Basic Lie Theory · their Lie algebras, subalgebras and ideals, the functorial relationship determined by the exponential map, the topology of the classical groups, the Iwasawa decomposition

Basic Lie Theory

Hossein AbbaspourMartin Moskowitz

ii

To Gerhard Hochschild

Contents

Preface and Acknowledgments ix

Notations xiii

0 Lie Groups and Lie Algebras; Introduction 1

0.1 Topological Groups . . . . . . . . . . . . . . . . . . . . . 1

0.2 Lie Groups . . . . . . . . . . . . . . . . . . . . . . . . . 6

0.3 Covering Maps and Groups . . . . . . . . . . . . . . . . 10

0.4 Group Actions and Homogeneous Spaces . . . . . . . . . 15

0.5 Lie Algebras . . . . . . . . . . . . . . . . . . . . . . . . . 25

1 Lie Groups 31

1.1 Elementary Properties of a Lie Group . . . . . . . . . . 31

1.2 Taylor’s Theorem and the Coefficients of expX expY . . 39

1.3 Correspondence between Lie Subgroups and Subalgebras 45

1.4 The Functorial Relationship . . . . . . . . . . . . . . . . 48

1.5 The Topology of Compact Classical Groups . . . . . . . 60

1.6 The Iwasawa Decompositions for GL(n,R) and GL(n,C) 67

1.7 The Baker-Campbell-Hausdorff Formula . . . . . . . . . 69

2 Haar Measure and its Applications 89

2.1 Haar Measure on a Locally Compact Group . . . . . . . 89

2.2 Properties of the Modular Function . . . . . . . . . . . . 100

2.3 Invariant Measures on Homogeneous Spaces . . . . . . . 101

2.4 Compact or Finite Volume Quotients . . . . . . . . . . . 106

v

vi

2.5 Applications . . . . . . . . . . . . . . . . . . . . . . . . . 112

2.6 Compact linear groups and Hilbert’s 14th problem . . . 121

3 Elements of the Theory of Lie Algebras 127

3.1 Basics of Lie Algebras . . . . . . . . . . . . . . . . . . . 127

3.1.1 Ideals and Related Concepts . . . . . . . . . . . 127

3.1.2 Semisimple Lie algebras . . . . . . . . . . . . . . 138

3.1.3 Complete Lie Algebras . . . . . . . . . . . . . . . 139

3.1.4 Lie Algebra Representations . . . . . . . . . . . . 140

3.1.5 The irreducible representations of sl(2, k) . . . . 142

3.1.6 Invariant Forms . . . . . . . . . . . . . . . . . . . 145

3.1.7 Complex and Real Lie Algebras . . . . . . . . . . 147

3.1.8 Rational Forms . . . . . . . . . . . . . . . . . . . 148

3.2 Engel and Lie’s Theorems . . . . . . . . . . . . . . . . . 150

3.2.1 Engel’s Theorem . . . . . . . . . . . . . . . . . . 150

3.2.2 Lie’s Theorem . . . . . . . . . . . . . . . . . . . 153

3.3 Cartan’s Criterion and Semisimple Lie algebras . . . . . 158

3.3.1 Some Algebra . . . . . . . . . . . . . . . . . . . . 158

3.3.2 Cartan’s Solvability Criterion . . . . . . . . . . . 162

3.3.3 Explicit Computations of Killing form . . . . . . 166

3.3.4 Further Results on Jordan Decomposition . . . . 170

3.4 Weyl’s Theorem on Complete Reducibility . . . . . . . 173

3.5 Levi-Malcev Decomposition . . . . . . . . . . . . . . . . 180

3.6 Reductive Lie Algebras . . . . . . . . . . . . . . . . . . . 188

3.7 The Jacobson-Morozov Theorem . . . . . . . . . . . . . 193

3.8 Low Dimensional Lie Algebrasover R and C . . . . . . . 198

3.9 Real Lie Algebras of Compact Type . . . . . . . . . . . 202

4 The Structure of Compact Connected Lie Groups 207

4.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 207

4.2 Maximal Tori in Compact Lie Groups . . . . . . . . . . 208

4.3 Maximal Tori in Compact Connected Lie Groups . . . . 210

4.4 The Weyl Group . . . . . . . . . . . . . . . . . . . . . . 217

4.5 What goes wrong if G is not compact . . . . . . . . . . 221

vii

5 Representations of Compact Lie Groups 223

5.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 224

5.2 The Schur Orthogonality Relations . . . . . . . . . . . . 226

5.3 Compact Integral Operators on a Hilbert Space . . . . . 228

5.4 The Peter-Weyl Theorem and its Consequences . . . . . 234

5.5 Characters and Central Functions . . . . . . . . . . . . . 243

5.6 Induced Representations . . . . . . . . . . . . . . . . . . 250

5.7 Some Consequences of Frobenius Reciprocity . . . . . . 255

6 Symmetric Spaces of Non-compact type 261

6.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 261

6.2 The Polar Decomposition . . . . . . . . . . . . . . . . . 264

6.3 The Cartan Decomposition . . . . . . . . . . . . . . . . 267

6.4 The Case of Hyperbolic Space and the Lorentz Group . 274

6.5 The G-invariant Metric Geometry of P . . . . . . . . . . 278

6.6 The Conjugacy of Maximal Compact Subgroups . . . . 289

6.7 The Rank and Two-Point Homogeneous Spaces . . . . . 294

6.8 The Disk Model for Spaces of Rank 1 . . . . . . . . . . 299

6.9 Exponentiality of Certain Rank 1 Groups . . . . . . . . 304

7 Semisimple Lie Algebras and Lie Groups 313

7.1 Root and Weight Space Decompositions . . . . . . . . . 313

7.2 Cartan Subalgebras . . . . . . . . . . . . . . . . . . . . . 316

7.3 Roots of Complex Semisimple Lie Algebras . . . . . . . 323

7.4 Real Forms of Complex Semisimple Lie Algebras . . . . 337

7.5 The Iwasawa Decomposition . . . . . . . . . . . . . . . . 343

8 Lattices in Lie Groups 355

8.1 Lattices in Euclidean Space . . . . . . . . . . . . . . . . 355

8.2 GL(n,R)/GL(n,Z) and SL(n,R)/SL(n,Z) . . . . . . . 360

8.3 Lattices in more general groups . . . . . . . . . . . . . . 371

8.4 Fundamental Domains . . . . . . . . . . . . . . . . . . . 374

9 Density results for cofinite Volume Subgroups 377

9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 377

9.2 A Density Theorem for cofinite Volume Subgroups . . . 379

viii

9.3 Consequences and Extensions of the Density Theorem . 389

A Vector Fields 397

B The Kronecker Approximation Theorem 403

C Properly discontinuous actions 407

D The Analyticity of Smooth Lie Groups 411

References 413

Index 421

Preface and

Acknowledgments

In the view of the authors, and as we hope to convince the reader, Lietheory, broadly understood, lies at the center of modern mathematics. Itis linked to algebra, analysis, algebraic and differential geometry, topol-ogy and even number theory, and applications of some of these othersubjects are crucial to many of the arguments we shall present here.This also holds true in the opposite direction: Lie theory can be usedto clarify or derive results in these other areas. In a philosophical senseLie groups are pervasive within much of mathematics, for whenever onehas some system, the “automorphisms” of it will frequently be a Liegroup. This even occurs in the oldest deductive system in mathematics,namely Euclidean geometry. Here the key issue is congruent figures,particularly triangles. Two such planar triangles are congruent if andonly if they differ by an element of the Lie group E(2) = O(2,R)⋉R2,the group of rigid motions of the Euclidean plane. The reader will findin these many interrelations a vast panorama well worth studying.

This book is the result of courses taught by one of the authors overmany years on various aspects of Lie theory at the City University ofNew York Graduate Center. The primary reader to which it is ad-dressed is a graduate student in mathematics, or perhaps physics, ora researcher in one of these subjects who wants a comprehensive ref-erence work in Lie theory. However, by a judicious selection of topics,some of this material could also be used to give an introduction to thesubject to well-grounded advanced undergraduate mathematics majors.

ix

x Preface and Acknowledgments

For example, Chapters 3 and most of 7 could form a semester’s coursein Lie algebras. Similarly, Chapters 0, 2 and 5 (respectively 0, 2 and8) could be a semester’s course in integration in topological groups andtheir homogeneous spaces (respectively lattices and their applications).For the reader’s convenience we have included a diagram of the inter-dependence of the chapters. We have also tried to make the text asself-contained as possible even at the cost of increased length. We shallassume the reader has some knowledge of basic group theory, topology,and linear algebra, and a general acquaintance with the grammar ofmathematics. While reading this book one may wish to consult some ofthe other books on the subject for clarification, or to see another view-point or treatment; especially useful books are listed in the bibliography.We have not attempted to detail the historical development of our sub-ject, nor to systematically give credit to the individual researchers whodiscovered these results.

The book’s organization is as follows:

Chapter 0 introduces the players; topological and Lie groups, cover-ings, group actions, homogeneous spaces, and Lie algebras.

Chapter 1 deals with the correspondence between Lie groups andtheir Lie algebras, subalgebras and ideals, the functorial relationshipdetermined by the exponential map, the topology of the classical groups,the Iwasawa decomposition in certain key cases, and the Baker CampbellHausdorff theorem, and local Lie groups.

Chapter 2 concerns Haar measure both on a group and on cocom-pact and finite volume homogeneous spaces together with a number ofapplications.

Chapter 3 gives the elements of Lie algebra theory in some consid-erable detail (except for the detailed structure of complex semisimpleLie algebras, which we defer until Chapter 7).

Chapter 4 deals with the structure of a compact connected Lie groupin terms of a maximal torus and the Weyl group.

Chapter 5 contains the representation theory of compact groups.

Chapter 6 concerns symmetric spaces of non-compact type.

Chapter 7 presents the detailed structure of complex semisimple Liealgebras.

Preface and Acknowledgments xi

Chapter 8 gives an introduction to lattices in Lie groups.Chapter 9 presents a “density theorem” for cofinite volume sub-

groups of certain Lie groups.Although we have included a rather detailed and extensive index, it

might be helpful to inform the reader of what we are not doing here.We do not deal extensively with the theory of algebraic groups, nor withtransformation groups, although each of these makes some appearance.Similarly, we do not prove that any connected Lie group is, as a man-ifold, the direct product of a Euclidean space and a maximal compactsubgroup, K; nor that any two such Ks are conjugate. But we do thisin important special cases. We do not deal, except by example, withthe theory of faithful representations. We have also omitted the Weylcharacter formula, the universal enveloping algebra, the classificationof complex simple algebras and, with the exception of one example,branching theorems.

We would like to thank Frederick Greenleaf, Adam Koranyi, KeivanMallahi and Grigory Margulis for reading various chapters of our bookand making a number of valuable suggestions. Of course any errors ormisstatements are the sole responsibility of the authors.

The authors would like to thank Richard Mosak for his help in thefinal preparation of this manuscript. We also thank Isabelle and Anitafor their extraordinary patience during the several years that it took tobring this project to completion.

Hossein Abbaspour Martin MoskowitzMax-Planck Institut fur Mathematik The Graduate School

Bonn The City University of New YorkNew York

xii Interdependency Chart

Chapter 0

Chapter 6 Chapter 4 Chapter 8 Chapter 5

Chapter 1 Chapter 3 Chapter 2

Chapter 7 Chapter 9

Notations xiii

Notations

For a complex number z, ℜz is the real part and ℑz is the imaginarypart.

The derivative of a differentiable map f : M → N at a point x ∈Mwill be denoted dxf : TxM → Tf(x)N and takes a tangent vector v ∈TxM to dxf(v) ∈ Tf(x)N .

We use capital letters G,H,K . . . for Lie groups and the correspond-ing german letters g, h, k . . . for Lie algebras. Lowercase letters such asa, b, c, . . . , g, h, k . . . are used for group elements and uppercase lettersX,Y,Z, . . . are reserved for vectors of Lie algebras. For a Lie group G,G0 is the connected component containing the identity element.

For a Lie group G with the Lie algebra g, Ad : G → GL(g) is theadjoint representation taking x ∈ G to Adx ∈ GL(g), and its image, theadjoint group, is denoted AdG. If H ⊂ G then AdG(H) is the imageof H under Ad and where is no risk of confusion we will simply writeAd(H). The Lie algebra representation ad : g → gl(g) takes a vector Xto adX ∈ gl(g). This is the map defined by the rule Y 7→ adX(Y ) =[X,Y ] and its image is ad g. To avoid any ambiguity we may writeadg to indicate that where the action is taking place. For a subalgebrah ⊂ g, the restriction of ad to h is denoted ad |h which is different fromadh.

For a real entries matrix A, At is the transpose of A and if A hascomplex entries then A∗ is the transpose conjugate of A.

For a vector space V over a field k and S ⊆ V then l.s.k(S) is the klinear span of S. Finally, Mn(k) is the set of n×n matrices with entriesin the field k. For T ∈ Endk(V ), define Spec(T ) to be the set of alleigenvalues of T in the algebraic closure of k.

Additional notations are introduced through the index at the end ofthe book.

xiv Notations

Chapter 0

Lie Groups and Lie

Algebras; Introduction

0.1 Topological Groups

Before dealing with the generalities concerning topological groups andLie groups which occupies next sections of this chapter, we provide somekey definitions and examples.

Definition 0.1.1. Let G be a group and at the same time a Hausdorfftopological space. Suppose in addition that the group operations

(1) (g, h) 7→ gh

(2) g 7→ g−1

are continuous, where in (1) we take the product topology on G × G.We then call G a topological group.

Exercise 0.1.2. Prove that continuity of (1) and (2) is equivalent tothat of (g, h) 7→ gh−1.

Exercise 0.1.3. Define the direct product of a finite number of topo-logical groups equipped with the product topology and show it is atopological group.

1

2 Chapter 0 Lie Groups and Lie Algebras; Introduction

We shall almost always be interested in locally compact groups. Notethat a closed subgroup of a locally compact topological group is againa locally compact topological group.

Examples of topological groups abound. Any abstract group is atopological group having discrete topology. So the theory of topologicalgroups includes abstract groups. The additive group of real numbers Ris also a topological group. The only point that needs to be checked isthe continuity of 1.

Exercise 0.1.4. Prove that the continuity of (x, y) 7→ x + y followsfrom the triangle inequality, |x+ y| ≤ |x| + |y|.

As indicated above further examples can be gotten by taking directproducts so Rn is a topological group. Of course, this also follows fromthe triangle inequality |x + y| ≤ |x| + |y|, where | · | denotes the norm.The multiplicative group of real numbers, R× = R \ 0, as well as themultiplicative group of complex numbers C× = C \ 0 (both with therelative topology) are also topological groups. These things follow fromthe triangle inequality together with |xy| = |x||y|.

Exercise 0.1.5. Prove that R× and C× with the usual multiplicationform groups. Show that a similar argument works for the multiplicativegroup of quaternions, H× = H \ 0.

Notice that R× is disconnected while C× is connected, but is notsimply connected whileH× is connected and simply connected. This lastexample is our first noncommutative group. Notice that all these groupsare locally compact by the Heine-Borel theorem (see [74]). The (closed)subgroup R×

+ consisting of positive reals however is connected. It hasindex 2 in R×. The subgroup of T of C× consisting of elements of norm1 is a closed and therefore also a locally compact topological group. Itis actually compact (and homeomorphic to the circle S1) as is the finitedirect product, Tn. Similarly, the subgroup of H× consisting of elementsof norm 1 is a compact topological group which is homeomorphic to the3 sphere S3.

Exercise 0.1.6. Prove that:

0.1 Topological Groups 3

(1) R× is the direct product of ±1 with R×+.

(2) C× is the direct product of T with R×+.

(3) H× is the direct product of S3 with R×+.

Exercise 0.1.7. Show that any closed subgroupH of G = Rn is isomor-phic to a subgroup of the form Zk ×Rj , where k+ j ≤ n. In particular,if H is connected it must be Rj, and if H is discrete it must be Zk. As aresult G/H is compact if and only if k+j = n in which case G/H = Tk,and the only closed subgroups of R are either 0, R itself or na : n ∈ Z,where a 6= 0.

Exercise 0.1.8. What are the closed subgroups of T?

Turning to an important source of noncommutative topologicalgroups we consider the real general linear group, GL(n,R). This isthe group of invertible n × n real matrices with the relative topologyfrom Mn(R) identified with Rn

2equipped its product topology. This

is a locally compact topological group because matrix multiplication isgiven by polynomial functions in the coordinates, inversion by rationalfunctions with non vanishing denominators and because an open set in aEuclidean space is locally compact by Heine-Borel. The same applies tothe complex general linear group, GL(n,C) ⊂Mn(C). Thus any closedsubgroup of either of these groups is again a locally compact group. Ofcourse when n = 1, these are nothing more than R× and C× and againGL(n,R) is disconnected while GL(n,C) is connected.

Other important examples are SL(n,R) and SL(n,C). These are thegroups of n× n real or complex matrices of determinant 1. Additionalexamples are O(n,R) and O(n,C), the real and complex orthogonalgroups which preserve the bilinear form

〈x, y〉 =

n∑

i=1

xiyi

for x = (x1, . . . , xn) and y = (y1, . . . , yn) ∈ Rn or Cn, respectively.

Then SO(n,R) = O(n,R) ∩ SL(n,R) and SO(n,C) = O(n,C) ∩SL(n,C). Further examples are the unitary group U(n,C), the sub-


group of GL(n,C) preserving the Hermitian form

〈x, y〉 =

n∑

i=1

xiyi,

for x = (x1, . . . , xn) and y = (y1, . . . , yn) ∈ Cn, and SU(n,C) =

U(n,C) ∩ SL(n,C). Finally we have the symplectic groups Sp(n,R)and Sp(n,C). These are respectively the subgroups of GL(2n,R) andGL(2n,C) which preserve the symplectic form

〈x, y〉 =n∑

i=1

(xiyn+i + xn+iyi).

A final example is SO(p, q). This is the subgroup of GL(p + q,R)which preserves the bilinear form

〈x, y〉 =

p∑

i=1

xiyi −p+q∑

i=p+1

xiyi.

This is also a nondegenerate bilinear form, making SO(p, q) a topologicalgroup.

Exercise 0.1.9. Prove each of these is a locally compact topologicalgroup.

This will give the reader some idea of what we have in mind whenwe refer to a topological group. Now as with any category one mustspecify the morphisms as well. Let G and H be topological groups. Weshall call f : G → H topological group homomorphism if it is a grouphomomorphism and continuous. So for example the exponential map,exp : R→ R×

+, or exp : C→ C× are topological group homomorphisms.The first of these is bijective and has a continuous inverse, namely log.Such a map is called a topological group isomorphism, while the second,although surjective, is not one-to-one since 2πni|n ∈ Z maps to 1.Evidently, isomorphic topological groups share the same properties astopological groups.

0.1 Topological Groups 5

Exercise 0.1.10. Show that the kernel of this map is exactly 2πni|n ∈Z.

Another example of a topological group homomorphism to keep inmind is given by the identity map from the discrete group of additivereals to R.

If f : G → H is a topological group homomorphism, then Ker f isa closed normal subgroup of G and f(G) is a subgroup of H. If G is atopological group and N is a closed normal subgroup, we can form thequotient group G/N equipped with the quotient topology.

Exercise 0.1.11. If G is is locally compact, show that G/N is a locallycompact topological group. Also demonstrate why the projection π :G→ G/N is a continuous, open, surjective homomorphism.

So, for example, taking t 7→ exp(2πit) shows that R/Z is a topolog-ical group that is isomorphic to T. More generally, if f : G → H is atopological group homomorphism then this induces an injective topo-logical group homomorphism f∗ : G/Ker f → H making a commutativediagram,

Gf //

$$HHHHHHHHH H

G/Ker f

f∗::uuuuuuuuu

.

This is called the first isomorphism theorem for topological groups.An important special case of this is when the induced map f∗ is actuallyan isomorphism (as was exp just above) which is treated in Corollary0.4.7.

Exercise 0.1.12. Prove that T is isomorphic with R/Z. The subgroupQ/Z of R/Z is the torsion subgroup. The other elements generate densecyclic subgroups.

Proposition 0.1.13. If G is a connected topological group and U isany symmetric neighborhood of 1, then G =

⋃∞n=1 U

n. Of course sinceU ⊇ U ∩ U−1, which is symmetric, the result actually holds for anyneighborhood U of 1 in G.


Proof. Note that⋃∞n=1 U

n is an open subgroup of G. Therefore it isclosed. Because of the connectedness of G, it must be G (see Exercise0.2.5).

0.2 Lie Groups

Analogously to the definition of a topological group we define a real Liegroup as a group G which is also a finite dimensional real differentiablemanifold whose operations are smooth, i.e. C∞. That is, the mapG×G→ G given by (g, h) 7→ gh−1 is C∞ (product manifold structure).We call the dimension n of the manifold the dimension of G. For Liegroups G and H, a Lie homomorphism f : G → H is a smooth grouphomomorphism. If in addition f is bijective and its inverse is smoothwe say f is an isomorphism. A Lie subgroup H of G is a submanifoldwhich is a subgroup as well.

Although the formal parts of this theory emulate those of topologicalgroups there are some not so obvious aspects. For example in the latterclosed subgroups are taken as a convenience to insure local compactnesswhile in the case of Lie groups it is an important theorem of Elie Cartanthat a closed subgroup of a Lie group has the structure of a Lie group.We mention some variants of the definition. One could consider realanalytic manifolds and real analytic maps instead of just C∞ ones. Thisis done in Hochschild [33]. In fact, it does not matter which one doesas the theory and the category of the real analytic and C∞ Lie groupscoincide (see Appendix D).

Another variant which does get one somewhere is to consider thenotion of a complex Lie group. Here one simply takes complex manifoldsand holomorphic maps. The result is called a complex Lie group.

Clearly a complex Lie group is a real Lie group. Another variantwould be to not limit the manifolds to be finite dimensional. Thisapproach has had only limited success and will not be pursued here.

Exercise 0.2.1. Prove that (g, h) 7→ gh−1 is C∞ if and only if multi-plication (g, h) 7→ gh and inversion g 7→ g−1 are C∞.

Given a Lie group G, the left translations Lg : G → G defined by

0.2 Lie Groups 7

Lg(h) = gh are global diffeomorphisms on G. Since they can take anypoint a to any other point b by taking g = ba−1 we see that all localtopological properties valid at a single point such as the identity arevalid at all other points. The same applies to right translations.

Of course a Lie group is a (very special kind of) locally compact topo-logical group. The converse question is Hilbert’s fifth problem whichwas solved in 1953 by Gleason-Montgomery-Zippin and Yamabe. Forthe details of this see [26]. As all manifolds, Lie groups are also locallyconnected. In particular the identity component1 G0 of G is open.

Lie groups arise in various ways. For example the isometry group ofa Riemannian manifold is always a Lie group (see [67, 32]). Similarly,the automorphism group of a Lie group is also a Lie group (see [33]).

As we shall see in Chapter 5, one way of studying Lie groups isthrough their representations. A representation of a Lie group G isa smooth homomorphism ρ : G → GL(V ), continuous complex rep-resentation of G. We call V = Vρ the representation space of ρ anddρ = dimVρ its degree and ρg : V → V is the map

ρg(v) = ρ(g)v.

A representation ρ is said to be faithful if it is injective.It is appropriate now to give a few simple examples. As usual,

when one has an open subset of Euclidean space the manifolds structureconsists of a one chart atlas. The various assertions concerning theseexamples should also be regarded as exercises.

Example 0.2.2.

(1) R, or more generally Rn, is a Lie group with the usual manifoldstructure.

(2) T, or more generally Tn, is a Lie group with the usual manifoldstructure and in fact the natural map π : Rn → Tn given byprojection in each coordinate is a smooth group homomorphism.It is the universal covering of Tn by Rn. Both Rn and Tn areconnected and Rn simply connected. As we saw above, R/Z is

1The connected component containing the identity element.


isomorphic as a topological group to the multiplicative group S1

of all complex numbers of modulus 1; the isomorphism is given byt 7→ e2πit, t ∈ R when regarded as a map R/Z → S1. Since thismap and its inverse are smooth they are isomorphic as Lie groups.

(3) Any discrete group G is a Lie group of dimension zero. In particu-lar Z or more generally Zn, is a Lie group. It is a closed subgroupof Rn which is the kernel of π above.

(4) The multiplicative group R× is a Lie group. It is not connected,but has two components. Its identity component R×

+ of positivereal numbers is also a Lie group. This group is isomorphic with Rvia the usual exponential map. Similarly, C× is a (2 dimensional)Lie group which is connected. It is actually a complex Lie group.As we saw here however the exponential map, exp : C → C× isnot an isomorphism. However, it is smooth (actually holomorphic)a homomorphism, surjective and a local diffeomorphism at eachpoint. The reader should prove this. Since C is simply connectedwe have just constructed the universal covering group of C× (seeSection 0.3 on covering spaces). As we saw previously, regardingC× as a real Lie group, it is isomorphic via the polar decompositionto R×

+ × T (direct product). Later in Chapter 6 we shall see thiscan be generalized considerably.

(5) More generally, GL(n,R), the group of invertible n × n real ma-trices is a (dense) open subset of Euclidean space Mn(R) and thusacquires a manifold structure in which multiplication is a polyno-mial function of the coordinates. Moreover, inversion is a rationalfunction of the coordinates with a nowhere vanishing denominator.As an exercise the reader should verify all of these facts includingthat GL(n,R) is open and dense. Hence GL(n,R) is a real Liegroup of dimension n2. When n = 1 we get R×. We shall see thatit has 2 components because GL+(n,R), the ones with positivedeterminant, is connected. To see that GL(n,R) is not connectedwe just observe that its image under the smooth (check!) mapA 7→ det(A) has two components.

(6) Similarly, GL(n,C), the group of invertible n×n complex matrices

0.2 Lie Groups 9

is a (dense) open subset of Euclidean space of n × n complexmatrices Mn(C) and thus acquires a complex manifold structurein which multiplication is a polynomial function of the coordinates.Moreover, inversion is a rational function of the coordinates witha nowhere vanishing denominator. Hence GL(n,C) is a complexLie group of complex dimension n2. When n = 1 we get C×. Laterwe shall see that, just as with n = 1, it is connected. Thus all theexamples given in Section 0.1 above are also Lie groups.

(7) Further examples are provided by the group of real Tn(R) or com-plex Tn(C) triangular matrices, or the group of strictly real Nn(R)or complex triangular matrices Nn(C). In all these cases we havean atlas with one coordinate patch. The latter three are connectedwhile Tn(R) has 2n components.

(8) One can form semidirect products to get additional examples.For instance G = GL(n,R) × Rn. This is a manifold withthe usual product manifold structure and with group operation(g, v)(g

′, v

′) = (gg

′, gv

′+ v). It is called the affine group of Rn

and is a Lie group of dimension n2+n. Similarly one can constructthe complex Lie group semidirect product GL(n,C)⋉Cn.

(9) Let G and H be Lie groups and η : G → Aut(H) be a smoothhomomorphism where Aut(H) takes on a natural Lie group struc-ture. Or more directly, we can give G × H the product mani-fold structure and just assume (g, h) 7→ η(g) · h is a smooth mapG×H → H. Then define (g, h)(g

′, h

′) = (gg

′, η(g)h

′ · h). This isa Lie group and is called the semidirect product G⋉H of G andH and contains a closed subgroup isomorphic to G and a closednormal subgroup isomorphic to H. Notice that in general G is notnormal.

(10) Of course when the action η is trivial we get the direct product.

Exercise 0.2.3. Show that a semidirect product G⋉H is direct if andonly if the G is normal.

Exercise 0.2.4. What are the smooth homomorphisms f : R→ T, andf : T → T? Suggestion: use Exercise 0.1.7 above to find Ker f . In thefirst case we have for t ∈ R, fx(t) = e2πixt, x ∈ R. In the second we


have for t ∈ R, fx(t) = e2πixt, x ∈ Z, where t is the image of t underthe covering.

Exercise 0.2.5. Show that:

(1) An open subgroup of a topological group is closed.

(2) If H is a closed subgroup then H is open if and only if G/H isdiscrete.

(3) Let G be a Lie group and G0 denote the identity component of 1.Then G0 is an open normal subgroup of G. In particular, G0 is aconnected Lie group of the same dimension as G.

We close our remarks on Lie groups with an example of a locallycompact (in fact compact and commutative) group which is not a Liegroup. Let p be a prime number and consider (Z, | · |p), where | · |p is thep-adic norm on Z which is defined as follows. We take |0|p = 0 and ifz 6= 0 ∈ Z the prime factorization of z = pns, where s is relatively primeto p. Then |z|p = p−n. It is easy to see that this gives a norm on Z withall the usual properties, but instead of the triangle inequality one hasthe stronger |x + y|p ≤ max(|x|p, |y|p). Then the p-adic integers Z(p)

is the completion of Z with respect to the metric dp(x, y) = |x − y|p.This group has a neighborhood basis at 0 of nested subgroups (pm),where m ∈ Z+. But as we shall see, a Lie group must have a sufficientlysmall neighborhood of 1 which contains no nontrivial subgroup. This isimpossible for Z(p).

0.3 Covering Maps and Groups

We begin this section by reviewing the notions of covering spaces andcovering maps. Then we will study the covering maps for Lie groupsand their relation to the group structure. We refer the reader to [40]for a detailed account of the covering theory. In this section all thetopological spaces are path connected and locally path connected.

Suppose that X and Y are two topological spaces and e : Y → X acontinuous map. We shall say e is a covering map if every point x ∈ Xhas an neighborhood set U such that e is a homeomorphism on each

0.3 Covering Maps and Groups 11

connected component of e−1(U) to U . Such an open set U is called anadmissible neighborhood. So by definition a covering map is surjective.We say that Y is a covering space for X. If e : Y → X and e′ : Y ′ → Xare two covering maps,

Y ′ f //

e′ AAA

AAAA

A Y

e~~~~

~~~~

~~

X

then a continuous map f : Y ′ → Y with e′ = e f is said to be fiberpreserving . Two covering spaces e : Y → X and e′ : Y ′ → X are saidto be equivalent if there is a fiber preserving map f : Y ′ → Y which isa homeomorphism.Lifting property: The most fundamental property of the covering isthe lifting property as follows. Suppose that e : (Y, y0) → (X,x0) is acovering map for the based spaces X and Y that is x0 ∈ X0, y0 ∈ Yand e(y0) = x0. Let g : (P, p0) → (X,x0) be a continuous map of basedspaces such that f∗(π1(P, p0)) ⊂ e∗(π1(Y, y0)), where f∗ and e∗ are theinduced maps on the fundamental groups.

Y

e

P

f>>~~~~~~~ g // X

Then there exists a unique continuous map f : P → Y such thatg = e f .

A universal cover of a topological space X is a covering space e :Y → X, where Y is a simply connected manifold.

Exercise 0.3.1. Prove that any two universal covers of a topologicalspace are equivalent.

Given a covering map e : Y → X with smooth base X, one can makeY into a smooth manifold such that the covering map is a smooth map.Let (U, φ) be a chart on X so that U is an admissible neighborhood.


Then the connected components of e−1(U) with the map φ e form anatlas for Y and by construction e is a smooth map.

Now we go one step further and consider the covering maps whosebase spaces are Lie groups.

Proposition 0.3.2. Suppose that e : G → G is a universal cover of aLie group G. Then there is a unique Lie group structure on G whichmakes e a group homomorphism.

Proof. Earlier we described the unique smooth structure on G whichmakes e smooth. We still need to introduce the group structure andverify that structural maps are smooth. We choose 1 ∈ e−1(1) where 1is the identity element of G. In what follows 1 is the base point of Gand 1 is the base point of G.

Suppose that (g, h) 7→ g.h denotes the multiplication of G, then wedefine the smooth map ψ : G× G→ G to be

ψ(x, y) = e(x).e(y) (1)

which sends the base point (1, 1) to the base point 1.

G

e

G× G

ψ<<yyyyyyyyy ψ // G

Because G is simply connected, by the lifting property of covering mapsψ lifts to a map ψ : G× G→ G such that

ψ = e ψ. (2)

Since ψ is the lift of a smooth map, therefore it is a smooth map withrespect to the natural smooth structure of G. Similarly, one can lift thesmooth map i : G→ G given by x 7→ (e(x))−1 to G

G

e

G

i@@

i // G

0.3 Covering Maps and Groups 13

to get the smooth map i : G→ G for which

e(x)−1 = e(i(x)). (3)

holds.We claim that G, with 1 as the identity element, ψ as the multipli-

cation, and i as the inverse element map is a group.1) Associativity : We must show that ψ(ψ(x, y), z) = ψ(x, ψ(y, z)). Notethat by the associativity of the multiplication for G, (2) and (1),

e(ψ(ψ(x, y), z)) = e(ψ(x, y)).e(z) = (e(x).e(y))e(z) = e(x).e(y).e(z)

and

e(ψ(x, ψ(y, z))) = e(x).e(ψ(y, z)) = e(x).(e(y).e(z)) = e(x).e(y).e(z).

Therefore, ψ(ψ(x, y), z) and ψ(x, ψ(y, z)) are both liftings of the map(x, y, z) 7→ e(x).e(y).e(z). Thus by the uniqueness of the lifting they areequal.2) Inverse: We prove that ψ(x, i(x)) = 1 and the proof is similar to theone above. By (3)

e(ψ(x, i(x))) = e(x).e(i(x)) = e(x).e(x)−1 = 1

which means that ψ(x, i(x)) is a lift of the constant map x 7→ 1 just likethe map x 7→ 1. Therefore ψ(x, i(x)) = 1.3) Identity element : It is a direct check that x 7→ ψ(x, 1) and the identitymaps are both liftings of x 7→ e(x). Therefore they are the same, provingthat 1 is the identity element for the multiplication ψ.

Proposition 0.3.3. If e : G → G is a universal covering homomor-phism then any group homomorphism f : H → G can be lifted to agroup homomorphism f : H → G.

G

e

H

f?? f // G


Proof. By the lifting property there is a unique lifting f : H → G forf . Consider the maps ψ1(x, y) = f(x).f (y) and ψ2(x, y) = f(xy), bothare a lifting of φ : H ×H → G given by

φ(x, y) = f(x)f(y).

Therefore, by uniqueness, we have ψ1 = ψ2 which implies that f is agroup homomorphism.

Proposition 0.3.4. Let Γ be a discrete subgroup of a connected Liegroup G. Then the natural projection map p : G → G/Γ is a coveringmap.

Proof. Let U be an open connected neighborhood of the identity withU ∩Γ = 1. This is possible as Γ is discrete. Because of the continuityof the multiplication we can choose an open neighborhood V of 1 suchthat V −1V ⊂ U . Therefore V −1V ∩ Γ = 1. Consider the set of theform V h ⊂ G for h ∈ Γ. These sets are disjoint because if v1h1 = v2h2

then v−12 v1 ∈ V −1V ∩ Γ = 1 which says that v1 = v2 and h1 = h2.

Moreover, p|V h is injective for the same reason as above. To get theadmissible opens we use the left translation. More precisely for a cosetgΓ the open set of cosets gV Γ = gvΓ|v ∈ V is an admissible open andp−1(gV Γ) = ∪h∈Γ(gV h) which is a disjoint union and the restriction ofp to each gV h is a diffeomorphism.

Remark 0.3.5. In the previous proposition, p is not necessarily a grouphomomorphism as Γ is not a normal subgroup. The next result addressesthis case, and proves something stronger namely that Γ has to be central.

Lemma 0.3.6. Let G be a connected topological group and Γ be a dis-crete normal subgroup. Then Γ is central in G.

Proof. To see that Γ is central, consider the continuous map G → Γgiven by g 7→ gγg−1 for a fixed γ ∈ Γ. This is well-defined as Γ isnormal, and constant as Γ is discrete and G connected, so γ is in thecenter.

0.4 Group Actions and Homogeneous Spaces 15

Corollary 0.3.7. Let G be a connected Lie group and f : G → H bea local isomorphism. Then f is a covering map and Γ = Ker f is adiscrete central subgroup of G. If G is a simply connected group, thenπ1(H) = Γ.

Proof. Since f is a local diffeomorphism then there is an open neighbor-hood U of the identity such that f |U is injective. Therefore, Γ∩U = 1and Γ is discrete. We have the isomorphism f : G/Γ → H and by Propo-sition 0.3.4, p : G → G/Γ is a covering map, and thus f is a coveringmap. Γ is central by the previous lemma.

0.4 Group Actions and Homogeneous Spaces

Here we discuss a notion which is central to many areas of mathematics,namely, that of a group action (either of a locally compact group on aspace, or of a Lie group on a manifold).

Let G be a locally compact group, X be a (usually locally compact)space and φ : G×X → X be a jointly continuous mapping, called theaction of G on X. Writing φ(g, x) as g · x or even gx we shall assumethat an action, (G,X), satisfies the following:

(1) For all x ∈ X, 1 · x = x

(2) For all g, h ∈ G and x ∈ X, (gh) · x = g · (h · x)We shall call X a G-space or equivalently, say that G acts as a

transformation group on X. An action is transitive if for all x, y ∈ Xthere exists g ∈ G such that g.x = y. If such g is unique, then it iscalled a simply transitive action.

If G is a Lie group, X a smooth manifold and the action is jointlysmooth, then we shall simply call this a smooth action,

Exercise 0.4.1. Show that each φ(g, .) is a homeomorphism ofX. ThusG operates on X by homeomorphisms.

In applications X could, for example, be some geometric space andG a group of transformations of X which preserves some geometricproperty such as length, angle, or area, etc. Such transformations alwaysform a group and it is precisely the properties of this group which is the


key to understanding length, angle, or area, respectively. This is theessential idea of Klein’s Erlanger Program, named after the late 19thcentury mathematician, Felix Klein. One can also turn this on itselfand use it as a tool to study group theory. The first example of a groupaction is provided by a group G acting on itself by left translation.Since G is a topological group this is evidently an action and one seesthat the conditions above are designed exactly to reflect this. Moregenerally we can consider a locally compact group G, a closed subgroupH and the space of left cosets X = G/H and let G act on G/H byleft translation, g1 · gH = g1gH, where g1 ∈ G and gH ∈ G/H. Itis an easy check to show that this is also an action. Another exampleis provided by taking a locally compact group G and letting Aut(G)the automorphism group (suitably topologized), or some subgroup ofAut(G) such as I(G), the inner automorphisms, act in the natural wayon G. Or one might take a real or complex (usually finite dimensional)representation ρ : G → GL(V ) of a locally compact group. Then weevidently have an action of G on V . Such actions are called linearactions. In particular, this would be the case if G ⊆ GL(V ), i.e. theidentity representation. Finally, let G, X be a G-space and F(X) besome real or complex vector space of functions defined on X. We candefine an action of G on F(X) by g · f = fg, the left translate of thefunction f , where fg(x) = f(g−1x). All that is required here is thatF(X) be G-stable. An easy check shows this is a linear action calledthe induced action on functions. We shall give other examples of actionsin the sequel.

Exercise 0.4.2. Let G be a locally compact group, H a closed subgroupand G/H the space of left cosets having quotient topology. Then theprojection π : G → G/H is a continuous and open map and G/H is alocally compact Hausdorff space. Show G/H is discrete if and only ifH is open in G. In particular, this means that every open subgroup isclosed. Also show that at the other extreme G/H is not discrete if andonly if H is dense in G.

We remark that there is a theorem of Pontrjagin [70] strengtheningthis considerably. Namely, that for a closed subgroup H of a topological


group G the quotient space, G/H, is actually T3 12. In particular, since

taking H to be trivial gives us the usual action of a topological groupon itself by left translation, we see that a locally compact group itselfis T3 1

2.

If G acts on X a subset Y of X is called G-invariant if G · Y ⊆ Y .When we have an invariant set we get a new action of G on Y . Similarly,if (G,X) is a group action and H is a closed subgroup of G, then (H,X)is also a group action. Let (G,X) be a group action. For x ∈ X wedefine the G-orbit of x, written OG(x), as g · x : g ∈ G. When theaction is transitive there is only one orbit. Choosing an x0 ∈ X givesrise to a map G→ OG(x) ⊆ X given by g 7→ g(x0) and called the orbitmap. Evidently each x ∈ X is in its OG(x), so that each orbit is non-empty. Also if OG(x) ∩ OG(y) 6= φ, then OG(x) = OG(y). For supposeg1 ·x = g2 ·y. Then g−1

2 g1 ·x = y and therefore g3g−12 g1 ·x = g3 ·y, proving

that OG(y) ⊂ OG(x). Similarly, OG(x) ⊂ OG(y) so OG(x) = OG(y).Thus X is the disjoint union of the orbits. (In the case that G andX are finite this gives a useful counting principle which plays a role inproving the Sylow theorems.) For example, for the standard action ofSO(n,R) on Rn the orbits are 0 and the various spheres are centered at0. Whereas if GL(n,R) acts on Rn, there are only two orbits 0 andRn \ 0. We observe that the orbits of an action need not be closed inX and can even be dense.

Exercise 0.4.3. Give an example of an action whose orbits are notclosed.

Just as for groups we must now also decide when two group actions(G,X) and (G,Y ) are essentially the same. We shall say that (G,X)and (G,Y ) are G-equivalent or equivariantly equivalent if there is abijective bi-continuous map π : X → Y which for all g ∈ G and x ∈ Xsatisfies

π(g · x) = g · π(x). (4)

We call such a π a G-equivariant isomorphism or equivalence. When π ismerely a continuous map X → Y satisfying (4), we say it is a morphismof G-spaces. Evidently each orbit map is a continuous morphism.


Let G be a group, H be a closed subgroup and G act on G/Hby left translation, as above. A moment’s reflection tells us this is atransitive action. As we shall see it is essentially the only one. WhenG acts transitively on X, since it acts by homeomorphisms, the localtopological properties of X are the same at every point. So, for example,if local compactness or local connectedness holds at one point, then itholds at all points. In particular, this is so for a topological group itself.

Proposition 0.4.4. Let (G,X) be an action and x ∈ X be fixed.Then the orbit, OG(x) is a G-invariant set; in fact it is the small-est G-invariant set containing x. Hence this gives a transitive action(G,OG(x)). If (G,X) is a transitive action and x ∈ X, then the stabi-lizer,

StabG(x) = g ∈ G : g · x = x

is a closed subgroup of G. If y is another element of X, then by tran-sitivity we can choose g ∈ G so that g · x = y. Then g StabG(x)g−1 =StabG(y).

Proof. The fact that OG(x) is G-invariant and G acts transitively onOG(x) is immediate. The same may be said as far as the stabilizer beinga subgroup of G. If gn is a net in G converging to g with gn(x) = x,then g(x) = x, by continuity of the action. Finally, if g′(x) = x, thengg′g−1(y) = gg′(x) = g(x) = y. This proves g StabG(x)g−1 ⊆ StabG(y).Hence StabG(x) ⊆ g−1 StabG(y)g which by the same reasoning is con-tained in StabG(x). Thus g StabG(x)g−1 = StabG(y).

We now come to a useful result which will show that when thereis a compact, or even locally compact σ-compact group2 G operatingtransitively and continuously on a locally compact space X then thetopology of X is determined by that of G. It is the quotient topologyon G/Stab(x) transferred to X. If G happens to be a Lie group thenas a closed subgroup H is also a Lie group (see Theorem 1.3.5) and Xgets a manifold structure as G/H where G is a Lie group and H is aLie subgroup.

2A group is said to be σ-compact if it is a countable union of compact subsets.


Theorem 0.4.5. Let G be a locally compact group, X a locally compactHausdorff space, (G,X) a transitive G-space and x ∈ X fixed. If G iscompact, or even σ-compact, then (G,X) is equivariantly equivalent tothe action of G on G/StabG(x) by left translation.

Proof. We first deal with the formal part. To see that we have anequivariant equivalence of actions let π : G → X be the orbit mapg 7→ g · x. By transitivity, π is onto. Moreover, π(g) = π(g′) if and onlyif g−1g′ ∈ StabG(x). Hence π induces a bijection π : G/StabG(x) →X. To check the commutativity of the diagram we must see that g ·π(g′ StabG(x)) = π(gg′ StabG(x)) for all g and g′ ∈ G. But the formeris just g · π(g′) while the latter is π(gg′). Since π is a G-map, π is G-equivariant. Because π is continuous so is π. Thus π is a continuous,bijective G-equivariant map. All that remains is to see that is open. Ofcourse, if G is compact, or even if G/StabG(x) is compact, then thisis so because π is continuous and bijective and X is Hausdorff. Thisalready suffices for many, but not all applications. For this reason wealso make the very mild assumption that G is σ-compact.

Exercise 0.4.6. Show this σ-compact result is rather general. Forinstance it shows that it applies whenever G is locally compact andsecond countable.

To deal with the σ-compact case we recall a version of the BaireCategory theorem [32], pp. 110.Baire’s Category Theorem: Let X be a σ-compact space. That isX = ∪Cn, where each Cn is compact. If X is locally compact, then oneof the Cn must have a non-void interior.

Continuing the proof in the σ-compact case, we have the following:Since the orbit map π is G-equivariant and the action of G on itself aswell as the action of G on X are both transitive, to show openness wemay consider a neighborhood of 1 in G. Let U be such a neighborhood.We show U · x contains a neighborhood of x in X. Choose a smallerneighborhood U1 of 1 in G which is compact and U−1

1 U1 ⊆ U . BecauseG is σ-compact and U1 is a neighborhood there is a countable numberof gn such that G =

⋃gnU1. Since π is continuous and surjective the

π(gnU1) are compact and cover X. By Baire’s theorem there is some


n for which π(gnU1) ⊇ V , a nontrivial open set in X. If u1 ∈ U1 andπ(gnu1) ∈ V , then (gnu1)

−1V is a neighborhood of x in X and

(gnu1)−1V = u−1

1 g−1n V ⊆ u−1

1 U1x ⊆ Ux.

Thus π, and of course also π, is open.

We now apply this to derive the Open mapping theorem for grouphomomorphisms.

Corollary 0.4.7. Suppose G and H are locally compact groups with Gσ-compact. Let f : G → H be a continuous surjective group homomor-phism. Then f is open.

Proof. Let G act on H by left multiplication through f . Thus if h =f(g′), g · h = f(g)h. This is clearly a transitive action whose stabilitygroup is Ker f . Moreover, taking h = f(1) = 1 the orbit map for thisaction given in the theorem above is f . Thus f is open.

We can now apply these methods to some important examples ofcompact homogeneous G-spaces, whereG is a compact Lie group. Theseare spheres, projective spaces, the Stiefel and Grassmann manifolds andthe Flag manifolds.

Spheres: We consider Rn and Cn to be inner product and Hermitianinner product spaces, respectively. Evidently, O(n,R) acts transitivelyon Sn−1, the unit sphere in Rn and U(n,C) acts transitively on S2n−1,the unit sphere in Cn, since these groups can take any orthonormal basisv1, . . . , vn to another one w1, . . . , wn. If it happens that det g 6= 1,where g ∈ O(n,R) or U(n,C), which we call G, replace v1, . . . , vn byv1, . . . , λvn, where |λ| = 1. Then this is again an orthonormal basisand there is an h ∈ G such that h(vi) = wi, for i < n and h(λvn) = wn.Then h(vn) = 1

λwn and deth = 1λ det g. Choosing λ = det g which has

absolute value 1 since g ∈ G, makes deth = 1.

Thus SO(n,R) and SU(n,C) act transitively on Sn−1 and S2n−1,respectively. What is the isotropy group of such an action? Since it istransitive we can choose any unit vector as the base point. Since G fixesv1 then it must stabilize its orthocomplement, the subspace W spanned


by v2, . . . , vn and the restriction of such a g to W is in SO(n − 1,R)or SU(n− 1,C), respectively.

Applying Theorem 0.4.5 to the various compact groups then yieldsthe following:

(1) O(n,R)/O(n− 1,R) = Sn−1

(2) SO(n,R)/SO(n− 1,R) = Sn−1

(3) U(n,C)/U(n− 1,C) = S2n−1

(4) SU(n,C)/SU(n− 1,C) = S2n−1

In a similar way one sees that Sp(n) acts transitively on S4n−1,n ≥ 1. Since Sp(1) = SU(2,C) = S3 it follows that all Sp(n) arecompact by Proposition 2.4.5. Also for similar reasons all Sp(n) areconnected and since all the spheres from n ≥ 3 are simply connected italso follows from the results of the next section that the Sp(n) are alsoall simply connected.

Projective space: This line of argument gives real RPn−1 andcomplex CPn−1 projective spaces as homogeneous spaces as follows.SU(n,C) acts transitively on CPn−1. Let p0 = Cvn, where v1, . . . , vn isan orthonormal basis of Cn, and let p = Cv, where v has norm 1. Then,as we showed above, by transitivity g(vn) = v for some g. Therefore forthe induced action on CPn−1 we see that g(p0) = p. Similarly, SO(n,R)acts transitively on RPn−1.

What is the isotropy group of p0? In the complex case it is

g ∈ SU(n,C) : g(vn) = λvn.

Clearly such a g is block diagonal and the upper block g′ is unitary. Theonly condition is that λ = 1

det g′ . Since this is no condition on g′, we see

that the isotropy group is U(n−1,C) so SU(n,C)/U(n−1,C) = CPn−1.Similarly, SO(n,R)/O(n− 1,R) = RPn−1.

Corollary 0.4.8. (1) SU(n,C)/U(n− 1,C) = CPn−1

(2) SO(n,R)/O(n− 1,R) = RPn−1

The next two corollaries depend on some results in the followingsection.


Corollary 0.4.9. CPn−1 is a compact, connected and simply connectedcomplex manifold. The simple connectivity follows from the fact thatSU(n,C) is simply connected and U(n− 1,C) is connected. RPn−1 is amerely a compact, connected real manifold.

Corollary 0.4.10. As real manifolds, S3/S1 = S2 (Hopf fibration).

Taking n = 2 in SU(n,C)/U(n − 1,C) = CPn−1 yields S3/S1 =CP 1. Since, as above, CP 1 is a compact, connected and simply con-nected manifold of real dimension 2 it is clearly the 2 sphere, S2.

Grassmann Space: Let V be a real or complex vector space of di-mension n. For each integer 1 ≤ r ≤ n consider the Grassmann spaceG(r, n), the set of all subspaces of V of dimension r. Of course, whenr = 1 we have a real or complex projective space. How can we topol-ogize this space in a convenient way and study it? Evidently GL(V )acts transitively and continuously on G(r, n) by natural action. HenceG(r, n) is a homogeneous space of this group. What is the isotropygroup? If g ∈ GL(V ) fixes an r-dimensional subspace W , i.e. a pointin the Grassmann space, must stabilize W . Hence

g =

(A B0 C

)

and obviously any such g will do. This gives a manifold structure onGrassmann space defined by the quotient structure from GL(V ). No-tice this shows that in the complex case, G(r, n) is actually a com-plex manifold. Also it will follow that G/StabG(W ) is actually com-pact even though G is not. However, we will prove the compactnessof G(r, n) in another way in a moment. But we can already see thatdim(G(r, n)) = r(n− r) (over R respectively C) and that G(r, n) is con-nected. The latter follows from the fact that GL(n,C) is connected andGL(n,R)0, the identity component of GL(n,R), also acts transitivelyin the real case. To see that G(r, n) is compact we need only showthat a compact group acts transitively. We consider a positive definitereal symmetric (respectively Hermitian symmetric) form on V . Choosean orthonormal basis of W and by Gram-Schmidt extend this to an or-thonormal basis of V . If W1 is another r-dimensional subspace of V and


we make a similar construction then there exists an orthogonal operator(respectively unitary operator) taking one orthonormal basis on V tothe other and taking W to W1. Since O(n,R) (respectively U(n,C)) iscompact it follows that G(r, n) is compact.

Exercise 0.4.11. Calculate the isotropy group when O(n,R) (respec-tively U(n,C)) acts.

Flag manifolds: Again let V be a finite dimensional real or complexvector space of dim = n and consider V0 = 0 < V1 . . . < Vn = V ,where dimVi = i. Such a thing is called a flag. Let F(V ) be the setof all flags on V . By choosing a basis for V1, extending this to a basisof V2, and eventually to V , we see that GL(V ) operates transitivelyand continuously on F(V ). The isotropy subgroup is easily seen to bethe group of upper triangular matrices B in GL(V ). Arguing, as in theGrassmann manifold we see that F(V ) is connected and its dimension isn(n−1)

2 . By using the Gram-Schmidt method to get an orthonormal basiscompatible with a flag we see that O(n,R) (respectively U(n,C)) actstransitively. Hence, as before, F(V ) is compact. Notice that this cutsboth ways. We have shown that GL(V )/B is compact without actuallyhaving looked at it. Just as in the case of the Grassmann manifold thisquotient gives a manifold structure on the flag manifold.

Exercise 0.4.12. Generalize both these constructions by considering apartition n1, . . . , ns of n. That is, n = n1 + . . . + ns, where all ni > 0and consider generalized flags made up of s subspaces of ni dimensions.Formulate and prove a result that is analogous to what we have justdone above.

Stiefel manifolds: This is defined as follows. Let V be a finitedimensional real Euclidean or complex Hermitian space and r be aninteger 1 ≤ r ≤ n = dimV . We consider the set SdimV

r of r-frames, by which we mean the set of all orthonormal r-tuples of vec-tors of V . Clearly, O(n,R) (respectively U(n,C)) acts transitively onSr. Since a g ∈ G which fixes an r-frame (it fixes each of the vec-tors which make up the r-frame) must also fix and hence stabilize


the space which they span. Therefore it also stabilizes the ortho-complement. Its restriction to the orthocomplement is something in

O(n−r,R) (respectively U(n−r,C)). Clearly anything of this type canoccur. Thus StabG(x0) = O(n− r,R) (respectively U(n− r,C)), so thatsince G is compact G/StabG(x0) = O(n,R)/O(n − r,R) (respectively

U(n,C)/U(n− r,C)). The dimension of SdimVr (R) as a real manifold is

n(n− 1)/2 − (n− r)(n− r− 1)/2 = r(2 dim V − r− 1)/2, while that ofthe real manifold SdimV

r (C) is n2 − (n− r)2 = r(2 dimV − r).We close this section with an example of a non-compact homoge-

neous space associated with the group G = SL(2,R). Let H+ = z =x+ yi|x ∈ R, y > 0 denote the Poincare upper half plane and

g =

(a bc d

),

with det g = 1. Then, as we show in Chapter 6, G acts transitively andcontinuously on H+ via fractional linear transformations,

g · z =az + b

cz + d,

with Stab(i) = SO(2). Since SL(2,R) is separable, SL(2,R)/SO(2) =H+. As we shall see in Section 1.5 since SO(2) and H+ are both con-nected we have:

Corollary 0.4.13. SL(2,R) is connected.

In the following example we will consider the R matrices, the com-plex case is identical. Evidently GL(n,R) acts transitively on Rn \ 0,by the natural action. However, certain subgroups also act transitivelyon Rn \ 0.Example 0.4.14. For n ≥ 2, SL(n,R) acts transitively on Rn \0 andfor n ≥ 1, Sp(n,R) acts transitively on R2n \ 0. Suppose v 6= 0 ∈ Rn.There are two possibilities. Either v, e1 are linearly independent, orthey are not. If they are, then v = λe1, where λ 6= 0. In this case let

g =

λ 0 00 1λ 0

0 0 I

.

0.5 Lie Algebras 25

Hence g(e1) = v. Otherwise, enlarge this to a basis,e1, v, v3, . . . , vn, and let

g =

0 −1 01 0 00 0 I

,

where I is the identity matrix of order n − 2. Then again, g(e1) = vand in either case det g = 1.

Now suppose G = Sp(n,R) and v ∈ R2n \ 0. Then making thesame choices, but this time with I being the identity matrix of order2n− 2, we see that in the first instance the diagonal matrix g preservesthe symplectic form. In the second instance g is also in the symplecticgroup. In fact, g lies in the compact subgroup U(n,C), of Sp(n,R) andin either case g(v) = e1.

Notice, however, that if G = SO(1, 1), a 1-parameter group of hy-perbolic rotations, then already for dimension reasons G cannot acttransitively on the 2-dimensional manifold, R2 \ 0. Moreover, in gen-eral, under the natural action G = SO(p, q), does not act transitivelyon Rp+q \ 0. This is because G preserves a (p, q)-form for all c ∈ R,and therefore it must preserve all the varieties

Vc = x ∈ Rp+q \ 0 : x21 + . . .+ x2

p − x2p+1 . . .− x2

p+q = c,

and these invariant subvarieties of Rp+q \ 0 are disjoint for differentc’s.

0.5 Lie Algebras

Here we define Lie algebras and give a few examples of them. The basictheory of Lie algebras will be presented in Chapter 3. Let g be a vectorspace over a field k having a zero characteristic.

Definition 0.5.1. We will say that g is a Lie algebra if it possesses ananti-symmetric bilinear product [·, ·] : g × g → g, called the Lie bracket, which satisfies the Jacobi identity,

[[X,Y ], Z] + [[Y,Z],X] + [[Z,X], Y ] = 0


for all x, y and z in g.

This is the moral equivalent of the associative law in the case ofassociative algebras.We shall call h a subalgebra of g if it is a subspace and is closed underthe bracket. Obviously any subalgebra of a Lie algebra is itself a Liealgebra. A Lie algebra with a trivial Lie bracket is called an abelian Liealgebra.

Example 0.5.2. Let V be a finite dimensional vector space over a fieldk as above and let gl(V ) denote the space of all k-Endomorphisms ofV . If n = dimV we usually denote gl(V ) be as gl(n, k).

Let [T, S] = T S − S T , where T and S are in gl(V ) and is thecomposition of maps. Then gl(V ) is a Lie algebra by virtue of the factthat gl(V ) is an associative algebra under .Thus gl(V ) and any of its subalgebras provide a wide class of examplesof Lie algebras. These are called linear Lie algebras.

Exercise 0.5.3. Any associative algebra can be made into a Lie algebraby a similar construction as above. We leave it to the reader to verifythis.

Let g be a Lie algebra andX1,X2, . . . ,Xn be a basis of g and considerthe expansion of [Xi,Xj ] in terms of the basis,

[Xi,Xj ] =n∑

k=1

ckijXk,

and ckij ’s are called structure constants of g with respect to the basisX1,X2, . . . ,Xn.

Exercise 0.5.4. Let g be a vector space of dimension n andX1,X2, . . . ,Xn be a basis. Define [Xi,Xj ] =

∑nk=1 c

kijXk, where ckij ’s are

in the field. Extend [·, ·] bilinearly to g× g. What are the requirementson the ckij ’s in order that g be a Lie algebra?

0.5 Lie Algebras 27

Example 0.5.5. Let g = gl(V ) and consider the basis consisting of thematrices eij . These are the matrices which are 1 in the (i, j) spot andzero elsewhere. A direct calculation shows that

[eij , ekl] = δjkeil − δilejk.

This shows that structure constants with respect to this basis are all 0or ±1. In particular notice that

[eij , eji] = eii − ejj. (5)

Example 0.5.6. A Lie algebra of dimension 1 is evidently abelian. Asfor Lie algebras g of a dimension 2 that is we first show that [g, g] hasdimension less than 1. If dim[g, g] = 2 and X,Y is a basis for g, onesees directly that [g, g] = g = l.s.[X,Y ] and this is a contradiction asl.s.[X,Y ], has dimension 1. Now if g is a Lie algebra of dimension 2then either g is abelian or [g, g] is of dimension 1 generated by a vectorU . In this case we can take U as the first element of a basis U, V .Then [V,U ] = λU where λ 6= 0. So [1/λV,U ] = U . Thus there is a basisX,Y with [X,Y ] = Y . This Lie algebra is called the ax+ b-Lie algebra.As we shall see it is solvable.

Example 0.5.7. Let V be a finite dimensional vector space over a fieldk which is equipped with a nondegenerate symmetric bilinear form (·, ·).For A ∈ End(V ) = gl(V ), one can consider the endomorphism At suchthat for all u and v in V ,

(Au, v) = (u,At).

Since the bilinear is nondegenerate At is well-defined. At is called theadjoint or transpose of A. The following properties can be easily veri-fied:

(1) (xA + yB)t = xAt + yBt or in other words taking adjoint is alinear operator.

(2) (At)t = A.

(3) (AB)t = BtAt.

(4) [A,B]t = −[At, Bt].


A ∈ End(V ) is said to be a symmetric operator if A = At. Similarly itis said to be a skew symmetric operator if A = −At. We denote the setof skew symmetric operators by k and that of symmetric operators by p.It follows from property (1) that k and p are linear subspaces of gl(V ).We have k ∩ p = 0, since A = At and A = −At implies that A = 0. Onthe other hand for any A ∈ gl(V ) one can write,

A =A−At

2+A+At

2.

It follows from (1) and (2) that A−At

2 ∈ k and A+At

2 ∈ p. Hence,gl(V ) = k ⊕ p. From (4) it follows easily that

(1) [k, k] ⊆ k

(2) [k, p] ⊆ p

(3) [p, p] ⊆ k

These relations are called the Cartan relations and will play an impor-tant role in symmetric spaces. In particular k is a subalgebra of gl(V ).When V has dimension n one also writes o(n, k) for k. Since in o(n, k),the trace of every element of o(n, k) is zero, we sometimes also writeso(n, k).

Similar arguments apply in the case of a vector space V = Cn overthe complex field and a Hermitian form 〈·, ·〉. Then we get the real Liealgebra of skew hermitian operators on V = (Cn, 〈·, ·〉) which we denoteby u(n). We leave it to the reader to check that this is not a complexLie algebra.

Exercise 0.5.8. For positive integers n, p and q where p + q = n,consider the matrix A = diag(1, . . . , 1︸︷︷︸

p-times

,−1, . . . ,−1︸︷︷︸q-times

). Prove that o(p, q) ⊂

gl(n,R) and consists of matrices X satisfying

XA+AXt = 0

form a subalgebra of gl(n,R). Prove that so(p, q) ⊂ o(p, q), matrices oftrace zero, form an ideal of o(p, q).

0.5 Lie Algebras 29

Definition 0.5.9. Let g and h be Lie algebras over the same field andf be a k-linear map from g to h. We shall say f is a Lie homomorphismif f([X,Y ]) = [f(X), f(Y )] for all X and Y in g. If the Lie homomor-phism is bijective we shall call it an isomorphism and say that g and h

are isomorphic Lie algebras. An isomorphism f : g → g is called auto-morphism and Aut(g) denotes the set of automorphisms of g. Clearlyunder composition Aut(g) forms a subgroup of GL(g)

Example 0.5.10. It follows from the Jacobi identity that for X ∈ g,adX : g → g, defined by

adX(Y ) = [X,Y ],

is a Lie homomorphism

Evidently isomorphic Lie algebras share all Lie algebra theoreticproperties. If h is a subspace of g, then the inclusion map from h to g

is a Lie homomorphism if and only if h is a subalgebra.

Definition 0.5.11. A Lie homomorphism ρ : g → gl(V ) is a calledLie algebra representation. The dimension of the representation is thedimension of V and V itself is called the representation space of ρ. TheLie algebra representation ρ : g → gl(V ) is said to be faithful if it isinjective. For every X ∈ g we denote the map ρ(X) : V → V by ρX .

Obviously the inclusion map of a linear subalgebra of gl(V ) is a Lierepresentation.

Example 0.5.12. Let g be a Lie algebra. The map ad : g → gl(g)defined as follows:

adX : g → g

adX(Y ) = [X,Y ]

is easily seen to be a Lie algebra representation, called the adjoint rep-resentation. The image of ad, denoted ad g, is called adjoint algebraand is clearly a linear Lie algebra. The kernel of ad is z(g), the centerof g.


Exercise 0.5.13. Let g ⊂ gl(V ) be a linear Lie algebra and X ∈ g.Then the eigenvalues of adX are all of the form λi − λj where λi andλj are eigenvalues of X. Hint: One may assume that the field is alge-braically closed. Then use the Jordan form.

Chapter 1

Lie Groups

In this chapter we define a Lie group and study its basic properties.

1.1 Elementary Properties of a Lie Group

We first give the definition of a smooth invariant vector field on a Liegroup. As in Appendix A, a smooth vector field X can be regarded asa first-order differential operator, Xg, each operating on the space ofsmooth functions on G. As in the page on Notation, TgG denotes thetangent space to G at g and dgf is the derivative of a map f at g ∈ G.

Definition 1.1.1. Let G is a Lie group and X a smooth vector field onG. We say X is left invariant if d(Lh)Xg = Xhg for all h and g ∈ G.Here Lg denotes left translation by g ∈ G and d(Lg) its derivative actingon tangent spaces.

We want to describe all such vector fields. Since the Lg act simplytransitively on G it follows that a left invariant vector field is completelydetermined by X1 ∈ T1G and, conversely, that any v ∈ T1G determinesa unique left invariant vector field defined by Xg = d(Lg)1v ∈ TgG. Thisvector field is evidently smooth. Also, since Lhg = LhLg, we have leftinvariance, dgLhXg = (dgLh)d1(Lg)v = d1(Lhg)v = Xhg.

Thus the linear map X 7→ X1 is a vector space isomorphism fromthe left invariant vector fields onto T1G which we call the Lie algebra

31

32 Chapter 1 Lie Groups

g of G, and hence g is a finite dimensional subspace of the space of allvector fields of dimension = dimG. Since G0, the identity componentof G, is open in G, it follows that T1G = T1G0. Therefore, because ofthis vector space isomorphism, the Lie algebras of G0 and G coincide.Here g is an invariant of G; that is, it depends intrinsically on G.

Now the left invariant vector fields form a subalgebra of the Liealgebra of all vector fields (see Appendix A). For let X and Y be leftinvariant vector fields on G.

d(Lh)[Xg, Yg] = d(Lh)(XgYg − YgXg) = d(Lh)(XgYg) − d(Lh)(YgXg)

= XhgYhg − YhgXhg = [Xhg, Yhg]

Hence [X,Y ] is again left invariant and a vector field. We now cometo a concept that is of fundamental importance to our subject, namely,the exponential map and 1-parameter subgroups.

Definition 1.1.2. Let G be a Lie group. We say φ : R → G is a1-parameter subgroup of G if φ is a smooth homomorphism.

X

exp(tX)

1

Figure 1.1: 1-parameter subgroups

For example if G = Rn and φ : R→ Rn is a smooth homomorphism,then φ takes 0 to 0. Hence its derivative d0φ is a linear map of the linearspaces d0φ : R → Rn. Therefore, there is a vector X = (x1, . . . xn) so

1.1 Elementary Properties of a Lie Group 33

that d0φ(t) = tX. Identifying the Lie groups R and Rn with theirrespective tangent spaces at 0 we get φ(t) = tX for all t ∈ R (see Figure1.1).

We can handle Tn similarly. If φ : R → Tn is a smooth homomor-phism and π : Rn → Tn is the universal covering (see Section 0.3),then φ lifts to a smooth homomorphism φ : R → Rn so that πφ = φ(see Proposition 0.3.3). Since, as above, φ(t) = tX, it follows thatφ(t) = π(tX) for all t ∈ R. This example shows that in order to findthe 1-parameter subgroups of a Lie group G it is sufficient to do it forthe universal covering group and apply the projection (see Section 0.3).

Finally, we consider 1-parameter subgroups of the non-abelian Liegroups, G = GL(n,R) or GL(n,C). As we shall see, this is more com-plicated than the examples given above. Let X be a tangent vector (i.e.an n × n real (respectively complex) matrices) to I = In×n of G. Thisis because G is open in the space of matrices. Then φ(t) = Exp tX isa 1 parameter group. Here Exp is the power series applied to matricesA ∈Mn(C).

ExpA =

∞∑

k=0

Ak

k!

Now this series is absolutely convergent and uniformly on compacta,therefore it defines an entire holomorphic functionMn(C) →Mn(C). To

see this consider the finite partial sums∑m

k=0Ak

k! . Calculating the normwe get

‖m∑

k=n

Ak

k!‖ ≤

m∑

k=n

‖A‖kk!

.

Since the series∑∞

k=0‖A||kk! converges, the sequence of partial sums is a

Cauchy sequence and by the completeness of Mn(C), we see that ExpAconverges absolutely and ‖ExpA‖ ≤ e||A||, for A ∈ Mn(C). Moreover,the same argument shows that the series for Exp converges uniformlyon compacta.

Moreover if A and B commute then ExpA · ExpB = Exp(A + B).Therefore, since A and −A commute ExpAExp(−A) = Exp 0 = Iso that ExpA is always invertible and its inverse is Exp(−A). Thus


Exp : Mn(C) → GL(n,C). Moreover, tX and sX commute for all t ands. It follows that φ(t) = Exp tX is a 1-parameter group.

The proof of this is quite similar to that of the functional equationfor the ordinary numerical exp. Because the convergence is absolute wemay perform rearrangements by the Weierstrass theorem,

ExpAExpB = (

∞∑

k=0

Ak

k!)(

∞∑

l=0

Bl

l!) =

∞∑

k,l=0

AkBl

k!l!.

On the other hand,

Exp(A+B) =

∞∑

p=0

(A+B)p

p!

and since A and B commute it follows from the binomial theorem that

(A+B)p

p!=

p∑

j=0

Aj

j!

Bp−j

(p − j)!.

Hence Exp(A+B) = ExpAExpB.Conversely, suppose φ(t) is a 1-parameter group. Then for all s

and t ∈ R, φ(s + t) = φ(s)φ(t). Differentiating with respect to s ats = 0 gives φ′(t) = φ′(0)φ(t), for all real t. Also φ(0) = I. This is afirst-order linear matrix differential equation with constant coefficients(or a system of such numerical equations) and hence has the uniqueglobal solution φ(t) = Exp tX where X is the tangent vector, φ′(0). Forclearly by absolute and uniform convergence we can differentiate φ(t) =Exp tX term by term and get φ′(t) = Xφ(t). Thus Exp tX satisfies thedifferential equation on all of R and φ(0) = I. If another function ψ didthis then d

dt(Exp−tX)ψ(t)) = Exp(−tX)ψ′(t)+−X(Exp(−tX))ψ(t) =Exp(−tX)(X+−X)ψ(t), since Exp−tX andX commute and ψ satisfiesthe differential equation. Since this is zero, Exp−tXψ(t) is a constant.Evaluating at t = 0 shows the constant is I. Hence ψ(t) = Exp(tX).Thus, once again, the 1-parameter groups are uniquely determined bya tangent vector at the origin, but this time by an ordinary differentialequation rather than a linear algebraic one. As we shall see, whenproperly understood, these examples are typical.


Exercise 1.1.3. Prove that for an n×n matrix X and for m a positiveinteger, one has ExpX = limm→∞(I + X

m )m.

Finally, notice also that the derivative of Exp at zero is the iden-tity, This is because the power series expression for Exp(X) shows that

Exp(X) = Exp(0) + (X − 0)(I) + (X − 0)(X2! + X2

3! + . . .). By the linear

approximation theorem if X2! +X2

3! +. . . tends to 0 as X → 0, we conclude

d0 Exp = I. But ||X2! + X2

3! + . . . || ≤ ||X||2! + ||X||2

3! + . . ., which definitelytends to zero since it is the tail of et, t ∈ R, which is differentiable at 0with derivative 1 so that limt→0(

t2! + t2

3! + . . .) = 0.

We include two other useful facts about Exp and linear Lie groupshere. Namely, for any complex n × n matrix A and P ∈ GL(n,C),P Exp(A)P−1 = Exp(PAP−1). That is, Exp commutes with conju-gation. This is because conjugation by P is an automorphism of theassociative algebra Mn(C). Hence for any positive integer j, and anyconstant c ∈ C we have PcAjP−1 = c(PAP−1)j . Hence for any polyno-mial p(A) =

∑cjA

j we get Pp(A)P−1 = p(A)(PAP−1). Taking limitsgives the result. Actually, we see that this holds for any absolutely con-vergent power series. That is, if f(z) is an entire function then f(A)commutes with conjugation.

Secondly, for any complex n × n matrix A, det(ExpA) = etr(A).When A is triangular with eigenvalues λ1, . . . , λn, then a direct calcu-lation shows that ExpA is also triangular with eigenvalues eλ1 , . . . , eλn .Hence det(ExpA) = eλ1 . . . eλn = eλ1+...+λn = etr(A). In general, we canapply the 3rd Jordan canonical form to get PAP−1 triangular. ThenP Exp(A)P−1 = Exp(PAP−1). The determinant and trace of this tri-angular matrix is the same as that of ExpA. Hence the result.

Let R be the additive group of real numbers, considered to be pa-rameterized by t and let its Lie algebra be generated by the tangentvector field Dτ = ∂

∂t |t=τ . In this way we can identify R with its Lie al-gebra. Let G be a Lie group, g be its Lie algebra, and suppose t 7→ f(t)is a smooth curve in G defined everywhere on R. From now on we shallwrite f ′(τ) = dτf(Dτ ). Thus by evaluating at τ = 0 we get a vectorf ′(0) ∈ T1G.

For X ∈ T1G consider the associated invariant vector field Xg on G.


Let φX(t) be an integral curve for this vector field, which passes through1 at t = 0. We check that t 7→ φ(t) is a group homomorphism, that is,φX(t+ s) = φX(t)φX(s). Because of the left invariance of X the curvess 7→ φX(t + s) and s 7→ φX(t)φX(s) are both integral curves passingthrough φX(t) at s = 0. Therefore, by, the uniqueness of solution ofODEs, they are equal in a neighborhood of s = 0. Then Proposition1.1.4 below shows that they are equal every where. Conversely if f wereany smooth homomorphism R → G and we take the derivative, thenf ′(0) = X ∈ g. Since f(0) = 1 = φX(0) and φX(0) = X = f ′(0).By uniqueness of local solutions to ODE’s, this means that f = φXin some neighborhood of zero. By Proposition 1.1.4 below, smoothhomomorphisms which agree in a neighborhood of zero must coincideon all of R.

In fact more generally one has,

Proposition 1.1.4. Let G be a connected Lie group, H be a Lie groupand f and g be globally defined smooth homomorphisms G → H whichcoincide in a neighborhood U of 1 in G. Then f ≡ g. (See a previousexercise on how U generates H.)

Proof. Because G is connected, the symmetric neighborhood V = U ∩U−1 generates G. Then by Proposition 0.1.13 G =

⋃∞n=1 V

n. Since fand g are homomorphisms which agree on V , they agree on V n for everyn.

Thus we get,

Proposition 1.1.5. There is a bijection X 7→ φX from g to the set ofall smooth 1-parameter subgroups of G, subject to the requirement thatφ′X(D) = X for X ∈ g.

Notice, however, that if s ∈ R is fixed and t 7→ φX(t) is a 1-parametergroup, then t 7→ φX(st) = f(t). Since f ′ = sX, we see by the injectivityof the correspondence that φsX(t) = φX(st), s, t ∈ G, X ∈ g. We cannow define the exponential map exp : g → G of a Lie group G.

Definition 1.1.6. For X ∈ g we define exp(X) = φX(1).


Since for all real t, exp(tX) = φX(t), by Proposition 1.1.5 above wecan identify all the 1-parameter subgroups of G. Also, exp(0) = 1. Byconnectedness all 1-parameter subgroups lie in G0 so the range of expis in G0. Therefore exp does not really help much in non-connected Liegroups and for this reason one often simply assumes one is working witha connected Lie group.

Corollary 1.1.7. The 1-parameter subgroups of G are precisely themaps t 7→ exp(tX), for X ∈ g.

This, together with Definition 1.1.6, enable us to determine the ex-ponential map in the case of the various examples discussed above:

The exponential map for Rn is the identity, for Tn it is π and forGL(n,C) or GL(n,R) it is Exp. This is the reason the exponential maphas its name.

Corollary 1.1.8. The exponential map is smooth and its derivative at0 is the identity, i.e. d0 exp = I. Since exp is smooth, the inversefunction theorem tells us that it is a local diffeomorphism of a ball about0 in g with a neighborhood of 1 in G. We shall call its local inverse log.

Proof. Let TG denote the tangent bundle of G. The map, (g,X) 7→d1Lg(X), going from G × g → TG is smooth. Now φX(t) = exp tXis the integral curve of the vector field Xg = d1Lg(X) with the initialdata φX(0) = 1 and exp(X) = φX(1). So it follows from the smoothdependence on initial data of solutions of ODEs that exp is itself smooth(see Appendix A). The directional derivative in the direction X ∈ g isddt(exp(tX))|t=0 = X. Therefore d0 exp = idg.

Notice that all of our discussion works equally well for complex Liegroups; just substitute complex Lie algebras for g, complex 1-parametergroups (namely, z 7→ exp(zX), z ∈ C) for φX , and the simply connectedgroup C for R. Here again G0 is also open in G so these groups havethe same Lie algebra.

Exercise 1.1.9. If the Lie algebra of a real Lie group G is a complexLie algebra then G is a complex Lie group.


Proposition 1.1.10. Let G be a complex connected Lie group and ρa holomorphic representation of G on the complex vector space V . Ifρ(G) is bounded, then it is trivial.

In particular, a compact, complex connected Lie group must beabelian. This follows by taking for ρ the adjoint representation. ThenAdG is trivial and hence G = Z(G) is abelian. (In fact it is a torus ofeven dimension.)

Proof. To prove this we may replace G by the complex connected sub-group of GL(V ) and show G is trivial. Consider the 1-parameter sub-group Exp(zX), where X ∈ g. This is a bounded entire function ofz ∈ C taking values in the finite dimensional vector space EndC(V ).Applying Liouville’s theorem to each of the finitely many numericalcoordinate functions, we conclude Exp(zX) is constant (see [55]). Eval-uating at z = 0 tells us this constant is I.) Taking the derivative d

dzat z = 0 shows X = 0. Since X was arbitrary g = 0 and since G isconnected G = I.

Exercise 1.1.11. Prove that if a homomorphism on a connected Liegroup is smooth in a neighborhood of 1 then it is smooth everywhere.

Proposition 1.1.12. Let X1, . . . ,Xn be a basis for g. Then for suitablysmall ti’s the map p : (t1, . . . , tn) 7→ expG(t1X1) · · · expG(tnXn) is adiffeomorphism onto an open neighborhood of 1 ∈ G.

Proof. This follows from Corollary 1.1.8 and the fact that the deriva-tive of p at (0, . . . , 0) is a block diagonal matrix of the derivatives ofexp(tiXi)’s, therefore it is the identity map and the conclusion followsfrom the inverse function theorem.

A similar argument proves

Corollary 1.1.13. Suppose G is a Lie group and its Lie algebra, g, isthe direct sum of subspaces, a1 ⊕ . . .⊕ aj. Then we can find small ballsUa1 . . . , Uaj

about 0 in a1 . . . , aj such that (a1, . . . aj) 7→ exp a1 . . . exp ajis a diffeomorphism.

1.2 Taylor’s Theorem and the Coefficients of expX expY 39

Corollary 1.1.14. Let G be a connected Lie group and H any Liegroup. Then a continuous homomorphism f : G→ H is smooth.

Proof. We note that this is true for a 1-parameter subgroup φ : R→ H.Since φX(t) = expH(tX), for X ∈ h and expH is smooth this istrue. Now let X1, . . . ,Xn be a basis of g. Then for each i, ti 7→f(expG(tiXi)) is a smooth 1-parameter subgroup ofH. Hence for each i,f(expG(tiXi)) = expH(tiYi), where Yi ∈ h. Since f is a homomorphismf(expG(t1X1) · · · expG(tnXn)) = expH(t1Y1) · · · expH(tnYn). By Corol-lary 1.1.12, for small ti’s p : (t1, . . . , tn) 7→ expG(t1X1) · · · expG(tnXn) isa diffeomorphism onto a small neighborhood U of 1 in G and we havef p(t1, . . . , tn) = expH(t1Y1) · · · expH(tnYn) which is smooth as expH issmooth. Therefore f is smooth in a neighborhood of 1 and by Exercise1.1.11 is smooth everywhere.

1.2 Taylor’s Theorem and the Coefficients of

expX expY

We first deal with Taylor’s theorem on a Lie group. Throughout thissection X, Y , . . . denotes the left invariant vector fields associated toX,Y . . . ∈ g. Consequently X, Y . . . act on C∞(G) as first-order differ-ential operators,

(X.f)(x) = dxf(X(x)).

Proposition 1.2.1. Let G be a Lie group with Lie algebra g. SupposeX ∈ g and f is a smooth function on G. Then for every positive integerm and g ∈ G,

Xmf(g exp(tX)) = (X · · · Xf)(g exp(tX)) =dm

dtm(f(g exp(tX)).

Moreover, for each positive integer m,

f(exp(X)) =

m∑

k=0

1

k!Xkf(1) +Rm(X),

where ||Rm(X)|| ≤ cm||X||n+1 and cm is a positive constant dependingonly on m.


In particular, taking m = 1 and t = 0 gives Xf(g) =ddt(f(g exp(tX))|t=0, for each g ∈ G.

Proof. To prove the first equation we may assume g = 1 by replacingthe C∞ function f by fg and using left invariance of X . So for the firstequation it remains to show that Xmf(exp(tX)) = dn

dtn (f(exp(tX)).

This is obvious from the definition of Xf .

Turning to the second equation, we consider the mth order Taylorexpansion of f(exp(tX)) about t = 0 with the integral remainder andevaluate at t = 1.

f(exp(X)) =m∑

k=0

1

k!Xmf(1)) +Rm(X),

where Rm(X) = 1(m+1)!

∫ 10 (1 − s)m dm+1

dsm+1 (f(exp(sX))ds. By the

first equation Rm(X) = 1(m+1)!

∫ 10 (1 − s)mXm+1f(exp(sX))ds. Let

X1, . . . ,Xn be a basis of g and write X =∑n

i=1 xiXi. ThenXm+1 = (

∑ni=1 xiXi)

m+1 which is a finite sum of products of Xi oforder m + 1 indexed by the various partitions of m + 1 into n partswith coefficients, the product of the corresponding xi’s. Now each ofthese coefficients is ≤ ||X||n+1. Since (1 − s)m ≥ 0, letting dn,m be thenumber of partitions and using the Banach algebra properties of || · ||and the fact that f(exp(sX)) is bounded, say, by c, on the interval [0.1]

we get ||Rm(X)|| ≤ ||X||m+1

(m+1)! cdn,m∫ 10 (1 − s)mds.

In what follows O(tk) indicates any smooth function of t in a sym-

metric interval about 0 with the property that O(tk)tk

remains boundedat t→ 0. We now come to a key lemma which can be considered as thesecond order approximation to the Baker-Campbell-Hausdorff (BCH)formula which reads

exp(X) exp(Y ) = exp(A+B + C2(A,B) + C3(A,B) + · · · ) (1.1)

where each Cn(A,B) is a finite linear combination of the expres-sions [X1, [X2[· · · [Xn−1,Xn] · · · ]]] = (adX1)(adX2) · · · (adXn−1)Xn,


for Xi = A or B and when A and B are sufficiently close to the identity.The remarkable fact is that Hn does not depend on G, A or B and itscoefficients are rational. For instance

C2(A,B) = 12 [A,B]

C3(A,B) = 112 [A, [A,B]] + 1

12 [B, [B,A]]

C4(A,B) = − 124 [A, [B, [A,B]]].

(1.2)

Since every Lie group is locally isomorphic to a subgroup of someGL(k,R) and the exponential map of a subgroup is the restriction ofthe exponential group of the group, one therefore only has to verify thisformula for GL(k,R); it turns out that the formula for exp(X) exp(Y )in GL(k,R) is independent of k.

Lemma 1.2.2. For X and Y ∈ g and t ∈ R we have

(1) exp(tX) exp(tY ) = exp(t(X + Y ) + 12 t

2[X,Y ] +O(t3))

(2) exp(tX) exp(tY ) exp(−tX) = exp(tY + t2[X,Y ] +O(t3))

(3) exp(tX) exp(tY ) exp(−tX) exp(−tY ) = exp(t2[X,Y ] +O(t3)).

We remark that the third relation gives a geometric interpretationof the bracket: [X,Y ] is the tangent vector at 1 to the curve t 7→exp(

√tX) exp(

√tY ) exp(−

√tX) exp(−

√tY ), t ≥ 0. It also shows that

if, for all small t, exp(tX) and exp(tY ) commute in G then [X,Y ] = 0.In particular, if for all small t, exp(tX) and exp(tY ) commute then by(1), we get,

exp(tX) exp(tY ) = exp(t(X + Y )). (1.3)

As a result exp(tX) and exp(−tX) commute and are mutual inversesof one another.

Another remark to be made is that the first relation impliesexp(tX) exp(tY ) = exp(t(X+Y )+O(t2)). This means that the tangentvector at 0 to the curve t 7→ exp(tX) exp(tY ) is X + Y .

Proof. Let f be a smooth function defined in a neighborhood of 1and X ∈ g. By Proposition 1.2.1 for n ≥ 0, (Xnf)(g exp tX) =


dn

dtn f(g exp tX). Hence (XnY mf)(1) = dn

dtndn

dsn f(exp tX exp sY )t=0,s=0.Therefore,

f(exp(tX) exp(sY )) =∑ ∑

n≥0,m≥0

tn

n!

sm

m!(XnY m)f(1).

On the other hand since exp is smooth and invertible in a neighborhoodof 0 and group multiplication is smooth we see that for |t| sufficientlysmall exp(tX) exp(tY ) = exp(Z(t)) where Z(t) is a smooth functionZ : U → g, and U is a symmetric interval about 0. Evidently Z(0) = 0.Taking the Taylor expansion of the second order of Z(t) about t = 0gives Z(t) = tZ1+t2Z2+O(t3), where Z1 and Z2 are constants in g. LetX1, . . . ,Xn be a basis of g and f be any of the coordinate functionsexp(x1X1+ · · ·+xnXn) 7→ xi. Then f(expZ(t)) = f(exp(tZ1+t2Z2))+O(t3). But f(exp(tZ1 + t2Z2)) =

∑∞n=0

1n!(tZ1 + t2Z2)f(1). Hence as

above,

f(expZ(t)) =

∞∑

n=0

1

n!(tZ1 + t2Z2)f(1) +O(t3). (1.4)

Whereas,

exp(tX) exp(sY ) =∑

n≥0,m≥0

tn

n!

sm

m!(XnY m)f(1). (1.5)

Letting s = t in (1.5) and comparing coefficients with (1.4) yields Z1 =X+Y and Z2 + 1

2Z21 = 1

2 (X2 +2XY +Y 2). Therefore 2Z2 +(X+Y )2 =X2 + 2XY + Y 2 and since (X + Y )2 = X2 + XY + Y X + Y 2 we seethat Z2 = 1

2 [X,Y ]. This proves (1).

Part (3) follows by applying (1) twice. To prove (2) observe thatsince exp(tX) exp(tY ) exp(−tX) exp(−tY ) = exp(t2[X,Y ] + O(t3)) weknow that exp(tX) exp(tY ) exp(−tX) = exp(t2[X,Y ] +O(t3)) exp(tY ).

Reasoning as before this is exp(tY + t2([X,Y ] + Y 2

2 ) +O(t3)). But alsoas before,

exp(tX) exp(tY ) exp(−tX) = expZ(t),

where Z(t) is again a smooth function with Z(0) = 0 and Z(t) = tZ1 +


t2Z2 +O(t3). Then as above,

exp(tZ1 + t2Z2)) +O(t3) = exp(tY + t2[X,Y ] +Y 2

2+O(t3))

and comparing coefficients we get Z1 = Y and Z2 + Y 2

2 = [X,Y ] + Y 2

2so Z2 = [X,Y ].

From (1) and (3) and the continuity of exp we conclude

Corollary 1.2.3. Let G be a Lie group. Then for X and Y ∈ g, andsufficiently small t ∈ R and n ∈ Z we have

(1) exp(t(X + Y )) = limn→∞(exp( 1n tX) exp( 1

n tY ))n

(2) exp(t[X,Y ]) = limn→∞[exp( 1n tX), exp( 1

n tY )]n2.

Corollary 1.2.4. Let G be a Lie group with Lie algebra g and H be a Liesubgroup. Then the Lie algebra of H is X ∈ g : exp(tX) ∈ H for all t.

Proof. Calling this set S we see immediately that h ⊆ S. On the otherhand if X ∈ g has the property that the whole curve is in H, then itstangent vector at t = 0 lies in T1(H). Thus h ⊆ S ⊆ T1(H). Hence theresult.

Exercise 1.2.5. Show that exp(tX) ∈ H for all small t then forexp(tX) ∈ H for all t ∈ R.

We now give an example of a general type of group which will havea certain importance.

Definition 1.2.6. One calls a subgroup G ⊆ GL(n,C) an algebraicgroup if it is the simultaneous zero set within gl(n,C) of a family of poly-nomials with complex coefficients in the xij coordinates of the matricesin gl(n,C). Clearly, such a group is a closed subgroup of GL(n,C) in theusual Euclidean topology and hence is a Lie group by a theorem of E.Cartan, Theorem 1.3.5. Furthermore, we shall call GR = G ∩ GL(n,R)the R-points or the real points of G. Similarly, GR of an algebraic groupG is also a Lie group. If the family of polynomials defining G happensto have all its coefficients lying in some subfield F of C, we then say


that G is defined over F . A group G is said to be essentially algebraicif it is either an algebraic group or it has finite index in the real pointsof an algebraic group.

Typical examples of algebraic groups are GL(n,C) itself (the emptyset of polynomials) and SL(n,C) itself (the single polynomial det−1 =0). These groups are defined over Q. The respective real points areGL(n,R) and SL(n,R). If V is a finite dimensional vector space overC and we have a nondegenerate bilinear form β : V × V → k. LetGβ = g ∈ GL(V ) : β(gv, gw) = β(v,w) for all v,w ∈ V . It is obviousthat Gβ is a subgroup of GL(V ) which is algebraic. These evidentlyinclude O(n,C) and Sp(n,C) with real points O(n,R) and Sp(n,R) (see[15] for the geometric significance of these groups.)

Exercise 1.2.7. Prove that in fact Gβ is an algebraic group definedover Q.

By Cartan’s theorem, Theorem 1.3.5, Gβ is a Lie subgroup. Wecompute its Lie algebra, gβ. By our criterion gβ is X ∈ End(V ) :β(Exp(tX)v,Exp(tX)w) = β(v,w). Calculating the derivative at t = 0tells us that β(v,Xw) + β(Xv,w) = 0. Because Exp is faithful on aneighborhood of the identity and ExpXt = (ExpX)t, the converse isalso true. We leave the details to the reader.

When β is the symplectic form, a 2n × 2n matrix g preserves β ifand only if gtJg = J , where J is the 2n × 2n matrix consisting of thefollowing n× n blocks:

J =

(0 I−I 0

).

This description makes it more convenient to calculate things. Forexample, it shows easily that Sp(1,C) = SL(2,C), but for higher n,Sp(n,C) 6= SL(2n,C). The Lie algebra, sp(n,C), of Sp(n,C) is thesubalgebra of the 2n × 2n matrices, M2n(C), consisting of

X =

(X1 X2

X3 X4

),

and satisfying XtJ + JX = 0. This means X2 and X3 are symmetricwhile X1 = −Xt

4. Thus X4 is arbitrary, X1 is determined by X4 and


X2 and X3 have n(n+1)2 free parameters each. Hence the Lie algebra

sp(n,C) of Sp(n,C) has complex dimension 2n2 +n. Similarly sp(n,R),the Lie algebra of Sp(n,R), has real dimension 2n2 + n.

1.3 Correspondence between Lie Subgroups

and Subalgebras

In this section we characterize the Lie subalgebras of the Lie algebra ofa Lie group. In fact we show that there is a one-to-one correspondencebetween Lie subalgebras of g = Lie G and connected Lie subgroups ofG.

A k-dimensional distribution D on a smooth manifold M is a choiceof k-dimensional subspace D(m) of TmM for each m ∈ M . We shallsay D is smooth if each m ∈M has a neighborhood U with vector fieldsX1, . . . ,Xk defined on U which span D at all points m ∈ U . A vectorfield X is in D if X(m) ∈ D(m) for all m ∈M .

A natural class of smooth distributions is given by foliations, moreprecisely if F is a foliation of M with leaves Fi then let D(m) = TxFi,the tangent space at the leaf containing m. These types of distributionsare called integrable i.e. for every m ∈ M there is a submanifold Nwhich passes through m and TmN = TmM |N = D(m); the submanifoldN is said to be an integral manifold for D.

One would like to characterize such distributions and Frobenius’theorem addresses this matter.

Definition 1.3.1. A smooth distribution D is called involutive if theset of vector fields in D is closed under Lie bracket.

Obviously integrable distributions are involutive as the vector fieldson a submanifold (or any manifold) form a Lie algebra.

Theorem 1.3.2. An involutive distribution on a smooth manifold Mis integrable. Moreover, for every m ∈ M there is a unique connectedmaximal integral manifold containing m and the set of maximal integralmanifolds of D gives rise to a foliation of M .


This is known as Frobenius’ theorem [80].

Here is one of the main results in this section

Theorem 1.3.3. Let G be a Lie group with Lie algebra g. Then thereis bijection between the connected subgroups of G and the subalgebras ofg.

Proof. Suppose that H is a connected Lie subgroup of G and X, Y aretwo left invariant vector fields on H which correspond to the vectorsX,Y ∈ h = T1H ⊂ g = T1G. By the definition of the Lie bracket in g

we have

[X, Y ](g) = dLg[X,Y ] (1.6)

for all g ∈ G which uniquely determines [X,Y ] ∈ g. By restricting (1.6)to H we observe that [X,Y ] has to be the Lie bracket of X and Y in h,in particular [X,Y ] ∈ h.

Conversely, a subalgebra h of g is indeed the Lie algebra of a uniqueconnected subgroup of G. Consider the smooth distribution D on Gwhich consists of all the left invariant vector fields generated by thevectors in h. Then D is involutive by (1.6) and due to the assumptionthat h is a subalgebra. Hence by Frobenius’ theorem there is a maximalconnected integral manifold H which contains 1 ∈ G. We claim that His subgroup and to prove that it suffices to show that xH = H for allx ∈ H. Note that xH is also an integral manifold since the distributionD is left invariant. On the other hand, since 1 ∈ H therefore x ∈ xHand by the uniqueness part of the Frobenius’ theorem we have H = xH.

Now we must show that H is a Lie group, or equivalently that τ :H ×H → H, the map given by τ(x, y) = x−1y is smooth. Consider thediagram

H ×Hτ

##HHH

HHHH

HH

τ ′ // G

H

i

OO

Obviously τ is continuous. Note that τ ′ : H ×H → G is smooth as theinclusion H × H → G × G is smooth. Now we are going to introducea chart for H which makes τ smooth. Suppose that G has dimension


k+n and k is the dimension of h. By Frobenius theorem, for each x ∈ Hthere is a chart φ : U → Rk+m

U ∩H = φ−1((x1, . . . , xk+m)|xk+1 = · · · = xk+m = 0).

Consider V = U ∩H and π : Rk+m → Rk the projection on the firstk coordinates, then (V,ψ = π φ i) is the desired chart. We have

ψ τ = π φ i τ = π φ τ ′

which is smooth as τ ′ and π are.For uniqueness, let K be another such connected Lie group. As K

is an integral manifold then there is smooth inclusion K ⊂ H. SinceT1H = T1K = h the inclusion is a local isomorphism thus surjectivewhich proves that K = H.

We now come to Cartan’s theorem mentioned earlier. For that weneed the following lemma.

Corollary 1.3.4. Let H be a closed subgroup of a Lie group G andh = X ∈ g : exp(tX) ∈ H for all t ∈ R. Then h is a Lie subalgebraof g.

Proof. Clearly h is closed under scalar multiplication. Since H is aclosed subgroup of G, Corollary 1.2.3 part 1, shows h is closed underaddition, while part 2 shows h is closed under bracketing.

Theorem 1.3.5. A closed subgroup H of a Lie group G is a Lie groupwith relative topology.

Proof. It is enough to prove the theorem when H is a connected sub-group. Consider the Lie subalgebra h as it is defined in Corollary1.3.4. Then by Theorem 1.3.3 there exists a Lie subgroup H ′ ⊂ Gsuch that T1H

′ = h. Let s be a complementary subspace for h in g

such that g = h ⊕ s. Let U and V be two sufficiently small neighbor-hoods of 0 in h and s such that the restriction of expG to U × V is adiffeomorphism onto its image (see Corollary 1.1.13). We prove thatH ∩ expG(U ×V ) = expU . If exp(X +Y ) ∈ H, X ∈ U and Y ∈ V then


by Corollary 1.2.3, expY = limn→∞(exp( 1n(X + Y )) exp(− 1

nX))n is inH as H is a closed subgroup, hence Y ∈ h ∩ s = 0. Consequently,expU is an open set in H. On the other hand expU is an open setin H ′ as expG |h = expH′ . Therefore, H =

⋃∞n=1 U

n = H ′ as both areconnected.

1.4 The Functorial Relationship

We next turn to functoriality questions. We will deal with the real case,but this works just as well in the complex case. For a Lie homomorphismf : G→ H we denote the derivative of f at the identity element 1 by

f ′ = d1f : g → h

which is a linear map. Since d0 exp = id, one can write

f ′(X) =d

dt|t=0f(expG(tX)).

Theorem 1.4.1. Let f : G → H be a smooth homomorphism betweenLie groups and f ′ : g → h be its derivative at 0. Then

(1) f ′ is a Lie algebra homomorphism.

(2) If expG and expH denote the respective exponential maps thenf expG = expH f ′.

gf ′ //

exp

h

exp

G

f // H

(3) f ′ is uniquely determined by (1) and (2).

(4) If e is any other smooth homomorphism e : H → L, then ef is asmooth homomorphism G→ L and (ef)′ = e′f ′.

Proof. Let X ∈ g and consider the corresponding 1-parameter groupt 7→ expG(tX). As we saw earlier, because f is a smooth homomor-phism t 7→ f(expG(tX)) is a 1-parameter subgroup of H and hence its

1.4 The Functorial Relationship 49

infinitesimal generator ddt |t=0f(expG(tX)) = f ′(X) ∈ h. Thus for all t,

expH(tf ′(X)) = f(expG(tX)). (1.7)

Taking t = 1 in (1.7) gives the commutativity of the diagram above. Wenow show f ′ is a Lie algebra homomorphism. Using (1.7) but replacingX by cX gives expH(tcf ′(X)) = expH(tf ′(cX)). Differentiating at t = 0shows cf ′(X)) = f ′(cX)). Let X and Y ∈ g. Now recall that, for anyLie group, the tangent vector at 0 to the curve t 7→ exp(tX) exp(tY ) isX + Y . Applying f and using (1.7) we get f ′(X + Y ) = f ′(X) + f ′(Y ).In a similar manner for t ≥ 0 applying

exp(√tX) exp(

√tY ) exp(−

√tX) exp(−

√tY ) = exp(t[X,Y ] +O(t

32 ))

we conclude that f ′([X,Y ]) = [f ′(X), f ′(Y )]. We leave the verificationof this last statement to the reader. The Chain rule also proves (4). Toprove (3). we suppose φ : g → h is any other Lie algebra homomorphismwhich commutes the diagram. Let B be a ball about 0 in g on whichexpG is a diffeomorphism with its image expG(B) = U . Then by thecommutativity of the diagram on log(U) = B, φ = f ′. Since these mapsare linear and B contains a basis of g they coincide.

Corollary 1.4.2. (1) f(G) is a Lie group whose Lie algebra is iso-morphic to f ′(g)

(2) If H is connected, then f is surjective if and only if f ′ is surjective.

(3) Ker f is Lie group whose Lie algebra is Ker f ′

(4) f is locally one-to-one if and only if f ′ injective.

Proof. Proof of (1). In Theorem 1.4.1 let H = f(G) which is a closedsubgroup. Therefore it is a Lie group by Cartan’s theorem, 1.3.5. Let h

be its Lie algebra. Then h ⊇ f ′(g). On the other hand by Proposition1.4.8,

dim f(G) = dimG− dim Ker f = dim g − dim h = dim g/h = dim f ′(g).

Thus h and f ′(g) have the same dimension and hence they are equal. Inparticular, f ′ is surjective if and only if h = f ′(g). Thus f(G) has thesame dimension as H. Therefore f(G) is open in H which proves (2).


Proof of (3). By our criterion for a Lie subalgebra, the Lie algebraof Ker is X ∈ g : expG(tX) ∈ Ker f, i.e f(expG(tX)) ≡ 1. Thatis, expH tf

′(X) ≡ 1. Differentiating gives f ′(X) = 0. Conversely iff ′(X) = 0, then f(expG(tX)) ≡ 1. Therefor the Lie algebra of Ker f isKer f ′. Finally, (4) follows from the inverse function theorem.

The proof of the following corollary uses Theorem 1.4.1 and Propo-sition 0.1.13 and is left to the reader.

Corollary 1.4.3. Let G be a connected Lie group and e and f be twosmooth homomorphisms G→ H, with e′ = f ′. Then e ≡ f .

As the next result, we show that for simply connected Lie groups,there is a one-to-one correspondence between Lie homomorphisms andLie algebra homomorphisms.

Theorem 1.4.4. For Lie groups G and H with G simply connected,every Lie algebra homomorphism φ : g → h is the derivative of a Liehomomorphism f : G→ H, i.e. φ = f ′.

This theorem follows from an important result called the monodromyprinciple. Here is a formulation of it as it appears in [15] pp.46.

Theorem 1.4.5. (Monodromy Principle) Let X be a simply connectedspace. Suppose that we are given a collection of sets Mp, p ∈ X, param-eterized by the elements of X. Assume that D ⊂ X ×X is a connectedsubset containing the diagonal such that for each (p, q) ∈ D there is amap φpq : Mp →Mq satisfying the following conditions:

(1) φpq is a one-to-one map and φpp = id.

(2) If φpq, φqr and φpr are all defined, then φpr = φqr φpq.Then there is a map ψ which assigns to each p ∈ X an element ψ(p) ∈Mp in such a way that

ψ(q) = φpq(ψ(p)),

whenever φpq is defined. If we required that ψ(p0) = m0 for some fixedelements p0 ∈ X and m0 ∈Mp0 , then ψ is unique.


Here is a consequence of the monodromy principle.

Lemma 1.4.6. Let G be a simply connected topological group and f alocal homomorphism from G to a topological group H. Then f can beextended to all of G. Local homomorphism means that f is defined onan open neighborhood of the identity element and f(ab) = f(a)f(b) onthe neighborhood.

Proof. Let U be the open neighborhood where f is defined. We mayassume that U is symmetric. If not, we can replace U by U ∩ U−1.Let D ⊂ G × G be the subset consisting of (p, q) such that qp−1 ∈ U .Evidently D contains the diagonal and is connected. To every p ∈ G,we associate the map φpq : x 7→ f(qp−1)x on H. One checks directlythat D, Mp = H and φpq satisfy the conditions of the previous theorem.Therefore, there is a unique map ψ : G→ H such that ψ(1) = 1 and

ψ(p) = f(qp−1)ψ(q), (1.8)

whenever qp−1 ∈ U . We shall prove that ψ is a group homomorphismwhich extends f . By taking q = 1 and p ∈ U in (1.8) we obtain f(p) =ψ(p), thus ψ extends f . Let r = qp−1, we get

ψ(rq) = ψ(r)ψ(q)

for r ∈ U . In particular we have,

ψ(r1r2 · · · rn) = ψ(r1) · · ·ψ(rn)

if ri ∈ U . Since G =⋃∞n=1 U

n, every element g ∈ G can be written asg = r1r2 · · · rn for some n. Therefore,

ψ(g) = ψ(r1r2 · · · rn) = ψ(r1) · · ·ψ(rn)

which implies that ψ is a group homomorphism.

Proof of Theorem 1.4.4: Given φ : g → h, and using the exponential onecan construct a local homomorphism f from G to H. More explicitly,let U ⊂ g and V ⊂ h be two neighborhoods of the origin where the


corresponding exponential maps are diffeomorphisms and φ(U) ⊂ V .Then for g = expX, X ∈ U , we define,

f(g) = exp(φ(X)).

It follows from the properties of the exponential map that f is a localhomomorphism and f ′ = ψ. By the previous lemma, f can be extendedto all of G which is unique by Corollary 1.4.3.

The next result follows immediately.

Corollary 1.4.7. Let G be a simply connected Lie group. Then thereis a one-to-one correspondence between the automorphims of G and theautomorphisms of g, the Lie algebra of G.

Here we give the Lie theoretic analogue of the first isomorphismtheorem whose proof is left to the reader.

Proposition 1.4.8. If f : G → H is a smooth surjective homomor-phism between connected Lie groups, then it induces a bijective Lie ho-momorphism f∗ : G/K → H. In particular, dimG = dimKer f +dimH.

Proof. Let K = Ker f which is a closed normal subgroup. Thereforea Lie group with Lie algebra k. Then we have the isomorphism f∗ :G/K → H as groups which is a smooth map by definition of the smoothstructure of the quotient. To prove that it is an isomorphism we mustshow that it is a local isomorphism at the identity. The derivative off∗ at the identity is basically the induced map on the Lie quotientf ′ : g/k → h. The latter is an isomorphism by the first isomorphismtheorem for Lie algebras, Theorem 3.1.6.

The other isomorphism theorems are formulated and proven simi-larly.

Corollary 1.4.9. (The second isomorphism theorem) If G is connectedLie group with connected subgroups K and H such that HK is closedand H is normalizing K in G, then

H/H ∩K ≃ HK/K.


Proof. Apply Corollary 1.4.8 to the natural projection map H →HK/H ∩K.

Corollary 1.4.10. (The third isomorphism theorem) If G is a con-nected Lie group with connected normal subgroups K ⊂ H then(G/K)/(H/K) ≃ G/H.

Proof. Apply the previous result to then natural map G/K → G/Hinduced by the inclusion K → H.

We now specialize Theorem 1.4.1 to finite dimensional representa-tions.

Proposition 1.4.11. Let ρ : G→ GL(V ) be a smooth representation ofthe Lie group G on V over k, where k = R or C, and let ρ′ : g → gl(V ),be its derivative. If G is connected, then a subspace W of V is ρ-invariant if and only if it is ρ′-invariant.

Proof. Here we use the following commutative diagram

g

exp

ρ′ // gl(V )

Exp

Gρ // GL(V )

Now W is ρ′-invariant if and only if it is stable under ρ′(B), where Bis a ball about 0 in g. This is because if X ∈ g, then X = cY , whereY ∈ B and ρ′ is linear. Choosing B small enough B = logU , where U isa canonical neighborhood of 1 in G we see, by the commutativity of thediagram, that this is equivalent to W being stable under ρ(U). Since Uis symmetric, G =

⋃∞n=1 U

n. Hence because ρ is a homomorphism, thiscondition is, in turn, equivalent to the invariance of W under ρ(G).

Corollary 1.4.12. If G is connected, then ρ is irreducible (respectivelycompletely reducible) if and only if ρ′ is irreducible (respectively com-pletely reducible).


Definition 1.4.13. If ρ and σ are representations of a Lie group actionon Vρ and Vσ, respectively, then Cρ,σ, the space of intertwining opera-tors, consists of all linear maps T : Vρ → Vσ, such that Tρg = σgT , forall g ∈ G.

If ρ : G → GL(V ) is a representation of G then ρ′ : g → gl(V )denotes the derivative of ρ at the identity element. Note that ρ′ is a Liealgebra representative. Similar reasoning, together with the fact thatExp commutes with conjugation, yields the following result whose proofis left to the reader.

Corollary 1.4.14. If G is connected and ρ and σ are finite dimensionalrepresentations of G, then ρ and σ are equivalent if and only if ρ′ andσ′ are equivalent. More generally, if ρ and σ are representations thenCρ,σ = Cρ′,σ′ .

The following corollary stems directly from Theorem 1.4.4.

Corollary 1.4.15. For a simply connected Lie group G, there is a one-to-one correspondence between the representations of G and its Lie al-gebra g.

We now need the following lemma.

Lemma 1.4.16. Let X ∈ Endk(V ), where k = R or C and W be ak-subspace of V . Then X(W ) ⊆W if and only if Exp tX(W ) = W forall t ∈ k.

Proof. Suppose Exp(tX)(W ) = W for all t ∈ k. Let w ∈ W and con-sider the smooth curve t 7→ Exp(tX)(w) in W . Then d

dt |t=0 of thiscurve is X(w) ∈ W because the tangent space of an open set in Eu-clidean space is the Euclidean space. Conversely, if X(W ) ⊆ W , thenXn(W ) ⊆W and sinceW is a subspace, any polynomial p(X)(W ) ⊆W .Because W is closed, for any entire function f , f(X)(W ) ∈W . Apply-ing this to tX as the X gives the conclusion.

Definition 1.4.17. Let ρ be a representation of G on V , ρ′ a representa-tion of a Lie algebra, g, and W be a subspace of V . We write StabG(W )


for the set of all g’s that stabilize W and similarly for Stabg(W ). Wealso shall write FixG(v) and Fixg(v) for the things that fix v ∈ V , ineach case.

Proposition 1.4.18. Let ρ be a representation of G on V and ρ′ be itsderivative and W a subspace of V . Then

(1) StabG(W ) = Stabg(W )

(2) FixG(v) = Fixg(v), v ∈ V

The proof of the first statement follows from Lemma 1.4.16, and thesecond can be seen directly. We leave the details to the reader.

Now we turn to the most important representation of a Lie group,namely the adjoint representation. If G is a Lie group (not neces-sarily connected) and α is a smooth automorphism of G then α′ isan automorphism of g and as usual we get a commutative diagram.Now take α = αg, the inner automorphism gotten throught conjuga-tion by g ∈ G. We call the differential Ad g. Since αgαh = αgh wesee that Ad(gh) = Ad gAdh. Thus we get a linear representationAd : G → GL(g) called the adjoint representation and the image ofG under Ad is denoted AdG. For g near 1, Ad g = log ·αg · exp, acomposition of smooth functions so Ad is smooth in a neighborhood of1 in G. Hence by the exercise below Ad is a smooth representation ofG in g. This works equally well over R or C.

Corollary 1.4.19. Let G be a connected Lie group. Then Z(G) =Ker Ad and its Lie algebra is z(g) = Ker ad.

Proof. Now because of connectedness, g ∈ Z(G) if and only if for allt ∈ k and X ∈ g we have

exp(tX) = g exp(tX)g−1 = exp(tAd g)(X).

Differentiating at t = 0, we see this is equivalent to Ad g(X) = X forall X. That is, g ∈ Ker Ad.

Also, the Lie algebra of Z(G) is X ∈ g : Ad(exp(tX)) ≡ for all t .So, again, differentiating at t = 0 gives adX = 0. Since all steps arereversible this proves the second statement.


Corollary 1.4.20. If G is a connected abelian Lie group, then g isabelian and exp is a surjective homomorphism.

Proof. It follows from Lemma 1.2.2, part 3, that the Lie algebra of[G,G] is [g, g]. Therefore since G is abelian, so is g and exp(X + Y ) =exp(X) exp(Y )+ higher order terms all involving commutators. Since g

is abelian, these all vanish.

Corollary 1.4.21. Any connected abelian Lie group is of the form Rj×Tk.

Proof. Since g is abelian we can regard it as Rn. Now the homomor-phism, exp is locally one-to-one since it is one-to-one in a neighborhoodof 0. Therefore, its kernel is discrete. Hence, its image, G, is Rj × Tk,where j + k = n.

It is sometimes convenient to have an explicit description of theadjoint representation which we will do in the next corollary.

Corollary 1.4.22. If G is a linear group, then Ad g(X) = gXg−1, forX ∈ g.

Proof.

exp t(Ad g(X)) = exp Ad g(tX) = αg(exp tX)

= g(exp tX)g−1 = exp(gtXg−1) = exp t(gXg−1)

Calculating the derivative at t = 0 gives the conclusion.

We now come to the question of what is Ad′?

Corollary 1.4.23. In a connected Lie group Ad expX = Exp(adX),X ∈ g.

Proof. Since Ad is a smooth representation of G on g its derivative Ad′

is a Lie algebra representation of g on g making the appropriate dia-gram commutative. For t ∈ k and X ∈ g we know Exp[tAd′(X)] =Exp Ad′(tX) = Ad exp(tX). Let Y ∈ g. Then since AdG is a linear Liegroup Exp[tAd′(X)](Y ) = Ad exp(tX)(Y ) = exp(tX)(Y ) exp(−tX).


By a fact proved earlier this last term is (1 + tX + O(t2))(Y )(1 −tX + O(t2)). But this is just Y + t[X,Y ] + O(t2). On the other hand,Exp tAd′(X) = (1 + tAd′X + O(t2))(Y ). Taking derivatives at t = 0yields Ad′(X)(Y ) = [X,Y ] for all Y . Thus Ad′ = ad and we have thecommutativity relation Exp(adX) = Ad exp(X), X ∈ g.

Corollary 1.4.24. Let G be a connected Lie group and H be a connectedLie subgroup with Lie algebras g and h. Then the following conditionsare equivalent.

(1) H is normal in G.

(2) h is an ideal in g.

(3) h is an Ad-invariant subspace of g.

Proof. Since h is a subspace of g, we know from the above that h isAd-invariant if and only if it is ad-invariant, proving that (2) and (3)are equivalent. Suppose H is normal in G, and U is a canonical neigh-borhood of 1 in H and B = logU is a ball about 0 in h. For g ∈ G,exp Ad g(B) = αg(U) ⊆ H. Then exp Ad g(tX) = exp tAd g(X) ∈ Hfor all X ∈ B and |t| ≤ 1. Hence Ad g(B) ⊆ h and, by linearity of Ad gand the fact that h is a vector space, Ad g(h) ⊆ h. Conversely, if eachAd g preserves h then reversing all these steps tells us that αg(U) ⊆ Hand since U generates H, αg(H) ⊆ H so H is normal. This proves parts(1) and (2) are equivalent and completes the proof.

We now identify the Lie algebras of some other commonly encoun-tered Lie subgroups. The following result shows, in particular, thata connected Lie group is solvable (respectively nilpotent) if and onlyif Lie algebra being solvable (respectively nilpotent). In the case of asemisimple (or reductive) Lie group we merely take as the definition thecorresponding property of the Lie algebra.

However, in contrast to all the other theorems in this book, Theorem1.4.25 requires the full statement of the BCH formula (1.9), not merelyits second order approximation.

Theorem 1.4.25. Let G be a connected Lie group and N a connectednormal Lie subgroup, with respective Lie algebras g and n. Then [G,N ]is a normal connected Lie subgroup whose Lie algebra is the ideal [g, n].


Proof. That [G,N ] is normal and [g, n] is an ideal follow directly fromthe fact that N is normal and n is an ideal, Corollary 1.4.24. We leavethe verification of this to the reader. By Lie’s theorem, Theorem 1.3.3,there is a connected Lie subgroup, H of G whose Lie algebra is [g, n]which by Corollary 1.4.24 is normal. We next show [G,N ] ⊆ H. SinceG and H are both connected in order to show [G,N ] ⊆ H it is sufficientto prove that for a small ball B about 0 in n that

[expX, exp Y ] ∈ H, for X ∈ B and Y ∈ B ∩ n.

Choose the ball small enough so that the BCH formula (1.9) is valid init. Then [expX, exp Y ] = exp([X,Y ] + η(X,Y )), where η is a conver-gent power series consisting of commutators of various orders in X andY (such as [X, [X,Y ]] or [X, [Y, [X,Y ]]]) with rational coefficients. Inparticular, since n is an ideal, [g, n] is closed, η(X,Y ) and [X,Y ] ∈ [g, n].Hence [expX, exp Y ] ∈ H. This means [G,N ] ⊆ H.

Now choose X1, . . . ,Xp from g and Y1, . . . , Yp from n so the [Xi, Yi]are a basis for [g, n]. Modify the Xi by scalar multiplication to insurethat they also lie in B. Let φi(t) = [expXi, exp tYi], for i = 1 . . . p,φi : R → [G,N ]. Let φ(t1, . . . , tp) = φ1(t1) . . . φp(tp) and φ be asmooth Rp → [G,N ]. We can identify Rp with [g, n] via the basis[X1, Y1], . . . , [Xp, Yp]. Choose t1, . . . , tp small enough so that the BCHformula applies. Now log(φ) : B1 → Rp, where B1 is a small ballabout 0 in [g, n] on which exp is invertible. Then, as above, for each i,log[expXi, exp tYi] = [Xi, tYi] + η(Xi, tYi) has a nonzero derivative att = 0. Hence d(log(φ))(0) is invertible. By the inverse function theoremthere is a ball B2 about 0 in Rp so that log(φ)(V ) contains a neighbor-hood of 0 in [g, n]. Hence φ(B) contains a neighborhood of 1 in H. Thisneighborhood is in [G,N ] since φ takes values there. Because H is con-nected and therefore the neighborhood generates H we get H ⊆ [G,N ].Therefore H = [G,N ] so the Lie algebra of [G,N ] is [g, n].

Corollary 1.4.26. A Lie group has no small subgroups.

Proof. Let U be a ball about 0 in g on which exp is a global diffeomor-phism and suppose that H were a subgroup of G contained in exp(U).


Let h ∈ H, h = expX, then for any positive integer n, hn = expnX.Since everything takes place in a neighborhood where exp is a globaldiffeomorphism, and hn ∈ H ⊂ exp(U) therefore expnX ∈ exp(U) sonX ∈ U for all n. This is impossible for a ball.

Proposition 1.4.27. Over C, Aut(g) is a complex algebraic group,while over R, Aut(g) is the real points of an algebraic group over R. Inany case, Aut(g) is closed and hence is either a complex Lie group, ora real Lie group, respectively.

Proof. Let X1, . . . ,Xn be a basis for g and [Xi,Xj ] =∑

k ckijXk be

structure constants for this basis. For an automorphism, α, α(Xk) =∑p αkpXp, where αkp are the matrix coefficients of α relative to this

basis. Thus

α([Xi,Xj ]) =∑

k

ckijα(Xk) =∑

k

ckij∑

p

αkpXp =∑

p

(∑

k

ckijαkp)Xp

These relations are determinative. On the other hand, α([Xi,Xj ]) =[α(Xi), α(Xj)]. Applying the same reasoning to α(Xi) and α(Xj) andthen using the linearity of the bracket and equating coefficients of thebasis gives a finite number of equations which on one side are (quadratic)polynomials in the matrix coefficients of α and on the other are linear.Thus α ∈ Aut(g) if and only if its coefficients satisfy this (finite) set ofpolynomial equations.

Definition 1.4.28. A map D : g → g is called a derivation of g if Dis linear and for all X,Y ∈ g, D[X,Y ] = [D(X), Y ] + [X,D(Y )]. Wedenote by Der(g) the derivations of g. Evidently Der(g) is a subspaceof the vector space End(g). As we show, under taking the commutatorof operators, Der(g) is actually a Lie subalgebra of End(g).

For example adX is a derivation for each X ∈ g. Such a derivation iscalled an inner derivation. We denote by Der(g) the set of all derivationsof g. It is not difficult to see that Der(g) is a subalgebra of gl(g) andthat ad g is a subalgebra of it, called the algebra of inner derivations.In fact ad g ⊂ Der(g) is an ideal: if D ∈ Der(g) and X ∈ g we get

[D, adX] = adD(X)


If g is an abelian Lie algebra then any endomorphism of g is a derivation.

Exercise 1.4.29. Let D be a derivation g and n a positive integer.Then,

Dn[X,Y ] =

n∑

i=0

(n

i

)[DiX,Dn−iY ]

Corollary 1.2.4 enables us to see the relationship of automorphismsand derivations.

Theorem 1.4.30. Let g be any real or complex Lie algebra. Then theLie algebra of Aut(g) is Der(g)

Proof. Let D ∈ Der(g). As we saw ExpD ∈ GL(g). If X,Y ∈ g, thenD[X,Y ] = [D(X), Y ]+[X,D(Y )]. It follows from the binomial theoremDn[X,Y ] =

∑ni=0

n!i!(n−i)! [D

i(X),Dn−iY ], for all n.Hence

ExpD[X,Y ] =∞∑

n=0

1

n!

n∑

i=0

n!

i!(n − i)![Di(X),Dn−iY ].

Arguing exactly as in the proof of Exp(A+B) = ExpAExpB when Aand B commute we see that ExpD[X,Y ] = [ExpD(X),ExpD(Y )] soExpD ∈ Aut(g).

Now suppose T ∈ End(g) and Exp tT ∈ Aut(g) for all t. Then weshow T ∈ Der(g). Since Exp tT [X,Y ] = [Exp tT (X),Exp tT (Y )] for allt we can differentiate both sides of the equation with respect to t at t = 0and get T [X,Y ] = [T (X), Y ] + [X,T (Y )]. Thus T is a derivation.

Remark 1.4.31. We remark that the Lie algebra of Haar measurepreserving automorphisms of a group consists of the derivations of theLie algebra of trace 0 (see [47]).

1.5 The Topology of Compact Classical Groups

Here we will determine a number of global topological properties ofimportant compact Lie groups. In Chapter 6 we shall see how theseglobal topological properties propagate to non-compact groups.

1.5 The Topology of Compact Classical Groups 61

Let G be a connected Lie group and H a closed subgroup. We shallshow that the smooth map π : G → G/H always has a smooth localcross-section which implies that π : G→ G/H is a fibration [76].

Let g be the Lie algebra of G. SinceH is a closed subgroup it is also aLie group. Let h be its Lie algebra and choose a vector space complementW to h in g. Then there is a local diffeomorphism φ : W → G/H suchthat π exp = φdπ, where exp is the exponential map of G and dπ is thedifferential of π at the identity. Also by the choice of W , dπ has a globalcross section on W , namely i, the injection of W into g.

Let σ = (exp i)φ−1. Then σ is a smooth map locally defined in aneighborhood of the coset H in G/H and πσ = π(exp i)φ−1. But sinceπ exp = φdπ, we have π(exp i) = φ so that πσ = π(exp i)φ−1 = I.

A fibration gives rise to a long exact sequence of homotopy groups(see [76])

. . . π2(H) → π2(G) → π2(G/H) → π1(H) → π1(G) →→ π1(G/H) → π0(H) → π0(G) → π0(G/H) → 0

From this we can draw a number of important conclusions. Forexample, by looking at the exactness of π0(H) → π0(G) → π0(G/H) wesee that if H and G/H are both connected then so is G.

We will prove this directly so that the reader will be confident thatthis is true. Let U and V be an open partition of G by non-emptysets. Then since π is open and surjective, π(U) and π(V ) are open non-empty sets whose union is G/H. But since G/H is connected these mustintersect at least in some coset gH. Thus there is a u ∈ U and a v ∈ Vwhich are each congruent to gH. But then U ∩gH and V ∩gH are non-empty and so give an open partition of gH. Since gH is homeomorphicto H, it is also connected so this is impossible.

Corollary 1.5.1. For all n ≥ 1, SO(n,R), U(n,C) and SU(n,C) areconnected. O(n,R) has 2 components.

This follows by induction from the fact that the spheres are con-nected and SO(1,R), U(1,C) and SU(1,C) are connected. SinceSO(n,R) is connected and det : O(n,R) → ±1 is a surjective and con-tinuous map O(n,R) has 2 components.


Now let us look at other parts of the long exact sequence. Considerπ1(H) → π1(G) → π1(G/H). This tells us that if H and G/H are bothsimply connected then so is G.

Corollary 1.5.2. For all n ≥ 1, SU(n,C) is simply connected.

To see this we must know that Sn is simply connected for n ≥ 3. Infact this is so for n ≥ 2 and follows from the Van Kampen theorem (see[40]).

The exactness of π1(G) → π1(G/H) → π0(H) → π0(G) tells us thatif G is connected and simply connected then π1(G/H) = H/H0. Inparticular, if H is also connected then G/H is simply connected. If Γis a discrete subgroup, then π1(G/Γ) = Γ. For example, this latter facttells us that if G is a simply connected Lie group and Γ is a discretecentral subgroup then the fundamental group of the quotient is Γ. Thusif (G, π) is the universal covering group of G then π1(G) = Kerπ.

The exactness of π2(G/H) → π1(H) → π1(G) → π1(G/H) tells usthat if π2(G/H) = 1 and π1(G/H) = 1 then π1(H) and π1(G) areisomorphic.

Corollary 1.5.3. For n ≥ 1, π1(U(n,C)) = Z and for n ≥ 3,π1(SO(n,R)) = Z2, π1(SO(2,R)) = Z (since U(1,C) = SO(2,R)).

To prove this we need only show in addition to what we know alreadythat π1(U(1,C)) = Z, π1(SO(3,R)) = Z2 and π2(S

n) = 1 for n ≥ 3.For this last fact see [76]. Concerning the first, since the universalcovering R → U(1,C) is t 7→ e2πit and R is simply connected then asabove π1(U(1,C)) = Z. The second fact will follow in a similar way byconstructing the universal covering group of SO(3,R) below.

We shall use this last principle to calculate both π1(SO(3,R)) andπ1(SO(4,R)). Consider the quaternions H and the nonzero quaternions,H×. Given a quaternion

q = a0 + a1i+ a2j + a3k,

we define its conjugate,

q = a0 − (a1i+ a2j + a3k)


and its norm,

N(q) = a20 + a2

1 + a22 + a2

3.

Then clearly N(q) = qq and the real number N(q) ≥ 0 and = 0 ifand only if q = 0. From this we see immediately that each nonzero

quaternion q is invertible with q−1 =q

N(q). Thus H× is a group. It

is easy to see from the formulas for multiplication and inversion thatH× is in fact a Lie group. Also − is an anti-automorphism of thisgroup; qr = rq and from this it follows that N is a homomorphism ofthis group to the multiplicative group R×; N(qr) = N(q)N(r). Let Gdenote the elements of unit norm in H. Then G is a subgroup of H×

which topologically is the 3-sphere S3. In particular, G is a compactconnected and simply connected Lie group which is noncommutative.Incidentally, like the 1-sphere this also shows that the 3-sphere carriesa Lie group structure. This is not so, for example, of the 2-sphere.

Now we define the left, right and two-sided regular representationsrespectively of the R-algebra H as follows: Lq(x) = qx, Rq(x) = xq−1

and T(q,r)(x) = qxr−1. It is easy to see that each of these is an R-linear representation H× or H× × H× on H. Restricting the two-sidedregular representation to G × G we get a smooth homomorphism ofthe latter to O(4,R) and by connectedness to SO(4,R). Since G × Gand SO(4,R) are both connected and have dimension 6 and the kernelof this map is ±(1, 1) which is finite we see that this map is ontoand a covering. Since G × G is simply connected (see [40]) it followsthat π1(SO(4,R)) = Z2. Now further restrict this representation to thediagonal subgroup of G × G. This gives a representation of G on H(by conjugation) which leaves the center fixed. Since G is compact itmust leave the orthocomplement stable and preserve the norm. Thus wehave a smooth homomorphism G → O(3,R). As above it takes valuesin SO(3,R) and its kernel is ±(1, 1). Since G and SO(3,R) are bothconnected and have dimension 3 and the kernel of this map is finite themap is onto and a covering. Because G is simply connected it followsas above that π1(SO(3,R)) is also Z2. Thus we have constructed theuniversal covering group G→ SO(3,R) and G×G→ SO(4,R). Finally,taking the differential of the first of these shows that the Lie algebra


su(2,C) of G is isomorphic with so(3,R). Taking the differential of thesecond yields an important decomposition of the Lie algebra of so(4,R)as the direct sum of ideals isomorphic with so(3,R) ⊕ so(3,R).

We note that the fundamental group of these compact semisimplegroups seem to be finite, while that of the non-semisimple ones arefinitely generated abelian, but infinite. We shall see later that this isnot an accident.

We now turn to the identification of the group G. First observe thatH contains the field C in the form a0+a1

√−1 and so is a vector space

over C and since H is a 4-dimensional algebra over R it has dimension2 over C. To be specific we shall take the scalar multiplication by C onthe right. The associative and distributive laws of H tell us that, in thismanner, H is a vector space over C. Here any q = a0+a1i+a2j+a3k ∈ Hcan be written as follows:

q = 1(a0 + a1

√−1) + j(a2 − a3

√−1)

In this way 1, j are a basis for H over C. Now consider the left regularrepresentation L of the R-algebra H. This representation is faithful. Forif Lq = 0, then qr = 0 for every r. But, if q 6= 0 taking r = q−1 wouldgive a contradiction. Hence q = 0 and so L is faithful. It is actually aC-representation since Lq(xc) = Lq(x)c, c ∈ C by associativity. Thuseach Tq can be represented by a 2×2 complex matrix whose coefficientsare determined by Tq(1) and Tq(j) as follows:

Tq(1) = 1z11(q) + jz21(q)

and

Tq(j) = 1z12(q) + jz22(q).

Now since q = 1(a0 + a1i) + j(a2 − a3i), Tq(1) = q1 = q = 1z11(q) +jz21(q). So

z11(q) = a0 + a1i and z21 = a2 − a3i,

while

Tq(j) = qj = (a0 + a1i)j + j(a2 − a3i)j = a0j + a1k − a2 − a3i.


Since Tq(j) = −a2 − a3i+ a0j+ a1k we see that z12(q) = −a2 − a3i andz22(q) = a0 − a1i. Denoting a0 + a1i by a and a2 − a3i by b, the matrixof Lq with respect to this basis is

Lq =

(a −bb a

)

and we have a faithful matrix representation of H. In particular,detLq = N(q). This gives an independent proof of the fact that thingswith nonzero norm are invertible and N(qr) = N(q)N(r). Also q ∈ Gif and only if |a|2 + |b|2 = 1. So G is isomorphic to this group of 2 × 2complex matrices.

Finally, we identify the latter. The set G of these matrices

g =

(a −bb a

)

clearly forms a subgroup of SU(2,C) which is homeomorphic to the 3sphere and in particular is connected and compact and hence closed.Thus G is a connected Lie subgroup of SU(2,C). Since the latter is alsoconnected and both these groups have dimension 3 they coincide andG = SU(2,C).

Turning to Sp(n), similar arguments show that Sp(1) = SU(2,C)and Sp(n)/Sp(n − 1) = S4n−1, n ≥ 2. Hence Sp(n) is a compactconnected and simply connected Lie group.

We conclude this section with the calculation of some coveringgroups of non-compact groups. Consider SO(2, 1)0 and SO(3, 1)0, theconnected components of the group of isometries of hyperbolic 2 and 3space, H2 and H3. That is, we consider the forms q12(X) = x2−y2−z2

on R3 and q13(X) = x2 − y2 − z2 − t2 on R4 and the correspondinggroups of isometries.

We let SL(2,R) act on the space of 2× 2 real symmetric matrices Sby ρg(S) = gSgt. Then S is a real vector space of dimension 3 and ρis a continuous real linear representation of SL(2,R). Since det g = 1,det(ρg(S)) = det(gSgt) = detS. Now if

S =

(a zz b

),


then detS = ab− z2. Consider the change variables with x = a+b2 and

y = a−b2 . This is an R linear change of variables and a = x + y and

b = x−y. Therefore, detS = x2−y2−z2. Thus ρ preserves a (1, 2) formon R3. Since SL(2,R) is connected so is its image ρ(G) in GL(3,R). Adirect calculation shows Ker ρ = ±I. Since this is finite and thereforediscrete ρ(G) has dimension 3 just as SL(2,R). But ρ(G) ⊆ SO(1, 2)0.Since this connected group also has dimension 3 ρ is onto and thereforea covering. It induces an isomorphism between SO(1, 2)0 = SO(1, 2)and SL(2,R)/±I = PSL(2,R).

Similarly, let SL(2,C) act on the space of 2 × 2 complex Hermitianmatrices H by ρg(H) = gHg∗. Then H is a real vector space of dimen-sion 4 and ρ is also a continuous real linear representation of SL(2,C).Since det g = 1, det(ρg(H)) = det(gHg∗) = detH. Now if

H =

(a zz b

),

then detH = ab − |z|2. Consider the change of variables x = a+b2 and

y = a−b2 . Then this is an R linear change of variables, a = x + y and

b = x − y. Therefore, detH = x2 − y2 − |z|2 = x2 − y2 − u2 − v2.Thus ρ preserves a (1, 3) form on R4. Since SL(2,C) is connected sois its image ρ(G) in GL(4,R). A direct calculation shows Ker ρ = ±I.Since this is finite and therefore discrete ρ(G) has dimension 6 justas SL(2,C) does. But ρ(G) ⊆ SO(1, 3)0. Since this connected groupalso has dimension 6 ρ is onto and therefore a covering. It induces anisomorphism between SO(1, 3)0 and SL(2,C)/±I = PSL(2,C). Sinceas we shall see SL(2,C) is simply connected (see Corollary 6.3.7), in factit is the universal cover of SO(1, 3)0.

Finally we apply the same method to SU(2,C) to get another wayof calculating the universal cover of SO(3,R).

Let SU(2,C) act on H, the space of 2 × 2 complex skew-Hermitianmatrices of trace 0 by ρg(H) = gHg∗ = gHg−1. Then H is a real vectorspace of dimension 3 and ρ is also a continuous real linear representationof SU(2,C). Then det(ρg(H)) = det(gHg−1) = detH. Now if

H =

(ia z−z −ia

),

1.6 The Iwasawa Decompositions for GL(n,R) and GL(n,C) 67

We see detH = −a2−|z|2. Thus ρ preserves a negative definite formand therefore also a positive definite form on R3. Thus ρ(G) ⊆ O(3,R).Since SU(2,C) is connected so is its image ρ(G). A direct calculationshows Ker ρ = ±I. Since this is finite and therefore discrete ρ(G) hasdimension 3 just as SU(2,C) does. But ρ(G) ⊆ SO(3,R). Since thisconnected group also has dimension 3, ρ is onto and therefore a cover-ing. It induces an isomorphism between SO(3) and SU(2,C)/±I =Ad(SU(2,C)). Since SU(2,C) is simply connected it is the universalcover of SO(3,R).

Exercise 1.5.4. Prove that the group SO(1, 1)0 is g : g =(cosh t sinh tsinh t cosh t

), where t ∈ R, and therefore SO(1, 1)0 is isomorphic

with R and is simply connected.

1.6 The Iwasawa Decompositions for GL(n,R)and GL(n,C)

Here we shall prove that the manifold, GL(n,R), is the direct productof three submanifolds K, A0 and N , where each of these is actually aclosed subgroup. K = O(n,R), the orthogonal group, A0 is the diagonalmatrix with positive entries, and N is the subgroup of unitriangularmatrices. Note that A0 is the identity component of A, the group ofdiagonal matrices with nonzero entries on the diagonal. Moreover, thediffeomorphism of K×A0×N with GL(n,R) is given by multiplication.

Similarly, for GL(n,C), we get a diffeomorphism of K×A0×N withGL(n,C) given by multiplication. Here K is the unitary group, Un(C),A0 is as before, and N is the unitriangular matrices on GL(n,C).

Exercise 1.6.1. Show that:

(1) Over R, A has 2n components, while over C, A is connected.

(2) On(R) has two components while Un(C) is connected.

(3) In both cases, over R and C, N is connected.

(4) In either case A0N is diffeomorphic to Euclidean space and deter-mine the dimension.


Proposition 1.6.2. G = KA0N where the diffeomorphism is given bymultiplication.

Before proving these facts we mention that they can be used totell much about the topology of the non-compact groups GL(n,R) andGL(n,C), if one knows something about the compact group, K, becausein both cases K is a deformation retract of G. In Section 1.5 we dealtwith the topology of compact Lie groups.

Another consequence is that in either case A0N is a subgroup (sinceA0 normalizes N . It consists of the triangular matrices with positivediagonal entries. Since G = KA0N it is also true that G = KB whereB is the group of all triangular matrices in G.

Since G = KB the second isomorphism theorem tells us that G/B =K/K ∩B and in particular that G/B is compact.

Proof. Our proof shall deal with both the real and complex cases simul-taneously. We let V stand for either Rn or Cn. Let ei : i = 1 . . . n bethe standard basis of V and g ∈ G. Then vi = g−1ei, i = 1 . . . n is alsoa basis of V . We apply the Gram-Schmidt orthogonalization process tovi : i = 1 . . . n. Let u1 = v1/||v1|| and for i = 2 . . . n,

ui =vi −

∑i−1j=2(vi, uj)uj

||vi −∑i−1

j=2(vi, uj)uj ||.

Then the ui’s form an orthonormal basis of V , depending smoothly ong ∈ G, and by the formulas above, one can write

ui(g) =∑

j≤iaji(g)vi(g),

where aii > 0. Let a(g) be the diagonal matrix with entries aii andn(g) = a(g)−1(aji(g)). Then a and n depend smoothly on g, a(g) ∈ A0

and n(g) ∈ N , all g ∈ G. Also, for all g ∈ G and i = 1 . . . n, a(g)n(g)vi =(aji(g))(vi(g)) = ui(g). Now since ei : i = 1 . . . n and ui : i = 1 . . . nare both orthonormal basis there exists a unique k(g) ∈ K so thatk(g)(ui(g)) = ei, for all i. k also depends smoothly on g. Moreover,k(g)a(g)n(g)(vi) = k(g)ui(g) = ei for all i. But also g(vi) = ei. Sincevi’s form a basis we get g = k(g)a(g)n(g).

1.7 The Baker-Campbell-Hausdorff Formula 69

We remark that similar reasoning applies to SL(n,R). Just replace

On(R) by SOn(R) and A0 by a ∈ A0 : det a = 1. Similarly, forSL(n,C). Just replace Un(C) by SUn(C) and A0 by a ∈ A0 : det a =1.

We also get corresponding decompositions of the respective Lie al-gebras, g = gl(n,R) or gl(n,C). We let a denote the diagonal elementsof either one, n denote the strictly triangular elements of either one andk denote the skew symmetric elements in the case of R and the skewHermitian symmetric elements in the case of C.

As the reader can easily check these are always real subspaces of g.Note, however, that in the complex case, k is not a complex subspace ofg. It follows immediately from the previous result that,

Corollary 1.6.3. For g = gl(n,R) or gl(n,C) we have g = k ⊕ a ⊕ n.

1.7 The Baker-Campbell-Hausdorff Formula

In order to prove the Baker-Campbell-Hausdorff formula, as well as forother purposes, we first calculate the derivative of the exponential map.

Let G be a connected Lie group, which for convenience we shallassume to be linear and g = T1(G) its Lie algebra. We now calculatethe derivative, dX exp, of the exponential map at a point X ∈ g. Sinced0 exp = I and in particular is nonsingular it follows by continuity thatfor small X, dX exp is also invertible. We can do much better thanthis with an explicit formula for dX exp. This will tell us how nearzero we have to be for dX exp to be invertible and will be importantfor other reasons as well. Under this identification, if f : G → H is asmooth homomorphism, X ∈ g, and f(expG(tX)) = expH(tf ′(X)) isthe corresponding 1-parameter subgroup of H, then

d1f(X) =d

dtexpH(tf ′(X))|t=0 = f ′(X).

So d1f = f ′. We shall also make the further convention that we iden-tify the tangent space Tg(G) of a point g of G with g by applying theleft translation by g−1 to Tg(G). This will enable us to normalize thesituation and for every X ∈ g view dX exp as an operator on g.


Before beginning the calculation of dX exp a word must be said aboutfunctional calculus. Let V be a finite dimensional complex vector spaceand gl(V ) be its endomorphism algebra. For each complex analyticfunction f(z) =

∑∞k=0 akz

k on some disk D about 0 in C and linearoperator L on V . We may assume

∑∞k=0 akz

k absolutely convergentby taking the radius of convergence smaller. We may form, for anyL with ‖L‖ < radius of D, f(L) =

∑∞k=0 akL

k. Each such f(L) is alinear operator on V and the series

∑∞k=0 akL

k is absolutely convergent.The resulting map (f, L) 7→ f(L) is called the operational calculus. Bylooking at the Jordan triangular form of such an L and taking intoaccount the fact that Pf(L)P−1 = f(PLP−1), we see easily that

Spec f(L) = f(λ) : λ ∈ SpecL.Now, for fixed L, the map f 7→ f(L) is clearly an algebra homomor-

phism from the holomorphic functions about 0 to gl(V ). In particular,if f is holomorphic and f(0) 6= 0 then 1/f is also holomorphic and

(1/f)(L) = f(L)−1.

For example, if f were exp or log then we have already applied thisfunctional calculus to study Exp and Log. Now let φ(z) = (ez − 1)/z =1 + z/2! + z2/3! + . . .. Then φ is an entire function with a removablesingularity at z = 0, φ(0) = 1.

Theorem 1.7.1. For each X ∈ g, dX exp = φ(− adX).

This formula will be important in proving the Baker-Campbell-Hausdorff formula. We shall also use it in studying the geometry ofthe symmetric spaces associated with certain Lie groups. As a corollarywe have

Corollary 1.7.2. dX exp is nonsingular if and only if adX has noeigenvalues of the form 2πin for some nonzero integer n.

Proof. dX exp is nonsingular if and only if φ(λ) 6= 0 for all λ ∈Spec(− adX) = − Spec(adX). Since φ(0) = 1, φ(z) = 0 if and only ifez = 1, z 6= 0 if and only if z = 2πin for some nonzero integer n.


We need the following lemma.

Lemma 1.7.3. For X and Y ∈ g and t ∈ R,

d

dtexp(X + tY )|t=0 = Y + 1/2!(XY + Y X)+

1/3!(X2Y +XYX + Y X2) + . . .

The terms of this convergent series are in gl(V ) in our case, or in theuniversal enveloping algebra in general.

Proof. Write exp(X + tY ) = c + t(. . .) + O(t2) where (...) = Y +1/2!(XY +Y X)+1/3!(X2Y +XY X+Y X2)+ . . . where c is a cosntant.Differentiating and evaluating at t = 0 gives the result.

Lemma 1.7.4. Let X and Y be fixed in g. Then

exp(−X) d

dt(exp(X + tY ))|t=0 = φ(− adX)(Y ).

Proof. In particular, from the previous lemma

∂

∂texp s(X + tY )t=0 = sY + s2/2!(XY + Y X)+

s3/3!(X2Y +XY X + Y X2) + . . . .

For s ∈ R, let Φ(s) = exp(−sX) ∂∂t(exp s(X + tY ))|t=0. Then Φ

is analytic and Φ(0) = 0. We shall show that Φ satisfies the globallinear nonhomogeneous differential equation with constant coefficientsdΦds = −[X,Φ(s)] + Y , for s ∈ R. Now

Φ′(s) = exp(−sX)(Y + s(XY + Y X)

+ s2/2!(X2Y +XYX + Y X2) + . . .)

+ exp(−sX)(−X)(sY + s2/2!(XY + Y X)

+ s3/3!(X2Y +XYX + Y X2) + . . .)

= exp(−sX)(Y + sY X + s2/2!(Y X2) + . . .)


On the other hand,

−[X,Φ(s)] + Y = Y + exp(−sX)(sY + s2/2!(XY + Y X)+

s3/3!(X2Y +XY X + Y X2) + . . .)X

−X exp(−sX)(sY + s2/2!(XY + Y X)+

s3/3!(X2Y +XY X + Y X2) + . . .)

= Y + exp(−sX)(s(Y X −XY )

+ s2/2!(Y X2 −X2Y ) + . . .).

Hence, exp(sX)Φ′(s) = Y + sY X + s2/2!(Y X2) + . . .. Whereas,

exp(sX)(−[X,Φ(s)] + Y ) = exp(sX)Y + s(Y X −XY )+

s2/2!(Y X2 −X2Y ) + . . .

= Y + sY X + s2/2!(Y X2) + . . . .

Since exp(sX)Φ′(s) = exp(sX)(−[X,Φ(s)]+Y ) we see that Φ′(s) =−[X,Φ(s)] + Y for all s ∈ R. Now the lemma follows from the nextproposition.

Proposition 1.7.5. Let X and Y be fixed in g and Φ : R → g be amap which satisfies the differential equation Φ′(s) = −[X,Φ(s)]+Y andinitial condition that Φ(0) = 0. Then for all s ∈ R,

Φ(s) = φ(−s adX)(sY ).

and, conversely this Φ does satisfy the differential equation with thisinitial condition. In particular, Φ(1) = φ(− adX)(Y ). (Notice thatφ(−s adX)(sY ) ∈ g for all s).

Proof. This is a system of nonhomogeneous equations with constantcoefficients so it has a global analytic solution where Xi ∈ g and s ∈ R.(See [69] for further detail)

Φ(s) = X0 + sX1 + s2X2 + s3X3 + . . .

Since Φ(0) = 0, we know X0 = 0. Now Φ′(s) = X1 +2sX2 +3s2X3 + . . .and −[X,Φ(s)]+Y = Y −s[X,X1]−s2[X,X2]−. . .. Therefore, X1 = Y ,


X2 = −12 [X,X1] = −1

2 adX(Y ), X3 = −13 [X,X2] = 1

6 ad2 Y and ingeneral

Xn = (−1)n−1 1

n!adXn−1(Y ).

So Φ(s) = sY − s2/2! adX(Y )+ s3/3! adX2(Y )+ . . .. Applying − adXwe get

− adX Φ(s) = −s adX(Y ) + s2/2! adX2(Y ) − s3/3! adX3(Y ) + . . .

= exp(−s adX − I)Y.

Hence,−s adX Φ(s) = (exp(−s adX) − I)(sY ).

So Φ(s) = φ(−s adX)(sY ). Conversely, if Φ(s) = φ(−s adX)(sY ), thenΦ(0) = 0 and Φ′(s) = −[X,Φ(s)] + Y .

We now deal with the proof of Theorem 1.7.1.

Proof. Since for t ∈ R, X+tY is a smooth curve passing throughX ∈ g,exp(X + tY ) is a smooth curve in G. Hence the directional derivativeof exp(X) in the direction Y is given by

d

dtexp |t=0 ∈ TexpX(G).

This means that according to our conventions

exp(−X) d

dtexp(X + tY )|t=0 ∈ exp(−X)TexpX(G) = T1(G) = g.

From the proposition above we see that for all s ∈ R

exp(−sX) d

dtexp(s(X + tY ))|t=0 = φ(−s adX)(sY ),

and so taking s = 1, we see that for each Y ∈ g

dX exp(Y ) = exp(−X) d

dtexp(X + tY )|t=0 = φ(− adX)(Y ).


We now turn to the Baker-Campbell-Hausdorff formula itself. LetX and Y be fixed and |t| be small. As every Lie group is analytic (seeAppendix D), exp(tX) exp(tY ) is an analytic function of t which tendsto 1 as t → 0. So by injectivity of exp near 0 we see that

exp(tX) exp(tY ) = expF (t,X, Y ).

where F (t,X, Y ) = F (t) is an analytic function of t. Expanding F in apower series for small |t| we get

F (t,X, Y ) =∑

n≥0

tnCn(X,Y ),

where Cn(X,Y ) = 1n!

dn

dtnF (t,X, Y ) at t = 0. Since we have alreadyseen that F (t,X, Y ) = t(X + Y ) + 1

2t2[X,Y ] + O(t3), it follows that

C0(X,Y ) = 0, C1(X,Y ) = X + Y , and C2(X,Y ) = 12 [X,Y ]. Our

objective now is to calculate the higher Cn(X,Y ). We shall see that foreach n, Cn(X,Y ) is a fixed homogeneous polynomial in the coordinatesof X and Y with rational coefficients consisting of brackets of degreen. They are universal for all linear Lie groups because it is really astatement about GL(n,C) alone. We would thus get

exp(tX) exp(tY ) = exp(t(X + Y ) +1

2t2[X,Y ] + . . .)

valid for small t. Replacing tX and tY by a new X and Y yields

exp(X) exp(Y ) = exp(X + Y +1

2[X,Y ] + . . .), (1.9)

an absolutely convergent series, involving higher-order brackets of Xand Y , valid for small X and Y . Or since log inverts exp locally at theorigin and X and Y are small, it follows that an absolutely convergentseries of these brackets is also small. Hence

log(exp(X) exp(Y )) = X + Y +1

2[X,Y ] + ...

We shall write this last expression as X Y . The Baker-Campbell-Hausdorff formula refers to any of these equations. Wherever valid it


generalizes a fact which we have already proven, namely, if X and Ycommute in g then

exp(X) exp(Y ) = exp(X + Y ).

To see this observe that since X and Y commute so do tX and tY forreal t. Taking t small enough so that tX and tY lie in an appropriateneighborhood and applying the BCH formula we get exp(tX) exp(tY ) =exp(t(X +Y )) for small t because all the other terms will be zero. Nowsince we have real analytic functions this must hold for all t by theidentity theorem (see [55]). Taking t = 1 yields the result. It mightbe worthwhile to mention here that the BCH formula shows in certaincases that the converse is also true (see [16]). Notice that the conversestatement does not follow from exp(tX) exp(tY ) = exp(t(X + Y ) +12t

2[X,Y ] + O(t3)), since if exp(X) exp(Y ) = exp(X + Y ) we cannotconclude that exp(tX) exp(tY ) = exp(t(X+Y )) for all small t and thenmerely take d2/dt2 at t = 0.

In fact, once we know the BCH formula is valid in the case of alinear group we will see that it is valid for an arbitrary Lie group. Thisis because everything in this chapter is local and it is a theorem ofSophus Lie that every Lie group is locally isomorphic to a linear Liegroup. This is due to the fact the Lie algebra of this group has afaithful representation by Ado’s theorem. It is for this reason we writeexp rather than Exp. The same remarks of course apply to the formuladX exp = φ(− adX).

It is also worth noting that if G is a connected nilpotent Lie group,then g is a nilpotent Lie algebra and so, for sufficiently high n, all com-mutators of order n equal zero. This means that the BCH formulabecomes a polynomial and hence converges everywhere on g, not justnear 0. It is also well known, that when G is a connected and simplyconnected nilpotent Lie group, the exponential map is an analytic dif-feomorphism. Together with the BCH formula our previous show that is a polynomial map giving an alternative way of defining a simply con-nected nilpotent Lie group, namely, one that is modeled on Euclideanspace with polynomial multiplication.

We now begin the proof of the BCH formula. Just as before it will


be desirable to blow things up by adding a new variable. We write

exp(uX) exp(vY ) = expZ(u, v,X, Y ) = expZ(u, v).

Then Z is an analytic function of (u, v) for small u and v. Lettingu = t = v we get F (t) = Z(t, t). We also let g(z) = (1 − e−z)/z =φ(−z). Since φ(0) = 1, g(0) = 1 and so g is an entire function. AlsodX exp = g(adX). The zeros of g are z = 2πin where n is a nonzerointeger. Hence g 6= 0 in a neighborhood of 0 so h = 1/g = z/(1 − e−z)is analytic there and h(0) = 1. Finally, let f(z) = z/(1 − e−z) − 1/2z.Then f is also analytic there and f(0) = 1.

Lemma 1.7.6. The function f is even and all its Taylor coefficientsare rational numbers.

Proof. To see that f(z) = f(−z) we show that

z

(1 − e−z)− 1

2z = − z

(1 − ez)+

1

2z.

That is, z( 11−e−z + 1

1−ez ) = z. Or, 11−e−z + 1

1−ez = 1. This last equationis clearly an identity. It follows that all odd Taylor coefficients equalzero. Hence f(z) = 1+

∑p≥1 k2pz

2p. Now all the Taylor coefficients of gare rational. We show the same is true of h. Since f = h+ a polynomialwith rational coefficients this will prove the lemma.

Now h(z)g(z) = 1. Differentiating n times we see that

hn(z)g(z) +∑

0≤i≤n−1

n!

i!(n− i)!hi(z)gn−i(z) = 0

so thathn(0)

n!= − 1

g(0)

∑

0≤i≤n−1

hi(0)

i!

gn−i(0)(n− i)!

By inductionhn(0)

n!∈ Q.


Proposition 1.7.7. For X and Y ∈ g and |t| small, F satisfies the(nonlinear) differential equation

dF

dt= f(adF )(X + Y ) +

1

2[X − Y, F ]

with initial condition F (0) = 0 and analytic coefficients.

Proof. We have already seen that F (0) = 0. Since for z near zero,

f(z) +1

2z =

1

g(z),

we see that for small Z ∈ g

f(adZ) +1

2adZ = g(adZ)−1

and alsof(adZ) = 1 +

∑

p≥1

k2p(adZ)2p.

Now∂

∂v(exp(uX) exp(vY )) =

∂

∂v(expZ(u, v))

so that

exp(uX) exp(vY )Y = dZ(u,v) exp∂Z

∂v.

Identifying Texp(uX) exp(vY )(G) with g by left translation by(exp(uX) exp(vY ))−1 we get

Y = dZ(u,v) exp∂Z

∂v.

Now for small u and v, g(adZ(u, v))−1 exists and equals

h(adZ(u, v)) = f(adZ(u, v)) +1

2adZ(u, v).

Therefore,

f(adZ)(Y ) +1

2adZ(Y ) =

∂Z

∂v.


On the other hand, exp(−vY ) exp(−uX) = exp(−Z(u, v)). Hencetaking ∂

∂u of both sides gives

exp(−vY ) exp(−uX)(−X) = d−Z(u,v) exp(−∂Z∂u

).

Since d−Z(u,v) exp = g(− adZ), after identification this gives

−X = g(− adZ)(−∂Z∂u

).

So

X = g(− adZ)(∂Z

∂u).

Inverting, as before, we get

f(− adZ)(X) − 1

2adZ(X) =

∂Z

∂u.

Now let u = t = v. Then F (t) = Z(u, v) so

F ′(t) =∂Z

∂uu′(t) +

∂Z

∂vv′(t) =

∂Z

∂u+∂Z

∂v.

This means that

dF

dt= f(− adF )(X) − 1

2[F,X] + f(adF )(Y ) +

1

2[F, Y ]

= f(adF )(X + Y ) +1

2[X − Y, F ].

(1.10)

We now complete the proof of the BCH formula. Let F (t) =∑n≥0 t

nCn(X,Y ) where F : (−ǫ, ǫ) → g is the local analytic solution

to the differential equation dFdt = f(adF )(X + Y ) + 1

2 [X − Y, F ] withinitial condition F (0) = 0. Then C0(X,Y ) = 0, and Cn+1(X,Y ) sat-isfies the following formula where S(n) = p ∈ Z+ : 2p ≤ n andT (n) = (a(1), . . . , a(2p)) : a(i) ∈ Z+,

∑a(i) = n.

(n+ 1)Cn+1(X,Y ) =1

2[X − Y,Cn(X,Y )]+ (1.11)


∑

S(n)

k2p

∑

T (n)

[Ca(1)(X,Y ), [. . . [Ca(2p)(X,Y ),X + Y ]] . . .].

These recursive relations clearly determine the Cn(X,Y ) (and hencealso F ) uniquely, since C1(X,Y ) = X + Y .

For example, if n = 1 then S = φ = T and hence

2C2(X,Y ) =1

2[X − Y,X + Y ] = [X,Y ]

so C2(X,Y ) = 12 [X,Y ]. If n = 2 then S = 1 i.e. p = 1 and T = (1, 1).

Hence 3C3(X,Y ) = 12 [X−Y, 1

2 [X,Y ]]+k2[X+Y,X+Y ]. Since k2 doesnot actually enter this formula we get

C3(X,Y ) =1

12[X, [X,Y ]] − 1

12[Y, [X,Y ]].

Proof of (1.1). Fix an integer n. Then

dF

dt(t) = C1 + 2tC2 + . . . (n+ 1)tnCn+1 +O(tn+1). (1.12)

Since ad is linear and continuous

adF (t) = t adC1 + t2 adC2 + tn adCn +O(tn+1).

Hence for any positive integer p with 2p ≤ n

(adF (t))2p =∑

S(n)

ts∑

T (s)

adCa(1) . . . adCa(2p) +O(tn+1).

But also adF (t) = O(t) so applying the power series definition of f wefind that

f(adF (t)) = I +∑

S(n)

k2p(adF (t))2p +O(tn+1).

Hence,

f(adF (t)) = I +∑

1≤s≤nts

∑

S(s)

k2p

∑

T (s)

adCa(1) . . . adCa(2p) +O(tn+1).

(1.13)


Substituting (1.12) and (1.13) into the differential equation for F andequating the coefficients of tn on both sides yields (1.11).

We have already observed that for small X and Y , that X Y is ananalytic function of X and Y . By induction from (1.11) it follows that.

Corollary 1.7.8. For each n, Cn(X,Y ) is a degree n homogeneouspolynomial in the coordinates of X and Y with rational coefficients thatconsisting of brackets.

Corollary 1.7.9. For any X and Y ∈ g we have

limn→∞(exp(1

nX) exp(

1

nY ))n = exp(X + Y ).

In fact, more generally, if Xn → X and Yn → Y , then

limn→∞(exp(1

nXn) exp(

1

nYn))

n = exp(X + Y ).

Using analyticity we now derive the functional equation for the ex-ponential map of a connected real Lie group G. Namely, if X and Y ∈ g

commute, i.e. and [X,Y ] = 0 then

expX · expY = exp(X + Y ). (1.14)

To see this, apply the BCH formula.

exp(X) · exp(Y ) = exp(X + Y +1

2[X,Y ] + . . .), (1.15)

where the right side is an absolutely convergent series, involving higher-order brackets of X and Y , and (1.15) is valid for all small X and Y .Now if X and Y commute, so do tX and tY for all real t. Take |t| smallenough so that the BCH formula applies to tX and tY (where X andY are arbitrarily large). Hence by (1.15), since all the other terms inthe formula are zero,

exp tX · exp tY = exp t(X + Y ), (1.16)

for all small t. Now the exponential function as well as multiplicationin the group are real analytic and exp tX is defined for all t. Also, by


the chain rule, the composition of analytic functions is again analytic.Hence, by the identity theorem for real analytic functions, this holds forall t. Taking t = 1 gives the result.

Definition 1.7.10. Let G be a Lie group and L be a manifold contain-ing a neighborhood U of 1 in G. Suppose V V −1 ⊆ U where V is itselfa neighborhood of 1 in G such that the map on V × V → U sending(g, h) 7→ gh−1 is C∞ (equivalently (g, h) 7→ gh and g 7→ g−1 take valuesin U and are C∞). Then L is called a local Lie subgroup of G and U iscalled a germ of L.

Proposition 1.7.11. Let L be a local Lie subgroup of G and U a germof L. If U is connected then there is a unique connected Lie subgroupH of G in which U is a neighborhood of 1. If (L′, U ′) is another suchpair, then H ′ = H if and only if U ′ ∩ U is open in both U and U ′.

Proof. Let H be the subgroup of the abstract group G generated byU i.e. H is the set of all finite products of elements of U togetherwith their inverses. It is easy to see that H is a submanifold of G andin fact a Lie subgroup of G (see pp. 45-46 of [31] for details). Weshow that H is connected. Now H0, its identity component, containsU . Hence U is a neighborhood of 1 in H0. Therefore, as a connectedLie group, H0 =

⋃n≥1(U ∩ U−1)n. But by definition this is H. H is

clearly determined by U . In fact, since H is connected if W were anyother connected neighborhood of 1 in G (say W ⊆ U), then since H isconnected W generates H. Finally, if H ′ = H, then U ′ and U are bothopen in H and hence U ′ ∩ U is open in H and therefore in both U andU ′. Conversely, if U ′ ∩ U is open in both U and U ′, then since theyeach contain 1 and L is a manifold, the identity component (U ′ ∩ U)0,is open in L and by the remark above generates both H and H ′. SoH = H ′.

Our next result, due to Sophus Lie, is usually proved by means ofthe Frobenius’ Theorem. Here we observe that it follows from the BCHformula.

Corollary 1.7.12. Let G be a Lie group and g its Lie algebra. If h

is any Lie subalgebra of g, then there is a unique connected subgroup


H of G with Lie algebra h. Since H is uniquely determined by U wehave a bijective correspondence between connected subgroups of G andLie subalgebras of g. In particular, if we knew that any Lie algebra hada faithful linear representation, then taking G = GL(n,R) we see thatany Lie algebra over R is the Lie algebra of some real Lie group.

Remark 1.7.13. In general, H need not be closed in G, but will be ifG is simply connected. In particular, this applies to [G,N ] in the resultbelow.

Proof. It is sufficient by the above to show that there is a local Liesubgroup L of G with its germ based on a connected neighborhood U of1 in L. Let L be exp h and V be h∩W , where W is a sufficiently smallspherical canonical neighborhood of 0 in g. Then V is open, contains 0and is connected, so U = exp(V ) and this is what we want.

Since V is symmetric, X 7→ −X is smooth and exp(X)−1 =exp(−X). Therefore, inversion is no problem. Now in V the groupmultiplication is for X and Y ∈ V given by

log(exp(X) · exp(Y )) = X Y = X + Y +1

2[X,Y ] + . . .

which is an analytic function of X and Y . These are called local loga-rithmic coordinates. Since h is closed under [·, ·] and all the terms inthe BCH formula involve brackets of various orders of X’s and Y ’s, eachterm and therefore each partial sum is in h. But h is a closed subspaceof g so X Y ∈ L. If X and Y are sufficiently small X Y ∈ V andso exp(X Y ) ∈ U . This proves that L is a local Lie subgroup of Gwith germ U and therefore the existence of H. The uniqueness of Halso follows from the Proposition 1.7.11.

We shall now derive some consequences of the BCH formula. To do thiswe need the following result which itself requires BCH.

Proposition 1.7.14. Let G be a Lie group, X1, . . . ,Xn be a basis forits Lie algebra g and φi(t) be a family of smooth curves in G such thatφi(0) = 1 and d

dtφi(t)t=0 = Xi. In particular, we could take for theφi(t) = exp(tXi). Since each element of g can be written uniquely in


the form∑

i tiXi, we show that a small enough neighborhood of 1 in Gcan be parameterized by

φ(∑

i

tiXi) = φ(t1, . . . , tn) = log(φ1(t1) . . . φn(tn).

That is, for small (t1, . . . , tn), φ is an analytic map of a neighborhoodof 0 in g → logG ⊆ g, and in fact is a local diffeomorphism at 0.

We say that φ is a set of canonical coordinates of the 2nd kind .

Proof. Since φi(ti) = exp(tiXi+ higher order terms), we see by BCHlog(φ1(t1) . . . φm(tm)) =

∑i tiXi + . . . . For purposes of calculating d0φ

we may assume the higher order terms is not present and hence thatφ(

∑i tiXi) =

∑i tiXi. From this it follows that φ is the identity map

near 0 so d0φ = I, and in particular is nonsingular. By the inversefunction theorem φ is a local diffeomorphism at 0.

A direct consequence of our next result is the fact that for a connectedLie group the notions of nilpotence and solvability for the group and itsLie algebra coincide.

Another consequence of the BCH formula and other facts is

Proposition 1.7.15. Let G be a connected Lie group and H and Nconnected Lie subgroups of G with N normal. Then G = HN if andonly if g = h + n, where g, h and n are the corresponding Lie algebrasof G, H and N respectively.

Proof. Suppose G = HN . Let H × N act on G by (h, n)g = hgn−1.Since OH×N (1) = HN = G, this action is transitive and hence byTheorem 0.4.5 G is H × N equivariantly homeomorphic with H ×N/StabH×N (1). In particular, the multiplication map H × N → Gis open. Let U be a canonical neighborhood of 1 in G and V smallenough so that V 2 ⊆ U . Let VH = H ∩ V and VN = N ∩ V . Thenthese are canonical neighborhoods in H and N , respectively, and by theabove VHVN contains a neighborhood W of 1 in G which is canonicalsince W ⊆ V 2 ⊆ U . If g = expX is in W then g = hn, where h ∈ VH


and n ∈ VN . Hence expX = expY expZ where Y ∈ h and Z ∈ n. Butthe latter is

exp(Y + Z +1

2[Y,Z] + . . .) = exp(Y + Z ′)

where Z ′ ∈ n since n is an ideal. By taking Y and Z small enough,exp(Y + Z ′) ∈ U . It follows that X = Y + Z ′. This proves the claimfor small X. By scaling we see that g = h + n.

Conversely, suppose g = h + n and g ∈ U . Then g = expX, whereX is near 0. By assumption X = Y + Z, where Y ∈ h and Z ∈ n

are near enough to 0 for the BCH series to converge. Accordingly,exp(−Y )g = exp(−Y ) exp(Y +Z) = exp(Z + 1

2 [−Y, Y +Z] + . . .). Now[−Y, Y + Z] = [Z, Y ] ∈ n and, similarly, all subsequent terms are inn, since n is an ideal. Thus exp(−Y )g = exp(Z + Z ′), where Z ′ ∈ n.This means g = expY exp(Z + Z ′) ∈ HN for each g ∈ U . Now since Ugenerates G and N is normal, G = HN .

Remark 1.7.16. An example of the use of Proposition 1.7.15 is itsapplication to compact connected Lie groups G = Z(G)0[G,G] where[G,G] is compact and semisimple.

We now turn to some results of Zassenhaus and Margulis concerningdiscrete subgroups, Γ, of a Lie group G. These were proved by Margulisusing the BCH formula. The original proof, due to Zassenhaus, which wegive here depends on the following lemma involving elementary matrixinequalities, and seems clearer. As usual, ‖·‖ denotes the operator normon Mn(C).

Lemma 1.7.17. Let A = I + α ∈ GL(n,C) where ‖α‖ < 1. ThenI+

∑n≥1(−1)nαn converges in M(n,C) and, equals A−1. Moreover, for

any X ∈ Mn(C), ‖A−1X‖ ≤ ‖X‖1−‖α‖ and similarly, ‖XA−1‖ ≤ ‖X‖

1−‖α‖ .

Finally, if A = I + α and B = I + β ∈ GL(n,C), where ‖α‖ and

‖β‖ < 1, then ‖[A,B] − I‖ ≤ 2 ‖α‖‖β‖(1−‖α‖)(1−‖β‖) .

Proof. That A−1 = I +∑

n≥1(−1)nαn is just a convergent geometric

series. To see that ‖A−1X‖ ≤ ‖X‖1−‖α‖ , simply estimate the ‖A−1‖ by the


geometric series and then apply the fact thatMn(C) is a Banach algebra.Finally, turning to our last inequality, we have ABA−1B−1−I = (AB−BA)A−1B−1 = [α, β]A−1B−1. Hence, applying the previous inequalitytwice and the fact that ‖[α, β]‖ ≤ 2‖α‖‖β‖, yields the result.

We now turn to a result which is usually called the Margulis Lemma.

Theorem 1.7.18. Any Lie group G has a neighborhood Ω of 1 such thatfor any sequence gn ∈ Ω, the sequence given by h1 = g1, h2 = [g2, g1],h3 = [g3, [g2, g1]], . . . converges to 1 in G.

Proof. Since G is a Lie group G0, its identity component, is open in Gso we may assume that G is connected and, as this is a local questionand any connected Lie group is locally isomorphic to a Lie subgroup ofGL(n,C), we may also assume that G is itself a linear group. Choosea neighborhood Ω of I so that for all g ∈ Ω, ‖g − I‖ < ǫ, where 0 <ǫ < 1

3 . We will prove by induction that, for all n, if C is an n-foldcommutator, then ‖C − I‖ < ǫ(3ǫ)n. To see this let C = [A,B], whereA = I +α, B = I +β and ‖α‖ < ǫ and ‖β‖ < ǫ(3ǫ)n−1. Since both ‖α‖and ‖β‖ < ǫ, then 1

(1−‖α‖) and 1(1−‖β‖) are each < 1

(1−ǫ) . Hence, since

‖[α, β]‖ ≤ 2‖α‖‖β‖, we see by Lemma 1.7.17 that,

‖[A,B] − I‖ ≤ 2ǫǫ(3ǫ)n−1

(1 − ǫ)2.

But since ǫ < 13 < 1 −

√32 , it follows that 1

(1−ǫ)2 < 32 and therefore,

that ‖C − I‖ ≤ ǫ(3ǫ)n, thereby proving the inductive statement. Now,for each n, hn is an n-fold commutator and since ǫ < 1, we see that‖hn − 1‖ < (3ǫ)n. But, since 3ǫ < 1, (3ǫ)n converges to 0 and sohn → 1.

In the following corollary what is important is that k may depend onΓ, but Ω depends only on G.

Corollary 1.7.19. Let G be a Lie group, Γ be any discrete subgroupof G and Ω as above. Then there is a fixed integer k such that for anyfinite set g1, g2, . . . , gk ∈ Ω ∩ Γ we have [gk, [gk−1, . . . , g1]] . . .] = 1.


Proof. Let gn be any sequence Ω ∩ Γ. By the Margulis lemma g1,[g2, g1], [g3, [g2, g1]], . . . converges to 1. But this sequence lies in Γ whichis discrete. Hence it is identically 1 from some term on. Thus there isan integer k such that [gk, [gk−1, . . . , g1] . . .] = 1.

We shall always denote by k the smallest such integer. The discrete partof our next result is also usually called the Margulis lemma. Howeverthis result was first proven by Zassenhaus in the 1930s.

Theorem 1.7.20. In any Lie group G there exists a neighborhood Ω of1 such that for any discrete subgroup Γ of G, Ω∩Γ generates a discretenilpotent subgroup N of G. In fact, N is contained in a connectednilpotent Lie subgroup of G.

Proof. Choose Ω smaller than the one in Theorem 1.7.18 and symmetric.Then Ω ∩ Γ is also symmetric. Since N ⊂ Γ it is discrete. Thereis a fixed integer k so that for any choice of g1, g2, . . . , gk ∈ Ω ∩ Γ,[gk, [gk−1, . . . g1] . . .] = 1. For each integer j ≥ 2, let Nj be the subgroupof G (actually of N) generated by the set of commutators Cj of length atleast j with gi ∈ Ω∩Γ. Since Cj+1 ⊆ Cj it follows that Nj+1 ⊆ Nj ⊆ N .We know that Nj = 1 for j ≥ k. We will prove by induction that eachNj is normal in N . For j ≥ k this is clearly so. Suppose inductivelythat Nj+1 is normal in N where j < k. Consider the exact sequence

1 → Nj+1 → Nπ→ N/Nj+1 → 1.

Then,

[π(Ω ∩ Γ), π(Cj)] = π[Ω ∩ Γ, Cj] = π(Cj+1) ⊆ π(Nj+1) = 1.

Since N is generated by Ω∩Γ, N/Nj+1 is generated by π(Ω∩Γ). Also,π(Nj) is generated by π(Cj). It follows that π(Nj) = Nj/Nj+1 is in thecenter of π(N) = N/Nj+1. Thus [π(N), π(Nj)] = π[N,Nj ] = 1 so

[N,Nj ] ⊆ Nj+1 ⊆ Nj.

This means that Nj is normal in N . Since we have also proven thatfor each j, [N,Nj ] ⊆ Nj+1 (for j ≥ k this is also clearly so) we see


that Nj/Nj+1 ⊆ Z(N/Nj+1). Now (N/Nj+1)/(Nj/Nj+1) = N/Nj soif N/Nj were nilpotent then N/Nj+1 would also be nilpotent. Since[N,N ] ⊆ N2, N/N2 is abelian and therefore nilpotent. This shows byinduction that N/Nj is nilpotent for all j and in particular N/Nk = Nis nilpotent.

We now strengthen our result by showing that N is actually con-tained in a connected nilpotent Lie subgroup of G. Let log be the inverseto exp on Ω. By taking Ω small enough we may assume in addition to itsother properties that it has compact closure. Then since Ω is compact,choose a neighborhood V of 0 in g small enough so that Ω ⊆ expV , andAd y(V ) ⊆ log(Ω) for all y ∈ Ω. Let t = log(expV ∩ Γ) (in other words,for this part of the argument we replace Ω by the smaller expV ) and h

by the subalgebra of g generated by t. We shall show by induction ondimG that h is nilpotent. Then the corresponding connected Lie groupH is also nilpotent. Since t ⊆ h, it follows that exp(t) = expV ∩Γ ⊆ H.Since H is a group and expV ∩ Γ generates N , N ⊆ H. Now by theestimates of Lemma 1.7.17

Ck−1 ⊆ (expV ∩ Γ) ⊆ expV,

so each g ∈ Ck−1 is of the form expX for some X ∈ h. Let

nk−1 = X ∈ n : expX ∈ Ck−1.

We show first that [n, nk−1] = 0. Let y = expY ∈ n and x = expX ∈nk−1. Since y ∈ Ω ∩ Γ and x ∈ Ck−1, [y, x] = 1 so yxy−1 = x. Butyxy−1 = exp Ad y(X) so exp Ad y(X) = x ∈ Ck−1 ⊆ Ω. On the otherhand, since y ∈ Ω, X ∈ n ⊆ V we know Ad y(X) ∈ log(Ω). But exp isone-to-one on log(Ω), X = log x and Ad y(X) ∈ log(Ω). We concludethat Ad y(X) = X. But Ad expY = Expad Y so Exp adY (X) = X andsince Exp adY is a linear operator on g, this means Exp adY (tX) =tX for all t and hence that adY (X) = 0. Thus [n, nk−1] = 0. Inparticular, [nk−1, nk−1] = 0. Let a be the abelian subalgebra of g

spanned by nk−1 over R, and A be the corresponding connected Liesubgroup, B its closure, and b the Lie algebra of B. Now the centralizer,z = zb(g) contains b since B and therefore also b is abelian. Let Z =


ZB(G)0 ⊇ B be the corresponding connected Lie subgroup of G. Zis evidently closed in G and so is a Lie group. Let π : Z → Z/Bbe the projection and π′ : z → z/zb(g) be its differential. Since zb(g)is central in z, B is central in Z. Because [n, nk−1] = 0 it followsthat exp(n) centralizes A and therefore also B. Hence n ⊆ z. NowCk−1 = exp nk−1 so since Ck−1 6= 1 it follows that nk−1 6= 0. Hence0 < dimA ≤ dimB so dimZ/B < dimZ ≤ dimG. We see by inductionthat the subalgebra of z/b generated by π′(z∩b) is nilpotent. Because b

is central in z the subalgebra of z (and g) generated by z∩b is nilpotent.Since b ⊆ z this is b.

Chapter 2

Haar Measure and its

Applications

2.1 Haar Measure on a Locally Compact Group

Given a locally compact Hausdorff space X and a continuous real(or complex) valued function f we denote by Supp(f) the setx|f(x) 6= 0.We shall denote by C0(X) the continuous real or com-plex valued functions on X with compact support and by C+

0 (X) theones with positive values. When X is a locally compact group G andf ∈ C0(G) and g ∈ G we define the left translate of f by g ∈ G to befg(x) = f(g−1x) for all x ∈ G.

On any locally compact topological group there is always a nontrivialand essentially unique left (or right) invariant measure dx, called Haarmeasure defined by µ(gE) = µ(E) for every measurable set E ⊆ G andg ∈ G. Alternatively,

∫G f(g−1x)dx =

∫G f(x)dx, for all continuous

functions f with compact support on G and g ∈ G. Here by a mea-sure we shall always mean a nontrivial positive regular measure, thatis one where the measure of a set E can be approximated by open setscontaining E and by compact sets contained in E. Such measures arepositive on non trivial open sets and finite on compact sets. For thedetails regarding regular measures see [65]. Since an invariant measurecan be modified by multiplying by a positive constant and still remain

89

90 Chapter 2 Haar Measure and its Applications

nontrivial positive and invariant there can be no uniqueness to such ameasure. However, if this is the worst that can happen we shall say themeasure is essentially unique.

Theorem 2.1.1. There is an essentially unique left (or right) invariantmeasure on any locally compact group G. This measure is called left (orright) Haar measure.

We first deal with the existence of Haar measure.

Let f and g be nonzero functions in C+0 (G). Then for some positive

integer n there are positive constants c1, · · · , cn and group elementsx1, · · · , xn so that for all x ∈ G

f(x) ≤n∑

i=1

cig(xix)

For example if Mf and Mg are the maximum values of f and g

respectively, then f(x) ≤ (Mf

Mg+ ǫ)gxi

(x) for any choice of n and the xi.

So consider the set of all possible such inequalities and let (f : g) standfor the inf

∑ni=1 ci over this set. Then evidently we have

(1) (fx : g) = (f : g), for every x ∈ G.

(2) (f1 + f2 : g) ≤ (f1 : g) + (f2 : g).

(3) (cf : g) = c(f : g), for c > 0.

(4) If f1 ≤ f2, then (f1 : g) ≤ (f2 : g).

(5) (f : h) ≤ (f : g)(g : h).

(6) (f : g) ≥ Mf

Mg.

This gives us a relative idea of the size of f as compared to g. In orderto have an absolute estimate of the size of f we must fix an f0 ∈ C+

0 (G)

for which (f0 : g) is positive. So we define Ig(f) = (f :g)(f0:g) . Now the

subadditivity of 2) will somehow have to be corrected to become closeto additivity. This will be done by taking g with smaller and smallersupport. Then we will take some kind of limit of the Ig(f) for smallg to get actual additivity of the integral I(f). To do so we need thefollowing lemma.

2.1 Haar Measure on a Locally Compact Group 91

Lemma 2.1.2. For f1 and f2 ∈ C+0 (G) and ǫ > 0 there is a sufficiently

small neighborhood U of 1 in G so that whenever Supp g ⊆ U , we have

Ig(f1) + Ig(f1) ≤ Ig(f1 + f1) + ǫ

Proof. Since f1+f2 ∈ C+0 (G) we know Supp(f1+f2) is compact. Choose

f ′ ∈ C+0 (G) which is ≡ 1 on Supp(f1 + f2). Let δ and ǫ′ > 0, f =

f1 + f2 + δf ′ and for i = 1, 2 let hi = fi

f , it being understood that

hi = 0 whenever f = 0. Then hi ∈ C+0 (G). By uniform continuity

choose a neighborhood U of 1 in G so that if x−1y ∈ U and i = 1, 2then |hi(x) − hi(y)| < ǫ′. Now choose g ∈ C+

0 (G) with Supp g ⊆ V .

Supposef(x) ≤ ∑nj=1 cjg(sjx). If some g(sjx) 6= 0, then |hi(x) −

hi(s−1j )| < ǫ′ for both i and

fi(x) = f(x)hi(x) ≤n∑

j=1

cjg(sjx)hi(x) ≤n∑

j=1

cjg(sjx)(hi(s−1j ) + ǫ′).

Hence (fi : g) ≤ ∑nj=1 cj(hi(s

−1j ) + ǫ′). So that (f1 : g) + (f2 : g) ≤∑n

j=1 cj(1+ 2ǫ′). Because∑n

j=1 cj approximates (f : g) we get Ig(f1)+Ig(f1) ≤ Ig(f1 + f1 + δf ′)(1+2ǫ′) ≤ (Ig(f1 + f1)+ δIg(f

′)(1+2ǫ′). Nowchoose first ǫ′ and then δ small enough so that 2ǫ′(f1 + f2 : f0) + δ(1 +2ǫ′)(f ′ : f0) < ǫ.

Now from 5) we see that Ig(f) always lies in a closed interval

1

(f0 : f)≤ Ig(f) ≤ (f : f0).

If we think of the space of functionals on C0(X) as a subspace of theproduct RC0(X) equipped with the product topology, then by the Ty-chonoff theorem Ig lies in the compact space which is the product ofthese intervals as f varies and the f component of Ig is Ig(f). Foreach neighborhood U of 1 denote by KU the closure of the set of all Igwhere Supp g ⊆ U . Now these closed sets KU have the finite intersec-tion property because for any finite number of Ui KU1∩...∩Un = ∩ni=1KUi

which, by Urysohn’s lemma, in non-empty. Hence we can find a point


I ∈ ∩KU the intersection of all of them. By the properties of theproduct topology, for any such U and any finite number of f1, . . . fn,there is a g with Supp g ⊆ U so that for all i, |I(fi) − Ig(fi)| < ǫ andmoreover

1

(f0 : f)≤ I(f) ≤ (f : f0).

The lemma now shows that I is additive and, of course, invariant. Fi-nally in the usual way one extends I from C+

0 (G) to C0(G) itself byI(f1 − f2) = I(f1) − I(f2) to get a left invariant Haar measure on G.

We now turn to the uniqueness of Haar measure. This is very im-portant because if one can find an invariant measure then it must beHaar measure, suitable normalized.

Proof. Let I = dx and J = dy be two positive left invariant measureson G and f ∈ C+

0 (G). Let C = Supp f and choose an open set Uabout C with compact closure. By Urysohn’s lemma choose a functionφ ∈ C+

0 (G) which is identically 1 on U . Let ǫ > 0 and V be a symmetricneighborhood of 1 in G such that CV ∪ V C ⊆ U . Since f ∈ C0(G) itis uniformly continuous and therefore |f(xy)− f(zx)| < ǫ, for all x ∈ Gand y, z ∈ V .

Then f(xy) = f(xy)φ(x) and f(yx) = f(yx)φ(x) for x ∈ G andy ∈ V . Hence for y ∈ V , |f(xy) − f(yx)| < ǫφ(x) everywhere on G.Now let h be any symmetric function in C+

0 (G) supported on V . Then,by invariance and the Fubini theorem,

I(h)J(f) =

∫ ∫f(x)h(y)dxdy =

∫ ∫h(y)f(yx)dxdy.

and

J(h)I(f) =

∫ ∫f(y)h(x)dxdy =

∫ ∫h(y−1x)f(y)dydx =

∫ ∫h(y)f(xy)dydx,


So that

|I(h)J(f) − J(h)I(f)| ≤∫ ∫

h(y)|f(yx) − f(xy)|dxdy ≤

ǫ

∫ ∫h(y)φ(x)dxdy = ǫI(h)J(φ).

Similarly, if g ∈ C+0 (G) and h is symmetric and suitably chosen

|I(h)J(g) − J(h)I(g)| ≤ ǫI(h)J(ψ).

Hence

|J(f)

I(f)− J(g)

I(g)| ≤ ǫ|J(φ)

I(f)− J(ψ)

I(g)|.

Since ǫ is arbitrary we see the left side of this equation is zero and henceJ(f)I(f) = J(g)

I(g) for any f and g satisfying the above conditions. Let g be

fixed. Then there is a positive c = J(g)I(g) for which J(f) = cI(f) for all

f ∈ C+0 (G).

We now know that on any locally compact group G there is anessentially unique left invariant Haar measure. The same reasoning alsoshows that there is an essentially unique right invariant measure.

Of course since Haar measure is regular, compact groups have fi-nite measure (which is usually normalized to have total mass 1). Theconverse is also true.

Corollary 2.1.3. A locally compact group G has a finite Haar measureif and only if G is compact.

Proof. Let U be a compact neighborhood of 1 in G and V be smallenough so that V V −1 ⊆ U . We consider finite subsets giV which arepairwise disjoint. Now for any such subset µ(G) ≥ µ(

⋃ni=1 giV ) =∑n

i=1 µ(giV ) = nµ(V ). Therefore n ≤ µ(G)µ(V ) . Since the number of such

subsets is bounded there must be a maximum number of them whichwe again call n. Let g ∈ G, but g 6= gi for any i = 1 . . . n. Then gVmust intersect one of the giV so that g ∈ giV V

−1 ⊂ giU . Clearly eachgi is also in giU . Thus G =

⋃ni=1 giU and the latter being a finite union

of compact sets is compact.


Using the uniqueness an obvious example of Haar measure on R isLebesgue measure since it is translation invariant. For the same reasonLebesgue measure on the circle T = S1 is Haar measure. Here sincewe have a compact group and therefore a finite measure it is customaryto normalize and divide by 2π. Later we shall see that normalizedLebesgue measure on S3 is Haar measure for SU(2,C) (= S3), but formore complicated reasons. Since on a finite direct product of groupsevidently left Haar measure is the product of the left Haar measureson the components we know that product measure is Haar measure onT n and Rn. Evidently, counting measure is Haar measure on a discretegroup.

In general in the case of a Lie group one can be somewhat moreexplicit concerning Haar measure. If the dimension of G is n we con-sider left invariant n forms on G. Such a form is determined on allof G by its value at 1. Also the space of all such forms has dimen-sion 1. Since G is orientable choose a nonzero left invariant n formω consistent with the orientation of G and then for each f ∈ C0(G)define I(f) =

∫G f(x)ω(x)dx, where dx is local Lebesgue measure

on a coordinate patch. A partition of unity argument together withthe change of variables formula for multiple integrals shows that I iswell-defined. Although I depends on the ω chosen, ω is uniquely de-termined up to a positive constant. Therefore the same is true ofI. Clearly I gives a measure on G. To see that it is left invariantobserve I(f) =

∫G fω =

∫G d(Lg)(fω) by the left invariance of the

form. But by the change of variables formula for multiple integralsthis is

∫G f(Lg)ω = I(f(Lg)). Thus for the Lie group G we have an

essentially unique left invariant Haar measure given by an invariantvolume form. That is locally on a chart U = (x1, . . . , xn) we havedx = ω(x1, . . . , xn)dx1 . . . dxn, where ω is a non-negative smooth func-tion on U and dx1 . . . dxn is Lebesgue measure on U . Thus dx is abso-lutely continuous with respect to Lebesgue measure. For the Lie groupscase (see [15]).

We will now see in a very explicit way how the change of variableformula can be used to identify Haar measure in many cases.

Proposition 2.1.4. Let G be a Lie group modeled on some open subset


of some Euclidean space, Rn and dg be Lebesgue measure on G inheritedfrom Rn. Let Lg and Rg denote left and right translations on G bythe element g and suppose for each g ∈ G, |det d(Lg)(x)| (respectively

|det d(Rg)(x)|) is independent of x. Then left Haar measure is dg|det d(Lg)|

(respectively right Haar measure is dg|det d(Rg)|). In particular, left and

right Haar measure are absolutely continuous with respect to Lebesguemeasure.

Proof. We prove this for left Haar measure. The case of right Haarmeasure is completely analogous. Since G is an open subset of Euclideanspace we can apply the change of variables formula for multiple integrals.

∫

Gf(x)dx =

∫

Gf(Tx)|det(dT )x|dx,

where T is a smooth global change of variables, dT is its derivative, dxis Lebesgue measure on G and f is a continuous function with compactsupport on G.

We specialize this to T = Lg for g ∈ G. By assumption|det d(Lg)(x)| = φ(g) is independent of x. The function φ is positiveeverywhere on G. Hence for all f ∈ C0(G) and g ∈ G,

∫G f(gx)φ(g)dx =∫

G f(x)dx. Now f(x)φ(x) is again such a function. Applying the last equa-

tion to these functions shows

∫

G

f(gx)

φ(gx)φ(g)dx =

∫

G

f(x)

φ(x)dx.

Taking into account the chain rule and the fact that |det | is multi-plicative we see that φ is a homomorphism on G and hence

∫

Gf(gx)

dx

φ(x)=

∫

Gf(x)

dx

φ(x).

Since f and g are arbitrary in this last equation, uniqueness tells usleft Haar measure is dx

φ(x) .

Now although left Haar measure is often right invariant, the nextexample shows this is not so in general.


Example 2.1.5. Let G be the affine group of the real line R. G consistsof all 2 × 2 real matrices

g =

(a b0 1

), (2.1)

where a 6= 0 ∈ R and b ∈ R. In this way G can be regarded as an openset in the (a, b) plane, R2. G is usually called the ax+ b-group.

A direct calculation shows dLg = aI. Hence |det dLg | = a2, which isindependent of the space variables (as well as b). A similar calculationshows |det dRg | = |a|, which is also independent of the space variables

(as well as b). Thus left Haar measure here is dadba2 while right Haar

measure is dadb|a| . Clearly neither of these measures is a constant multiple

of the other. In fact, they do not even have the same L1 functions!

G is said to be unimodular if left invariant Haar measure is also rightinvariant. Of course abelian groups are unimodular, but as we just sawsolvable ones need not be. Clearly discrete groups are unimodular.

Exercise 2.1.6. Use the proposition 2.1.4 to calculate Haar measureon the following examples which are all unimodular.

(1) Haar measure on GL(n,R) is dx|det x|n , where dx is Lebesgue mea-

sure on Mn(R). This is because |detLg(x)| = |detRg(x)| =|det g|n.

(2) Haar measure on GL(n,C) is dx|det x|2n , where dx is Lebesgue mea-

sure on Mn(C). This is because |detLg(x)| = |detRg(x)| =|det g|2n.

(3) Haar measure on Nn(R), the n × n real unitriangular matri-ces, is just Lebesgue measure. This is because |detLg(x)| =|detRg(x)| = 1. This is a special case of the fact that nilpotentgroups are always unimodular.

On the other hand as we know in the case of the affine group ofthe line, the group GL(n,R) ×η Rn of all affine motions of Rn is notunimodular.


Indeed calculations similar to the ones we have made show that leftHaar measure is dxdy

| det x|n+1 , where dx is Lebesgue measure on GL(n,R)

and dy is Lebesgue measure on Rn. Thus I(f) =∫GL(n,R)

∫Rn

f(x,y)dxdy|det x|n+1 .

This is because |detLg(x)| = |det g| together with what we know about

GL(n,R) itself. Similarly, right Haar measure is dxdy|det x|n , where dx is

Lebesgue measure on GL(n,R) and dy is Lebesgue measure on Rn.

Exercise 2.1.7. Generalize these facts concerning the group of affinemotions to semi-direct products as follows:

Let G×ηH be a semidirect product of unimodular groups, where Gacts on H and dg and dh are Haar measure on G and H respectively.Then right Haar measure on G ×η H is dgdh, while left Haar measure

is dgdh∆(η(g)) , where ∆(η(g)) is the amount that the automorphism η(g)

acting on H distorts Haar measure on H.

In particular, G ×η H is unimodular if and only if G acts on H bymeasure preserving automorphisms and in this case Haar measure is theproduct measure. So for example this is the case for SL(n,R) ×η Rn or

O(n,R) ×η Rn.

Exercise 2.1.8. We now consider the solvable, but not nilpotent, fulltriangular subgroup, B of GL(n,R). This is evidently an open set in aEuclidean space, X. Prove for g = (gij),

|det d(Lg)(x)| = |gn11gn−122 . . . g1

nn|,

which is independent of x. Therefore

d(µl) =dx

|xn1xn−12 . . . x1

n|,

where dx is Lebesgue measure on X. Similarly,

d(µr) =dx

|xnnxn−1n−1 . . . x

11|.

Therefore these groups are not unimodular.


Exercise 2.1.9. Let n1, . . . nr be a partition of n. Thus each ni is apositive integer and

∑ri=1 ni = n. Let P be the subgroup of GL(n,R)

consisting of block triangular matrices with diagonal blocks gi corre-sponding to the ni. Notice that this includes GL(n,R) itself as well asB. Calculate Haar measure.

An example we have not seen before is that of Haar measure ona compact non-abelian group. We will consider the most importantnon-abelian compact group, namely G = SU(2,C). Since as we shallsee in the next section, compact groups are unimodular we need onlyconsider left invariant Haar measure. Once we determine normalizedHaar measure µ on G this automatically gives normalized Haar measureν on the quotient group, SO(3,R). For if π is the universal covering mapand A is a Borel set in SO(3,R), then ν(A) = µ(π−1)(A) is normalizedand invariant.

As we saw (see Section 1.5) each g ∈ SU(2,C) is of the form

g =

(α β−β α

),

where |α|2 + |β|2 = 1. In this way our group can be regarded as the 3-sphere, S3. We go further and give a 4-dimensional real linear realizationof the transformation group G × G → G acting by left translation.This is actually a special case of the equivariant embedding theorem ofMostow-Palais (see [58] ). Since

(α β−β α

)(γ δ−δ γ

)=

(αγ − βδ αδ + βγ−βγ + αδ −βδ + αγ,

)

if we write α = α1 + iα2 and similarly for β, γ and δ, this says

X(γ1, γ2, δ1, δ2) = (Re(αγ−βδ), Im(αγ−βδ), Re(αδ+βγ), Im(αδ+βγ)),

where X = X(g) is the linear transformation on R4 given by

α1 −α2 −β1 −β2

α2 α1 −β2 β1

β1 β2 α1 −α2

β2 −β1 α2 α1.


Now by definition each X(g) preserves S3. Since X(g) is linear itis therefore orthogonal and in particular it is invertible. The subset Xof O(4,R) consisting of these X(g) when acting on the invariant set S3

acts equivariantly with G × S3 → S3 under translation. The fact thatthe map g 7→ X(g) is a homomorphism and X is a subgroup of O(4,R)is immaterial. In any case, since ordinary Lebesgue measure λ on S3 (λis the measure on S3 such that if dx is Lebesgue measure on R4 thendx = r3drdλ is invariant under all of O(4,R) and by the propositionbelow λ(S3) = 2π2, it follows that µ = λ

2π2 is normalized Haar measureon G.

Proposition 2.1.10. In R4 let B4(r) stand for the ball centered at theorigin of radius r > 0 and S3(r) the surface of the corresponding sphere.We denote the Lebesgue measures on Rn and Sn−1(r) by voln and voln−1

respectively.

Then vol4(B4(r)) = π2

2 r4 and vol3(S

3(r)) = 2π2r3.

Proof. As is well known the improper integral,∫ ∞−∞ e−t

2dt =

√π.

Hence by Fubini’s theorem and the functional equation for exp we get∫. . .

∫Rn e

−||x||2dx1 . . . dxn = πn2 . Writing this integral in polar coordi-

nates gives πn2 =

∫ ∞0 e−ρ

2ρn−1dρ

∫Sn−1 dΘ, where dρ is Lebesgue mea-

sure on (0,∞) and dΘ is voln−1 on Sn−1(1) = Sn−1. Now consideringthe volumes of two concentric spherical balls of radius r and r + drcentered at 0 we see that d voln

dr = voln−1(r). Since voln(Bn(r)) =cnr

n, where cn is some constant (to be determined) it follows thatvoln−1(S

n−1(r)) = ncnrn−1 so that

cn =π

n2

n∫ ∞0 e−ρ2ρn−1dρ

,

this latter integral being the value of the gamma function at some halfintegral point. We shall calculate this integral when n = 4 using inte-gration by parts.

∫ ∞

0udv = (uv)|∞0 −

∫ ∞

0vdu


Letting dv = e−ρ2dρ and u = ρ2 we get du = 2ρdρ and v = −1

2e−ρ2 .

Since the evaluative term is zero we conclude∫ ∞0 e−ρ

2ρ3dρ = 1

2 so that

c4 = π2

2 and hence the conclusions.

Exercise 2.1.11. Show that λ(S3) = 2π2 as follows:∫

R4 e−||x||2dx =

(∫ ∞−∞ e−t

2dt)4 = π2. On the other hand this is

∫R4−0 e

−||x||2dx =∫ ∞0 e−r

2r3dr

∫S3 dλ. Thus λ(S3) = π2

R ∞0 e−r2r3dr

. Then calculate the

denominator using integration by parts.

2.2 Properties of the Modular Function

In general left and right Haar measures on a group are connected by themodular function ∆G. This is a continuous map ∆G : G → R×

+ whichmeasures the deviation from right invariance of left Haar measure dgand is defined as follows

∆G(g)

∫

Gf(x)dx =

∫

Gf(xg)dx (2.2)

for all f ∈ Cc(G).

Lemma 2.2.1. (1) ∆G : G→ R×+ is a homomorphism.

(2) ∫

Gf(x−1)∆(x−1)dx =

∫

Gf(x)dx.

Proof. The proof of 1) is a direct check of the definition of the mod-ular function. For 2) note that f 7→

∫G f(x−1)∆(x−1)dx define a left

invariant measure, therefore by uniqueness has to be∫G f(x)dx.

For example we see that the modular function of the affine group of

the real line is given by ∆G(

(a b0 1

)) = 1

|a|

From these properties of the modular function we can immediatelysee that certain groups must be unimodular:

Corollary 2.2.2. Compact and semisimple groups are unimodular.

2.3 Invariant Measures on Homogeneous Spaces 101

This is because ∆G is a continuous homomorphism from G → R×+

and such groups have no nontrivial homomorphisms into R×+.

Suppose for example G has a compact invariant neighborhood ofthe identity, U . Then for each x ∈ G, µ(xUx−1) = µ(U). But by leftinvariance µ(xUx−1) = µ(Ux−1) while the latter is ∆G(x)µ(U). Thus∆G(x)µ(U) = µ(U). Since µ(U) is finite and positive we see ∆G isidentically 1. In particular, compact, discrete and of course abeliangroups are unimodular.

Another example of a class of unimodular groups are the connectednilpotent ones. Since this fact will have little bearing on our work wewill just sketch the proof. In such a group the center Z(G) is alwaysnontrivial by Corollary 3.2.9. So by induction on the dimension G/Z(G)is unimodular. Let dµ be the left and right invariant measure on thisquotient group and dz be (left and right) Haar measure on Z(G). Thenby Theorem 2.3.5 below, I(f) =

∫G/Z(G)

∫Z f(zx)dzdµx is both left and

right invariant on G.

2.3 Invariant Measures on Homogeneous

Spaces

A natural step after studying Haar measure is to find conditions thatguarantee a homogeneous G-space has a G-invariant measure and par-ticularly a finite G-invariant measure. That is the main purpose of thissection. Of course if H is normal in G then since G/H is a group it hasan G-invariant that is G/H-invariant measure, namely Haar measure.

Definition 2.3.1. LetG be a locally compact group acting continuouslyon a locally compact space X (all spaces considered being Hausdorff).We shall call a nontrivial positive (regular) measure dµ(x) on X in-variant if for each g ∈ G, and each measurable set E ⊆ X, we haveµ(gE) = µ(E). Alternatively

∫f(g · x)dµ(x) =

∫f(x)dµ(x) for every

continuous function f on X with compact support.

Just as with Haar measure there can be no uniqueness to invariantmeasures.


An important special of the definition is when G acts transitively onX. When X is G itself and the action is by left translation we have Haarmeasure. When G merely acts transitively, we know X = G/H, whereH is a closed subgroup of G and G operates on G/H by left translation.As we shall see G/H has an (essentially) unique invariant measure ifand only if

∆G|H = ∆H .

So for example, if G is unimodular (and non-compact) and Γ isdiscrete then both sides of this equation would be identically one and soG/Γ would always have a G-invariant measure (which may be infinite).If G/H were compact where H is a closed subgroup and G/H had aninvariant measure then by regularity µ would have to be finite. Thus µ isfinite and invariant. Hence if G = GL(n,R) and B is the full triangulargroup then G/B can have no invariant measure, because it would haveto be finite. This cannot happen because the modular functions do notagree as G is unimodular and B is not. As we shall see in Chapter7 there are other reasons why this cannot happen). Another way tothink of this situation without considering the finiteness of the measureis that since G is unimodular and B is not the nontrivial character 1

∆B

which must extend to G get a invariant measure cannot do it. Such anextension restricted to SL(n,R) must be trivial. Hence, 1

∆Bmust be

trivial on B∗ = B ∩ SL(n,R) which it clearly is not.The following fact is basic. Let G be a locally compact group and

H be a closed subgroup with respective left Haar measures dg and dhand let π : G→ G/H be the natural map.

For f ∈ C0(G) and g ∈ G consider fg|H , the left translate of f by grestricted to H. This is in C0(H) so F (g) =

∫H f(gh)dh exists for each

g ∈ G. Moreover, if h1 ∈ H, then∫H f(gh1h)dh =

∫H f(gh)dh so that

F (gh1) = F (g). Hence F is constant on left cosets and gives a functionon G/H.

Lemma 2.3.2. F ∈ C0(G/H).

Proof. Let ǫ > 0. Since f is uniformly continuous choose a neighbor-hood U(ǫ) of 1 in G so that if xy−1 ∈ U(ǫ), then |f(x)− f(y)| < ǫ. Alsolet U0 be a fixed neighborhood of 1 in G. If gν → g, then eventually


gν ∈ (U0 ∩ U(ǫ))g. Now |F (gν) − F (g)| ≤∫H |f(gνh) − f(gh)|dh. Since

gνh(gh)−1 = gνg

−1 ∈ U(ǫ) we see |f(gνh) − f(gh)| < ǫ. We will showthat function on H is 0 whenever h ∈ H is outside the fixed compactset H ∩g−1U0 Suppf . This is because f(gνh) = 0 if h /∈ g−1

ν Suppf andsimilarly f(gh) = 0 if h /∈ g−1 Supp f . So if h /∈ g−1

ν Supp∪g−1 Suppf ,then |f(gνh) − f(gh)| = 0. But g−1

ν ∈ g−1U0 so if h /∈ g−1U0 Suppf ∪g−1 Suppf = g−1U0 Suppf we get |f(gνh) − f(gh)| = 0. Since this sethas finite H measure and |F (gν) − F (g)| ≤ ǫµH(H ∩ g−1U0 Suppf) itfollows that F is continuous at each g ∈ G. Thus F ∈ C(G/H).

Finally, suppose g /∈ π(Supp f). Then f(gh) = 0 for all h ∈ H.Hence

∫H f(gh)dh = F (g) = 0 so g /∈ SuppF . Thus SuppF ⊆

π(Supp f) and so is compact.

This enables us to define I : C0(G) → C0(G/H) by f 7→ F . Evi-dently I is a linear map taking positive functions to positive functions.

Lemma 2.3.3. I is surjective.

Proof. Let v ∈ C0(G/H) and denote its lift back toG by v. Then for anyψ ∈ C0(G) we have I(ψ·v) = I(ψ)·v. This is because

∫H φ(gh)v(gh)dh =∫

H φ(gh)v(g)dh = v(g)∫H φ(gh). Now let u ∈ C0(G/H). Since u has

compact support choose an open set Ω in G with compact closure sothat u vanishes outside π(Ω). By Urysohn’s lemma choose ψ ∈ C0(G)so that ψ ≥ 0 and ψ|Ω ≡ 1. Since g = gH where g ∈ Ω we haveψ(g) = 1. Thus ψ(xh) ≥ 0 for all h and ψ(g1) > 0. Hence for g ∈ π(Ω),I(ψ)(g) > 0.

Define v ∈ C0(G/H) by v(g) = u(g)I(ψ)(g) , if g ∈ π(Ω) and v(g) = 0

otherwise. Then v has compact support and is continuous on the openset π(Ω). Since u vanishes outside π(Ω) and is continuous, v is alsocontinuous on the boundary so v ∈ C0(G/H). Also U(g) = I(ψ)v(g)everywhere on G/H. Hence u = I(ψ)v = I(ψ · v).

We keep the same notation. Let the modular functions on G and Hbe ∆G and ∆H .

Lemma 2.3.4. Suppose∆G|H ≡ ∆H . Let f ∈ C0(G). If I(f) = 0, then∫G f(g)dg = 0.


Proof. Let φ ∈ C0(G) be a function such that I(φ) ≡ 1 on Supp f . Now∫H

∫G φ(g)f(gh)dgdh =

∫G φ(g)dg

∫H f(gh)dh = 0.

Also since∫G |φ(g)|dg

∫H |f(gh)|dh < ∞, Fubini’s theorem applies.

Hence

0 =

∫

H

∫

Gφ(g)f(gh)dgdh =

∫

H

∫

Gφ(gh−1)f(g)∆G(h)dgdh.

But this is∫

Gf(g)

∫

Hφ(gh−1)∆G(h)dhdg =

∫

Gf(g)

∫

Hφ(gh)∆H (h−1)∆G(h)dhdg

=

∫

Gf(g)

∫

Hφ(gh)dhdg =

∫

Gf(g)

by the choice of φ.

Theorem 2.3.5. Let G be a locally compact group and H be a closedsubgroup. Then there exists an essentially unique invariant measure dgon G/H satisfying

∫

Gf(g)dg =

∫

G/H

∫

Hfg(h)dhdg,

f ∈ C0(G) if and only if ∆G|H = ∆H .

Proof. Suppose G/H has a G-invariant measure d(g). Let f ∈ C0(G).Consider F ∈ C0(G/H) as above. Then

∫G/H F (g)d(g) = J(f) is a

positive linear functional on C0(G).

J(f) =

∫

G/H

∫

Hf(gh)dhd(g)

For g1 ∈ G,

J(fg1) =

∫

G/H

∫

Hf(g1gh)dhd(g) =

∫

G/HF (g1g)d(g)

=

∫

G/HF (g)d(g) = J(f).


So that J is G-invariant. Since I is surjective J is also nontrivial.This means by uniqueness of Haar measure J must be Haar measure onG with some normalization. Any two such measures on G/H give Haarmeasure after normalization. Since I is surjective these measures mustcoincide. Furthermore

∫G f(g)dg =

∫G/H

∫H f(gh)dhd(g). Let h1 ∈ H,

then∫G f(gh1)dg =

∫GRh1f(g)dg = ∆G(h1)

∫G f(g)dg. On the other

hand this is∫G/H

∫H Rh1f(gh)dhd(g) =

∫G/H

∫H ∆H(h1)f(gh)dhd(g) =

∆H(h1)∫G f(g)dg. Hence for every f ∈ C0(G), ∆G(h1)

∫G f(g)dg =

∆H(h1)∫G f(g)dg. This means ∆G|H ≡ ∆H .

Conversely, suppose ∆G|H ≡ ∆H . Then by Lemma 2.3.4 the linearform f 7→

∫G f factors through I and defines an invariant measure on

G/H. Uniqueness follows from that of Haar measure

Our next result is usually referred to as pushing a measure forward.Its proof is straight forward and is left to the reader.

Proposition 2.3.6. Suppose X and Y are locally compact G-spaces, µis a regular, G-invariant measure on X and π : X → Y is a continuous,surjective G-equivariant map, then for any measurable set S ⊆ Y wedefine ν(S) = µ(π−1(S)). It is easy to see that ν is a regular G-invariantmeasure on Y which is finite if µ is.

We can apply some of this to calculate Haar measure on SL(n,R).(As with SU(2,C) this group is also not diffeomorphic to a single openset in some Euclidean space). Write the Iwasawa decomposition ofSL(n,R) = G = KAN , where here K = SO(n) and AN = B+, the realtriangular matrices all of whose eigenvalues are positive and of det = 1.Let dk, da+, and dn be Haar measures on K, A+ and N respectively.Because these groups are compact, abelian, or nilpotent they are all uni-modular (For the compact case see Corollary 2.2.2. Here N is actuallythe nilpotent group of Exercise 2.1.6). Write G = B+K. Hence the mapG/K → B+ is a B+-equivariant diffeomorphism. Since both G and Kare unimodular G/K has a G-invariant and therefore B+-invariant mea-sure, which by Proposition 2.3.6 can be pushed forward to give left Haarmeasure db+ on B+. Hence by Theorem 2.3.5 f 7→

∫B+ db

+∫K f(b+k)dk

is (left) Haar measure on the unimodular group G. So we are reduced


the question of what Haar measure is on B+? Since B+ is the semidi-rect product of A+ and N using the semi-direct product result one getsdb+ = Πi<j

aii

ajjda+dn and so dg = Πi<j

aii

ajjdkda+dn.

2.4 Compact or Finite Volume Quotients

Definition 2.4.1. We say a closed subgroup H of G has cofinite vol-ume in G if G/H has a finite G-invariant measure. We shall say H iscocompact or a uniform subgroup of G if G/H is compact. In particular,if Γ is discrete and of cofinite volume then we say Γ is a lattice in G; ifΓ is a discrete subgroup and G/Γ is compact then we say Γ is a uniformlattice in G.

Notice that if Γ is a lattice or a uniform lattice in a connected Liegroup, G, we can always pull this back to such a thing in G, its universalcovering group. This means that in many situations we may as wellassume G itself is simply connected.

Proposition 2.4.2. If a locally compact group G admits a lattice itmust be unimodular.

Proof. Observe that ∆G|Γ = ∆Γ = 1. Hence Γ ⊆ Ker ∆G. So thatthe finite measure on G/Γ pushes forward to give a finite G-invariantmeasure on G/Ker ∆G ⊆ R×

+. As Ker ∆G is normal in G, G/Ker ∆G

is a group so by uniqueness this must be left Haar measure. Since themeasure is finite G/Ker ∆G is compact. On the other hand R×

+ has nonontrivial compact subgroups and so G is unimodular.

Thus for example the group of affine motions of the real line notbeing unimodular cannot have lattices. There are also other necessaryconditions, but there are no known necessary and sufficient conditions,in general, for a locally compact group or even a Lie group to possess alattice. However, Borel has shown [7] that any connected semisimple Liegroup of non-compact type has both uniform and non uniform lattices.It is a theorem of Mostow [71] that in a connected solvable Lie groupG and a closed subgroup H, then G/H is compact if and only if it

2.4 Compact or Finite Volume Quotients 107

carries a finite invariant measure. In particular, this holds for nilpotentgroups and discrete subgroups. A theorem of Malcev [71] tells us asimply connected nilpotent Lie group has a lattice if and only if theLie algebra has a basis in which all structure constants are rational.Proposition 3.1.69 now shows that there are simply connected 2-stepnilpotent groups which have no lattices. The next few Propositions willbe useful.

Proposition 2.4.3. Let G be a locally compact group, Γ a discrete sub-group. If Ω is a measurable set in G of finite Haar measurable satisfyingΩΓ = G, then Γ is a lattice in G. That is, G/Γ has a finite invariantmeasure.

Proof. Choose measures dg, dg and dγ, with dg invariant as in Theorem2.3.5 so that ∫

Gdg =

∫

G/Γ

∫

Γdγdg

and apply this to χΩ, the characteristic function of Ω. Then

∫

G/Γ

∫

ΓχΩ(gγ)dγdg = µ(Ω) <∞.

Since each g ∈ G is of the form g = ω1γ1 and dγ is left invariant we seethat because Γ is discrete

∫

ΓχΩ(gγ)dγ =

∫

ΓχΩ(ω1γ)dγ =

∑

γ∈Γ

χΩ(ω1γ).

Now this last term is everywhere ≥ 1. If not, χΩ(ω1γ) = 0 for all γ ∈Γ; that is for all γ, ω1γ lies outside of Ω. Thus Γ∩ω−1

1 Ω is empty. Thisis impossible since 1 ∈ Γ and 1 = ω−1

1 ω1. Thus ∞ > µ(Ω) ≥∫G/Γ 1dg.

An application of Corollary 2.3.6 completes the proof.

Conversely we have

Proposition 2.4.4. Let G be a Lie group and Γ a lattice. Then thereexists a measurable set Ω in G of finite measure satisfying ΩΓ = G.


Actually, more is true as we shall see in Chapter 8. For any discretesubgroup Γ, there exists an open set Ω ⊆ G satisfying

(1) For any two distinct γ1 and γ2 ∈ Γ the sets γ1Ω and γ2Ω aredisjoint.

(2)⋃γ∈Γ γΩ = G.

Here we need the fact that the space on which Γ acts is a completeRiemannian manifold.

In particular from 2), taking inverses, we have G = (Ω)−1Γ so π :Ω−1 → G/Γ is surjective and injective. Since G/Γ has finite volumewith respect to the push forward measure therefore Ω−1 has a finitemeasure and hence also Ω

The sister result to Propositions 2.4.3 and 2.4.4 above is the follow-ing:

Proposition 2.4.5. Let G be a locally compact group and H be a closedsubgroup. Then G/H is compact if and only if there is a compact sym-metric neighborhood Ω of 1 in G satisfying ΩH = G.

In particular, if H and G/H are both compact, then so is G. Animportant special case of this is the situation where G is a compactconnected Lie group and G is its universal cover. Then G is compact ifand only if π1(G) is finite.

Proof. If there is such an Ω, then π : G → G/H is surjective whenrestricted to Ω. Since π is continuous and Ω is compact so is G/H.

Conversely, choose a compact neighborhood U of 1 in G. Then sinceπ is both continuous and open π(U) is a compact neighborhood of π(1)in G/H. Cover G/H by a finite number of its G translates giπ(U) wherei = 1, . . . , n. Then Ω =

⋃ni=1 gi(U) is compact and ΩH = G. Since, we

can always include 1 as one of the translates, Ω is a neighborhood of 1.By replacing Ω by Ω ∪ Ω−1 we may assume Ω is symmetric.

Proposition 2.4.6. If G/H is compact and dg and dh are the respectiveleft Haar measures then there is a non-negative function ω in C+

0 (G)such that

∫H ωg|H ≡ 1. Hence if f the lift back to G of a continuous

function f on G/H, then∫G ω(g)f(g)dg =

∫G/H f(g)dg.


Proof. Since G/H is compact and I is surjective by Lemma 2.3.3 theconstant function 1 has an inverse image. For the second statement,

∫

Gω(g)f(g)dg =

∫

G/H

∫

Hω(gh)f(gh)dhdg.

But this last term is∫G/H f(g)(

∫H ω(hg)dh)dg =

∫G/H f(g)dg

We now formulate two general propositions concerning subgroupsof cofinite volume which are analogous to the second and third isomor-phism theorems for topological groups (see Corollaries 1.4.10 and 1.4.9).

Proposition 2.4.7. Let G be a locally compact σ-compact group andL and H closed subgroups with H normalizing L and HL closed in G.Then HL/H has a finite HL-invariant measure if and only if L/H ∩Lhas finite L-invariant measure.

Proof. Consider the natural map L/H ∩ L→ HL/H. As we saw therethis map is a homeomorphism which intertwines the actions L on thefirst and HL on the second. By Proposition 2.3.6 (perhaps slightlygeneralized to two different groups acting) applied to the inverse of themap if HL/H has a finite HL-invariant measure then L/H∩L has finiteL-invariant measure.

Conversely, if L/H ∩ L has finite L-invariant measure then by thesame reasoning as above we see that HL/H also has a finite L-invariantmeasure µ. For h ∈ H let νh be the measure on HL/H defined byµh(E) = µ(hE), where E is a measurable set in HL/H. Now for l ∈ L,we have by L-invariance,

µh(lE) = µ(hlE) = µ(hlh−1hE) = µ(l′hE) = µ(hE) = µh(E).

Thus each µh is also an L-invariant measure on HL/H. By unique-ness µh = λ(h)µ, where λ : H → R×

+ is a character. But µ is a finitemeasure. Therefore letting E = HL/H we see λ(h)µ(E) = µh(E) =µ(hE) = µ(E) so that λ(h) ≡ 1 and µh = µ for all h and µ is H-invariant. This means µ is HL -invariant.


Proposition 2.4.8. Let G be a locally compact group and H1 and H2

closed subgroups with H1 ⊇ H2. Then G/H2 has a finite G-invariantmeasure if and only if G/H1 and H1/H2 each have a finite G-invariantmeasure.

Proof. If G/H2 has a finite G-invariant measure then so does G/H1

since π : G/H2 → G/H1 can be used to push this measure forward(see Proposition 2.3.6). Since both G/H2 and G/H1 carry invariantmeasures we know from Theorem 2.3.5 that ∆G|Hi = ∆Hi

for i = 1, 2.Hence ∆H1 |H2 = ∆H2 and therefore again by Theorem 2.3.5 H1/H2 alsocarries an H1-invariant measure. Let dx, dy and dz be these measures.Consider the linear functional I defined on C0(G/H2) by

I(f) =

∫

G/H1

(

∫

H1/H2

f(ghH2)dz)dy (2.3)

This is a G invariant measure on G/H2 so by uniqueness it is dxafter normalization. Applying (2.3) to the constant function 1 tells∫G/H1

dy∫H1/H2

dz = 1. By Fubini’s theorem∫H1/H2

dz <∞.

Conversely if G/H1 and H1/H2 each carries a finite G-invariantmeasure, then (2.3) defines a finite G-invariant measure on G/H2.

Next we come to a general result which is useful in distinguishinguniform from non-uniform lattices. Refinements of this result play animportant role in arithmetic groups.

Theorem 2.4.9. Let G be a connected Lie group, Γ be a lattice in Gand π : G→ G/Γ the natural projection. For a sequence xn in G, π(xn)has no convergent subsequence if and only if there exists a sequenceγn 6= 1 in Γ so that xnγnx−1

n converges to 1.

So for example if Γ is a uniform lattice, then given any sequencexn ∈ G, the only sequence γn ∈ Γ for which xnγnx

−1n → 1 is one where

eventually all γn = 1.

Proof. Since G is connected and locally compact it is σ-compact andbecause G has a lattice it is unimodular Proposition 2.4.2. As a σ-compact group choose an increasing sequence Fn of compact subsets


which fill out G. Let µ be Haar measure on G and µ a finite invariantmeasure on G/Γ. Since π is surjective π(∪Fn) = G/Γ. Now π(Fn)is compact and measurable and µ is finite so by Ergoroff’s theoremµ(π(Fn)) ↑ µ(G/Γ). Letting ǫn = µ(G/Γ)−µ(π(Fn)) it follows that ǫn ↓0. Since G obeys the first axiom of countability, choose a fundamentalsequence Vn of compact neighborhoods of 1 in G with µ(Vn) > ǫn.This can be done by considering balls B(r) of radius r > 0 centered at0 in g, the Lie algebra. Now r 7→ µ(exp(B(r))) is a continuous functionon some neighborhood 0 < r < δ of 0 and takes on all positive valuesin some interval. Therefore, there is a sequence B(rn) with rn ↓ 0, withµ(exp(B(rn))) = 2ǫn for each n. Letting Vn = exp(B(rn)) gives such asequence.

Now VnV−1n is also a fundamental sequence of compact neighbor-

hoods of 1 in G. Suppose π(xn) has no limit point in G/Γ. Sincefor each n, π(VnV

−1n Fn) is compact for any n there must be an inte-

ger kn so that π(xm) /∈ π(VnV−1n Fn) if m ≥ kn. As a consequence

π(Vnxm)∩π(VnFn) = φ for all m ≥ kn because if for some γ andm ≥ kn,vnxmγ = v′nfnγ

′, then xm = v−1n v′nfnγ

′′so π(xm) ∈ π(V −1

n V ′nFn), a con-

tradiction.

Therefore π(Vnxm) ⊆ G/Γ − π(VnFn) ⊆ G/Γ − π(Fn) and soµ(π(Vnxm)) ≤ µ(G/Γ − π(Fn)) = ǫn. But µ(Vnxm) = µ(Vn)) > ǫn.Since Γ is discrete so measure on Γ is given by counting, therefore fora measurable set S ⊂ G which intersects Γ in at most one point wehave µ(S) = µ(π(S)). Since Vn is a neighborhood basis at 1 this is acontradiction unless (taking Vnxm for S) there is a γm and γ′m ∈ Γ,where γm 6= γ′m so that vxmγm = v′xmγ′m. But then xmγ

′′

mx−1m = v−1v′.

Therefore for each n there exist a large enough m and a γ′′m ∈ Γ suchthat xmγ

′′

mx−1m ∈ VnV

−1n ; hence xmγ

′′

mx−1m converges to 1.

Conversely, let xn ∈ G be a sequence and suppose there is a sequenceγn ∈ Γ eventually 6= 1 with xnγnx

−1n converging to 1 as n → ∞. We

show the image π of such a sequence cannot have a limit point. Forsuppose it did, say π(xn) → π(x). Since Γ is discrete π is a localhomeomorphism. So in some neighborhood of x in G we can find θn ∈ Γso that xnθn → x. But then xnγnx

−1n = xnθnθ

−1n γnθnθ

−1n x−1

n so thelimiting value of this is xθ−1

n γnθnx−1 → 1. But then θ−1

n γnθn must also


converge to 1. Since Γ is discrete θ−1n γnθn must eventually be 1. Hence

γn is also eventually 1, a contradiction.

We now apply Theorem 2.4.9 to the following question. Let G be aLie group H a closed subgroup and Γ a lattice in G. When is Γ ∩H alattice in H? As we shall see this is a rare occurrence.

Corollary 2.4.10. Let G be a connected Lie group, H be a closed sub-group and Γ be a lattice in G. If H ∩ Γ is a lattice in H if and only ifHΓ is closed in G. Equivalently the injection ι : H/H ∩ Γ → G/Γ isproper.

If H is normal then these conditions are equivalent by Proposition2.4.7.

Proof. Suppose H ∩ Γ is a lattice in H and πH : H → H/H ∩ Γ andπG : G → G/Γ be the natural projections. To show that ι is properit is enough to prove that for a sequence hn ∈ H, πH(hn) has a limitpoint if and only if πG(hn) has one. If πH(hn) converges so does πG(hn)because ι is continuous. Suppose πG(hn) converges, but πH(hn) has nolimit point. Then by Theorem 2.4.9 there are elements γn ∈ H ∩Γ suchthat hnγnh

−1n converges to 1. Then by Theorem 2.4.9 again πG(hn) has

no limit point, a contradiction. Consider the commutative diagram

H

πH

// G

πG

H/H ∩ Γ

ι // G/Γ

(2.4)

We have HΓ = π−1G (ι(H/Γ ∩ H)), by the argument above ι sends the

closed sets to closed sets which shows ι(H/Γ ∩ H) is closed and sinceπG is continuous therefore HΓ is closed.

2.5 Applications

In this section we shall give a number of applications of Haar measureon a compact group to derive various algebraic and geometric propertiesof compact Lie groups.

2.5 Applications 113

A first application is the fact that compact linear groups are com-pletely reducible.

Theorem 2.5.1. ρ : G → GL(V ) be a finite dimensional continuousrepresentation of a compact group on a real or complex linear space V .Then any invariant subspace W of V has an invariant complement.

Proof. By replacing G by ρ(G) we may assume the compact group isa subgroup of GL(V ). Let (·, ·) be any positive definite symmetric (re-spectively Hermitian) form on V and let dg be normalized right Haarmeasure on G. Then the form 〈·, ·〉 is defined on V as follows:

〈v,w〉 =

∫

G(gv, gw)dg.

Since∫G is linear and (·, ·) is bilinear symmetric (respectively Hermi-

tian conjugate linear) it follows 〈·, ·〉 is also bilinear symmetric (respec-tively Hermitian conjugate linear). In addition 〈·, ·〉 is positive definite.For 〈v, v〉 =

∫G(gv, gv)dg ≥ 0 because the integrand (gv, gv) ≥ 0 every-

where since (·, ·) itself is positive definite. If 〈v, v〉 = 0 then v = 0. Thisis because the integrand (gv, gv) ≥ 0 and is positive at g=1 unless ofcourse v itself is 0. Finally, 〈·, ·〉 is G-invariant. For any h ∈ G, becausedg is right invariant

〈hv, hw〉 =

∫

G(ghv, ghw)dg =

∫

G(gv, gw)dg = 〈v,w〉.

Thus we have V has a positive definite invariant symmetric (respec-tively Hermitian) form. It follows from this that W⊥ with respect tothis form is also G-invariant. For if w⊥ ∈ W⊥, g ∈ G and w ∈ W〈(gw⊥), w〉 = 〈w⊥, g−1w〉 Since W is G-invariant and w⊥ ∈ W⊥ thislast term is zero so that gw⊥ ∈ W⊥. Thus W⊥ is G-invariant. Choos-ing an orthonormal basis for V by putting together two such forms Wand W⊥ shows G acts completely reducibly.

Proposition 2.5.2. Let G be a compactly generated locally compactgroup and H a closed subgroup with G/H compact. Then H is alsocompactly generated.


Proof. By Proposition 2.4.5 choose a compact symmetric neighborhoodU0 of 1 in G with G = U0H and large enough so that it generates G.Then U2

0 is compact and is contained in G = U0H. Therefore U20 ⊆⋃n

i=1 U0hi, where hi ∈ H. Let F = h1, . . . hn and 〈F 〉 be the (finitelygenerated) subgroup of H generated by F . Since U2

0 ⊆ U0F ⊆ U0〈F 〉we see that U3

0 ⊆ U20 〈F 〉 = U0〈F 〉2 = U0〈F 〉. Continuing in this way it

follows that Un0 ⊆ U0〈F 〉 for every n ≥ 1. Since U0 generates G we getG ⊆ U0〈F 〉 and in particular H ⊆ U0〈F 〉. Let U0,H = U0 ∩ H. Thenthis is a compact neighborhood of 1 in H and H = U0,H〈F 〉. Thus His compactly generated.

This has as a consequence

Corollary 2.5.3. (1) Let G be a connected locally compact group andΓ a discrete cocompact subgroup. Then Γ is finitely generated.

(2) If X is a compact space on which a connected Lie group acts tran-sitively, then π1(X) is finitely generated.

(3) If G be a compact connected Lie group, then π1(G) is finitely gen-erated.

Proof. (1) The first statement follows from the fact that a connectedlocally compact group is compactly generated.

(2) Let X = G/H and (G, π) be the universal covering group ofG. Then G/π−1(H) is homeomorphic and G-equivariantly equiv-alent with G/H = X. So these spaces have the same funda-mental groups. But π1(G/π

−1(H)) = π−1(H)/π−1(H)0. SinceG/π−1(H) is compact and G is connected π−1(H) is compactlygenerated. Hence so is π−1(H)/π−1(H)0. It is also discrete be-cause π−1(H) is closed in G and therefore is also a Lie group.Hence π−1(H)/π−1(H)0 is discrete and therefore finitely gener-ated.

(3) This follows from 2 by letting G = X and H = 1.

Proposition 2.5.4. The fundamental group of a connected Lie groupis abelian.


Proof. Let (G, π) be the universal covering group of G. Then Kerπis the fundamental group π1(G) and it is normal discrete subgroup ofG. Thus all we need to know is that a discrete normal subgroup ofa connected group is central and therefore abelian and for that seeProposition 0.3.7.

The following is a modification of an important observation of PierreCartier but before that we remark that if H is a locally compact groupand dh is right Haar measure we can uniquely extend it vector valuedfunctions i.e. to C0(H,V ) where V is a finite dimensional real vectorspace as follows: For λ ∈ V ∗, the dual space and φ ∈ C0(H,V ) we define∫φdh by

λ(

∫

Hφdh) =

∫

Hλφdh

Thus we integrate in each coordinate. When we do this we again getan H-invariant linear vector valued function. Now if we take as ourvector space EndR(V ) and if T is a fixed linear operator on V , then fora continuous φ : H → EndR(V ) with compact support we have

T (

∫

Hφdh) =

∫

HTφdh.

Theorem 2.5.5. Let G be a locally compact group and H a closed nor-mal subgroup with G/H compact and ρ be a continuous finite dimen-sional representation of G on the real vector space V whose restrictionto H is trivial. If f : H → V is a group homomorphism i.e.

f(hh′) = f(h) + f(h′)

satisfying the invariance condition

ρ(g)(f(h)) = f(ghg−1)

for all g ∈ G and h ∈ H, then f extends to a continuous f∗ : G → Vsatisfying1 f∗(xy) = ρ(x)(f∗(y)) + f∗(x) for all x, y ∈ G.

1This is 1-cocycle condition and leads to a degree one class in the cohomology ofgroup G.


Proof. Since G/H is compact by Proposition 2.4.5 there exits a weight-ing function ω ∈ C0(G,R) such that for all x ∈ G

∫

Hω(hx)dh ≡ 1.

Since f(h′h) = f(h′) + f(h) for h, h′ ∈ H we have,

∫

Hω(h′x)f(h′h)dh′ −

∫

Hω(h′x)f(h′)dh′ = f(h)

∫

Hω(h′h)dh′ = f(h).

Now translate h′ 7→ h′h−1 and apply right invariance giving

∫

Hω(h′h−1x)f(h′)dh′ −

∫

Hω(h′h−1x)f(h′h−1)dh′ = f(h).

So that ∫

H(h−1x)ω · fdh′ −

∫

H(x)ω · fdh′ = f(h), (2.5)

here y.ω means the right translation action. Define f1 : G→ V by

f1(x) =

∫

Hxω · fdh′ − x ·

∫

Hω · fdh′,

for x ∈ G. Since ρ is continuous f1 is a continuous V valued functionand

f1(h−1x)−f1(x) =

∫

H(h−1x·ω)f−h−1x

∫

H(ω)f−(

∫

H(x·ω)f−x·

∫

Hωf).

By (2.5) get f1(h−1x) − f1(x) = f(h) − h−1x

∫H ωf + x

∫H ωf . Let

v0 = x∫H ωf ∈ V . Then for all h ∈ H and x ∈ G, f1(h

−1x) − f1(x) =f(h) − h−1v0 + v0. But since H acts trivially on V we get

f1(h−1x) − f1(x) = f(h). (2.6)

For each x ∈ G let gx : G→ V be defined by

gx(t) = t−1(f1(t) − f1(tx)),


for t ∈ G. Since ρ is continuous and right translation is also each gx is acontinuous function of t. We prove that gx is constant on right cosets ofH in G. We have gx(ht) = (ht)−1(f1(ht) − f1(htx)). By (2.6) we knowf1(htx) − f1(tx) = f(h−1) and f1(ht) − f1(t) = f(h−1), therefore

f1(ht) − f1(htx) = f1(t) − f1(xt).

Since H acts trivially we gx(ht) = t−1h−1(f1(t) − f1(xt)) = gx(t). Letgx be the induced function on cosets and then consider the V -valuedintegral

f∗(x) =

∫

G/Hgx(t)dt,

where dt is the normalized Haar measure on G/H. Then f∗ : G→ V iscontinuous. For h ∈ H we have gh(t) = t−1(f1(t)−f1(th)) = t−1(f1(t)−f1(tht

−1t)). Since h′ = tht−1 ∈ H, using (2.6) we get −t−1(f1(h′t) −

f1(t)) = −t−1f((tht−1)−1) = −t−1f(th−1t−1) = −f(h−1) = f(h). Thusfor all h ∈ H and t ∈ G we get gh(t) = f(h). This means f∗(h) =∫G/H gh(t)dt =

∫G/H f(h)dt = f(h), so that f∗ is an extension of f .

For x, t, u ∈ G, a direct calculation similar to one just above tells us

u−1((ux)gt(ux)) = u−1(f1(ux) − f1(uxt))

and so x · gt(ux) = u−1(f1(ux)− f1(uxt)). This means gxt(u) = gx(u)+x · gt(ux) and therefore. gxt = gx + x · gt(Rx(u)) (right translation).Integrating this last equation over G/H yields

f∗(xt) = f∗(x) + x ·∫

G/Hgt(Rx(u))du

But since∫G/H gt(Rx(u))du =

∫G/H gt(u)du by invariance of the integral

we getf∗(xt) = f∗(x) + x · f∗(t).

Corollary 2.5.6. Let G be a locally compact group and V a normalvector subgroup with G/V compact. Then G is a semidirect product ofa compact subgroup K of G with V .


Proof. Now conjugation on G leaves V stable and gives a continuousrepresentation g → αg|V which we denote by α of G on V . SinceV is abelian α restricted to V is trivial. Evidently i is a continuoushomomorphism V → V satisfying the invariance condition with respectto α. Hence i extends to a continuous map i∗ : G → V as in Theorem2.5.5 satisfying

i∗(xy) = xi∗(y)x−1 · i∗(x),here we write the group V multiplicatively. We show the sequence

1 → Vi→ G

π→ G/V → 1

has a continuous global cross section.For x ∈ G let Φ(x) = i∗(x)−1x. Then Φ is a continuous map G→ G.

Moreover

Φ(xy) = i∗(xy)−1xy = (xi∗(y)x−1i∗(x))−1xy = i∗(x)−1xi∗(y)−1x−1xy

= i∗(x)−1xi∗(y)−1y = Φ(x)Φ(y)

Thus Φ : G→ G is a continuous homomorphism. If v ∈ V then Φ(v) =i∗(v)v−1 = 1 since i∗ extends i. Thus V ⊆ Ker Φ and therefore Φ inducesa continuous homomorphism Φ : G/V → G with Φ(G/V ) = Φ(G) = K,a compact subgroup of G. Finally we show π Φ = idG/V . Since Vis normal g−1i∗(g)−1g ∈ V for all g ∈ G. Thus gV = i∗(g)−1gV fromwhich π Φ = idG/V follows.

Using Corollary 2.5.6 and induction on the length of the derived se-ries together with the fact that a connected solvable Lie group is simplyconnected if and only if it has no nontrivial compact subgroup, one canextend this result to arbitrary simply connected solvable subgroups asfollows. We leave this as an exercise for the reader.

Corollary 2.5.7. Let G be a locally compact group and S a normalsimply connected solvable subgroup with G/S compact. Then G is asemidirect product of a compact subgroup K of G with S.

We now turn to a fundamental result, namely Weyl’s finiteness the-orem.


Corollary 2.5.8. If G is a connected compact semisimple Lie groupthen π1(G) is finite. Alternatively G or indeed any group locally iso-morphic to G is compact.

Proof. The property we need concerning semisimple groups L is that[L,L] = L. (Actually, [L,L] = L). We prove that if D is a discretecentral subgroup of H with H/D compact and [H/D,H/D] = H/D,then D is finite.

Let G = H be the universal covering of G and D be its fundamentalgroup. Then H/D = G is compact and D is a finitely generated, dis-crete, abelian group. Hence D = Zr × F , where F is finite and r ≥ 0is an integer. If D is not finite then r ≥ 1 so there is a surjective con-tinuous homomorphism φ : D → Z. Injecting Z → R and composinggives a nontrivial homomorphism f : D → R. Consider the extensionf∗, where ρ is the 1-dimensional trivial representation of H on R. Thenevidently all the requirements of Theorem 2.5.5 are satisfied. If f∗ isthe extension to H → R we see that f∗ is actually a continuous ho-momorphism and f∗(D) = f(D) = Z. Since f∗ is continuous f∗(H)is connected so this must be larger then Z. Compose f∗ with the pro-jection π : R → T. Then this is a nontrivial homomorphism H → T .but π f∗(D) = π(Z) = 1. Therefore this drops down to a nontriv-ial continuous homomorphism χ : H/D → T . This is impossible sinceχ([H/D,H/D]) = 1 = χ(H/D). Hence D is finite so G is compact.Therefore so are all locally isomorphic groups since they are covered byG.

As a further application of the use of Haar measure we prove theBochner linearization theorem which says that when a compact groupacts on a manifold, near a fixed point the action is locally linear (and,of course, orthogonal).

Theorem 2.5.9. Let G×M →M denote the smooth action of a com-pact Lie group G on a smooth manifold M and let p be a G-fixed pointof M . Then there is a G-invariant neighborhood U of p in M and a Gequivariant diffeomorphism F : U → B, where B is an open ball about0 in Tp(M).


Proof. First we show that around p there is a neighborhood basisconsisting of G-invariant neighborhoods. This actually follows as theequicontinuity from a general Ascoli type theorem for group actions,but we will give a direct proof. Let U be a neighborhood of p ∈ M .By continuity of the action together with the fact that p is G-fixed, wecan find for each g ∈ G neighborhoods Wg of g and Ug of p so thatWgUg ⊆ U . By compactness G =

⋃ni=1Wgi

. Let U∗ = ∩ni=1Ugi. Then

U∗ is a neighborhood of p ∈ M and U∗ ⊆ GU∗ ⊆ U . Hence GU isalso neighborhood of p ∈ M . It is clearly G-invariant and since U wasarbitrary we have a neighborhood basis about p ∈M .

Now let U be a G-invariant neighborhood of p ∈ M small enoughto be in a chart f around p. Then f can be regarded as mapping Udiffeomorphically to Tp(M), taking p to 0. Hence dpf is invertible. Byinvariance of U , u 7→ f(g−1u), also takes values in Tp(M). Because pis G fixed and U is G invariant, since Tp(M) = Tp(U), each g ∈ G hasderivative d(g) ∈ GL(Tp(M)). Hence g 7→ d(g)(f(g−1u)) also takes val-ues in Tp(M). It follows that the integral

∫G d(g)(f(g−1u))dg (normal-

ized Haar measure) is a tangent vector and so defines a function F : U →Tp(M) which by differentiation under the integral is a smooth functionsince f is. Now f(g−1p) = f(p) = 0 so F (p) =

∫G d(g)(f(g−1p))dg = 0.

To calculate dpF let ǫ > 0. Since for u near p,

f(g−1u) − f(p) = dpf ||f(u) − f(p)|| + ǫ||f(u) − f(p)||,

taking into account f(p) = 0, applying d(g) and integrating we get∫

Gd(g)f(g−1u)dg = dpf

∫

Gd(g)||f(u) − f(p)||dg

+ ǫ

∫

Gd(g)||f(u) − f(p)||dg.

Since we have normalized Haar measure

F (u) = F (u) − F (p) = (dpf + ǫ)||f(u) − f(p)||.

Hence dpF = dpf 6= 0. By the inverse function theorem F is a localdiffeomorphism on some neighborhood of p within U with a ball about0 in Tp(M). Finally by invariance of Haar measure, for h ∈ G,

2.6 Compact linear groups and Hilbert’s 14th problem 121

F (hu) =

∫

Gd(g)(f(g−1hu))dg =

∫

Gd(hg)(f((hg)−1hu))dg

= d(h)

∫

Gd(g)(f((g)−1u))dg = d(h)F (u).

2.6 Compact linear groups and Hilbert’s 14th

problem

Here we will prove a theorem of Chevalley which states that a compactlinear group is the set of real points of an algebraic group defined overR. We shall do this by means of the study of invariant polynomials.This method leads in a natural way to the classical solution of Hilbert’s14th problem in the case of compact linear groups. Namely, that thealgebra of invariants, P (V )G, is finitely generated. (Something of thesemethods also can be made to work in the case of non-compact reductivegroups, but that is another story).

Suppose k is a field and V a vector space of dimension n over k.We shall call k[x1, . . . , xn], the k-algebra of polynomials in n indeter-menants, P (V ). Elements p ∈ P (V ) are finite sums

p(x) = Σa(e1,...,en)xe11 . . . xen

n

where the monomials are formed with coefficients from k and (e1, . . . , en)are n-tuples of non negative integers. The degree of the monomial is∑

i ei and the degree of p is the maximum of the degrees of the monomi-als of which p is composed. A polynomial is called homogenous if all itsmonomials have the same degree. Now, P (V ) can be regarded as a k-space (in fact a k-algebra) of k-valued functions on V as follows. Choosea basis v1, . . . , vn of V . If x =

∑xivi, then the value of p(x) is given

by the equation above. As a result, we get an action of GL(V ) on P (V )by left translation, namely, (g, p) → pg where pg(x) = p(g−1x). Clearly,pg ∈ P (V ) and GL(V ) × P (V ) → P (V ) is a k-linear (infinite dimen-sional) representation of GL(V ) on P (V ). Note also that GL(V ) acts by


k-algebra automorphisms; that is, (pq)g = pgqg for all p and q ∈ P (V ).Now, deg pg = deg p for all p and g. So, if we consider the filtration ofP (V ) =

⋃n P (V )m by degree where P (V )m = p ∈ P (V ) : deg p ≤ m,

then each P (V )m is a finite dimensional GL(V )-invariant subalgebra ofP (V ). So, for each integer m, we get a finite dimensional representationon P (V )m as well as on P (V )m \ P (V )m−1, the space of homogenouspolynomials of degree m. Now let G×V → V be a linear representationof G on V . Then by restriction from GL(V ) to the image of G underthe representation, we get an action of G on P (V ), P (V )m etc. A G-invariant polynomial p is one for which pg = p for all g ∈ G. The set ofG-invariant polynomials will be denoted by P (V )G. Clearly, P (V )G is aG-invariant k-subalgebra of P (V ). It is called the algebra of invariants.

As an illustration, let k = R and G be any subgroup of GL(n,R)which acts transitively on the unit sphere Sn−1 of V , such as O(n,R)or SO(n,R). Then

P (V )G = q(x21 + · · · + x2

n) : q(t) ∈ R[t],

i.e., the algebra of invariant polynomials has a single generator.

To see this let x 6= 0 ∈ V and write x = ‖x‖ · v, where v ∈ Sn−1.If p is homogeneous, then p(x) = p(‖x‖ · v) = ‖x‖deg p · v. Now if, inaddition, p is G-invariant, then since G operates transitively on Sn−1,p(v) = c, a constant. Let p ∈ P (V )G and write p =

∑pi where each

pi is homogeneous and, by Lemma 2.6.1 below, is also G-invariant. Bythe above, pi(x) = ‖x‖ici, for i = 1, . . . ,deg p and x 6= 0 ∈ V . Clearly,this also holds for i = 0 and for x = 0 since pi(0) = 0 if i > 0 and p0(x)is constant. If any ci = 0, then pi = 0 on Sn−1 and therefore in all ofV , by homogeneity. Hence we may assume all ci 6= 0. Thus

pi(x)

ci= (x2

1 + · · · + x2n)

i2 .

Now, the left side is a polynomial so i must be even. Let q(t) =∑c2jtj

where j = i2 . Then p(x) =

∑pi(x) = q(x2

1 + · · · + x2n). Conversely, any

q(x21 + · · · + x2

n) ∈ P (V )G.


We shall presently see that the algebra P (V )G always has a finitenumber of generators; where by this we mean that an algebra A haselements α1, . . . , αr such that

A = q(α1, . . . , αr) : q ∈ k[x1, . . . , xr].

We require three lemmas.

Lemma 2.6.1. Let the polynomial p be written p(x) =∑pi(x) where

pi is homogenous of degree i. If p is G-invariant, then each pi is alsoG-invariant.

Proof. Clearly p(x) can be expressed in terms of pi(x). Now, suppose qiare homogenous polynomials of distinct degrees and

∑ciqi = 0 where

ci ∈ k. Then for each i, either ci = 0 or qi = 0. To see this, wemay assume all qi 6= 0 and show all ci = 0. But this is clear sincethe distinct monomials are linearly independant over k. Now, if p(x) =∑pi(x) =

∑qi(x) where pi and qi are homogeneous to degree i, then∑

1(qi − pi) = 0. Since qi − pi is homogeneous of degree i, we see thatqi − pi = 0 for all i, by the above. For g ∈ G if p(x) =

∑pi(x), then

p(g−1x) =∑pi(g

−1x) = p(x) since p is G-invariant. Now pi(g) is ahomogeneous polynomial of degree i for each g ∈ G. By uniqueness,pi(x) = pi(gx) for all i.

Lemma 2.6.2. Let I+ denote the homogeneous G-invariant polynomi-als of positive degree and I the ideal in P (V ) generated by I+. Then Ias an ideal, has a finite number of G-invariant homogeneous generators.

Proof. By the Hilbert’s basis theorem [82], I = (p1, . . . , pr). Hence, eachpi =

∑qji s

ji where qji ∈ P (V ) and sji ∈ I+. Now, the ideal (sji ) ⊆ I.

If p ∈ I, then p =∑tipi where ti ∈ P (V ) so p =

∑ti

∑qji s

ji ∈ (sji ).

Thus I = (sji ).

Lemma 2.6.3. Let G be a compact subgroup of GL(V ). Then thereexists a map # : P (V ) → P (V )G such that

(1) # is R (respectively C) linear


(2) p = p# if and only if p ∈ P (V )G

(3) (pg)# = p#q if p ∈ P (V ) and q ∈ P (V )G

Proof. Let dg be normalized Haar measure on G. For p ∈ P (V ), de-fine p#(x) =

∫G p(g

−1x)dg. Then p# ∈ P (V ) and # is k-linear. Byinvariance of dg, p# ∈ P (V )G. Clearly, if p ∈ P (V )G, then p# = p.Finally, (pq)#(x) =

∫p(g−1x)q(g−1x)dg = q(x)

∫p(g−1x)dx, since q is

G-invariant. Thus (pq)# = p#q.

We now come to Hilberts 14th problem in the case of compact groups.Such results are also known as the fundamental theorem of invarianttheory.

Theorem 2.6.4. Let G be a compact group and G × V → V be acontinuous real or complex linear action of G on V . Then P (V )G is afinitely generated algebra over R or C. In fact, P (V )G is generated asan algebra by a finite number of homogeneous, G-invariant polynomials.

Proof. Let J+ = p ∈ P (V )G : deg p > 0 and I+, as above, be thoseelements in J+ which are homogeneous. Let J and I be the respectiveideals generated by J+ and I+. By Lemma 2.6.2, I has a finite number ofideal generators belonging to I+. We shall show J = I. Since 1 ∈ P (V ),J+ ⊂ J , so from I+ ⊂ J+ we know I ⊂ J . Let p ∈ J . Then p =

∑qipi

where qi ∈ P (V ) and pi ∈ J+. By Lemma 2.6.1, each pi =∑pij, where

pij ∈ I+. Hence, p =∑qi

∑pij ∈ I so J = I and J as an ideal has a

finite number of G-invariant homogeneous generators; J = (p1, . . . , pr).Let p ∈ J+; we show p = q(p1, . . . , pr) for some q ∈ k[x1, . . . , xr] byinduction on deg p. This would complete the proof, since if p ∈ P (V )G,then either p ∈ J+ or p is constant. In the latter case, we would takeq to be the constant polynomial. Of course, conversely all q(p1, . . . , pr)are in P (V )G. Now, take p ∈ J+. As an element of J , p =

∑qipi. By

Lemma 2.6.3,

p = p# = (∑

qipi)# =∑

q#i pi

It follows that deg p ≥ deg q#i pi = deg q#i + deg pi for each i. If deg pi =0, then pi is a constant and does not generate any more subalgebra thanif it were not there at all. We can therefore assume all deg pi > 0. The


deg q#i < deg p. If deg q#i > 0, then deg q#i ∈ J+. By induction, deg q#iis in the algebra generated by p1, . . . , pr. If deg q#i = 0, then q#i is

constant and is also in this algebra. This means∑q#i pi = p is in the

algebra.

Now we shall see that there are sufficiently many G-invariant poly-nomials on V .

Theorem 2.6.5. Let G be a compact subgroup of GL(V ). Then the G-invariant polynomials on V (with real coeficients) separate the disjointcompact G-invariant subsets of V . In particular, they separate the G-orbits.

Proof. Suppose A and B are disjoint compact G-invariant subsets ofV . Let φ(x) = d(x,A) − d(x,B) where x ∈ V and d is the distancefunction on V . Then φ is continuous and is < 0 on A and > 0 on B.By compactness, there is a δ > 0 so that φ > δ on B and φ < −δ on A.By the Weierstrass approximation theorem, φ can be approximated, towithin δ

2 , by a polynomial on the compact set A∪B. Then p# ∈ P (V )G

and is p# > 0 on B and < 0 on A.

Finally, we come to Chevalley’s theorem.

Corollary 2.6.6. A compact linear Lie group is the set of real pointsof an algebraic group defined over R.

Proof. G acts on EndR(V ) by (g, T ) → g · T . This is a linear represen-tation of G. Now, G = G · 1, the G-orbit of 1. If T ∈ EndR(V ), butnot in G, then there is a p ∈ P (EndR(V ))G so that p(T ) 6= p(1). Thus,G = ∩p∈P (EndR(V ))GT ∈ EndR(V ) : p(T ) − p(1) = 0.


Chapter 3

Elements of the Theory of

Lie Algebras

3.1 Basics of Lie Algebras

3.1.1 Ideals and Related Concepts

Definition 3.1.1. A subspace h of g is a called an ideal in g if [X,Y ]is in h whenever X is h. Evidently an ideal is a subalgebra.

Example 3.1.2. The center of g, z(g) = X ∈ g : [X,Y ] = 0,∀Y ∈ g,is an ideal in g.

Example 3.1.3. In g = gl(n, k) we consider the center, z(g) namelyall linear operators commuting with g. This is evidently just the scalarmatrices. Consider s = sl(n, k) which is the set of matrices of tracezero. This is a subalgebra since the trace is of any commutator is zero.In fact s is an ideal in g. Also s ∩ z(g) = 0 since the characteristic ofk is zero. Since dim z(g) + dim s = n2 = dim g this is a direct sum ofideals. For X ∈ g then X = trX

n In×n + Y , where Y is in s, implementsthe decomposition.For the particular case of n = 2 we see that a basis of sl(2, k) is given by

H =

(1 00 −1

), X+ =

(0 10 0

)and X− =

(0 01 0

). The relations (structure

127

128 Chapter 3 Elements of the Theory of Lie Algebras

constants) are [H,X+] = 2X+, [H,X−] = −2X− and [X+,X−] = H.Notice that adH with respect to the basis X+,H,X− is diagonalwith (distinct) eigenvalues 2, 0,−2.

Exercise 3.1.4. Prove that sl(2,C) has a basis X,Y,Z such that[X,Y ] = Z, [Y,Z] = X and [Z,X] = Y .

Hint: Let, X = H2i , Y = X++X−

2 and Z = X+−X−

2i .

Exercise 3.1.5. Prove that so(2, 1) ∼= sl(2,R) as Lie algebras.

Let g be a Lie algebra and h and ideal in g; then we can equip g/hwith a Lie bracket, making it a Lie algebra. For X ∈ g, X denotes itsimage in g/h. The Lie bracket on g/h is defined by setting

[X,Y ] = [X,Y ].

To see that it is well-defined, replace X by X +X1 and Y + Y1 whereX1 and Y1 are in h. Then,

[X +X1, Y + Y1] = [X +X1, Y + Y1]

= [X,Y ] + [X1, Y ] + [X,Y1] + [X1, Y1]

= [X,Y ].

Here [X1, Y ], [X,Y1] and [X1, Y1] are 0 because h is an ideal. It followsimmediately from the definition that the projection map π : g → g/h isa Lie algebra homomorphism with kernel h.

Proposition 3.1.6. (The First Isomorphism Theorem) Let f : g → h bea Lie homomorphism, then Ker f is an ideal in g and f(g) is subalgebraof h. Moreover, f induces a Lie isomorphism from g/Ker f to f(h)making the following diagram commutative:

gf //

π##F

FFFFFFFF h

g/Ker f

f;;xxxxxxxxx

(3.1)

3.1 Basics of Lie Algebras 129

Proof. It is obvious that Ker f is a linear subspace. If X ∈ Ker f andY ∈ g then we have

f([X,Y ]) = [f(X), f(Y )] = [0, f(Y )] = 0.

Therefore [X,Y ] ∈ Ker f which means that Ker f is an ideal. Since[f(X), f(Y )] = f([X,Y ]), f(g) is closed under bracketing so it is a sub-algebra of h. Then f induces a map f : g/Ker f → h defined as follows,

f(X) = f(X)

where X is a representative of the equivalence class X. One has to checkthat f does not depend on the representative X, but If X and Y aretwo representatives for an equivalence class we have X−Y = Z ∈ Ker fso f(X) − f(Y ) = f(Z) = 0. This shows that f is well-defined. f isinjective since if f(X) = f(X) = 0 means that X ∈ Ker f or in otherwords X = 0 ∈ g/Ker f . It is obvious that f : g/Ker f → f(g) issurjective since that f(X) = f(X) for any X ∈ g. Therefore f is anisomorphism. Also f(X) = f(X) which means that the above diagramcommutes.

Definition 3.1.7. Given two subalgebras a and b of a Lie algebra g

one can consider the subalgebra generated by a and b. If a and b areideals then a + b is also an ideal.

Exercise 3.1.8. Prove that if a and b are ideals in g then a ∩ b, [a, b]are ideals in a + b.

Definition 3.1.9. Let h and l be two Lie algebras and g = h⊕ l be thedirect sum of the two vector spaces. This vector space can be equippedwith a Lie bracket such that h and l are subalgebras and [h, l] = 0. Wecall this Lie algebra the external direct sum of h and l. Evidently h andl are ideals in g. A similar construction can be made with any finitenumber of factors.

Remark 3.1.10. Let g be a Lie algebra and h and l two ideals in g

with trivial intersection. If g = h + l then g ≃ h ⊕ l.


Exercise 3.1.11. Let g be a Lie algebra and h and l be two subalgebraswith h an ideal. Then,

(1) (The Second Isomorphism Theorem) Then the subalgebra h + l

generated by h and l contains h as an ideal. Moreover h ∩ l is anideal in l and

h + l/h ≃ l/h ∩ l

(2) (The Third Isomorphism Theorem) If h ⊆ l ⊆ g are ideals in g

then l/h can be regarded as an ideal in g/h (using the map inducedby inclusion), and then

g/h

l/h≃ g

l

Clearly a subspace of the center z(g) is an ideal in g. Such an idealis called a central ideal .

Definition 3.1.12. Let g be Lie algebra and h an ideal in g. We denoteby [g, h] the linear span of the set of all [X,Y ] where X ∈ g and Y ∈ h.It is easy to see, using Jacobi identity, that [g, h] is also an ideal. Animportant special case of this is [g, g]. This ideal is called the derivedsubalgebra.

Proposition 3.1.13. If h is a subalgebra of Lie algebra g and [g, g] ⊆ h

then h is an ideal.

Proof. We have[g, h] ⊆ [g, g] ⊆ h.

Proposition 3.1.14. For a Lie algebra g, g/[g, g] is abelian. Moreoverany ideal h ⊂ g for which g/h is abelian contains the derived subalgebra[g, g].

Proof. If X and Y ∈ g/[g, g] then [X,Y ] = [X,Y ] = 0 ∈ g/[g, g]. Sothe derived subalgebra is abelian. Conversely if g/h is abelian, X andY ∈ g/h commute, for X and Y so

[X,Y ] = [X,Y ] = 0 ∈ g/h.


This implies that [X,Y ] ∈ h and since this is true for all X and Y , wesee that h contains the derived subalgebra.

Definition 3.1.15. Let g be a Lie algebra. A subspace a of g is saidto be a characteristic ideal if it is invariant under every derivation of g.

A characteristic ideal is an ideal since it is invariant under adX forall X ∈ g.

Example 3.1.16. Typical examples of characteristic ideals in a Liealgebra g are the center z(g) and the derived subalgebra [g, g].

Proposition 3.1.17. Let a and b be two characteristic ideals of g. Thena + b, a ∩ b and [a, b] are also characteristic ideals of g.

Proof. We check this for [a, b]. For X ∈ a and Y ∈ b and D ∈ Der(g)we have

D[X,Y ] = [DX,Y ] + [X,DY ] ∈ [a, b]

since a and b characteristic ideals.

Proposition 3.1.18. Let g be a Lie algebra, h an ideal in g, and l acharacteristic ideal in h. Then l is an ideal of g.

Proof. If X ∈ g then adX is a derivation of h, therefore l is stable underadX and this means that l is an ideal in g.

Definition 3.1.19. Let g and h be two Lie algebras and η : g → Der(h)be a Lie homomorphism. The semi direct sum g⊕η h which is the vectorspace g ⊕ h equipped with the Lie bracket is

[(X,Y ), (X ′, Y ′)] = ([X,X ′], ηX(Y ′) − ηX′(Y ) + [Y, Y ′]).

Making the obvious identification h is an ideal and g is a subalgebra ofg ⊕η h.

Alternatively, let l be a Lie algebra, h an ideal and g a subalgebraof l such that l = h ⊕ g as vector spaces. For X ∈ g let ηX = adX|h.Then η is a Lie algebra homomorphism g → Der(h) and the resultingsemi direct product is isomorphic to l.


Example 3.1.20. Let hn be the space of matrices in gl(n+ 2, k) of theform

(X,Y, z) =

0 x1 . . . xn z0 0 . . . 0 y1...

......

0 0 . . . 0 yn0 0 . . . 0 0

where X = (x1, x2, . . . , xn) and Y = (y1, y2, . . . , yn) and n ≥ 1. In thisnotation

[(X,Y, z), (X′,Y′, z′)] = (0, 0,XY′t − X′Yt).

In particular this shows hn is a Lie algebra called the Heisenberg Liealgebra.. Here [hn, hn] = z(hn) which has dimension one.

It is of interest to identify the derivation algebra of hn. Consider thebasis

Xi = ((0, 0, . . . , 1, . . . , 0), (0, 0, . . . , 0), 0), 1 ≤ i ≤ nYi = ((0, 0, . . . , 0), (0, 0, . . . , 0, 1, 0, . . . , 0), 0), 1 ≤ i ≤ n

Z = ((0, 0, . . . , 0), (0, 0, . . . , 0), 1)

for hn and let D be a derivation for hn. Then

DXi =∑aijXj +

∑bijYj + λiZ,

DYi =∑cijXj +

∑dijYj + µiZ,

DZ = λZ.(3.2)

The last equation follows because the center is a characteristic ideal.By applying D to the identities

[Xi, Yj ] = δijZ

[Xi,Xj ] = 0

[Yi, Yj ] = 0

and inserting the values from (3.2) we get,

aij + dji = δijλbij + bji = 0cij + cji = 0.


In particular we have λ = 1n(

∑aii+

∑dii). Letting A = (aij), B = (bij),

C = (cij) and D = (dij), we have

A+Dt = ρInB = −Bt

C = −Ct.The matrix representation of D is

D =

A C | 0

B D | ...−−− −−− | −−λ1 . . . λn µ1 . . . µn | λ

When λ = 0 or equivalently trD = 0, we get

D =

A C | 0

B −At | ...−−− −−− | −−λ1 . . . λn µ1 . . . µn | 0

.

Notice that H =

(A CB −At

)lies in Sp(n, k). Indeed, the set of

derivations of hn of trace zero is isomorphic to the semidirect sumSp(n, k) ⊕η k

2n where η is the inclusion of Sp(n, k) in GL(2n, k) andk2n is regarded as an abelian Lie algebra. Each such derivation

D =

A C 0B −At 0

λ1 . . . λn µ1 . . . µn, 0

is mapped to (

(A CB −At

), (λ1 . . . λn, µ1 . . . µn)).

We now turn to the concepts of nilpotence and solvability.

Definition 3.1.21. Let g be a Lie algebra. We say g is nilpotent if itslower central series, g0 = g, g1 = [g, g], . . . , gk = [g, gk−1] . . ., eventuallyhits 0. The first index where this occurs is called the index of nilpotenceof g. We call an ideal of g nilpotent if it is nilpotent as a Lie algebra.


Exercise 3.1.22. Show that each term in the lower central series is acharacteristic ideal in g.

Example 3.1.23. The Heisenberg Lie algebra hn, is a 2-step nilpotentLie algebra. This follows from the fact that [g, g] = z(g).

We now construct two other important examples of 2-step nilpotentLie algebras. Regard gn(C) = Cn⊕iR as a real vector space of dimension2n + 1 and let 〈·, ·〉 be the Hermitian form on Cn given by 〈X,Y〉 =∑n

i=1 xiyi. Now we define the bracket [·, ·] by

[(X, it), (Y, is)] = (0, iℑ〈X,Y〉)

We the reader to check that this is a 2-step nilpotent Lie algebra overR with 1-dimensional center, and in fact is isomorphic to hn.

Exercise 3.1.24. Prove that any 2-step nilpotent Lie algebra over afield k is isomorphic to hn(k) if its center is 1-dimensional.

We now consider the quaternionic analogue of the example justabove. Let H be the quaternions and X 7→ X be quaternionic con-jugation. Then gn(H) = Hn ⊕ ℑ(H) where ℑ(H) is the 3-dimensionalreal vector space spanned by i, j and k. Then gn(H) is a real vectorspace of dimension 4n+ 3. We make this into a Lie algebra by

[(X,V), (Y,W)] = (0,ℑ〈X,Y〉)

where 〈X,Y〉 =∑n

i=1 xiyi is the Hermitian form on Hn (usingquoternionic conjugation) and ℑ〈X,Y〉 is the imaginary part of 〈X,Y 〉.Direct calculations show that gn(H) is a 2-step nilpotent Lie algebrawith a 3-dimensional center1. These Lie algebras play an importantrole in the study of rank one simple Lie algebras and groups which willbe given in Chapter 6.

Proposition 3.1.25. The sum of two nilpotent ideals, a and b, is anilpotent ideal.

1Similar constructions can be made using the Cayley numbers.


Proof. Let an (respectively bn) be the nth term of the lower central seriesof a (respectively b). Now for any sequence h1, h2, . . . , hk of subalgebrasof g where at least n of them are equal to a (respectively b), thenany bracketing [[[. . . [hi1 , hi2 ], hi3 ] . . .]]] of these hi is in an (respectivelybn). Let m = max(m1,m2) where m1 (respectively m2) is the index ofnilpotence of a (respectively b). Then

(a + b)2m ⊂ am + bm = 0.

Hence a + b is nilpotent. We already know it’s an ideal.

Definition 3.1.26. It follows from Proposition 3.1.25 and finite dimen-sionality that every Lie algebra g has a unique maximal nilpotent idealnil(g), called nilradical , which contains any nilpotent ideal.

Definition 3.1.27. A Lie algebra g is said to solvable if the derivedseries of g, g0 = g, g1 = [g, g], . . . , gk = [g(k−1), g(k−1)], . . . hits 0. Thefirst index where this occurs is called the index of solvability of g. Wecall an ideal of g solvable if it is solvable as a Lie algebra.

Exercise 3.1.28. Show that each term in the derived series is a char-acteristic ideal in g.

Evidently since [gk−1, gk−1] ⊆ [g, gk−1] for every k, a nilpotent al-gebra is always solvable. Also it is clear that a subalgebra or quotientalgebra of a nilpotent (respectively solvable) algebra is nilpotent (re-spectively solvable).

Example 3.1.29. Let V be a finite dimensional vector space and lets(V ) be the set of upper triangular matrices of gl(V ) and n(V ) be theset of upper triangular matrices with all zero entries on the diagonal.The latter is niltriangular matrices. We leave as an exercise that theseare both subalgebras of gl(V ) and that [s(V ), s(V )] = n(V ). A directcalculation shows that the series nk(V ) = [n(V ), nk−1(V )] introducesanother row of off diagonal zeros. This means that n(V ) is nilpotent ofindex dimV − 1 and also that s(V ) is solvable of index dimV .


Remark 3.1.30. As we shall see later, by Lie’s theorem, Theorem3.2.18, a Lie algebra is solvable if and only if its derived subalgebra isnilpotent.

Remark 3.1.31. Notice that (5) in Section 0.5 tells us that[n(V ), n(V )t] is contained in the diagonal subalgebra, where n(V )t con-sists of the transpose of the elements of n(V ).

Definition 3.1.32. Let f : g → h be a surjective Lie algebra homo-morphism. If Ker f ⊂ z(g), then g is called a central extension of h.

Proposition 3.1.33. Let g be a Lie algebra and h be an ideal andconsider the short exact sequence,

0 → h → gπ→ g/h → 0

then

(1) g is solvable if h and g/h are solvable.

(2) If h is a central ideal and g/h is nilpotent, then so is g.

Proof. Because π is a homomorphism it takes the derived series of g tothe derived series of g/h. Since g/h is solvable, [gk, gk] ⊆ h for some k.It follows from the solvability of h that g is solvable. For the secondpart we observe that since g/h is nilpotent and π is a homomorphismgk+1 = [g, gk] ⊆ h. Hence [g, gk+1] ⊂ [g, h] = (0).

Proposition 3.1.34. Let g be a Lie algebra and a and b two solvableideals of g, then a + b is also solvable.

Proof. By the second isomorphism theorem we have (a + b)/a ∼= b/(a∩b). The latter is solvable since it is homomorphic image of a solvablealgebra. Therefore a + b is solvable by Proposition 3.1.33.

Definition 3.1.35. By virtue of Proposition 3.1.34 and finite dimen-sionality of g, every Lie algebra has a unique maximal solvable idealrad(g), called the radical .


Notice that by their very definitions rad(g) and nil(g) are stableunder all automorphisms of g. Let h stand for either of these ideals.Now suppose we are over the real, or complex field and D is a derivationof g. Then for all real t, Exp tD is an automorphism of g (1.4.29) andso Exp tD(h) = h. Differentiating at t = 0 shows D(h) = h. Hence bothrad(g) and nil(g) are characteristic ideals in g.

Remark 3.1.36. Evidently rad(g) ⊃ nil(g). (We shall see later thatrad(g)/nil(g) is abelian (see Corollary 3.2.19)).

Proposition 3.1.37. For every Lie algebra g,

nil(rad(g)) = nil(g).

Proof. Now since, as we saw, nil(rad(g)) is a characteristic ideal inrad(g) which is itself an ideal in g, it follows that nil(rad(g)) is anideal in g. Since its nilpotent it’s contained in the largest such ideal ofg; nil(rad(g)) ⊂ nil(g). Conversely, let a any nilpotent ideal in g. Thenit is also a solvable ideal so a ⊂ nil(rad(g)). Taking a = nil(g)) we getnil(g)) ⊂ nil(rad(g)).

Example 3.1.38. Here we introduce the affine Lie algebra. Let V bea vector space of dimension n over k, A a linear operator on V and b avector in V . We consider the linear Lie algebra g consisting of matricesof order n+ 1, (

A b0 0

)

It is easy to check that the matrices with A = 0 form an ideal in g. TheLie bracket is given by

[(A, b), (A′, b′)] = ([A,A′], Ab′ −A′b)

This Lie algebra is called the ax+ b-Lie algebra and any of its sub-algebras is called an affine Lie algebra. It is the semi direct sum ofgl(n, k) with V . This is a subalgebra of gl(n+ 1, k). It is only solvableif n = 1.


3.1.2 Semisimple Lie algebras

Definition 3.1.39. Let g be a Lie algebra.

(1) g is said to be simple if it has no nontrivial proper ideal.

(2) g is said to be semisimple if it has no nontrivial solvable ideal.

It is clear that:

Proposition 3.1.40. A Lie algebra is semisimple if and only if itsradical is trivial.

Proposition 3.1.41. A Lie algebra is semisimple if and only if it hasno nontrivial abelian ideal. In particular the center of a semisimple Liealgebra is trivial.

Proof. Clearly a semisimple Lie algebra has no nontrivial abelian idealas abelian ideals are solvable. Conversely, consider the derived seriesof rad(g). By assumption that has no nontrivial abelian ideal. This isbecause the ideals in this derived series are characteristic ideals in rad(g)and therefore are ideals in g. But since rad(g) is solvable and non trivialthe last term in this series is abelian and non trivial. Therefore rad(g)must be (0) and g is semisimple.

Remark 3.1.42. Let g be a Lie algebra and rad(g) its radical. Theng/ rad(g) is semisimple. This is an immediate consequence of Proposi-tion 3.1.40.Examining the following exact sequence,

0 → rad(g) → g → g/ rad(g) → 0

we see that any Lie algebra g is the middle term of a short exact sequencewhose other two terms are solvable and semisimple. So it seems plausiblethat studying solvable Lie algebras and semisimple Lie algebras is anappropriate guide line in the theory of representations of Lie algebrasin general.

Definition 3.1.43. For a Lie algebra g and h a subalgebra of it, wedefine the normalizer ng(h) of h in g to be the set of all X ∈ g such that


[X, h] ⊆ h. It is easy to see that ng(h) is a subalgebra of g containingh as an ideal and is the largest such subalgebra. We also define thecentralizer zg(h) of h in g to be the set of all X ∈ g such that [X, h] = 0.Evidently zg(h) is a subalgebra of g. If h = g we get the center. If h

is abelian then the centralizer contains h. In any case the centralizeris a subalgebra of the normalizer. If X ∈ g then zg(X) will mean thecentralizer of the subalgebra generated by X.

Proposition 3.1.44. If h is an ideal in g then its centralizer is also anideal.

Proof. Let X ∈ zg(h), Y ∈ g and H ∈ h then

[[Y,X],H] + [[H,Y ],X] + [[X,H], Y ] = 0.

Since [X,H] = 0, the last term is trivial and since h is an ideal [H,Y ] ∈ h

and therefore the middle term is also zero. Hence the first term is zero,so [Y,X] ∈ zg(h).

3.1.3 Complete Lie Algebras

Definition 3.1.45. A Lie algebra g is called complete if its center istrivial and every derivation is inner.

Some examples of complete Lie algebras are semisimple Lie algebrasas we shall see in next section, and the affine Lie algebra of dimension2 given just above.

Proposition 3.1.46. The ax+ b Lie algebra g is complete.

Proof. Let X,Y be a basis of g with [X,Y ] = Y and let cX + dY bea generic element of the Lie algebra. If cX + dY is in the center thenit must bracket trivially with both X and Y . It follows that c = d = 0and hence z(g) = 0.

Now let D be a derivation, so that

DY = D[X,Y ] = [DX,Y ] + [X,DY ]. (3.3)


Let DX = aX + bY and DY = cX + dY . By substituting in (3.3) weget c = a = 0. Hence DX = bY and DY = dY . Then it can be easilychecked that D = adU where U = dX − bY .

Proposition 3.1.47. Let g be a Lie algebra and h an ideal. If h iscomplete then it is a direct summand. In fact g = h ⊕ zg(h).

Proof. Notice that h ∩ zg(h) = z(h) = 0. If X ∈ g then adX restrictedto h is a derivation of h and therefore it is an inner derivation of h,

adX|h = adH

for some H ∈ h. Hence X −H is in the centralizer of h, therefore

g = h ⊕ zg(h).

3.1.4 Lie Algebra Representations

Definition 3.1.48. Let ρ : g → gl(V ) be Lie algebra representation ofg. A subspace W of V is called an invariant subspace if ρX(W ) ⊂ Wfor every X ∈ g. We shall say that a representation is reducible if it hasa nontrivial proper invariant subspace, otherwise we call it irreducible.

Definition 3.1.49. Let ρ1 : g → gl(V1) and ρ2 : g → gl(V2) be tworepresentations of a Lie algebra g. A intertwining operator from (ρ1, V1)to (ρ2, V2) is a linear map T : V1 → V2 such that for every X ∈ g,

T ρ1(X) = ρ2(X) T.

Two representations are equivalent if there exists an invertible inter-twining operator between them.

On can do certain operations on the space of representations.

Definition 3.1.50. Let ρ1 : g → gl(V1) and ρ2 : g → gl(V2) be twoLie representations of a Lie algebra g. Then one can consider the map


ρ1 ⊕ ρ2 : g → gl(V1) ⊕ gl(V2) ⊂ gl(V1 ⊕ V2) defined as following: Forevery X ∈ g ,

(ρ1 ⊕ ρ2)(X) = (ρ1(X), ρ2(X)) : V1 ⊕ V2 → V1 ⊕ V2

We call (ρ1 ⊕ ρ2, V1 ⊕ V2) the direct sum of ρ1 and ρ2.

A Lie representation is said to be completely reducible if it is isomor-phic to a direct sum of irreducible representations.

Definition 3.1.51. A family of operators Υ ⊂ gl(V ) is called irreducibleif there is no nontrivial subspace of V stable under all X ∈ Υ.

Lemma 3.1.52. (Schur’s Lemma) Let Υi ⊂ gl(Vi) be an irreducibleset of operators for i = 1, 2 and T : V1 → V2 be an operator such thatTΥ1 = Υ2T . Then either T = 0 or dimV1 = dimV2 and T is anisomorphism.

Proof. KerT is a Υ1 invariant subspace of V1 and T (V1) is a Υ2 invariantsubspace of V2. Hence KerT = V1 or 0. If KerT = V1 then T = 0.Otherwise T is injective. Also T (V1) = V2 or 0. Since we have disposedof the case T = 0 we see that T is an isomorphism.

Corollary 3.1.53. (Schur’s Lemma over an algebraically closed field)Let Υ ⊂ gl(V ) be an irreducible set of operators and T : V → V be anoperator such that TΥ = ΥT . Then T = cI. In particular this is so ifTX = XT for all X ∈ Υ.

Proof. Let c be an eigenvalue of T . Since TΥ = ΥT it follows that(T − cI)Υ = Υ(T − cI). But det(T − cI) = 0 so T = cI by Schur’slemma.

The converse of the Schur’s lemma is also true and does not dependon the field being algebraically closed.

Proposition 3.1.54. Let ρ be a completely reducible representation ofg on a finite dimensional vector space V over k. If the intertwinersconsist only of scalars, then ρ is irreducible.


Proof. Let W be an invariant subspace of V . By complete reducibilitychoose an invariant complementary and subspace U to W . Let PW bethe projection of V onto W . Since both W and U are invariant aneasy calculation shows that for each X ∈ g, PW ρ(X) = ρ(X)PW whenrestricted to W and to U . Hence they also agree on V . Thus PW isan intertwiner. By hypothesis PW is a scalar. But the eigenvalues ofa projection consist of 0 and 1. Therefore PW = 0 or PW = I. HenceW = 0 or W = V so ρ is irreducible.

3.1.5 The irreducible representations of sl(2, k)

We now consider the “simplest” non solvable Lie algebra, namely sl(2, k)(see 3.1.3 for the definition). This algebra is of dimension 3 over k.

Proposition 3.1.55. sl(2, k) is a simple Lie algebra.

Proof. If not, then it has a proper ideal which would be of dimension 1or 2. But all Lie algebras of dimension ≤ 2 are solvable (see Example3.1.38) so by Proposition 3.1.33 it would follow that g = sl(2, k) is itselfsolvable. On the other hand, from the relations (see Example 3.1.3) wesee that [g, g] = g so g cannot be solvable.

Corollary 3.1.56. If ρ is a representation of sl(2, k) on V and theintertwiners consist of scalars, then ρ is irreducible.

Proof. Since sl(2, k) is simple by Proposition 3.1.55 it is thereforesemisimple. This means that ρ is completely reducible by Weyl’s theo-rem so the proposition above applies.

This just means we have to refer to Weyl’s theorem, Theorem 3.4.3,out of order. The proof we give of Weyl’s theorem is independent offacts concerning sl(2, k).

Corollary 3.1.57. Any representation ρ : sl(2, k) → gl(n, k) is eithertrivial or faithful. In particular, any nontrivial representation of g takesa basis of sl(2, k) to a linearly independent family of operators in gl(n, k).

Proof. Let ρ be a nontrivial representation of g. Then Ker ρ is a properideal in g. Hence Ker ρ = 0 so ρ is faithful.


We shall now find all equivalence classes of finite dimensional irre-ducible representations of g = sl(2, k), where k is algebraically closed.The set of all equivalence classes of finite dimensional representationsof g is called the finite dimensional irreducible dual.

Let Vn+1 be a vector space of dimension n + 1 over k with basisv0, . . . vn and n ≥ 1. For each n and X ∈ g we define operators ρ(X)on V = Vn+1 as follows (following the notation in Example 3.1.3):

(1) ρ(X+)vi = i(n− i+ 1)vi−1 for i = 1, . . . , n and ρ(X+)v0 = 0.

(2) ρ(H)vi = (n − 2i)vi for i = 0, . . . , n.

(3) ρ(X−)vi = vi+1 for i = 0, . . . , n − 1 and ρ(X−)vn = 0.

Extending by linearity, this gives a well-defined linear operator ρ(X)on V for each X ∈ g. With respect to the given basis ρ(X+) is lowertriangular with integer entries, ρ(X−) is upper triangular with integerentries, and ρ(H) diagonal with integer entries symmetric about 0.

To see that ρ is a representation, by linearity it is sufficient to verifythat

[ρ(H), ρ(X+)] = ρ([H,X+]),

[ρ(X+), ρ(X−)] = ρ([X+,X−]),

[ρ(H), ρ(X−)] = ρ([H,X−]).

We leave this to the reader to check by using the definitions. Theserepresentations are all inequivalent since they all have different degrees.Thus for each integer n we have a representation of sl(2, k) of degreen+ 1.

We will now show that ρ is an irreducible representation of g. If Tis an intertwining operator on V , then in particular Tρ(H) = ρ(H)T .Since ρ(H) has n+1 distinct eigenvalues this means T is diagonal. Thenapplying Tρ(X+) = ρ(X+)T tells us that all these diagonal entries areequal. Since k is algebraically closed it follows from the converse ofSchur’s Lemma 3.1.54 that ρ is irreducible.

We now show that if k is algebraically closed, then up to equivalencewe have identified all finite dimensional irreducible representations ofsl(2, k).


Proof. To see this let σ be such a representation of degree n+1. Supposeσ(h)v0 = λv0, where λ ∈ k and v0 6= 0 in V . Then since σ is arepresentation,

σ(H)σ(X+)v0 = σ(X+)σ(H)v0 + σ([X+,H])v0 = (λ+ 2)σ(X+)v0

But then λ, λ + 2, λ + 4 etc. is an infinite sequence (of distinct)eigenvalues of σ(h) and dimk V < ∞, it follows that by replacing λ byone of the succeeding terms in the sequence we get σ(X+)v0 = 0 as wellas σ(H)v0 = λv0 for some v0 6= 0.

Define vi = σ(X−)iv0, for i ≥ 0. Then arguing as above, we see thatσ(H)σ(X−)v0 = (λ−2)σ(X−)v0, so that σ(H)v1 = (λ−2)v1. We showby induction that σ(H)vi = (λ− 2i)vi, the case i = 1 having just beendone. Indeed suppose σ(H)vi−1 = (λ− 2(i− 1))vi−1. Then

σ(H)σ(X−)vi−1 = σ(X−)σ(H)vi−1 + σ([H,X−])vi−1

= σ(X−)(λ− 2(i − 1))vi−1 − 2σ(X−)vi−1

= σ(X−)(λ− 2i)vi−1.

That is, σ(H)vi = (λ − 2i)vi. It follows from σ(H)σ(X−)vi−1 = (λ −2i)σ(X−)vi−1 that each σ(X−)iv0 is an eigenvalue of σ(H), or is zero.But just as before since the set of λ, λ − 2, λ − 4 etc. is an infinitesequence of distinct eigenvalues of σ(H), σ(X−)n+1 = 0 for some n+ 1,where n+ 1 is the smallest such integer. Now v0, . . . , vn is a linearlyindependent set since v0, . . . , vn are eigenvectors of σ(H) with distincteigenvalues. To summarize, σ(H)vi = (λ − 2i)vi, σ(X+)v0 = 0 andσ(X−)vi = vi+1, σ(X−)vn+1 = 0. We next prove v0, . . . , vn spansV . To see that this is so, we prove v0, . . . , vn is stable under σ(X+).Then since it is stable under σ(H) and σ(X−) and these generate g, thelinear span of v0, . . . , vn would be a nontrivial invariant subspace andbecause σ is irreducible this must be all of V . We prove our claim byactually showing σ(X+)vi = i(λ− i+ 1)vi−1 for each i.

First observe that

σ(X+)vi+1 = σ(X+)σ(X−)vi = σ([X+,X−])vi + σ(X−)σ(X+)vi

= σ(H)vi + σ(X−)σ(X+)vi = (λ− 2i)vi + σ(X−)σ(X+)vi


By inductive hypothesis this is

(λ− 2i)vi + i(λ− i+ 1)σ(X−)vi−1 = (λ− 2i)vi + i(λ− i+ 1)vi

= (i+ 1)(λ − i)vi.

Thus the vi’s form a basis for V and in addition to the other relations,we have σ(X+)vi+1 = (i+ 1)(λ − i)vi for each i.

We conclude the proof showing σ is equivalent to ρn, the represen-tation defined above of degree n + 1, by proving λ = n and hence therelations are the same. But

tr(σ(H)) = tr(σ(X+)σ(X−) − σ(X−)σ(X+)) = 0.

This means∑n

i=0(λ − 2i) = 0 so that (n + 1)λ = 2∑n

i=0 i = n(n + 1).Thus λ = n.

We remark that when n = 0 we get the trivial representation, whenn = 1 we get the identity representation and when n = 2 we get theadjoint representation. We leave this as an exercise for the reader.

3.1.6 Invariant Forms

One can consider further structures on a Lie algebra, for example bilin-ear forms. But we must tie these to the Lie algebra structure (otherwisewe are merely talking about the vector space). Therefore we require aninvariance condition to be described below.

Definition 3.1.58. Let g be a Lie algebra over a field k. A bilinearform β : g × g → k is said to invariant if for any triple X, Y , and Z ing,

β([X,Y ], Z) = β(X, [Y,Z])

Example 3.1.59. gl(V ) is equipped with a natural invariant bilinearform defined as follows:

β(X,Y ) = tr(XY ).


It is easy to see that β is invariant,

β([X,Y ], Z) = tr([X,Y ]Z) = tr(XY Z − Y XZ)

= tr(XY Z) − tr(Y XZ)

= tr(XY Z) − tr(XZY )

= tr(XY Z −XZY ) = tr(X(Y Z − ZY ))

= tr(X[Y,Z])

= β(X, [Y,Z]).

Here we use the fact that tr(AB) = tr(BA), taking A = Y , B = XZ.

Example 3.1.60. Let φ : g → h be a Lie homomorphism, and β aninvariant form on h then one can pull β back to g using φ in followingmanner,

βφ(X,Y ) = β(φ(X), φ(Y )) for X,Y ∈ g.

Since φ is a Lie homomorphism and β is an invariant form on h itfollows that βφ is an invariant form on g.

Let ρ : g → gl(V ) be a Lie representation. It follows from twoprevious Examples 3.1.59 and 3.1.60 that the trace on gl(V ) induces aninvariant form on g given by the following formula,

βρ(X,Y ) = tr(ρ(X)ρ(Y )).

If ρ is adjoint representation this construction gives rise to an invariantwhich is called Killing form.

Remark 3.1.61. There are invariant forms which are not the traceforms of any representation.

Lemma 3.1.62. Let g be a Lie algebra, h an ideal and β the Killingform on g. Then the restriction of β to h × h is the Killing form of h.

Proof. Let X ∈ h then since [X, g] ⊂ h, adX is of the form

adX =

adh | ∗

0 | 0

(3.4)


Hence for X and Y ∈ h it follows that

tr(adX adY ) = trh(adh(X) adh(Y )).

Let g be a Lie algebra endowed with an invariant form β and letW be a subset of g. One can consider the orthocomplement of W withrespect to β,

W⊥ = X ∈ g| β(X,Y ) = 0, ∀ Y ∈W.

Corollary 3.1.63. Let g be a Lie algebra with an invariant form β andlet h be an ideal in g. Then h⊥ is also an ideal.

Proof. If X ∈ h⊥, Y ∈ g and Z ∈ h, then

β([X,Y ], Z) = β(X, [Y,Z]) = 0

since [Y,Z] ∈ h. Therefore [X,Y ] ∈ h⊥ and h⊥ is an ideal.

3.1.7 Complex and Real Lie Algebras

Let g be a Lie algebra over a field k and k′ be a field extension of k.The tensor product gk′ = k′ ⊗k g, a vector space over the field k′, canbe equipped with a Lie algebra structure induced by that of g. The Liebracket on the generators of gk′ is defined as follows,

[a⊗X, b⊗ Y ]gk′= ab⊗ [X,Y ]g

extending to all of gk′ by linearity.

In other words gk′ has the same structure constants as gk.In particular when k = R and k′ = C, we call gC complexification of

g. It is obvious that:

Proposition 3.1.64. Let g and k and k′ be as above. If h is a subalgebra(respectively ideal) of g then hk′ is also subalgebra (respectively ideal) ofgk′


Proposition 3.1.65. Let g be a Lie algebra over k and k′ a field ex-tension of k. Then for any two ideals a, b in g,

[a, b]k′ = [ak′ , bk′ ]

Proof is left to the reader as an exercise.

Proposition 3.1.66. Let g, k and k′ be as above. Then

(1) g is nilpotent if and only if gk′ is nilpotent.

(2) g is solvable if and only if gk′ is solvable.

(3) g is semisimple if and only if gk′ is semisimple.

Proof. (i) and (ii) follow from the fact that lower central series andderived series of gk′ are gi ⊗ k′i=1,2.. and gi ⊗ k′i=1,2..; part (iii)follows from Cartan’s criterion which we prove later in the chapter.

Proposition 3.1.67. Let g be a Lie algebra over a field k and let k′ bea field extension of k. If βk denotes the Killing form of g and βk′ theKilling form of gk′, then

βk′ |g⊗g = βk

Proof. This follows from the simple fact that the trace is independentof field extension.

3.1.8 Rational Forms

Let g be a Lie algebra over k and let k0 be a subfield of k. We say g hasa k0-rational form if there is a basis for g whose structure constants liein k0. If k = R and k0 = Q then we say g has a rational form.

Lemma 3.1.68. A Lie algebra g over k has a ko-rational form if andonly if there is a Lie algebra h over k0 such that g = hk.

Proof. If g = hk then any basis for h over k0 is a basis for g over k andobviously the structure constants are independent of the field extension.

To prove the converse, let vii∈I be a basis for g such that thestructure constants with respect to this basis are rational. Considerh, the vector space over k0 spanned by vii∈I . The Lie bracket of g


induces a Lie bracket on h, since the structure constants of the bracketof g with respect to vii∈I are in k0. Now it is clear that g = hk as Liealgebras over k.

Lie algebras do not always have rational forms. In fact,

Proposition 3.1.69. There exists a real (2-step) nilpotent Lie algebrawithout a rational form.

As we shall see in Chapter 8, this fact will prove the well-knownresult of Malcev [39] that even simply connected 2-step nilpotent groupsdo not always contain lattices.

We shall need a lemma about families of surjective linear maps.

Lemma 3.1.70. Let V and W be vector spaces of dimension n and mrespectively where n ≥ m, over a field k. Suppose that T is the subset ofmaps in Homk(V,W ) consisting of the linear maps which are surjective.Then T is an open set in Homk(V,W ).

Proof. T is the subset of Homk(V,W ) = Matn,m(k) consisting of oper-ators of maximal rank. That is to say the T ’s with some m×m-minorwith nonzero determinant. Therefore the complement of T consists ofthose matrices all of whose minors have determinant zero. This is the in-tersection of a finite number of (Zariski) closed sets which are Euclideanclosed. Hence T is open.

Proof of Proposition 3.1.69: We want to construct a Lie algebra g = E⊕V where E and V are real vector spaces such that [g, g] = V and [g, V ] =0. With these relations the Jacobi identity is automatically satisfied.Such a Lie algebra will be 2-step nilpotent and there is bijection betweenthe set of such Lie algebra structures Φ and the surjective R-linear mapsφ :

∧2E → V ; this is a manifold by the previous lemma and hence has

a dimension. Let n = dimE and m = dimV ; then dim Φ is m · n(n−1)2 .

Existence of a rational structure, φ0, on g means that there is a basise01, . . . , e

0n for E and a basis v0

1, . . . , v0m for V such that the matrix of

φ0 has all rational entries. Note that e0i ∧ e0j , i < j is a basis for∧2E. Suppose that e1, . . . , en and v1, . . . , vm are another such basis.


Let TE : E → E be an element of GL(E) taking the one basis of Eto the other and let T∧E :

∧2E → ∧2E be the map induced by TE .Similarly TV : V → V is an element of GL(V ) taking one basis of V tothe other. Then ΦQ, the space of all rational structures on g, equals theset of all maps of the form TV φ0 T−1

∧E . There are countably manysuch φ0 and the dimension of the set of all TV φ0 T∧E for a fixed φ0 isn2 +m2. For m ≥ 3 and sufficiently large n, this is < m · n(n−1)

2 whichis the dimension of the ambient space. Hence each of these sets hasLebesgue measure zero. By countable subadditivity the union also hasmeasure zero. Thus the complement is very large. In fact it is dense.

3.2 Engel and Lie’s Theorems

Here we shall deal with certain results concerning representations ofsolvable and nilpotent Lie algebras. As we shall see each of these isa generalization of a familiar theorem of linear algebra concerning asingle operator. These results are also important for non-solvable Liealgebras because those more general algebras have interesting solvableand nilpotent subalgebras. Later we shall consider the Lie group ana-logues of these results.

3.2.1 Engel’s Theorem

Definition 3.2.1. An operator T on a finite dimensional vector spaceV over a field k is called nilpotent if T j = 0 for some integer j.

Lemma 3.2.2. Let N1 and N2 be commuting nilpotent operators on avector space V of dimension n. Then N1 ±N2 is nilpotent.

Proof. By the binomial formula

(N1 ±N2)2n =

2n∑

i=0

(±1)i(

2n

i

)N i

1N2n−i2 .

Thus (N1±N2)2n = 0 since N i

1 = 0, if i ≥ n and N2n−i2 = 0, if i ≤ n.

3.2 Engel and Lie’s Theorems 151

Corollary 3.2.3. Let X ∈ gl(n, k) be nilpotent. Then adX ∈gl(gl(n, k)) is also nilpotent.

Proof. For any X ∈ g(n, k) define the left and right translation oper-ators, lX and rX by lX(T ) = XT and rX(T ) = TX. These operatorscommute because of the associativity of multiplication in gl(n, k) andalso lX − rX = adX. Hence by Lemma 3.2.2, adX is nilpotent.

Proposition 3.2.4. Let g be a subalgebra consisting of nilpotent oper-ators in the Lie algebra gl(n, k). Then there exists a v0 6= 0 ∈ V suchthat X(v0) = 0 for all X ∈ g.

Such a vector is called an invariant vector of g.

Proof. We prove this by induction on the dimension of g. If the di-mension is one we are really just talking about the line through X andso everything is determined by X itself. As is well known from linearalgebra a nilpotent operator always annihilates some nonzero vector.

Now suppose inductively that the proposition holds for all linear Liealgebras of dimension strictly lower than dimk(g). Let h be a proper sub-algebra of g of maximal dimension. Such an h exists since 1-dimensionalsubspaces are (abelian) subalgebras. Let H ∈ h. By 3.2.3 adH actingon gl(V ) is nilpotent. Hence adgl(V )H restricted to g, which is justadgH is also nilpotent. This operator stabilizes the subalgebra h and

induces a linear endomorphism adg on g/h, which is also nilpotent, de-

fined by adg(X + h) = adgH(X)+ h = [X,H]+ h. Now these operatorsform a Lie subalgebra of gl(g/h) and since h is a proper subalgebra, weget

dim adg h = dim(h + z(g)/z(g)) = dim(h/h ∩ z(g)) ≤ dim h < dim(g).

Therefore by induction there is an X ∈ g\h such that [X,H] ∈ h forall H ∈ h. This means that the linear span of h and X is a subalgebraof g strictly containing h. By maximality this subalgebra is g. Butsince [X, h] ⊆ h we see that h is actually an ideal in g. Let W = w ∈V : Hw = 0 for all H ∈ h. W is clearly a subspace of V . Now h

is a subalgebra of nilpotent operators on V of dimension strictly lower


than that of g. Hence there is some w 6= 0 ∈ W with Hw = 0 for allH ∈ h and in particular, W 6= 0. For w ∈ W and H ∈ h we haveHX(w) = [H,X]w +XH(w). Hence HX(w) = 0 for all H ∈ h. ThusW is an X stable subspace of V , and since X is nilpotent on V it isnilpotent on W . By the theory of a single nilpotent operator there issome w0 6= 0 ∈ W such that X(w0) = 0. Since h kills everything in Wand X kills w0 it follows that g kills w0.

We now come to Engel’s theorem itself which is just an amplifica-tion of the previous result. This is easily proved by induction on thedimension of V using the previous proposition and is left to the reader.

Theorem 3.2.5. Let g be a Lie subalgebra of the Lie algebra gl(n, k)consisting of nilpotent operators. Then there exists a basis v1, . . . vn ofV with respect to which g is simultaneously nil-triangular.

Corollary 3.2.6. A subalgebra g of gl(n, k) consisting of nilpotent op-erators is a nilpotent Lie algebra.

This is so because the full algebra of nil-triangular operators is nilpo-tent and hence so is any subalgebra. For similar reasons we get

Corollary 3.2.7. A Lie subalgebra g of gl(n, k) consisting of nilpotentoperators has the property that the associative product of any n elementsis zero.

The following variant is sometimes also called Engel’s theorem.

Corollary 3.2.8. A Lie algebra g is nilpotent if and only if adX is anilpotent operator on g for all X ∈ g.

Proof. If g is nilpotent, let n = 1+index of nilpotence. Then[X1, [X2, [X3 . . .]]] = 0. Taking all X1 = X2 . . . = Xn−1 = X andXn = Y we get (adX)n−1(Y ) = 0. Since Y is arbitrary each adX isnilpotent. On the other hand if each adX is nilpotent, then by a corol-lary to Engel’s theorem the linear Lie algebra ad g is nilpotent. Butthen so is g since it is a central extension of ad g.


Corollary 3.2.9. If g is a nilpotent Lie algebra and h is an ideal thenh ∩ z(g) is nonzero. In particular the center z(g) itself is nonzero.

Proof. ad g is a Lie subalgebra of gl(g) consisting of nilpotent operatorsand h is an invariant subspace under this subalgebra. Therefore ad g|his a Lie subalgebra of gl(h) consisting of nilpotent operators. Hence,there is a nonzero H ∈ h such that adX(H) = 0 for all X ∈ g and thusH ∈ h ∩ z.

Engel’s theorem can be formulated in terms of representations asfollows:

Theorem 3.2.10. Let g be a Lie algebra and ρ a representation of g onV . If ρ(g) consists of nilpotent operators, then there exists an operatorP0 ∈ GL(V ) such that P0(ρ(g))P−1

0 is in nil-triangular form.

Corollary 3.2.11. If g be a nilpotent Lie algebra and h is a propersubalgebra of g, then h ng(h).

Proof. Consider the adjoint representation of h on g. Since g is nilpotentthese operators are nilpotent. Because h is a subalgebra this inducesan action of h on g/h by nilpotent operators. Proposition 3.2.4 tells usthat there has to be X ∈ g \ h such that takes [X, h] ⊆ h. ThereforeX ∈ ng(h) \ h.

3.2.2 Lie’s Theorem

Before turning to Lie’s theorem we make some convenient definitionsamplifying the notion of invariant in the previous section.

Definition 3.2.12. Let g be a Lie algebra and ρ a representation of g

on V . If there is a nonzero vector v ∈ V and a function χ : g → k suchthat ρX(v) = χ(X)v for all X ∈ g we shall call χ a semi-invariant or aweight of ρ and v a weight vector .

Evidently a semi-invariant is k-linear and the χ(X)’s are simulta-neously eigenvalues of the ρX ’s. When χ is identically zero we obtaininvariants as before. Also, if ρ = ad, we shall say χ is a root and v a


root vector . Since each semi invariant χ kills the derived subalgebra,χ([g, g]) = 0, there will not be any nontrivial semi invariants at all ifg = [g, g]. This is the situation, for example, when g is a semisimple Liealgebra. At the opposite extreme is the case of a solvable Lie algebra.

Theorem 3.2.13. Let g be a solvable Lie algebra and ρ a representationof g on V over an algebraically closed field of characteristic zero. Thenthere exists an operator in P0 ∈ GL(V ) such that P0(ρ(g))P−1

0 is intriangular form.

Before turning to the proof of Lie’s theorem we make a few remarks.This result can be stated in several different ways. For example, it as-serts that there exists a basis v1, . . . vn of V with respect to which ρ(g)is in simultaneously upper triangular form. Evidently, Lie’s theoremgeneralizes the fact that, over an algebraically closed field (of character-istic zero), any operator can be put in triangular. Indeed, in the caseof a 1-dimensional (abelian) Lie algebra this is the content of Lie’s the-orem. It should also be remarked that even in the 1-dimensional case,the results fails if the field is not algebraically closed. For example, ifk = R, g is the 1-dimensional abelian Lie algebra of 2 × 2 skew sym-metric matrices and ρ is the identity representation, then there are nosimultaneous eigenvectors in R. Indeed, they are all in iR.

It should also be mentioned that Lie’s theorem is a generalizationof the following result. Let T be a family of commuting operators on avector space V over an algebraically closed field of characteristic zero,then these operators can be simultaneously put in triangular form. Thisis because the linear span of T is also a commuting family and thereforean abelian and hence solvable Lie algebra of operators on V .

Finally just as in the previous section it is sufficient to prove thefollowing result by making an induction on dimk V .

Proposition 3.2.14. Let g be a solvable Lie algebra and ρ a represen-tation of g on V over an algebraically closed field of characteristic zero.Then there exists a nonzero weight χ and weight vector v.

Proof. We shall argue by induction on dimk(g). Since [g, g] ⊆ g choosea subspace h of g such that [g, g] ⊆ h <( g with dim(g/h) = 1, i.e. h is


maximal. Then h is an ideal in g ([g, h] ⊆ [g, g] ⊆ h. In particular, it isa subalgebra and hence solvable. Consider the restriction of ρ to h. Byinductive hypothesis there is a v 6= 0 ∈ V satisfying ρHv = χ(H)v forall H ∈ h. Let X0 ∈ g \ h. Then g = h + cX0; c ∈ k. We complete theproof using the following lemma.

Lemma 3.2.15. Let ρ : g → gl(V ) be a representation of a Lie algebrag with an ideal h. Suppose there is a vector v 6= 0 ∈ V and χ : h → ksuch that ρ(H)v = χ(H)v for all H ∈ h. Then χ([X0,H]) = 0 for allX0 ∈ g and H ∈ h.

Proof. Let Vi be the subspace of V generated by v, ρX0v, . . . ρi−1X0v,

with V0 = 0. We get an increasing sequence of subspaces of V . Byfinite dimensionality there must be a place where Vi = Vi+1. Let n bethe smallest such index. But because Vi = Vi+1 if and only if ρiX0

v is

a linear combination of v, ρX0v, . . . , ρi−1X0

v, we see that dimVi = i, fori = 0, . . . , n. In particular, dimVn = n. Now ρX0(Vn) is spanned byρX0v, . . . , ρ

iX0v ⊆ Vn+1 = Vn. Hence ρX0(Vn) ⊆ Vn. From this it also

follows that Vn = Vi for all i ≥ n.Next we will prove by induction on i that for all H ∈ h,

ρHρiX0v ≡ χ(H)ρiX0

vmodVi.

When i = 0 this is just the statement ρHv = χ(H)v for all H. In theinductive step

ρHρi+1X0

v = ρHρX0ρiX0

(v)

= ρX0ρHρiX0

(v) + ρ[H,X0]ρiX0v

= ρX0(χ(H)ρiX0(v) + vi) + χ([H,X0])ρ

iX0

(v) + v′

i,

(3.5)

where vi and v′

i ∈ Vi.But this in turn is χ(H)ρi+1

X0(v) + ρX0(vi) + χ([H,X0])ρ

iX0

(v) + v′

i,

Since each of these terms is in Vi+1 we see that indeed ρHρi+1X0v ≡

χ(H)ρi+1X0

vmodVi+1. Thus with respect to a basis compatible with ourflag on Vn, for each H ∈ h, ρH is in simultaneous triangular form, withall diagonal terms equal to χ(H), Hence trVn(ρH) = (dimk Vn)χ(H),


for all H ∈ h. In particular, trVn(ρ[H,X0]) = 0 = dimk Vnχ([H,X0])and since k has characteristic zero it follows that χ([H,X0]) = 0 for allH ∈ h.

Let W =⋂H∈h

Ker(ρH −χ(H)I). Then W is a nonzero subspace of

V since v ∈W , and W is clearly h-invariant. For H ∈ h,

ρHρX0(w) = ρX0ρH(w) + ρ[H,X0](w) = ρH0χ(H)(w) + χ([H,X0])(w)

= χ(H)ρX0(w)

Since H is arbitrary we see that W is ρX0-invariant and hence g

invariant. Choose an eigenvector w0 for ρX0 in W with eigenvalue λ.Then since ρX = ρH + c(X)ρX0 everywhere on V we see that ρX(w0) =(χ(H) + λc(X))w0.

Our next corollary is actually equivalent to Lie’s theorem. No-tice also that if all finite dimensional irreducible representations are1-dimensional, this also implies solvability. This is because, as just re-marked, ρ(g) is a subalgebra of the full triangular algebra and hence issolvable. In particular, ad g is solvable. But then so is g itself.

Corollary 3.2.16. In a solvable Lie algebra each finite dimensionalirreducible representation over an algebraically closed field of character-istic zero is 1-dimensional.

Proof. Suppose ρ is the irreducible representation. By the propositionabove there is a weight χ and a weight vector v in V , the representationspace. The line through v is an invariant subspace and, by irreducibility,this must be all of V .

Applying Lie’s theorem to the adjoint representation we get,

Corollary 3.2.17. In a solvable Lie algebra over an algebraically closedfield of characteristic zero there is always a flag of ideals .

Here is a version of Lie’s theorem which does not require the fieldto be algebraically closed.


Corollary 3.2.18. Let g be a Lie algebra over a field of characteristiczero. g is solvable if and only if [g, g] is nilpotent.

Proof. If [g, g] is nilpotent and hence solvable, then since g/[g, g] isabelian and therefore also solvable so is g. Conversely, suppose g issolvable and the field is algebraically closed of characteristic zero. Ap-plying Lie’s theorem to the adjoint representation we see that ad g isa subalgebra of upper triangular operators. Hence its derived algebra,[ad g, ad g] = ad [g, g] consists of nil-triangular operators and is thereforenilpotent. Since Ker ad = z(g), it follows that [g, g] itself is nilpotent.This completes the proof when k is algebraically closed. In general weproceed by complexifying everything. Now suppose we have a solvableLie algebra over a field k of characteristic zero. Then, by Proposition3.1.65 and 3.1.66 [gk, gk] is nilpotent, where k is algebraic closure of k.Therefore gk is solvable and then by Proposition 3.1.66, g is solvable.

From this follows:

Corollary 3.2.19. For a Lie algebra g,

[rad(g), rad(g)] ⊂ nil(g).

Corollary 3.2.20. Let g be a solvable real Lie algebra and ρ a realrepresentation of g on V . Then there exists an increasing family ofinvariant subspaces which are each of codimension 1 or 2 in the next.Choosing a natural basis for these quotients puts ρ(g) in simultaneousblock triangular form over R, where the 2 × 2 blocks are of the type.

(aj(X) bj(X)−bj(X) aj(X)

)

for X ∈ g.

Notice that when aj(X) = 0 this is the form of the skew symmetricmatrices mentioned earlier.


3.3 Cartan’s Criterion and Semisimple Lie al-

gebras

3.3.1 Some Algebra

Definition 3.3.1. An operator T on a finite dimensional vector spaceV over k is called semisimple if T is diagonalizable over the algebraicclosure of k.

Theorem 3.3.2. (Jordan Decomposition) Let V be a finite dimensionalspace over an algebraically closed field and T ∈ Endk(V ). Then T =S +N where S is diagonalizable (semisimple), N is nilpotent and theycommute. These conditions uniquely characterize S and N . Moreoverthere exist polynomials p and q without constant term in k[x] such thatS = p(T ) and N = q(T ). Hence not only do S and N commute with Tbut they commute with any operator which commutes with T. If A ⊂ Bare subspaces of V and T (B) ⊂ A then S(B) ⊂ A and N(B) ⊂ A. Inparticular if A is a T invariant subspace then it is S and N invariant.If T (B) = 0 then S(B) = 0 and N(B) = 0.

The Jordan form of T consists of blocks each of which is the sumof a scalar and a nilpotent operator. This proves the first statement.Regarding the uniqueness, we first note that

Lemma 3.3.3. An operator S is semisimple if and only if every S-invariant W space has a S-invariant complement.

Proof. Suppose that S is semisimple and k is algebraically closed, thenwe can write V =

⊕α∈A Vα where Vα are 1-dimensional S-invariant

vector spaces and A is a finite set. Consider the family of sets of theform

⋃β∈BVβ ∪ W where B ⊂ A and Vβ’s and W are linearly

independent. This family is nonempty as it contains N. Since it isfinite, it has a maximal K =

⋃β∈BVβ∪W. LetM ′ =

⊕β∈B Vβ⊕W .

We prove that M ′ = M , otherwise there exists α ∈ A such that Vα *M ′. Since Vα is 1-dimensional therefore Vα ∩M ′ = (0). Therefore L =⋃β∈BVβ ∪ W ∪ Vα is in the family and K ⊂ L which contradicts

the maximality of K, hence M ′ = M . Now take W ′ =⊕

α∈A Vα. ThenM = W ⊕W ′ and W ′ is S invariant.

3.3 Cartan’s Criterion and Semisimple Lie algebras 159

The converse can be proved by an induction, and by noticing thatthere is always an eigenvector for S over an algebraically closed field.

Lemma 3.3.4. The restriction of a semisimple operator S to an in-variant subspace W is still a semisimple operator.

Proof. Let U be an S-invariant subspace ofW . Then U is an S-invariantsubspace of V . By the previous lemma there is an S-invariant subspaceU ′ of V which complements U in V . Then U ′ ∩W is an S-invariantsubspace of W which complements U in W .

Lemma 3.3.5. If S and S′ are diagonalizable and commute, then S±S′

is diagonalizable. If N and N ′ are nilpotent and commute, then N ±N ′

is nilpotent.

Proof. For α ∈ Spec(S) let Vα be the eigenspace of α. Then V is thedirect sum of the Vα. If v ∈ Vα then S(v) = αv so S′S(v) = αS′(v) =SS′(v) so that S′(Vα) ⊂ Vα. Now the restriction of S′ to each Vα isstill semisimple. Choose a basis in each Vα in which the restriction isdiagonal and in this way get a basis of V . Since, on each Vα, S = αI,it follows that S + S′ is diagonal. Moreover, Nn = 0 = (N ′)m so by thebinomial theorem (N +N ′)n+m = 0.

Now suppose S + N = T = S′ + N ′, where [S,N ] = 0 = [S′, N ′].Then S′ and N ′ each commutes with T . Hence each commutes withS and N . In particular, S and S′ and N and N ′ commute. But thenby the lemma S′ − S is semisimple and N − N ′ is nilpotent. SinceS′ − S = N −N ′ each of these is zero. This proves the uniqueness partof Theorem 3.3.2

Completion of proof of 3.3.2: Let χT (x) =∏

(x−αi)ni be the factor-ization of the characteristic polynomial of T into distinct linear factorsover k. Since the αi are distinct the (x − αi)

ni are pairwise relativelyprime. Consider the following

p(x) ≡ αi mod (x− αi)ni

p(x) ≡ 0 mod (x)


If no αi = 0 then x together with the (x − αi)ni is also relatively

prime. If some αi = 0 then the last congruence follows from the others.In either case by the Chinese Remainder Theorem there is a polynomialp satisfying them. Then p(T ) − αiI = φi(T )(T − αiI)

ni . So, on eachVαi

, p(T ) = αiI. This is equal to S on Vαiso S = p(T ). Taking

q(x) = x−p(x) we see that q(0) = 0 and q(T ) = T −p(T ) = T −S = N .Suppose A ⊂ B are subspaces of V and T (B) ⊂ A. Since S =

∑αiT

i,we get S(B) ⊂ A if we can show that T i(B) ⊂ A for i ≥ 1. NowT 2(B) = T (T (B)) ⊂ T (A) ⊂ T (B) ⊂ A and proceed by induction.

Theorem 3.3.6. (Lagrange Interpolation Theorem) Let c0, c1, ..., cn bedistinct elements of a field, k, and a0, a1, ..., an lie in k. Then there isa unique polynomial p in k[x] of degree n such that p(ci) = ai for all i.

Proof. For each i = 1, ..., n let φi(x) =∏j 6=i(x − cj). Then φi is a

polynomial in k[x] of degree n, φi(ci) 6= 0 and if k 6= i, φi(ck) = 0.Now fi = φi/φi(ci) is also a polynomial of degree n and fi(cj) = δijfor all i, j. Let a0, a1, ..., an be given and p =

∑n0 aifi. Then p(cj) =∑

aifi(cj) =∑aiδij = aj for all j. If q is another such polynomial of

degree ≤ n then (p − q)(cj) = 0. Since p − q has degree ≤ n and hasn+ 1 distinct roots p− q = 0 so p = q.

Corollary 3.3.7. Let a1, ..., an be in a field of characteristic 0, andE = l.s.Qa1, ..., an. If f ∈ E∗, the Q-dual space of E, then there isa polynomial p in k[x], without constant term, such that p(ai − aj) =f(ai) − f(aj) for all pairs i, j.

Proof. Let S and T denote the following finite sets S = ai − aj :i, j = 1, ..., n and T = f(ai) − f(aj) for i, j = 1, ..., n. Consider themap S → T given by ai− aj → f(ai)− f(aj). This is well-defined sinceif ai− aj = ak − al then f(ai− aj) = f(ak − al) and since f is Q-linear,f(ai) − f(aj) = f(ak) − f(al). If ai = aj then f(ai) = f(aj) so thismap takes 0 to 0. By Lagrange interpolation there is a polynomial psuch that p(ai − aj) = f(ai) − f(aj) for all pairs i, j and in particularp(0) = 0.


In what follows k is an algebraically closed field of characteristic 0,gl(V ) stands for Endk(V ) and Eij : i, j = 1, ..., n denotes its matrixunits.

Lemma 3.3.8. If X is a diagonal matrix with entries a1, ..., an thenadX(Eij) = (ai − aj)(Eij). Thus if X is semisimple on V then adX issemisimple on Endk(V ).

Proof. We have X =∑k

akEkk so

adX(Eij) =∑

k

ak[Ekk, Eij ] =∑

k

ak(EkkEij − EijEkk)

=∑

k

akδikEkj −∑

k

akEikδkj

= aiEij − ajEij = (ai − aj)Eij .

Lemma 3.3.9. Let X ∈ gl(V ) and let X = S + N be its Jordan de-composition. Then adX = adS + adN is the Jordan decomposition ofadX.

Proof. Since ad is linear adX = adS + adN . Moreover [adS, adN ] =ad [S,N ] = ad 0 = 0 so that adS and adN commute. By Lemma 3.3.8,adS is semisimple. By Lemma 3.2.3, adN is nilpotent. The uniquenessof the Jordan decomposition gives the result.

Lemma 3.3.10. Let p be a polynomial without constant term and letT, S ∈ gl(V ) for some V . Suppose that, relative to a basis v1, ..., vn,T and S are diagonal with entries a1, ..., an and p(a1), ..., p(an) re-spectively. Then clearly S = p(T ). In particular, by the argument inthe Jordan decomposition theorem, if A ⊂ B are subspaces of V andT (B) ⊂ A then S(B) ⊂ A.

In order to formulate and prove Cartan’s criterion for an arbitraryfield of characteristic zero, it will be necessary to deal with certain finitedimensional rational vector spaces in the following lemma.


Lemma 3.3.11. Let A ⊂ B be subspaces of gl(V ) and M = X ∈gl(V ) : [X,B] ⊂ A. If X ∈ M and tr(XY ) = 0 for all Y ∈ M then Xis nilpotent.

Proof. Let X = S +N be the Jordan decomposition of X and v1, ..., vnbe a basis such that S is diagonal with entries a1, ..., an. Let E bethe subspace of k spanned by by a1, ..., an as a Q-vector space. IfE = 0 then all ai = 0 and hence S = 0 and X is nilpotent. Since Eis finite dimensional space to see E = 0 it suffices to show its dual E∗

over Q, is trivial. If f ∈ E∗ let Y be the diagonal matrix with entriesf(a1), ..., f(an) with respect to v1, ..., vn. By a previous lemma,adS(Eij) = (ai − aj)Eij and adY (Eij) = (f(ai) − f(aj))Eij . By thecorollary to the Lagrange interpolation Theorem, Corollary 3.3.7, thereis a polynomial p such that p(0) = 0 and p(ai − aj) = f(ai) − f(aj)for all pairs i, j. This means adY (Eij) = p(ai − aj)Eij . Since X ∈ M ,adX(B) ⊂ A. As above adS is a polynomial in adX without constantterm; it follows that adS(B) ⊂ A. By the last lemma ad Y (B) ⊂A and hence Y ∈ M . Now XY is triangular with diagonal entriesa1f(a1), ..., anf(an) and so tr(XY ) =

∑aif(ai) = 0. This is a Q-

linear combination of elements of E. Applying f yields∑f(ai)

2 = 0.Since the f(ai) are in Q they are all 0. We conclude that f kills thegenerators for E so f = 0 and since f is arbitrary E∗ = (0).

3.3.2 Cartan’s Solvability Criterion

Cartan’s solvability criterion is the following:

Proposition 3.3.12. (Cartan’s Criterion) Let g be a subalgebra ofgl(V ) over any field of characteristic zero such that tr(XY ) = 0 forall X ∈ [g, g] and Y ∈ g. Then g is solvable.

Proof. By a corollary to Lie’s theorem it suffices to show that [g, g] isnilpotent; and by Engel’s theorem that each element of [g, g] is nilpotent.Let A = [g, g], B = g, X and Y ∈ g and Z ∈ M where M is defined asin Lemma 3.3.11. Then [Z, g] ⊂ [g, g]. In particular [Z, Y ] ∈ [g, g] so ourhypothesis tells us that tr(X[Y,Z]) = 0 or by invariance tr([X,Y ]Z) =


0. By linearity tr(UZ) = 0 for any U ∈ [g, g]. Since g ⊂ M Lemma3.3.11 tells us that each such U is nilpotent.

Corollary 3.3.13. (Also called Cartan’s criterion) Let g be a Lie alge-bra (any field of characteristic zero) such that tr(adX adY ) = 0 for allX ∈ [g, g] and Y ∈ g then g is solvable.

Proof. By Cartan’s criterion ad g is solvable. Since Ker ad = z(g) is alsosolvable so is g.

Let ρ : g → gl(V ) be a representation of a Lie algebra g on V (anyfield of characteristic zero). Then, as above, the trace form

βρ : g × g → k

is given by βρ(X,Y ) = tr(ρ(X)ρ(Y )) and βρ is a symmetric bilinearform on g. In particular in the case of the Killing form we we have aninvariant symmetric bilinear form β on g.

Another way of expressing Cartan’s criterion is the first half of thefollowing:

Corollary 3.3.14. Let g be Lie algebra over any field k such that [g, g]is orthogonal to g with respect to the Killing form then g is solvable.Conversely if g is a solvable then [g, g] is orthogonal to g with respect toKilling form.

Proof of Converse: Let k be the algebraic closure of k and regardad g as a subalgebra of gl(gk) rather than of gl(g). Since ad g is solvableby Lie’s theorem we know that ad g can be triangularized over k. Henceits derived algebra has zeros on the diagonal and so trgC

(adX adY ) = 0for all X ∈ [g, g] and Y ∈ g. Since the trace is independent of fieldextensions, [g, g] is orthogonal to g with respect to the Killing form.

Remark 3.3.15. The same argument shows that even without consid-ering field extensions, the Killing form is identically zero if g is nilpotent.For then ad g can be nil-triangularized, therefore adX adY is niltrian-gular and so has trace 0. Exercise: What about the converse?


Corollary 3.3.16. A Lie algebra g is semisimple if and only if itsKilling form is nondegenerate. In fact (since the adjoint representationof a semisimple Lie algebra is faithful) more generally if g is semisimpleand ρ is any faithful representation of g then βρ is nondegenerate.

Proof. Suppose g is semisimple. Since βρ is invariant, h = g⊥ is an ideal.Because h is orthogonal to g it is orthogonal to [h, h]. Hence by Cartan’scriterion ρ(h) is solvable and since ρ is faithful this means h is itselfsolvable and so is trivial. Thus βρ is nondegenerate. Conversely supposethe Killing form β is nondegenerate and h is an abelian ideal. LetX ∈ h and Y ∈ g. Then adX ad Y (g) ⊂ h. So that (adX ad Y )2(g) ⊂adX adY (h). Since h is an ideal, adX adY (h) ⊂ [X, h] = 0 and h isabelian. This means adX adY is nilpotent and so has trace 0. As β isnondegenerate we must have X = 0 so h = 0.

Proposition 3.3.17. If h is an ideal in a semisimple Lie algebra g anda is an ideal in h then a is an ideal in g.

Proof. Let h⊥ be the orthocomplement of h in g with respect to Killingform. Since Killing form is nondegenerate we have h ⊕ h⊥ = g and h⊥

is also an ideal in g. We have [a, h⊥] ⊂ [h, h⊥] ⊂ h∩ h⊥ = 0, therefore[a, g] ⊂ [a, h] ⊂ a.

Lemma 3.3.18. Let V be a finite dimensional k-vector space and letβ : V ×V → k be a bilinear form. Given a subspace W of V write W⊥ =v ∈ V : β(v,W ) = 0. Then W⊥ is a subspace of V and dimW +dimW⊥ ≥ dimV . If β is nondegenerate then dimW + dimW⊥ =dimV .

Proof. Choose a basis v1, ..., vn of V so that v1, ..., vk is a basis ofW .For x ∈ V and i = 1, ..., k let αi(x) = β(vi, x) and S = αi : i = 1, ..., k.Then each αi ∈ V ∗ (the dual space of V ) and W⊥ =

⋂S Kerαi. If j

is the maximum number of linearly independent elements in the spanof S then j = k and dimW⊥ = n − j so k + dimW⊥ ≥ n. If βis nondegenerate and

∑ciαi = 0 is a dependence relation among the

elements of S then β(∑civi, x) = 0, for all x ∈ V and hence

∑civi = 0,

and so each ci = 0. This means that j = k and so dimW + dimW⊥ =dimV .


Corollary 3.3.19. If g is semisimple then adjoint representation iscompletely reducible i.e. g is the direct sum of simple ideals gi. Further-more every simple ideal in g coincides with one of the gi. In particularthe simple ideals are absolutely unique (not just up to equivalence ofrepresentations). Conversely, a direct sum of simple (or semisimple)algebras is semisimple.

Proof. If h is an ideal in g then h is an ad-invariant subspace. But h⊥

is also an ideal. Hence h ∩ h⊥ is an ideal and β restricted to h ∩ h⊥ isidentically 0. But β restricted to h∩ h⊥ coincides with the Killing formof h∩h⊥. By Cartan’s criterion h∩h⊥ is solvable and so is trivial. Sinceβ is nondegenerate dim h + dim h⊥ = dim g. Hence h ⊕ h⊥ = g. Thusby induction on dimension ad is completely reducible and g = ⊕gi, thedirect sum of simple ideals. If h is any simple ideal of g then [h, g] isan ideal in h and therefore is either trivial or h. In the former caseh ⊂ z(g) = 0. In the latter, h = [h, g] = ⊕[h, gi] a direct sum of ideals.Since h is simple h = [h, gi] for some i. But [h, gi] ⊂ gi. Since h ⊂ giand the latter is also simple, and h is not the zero ideal we must haveh = gi.

For the converse it is sufficient to show that a direct sum of twosemisimple algebras is semisimple. Let g = h⊕ l. If a is an abelian idealin g then the π(a) is an abelian ideal in h where π is the projection onh. Therefore π(a) = 0 and a ⊂ l. Similarly a ⊂ h. Hence a = 0.

Corollary 3.3.20. If g is semisimple then [g, g] = g.

Proof. Since g = ⊕gi, the direct sum of simple ideals we see that [g, g] =∑i,j[gi, gj ]. If i 6= j then [gi, gj ] ⊂ gi ∩ gj = 0 and if i = j then [gi, gi]

is a nontrivial ideal in the simple algebra gi. Hence [gi, gi] = gi and so[g, g] = g.

Corollary 3.3.21. If g is semisimple Lie algebra so is any ideal of g,as is any homomorphic image of g. Any ideal in g is the direct sum ofsome of the gi.

Proof. If h is an ideal in g then g = h⊕h′. Continue decomposing thesetwo summands. Then h (and h′) is the direct sum of certain of the gi.


In particular h is semisimple. Since g/h ∼= h′ and h′ is semisimple thiscompletes the proof.

Corollary 3.3.22. If g is an arbitrary Lie algebra and h is an ideal ing which as a Lie algebra is semisimple then h is a direct summand.

Proof. Let β be the Killing form of g and γ the restriction of β to h×h.Then since h is an ideal γ is the Killing form of h and is non degeneratesince h is semisimple. If X ∈ h ∩ h⊥ then β(X,H) = 0 for all H ∈ h,but since X is itself in h this means γ(X,H) = 0 and so X = 0. Thush∩ h⊥ = 0. Since dimh + dim h⊥ ≥ dim g and both h and h⊥ are idealsin g this completes the proof.

Corollary 3.3.23. In a semisimple algebra each derivation is inner.

Proof. ad g is a subalgebra of Der(g); in fact if X ∈ g and D ∈ Der(g),then [D, adX] = adD(X), hence ad g is an ideal in Der(g). Being ahomomorphic image of a semisimple algebra ad g is semisimple. Henceit is a direct summand: Der(g) = ad g ⊕ h. Let D ∈ h. Then for all X,[D, adX] = 0 = adD(X). Since g is semisimple ad is faithful and soD = 0 and Der(g) = ad g.

Corollary 3.3.24. Let g and g′ be an arbitrary Lie algebras with radi-cals r and r′ respectively and let f : g → g′ be a Lie algebra epimorphism.Then f(r) = r′.

Proof. Clearly f(r) ⊂ r′. If f : g/r → g′/f(r) denotes the induced Liealgebra epimorphism then since g/r is semisimple so is g′/f(r). Sincef(r) is an ideal in g′, we have r′ ⊂ f(r).

3.3.3 Explicit Computations of Killing form

We now compute the Killing form of certain Lie algebras explicitly.Although we do this over R or C our method works without change overfields k of characteristic 0. In order to do this we realize gl(V ) in anotherway. Let v1, ..., vn be a basis of V . Then vi ⊗ vj : i, j = 1, ..., n is abasis of V ⊗V and so this space has dimension n2. This means that thek-linear map φ : gl(V ) → V ⊗V given by Y → ∑

i,j yijvi⊗vj is a k-linear


isomorphism. If X ∈ gl(V ) then X(vi) =∑k,i

xkivk and Xt(vj) =∑j,l

xjlvl.

The question is, what does ad look like on V ⊗ V ? For each X ∈ gl(V )the diagram

gl(V )φ−−−−→ V ⊗ V

yadX

yX⊗I−I⊗Xt

gl(V )φ−−−−→ V ⊗ V

(3.6)

is commutative. We have

φ(adX(Y )) = φ([X,Y ]) =∑

i,j

(XY − Y X)ijvi ⊗ vj ,

whereas

(X ⊗ I − I ⊗Xt)φ(Y ) = (X ⊗ I − I ⊗Xt)(∑

i,j

yijvi ⊗ vj)

=∑

i,j

xij(X ⊗ I − I ⊗Xt)(vi ⊗ vj)

=∑

i,j

xij(X(vi) ⊗ vj − vi ⊗Xt(vj))

=∑

i,j

xij(∑

k

xkivk ⊗ vi −∑

1

xjlvi ⊗ vl)

=∑

k,j,i

xkiyijvk ⊗ vj −∑

l,i,j

yijxjlvi ⊗ vl

=∑

k,j

(XY )kjvk ⊗ vj −∑

i,l

(Y X)ilvi ⊗ vl

=∑

s,t

(XY − Y X)stvs ⊗ vt.

Since this holds for all Y ∈ gl(V ), we conclude

φ(adX) = (X ⊗ I − I ⊗Xt)φ


for all X ∈ gl(V ). Hence

adX = φ−1(X ⊗ I − I ⊗Xt)φ

and so

trgl(V )(adX2) = trV⊗V ((X ⊗ I − I ⊗Xt)2)

But (X ⊗ I − I ⊗ Xt)2 = (X2 ⊗ I − 2(X ⊗ Xt) + I ⊗ (Xt)2). Sincetr(X) = tr(Xt) and tr(X2) = tr((Xt)2) we see

trV⊗V ((X ⊗ I − I ⊗Xt)2) = 2n tr(X2) − 2 tr(X)2.

Lemma 3.3.25. (Polarization Lemma) Let α and β be symmetric bi-linear forms W ×W → k where char k 6= 2. If α(X,X) = β(X,X) forall X ∈W then α = β.

Proof. α(X +Y,X +Y ) = α(X,X) + 2α(X,Y ) +α(Y, Y ) and similarlyfor β. Therefore α(X,Y ) = β(X,Y ).

Since the Killing form satisfies β(X,X) = 2n tr(X2) − 2 tr(X)2 andα(X,Y ) = 2n tr(XY ) − 2 tr(X) tr(Y ) is a symmetric bilinear form ongl(V ) it follows, by polarization, that for gl(V )

β(X,Y ) = 2n tr(XY ) − 2 tr(X) tr(Y ).

Corollary 3.3.26. For sl(V ) (any k of char 6= 2)the Killing form β isgiven by β(X,Y ) = 2n tr(XY ).

Proof. sl(V ) is an ideal in gl(V ).

Corollary 3.3.27. For k = R or C and n ≥ 2, gl(V ) is not semisimplewhereas sl(V ) is semisimple.

Proof. If g = gl(V ) and X = αI then β(αI, Y ) = 0 so β is degenerate.

For sl(V ), β(X,Y ) = 2n tr(XY ) so if β(X,Y ) = 0 for all Y ∈ g thenβ(X,Xt) = 0 because Xt ∈ sl(V ). Therefore

∑i,j |xij |2 = 0 where xij

are the entries of X. This means X = 0.


Let

X =

0 ∗ ∗ ... ∗0 0 ∗ ... ... ∗0 0 0 ∗ ... ∗... ... ... ... ... ...... ... ... ... ... ...0 0 0 0 0 0

Y =

0 1 0 ... ... 0−1 0 0 ... ... 00 0 0 ... ... 0... ... ... ... ... ...... ... ... ... ... ...0 0 0 0 0 0

H =

−1 0 0 ... ... 00 1 0 ... ... 00 0 0 0 ... 0... ... ... ... ... ...... ... ... ... ... ...0 0 0 0 0 0

Then X, H, Y ∈ sl(n,R) and 2n tr(X2) = 0, 2n tr(H2) = 4n,2n tr(Y 2) = −4n. Thus β is neither positive nor negative definite.Whereas for so(n,R) we know β is negative definite (see Section 3.9).This shows that sl(2,R) is not isomorphic to so(3,R).

Now let k = R or C, n ≥ 2 and so(V ) = X ∈ gl(V ) : Xt = −X.We wish to compute the Killing form of h = so(V ). Let S = X ∈ gl(V ) :Xt = X and σ : V ⊗ V → V ⊗ V be defined by σ(v ⊗ w) = w ⊗ v.Then h is a subalgebra of gl(V ) and gl(V ) is the direct sum of h and Sas k-spaces.

Lemma 3.3.28. trV⊗V (σ(A⊗B)) = tr(AB).

Proof. Both sides are bilinear maps gl(V )× gl(V ) → k. By polarizationit suffices to show that trV⊗V (σ(A ⊗ A)) = trV (A2). If A is a matrixthen

σ(A⊗A)(vi ⊗ vj) = A(vj) ⊗A(vi) =∑

k

ajkvk ⊗∑

l

ailvl

=∑

k,l

ajkailvk ⊗ vl.

Therefore trV⊗V (σ(A⊗A)) =∑

i,j ajiaij = trV (A2).


Decompose gl(V ) = h⊕S under Y → (Y −Y t)/2+(Y +Y t)/2. Applyφ and get V ⊗ V = φ(h) ⊕ φ(S) where φ(h) = ∑

i,jyijvi ⊗ vj|yij = −yji

and φ(S) = ∑i,jyijvi ⊗ vj : yij = yji. Let π : V ⊗ V → φ(h) be the

projection onto φ(h). Then clearly π = (I − σ)/2. Also notice thatif X ∈ h, Y ∈ h and Z ∈ S then [X,Y ] ∈ h (h is a subalgebra) and[X,Z] ∈ S; that is for X ∈ h, adX leaves both h and S stable. We areinterested in

trh(adX2 :h) = tr((X2 ⊗ I − 2X ⊗Xt + I ⊗ (Xt)2)|φ(h))

= trV⊗V (1

2(I − σ)(X2 ⊗ I − 2X ⊗Xt + I ⊗ (Xt)2)).

But since X ∈ h, this is trV⊗V (12(I −σ)(X2 ⊗ I + 2(X ⊗X) + I ⊗X2)).

Applying the lemma above to calculate this, we get (n− 2) tr(X2).

Corollary 3.3.29. For k = R or C the Killing form β of so(V ) is givenby β(X,Y ) = (n − 2) trV (XY ).

Proof. Apply the polarization lemma.

Corollary 3.3.30. For k = R, or C and n ≥ 3, so(V ) is semisimple.

Proof. We will show β is nondegenerate. Let X ∈ h and supposetr(XY ) = 0 for all Y ∈ h. If Z ∈ gl(V ) and Y = Z − Zt. ThenY ∈ h and so 0 = tr(X(Z − Zt)). Since tr(XZt) = tr(ZXt) = tr(XtZ),we see that 0 = tr((X − Xt)Z). But Z is arbitrary and tr(XY ) isnondegenerate on gl(V ). Therefore X ∈ S and since X is also in h,X = 0.

3.3.4 Further Results on Jordan Decomposition

Corollary 3.3.31. Let g be a Lie algebra an algebraically closed field kof characteristic 0, and let D ∈ Der(g). Then the Jordan componentsof D are also derivations.


Proof. If D = S +N it suffices to show that S ∈ Der(g). For α ∈ k letgα = X ∈ g : (D − αI)iX = 0 for some integer i. Then Spec(D) =α : gα 6= 0 and the gα’s are D (and S) invariant subspaces of g

on which S acts as scalars by αI; hence g = ⊕gα|α ∈ SpecD. Aninductive calculation shows that for all n

(D − (α+ β)I)n[X,Y ] =n∑

i≥0

(n

i

)[(D − αI)n−iX, (D − βI)iY ]

In particular if α and β ∈ SpecD then [gα, gβ] ⊂ gα+β, and if α + β isnot in SpecD then [gα, gβ ] = 0. Let X =

∑Xα and Y =

∑Yβ. Then

[X,Y ] =∑

α,β [Xα, Yβ], so

S[X,Y ] =∑

α,β

S[Xα, Yβ] =∑

α,β

(α+ β)[Xα, Yβ ].

On the other hand [SX, Y ] + [X,SY ] clearly also equals∑α,β

(α +

β)[Xα, Yβ].

In general, if k is an algebraically closed field of characteristic zero,for a derivation in a Lie algebra over k, the Jordan components compo-nents of Lie derivation are also derivations. In a semisimple algebra g

each derivation is inner. Hence for X ∈ g, adX = adY + adZ whereadY is semisimple and adZ is nilpotent. We now derive a result whichimplies this in a stronger form.

Theorem 3.3.32. Let g ⊂ gl(V ) be a semisimple Lie algebra over analgebraically closed field k of characteristic 0. Then g contains thesemisimple and nilpotent parts of each of its elements. In particu-lar, for each X ∈ g, we have X = S + N , and hence we know thatadX = adS + adN is the Jordan decomposition of adX ∈ gl(gl(V )).

Proof. For a subspace W of V let

gW = X ∈ gl(V ) : X(W ) ⊂W and tr(X|W ) = 0

For each W , gW is a subalgebra of gl(V ). Now in general ngl(V )(g)is a subalgebra of gl(V ) containing g. (If [X, g] ⊂ g and [Y, g] ⊂ g


then ad[X,Y ] stabilizes g by Jacobi). Let s = W : W is g − stableand g# =

⋂W∈s

gW ∩ ngl(V )(g). If W ∈ s then g(W ) ⊂ W . The map

X → X|W is clearly a Lie algebra homomorphism so X|W : X ∈ gis semisimple. In particular if X ∈ g, tr(X|W ) = 0 so X ∈ gW . Thismeans that g ⊂ ⋂

W∈s

gW so that g# is a subalgebra of gl(V ) containing

g as an ideal.We show g# = g: Since g is a semisimple ideal in g# then by Corol-

lary 3.3.21 it is a direct summand so g# = g⊕ h. Let H ∈ h and let W0

be a minimal g-stable subspace of V . Then W0 ∈ s so g# ⊂ gW0 andeach element of g# leaves W0 stable. In particular H(W0) ⊂W0. Since[g, h] = 0 on V , and therefore also on W0, each H is an intertwiningoperator on the irreducible subspace W0 so H = cI on W0 by Schur’slemma. Hence trH = cdimW0 = 0 so c = 0 and H = 0 on W0. ByWeyl’s theorem, to be proven in Section 3.4, V is the direct sum ofirreducible g-subspaces so H = 0 on V . Thus h = 0 and g# = g.

Now let X ∈ g and X = S +N be its Jordan decomposition. ThenadX = adS + adN is the Jordan decomposition of adX in gl(gl(V )).In particular, adS = s(adX) where s is a polynomial without constantterm and since adX stabilizes g so does adS. On the other hand S =p(X) where p is also a polynomial without constant term. We provep(X) ∈ g#(= g); hence also N ∈ g. Since adS ∈ Ngl(V )(g) we mustshow that p(X) ∈ gW for all W ∈ s. But for W ∈ s, X(W ) ⊂ Wand hence p(X)(W ) ⊂ W . Similarly N stabilizes W . Since X ∈ g

and N is nilpotent so is its restriction to W ; hence tr(N |W ) = 0. ButX|W = N |W + S|W so tr(X|W ) = tr(N |W ) + tr(S|W ) = tr(S|W ). Thensince g ⊂ ⋂

W∈s

gW and X ∈ g, tr(X|W ) = 0 we get S ∈ gW .

Corollary 3.3.33. Let g ⊂ gl(V ) be a semisimple Lie algebra over analgebraically closed field and X ∈ g. Then X is semisimple (respectivelynilpotent) if and only if adX is semisimple (respectively nilpotent). Infact, adgX = adgS + adgN is the Jordan decomposition of adgX ingl(g).

Proof. Let S and N be the semisimple and nilpotent parts of X. By thetheorem we know S and N ∈ g. We also know that adX = adS+adN

3.4 Weyl’s Theorem on Complete Reducibility 173

is the Jordan decomposition of adX ∈ gl(gl(V )). Since the restrictionof a semisimple or nilpotent operator to an invariant subspace is respec-tively semisimple or nilpotent we see that adgS is the semisimple partof adgX and adgN is the nilpotent part of adgX. In particular if Xis semisimple (respectively nilpotent) then adgX is semisimple (respec-tively nilpotent). Conversely if adgX is semisimple i.e. adgX = adgSthen X = S since ad is faithful, similarly for nilpotent case.

3.4 Weyl’s Theorem on Complete Reducibility

We prove Weyl’s complete reducibility theorem. Our first step is todefine the Casimir element associated with a representation.

Lemma 3.4.1. Let V be a finite dimensional vector space over a fieldk and β : V × V → k be a nondegenerate symmetric bilinear form. IfX1, . . . ,Xn is a basis of V then there exists a dual basis Y1, . . . , Ynthat satisfies β(Xi, Yj) = δij .

Proof. For fixed X, the map Y 7→ β(X,Y ) is in V ∗, so we get a map µfrom V → V ∗ which, by assumption is injective. Comparing dimensionswe see that µ is an isomorphism. If X1, . . . ,Xn is a basis of V thereis a corresponding dual basis X∗

1 , . . . ,X∗n of V ∗ satisfying X∗

j (Xi) =δij . Taking Y1, . . . , Yn as the µ pre-image of X∗

1 , . . . ,X∗n yields the

result.

In particular, if g is a semisimple Lie algebra, ρ : g → gl(Vρ) is afaithful representation of g on Vρ and βρ : g×g → k is the trace form ofρ then βρ is nondegenerate and for each basis X1, . . . ,Xn of g thereis a basis Y1, . . . , Yn satisfying βρ(Xi, Yj) = δij .

We now define the Casimir operator Cρ of ρ to be the element of theassociative algebra Endk(Vρ) given by

Cρ =∑

i

ρ(Xi) · ρ(Yi).

Our next result shows that the Casimir operator is an invariant ofthe particular representation, (g, ρ, V ), we are lo.


Proposition 3.4.2. Let g is a semisimple Lie algebra and ρ : g → gl(V )is a faithful representation of g.

(1) The operator Cρ, is independent of the choice of the basisX1, . . . ,Xn.

(2) If X1, . . . ,Xn is a basis and Y1, . . . , Yn is the dual basis X ∈ g

we write [X,Xi] =∑

j aij(X)Xj and [X,Yi] =∑

j bij(X)Yj . Thenaik(X) = −bki(X) for all i, k ≤ dim g and X ∈ g.

(3) trCρ = dim g.

(4) Cρ is an intertwining operator on V. In particular, if ρ is irre-

ducible and k is algebraically closed then Cρ = dim gdimV · I

Proof of 1: Let X ′

1, . . . ,X′

n be another basis and Y ′

1 , . . . , Y′

n itsdual basis. Then X

′

i =∑jαijXj and Y

′

k =∑l

βklYl. This means

δik = βρ(X′

i , Y′

k) =∑

j,l

αijβklβρ(Xj , Yl)

But this is∑j,l

δjlαijβkl =∑jαijβkj, so α · βt = I. Taking transposes,

αt · β = I. Now∑

i

ρ(X′

i) · ρ(Y′

i ) =∑

i

(∑

j

αijρ(Xj))(∑

l

βilρ(Yl))

=∑

i

∑

j,l

αijβilρ(Xj)ρ(Yl)

=∑

j,l

∑

i

αtjiβilρ(Xj)ρ(Yl)

(3.7)

Since β · αt = I the last term is∑

j,l δjlρ(Xj)ρ(Yl) = Cρ.

Proof of 2: aik(X) =∑jaij(X)βρ(Xj , Yk) = βρ(

∑jaij(X)Xj , Yk). But

this is βρ([X,Xi], Yk) = −βρ([Xi,X], Yk) = −βρ(Xi, [X,Yk]), which inturn equals

−βρ(Xi,∑

j

bkj(X)Yj) = −∑

j

bkj(X)βρ(Xi, Yj) = −∑

j

bkj(X)δij

= −bki(X).


Proof of 3:

tr(Cρ) =∑

i

tr(ρ(Xi)ρ(Yi)) =∑

i

βρ(Xi, Yi) =∑

i

δii = dim g.

Proof of 4: For X ∈ g, we have

[ρ(X), Cρ] = [ρ(X),∑

i

ρ(Xi)ρ(Yi)] =∑

i

[ρ(X), ρ(Xi)ρ(Yi)].

But this is∑

i[ρ(X), ρ(Xi)]ρ(Yi) + ρ(Xi)[ρ(X), ρ(Yi)] (we invoke thematrix identity [A,BC] = [A,B]C +B[A,C]). This last expression canbe rewritten as

∑

i

ρ([X,Xi])ρ(Yi) + ρ(Xi)ρ([X,Yi])

=∑

i

ρ(∑

j

aij(X)Xj)ρ(Yi) + ρ(Xi)ρ(∑

j

bij(X)Yj)

=∑

i

∑

j

aijρ(Xj)ρ(Yi) + bijρ(Xi)ρ(Yj)

This is zero aij(X) = −bji(X) for all i, j. If ρ is irreducible and k isalgebraically closed then by Schur’s lemma Cρ = cI. But,

tr(Cρ) = cdimV = dim g. Hence Cρ = dimgdimV · I.

Let ρ and σ be representations over k of g on Vρ and Vσ, respectively.We define a new representation of g on the k space Homk(Vρ, Vσ) asfollows. For X ∈ g and T ∈ Homk(Vρ, Vσ) take X(T ) = T ρX −σX T .One checks immediately that this is a Lie representation of g.

Theorem 3.4.3. (Weyl) If ρ : g → gl(V ) is a representation of asemisimple Lie algebra over k, then ρ is completely reducible.

Proof. We may assume that g acts faithfully since this cannot affectcomplete reducibility and still keeps semisimplicity. Using the remarksabove we may also assume the field k is algebraically closed. We firstdeal with the case in which there is a g-invariant subspace W of V ofcodimension 1. Our proof, in this case, goes by induction on dimV . If


W has a proper g-invariant subspace then it has a minimal proper one,say W0. Then we have an exact sequence of g modules

0 → W/W0 → V/W0 → V/W → 0

Since W/W0 is a submodule of codimension 1 of V/W0 and dimV/W0 <dimV , there is a complementary 1-dimensional g-invariant subspaceU/W0 in V/W0 to W/W0. But since U/W0 is ρ-invariant and 1-dimensional, and g is semisimple, ρ acts trivially on U/W0. Thus,ρX(U/W0) = 0 for all X ∈ g. Hence ρX(U) ⊂ W0, and since W0 ⊆ U ,this means that U is ρ-invariant. But dimU = 1 + dimW0 and if W isreducible this is < 1+dimW = dimV . Hence, again by induction, thereis a ρ-invariant 1-dimensional subspace of U complementary to W0. Letu0 ∈ U generate this line L over k. Then u0 /∈ W and U +W = V , soL+W = V . Since L∩W = 0 and dimV/W = 1, L⊕

W = V and Whas a complementary 1 dimensional g-invariant subspace. This meanswe may assume that W is irreducible.

Because W is a g-invariant subspace of V of codimension 1, ρ inducesρ of g on V/W which, as above, is trivial. So ρX(V ) ⊂ W for allX ∈ g. If Cρ is the Casimir operator, then Cρ(v) =

∑i ρ(Xi)ρ(Yi)(v) so

Cρ(V ) ⊂W . In particular W is Cρ-invariant. Because W is irreducible,Cρ restricted to W is cI. If this c = 0 then C2

ρ = 0. But since trV Cρ =dim g = 0, this is a contradiction. Now ρ is a faithful representationof a semisimple Lie algebra, so we also know Cρ is an intertwiningoperator on V . This means that KerCρ is an g-invariant subspace.Suppose w ∈ KerCρ ∩ W . Then Cρ(w) = cw = 0 and since c 6= 0,w = 0 so that KerCρ ∩W = 0. On the other hand, dimKerCρ +dimCρ(V ) = dimV and dimCρ(V ) ≤ dimW so dim KerCρ ≥ 1. Thusdim KerCρ + dimW ≥ dimV . Together with disjointness from W thisshows KerCρ

⊕W = V , completing the proof whenW has codimension

1.

Finally let W be an arbitrary g-invariant subspace of V and considerthe representation of g on Homk(V,W ) defined above where for σ wetake ρ restricted to W . Let V = T ∈ Homk(V,W ) : T |W = λIW andW = T ∈ Homk(V,W ) : T |W = 0.


Lemma 3.4.4. V and W are subspaces of Homk(V,W ); W is codimen-sion 1 in V and g(V) ⊆ W. In particular V and W are g-invariant.

Proof. Clearly W ⊆ V and V and W are subspaces of Homk(V,W ).Let w1, . . . , wk be a basis of W and extend this to a basisw1, . . . , wk, v1, . . . , vj of V. Define T0 ∈ Homk(V,W ) by T0(wi) = wiand T0(vi) = 0. Then T0 ∈ V − W. If T ∈ V, then T |W = λIW .So for w ∈ W , T − λT0(w) = 0 and dimV/W = 1. For X ∈ g andT ∈ Homk(V,W ) we have

X(T ) = ρX |W T − T ρX .

If T ∈ V then since T |W = λIW , and W is invariant we have

X(T )(w) = ρX T (w) − T ρX(w)

= ρX(λw) − λρX(w) = 0.

This means X(T ) ∈ W, so g(V) ⊆ W.

Continuing the proof, we see that by the lemma and the codimension1 case there is some T0 ∈ V − W such that V = W + cT0c∈k as g-modules. Hence T0|W = λIW , where λ 6= 0. Normalizing, we mayassume T0|W = I. Since g is semisimple and the invariant subspace hascodimension 1, X T0 = 0 for all X ∈ g. Thus T0 is an intertwiningoperator and so KerT0 is an g-invariant subspace of V . If w ∈ W ∩KerT0, then T0(w) = w = 0 so W ∩ KerT0 = 0. Since T0 maps ontoW we have dimV = dimW + dim KerT0; it follows that KerT0 is thedesired complementary subspace.

Corollary 3.4.5. A semisimple subgroup of GL(n,C) is closed.

Proof. Let H = G, the closure of G in GL(n,C) = GL(V ) and h and g

the respective Lie algebras. Since G normalizes itself, H also normalizesG. Therefore g is an ideal in h. Since g is semisimple it is a direct factor,so h = g⊕ l. Let L be the corresponding normal analytic subgroup of H.Then H = GL, and L commutes with G (and therefore also with H);thus L ⊆ Z(H)0. By Weyl’s theorem, the action of G on V is completely


reducible, so V =∑r

i=1 Vi where each Vi is G invariant. Since G issemisimple each of these irreducible representations lies in SL(Vi) andsince L commutes with G each l ∈ L consists of diagonal operatorswhich are scalars on each Vi. Hence all eigenvalues of each l ∈ L areroots of unity with bounded order, namely the product

∏ri=1 dimVi.

This is less than or equal to nr ≤ nn, hence L is finite. Since it is alsoconnected L is trivial and so G = H.

Corollary 3.4.6. sl(n,C) and sl(n,R) are simple.

Proof. We first deal with the case of sl(n,C) = sl(V ). We know fromthe calculation of the Killing form, or as a corollary of Lie’s theorem,that sl(n,C) is semisimple and so is the direct sum of simple ideals aj . Ifa1, for example, acts irreducibly on V , then since aj commutes with a1

for each j ≥ 2, each aj consists of scalars by Schur’s lemma. But thesescalars are of trace 0. Hence aj = 0 for each j ≥ 2 and sl(V ) = a1 issimple. Therefore we may assume each aj acts reducibly on V . Choosean X 6= 0 ∈ sl(n,C) which is diagonal with distinct eigenvalues andwrite X =

∑Xj according to the decomposition of g given above.

Relabeling the indices we may assume X1 6= 0. Clearly [X,X1] = 0and hence X1 is also diagonal. Since a1 does not act irreducibly thereis a proper a1-invariant subspace W ⊆ V . Let W

′be a complementary

a1-invariant subspace as in Weyl’s theorem. Since X1 is diagonalizable,its restrictions to W and W

′are also diagonalizable. Choose a basis

of each so that both restrictions are diagonal. Together these form abasis v1, . . . , vn of V . If X1 had only one eigenvalue it is a scalarof trace 0, and X1 = 0, a contradiction. Let ai and aj, i 6= j bedistinct eigenvalues of X1 corresponding to eigenvectors vi and vj. Byinterchanging the roles of W and W

′, if necessary, we may assume at

least one of them, say vi lies in W . Let T be a linear transformationdefined by T (vi) = vj , and T (vk) = 0 for all other k. Then T ∈ sl(V )and [T,X1](vi) = TX1(vi) −X1T (vi) = (ai − aj)vj 6= 0. Since a1 is anideal and W is invariant, we have a contradiction if vj ∈W

′. The other

possibility is that all eigenvalues of X1 on W′have equal value, say a,

and the vj are in W . But then, a must differ from either ai, or aj , orboth, and we argue as above replacing aj by a.


Now since the complexification sl(n,R)C = sl(n,C) is simple, to seethat sl(n,R) itself is simple we show in general that if g is semisimpleand gC is simple then so is g. For let a be a nonzero ideal in g. ThenaC is a nonzero ideal in gC and so equals gC. Hence

dimR a = dimC aC = dimC gC = dimR g,

and so a = g.

Another corollary of Weyl’s complete reducibility theorem is White-head’s lemma, which is actually equivalent to Weyl’s theorem.

Lemma 3.4.7. (Whitehead’s lemma) Let ρ : g → gl(V ) be a representa-tion of the semisimple Lie algebra g on V and φ : g → V be a 1-cocycle,that is, a linear function satisfying

φ([X,Y ]) = ρX(φ(Y )) − ρY (φ(X)).

Then φ is a coboundary. That is, φ(X) = ρX(v0) for some v0 ∈ V .

Proof. Let U = V ⊕ (t) be a space of dimension 1 more than the dimen-sion of V and let σ be defined by σX(v, t) = (ρX(v) + tφ(X), 0). Sinceρ is a representation, one sees easily that σ is a representation on U ifand only if φ is a 1-cocycle with values in V . Evidently σX(U) ⊆ Vfor all X ∈ g, and in particular V is a σ-invariant subspace of U . Bycomplete reducibility there is a vector u0 = (v0, t0) ∈ U − V , witht0 6= 0 and σ(u0) = 0. Normalizing by taking t0 = −1, we see thatφ(X) = ρX(v0).

We remark that even if g were not semisimple, but merely reductive(i.e. g always acts completely reducibly in any finite dimensional linearrepresentation ρ over k) then again H1(g, V, ρ) = 0. The argument isthe same except now we get a vector u0 = (v0, t0) ∈ U − V with t0 6= 0,such that (ρX(v0) + t0φ(X), 0) = (λXv0, λX t0), where λ is a k valuedlinear functional on g. Since λXt0 = 0 for all X and t0 6= 0 it followsthat λ = 0, so u0 is killed by σ and we can proceed as above.


3.5 Levi-Malcev Decomposition

We now turn to the theorem of Levi-Malcev.

Theorem 3.5.1. Let π : g → s be a Lie algebra homomorphism onto asemisimple Lie algebra s. Then there exists a Lie algebra homomorphismǫ which gives a global cross section to π. That is to say, ǫ : s → g andπ ǫ = Is.

The uniqueness statement whose proof will be given below is due toMalcev. Before proving this theorem we give two of its corollaries.

Corollary 3.5.2. Let r be the radical of a Lie algebra g. Then thereexists a semisimple subalgebra s of g such that g = r

⊕s. In fact, g is

the semidirect sum of the ideal r with s. This is what is usually calledLevi’s splitting theorem. In addition, s is unique up to conjugation byExp(ad Y ) where Y ∈ n and n is the nilradical of r (actually Y ∈ [g, r]).Notice that Exp(adX) is well-defined over any field of characteristiczero and that it is an inner automorphism of g.

Proof. Let π : g → g/r be the projection mod r. Then g/r is semisimpleso there exists ǫ : g/r → g such that π ǫ = Ig/r. In particular ǫ isinjective and so ǫ(g/r) is a semisimple Lie subalgebra s of g isomorphicwith g/r and hence is of dimension dim g− dim r. Thus dim s + dim r =dim g. Now if X ∈ s ∩ r, then X = ǫ(Y ) for a unique Y ∈ g/r andtherefore π(X) = πǫ(Y ) = Y . But since π(X) = 0 we see that Y = 0and therefore X = ǫ(Y ) is also 0. Thus s ∩ r = 0, and it follows thatg = r

⊕s.

In general if g = a⊕

b, where a is an ideal and b is a subalgebrathen g is the semidirect sum of a with b. For if A and A′ ∈ a and Band B′ ∈ b then, [A+B,A′ + B′] = [A,A′] + [B,A′] + [A,B′] + [A,B′]and since a is an ideal, the first three terms are ∈ a. By uniqueness ofthe decomposition we therefore get

[(A,B), (A′, B′)] = ([A,A′] + adB(A′) − adB′(A), [B,B′]).

Since ad b is an algebra of derivations of a, this is a semidirect sum.

3.5 Levi-Malcev Decomposition 181

Corollary 3.5.3. Any finite dimensional Lie algebra g that is not sim-ple and not 1 dimensional is a semidirect sum of lower dimensional Liesubalgebras.

Proof. We may clearly assume that g is not semisimple for if it were wecould decompose it as a sum of simple ideals. Thus we may assume itsradical r 6= 0. If g is not solvable then by the Levi theorem g is thesemidirect sum of the ideal r with a Levi factor s. Thus we may assumeg is solvable. In particular, g 6= [g, g] and we can find an ideal a in g

of codimension 1. Thus g = a⊕kX0. Since kX0 is a subspace this

completes the proof.

We now turn to the proof of Theorem 3.5.1. It clearly suffices toshow that there exists a subalgebra t of g such that g = Kerπ

⊕t, as k-

spaces. For if this were so π would give rise to a Lie algebra isomorphismπ : t → s and we could than take ǫ = (π)−1. Then ǫ : s → g, andπ ǫ = Is.

Now let a = Kerπ and write s = g/a. We shall prove our resultby induction on dim a. Suppose there is a g-ideal a0 lying within a.Then by inductive hypothesis we can find a supplementary subalgebras0 = g0/a0 to a/a0 in g/a0 and also a supplementary subalgebra s

′to

a0 in g0. Then s′

is a supplementary subalgebra to a in g, so we mayassume that a itself is a g-simple ideal. Now since s is semisimple, a

must contain the radical r of g. If r = 0, then g would be semisimpleand then we would be done, since the ideal a would then be a directsummand. Otherwise, by irreducibility of a under g, we have r = a. Butthen by solvability [a, a] is a proper g-ideal in a and hence is 0. Thuswe may assume a is an abelian ideal (on which g and therefore also s

act irreducibly).

Lemma 3.5.4. Let ρ : g → gl(V ) be a representation of the Lie algebrag over k, and suppose that there is a vector v0 ∈ V such that the mapX 7→ ρ(X)v0 is a bijection of a with the orbit ρ(a)v0. Assume alsothat ρ(g)v0 = ρ(a)v0. Then Stabg(v0) = X ∈ g : ρ(X)v0 = 0 is asubalgebra of g and g = a

⊕Stabg(v0).


Proof. If ρ(X)v0 = 0 and ρ(Y )v0 = 0, then ρ([X,Y ])v0 = ρ(X)ρ(Y )v0−ρ(Y )ρ(X)v0 = 0 so Stabg(v0) is a subalgebra of g. By assumption,the map X 7→ ρ(X)v0 is a linear bijection of a with the orbit g(v0).On the other hand the orbit ρ(g)(v0) is isomorphic as a k-space withg/Stabg(v0). Thus dim g = dim a + dimStabg(v0). If X ∈ a andρ(X)v0 = 0, then X = 0 (by assumption), so a ∩ Stabg(v0) = 0.Hence g = a

⊕Stabg(v0).

Continuing the proof of Theorem 3.5.1, let V = Endk(g) and ρ be therepresentation of g on V defined by ρ(X)T = [adX,T ]. Define threesubspaces of V as follows: P = adg(a), Q = T ∈ Endk(g) : T (g) ⊆a, T |a = 0, and finally, R = T ∈ Endk(g) : T (g) ⊆ a, T |a = λIa.Then P ⊆ Q ⊆ R ⊆ V, and Q has codimension 1 in R. We firstshow that P, Q and R are sub g-modules of V. In fact, if X ∈ g andY ∈ a, then [adX, ad Y ] = ad [X,Y ] ∈ P, since a is an ideal, thus P isa submodule of V. If X ∈ g and T (g) ⊆ a, then for Y ∈ a, we have

[adX,T ](Y ) = adX T (Y ) − T adX(Y ) ∈ a

since a is an ideal. Finally, if Y ∈ a and T (a) is a homothety, thenT (Y ) = λ(Y ) so

adX T (Y ) − T adX(Y ) = λ[X,Y ] − λ[X,Y ] = 0.

This shows that g(R) ⊆ Q; in particular, both R and Q are submodulesof V. Since a is an ideal in g, P is also an g-submodule of V.

Taking quotients by P we get an exact sequence of g modules

0 → Q/P → R/P → R/Q → 0.

We show that a acts trivially on R/P, that is a(R) ⊆ P. Let Y ∈ a,T ∈ R and X ∈ g. Then adY T (X) = 0 since T ∈ R, Y ∈ a and a

is abelian. On the other hand since a is an ideal and adY (X) ∈ a, wehave T ad Y (X) = [λY,X]. Thus ρY (T ) = adλY ∈ P for each Y ∈ a.

Because a acts trivially on R/P these are actually all representationsof the semisimple algebra s, and this sequence splits by Weyl’s theorem.Because Q has codimension 1 in R there exists T0 ∈ R \ Q, such that


[ad g, T0] ∈ P, that is [ad g, T0] ∈ adg(a). This T0 is our “v0”. We caneven normalize T0 so that the homothety has λ = 1

We show that ρ(A)T0 = − adgA. That is for every X ∈ g,[adA,T0](X) = adA T0(X) − T0 adA(X). But since T0(X) ∈ a,adA T0(X) = [A,T0(X)] = 0. Hence by our normalization of T0,ρ(A)T0(X) = −T0([A,X]) = −[A,X] and ρ(A)T0 = − adg(A).

For A ∈ a the map A 7→ ρ(A)T0 is bijective (injective). Since thismap is linear in A this means that if ρ(A)T0 = − adg(A) = 0, thenA = 0. But this condition says that [X,A] = 0 for all X ∈ g. ThusA would be fixed under the original irreducible action, a contradictionunless A = 0. Finally, let X ∈ g. We must show that ρ(X)T0 = ρ(A)T0,for some A ∈ a i.e. ρ(X)T0 = − ad g(A) for that A. But ρ(X)T0 =[adX,T0] is in ad g(a). This completes the proof.

We now turn to the Malcev uniqueness part of the Levi theorem.For the convenience of the reader we recall the uniqueness statement.

Malcev uniqueness: If t is any semisimple subalgebra of a finite di-mensional Lie algebra g there is an inner automorphism α such thatα(t) ⊂ s. In fact α = Exp(adX) for some ad-nilpotent X ∈ rad(g). Inparticular, a Levi factor of g is unique up to inner automorphism of theform α = Exp(adX), for some ad nilpotent X ∈ rad(g).

Proof of uniqueness: By Levi splitting for X ∈ g can write X =r(X) + s(X), the unique r and s components of X. If Y = r(Y ) + s(Y )is another such element, then since r is an ideal,

r[X,Y ] = [r(X), r(Y )] + [r(X), s(Y )] + [s(X), r(Y )]

and s[X,Y ] = [s(X), s(Y )]. Now [r, r] is an ideal in g and hence also in[g, r]. Let π be the projection [g, r] → [g, r]/[r, r] and let φ = π ·r|t. Then

φ is a linear map and if ρX = ˜(adX|[g,r]), where˜means the inducedmap on [g, r]/[r, r], then ρ is a representation of g on [g, r]/[r, r]. Weconsider the restriction of the representation to t which we again call ρ.We will show

φ([H1,H2]) = ρH1(φ(H2)) − ρH2(φ(H1)),


that is, ρ is a 1 cocycle. One shows by direct calculation that for allH1, h2 ∈ t,

φ([H1,H2]) − ρH1(φ(H2)) + ρH2(φ(H1)) ∈ [r, r]. (3.8)

This means

r[H1,H2] − [H1, r(H2)] + [H2, r(H1)] ∈ [r, r].

But

r[H1,H2] = [r(H1), r(H2)] + [r(H1), s(H2)] − [r(H2), s(H1)],

while,[H1, r(H2)] = [r(H1), r(H2)] + [s(H1), r(H2)],

and[H2, r(H1)] = [r(H2), r(H1)] + [s(H2), r(H1)].

Hence (3.8) is just [r(H1), r(H2)] ∈ [r, r]. Since t is semisimple, byWhitehead’s lemma there is some v0 = X0 +[r, r] where X0 ∈ [g, r] suchthat ρ(H)v0 = [H,X0] + [r, r] = φ(H). But since φ(H) = r(H) + [r, r]we see that [H,X0] − r(H) ∈ [r, r] for all H ∈ t. In other words,H + [X0,H] = H + adX0(H) ∈ s(H) + [r, r]. On the other hand, [r, r]is an ideal in g and therefore normalized by s. So [r, r]

⊕s = g1 is a

subalgebra of g, containing [r, r] as a solvable ideal. Since s is semisimple,its radical is [r, r] and therefore s is a Levi factor.

We show Exp(adX0)(t) ⊆ g1. Now (adX0)2(H) = [X0, [X0,H]]

and since X0 ∈ [g, r] and this is an ideal, [X0,H] ∈ [g, r]. Hencealso [X0, [X0,H]] ∈ [[g, r], [g, r]] ⊆ [r, r], since [g, r]] ⊆ r. Thus(adX0)

2(H)/2! ∈ [r, r]. Similarly, (adX0)3(H) = [X0, [X0, [X0, t]]

and since [[g, r], [r, r]] ⊆ [[g, r], [g, r]] ⊆ [r, r], we see that by induction(adX0)

n(H)/n! ∈ [r, r] for all n ≥ 2. Since t + adX0(H) ∈ g1, it fol-lows that the automorphism Exp(adX0) also takes t into g1 and henceExp(adX0)(t) is a semisimple subalgebra of the latter. Since [r, r] 6= r,by solvability, we see by induction on the dimension of g that thereexists an X1 ∈ rad(g1), so that Exp(adX1) Exp(adX0)(t) ⊆ s. Thusif α = Exp(adX1) · Exp(adX0), then α(t) ⊆ s. Since X0 ∈ [g, r] and


X1 ∈ [r, r] ⊆ [g, r], once we know that the operators adX are nilpotentfor X ∈ [g, r] we would then argue as follows:

Exp(adX1) · Exp(adX0) = Exp(adX1 + adX0 +1

2[adX1, adX0] + . . .)

But this is Exp(adY ) where

Y = X1 +X0 +1

2[X1,X0] + . . . .

Since [g, r] is a subalgebra and we have a finite sum, Y ∈ [g, r]. Thatconcludes the uniqueness proof.

As a corollary of Malcev uniqueness we have:

Corollary 3.5.5. In a finite dimensional Lie algebra g, a Levi factoris a maximal semisimple subalgebra, and conversely. In particular, anysemisimple subalgebra is contained in the same Levi factor.

Proof. A Levi factor is a maximal semisimple subalgebra. For if it wereproperly contained in a larger semisimple subalgebra, the larger onewould have to intersect the radical nontrivially in a solvable ideal init, therefore violating semisimplicity. Conversely, if t were a maximalsemisimple subalgebra of g, then α(t) ⊂ s. So that t ⊂ α−1(s). Sincethe latter is also semisimple, by maximality t = α−1(s), must be a Levifactor.

Proposition 1.7.15 enables one to transform decompositions of theLie algebra usually gotten by linear algebra to the group. For example,in this way we get the Levi decomposition of a a connected Lie groupG with Lie algebra g. Let g = r ⊕ s be a Levi decomposition of g. LetR and S be the unique connected Lie subgroups of G corresponding tothe Lie subalgebras r and s (see Theorem 1.3.3). Then R∩S is discrete,since R is normal in G, and G = RS. Here R is the radical of G andS is a Levi factor, a maximal semi-simple connected subgroup. This isthe Levi decomposition of G. For each such global decomposition thereis a uniqueness statement concerning S. Notice R ∩ S is also central inS normal in S and S is connected.


We now make a few remarks about faithful representations of Liegroups. This works equally well in the real or complex cases and im-plies Ado’s theorem, which we mentioned in Section 1.7. Let G be aconnected Lie group and G = RS be a Levi decomposition. A theoremof Hochschild and Mostow [33, 34] states that if R and S each havefaithful representations, then G has a faithful representation and con-versely. Any semisimple group always has a locally isomorphic groupwith a faithful representation namely the adjoint group and the univer-sal covering group always has a faithful representation [33, 34]. Hence,G is locally isomorphic to a faithfully represented group. It follows,therefore, that any connected Lie group is locally isomorphic to a faith-fully represented Lie group. Taking the derivative and using Lie’s thirdtheorem (that Lie algebras of Lie groups comprise all Lie algebras), weget a proof of Ado’s theorem for Lie algebras.

To inject a note of reality into our brief discussion of faithful repre-sentations we now give two examples of classes of connected Lie groups,one nilpotent and one semisimple which have no faithful representations.For the general situation see [48] and [49].

Let G be any simply connected 2-step nilpotent group, G (for ex-ample, G could be Nn, the Heisenberg group of dimension (2n + 1).Since the center, Z(G), is nontrivial abelian and simply connected, letD be a discrete subgroup of Z(G) with K = Z(G)/D compact. Thenthe locally isomorphic group H = G/D has no faithful linear repre-sentation. We denote by π : G → H the canonical map. For sup-pose ρ : H → GL(n,C) were such a representation. Then ρ(K) is acompact and therefore completely reducible subgroup of GL(n,C). Onthe other hand, since D is discrete π(Z(G)) = Z(H), so K = Z(H).Since G is 2-step nilpotent and H has the same Lie algebra H is also2-step nilotent. Therefore [H,H] ⊆ Z(H) = K. By Lie’s theorem(see Theorem 3.2.13), ρ(H) is contained in the triangular matrices andhence ρ([H,H]) = [ρ(H), ρ(H)] acts by unipotent operators. But sinceρ([H,H]) also completely reducible ρ([H,H]) = I. Hence since ρ isfaithful [H,H] = 1, a contradicion because H is non abelian being 2-step nilpotent.

Now consider ˜SL(2,R), or more generally G = ˜Sp(n,R). Since here


K = U(n,C) is a maximal compact subgroup ofG and π1(K) = π1(G) =Z, we see that the center of G is infinite (its actually Z). Since ρ isfaithful ρ(G) also has infinite center. But by Theorem 7.5.17 a linearsemisimple group must have a finite center, a contradiction.

Evidently this last example works whenever G a non-compactsemisimple group and the maximal compact subgroups are not semisim-ple (but are merely reductive).

The following is a useful result.

Theorem 3.5.6. IF g is a Lie algebra and r is its radical then [g, g]∩r =[g, r]. Moreover, if ρ : g → V is a representation of g then [g, r] acts onV by nilpotent operators.

Proof. By the Levi decomposition, g = r + s where s is a semisimplesubalgebra of g and r ∩ s = 0 (s being a Levi factor). Since r is anideal and [s, s] = s,

[g, g] = [r, r] + [r, s] + [s, s] ⊆ [r, g] + s.

This means that [g, g] = [r, g] + s and hence that [g, g] ∩ r = [g, r].Weknow from Lie’s theorem that [r, r] acts on V by nilpotent operators.If m is a maximal subspace of [g, r] which acts nilpotently on V . Wewill show that m = [g, r]. Suppose this is not so and m is a propersubspace. Then there exists an X ∈ g and R ∈ r so that [X,R] does notact nilpotently on V . The subalgebra generated by X and r consistingof cX +R : c ∈ k,R ∈ r is clearly solvable since it contains a solvableideal of codimension ≤ 1. Again by Lie’s theorem, its derived alsoacts nilpotently on V . In particular, [X,R] acts nilpotently on V , acontradiction.

Corollary 3.5.7. The subalgebra [g, r] is in the nilradical n of g. Inparticular, if r is solvable then r/n is abelian. In general, the radical ofg/n is abelian.

Proof. Taking ρ to be the adjoint representation, we see by the presentresult together with Engel’s theorem that ad[g, r] = [ad g, ad r] acts as anilpotent Lie algebra. Hence [g, r] is nilpotent ideal. Therefore [g, r] ⊆


n, the nilradical of g. Taking g to be solvable we obtain the secondstatement. Finally, if g = r ⊕ s is the Levi decomposition of g, dividingby n ⊆ r gives g/n = r/n⊕ s, a Levi decomposition of g/n. Hence r/n isits radical which is abelian.

3.6 Reductive Lie Algebras

Definition 3.6.1. We say a Lie algebra g is reductive if it is non-abelianand the adjoint representation is completely reducible.

Lemma 3.6.2. If ρ : g → gl(V ) is a representation of a semisimpleLie algebra then ρ(g) ⊂ sl(V ). In particular g acts trivially on any1-dimensional space.

Proof. g = [g, g] so ρ(g) = ρ[g, g] = [ρ(g), ρ(g)] ⊂ [gl(V ), gl(V )] ⊂sl(V ).

Proposition 3.6.3. The following conditions are equivalent.

(1) g is reductive.

(2) [g, g] is semisimple.

(3) g = z(g) ⊕ [g, g] where [g, g] is semisimple.

(4) z(g) = rad(g).

Proof. If the adjoint representation is completely reducible then, asabove (even if g is not semisimple) g =

⊕gi, the direct sum of sim-

ple ideals, except that now some of them may be 1-dimensional abelian.Thus g is the direct sum of an abelian and a semisimple algebra h. Thismeans that [g, g] = [h, h] = h. If [g, g] is semisimple then g = [g, g] ⊕ h

for some ideal h. Since g/[g, g] = h and the former is abelian thisproves (iii) because z([g, g]) = 0. If g = z(g) ⊕ [g, g] where [g, g] issemisimple then rad(g) = rad(z(g)) ⊕ rad([g, g]) = z(g). Finally, sup-pose z(g) = rad(g). Then ad induces a map g/z = g/ rad(g) → ad g.The algebra g/r is semisimple, hence is ad g. By Weyl’s theorem ad iscompletely reducible.

An example of a reductive Lie algebra is a Lie algebra of compacttype. Another example is provided by

3.6 Reductive Lie Algebras 189

Corollary 3.6.4. If g is a Lie algebra and n is its nil radical then g/nis reductive.

Proof. Let r = rad(g) then r/n is an ideal in g/n and the quotient, g/r issemisimple. By Corollary 3.2.19, [r, r] ⊂ n so r/n is abelian. By the Levitheorem g/n = r/n ⊕ g/r the semi-direct sum of an abelian ideal anda semisimple algebra. To see that this is a direct sum, i.e. that r/n iscentral in g/n, we must show that [g/n, r/n] = 0 i.e. that [g, r] ⊂ n.Thisis so because [g, r] is a nilpotent ideal.

Definition 3.6.5. Let g be a subalgebra of gl(V ) where V is a finitedimensional k-vector space and k is a field of characteristic 0. We sayg is to be linearly reductive if g acts completely reducibly V .

We now study linearly reductive subalgebras of gl(V ).

Proposition 3.6.6. A representation ρ : g → gl(V ) of a Lie algebra g

is completely reducible if and only if V is the direct sum of nontrivialg-invariant subspaces Wi of V on each of which g acts irreducibly.

Proof. The restriction of a completely reducible representation to aninvariant subspace W is still a completely reducible representation. Forif U is a g-invariant subspace of W , then U is a g-invariant subspaceof V . If U ′ is a complementary g-invariant subspace of V then U ′ ∩Wis a g-invariant subspace of W which complements U in W . Henceby induction on the finite dimension of V , if ρ is completely reduciblethen V is the direct sum of such subspaces Wi of V on which g actsirreducibly. Conversely, suppose V is the direct sum of nontrivial g-invariant irreducible subspaces Vi. Let W be a g-invariant subspaceof V . We argue by induction on the codimension of such W . SinceVi ∩W is a g-invariant subspace and Vi is irreducible we have eitherVi ∩W = Vi or Vi ∩W = 0 for each i. If Vi ∩W = Vi for all i thenW = V . Otherwise choose i so that Vi ∩W = 0 and let U = Vi +W .Then U is a g-invariant subspace of V of codimension less than thatof W . Hence there is a complementary g-invariant subspace U ′ of V ,which means that U ′ + Vi complements W .


Proposition 3.6.7. If a representation ρ : g → gl(V ) of a Lie algebrag, and if V is the direct sum of nontrivial g-invariant subspaces Wi

of V on which g acts irreducibly, then this decomposition is unique upto equivalence. That is if ρ =

∑n1 niρi where ni and q are integers

and the ρi’s are irreducible, the equivalence of classes of ρi’s and theirmultiplicities ni’s are uniquely determined by ρ.

Proof. Apply the Jordan-Holder theorem.

We now deal with some questions about extension of the base field.If ρ : g → gl(V ) is a representation over R of a real Lie algebra let ρC, gC,and V C denote the respective complexifications. Then gC is a complexLie algebra, V C is a complex vector space and ρC : gC → gl(V C) is arepresentation overC where ρC is defined by

ρC(X + iY )(v + iw) = ρX(v) − ρY (w) + i(ρX(w) + ρY (v)). (3.9)

for X + iY ∈ gC, v + iw ∈ V C.In particular if ρ = ad then ρC(X + iY )(X ′ + iY ′) = ρX(X ′) −

ρY (Y ′) + i(ρX(Y ′) + ρY (X ′)). The latter clearly equals [X + iY,X ′ +iY ′] = adX + iY (X ′ + iY ′). We now show that the trace form βρC :gC × gC → C of ρC is given by

βρC(X + iY,X ′ + iY ′) = trV C(ρC(X + iY )ρC(X ′ + iY ′)).

The latter term equals

trV (ρ(X)ρ(X ′) − ρ(Y )ρ(Y ′)) + i(trV (ρ(X)ρ(Y ′) + trV (ρ(Y )ρ(X ′)).

In particular if Y and Y ′ = 0, we have βρC(X,X ′) = βρ(X,X′). Now

βρ is nondegenerate if and only if there exists a R-basis X1, ...,Xnof g (also a C-basis for gC) such that detβρ(Xi,Xj) 6= 0. But thendetβρC(Xi,Xj) is also 6= 0. We have proved that if βρ is nondegenerateso is βρC. Applying this to the adjoint representation of g we see thatif g semisimple Lie algebra then gC is also semisimple.

Conversely, suppose gC is semisimple and h is an abelian ideal in g.From the identity [h, k]C = [hC, kC] we see that hC is an abelian ideal ingC and hence is trivial. But h ⊂ hC. Thus g is also semisimple. Wehave proved:

3.6 Reductive Lie Algebras 191

Proposition 3.6.8. A real Lie algebra g semisimple if and only if gC

is also semisimple.

Lemma 3.6.9. (1) Any C-subspace S of V C is of the form S = WC

for some R-subspace W of V .

(2) If W is a subspace of V then W = WC ∩ V .

(3) If W is a subspace of V then W is g-invariant if and only if WC

is gC-invariant.

Proof. (i) is clear. Let w1, ..., wj be a basis of W and extend it toa basis w1, ..., wj , v1, ..., vk of V over R (also a basis of WC over C).Let

∑i ciwi ∈WC (where ci ∈ C). If this also lies in V then

∑i ciwi =∑

i aiwi +∑

j bjvj. Hence∑

i(ci − ai)wi +∑

j bjvj = 0. It follows that

ci = ai and b1 = 0, and in particular∑

i ciwi ∈W . Since W ⊂WC ∩ Vthis proves (ii).

As for (iii), if W is g-invariant then for w + iw′ ∈ WC we knowρX(w) − ρY (w′) and ρX(w′) + ρY (w) ∈ W . By (3.9) ρC(X + iY )(w +iw′) ∈ WC. Conversely, if ρC(X + iY )(w + iw′) ∈ WC for all X, Y , wand w′ then by (3.9) ρX(w) ∈WC ∩ V = W for all X and w.

Corollary 3.6.10. A linear Lie algebra g ⊂ gl(V ) is completely re-ducible if and only if gC ⊂ gl(V C) is also completely reducible.

Proof. If gC is completely reducible and W is a g-invariant subspaceof V then WC is gC-invariant. Hence V C = WC ⊕ S where S is gC-invariant. But S = UC for some U which is g-invariant by (ii). FromV C = WC⊕UC it follows that V = W⊕U . Now suppose g is completelyreducible and let S be a gC-invariant subspace of V C. Then S = WC

for some W ⊂ V and W is g-invariant. Hence W has a complementaryg-invariant subspace U and V = W ⊕ U . But then UC is gC-invariantand V C = WC ⊕ UC.

Theorem 3.6.11. A linear Lie algebra g is linearly reductive if andonly if

(1) g is reductive and

(2) the elements of z(g) are simultaneously diagonalizable.


Proof. Using the remarks above we may assume the field k is alge-braically closed. Let r = rad(g). By Lie’s theorem there exists asemi-invariant χ ∈ r∗ with nonzero semi-invariant vector w. ThenVχ = v ∈ V : Xv = χ(X)v for all X ∈ r is a nonzero subspace ofV . By Lemma 3.2.15, which was used in the proof of Lie’s theorem,χ([g, r]) = 0. If v ∈ Vχ, Y ∈ r and X ∈ g then Y Xv = XY v+ [Y,X]v =X(χ(Y )v) + 0v = χ(Y )X(v) so that Vχ is a g-invariant subspace of V .By complete reducibility (and the fact that a submodule of a completelyreducible module is itself completely reducible) there is a finite set ofχ such that V =

⊕Vχ and each Vχ is g-invariant. Choose a basis in

each Vχ and put these together to get a basis of V . On each Vχ, r actsby Y → χ(Y )I so on V , r acts diagonally and in particular z acts di-agonally, proving (ii). Moreover for v ∈ Vχ, Y ∈ r and X ∈ g we haveXY v = X(χ(Y )v) = χ(Y )(X(v)), while Y X(v) = χ(Y )(X(v)) becauseVχ is X-stable. Thus Y X = XY on each Vχ, [X,Y ] acts trivially V andr ⊂ z. Hence r = z, proving (i) by Proposition 3.6.3. Conversely suppose(i) and (ii) hold. By (ii) if Z ∈ z(g) we have

Z =

χ1(Z) 0 0 0 00 χ2(Z) 0 0 0... ... ... ... ...... ... ... ... ...0 0 0 0 χn(Z)

, (3.10)

where χi ∈ z(g)∗. Here we use the well known fact from linear algebrathat a commuting family of diagonalizable operators can be simultane-ously diagonalized. Let Vχi

= v ∈ V : Zv = χi(Z)v for all Z ∈ z(g).Then each Vχi

is g-stable. For if X ∈ g and Zv = χi(Z)v for all Z ∈ z(g)then ZXv = XZv = X(χi(Z)v) = χi(Z)Xv. Since this holds for allZ ∈ z(g) we conclude that Xv ∈ Vχi

. To prove that g acts completelyreducibly on V we may assume V = Vχi

. Then z(g) acts by scalars so by(i) g-submodules are the same as [g, g]-submodules. By Weyl’s theorem[g, g] acts completely reducibly.

3.7 The Jacobson-Morozov Theorem 193

3.7 The Jacobson-Morozov Theorem

We recall (see Example 3.1.3) that if X+,H,X− are the usual gen-erators of sl(2), then they form a basis for sl(2) and [H,X+] = 2X+,[H,X−] = −2X− and H = [X+,X−].

Definition 3.7.1. Let g be a Lie algebra containing linearly indepen-dent elements, X+, H and X+. We shall say X+,H,X− is an sl(2)triple if they satisfy the sl(2) relations.

Our objective here is to prove the following result which provides usanother criterion (see Theorem 3.6.11) for complete reducibility of a Liealgebra of operators in characteristic 0.

Theorem 3.7.2. Let g be a completely reducible Lie subalgebra of gl(V )and let N 6= 0 ∈ g. If adN is nilpotent then N can be imbedded in ansl(2) triple, in a way that N = X+. We shall call this condition J-M.Moreover, g contains the nilpotent and semisimple parts of each of itselements.

Conversely, if J-M holds and g contains the nilpotent and semisimpleparts of each of its elements, then g is completely reducible.

In particular, if J-M holds and g has a trivial center, then g is com-pletely reducible. Of course if g is semisimple, we already know thisby Weyl’s theorem, which suggests that Weyl’s theorem will have toplay a role in our proof. Moreover, by Weyl’s theorem, if g is a linearsemisimple algebra then it contains the nilpotent and semisimple partsof each of its elements.

We begin with a result known as Morozov’s lemma.

Lemma 3.7.3. Let g be a Lie algebra containing elements X+ and Hsuch that [H,X+] = 2X+ and H = [Z,X+], for some Z ∈ g. Then thereexists an X− ∈ g such that X+,H,X− form an sl(2)-triple.

Proof. Let X ′+ = adX+, H ′ = adH and Z ′ = adZ. Since ad is a rep-

resentation we know [H ′,X ′+] = 2X ′

+ and H ′ = [Z ′,X ′+]. From the first

of these relations we see by Lemma 3.7.6 that X ′+ is a nilpotent opera-

tor on g. Moreover, [[Z,H] − 2Z,X+] = [[Z,H],X+] − 2[Z,X+] which,


by the Jacobi identity, equals −[[X+, Z],H]− [[H,X+], Z]− 2[Z,X+] =[H,H] − 2[X+, Z] + 2[X+, Z] = 0. In other words, [Z,H] − 2Z is inzg(X+), the centralizer of X+. We can therefore write [Z,H] = 2Z+C,where C ∈ zg(X+).

Now let U ∈ zg(X+). Because [H ′,X ′+] = 2X ′

+ and X ′+(U) = 0,

X ′+H

′(U) = [X ′+,H

′](U) +H ′X ′+(U)

= −2X ′+(U) +H ′X ′

+(U) = 0,

and hence H ′(U) ∈ zg(X+). Thus H ′ leaves zg(X+) stable. Moreoverfor a positive integers i, we have

[Z ′,X ′i+] = X ′i−1

+ [Z ′,X ′+] +X ′i−2

+ [Z ′,X ′+]X ′

+ + · · · + [Z ′,X ′+]X ′i−1

+

= X ′i−1+ H ′ +X ′i−2

+ H ′X ′+ + · · · +H ′X ′i−1

+

By an easy induction, we see that for every positive integer i, H ′X ′i+ =

X ′k+H

′ + 2iX ′i+,

[Z ′,X ′i+] = (H ′ − 2(i − 1)I +H ′ − 2(i− 2)I + · · · +H ′)X ′i−1

+

= i(H ′ − (i− 1)I)X ′i−1+ .

Now suppose U ∈ zg(X+) ∩X ′i−1+ (g). Then U = X ′i−1

+ (V ) for some V

and so X ′+(U) = X ′i

+(V ) = 0. But then,

i(H ′ − (i− 1)I)X ′i−1+ (V ) = Z ′X ′i

+(V ) −X ′i+Z

′(V )

= −X ′i+Z

′(V ) ∈ X ′i+(g),

so that (H ′ − (i − 1)I)(U) ∈ X ′i+(g). This means that for every i,

H ′ − (i − 1)I sends zg(X+) ∩ X ′i−1+ (g) to X ′i

+(g). But since X ′+ is

nilpotent it follows that there is some m for which (H ′ −mI) · · · (H ′ −2I)(H ′ − I)H ′(U) = 0 for all U ∈ zg(X+).

Consider the restriction of H ′ to zg(X+) and put it in upper triangu-lar form over the algebraic closure of our field. From this and the aboveequation, it follows that each eigenvalue of this restriction is a nonpos-itive integer, so the restriction of H ′ + 2I to zg(X+) is invertible. Since


[Z,H] = 2Z+C where C ∈ zg(X+), and since the restriction of H+2I tozg(X+) is onto, there must be a Y ∈ zg(X+) for which (H ′+2I)(Y ) = C.But then, [H,Y ] = −2Y + C. Letting X− = −(Y + Z), we see that

[H,X−] = −[H,Y + Z] = −[H,Y ] − [H,Z]

= −(−2Y + C) + (2Z + C) = 2(Y + Z) = −2X−.

Since Y ∈ zg(X+), we also get

[X+,X−] = [X+,−(Y + Z)] = [X+,−Z] = [Z,X+] = H.

Before turning to the J-M theorem itself we need the followinglemma.

Lemma 3.7.4. Suppose g is a Lie subalgebra of gl(V ) with the propertythat every nonzero nilpotent element can be imbedded in an sl(2) triple.Let h be a subalgebra of g satisfying

(1) g = h ⊕ l, where l is a subspace of g.

(2) [h, l] ⊆ l.

Then h also has the property that every nonzero nilpotent element canbe imbedded in an sl(2) triple lying in h.

Proof. Suppose F is a nonzero nilpotent operator in h. Choose Eand H ∈ g so that E,H,F form an sl(2) triple in g. Using thedecomposition above write H = Hh + Hl and E = Eh + El. Then−2F = [F,H] = [F,Hh] + [F,Hl], where [F,Hh] ∈ h and [F,Hl] ∈ l.Since we have a direct sum decomposition, −2F = [F,Hh]. Also,H = [E,F ] = [Eh, F ] + [El, F ] where the components belong to h and l,respectively, hence Hh = [Eh, F ]. Thus by Morozov’s lemma applied toHh and F , both of which are in h, we get an E− ∈ h so that E−,Hh, Fsatisfy the relations of an sl(2) triple. The subalgebra of h generatedby these elements is a homomorphic image of sl(2) and is therefore asimple Lie algebra of dimension ≤ 3. Therefore, it is either trivial orisomorphic to sl(2) since Lie algebras of dimension 1 or 2 are alwayssolvable. But since F 6= 0 this means it is isomorphic to sl(2), that is,E−,Hh, F are linearly independent.


To prove the J-M theorem we need following lemmas.

Lemma 3.7.5. Let T ∈ Endk(V ) where V is a k-vector space of di-mension n. If tr(T j) = 0 for all j = 1, ..., n then T is nilpotent.

Proof. We may clearly assume k is algebraically closed. Hence T istriangular with diagonal entries α1, ..., αn. This means that for all j, T j

is triangular with diagonal entries αj1, ..., αjn. Thus our hypothesis says

αj1 + ... + αjn = 0 for j = 1, ..., n and we must show that each αi = 0.If we knew one of the αi say αn = 0 then we would have (by throwingaway the last equation) a system of n−1 equations, which by inductionwould have only the trivial solution. This would complete the proof.

Now let χT (x) = xn− tr(T )xn−1 + ...+ det(T ) be the characteristicpolynomial of T . By the Cayley-Hamilton theorem T n − tr(T )T n−1 +...+det(T ) = 0. Taking traces and using our hypothesis we get det(T ) =0. Thus one of the αi = 0.

Lemma 3.7.6. Let X ∈ gl(V ) and assume X =∑i[Xi, Yi] where

[X,Xi] = 0 for all i. Then X is nilpotent.

Proof. We first show that Xj =∑

i[Xi,Xj−1Yi] for j ≥ 1. Now Xj =∑

iXj−1(XiYi−YiXi) =

∑iX

j−1XiYi−Xj−1YiXi. Since X commuteswith all the Xi so does any power Xj . Hence this last term equals∑

iXiXj−1Yi−Xj−1YiXi =

∑i[Xi,X

j−1Yi]. Since tr is linear and takesthe value 0 on a commutator it follows that tr(Xj) = 0 for all j. ByLemma 3.7.5, X is nilpotent.

The proof the of J-M theorem, is

Proof. Let g be a completely reducible Lie subalgebra of gl(V ) andF 6= 0 be a nilpotent element in it. Let V =

⊕Vi be the decomposi-

tion of into Jordan blocks relative to F , so in each Vi we have a basis,v0, . . . vri such that Fvi = vi+1, when i < ri and Fvri = 0. We defineH and E to be the linear transformations on V which leave each Vi-invariant and on Vi we define for i = 0, . . . ri, Hvi = (ri−2i)vi, E(v0) = 0and for i > 0, E(vi) = (−iri + i(i − 1))vi+1. Then, [E,H] = 2E,


[F,H] = −2F , [E,F ] = H and E,H,F are linearly independent.Therefore they form a subalgebra of g isomorphic to sl(2) containing F .

Next let X0 ∈ g and X0 = N + S be its Jordan decomposition ingl(V ). We denote by ad the adjoint representation of gl(V ) on itself.Then adX0 = adN + adS and moreover adN is nilpotent and adSis semisimple (see Lemma 3.3.9) and they commute since N and Scommute. Thus by uniqueness of the additive Jordan decompositionadX0 = adN +adS is the Jordan decomposition of adX0. This meansthey are polynomials in adX0 without constant term. Since adX0 leavesg stable, the same is true of adN and adS. Hence the maps g → g givenby X 7→ [X,N ] and X 7→ [X,S], are both derivations of g. Since g actscompletely reducibly on V we know by Proposition 3.6.3 that g = s⊕ z,where s is a Levi factor and z is the center. But the derivations of asemisimple Lie algebra are all inner by Corollary 3.3.23, and hence anyderivation of g which maps z to 0 is also an inner derivation determinedby an element of s. Because [X0, Z] = 0 for all Z ∈ z and N is apolynomial without constant term in X0 it follows that [N,Z] = 0 forall Z ∈ z, which means the derivation X 7→ [X,N ] maps z to zero. Thusthe derivation X 7→ [X,N ] of g is inner and determined by an element,say N1 in s. Thus [X,N ] = [X,N1] for all X ∈ g, or alternatively,adN coincides with adN1 as operators on g. But adN is nilpotent asan operator on gl(V ) and therefore is also nilpotent on g. Hence sois adN1. Therefore the result proved just above applied to ad s showsthat there is some X0 ∈ s so that [adN1, adX0] = 2 adN1 (on s). Buts is semisimple and therefore centerless this means [N1,X] = 2N1. ByLemma 3.7.6, N1 is a nilpotent operator on V . Since [X0, N ] = 0 weknow [X0, N1] = 0 and hence [N1, N ] = 0. Because both N and N1 arenilpotent operators N −N1 is also nilpotent (see Lemma 3.2.2). But wealso showed [X,N ] = [X,N1] for all X ∈ g. Therefore [X,N −N1] = 0for all X ∈ g. Now consider the associative algebra g∗ generated by g

in gl(V ). It contains N as a polynomial in X0. It also contains N1 ∈ s,so N − N1 ∈ g∗. On the other hand g∗ ⊇ g so it also acts completelyreducibly. It follows that N − N1 is diagonalizable. Since it is alsonilpotent it must be zero and so N = N1. Therefore N ∈ g. SinceS = X0 −N it is also in g.


Conversely, suppose g ⊆ gl(V ) satisfies the J-M condition and z

contains the nilpotent and semisimple parts of each of its elements. Wewill show g acts completely reducibly on V . Let r be the radical of g. IfF ∈ [g, r], then by Theorem 3.5.6, F is nilpotent. If F were non zero, itcould be imbedded in a 3-dimensional simple subalgebra s of g. Hences ∩ r 6= 0. But s is simple so s ∩ r = 0. Thus F = 0 and therefore[g, r] = 0. This means r = z, and therefore by Proposition3.6.3 g isreductive, and acts completely reducibly on V .

3.8 Low Dimensional Lie Algebras

over R and C

In this section we indicate a classification of Lie algebras over R and C,up to dimension 3. Having done dimensions 1 and 2 already we now dealwith dimension 3. The classification is done by dimension of the derivedsubalgebra. We consider four subcases corresponding to dimension of[g, g] equal to 0, 1, 2 and 3.

(a) dimk[g, g] = 0. Here g is abelian so in the complex case g = C3

and in the real case g = R3.

(b) dimk[g, g] = 1. In the complex case g = h(C)⊕C, where h(C) is theax + b-Lie algebra over C, or g = n1(C), the complex Heisenberg(see Example 3.1.20). In the real case, g = h(R) ⊕ R, where h(R)is the real ax+ b-Lie algebra, or g = n1(R) the real Heisenberg.

Proof. There are two possibilities depending on whether [g, g] ⊆z(g) or not. In the former case let Z 6= 0 ∈ [g, g]∩ z(g) and extendthis to a basis X,Y,Z of g. Then [X,Y ] = λZ. Now λ 6= 0 for ifit were otherwise [X,Y ] = 0 and since everything commutes withZ, g would be abelian, whereas here we are assuming [g, g] hasexactly dimension 1. Since λ 6= 0 we can absorb it into X or Yand then [X,Y ] = Z and all other brackets are zero. This is thereal or complex Heisenberg Lie algebra.On the other hand suppose [g, g] has dimension 1, but is not con-tained in the center. Let X 6= 0 ∈ [g, g], X not in z(g). Then there

3.8 Low Dimensional Lie Algebras over R and C 199

is some Y ∈ g with [X,Y ] 6= 0 and so [X,Y ] = λX. Since λ 6= 0we can absorb it into y and get [X,Y ] = X, so X,Y generate theax+b-Lie algebra, h. Since h ⊃ [g, g] it is an ideal (see Proposition3.1.13). By Propositions 3.1.46 and 3.1.47 h is a direct summandg = h ⊕ a, where a is abelian since it has dimension 1.

For both cases there are only finitely many non isomorphic Liealgebras and they are all solvable (both real and complex).

(c) dimk[g, g] = 2 we shall see that we get two continuous families forthe complex field and three continuous families over the real field.

Proof. Suppose dimk[g, g] = 2. Then [g, g] cannot be the ax + b-algebra since it would then be a direct summand by Corollary3.3.22. But if g = h⊕a where h = [g, g], then since a is abelian wehave h = [g, g] = [h, h] and this is a contradiction. Thus [g, g] isabelian. Choose a basis X,Y for [g, g] and extend this to a basisX,Y,U of g. Then [U,X] = aX + bY , [U, Y ] = cX + dY and[X,Y ] = 0, where a, b, c, d ∈ k. This gives a matrix

A =

(a cb d

)

which determines the Lie algebra. Since [X,Y ] = 0, [g, g] is gener-ated by [U,X] = adU(X) and [U, Y ] = adU(Y ). Hence adu|[g,g]

is one-to-one. This means that A is nonsingular and we may takeany nonsingular A because the skew symmetry and Jacobi identityare automatically satisfied with these structure constants. Thuswe get many different Lie algebras in this way. Now the questionis exactly which of these are non isomorphic? Evidently, we canchange the basis X,Y of [g, g] and also change the U . The firstresults in changing A to a conjugate PAP−1 by P ∈ GL(2, k).We can also change the U to λU +W , where w ∈ [g, g]. But then[λU+W,X] = λ[U,X]+[W,X] = λ[U,X], since [W,X] = 0. Simi-larly, [λU+W,Y ] = λ[U, Y ]. Thus the effect of this is to change Ato λA. Thus the invariants are those of A 7→ λPAP−1. If k = Cwe can choose


A =

(1 00 α

),

α 6= 0 ∈ C, or

A =

(1 β0 1

),

β 6= 0 ∈ C, denoting these Lie algebras g3,α and gβ,3.Whe k = R we get

A =

(1 00 α

),

α 6= 0 ∈ R, or

A =

(α β−β α

)

where both α and β are real and β 6= 0,or

A =

(1 β0 1

)

β 6= 0 ∈ R These Lie algebras are denoted gC3,α, gC

α,β and gCβ,3

(d) dimk[g, g] = 3. Here g is simple and g = sl(2,C) in the complexcase, while in the real case g = sl(2,R) or g = so(3,R).

The proof of this requires two lemmas.

Lemma 3.8.1. If g is a 3-dimensional Lie algebra over any field k andg = [g, g], then g is simple.

Proof. Let a be a nontrivial abelian ideal in g then dim a = 1, 2 or 3.In the latter case g is abelian and so solvable. If the dimension of a

is 1 or 2, then dim g/a is 2 or 1. Hence g/a is solvable (see Example3.1.38), therefore g is also solvable by Proposition 3.1.33. Thus whena is nontrivial g is solvable so [g, g] 6= g, a contradiction. This shows g

is semisimple. Now the same argument as in the proof of Proposition3.1.55 shows that g is simple.

3.8 Low Dimensional Lie Algebras over R and C 201

Lemma 3.8.2. Let g be a simple Lie algebra of dimension 3. If k = Cthen g = sl(2,C) and if k = R then g = sl(2,R) or so(3,R).

Proof. We first consider the complex case. let X,Y,Z be a basis for g.It is easy to see that we can choose this basis such that [X,Y ] = Z.Then [Z, Y ] = aX + bY + cZ for some complex numbers a, b, c. Notethat a 6= 0 otherwise the vector space generated by Y and Z would bea nontrivial ideal of g contradicting the fact that g is simple.

Let d =√a then [Y/d,Z/d] = X + a/d.Y/d + c/d.Z/d, therefore

the new basis X ′ = X + a/d.Y/d, Y ′ = Y/d and X ′ = Z/d satisfies therelations

[X ′, Y ′] = Z ′

[Y ′, Z ′] = X ′ + tZ ′ (3.11)

where t ∈ C. If t = 0 then it immediately follows from the Jacobiidentity that g has a basis X ′′, Y ′′, Z ′′ in which the Lie bracket is definedby

[X ′′, Y ′′] = Z ′′

[Y ′′, Z ′′] = X ′′

[Z ′′,X ′′] = Y ′′,

therefore by Exercise 3.1.4 g is sl(2,C). If t 6= 0 then it follows from(3.11),

−[[Z ′,X ′], Y ′] = t[Z ′,X ′],

therefore by Lemma 3.7.6 [Z ′,X ′] is nilpotent. Note that [Z ′′,X ′′] 6= 0as [g, g] = g. Now that there is a nilpotent element in g, by Jacobson-Morozov there must be a sl(2) triple within g. By dimension we seethat g is sl(2,C). This takes care of the complex case.

Turning to the real case, consider the Killing form β of g. β isnondegenerate. If it is indefinite then its signature is either (2, 1) or(1, 2). These are really the same because one is the negative of theother, so we shall consider only the former. Now in any case since g issemisimple it is a linear Lie algebra because the adjoint representationis faithful; also by semisimplicity tr(g) = 0. If β is of type (2, 1) then


g (or rather its adjoint algebra) preserves the Killing form, so we haveg ⊆ o(2, 1), and then because the trace is identically zero g ⊆ so(2, 1).As both these Lie algebras are of dimension 3 they are equal: g =so(2, 1) (see Exercise 3.1.5). The proof of this part can be completed byobserving that so(2, 1) = sl(2,R). On the other hand, if β is negativedefinite then −β is positive definite and g preserves these forms similarly,we have g ⊆ so(3,R) and by dimension counting g = so(3,R).

dim[g, g] C Ra 0 C3 R3

b 1 h(C) ⊕ C, n1(C) h(R) ⊕ R, n1(R)

c 2 g3,α, gβ,3 gC3,α, gC

α,β, gCβ,3

d 3 sl(2,C) sl(2,R), so(3,R)

Figure 3.1: The 3-dimensional Lie algebras over R and C

Corollary 3.8.3. A 3-dimensional simple Lie algebra over R or C musthave rank one (Theorem 6.7.1 for the definition of rank).

3.9 Real Lie Algebras of Compact Type

In this section we will find out which real lie algebras are the Lie algebrasof compact real Lie groups.

Proposition 3.9.1. Let g be a real Lie algebra and ρ any faithful (real)representation of g on V such that the matrices ρ(X) are skew sym-metric with respect to some positive definite symmetric form 〈·, ·〉 on V .Then the trace form, βρ, is negative definite on g.

Proof. Relative to some orthonormal basis v1, . . . , vn of V we haveρij(X) = −ρji(X) where (ρij(X)) is the matrix of ρ(X) with respect tov1, . . . , vn. Hence for each i, ρ(X)(vi) =

∑j ρji(X)vj and so

ρ(X)2(vi) =∑

j

ρji(X)ρ(X)(vj) =∑

j,k

ρji(X)ρkj(X)vk.

3.9 Real Lie Algebras of Compact Type 203

Thustr(ρ(X)2) =

∑

i,j

ρij(X)ρji(X) = −∑

i,j

ρij(X)2 ≤ 0.

Since ρ is faithful, βρ is negative definite.

We recall the definition of an invariant form β on a Lie algebra g:for all X, Y and Z ∈ g,

β(adX(Y ), Z) + β(X, ad Y (Z)) = 0.

Definition 3.9.2. A real Lie algebra g is said to be of compact type ifit has a positive definite invariant form, β.

An obvious consequence of the definition is that subalgebras of Liealgebra of compact type are themselves of compact type.

Some examples of Lie algebras of compact type:so(n,R) = X ∈ gl(V ) : Xt = −X is of compact type since βρ,

where ρ is the inclusion in gl(V ), is positive definite by the previousresult.

Every abelian Lie algebra g is of compact type since any form isautomatically invariant. So here we may simply choose any positivedefinite form on g.

As a final example, let G be a compact connected Lie group and g beits Lie algebra. Then g is of compact type. To see this observe that bythe proof of Theorem 2.5.1 there is an AdG-invariant inner product ong. Relative to this inner product the operators of AdG are all orthogonaland hence those of ad g are skew symmetric. By Proposition 3.9.1, thenegative of the trace form is bilinear, symmetric, invariant and positivedefinite.

Remark 3.9.3. Let g be a semisimple Lie algebra over R. Then theKilling form, β, is nondegenerate and invariant. Hence for each Y ∈ g,adY is skew symmetric with respect to β. This, however, does not meanthat β is definite. The result above requires a positive definite form onV (= g) and in general (for non-compact semisimple Lie algebra) theKilling form is of mixed type. For example when g = sl(2,R) it is a(1, 2) form.


We now come to the following result which characterizes real Liealgebras of compact type.

Theorem 3.9.4. If g is of compact type then g = z(g) ⊕ [g, g] where[g, g] is semisimple (and of compact type as remarked above). If g is asemisimple Lie algebra of compact type then its Killing form is negativedefinite. Conversely, if g = z(g) ⊕ [g, g], where [g, g] is semisimple andof compact type then g is of compact type.

Proof. Since g has a positive definite invariant form β, the invariancetells us that for each X ∈ g, adX is skew symmetric with respect to β.By the argument given for semisimple algebras, the orthocomplement ofany ideal h in g is also an ideal and g is the direct sum of these two ideals.In particular since z(g) is an ideal g = z(g) ⊕ l. If a is an abelian idealin l, then since l is of compact type, a is a direct summand; l = a ⊕ b.Therefore a commutes with b. Since a is itself abelian a commutes withall of l. On the other hand a ⊆ l so a commutes with z(g). But then a

commutes with all of g. Therefore a ⊆ z(g). But a is a subset of l soa = 0 and l is semisimple. Now from g = z(g)⊕ l it follows directly that[g, g] = [l, l] and since l is semisimple, [l, l] = l. Because l is semisimple,the adjoint representation is faithful and the Killing form is negativedefinite by Proposition 3.9.1.

Before turning to the converse we need the following lemma whichshows that the direct sum of Lie algebras of compact type is again ofcompact type.

Lemma 3.9.5. Suppose g = u⊕ v is a direct sum of ideals each havinga positive definite invariant form 〈·, ·〉u and 〈·, ·〉u. Then g has a positivedefinite invariant form.

Proof. Putting these two forms together we get a positive definite formon g with the summands orthogonal.

〈(U, V ), (U′, V

′)〉g = 〈U,U ′〉u + 〈V, V ′〉v.

3.9 Real Lie Algebras of Compact Type 205

Now

〈ad g(U, V )(U1, V1), (U2, V2)〉g = 〈(adU(U1), ad V (V1)), (U2, V2)〉g= 〈adU(U1), U2〉u + 〈adV (V1), V2〉v= −〈U1, adU(U2)〉u − 〈V1, ad(V )V2〉v.

Since the forms on u and v are invariant, the form on g is also invariant.

Now for the converse, if g = z(g) ⊕ [g, g] where [g, g] is semisimpleand of compact type, then the negative of the Killing form of [g, g] ispositive definite (and invariant) and hence [g, g] is of compact type. Asabove the abelian Lie algebra z(g) is also of compact type and thereforeso is g.

Corollary 3.9.6. In a Lie algebra of compact type the Killing form isnegative semidefinite. It is negative definite if and only if g is semisim-ple.

Corollary 3.9.7. A Lie algebra g is of compact type if and only if it isthe Lie algebra of a compact connected Lie group.

Proof. We know the Lie algebra of a compact connected Lie group isof compact type. Conversely let g be a Lie algebra of compact type.Then g = z(g) ⊕ [g, g], where [g, g] is semisimple and of compact type.Therefore ad g is a semisimple Lie algebra of compact type. Let G bethe simply connected real Lie group whose Lie algebra is g. Then G is adirect product Rn×H, where H is a subgroup of G whose Lie algebra is[g, g]. Clearly, Tn×H is also a Lie group whose Lie algebra is g which willbe compact if and only ifH is. Since the Lie algebra ofH is compact andsemisimple we may assume these properties for g. Thus we may assumeg is semisimple and of compact type. In particular, g is isomorphic toad g. Since AdG is a Lie group whose Lie algebra is ad g we see thatits Lie algebra is g. We need only show that AdG is compact. Howeversince g is semisimple ad g = Der(g). It follows that AdG = Aut0(g).Since the latter is an algebraic group AdG it is a closed subgroup ofGL(g). On the other hand since g is semisimple and of compact type the


negative of its Killing form is positive definite. Therefore by invarianceof this form ad g consists of skew symmetric operators. This means, atleast near the identity, AdG consists of orthogonal operators. BecauseAdG is a group it is a subgroup of the orthogonal group on g. Inparticular, AdG is bounded. Since it is also closed it is compact.

Corollary 3.9.8. Any compact connected Lie group G is isomorphic to

(Z(G)0 × [G,G])/F,

the quotient group of a direct product of a torus, Z(G)0, and a compactsemisimple group by a finite central subgroup F .

Proof. If G is compact then, as we showed, its Lie algebra, g, is ofcompact type. By the argument just above G is a direct product Rn ×H, where H is as above. But since this H is locally isomorphic withAdG (has the same Lie algebra) and the latter is compact, so is H, byTheorem 2.5.8. The covering π : G → G maps onto a compact groupso its kernel must contain a lattice of maximal rank in Rn. Hence G iscovered by a direct product T n × H. Then H = [H,H] since the Liealgebra of H is [g, g], (see Theorem 3.9.4). Moreover this product groupis compact since both factors are. We have shown G is covered by Tn×[H,H]. But the image of [H,H] is clearly [G,G]. Therefore G = (Tn ×[G,G])/F where F is a discrete central subgroup, which must be finitesince Tn × [G,G] is compact. Since we have a direct product upstairs,G is the commuting product of Tn and [G,G] downstairs. This impliesits center Z(G) = Tn ·Z[G,G]. As [G,G] is compact and semisimple itscenter is finite so Tn = Z(G)0.

Chapter 4

The Structure of Compact

Connected Lie Groups

4.1 Introduction

In this chapter we deal with the important role a maximal torus playsin the structure and representation theory of a compact connected Liegroup.

As we know, if H is a connected abelian Lie group then it is isomor-phic as a Lie group to Rm × Tn. In particular, a compact connectedabelian Lie group is isomorphic with Tn. A maximal connected abeliansubgroup H of a connected Lie group means one which is contained inno strictly larger such subgroup. These clearly exist in any connectedLie group G for dimension reasons. Similarly, maximal tori exist in anyconnected Lie group. If H is a maximal connected abelian subgroup ofa connected Lie group G, then H must be closed, for its closure is apossible larger connected abelian Lie subgroup of G. If G is also com-pact then H, being closed, is also compact and hence is a torus. So inthe compact case maximal abelian connected is the same as a maximaltorus. Similarly, a maximal abelian subalgebra in a Lie algebra is anabelian subalgebra which is not properly contained in a larger abeliansubalgebra. These also exist for dimension reasons.

For example supposeG = U(n,C), the unitary group. This compact,

207

208 Chapter 4 The Structure of Compact Connected Lie Groups

connected Lie group and is a good place to start. Consider the subgroupD of diagonal matrices which is clearly isomorphic to Tn. Since it iscompact D must be closed. Now D is actually a maximal torus. Forif there were a strictly larger one, then there would be some element gwhich commutes with each point of D. But D contains elements with ndistinct eigenvalues. Since g commutes with such an element g is itselfdiagonal, a contradiction. At the same time this shows D is its owncentralizer ZG(D) = D. Another observation is that every point of Gis conjugate to something in D. That is to say, G =

⋃g∈G gDg

−1. Thisfollows by finite dimensional spectral theorem, which says that any uni-tary operator is similar under a unitary operator to a diagonal unitaryoperator. We shall see that this holds for any compact connected Liegroup G so that G =

⋃g∈G gTg

−1, where T is any maximal torus in G.However, this is a much deeper fact than the finite dimensional spectraltheorem because of the profusion of compact connected Lie groups. Aswe shall see in the next chapter, such groups are linear, but they may bemuch smaller than the ambient unitary group. Hence the nature of themaximal torus may also be different because of the relations definingG. Here the example to keep in mind is the most important compact,connected, nonabelian, simply connected Lie group G = SU(2). As wehave seen above, here the maximal torus has dimension 1.

4.2 Maximal Tori in Compact Lie Groups

The point of looking at maximal tori in a compact Lie group is to reducea more complicated non-abelian situation to an abelian one. We firstdeal with some issues of abelian groups.

Abelian Lie groups

If an abelian topological group A has a an element a0 which gener-ates a dense cyclic subgroup, we call a0 a quasi generator of A. Bythe Kronecker approximation theorem (Appendix B), a torus Tn has aquasi-generator.

4.2 Maximal Tori in Compact Lie Groups 209

Definition 4.2.1. Let G be a group and let k be a fixed a positiveinteger. Consider the map πk : g 7→ gk. We say G is divisible if πk issurjective for all k ∈ Z+.

This notion is particularly important when G is abelian. For exam-ple, if Tn is a torus and π : Rn → Tn is its universal covering, thensince R is a field it, and therefore also Rn, is divisible. Hence by usingthe homomorphism π so is Tn. As to the significance of divisibility inabelian groups we mention the following well known fact (see [36]).

Proposition 4.2.2. Let H be a subgroup of the abelian group G andf : H → D be a homomorphism where D is a divisible group. Then fextends to a homomorphism f : G→ D.

Corollary 4.2.3. A divisible subgroup H of an abelian group G is adirect summand of G.

Proof. Take the identity map i : H → H and extend it to a homomor-phism, i : G→ H. Then as is easily seen G = H ⊕ Ker i.

Corollary 4.2.4. A compact abelian Lie group G is the direct productof a torus and a finite group.

Proof. Clearly the identity component G0 being a compact connectedabelian Lie group is a torus, Tn (which is normal). Because G is a Liegroup Tn is open in G. Hence G/Tn is discrete and compact and sofinite. Now as remarked earlier Tn is divisible. Therefore the identityhomomorphism i : Tn → Tn extends to a homomorphism i : G → Tn.If i were continuous the proof of the previous results would show G =Tn ⊕ G/Tn as topological groups. But a homomorphism is continuousif it is so at the identity. Since Tn is an open subgroup containing theidentity element and i|Tn = i which is continuous, i is indeed continuouson G.

It now makes sense to ask when does a compact abelian Lie groupG have a quasi-generator?

Corollary 4.2.5. A compact abelian Lie group G has a quasi-generatorif and only if G = Tn ⊕ Zm. That is if and only if G/Tn is cyclic.


Proof. Since the projection is a continuous homomorphism if G has aquasi-generator then so does G/H for any subgroup H. Because G/Tn

is discrete this quotient must be cyclic. More generally let Tn be a toralsubgroup and suppose G/Tn is a finite cyclic group, then G will have aquasi-generator.

To see this choose a quasi-generator of t0 ∈ Tn and a g0 ∈ G whichprojects onto the generator of G/Tn = Zm. Then mg0 ∈ Tn. BecauseTn is divisible there is a t ∈ Tn satisfying mt = t0 − mg0, and theng = t+ g0 is a quasi-generator of G because mg = m(t+ g0) = t0. Thusthe m-fold powers of g which lie in Tn, are dense in Tn. As to the otherpowers, ng = nt + ng0. Write n = qm + r, where 0 ≤ r < m. Thenng = (qm+r)t+rg0. This gives a dense set in the coset Tn+rg0. Sincewe have a finite coset decomposition, G =

⋃m−1r=0 T

n + rg0, we see thatg is a quasi-generator for G.

Another important feature of tori is the fact that their automor-phism groups are discrete. When the group is an abelian Lie group itis easy to see this by duality, because Aut(G) is isomorphic to Aut(G)where G is character group of G. Since G is a finitely generated discretegroup its automorphism group is surely discrete. Actually in the caseof a torus Tn we can calculate Aut(G) explicitly: Aut(Zn) = GL(n,Z).

Exercise 4.2.6. Work out the details of the paragraph above. In par-ticular, show that Aut(G) is discrete for a compact abelian Lie groupG. Show that Aut(G) is not discrete when G = SO(3,R).

4.3 Maximal Tori in Compact Connected Lie

Groups

We now turn to the following basic result. This argument is due to G.Hunt [35].

Theorem 4.3.1. Let G be a compact connected Lie group and g its Liealgebra. Then

(1) G has a maximal torus, T .

4.3 Maximal Tori in Compact Connected Lie Groups 211

(2) g has a maximal abelian subalgebra, t.

(3) If h is a maximal abelian subalgebra, then the connected subgroupH of G with Lie algebra h is a maximal torus of G.

(4) Any two maximal abelian subalgebras h1 and h2 are conjugate;Ad g(h1) = h2 for some g ∈ G.

(5) Similarly, any two maximal tori T1 and T2 of G are conjugate;gT1g

−1 = T2.

In particular, any two maximal tori of G have the same dimension,and any two maximal abelian subalgebras of g have the same dimension.This number, called the rank of G, is an important invariant. As weshall see later, an analogue of the conjugacy theorem does hold in thecase of connected complex semisimple (or reductive) Lie groups.

Proof. We know G has a torus and each torus is contained in a maximalone; and similarly for abelian subalgebras of g. The correspondencebetween Lie subgroups and subalgebras, Theorem 1.3.3, proves the first3 items and shows that 4 and 5 implies each other. We shall prove 4.

Let X1 ∈ h1 and X2 ∈ h2 be fixed and f(g) = 〈Ad g(X1),X2〉,where by compactness 〈·, ·〉 can be taken to be an Ad-invariant in-ner product on g (see Chapter 2). Again by compactness thissmooth function has a minimum value at say g0 ∈ G. Let X ∈ g.Then d

dt |t=0〈Ad exp tXg0X1,X2〉 = 0. But 〈Ad exp tXg0X1,X2〉 =〈Ad exp tX Ad g0X1,X2〉 = 〈Exp(ad tX)Ad g0X1,X2〉 so taking thederivative at t = 0 tells us 〈adX Ad g0X1,X2〉 = 0. Since Ad leaves〈·, ·〉 invariant, ad leaves it infinitesimally invariant (i.e. operates byskew symmetric matrices). Hence 〈X, [Ad g0X1,X2]〉 = 0 and since Xis arbitrary and 〈·, ·〉 is positive definite, we get [Ad g0X1,X2] = 0. ButXi ∈ hi are also arbitrary hence Ad g0(h1) and h2 commute pointwise.Since h1 is a maximal abelian subalgebra and Ad g0 is an automorphismof g, so is Ad g0(h1). By maximality Ad g0(h1) + h2 = h2 from which itfollows easily that Ad g0(h1) = h2.

Exercise 4.3.2. Consider the compact connected Lie group SO(3,R).It contains a subgroup A (isomorphic to the Klein 4 group) consisting ofthe identity together with diagonal elements with two −1 and one 1 in


the various possible places. Show that A is maximal abelian. This doesnot contradict the Theorem 4.3.1 above because A is not connected;rather it is discrete.

We will illustrate the importance of the rank of G by finding the(non-abelian) compact connected Lie groups of rank 1.

Theorem 4.3.3. Let G be a non-abelian compact connected Lie groupof rank 1. Then G is either SU(2,C) or SO(3,R).

Proof. Since G is compact we know that g is the direct sum of its centerand its derived subalgebras (Theorem 3.9.4). But [G,G] has positiverank and G has rank 1 so this means the center is trivial and G issemisimple. Next we prove dimG = 3. To do so we may assume Gis simply connected. This is because G is semisimple so its universalcovering group is compact (by Theorem 2.5.8) . As they have the sameLie algebra Theorem 4.3.1 tells us the ranks are the same.

As above, let 〈·, ·〉 be an Ad-invariant inner product on g and letX0 ∈ t be a unit vector in the Lie algebra of a maximal torus T . Definef(g) = Ad g(X0). This is a smooth function f : G→ g which is constanton cosets. The induced map f : G/T → g is injective. This is because ifAd g(X0) = Adh(X0), then Adh−1g(X0) = X0 and since the group hasrank is 1, Ad(h−1g) fixes all of t. By maximality h−1g ∈ T so hT = gT .By compactness of G/T , f is a homeomorphism onto its image. Whatis this image? If n = dimG, f(G/T ) = f(G) is the orbit of X0 so it iscontained in Sn−1. On the other hand dimG/T = n− 1 because G hasrank 1. Since this is also the dimension of the sphere and everythingis connected we see G/T and Sn−1 are homeomorphic. Now apply thelong exact homotopy sequence to the fibration T → G → G/T = Sn−1

to get π2(Sn−1) → π1(T ) → π1(G). If n ≥ 3 then π2(S

n−1) = 1,π1(T ) → π1(G) is an injection. This is impossible since G is simplyconnected and T is not. Hence π2(S

n−1) 6= 1. Therefore n − 1 = 2and n = 3.

It remains to prove G is either SU(2,C), or SO(3,R). We continueto assume G is simply connected (and semisimple). Then Z(G) is finite,hence dimAdG = 3. But by invariance of the form we have AdG ⊆SO(3) and since both these are connected and SO(3) also has dimension


3, we conclude that AdG = SO(3). This means the Lie algebra is thatof SU(2,C), proving the theorem (see Lemma 3.8.2).

Remark 4.3.4. As a corollary of the proof (since, as we saw above,G = SU(2,C) has rank 1) we know G/T = S2. That is, S3/S1 = S2.This is the Hopf fibration.

In order to deal with many of the properties of a compact connectedLie group we shall require the concept of an exponential Lie group.

Definition 4.3.5. Let G be connected Lie group. We say G is expo-nential if exp : g → G is surjective. Alternatively, every point of G lieson a 1-parameter subgroup.

Remark 4.3.6. A few comments about this notion are in order. As weknow, the exponential map is surjective to sufficiently small neighbor-hoods of 1. The question is, can this be extended to the entire group.Of course, an exponential group would have to be connected. That iswhy we restricted ourselves to connected groups in the definition above.Non-compact connected Lie groups are rarely exponential, for exampleSL(2,R) is not exponential. In fact, here the range of exp is not evendense in G.

We now prove a lemma due to H. Hopf. Because the theorem onthe degree of a mapping requires compactness, the proof of this lemmabreaks down if the group is not compact. In fact this statement, aswell as the following two (or their analogues), are false for non-compactsemisimple Lie groups.

Lemma 4.3.7. Compact connected Lie groups are divisible.

Proof. Our proof relies on the concept of the degree of a mapping. Werefer the reader to [18] for the definition of the degree of a map. LetX be a smooth oriented compact connected manifold and f : X → Xa smooth map. It follows from the very definition of degree that if thedegree of f is different from zero, then f is surjective [18]. Here wetake G itself as the manifold, and for the mapping, πk : g 7→ gk and


we determine the sign deg(πk). First we fix a left invariant orientationand volume form on G. Let Rg and Lg denote the right and the lefttranslation by g. We shall prove that deg(πk) is positive. For that wemust determine the sign of the Jacobian of πk, i.e. Ja(πk) = det(daπk)and for a ∈ G, we know that TaG is generated by left invariant vectorfields, so let Xa be the invariant vector field is generated by X ∈ g.Note that πk La is given by g 7→ a · g · a · · · a · g and its derivative atthe identity element is

daπkd1La = d1(πkLa) =∑

i+j=k

dLaidRaj

= dLak(In + T + T 2 + · · · + T k−1)

where T = Ad(a−1). Therefore

daπk(Xa) = daπkd1La(X) = d1Lak(In + T + T 2 + · · · + T k−1)(X),

and since dLak is an orientation preserving diffeomorphism of G, thesign of the determinant of dπk and of T ka = Id + T + T 2 + · · · + T k−1

are the same.As G is compact, we can choose an Ad-invariant measure (see Chap-

ter 2) so AdG is contained in some unitary group U(n,C), hence theeigenvalues of T , we have absolute value 1. Now consider the character-istic polynomial of Ad(g−1),

Pg(t) = det(tIn − Ad(g−1))

which is positive for large enough t. Its roots have absolute value one,therefore Pg(t) is positive for t > 1.

LetT ka (t) = tk−1In + tk−2T + T 2 + · · · + tT k−1,

then T ka (1) = T ka so we have

(tIn − T )T ka (t) = T ka (t)(tIn − T ) = tkIn −Ak,

and Pa(t) detT ka (t) = Pa(tk). Since Pa(t) > 0 and Pa(t

k) > 0 for t > 1,we get

detT ka (t) > 0


for t > 1 therefore detT ka = detT ka (1) ≥ 0. One can calculate the degreeof πk by ∫

Gπ∗k(dg) = deg(πk)

∫

Gdg,

where dg is the volume form on G. We have

∫

Gπ∗k(dg) =

∫

GJ(πk)(g)dg.

Since J(πk)(g) ≥ 0 for all g ∈ G, and J(πk)(1) = kn, we get∫G π

∗k(dg) >

0 which proves that deg(πk) > 0.

We are now in a position to prove:

Theorem 4.3.8. Compact connected Lie groups are exponential.

Proof. Let g ∈ G, let U be a canonical neighborhood of 1 in G andconsider the closed subgroup H generated by g. Since a compact Liegroup is linear by Peter-Weyl theorem, Chapter 5, it is also secondcountable, therefore gn : Z+, has a convergent subsequence gni con-verging to g0, say. Hence limi→∞ gni−ni−1 = 1. This means that somepower gk lies in U . By continuity of πk, there is a neighborhood V (g)of g so that πk(V (g)) ⊆ U . Hence G =

⋃g∈G V (g) and by compactness

G =⋃gi∈S V (gi) for a finite number of gi where πi(V (gi)) ⊂ U . Let

m be the product of these i′s. For each i, we have i · i = m, where iis the product of all the others. Then for all i, πi(V (gi)) ⊆ U . Hencefor each i, πm(V (gi)) ⊆ πi(U). But since every point of U does lie on a1-parameter group of G therefore U ⊆ πi(U) for each i. Applying πi weget πi(U) ⊆ πm(U) and therefore πm(V (gi)) ⊆ πm(U). But the V (gi)cover G so πm(G) ⊆ πm(U) and since G is divisible, Lemma 4.3.7 wefinally get G = πm(U). Thus some power of each group element lieson a 1-parameter subgroup of G and this means the same is true of thegroup element itself.

Corollary 4.3.9. For a compact connected Lie group, G, the conjugatesa maximal torus T fill out all of G:


G =⋃

g∈GgTg−1.

Proof. Since each point of G lies on a 1-parameter subgroup which isitself contained in a maximal torus we conclude that G is a union of itsmaximal tori. The result now follows from the fact that the maximaltori are all conjugate, Theorem 4.3.1.

We remark that actually the conclusion of Corollary 4.3.9 impliesexponentiality which in turn implies divisibility. Thus all three notionsare equivalent in the case of compact connected Lie groups. To seethis, let g ∈ G. Then g = g1tg

−11 where t ∈ T . Since G is linear,

because it is compact, we have exp = Exp and therefore exp(PXP−1) =P exp(X)P−1. Since T is abelian here exp is the universal covering ofT by Rn, where n = dimT . In particular exp |T is surjective. Thereforet = expX and g = exp(g1Xg

−11 ). Conversely If g = exp(X) and k is a

positive integer, then g = [exp 1kX]k.

Corollary 4.3.10. If G is a compact connected Lie group, then Z(G)is the intersection of all maximal tori of G.

Proof. Let g ∈ Z(G). Then g ∈ T for some maximal torus. Thereforeg = hgh−1 ∈ hTh−1. Hence g lies in every maximal torus of G. Onthe other hand suppose g lies in every maximal torus of G. Then gcommutes pointwise with every maximal torus. Since G is the union ofall maximal tori, g has to be in the center.

Corollary 4.3.11. If A ⊆ G is a connected abelian subgroup of G itscentralizer ZG(A) is the union of the maximal tori in G containing A.

Proof. Since A is a connected abelian subgroup of G so is its closure,A. Therefore A is a torus. If g centralizes A, then it also centralizesA (and vice versa). Thus we can assume A is a torus and must showZG(A) = ∪T ′, where T ′ ⊇ A. We have T ′ ⊆ ZG(A), for any toruscontaining A hence ∪T ′ ⊆ ZG(A).

Conversely, let g ∈ ZG(A) and let B be the closure of the (abelian)subgroup of G generated by g and A. In particular, B ⊇ A. Also, B is

4.4 The Weyl Group 217

compact and abelian. Hence its identity component B0 is a torus. SinceB0 ⊇ A and g ∈ B, gB0 generates B/B0, which is finite and hence finitecyclic. By the result Corollary 4.2.5, since B0 has a quasi-generator, sodoes B. Let b be a quasi-generator of B ⊆ G. Hence by Corollary 4.3.9b ∈ T ′, some maximal torus of G. Therefore B must also be containedin T ′ and g ∈ T ′. Since T ′ ⊇ A, this completes the proof.

We now come to a criterion for maximality which made an appear-ance in many previous examples.

Corollary 4.3.12. Let G be a compact connected Lie group, T be atorus and g and t be the respective Lie algebras. Then T is maximal ifand only if ZG(T ) = T .

Proof. We know from the previous proposition that if T is maximal,then ZG(T ) = T . Suppose ZG(T ) = T and T ′ is a torus in G containingT . Then T ′ ⊆ ZG(T ). Therefore T ′ = T .

4.4 The Weyl Group

In this section we introduce Weyl group, an invariant for compact con-nected Lie groups.

Proposition 4.4.1. Let G be a compact connected Lie group and Tmaximal torus. Then G/T is simply connected.

Proof. First assume G is compact semisimple. Consider the universalcovers, G and T of G and T respectively. Since G is semisimple G iscompact and T is a maximal torus in G. Now G/T = G/T and since Gis simply connected we have Π1(G/T ) = T /T0 which is trivial becauseT as a maximal torus is connected. Therefore G/T is simply connected.Now in general G = Z(G)0[G,G] where [G,G] is semisimple. SinceG is compact Z(G)0 is a torus and hence Z(G)0T is a maximal torusof G where T is a maximal torus of [G,G]. Therefore by the secondisomorphism theorem G/Z(G)0T = [G,G]/T .

Definition 4.4.2. We denote the centralizer and normalizer of T byZG(T ) and NG(T ) respectively.


Proposition 4.4.3. Let G be a compact connected Lie group and T amaximal torus. Then NG(T ) contains T as a subgroup of finite index.

Proof. Using joint continuity Aut(G) × G → G one sees easily thatNG(T ) is a closed subgroup of G, so NG(T ) and NG(T )/NG(T )0 bothare compact. Moreover since NG(T ) is a Lie group, NG(T )0 is openin NG(T ) so NG(T )/NG(T )0 is discrete. Therefore NG(T )/NG(T )0 isfinite. We complete the proof by showing NG(T )0 = T . EvidentlyNG(T )0 ⊇ T . Consider αn0 |T is in Aut(T ), the conjugation by n0 ∈Ng(T ). This gives a connected subgroup of Aut(T ). On the other hand,as we saw earlier, Aut(T ) is discrete. Therefore this subgroup is trivialand n0 centralizes T . Therefore by the previous paragraph n0 ∈ T .

Definition 4.4.4. Since NG(T ) contains T as a closed normal subgroupof finite index, the quotient NG(T )/T is a finite group called the Weylgroup, W(G).

In principle W could depend on the choice of maximal torus T .However, suppose gTg−1 = T ′ were another maximal torus, withW ′ = NG(gTg−1)/gTg−1. An easy calculation shows gNG(T )g−1 =NG(gTg−1). This means that W ′ is naturally isomorphic with W. Thusthe Weyl group, and in particular its order, is another important invari-ant of G.

Exercise 4.4.5. SupposeG = U(n,C). Its Weyl group is the symmetricgroup W = Sn. In particular, here |W| = n!.

Now we turn to the mapping φ : G/T × T → G given by (gT, t) 7→gtg−1 which is of some importance in the representation theory of com-pact connected Lie groups (see e.g. [1]), whereG is a compact connectedLie group, and is T a maximal torus. The map is well-defined since ifh−1g ∈ T , then hth−1 = gtg−1. It is also surjective by Corollary 4.3.9.Evidently here we have a smooth map between compact connected mani-folds of the same dimension. What is the degeneracy of φ in some genericsense? Let t0 be a quasi-generator of T . Then φ(gT, s) = gsg−1 = t0if and only if g normalizes T . Therefore |φ−1(t0)| = |NG(T )/T | = |W|.We want to see what this map looks like locally near (T, 1), to do so,consider g and its subalgebra t.

4.4 The Weyl Group 219

Now the significance of the Weyl group is that it operates on Tby inner automorphisms W × T → T is a (nT, t) 7→ ntn−1. Sincen ∈ NG(T ) and t ∈ T , ntn−1 ∈ T . This action is effective, i.e. themap W → Aut(T ) is 1 : 1, since if w = nT ∈ W and ntn−1 = t, for allt ∈ T then n ∈ ZG(T ) = T so nT = w is the identity. We shall see ita moment that the action of the Weyl group on T reflects exactly theaction of G on itself by conjugation.

Lemma 4.4.6. Let t1 and t2 lies in T , a maximal torus. Then there isa g ∈ G with gt1g

−1 = t2 if and only if w(t1) = t2 for some w ∈ W.

Proof. Suppose gt1g−1 = t2. A direct calculation shows gZG(t1)g

−1 =ZG(t2). Since T ⊂ ZG(t1) we get gTg−1 ⊆ ZG(t2). Thus T andgTg−1 are maximal tori in ZG(t2)0. They are conjugate in this com-pact connected Lie group, so there is some h ∈ ZG(t2)0 satisfyingh(gTg−1)h−1 = T . Hence hg ∈ NG(T ). Also hgt1(hg)

−1 = ht2h−1 = t2.

Hence hgT = w ∈ W and w(t1) = t2. The converse statement is obvi-ous.

Let O be the space of conjugacy classes of G with the quotienttopology. Since G is compact one defines a compact Hausdorff topologythis orbit space. Let T/W be the orbit space of the action of W onT . Then there is a canonical homeomorphism T/W → O given byW(t) 7→ O(t), t ∈ T , which is surjective because of Corollary 4.3.9 andinjective because of Lemma 4.4.6 above. Since T/W is also compactand Hausdorff we have a homeomorphism.

Let C(X) be the complex valued continuous functions on the com-pact space, X. If a compact group G acts continuously on X, this givesrise to a linear representation of G on C(X) via fg(x) = f(g−1x). Weleave it to the reader to check this is a linear representation. We denotethe G-fixed functions by C(X)G. For example, if G acts on itself byconjugation, then C(G)G indicates the continuous class functions on G.In particular conjugation also gives rise to an action of W on C(T ).Here C(T )W = C(OW).

For a compact connected Lie group G and maximal torus T weconsider the restriction map C(G) → C(T ). Since conjugation gives


rise to an action of G on G and W on T we see that the restriction maptakes C(G)G → C(T/W) = C(T )W . If f |T = f1|T , where f1 is anothercontinuous class function, then since these are class functions we havef(gtg−1) = f1(gtg

−1) for all g ∈ G. If h ∈ G, then h ∈ gTg−1 for someg so f(h) = f1(h), thus f = f1 and the restriction map is 1 : 1. Thismap is also onto. Suppose f is a continuous function on T invariantunder the Weyl group. Define f on G by f(gtg−1) = f(t). Its easy tosee that f is well defined, continuous and a class function on G. This weleave as an exercise to the reader. Evidently f |T = f . The restrictionmap is a complex algebra homomorphism, hence an isomorphism.

Now let K(G) denote the ring of isomorphism classes of repre-sentations of G. The basic operations are the tensor product andthe direct sum. Note that K(G) is not quite a ring but to obtaina ring we must take all formal finite Z-linear combinations of repre-sentations (including those with possibly negative coefficients), then(−nρ) ⊕ σ = −(nρ) ⊕ σ = −n(ρ⊕ σ) and (−nρ) ⊗ σ = −n(ρ⊗ σ), etcfor n ∈ Z+. Now we really do have a ring with identity called the ringof virtual representations. Since the character of a finite dimensionalrepresentation determines the representation (see Chapter 5), K(G) isbasically the set of all the characters of the finite dimensional represen-tations of G, together with 0 under pointwise operations. We recall thatχρ⊗ρ′ = χρχρ′ and χρ⊕ρ′ = χρ + χρ′ .

Now we consider a character χρ of the finite dimensional continuousrepresentation ρ of G on V ; ρ is determined by its character χρ andsince χρ is a class function it is determined by its restriction to T .

Corollary 4.4.7. K(G) is an integral domain.

Proof. The paragraph above tells us that restriction K(G) → K(T )W

is an injective C-algebra homomorphism. Suppose χρχσ = 0, or even ifχρ|Tχσ|T = 0. We will show either χρ = 0 or χσ = 0. Let t0 be a quasi-generator of T and let χρ =

∑i niχρi

, χσ =∑

jmiχσj∈ K(G) where

ni and mj ∈ Z. Then since χρ(t0)χσ(t0) = 0 and C is a field we getχρ(t0) = 0, or χσ(t0) = 0. Let us assume its the former. Ignoring, as wemay, the zero coefficients, reorder the irreducibles of ρ so that n1, . . . , nkare positive and nk+1, . . . , nr are negative. Let ρ+ = n1ρ1 ⊕ · · · ⊕ nkρk

4.5 What goes wrong if G is not compact 221

and ρ− = (−nk+1ρk+1) ⊕ · · · ⊕ (−nrρr). Then ρ+ and ρ− are finitedimensional representations of G and χρ+(t0) = χρ−(t0). Since t0 is aquasi-generator of T , they agree on all of T . Hence χρ+ and χρ− agreeon T . Therefore they agree on all of G so that χρ = 0.

We conclude with an application of representation theory.

Proposition 4.4.8. Let G be a compact connected Lie group and Tmaximal torus. Then G/T has even dimension.

Proof. Consider the adjoint representation of G on its Lie algebra g andlet ρ be the restriction to a maximal torus.T . Since T is compact andabelian the continuous representation ρ decomposes into the direct sumof irreducibles over R. Thus we have some 1-dimensional representationsof the form ρ(t) = λ(t)X, t ∈ T , where λ is a continuous homomorphismT → R×, or we have 2-dimensional real representations. In the formercase since T is compact λ(t) = ±1, and since T is connected λ(t) ≡1. On the other hand suppose we have a 2-dimensional invariant realsubspace, V , of g. Then ρt|V = Rt of rotations. Thus g is the directsum of g0, the space on which ρ acts trivially, with

∑kj=1 gj, of rotations.

Hence dim g = dim g0 + 2k. But g0 is actually the Lie algebra of T .Hence dim(G/T ) = dimG− dimT = 2k.

To complete the proof of even dimensionality we must showAd t(X) = X for all t ∈ T if and only if Exp(sX) ∈ T for alls ∈ R. Suppose Exp(sX) ∈ T for all s ∈ R. Then Ad t(Exp(sX)) =tExp(sX)t−1 = Exp(sX) for all s and t ∈ R. Hence Exp(tsXt−1) =Exp(sX) for all s and t ∈ R. Taking d

ds |s=0 of both sides tells ustXt−1 = Ad t(X) = 0. Conversely, if tXt−1 = 0, then reversing ourargument we get Ad t(Exp sX) = Exp sX for all s and t so that for alls, Exp(sX) ∈ ZG(T ) = T .

4.5 What goes wrong if G is not compact

Is there any hope of finding a connected abelian subgroup whose con-jugates fill our the group. Let us take an important and familiar non-compact, but simple Lie group, namely G = SL(2,R). Write G = KAN


as in Section 1.6. We claim that both A and N are maximal connectedabelian subgroups of G. First, they are both abelian connected Lie sub-groups. Suppose H was a such a subgroup of G containing A. Everyh ∈ H centralizes A. But a matrix commuting with a diagonal matrixthat has distinct eigenvalues must itself be diagonal, and since H is con-nected the diagonal entries of h are positive. Therefore h ∈ A, whichshows A is maximal. To see that N is also maximal, let h ∈ H which isa connected abelian subgroup of G containing N . A direct calculationshows that ZSL(2,R)(N) consists of ±N . Since H is connected, h ∈ N .Thus N is also a maximal abelian subgroup of G.

Now is it possible that G =⋃g∈G gAg

−1, or G =⋃g∈G gNg

−1?The answer to each of these questions is no, both for the same reason.If either were true then each element of A would be conjugate to anelement of N and vice versa. But this cannot be because trh = a+ 1

a ,where h ∈ A, while if h ∈ N , then trh = 2. However, a+ 1

a > 2, unlessof course a = 1.

Another possibility for a connected abelian subgroup (which likeA acts completely reducibly) is K, which after all is a torus. CouldG =

⋃g∈G gKg

−1? (Even if K were not a maximal torus it would becontained in one and so the conjugates of it would also fill out G andso we could argue as below.) If G = ∪g∈GgKg−1, then each element ofG would be diaganolizable with eigenvalues on the unit circle. But A isdiagonal with positive eigenvalues. This means A = 1, a contradic-tion.

Moreover the conjugacy relation between maximal abelian sub-groups is also false. For K cannot be conjugate to either A or N sinceK is compact and the other two are not. Nor can A and N be conju-gate since the eigenvalues of elements of A give all positive reals whilethose of N are always 1. It is worth noting that these groups are notconjugate in spite of the fact that they are isomorphic.

Chapter 5

Representations of

Compact Lie Groups

In this chapter we deal with the classical representation theory of a com-pact group. We shall see that compact Lie groups play a special rolealthough the case of finite or abelian groups are also of great interest.The central object of study here is the set of all finite dimensional, con-tinuous, irreducible unitary representations. Although we will not needthis, actually every continuous, irreducible, unitary representation of acompact group on a Hilbert space is automatically finite dimensional.In Section 1 we introduce the players, in Section 2 we prove the Schurorthogonality relations. In Section 3 we develop what we need fromfunctional analysis and in Section 4 we prove the Peter-Weyl theoremand its many consequences. Section 5 deals with characters and classfunctions. In our final section we study induced representations andthe Frobenius reciprocity theorem as well as a number of related ideaswhich have proven to be quite useful in geometric questions (such as theMostow-Palais equivariant embedding theorem) and spherical harmon-ics. It might also be mentioned that the results on representations andharmonic analysis have been generalized from compact to other classesof groups. The most direct analogies have been found in the case ofcentral groups, those which are compact modulo their center (see [27]and [28]).

223

224 Chapter 5 Representations of Compact Lie Groups

5.1 Introduction

Unless otherwise stated, throughout this chapter G will denote a com-pact topological group and dg will be the normalized Haar measure onG (recall G is unimodular by Corollary 2.2.2), L1(G) and L2(G) willdenote the integrable, respectively square integrable, complex valuedmeasurable functions on G with respect to the Haar measure and C(G)the continuous complex valued functions on G. Likewise if G operateson a space X with a measure dx preserved by G we denote by L1(X)and L2(X) the integrable, respectively square integrable, complex val-ued measurable functions on X and C(X) the continuous ones.

Even though compact groups have many interesting infinite dimen-sional representations, we shall also see why we concentrate on finitedimensional representations.

The following definitions are fundamental and actually do not re-quire compactness of G, but merely that the representations are contin-uous and finite dimensional.

Definition 5.1.1. (1) Given two such representations of ρ and σ ofG we shall call an operator T : Vρ → Vσ an intertwining operatorif we have a commutative diagram Tρg = σgT , g ∈ G.

(2) ρ and σ are said to be equivalent if there exists an invertible in-tertwining operator between them.

(3) We shall say ρ is a unitary representation if ρ(G) ⊆ U(n,C) forsome n.

(4) A Hermitian inner product 〈·, ·〉 on Vρ is called invariant if〈ρg(v), ρg(w)〉 = 〈v,w〉 for all g ∈ G and v,w ∈ Vρ.

(5) A subspace W of Vρ is called an invariant subspace if ρg(W ) ⊆Wfor all g ∈ G.

(6) ρ is called completely reducible if every invariant subspace has acomplementary invariant subspace.

(7) ρ is called irreducible if it has no nontrivial invariant subspaces.

(8) Finally, we denote by R(G) the equivalence classes of finite di-mensional, continuous, irreducible unitary representations of G.As we shall see this set is quite interesting even when G is finite

5.1 Introduction 225

or abelian.

The following proposition and its corollary was proved in Chapter 2(see the proof of Theorem 2.5.1).

Proposition 5.1.2. Any finite dimensional representation of a compactgroup G has a G-invariant inner product and hence is equivalent to aunitary representation.

In particular,

Corollary 5.1.3. Any finite dimensional continuous representation ofa compact group G is completely reducible.

We leave the following important exercise to the reader.

Exercise 5.1.4. (1) Show that equivalence of representations is anequivalence relation.

(2) Given a single representation ρ, show the set of intertwining op-erators forms a subalgebra of End(Vρ).

(3) Give an example to show that the proposition and corollary aboveis false if G is not compact e.g consider a unipotent representationof R.

(4) Show that two representations are equivalent if and only if themodules (G,Vρ, ρ) and (G,Vσ , σ) are isomorphic. Thus they shareall module theoretic properties such as a composition series forone corresponds to such a series for the other etc.

(5) Show that a finite dimensional continuous representation ρ iscompletely reducible if and only if the corresponding module issemisimple.

(6) Show that a finite dimensional continuous representation ρ is ir-reducible if and only if the corresponding module is simple.

(7) Define a unitary representation (not necessarily continuous) of agroup (not necessarily compact) on a Hilbert space V and showthat it is completely reducible in the sense that any closedG invari-ant subspace of V has a complimentary closed invariant subspace.


We conclude this section with an important example. We shall findall the finite dimensional irreducible unitary representation of SU(2,C).Now we know all the complex irreducible representations of the Liealgebra sl(2,C) (see Section 3.1.5). Since the Lie group SL(2,C) issimply connected (Corollary 6.3.7) and has sl(2,C) as its Lie algebra itsirreducible representations are in bijective correspondence with thoseof the Lie algebra by ρ 7→ ρ′ (Corollary 1.4.15). Similarly, SU(2,C) isalso simply connected (Corollary 1.5.2), so its real continuous (smooth)irreducible representations are in bijective correspondence with those ofits Lie algebra, su(2,C). Finally, as we will see in Chapter 7, su(2,C)is a compact real form of sl(2,C). Hence complex irreducibles of thelatter bijectively correspond with the real irreducibles of the former.

Corollary 5.1.5. There are an infinite number of continuous, finitedimensional, irreducible, unitary representations of SU(2,C), one foreach degree.

Exercise 5.1.6. Show that, within these, the representations ofSO(3,R) are the ones of odd degree.

5.2 The Schur Orthogonality Relations

The Schur orthogonality relations are the following.

Theorem 5.2.1. Let G be a compact group, dg be the normalized Haarmeasure and ρ and σ finite dimensional continuous irreducible unitaryrepresentations of G. Then

(1) If ρ and σ are inequivalent, then∫G ρij(g)σlk(g)dg = 0 for all

i, j = 1 . . . dρ and k, l = 1 . . . dσ.

(2)∫G ρij(g)ρkl(g)dg =

δikδjl

dρ.

Proof. Let Vρ and Vσ be the respective representation spaces andB(Vσ, Vρ) be the (finite dimensional) C-vector space of linear operatorsbetween them. Let T ∈ B(Vσ, Vρ) and consider the map G→ B(Vσ, Vρ)given by g 7→ ρ(g)Tσ(g−1). This is a continuous operator valued func-tion on G and so

∫G ρ(g)Tσ(g−1)dg is also an operator in B(Vσ, Vρ). For

h ∈ G we have

5.2 The Schur Orthogonality Relations 227

ρ(h)

∫

Gρ(g)Tσ(g−1)dgσ(h)−1 =

∫

Gρ(h)ρ(g)Tσ(g−1)σ(h)−1dg

=

∫

Gρ(hg)Tσ(hg)−1dg

=

∫

Gρ(g)Tσ(g)−1dg

Letting T0 =∫G ρ(g)Tσ(g)−1dg we see ρ(g)T0 = T0σ(g) for ev-

ery g ∈ G. That is, T0 is an intertwining operator. By Schur’slemma, Lemma 3.1.52, there are only two possibilities. Either ρ andσ are equivalent and T0 is invertible (and implements the equiva-lence), or ρ and σ are inequivalent and T0 = 0. In the latter case∫G ρ(g)Tσ(g)−1dg = 0, where T is arbitrary. Let T = (tjk). Then

(ρ(g)Tσ(g)−1)il =∑

jk ρij(g)tjkσkl(g−1). Since (tjk) are arbitrary we

get∫G ρij(g)σkl(g

−1)dg = 0 for all i, j = 1 . . . dρ and k, l = 1 . . . dσ.Because σ is a unitary representation σ(g−1) = σ(g)−1 = σ(g)∗. Thus∫G ρij(g)σlk(g)dg = 0 for all i, j = 1 . . . dρ and k, l = 1 . . . dσ.

We now consider the case when we have equivalence. Here we mayas well just take σ to be ρ. In this case Schur’s lemma tells us T0 is ascalar multiple of the identity. Thus

∫G ρ(g)Tρ(g)

−1dg = λ(T )I. Takingthe trace of each side yields

tr(

∫

Gρ(g)Tρ(g)−1dg) =

∫

Gtr(ρ(g)Tρ(g)−1)dg =

∫

Gtr(T )dg = tr(T ),

while tr(λ(T )I) = λ(T )dρ. We conclude λ(T ) = tr(T )dρ

and so∫G ρ(g)Tρ(g)

−1dg = tr(T )dρ

I. Using reasoning similar to the earlier case

one finds∫G ρij(g)ρlk(g)dg = 0 for all i, j, k, l = 1, . . . , dρ whenever i 6= l,

or j 6= k. Now we consider the case when i = l and j = k. By taking Tto be diagonal with all zero entries except for one we get i, j = 1, . . . , dρ.

Hence in general one has∫G ρij(g)ρkl(g)dg =

δikδjl

dρ.


5.3 Compact Integral Operators on a Hilbert

Space

Before proceeding further we must now prove the spectral theorem forcompact self-adjoint operators on a Hilbert space. Then we will applythis result to compact self-adjoint integral (Fredholm) operators to con-clude that the range of such an operator always has an eigenfunctionexpansion. It is this fact which is behind the Peter-Weyl theorem.

Definition 5.3.1. Let V and W be real or complex Hilbert spaces. Abounded linear operator T : V → W is called a compact operator ifT (B1(0)) is compact where B1(0) is the unit ball in V .

For an operator T , the norm, if it exists, is defined to

‖T‖ = sup‖Tv‖ : ‖v‖ = 1.

Exercise 5.3.2. Evidently, if this were so it would be true of every ballabout 0. In fact, T is compact if and only if it takes bounded sets tocompact ones. Notice that when V = W and is infinite dimensional,then the identity map I or, more generally λI, λ 6= 0 is not compactwhile if W is finite dimensional all bounded linear operators are com-pact. Such operators are called finite rank operators. Observe also thatthe restriction of a compact operator to a closed invariant subspace isagain compact.

What would be an nontrivial example of a compact operator? Sup-pose V = L2(X), where X is a compact Hausdorff space, dx is a (finitewhich we may as well normalize to have total mass 1) regular mea-sure on X and k is a continuous function on X × X. We can use kto define an operator Tk : V → V , by Tk(f)(x) =

∫X k(x, y)f(y)dy.

Tk is well defined since by compactness k is bounded and by theSchwarz inequality together with compactness tells us L2(X) ⊆ L1(X).Tk is evidently linear. Applying the Schwarz inequality again shows∫X |k(x, y)f(y)|2dy ≤ ||k||2X×X

∫X |f(y)|2dy ≤ ||k||2X×X ||f ||22. Hence Tk

is a bounded operator (whose operator norm is ≤ ||k||2X×X ).

5.3 Compact Integral Operators on a Hilbert Space 229

Definition 5.3.3. In this context the k above is called a kernel functionand Tk a Fredholm operator.

So for example. we can let φi and ψi be two sets of n of continuousfunctions on X, where n is any integer and k(x, y) =

∑ni=1 φi(x)ψi(y).

Then Tk is a compact operator for a very simple reason. Tk(V ) ⊆ W ,where W is the linear span of φi and hence is finite dimensional.

Now this conclusion actually holds for any jointly continuous k. Weshall see that it will be sufficient for our purposes to understand Fred-holm integral operators.

Theorem 5.3.4. For jointly continuous k, Tk is a compact operator.Furthermore, Tk(L2(X)) ⊆ C(X).

Proof. We first prove the second statement. Since X ×X is compact kis uniformly continuous. That is given x0 ∈ X and ǫ > 0 there existsa neighborhood Ux0 of x0 so that |k(x, y) − k(x0, y)| < ǫ, whenevery ∈ X and x ∈ Ux0 . Therefore |Tk(f)(x) − Tk(f)(x0)| ≤

∫X |k(x, y) −

k(x0, y)|‖f(y)‖dy ≤ ǫ‖f‖1 and since L2 ⊆ L1, ||f ||1 < ∞. Thus Tk(f)is always a continuous function.

We now show Tk is a compact operator. Notice that on C(X) wehave two norms, the sup norm || − ||X and the restricted L2 norm.But since we have normalized the measure, ||f ||2 ≤ ||f ||X . Let B bea bounded set in L2. If we can prove Tk(B) is compact in C(X), thenby continuity of the injection C(X) → L2(X) we will be done. To dothis we apply the Ascoli theorem. Now again by Schwarz, for x ∈ X,|Tkf(x)| ≤

∫X |k(x, y)||f(y)|dy ≤ ||k||2X×X ||f ||22. Thus ||Tkf ||X < ∞

and Tk(B) is uniformly bounded. Moreover |Tk(f)(x) − Tk(f)(x0)|2 ≤∫X |k(x, y) − k(x0, y)|2|f(y)|2dy ≤ ǫ2||f ||22, if x ∈ Ux0 . Thus Tk(B) is

equicontinuous at every point of X. By Ascoli, Tk(B) has compactclosure in C(X).

Definition 5.3.5. (1) A linear operator T : V → V on a Hilbertspace is called self adjoint if for all v,w ∈ V , 〈Tv,w〉 = 〈v, Tw〉.

(2) For such an operator if λ ∈ C, we define Vλ = v ∈ V : Tv =λv. Here λ is called an eigenvalue of T and Vλ the correspondingeigenspace.


Exercise 5.3.6. (1) So for example, a Fredholm operator Tk is self-adjoint if and only if k(x, y) = k(y, x) for all x, y ∈ X.

(2) If a linear operator T : V → V on a Hilbert space is self adjoint,then all its eigenvalues are real.

Now we wish to prove the following spectral theorem for compactself adjoint operators on a Hilbert space.

Theorem 5.3.7. Let T be a compact self-adjoint operator on a Hilbertspace V . Then

(1) These are all real.

(2) If λ 6= µ are distinct eigenvalues then Vλ and Vµ are orthogonal.

(3) If λ 6= 0 then Vλ is finite dimensional.

(4) T has at most a countable number of nonzero eigenvalues.

(5) KerT = V0.

(6) V = V0 ⊕ (∑

λi 6=0 Vλi) (orthogonal direct sum).

The main point being the last item which says, in particular, thatthe range of T can be expanded in a convergent series of eigenvectorsi.e. given ǫ > 0, for any v ∈ V , there exists a positive integer n(v) sothat ||T (v) − ∑n

i=1 λiT (vi)|| < ǫ.Here are some consequences of the spectral theorem.

Exercise 5.3.8. Prove that:

∞∑

i=1

λ2i dimVλi

= ||T ||.

In particular,for any nonzero eigenvalue, λ2i dimVλi

≤ ||T ||. Hence

dimVλi≤ ||k||22

λ2i

.

Exercise 5.3.9. Notice that in the case of a finite dimensional operatorthis just amounts to the fact that a self-adjoint operator is unitarilydiagonalizable with real eigenvalues.

Before turning to the proof of Theorem 5.3.7 we need some prepara-tory results.


Lemma 5.3.10. Let T : V → V be a compact operator and δ > 0, thenthe number of eigenvectors of norm 1 with eigenvalues λ > δ is finite.

In particular the number of such distinct (i.e. orthogonal)eigenspaces is finite. In particular, the total number of such distincteigenspaces associated with positive eigenvalues is countable. (even ifV has an uncountable orthonormal basis!). Moreover, (using the pos-itive integers) if we order the positive eigenvalues, λn, in decreasingorder, then λn → 0.

Proof. Let λ and µ be distinct eigenvalues of T both bigger than δ andv and w be the respective eigenvectors of norm 1. Then since theseeigenspaces are orthogonal (see item 2), ||Tv − Tw||2 = ||λv − µw||2 =λ2 +µ2 ≥ 2δ2. Thus ||Tv−Tw|| ≥

√2δ. Clearly if there were an infinite

number of such eigenvalues there could be no convergent subsequencecontradicting the fact that T is compact.

Lemma 5.3.11. Let T : V → V be a bounded self-adjoint operator andW a T -invariant subspace of V . Then W⊥ is also T -invariant.

In particular if T is compact self-adjoint and W is a closed T -invariant subspace then T restricted to W⊥ is again a compact selfadjoint operator.

Proof. Let w ∈ W and w⊥ ∈ W⊥. Then 〈Tw⊥, w〉 = 〈w⊥, Tw〉 = 0since W is T -invariant. Thus since w is arbitrary Tw⊥ ∈W⊥.

Lemma 5.3.12. For a self-adjoint operator T ,

‖T‖ = sup|〈T (v), v〉 : ‖v‖ = 1.

Proof. Let M = sup|〈T (v), v〉 : ‖v‖ = 1. By the Cauchy-Schwarzinequality it is obvious that

‖〈Tv, v〉‖ ≤ ‖T (v)‖ · ‖v‖ ≤ ‖v‖ = ‖T‖

if ‖v‖ = 1, therefore M exists and M ≤ ‖T‖. It remains to prove that‖T‖ ≤M for which it suffices to show that ‖T (v)‖ ≤M if ‖v‖ = 1. We


assume that T (v) 6= 0 and let w = Tv/‖Tv‖. Then 〈Tv,w〉 = 〈v, Tw〉 =‖Tv‖ and

4‖Tv‖ = 〈T (v + w), v + w〉 − 〈T (v − w), v − w〉≤M‖v + w‖2 +M‖v − w‖2 = 4M.

Proposition 5.3.13. Let T : V → V be a compact self-adjoint operatoron a Hilbert space. Then there is some w of norm 1 with T (w) =±||T ||w.

Proof. By previous result there is sequence of vectors vn of norm 1 suchthat ||T || = limn→∞ |〈T (vn), vn〉|. By passing to a subsequence, we mayassume that 〈T (vn), vn〉 converges to r which is ||T || or −||T ||, and T (vn)converges to some vector v, as T is compact. Then

0 ≤ ‖T (vn) − rvn‖ = ‖T (vn)‖2 − 2r〈T (vn), vn〉 + r2‖vn‖2

≤ 2r2 − 2r〈T (vn), vn〉.

As the right side of the inequality converges to zero thereforelimn→∞ ‖T (vn) − rvn‖ = 0. On the other hand w = limn→∞ T (vn)hence limn→∞ rvn = w. For w = r−1v, we have T (w) = rw.

Proof of the spectral theorem.

(1) This is the exercise above.

(2) Suppose Tv = λv and Tw = µw. Then 〈Tv,w〉 = λ〈v,w〉. But itsalso 〈v, Tw〉 = µ〈v,w〉. Therefore either 〈v,w〉 = 0 or λ = µ, butsince µ is real this would mean λ = µ.If λ 6= 0, then since T acts on Vλ as a nonzero scalar multipleof the identity, Vλ is finite dimensional by a remark made earlier.By Lemma 5.3.10 T has at most a countable number of positiveeigenvalues. Since −T is also a compact operator, T must alsohave at most a countable number of negative eigenvalues hence acountable number of nonzero eigenvalues.

(3) Clearly V0 = KerT .


(4) Finally, let W = (∑

λi 6=0 Vλi). Then we have V = W ⊕ W⊥.

We prove that V0 = W⊥ which complete the proof. Since eachVλi

⊆ V ⊥0 , hence V0 ⊆W⊥.

Since W is clearly T -invariant the same is true of W⊥ by Lemma5.3.11 and moreover T restricted to W⊥ is a compact self-adjointoperator. By Proposition 5.3.13 choose w⊥

0 of norm 1 in W⊥ sothat T (w⊥

0 ) = ±||S||w⊥0 , where S is the restriction of T to W⊥.

If ||S|| > 0, then w⊥0 ∈ Vλi

for some i. Since Vλi⊆ W this means

w⊥0 = 0 which is impossible as its norm is 1. Thus ||S|| = 0 so T

restricted to W⊥ is zero or in other words W⊥ ⊆ KerT = V0.

Since we were within C(X) and estimated by the sup norm we get

Corollary 5.3.14. The range, Tk(L2(X)), can be expanded in a uni-formly convergent series of eigenfunctions of Tk with λi 6= 0.

Tk(f) =∞∑

i=1

fλi.

Moreover each eigenfunction φ of Tk associated with a nonzero eigen-value is continuous because T (φ) = λφ and T (φ) is continuous, henceso is φ.

This completes our study of compact operators. We remark that us-ing the spectral theorem for compact integral operators proven above,one can also get the following theorem which is important in the studyof compact Riemann surfaces or, more generally, compact hyperbolicmanifolds of higher dimension. Here G is a non-compact simple Liegroup and H is the (discrete) fundamental group of the compact man-ifold. For the details the reader is referred to Representation Theoryand Automorphic Functions by I.M. Gelfand et.al. The definition ofinduced representations is given in the last section of this chapter.

Theorem 5.3.15. Let G be a locally compact group, H a closed sub-group with G/H compact and having a finite G-invariant measure. Let σbe a finite dimensional, continuous, unitary representation of H. Thenthe induced representation, Ind(H ↑ G,σ), decomposes into a count-able orthogonal direct sum of irreducible unitary representations, eachof finite multiplicity (but usually of infinite dimension).


5.4 The Peter-Weyl Theorem and its Conse-

quences

In order to prove the Peter-Weyl theorem it will be necessary to studya certain infinite dimensional representation called the left regular rep-resentation L which is defined as follows.

The representation space of L is L2(G) and the action is given byleft translation, Lg(f)(x) = f(g−1x), where g, x ∈ G and f ∈ L2(G).We leave it to the reader to check that this is well-defined on L2 andis a linear action. It is actually a unitary representation; that is eachLg is a unitary operator since 〈Lg(f1), Lg(f2)〉 = 〈f1, f2〉 for all g ∈ G.f1, f2 ∈ L2(G) because of the invariance of Haar measure.

L has another important feature. Namely it is continuous in thefollowing sense (called strong continuity). If gν → g inG and f ∈ L2(G),then Lgν (f) → Lg(f). First let f ∈ C(G). Then by compactness, f isuniformly continuous. So if ǫ > 0 then |f(g−1

ν x)−f(g−1x)| < ǫ wheneverg−1ν x(g−1x)−1 = g−1

ν xx−1g = g−1ν g ∈ U , where U is a sufficiently small

neighborhood of 1 in G. Hence ||Lgν (f) − Lg(f)||G ≤ ǫ and so also .||Lgν (f) − Lg(f)||2 ≤ ǫ. Now since Haar measure is regular, C(G) isdense in L2. So if f ∈ L2 we can choose f1 ∈ C(G) with ||f − f1|| < ǫ.Then

||Lgν (f) − Lg(f)||2 ≤ ||Lgν (f) − Lgν (f1)||2 + ||Lgν (f1) − Lg(f1)||2+ ||Lg(f1) − Lg(f)||2 ≤ 3ǫ

if g−1ν g ∈ U .Now let φ be a continuous non-negative function on G which is not

identically zero. We can make it symmetric (that is φ(x) = φ(x−1))by replacing φ by φ(x) + φ(x−1). We can also have

∫φdx = 1 by

normalizing i.e. replacing φ by φR

φdx. Let k(x, y) = φ(x−1y). Then

k is continuous and since φ is symmetric k(x, y) = k(y, x). Because φis real we see Tk is a compact self-adjoint Fredholm operator. In facthere Tk(f) is called the convolution of φ and f and this is precisely thetype of integral operator we are interested in. Let Ω be the set of allthe eigenfunctions associated with nonzero eigenvalues of all such Tk.Then Ω and therefore also its complex linear span, l.s.C(Ω), is contained

5.4 The Peter-Weyl Theorem and its Consequences 235

in C(G). We will now show that any continuous function on G is theuniform limit of some complex linear combination of such Tk. To do sorequires a lemma, sometimes called the approximate identity lemma.

Lemma 5.4.1. Let f ∈ C(G), ǫ > 0 and U be a symmetric neighborhoodof 1 in G so that |f(x) − f(y)| < ǫ, if x−1y ∈ U . Let φ be a functionas described above with a support contained in U . Then for all x ∈ G|f(x) − Tk(f)(x)| < ǫ.

Proof. |f(x) − Tk(f)(x)| = |f(x)∫G φ(y) −

∫G φ(x−1y)f(y)|. But

by invariance of Haar measure and the fact that φ non-negativethis is |f(x)

∫G φ(x−1y)dy −

∫G φ(x−1y)f(y)dy| ≤

∫G φ(x−1y)|f(x) −

f(y)|dy. Now because Suppφx−1 ⊆ xU , we see |f(x) − Tk(f)(x)| ≤ǫ∫Supp φx

φ(x−1y) = ǫ∫G φ(x−1y) = ǫ.

Proposition 5.4.2. l.s.C(Ω) is dense in C(G).

Proof. Let f ∈ C(G), ǫ > 0 and U be a symmetric neighborhood of 1in G sufficiently small so that |f(x) − f(y)| < ǫ, if x−1y ∈ U . Choosea neighborhood U1 of 1 so that U1 ⊆ U and by Urysohn’s lemma acontinuous function h : G → [0, 1] with Supph ⊆ U and h ≡ 1 onU1. Then this gives rise to a function φ as above with Suppφ ⊆ U . ByLemma 5.4.1 |f(x)−Tk(f)(x)| < ǫ for all x ∈ G so that ||f−Tk(f)||G ≤ ǫ.This proves the proposition since by spectral theorem Tk(f) itself is theuniform limit of a finite linear combination of nonzero eigenfunctionsassociated with Tk.

Our next lemma shows that L restricted to Vλ gives a finite dimen-sional continuous unitary representation of G.

Lemma 5.4.3. If Vλ is an eigenspace of such a Tk, where λ 6= 0, ThenVλ (which we know is a finite dimensional ⊆ C(G) ⊆ L2) is invariantunder L.

Proof. We know∫G φ(x−1y)ψ(y)dy = λψ(x), x ∈ G. we apply L and

replace x by g−1x. Then∫

Gφ((g−1x)−1y)ψ(y)dy = λψ(g−1x).


That is ∫

Gφ(x−1gy)ψ(y)dy = λψ(g−1x).

Applying left invariance, the first term is

∫

Gφ(x−1gy)ψ(y)dy =

∫

Gφ(x−1y)ψ(g−1y)dy,

therefore ∫

Gφ(x−1y)ψg(y)dy = λψg(x).

Now let ∆ be the set of all matrix coefficients of all finite dimensionalcontinuous unitary representations of G and R(G) be l.s.C∆. R(G) iscalled the space of representative functions on G. We leave it to thereader to check that R(G) is intrinsic to G and does not depend on thechoice of basis needed to get these matrices. Notice that R(G) is stableunder conjugation since if ρ is a finite dimensional continuous unitaryrepresentations of G so is ρ, its conjugate. If ρ is irreducible so is ρ.

Exercise 5.4.4. The reader should verify these statements. Particu-larly the irreducibility of ρ. Hint use Schur’s lemma.

We now show that l.s.CΩ ⊆ R(G) and R(G) is uniformly dense inC(G). To do so only requires the following.

Lemma 5.4.5. Ω ⊆ R(G)

Proof. Let f ∈ Ω. Then for some appropriate k, Tk(f) = λf , whereλ 6= 0. Choose an orthonormal basis, φ1, . . . , φn for Vλ. Then f =∑n

i=1 ciφi. Since Vλ is invariant under L we get Lg(φi) =∑n

i=1 ρij(g)φj .That is φi(g

−1x) =∑n

i=1 ρij(g)φj(x) for all g, x ∈ G. Taking x = 1and replacing g by its inverse tell us φi(g) =

∑ni=1 ρij(g

−1)φj(1) forall g ∈ G. But since ρ(g−1) = ρ(g)−1 = ρ(g)∗ and ρ∗ij = ρji we see

φi(g) =∑n

i=1 ρji(g)φj(1). Thus each φi ∈ R(G) and since this a linearspace so is f .


Corollary 5.4.6. For a compact topological group G, R(G) is uniformlydense in C(G). Also R(G) is dense in L2 (with respect to ‖ · ‖ norm.)

This yields the following which is also called the Peter-Weyl theorem.

Corollary 5.4.7. For a compact topological group G, R(G) separatesthe points of G

Proof. Let g 6= h ∈ G and suppose r(g) = r(h) for all r ∈ R(G).Choose a continuous real valued function f such that f(g) 6= f(h).Since R(G) is uniformly dense in C(G) we can choose a representativefunction r so that ||r − f ||G < 1

2 |f(g) − f(h)|. Since |r(g) − f(g)|and |r(h) − f(h)| ≤ 1

2 |f(g) − f(h)|, it follows that |f(g) − f(h)| ≤|f(g)−r(g)|+|r(g)−r(h)|+|r(h)−f(h)| < |f(g)−f(h)|, a contradiction.Now there must be an irreducible representation ρ ∈ R(G) satisfyingρ(g) 6= ρ(h). For otherwise by complete reducibility all continuous finitedimensional unitary representations would take the same value on g andh. Hence r(g) = r(h) for all r ∈ R(G), a contradiction.

Corollary 5.4.8. A compact topological group G is isomorphic to aclosed subgroup of a product of unitary groups. Conversely, such a groupis compact.

Proof. For each ρ ∈ R(G) we get a unitary representation ρ : G →Uρ. Putting them together gives a continuous homomorphism G →Πρ∈R(G) Uρ, a product of unitary groups. Since R(G) separates thepoints of G this map is injective. By compactness G is homeomorphic(and isomorphic) to its image which is closed. The converse is obvious.

Corollary 5.4.9. Given a compact topological group G and a neighbor-hood U of 1 there is a closed normal subgroup HU of G contained in Usuch that G/HU is isomorphic to a closed subgroup of some U(n,C).

Proof. Now G \ U is closed and therefore compact. By Corollary 5.4.7each g ∈ G\U has ρg ∈ R(G) so that ρg(g) 6= I. By continuity of ρg thereis a neighborhood, Vg of g where ρg is never the identity anywhere on Vg.Since these Vg cover G \ U we have by throwing in U an open covering


of G itself. By compactness G = U ∪ Vgi, the union of a finite number

of these. Consider the corresponding ρgi. Let HU = ∩iKer ρgi

. ThenHU is a closed normal subgroup of G. Let ρ = ⊕iρgi

. Then ρ is a finitedimensional unitary representation, Ker ρ = HU and G/HU = ρ(G) isa closed subgroup of some unitary group. Let g ∈ HU . If g is not inU then g ∈ Vgi

for some i. But then ρgi(g) 6= Iρgi

, On the other handsince g ∈ HU , ρ(g) = Iρ and hence ρgi

(g) = Iρgi, a contradiction.

Corollary 5.4.10. A compact Lie group G is isomorphic to a closedsubgroup of some U(n,C) and conversely.

Proof. This follows immediately from Corollary 5.4.9 since G has nosmall subgroups. That is, there is some U containing only the subgroup1. Hence the HU is trivial and so G is isomorphic to a closed subgroupof some U(n,C). The converse follows from Cartan’s theorem.

We make a few final remarks concerning the abelian case. Herethe irreducibles are all 1-dimensional. That is, they are multiplicativecharacters χ : G → T and enables us to sharpen the conclusions of thePeter-Weyl theorem in this case.

The next corollary shows that if a compact group has only 1-dimensional irreducible representations in must be abelian since it isembedded in an abelian group.

Corollary 5.4.11. For a compact abelian topological group G, Thecharacters separate the points and the linear span of the characters isuniformly dense in C(G). G is isomorphic to a closed subgroup of aproduct of tori and if G is a Lie group it is isomorphic to a closedsubgroup of a torus.

We can now study the (in general infinite dimensional) left regularrepresentation, L.

Definition 5.4.12. If ρ is a finite dimensional irreducible representa-tion of the compact group G we denote by R(ρ) the linear span of thecoefficients of ρ i.e. the representative functions associated with ρ.


R(ρ) is a subspace of R(G) ⊆ C(G) ⊆ L2(G). By Section 5.2 di-mension is d2

ρ. If ρ and σ are distinct in R(G) then R(ρ) and R(σ)are orthogonal (see Section 5.2). Now the linear span of all R(ρ), asρ ∈ R(G), is R(G). Hence by the Peter-Weyl theorem we have

Corollary 5.4.13. L2(G) =⊕

ρ∈R(G)R(ρ) with d12ρ ρij as an or-

thonormal basis.

We can look at R(ρ) in another way as follows.

Proposition 5.4.14. R(ρ) is both a left and right invariant subspace ofL2. In particular, R(ρ) is also invariant under inner automorphisms.

Proof. We prove left invariance. Right invariance is done similarly. Letr(x) =

∑cijρij(x), cij ∈ C, be a generic element of R(ρ). Since L is a

linear representation its sufficient to show Lgρij ∈ R(ρ). But ρ(g−1x) =ρ(g−1)ρ(x) so ρij(g

−1x) =∑

k ρik(g−1)ρk,j(x) ∈ R(ρ).

Thus we have decomposed L2 into the orthogonal direct sum ofperhaps a large number of finite dimensional (closed) left invariant sub-spaces. In order to completely analyze L we merely need to know whichirreducibles occur in each of the R(ρ).

Now if τ is a finite dimensional representation of G on V and ρis an irreducible representation of G, then [τ : ρ], the multiplicitythat ρ occurs in τ , is given in Corollary 5.5.5 below by 〈χτ , χρ〉 =∫G χτ (x)χρ(x)dx. In our case τ = L|R(ρ). It can be easily checked

that χL|R(ρ)(g) = dρχρ. We also saw that ρ ∈ R(G) if ρ is. Thus the

multiplicity of ρ in L|R(ρ) is dρ = dρ. Since dimCR(ρ) = d2ρ it follows

that L|R(ρ) contains only ρ with multiplicity dρ and nothing else. Sinceρ then occurs in L|R(ρ) with multiplicity dρ = dρ we see

Corollary 5.4.15. Each irreducible of G occurs in L with a multiplicityequal to its degree.

In particular, if the group is finite one has

Corollary 5.4.16. A finite group has a finite number of inequivalentfinite dimensional irreducible representation ρ1, . . . ρr. These are con-strained by the requirement

∑ri=1 d

2ρi

= |G|


Example 5.4.17. Let G = S3, the symmetric group on 3 letters. Thisgroup has two 1-dimensional characters. These are the characters ofS3/A3 = Z2 lifted to G. It has no others since [S3, S3] = A3. S3 musthave an irreducible of degree d > 1 for otherwise it would be abelian see1.4.20. Since 12 + 12 + 22 = 6, the order of S3, we see |R(S3)| = 3 andthe higher dimensional representation has degree 2.

If we consider the two-sided regular representation of G on L2 (theHaar measure is both left and right invariant and left and right trans-lations commute) then a similar analysis shows that this representationon R(ρ) is now actually irreducible and equivalent to ρ ⊗ ρ. This canbe done by calculating the character of this representation (see begin-ning of the next section). Here R(ρ) is identified with Vρ ⊗ Vρ andχρ⊗ρ(g, h) = χρ(g)χρ(h), g, h ∈ G. Applying Proposition 5.5.4 showsthese representations are equivalent and Corollary 5.5.6 that they areirreducible. We leave this verification to the reader an exercise.

We now turn to the Plancherel theorem. Let ρ ∈ R(G) and φ be anL1 function. We define Tφ(ρ) =

∫G φ(g)ρ(g)dg. Thus Tφ(ρ) is a linear

operator on Vρ. It is called the Fourier transform of φ at ρ and so eachfixed φ gives an operator valued function on R(G), but always taking itsvalue in a different space of operators. Since φ ∈ L1 and the coefficientsof ρ are bounded Tφ(ρ) always exists.

Now let φ and ψ ∈ L2(G). We want to calculate 〈φ,ψ〉L2by Fourier

analysis. This is exactly what the Plancherel theorem does

〈φ,ψ〉L2 =∑

ρ∈R(G)

dρ tr(Tφ(ρ)Tψ(ρ)∗),

where ∗ means the adjoint operator. Since L2 ⊆ L1 the Fourier trans-form applies. To prove this using polarization we may take ψ = φ. Thenwe get the following formula involving the Hilbert Schmidt norm of anoperator.

||φ||22 =∑

ρ∈R(G)

dρ tr(Tφ(ρ)Tφ(ρ)∗).


Matrix calculations similar to those involved in the orthogonalityrelations themselves yield

dρ tr(Tφ(ρ)Tφ(ρ)∗) =

dρ∑

i,j=1

|d12ρ

∫

Gφ(g)ρij(g)dg|2.

Since by Corollary 5.4.13 d12ρ ρij form an orthonormal basis, the

claim follows from the identity

||φ||22 =∑

〈φ, d12ρ ρij〉2.

Corollary 5.4.18. Let G be a compact Lie group and ρ0 be a faith-ful finite dimensional unitary representation as guaranteed by Corollary5.4.10. Then each irreducible representation ρ of G is an irreduciblecomponent of ⊗nρ0 ⊗m ρ0 for some n and m non-negative integers.

Proof. Consider the representative functions F associated with irre-ducible subrepresentations of ⊗nρ0 ⊗m ρ0 as n and m vary. Since ρ0 isfaithful F separates the points. It is clearly stable under conjugation,contains the constants and is a subalgebra of C(G). By the Stone-Weierstrass theorem F is dense in C(G). Let σ ∈ R(G). If σ is notequivalent to some irreducible component of ⊗nρ0 ⊗m ρ0 for some nand m then R(σ) ⊥ F . On the other hand given σij we can choosefν → σij uniformly on G. Therefore 〈fν , σij〉 = 0 → 〈σij, σij〉 6= 0, acontradiction.

We conclude this section with the following result concerning infinitedimensional representations of a compact group

Theorem 5.4.19. Let γ be a strongly continuous unitary representationof a compact group, G, on a complex Hilbert space, V . Then γ is thedirect sum of finite dimensional, continuous, irreducible unitary sub-representations.

Notice that here the multiplicities need not be finite. Also observethat it follows from Theorem 5.4.19 that irreducible unitary represen-tations of a compact group on a Hilbert space are finite dimensional.


Before turning to the proof of this result we extend the definitionof the Fourier transform to the case of a strongly continuous unitaryrepresentation, γ on a complex Hilbert space V . For f ∈ L1(G) defineTf (γ) =

∫G f(x)γxdx. This is the integral of an operator valued func-

tion. Hence if the integral exists, the result is an operator. This integraldoes indeed exist since f ∈ L1 and the coefficients, x 7→ 〈γx(v), w〉,(v,w ∈ V ) are all bounded. Hence this operator has the propertythat 〈Tf (γ)(v), w〉 =

∫G f(x)〈γx(v), w〉dx. Also, since γ is unitary and

Tf (γ)(v) =∫G f(x)γx(v)dx. It follows that for v ∈ V ,

||Tf (γ)(v)|| ≤ ||f ||1||v||. (5.1)

Lemma 5.4.20. For each f ∈ L1, Tf preserves all γ-invariant sub-spaces of V .

Proof. Let W be an invariant subspace and W⊥ be its orthocomple-ment. We want to prove that if w ∈ W , then Tf (γ)(w) ∈ W . Thatis, 〈Tf (γ)(w), w⊥〉 = 0 for all w⊥ ∈ W⊥. But 〈Tf (γ)(w), w⊥〉 =∫G f(x)〈γx(w), w⊥〉dx. Since γx(w) ∈ W for all x ∈ G and w⊥ ∈ W⊥

the integrand is zero.

Proof of Theorem 5.4.19. Ordering the set of all orthonormal, finitedimensional, irreducible, γ-invariant subspaces of V by inclusion andapplying Zorn’s lemma shows there is a maximal such set. Let W bethe closure of the subspace generated by the subspaces in this maximalset. We want to show that W = V . In any case W is a γ-invariantsubspace. Hence since γ is unitary the orthocomplement W⊥ is alsoa γ-invariant subspace (by the same argument we used for the finitedimensional case). Choose an family of functions (approximate iden-tity), fU , consisting of continuous non-negative functions on G withSuppfU ⊆ U and

∫G fUdx = 1, as in the beginning of Section 5.4.

By Lemma 5.4.20 for each U , TfUpreserves W⊥. Hence if v ∈ W⊥,

TfUv ∈W⊥ for all U . Let us assume there is such a nonzero v.

Now let v1 ∈ V . By the Schwarz inequality |〈TfU(v) − v, v1〉| ≤∫

G |fUγx(v) − v)|dx||v1||. Since fU is non-negative, is supported on Uand has integral 1 we get |〈TfU

v − v, v1〉| ≤ supx∈U ||γx(v) − v||||v1||.

5.5 Characters and Central Functions 243

Taking the sup over all v1 with ||v1|| ≤ 1 we conclude that ||TfUv−v|| ≤

supx∈U ||γx(v)−v||. Hence by strong continuity of γ, ||TfUv−v|| → 0 as

U shrinks to 1. (The reader will notice the similarity with Lemma 5.4.1).Finally, since v 6= 0, TfU

v 6= 0 for some small U . Next we apply thePeter-Weyl theorem to uniformly approximate fU by a representativefunction r ∈ R(G) to within ǫ

||v|| . Since this is also an approximation in

L2 we see ||TfUv−Tr(v)|| = ||TfU−rv||. The latter is ≤ ||fU − r||1||v|| ≤

||fU − r||2||v|| < ǫ. Thus also Tr(v) 6= 0.Now the linear span of all left translates by x ∈ G of r lies in a

finite dimensional subspace F of L2(G) and so gives a finite dimensionalcontinuous unitary representation of G. Let f1, . . . fn be an orthonormalbasis of F . Then Lg(fi) =

∑nj=1 cij(g)fj . These functions all being in

L2(G) and hence also in L1(G). Therefore for g ∈ G and i = 1 . . . n,

γgTfi(v) = γg

∫

Gfi(x)γx(v)dx =

∫

Gfi(x)γgγx(v)dx =

∫

Gfi(x)γgx(v)dx.

Now by invariance of the integral under translation this is

∫

Gfi(g

−1x)γx(v)dx = TLg(fi)(v) = TPnj=1 cij(g)fj

(v) =

n∑

j=1

cij(g)Tfj(v).

Hence, TF (v) is a finite dimensional γ-invariant subspace of V . whichlies in W⊥ by Lemma 5.4.20. Moreover, it is nontrivial since Tr(v) 6= 0.This means it lies in W , a contradiction.

5.5 Characters and Central Functions

Definition 5.5.1. Let ρ be a finite dimensional, continuous, unitaryrepresentation of G. We shall call χρ(g) = tr(ρ(g)), g ∈ G the characterof ρ. Then χρ : G→ C is a bounded continuous function on G.

Exercise 5.5.2. Why is χρ bounded? Where does it take its largestabsolute value?

Corollary 5.5.3. Let ρ and σ be a finite dimensional, continuous, ir-reducible, unitary representation of G. Then 〈χρ, χσ〉 = 0 if ρ and σ areinequivalent and 〈χρ, χρ〉 = 1.


Thus the set X (G) consisting of the characters of the irreducibleunitary representations of G form an orthonormal family of functionsin L2(G). In particular they are linearly independent.

Proof. We have,

dρ∑

i=1

ρii(g)(

dσ∑

j=1

σjj(g) =

dρ∑

i=1

dσ∑

j=1

ρii(g)σjj(g).

Hence ∫

Gχρχσdg =

dρ∑

i=1

dσ∑

j=1

∫ρii(g)σjj(g)dg.

This is clearly 0 if ρ and σ are inequivalent. Now if σ = ρ, then

||χρ||22 =

dρ∑

i=1

dρ∑

j=1

∫

Gρii(g)ρjj(g)dg.

If i 6= j we get 0. Hence ||χρ||22 =∑dρ

i=1

∫G ρii(g)ρii(g)dg = 1 by Theorem

5.2.1.

Proposition 5.5.4. Let ρ and σ be finite dimensional, continuous, uni-tary representation of G. Then ρ and σ are equivalent if and only if theyhave the same character.

Proof. Evidently, equivalent representations have the same character.We now suppose χρ ≡ χσ. Decompose ρ and σ into irreducibles. ρ =∑r

i=1 niρi, ni > 0 and σ =∑s

i=kmiρi, mi > 0.After renumbering we can consider the overlap to be from ρk . . . ρr.

Then

0 = χρ − χσ =k−1∑

i=1

niχρi+

r∑

i=k

(ni −mi)χρi+

s∑

i=r+1

−miχρi.

Since the χρiare linearly independent we conclude ni = 0 for i =

1 . . . k−1, ni = mi for i = k . . . r and −mi = 0, i = r+1 . . . s. But since


the ni and mi are positive k = 1, r = s and ni = mi for all i = 1 . . . r.That is, ρ and σ are equivalent.

The next result follows from the orthonormality of X (G) in a similarmanner. We leave its proof to the reader an exercise.

Corollary 5.5.5. Let ρ be a finite dimensional, continuous, unitary rep-resentation of G whose decomposition into irreducibles is ρ =

∑ri=1 niρi,

ni > 0. Then ||χρ||22 =∑r

i=1 n2i . In particular ρ is irreducible if and

only if ||χρ||22 = 1. Moreover the multiplicity of an irreducible ρi in ρ is〈χρ, χρi

〉.

We can use our irreducibility criterion to study tensor product rep-resentations. Let G and H be compact groups and ρ and σ are finite di-mensional continuous representations of G andH respectively. Form therepresentation ρ⊗σ of G×H on Vρ⊗Vσ by defining ρ⊗σ(g, h) = ρg⊗σh.

We leave it to the reader to check that this is a continuous finitedimensional representation of G×H. It is actually unitary, but this doesnot matter since everything is equivalent to a unitary representationanyway.

Corollary 5.5.6. If ρ ∈ R(G) and σ ∈ R(H), then ρ⊗σ ∈ R(G×H).Conversely, all irreducibles of G×H arise in this way.

Proof. If dg and dh are normalized Haar measures on G and H respec-tively then dgdh is normalized Haar measure on the compact groupG×H. Now χρ⊗σ(g, h) = χρ(g)χσ(h). Hence

||χρ⊗σ||22 =

∫

G

∫

Hχρ(g)χσ(h)χρ(g)χσ(h)dgdh = ||χρ||22||χσ||22 = 1·1 = 1,

proving the irreducibility.

Now let τ ∈ R(G × H) and consider, as before, all ρ ⊗ σ whereρ ∈ R(G) and σ ∈ R(H). Let f ∈ C(G × H). By the Stone-Weierstrass theorem f can be uniformly approximates by the functionsof the form

∑ni=1 gi(x)hi(y), where gi ∈ C(G) and hi ∈ C(H). But by

the Peter-Weyl theorem these in turn can be uniformly approximated


by φ(x, y) =∑n

i=1 ri(x)si(y), where ri ∈ R(G) and si ∈ R(H). HenceΦ the collection of these φ’ are the representative functions of the ir-reducibles representations of G ×H which are of the form ρ⊗ σ, forma uniformly dense linear subspace of C(G ×H). If an irreducible rep-resentation τ is not of the form ρ ⊗ σ, then its coefficients must beperpendicular to Φ and therefore to all of L2(G ×H). In particular itmust be orthogonal to itself, a contradiction.

Definition 5.5.7. A function f : G→ C is called a central function ora class function if f(xy) = f(yx) for all x, y ∈ G. Equivalently, f is aclass function if and only if f(gxg−1) = f(x) for all g, x ∈ G. That is, fis constant on conjugacy classes of G. We denote the central functionsby C(G)G.

Exercise 5.5.8. Show these two definitions are equivalent.

Obvious examples of class functions are characters of finite dimen-sional, continuous representations ρ of G and since a linear combinationof a class function is again such a function, the linear span of all char-acters is a class function. Pursuing this idea somewhat further its quiteclear that the uniform limit (even the pointwise limit if the limitingfunction is continuous) of class functions is again a class function. Thuswe know that the elements of X (G) are class functions. It turns outthat the converse is also true. Namely,

Theorem 5.5.9. Every central function is a uniform limit of functionsin X (G) and conversely.

Before turning to the proof we need a pair of lemmas.

Lemma 5.5.10. Let ρ ∈ R(G) and r ∈ R(ρ). If r is central, thenr = λχρ, where λ ∈ C.

Proof. Here r(x) =∑dρ

i,j=1 cijρij(x). Since r(x) = r(gxg−1) we concludeupon substituting and taking into account the linear independence ofρij that C = ρ(g)tCρ(g). Alternatively, ρ(g)C = Cρ(g) for all g ∈ G,where C is the matrix of cij . By Schur’s lemma C is a scalar multipleof the identity and hence so is C.


Lemma 5.5.11. Suppose that f ∈ R(G) is central then f is in thelinear span of X (G).

Proof. f =∑n

i=1 ri, where ri ∈ R(ρi) and the ρi are distinct in R(G).Applying the assumption f(gxg−1) = f(x) and taking into account thelinear independence of the ri and Proposition 5.4.14 tells us each ri isitself a class function. Hence by Lemma 5.5.10 each ri = λiχρi

andf =

∑ni=1 λiχρi

.

Proof of Theorem 5.5.9. We first define a projection operator # :C(G) → C(G)G via the formula f#(x) =

∫G f(gxg−1)dg. It is easy

to see that this gives a continuous function f#, the operator is normdecreasing ||f#||G ≤ ||f ||G and f is central if and only if f = f#. Weleave these details to the reader to check.

Let f ∈ C(G)G. Then by the Peter-Weyl theorem f can be uniformlyapproximated on all of G by representative functions, φ, ||f − φ||G < ǫ.Apply the # operator and get ||f# − φ#||G = ||(f − φ)#||G ≤ ||f −φ||G < ǫ. On the other hand f = f# and φ# is a central representativefunction which by Lemma 5.5.11 is a linear combination of characters ofR(G). Thus f is the uniform limit of a linear combination of irreduciblecharacters.

Exercise 5.5.12. The reader should verify the various properties of# mentioned above as these are necessary to complete the proof ofTheorem 5.5.9.

Corollary 5.5.13. X (G) separates the conjugacy classes of G.

Proof. Let Cx and Cy be disjoint conjugacy classes. Since these aredisjoint compact sets Urysohn’s lemma tells us there is f ∈ C(G) withf |Cx = 0 and f |Cy = 1. Applying # yields f#|Cx = 0 and f#|Cy = 1.Now approximate f# by a linear combination of characters to within 1

2by Theorem 5.5.9. If χρ(x) = χρ(y) for every ρ ∈ R(G) this would givea contradiction. Hence the conclusion.

Thus the irreducible representations of a compact group are in bi-jective correspondence with the irreducible characters and, if the group


is finite, the characters are in bijective correspondence with the set ofconjugacy classes. This is the basis of so called character tables of afinite group. Vertically the characters are listed and horizontally theconjugacy classes are listed. Then the table must be filled in with thevalue of that character on that particular conjugacy class. For exam-ple as we saw above, S3 has exactly 3 characters and therefore also 3conjugacy classes.

We conclude this section with the functional equation for a characterof a representation in R(G).

Theorem 5.5.14. Let f = χρ be the character of a finite dimensional,continuous, irreducible, unitary representation ρ. Then for all x, y ∈ G,

f(x)f(y) = f(1)

∫

Gf(gxg−1y)dg.

Conversely, if f is a continuous function G → C, not identically zerosatisfying this equation, then f

f(1) =χρ

χρ(1) , for a unique ρ ∈ R(G).

Proof. We extend the # operator defined earlier on functions to rep-resentations. For any finite dimensional continuous unitary representa-tion, ρ, let ρ#(x) =

∫G ρ(gxg

−1)dg, giving an operator valued functionon G. For y ∈ G using invariance of dg we get

ρ(y)ρ#(x)ρ(y)−1 =

∫

Gρ(y)ρ(gxg−1)ρ(y)−1dg =

∫

Gρ((gy)x(gy)−1)dg

=

∫

Gρ(gxg−1)dg = ρ#(x).

Thus ρ#(x) is an intertwining operator. If ρ is irreducible ρ#(x) =

λ(x)I and taking traces shows λ(x) = tr(ρ#(x))dρ

. On the other hand,

tr(ρ#(x)) =

∫

Gtr(ρ(gxg−1))dg = χρ(x),

so that for all x ∈ G,∫

Gρ(gxg−1)dg =

χρ(x)

dρI.


Hence ∫

Gρ(gxg−1y)dg =

χρ(x)

dρρ(y).

Taking traces yields the functional equation

∫

Gχρ(gxg

−1y)dg =χρ(x)χρ(y)

dρ.

Conversely, let f be an arbitrary continuous function satisfying thefunctional equation. From it we see f(1) 6= 0 for otherwise f ≡ 0.Let y = 1 in the equation. Then f(x)f(1) = f(1)

∫G f(gxg−1)dg =

f(1)f#(x). Since f(1) 6= 0 f(x) = f#(x) so f is central. We will showthat for every ρ ∈ R(G) and every x ∈ G,

f(x)

f(1)〈f, χρ〉 =

χρ(x)

χρ(1)〈f, χρ〉. (5.2)

Having done so we complete the proof by choosing a ρ ∈ R(G) suchthat 〈f, χρ〉 6= 0. For then we can cancel and conclude from (5.2) that

f(x)

f(1)=χρ(x)

χρ(1). (5.3)

Since f is central such a ρ must exist by Theorem 5.5.9. The ρ satisfying(5.3) is unique because the characters of distinct representations arelinearly independent. It remains only to prove (5.2). To do so consider∫G

∫G f(gxg−1y)χρ(y)dgdy. By hypothesis this is

∫

G

∫

Gχρ(y)

f(x)f(y)

f(1)dgdy =

f(x)

f(1)

∫

Gf(y)χρ(y)dy =

f(x)

f(1)〈f, χρ〉.

On the other hand by Fubini’s theorem, left translating

∫

G

∫

Gf(gxg−1y)χρ(y)dgdy =

∫

G(

∫

Gf(gxg−1y)χρ(y)dy)dg

=

∫

G(

∫

Gf(y)χρ(gx−1g−1y)dy)dg.


Using χρ(t−1) = χρ(t), the latter is

∫

G

∫

Gf(y)χρ(y

−1gxg−1)dydg =

∫

Gf(y)(

∫

Gχρ(y

−1gxg−1)dg)dy

=

∫

Gf(y)(

∫

Gχρ(gxg

−1y−1)dg)dy

By the part of the theorem already proved this is just

∫

Gf(y)

χρ(x)χρ(y−1)

χρ(1)dy =

χρ(x)

χρ(1)

∫

Gf(y)χρ(y)dy

orχρ(x)

χρ(1)〈f, χρ〉.

5.6 Induced Representations

We now study induced representations of a compact group, G. Recall(see Theorem 2.3.5) that if H is a closed subgroup of G and dg anddh are the respective normalized Haar measures, then there is a (finite)G-invariant measure µ on the homogeneous space G/H satisfying

∫

Gf(g)dg =

∫

G/H

∫

Hf(gh)dhd(µ).

Now suppose we have a finite dimensional representation σ of Hon Vσ. We now define the induced representation of σ to G. Thisrepresentation, written Ind(H ↑ G,σ), will be infinite dimensional unlessH has finite index in G. We consider only finite dimensional, σ, toavoid technical difficulties and because most of the applications we areinterested in are in this situation.

Consider the vector spacecW consisting of functions F : G→ Vσ satisfying

(1) F is measurable,

(2) F (gh) = σ(h)−1F (g), for h ∈ H and g ∈ G,

5.6 Induced Representations 251

(3)∫G/H ||F (g)||2Vσ

d(µ)(g) <∞.

Such functions clearly form a complex vector space under pointwiseoperations. If 〈·, ·〉Vσ

denotes the Hermitian inner product on Vσ we canuse this to define an inner product on this space as follows. For F1 andF2 here, the function g 7→ 〈F1(g), F2(g)〉Vσ

is a continuous function onG, which by condition 2 descends to a function on G/H. In particular,g 7→ ||F (g)||2Vσ

is a non-negative measurable function on G/H.

Now W is actually a Hilbert space whose inner product is given by〈F1, F2〉 =

∫G/H 〈F1(g), F2(g)〉Vσ

d(µ)(g). This inner product convergesby the Schwarz inequality which comes built in.

∫

G/H〈F1(g), F2(g)〉Vσ

d(µ)(g) ≤∫

G/H||F1(g)||2Vσ

d(µ)(g)

∫

G/H||F2(g)||2Vσ

d(µ)(g)

Now let G act on W by left translation (x · F )(g) = F (x−1g), whereF ∈ W and x, g ∈ G.

Proposition 5.6.1. Ind(H ↑ G,σ) is a unitary representation of G onW

Exercise 5.6.2. The proof of this is routine and is left it to the reader.We also leave to the reader to check that the left regular representation,L acting on L2, is an induced representation. Here σ is the trivial 1-dimensional representation of H = 1. (This is a good example of aninduced representation to keep in mind).

We now show W contains a certain dense set of functions to bedescribed below. Let f ∈ C(G,Vσ), the continuous vector valued func-tions on G and define for x ∈ G, Ff (x) =

∫H σ(h)f(xh)dh. Since the

integrand is a Vσ valued continuous function on H which is compact theintegral exists and is a Vσ valued function Ff : G→ Vσ on G.

Lemma 5.6.3. The Ff are continuous and in W.


Proof. We prove 2). After we show Ff is continuous, 1) and 3) followautomatically.

Ff (gh1) =

∫

Hσ(h)f(gH1h)dh =

∫

Hσ(h−1

1 h)f(gh)dh

=

∫

Hσ(h−1

1 )σ(h)f(gh)dh = σ(h−11 )

∫

Hσ(h)f(gh)dh.

Thus Ff (gh1) = σ(h−11 )Ff (g) proving 2).

Now since f is uniformly continuous given ǫ > 0 there is a neighbor-hood U of 1 in G so that ||f(xh) − f(yh)||Vσ < ǫ, whenever h ∈ H andxy−1 ∈ U . Therefore

||Ff (x) − Ff (y)||Vσ ≤∫

H||σ(h)|||f(xh) − f(yh)||Vσdh.

Since σ is unitary we see if xy−1 ∈ U , then ||Ff (x)− Ff (y)||Vσ < ǫ.

Lemma 5.6.4. The Ff are dense in W.

Proof. Clearly the continuous functions in W form a dense subspace ofW. We will actually show that the Ff are not only dense, but actu-ally comprise all continuous functions in W. Let F1 be any continuousfunction satisfying 2). We want to find an f ∈ C(G,Vσ) so that

||F1 − Ff ||2 =

∫

G/H〈Ff − F1, Ff − F1〉dµ(g)

is small. Now F1(g) = σ(h)F1(gh), so

F1(g) =

∫

HF1(g)dh =

∫

Hσ(h)F1(gh)dh.

On the other hand, Ff (g) =∫H σ(h)f(gh)dh. Therefore since σ is

unitary ||F1 − Ff ||2 =∫G/H

∫H ||f(gh) − F1(gh)||2Vσ

dhdµ(g). Takingf = F1, then this last integral is zero so Ff = F1. Since they are bothcontinuous they are identically equal on G. Therefore the Ff consist ofall continuous functions in W. and hence they are dense in W.

5.6 Induced Representations 253

Corollary 5.6.5. Ind(H ↑ G,σ) is a strongly continuous unitary rep-resentation of G on W.

Proof. We will show that if gν → g, and F ∈ W is fixed, then ||Lgν (F )−Lg(F )|| → 0. Since Ind(H ↑ G,σ) is a unitary representation we have||Lgν (F )−Lg(F )|| = ||Lg−1gν

(F )−F || so we may as well assume g = 1.Also if we were to prove this for all Ff , then by density it would hold forall F ∈ W. We leave this to be checked by the reader. (It is essentiallythe same argument as in the one for the regular representation in thethird paragraph of section 4). Thus we may assume F is continuous.Hence F is uniformly continuous and ||F (x−1g)−F (g)||2Vσ

< ǫ2, if x ∈ Ua neighborhood of 1 and g ∈ G. Then if x ∈ U ,

||Lx(F ) − F ||2 =

∫

G/H||LxF (g) − F (g)||2Vσ

< ǫ2µ(G/H).

So ||Lx(F ) − F || < ǫ.

Since by Theorem 5.4.19 any continuous unitary representation of acompact group on a Hilbert space is the direct sum of finite dimensionalcontinuous irreducible unitary representations. In particular,

Corollary 5.6.6. Ind(H ↑ G,σ) is a direct sum of finite dimensional,continuous, irreducible unitary representations of G on W

Exercise 5.6.7. The following is a consequence of the fact that a con-tinuous function F : G → Vσ which satisfies condition 2) is determinedby its values on coset representatives of H in G and its proof is left tothe reader.

Corollary 5.6.8. (1) dimC W is infinite unless [G : H] is finite.

(2) If [G : H] is finite, then dimC W = [G : H] dimC Vσ.

The next result includes the possibility that γ could be induced, orfinite dimensional.

Proposition 5.6.9. Let G be a compact group and γ a strongly con-tinuous representation of G on a Hilbert space, V and ρ ∈ R(G). Then[γ : ρ] = dimC HomG(Vρ, V ).


Proof. By 5.4.19 we can write V as the orthogonal direct sum of finite di-mensional irreducible continuous unitary subrepresentations, (Vi, γ|Vi

),where i ∈ I and V =

⊕Vi. Let πi be the orthogonal projection of V

onto Vi. Partition I = I1 ∪ I2, where I1 contains those representationsequivalent to ρ, while I2 contains those representations which are notequivalent to ρ. For T ∈ HomG(Vρ, V ), πi T ∈ HomG(Vρ, Vi). So ifi ∈ I2, Schur’s lemma tells us πi T = 0, while if i ∈ I1, πi T is a scalarmultiple of the identity. Thus the former components have dimension 0while the latter have dimension 1. Let W be the closure of the sum ofthe Vi for i ∈ I1. Hence the dimension of HomG(Vρ, V ) is the same asthat of HomG(Vρ,W ) which is the cardinality of I1.

We now come to the Frobenius reciprocity theorem a particular caseof which stats that each irreducible representation ρ of G is containedin the induced from σ with the same multiplicity that its restrictioncontains the irreducible σ of H. In particular the multiplicity of ρ inthe induced is always finite.

Theorem 5.6.10. Let G be a compact group, H be a closed subgroup,σ be any finite dimensional continuous unitary representation of H andρ a finite dimensional continuous unitary representation of G. Then

HomG(ρ, Ind(H ↑ G,σ)) ≃ HomH(ρ|H , σ). (5.4)

In particular, these have the same dimension. Hence by Proposition5.6.9 if ρ and σ are each irreducible, then [Ind(H ↑ G,σ) : ρ] = [ρ|H : σ].

Proof. Our proof of this result is functorial, In this way it does notreally depend on compactness of G at all. For example it also works forany (not necessarily unitary, but) finite dimensional representations if[G : H] <∞. Nor does it depend on irreducibility!

We will prove (5.4) by constructing a vector space isomorphism be-tween them. Let T be a G-linear map T : Vρ → W. Then for eachvρ ∈ Vρ we know T (vρ) ∈ W and so T (vρ)(1) ∈ Vσ. This gives us a lin-ear map T ∗ from Vρ to Vσ. So T ∗ ∈ HomC(Vρ, Vσ). Moreover, T 7→ T ∗

is itself a C-linear map.

5.7 Some Consequences of Frobenius Reciprocity 255

Now let us consider how the action of H fits into this picture. Leth ∈ H. Then T ∗(ρh(vρ)) = T (ρh(vρ))(1) and since by 2) F (h−1) =σ(h)F (1) we get

T (ρh(vρ))(1) = LhT (vρ)(1) = T (vρ)(h−1) = σ(h)T (vρ)(1) = σ(h)T ∗(vρ).

This says T ∗ρh = σhT∗, for all h ∈ H. Thus T ∗ ∈ HomH(ρ|H , σ). We

now construct the inverse of this map.Let S ∈ HomH(ρ|H , σ) and define S∗ : Vρ → W by S∗(vρ)(g) =

S(ρ−1g (vρ)) ∈ Vσ. Since S∗(vρ) is a mapping from G to Vσ, it has a

chance of being in W. Because ρ is continuous as is S, we see S∗(vρ) iscontinuous and so measurable.

It also satisfies 2).

S∗(vρ)(gh) = S(ρ(gh)−1(vρ)) = S(ρ(h)−1ρ(g)−1(vρ))

= σ(h)−1S(ρ(g)−1(vρ)) = σ(h)−1S∗(vρ)(g).

Since G/H is compact and this function is continuous, it has finitesquare integrable norm. Thus S∗(vρ) ∈ W and so we have a linear mapS∗ : Vρ → W.

For g ∈ G, one checks easily that S 7→ S∗ is linear and LgS∗ = S∗ρg.Hence S∗ ∈ HomG(Vρ,W). It remains only to see that these maps invertone another.

Now

T ∗(vρ)∗(g) = T ∗(ρ(g)−1)(vρ) = Tρ(g)−1(vρ)(1)

= Lg−1T (vρ)(1) = T (vρ)(g).

Since this holds for all g ∈ G and vρ ∈ Vρ we conclude T ∗∗ = T . Also,

S∗(vρ)(1) = S(vρ). Hence (S∗)∗(vρ) = S(vρ) so (S∗)∗ = S.

5.7 Some Consequences of Frobenius Reci-

procity

Let SO(3,R) act on S2 with isotropy group SO(2,R). Then SO(3,R)also acts on C(S2) and therefore on L2(S

2). Hence (see 5.1.3) we can de-


compose this representation into irreducible components, called spheri-cal harmonics. An interesting question is then which spherical harmon-ics occur and with what multiplicities?

It is easy to verify directly that SU(2,C) is a compact real formof SL(2,C), that is the complexification of su(2,C) is sl(2,C). Thegeneral fact following from Corollary 7.4.10. Since SU(2,C) is simplyconnected its finite dimensional irreducible representations are the sameas those of su(2,C) by Corollary 1.4.15. Hence the finite dimensionalirreducible representations of SU(2,C) are in bijective correspondencewith those of sl(2,C), that is to say the positive integers by the degreeof the representation (see Section 3.1.5). Since SU(2,C) is two-sheetedcovering of SO(3,R), its irreducibles are those of odd degree.

Exercise 5.7.1. Show that the irreducibles of SU(2,C) which are trivialon ±id are those of odd degree.

Theorem 5.7.2. In the action of SO(3,R) on L2(S2) each irreducible

representation of SO(3,R) occurs and with multiplicity 1.

Proof. We know S2 = SO(3,R)/SO(2,R). Consider the trivial ir-reducible representation σ of SO(2,R). Then the representation ofSO(3,R) on L2(S

2) is Ind(SO(2,R) ↑ SO(3), σ). If ρ is an irreduciblerepresentation of SO(3), then [Ind(SO(2) ↑ SO(3), σ) : ρ] is the sameas [ρ|SO(2,R) : 1SO(2,R)]. But we know what the irreducibles of SO(3)are; these are just the irreducibles of SU(2,C) of odd degree, or whatis the same thing, the complex Lie algebra irreducibles of sl(2,C) ofodd degree. So the question is given an irreducible representation ofsl(2,C) of odd degree, how many times does its restriction to h (theline through H) contain the 0 representation of h? Our study of theserepresentations tells us the answer is 1.

We now study the relationship between the representations of G andthose of a proper subgroup H.

Proposition 5.7.3. Let G be a compact group and H a proper closedsubgroup. Then there exists ρ ∈ R(G) \ 1 whose restriction to Hcontains the 1-dimensional trivial representation. That is, there exists


v0 ∈ Vρ with ρh(v0) = v0 for all h ∈ H.

Proof. If ρ ∈ R(G) \ 1 the orthogonality relations show∫G ρij(x)1(x)dx = 0 for all i, j = 1 . . . dρ. Hence

∫G r(x)dx = 0 for

all r ∈ R(ρ) and all such ρ. For each such ρ, ρ|H is a direct sum ofirreducibles, σ1, . . . σm of H. If the statement of the proposition werefalse none of the σi would be 1H . If r ∈ R(ρ), then r|H is a linearcombination of the coefficients of the σi. By the orthogonality relationson H,

∫H σ

ikl(h)1(h)dh = 0 for all i. Hence

∫H r(h)dh = 0. Thus

∫

Gr(x)dx =

∫

Hr(h)dh for all r ∈ R(ρ), ρ ∈ R(G) \ 1. (5.5)

On the other hand consider ρ = 1G ∈ R(G). Here ρ|H = 1 and therepresentative functions associated with this are r(x) = λ · 1 = λ so∫G r(x)dx = λ =

∫H r(h)dh. Hence (5.5) holds for all ρ ∈ R(G). Now

let f ∈ C(G) and ǫ > 0. Choose r ∈ R(G) so that ||f − r||G < ǫ. Then

|∫

Hf(h)dh−

∫

Gf(g)dg| ≤ |

∫

Hf(h)dh−

∫

Hr(h)dh|+

|∫

Hr(h)dh−

∫

Gr(g)dg|+

|∫

Gr(g)dg −

∫

Gf(g)dg|

≤ 2ǫ

and since ǫ is arbitrary,∫G f(g)dg =

∫H f(h)dh for all f ∈ C(G). Now

H 6= G so there must be another coset, x0H. choose a neighborhood Uof x0H which is disjoint from H and a continuous non-negative real val-ued function, f , which is ≡ 1 on x0H and ≡ 0 on H. Then

∫G f(g)dg > 0

and∫H f(h)dh = 0, a contradiction.

Because of Frobenius reciprocity and the fact that [ρ|H : 1] ≥ 1 thisproposition has the following corollary.


Corollary 5.7.4. Let G be a compact group and H a proper closedsubgroup. Then there exists ρ ∈ R(G) \ 1 for which [Ind(H ↑ G, 1) :ρ] ≥ 1.

We will prove the following

Theorem 5.7.5. Let H be a closed subgroup of the compact group G.For each σ ∈ R(H) there is some ρ ∈ R(G) with [ρ|H : σ] ≥ 1.

Hence by Frobenius reciprocity we would get

Corollary 5.7.6. Let H be a closed subgroup of the compact Lie groupG. For each σ ∈ R(H) there is some ρ ∈ R(G) with [Ind(H ↑ G,σ) :ρ] ≥ 1.

Proof of Theorem 5.7.5. Let ρ0 be faithful representation of G (and

also H). Earlier 5.4.18 we proved that σ is a subrepresentation of ρ(n)0 ⊗

ρ−(m)0 |H for some choice of n and m. Therefore σ is a subrepresentation

of some irreducible component of ρ since ρ is an irreducible component

of some ρ(n)0 ⊗ ρ

−(m)0 . Therefore some irreducible component ρ of ρ

(n)0 ⊗

ρ−(m)0 must restrict to σ.

Exercise 5.7.7. Let G be a compact group and H be a closed sub-group.

(1) Each σ ∈ R(H) is an irreducible component of the restriction ρ|Hof some ρ ∈ R(G).

(2) If H happens to be a Lie group, the there is a finite dimensionalcontinuous representation ρ of G whose restriction to H is faithful.

(3) The restriction map R(G) → R(H) is surjective.

We conclude this chapter with the following result connected withequivariant imbeddings of compact G-spaces.

Theorem 5.7.8. Let H be a closed subgroup of the compact Lie groupG. Then there exists a finite dimensional continuous unitary represen-tation ρ of G on Vρ and a nonzero vector v0 ∈ Vρ so that H = StabG(v0).


We first need a lemma which tells us that the dimension togetherwith the number of components determines the size of a compact Liegroup.

Lemma 5.7.9. Let G be a compact Lie group and G ⊇ G1 ⊇ G2 . . . bea chain closed subgroups. Then this chain must eventually stabilize.

Proof. Since dimGi ≥ dimGi+1 . . . and all these dimensions are finite,then for i ≥ n0 all the dimensions must be constant. Hence for i ≥ n0

each Gi+1 is open in Gi which is itself open in Gn0 , the number ofcomponents of Gi+1 is ≤ the number of components of Gi which is ≤ thenumber of components of Gn0 Since Gn0 is closed in G it is compact andtherefore has a finite number of components. It follows that eventuallythese must also stabilize and hence the conclusion.

Proof of Theorem 5.7.8. We may assume H < G since if H = G wemay take ρ to be the trivial 1-dimensional representation and v0 anynonzero vector. We will now prove

(**) If g0 ∈ G−H, then there exists a representation ρ of G on V andv0 6= 0 ∈ Vρ such that ρg0(v0) 6= 0 and ρh(v0) = 0 for all h ∈ H.

Suppose we can do this. Then G ⊇ StabG(v0) ⊇ H and g0 is not inStabG(v0). Replacing G by the closed and therefore compact groupStabG(v0) we can apply (**) again to this subgroup. In this way we geta descending chain of closed subgroups terminating in H which mustterminate by Lemma 5.7.9. Therefore they must terminate in H. Thiswould prove the Theorem.

Proof of (**). Since H and Hg−10 are disjoint compact sets we can

find a continuous function f on G for which f |H = α and f |Hg−10

= β,

where α < β. Approximate f by r ∈ R(G) to within ǫ = β−α2 . Let

F (g) =∫H r(hg)dh. Then F is continuous and therefore in L2(G).

Since r ∈ R(G), F ∈ R(G) also. For h1 ∈ H,

F (h1) =

∫

Hr(hh1)dh =

∫

Hr(h)dh ≤ ǫ+ α.


So F |H ≤ ǫ+ α. On the other hand,

F (h1g−10 ) =

∫

Hr(hh1g

−10 )dh =

∫

Hr(hg−1

0 )dh ≥ β − ǫ.

So F |Hg−10

≥ β − ǫ. In particular, F (1) ≤ ǫ+ α and F (g−10 ) ≥ β − ǫ so

F (1) 6= F (g−10 ). Now apply L, the left regular representation of G on

L2. Hence because Lg0F (1) = F (g−10 ) we see Lg0F 6= F . On the other

hand,

Lh1(F )(g) = F (h−11 g) =

∫

Hr(hh−1

1 g)dh =

∫

Hr(hg)dh = F (g)

Thus Lh1(F ) = F for all h1 ∈ H. Since F ∈ R(G) it lies in a finitedimensional L-invariant subspace Vρ of C(G). So there is a finite dimen-sional continuous unitary representation ρ of G and a nonzero vector Fin it with ρh(F ) = F for all h ∈ H and ρg0(F ) 6= F , proving (**).

Chapter 6

Symmetric Spaces of

Non-compact type

6.1 Introduction

In this chapter we shall give an introduction to symmetric spaces ofnon-compact type. This subject, largely the creation of Elie Cartan(1869-1951), is of fundamental importance both to geometry and Lietheory. Indeed, one of the great achievements of the mathematics ofthe first half of the twentieth century was E. Cartan’s discovery of thefact that these two categories correspond exactly. Namely, given a con-nected, centerless, real semisimple Lie group G without compact factorsthere is associated to it a unique symmetric space of non-compact type.This is G/K, where K is a maximal compact subgroup of G and G/Ktakes the Riemannian metric induced from the Killing form of G. Con-versely, if one starts with an arbitrary symmetric space, X, none ofwhose irreducible constituents is either compact or Rn, then X = G/K,where G is the identity component of the isometry group of X. HereG is a centerless, real semisimple Lie group without compact factors.Thus, we have a bijective correspondence between the two categoriesand this fact underlies an important reason why differential geometryand Lie theory are so closely bound. As one might expect, this closerelationship between the two will show up in some of the proofs. For the

261

262 Chapter 6 Symmetric Spaces of Non-compact type

details of all this, see [32] and [61]. Also, [32] has a particularly conve-nient and useful early chapter on differential geometry. Concerning thiscorrespondence, the same may be said of Euclidean space and its groupof isometries, or of compact semisimple groups and symmetric spaces ofcompact type, which were also studied by E. Cartan. However, we shallnot deal with these here. Taken as a whole, Cartan’s work on symmetricspaces can be considered as the completion of the well-known “ErlangerProgram” first formulated by F. Klein in 1872. In particular, it tiestogether Euclidean, elliptic and hyperbolic geometry in any dimension.

Before turning to our subject proper it might be helpful to considera most important example, namely that of G = SL(2,R) and X thehyperbolic plane, which we view here as the Poincare upper half plane,H+, consisting of all complex numbers z = x+ iy, where y > 0. We letG act on H+ by fractional linear transformations, g · z = az+b

cz+d ,

g =

(a bc d

)

where a, b, c and d are real and det g = 1. Since I(az+bcz+d ) = I(z)|cz+d|2 > 0,

we see that g · z ∈ H+. It is easy to verify that this is an action. Nowthis action is transitive. Let c = 0, then a 6= 0 and d = 1

a . Theng · i = a2i+ ab. Evidently, by varying a > 0 and b ∈ R this gives all ofH+. A moment’s reflection tells us that the isotropy group, StabG(i),is given by a = d and c = −b. Since det g = a2 + b2 = 1, we see

StabG(i) = g : g =

(cos t sin t− sin t cos t

): t ∈ R.

On H+ we place the Riemannian metric ds2 = dx2+dy2

y2(meaning the

hyperbolic metric ds = dsEuc

I(z) ) and check that G acts by isometries on

H+ (for this see, for example, p. 118 of [55]). Since G is connected, itsimage, PSL(2,R), is contained in Isom0(H

+). Actually it is Isom0(H+)

but that will not matter. From the point of view of the symmetricspace it does not even matter whether we take SL(2,R) or PSL(2,R).However, we note that PSL(2,R), the group that is really acting, is thecenterless version.

6.1 Introduction 263

Another model for this symmetric space is the unit disk, D ⊆ C,

called the disk model. It takes the metric ds2 = 4dx2+dy2

(1−r2)2and has the

advantage of radial symmetry about the origin, 0. Here r is the usualradial distance from 0. The quantity 4, as we shall see, makes D isomet-ric with H+, or put another way, it normalizes the curvature on D to be−1. Now the Cayley transform c(z) = z−i

z+i maps H+ diffeomorphically

onto D. Its derivative is c′(z) = 2i

(z+i)2. A direct calculation shows that

for z ∈ H+

2|c′(z)|1 − |c(z)|2 =

1

I(z).

Using this we see that if w = c(z), then |dw| = |c′(z)||dz| and so

2|dw|1 − |w|2 =

2|c′(z)|1 − |c(z)|2 |dz| =

|dz||I(z)| .

Thus c is an isometry. Of course in the form of the disk, the group ofisometries and its connected component will superficially look different.

Example 6.1.1. The action of SL(2,R) on the upper half plane can begeneralized in two different ways. One is SO(n, 1) acting on hyperbolicn-space which will be discussed in detail in Section 6.4. The other is theSiegel generalized upper half space consisting of z = x + iy, where x ∈Symm(n,R), n×n the real symmetric matrices, and y ∈ Symm(n,R)+,the positive definite real symmetric matrices. The action of Sp(n,R) onZ is given by g · z = az+b

cz+d .Notice when n = 1, just as Sp(1,R) = SL(2,R), Z is the usual upper

half plane and the action is also the usual one. In general, since as weshall see Symm(n,R) and Symm(n,R)+ are diffeomorphic and each has

dimension n(n+1)2 , so Z has dimension n(n + 1). Finally, because a

maximal compact subgroup of Sp(n,R) is U(n,C), Sp(n,R)/U(n,C) =Z.

Exercise 6.1.2. Prove:

(1) g · z = az+bcz+d defines a transitive action of Sp(n,R) on Z.

(2) The isotropy group of i1n×n is U(n,C).

(3) U(n,C) is a maximal compact subgroup of Sp(n,R).


6.2 The Polar Decomposition

We shall begin by studying the exponential map on certain specificmanifolds.

As usual n×n complex matrices will be denoted by gl(n,C) and thereal ones by gl(n,R). Denote by H the set of all Hermitian matricesin gl(n,C) and by H the positive definite ones. It is easy to see thatH is a real (but not a complex!) vector space of dimn2. Similarly, wedenote by P the symmetric matrices in gl(n,R) and by P those that

are positive definite. P is a real vector space of dim n(n+1)2 . As we shall

see, H and P and certain of their subspaces will actually comprise allsymmetric spaces of non-compact type.

Proposition 6.2.1. P and H are open in P and H, respectively. Asopen sets in a real vector space each is, in a natural way, a smoothmanifold of the appropriate dimension.

Proof. Let p(z) =∑

i pizi and q(z) =

∑i qiz

i be polynomials of degreen with complex coefficients, let z1 . . . zn and w1 . . . wn denote their re-spective roots counted according to multiplicity and let ǫ > 0. It followsfrom Rouche’s theorem (see [55]) that there exists a sufficiently smallδ > 0 so that if for all i = 0, . . . , n, |pi − qi| < δ, then after a possiblereordering of the wi’s, |zi − wi| < ǫ for all i. Suppose H were not openin H. Then there would be an h ∈ H and a sequence xj ∈ H −H con-verging to h in gl(n,C). Since h is positive definite, all its eigenvaluesare positive. Choose ǫ so small that the union of the ǫ balls about theeigenvalues of h lies in the right half plane. Since the coefficients of thecharacteristic polynomial of an operator are polynomials and thereforecontinuous functions of the matrix coefficients and xj converges to h,for j sufficiently large, the coefficients of the characteristic polynomialof xj are in a δ- neighborhood of the corresponding coefficient of thecharacteristic polynomial of h. Hence all the eigenvalues of such an xjare positive. This contradicts the fact that none of the xj are in H,proving H is open in H. Intersecting everything in sight with gl(n,R)shows that P is also open in P.

6.2 The Polar Decomposition 265

Proposition 6.2.2. Upon restriction, the exponential map of gl(n,C)is a diffeomorphism between H and H. Its inverse is given by

Log h = log(tr h)I −∞∑

i=1

(I − h

trh)i/i,

which is a smooth function on H.

As a consequence we see that the restriction of Exp to any realsubspace of H gives a diffeomorphism of the subspace with its image.In particular, Exp is a diffeomorphism between P and P . In particular,in all these cases Exp is a bijection.

Proof. We shall do this for H, the real case being completely analogous.Suppose h ∈ H is diagonal with eigenvalues hi > 0. Then tr(h) > 0 and0 < hi

tr(h) so log(tr(h)) is well-defined and log( hi

tr(h)) is defined for all i.

But since 0 < hi

tr(h) < 1, we see that 0 < (1 − hi

tr(h)) < 1 for all positive

integers k. Hence Log( htr(h)) is given by an absolutely convergent power

series −∑∞i=1(I − h

tr h)i/i. If u is a unitary operator so that uhu−1 isdiagonal, then tr(uhu−1) = tr(h) and since conjugation by u commuteswith any convergent power series, this series actually converges for allh ∈ H and is a smooth function Log on H. Because on the diagonalpart of H this function inverts Exp, and both Exp and this power seriescommute with conjugation, it inverts Exp everywhere on H. Finally,log(tr(h))I and Log( h

tr h) commute and Exp of a sum of commutingmatrices is the product of the Exp’s. Since Log inverts Exp on thediagonal part of H it follows that

Log(h) = log(tr(h))I + Log(h

trh) = log(tr(h))I −

∞∑

i=1

(I − h

trh)i/i.

We shall need the following elementary fact whose proof is left tothe reader.

Lemma 6.2.3. For any g ∈ GL(n,C), g∗g ∈ H.


It follows that for all g ∈ GL(n,C), Log(g∗g) ∈ H and since this isa real linear space also 1

2 Log(g∗g) ∈ H. This means we can apply Expand conclude the following:

Corollary 6.2.4. h(g) = Exp(12 Log(g∗g)) ∈ H is a smooth map from

GL(n,C) → H.

Hence h(g)n = Exp(n2 Log(g∗g)) ∈ H for every n ∈ Z. In particular,h(g)−2 = Exp(2

2 Log(g∗g)) = g∗g. So that

gh(g)−1(gh(g)−1)∗ = gh(g)−1h(g)−1∗g∗ = gh(g)−2g∗

and, since h(g)−1 ∈ H, g(g∗g)−1g∗ = I. Thus, gh(g)−1 = u(g) isunitary for each g ∈ GL(n,C). Since group multiplication and inversionare smooth, g 7→ u(g) is also a smooth function on GL(n,C) (as ish(g)). Now this decomposition g = uh, where u ∈ U(n,C) and h ∈ H isactually unique. To see this, suppose u1h1 = g = u2h2. Then u−1

2 u1 =h2h

−11 so that h2h

−11 is unitary. This means (h2h

−11 )∗ = (h2h

−11 )−1 and

hence h21 = h2

2. But since h1 and h2 ∈ H, each is an exponential ofsomething in H; hi = expxi. But then h2

i = exp 2xi and since exp is1 : 1 on H, we get 2x1 = 2x2 so x1 = x2 and therefore h1 = h2 and u1 =u2. The upshot of all this is that we have a smooth map GL(n,C) →U(n,C) ×H given by g 7→ (u(g), h(g)). Since g = u(g)h(g) for every g(multiplication in the Lie group GL(n,C)), this map is surjective andhas a smooth inverse. We summarize these facts as following :

Theorem 6.2.5. (polar decomposition) The map g 7→ (u(g), h(g)) givesa real analytic diffeomorphism GL(n,C) → U(n,C) ×H. Identical rea-soning also shows that as a smooth manifold GL(n,R) is diffeomorphicto O(n,R) × P .

From this it follows that, since H and P are each diffeomorphic witha Euclidean space, and therefore are topologically trivial, in each casethe topology of the non-compact group is completely determined bythat of the compact one. In this situation, one calls the compact groupa deformation retract of the non-compact group. Since P and H arediffeomorphic images under Exp of some Euclidean space, one calls them

6.3 The Cartan Decomposition 267

exponential submanifolds. For example, connectedness, the number ofcomponents, simple connectedness and the fundamental group of thenon-compact group are each the same as that of the compact one. Thusfor all n ≥ 1, GL(n,C) is connected and its fundamental group is Z,while for all n, GL(n,R) has 2 components and the fundamental groupof its identity component is Z2 for n ≥ 3 and Z for n = 2 (see Section1.5)

As a final application of the polar decomposition theorem we havethe following inequality which is a variant of one in Margulis [41] p. 169

Corollary 6.2.6. Let T be a linear transformation on a finite dimen-sional real or complex vector space V of dimension n and || · || be theHilbert-Schmidt norm on End(V ). Then |detT | ≤ ||T ||n. See Section6.5 for the definition of the Hilbert-Schmidt norm.

Proof. Clearly we may assume T is invertible, since otherwise |detT | =0. For T ∈ GL(V ) write the polar decomposition T = kp. Then since|det k| = 1, |detT | = |det p| and

||T ||2 = tr(kp(kp)∗) = tr(kpp∗k−1) = tr(pp∗) = ||p||2.Thus we may assume T is positive definite symmetric, or Hermi-

tian. As such it is diagonalizable T = kDk−1. Thus |detT | = |detD|and ||T ||2 = ||D||2 so we may actually assume T is diagonal with posi-

tive eigenvalues, d1, . . . , dn. We have to show (d1 . . . dn)1n ≤

√∑ni=1 d

2i .

Now the geometric mean is less than or equal to the arithmetic mean

(d1 . . . dn)1n ≤ 1

n

∑ni=1 di, so we show

∑ni=1 di ≤ n

√∑ni=1 d

2i , or

(∑n

i=1 di)2 ≤ n2

∑ni=1 d

2i . By the Schwarz inequality (

∑ni=1 di)

2 ≤n

∑ni=1 d

2i . Thus, the question is just n

∑ni=1 d

2i ≤ n2

∑ni=1 d

2i which

is true since∑n

i=1 d2i > 0 and n ≥ 1.

6.3 The Cartan Decomposition

We now turn to more general groupsG and also streamline our notation.Instead of H, we shall consider certain real subspaces of H denoted byp whose exponential image will be P and make the following definition.


Definition 6.3.1. Let G be a Lie subgroup of GL(n,R) with Lie algebrag. We denote by K = O(n,R)∩G, by P the positive definite symmetricmatrices in G, by p the symmetric matrices of g, and by k the skewsymmetric matrices in g. In the case that G be a Lie subgroup ofGL(n,C) we again denote its Lie algebra by g, but now K = U(n,C) ∩G, P is the positive definite Hermitian matrices of G, p is Hermitianmatrices of g and k the skew Hermitian matrices in g.

Lemma 6.3.2. Let q(t) =∑n

j=1 cj exp(bjt) be a trigonometric polyno-mial, where cj ∈ C, and bj and t ∈ R. If q vanishes for an unboundedset of real t’s, then q ≡ 0.

An immediate consequence is that for a polynomial p ∈ C[z1, . . . , zn]in n complex variables with complex coefficients and (x1, . . . , xn) ∈ Rn,if p(exp(tx1), . . . , exp(txn)) vanishes for an unbounded set of real t’s,then it vanishes identically in t.

Proof. First we can assume that the t’s for which q vanishes tend to+∞. Otherwise, they would have to tend to −∞ and in this case wejust let p(t) = q(−t). Then p is also a trigonometric polynomial andif p = 0, then so is q. Reorder the bj’s, if necessary, so that they arestrictly increasing by combining terms by adding the corresponding cj ’s.Of course, we can now assume that all the cj ’s are nonzero. Let tk bea sequence tending to +∞ on which q vanishes. Suppose there are twoor more bj’s. Since

q(t)

cn exp(bnt)=

n−1∑

j=1

cjcn

exp((bj − bn)t) + 1,

it follows that q(t)cn exp(bnt)

→ 1 as k → ∞. But since q is identically 0 ink so is this quotient, a contradiction. This means that all the bj’s areequal and so q(t) = c exp(bt) for some c ∈ C and b ∈ R. This functioncannot have an infinite number of zeros unless c = 0, that is q = 0.


Proposition 6.3.3. Suppose M is an algebraic subgroup of GL(n,C)and G be a Lie subgroup of GL(n,R) (or GL(n,C)) with Lie algebrag. Let G have finite index in MR (respectively M). If X ∈ H andexpX ∈ G, then exp tX ∈ P for all real t. In particular, X ∈ g andhence X ∈ p.

Proof. To avoid circumlocutions we shall prove the complex case, thereal case being completely analogous. Choose u ∈ U(n,C) so thatuXu−1 is diagonal with real eigenvalues λj . Replace G by uGu−1, a Liesubgroup of GL(n,C) which is contained in uMu−1 with finite index.Now uMu−1 is an algebraic subgroup of GL(n,C) (and in the real caseuMRu

−1 = (uMu−1)R). Hence we can assume X is diagonal. Let p(zij)be one of the complex polynomials defining M . Since expX ∈ G and Gis a group, exp kX ∈ G ⊆M for all k ∈ Z. But exp kX is diagonal withdiagonal entries exp(kλj). Applying p to exp kX, we get p(exp kX) = 0for all k. By the corollary, p(exp tX) = 0 for all t. Because p wasan arbitrary polynomial defining M , it follows that exp tX ∈ M forall real t. Since G has finite index in M and the 1-parameter groupexp tX is connected, it must lie entirely in G and therefore in P . HenceX ∈ g.

Definition 6.3.4. A subgroup G of GL(n,R) (or GL(n,C)) is calledself-adjoint if it is stable under taking transpose (respectively ∗). Heretranspose and ∗ refer to any linear involution (respectively conjugatelinear involution) on Rn (respectively Cn).

For example, SL(n,R) and SL(n,C) are self-adjoint since det gt =det g (det g∗ = det(g)). The routine calculations showing O(n,C),SO(n,C), O(p, q) and SO(p, q) are also self-adjoint are left to the reader.In fact, the reader can check that any classical non-compact simplegroup in E. Cartan’s list (see [32]) is self-adjoint. Clearly by their verydefinition these groups are either algebraic or have finite index in thereal points of an algebraic group (essentially algebraic). Now it is an im-portant insight of Mostow [57] that any linear real semisimple Lie groupis self-adjoint under an appropriate involution. Moreover, by the rootspace decomposition, Section 7.3, the adjoint group of any semisimple


group without compact factors is algebraic (actually over Q). Thus herewe are really talking about all the semisimple groups without compactfactors and, of course, this means our construction actually gives allsymmetric spaces of non-compact type. But even if we did not knowthis, since any classical non-compact simple group is easily seen to beself-adjoint as well as essentially algebraic, we already get a plethora ofsymmetric spaces from them.

Particular cases of Theorem 6.3.5 below are the following. We shallleave their routine verification to the reader. SL(n,R) is diffeomorphicwith SO(n)×P1, where P1 is the positive definite symmetric matrices ofdet 1, which in turn is diffeomorphic under exp with the linear space ofreal symmetric matrices of trace 0. Similarly, SL(n,C) is diffeomorphicwith SU(n)×H1, where H1 is the positive definite Hermitian matrices ofdet 1, which in turn is diffeomorphic with the linear space of Hermitianmatrices trace 0. As deformation retracts, similar conclusions can bedrawn about the topology of these, as well as the other groups mentionedearlier.

The following result is a special case of the general Iwasawa decom-position theorem which holds for an arbitrary Lie group with a finitenumber of components, but with a somewhat more elaborate formula-tion (see G.P. Hochschild [33]). Here, we content ourselves with the mat-ter at hand, namely self-adjoint algebraic groups, or their real points.In this context, it is called the Cartan decomposition. By a maximalcompact subgroup of G we mean one not properly contained in a largercompact subgroup of G. Our next result is the Cartan decomposition.

Theorem 6.3.5. Let G be a self-adjoint subgroup of GL(n,C) (orGL(n,R)) with Lie algebra g. Suppose that G has finite index in analgebraic subgroup M of GL(n,C) (G has finite index in MR, the realpoints of M). Then

(1) G = K × P as smooth manifolds.

(2) g = k ⊕ p as a direct sum of R-vector spaces.

(3) exp : p → P is a diffeomorphism whose inverse is given by theglobal power series of Proposition 6.2.2.

(4) K is a maximal subgroup of G. In particular, P is simply con-


nected and K is a deformation retract of G.

Proof. Here again we deal with the complex case, the real case beingsimilar. First we show each g ∈ G can be written uniquely as g =u expX, where u ∈ K and X ∈ p. By Theorem 6.2.5, g = up, whereu ∈ U(n,C) and p ∈ H. Now g∗ = (up)∗ = p∗u∗ = pu−1, so g∗g =pu−1up = p2. Since G is self-adjoint, p2 ∈ G. Now p = expX forsome Hermitian X, then p2 = exp 2X where 2X is also Hermitian. ByProposition 6.3.3, exp t2X ∈ P for all real t, in particular for t = 1

2 forwhich we get expX = p ∈ P ⊆ G and X ∈ p. But then gp−1 = u ∈ G,therefore u ∈ K. Also, since exp tX ∈ P for all real t, X ∈ p. Thusg = up, where u ∈ K and p ∈ P . Thus we have a map g 7→ (u, p) fromG to K ×P . As above, if we can show uniqueness of the representationg = up, then the map is onto. But since K ⊆ U(n,C) and P ⊆ thepositive definite Hermitian matrices, this follows from the uniquenessresult proven earlier. Since multiplication inverts this map it is one-to-one and has a smooth inverse. The formula, p(g) = exp(1

2 log(g∗g)) ∈ Pderived in the case of GL(n,C) is still valid, if suitably interpreted, andgives a smooth map G→ P . Arguing exactly as in the case of GL(n,C)we see that part 1 is true. Part 3 follows immediately from the case ofGL(n,C) treated earlier.

For part 2, write X = X−X∗

2 + X+X∗

2 . Since the first term is skewHermitian, the second is Hermitian and each is an R-linear functionof X ∈ gl(n,C), this proves part 2 for the case gl(n,C). To prove itin general we need only show that X−X∗

2 ∈ k and X+X∗

2 ∈ p and forthis it suffices to show that g is stable under map X 7→ X∗. Notethat for X ∈ g, exp tX ∈ G for all t. Since G is self-adjoint and(exp tX)∗ = exp t(X∗), it follows that X∗ ∈ g.

To prove part 4, we first consider the basic cases, GL(n,R) andGL(n,C).

Proposition 6.3.6. Let L be a compact subgroup of GL(n,C) (orGL(n,R)). Then some conjugate gLg−1, g ∈ GL(n,C) (respectivelyin GL(n,R)) is contained in U(n,C) (respectively O(n,R)). In particu-lar, U(n,C) is a maximal compact subgroup of GL(n,C) and O(n,R) amaximal compact subgroup of GL(n,R). In GL(n,C) and GL(n,R) any


two maximal compact subgroups are conjugate.

Proof. We deal with the complex case, the other being completely anal-ogous. If (·, ·) is a Hermitian inner product on Cn, using (finite) Haarmeasure dl on L we can form an L-invariant Hermitian inner product onCn given by 〈v,w〉 =

∫L(lv, lw)dl. Thus for some g ∈ GL(n,C), gLg−1

is contained in U(n,C).

If L ⊃ U(n,C), where L is a compact subgroup of GL(n,C), thenby the previous discussion gLg−1 ⊂ U(n,C) for some g, so both havethe same dimension. Therefore U(n,C) is an open subgroup of L. Since

U(n,C) is connected, we conclude that U(n,C) = L0, the identity com-ponent of L. On the other hand gL0g

−1 ⊂ gLg−1 ⊂ U(n,C) = L0,since L0 is connected, therefore they all are equal, in particular L isconnected and L = L0 = U(n,C).

In the real case we just work with the compact connected groupSO(n,R) instead of U(n,C). Thus U(n,C) and O(n,R) are maximalcompact subgroups of GL(n,C) and GL(n,R), respectively. That anyother maximal compact subgroup is conjugate to one of these now fol-lows from the first statement of the proposition.

In particular, if L is any compact subgroup of GL(n,C), all its ele-ments have their eigenvalues on the unit circle. From this we see thatif an element l ∈ L has all its eigenvalues equal to 1, then l = I. Thisis because gLg−1 is unitary for some g. Hence for some u we knowuglg−1u−1 is diagonal and also has all eigenvalues equal to 1. Thusuglg−1u−1 = I and hence l itself equals I.

Finally, we turn to the proof of part 4 itself. First suppose L is anycompact subgroup of G. Then L ∩ P = 1. To see this just observethat, by the previous result, since L is compact, all its elements haveall their eigenvalues on the unit circle. But the eigenvalues of elementsof P are all positive. Hence all the elements of L ∩ P have all theireigenvalues equal to 1 as above, so L ∩ P = 1. Now we prove that Kis a maximal compact subgroup. Suppose that L ⊇ K, then each l ∈ Lcan be written l = up, where u ∈ K ⊂ L and p ∈ P . But since u ∈ L,so is p. Hence by the above p = I and l = k. Hence L ⊆ K, so thatactually L = K.


We have essentially used the conjugacy of maximal compact sub-groups in GL(n,C) and GL(n,R) to show that K is a maximal compactsubgroup of G, in general. However to prove, in general, that any twomaximal compact subgroups of G are conjugate will require somethingmore. For this we will rely on the important differential geometric fact,called Cartan’s fixed point theorem, that a compact group of isometriesacting on a complete simply connected Riemannian manifold of nonposi-tive sectional curvature at every point (Hadamard manifold) always hasa unique fixed point and, for the reader’s convenience, we will proveCartan’s result as well in the next section. However, we will only proveit for symmetric spaces of non-compact type. This will also establishthe fact that for each p ∈ P , StabG(p) is a maximal compact subgroupof G.

We note that the Cartan involution of g is given by k + p 7→ k − p.It is an automorphism of g whose fixed point set is k. We also mentionthe Cartan relations, which were also proved earlier. If the Cartandecomposition of g is g = k ⊕ p, since k is a subalgebra and [x∗, y∗] =−[x, y]∗ and [xt, yt] = −[x, y]t it follows that

(1) [k, k] ⊆ k,

(2) [k, p] ⊆ p,

(3) [p, p] ⊆ k.

We conclude this section by observing that for all the Lie group Gconsidered in this section, there is a natural smooth action of G onP given by (g, p) 7→ gtpg. Now this action is transitive. To see this,consider the G orbit of I ∈ P , OG(I) = gtg : g ∈ G. As we sawearlier, this is p2; p ∈ P. But since everything in P is exp of a uniqueelement X of p, it follows that everything in P has a unique squareroot in P , namely exp 1

2X. This means the action is transitive. Whatis the isotropy group StabG(I) of I? This is g ∈ G : gtg = I =G ∩ O(n,R) = K. Hence, by general principles, P ≃ OG(I) is G-equivariantly diffeomorphic with G/K, endowed with the action G byright translation. As we shall see, this transitive action will be of greatimportance in what follows.

Observe that this action does not have the two-point homogeneity


property. That is, given p, q and p′, q′, all in P , there may not be ag ∈ G so that g(p) = p′ and g(q) = q′, even when dimP = 1. Note alsothat gt(expX)g is not equal to exp(gtXg), so this is not G-equivalentwith the R-linear representation of G acting on p by (g,X) 7→ gtXg,X ∈ p. Concomitantly, the latter is not a transitive action because itis linear, so 0 is a single orbit. In fact, here the orbit space can beparameterized by the number of positive, negative and zero eigenvaluesof a representative.

Corollary 6.3.7. For all n ≥ 1, SL(n,C) is simply connected.

Proof. This follow from the Cartan decomposition that the homotopytype of SL(n,C) is that of its maximal compact subgroup SU(n,C),which is simply connected (see Corollary 1.5.2)

Exercise 6.3.8. Find the Cartan decompositions of Sp(n,R) andsp(n,R).

6.4 The Case of Hyperbolic Space and the

Lorentz Group

We now make explicit the Cartan decomposition in an important spe-cial case and give the Lorentz model for hyperbolic n space, Hn. Weconsider O(n, 1) the subgroup of GL(n+1,R) leaving invariant the non-degenerate quadratic form q(v, t) = v2

1 + . . . + v2n − t2, where v ∈ Rn

and t ∈ R. Equivalently, by polarization, this means leaving invariantthe nondegenerate symmetric bilinear form 〈(v, t), (w, s)〉 = (v,w) − ts,where (v,w) is the usual (positive definite) inner product in Rn. ThusG is defined by the condition g−1 = gt (transpose with respect to 〈·, ·〉).It is easy to check that G is the set of R-points of a self-adjoint algebraicgroup and, in particular, is a Lie group. Now G is not compact. Forexample, SO(1, 1) ⊆ O(1, 1), which sits inside O(n, 1), consists of allmatrices

g =

(a bc d

)

6.4 The Case of Hyperbolic Space and the Lorentz Group 275

with a2 − c2 = 1, ab − cd = 0 and b2 − d2 = −1. In particular, takingan arbitrary a and c = (a2 − 1)

12 , where a2 − 1 = c2 > 0 and letting

b and d be determined by the remaining two equations we see thatb = (a2 − 1)

12 = c and d = a. Now consider the identity component

SO(1, 1)0. Since the locus a2 − c2 = 1 has two connected components,if g ∈ SO(1, 1)0, then a > 0 and so there is a unique t ∈ R for whicha = cosh t and b = sinh t. Thus

g(t) =

(cosh t sinh tsinh t cosh t

).

Because these hyperbolic functions are unbounded, we see evenSO(1, 1)0 is not compact. The identities satisfied by the hyperbolicfunctions show that this is an abelian subgroup. However, we shall seethis without these identities; in fact, we will derive the identities. Let

X =

(0 11 0

).

A direct calculation using the fact that X2 = I shows that exp tX =I cosh t+X sinh t = g(t), from which it follows that g(s+ t) = g(s)g(t).This equation gives all the identities satisfied by the hyperbolic functionssinh and cosh and g is a smooth isomorphism of SO(1, 1)0 with R. Thegeometric importance of such 1-parameter subgroups will be seen in amoment.

By Theorem 6.3.5 a maximal compact subgroup of G is given by

O(n+ 1,R) ∩ O(n, 1). Because subgroups of GL(n,R) can be regardedas subgroups of GL(n+1,R) via the imbedding g 7→ diag(g, 1), we maythink of O(n,R) as a subgroup of GL(n+ 1,R) and, in fact, of O(n, 1).Thus O(n,R) ⊆ O(n + 1,R) ∩ O(n, 1). Clearly these are equal. Since

O(n,R) has two components, so does O(n, 1) which equals O(n,R)×P ,where P an exponential submanifold. Therefore, O(n, 1)0 = SO(n,R)×P .

Note that for g ∈ O(n, 1) we have ggt = I, so (det g)2 = 1, Thusdet g = ±1, a discrete set. It follows that SO(n, 1) is open in O(n, 1)and hence has the same P . The same is true of SO(n, 1)0 because we aredealing with Lie groups. Thus SO(n, 1)0 = SO(n,R) × P and we now


work with this connected group G = SO(n,R)× P 1. The Lie algebra g

of G = SO(n, 1)0 is

X ∈ gl(n+ 1,R) : Xt = −X

which has dimension (n+1)n2 . Now consider the subspace of gl(n+ 1,R)

consisting of (X vv 0

),

where X ∈ so(n,R) the Lie algebra of SO(n,R) and v ∈ Rn. It is clearly

a subspace and has dimension (n−1)n2 + n = (n+1)n

2 and it consists ofskew symmetric matrices with respect to 〈·, ·〉. Hence it must coincidewith g. Here the Cartan decomposition is perfectly clear. The k part is

(X 00 0

),

for X ∈ so(n,R), while the p part is

(0 vv 0

),

for v ∈ Rn. Consider the locus of points,

H = (v, t) ∈ Rn+1 : q(v, t) = −1.

For g ∈ O(n, 1), q(g(v, t)) = q(v, t). In particular, if q(v, t) = −1,then q(g(v, t)) = −1. Thus H is invariant under O(n, 1). Now H is a

hyperboloid of two sheets: 1 + ‖v‖2 = t2. So t = ±(1 + ‖v‖2)12 . Write

H = H+ ∪ H−, a disjoint union of the upper and lower sheet. Bothsheets are open subsets of H since they are the intersection of H witha half space. Each is diffeomorphic with Rn. In particular, each isconnected and simply connected. We show that G = SO(n, 1)0 leavesboth H+ and H− invariant. Note that

g(H+) = (g(H+) ∩H+) ∪ (g(H+) ∩H−),

1Actually, SO(n, 1) is connected if n is even, and has two components if n is odd.

6.4 The Case of Hyperbolic Space and the Lorentz Group 277

and g(H+) is connected. Therefore g(H+) ⊆ H+ or g(H+) ⊆ H−.Since g is a diffeomorphism of H, g(H+) = H+ or g(H+) = H−. Weshow that the former must hold. Since G is arcwise connected, theremust be a smooth path gt in G joining g = g1 to I = g0. Considerthe disjoint sets T+ = t ∈ [0, 1] : gt(H

+) = H+ and T− = t ∈[0, 1] : gt(H

+) = H−. Note that [0, 1] = T+ ∪ T− and T+ 6= ∅ ast = 0 ∈ T+. We prove that T+ and T− are closed. For if tk → t andsay gtk(H+) = H+, for all k, but gt(H

+) = H−, then for x ∈ H+,gtk(x) → gt(x). This is impossible as the distance between H+ and H−

is 2. Therefore [0, 1] = T+ and g(H+) = g1(H+) = H+.

We now know G operates on H+ which we shall call Hn, theLorentz model of hyperbolic n-space. Consider the lowest point,p0 = (0, . . . , 0, 1) ∈ Hn. What is StabG(p0)? This is clearly a sub-group which does not change the t coordinate and is arbitrary inthe other coordinates since it is linear and so always fixes 0. Hence,StabG(p0) = SO(n,R), a maximal compact subgroup of G. Next welook at the G-orbit O(p0) and show G acts transitively on Hn. Let

p = (v, t), where t = (1+‖v‖2)12 , be any point inHn and apply SO(n,R)

on the first n coordinates to bring p to (‖v‖, 0, ...0, t). Now the prob-lem is reduced we are to a two-dimensional situation, let us consider(x, y) = (‖v‖, t), where y2 − x2 = 1. We want to transform (0, 1) to(x, y) by something on the 1-parameter group

g(s) =

(cosh s sinh ssinh s cosh s

).

But this is just the fundamental property of the right hand branchof the hyperbola mentioned earlier. Therefore, G acting transitivelyon Hn is equivariantly equivalent to the action by left translation onSO0(n, 1)/SO(n,R).

Now consider the hyperplane t = 1 in Rn+1. This is the tangentspace Tp0 to Hn at p0. Consider (·, ·) the standard Euclidean metric onTp0. If p is another point of Hn, choose g ∈ G such that g(p) = p0.Then its derivative dpg at p maps Tp to Tp0 bijectively. Use this totransfer the inner product from Tp0 to Tp. Now if h(p) also equals p0,then gh−1 ∈ StabG(p0) = SO(n,R). Therefore dp0(gh

−1) = dpgdp0h−1


is a linear isometry of Tp0 . This shows the inner product on Tp isindependent of g and is well defined. Hence we get a Riemannian metricon Hn because G is a Lie group acting smoothly on Hn. Evidently, Gacts by isometries, the action is transitive and Hn can be identified withG/StabG(p0) = SO(n, 1)0/SO(n,R).

Notice that SO(n,R) = StabG(p0) acts transitively on k-dimensionalsubspaces for all 1 ≤ k ≤ n. In particular, this is so for 2-planes inRn = Tp0(H

n). Since it acts by isometries, this means the sectionalcurvature is constant as both the point and the plane section vary.

6.5 The G-invariant Metric Geometry of P

Here we introduce a Riemannian metric on any P and study its mostbasic differential geometric properties. From now on we will write expand log instead of Exp and Log. Much if this section is an elaborationof results in [61].

Lemma 6.5.1. If A and B are n×n complex matrices, then tr(AB) =tr(BA). Also tr(B∗B) ≥ 0 and equals 0 if and only if B = 0. Evidently,tr(B)− = tr(B∗).

Proof. Suppose A = (aij) and B = (bkl). Then (AB)il =∑

j aijbjl.Therefore tr(AB) =

∑i,j aijbji. But then tr(BA) =

∑i,j bijaji =∑

i,j ajibij =∑

j,i aijbji = tr(AB). Taking B∗ for A we get tr(B∗B) =∑i,j bjibji ≥ 0 and equals 0 if and only if B = 0.

This enables us to put a Hermitian inner product on gl(n,C) calledthe Hilbert-Schmidt inner product and a symmetric inner product ongl(n,R) by defining

〈Y,X〉 = tr(Y ∗X).

For X Hermitian (or symmetric), we now study the linear operator adXon gl(n,C) (respectively gl(n,R)). As we saw from the Cartan relationsfor T ∈ gl(n,C) and X Hermitian, [X,T ]∗ = [T ∗,X] = −[X,T ∗].

Lemma 6.5.2. If X is Hermitian, 〈adX(T ), S〉 = 〈T, adX(S)〉 for allS and T ; that is, adX is self-adjoint. In particular, the eigenvalues ofsuch an adX are all real.

6.5 The G-invariant Metric Geometry of P 279

Proof. We calculate tr([X,T ]∗S) = tr(−[X,T ∗]S) = − tr((XT ∗ −T ∗X)S) = tr(T ∗XS) − tr(XT ∗S). On the other hand, tr(T ∗[X,S]) =tr(T ∗XS)−tr(T ∗SX). Thus we must show that tr(XT ∗S) = tr(T ∗SX).But this follows from the lemma above.

A formal calculation, which we leave to the reader, proves the fol-lowing:

Lemma 6.5.3. For each U ∈ gl(n,C), Lexp(U) = exp(LU ) andRexp(U) = exp(RU ).

Definition 6.5.4. For X and Y ∈ gl(n,C) let

DX(Y ) =d

dtexp(−X/2) exp(X + tY ) exp(−X/2)|t=0.

Proposition 6.5.5. For X ∈ p, the operator dX is self-adjoint ongl(n,C). Using functional calculus, this operator is given by the formula

DX = sinh(adX

2)/(

adX

2).

Proof. Let t ∈ R, X, Y ∈ gl(n,C) and X(t) = X + tY . Then

DX(Y ) = exp(−X/2) ddt

exp(X(t))|t=0 exp(−X/2).

Now for all t,

X(t) · exp(X(t)) = exp(X(t)) ·X(t).

Differentiating we get

X ′(t) · exp(X(t)) +X(t) · ddt

exp(X(t)) =d

dtexp(X(t)) ·X(t)

+ exp(X(t)) ·X ′(t).

Evaluating at t = 0 and subtracting gives X · ddt exp(X(t))|t=0 −

ddt exp(X(t))|t=0 · X = exp(X)Y − Y exp(X). Multiplying on both


the left and right by exp(−X/2) and taking into account the fact thatexp(−X/2) and X commute, we get

X · exp(−X/2) ddt

exp(X(t))|t=0 exp(−X/2)

− exp(−X/2) ddt

exp(X(t))|t=0 exp(−X/2)X= exp(X/2)Y exp(−X/2) − exp(−X/2)Y exp(X/2).

Substituting for DX(Y ), the left hand side becomes

XDX (Y ) −DX(Y )X = adXDX (Y ),

while the right hand side is

Lexp(X/2)Rexp(−X/2)(Y ) − Lexp(−X/2)Rexp(X/2)(Y ).

But by the lemma above

Lexp(U) = exp(LU ) and Rexp(U) = exp(RU ).

Substituting we get

adXDX (Y ) = exp(LX/2) exp(R−X/2)(Y ) − exp(L−X/2) exp(RX/2)(Y ).

Since LU and RU ′ commute for all U and U ′, we see that

exp(LX/2) exp(R−X/2) = exp(LX/2 +R−X/2) = exp(LX/2 −RX/2)

= exp(adX/2).

Similarly,

exp(L−X/2) exp(RX/2) = exp(L−X/2 +RX/2) = exp(− adX/2).

So for all Y ,

adX ·DX(Y ) = (exp(adX/2) − exp(− adX/2)) (Y ).

Now let

f(z) = ez/2 − e−z/2 = z + 2(z/2)3/3! + 2(z/2)5/5! + · · · .


Then f is an entire function and f(0) = 0. In terms of f , the equationabove says

adXDX = f(adX).

This means if we let

g(z) = f(z)/z = 1 + (z/2)2/3! + (z/2)4/5! + · · · ,with g(0) = 1, then g is also entire and DX = g(adX). Now sinh z = z+z3/3! + z5/5! + · · · so g(z) = sinh(z/2)/(z/2) and hence the conclusion.Finally, because DX = g(adX), adX is self-adjoint and the Taylorcoefficients of g are real, DX is also self-adjoint.

Exercise 6.5.6. Using the same method in the proof of Proposition6.5.5, prove that

dX exp(Y ) = φ(− adX)(Y )

where φ(z) =∑∞

n=0zn

(n+1)! .

Corollary 6.5.7. For X ∈ p, Spec( sinh(adX)adX ) consists of real numbers

greater than or equal to 1. The same is true for the operator DX .

Proof. Since for t ∈ R, sinh tt = 1+t2/3!+t4/5!+· · · , we see that sinh t

t > 1unless t = 0. Now by Exercise 0.5.13, Spec(adX) ⊆ λi − λj |λi, λj ∈SpecX, therefore

Spec(sinh(adX)

adX) = sinh(λ)

λ: λ ∈ Spec adX

⊆ sinh(λi − λj)

λi − λj: λi, λj ∈ SpecX.

If λ = λi − λj for distinct eigenvalues of X, then sinh(λ)λ > 1. If λi and

λj are equal, then λ = 0 and sinh(λ)λ = 1.

We now work exclusively over R. The same type of arguments alsowork just as well over C.

Corollary 6.5.8. For X ∈ p and Y ∈ gl(n,R), tr(Y 2) ≤ tr(DX(Y ))2).Equality occurs if and only if [X,Y ] = 0.


Proof. Because adX is self-adjoint, we can choose an orthonormal basisof real eigenvectors of adX, Y1, . . . Yj ∈ gl(n,R) which, since DX =g(adX) are also eigenvectors forDX with corresponding real eigenvaluesµ1, . . . µj. Then DX(Yk) = µkYk for all k. If Y =

∑k ak(Y )Yk, then

DX(Y ) =∑

k ak(Y )DX(Yk) =∑

k ak(Y )µkYk. Since the Yk form anorthonormal basis, we see tr(DX(Y )2) =

∑k ak(Y )2µ2

k, while tr(Y 2) =∑k ak(Y )2. Thus we are asking whether

∑k ak(Y )2 ≤ ∑

k ak(Y )2µ2k.

Since each µk ≥ 1, this is clearly so and equality occurs only if µk = 1whenever ak(Y ) 6= 0. Rearrange the eigenvectors so that the µk = 1come first and for k ≥ k0, µk > 1. Hence gl(n,R) = W1 ⊕W∞ is theorthogonal direct sum of two adX-invariant subspaces. Here W1 is the1-eigenspace, and W∞ the sum of all the others. But since ak(Y ) = 0for k ≥ k0, Y ∈ W1. But on W1 all eigenvalues of g(adX) = DX are1, and the eigenvalues of adX are 0 so adX = 0 on W1 and hence[X,Y ] = 0.

Conversely, if [X,Y ] = 0, then adX(Y ) = 0. Therefore DX =g(adX) = I.

Theorem 6.5.9. Along any smooth path p(t) in P we have

tr[(d

dtlog p(t))2] ≤ tr[(p−1(t)p′(t))2].

with equality if and only if p(t) and p′(t) commute for that t.

Proof. For each t, it is easy to see that

p1/2(p−1p′)2p−1/2 = (p−1/2p′p−1/2)2.

It follows that tr[(p−1p′)2] = tr[(p−1/2p′p−1/2)2]). Set X(t) = log p(t).

Then X(t) is a smooth path in p and p(t)−12 = exp(−X(t)/2). Let

t be fixed and Y = X ′(t). Since DX(Y ) = exp(−X/2) dds exp(X +

sY ))|s=0 exp(−X/2), this is p−12p′p−

12 , where p′ = d

ds exp(X +sY )|s=0 (the tangent vector to curve p(t) at p = expX). Hence

tr[(p−12 p′p−

12 )2] = tr[(DX(X ′))2]. Also tr[( ddt log p(t))2] = tr[X ′(t)2].

Now by the corollary, for each t,

tr[X ′(t)2] ≤ tr[(dX(t)(X′(t))2]


with equality if and only if X(t) and X ′(t) commute for that t.

Finally we show that for all fixed t, X(t) and X ′(t) commute if andonly if p(t) and p′(t) commute. For by the chain rule and the formulaExercise 6.5.6,

p′(t) = dX(t) exp(X ′(t)) = φ(− adX(t))X ′(t),

where φ is the entire function given by φ(z) =∑∞

n=0zn

(n+1)! . If X ′

commutes with X for fixed t, then since φ(0) = 1, we see thatφ(− adX)X ′ = X ′ so that p′ = X ′. In particular, p′ commutes with Xand therefore with expX = p. On the other hand, if φ(− adX)X ′ com-mutes with expX = p, then since log : P → p is given by a convergentpower series in p (see Theorem 6.3.5, part 3), it must also commute withlog p = X. Looking at the specific form of the function φ, it follows that

[X,X ′ − adX(X ′)/2! + ad2X(X ′)/3! + · · · ] = 0.

That is, adX(X ′) − ad2X(X ′)/2! + ad3

X(X ′)/3! + · · · = 0. Henceexp(− adX(X ′)) = X ′. Taking exp(adX) of both sides tells usexp(adX)(X ′) = X ′. Therefore Ad(expX)(X ′) = X ′ so X ′ commuteswith expX. But then, reasoning as above, X ′ must commute withlog(expX) = X.

Since what is inside the square root is real and positive, we makethe following definition.

Definition 6.5.10. Let p(t) be a smooth path in P , where a ≤ t ≤b. Then its length l(p) equals

∫ ba [tr((p−1p′(t))2)]

12 dt. The Riemannian

metric is given by ds2 = tr((p−1p′)2)dt2. We call this metric d.

Proposition 6.5.11. G acts isometrically on P .

Proof. We calculate that

(gtpg)−1(gtpg)′ = g−1p−1(gt)−1gtp′g = g−1p−1p′g.

Hence ((gtpg)−1(gtpg)′)2 = g−1(p−1p′)2g. Taking traces we get

tr[(gtpg)−1(gtpg)′)2] = tr[(p−1p′)2].


On p we place the metric given infinitesimally by ds2 =tr[( ddt log p(t))2]dt2, that is, if X(t) is a smooth path in p, then ds2 =tr(X ′(t)2)dt2. We call this metric dp. Earlier we defined an inner prod-uct on gl(n,R) by 〈Y,X〉 = tr(Y tX). Hence the linear subspace p hasan inner product on it by restriction, namely 〈Y,X〉 = tr(Y X). Theassociated norm is ‖Y ‖2 = tr(Y 2). This, together with the formulaabove, shows dp is the Euclidean metric. If we transfer dp to P , thendp(p, q) = ‖ log p− log q‖. This will give us the opportunity to comparedp and d on P . Since by Theorem 6.5.9 along any smooth path p(t) inP we have,

tr[(d

dtlog p(t))2] ≤ tr[(p(t)−1p′(t))2],

we see that infinitesimally and hence globally dp ≤ d.

Now for X ∈ p, DX = sinh(adX/2)adX/2 . By Corollary 6.5.7 we have,

SpecDX = sinh(λ/2)

λ/2: λ ∈ Spec adX.

As sinh tt is analytic, by continuity sinh t

t → 1 as t → 0. This tells usthat from the formulas for tr[(DX(Y ))2] and tr(Y 2), if X → 0, thenindependently of Y , tr[(DX (Y ))2] can be made as near as we want totr(Y 2). This last statement implies that for p and q in a sufficientlysmall neighborhood of a point p0, which by transitivity of G we mayassume to be I, the nonpositively curved symmetric space and Euclideandistances approach one another.

limp,q→p0

d(p, q)

dp(log p, log q)= 1.

This has the interesting philosophical consequence that in the nearbypart of the universe that man inhabits, because of experimental error inmaking measurements, nonpositively curved symmetric space distancesand Euclidean ones are (locally) indistinguishable. As we shall showbelow, angles at I are in any case identical. This means no experimentcan tell us if we “really” live in a hyperbolic or Euclidean world.


Corollary 6.5.12. If p = logX ∈ P , the 1-parameter subgroup exp tXis the unique geodesic in (P, d) joining I with p. Moreover, any twopoints of P can be joined by a unique geodesic.

Proof. Consider a path p(t) in P which happens to be a 1-parametersubgroup. Since p(t) = exp tX, log p(t) = tX and its derivative is X.Thus for each t, log p(t) and its derivative commute. Hence, as weshowed, p(t) and p′(t) also commute. This tells us that all along p(t),dp and d coincide. But the 1-dimensional subspaces of p are geodesicsfor dp. Hence if p = logX ∈ P , the 1-parameter subgroup exp tX isthe unique geodesic in (P, d) joining I with p. Let p and q be distinctpoints of P . Since G acts transitively on P , we can choose g so thatg(q) = I. Connect I with g(p) by its unique geodesic γ. Since Gacts isometrically, g−1(I) = q, g−1(g(p)) = p and g−1(γ) is the uniquegeodesic joining them.

This corollary also follows from more general facts in differential ge-ometry. This is because as a 1-parameter subgroup every geodesic ema-nating from I has infinite length. SinceG acts transitively by isometries,this is true at every point. Hence by the Hopf-Rinow theorem (see [23])P is complete. In particular, any two points can be joined by a shortestgeodesic (also Hopf-Rinow). Being diffeomorphic to Euclidean space, Pis simply connected. If P had nonpositive sectional curvature in everysection and at every point, then this geodesic would be unique. Thislast fact is actually valid for any Hadamard manifold and is called theCartan-Hadamard theorem. We will give a direct proof of completenessof P shortly.

Corollary 6.5.13. A curve p(t) in P is a geodesic through p0 ∈ P ifand only if p(t) = g(exp tX)gt, where X ∈ p and g ∈ G.

Proof. Since G acts transitively by isometries on P , choose g ∈ G sothat gIgt = p0. The result follows from the above since the 1-parametersubgroup exp tX is the unique geodesic in (P, d) beginning at I in thedirection X.

Corollary 6.5.14. At I the angles in the two metrics coincide.


Proof. Let X and Y be two vectors in p and p(t) and q(t) be curvesin P passing through I with tangent vectors X and Y , respectively,and let p0(t) = exp tX and q0(t) = exp tY be two 1-parameter groupsin P . Then since X and Y are also the tangent vectors of p0 and q0,respectively, the angle between p and q equals that between p0 and q0.We may therefore replace p and q by p0 and q0. Now p−1p′q−1q′(0) isjust XY so that tr(p−1p′q−1q′(0)) = tr(XY ).

Corollary 6.5.15. For X ∈ p, d(I, expX) = [tr(X2)]12 .

Proof. The 1-parameter group exp tX is a geodesic in P passing throughI at t = 0. Hence, infinitesimally along this curve, d = dp. Thisimplies the same is true globally along it. Put another way, at eachpoint of exp tX, for 0 ≤ t ≤ 1, the theorem tells us the metric istr[( ddt(tX))2] = tr(X2). Since this is independent of t, integrating from0 to 1 gives tr(X2).

Corollary 6.5.16. For X and Y ∈ p,

d(expX, exp Y ) ≥ (tr[(X − Y )2])12 .

Corollary 6.5.17. P is complete.

Proof. Let pk be a Cauchy sequence in (P, d). By the inequality above,Xk = log pk is a Cauchy sequence in (p, dp) which must converge toX since Euclidean space is complete. By continuity, pk converges toexpX = p.

Corollary 6.5.18. (Law of Cosines). Let a, b and c be the lengths of thesides of a geodesic triangle in P and A, B and C be the correspondingvertices. Then

c2 ≥ a2 + b2 − 2ab cosC

and the sum of the angles A+B +C ≤ π. Moreover, if the vertex C isat I then the equality holds if and only if

(1) The triangle lies in a connected abelian subgroup of P , or equiva-lently,


(2) A+B + C = π.

Proof. Put C at the identity via an isometry from G. Then the Eu-clidean angle at C equals the angle in the metric d. Also, lp(c) ≤ c andlp(a) = a and lp(b) = b. The inequality now follows from the EuclideanLaw of Cosines.

The equality holds if and only if lp(c) = c. This occurs if and only iflog takes the side c to a geodesic in p (i.e. a straight line) of the samelength. This is also equivalent to tr[( ddt log p(t))2] = tr[(p−1p′(t))2], forall t, where p(t) denotes the geodesic side of length c. This occurs if andonly if p(t) satisfies the condition that p(t) and p′(t) commute for all twhich, as we showed, is equivalent to [X,Y ] = 0, where X and Y arethe infinitesimal generators of the sides a and b. Thus the equality inthe Law of Cosines holds if and only if the Euclidean triangle lies in a2-dimensional abelian subalgebra of g contained in p. Equivalently, thegeodesic triangle lies in a 2-dimensional abelian subgroup of G containedin P .

Next we show that in general the sum of the angles is at most π.Since d is a metric and c = d(A,B), etc., it follows that each lengtha, b, or c is less than the sum of the other two. Therefore there is anordinary plane triangle with sides a, b and c. Denote its angles by A′, B′

and C ′. Then A ≤ A′, B ≤ B′ and C ≤ C ′. For by the Law of Cosinesc2 ≥ a2 + b2 − 2ab cosC and c2 = a2 + b2 − 2ab cosC ′. This meanscosC ′ ≤ cosC. But then because C and C ′ are between 0 and π andcos is monotone decreasing there, we see C ≤ C ′. Similarly, this holdsfor the others. Since A′ +B′ +C ′ = π, it follows that A+B + C ≤ π.

If c2 > a2 + b2 − 2ab cosC, then, as above, construct an ordinaryplane triangle with sides a, b and c and angles A′, B′ and C ′. Then sincehere we have a strict inequality, it follows as above that C < C ′. But itis always the case that A ≤ A′ and B ≤ B′. Hence A+ B + C < A′ +B′+C ′ = π. Conversely, if A+B+C = π, then c2 = a2 +b2−2ab cosCand [X,Y ] = 0. Therefore X and Y generate an abelian subalgebra,and the triangle lies in a flat.

Our next result is of fundamental importance. Nonpositive andpositive sectional curvature distinguish the symmetric spaces of non-


compact type from those of compact type.

Corollary 6.5.19. The sectional curvature of P is nonpositive andstrictly negative off the flats. In particular, P is a Hadamard manifold.

Proof. Each geodesic triangle lies in a plane section. We have just shownthat each geodesic triangle in each such section has the sum of the angles≤ π and the sum of the angles < π if we are off a flat. It is a standardresult of 2-dimensional Riemannian geometry (Gauss-Bonnet theorem)that these conditions are equivalent to K ≤ 0 and K < 0, respectively,where K denotes the Gaussian curvature of the section, that is, thesectional curvature.

Remark 6.5.20. We remark that when X,Y ∈ p and are orthonor-mal with respect to the Killing form, one actually has K(X,Y ) =−‖[X,Y ]‖2. See [14] for more details.

Definition 6.5.21. A submanifold N of a Riemannian manifold M iscalled totally geodesic if given any two points of N and a geodesic γ inM joining them, γ lies entirely in N .

Corollary 6.5.22. P is a totally geodesic submanifold in the set of allpositive definite symmetric matrices.

Proof. Let p and q ∈ P be two arbitrary points. Since p12 and p−

12

are self-adjoint, p−12 qp−

12 is positive definite and symmetric. But as we

showed earlier, p−12 ∈ G. Hence p−

12 qp−

12 ∈ G. Because p−

12 is self-

adjoint we see that p−12 qp−

12 ∈ P . Let X ∈ p be its log. Then exp tX

lies in P , for all real t. Therefore γ(t) = p12 (exp tX)p

12 is a geodesic in P .

Clearly, γ(0) = p and γ(1) = q. Therefore there is a unique geodesic inP joining p and q, which means P is a totally geodesic submanifold.

We conclude this section with the standard definition of a symmetricspace.

Definition 6.5.23. A Riemannian manifold M is called a symmetricspace if for each point p ∈ M there is an isometry σp of M satisfyingthe following conditions.

6.6 The Conjugacy of Maximal Compact Subgroups 289

(1) σ2p = I, but σp 6= I,

(2) σp has only isolated fixed points among which is p,

(3) dpσ = −idTpM .

Thus the main feature of the definition is that for each point p there isan isometry which leaves p fixed and reverses geodesics through p.

Corollary 6.5.24. P is a symmetric space.

Proof. Since G acts transitively and by isometries, we may restrict our-selves to the case p = I. Take σI = σ(p) = p−1, for each p ∈ P . Thismap is clearly of order 2. If p is σ fixed, then p2 = I. Since p ∈ P ,pt = p = p−1, hence p ∈ K ∩ P which is trivial. Thus I is the onlyfixed point. Let p = expX, then σ(p) = exp(−X) so that d(σp)p = −I,where here we identify TI(P ) with p.

It remains to see that σ is an isometry. For a curve p(t) in P , sincep(t)p(t)−1 = I, differentiating tells us

p(t)d

dt(p(t)−1) +

dp

dtp(t)−1 = 0,

hence,

p(t)d

dt(p(t)−1) = −dp

dtp(t)−1.

By taking the trace we obtain

tr[(p(t)d

dt(p(t)−1))2] = tr[(

dp

dtp(t)−1)2] = tr[

dp

dtp(t)−1dp

dtp(t)−1]

= tr[(p(t)−1 dp

dt)2]

Hence σ is an isometry of P and the latter is a symmetric space.

6.6 The Conjugacy of Maximal Compact Sub-

groups

The theorem on the conjugacy of maximal compact subgroups of G inthe present context is due to E. Cartan. Actually, the result is true for


an arbitrary connected Lie group, and due to K. Iwasawa, and in thecase of a Lie group with finitely many components, to G.D. Mostow.In this more general context see [33]. We shall deal with this problemin the present context by means of Cartan’s fixed point theorem whichstates that a compact group of isometries acting on a complete, sim-ply connected Riemannian manifold of nonpositive sectional curvature(Hadamard manifold) has a unique fixed point. However, here we willprove the fixed point theorem where we need it, namely, in the specialcase when the manifold is a symmetric space of non-compact type.

Theorem 6.6.1. Let f : C → (P, d) be a continuous map where ddenotes the distance on a symmetric space P of non-compact type andC is a compact space with a positive finite regular measure, µ. Then thefunctional

J(p) =

∫

Cd2(p, f(c))dµ(c), p ∈ P

attains its minimum value at a unique point of P called the center ofgravity of f(C) with respect to µ.

Proof. Fix a point p0 ∈ P . Since C is compact, there is a ball Br(p0)centered at p0 such that if p /∈ Br(p0) then J(p) > J(p0). As the closureof Br(p0) is compact, J takes its minimum at some point q0 ∈ Br(p0).To prove that q0 is unique, it suffices to show that

J(q) > J(q0), if q 6= q0.

Let q(t) be the geodesic joining q and q0, q(0) = q0 and q(1) = q. ByLemma 6.6.2 below

d

dtd2(q(t), f(c)) =

‖q′(t)‖d(q(t), f(c)) cos αt(c) if f(c) 6= q(t),

0 otherwise.

where αt(c) is the angle between the unique geodesic f(c)q(t) and q(t)q.One can prove that the map (t, c) 7→ d

dtd2(q(t), f(c)) is continuous, we

leave this as an exercise to the reader. So t 7→ J(q(t)) is differentiableand since t = 0 is a minimal point for J(qt), by differentiating we obtain

‖q(0)′‖∫

Cd(q0, f(c)) cosα0(c)dµ(c) = 0.


which implies ∫

Cd(q0, f(c)) cosα0(c)dµ(c) = 0. (6.1)

Since the curvature is non-positive, by cosine inequality, if f(c) 6= q0then

d2(q, f(c)) ≥ d2(q0, f(c))+ d2(q0, q)− 2d(q0, q)d(q0, f(c)) cos(π−α0(c)).

A similar inequality trivially holds if f(c) = q0. After integrating bothsides and using (6.1) we get,

J(q) ≥ J(q0) + d2(q0, q),

which proves that J(q) > J(q0).

Lemma 6.6.2. Let q(t) be a curve not passing through p ∈ P . Then

d

dtd(q(t), p)|t=0 = ‖q′(0)‖ cos α

where α is the angle between the geodesic pq(0) and q(0)q(1).

Proof. Let Q(t) be the a curve in the tangent space TpP such thatexpp(Q(t)) = q(t), where expp : TpP → P is the exponential map at thep. We also think of p as the origin of TpP . Then

d

dtd(q(t), p)|t=0 = limt→0

1

t(d(q(t), p) − d(q(0), p))

= limt→01

2d(q(0), p)t(d2(q(t), p) − d2(q(0), p))

= limt→01

2dp(Q(0), p)t(d2p(Q(t), p) − d2

p(Q(0), p)),

(6.2)

where dp is the metric of the tangent space TpP . By the cosine law inthe Euclidean space TpP we have

dp(Q(t), p)2 − dp(Q(0), p)2 = dp(Q(t), Q(0))2

+ 2dp(Q(0), p)dp(Q(0), Q(t)) cos β(t)


where β(t) is the angle between the lines pQ(0) and Q(0)Q(t). Let Ltbe the arc length from Q(0) to Q(t). Then we have

limt→0dp(Q(t), Q(0))

Lt= 1

and

limt→0Ltt

= ‖Q′(0)‖,Combining these we get

limt→0dp(Q(t), Q(0))2

t= 0.

Continuing with (6.2), we have

q

q(t )

q(t)

q(0)

α

d

dtd(q(t), p)|t=0 =

dp(Q(0), Q(t))

tcosβ(t) = ‖Q′(0)‖ cos β(0) (6.3)

Writing Q′(0) = X1+X2 whereX1 is in the direction of the line pQ0 andY1 is perpendicular to it. This orthogonal decomposition is preservedunder the map dQ0 exp and ‖dQ0 exp(X1)‖ = ‖X1‖, therefore,

‖Q′(0)‖ cos β(0) = ‖X1‖ = ‖dQ0 expp(X1)‖ = ‖dQ0 expp(Q′(0))‖ cos α

= ‖q′(0)‖ cos α


As usual, G is a self-adjoint essentially algebraic subgroup ofGL(n,R), or GL(n,C) acting on P by (g, p) 7→ gtpg. The followingis Cartan’s fixed point theorem for symmetric spaces of non-compacttype.

Corollary 6.6.3. If C is a compact subgroup of G, then C has a si-multaneous fixed point acting on P .

Proof. Let µ = dc be the normalized Haar measure on C, p0 a point ofP and f : C → P be the continuous function given by f(c) = c · p0.Then J(p) =

∫C d

2(p, c · p0)dc. Now for c′ ∈ C, J(c′p) =∫C d

2(c′p, c ·p0)dc. Since C acts by isometries this is

∫C d

2(p, (c′)−1c · p0)dc. By leftinvariance of dc we get

∫C d

2(p, c · p0)dc. Thus J(p) = J(c · p) for allc ∈ C and p ∈ P . But by Theorem 6.6.1, J has a unique minimum valueat some p ∈ P . This means c.p = p for all c ∈ C since J(p) = J(c · p).Therefore p is a simultaneous fixed point.

We now prove the conjugacy theorem for maximal compact sub-groups of G. The proof in [33] is similar to the one given here, butrather than involving differential geometry itself, it uses a convexityargument and a function which mimics the metric.

Theorem 6.6.4. Let G be a self-adjoint essentially algebraic subgroupof GL(n,R), or GL(n,C). Then all maximal compact subgroups of Gare conjugate. Any compact subgroup of G is contained in a maximalone.

Proof. Let C be a compact subgroup of G. By Corollary 6.6.3 there isa point p0 ∈ P fixed under the action of C. Thus C ⊆ StabG(p0). SinceG acts transitively so StabG(p0) = gKg−1 for some g ∈ G. Since Kis a maximal compact subgroup by Theorem 6.3.5, so is the conjugategKg−1. This proves the second statement. If C is itself maximal thenC = gKg−1.


6.7 The Rank and Two-Point Homogeneous

Spaces

Let g be the Lie algebra of G a self-adjoint algebraic subgroup ofGL(n,R) or GL(n,C), as discussed earlier in this chapter. Let g = k⊕p

be a Cartan decomposition. By abuse of notation we shall call a sub-algebra of g contained in p a subalgebra of p. Such subalgebras areabelian since [p, p] ⊂ k and they will play an important role in what fol-lows. By finite dimensionality, maximal abelian subalgebras of p clearlyexist. In fact, any abelian subset of p is contained in a maximal abeliansubalgebra of p.

Consider the adjoint representation of K on g. Then the subspace p

is invariant under this action. Since Ad k(p) ⊆ g for k ∈ K, to see thiswe only need to check that Ad k(p) is symmetric (Hermitian). We shallalways deal with the symmetric case except when the Hermitian one isharder. So for X ∈ p and k ∈ K we have Ad k(X) = kXk−1 = kXkt.Hence the transpose is (kXkt)t = kXkt = Ad k(X).

Theorem 6.7.1. In g any two maximal abelian subalgebras a and a′

of p are conjugate by some element of K. In particular, their commondimension is an invariant of g called r = rank(g).

This theorem was originally proved by E. Cartan. Here we adaptthe argument of Theorem 4.3.1.

Proof. Let 〈·, ·〉 be the Killing form on g. This is positive definite onp and negative definite on k. Since K is compact and acts on g, byaveraging with respect to Haar measure on K we can, in addition,assume this form to be K-invariant. That is, each Ad k preservesthe form. Let A ∈ a and A

′ ∈ a′

and consider the smooth numer-ical function on K given by f(k) = 〈Ad k(A), A

′〉. By compactnessof K, this continuous function has a minimum value at k0 and bycalculus, at this point the derivative is zero. Thus for each X ∈ k,ddt〈Ad(exp tX · k0)(A), A

′ 〉|t=0 = 0. But 〈Ad(exp tX · k0)(A), A′〉 =

〈Ad(exp tX)Ad k0(A), A′〉 = 〈Exp(t adX)Ad k0(A), A

′〉. Hence dif-ferentiating with respect to t at t = 0 gives 〈adX Ad k0A,A

′〉 = 0

6.7 The Rank and Two-Point Homogeneous Spaces 295

for all X ∈ k. A calculation similar to the one just given showsthat the K-invariance of the form on k has an infinitesimal version,〈[X,Y ], Z〉 + 〈Y, [X,Z]〉 = 0, valid for all X ∈ k and Y,Z ∈ p. Hence,also for all X ∈ k, we get 〈x, [Ad k0(A), A

′]〉 = 0. Now Ad k0(A)

and A′ ∈ p and [p, p] ⊆ k. Hence [Ad k0(A), A

′] ∈ k and because

〈X, [Ad k0(A), A′]〉 = 0 for all X ∈ k and 〈·, ·〉 is nondegenerate on

k, it follows that [Ad k0(A), A′] = 0. Now hold A ∈ a fixed. Because

[Ad k0(A), a′] = 0 we see by maximality of a

′that Ad k0(A) ∈ a

′and

since A is arbitrary Ad k0a ⊆ a′. Thus a ⊆ Ad k−1

0 (a′). The latter is

an abelian subalgebra of p and by maximality of a they coincide. ThusAd k0(a) = a

′.

It might be helpful to mention the significance of this theorem in themost elementary situation, namely, when G = GL(n,R), or GL(n,C).As usual, we restrict our remarks to the real case. Here p is the set ofall symmetric matrices of order n. Let d denote the diagonal matrices.These evidently form an abelian subalgebra of p. Now d is actually max-imal abelian. To see this, suppose there were a possibly larger abeliansubalgebra a. Each element of a is diagonalizable being symmetric.Since all these elements commute they are simultaneously diagonaliz-able. This means, in effect, that a = d. Thus d is a maximal abeliansubalgebra of p. Similarly, over C it says any commuting family of Her-mitian matrices is simultaneously conjugate by a unitary matrix to thediagonal matrices. This is exactly the content of the theorem in thesetwo cases. Thus Theorem 6.7.1 is a generalization of the classic resulton simultaneous diagonalization of commuting families of quadratic orHermitian forms.

We also note that the statement of Theorem 6.7.1 without the stipu-lation that the subalgebras are in p is false. That is, in general, maximalabelian subalgebras of g are not conjugate. For example, in g = sl(2,R),the diagonal elements, the skew symmetric elements and the unitrian-gular elements are each maximal abelian subalgebras of g, but no twoof them are conjugate (by an element of K or anything else, see Section4.5).


Corollary 6.7.2. In g let a be a maximal abelian subalgebra of p. Thenthe conjugates of a by K fill out p, that is,

⋃k∈K Ad k(a) = p. Of

course, exponentiating and taking into account that exp commutes withconjugation, this translates on the group level to P =

⋃k∈K kAk

−1,where A is the connected abelian subgroup of G with Lie algebra a.

This corollary is the analogue for non-compact groups of Corollary4.3.9.

Proof. Let X ∈ p and choose a maximal abelian subalgebra a′contain-

ing it. By our theorem there is some k ∈ K conjugating a′

to a. Inparticular, Ad k(X) ∈ a for some k ∈ K and so X ∈ Ad k−1(a).

Our next corollary, also called the Cartan decomposition, followsfrom this last fact together with the usual Cartan decomposition, The-orem 6.3.5.

Corollary 6.7.3. Under the same hypothesis G = KAK.

Proof. G = KP ⊆ KKAK = KAK ⊆ G.

Remark 6.7.4. An important use of this form of the Cartan decom-position is that it reduces the study of the asymptotic at ∞ on G to A.That is, suppose gi is a sequence in G tending to ∞. Now gi = kiaili,where ki and li ∈ K and ai ∈ A. Since both ki and li have conver-gent subsequences, again denoted by ki and li, which converge to k andl, respectively, the sequence ai must also tend to ∞. Thus in certainsituations we can assume the original sequence started out in A.

We now make explicit the notions of a homogeneous space and two-fold transitivity from differential geometry mentioned earlier. If X is aconnected Riemannian manifold, we shall say X is a homogeneous spaceif the isometry group Isom(X) acts transitively on X. Now even whenthe action may not be transitive it is a theorem of Myers and Steenrod(see [32]) that Isom(X) is a Lie group and the stabilizer Kp of any pointp is a compact subgroup. In the case of a transitive action it followsfrom general facts about actions that X is equivariantly equivalent as aRiemannian manifold to Isom(X)/Kp with the quotient structure. Of

6.7 The Rank and Two-Point Homogeneous Spaces 297

course, if some subgroup of the isometry group acted transitively thenthese same conclusions could be drawn replacing the isometry group bythe subgroup. Clearly, by its very construction, every symmetric spaceof non-compact type is a homogeneous space.

Now suppose in our symmetric space P we are given points p and qand p

′and q

′of P with d(p, q) = d(p

′, q

′). We shall say that a subgroup

of the isometry group acts two-fold transitively if there is always anisometry g in the subgroup taking p to p

′and q to q

′for any choices of

such points. When this occurs we shall say P is a two-point homogeneousspace. Clearly, every two-point homogeneous space is a homogeneousspace. As we shall see the converse is not true and we will learn which ofour symmetric spaces is actually a two-point homogeneous space. Beforedoing so, we make a simple observation which follows immediately fromtransitivity.

Proposition 6.7.5. Let G be as above and K be a maximal compactsubgroup. Then G/K = P is a two-point homogeneous space if and onlyif K acts transitively on the unit geodesic sphere U of P .

For example, whenG = SO(n, 1)0 andK = SO(n), thenG/K = Hn,hyperbolic n-space. Here K acts transitively on U . Hence SO(n, 1)0acts two-fold transitively on Hn. As we shall see in Theorem 6.7.6,this fact is a special case of a more general result. We also remark thatthis definition can be given for any connected Riemannian manifold andindeed such a manifold is of necessity, a symmetric space (see [32]).

Our last result tells us the significance of the rank in this connection.Before proving it we observe that for all semisimple or reductive groupsunder consideration dim p ≥ 2. The lowest dimension arising, is thecase of the upper half plane introduced at Section 6.4. Indeed, supposedim p = 1. Then since p is abelian and exp is a global diffeomorphismp → P , it follows easily from exp(X + Y ) = expX expY , where X,Y ∈p, that P is a connected 1-dimensional abelian Lie group. Now sinceK acts on P by conjugation, and in this case these form a connectedgroup of automorphisms of P we see that this action is trivial becauseAut(P )0 = 1. Thus K centralizes P and we have a direct product ofgroups. Such a group is not semisimple. It is clearly also not GL(n,R)


or GL(n,C) for n ≥ 2.We now characterize two-point homogeneous symmetric spaces.

Theorem 6.7.6. Let G be as above, g be its Lie algebra and K bea maximal compact subgroup. Then G/K is a two-point homogeneousspace if and only if rank(g) = 1.

Proof. We first assume rank(g) = 1. By Proposition 6.7.5, to see thatG/K is a two-point homogeneous space, it is sufficient to show K actstransitively on geodesic spheres of P . Of course, we know AdK actslinearly and isometrically on p. Now by Corollary 6.7.2

⋃k∈K Ad k(a) =

p. Hence each point p ∈ U is a conjugate by something in K to a pointon the unit sphere of a. Since the dimension of this sphere is zero, itconsists of two points, ±a0. Hence U = AdK(a0)∪AdK(−a0). In anycase, U is a union of a finite number of orbits all of which are compactand therefore closed since K itself is compact. Since these are closed,so is the union of all but one of them. Hence U is the disjoint unionof two nonempty closed sets. This is impossible since U is connectedbecause dim p ≥ 2. Thus there is only one orbit and therefore K actstransitively on U .

Before proving the converse, the following generic example will beinstructive. Let G = SL(n,R), n ≥ 2. We shall see SL(n,R)/SO(n)is a two-point homogeneous space if and only if n = 2. This suggeststhat unless the rank = 1, one can never have a two-point homogeneoussymmetric space.

To see this, observe that since G/K = P is the set of positive definite

n × n symmetric matrices of det 1, it follows that dimP = n(n+1)2 − 1.

Also dimK = n(n−1)2 . Hence if U denotes the geodesic unit sphere in P ,

its dimension is n(n+1)2 − 2. Let K act on P and U by (k, p) 7→ kpk−1 =

kpkt. For p ∈ U the dimension of OK(p), the K-orbit of p, is

dimOK(p) =n(n− 1)

2− (n− 1)(n − 2)

2= n− 1.

Now if K were to act transitively on U , then dimOK(p) = dimU . That

is, n − 1 = n(n+1)2 − 2. Alternatively, (n − 2)(n + 1) = 0. Since n ≥ 2,

this holds if and only if n = 2.

6.8 The Disk Model for Spaces of Rank 1 299

We conclude by proving the converse. Suppose (P,G) is a two-pointhomogeneous space and hence K acts transitively (by conjugation) onthe unit geodesic sphere U in p. Then U = OK(A0), where A0 ∈ p and‖A0‖ = 1. Since A0 is conjugate to something in a, we may assumeA0 ∈ a. In particular, everything in U ∩ a is K-conjugate to everythingelse. Because these matrices commute, they can be simultaneously di-agonalized by some u0 (which may not be in K). By replacing theseA0’s by their u0 conjugates we may assume they are all diagonal. Beingconjugate under K these matrices have the same spectrum S. Since Sis finite and K is connected, K cannot permute this finite set. Thus theaction of K leaves each of these matrices fixed. But K acts transitivelyon U ∩a so U∩a must be a point. Hence it has dim 0 and dim a = 1.

Exercise 6.7.7. Show the rank of Sp(n,R) is n.

6.8 The Disk Model for Spaces of Rank 1

In this section we focus on the classical simple groups of rank 1 andtheir associated irreducible symmetric spaces which we view in the diskmodel. We will then use the geometry of the latter to indicate thatthe exponential map of the corresponding centerless simple group issurjective.

In this section, whose material is mostly taken from [19], we unifythe study of the three infinite families of classical simple rank 1 groupsor classical non-compact irreducible rank 1 symmetric spaces, Hn(F ) byconsidering a field F , where F = R, C, or H, the quaternions and defineG = U(n, 1, F ) as follows: Let Fn+1 be the right vector space over Fconsisting of (n + 1)-tuples of points from F . For such (n + 1)-tuplesx = (x0, . . . xn) and y = (y0, . . . yn) ∈ Fn+1, consider 〈·, ·〉 defined

〈x, y〉 = x0y0 −n∑

i=1

xiyi.

This is a nondegenerate form over F which is linear in x and conjugatelinear in y, where the conjugation is the natural one coming from F .(In the case F = R conjugation is the identity). G = U(n, 1, F ) is then


defined to be those g ∈ GL(n + 1, F ) which preserve this form. G isevidently a Lie group, we denote its Lie algebra by u(n, 1, F ). Just asin Section 6.3 above, G is actually a self-adjoint algebraic subgroup ofGL(n+ 1, F ).

When F = R taking the identity component we get SO(n, 1)0 whichis centerless and simple, When F = C, U(n, 1,C) is merely reductive andnon-semisimple. But, of course, Ad(U(n, 1,C)) = PSU(n, 1) is center-less and simple as is Ad(U(n, 1,H)) = Ad(Sp(n, 1)). A direct calculationtells us that respectively K = SO(n), U(n), and Sp(n), operating on Rn,Cn, Hn in the usual manner. The unit sphere, U , being Sn−1, S2n−1,S4n−1, respectively. Thus in all three cases K operates transitively onU . From this it follows that AdG operates 2-fold transitively on G/Kand hence AdG always has rank 1. Using classification, these are theclassical irreducible non-compact rank 1 symmetric spaces and exceptfor one irreducible rank 1 symmetric space of non-compact type (or realsimple non-compact Lie group) this accounts for all non-compact, cen-terless, simple groups of rank 1 (see [32]). The missing one is calledthe exceptional group is related to the Cayley numbers which we willnot deal with here. However all the properties of the disk model of theexceptional rank 1 symmetric space are actually the same as those ofthe classical ones.

Let P (Fn+1) be the projective space corresponding to Fn+1 andπ : Fn+1 \ 0 → P (Fn+1) the canonical map taking x 7→ [x]. NowGL(n + 1, F ) operates of P (Fn+1) through Fn+1. Thus if g ∈ GL(n +1, F ) and x ∈ Fn+1−(0) we take g[x] = [g(x)]. This map is well-definedand gives an action of GL(n + 1, F ) on P (Fn+1) and upon restrictionthe same is true of any subgroup of GL(n+ 1, F ). Thus U(n, 1, F ) = Gacts on P (Fn+1). Let

Ω = πx ∈ Fn+1 \ 0 : 〈x, x〉 > 0,

be the projective image of the interior of the light cone. Since Gpreserves 〈·, ·〉, it leaves Ω invariant. Now let x = (x0, . . . xn) be avector with [x] ∈ Ω. Then |x0|2 >

∑ni=1 |xi|2. Next we show that

G operates transitively on Ω. Let y = (y0, . . . yn) ∈ Fn+1 and as-sume |y0|2 − ∑n

i=1 |yi|2 = 1. Chose t ≥ 0 so that |y0| = cosh t and


√∑ni=1 |yi|2 = sinh t. Choose u ∈ U(1, F ) and v ∈ U(n,F ) so that

y0 = u cosh t and (y1, . . . , yn) = v(0, . . . , 0, sinh t). This is possible since

U(n,F ) operates transitively on spheres in Fn for all n. Let

k =

(u 00 v

),

Then k ∈ K. If x0 is the point in P (Fn+1) represented by (1, 0 . . . , 0),then x0 ∈ Ω. We show that y = katx0, where at = Exp tX0 and thematrix X0 of order n+ 1 given by

X0 =

0 v0 1w0 O w0

1 v0 0

,

where v0 = (0, . . . , 0) of order n − 1, w0 = vt0 and O is the zero matrixof order n − 1. Hence X0 ∈ P as in Section 6.2. Since the rank of Gis 1 because it operates two fold transitively, therefore at : t ∈ R is amaximal abelian subgroup of P . A direct calculation shows that

at =

cosh t v0 sinh tw0 I w0

sinh t v0 cosh t

,

where v0 = (0, . . . , 0) of order n − 1, w0 = vt0 and I is the identitymatrix of order n− 1. Hence at(1, 0, . . . , 0) = (cosh t, 0, . . . 0, sinh t) andtherefore

kat(1, 0, . . . , 0) = k(cosh t, 0, . . . 0, sinh t) = (u cosh t, v(0, . . . , 0, sinh t))

= (y0, y1, . . . , yn).

Since K and at are connected we actually get G0 acts transitively onΩ. Here we take the upper component of the hyperboloid y2

0 = 1 +∑ni=1 |yi|2, y0 > 0, projectivizing this, gives everything.We now calculate the isotropy group of [(1, 0, . . . , 0)] within G0.

Since this group is connected we can calculate the isotropy group withinthe Lie algebra and exponentiate. Here we take

X =

(A BB∗ C

),


where A = −A ∈ F , B ∈ Fn, B∗ is B conjugate transposed and C is ann×n matrix from F with C∗ = −C. Since [(1, 0, . . . , 0)] = [(λ, 0, . . . , 0)]for any λ 6= 0 ∈ F , if X ∈ Stabg[(1, 0, . . . , 0)], then X(λ, 0, . . . , 0) = 0so B∗ = 0. Hence also B = 0. Therefore A ∈ u(1, F ) and C ∈ u(n,F ),and StabG0 [(1, 0, . . . , 0)] = K (see Corollary 1.4.15). Thus Ω = G/K,where G is the connected component of the identity in G and K is amaximal compact subgroup of G0.

Finally, we come to the ball model of Hn(F ). We will denote thisby

B(Fn) = (x1, . . . , xn) ∈ Fn :

n∑

i=1

|xi|2 < 1.

Define φ : Ω → B(Fn) as follows. If x ∈ Ω, then |x0|2 > 0. Since x0 6= 0we can form xix

−10 for each i. Let φ be defined by (x0, . . . , xn) 7→

[(x1x−10 , . . . , xnx

−10 )]. Then for x ∈ Ω, |x1x

−10 |2 + . . .+ |xnx−1

0 |2 < 1. Soφ(x) ∈ B(Fn). Conversely, let (y1, . . . , yn) ∈ B(Fn). Then

∑ni=1 |yi|2 <

1. Therefore, (1, y1, . . . , yn) ∈ Ω and φ(1, y1, . . . , yn) = (y1, . . . , yn).Evidently, φ−1(y1, . . . , yn) = x0, y1x0, . . . , ynx0 : x0 6= 0. Thus φmaps Hn(F ) bijectively to B(Fn). How does G operate on B(Fn)? Letg ∈ G and y ∈ B(Fn). Then

(gy)i = (gi0 +

n∑

j=1

gijyj)(g00 +

n∑

j=1

g0jyj)−1.

Thus,

1 − ‖gy‖2 = (1 − ‖y‖2)(g00 +n∑

j=1

g0jyj)−2.

It follows that g operates by the same formula on the boundary,∂(B(Fn)) = y ∈ Fn : ‖y‖2 = 1, as well. We conclude this sec-tion with introduction the Lorentz model. and its connection of the ballor disk model.

We show [x] = [y], for some y = (y0, · · · , yn) such that |y0|2 −∑ni=1 |yi|2 = 1. That is, we can change coordinates and sharpen the

inequality to an equality. We can assume that y0 ≥ 0, otherwise wereplace (y0, . . . yn) by (−y0, . . . − yn). Therefore Ω is diffeomorphic to


L = (y0, . . . yn) ∈ Fn+1 : y0 =

√√√√1 +

n∑

i=1

|yi|2, y0 > 0.

which is called the hyperboloid or Lorentz model .

L

0x

n B( F )

light cone

Figure 6.1: Lorentz model and disk model

To see this just choose λ 6= 0 ∈ F . Then xλ = (x0λ, . . . xnλ) =y. Writing down the equation we want for y we see that (|x0|2 −∑n

i=1 |xi|2)|λ|2 = 1. Since |x0|2 >∑n

i=1 |xi|2 the first term is nonzeroforcing

λ =

√1

|x0|2 −∑n

i=1 |xi|2∈ R ⊆ F.

Now the projection from the hyperboloid with respect to vertex of the


light cone (1, 0 . . . , 0) to the unit disk in 0 × Rn ⊂ Rn+1 identifiesthese two models.

6.9 Exponentiality of Certain Rank 1 Groups

As above we call a connected Lie group exponential if exp is surjective,or alternatively if every point lies on a 1-parameter subgroup. For ex-ample, compact connected groups are exponential (see Theorem 4.3.8),but often non-compact simple groups are not. The purpose of this sec-tion is to indicate that the groups SO(n, 1)0, n ≥ 2, PSU(n, 1), n ≥ 1and Ad Sp(n, 1), n ≥ 1 are all exponential [52]. We remark that theexceptional non compact, centerless, rank 1 simple group, AdF(4,−20),was proved to be non exponential by D. Djokovic and N. Thang in [17]and we will see where our line of argument breaks down in this case.

Theorem 6.9.1. The exponential map is surjective for all classical,connected, non-compact, centerless, rank 1 simple Lie groups.

This will be proven by means of geometry; that is, studying theaction of G as the connected component of the isometry group of thesymmetric space X = G/K. A self evident principle which we shallemploy here is the fact that if a Lie group is a union of exponential Liesubgroups, it must itself be exponential.

Now G/K is diffeomorphic to the interior of the closed unit ballBn of dimension n, where K is a maximal compact subgroup of G andn = dim(G/K). As we noted above each isometry g of G extendscontinuously to the boundary of G/K. Thus G acts on Bn and since Gis arcwise connected each isometry g is homotopic to the identity. Bythe Brouwer fixed point theorem each g has a fixed point in Bn. If this isin the interior of the ball, that is in G/K, then since G acts transitivelywith K as the isotropy group of 0, g lies in some conjugate of K. Butbecause K is compact and connected, g lies on a 1-parameter subgroupof this conjugate of K and therefore of G. The other possibility is thatthere is a g-fixed point p on the boundary. Let Kp be the compact Liesubgroup of K consisting of the isometries leaving p fixed. We shall firstprove

6.9 Exponentiality of Certain Rank 1 Groups 305

Proposition 6.9.2. Kp is connected.

Proof. By Proposition 6.7.5 and Theorem 6.7.6, K acts transitively onall geodesic spheres centered at 0. Now each point on the boundary ofBn lies on a unique geodesic emanating from 0. Let q and r be boundarypoints and γq(t) and γr(t), t ∈ R be the corresponding geodesics. Thenfor n ∈ Z+, γq(n) = qn and γr(n) = rn converge respectively to q andr and for each such n there is a kn ∈ K such that kn(qn) = rn. Bycompactness some subsequence, which we again call kn → k ∈ K. Sincealso the corresponding subsequences qn → q and rn → r, and isometriesextend continuously to the boundary we see that k(q) = r. Thus Kalso acts transitively on the boundary of Bn and therefore K/Kp is(K-equivariantly) homeomorphic with Sn−1. But the latter is simplyconnected for n > 2 and K is connected so we conclude from the longexact sequence for homotopy of a fibration that Kp is also connected.This leaves only the case n = 2, i.e. the hyperbolic disk. Here, by directcalculation, one sees easily that Kp = 1.

Continuing the proof of Theorem 6.9.1, let Gp denote the subgroupof all elements of G fixing p. Then according to p.154 of [61]

Gp = KpAUp, (6.4)

where Up is the unipotent radical of Gp. In particular, Up is normalin Gp and connected and simply connected. This means, in particular,that Hp = KpUp is a subgroup of Gp and that Up contains no nontrivialcompact subgroups. Thus Up∩Kp = 1 and Hp = Kp×ηUp (semidirectproduct).

Now suppose that there is another g-fixed point q on the boundary.In this case p and q can be joined by a unique geodesic γ of G/K. Sinceg is an isometry, g(γ) is also a geodesic joining p and q and thereforeit must coincide with γ so that γ is stabilized by g. By conjugation byan isometry in G taking a point on the interior of γ to 0 we can assumethat p and q are opposite to one another. Since as we saw earlier Kacts transitively on the boundary we can also assume that q is anyparticular boundary point. Passing to the upper half space model wecan therefore take q = ∞, p = 0 then γ is a vertical half line. In this


model if g(∞) = ∞ then using (6.4), g = θλu. where λ > 0, θ ∈ Kp

and u ∈ Up. Since u is unipotent,

g(x, z) = (λθ(x) + b(u), λz) (6.5)

for x ∈ Rn and z > 0 where b(u) ∈ Rn. Now if in addition g(0) = 0then b(u) = 0 and

g(x, z) = (λθ(x), λz) (6.6)

and so on γ our map is z 7→ λz. Now compose with the isometryh : (x, z) 7→ λ−1(x, z). By (6.6), h fixes 0 and ∞ and the compositionleaves γ pointwise fixed and has many fixed points in the interior, so∈ K. In fact the composition is in Kp and hence g ∈ R×

+ ×Kp (directproduct). Thus the original g lies in H, a conjugate by something in Gof this direct product subgroup. (Notice that since A commutes withKp and hg = θ leaves γ fixed it must leave the orthogonal hyperplanein the disk model i.e. the boundary in the upper half space invariant.Hence by (6.6) so does g. In any case, since Kp is connected, such agroup is clearly of exponential type (this will also be included in theresult below), so g = expY , where Y ∈ the Lie algebra of H.

If g has more than two fixed points say p, q and r on the boundary,reasoning as above we would then have, in the upper half space modelq = ∞, p = 0, γ is a vertical half line and r is a boundary point 6= p.Hence, by (6.6), θ = 1. But then, after applying g, the distance d(p, r)is multiplied by λ. Since both points are g-fixed this means that λ = 1and g = I and so every interior point is also g-fixed.

The only other possibility is that p is the unique g-fixed point onthe boundary. Passing again to the upper half space model we cantake p = ∞. Since Gp is a group g−1 also leaves p fixed. Clearly1/λ(g) = λ(g−1) and so if λ(g) 6= 1 then we may assume that each ofthese has norm 6= 1. But then applying (6.5)to a boundary point (x, 0)we have g(x, 0) = (λθ(x) + b(u), 0). This means that x is g-fixed if andonly if x = λθ(x) + b(u) or alternatively (θ − λ−1I)x = −b(u)/λ. Sinceall the eigenvalues of θ are of absolute value 1 and |λ−1| 6= 1 it followsthat (θ − λ−1I) is invertible and this equation can be solved giving ag-fixed point on the (finite) boundary. Hence x 6= ∞ and we are back


in the case of a screw motion (g ∈ R×+ ×Kpe) . Thus we may assume

that λ = 1 and g has no A part and this means g lies in Hp = KpUp(semidirect product)2.

We will complete the proof of Theorem 6.9.1 by showing that in allthree cases, Hp = H is exponential. Now for hyperbolic space since Nis abelian, this can be done directly as follows.

Theorem 6.9.3. The identity component, SO(n) ×η Rn = H, of theEuclidean motion group is exponential.

That is, we will show the exponential map is surjective for the con-nected group of isometries of a (simply connected) space form of zerocurvature,

Proof. Take faithful matrix representations of H and its Lie algebra h

of order n+ 1 as follows.

H = (α v0 1

): α ∈ SO(n), v ∈ Rn

while

h = (X w0 0

): X ∈Mn(R),Xt = −X,w ∈ Rn.

If we denote the elements of h by (X,w), then by a direct calculation

exp(X,w) =

(expX expX−I

X (w)0 1

)

where, by expX−IX we understand the functional calculus, i.e. this matrix

valued function of a matrix argument, has a removable singularity at 0with value I.

2It should be remarked that another way of looking at the three possibilitieswhich arise in the proof below is by means of the following classification scheme forisometries of spaces of negative curvature (X, d), due to M. Gromov (see p.77 of[3]), which, in our situation, gives the following trichotomy for an isometry g. Ifinfx∈X d(x, gx) is 0 and is assumed, or is 0 and is not assumed, or is positive (whenit must be assumed), then g is elliptic, parabolic, or hyperbolic, respectively.


Let (α, v) ∈ H. Since SO(n) is a compact connected Lie group wecan choose X ∈ End(V ), so that expX = α. Then exp(X,w) = (α, v)for some w if and only if expX−I

X (w) = v. Since v is arbitrary this

amounts to knowing that the linear transformation expX−IX is onto, i.e.

all its eigenvalues are 6= 0. But by functional calculus the eigenvalues ofthis operator are either 1, or are of the form eλ−1

λ , where λ is a nonzeroeigenvalue of X. Clearly such an eigenvalue is 0 if and only ifλ = 2πimfor some integer m. Hence this operator is invertible if and only if Xhas no eigenvalues of the form 2πim for some m 6= 0 ∈ Z. Now by anorthonormal change of basis (which affects nothing since exp is constanton conjugacy classes), α and X are respectively given by

α = diag(R(t1), . . . , R(tj), 1, . . . 1),

and

X = diag(S(t1), . . . , S(tj), 0, . . . 0),

where R(tk) and S(tk) are the planar rotation and infinitesimal rotationdetermined by tk:

R(tk) =

(cos tk sin tk− sin tk cos tk

)

S(tk) =

(0 tk

−tk 0

).

We may assume none of the tk is an integral multiple of 2π. For ifthere were such a tk it would just produce additional 1’s in the blockdiagonalization of α, above and we exponentiate onto I by O. Hencewe may assume that each tk 6= 0 and satisfies −π ≤ tk ≤ π. But then,X has no eigenvalues of the form 2πim, where m 6= 0 ∈ Z.

Turning to the other two cases, in general since Kp and Up are bothconnected so is Hp. Denote Up by U and Kp by C. If U were trivial,then H would be compact and connected and so would be of exponentialtype. Otherwise we shall show that ZC(U) = v. In the disk modellet q be the point opposite p on the boundary, k0 6= 1 ∈ C and supposethat k0u = uk0 for all u ∈ U . Then k0u(q) = uk0(q). Since k0 ∈ C and


k0(p) = p it follows that k0(q) also equals q and therefore each U(q) isk0 fixed. On the other hand each u stabilizes each horosphere and inparticular the boundary of Bn. Since the only points on Bn which k0

leaves fixed are p and q, either U(q) = p or q. But if U(q) = p for someu then u−1(u(q)) = q = u−1(p) = p, a contradiction. Thus for each u,

U(q) = q and U(p) = p. Since p and q are also C fixed and U and Cgenerate H, we see that H itself leaves p and q fixed so H is contained inthe a group isomorphic with the direct product of a compact connectedgroup with R×

+ and therefore is also exponential. The remaining case isthat only the identity of C centralizes U so C acts faithfully as a groupof automorphisms of U .

In order to complete the proof of Theorem 6.9.1 in the remaining twocases we must now investigate the exponentiality of certain semidirectproducts where N is non-abelian. In these cases although H is notsolvable, the solvable methods of [56] can be made to apply. Namely, weconsider a compact connected group, L, of automorphisms of a simplyconnected nilpotent group, N with Lie algebra, n, and H = L×ηN , thenatural semi-direct product. Since N is simply connected and nilpotentwe can identify Aut(N)0 with Aut(n)0.

In [56] the following result is proved. Here T (X) is the subgroup ofT fixing X ∈ n.

Theorem 6.9.4. Let N be a connected nilpotent Lie group and G =T×ηN be a semi-direct product of N with a torus. Then G is exponentialif and only if T (X) is connected for each X ∈ n.

Using Theorem 6.9.4 we can get at exponentiality of H as follows.

Corollary 6.9.5. Suppose a maximal torus of T of L is the set ofdiagonal matrices in L whose coefficients vary independently. Then H =L×η N is exponential.

Proof. Let X = (x1, . . . , xn) ∈ n and t = (t1, . . . , tn) ∈ T , we have

T (X) = X ∈ n : t ·X = (t1x1, . . . , tnxn) = X, t ∈ T.

If xi = 0, then there is no condition on ti and T (X)i = T1. Whereasif xi 6= 0, then tixi = xi if and only ti = 1 so T (X)i = 1. In any


case, for each i, T (X)i is connected. As T (X) is a direct product ofits components and these are all connected so is T (X). Since this istrue for every X ∈ n, it follows from Theorem 3 of [56] that T · N isexponential. Now N is L-invariant, and the conjugates of T fill out Lso ⋃

l∈Ll(T ·N)l−1 =

⋃

l∈Ll(T )l−1 ·N = L ·N.

Thus the conjugates of T ·N fill out H. Since exp is constant on con-jugacy classes of H, the latter is also exponential.

Now consider n the Heisenberg Lie algebra of dimension 2n + 1,viewed as Cn⊕ iR and N is the Heisenberg group. Here the bracketingrelations are [v,w] = Im〈v,w〉, where v and w ∈ Cn, 〈−,−〉 is the stan-dard Hermitian form on Cn and all other brackets are zero. Then thenatural action of U(n) on Cn, leaving the center, iR, pointwise fixed isevidently by Lie algebra automorphisms. To identify a maximal com-pact subgroup, L, of Aut(n)0, we proceed as follows: By a calculation(see [47]) one sees that the identity component of the group of measurepreserving automorphisms is Sp(n,R). Since L must be contained inthis group, L is a maximal compact subgroup of Sp(n,R). Thus byTheorem 6.3.5 L is conjugate to Sp(n,R) ∩ SO(2n,R). Its Lie algebrais therefore conjugate to

l = (A B−B A

): At = −A,Bt = B,

and so dim l = n2. Since L is connected and U(n) is a compact con-nected subgroup of Aut(n)0 of the same dimension, by the conjugacy ofmaximal compact subgroups L and U(n) are conjugate. Hence U(n) isa maximal compact subgroup of Aut(n)0. Now a maximal torus of T of

U(n) is the set of diagonal matrices in U(n). Since the action of U(n) onn is linear on Cn and leaves z(n) fixed, it follows that a maximal torusis Tn of U(n) consisting of the diagonal matrices in U(n) together witha 1 in the z(n) component. By Corollary 6.9.5 we see that U(n)×ηN isexponential.


Our last case is n, the Lie algebra of dimension 4n + 3, defined asfollows. n = Hn⊕Im(H), where H is the quaternions and Im(H) is the3-dimensional subspace of pure quaternions. Here we take for bracketingrelations [v,w] = Im〈v,w〉, where v and w ∈ Hn, 〈·, ·〉 is the standardH-Hermitian form on Hn, all other brackets are zero, and Im is theprojection from H onto Im(H). This construction gives us a 2-stepnilpotent Lie algebra with 3index2-step nilpotent dimensional center.Let N be the corresponding simply connected group. Calculations ofthe automorphism group of n in the dissertation of P. Barbano, [4] showthat the natural action of Sp(n) on Hn together with a 3-dimensionalsurjective representation, ρ, acting on the center, Im(H) is a maximalcompact subgroup of Aut(n)0. We shall restrict our attention to Sp(n).A maximal torus of T of Sp(n) is the set of block diagonal matrices(D,D−), where D ∈ U(n) and − is complex conjugation. As above, ateach Hn component either we get a circle, if that component is zero,or a point, if it is nonzero (complex conjugation leaving this situationunchanged). It follows that T (X) is connected for all X ∈ n. Hence, byCorollary 6.9.5, Sp(n)×η N is also exponential. Thus H is exponentialin the remaining two cases and this completes the proof of Theorem6.9.1.

Theorem 6.9.6. Let L be a maximal compact subgroup of Aut(N)0,where N is Heisenberg group, or let L be Sp(n), where N is the sim-ply connected group of Heisenberg type, based on the quaternions givenabove. Then H = L×η N is exponential.

We remark that this type of argument also works for the Euclideanmotion group, SO(n) ×η Rn. However, if one attempts to apply thereasoning above to the exceptional rank 1 group, one must look at theanalogous 15-dimensional Heisenberg type Lie algebra, n, based on theCayley numbers, with 7 dimensional center, z. In this case K = Spin(9),Kp = Spin(7), and the action on z is by the full rotation group, SO(7) =Ad(Spin(7)). Since a maximal torus, T3, of Spin(7) is a two fold coveringof a maximal torus of SO(7) one sees, for suitable X ∈ n, that T3(X)can be finite and nontrivial. This means Kp ×η N is not exponential.Indeed, it could not be, for if it were then AdF(4,−20) would also be


exponential.

Chapter 7

Semisimple Lie Algebras

and Lie Groups

7.1 Root and Weight Space Decompositions

Here we will discuss the root space decomposition and the existenceand fundamental properties of Cartan subalgebras, particularly whenthe complex Lie algebra is semisimple. Let h be a (finite dimensional)complex Lie algebra and ρ : h → gl(V ) a finite dimensional complex Liealgebra representation of h on V . For λ ∈ h∗, its dual, we consider

Vλ = v ∈ V : (ρH − λ(H)I)k(v) = 0, for some k,for all H ∈ h and some integer k. Here k could, in principle, dependon H and v. However, by the Jordan canonical form, for fixed H andv 6= 0 ∈ V , (ρH−λ(H)I)k(v) only takes value 0 for an eigenvalue. Hencek can always be taken to be dimV .

We call Vλ a weight space, its vectors weight vectors and λ is calleda weight . Of course there are at most dimV nonzero weight spaces.

For a fixed H ∈ h it is also convenient to write

Vλ,H = v ∈ V : (ρH − λ(H)I)k(v) = 0,so that

Vλ = ∩H∈hVλ,H .

313

314 Chapter 7 Semisimple Lie Algebras and Lie Groups

The following result shows that complex representations of nilpotentalgebras are rather special.

Theorem 7.1.1. Suppose h is nilpotent. Then

(1) V is the direct sum of the nonzero weight spaces.

(2) Each weight space is ρ-invariant.

Proof. We first show that each Vλ is invariant under h. To do so it issufficient to show each Vλ,H is invariant. Since h is nilpotent we knowby Engel’s theorem adH is nilpotent for all H. Let H 6= 0 ∈ h be fixedand define

hk = Y ∈ h : adHk(Y ) = 0.

Then hk’s form an increasing sequence of sets whose union is h. We firstshow by induction on k that for Y ∈ hk we have ρ(Y )Vλ,H ⊆ Vλ,H . Ifk = 0, then h0 = Y ∈ h : adH0(Y ) = Y = 0 so ρ(0) = 0 which leaveseverything invariant. Now suppose Y ∈ hk. Then [H,Y ] ∈ hk−1 andsince ρ is a representation,

(ρ(H) − λ(H)I)ρ(Y ) = ρ([H,Y ]) + ρ(Y )ρ(H)−λ(H)ρ(Y )

= ρ(Y )(ρ(H) − λ(H)I) + ρ([H,Y ]).

Iterating this several times we get for each integer k,

(ρ(H) − λ(H)I)kρ(Y ) = ρ(Y )(ρ(H) − λ(H)I)k+

k−1∑

j=0

(ρ(H) − λ(H)I)k−1−jρ([H,Y ])(ρ(H) − λ(H)I)j .

(7.1)

Now let v ∈ Vλ,H and (ρ(H) − λ(H)I)n(v) = 0. Take k ≥ 2n. Thenif j ≥ n the right side of (7.1) gives zero. so we can assume j < n. Butthen k− 1− j ≥ 2n− 1− j ≥ n. Note that (ρ(H)− λ(H)I)j(v) ∈ Vλ,H ,and since [H,Y ] ∈ hk−1 we know by induction that ρ([H,Y ]) preservesVλ,H . This means for large enough k

7.1 Root and Weight Space Decompositions 315

(ρ(H) − λ(H)I)k−1−jρ([H,Y ])(ρ(H) − λ(H)I)j(v) = 0

and by (7.1), (ρ(H) − λ(H)I)kρ(Y )(v) = 0. Thus ρ(Y ) stabilizes Vλ,H .This completes the induction and shows each Vλ is invariant under h.

We now turn to the direct sum decomposition. Let H1, . . . ,Hn bea basis of h. By the Jordan canonical form applied to ρ(H1) we getV =

⊕Vci,H1 where the ci are the eigenvalues of ρH1. Here, since we

have a single operator, ci can be regarded as λi(H1) so that Vλi,H1 isthe generalized eigenspace of a single operator defined earlier, which isstable under ρ(h). Hence each of these spaces can be further decomposedunder ρ(H2) so that V =

⊕i,j Vλi,H1 ∩ Vλj ,H2 etc. until we finally get

V =⊕

λ(H1),...,λ(Hn)

∩ni=1Vλ(Hi),Hi.

with each of these spaces stable under ρ(h). But since h is nilpotentand therefore solvable, Lie’s theorem 3.2.18 tells us that the whole Liealgebra h and in particular each of the ρ(Hi), i = 1 . . . , n acts as simul-taneous triangular operators on ∩ni=1Vλ(Hi),Hi

with diagonal entries theλ(Hi). Hence the linear span, ρ(h) must also do this. Taking for ourlinear functional λ(

∑ziHi) =

∑ziλ(Hi) we get our result.

Now let g be a finite dimensional Lie algebra, h a nilpotent subal-gebra and ρ be the adjoint representation of g restricted to h. Then wewrite gλ instead of Vλ and call this a root space, its elements root vectorsand λ a root if λ 6= 0. Similarly gλ,X replaces Vλ,X for a fixed elementX ∈ g.

Before turning to our next result we need the following lemma which,just as the binomial theorem, can easily be proved by induction on k.We leave it to the reader as an exercise.

Lemma 7.1.2. Let g be a complex Lie algebra, D be a derivation, Yand Z ∈ g and a and b ∈ C. Then for each positive integer, k,

(D − (a+ b)I)k[Y,Z] =

k∑

r=0

(k

r

)[(D − aI)r(Y ), (D − bI)k−r(Z)].


In particular, for X ∈ g,

(adX−(a+b)I)k[Y,Z] =

k∑

r=0

(k

r

)[(adX−aI)r(Y ), (adX−bI)k−r(Z)].

Corollary 7.1.3. Suppose h is nilpotent. Then g is the direct sum ofthe nonzero root spaces each of which is ad-invariant. Moreover, h ⊆ g0

and [gλ, gµ] ⊆ gλ+µ, where the latter is understood to be zero if λ+ µ isnot a root. Finally, g0 is a subalgebra of g.

Proof. Now g0 = H ∈ h : adHk = 0, for some k. As we alreadysaw this includes ad h since this is nilpotent. [gλ, gµ] ⊆ gλ+µ followsimmediately from Lemma 7.1.2. Finally, The last statement followsfrom this relation by taking λ and µ both zero.

7.2 Cartan Subalgebras

Definition 7.2.1. A nilpotent subalgebra, h is called a Cartan subal-gebra of g if h = g0.

We now come to Chevalley’s characterization of Cartan subalgebras.

Proposition 7.2.2. A nilpotent subalgebra, h is a Cartan subalgebra ifand only if it equals its own normalizer. That is h = ng(h).

Proof. Now, in general, h ⊆ ng(h) since its a subalgebra. Moreover, if[H,X] ∈ h for every H ∈ h, then since adHk(X) = adHk−1[H,X] andadH is nilpotent we see ng(h) ⊆ g0. If h were a Cartan subalgebrathen it would equal g0 and hence all three subalgebras would coincide.Thus h would equal its normalizer. Suppose h were not a Cartan, i.e.were strictly smaller than g0. Then ad h acts as a Lie algebra of linearoperators on the nonzero vector space g0/h. Since ad h is nilpotent andtherefore solvaheorem gives us a vector X+h ∈ g0/h, where X is not inh satisfying adH(X)−λ(H)X ∈ h for all H. But since, as we know fromEngel’s theorem, ad h consists of nilpotent operators, λ(H) = 0. HenceX normalizes h. This means h is strictly smaller than its normalizer.

7.2 Cartan Subalgebras 317

Lemma 7.2.3. A Cartan subalgebra h of a Lie algebra g is a maximalnilpotent subalgebra.

Proof. Suppose n is a nilpotent subalgebra of g strictly containing h.Consider the representation ad h : h → gl(n/h) induced by the adjointrepresentation of n. Since h is nilpotent, there is a nontrivial class[U ] ∈ n/h such that ad |h(U) = [0]. This means ad h(U) ∈ h. HenceU normalizes of h. Because the latter is a Cartan subalgebra U ∈ h, acontradiction.

Notice that we do not yet know whether Cartan subalgebras ex-ist. As before let ρ be a representation of g on V . For X ∈ g weconsider the algebraic eigenspace, V0,X , of ρ(X) with eigenvalue 0. De-fine min(ρ, g, V ) to be the smallest dimV0,X , as X varies over g andreg(ρ, g, V ) to be those X ∈ g which achieve this minimal dimension.Let n = dimV and consider the characteristic polynomial

det(λI − ρ(X)) = λn +n−1∑

j=0

dj(X)λj

of X. Here the dj(X) are polynomial functions in the coefficients ofρ(X). For instance dn−1(X) = − tr(ρ(X)) and d0(X) = ± det(ρ(X)).(In fact, each dj(X) is a homogeneous polynomial of degree n−j). Sinceρ is linear these are polynomial functions on g. Let X be fixed and con-sider the smallest j where dj(X) 6= 0. This j must be dimV0,X becausethe multiplicity of 0 in the characteristic polynomial as an algebraiceigenvalue is dimV0,X . Hence min(ρ, g, V ) is the smallest j for whichdj is not identically zero and reg(ρ, g, V ) consists of those X ∈ g wheredmin(X) 6= 0. We call such elements regular elements for (ρ, g, V ). Theminimum value of j is called the rank , which we denote by r. Clearly,regular elements always exist. When ρ is the adjoint representation wejust say regular elements.

We will need the following lemma which shows that (ρ, g, V ) is anopen, dense and connected subset of g. Compare Lemma 7.4.12.

Lemma 7.2.4. Let p be a non-identically zero polynomial function de-fined on Cn. If p vanishes on a non-empty open set in U in Cn, then


it vanishes everywhere. The set where p 6= 0 is Euclidean dense in Cn

and is also connected.

Proof. The first statement follows from the identity theorem since p isentire. Let V = x ∈ Cn : p(x) 6= 0. If V were not dense there wouldbe a diskD ⊆ Cn with positive radius on which p vanishes. Hence p ≡ 0,a contradiction. To see that V is connected, let x and y be two distinctelements of V . These can be joined by a (complex) line segment, L.Since L is compact and is has dim = 2 over R it can only hit the zeroset in at most a finite number of points. For otherwise we would have alimit point and again by the identity theorem p ≡ 0. Since removal of afinite number of points cannot disconnect L we see that L is connected.Hence so is V .

Theorem 7.2.5. Every complex Lie algebra has a Cartan subalgebra.This is because g0,X is a Cartan subalgebra for any regular element,X ∈ g. Its dimension is r, the rank.

Proof. To prove this we must show g0,X is nilpotent and coincides withits own normalizer. To see that g0,X is nilpotent it suffices by Engel’stheorem to show for each Y ∈ g0,X that adY restricted to g0,X is anilpotent operator. Call this restriction adY 1 and denote by ad Y 2 theinduced endomorphism on g/g0,X . Let d = dim(g0,X) = r which is therank as X is regular.

Let U = Y ∈ g0,X : (ad Y 1)d 6= 0. In other words, U = Y ∈ g0,X :adY 1 is not nilpotent . Let V = Y ∈ g0,X : adY 2 is invertible.Both U and V are open in g0,X . Then V is non-empty as it contains X.This is because (adX2)(U +g0,X) = [X,U ]+g0,X . If this operator weresingular there would be U ∈ g \ g0,X with [X,U ] ∈ g0,X . But then thiswould force U ∈ g0,X , a contradiction. Since V is the complement of thezero set of a polynomial, namely det, we see by the lemma just abovethat V is dense in g0,X . Suppose U were non empty. Since U is openit would have to intersect V . Let Y ∈ U ∩ V . Because Y ∈ U , ad Y 1

has 0 as an eigenvalue with multiplicity strictly smaller than d. On theother hand since Y ∈ V , 0 is not an eigenvalue of adY 2. Thereforethe multiplicity of the eigenvalue 0 of adY on g is strictly < r. This


contradicts the definition of r. We conclude U must be empty andtherefore adY 1 is nilpotent for every Y ∈ g0,X .

Now we show g0,X is its own normalizer. Suppose adY preservesg0,X . Then [Y,X] ∈ g0,X . Hence by definition there is some integer k sothat (adX)k[X,Y ] = 0. Hence (adX)k+1Y = 0 so indeed Y ∈ g0,X .

In the next section we shall see that the Cartan subalgebras of a com-plex semisimple Lie algebra are all conjugate and therefore this methodof constructing Cartan subalgebras gives them all.

Proposition 7.2.6. If g is a complex semisimple Lie algebra then anyCartan subalgebra, h, is abelian.

Proof. We know h is nilpotent. Hence so is ad h. Therefore this algebrais solvable, so Lie’s theorem 3.2.13 tells us that these operators on g arein simultaneous triangular form. Let H1 and H2 ∈ h, X ∈ g and Bdenote the Killing form. We want to calculate

B([H1,H2],X) = tr(ad [H1,H2] adX)

= tr(adH1 adH2 adX) − tr(adH2 adH1 adX).

When X ∈ h, and therefore all three matrices are triangular wesee B([H1,H2],X) = 0. Now let X ∈ g =

∑gλ as in Corollary 7.1.3.

In fact suppose X ∈ gλ, for a root, λ and let H ∈ h. We know byCorollary 7.1.3, adH adX(gµ) ⊆ gλ+µ. Since g is a direct sum of theseroot spaces and tr is linear, tr(adH adX) = 0. Now take H = [H1,H2]in the calculation above. Since B([H1,H2],X) = 0 for all X and B isnondegenerate we conclude [H1,H2] = 0.

Corollary 7.2.7. Suppose g is a complex semisimple Lie algebra. LetX ∈ g and adX = S+N be its Jordan decomposition. Then there existXs and Xn ∈ g so that adXs = S and adXn = N .

Proof. We first note that on the root space, gλ, S = λI. This is becauseS and adX share the same invariant subspaces and on such an invariantsubspace they share the same eigenvalues (see Jordan decomposition,Theorem 3.3.2). But on gλ, adX has only λ as an eigenvalue. Since Sis semisimple on g and hence also on gλ we see S = λI.


On the other hand because [gλ, gµ] ⊆ gλ+µ, for X ∈ gλ and Y ∈ gµwe get S[X,Y ] = (λ + µ)[X,Y ], while S(X) = λX and S(Y ) = µY .This means S[X,Y ] = [S(X), Y ] + [X,S(Y )] so S is a derivation. ByCorollary 3.3.23, it must be inner; S = adXs. Hence N = adX−adXs

so N = adXn, where Xn = X −Xs.

In the case of a semisimple algebra we get another useful character-ization of a Cartan subalgebra.

Proposition 7.2.8. Let g be a complex semisimple Lie algebra. Thena subalgebra, h, is a Cartan subalgebra if (and only if) it is a maximalabelian diagonalizable subalgebra.

We will complete the proof of this by dealing with “only if” part inProposition 7.3.3.

Proof. Suppose h is abelian, ad h is simultaneously1 diagonalizable andthere is no larger such subalgebra of g. We will show that h is a Cartanby proving it coincides with its normalizer.

Since h is abelian and therefore nilpotent we can apply Corollary7.1.3 to get g = g0 ⊕λ6=0 gλ, where these weight spaces are invariantunder ad h. Since the latter is simultaneously diagonalizable and h is asubspace of g0 (because h is abelian) which is clearly ad h-invariant, wesee g0 = h ⊕ l with [h, l] = 0. We know h ⊆ ng(h) ⊆ g0 so all we needto do is to show l = 0. Suppose X 6= 0 ∈ l. Then h + CX is a properlylarger abelian subalgebra of g so adX cannot be diagonalizable. Butwe can still decompose g into weight spaces according to this abeliansubalgebra and get a (perhaps) more refined decomposition.

By Corollary 7.2.7 there exist Xs and Xn ∈ g so that adXs = S.Since S is a polynomial in adX without constant term andX centralizesh so does Xs. Hence by maximality Xs ∈ h. But then Xn = X − Xs

must be in h + CX. Hence we may assume adX is actually nilpotenton g. Because g0 is a subalgebra and X is in it, g0 is adX invariant andacts as a nilpotent operator on it. As X was taken arbitrarily from l we

1As we know, an abelian family of operators which is individually diagonalizableis simultaneously so


find l and therefore all of l + h = g0 acts nilpotently on g0. By Engel’stheorem g0 is nilpotent. Again using the weight space decomposition,this time under all of g0, we find that the zero weight space cannotpossibly be any bigger than what we got under h itself. This means g0

is a Cartan. So we have

g = g0 ⊕λ6=0 gλ, (7.2)

where the λ’s are weights of g0. Let X0 ∈ g0 and Xλ ∈ gλ, λ 6= 0.For our original X we calculate B(X,X0) and B(X,Xλ) where B is theKilling form. By bilinearity of B we get

B(X,X0) =∑

λ

(dim gλ)λ(X)λ(X0).

Since adX is nilpotent on all of g0, λ(X) = 0 for every λ. HenceB(X,X0) = 0. As for B(X,Xλ), this is also zero since we are justshifting these weight spaces. By (7.2) it follows that B(X,Y ) = 0 forall Y ∈ g. Since B is nondegenerate X = 0 and this contradictioncompletes the proof.

Let Aut(g) denote the full group of Lie algebra automorphisms, whileInn(g) be the inner automorphisms; that is the subgroup of Aut(g)generated by Exp(adX), for X ∈ g.

Before turning to the proof of the conjugacy of Cartan subalgebrasof a complex semisimple Lie algebra we need two lemmas.

Lemma 7.2.9. For a Cartan subalgebra h of g, reg(ad |h, h, g) consistsof those X ∈ h for which λ(X) = 0 for all λ 6= 0. Alternatively, itconsists of all X ∈ h for which g0,X = h.

Proof. For a Cartan subalgebra h of g we know reg(ad |h, h, g) consistsof those X ∈ h where dmin(X) 6= 0, or alternatively, where dim g0,X isminimal. Now g = h⊕λ6=0 gλ. Let X ∈ h then g0,X consists of those Y ∈g such that adXdim(g)(Y ) = 0. Hence g0,X = h⊕λ6=0,λ(X)=0 gλ. Now wehave a finite number of nonzero linear functionals λ on h each of whichhas for its zero set a hyperplane in h. The union of finitely many (orby countable subadditivity of Lebesgue measure even countably many)


of such hyperplanes cannot exhaust h. Hence g0,X is smallest when itis h.

Consider the natural action φ : Inn(g)× reg(ad |h, h, g) → g given by(α,X) 7→ α(X).

Lemma 7.2.10. φ is an open map. Its image is contained inreg(ad g, h, g).

Proof. Since we have a transitive group action it is sufficient to proveeach Y ∈ reg(ad |h, h, g) has a neighborhood contained in the orbit ofthe action φ. To this end we linearize and calculate the derivative ofφ at (I, Y ). As reg(ad |h, h, g) is open in h, the latter is the tangentspace at any point. Similarly, because reg(ad g, h, g) is open in g thetangent space at φ(I, Y ) can be regarded as g itself. The tangent spaceto I in Inn(g) is of course its Lie algebra, ad g (see Corollary 1.4.30 ).Thus d(I,Y )φ : ad g × h → g. Evidently, d(I,Y )φ(adX, 0) = [X,Y ] andd(I,Y )φ(0,H) = H so because the derivative is linear we get,

d(I,Y )φ(adX,H) = d(I,Y )φ(adX, 0) + d(I,Y )φ(0,H) = H + [X,Y ].

For X ∈ reg(ad |h, h, g), adX is non singular when restricted to thesum of the nonzero weight spaces. This because it acts on each gλ witheigenvalue λ(X), which is nonzero by definition. Hence d(I,Y )φ mapsonto g and so has maximal rank. Hence the image of φ itself containsa neighborhood in g by the implicit function theorem. This proves thefirst statement.

Since reg(ad g, g, g) is dense in g it must meet this neighborhoodin say Z. Hence there is some α ∈ Inn(g) (and recall Y ∈ h) withα(Y ) = Z. Since α is an automorphism of g we have α(g0,Y ) = g0,Z .In particular these have the same dimension and since dim g0,Z =min dim g0,X for X ∈ g and dim g0,Y = min dim g0,H for H ∈ h, weget reg(ad |h, h, g) ⊆ reg(ad g, g, g). Because reg(ad |h, h, g) is stable un-der every automorphism of g, φ(Inn(g) × reg(ad g, h, g)) is contained inreg(ad g, h, g).

7.3 Roots of Complex Semisimple Lie Algebras 323

We can now prove the Cartan subalgebras of g are conjugate. Thisgives another proof that the rank is an invariant.

Theorem 7.2.11. Let g be a complex semisimple Lie algebra. Thenany two Cartan subalgebras are conjugate.

Proof. We shall prove this theorem by contradiction by assuming therewere two non-conjugate Cartan subalgebra h1 and h2.

We first show their images under φ must be disjoint. For supposeφ(X1) = φ(X2), where each Xi ∈ reg(ad |hi

, hi, g). Then α(X1) = β(X2)so β−1α(X1) = X2. Hence, we see as above there is some γ ∈ Inn(g)with γ(g0,X1) = g0,X2 . By Lemma 7.2.9, g0,Xi

= hi. So γ(h1) = h2, acontradiction.

Thus by Lemma 7.2.10, reg(ad g, g, g) is a nontrivial disjoint unionof open sets. On the other hand by Lemma 7.2.4, we also know it isconnected. This contradiction completes the proof.

7.3 Roots of Complex Semisimple Lie Algebras

Throughout this section g will stand for a complex semisimple Lie alge-bra, h a fixed Cartan subalgebra and g = h ⊕λ6=0 gλ the correspondingroot space decomposition and Λ will denote the set of nonzero roots andB the Killing form of g.

Proposition 7.3.1. (1) If λ and µ ∈ Λ∪0 and λ+µ 6= 0, then gλand gµ are orthogonal under B.

(2) For λ ∈ Λ ∪ 0, B induces a nondegenerate form on gλ × g−λ.

(3) If λ ∈ Λ, then −λ ∈ Λ.

(4) B restricted to h × h is nondegenerate.

(5) For each λ ∈ Λ there exists a unique Hλ ∈ h so that λ(H) =B(H,Hλ).

(6) Λ spans h∗, the dual space.

Proof. 1. Choose Xλ and Xµ in each of the corresponding root spaces.Then adX adY (gν) ⊆ gλ+µ+ν . Since λ + µ 6= 0, gν ∩ gλ+µ+ν = 0.


Taking a basis of each root space including h one sees that the matrixof adX adY has trace 0 because all diagonal entries are zero.

2. Let Xλ ∈ gλ and suppose B(Xλ, g−λ) = 0. Then since gλ isorthogonal to everything else by 1, and therefore by the root spacedecomposition to everything in g, it follows that Xλ = 0 since B isnondegenerate.

3. Suppose to the contrary that g−λ = 0. Then B(gλ,X) = 0 forall X ∈ g−λ which is impossible by 2. Hence −λ is also a root.

4. Follows from 2 by taking λ = 0.5. Follows from 4.6. We first show that if H ∈ h and λ(H) = 0 for all λ ∈ Λ, then

H = 0. This is because the root space decomposition Corollary 7.1.3then forces [H,X] = 0 for all X ∈ ∑

λ∈Λ gλ and since h is abelian, byCorollary 7.2.6, it follows that [H,X] = 0 for all X ∈ g. But thenH ∈ z(g) = 0, again by semisimplicity. Therefore Λ separates thepoints of h and so spans the dual.

For each root λ, consider the adjoint representation of h on gλ andapply Corollary 7.1.3 to choose a fixed nonzero Eλ ∈ gλ satisfying[H,Eλ] = λ(H)Eλ on h.

Proposition 7.3.2. (1) If λ is a root and X ∈ g−λ, then [Eλ,X] =B(Eλ,X)Hλ.

(2) If λ and µ ∈ Λ, then µ(Hλ) is a rational multiple of λ(Hλ).

(3) For λ ∈ Λ, λ(Hλ) 6= 0.

Proof. 1. By Corollary 7.1.3, [gλ, g−λ] ⊆ g0 = h. Hence [Eλ,X] ∈ h.Let H ∈ h. Then, by invariance and skew symmetry,

B([Eλ,X],H) = −B(X, [Eλ,H]) = B(X, [H,Eλ]) = λ(H)B(X,Eλ).

This in turn is

λ(H)B(X,Eλ) = B(Hλ,H)B(Eλ,X) = B(B(Eλ,X)Hλ,H).

Since this holds for all H and B is nondegenerate on h, we concludeB(Eλ,X)Hλ = [Eλ,X].


2. By the above choose X−λ ∈ g−λ so that B(Eλ,X−λ) = 1. By 1we see

Hλ = [Eλ,X−λ]. (7.3)

Let µ ∈ Λ be fixed and considerW = ⊕n∈Zgµ+nλ. ThenW is a subspaceof g which is invariant under adHλ because adHλ leaves gµ+nλ invariantand acts on it by a single algebraic eigenvalue, µ + nλ. Since the sumis direct we see the trace of adHλ on W is

∑

n∈Z

(µ(Hλ) + nλ(Hλ)) dim gµ+nλ.

On the other hand W is invariant under adEλ and adX−λ andtherefore also their bracket. Since ad is a representation this is adHλ

so by (7.3) this trace is zero and hence the conclusion.3. If λ(Hλ) = 0, then µ(Hλ) being a multiple of it would also

be zero for every µ. Since Λ spans h∗ this forces Hλ = 0. But thenby 5. of the previous proposition we would then have λ itself zero, acontradiction.

Proposition 7.3.3. (1) For λ ∈ Λ, dim gλ = 1.

(2) If λ ∈ Λ, the only multiples of it which are also in Λ are ±λ.(3) ad h acts simultaneously diagonally on g.

(4) On h × h the Killing form is given by B(H,H ′) =∑λ∈Λ λ(H)λ(H ′).

(5) The pair Eλ, E−λ can be normalized so B(Eλ, E−λ) = 1.

Proof. As above, choose X−λ ∈ g−λ so that B(Eλ,X−λ) = 1 so that[Eλ,X−λ] = Hλ. Now consider the (complex) subspace U of g spannedby Eλ,Hλ together with gnλ, where n < 0. Because [gλ, gµ] ⊆ gλ+µ

we see U is invariant under bracketing by Eλ and Hλ. By part (1)of Proposition 7.3.2 it is also invariant under bracketing by X−λ. Since[Eλ,X−λ] = Hλ we know adHλ has trace zero on U . Since this operatoracts with a single algebraic eigenvalue on each summand we conclude

λ(Hλ) + 0 +∑

n<0

nλ(Hλ) dim(gnλ) = 0.


Since λ(Hλ) 6= 0 this tells us

∑

n>0

n dim(gnλ) = 1.

Hence dim(g−λ) = 1 and dim(g−nλ) = 0 for n ≥ 2. Since we alreadyknow that Λ is symmetric under taking negatives this proves (1) and(2).

(3) Using the root space decomposition and the fact that h is abelianand so acts trivially on itself we see (3) follows immediately from (1).

(4) Follows from (3).

(5) Follows from the fact that B is nondegenerate on gλ × g−λ andthese spaces are 1-dimensional.

Remark 7.3.4. We see here that for each λ ∈ Λ we get a 3-dimensionalsubalgebra of g. This is because we now know each gλ is 1-dimensional.Hence gλ is generated by Eλ and g−λ by E−λ. Now [Hλ, Eλ] = λ(Hλ)Eλand [Hλ, E−λ] = −λ(Hλ)E−λ. Since B(Eλ, E−λ) = 1 we also have[Eλ, E−λ] = Hλ. Thus Eλ, Hλ and E−λ generate a 3-dimensional sub-algebra. But what subalgebra is this?

Since λ(Hλ) 6= 0 we can further normalize the basis (and there-fore the structure constants) to get H ′

λ = 2λ(Hλ)Hλ, E

′λ = 2

λ(Eλ)Eλ and

E′−λ = E−λ. Then

[H ′λ, E

′λ] = 2E′

λ,

[H ′λ, E

′−λ] = −2E′

−λ,

and

[E′λ, E

′−λ] = H ′

λ.

So this is just sl(2,C).

We now come to the definition of a root string.

Definition 7.3.5. Let λ ∈ Λ and µ ∈ Λ∪0. We shall call all elementsof Λ ∪ 0 of the form µ+ nλ, where n ∈ Z, the λ root string throughµ.


Before stating our next result it will be convenient to transfer B|h×h

to h∗ × h∗ as follows: For φ and ψ ∈ h∗we take 〈φ,ψ〉 = B(Hφ,Hψ),where the isomorphism between h and h∗ is given Proposition 7.3.1, part5. This gives us a nondegenerate symmetric form on h∗ × h∗. Noticethat because of the way we identify h and h∗, B(Hφ,Hψ) = φ(Hψ), orψ(Hφ).

We now turn to a proposition which will play an important role infinding compact real forms of complex semisimple algebras. As we shallsee, our previous work on the representation theory of sl(2,C) will playan important role here.

Proposition 7.3.6. Let λ ∈ Λ and µ ∈ Λ ∪ 0. Then

(1) The λ string through µ has the form µ+nλ (−p ≤ n ≤ q) i.e. thestring is uninterrupted. Here p and q ≥ 0.

(2) p− q = 2〈µ,λ〉〈λ,λ〉 . In particular, the latter quantity is in Z.

(3) If µ + nλ is never zero, then the adjoint action of sl(2,C) onW = ⊕n∈Zgµ+nλ is irreducible.

Proof. If µ + nλ = 0 for some n then by Proposition 7.3.3, µ = 0, orµ = ±λ. In either case there are no gaps and p − q = 2〈µ,λ〉

〈λ,λ〉 while thethird conclusion has no content since its hypothesis is false.

We may therefore assume µ + nλ is never 0 and will prove all con-clusions simultaneously.

By Proposition 7.3.3, adH ′λ is diagonal on W with distinct eigenval-

ues. The eigenvalues are (µ + nλ)(H ′λ) = 2

〈λ,λ〉(µ + nλ)(Hλ). But this

is 2〈µ,λ〉+n〈λ,λ〉〈λ,λ〉 so

(µ+ nλ)(H ′λ) =

2〈µ, λ〉〈λ, λ〉 + 2n. (7.4)

This tells us that an adH ′λ-invariant subspace ofW is a sum of certain of

the gµ+nλ. In particular this must also be true of any sl(2,C) invariantsubspace of W . To show sl(2,C) acts irreducibly on W , suppose Vis any sl(2,C) invariant (and therefore by Weyl’s theorem 3.4.3, wemay assume to be an irreducible) subspace and let −p and q be thesmallest and largest integers n appearing in this representation. Our


study of irreducible representations sl(2,C) tells us that the eigenvaluesof ad h = adH ′

λ on V are N − 2i that is they drop by 2, where hereN = dim(V ) − 1. This tells us that all n between −p and q come up.

Furthermore by (7.4), we get N = 2〈µ,λ〉〈λ,λ〉 + 2q, and −N = 2〈µ,λ〉

〈λ,λ〉 − 2p soadding gives

p− q =2〈µ, λ〉〈λ, λ〉 . (7.5)

But the theorem on representations of sl(2,C) tell us that W isthe direct sum of its irreducible components. If W0 is another suchirreducible subspace and −p0 and q0 the corresponding smallest andlargest integers appearing, then (7.5) tells us p0 − q0 = 2〈µ,λ〉

〈λ,λ〉 . Hencep0 − q0 = p − q. But all the n’s between −p and q are accounted forby W0 so either q0 < −p or −p0 > q. Using symmetry we may assumethe former. Hence p0 < −q. But then q0 ≥ −p0 > q ≥ −p. So we findp0 − q0 < p− q, a contradiction.

Definition 7.3.7. Let V be a complex vector space and U be a realsubspace of VR (this means considering V as a real vector space). IfUC = V , then we shall call U a real form of V .

Put more simply, U is a real form of V if, U ⊕ iU = V .

Corollary 7.3.8. (1) Let λ and µ ∈ Λ ∪ 0 with λ + µ 6= 0. Then[gλ, gµ] = gλ+µ.

(2) Let Xλ ∈ gλ, X−λ ∈ g−λ and Xµ ∈ gµ, where µ+nλ is never zerofor any n ∈ Z. Then

[Xλ, [X−λ,Xµ]] =p(1 + q)

2〈λ, λ〉B(Xλ,X−λ)Xµ.

(3) Let λ ∈ Λ and µ ∈ Λ ∪ 0. Then µ(Hλ) is rational.

(4) Let V be the R-linear span of Λ in h∗. Then V is a real form ofh∗ and the restriction of our symmetric bilinear form to V × V ispositive definite.

(5) Let h0 be the R-linear span of Hλ : λ ∈ Λ. Then h0 is a realform of h and V consists of those linear functionals on h whichtake real values on h0.


Proof. 1. Let λ and µ ∈ Λ ∪ 0 with λ + µ 6= 0. We may evidentlyassume λ 6= 0. Also, we know by Corollary 7.1.3 that [gλ, gµ] ⊆ gλ+µ.If µ is an integral multiple of λ, then we know µ = 0, or ±λ. If µ = 0,then [h, gλ] = gλ because ad h acts diagonally with a nonzero eigenvalue.Since our hypothesis rules out µ = −λ, we are left with µ = λ. Butthen gλ+µ = 0 so the statement is trivially true.

Thus we can assume that µ is not an integral multiple of λ. Henceby the previous proposition a copy of sl(2,C) acts irreducibly on theinvariant subspace W = ⊕n∈Zgµ+nλ. Now we have classified the fi-nite dimensional irreducible representations of sl(2,C) in Section 3.1.5.When we match this up with X+,H and X− in Section 3.1.5, ignoringconstant factors, the vectors Eµ+nλ correspond to the vi where the onlyi for which e maps vi to zero is i = 0 and v0 corresponds to Eµ+qλ.Hence if [gλ, gµ] = 0 we must have q = 0 which means µ+ λ is not aroot. Hence gλ+µ is also zero.

2. Suppose Xλ ∈ gλ, X−λ ∈ g−λ and Xµ ∈ gµ, where µ+nλ is never

0. We want to prove [Xλ, [X−λ,Xµ]] = p(1+q)2 〈λ, λ〉B(Xλ,X−λ)Xµ,

where p and q are the positive integers defined earlier. Now both sidesare linear in Xλ and X−λ. We can therefore normalize them as inRemark 7.3.4 so that B(Xλ,X−λ) = 1. Therefore we can make theidentification of the linear span of Xλ, Hλ and X−λ with X+, H andX− in sl(2,C) as in Remark 7.3.4. Using the fact that

B(Xλ,X−λ) =2

〈λ, λ〉

the formula we wish to prove in these terms yields [Xλ, [X−λ,Xµ]] =p(1 + q)Xµ. So the question is, is this true? By Proposition 7.3.6,sl(2,C) acts irreducibly on W =

∑n∈Z gµ+nλ. In these identifications

Xµ+qλ corresponds to a multiple of v0 in Section 3.1.5. Because Xµ is amultiple of adX−λ

q(Xµ+qλ), it follows from our work on the irreduciblerepresentations of sl(2,C) that

[Xλ, [X−λ,Xµ]] = adXλ adX−λ adX−λq(Xµ+qλ)

= (q + 1)(N − q − 1 + 1)Xµ,


where N = dimW −1; that is N = q+p+1−1. Hence (q+1)(N−q) =p(q + 1).

3. For φ, ψ ∈ h∗ we have 〈φ,ψ〉 = B(Hφ,Hψ). Hence by Proposition7.3.3, part 4, we get

∑λ∈Λ λ(Hφ)λ(Hψ), that is

〈φ,ψ〉 =∑

λ∈Λ

〈λ, φ〉〈λ, ψ〉. (7.6)

Let pλ,µ and qλ,µ be the integers associated with the λ root stringcontaining µ. Setting φ = λ = ψ we get 〈λ, λ〉 =

∑µ∈Λ 〈µ, λ〉2 which

by Proposition 7.3.6, part 2, is∑

µ∈Λ[(pλ,µ − qλ,µ)〈λ,λ〉

2 ]2. Since as wesaw in Proposition 7.3.2, part 3, λ(Hλ) = 〈λ, λ〉 6= 0, dividing by it andsolving for 〈λ, λ〉 we get

|λ|2 = 〈λ, λ〉 =4∑

µ∈Λ (pλ,µ − qλ,µ)2. (7.7)

This shows that 〈λ, λ〉 is positive rational. By Proposition 7.3.2, part2, µ(Hλ) is always rational.

4. Suppose dimC h = r. Since by Proposition 7.3.1, part 6, Λ spansh∗ we can choose roots λ1, · · · , λr so that Hλ1, · · · ,Hλr

is a basis forh. Let φ1, · · ·φr be the dual basis of h∗. Thus φi(Hλj

) = δij . Let Vbe the R-subspace of all functionals in h∗ which take real values on allHλj

. Then V is the direct sum of the R-lines through the φi. HenceV is a real form of h∗. Since λ1, · · · , λr are linearly independent overC and hence also over R we see that V is the R span of Λ. Finally,for φ ∈ V since φ(Hµ) is real for each µ ∈ Λ so by (7.6) we have〈φ, φ〉 =

∑µ∈Λ 〈µ, φ〉2 =

∑µ∈Λ φ(Hµ)

2. As a sum of squares of realnumbers we see 〈·, ·〉 is positive definite on V × V .

5. φ 7→ Hφ is an isomorphism of V with h0. Since V has realdimension r and is the linear span of Λ, hence the real span of the Hλ

must also have real dimension r. But they are independent over C andso also over R. Thus they form an R-basis of h0. This means V is theset of functionals in h∗ that are real on h0 and the restriction of theisomorphism above gives an isomorphism of V onto h0.


We now define new structure constants and study their relations. Ifλ + µ ∈ Λ, define [Xλ,Xµ] = Cλ,µXλ+µ. If λ + µ is not in Λ, defineCλ,µ = 0.

Lemma 7.3.9. (1) Cλ,µ = −Cµ,λ.(2) If λ, µ and ν ∈ Λ and λ+ µ+ ν = 0, then Cλ,µ = Cµ,ν = Cν,λ.

(3) If λ, µ, ν and ξ ∈ Λ, λ+ µ+ ν + ξ = 0, and ξ is not one of −λ,−µ and −ν, then

Cλ,µCν,ξ + Cµ,νCλ,ξ + Cν,λCµ,ξ = 0.

Proof. The first relation follows immediately from the skew symmetryof the bracket. As to the second, by the Jacobi identity

[[Xλ,Xµ],Xν ] + [[Xµ,Xν ],Xλ] + [[Xν ,Xλ],Xµ] = 0.

Hence Cλ,µHν + Cµ,νHλ + Cν,λHµ = 0. Note that Hν = −Hλ − Hµ.The linear independence of Hλ and Hµ yields the conclusion.

The third relation follows from the Jacobi identity once we provethat

Cλ,µCν,ξ = B([[Xλ,Xµ]Xν ],Xξ).

Since B is the Killing form,

B([[Xλ,Xµ]Xν ],Xξ) = B([Xλ,Xµ], [Xν ,Xξ ]).

First suppose that λ+µ ∈ Λ. Then since λ+µ = −(ν+ ξ) 6= 0 we have

B([Xλ,Xµ], [Xν ,Xξ]) = Cλ,µCν,ξB(Xλ+µ,Xν+ξ)

= Cλ,µCν,ξB(Xλ+µ,X−(λ+µ)) = Cλ,µCν,ξ(7.8)

If λ+ µ /∈ Λ then the identity holds as Cλ,µ = 0 and [Xλ,Xµ] = 0.

Lemma 7.3.10. Let λ, µ and λ + µ ∈ Λ and let µ+ nλ, where −p ≤n ≤ q be the λ string containing µ. Then Cλ,µC−λ,−µ = −p(1+q)

2 |λ|2.


Proof. By Corollary 7.3.8, taking into account the fact that 〈·, ·〉 is pos-itive definite on a real form of h, we know

[X−λ, [Xλ,Xµ]] =p(1 + q)

2|λ|2B(Xλ,X−λ)Xµ.

Taking our normalizations into account as well gives,

[X−λ, [Xλ,Xµ]] =p(1 + q)

2|λ|2Xµ.

Since the left side of this equation is C−λ,λ+µCλ,µXµ we see

C−λ,λ+µCλ,µ =p(1 + q)

2|λ|2. (7.9)

But because −λ+(λ+µ)+−µ = 0 the two lemmas just above showC−λ,λ+µ = C−µ,−λ = −C−λ,−µ. Substituting into (7.9) completes theproof.

We conclude this section with the Chevalley normalization theoremwhose proof requires the Serre’s isomorphism theorem which follows.

Theorem 7.3.11. Let g and g′ be two complex semisimple Lie algebraswith Cartan subalgebras h and h′. Suppose that µ 7→ µ′ is a bijectionbetween the roots of g and g′ such that µ′ +λ′ is a root for g′ if and onlyif µ+ λ is a root for g and that µ′ + λ′ = (µ+ λ)′. Then there is a Liealgebra isomorphism f : g → g′ such that f(h) = h′ and for every rootµ of g we have

µ′ f |h = µ.

Proof. It follows from the assumption that for any two roots µ and λof g, the length of the root string of µ through λ and µ′ through λ′ areidentical. Therefore by (7.7),

λ(Hλ) = λ′(Hλ′) (7.10)

and by Proposition 7.3.6, part 2,

λ(Hµ) = λ′(Hµ′)


for all µ and λ. Let µ1, . . . , µn be a maximal set of linearly independentroots of g with the corresponding Hµi

∈ h, 1 ≤ i ≤ n. Since µ1, . . . , µnis a basis for h∗, then Hµ1 , . . . ,Hµn is a basis for h and the determinantdet((µi(Hµj

))i,j) is nonzero.

Now consider µ′1, . . . , µ′n with their corresponding Hµ′i

∈ h′, 1 ≤ i ≤n. Since µi(hµj

) = µ′i(h′µj

), µ′i are also linearly independent. Inter-changing the role of g and g′, we conclude that µ′1, . . . , µ

′n is a maximal

linearly independent set.

Define the isomorphism fH : h → h′ by setting fH(Hµi) = Hµ′i

, thenby (7.10) we get

µ′ fH = µ (7.11)

for all i. So to prove the theorem we must show that fH extends to aLie algebra isomorphism from f : g → g′. For each root µi of g considervectors Xµ andX−µ ∈ g such thatB(Xµ,X−µ) = 1 and [Xµ,X−µ] = Hµ

(see Remark 7.3.4). We make a similar choice of Xµ′ and X−µ′ for g′.Then f is defined once we have determined f(Xµ) and f(X−µ). Theseare defined by f(Xµ) = cµXµ′ and f(X−µ) = c−µX−µ′ for a suitablechoice of cµ and c−µ. The identity f([X,Y ]) = [f(X), f(Y )] requiresthat

(1) cµc−µ = 1, obtained for X = Xµ and Y = X−µ,

(2) Cµ,λcµ+λ = Cµ′,λ′cµcλ if µ+λ is also a root, obtained for X = Xµ

and Y = Xλ.

Here Cµ,λ’s and Cµ′,λ′ ’s are the structure constants relative to the basisXµ and Xµ′ for g and g′ respectively.

In fact these conditions are sufficient because f([Xµ,Xλ]) =[f(Xµ), f(Xλ)] is basically conditions 1 and 2,

f([Hµ,Xλ]) = f(λ(Hµ)Xλ) = λ(Hµ)f(Xλ) = λ′(Hµ′)cλXλ′

= [Hµ′ , cλXλ′ ] = [f(Hµ), f(Xλ)]

follows from (7.11), and f([X,Y ]) = [f(X), f(Y )] = 0 if X,Y ∈ h.

Now we construct cµ’s, this is done inductively. Note that the ra-tional linear combinations of all µi give all roots of µ ∈ g, because if


µ =∑n

i=1 aiµi then

µ(Hµj) =

n∑

i=1

aiµi(Hµj),

since all µi(Hµj)’s and µ(Hµj

)’s are rational (see Proposition 7.3.2)therefore all ai are rational. We order lexicographically all rationallinear combinations of µi’s. That is

∑i aiµi >

∑i biµi if and only if the

first nonzero ai−bi is positive. Once we have defined cλ for λ > 0, we setc−λ = c−1

λ to assure condition 1. Let λ be a root of g and suppose thatcµ has been defined for all −λ < µ < λ. We show that we can definecλ while maintaining condition 2. If we cannot write λ = µ + ν where−λ < µ, ν < λ, we can simply define cλ = c−λ = 1. If λ = µ+ ν where−λ < µ, ν < λ, guided by condition 2 we define cλ = C−1

µ,νCµ′,ν′cµcν ,this is well-defined since Lemma 7.3.10 guarantees that Cµ,ν is nonzero.Then cλ is nonzero as Cµ′,ν′ is nonzero again by Lemma 7.3.10. We haveto check that condition 2 also holds for the pair (−µ,−ν). By Lemma7.3.10 we have

Cµ,νC−µ,−ν = Cµ′,ν′C−µ′,−ν′ ,

therefore

C−µ,−νc−µ−ν = C−µ,−νc−1µ+ν = C−µ,−νCµ,νC

−1µ′,ν′c

−1µ c−1

ν

= Cµ′,ν′C−µ′,−ν′C−1µ′,ν′c

−1µ c−1

ν = C−µ′,−ν′c−1µ c−1

ν

= C−µ′,−ν′c−µc−ν

To complete the proof we prove that condition 2 holds for any otherpair (µ1, ν1) such that λ = µ1 + ν1 and −λ < µ1, ν1 < λ. Since ν1 isdifferent from µ, ν and −µ1 by Lemma 7.3.9 applied to the quadruple−µ,−ν, µ1 and ν1 we

C−µ,−νCµ1,ν1 + C−ν,µ1C−µ,ν1 + Cµ1,−µC−ν,ν1 = 0 (7.12)

Because of the orderings we have that µ, ν, µ1and ν1 are all positivetherefore the difference of any two of them is between −λ and λ. By


condition 2 we have

C−ν,µ1c−ν+µ1 = C−ν′,µ′1c−νcµ1 , C−µ,ν1c−µ+ν1 = C−µ′,ν′1c−µcν1,

Cµ1,−µcµ1−µ = Cµ′1,−µ′cµ1c−µ, C−ν,−ν1c−ν+ν1 = C−ν′,−ν′1c−νcν1 .

These relations are true even if −ν+µ1 etc. is not a root, with conven-tion that cρ = 1 if ρ is not a root. These last fours relations imply that

C−ν,µ1C−µ,ν1 = C−ν′,µ′1C−µ′,ν′1c−νcµ1c−µcν1 (7.13)

Cµ1,−µC−ν,−ν1 = Cµ′1,−µ′C−ν′,ν′1cµ1c−µc−νcν1 (7.14)

because c−ν+µ1c−µ+ν1 = 1 = cµ1−µc−ν+ν1 as −ν+µ1 = −(−µ+ν1) andµ1 − µ = −(−ν + ν1).

By inserting the left-hand side of (7.13) and (7.14) in (7.12) andmultiplying by cνc−µ1cµc−ν1 = (c−νcµ1c−µcν1)

−1 we obtain,

C−µ,−νCµ1,ν1cνc−µ1cµc−ν1+C−ν′,µ′1C−µ′,ν′1+Cµ′1,−µ′C−ν′,ν′1 = 0. (7.15)

Since (7.12) hold for µ′, ν ′, µ′1, ν′1 in the place of µ, ν, µ1, ν1 and compar-

ing with (7.15) we get

C−µ′,−ν′Cµ′1,ν′1 = C−µ,−νCµ1,ν1cνc−µ1cµc−ν1. (7.16)

Condition 2 holds for the pair (−µ,−ν), that is

C−µ′,−ν′ = C−µ,−νc−ν−µc−1−νc

−1−µ = C−µ,−νcνcµc

−1ν+µ

= C−µ,−νcνcµc−1ν1+µ1

,

which together with (7.16) imply that

Cµ′1,ν′1 = Cµ1,ν1cν1+µ1c−µ1c−ν1 = Cµ1,ν1cν1+µ1c−1µ1c−1ν1

orCµ′1,ν′1cµ1cν1 = Cµ1,ν1cν1+µ1

So condition 2 holds for the pair (µ1, ν1) as well.


Theorem 7.3.12. Let g be a complex semisimple Lie algebra, h be aCartan subalgebra and Λ be the corresponding roots. For each λ ∈ Λ wecan choose Xλ ∈ gλ so that

(1) [Xλ,X−λ] = Hλ,

(2) [Xλ,Xµ] = Nλ,µXλ+µ, if λ+ µ ∈ Λ and

(3) [Xλ,Xµ] = 0, if λ+ µ 6= 0 and is not in Λ.

Moreover, these constants satisfy Nλ,µ = −N−λ,−µ and N2λ,µ =

p(1+q)2 |λ|2, where p and q are the integers associated with the λ string

containing µ. In particular since N2λ,µ ≥ 0, Nλ,µ is real.

Proof. Consider the linear map −I : h → h which is a linear iso-morphism. Its transpose (also −I) carries Λ bijectively onto itselfand hence by Theorem 7.3.11, extends to an automorphism α of g.Now by our choice of α, α(Xλ) ∈ g−λ. Hence there is some con-stant c−λ so that α(Xλ) = c−λX−λ. As we shall see below in Lemma7.5.2, the Killing form is preserved by all automorphisms of g. HenceB(α(X), α(Y )) = B(X,Y ) for all X, Y ∈ g. Taking X = Xλ andY = X−λ we get

c−λcλ = c−λcλB(Xλ,X−λ) = B(α(Xλ), α(X−λ)) = B(Xλ,X−λ) = 1.

Thus c−λ = 1cλ

. Now for each λ ∈ Λ choose zλ ∈ C so that z−λ = 1zλ

and z2λ = −cλ. This can be done because c−λcλ = 1. For instance, if

cλ = reiθ and c−λ = r−1e−iθ then let zλ = ir12 e

12iθ and z−λ = ir−

12 e−

12iθ.

This consistent choice of the zλ and z−λ gives us multiples Zλ = zλXλ ∈gλ which satisfy

[Zλ, Z−λ] = [zλXλ, zλX−λ] = zλz−λ[Xλ,X−λ] = [Xλ,X−λ] = Hλ.

Also,

α(Zλ) = α(zλXλ) = zλα(Xλ) = zλc−λX−λ = −z−1λ X−λ = −z−λX−λ.

But this last term is −z−λX−λ = −Z−λ. Hence

α(Zλ) = −Z−λ. (7.17)

7.4 Real Forms of Complex Semisimple Lie Algebras 337

Now define the constants Nλ,µ just the way we defined Cλ,µ earlier.Then

−Nλ,µZ−λ−µ = α(Nλ,µZλ+µ) = [α(Zλ), α(Zµ)]

= [−Z−λ,−Z−µ] = N−λ,−µZ−λ−µ

Therefore −Nλ,µ = N−λ,−µ. The relation N2λ,µ = p(1+q)

2 |λ|2 followsimmediately from Lemma 7.3.10.

7.4 Real Forms of Complex Semisimple Lie Al-

gebras

The main purpose of this section is to prove E. Cartan’s theorem onthe existence of a compact real form for any complex semisimple Liealgebra.

We now extend our definition of a real form from vector spaces toLie algebras.

Definition 7.4.1. Let g be a complex Lie algebra and h be a realsubalgebra of gR (this means considering g as a real Lie algebra). IfhC = g, then we shall call h a real form of g.

Just as before h is a real form of g if, h ⊕ ih = g.

Exercise 7.4.2. For example, sl(n,R) is a real form of sl(n,C). Thereader should check that su(n,C) is also a real form of sl(n,C).

Definition 7.4.3. If in addition h is of compact type (see Definition3.9.2), we shall call it a compact real form of g.

We leave the following observation to the reader.

Lemma 7.4.4. Let h be a real form of g. For X and Y ∈ h we defineα(X + iY ) = X − iY . Then α is an automorphism of the real Liealgebra, g. Moreover, for any X ∈ g and c ∈ C, α(cX) = cα(X). Also,α2 = I.


Definition 7.4.5. Whenever we have a complex Lie algebra, g, and anautomorphism α of it as a real Lie algebra satisfying

(1) α(cX) = cα(X), X ∈ g, c ∈ C(2) α2 = I.

We call α a conjugation.The conjugation given in Lemma 7.4.4 is calledthe conjugation relative to the real form h

Lemma 7.4.6. Let g be a complex Lie algebra and α be a conjugation.Then the α fixed points, gα, of g is a real form of g and the conjugationrelative to this real form is α.

Proof. Let X and Y ∈ gα. Then α(X) = X and α(Y ) = Y . Henceα[X,Y ] = [α(X), α(Y )] = [X,Y ] and similarly for the linear combina-tions. Thus gα is a real subalgebra of g. Notice that if α(Z) = −Z, thenZ ∈ igα. That is −iZ ∈ gα. To see that it is a real form, let X ∈ g andwrite X = 1

2(X+α(X))+ 12(X−α(X)). Now α(X+α(X)) = α(X)+X

and α(X−α(X)) = α(X)−X = −(X−α(X)). Hence 12(X+α(X)) ∈ gα

and 12 (X−α(X)) ∈ igα. This shows g = gα+igα. Clearly, gα∩igα = 0

so gα is a real form. The last statement is left to the reader.

Theorem 7.4.7. Any complex semisimple Lie algebra has a compactreal form.

Before turning to the proof of this important fact we mention thatfor classical groups and their Lie algebras one can actually verify theresult by inspection.

Example 7.4.8. In the case of classical Lie algebras we can verifyCartan’s theorem by hand. We have the following examples.

gl(n,C) = u(n,C)C, (7.18)

sl(n,C) = su(n,C)C (7.19)

sp(n,C) = sp(n)C (7.20)

so(n,C) = so(n,R)C (7.21)


To see that the real Lie algebras u(n,C), su(n,C), sp(n) and so(n,R)are of compact type it is sufficient to show that they are respectivelythe Lie algebras of appropriate compact Lie groups. Indeed,

U(n,C)• = u(n,C)

SU(n,C)• = su(n,C)

Sp(n)• = sp(n)

SO(n,R)• = so(n,R),

where the • signifies the associated Lie algebra.As a result, since all the Lie groups above are compact we conclude

by Corollary 3.9.7 that the corresponding Lie algebras are all of compacttype. On the other hand the dimension of the real Lie algebras arerespectively n2, n2 − 1, 2n2 + n and n(n−1)

2 which equal the complexdimension of the corresponding complex Lie algebra on the left handside of (7.18).

We now show this is true in general.Proof of Theorem 7.4.7: Let h be a Cartan subalgebra. Define rootvectors as in Theorem 7.3.12 and gk as follows:

gk =∑

λ∈Λ

R(iHλ) +∑

λ∈Λ

R(Xλ −X−λ) +∑

λ∈Λ

Ri(Xλ +X−λ).

Evidently gk is a real subspace of gR. Moreover, gk + igk containsthe C span of the Hλ, the Xλ −X−λ and the Xλ +X−λ. Therefore itcontains the C span of the Hλ, the Xλ and X−λ and by the root spacedecomposition this is g.

To show gk is a Lie algebra, we write it as gk = I + II + III. Wefirst consider the case when we are in the same root space.

Since [iHλ, (Xλ −X−λ)] = i|λ|2(Xλ +X−λ) we see that [Iλ, IIλ] ⊆IIIλ. On the other hand [iHλ, i(Xλ +X−λ)] = −|λ|2(Xλ −X−λ) so weget [Iλ, IIIλ] ⊆ IIλ, and

[(Xλ −X−λ), i(Xλ +X−λ)] = 2iHλ


says that [IIIλ, IIλ] ⊆ Iλ.Now suppose λ 6= ±µ. Then

[(Xλ −X−λ), (Xµ −X−µ)] = Nλ,µXλ+µ +N−λ,−µX−λ−µ−N−λ,µX−λ+µ −Nλ,−µXλ−µ= Nλ,µ(Xλ+µ −X−(λ+µ))

−N−λ,µ(X−λ+µ −X−(−λ+µ)).

Using the relations Nλ,µ = −N−λ,−µ we see [II, II] ⊆ II. Similarlywe get, [II, III] ⊆ III and of course [I, I] = 0 since h is abelian. Thusgk is closed under bracketing and so is a Lie subalgebra of gR. Hencegk is a real form of g.

Finally we will show the Killing form of gk, which is the restrictionof B to gk × gk (see Lemma 3.1.62) is negative definite. Hence byTheorem 3.9.4, gk is of compact type. Now by Proposition 7.3.1, part1, B(I, II + III) = 0. Also, B is positive definite on

∑λ∈Λ R(Hλ)

by Corollary 7.3.8, part 4. Hence B is negative definite on I. Now ifλ 6= ±µ, Proposition 7.3.1 again shows

B(Xλ −X−λ,Xµ −X−µ) = 0,

B(Xλ −X−λ, i(Xµ +X−µ)) = 0,

andB(i(Xλ +X−λ), i(Xµ +X−µ)) = 0.

Finally,

B(Xλ −X−λ,Xλ −X−λ) = −2B(Xλ,X−λ) = −2,

and

B(i(Xλ +X−λ), i(Xλ +X−λ)) = −2B(Xλ,X−λ) = −2,

showing B is negative on gk.

Exercise 7.4.9. Let g = sl(2,C) with the usual basis X+,H,X−(which gives roots). Show Theorem 7.4.7 gives gk = su(2,C).


The examples given just above also suggest the following

Corollary 7.4.10. Let G be a connected complex semisimple Lie group,k be a compact real form of g, and K be the connected Lie subgroup withLie algebra k. Then K is a maximal compact subgroup of G.

Proof. We know K is compact by Theorem 3.9.4 since its Killing form isnegative definite. Let L be a maximal connected compact subgroup of Gcontaining K. If x ∈ Z(L) since x then commutes with all ofK, AdG(x)leaves k pointwise fixed. As a complex linear automorphism it must thenleave g pointwise fixed. Therefore AdG(x) = I and x ∈ Z(G). ThusZ(L) ⊆ Z(G). In particular Z(L) is discrete. It follows from Theorem3.9.4 that L is semisimple. Since g = k⊕ ik we know l = k⊕ is, where s

is a vector subspace of k. But

i[s, k] = [is, k] ⊆ l ∩ ik = is.

Hence [s, k] ⊆ s so s is an ideal in k. This means s+is is an ideal in g. Inparticular s+ is is itself semisimple (see Corollary 3.3.21). Let P be thecorresponding complex semisimple subgroup of G. On the other hands + is is also a real subalgebra of l. Therefore P is compact semisimplegroup. But a complex connected Lie group which is compact must beabelian by Proposition 1.1.10. Therefore P is abelian. But it is alsosemisimple therefore P = 1. This means s = 0 and hence l = k soL = K.

We will now give an alternative proof of the theorem of HermannWeyl concerning complete reducibility of representations of a complexsemisimple Lie group using the so called unitarian trick. This was ac-tually the first proof of this result. Of course, Theorem 3.4.3 is moregeneral that Theorem 7.4.11, but the present one retains great appeal tothe authors. Here is its statement. Because of Theorem 7.4.7 it appliesto any complex semisimple (or reductive) Lie group.

Theorem 7.4.11. Let G be a complex connected Lie group whose Liealgebra has a compact real form, k. Then every finite dimensional holo-morphic representation is completely reducible.


Of course just as in Theorem 3.4.3 once we know the complete re-ducibility for holomorphic representations of complex groups we also getcomplete reducibility for real groups.

The proof of Theorem 7.4.11 below requires the following simplelemma.

Lemma 7.4.12. Let φ : Cn → C be an entire function which vanishesidentically on Rn. Then φ ≡ 0.

Proof. φ(z1, . . . , zn) vanishes when all zi are real, so considerφ(z1, x2 . . . , xn) where the xi are real. This is an entire function of z1and vanishes on the real axis. By the identity theorem, it vanishes iden-tically. Let z1 be fixed, but arbitrary and consider φ(z1, z2, x3 . . . , xn)where the xi are real. This is an entire function of z2 which vanisheswhen z2 is real and, therefore, identically in z2. Continuing by induc-tion, we see that φ(z1, . . . , zn) ≡ 0.

It also requires the following results:

(1) Let ρ be a representation of a connected Lie group H on V andρ′ its differential representation on the Lie algebra h. Then asubspace W of V is H-invariant if and only if it is h-invariant.

(2) The Lie algebra g of a complex semisimple Lie group G has acompact real form k and the Lie subgroup K of G with Lie algebrak is compact. (In fact its a maximal compact subgroup of G).

Proof. Let ρ be a holomorphic representation of G on V . We will showthat if W is a K-invariant subspace of V , then it is actually G-invariant.This would imply

(1) If ρ is irreducible, then so is ρK .

(2) ρ is completely reducible since ρK is.

Proof of the first statement. Suppose not, then since K is compact,ρK = Σρi, a direct sum of irreducibles. Each of the correspondingsubspaces Vi is K and, therefore, G-invariant. Hence ρ is reducible, acontradiction.

7.5 The Iwasawa Decomposition 343

Proof of the second statement. Let W be a G-invariant subspace ofV . Then W is K-invariant. Since K is compact, there is a complemen-tary K-invariant subspace W ′ which would be therefore G-invariant.Thus ρ would be completely reducible.

Therefore, it only remains to show that ifW is a k-stable, C-subspaceof V , then it is g-stable. Let λ ∈ (V/W )∗, the C-dual of V/W , andw ∈ W . Then for k ∈ k, we know ρk(w) ∈ W and hence λ(ρk(w)) = 0.For X ∈ g let φ(X) = λ(ρX(w)), where w ∈ W and λ ∈ (V/W )∗.Then φ : g → C is an entire function which vanishes on k and hence, byLemma 7.4.12, it vanishes on all of g. Since this is true for all w ∈ Wand all λ ∈ (V/W )∗, it follows that W is g-stable.

7.5 The Iwasawa Decomposition

Definition 7.5.1. Let g be a real semisimple Lie algebra. An automor-phism, α, of g is called an involution if α2 = I. Now let B be the Killingform of g and θ be an involution. We call θ a Cartan involution if thebilinear form Bθ(X,Y ) := −B(X, θY ) on g is symmetric and positivedefinite. .

Actually, by the following Lemma Bθ is always symmetric.

Lemma 7.5.2. (1) Let α be an automorphism of any Lie algebra g.Then for each X ∈ g we have adα(X) = α(adX)α−1.

(2) The Killing form B of g is preserved by Aut(g). That is,B(α(X), α(Y )) = B(X,Y ) for all α ∈ Aut(g).

(3) If α is an involution of g, then Bα(X,Y ) = −B(α(X), Y ) is al-ways symmetric.

Proof. (1) For X and Y ∈ g,

adα(X)(Y ) = [α(X), Y ] = α[X,α−1(Y )] = α(adX)α−1(Y ).

(2) B(α(X), α(Y )) = tr(adα(X) adα(Y )). But by 1) this is,

tr(α(adX)α−1α(ad Y )α−1) = tr(α(adX)(ad Y )α−1)

= tr(adX adY ) = B(X,Y ).


(3) By 2) B(α(X), Y ) = B(α2(X), α(Y )) = B(X,α(Y )). But becauseB is symmetric B(X,α(Y )) = B(α(Y ),X). Hence

Bα(X,Y ) = −B(X,α(Y )) = −B(α(Y ),X) = Bα(Y,X).

Example 7.5.3. Suppose g is a linear semisimple Lie algebra which isstable under taking transpose. Let θ(X) = −Xt. Then θ is a Cartaninvolution. Evidently, θ is a linear operator and θ(θ(X)) = −((−Xt))t =X. Also,

θ[X,Y ] = −[X,Y ]t = −[Y t,Xt] = [−Xt,−Y t] = [θ(X), θ(Y )],

so θ is an involution. To see that it is a Cartan involution since sym-metry is automatic we show Bθ(X,Y ) = −B(X, θY ) is positive defi-nite. i.e -tr(adX ad θ(X)) ≥ 0 and positive unless X = 0. But thisis − tr(adX ad−Xt) = tr(adX adXt) ≥ 0. If it were zero then adXwould be zero, since this is the Hilbert-Schmidt norm. By semisimplicityX = 0.

Remark 7.5.4. Notice that in these examples (see Chapter 6) theCartan decomposition g = k ⊕ p is given by X = X−Xt

2 + X+Xt

2 ,where the first factor is in k and the second in p. Then we havek = X ∈ g : θ(X) = X and p = X ∈ g : θ(X) = −X. Thus θ|k = Iand θ|p = −I. Since g = k ⊕ p this means that θ is diagonalizable witheigenvalues ±1. k is the 1-eigenspace and p the −1-eigenspace.

One further observation is that for X and Y ∈ k, Bθ(X,Y ) =−B(X, θ(Y )) = −B(X,Y ) so that Bθ(X,X) = −B(X,X). Similarlyfor X and Y ∈ p we have Bθ(X,X) = B(X,X). Since (see Chapter6) Bk×k is negative definite and Bp×p is positive definite, we see thatBθ is positive definite on k and negative definite on p. All these arecharacteristic properties of a Cartan involution.

We do not yet know that Cartan involutions exist. To this end weneed the following.


Proposition 7.5.5. Let g0 is a real Lie algebra, g its complexification,and gR be g regarded as a real Lie algebra. Then the Killing forms arerelated as follows: B0(X,Y ) = B(X,Y ), X and Y ∈ g0 and BR(X,Y ) =2ℜ(B(X,Y )), X and Y ∈ g. If any of these algebras is semisimple soare all the others.

Proof. We first show if g0 is semisimple then so are the other two. Sup-pose g0 is semisimple. Then B0 is nondegenerate. Let B(Z1, Z2) =0 for all Z2 ∈ g. Then for j = 1, 2, Zj = Xj + iYj andB(Z1, Z2) = B(X1,X2) + iB(X1, Y2) + iB(Y1,X2) − B(Y1, Y2) = 0.Hence B0(X1,X2) = B0(Y1, Y2) and B0(X1, Y2) = −B0(Y1,X2). TakeZ2 = X2 (and Y2 = 0) and get X1 and Y1 = 0 i.e. Z1 = 0. Hence g issemisimple.

Since g is semisimple B is nondegenerate. Suppose BR(Z1, Z2) = 0for all Z2. Then B0(X1,X2) = B0(Y1, Y2) for all X2 and Y2. Takingeach of these in turn to be zero shows X1 and Y1 are both zero. HenceZ1 = 0.

Now to the computation of the Killing forms. The first of these isobvious since g0 is a real subalgebra of gR. If A + iB is the matrix ofadX adY with respect to a basis X1 + iY1, . . . ,Xn + iYn of g (here Aand B are real), then the matrix of this same operator on gR is the

2n× 2n matrix

(A B−B A

), so the second relation follows.

Corollary 7.5.6. Let g be a complex semisimple Lie algebra and gR

be g considered as a real (semisimple) Lie algebra. Let u be a compactreal form of g and τ be the associated conjugation. Then τ is a Cartaninvolution of gR.

Proof. τ is evidently an involution. To see that τ is a Cartan involu-tion of gR we must show BgRτ is positive definite. But BgR

(Z1, Z2) =2ℜBg(Z1, Z2). Writing Z ∈ g as X + iY , where X and Y ∈ u weget BgR

(Z, τ(Z)) = BgR(X + iY,X − iY ) = Bg(X,X) + Bg(Y, Y ) =

Bu(X,X) + Bu(Y, Y ), which is ≥ 0 and > 0 unless X = 0 = Y i.eZ = 0.


Proposition 7.5.7. Let g be a real semisimple Lie algebra and θ be aCartan involution. For any involution σ of g there exists α ∈ Inn g sothat αθα−1 commutes with σ.

Proof. Let Bθ be the associated positive definite inner product on g.Then η = σθ ∈ Aut(g). Now because θ2 = I we see ηθ = σθ2 = σ.Taking inverses we get θ−1η−1 = σ−1. But because σ and θ are each oforder 2 we see. θη−1 = σ. Hence ηθ = θη−1. Alternatively, η−1θ = θη.

Now Lemma 7.5.2 tells us that Aut(g) leaves B invariant. HenceB(η(X), θ(Y )) = B(X, η−1θ(Y )) = B(X, η−1θ(Y )) = B(X, θη(Y )).Thus Bθ(η(X), Y ) = Bθ(X, η(Y )) so η is self adjoint. Taking X = η(Y )we get Bθ(η

2(X),X) = Bθ(η(X), η(X)) which tells us that since η isan automorphism, η2 is positive definite and hence is diagonalizablewith positive real eigenvalues. Now we make use of the fact that Expis a diffeomorphism between P and P in Chapter 6. In particular,η2 = Exp(W ) for some self-adjoint operator W with respect to Bθ. HereW is in the Lie algebra of Aut(g). Hence Exp(tW ) ∈ Aut(g)0 = Inn(g)for all real t. The latter because g is semisimple and so all derivationsare inner. Because W is diagonal so is Exp(tW ) for all real t. Henceeach of the Exp(tW ) commutes with η. The relation η−1θ = θη thenpropagates to the whole 1-parameter group Exp(−tW )θ = θExp(tW ).Hence

Exp(1

4W )θExp(−1

4W )σ = Exp(

1

2W )θσ = Exp(

1

2W )η−1

= ηExp(−1

2W ) = σθExp(−1

2W )

= σExp(1

4W )θExp(−1

4W ).

Taking α = Exp(14W ), we see σ commutes with αθα−1 for some

α ∈ Inn(g).

Corollary 7.5.8. Any real semisimple Lie algebra, g, has a Cartaninvolution.


Proof. Let gC be the complexification of g. Then by Proposition 7.5.5,gC is a complex semisimple Lie algebra and hence by Theorem 7.4.7 has acompact real form u. Let σ and τ be conjugations of gC with respect to g

and u, respectively (see Lemma 7.4.4). Then they are each involutionsof gC regarded as a real Lie algebra. Here we write l for uC = gC.By Proposition 7.5.5 l is semisimple. Hence by Corollary 7.5.6, τ is aCartan involution of l. By Proposition 7.5.7, we can find α ∈ Inn(l)so that ατα−1 commutes with σ. Now ατα−1 is the conjugation of l

with respect to α(u), which is also a compact real form of g. Hence,Bατα−1(Z1, Z2) = −2ℜBgC

(Z1, ατα−1Z2) is positive definite on l.

Now g is precisely the fixed set under σ. But if σ(X) = X,then σατα−1(X) = ατα−1σ(X) = ατα−1(X) so that ατα−1 re-stricts to an involution, θ, of g and Bθ(X,Y ) = −Bg(X, θ(Y )) =−Bg(X,ατα

−1(Y )) = 12Bατα−1(X,Y ) so that Bθ is positive definite

and θ is a Cartan involution.

Before turning to the Iwasawa decomposition we need the followinglemma. We denote by (·, ·) the inner product on gC associated with aCartan involution (by Corollary 7.5.8).

Lemma 7.5.9. Let g be a real semisimple Lie algebra and θ be aCartan involution. Then for each X ∈ g as an operator on gC,adX∗ = − ad θ(X). In particular ad k acts on gC by skew Hermitianoperators while ad p acts by Hermitian operators. Hence each of theseis diagonalizable with purely imaginary eigenvalues, or real eigenvalues,respectively.

Proof. Let Y and Z ∈ gC. Then (adXY,Z) = −B(adX(Y ), θ(Z)). Bythe invariance of the Killing form (see 3.1.60), this is B(Y, adX(θ(Z))).Because θ is an involution this last term is just B(Y, θ[θ(X), Z]) =−Bθ(X, ad θ(X)Z). Thus adX∗ = − ad θ(X).

We first formulate the Iwasawa decomposition for a (non-compact)real semisimple Lie algebra. Let g = k⊕ p be the Cartan decompositionof g (see Chapter 6). Let a be a maximal abelian subspace of p. Thenby the Lemma 7.5.9 just above, the elements of ad a are simultaneously


diagonalizable with real eigenvalues. Let θ be the corresponding Cartaninvolution of g.

This leads to the following

Definition 7.5.10. For λ ∈ a∗, the real dual space of a, λ 6= 0 we formthe restricted root space,

gλ = X ∈ g : adHX = λ(H)X for all H ∈ a.

We write Λ for the set of restricted roots with gλ 6= 0. We also writem = zk(a), the centralizer of a in k.

Proposition 7.5.11. g is a direct sum of subspaces, a⊕m⊕∑λ∈Λ gλ.

Proof. Let g0 = X ∈ g : adHX = 0 for all H ∈ a. Then g =g0⊕

∑λ∈Λ gλ. Now θ is an automorphism of g which sends each element

of p and hence of a to its negative. So if [H,X] = 0 for all H ∈ a, thenθ[H,X] = [−H, θ(X)] = 0. So [θ(X),H] is also 0 for all H ∈ a. Henceg0 is θ-stable. Applying this to the Cartan decomposition and takinginto account that θ also preserves k and p tells us g0 = g0 ∩ k ⊕ g0 ∩ p.But g0 ∩ k = m and by maximality of a, g0 ∩ p = a.

Now let H1, . . . ,Hr be a basis of a. Order a∗ lexicographically rela-tive to this ordered basis. If Λ+ is the positive roots and Λ− the negativeroots, Then Λ is the disjoint union of Λ+ and Λ−. Also, if λ and µ ∈ Λ+

and λ+ µ ∈ Λ, then λ+ µ ∈ Λ+ and finally −Λ+ = Λ−.Let n+ =

∑λ∈Λ+ gλ. Then n+ is a subalgebra of g. Since Λ+ is

finite and if λ and µ ∈ Λ+, then λ+ µ is larger then either of them wesee that n+ is nilpotent. Similarly let n− =

∑λ∈Λ− gλ. i.e. n− = θ(n+).

Then we get another nilpotent subalgebra and g = n− ⊕ m ⊕ a ⊕ n+.We now come to the Iwasawa decomposition of a real semisimple Lie

algebra.

Theorem 7.5.12. g = k ⊕ a ⊕ n+ (direct sum of subspaces).

Proof. Let N− ∈ n−, then N− = N− + θ(N−) − θ(N−) ∈ k + n+ asθ(N−) ∈ n+. Hence n− ⊂ n++k. Since m ⊂ k and g = a⊕m⊕n−⊕n+ ⊂a + n+ + k, g = a + n+ + k.


Let X ∈ k, H ∈ a and N+ ∈ n+ and assume X + H + N+ = 0.Then applying θ we get 0 = X − H + θ(N+). Subtracting we find,2H +N+ − θ(N+) = 0, where θ(N+) ∈ n−. But a∩ (n+ + n−) = 0 sothat H = 0 and N+ = N−. Since n+ ∩ n− is also 0, N+ = 0 = N−.Therefore X = 0 and the sum is direct.

Before turning to the Iwasawa decomposition of a real semisimpleLie group G, we first deal with a few preliminaries.

Proposition 7.5.13. Let G be a real semisimple Lie group. Then

(1) AdG is closed in GL(g).

(2) If G is linear, then Z(G) is finite and Z(G) ⊆ K.

Proof. 1. This is because Aut(g), as the real points of an algebraic group(see Proposition 1.4.27), is definitely closed in GL(g), as is its identitycomponent. Alternatively see Corollary 3.4.5. Since every derivation ofg is inner and AdG is connected we see (Aut(g))0 = AdG.

2. Suppose G ⊆ GL(n,C) = GL(V ). Since G is semisimple, byTheorem 3.4.3 V is the direct sum of invariant irreducible subspaces Vi.For each i, the map g 7→ g|Vi

is a smooth homomorphism. Hence G|Vi

is also a semisimple group. By Schur’s lemma for each i, Z(G) acts byscalars λi on Vi. Since g 7→ (g|V1 , . . . , g|Vr ) is injective it suffices to proveeach Z(G)|Vi

is finite. Thus we have replaced G by G|Vi. In other words

we may assume Z(G) acts by scalars on V . Since G is a semisimplegroup it has no characters. In particular, det g ≡ 1. Restricting to thecenter we see λn ≡ 1 so Z(G) is finite. Also, K is compact (Chapter6). Therefore, Z(G)K is a compact subgroup of G containing K. Butsince K is actually a maximal compact subgroup of G, Z(G)K = K soZ(G) ⊆ K.

Exercise 7.5.14. Under the hypothesis of 2) show that the order ofZ(G) can be estimated by |Z(G)| ≤ n.

We now set notation and some preparatory ideas for the Iwasawadecomposition.


In GL(n,R) we let Dn and Nn stand for the diagonal and strictlyupper triangular matrices. If g is a real semisimple Lie algebra, let(X,Y ) = −B(X, θY ), where B is the Killing form. Then (·, ·) is a posi-tive definite inner product on g. Let Λ be the restricted roots relative toa. Choose a linear ordering on a∗ as before. Then we know by Proposi-tion 7.5.11 that g = a⊕m⊕n−⊕n+. Let Λ+ = λ1 < . . . < λr. Supposethe dimension of gλi

is pi, i = 1 . . . r. Choose an orthonormal basis foreach gλi

putting them together in reverse order. Set q = p1 + . . . pr,the dimension of n+. Let Xq+1 . . . Xq+m be an orthonormal basis ofa ⊕ m. Finally let Xq+m+j = θ(Xq−j+1) giving an orthonormal basis ofn−. Then X1, . . . Xn is an orthonormal basis of g, where n = 2q + m.Relative to this basis ad k is skew symmetric ad a is diagonal and ad n+

is strictly upper triangular.In our formulation of the Iwasawa decomposition, just below, one

must assume G is linear (or at least has a finite center) in order to besure that K is compact. Examples of when this difficulty can arise areprovided by the universal covering group of SL(2,R), or more generallyby the universal covering group of Sp(n,R).

Theorem 7.5.15. Let G be a linear real semisimple Lie group withLie algebra g = k ⊕ a ⊕ n+ and let K, A and N be the correspondingconnected Lie subgroups (see Section 1.6). Then

(1) exp : a → A is a Lie isomorphism. So A is a simply connectedabelian group.

(2) exp : n+ → N is a (surjective) diffeomorphism. N is a simplyconnected nilpotent group.

(3) The multiplication map (k, a, n) 7→ kan is a diffeomorphism fromK ×A×N → G.

Proof. We first prove 1) and 2) in general.Now ad : g → End(g). Here ad is injective since g is semisimple.

Hence for X ∈ a or n+, respectively, we can regard X as a diagonal,respectively upper nil-triangular operator on g. (However, in doingso we are identifying Exp with exp which is not strictly correct sinceAd expX = Exp(adX). Thus we must keep in mind that Exp actuallytakes us into AdG.) When X ∈ a, respectively, n+ then exp(X) is


diagonal with positive entries, respectively upper unitriangular with realentries and in each case is a diffeomorphism onto A, or N respectively.In the former case this is essentially because exp : R → R×

+ is a globaldiffeomorphism and since a is abelian exp is also a homomorphism,proving 1).

Now in the latter case as we saw in Proposition 7.5.13 AdG is closedin GL(g) and as a linear subspace ad n+ is closed in the nil-triangularmatrices of End(g). Therefore Exp(n+) is closed in GL(g) and thereforealso in AdG. Because on n+ we know exp and log are inverses of oneanother (see Proposition 6.2.2) this proves 2). In order to prove 3) wefirst consider the case when G is the adjoint group, i.e. when Z(G) istrivial. Now Ad(A) ⊆ Dn, Ad(N) ⊆ Nn and each is closed in the re-spective linear group. By Proposition 1.6.2, GL(n,R) the multiplicationmap there is a global diffeomorphism. It follows that Ad(A)Ad(N) isclosed in DnNn and hence in GL(n,R). But Ad(K) is compact henceAd(K)Ad(A)Ad(N) = Ad(KAN) is closed in GL(n,R) and hence inAdG.

By the GL(n,R) case (see Proposition 1.6.2) the multiplication map,φ, taking Ad(k)Ad(a)Ad(n) 7→ Ad(kan) is injective. We calculate itsderivative at a general point kan. LetX ∈ ad k, Y ∈ ad a and Z ∈ ad n+.Then we will show dkanφ(X,Y,Z) = Ad(an)−1X+Ad(n)−1Y +Z. Thereader will notice how similar this calculation is to that in Lemma 7.2.10.We have

dkanφ(X, 0, 0)f =d

dtf(k exp(tX)an)|t=0

=d

dtf(kan exp(Ad(an)−1tX)|t=0

= Ad(an)−1Xf,

dkanφ(0, Y, 0)f =d

dtf(ka exp(tY )n)|t=0

=d

dtf(kan exp(tAd(n)−1Y ))|t=0 = Ad(n)−1Y f,


and

dkanφ(0, 0, Z)f = Zf.

Hence using the linearity of d0φ we see

dkanφ(X,Y,Z) = Ad(an)−1X + Ad(n)−1Y + Z.

Hence if dkanφ = 0 we get Ad(an)−1X + Ad(n)−1Y +Z = 0 so thatX = −Ad(a)Y −Ad(an)Z. Since Y ∈ ad a, Z ∈ ad n+ and N is normalin AN (i.e. N is normalized by A), the same is true of −Ad(a)Y and−Ad(an)Z respectively. By Theorem 7.5.12, we see X = 0 and henceAd(a)Y +Ad(an)Z = 0. But then Ad(a)Y and Ad(an)Z are each zero.Finally, this yields Y = 0 = Z so that dkanφ is non singular at everypoint. By the inverse function theorem φ is a global diffeomorphism.In particular the image is open in AdG. But by connectedness this hasno open subgroups so φ is also surjective. This proves 3) when G is theadjoint group.

Now in general for g ∈ G, AdG = Ad(k)Ad(a)Ad(n). Thereforeg = zkan, where z ∈ Z(G). But Z(G) is itself in K so any g ∈ G can bewritten as g = kan and the multiplication map, φ, is surjective here aswell. If k′a′n′ = kan, then taking Ad of everything in sight and applyingwhat we already know tells us that Ad(k′) = Ad(k), Ad(a′) = Ad(a)and Ad(n′) = Ad(n). But then by 1) and 2) of the theorem which wehave already proved we get a′ = a and n′ = n. Hence also k′ = k soφ is injective here as well. Thus φ is a bijection. Using this φ in thederivative calculations just above shows it too is a diffeomorphism atevery point proving 3) and with it the theorem.

We conclude this chapter with the following important global resultwhich shows, for example, that the Iwasawa decomposition theoremapplies to all complex semisimple groups. It is important for otherreasons as well. Our proof will use a theorem of algebraic groups dueto Chevalley. The original proof due to Goto was different. Chevalley’stheorem had not yet been discovered. However, before turning to thistheorem we need the following lemma.


Lemma 7.5.16. Let G any complex Lie group with a faithful represen-tation ρ : G→ GL(n,C) and F be a finite central subgroup. Then G/Falso has a faithful representation.

Proof. The finite group ρ(F ) is Zariski closed in GL(V ) as is its nor-malizer N in GL(V ). Therefore by the theorem of Chevalley [9]N/ρ(F ) is an algebraic group (in say GL(W )) and the projection,π : N → N/ρ(F ), is a rational morphism. In particular π is holo-morphic. Now consider π ρ which is a holomorphic representation ofG on W . Its kernel is exactly F .

Theorem 7.5.17. A complex semisimple Lie group G always has afaithful holomorphic linear representation. In particular, by Proposition7.5.13 a complex semisimple Lie group always has a finite center.

Proof. Let (G, π) be the universal covering of G. Then G is also a com-plex semisimple group. If we can show it has a faithful representationthen Z(G) must be finite by Proposition 7.5.13. Since Kerπ = F is a(discrete) central subgroup of G, we may assume by Lemma 7.5.16 thatG itself is simply connected.

Let k be a compact real form of g. Then the corresponding group, K,is a maximal compact subgroup (see Corollary 7.4.10). By a corollaryto the Peter-Weyl theorem, K has a faithful smooth representation onU . Its derivative gives a faithful representation of k on U which extendscanonically to a complex representation of g = k⊕ ik on UC. By simpleconnectivity of G there is a holomorphic representation σ of G on UC

whose derivative, dσ is this representation of g. Now KerAd = Z(G) ⊆K (Proposition 7.5.13). Thus σ⊕Ad is a holomorphic representation ofG on UC⊕g. If AdG = I, then g ∈ Z(G) ⊆ K so if in addition σ(g) = Iwe get g = 1. Hence Ker(σ ⊕ Ad) is trivial and σ ⊕ Ad is faithful.

Theorem 7.5.17 is not true, in general, for real groups for a verysimple reason. Namely, if such a group were linear it would have tohave finite center. But we have seen many examples of real semisimplegroups with an infinite center.


Exercise 7.5.18. Show the intermediate coverings of SL(2,R) also haveno faithful representations (in spite of the fact that they have finitecenters).

Chapter 8

Lattices in Lie Groups

In this chapter we will consider a Lie group G and a lattice (or a uniformlattice) Γ and we will ask how much of G can be recovered, or is deter-mined by Γ? Another perhaps even more fundamental question is whenis there such a Γ, or how can one construct a Γ? A third might be toinvestigate the properties of such Γ’s and to distinguish between latticesand uniform lattices in G. Finally we should ask, just how different isa lattice in a general Lie group in relation to that group in comparisonto a lattice in Euclidean space as compared to Rn? These are all as-pects of a fundamental issue in mathematics. Namely, the comparisonof the discrete to the continuous. We begin with the progenitor, namelyEuclidean space.

8.1 Lattices in Euclidean Space

In this section we discuss some results concerning lattices in Euclideanspace. These are fundamental to further developments and, as thereader will see, are of considerable interest in their own right. Hereby a lattice we mean a discrete subgroup Γ of Rn with finite volumequotient; in other words a subgroup of Rn with n linearly independentgenerators.

Exercise 8.1.1. Show that in Rn a closed subgroupH has finite volume

355

356 Chapter 8 Lattices in Lie Groups

quotient if and only if the quotient is compact. Hint: Consult Exercise0.1.7.

A typical example of a lattice is Zn. But of course if g ∈ GL(n,R)and Γ is a lattice then so is gΓ. In fact a moments reflection tells us thatwe get all lattices in this way. Thus GL(n,R) acts transitively on theset L of lattices. Therefore we can choose any lattice as a base point forthis orbit. Choosing the standard lattice, Zn, we see that the isotropygroup is GL(n,Z). Thus L can be identified in a natural way with thehomogeneous space GL(n,R)/GL(n,Z). To topologize L we take thatnatural topology from this coset space. It does not depend on a choiceof generators in the lattice and makes L into a locally compact, secondcountable, Hausdorff manifold. In this way the lattices in Rn and thehomogeneous space GL(n,R)/GL(n,Z) (as well as SL(n,R)/SL(n,Z))are very closely related.

Let Γ be a lattice in Rn, dµ be Lebesgue measure on Rn and π :Rn → Rn/Γ be the natural projection. Then (see Theorem 2.3.5 ) thereis a invariant finite regular measure dµ on the torus, Rn/Γ such that fora continuous function on Rn with compact support one has

∫

Rn

f(x)dµ(x) =

∫

Rn/Γ(∑

γ∈Γ

f(γ + x))dµ(x).

Thus the three measures are related and normalizing any two of them(say Lebesgue measure and counting measure) determines the third, dµ.So for example if Γ = gZn, then µ(Rn/Γ) = |det g|. (Notice that thisstatement is independent of the g doing this since if h leaves Zn stable,then |deth| = 1).

Our study of lattices in Rn begins with Minkowski’s theorem.

Theorem 8.1.2. Let Γ be a lattice in Rn and Ω be an open convex setwhich is symmetric about the origin. If vol(Ω) ≥ 2n vol(Rn/Γ), then Ωmeets Γ is a nontrivial lattice point.

Proof. Let π : Rn → Rn/Γ be the natural projection. This map is eitherinjective on 1

2Ω, or it is not. In the latter case there must be a γ 6= 0 ∈ Γso that γ + x ∈ 1

2Ω and x ∈ 12Ω. But then by symmetry and convexity

8.1 Lattices in Euclidean Space 357

of 12Ω we get 1

2(−x) + 12(γ + x) = 1

2γ ∈ 12Ω. Hence γ ∈ Ω and we would

be done.We will show that the other alternative, namely that π is injective

on 12Ω leads to a contradiction. Suppose π is injective on 1

2Ω, then πwould also be injective on 1

2Ω. Now vol(12Ω) = vol(π(1

2Ω)) ≤ vol(Rn/Γ).But since vol(Ω) ≥ 2n vol(Rn/Γ) we know vol(π(1

2Ω)) ≥ vol(Rn/Γ). Itfollows that π restricted to this set is surjective. For if the image weresmaller, since Rn/Γ is of finite (regular) measure there would be an openset of positive measure left out, a contradiction. Because π(1

2Ω) = Rn/Γit follows Rn =

⋃γ∈Γ γ + 1

2Ω, and since π is injective here this union is

disjoint. Let U = 12Ω and V =

⋃γ 6=0∈Γ(γ + 1

2Ω). Then U and V areboth closed (and open) and disjoint and U ∪ V = Rn which contradictsthe connectivity of Rn.

Applications of Minkowski’s theorem:

Consider an n × n nonsingular real matrix (aij) and use it to definelinear functionals λi where i = 1, . . . , n on Rn by

λi(x) =∑

j

aijxj ,

where x = (x1, . . . , xn). Then A : Rn → Rn defined by

A(x) = (λ1(x), . . . , λn(x))

is an invertible linear transformation whose determinant is det(aij).Choose positive constants ci, i = 1, . . . , n so that c1 . . . cn ≥ |detA|and let

Ω = x ∈ Rn : |λi(x)| ≤ ci for all i.Corollary 8.1.3. Ω meets Zn is a nontrivial lattice point.

Proof. It is easy to see that Ω is a closed convex set which is symmetricabout the origin. Now

vol(Ω) = (detA)−1∏

(2ci) ≥ 2n = 2n vol(Rn/Zn).

Hence by Minkowski’s theorem Ω meets Γ nontrivially.


Minkowski’s theorem can also be applied to positive definite, sym-metric bilinear forms on Zn. Indeed this was its original purpose. Wedenote the ball in Rn centered at 0 of radius 1 by B1(0).

Exercise 8.1.4. Let β(x, y) = (Bx, y), where (·, ·) is the usual innerproduct, x, y ∈ Zn and B is a positive definite symmetric n× n matrixwith integer coefficients. Prove there exists a non zero x ∈ Zn such that

β(x, x) ≤ 4 det1n (B). vol(B1(0))

−2n .

Suggestion: Orthogonally reduce β to diagonal form and use thefact that the volume of an ellipsoid is volB1(0) times the product of itsvarious semiaxes.

Let q(x) =∑n

i,j=1 aijxixj be a real quadratic form, where x =(x1, . . . , xn) ∈ Rn and (aij) = A is a positive definite real symmetricmatrix. For c > 0 let

Xc = x ∈ Rn : q(x) ≤ cdet1n (aij).

Corollary 8.1.5. Let q(x) =∑n

i,j=1 aijxixj be a positive definite sym-metric form. Given a lattice Γ for c sufficiently large Xc must meetΓ.

Proof. We first prove that vol(Xc) = volB1(0)cn. Since A is positive

definite it can be diagonalized i.e. there is an orthonormal matrix Bsuch that BtAB is diagonal with λi‘s as eigenvalues. Consider x =By ∈ Xc, we have q(x) =

∑i λiy

2i , therefore

∑i λiy

2i ≤ c(Πλi)

1n . Hence

for positive numbers µi = λi

c(Πλi)1n

we see that B−1(Xc) is defined by∑µix

2i ≤ 1. Since this is an ellipsoid centered at 0 it is clearly is closed

and convex. Its volume is given by

volB1(0)Πµi = volB1(0)Πλi

c(Πλi)1n

= volB1(0)cn.

On the other hand vol(Xc) = vol(B−1(Xc)) = volB1(0)cn since B is

orthonormal. Now Xc is a closed symmetric convex set centered at 0in Rn satisfying vol(Xc) = volB1(0)c

n. In particular, this volume isindependent of q, and for a lattice Γ, c can be chosen large enough sothat vol(Xc) ≥ 2n vol(Rn/Γ).

8.1 Lattices in Euclidean Space 359

An interesting application of Minkowski’s theorem to number theoryis the four square theorem which was first proved by Lagrange 100 yearsprior to Minkowski by other methods.

Corollary 8.1.6. Any positive integer is the sum of at most 4 squares.

Before turning to a sketch of the proof we mention that as it canbe easily checked 7, for example, cannot be written as the sum of 3 orfewer squares.

Proof. If x and y are quaternions and N(x) = x21 + · · · + x2

4 is its normthen it is well known that N(x)N(y) = N(xy). Thus if one has aproduct of a sum of four squares by another sum of four squares theresult is again a sum of four squares. It shows that it is sufficient toprove the 4 square theorem for primes, p and we may evidently assume pis odd. For such a prime there exist integers r and s such that r2+s2+1is divisible by p.

Proof of this: Let S+ = 02, 12, . . . (p−12 )2. Then |S+| = p−1

2 + 1 =p+12 . Similarly, S− = 02 − 1,−12 − 1, . . .− (p−1

2 )2 − 1 and |S−| = p+12 .

Now if x2 ≡ y2 mod(p), p divides x2 − y2 so p divides x − y or x + y.That is x ≡ ymod(p) or x ≡ −ymod(p). Therefore none of the elementsof S+ are congruent mod(p) and similarly none of the elements of S−are congruent mod(p). But there are only p residue classes mod(p)and |S+ ∪ S−| = |S+| + |S−| = p + 1, hence there exist r2 ∈ S+ and−s2 − 1 ∈ S− so that r2 ≡ −s2 − 1mod(p).

Now consider the matrix

T =

p 0 r s0 p s −r0 0 1 00 0 0 1

.

Then T is nonsingular so Γ = T (Z4) is a lattice. In fact since|detT | = p2 we see that µ(R4/Γ) = p2. The volume of a ball B4(r) in

R4 of radius r > 0 is π2

2 r4 (see Proposition 2.1.10). Therefore the ball

of radius√

2p which is a convex body symmetric about the origin hasπ2

2 4p2 = 2π2p2 > 24p2. Therefore by Minkowski there is a nonzero


lattice point γ in this ball and the sum of the squares of the γi is< r2 = 2p. But a direct calculation using the matrix T , where γ = T (x)shows that p divides N(γ). So N(γ) ≥ p. Hence N(γ) = p.

We shall return to questions concerning families of lattices in Eu-clidean space shortly.

8.2 GL(n,R)/ GL(n,Z) and SL(n,R)/ SL(n,Z)

As we saw GL(n,R)/GL(n,Z) can be identified with L, the space ofall lattices in Rn and in this way we can put a manifold structure onL. This cuts both ways, we can also use our knowledge of L to learnsomething about the coset space. Given a lattice, Γ in Euclidean space,we take a basis x1, . . . , xn for it. Then consider the parallelepipedspanned by this basis. By abuse of notation we say vol(Rn/Γ) is theEuclidean volume of the parallelepiped. So vol(g(Zn)) = |det(g)| givesa well-defined map ∆ : L → R. Now consider L0 the space of latticeswhose parallelepiped has volume 1. Clearly SL(n,R) operates transi-tively on L0 with isotropy group SL(n,Z). Hence SL(n,R)/SL(n,Z)can be identified with the space of lattices, L0. As we shall see thishomogenous space has finite volume, but is non-compact. Whereas Litself does not even have finite volume.

To see that GL(n,R)/GL(n,Z) cannot support a finite GL(n,R)-invariant measure. Suppose µ was such a measure, consider thedet : GL(n,R) → R×. It induces an onto map GL(n,R)/GL(n,Z) →R×/(±1) ∼= R×

+. Push µ forward (Proposition 2.3.6) with the latter mapto get a finite invariant measure on R×

+ which must be Haar measure.Hence this group would have to be compact, a contradiction.

We now define a Siegel domain for GL(n,Z) within GL(n,R). Thiswas actually done by C.L. Siegel for all the classical non compact simplegroups ([8]). We shall only deal with GL(n,R) and SL(n,R).

Recall the Iwasawa decomposition for G = GL(n,R) which says thatas a manifold G = KAN where K = O(n,R), A consists of diagonalmatrices with positive entries and N is all upper triangular matriceswith eigenvalues 1. This is a direct product K × A × N where the

8.2 GL(n,R)/GL(n,Z) and SL(n,R)/SL(n,Z) 361

inverse map is given by group multiplication in G. For t, u > 0 wedenote by

At = a ∈ A :aii

a(i+1)(i+1)≤ t, i = 1, . . . n− 1

and by

Nu = n ∈ N : |nij| ≤ u, 1 ≤ i < j ≤ n.

Since N is diffeomorphic to Rn(n−1)

2 we know Nu is compact. We definethe Siegel domain St,u = KAtNu. Evidently Siegel domains are stableunder left translation by K and by scalar multiples of the identity andare compact if and only if At is compact.

We will prove the following using a sequence of lemmas.

Theorem 8.2.1. GL(n,R) = S 2√3, 12GL(n,Z).

Before proving this we first deal with N .

Lemma 8.2.2. N = N 12NZ, where NZ = N ∩ GL(n,Z).

Proof. Suppose that u = (uij) ∈ N . We shall find z = (zij) ∈ NZ suchthat

|(u.z)ij | ≤ 1/2, i < j. (8.1)

As uik = 0 for k < i, zjk = 0 for k < j and uii = zii = 1 for all i, (8.1)reads,

|zij + ui,i+1zi+1,j + ui,i+2zi+2,j + . . . + uij| ≤ 1/2, i < j (8.2)

We find zij recursively starting by j = n and i = n−1. For these values(8.2) is

|zn−1,n + un−1,n| ≤ 1/2

where we can find zn−1,n such that the inequality (8.2) holds. Now byfixing j = n and varying i we can recursively find all zi,n for all i ≤ nsuch that the inequality is satisfied. By the same process we can findall zi,n−1 for i ≤ n− 1 and eventually zi,j for i ≤ j, for all j’s.


Let e1, e2, ..., en be the standard basis for Rn and let Φ(g) = ||ge1||for g ∈ GL(n,R). Then Φ : GL(n,R) → R× is a continuous map.Φ(g) = a11 = Φ(a) where g = kan is the Iwasawa decomposition of g.

Lemma 8.2.3. Let g ∈ GL(n,R) be fixed and consider γ 7→ Φ(gγ).Then this function has a positive minimum.

Proof. gGL(n,Z)(e1) ⊆ g(Zn \ 0) which is the nonzero elements ofsome lattice in Rn, hence ||gγ(e1)|| has a positive minimum as γ variesover GL(n,Z).

Lemma 8.2.4. Let g = kan ∈ GL(n,R) and suppose that Φ(g) ≤ Φ(gγ)for all γ ∈ GL(n,Z). Then a11 ≤ 2√

3a22

Proof. Let n0 ∈ N then gn0 = kann0. Since n, n0 ∈ N so Φ(gn0) =a11 = Φ(g). By Lemma 8.2.2 there is an n0 ∈ NZ so that |(nn0)ij | ≤ 1

2for all i, j, so we can assume that |nij | ≤ 1/2. Now we let γ0 ∈ GL(n,Z)be the following element:

0 −1 01 0 00 0 In−2

.

Then γ0(e1) = −e2, γ0(e2) = −e1 and gγ0(e1) = ge2 = kan(e2) =ka(e2 + e21e1) = k(a22e2 + a11n12e1). So ||gγ0(e1)||2 = a2

22 + a211n

212 ≤

a222 + 1

4a211. By the assumption a2

11 ≤ a222 + 1

4a211 from which the conclu-

sion follows.

Proof of Theorem 8.2.1: We prove Siegel’s theorem by induction on n.When n = 1, GL(n,R) = St,u = R×, therefore there is nothing to prove.Now let g ∈ GL(n,R) and y ∈ gGL(n,Z) so that Φ(y) ≤ Φ(gγ) for allγ ∈ GL(n,Z). Hence also Φ(y) ≤ Φ(y.γ) for all γ ∈ GL(n,Z). One canwrite

k−1y y =

(a11 ∗0 b

)

where b is in GL(n − 1,R). So by inductive hypothesis there is z′ ∈GL(n−1,Z) such that bz′ ∈ S 2√

3, 12. Consider the Iwasawa decomposition


of bz′ = k′a′n′ and let

z =

(1 00 z′

)∈ GL(n,Z).

Then

k−1y yz =

(a11 ∗0 k′a′n′

),

and this has an Iwasawa decomposition k′′a′′n′′, where

k′′ = ky

(1 00 k′

)∈ K, a′′ =

(a11 00 a′

)∈ A, n′′ =

(1 00 n′

)∈ N.

By induction a′′ii ≤ a′′(i+1)(i+1) for 2 ≤ i. Since z leaves e1 fixed

therefore Φ(yz) = Φ(y) and consequently Φ(yz) = Φ(y) ≤ Φ(yzγ) for allγ ∈ GL(n,Z). By Lemma 8.2.4 we a′′11 ≤ 2√

3a′′22 therefore yz ∈ KA 2√

3

N

and hence g ∈ yGL(n,Z) = yzGL(n,Z) ⊂ KA 2√3N GL(n,Z). By

Lemma 8.2.2 N = N1/2NZ and therefore KA 2√3N = KA 2√

3N1/2NZ ⊂

S 2√3, 12GL(n,Z).

We now turn to Mahler’s compactness criterion. For a lattice Γ in Rn

we know Γ = g.Zn for some g ∈ GL(n,R) and g is uniquely determinedup to an element of GL(n,Z). Since |det | ≡ 1 on GL(n,Z) we get awell defined function ∆(Γ) = |det g|.

An important result concerning subsets S ⊆ L, the family of alllattices in Rn is Mahler’s theorem first proved in 1946 in [38]. A veryefficient proof of this result can be given by means of Siegel domains inGL(n,R). Mahler’s theorem, which bears a striking resemblance to theclassical theorem of Ascoli, is the following:

Theorem 8.2.5. A subset S ⊆ L has compact closure if and only if:

(1) ∆ is bounded on S.

(2) There exists a neighborhood U of 0 in Rn so that Γ∩U = 0 forall Γ ∈ S.

The first condition is analogous to uniform boundedness while thesecond (often described as S being uniformly discrete) is analogous toequicontinuity in Ascoli’s theorem.


Proof. Because of Theorem 8.2.1 we get all the lattices in Rn alreadyfrom Siegel set. Hence the statement that is compact is equivalent tohaving a subset S of the Siegel set with compact closure and S(Zn) = S.This is equivalent to having the A part of the Siegel set compact. Thatis to say that there should be α, β with

0 < α ≤ aii ≤ β (8.3)

where g varies over the Siegel set and 1 ≤ i ≤ n. We will prove that(8.3) is equivalent to the following two statements:

(a) |det | is bounded on S.

(b) There exists c > 0 such that ||g(x)|| > c for each x ∈ Zn \ 0 andg ∈ S.

These two conditions are exact reformulation of the two conditions inthe theorem.

Suppose that (a) and (b) hold and (aii) is the A part of the Iwasawadecomposition of g = kan. By (b) ||(g(e1)|| = a11 ≥ c > 0 for everyg ∈ S. Since S is a subset of Siegel set we know that a ∈ A 2√

3so we

have c ≤ a11 ≤ ta22 where t = 2√3. So a22 ≥ c

t and a33 ≥ ct2

etc.

By taking the minimum of this finite number of positive quantity wehave aii ≥ α > 0 for all 1 ≤ i ≤ n and g ∈ S. By (a), |det g| =∏ni=1 aii ≤ M for some constant M . Let j be a fixed index and since

αn−1ajj ≤ a11 . . . ajj . . . ann ≤M thus ajj ≤ Mαn−1 = β

Turning to the converse, suppose that (8.3) holds then |det g| ≤ βn

for all g ∈ S proving (a). Let x ∈ Zn\0 and write x =∑miei where for

some i, mi 6= 0. Let k be the first of such i’s, then ||g(x)|| = ||an(x)||and the kth coordinate of an(x) is akkmk. Hence ||g(x)|| ≥ akk|mk| ≥akk ≥ α > 0 for all g ∈ S proving the second condition.

Since L0 is closed in L (prove!). A direct corollary of Mahler’stheorem is then: A subset S of L0 has compact closure if and only ifthere exists some neighborhood of 0 in Rn so that Γ ∩ U = 0 for allΓ ∈ S.

Corollary 8.2.6. For n ≥ 2, SL(n,R)/SL(n,Z) is non-compact.


Proof. Now L0, the space of lattices with parallelepiped of volume 1 andis identified with SL(n,R)/SL(n,Z). If the latter is compact so is L0.Since L0 has compact closure in L, Mahler’s criterion must be satisfied.However the second condition cannot be satisfied. Consider the matrix

gk =

1/k 0 00 k 00 0 In−2

.

which is an element of SL(n,R) and ||gk(e1)|| = 1/k. Since this tendsto zero as k → ∞ this violates the second condition.

We now make a brief digression to give the reader a longer view ofthe terrain. We first define the concept of the unipotent radical of aconnected algebraic group.

Definition 8.2.7. Let G be a connected algebraic a group. Its unipo-tent radical, Gu, is the largest normal, connected, unipotent subgroupof G.

The basic examples just above illustrate an important result ofBorel-Harish Chandra 8.2.8 whose proof is beyond the scope of thisbook [6].

Theorem 8.2.8. If G is a connected algebraic group defined over Q,then GR/GZ has a finite invariant measure if and only if G has nonon-trivial Q-characters.

Here GR and GZ are respectively the real and integer points of G.Moreover the results of both Mostow-Tamagawa [64] and Borel-HarishChandra each tell us that under the same conditions GR/GZ is compactif and only if G has no nontrivial Q characters and every unipotentelement is in the unipotent radical of G.

To illustrate these results in a simple situation, let G = GL(1,C) =C×. For every n ∈ Z, z 7→ zn is a Q-character. GR = R× and GZ = Z2.So GR/GZ = R×

+. Since this is non-compact and everything is abelianGR/GZ does not have finite volume. On the other hand if G is theabelian subgroup (= (C,+)) of unitriangular matrices in GL(2,C), then


GR/GZ is compact and has finite volume. Concomitantly, as a unipotentQ-group G has no nontrivial characters.

Since det is such a Q character for GL(n,C), the Borel-Harish Chan-dra theorem gives an alternative proof that GL(n,R)/GL(n,Z) cannothave a finite invariant measure. On the other hand, for a semisim-ple group G (such as SL(n,C)) there are no Q characters. Hencehere GR/GZ always has a finite invariant measure. In particular, fora semisimple group G (which also has no unipotent radical) the condi-tion for compactness means G has no non-trivial unipotent elements atall.

It should be noted that a close look at the exposition of this com-pactness criterion (see [71]) shows that SL(n,R)/SL(n,Z) is the crucialcase of the compactness result. Later we shall deal with it directly.

In the case of semisimple Lie groups without compact factors andtheir lattices this has been further generalized by Kazdan and Margulis(see [71]) proving Selberg’s conjecture (2. below).

Let G be connected linear semisimple Lie group without compactfactors and µ be a fixed Haar measure on G. Then

(1) There is a constant c(G) > 0 such that for all lattices, Γ, in G themeasure induced on G/Γ ≥ c(G).

(2) If Γ is a non uniform lattice in G, then Γ has a non-trivial unipo-tent element.

(3) If Γ is a uniform lattice in G, then every element in it is Adsemisimple.

Another important general result is Mostow’s rigidity theorem (orthe Mostow-Margulis rigidity theorem).

Mostow’s theorem is the following:

Theorem 8.2.9. Let G and G′ be connected semisimple linear groupswithout compact factors, or factors locally isomorphic with SL(2,R) andlet Γ and Γ′ be uniform lattices in G and G′, respectively. Then anyisomorphism Γ → Γ′ extends to a smooth isomorphism of G→ G′.

This was proven by Mostow in stages starting with the group ofhyperbolic motions G = G′ = SO(n, 1)0, n ≥ 3 and then extending it


to the general case. When G = G′ the algebraic formulation can bereplaced by a more geometric one. Consider the associated symmetricspace of non-compact type G/K = P . Then P/Γ = X and P/Γ′ = X ′

are compact manifolds of the same dimension. (Since any two maximalcompact subgroups of G are conjugate the dimension of G/K is an in-variant of G called its characteristic index ). Since P is simply connectedΓ and Γ′ are the respective fundamental groups. Hence our hypothesis isthat these two compact manifolds have isomorphic fundamental groups.On the other hand, G is essentially the connected isometry group of Pso X and X ′ are isometric.

Why is SL(2,R) excluded? Taking G = G′ = PSL(2,R). HereG/K is the Poincare upper half plane, the universal covering surfaceof all compact oriented Riemann surfaces of genus, g ≥ 2. Let X andX ′ be two such Riemann surfaces of the same genus g ≥ 2. ThenX and X ′ are homeomorphic and hence have isomorphic fundamentalgroups. But they need not be analytically equivalent because there are6g − 6 analytically inequivalent such surfaces [20]. Since for PSL(2,R)analytic equivalence is the same as being isometric [20] this gives acounterexample.

For non-compact simple groups of real rank ≥ 2, Margulis has ex-tended the rigidity theorem to non-uniform lattices. Closely connectedwith the Mostow-Margulis rigidity theorem is the following result ofPrasad: Let G be a non-compact simple linear Lie group, not locallyisomorphic with SL(2,R) and Γ be a lattice in G. If G′ is another suchsimple group and Γ′ is a discrete subgroup isomorphic with Γ, then Γ′

is a lattice in G′ if and only if the characteristic index of G equals thatof G′. So for example, Γ cannot be isomorphic to any of its subgroupsof infinite index since such a subgroup cannot be a lattice. Also thisrecaptures the result of Furstenberg [21] that no lattice in SL(n,R) canbe isomorphic with a lattice in SL(m,R), if n 6= m and both are greaterthan 2. Since these matters are also beyond the scope of this book wewill not pursue them further.

Returning to Siegel domains in GL(n,R) another interesting conse-quence is Hermite’s inequality. This tells us given a lattice Γ = gZn

there is a universal constant cn > 0 such that cn|det g| 1n dominates


the smallest non zero length of all the lattice points in Γ = gZn. Inparticular, if Γ ∈ L′, then cn itself dominates these lengths.

Corollary 8.2.10. Let g ∈ GL(n,R). Then

minγ∈Zn\0

||g(γ)|| ≤ 2√3

n−12 |det g| 1

n .

Proof. Let Γ = g(Zn) and choose g′ ∈ gΓ∩S 2√3, 12

so that Φ(g′) ≤ Φ(gγ),

for all γ ∈ GL(n,Z). Hence

minγ∈Zn\0

||g(γ)|| ≤ minγ∈GL(n,Z)

||gγ(e1)|| = minγ∈GL(n,Z)

Φ(g(γ)) = Φ(g′) = a′11,

where a′11 is the first component of a′, the a part of g′. But g′ ∈S 2√

3, 12. Hence a′11 ≤ 2√

3a′22, . . . a

′(n−1)(n−1) ≤ 2√

3a′nn. Therefore

a′11n ≤ 2√

3

1+2+...+n−1

a′11 . . . a′nn.

Thus

a′11n ≤ 2

√3

n(n−1)2

|det a′|.

But |det g′| = |det a′| since |det k′| = 1 = |detn′| and |det g′| =|det g| since g′ = gγ and |det γ| = 1. This means

( minγ∈Zn\0

||g(γ)||)n ≤ a′11n ≤ 2√

3

n(n−1)2 |det g|.

Taking nth roots proves the result.

We now apply some of the results above to the so called “reductiontheory” of quadratic forms. Let p be a positive definite symmetricmatrix and q(x) = (px, x) be the associated positive definite quadraticform on Rn. As in Chapter 6 G = GL(n,R) operates transitively in anatural way on the space P of such forms via p 7→ gpgt, g ∈ GL(n,R)with isotropy group StabG(I) = O(n,R). The projection, π : G →


P given by g 7→ ggt commutes with the action of G on itself by lefttranslation. We also have the usual Siegel set S(t,u) and also S′

(t,u) =

nant : a ∈ At, n ∈ Nu. Since (nak)(nak)t = nakk−1ant = na2nt wesee that if g = nak ∈ S(t,u), then ggt ∈ S′

(t2,u). Hence π(S(t,u)) = S′(t2,u)

and π−1(S′(t2,u)) = S(t,u). Hence we have,

Corollary 8.2.11. (1) P = S′(t,u)(GL(n,Z)) whenever t ≥ 4

3 and u ≥12 .

(2) minx∈Zn\0 q(x) ≤ (43 )

n−12 (det p)

1n .

Similarly SL(n,R) acts transitively on P ∗, the positive definite sym-metric matrices of determinant 1 with isotropy group SO(n,R). Denot-ing the corresponding Siegel domains here by S∗

(t,u) and S∗′(t,u), using the

same actions restricted to SL(n,R), we get

Corollary 8.2.12. P ∗ = S∗′(t,u)(SL(n,Z)) whenever t ≥ 4

3 and u ≥ 12

and (by the result just below) since SO(n,R) is compact P ∗/SL(n,Z)has finite volume.

In the crucial case of SL(n,R)/SL(n,Z), finiteness of volume can beproved ”by hand” using the method of Siegel as follows.

Theorem 8.2.13. SL(n,R)/SL(n,Z) has finite volume.

Proof. We intersect all the elements in Siegel’s theorem, Theorem 8.2.1,with SL(n,R). By Iwasawa decomposition SL(n,R) = SO(n,R)×A∗×Nwhere A∗ = a ∈ A|det a = 1. Hence here

S∗t,u = SO(n,R) ×A∗

t ×Nu.

Let µ be left Haar measure on SL(n,R). We will prove thatµ(S∗

t,u) is finite. From this it will follow from Proposition 2.4.3 thatSL(n,R)/SL(n,Z) has finite volume.

dµ = dkdb∗ where dk is left Haar measure on K = SO(n,R) and db∗

is left Haar measure on B∗ = A∗N . Here db∗ = ρ(a∗)da∗dn where ρ(a∗)is the distortion of the Euclidean volume of N by the automorphismia∗ |N where ia∗ is the inner automorphism determined by a∗ (see Section


2.3). ρ(a∗) =∏i<j

a∗iia∗jj

. So dµ =∏i<j

a∗iia∗jjdkda∗dn. Now we calculate

µ(S∗t,u).

µ(S∗t,u) =

∫ ∫ ∫KA∗

tNu

∏

i<j

a∗iia∗jj

dkda∗dnFubini

=

∫

Kdk

∫

A∗t

∏

i<j

a∗iia∗jj

∫

Nu

dn.

Since K and Nu are compact∫K dk and

∫Nudn are finite. It remains

to show that∫A∗

t

∏i<j

a∗iia∗jj

is finite. We use a change of coordinate to

compute this integral. Our new coordinates are bi =a∗ii

a∗i+1,i+1, i = 1, . . . n.

On A∗t , bi ≤ t for all i. One sees directly that ρ(a∗) =

∏n−1i=1 b

rii where

are ri are certain positive integers combinatorially dependent on n. So

∫

A∗t

ρ(a∗)da∗Fubini

=

n−1∏

i=1

∫

bi≤tbrii dbi.

If a∗ = y|y real diagonal matrix of trace zero and the exponen-tial map defines a global diffeomorphism from a∗ to A∗ and it takes theglobal measure to the Haar measure. Choose yi ∈ a∗, i = 1, ...n − 1 sothat exp yi = bi for all i. Hence exp ryi = brii . Therefore

∫

bi≤tbrii dbi =

∫ log t

−∞(exp riyi)dyi.

For λ > 0 , ∫ log t

−∞(exp λy)dy =

eλy

λ|log t−∞ =

tλ

λ.

Thus∫A∗

tρ(a∗)da∗ =

∏n−1i=1

tri

ri<∞.

Remark 8.2.14. Here we again make contact with Riemann surfacetheory by showing that although we have constructed non-uniform lat-tices in, for example SL(2,R), we can construct an infinite family ofexamples of uniform lattices in SL(2,R) as well. Let S be a compactRiemann surface of genus g ≥ 2. By uniformization theorem S = H+

2 /Γwhere H+

2 is the upper half plane and the universal cover of S and Γ is

8.3 Lattices in more general groups 371

the fundamental group of S which is a discrete group of SL(2,R). NowH+

2 is K\SL(2,R), the right cosets K in SL(2,R) where is K = SO(2,R)is a maximal compact subgroup of SL(2,R). Therefore SL(2,R)/Γ iscompact. Another way of constructing uniform lattices in SL(2,R) isby means of quaternions [25].

8.3 Lattices in more general groups

In this section we only sketch the results. It is now natural to turnfrom Rn to simply connected nilpotent Lie groups where the results aredue to Malcev [71]. As with Rn these groups also have faithful linear(unipotent) representations. As mentioned earlier there in no distinc-tion between lattices and uniform lattices. However, there are strictrequirements for a discrete subgroup to be a lattice and in particularthere are simply connected nilpotent Lie groups which have no lattices.In fact, G has a lattice if and only if its Lie algebra has a basis withrespect to which all of the structure constants are rational. Hence byProposition 3.1.69 there exist simply connected 2-step nilpotent groupswhich have no lattices at all. Further, an abstract group is isomorphicto a lattice in some simply connected nilpotent group if and only if it isfinitely generated, nilpotent and torsion free. Thus, in this regard thesituation here is similar to the abelian case.

If the simply connected nilpotent group is the full strictly triangulargroup i.e. the N of Proposition 8.2.2 then since N = N 1

2NZ and N 1

2is

compact we see that NZ is a uniform lattice in N . Actually, N/NZ iscompact for any unipotent (hence simply connected, nilpotent) groupwhose Lie algebra has rational structure constants. This follows fromthe Borel-Harish Chandra and Mostow-Tamagawa theorems since sucha group has no nontrivial Q characters and, of course, every unipotentelement lies in the unipotent radical. A lattice is provided by takingNZ, the matrices of N with integer coordinates.

Two other interesting things happen here. First the integer param-eters for the set of all lattices are not arbitrary, but are governed bycertain divisibility conditions. Secondly, in general the log of a lattice isnot a lattice (or even a subgroup) of the additive group of g. When it is


such a lattice is called a log lattice. C. Moore has shown [43] Any latticein a simply connected nilpotent group is always bracketed between twolattices both of which are log lattices.

Exercise 8.3.1. Find all lattices in the Heisenberg group G up to au-tomorphisms, Aut(G). See which ones have log(Γ) a lattice in g, theLie algebra of G. Notice that here Aut(G) does not act transitively onL(G).

If Γ is a lattice in a simply connected solvable group, G, a theoremof Mostow [61] Γ∩Nil(G) is a lattice in Nil(G), the nilradical. Moreoveras mentioned earlier, for a connected solvable Lie group cofinite volumeand cocompactness of a closed subgroup are the same. This result isalso due to Mostow [57].

Exercise 8.3.2. In connection with Mostow’s theorem mentioned justabove, the reader should construct an example of a lattice Γ in R2 anda closed connected subgroup H of R2 with the property that H ∩ Γ isnot a lattice in H.

Let G be a connected Lie group with Levi decomposition G = SR,where S is a Levi factor and R is the radical. Since S is semisimplewe can further decompose S = S0C, where S0 is semisimple withoutcompact factors (the product of all non-compact simple subgroups ofS) and C is compact semisimple (the product of all compact simplesubgroups of S). Then S0 and C commute pointwise. For all this seeCorollary 3.3.19. Since R is characteristic and therefore normalized byC, CR is a subgroup of G which is connected. It is closed since C iscompact because of Weyl’s theorem, Theorem 2.5.8, and R is closedbecause it is the radical. It is evidently also normal since R is and Ccommutes with S0. Also CR while not solvable is almost as good; it isamenable and G/CR is semisimple without compact factors. Evidentlyno larger connected subgroup can be amenable so CR is itself a kindof radical. In this way we have separated the parts of G which aresemisimple without compact factors from the rest.

Similarly to the result of Mostow mentioned above, a theorem ofH.C. Wang [79], Garland and Goto [24] states that if Γ is a lattice in a

8.3 Lattices in more general groups 373

Lie group G, then Γ∩CR is a uniform lattice in CR. This fact providesa method for proving the folk theorem that lattices in a Lie group arealways finitely generated. (In particular lattices are always countable).For we have already proved, Proposition 2.5.2, that a uniform latticeis finitely generated. Since the image of Γ mod CR is a lattice in asemisimple group without compact factors one is reduced to showingthat a lattice in such a group is finitely generated. This is done bydifferent methods in the rank one and higher rank cases.

One final result along these lines (see [29]) is the following. Let Gbe a connected Lie group containing a lattice Γ and B(G) the boundedpart of the group G, namely the elements g ∈ G whose conjugacy classhas compact closure. If B(G) = Z(G) or even more generally if B(G) =Z(G) (the universal coverings), then Γ ∩ Z(G) is a (uniform) lattice inZ(G).

For a connected Lie group G and a smooth automorphism α, bytaking is derivative d1α at 1 we get a linear automorphism of its Liealgebra g. This map preserves composition and is injective. If G is sim-ply connected the map α 7→ d1α is an isomorphism Aut(G) → Aut(g).In this way we can regard Aut(G) as a the real points of a real linearalgebraic group. If g has a basis whose structure constants are ratio-nal this linear algebraic group is defined over Q. Its Lie algebra is thederivations Der(g) of g. If we consider the subgroup of Haar measurepreserving automorphisms, the Lie algebra of this subgroup consists ofDer0(g), the derivations of trace zero. Hence, if N is a simply connectednilpotent group which has a lattice, then its automorphism group is analgebraic Q-group. It is follows that the same is true of the group ofHaar measure preserving automorphisms M(N), as well as its identitycomponent M(N)0.

The following gives a new construction of both lattices and uniformlattices. Of course, it depends on the theorems of Borel-Harish Chandra[6] and Mostow-Tamagawa [64]. Let Γ be a lattice in N where N isnilpotent part of the Iwasawa decomposition G = KAN of a real rank 1simple group G. Such N ’s always possesses lattices. If G = SO0(n, 1) orSU(n, 1), StabM(N)0(Γ) is a non uniform lattices in M(N)0, the identitycomponent of the group of measure preserving automorphisms of N .


However, if G = Sp(n, 1), or the exceptional group our constructiongives a uniform lattice in M(N)0 (see [47] and [4]).

8.4 Fundamental Domains

In our final section of this chapter we define and then construct a fun-damental domain for a discrete subgroup Γ of a connected unimodularLie group G. Although lattices are the most interesting case, here G/Γneed not have finite volume.

Let the Lie group G act smoothly on a manifold X and Γ be adiscrete subgroup of G. A fundamental domain for Γ with respect tothis action is a closed set D ⊆ X satisfying the conditions listed below.

(1) If γ and γ′ are distinct points of Γ then γD and γ′D are disjoint.

(2) The union⋃γ∈Γ γD = X.

The idea here is to have exactly one representative from each orbit,but we will have to compromise about boundary points. If we werewilling to have a measurable fundamental domain we could just take ameasurable cross section (measurable axiom of choice). Of course, themeasure of a fundamental domain will be unaffected by what we chooseto do on the boundary since this has measure zero.

This representation x ∈ X as γd is essentially unique. That is,except for the Γ orbit of the boundary, ∂(D). If X has a G-invariantvolume it is usually the case that ∂(D) is lower dimensional and thereforevol(∂(D)) = 0.

Now suppose G is a unimodular Lie group and Γ is a discrete sub-group of G. Then G/Γ has an essentially unique G-invariant measure µand Γ is a lattice if and only if µ(D) <∞.

An example of a fundamental domain for the subgroup SL(2,Z) = Γin SL(2,R) = G under the action of G on the upper half planeH+ = G/K. Now since K is compact the question of compactness,or finite volume of G/Γ is the same as compactness, or finite volumeof a fundamental domain in H+. Such a fundamental domain for themodular group has been known for a long time. Since G operates by

8.4 Fundamental Domains 375

isometries in the Poincare metric and the curvature is constant −1. ByGauss-Bonnet, or rather just Gauss, Area = π

3 . Or one could use the in-

variant area dA = dxdyy2

coming from the invariant metric ds2 = dx2+dy2

y2

and estimate the area of a strip bounded away from zero in the verticaldirection.

We let G be a Lie group and Γ be a discrete subgroup. We shallsay Ω, an open subset of G, is a fundamental domain for Γ if Ω is afundamental domain for the standard of the action of Γ on G by lefttraslation i.e.

(1) If for γ1 and γ2 different elements of Γ, γ1Ω and γ2Ω are disjoint.

(2)⋃γ∈Γ γΩ = G.

This definition is equivalent to the statement that for any x ∈ G wecan find γ ∈ Γ and ω ∈ Ω such that x = γω. This representation isessentially unique except for points in Ω \ Ω.

In order to proceed we first construct a left invariant Riemannianmetric. This can be done by choosing an inner product for the tangentspace of the Lie algebra and transferring by left translation to the tan-gent spaces of any other point in G. The metric d on G and the resultingtopology determined by this left G-invariant Riemannian metric is thesame as that of the Lie topology. This is because the equivalence of thetwo topologies is a local question and in a neighborhood of the identityis equivalent to the fact that exponential map is a local diffeomorphism.By the Hopf-Rinow theorem (see[23]) and the fact that the Lie topologyis always complete (because it is locally compact) any two points of Gcan be joined by a minimal geodesic.

We now construct a fundamental domain for Γ in G containing 1.We let Ω the set of all points in g ∈ G such that

d(g, 1) < d(g,Γ \ 1).First of all Ω is open. If Ω is not open at g ∈ Ω then there would be

a sequence gn ∈ G and γn ∈ Γ such that

d(gn, 1) ≥ d(gn, γn) (8.4)

for all n and gn → g. Since gn → g and by (8.4) we have d(gn, γn) isbounded and therefore d(g, γn) is itself bounded. Since Γ is discrete the


only way that can happen is that for that sequence γn is a finite set. Inparticular there is subsequence γni

which constant γ0. By inserting thissubsequence in (8.4) we get

d(gni, 1) ≥ d(gni

, γ0) (8.5)

and then by taking limits we have d(g, 1) ≥ d(g, γ0) which is contradic-tion.

We now verify the conditions (i) and (ii). So for condition (i) supposethat γ0Ω ∩ Ω 6= ∅ for some γ0 6= 1. Then there are ω0 and ω1 in Ω suchthat γ0ω0 = ω1. Since we have

d(ω0, 1) < d(ω0, γ) and d(ω1, 1) < d(ω1, γ) (8.6)

for all γ ∈ Γ \ 1 we know that d(ω1, 1) < d(ω1, γ0) or d(γ0ω0, 1) <d(γ0ω0, γ0). Hence by left invariance we have that d(ω0, γ

−10 ) < d(ω0, 1)

which contradicts (8.6) for γ = γ−10 .

As for condition (ii), first note that Ω is set of all g ∈ G such thatd(g, 1) ≤ d(γ, g) for all γ ∈ Γ. Let g be an arbitrary point in G thenγ 7→ d(g, γ), γ ∈ Γ, has a minimum since Γ is discrete. Therefore thereis γ0 ∈ Γ such that d(γ0, g) ≤ d(γ, g) for all γ. By left invariance we getd(1, g) ≤ d(γ−1

0 γ, γ−10 g) for all the γ. Hence d(1, γ−1

0 g) ≤ d(γ, γ−10 g) for

all γ ∈ Γ. Therefore γ−10 g ∈ Ω. So g ∈ ΓΩ.

We remark that similar arguments work for a discrete group act-ing properly discontinuously by isometries on a complete Riemannianmanifold. For example if G is a semisimple linear Lie group withoutcompact factor, X = G/K where K is a maximal compact subgroupof G and Γ is a torsion free subgroup of G, then X/Γ is a completeRiemannian manifold on which Γ operates properly discontinuously byisometries. The case of Γ acting properly discontinuously on a compactmetric space is treated in Appendix C.

Chapter 9

Density results for cofinite

Volume Subgroups

9.1 Introduction

n this chapter we will study the situation of a connected Lie group Gand a closed subgroup H where G/H has finite volume (see Section 2.3)Often, but not always, H will actually be discrete. We shall study theextent to which features of G are determined by those of H. To doso we will occasionally have to use notions of algebraic groups. So forexample, in some appropriate context, we might say that H is Zariskidense in G as defined in Section 9.3. The earliest result along these,is the well known Borel density theorem [8], which can be stated asfollows:

Theorem 9.1.1. Let G be a connected semisimple Lie group withoutcompact factors, H a cofinite volume subgroup and ρ a smooth finitedimensional representation of G on V . Then every subspace of V in-variant under H must be invariant under G.

Actually, in [8] H was discrete.

Exercise 9.1.2. Show that the hypothesis that G have no compactfactors is necessary.

377

378 Chapter 9 Density results for cofinite Volume Subgroups

We shall provisionally take the conclusion of Theorem 9.1.1 as ourprincipal goal in extending this result. It should be mentioned thatthe conclusion of Theorem 9.1.1 need not hold when G/H is merelycompact, but does not have finite volume. Thus for example if B = ANis a Borel subgroup of G = KAN , thenG/B is compact. But B does nothave cofinite volume (see Theorem 2.3.5 and Exercise 2.1.8). Moreovertaking a nontrivial character of B gives a B-invariant line which cannotbe G-invariant since G = [G,G] and so G has no nontrivial characters.G/B is a typical example of a compact homogeneous space with nofinite G-invariant measure.

An interesting consequence of Borel density is the theorem of Hur-witz on finiteness of automorphism group of a compact (or finite volume)Riemann surface, S. Moreover, it also shows the only cofinite volumesubgroups in a non-compact simple group are lattices. The proof ofthese statements is as follows:

Proof. Let G be a non-compact simple group and H a cofinite volumesubgroup. A direct calculation shows that H normalizes NG(H) andtherefore by continuity H normalizes NG(H)0, its identity component.Taking differentials we see that Ad(H) acts on ng(h) ⊆ g. By the Boreldensity theorem AdG leaves this subspace invariant. Thus ng(h) is anideal in g. Since g is simple this ideal is either trivial or g itself. In thelatter case since g normalizes h hence H and therefore H0 are normalin G. Since G is simple and H0 is connected, H0 must be trivial andtherefore H is discrete. Hence H, as a discrete normal subgroup of G, iscentral by Lemma 0.3.6. Since G/H is finite volume group therefore byCorollary 2.1.3 it is compact. This means G/Z(G) is also compact andtherefore G is of compact type, a contradiction. Thus ng(h) = 0 andNG(H) is discrete and therefore so is H. Moreover, NG(H)/H is finite(Proposition 2.4.8). When G = SL(2,R) and H is the fundamentalgroup of the Riemann surface, S, this quotient is the automorphismgroup of S.

In the next section we will prove a generalization of Theorem 9.1.1due to one of the present authors.

9.2 A Density Theorem for cofinite Volume Subgroups 379

9.2 A Density Theorem for cofinite Volume

Subgroups

We now turn to a series of results ([51] and [45]) which generalize theBorel density theorem. This is based on an extension of the basicmethod of Furstenberg [22] together with a number of additional ob-servations.

In what follows, V will denote a vector space over k of finite dimen-sion, n, where k = R, or actually any subfield of C. GL(V ) denotes, asusual, the general linear group of V and P (V ) its projective space. If ris an integer between 1 and n, then ∧rV is the r-fold exterior productand Gr(V ) the Grassmann space of r-dimensional subspaces of V . Ofcourse, P (V ) is G1(V ). Each Gr(V ) is a compact manifold (see Section0.4) and G(V ), the Grassmann space of V , is a disjoint union of theseopen submanifolds. We let π : V \ 0 → P (V ) denote the canonicalmap v 7→ v. If W is a subspace of V of dimension ≥ 1, then W de-notes the corresponding subvariety of P (V ). A finite union, ∪Wi, willbe called a quasi-linear variety (qlv). Since each W is compact, a qlv isa closed subspace of P (V ).

We begin with some lemmas needed to prove Proposition 9.2.6 be-low.

Lemma 9.2.1. If A ⊆ P (V ), then there exists a unique minimal qlv Scontaining A.

Proof. By considering π−1(A) it is enough to show that any subset B ⊆V is contained in a unique minimal set of the form ∪Wi : i = 1, . . . , r.Now, B is contained in one such set, namely V . If we show there existsa smallest such set this will also imply uniqueness. Since each Wi isa linear subspace and hence is algebraic, a finite union of such setsis also algebraic. But an infinitely descending chain of algebraic setsin V would correspond to an infinitely ascending sequence of ideals ink[x1, . . . , xn] which is impossible by the Hilbert basis theorem [82].

Now, for g ∈ GL(V ) define g : P (V ) → P (V ) by g(v) = g(v). Rou-tine calculations prove that g is well-defined, gπ = πg, (gh) = gh and if


λ 6= 0, then (λg) = g. Now, suppose one has a linear representation, ormodule G× V → V of G on V . Then this induces a compatible actionof G on P (V ), making the diagram below commutative.

G× (V \ 0) −−−−→ V \ 0y(id,π)

yπ

G× P (V ) −−−−→ P (V )

(9.1)

Lemma 9.2.2. Let A be a G-invariant subset of P (V ) and ∪Wi : i =1, . . . , r be the minimal qlv containing A. Then G permutes Wi : i =1, . . . , r.

Proof. Since gπ = πg for g ∈ GL(V ), we know π−1(A) is also G-invariant, π−1(A) ⊆ ∪Wi : i = 1, . . . , r and this is the minimal linearvariety containing it. But then

g(π−1(A)) = π−1(A) ⊆ g.(∪Wi : i = 1, . . . , r) = ∪gWi : i = 1, . . . , r.

The latter is a linear variety for each g ∈ G. By minimality

∪Wi : i = 1, . . . , r ⊆ ∪gWi : i = 1, . . . , r.But this means the two sets are equal for each g ∈ G. The spaces

involved in the unique linear variety containing a set are clearly alsounique and gWi is one of them. Therefore, gWi = Wj for some j.

Lemma 9.2.3. Let gk be a sequence in GL(V ) and suppose det gk

‖gk‖n→0 ,

where ‖ · ‖ is any convenient Banach algebra norm on End(V ). Thenthere exists a map φ : P (V ) → P (V ) such that φ(P (V )) is a proper qlvof P (V ) and a subsequence of gk which converges to φ pointwise onP (V ).

Proof. Let W be a nonzero subspace of V and consider gk|W : W →V . Denote 1

‖gk|W ‖ by γk,W . Then ‖γk,W gk|W ‖ = 1 for all k. Since

A : AW : W → V, ‖A‖ = 1 is a compact set, there is a subsequence,which we again call γk,W gk|W , such that γk,W gk|W converges in norm


and, therefore, pointwise on W to σW . Here σW is a linear map W →V . Since ‖σW‖ = 1, σW 6= 0. Now, since π is continuous and forw ∈ W , γk,W gk|W (w) → σW (w), we have γk,W gk|W (w) → σW (w).

But γk,W gk|W (w) = gk(w) so gk(w) → σW (w) pointwise on W , and inparticular for w outside of KerσW .

In particular, if W = V , we have γkgk → σV and so

det(γkgk) = (γk)n det gk =

det gk‖gk‖n

→ det σV .

Since this sequence tends to 0, σV is singular. Now, inductively definesubspaces W0,W1, . . . , of V by W0 = V , Wi+1 = KerσWi

, i ≥ 0. ThenKerσV < V since σV 6= 0. Similarly, Wi+1 < Wi since σWi

6= 0. Thus,the sequence V = W0 > W1 > . . . must terminate at 0 after a certainnumber of steps; Wi0 = 0 for some i0. For each i and finer and finersubsequences, which are again called gk, we have gk(wi) → σWi

(wi)pointwise for wi ∈ Wi. Define φ : P (V ) → P (V ) by φ(v) = σWi

(v), ifv ∈Wi, but not in Wi+1, i = 0, . . . , i0 − 1. If v = u, then v = γu, γ 6= 0.If v ∈Wi −Wi+1, the same is true of u, so

σWi(v) = σWi

(γu) = γσWi(u) = σWi

(u),

since σWiis linear. Thus φ(v) = φ(u), φ is well-defined and gk converges

to φ pointwise on P (V ). Moreover,

φ(P (V )) = ∪σWi(Wi) : i = 0, . . . i0 − 1,

so the range of φ is a qlv. Since σV is singular, σV (V ) < V . For i > 0,σWi

: Wi → V so dimσWi(Wi) ≤ dimWi < dimV and σWi

(Wi) < Vfor i ≥ 0. Now the union of a finite (or even countable) number ofsubspaces each of strictly lower dimension cannot equal V . To see thisit is clearly sufficient to take k = R. Now this follows from the Bairecategory theorem, but it is more in the spirit of our subject to argueas follows: If V = ∪Vi : i ∈ Z, then take a finite positive measure µon V which is absolutely continuous with respect to Lebesgue measure,e.g., dµ = exp(−‖x‖2)dx. Then, by countable subadditivity, since eachµ(Vi) = 0 we see that µ(V ) = 0, a contradiction. This means thatP (V ) 6= ∪σWi

(Wi) : i = 0, . . . i0 − 1 and the range of φ is proper.


Lemma 9.2.4. Let G×X → X be an action of a topological group, G,on a metric space X. Suppose there exists a sequence gk and a closedsubspace Y of X such that for each x ∈ X, gk(x) converges to y(x) ∈ Ypointwise on X. Then each finite G-invariant measure µ on X hasSuppµ ⊆ Y .

Proof. Let D(x) = dist(x, Y ), where dist is an equivalent bounded met-ric on X. Then D is a bounded continuous nonnegative function on Xand D(x) = 0 if and only if x ∈ Y . Now, for all k,

∫

XD(gkx)dµ(x) =

∫

XD(x)dµ(x).

Since gkx → y(x), by continuity of D we have D(gkx) → D(y(x))pointwise on X. Because D is bounded, there is a c such that for allk ∈ Z and x ∈ X, |D(gkx)| ≤ c. The finiteness of µ together with thedominated convergence theorem [73] shows

∫X D(gkx)dµ(x) tends to 0.

Therefore,∫X D(x)dµ(x) = 0, so D ≡ 0 on Suppµ. Since D = 0 exactly

on Y , Suppµ ⊆ Y .

The following definition will play an important role in what follows.

Definition 9.2.5. Let G be a topological group and ρ : G → GL(V )be a continuous representation. We shall say ρ is admissible if thereis a family Hi of subgroups of G which together generate G andeach restriction has the following properties, where here again every-thing is done with respect to some convenient Banach algebra norm onEndk(W ).

(1) For each i, Hi has no closed subgroup of finite index.

(2) For each i and a Hi-invariant subspace W of V , either Hi acts onW by scalars, or else there is a sequence gk ∈ ρ(Hi) such that

det(gk|W )

‖(gk|W )‖dimW→ 0.

Furthermore we shall say ρ is strongly admissible if each rth exteriorpower ∧rρ acting on ∧rV is admissible for r = 1, . . . , n = dimV .


Proposition 9.2.6. Let ρ be an admissible representation of G on Vand G × P (V ) → P (V ) be the associated action on projective space.Then each finite G-invariant measure µ on P (V ) has Suppµ ⊆ P (V )G,the G-fixed points.

Proof. First assume that G satisfies the two conditions above. That is,G is one of the Hi. If G acts on V by scalars, then P (V ) = P (V )G

and we are done. Otherwise, by the second condition there exists asequence gk in G with the property that det(gk)

‖gk‖n → 0. By Lemma 9.2.3,

there exists φ : P (V ) → P (V ) such that φ(P (V )) = Q is a proper qlv ofP (V ) and we can assume gk converges to φ pointwise on P (V ). Since Qis closed, Suppµ ⊆ Q, by Lemma 9.2.4. By Lemma 9.2.1, there exists asmallest qlv, which we shall call S = ∪Wi : i = 1, . . . ,m, containingSuppµ. Thus

Suppµ ⊆ S ⊆ Q < P (V ).

Since µ is G-invariant, so is Suppµ. By Lemma 9.2.2, G permutes Wi.But there are only a finite number of Wi so each has a stability groupof finite index. Moreover, since G × V → V is continuous and the Wi

are closed, the stability groups are also closed. By the first condition Gmust leave each Wi stable. Let W be any one of the Wi and considerthe action of G on W . The two conditions above are clearly satisfiedfor this action. If we let µ′ = µW , then we get a G-invariant measureon P (W ) and argue as before. Unless G acts on W by scalars, we knowthere exists a proper qlv T of W = P (W ) such that Suppµ′ ⊆ T . Thiscontradicts the minimality of S. Otherwise G acts on Wi by scalars foreach i. But then each Wi is G-fixed. This means

Suppµ ⊆ S ⊆ ∪Wi : i = 1, . . . ,m ⊆ P (V )G.

We have just shown that for each Hi, Suppµ ⊆ P (V )Hi . HenceSuppµ ⊆ ∩P (V )Hi = P (V )G.

We now pass from projective space to the Grassmann space, Gr(V ).There is a canonical map φ : Gr(V ) → P (∧rV ), defined as follows: Foran r-dimensional subspace W of V , choose a basis w1, . . . , wr. Thenw1 ∧ . . . ∧ wr is a nonzero element of ∧rV and so the line through it


gives a point in P (∧rV ). This is a well-defined map because if u1, ..., uris another basis for W then u1 ∧ . . . ur is a multiple of w1 ∧ . . . ∧ wr.Moreover this map is injective. To see this suppose that for w1∧. . . wr =λu1 ∧ . . . ur where wi is a basis for W and ui is a basis for U .Then consider the subspace TW = v ∈ V |v ∧ w1 ∧ · · ·wr = 0. IndeedTW = W and similarly TU = U and by assumption TU = TW so U = W .Since this map is clearly smooth with respect to quotient structure (seeSection 0.4) we get:

Proposition 9.2.7. The map φ : Gr(V ) → P (∧rV ) is well-defined,smooth and injective.

We now come to our first theorem.

Theorem 9.2.8. Let ρ be a strongly admissible representation of G onV . Then under the induced action of G on the Grassmann space, G(V ),each finite G-invariant measure µ on G(V ), has Suppµ ⊆ G(V )G, theG fixed points.

Proof. Since G(V ) = ∪Gr(V ) a disjoint union of open G-invariant sets,it clearly suffices, by restricting the measure and the action to Gr(V ),to prove the theorem for that case. Now, since, as explained above,GL(V ) acts transitively and continuously on Gr(V ), the latter is a quo-tient space GL(V )/StabGL(V )(W ) where W is some fixed r-dimensionalsubspace of V . If γ : GL(V ) → Gr(V ) denotes the corresponding pro-jection and w1, . . . , wr is a basis of W , then since gw1, . . . , gwrare linearly independent for each g ∈ G, g 7→ gw1∧, . . . ,∧gwr is amap ψ : GL(V ) → ∧rV \ 0. Clearly, φ factors as φ1π whereφ1 : Gr(V ) → ∧rV \ 0 and π : ∧rV \ 0 → P (∧rV ) is the nat-ural map. The diagram below, including the map φ is commutative,since φγ(g) = φ(gW ) = gw1∧, . . . ,∧gwr = ψ(g).

GL(V ) −−−−→ψ

∧rV \ 0yγ

yπ

Gr(V ) −−−−→φ

P (∧rV )


To see that φ is continuous, note that γ is continuous, open andsurjective and that ψ and π are both continuous. Then it follows fromthe commutativity of the diagram above that φ is continuous.

For each g ∈ G we get a commutative diagram as follows:

Gr(V )φ−−−−→ P (∧rV )

g

y (∧rg)

y

Gr(V ) −−−−→φ

P (∧rV )

where g : Gr(V ) → Gr(V ) is the induced map Gr(V ) by g ∈ GL(V ).For, let W = l.s.wi be any point of Gr(V ) and g ∈ G. Then

φ(g(W )) = (gw1∧, . . . ,∧wr). While

(g ∧ . . . ∧ g)(φ(W )) = (g ∧ . . . ∧ g)(w1∧, . . . ,∧wr)= [(g ∧ . . . ∧ g)(w1∧, . . . ,∧wr)]= (gw1∧, . . . ,∧gwr).

Because φ is a G-equivariant measurable function, the measure µcan be pushed forward, by Proposition 2.3.6, and be regarded as a fi-nite G-invariant measure on P (∧rV ), supported on the image of Gr(V ).Since ρ is strongly admissible, ∧rρ is admissible. By Proposition 9.2.6Suppµ ⊆ P (∧rV )G. Therefore, by G-equivariance and Proposition9.2.7, Suppµ ⊆ Gr(V )G.

We now turn cofinite volume subgroups.

Theorem 9.2.9. Let G be a locally compact group and ρ a strongly ad-missible representation of G on V . If H is a closed subgroup with G/Hof finite volume, then each H-invariant subspace of V is G-invariant.

Proof. If the dimension of the subspaceW is r, form Gr(V ) and considerthe action G × Gr(V ) → Gr(V ). Here W corresponds to a point p ∈Gr(V ). Because H leaves W stable, the point p is H-fixed. So H ⊆StabG(p). Now since G/H has a finite G-invariant measure, by pushing


the measure forward the same is true of G/StabG(p). This means thatp ∈ Suppµ for an appropriate G-invariant measure µ on Gr(V ). ByTheorem 9.2.8, p is fixed under G. This means W is G-stable.

As an immediate corollary we get:

Corollary 9.2.10. Under the assumptions of Theorem 9.2.9, if ρ isirreducible, then so is ρH .

We now seek conditions for a representation to be admissible, orstrongly admissible so we can apply our results. To state one of thesewe define minimally almost periodic groups. This definition is due to J.von Neumann (spelling).

Definition 9.2.11. A minimally almost periodic group G is a locallycompact group which has no nontrivial finite dimensional continuousunitary representations.

Exercise 9.2.12. In particular, minimally almost periodic groups in-clude semisimple Lie groups having no compact factors. (In this wayTheorem 9.2.14 contains a generalization of the Borel density theorem).

Remark 9.2.13. Also a minimally almost periodic group has no closedsubgroups of finite index. This is because if had it would also have anormal subgroup of finite index and this finite quotient would have afaithful unitary representation.

Theorem 9.2.14. Let G be a locally compact group and ρ : G→ GL(V )be a continuous finite dimensional linear representation. Suppose thateither

(1) G is minimally almost periodic and ρ is arbitrary or

(2) G is a complex connected Lie group and ρ is holomorphic or

(3) G is a connected Lie group with G/R having no compact factorsand the radical R acts under ρ with only real eigenvalues.,

then ρ is strongly admissible. In particular, if H is a closed subgroupwith G/H of finite volume, then each H-invariant subspace of V is G-invariant.


Proof. Notice that in all three cases it is sufficient to show ρ is admissi-ble. This is because ∧rρ would be continuous (respectively holomorphic)if ρ were. So in cases 1 and 2, ρ would be strongly admissible. In case3, if R acts with only real eigenvalues under ρ, then the same is truefor ∧rρ. This is because the tensor product of operators has as its spec-trum the set of products of elements from the spectra of the individualoperators and, hence, is real, and since the wedge product of operatorsis induced by their tensor product, the spectrum here is a subset of thatof the tensor product and so is also real. Thus, in case 3, as well, ρwould be strongly admissible.Proof of case (i) (Furstenberg’s case): Here we need take only one Hi,namely, G itself. By Remark 9.2.13, G has no closed subgroup of finiteindex. Regarding the second condition, since g 7→ det g|W is a homomor-phism into an abelian group and such groups many finite dimensionalunitary characters, it is clear that this homomorphism is trivial. Weare therefore looking at 1

‖g|W ‖dim W , which, of course, tends to zero if we

merely select g’s so ‖g|W ‖ tends to ∞. This can be done, since G|Wis not a bounded group, since G is minimally almost periodic. As weshall see, case 1 is simpler than the others because there is just one Hi

and the alternative that G acts on W by scalars in condition 2 does notarise.Case(ii): Let Hi’s be all the 1-parameter subgroups exp zX : z ∈ Cwhere X ∈ g. Since the Hi are connected, condition 1 is automatic. LetW be an exp zX-invariant subspace of V . Then W is X-invariant andwe may as well assume W = V . We show det g

‖g‖n → 0 for some sequence

of the g’s in ρ(exp zX) = Exp zρ′(X), which we write henceforthas Exp z(X). But

‖ gn

det(g)‖ ≤ ‖g‖n

|det(g)|so it suffices to show ‖ gn

det(g)‖ → ∞. Now g = Exp zX so

gn

det(g)=

ExpnzX

det(Exp zX).

This is a vector valued holomorphic function of z ∈ C. By the maximum


principle, it tends to ∞ as |z| does, or it is constant. In the lattercase, gn

det(g) = A ∈ EndC(V ). Taking g = 1, we see that A = I and

gn = det(g)I. Since each (fixed) n-th power of every element on the1-parameter group acts as a scaler and every such element has an n-throot, the entire 1-parameter group acts as scalars.Case (iii): We let G = RS be a Levi decomposition and take for the Hi

the various 1-parameter groups of R together with S itself. If Hi = S,then we are done, by case (i). Thus we may assume we have some 1-parameter group, which we write Exp tX as above, acting with only realeigenvalues. As before, by connectedness, condition 1 is satisfied. Now

X =

λ1 0 · · · 0∗ λ2 · · · 0∗ · · · ∗ λn

also has only real eigenvalues. This is because if λ = a + bi and eλt =eateibt is real for all t, then bt is an integer multiple of π for all t. Thismeans b = 0, for otherwise t would lie in a discrete set. As above, wemay assume W = V and that |t| → ∞. We show

det(Exp tX)

‖Exp t(X)‖n → 0.

Now since

Exp t(X) =

exp tλ1 0 · · · 0∗ exp tλ2 · · · 0∗ · · · ∗ exp tλn

we see that

‖Exp t(X)‖ ≥ ||diag(exp tλ1, · · · , exp tλn)|| = max(| exp tλi|),

while det Exp t(X) = et trX .Let λi to be the largest eigenvalue. Then nλi ≥ trX and exp t(nλi−

trX) → ∞ as t → ∞ unless all λi’s are equal. Similarly, if −t → ∞,choose λi to be the smallest eigenvalue. If all λi are equal, then Exp tXacts as scalars.

9.3 Consequences and Extensions of the Density Theorem 389

Remark 9.2.15. Case 3 applies, in particular, to a solvable group act-ing with real eigenvalues and in particular to a unipotent action.

This concludes the proof of our generalization of the Borel densitytheorem.

9.3 Consequences and Extensions of the Den-

sity Theorem

Various other generalizations and extensions of Theorem 9.1.1 havebeen proved by Mostow, S.P. Wang, Rothman, Mosak and Moskowitz,Moskowitz, and Dani some of which we will describe, but withoutproofs. As Theorem 9.2.14 these all also remove the hypothesis that G issemisimple without compact factors and consider more general groups.We will also derive a number of consequences of Theorem 9.2.14 (mostlywith proofs).

We now define the algebraic hull, G#, of a linear group G ⊆ GL(V ).This is the smallest algebraic subgroup of GL(V ) containing G. Equiva-lently its the closure of G in GL(V ) with respect to the Zariski topology.All this with respect to the field k of definition.

The density Theorem 9.3.3 [45] below extended and unified the re-sults we have gotten so far. To state this version of the result we needthe following two definitions. If G ⊆ GL(V ) is a linear Lie group, arepresentation ρ of G on W is called k-rational, if G# is an algebraic k-group and ρ is the restriction of a k-rational morphism G# → GL(WC).In [45] such a linear group G is called k-minimally almost periodic if,for each k-rational representation, ρ of G, if ρ(G) is bounded, then itmust be trivial. In effect, this is what is being verified in the three casesof Theorem 9.3.3.

Corollary 9.3.1. Let G be a connected subgroup of GL(V ) which iseither

(1) minimally almost periodic, or

(2) complex connected Lie group, or


(3) a group with G/R having no compact factors and R acts on Vwith real eigenvalues.

Then

(1) If ρ is a k-rational representation of its algebraic hull G#, thenρ|G is strongly admissible.

(2) If G/H has finite volume, then l.s.kG = l.s.kH and ZEndV (H) =ZEndV (G).

(3) Any connected subgroup of G normalized by H is normal in G.

Proof. The first statement is clear from Theorem 9.2.14. As for thesecond, let W = l.s.kH. We first show that G ⊆ W . Then l.s.kG ⊆W . Consider the k-rational representation ρ of G# on EndV givenby (g, T ) 7→ gT . Since ρh(

∑i cihi) =

∑cihhi, we see that W is H-

invariant. By the first statement together with Theorem 9.2.14, Wis G-invariant. Since I ∈ W , we see that G ⊆ W and this provesthe second statement. If T ∈ EndV and Th = hT for all h, thenT commutes with any linear combination of h’s and hence with anyg ∈ G.

Finally, if L is a connected subgroup of G normalized by H andl is its Lie algebra, let ρ be the adjoint representation of G# on itsLie algebra. Then ρG = AdG is strongly admissible and, since L isnormalized by H, AdG(H) leaves l stable. By Theorem 9.2.14, l is alsoAdG stable, so L is normal in G.

In fact, we can extend Corollary 9.3.1 to nonlinear groups. To do sorequires the observation that if G acts on V with real eigenvalues, thenAdG must act on g also with real eigenvalues as its eigenvalues are theexponentials of the eigenvalues of g which are real by Exercise 0.5.13.We recall that the radical Rad(G) of a Lie group G largest connectednormal solvable subgroup of G. It is the connected Lie subgroup whoseLie algebra is r the radical of g.

Corollary 9.3.2. Let G be a connected group and H be a closed sub-group with G/H of finite volume. Suppose that either

(1) G is minimally almost periodic or,


(2) G is a complex connected group or,

(3) G/R has no compact factors and AdG(Rad(G)) acts on g with realeigenvalues.

Then any connected subgroup L of G normalized by H is normal in G.In particular, if A is a closed subgroup of G containing H, then A isnormal. Also, if G/NG(L) has finite volume where L is a connectedsubgroup of G, then L is normal.

Proof. L is normalized by H if and only if l, its lie algebra, is AdG(H)-stable, that is, if and only if l is stable under AdG(H), the Euclideanclosure in GL(g). Since AdG/Ad(H) has finite volume by pushing themeasure forward (Proposition 2.3.6), the result follows from Corollary9.3.2.

We now turn to the density theorem in the context of algebraicgroups.

Theorem 9.3.3. Let G be a Lie subgroup of GL(V ) and H be a closedsubgroup with G/H of finite volume. Suppose that either

(1) G is minimally almost periodic or,

(2) G is a complex connected Lie group or,

(3) G is a real connected Lie group with G/R having no compact fac-tors and R acts on V with real eigenvalues.

Then H# = G#.

Proof. Since H# is an algebraic group defined over C, we know by atheorem of Chevalley (see [8]) that there is a C-space WC, a line lCdefined over C in it and a C-morphism ρ : G# → GL(WC) such thatH# = g ∈ G# : ρ(g)lC = lC. Then ρG is a rational morphism of Gon W = WC and the line lC is ρ(H)-stable. By Theorem 9.2.9, it is alsoρ(G)-stable so G ⊆ H# and hence G# = H#.

Example 9.3.4. To appreciate the significance of a subgroup beingmerely Zariski dense we now give an example of a simply connectedabelian, in fact diagonal subgroup G of GL(2,R) and a connected Liesubgroup which is therefore closed subgroup H of G. Here H is Zariski


dense, but not of cofinite volume. Let G = diag(λ, µ) : λ, µ > 0 andH be the 1-parameter subgroup diag(et, eαt) : t ∈ R where α is anirrational number. Then G is the Euclidean identity component of thereal points of an algebraic group defined over Q. Let

p(λ, µ) =∑

i,j

ai,jλiµj

be one of the polynomials defining H#. Since λα = µ on H,∑i,j ai,jλ

i+αj ≡ 0. Now, the exponents i + αj are all distinct becauseα is irrational. If

∑mk βkλ

αk ≡ 0 for all λ > 0 where αk and βk are realand α1 < . . . < αm, then all βk must be 0. For the latter equals

λα1(β1 + β2λα2−α1 + . . . + βmλ

αm−α1).

Since the first factor is positive, the second must be identically 0. Let-ting λ → 0, we see that β1 = 0 and then reason by induction on m.We conclude that ai,j = 0 for all i, j. Since p = 0 and p was arbitrary,H# = G#. On the other hand, H is a Lie subgroup of lower dimensionthen that of G and so is proper, and since G is simply connected andsolvable, G/H is non-compact and has no finite invariant measure.

Definition 9.3.5. We shall say a subgroup H of a Lie group G is ana-lytically dense in G if the only connected Lie subgroup of G containingH is G itself.

Theorem 9.3.6. ([45]) Let G be a connected linear Lie group whoseradical is simply connected and whose Levi factor has no compact part.If Rad(G) acts with real eigenvalues, then any closed subgroup H withG/H of finite volume is analytically dense.

Closely related to the previous theorem is the following:

Theorem 9.3.7. ([45]) Let G be a non-compact exponential Lie group(such as the adjoint group of a classical real rank 1 simple group, oran almost direct product of such things) with Lie algebra g and supposeG/H has finite volume. Then every X ∈ g is a finite linear combinationof elements of g that exponentiate into H.


We shall say that a representation ρ of a group G on a vector spaceV is completely reducible if every G-invariant subspace of V has a com-plementary G-invariant subspace.

Theorem 9.3.8. ([54]) Let G be a locally compact group, H be a closedsubgroup such that G/H is compact, or has finite volume and ρ be acontinuous finite dimensional real or complex representation of G on V .If its restriction to H is completely reducible, then ρ itself is completelyreducible.

Theorem 9.3.8 can even be extended to infinite dimensional repre-sentations on a Hilbert space, but for that one needs G/H to be bothcompact and of finite volume.

We shall now make the following provisional definition: A subgroupA of an algebraic Q-group G is called arithmetic if it is commensurablewith GZ

Theorem 9.3.8 above can be freed of these assumptions entirely ifthe group is an algebraic Q-group, the representation is rational and thesubgroup is arithmetic.

Theorem 9.3.9. ([54]) Let ρ be a rational representation of a linearalgebraic group G defined over Q and A be an arithmetic subgroup ofG. Then ρ is completely reducible if and only if its restriction to A iscompletely reducible.

By combining Theorem 9.3.3 with a result of S. Rothman [72], weget a generalization of the density theorem of Mostow [60], where itwas assumed that G = [G,G], G/Rad(G) has no compact factors andRad(G) is abelian.

Corollary 9.3.10. Let G be a connected Lie subgroup of GL(n,R) andH be a closed subgroup of G with G/H of finite volume. If G/Rad(G)has no compact factors and G = [G,G], then G# = H#.

This is because these conditions on G characterize connected Liegroups which are map (see [72]).

We can formulate a more general form of our density theorem asTheorem 9.3.11 below. The reader who is not comfortable with this canusually just work with Theorem 9.3.3 instead.


Theorem 9.3.11. Let G be a connected Lie subgroup of GL(V ) which isk-minimally almost periodic and H be a closed subgroup of G of cofinitevolume. Then H# = G#.

We conclude with three further applications of the density theorem.This result comes from [45] and can be used to prove that under thesehypotheses, for a lattice Γ in G, the orbit Aut(G) Γ is locally compactin the Chabauty topology. For details see [45].

Proposition 9.3.12. Let G be a solvable connected Lie subgroup ofGL(n,R) having only real eigenvalues, H be a closed uniform subgroupof G and ρ : G → GL(W ) be an R-rational representation. Then thecohomology restriction maps Hp(G,W ) → Hp(H,W ) are isomorphismsfor all p ≥ 0.

Proof. Let K be a maximal compact subgroup of G. Since K is con-nected, it is contained in SO(n,R) (in appropriate coordinates) andeach element of K lies on a 1-parameter group of K. It follows thateach element of K can be put into block diagonal form with rotations(or I) in the blocks. Since the eigenvalues of the elements of K are real,K = 1. This means that G is simply connected. The result will followfrom [59], Theorem 8.1, if we can show that H is ρ-ample in G, that is,(ρ AdG)(H) is Zariski dense in (ρ AdG)(G). Since H is a uniformsubgroup of G, it follows that G/H carries a finite invariant measureand so this follows from Theorem 9.3.3.

Our second result deals with non-amenability of lattices. It relies onthe well known Tits alternative for linear groups.

Definition 9.3.13. For our purposes we shall say an abstract group isamenable if it does not contain a non abelian free group.

Corollary 9.3.14. A lattice Γ in a connected minimally almost periodicLie group G is never amenable.

Proof. Let G be such a group. Then B(G) = Z(G) by [72]. Hence(see [29]) Ad(Γ) is a lattice in AdG. As Ad(Γ) is a linear group theTits alternative [77] tells us that either it contains a free group on 2


generators, or it has a solvable subgroup H of finite index. In the lattercase by transitivity H is also a lattice in AdG. Since AdG is also mapwe see by Theorem 9.3.3 that H# = AdG#. But H# is solvable sinceH is. It follows that AdG is itself solvable and hence so is G. HenceG > [G,G]−. This is impossible since G is a minimally almost periodicgroup. Thus Ad(Γ) contains a free group and so is not amenable andso neither is Γ.

Our final application of the density theorem requires the BorelHarish-Chandra theorem, Theorem 8.2.8. Corollary 9.3.15 applieswhen G is the complexification of a non-compact simple group suchas SL(n,C), or Sp(n,C), but not when it is the complexification of acompact simple group such as SO(n,C). For a proof we refer the readerto [54].

Corollary 9.3.15. ([54]) Let G be a Zariski connected linear algebraicgroup defined over Q. Then GZ is Zariski dense in G if and only ifXQ(G) is trivial and GR is Q-minimally almost periodic.


Appendix A

Vector Fields

Here we recall some basic notions in differential topology, a full accountof the subject can be found in [30]. We begin with the definition ofTp(M), the tangent space of a smooth manifold M at a point p. Foran open U ⊂ Rn the tangent space at x ∈ U is defined to be TxU =x × Rn. Let (Ui, φi)i be an atlas for M where each φi : U →φ(U) ⊂ M is a homeomorphism. Then TpM is defined to be the set ofequivalence classes [p, v, i] where v ∈ Tφ−1

i (p)Ui and [p, v, i] = [p, v′, j] if

vi = dφ−1j (p)(φ

−1i φj)(vj). Then TpM can be made into a linear vector

space by defining

(1) [v, i] + [u, i] = [v + u, i],

(2) k[v, i] = [kv, i] for k ∈ R.

The tangent space TM is the union⋃x∈M TxM and can be made into a

manifold. One can give an atlas for TM by declaring (TUi, dφi) a chartwhere

dφi(x, v) = [φ(x), v, i]

and the change of coordinates are (φ−1i φj, d(φ−1

i φj)). A vectorfield is a smooth map X : M → TM such that X(p) ∈ TpM or, asis customary, we say that X is a smooth section of the vector bundleπ : TM → M where π([p, v, i]) = p. Notice that the space of vectorfields χ(M) on M is a module over C∞(M) where the scalar product is

397

398 Appendix A: Vector Fields

defined by pointwise multiplication i.e

(f ·X)(p) = f(p)X(p),

and using (2) above. One can consider the derivative of a smooth mapf : M → N at a point p, which is a linear map

dpf : TpM → Tf(p)N

defined using charts (Ui, φi) and (Vj , ψj) for a neighborhood of p andf(p) respectively,

dpf([p, v, i]) = [f(p), dφ−1(p)(ψ−1j f φj)(v), j].

The vector fields on M act on C∞(M) as first-order differentialoperators by

(Xf)(p) = dpf(X(p)).

for f ∈ C∞(M) and p ∈M . It is a direct check that

(1) X(fg) = fXg + gXf for all f, g ∈ C∞(M).

(2) X(f + λg) = Xf + λXg for all λ ∈ R.

which says that a vector field on M defines a first-order differentialoperator on C∞(M). In fact one can prove that any first-order differ-ential operator is given by a vector field. Given an atlas (Ui, φi) onM , then a vector field X on φi(Ui) has the local expression X(p) =∑n

i=1 ζi(p)∂/∂xi where ∂/∂xi(p)ni=1 is thought of as a basis for thetangent space TpM induced by the trivialization TUi = Ui×Rn, and ζiare smooth functions defined on φi(Ui). We have

(Xf)(p) =

n∑

i=1

ζi(p)∂f

∂xi(p),

which gives a local expression for the operator defined by X. One cancompose two such operators (vector fields) but of course the result isnot a first-order differential operator. We now define the bracket of twovector fields X and Y as

Appendix A: Vector Fields 399

[X,Y ] = XY − Y X,

the commutator of differential operators X, Y . Since [·, ·] is clearly skewsymmetric and the Jacobi identity, [[X,Y ], Z]+[[Y,Z],X]+[[Z,X], Y ] =0 is a formal verification, we see that the set of all vector fields is aninfinite dimensional real Lie algebra if the bracket of two vector fields isalso a vector field. The miracle is:

Proposition A.0.16. [X,Y ] is a vector field.

For purposes of comparison we give two proofs for this, one of whichis classical and one modern in spirit. Typically, the classical one is moreelaborate. It is full of sturm und drang. But, in recompense, it givesmore insight. The modern one is quick and machine-like and has littleinsight. It is merely the verification of a previously established criterion.

Proof. Modern Proof. We use the isomorphism established betweentangent vectors and vector fields. Clearly, [X,Y ] is a linear operatoron functions. We therefore need only verify [X,Y ](fg) = f [X,Y ](g) +[X,Y ](f)g. Indeed,

[X,Y ](fg) = (XY − Y X)(fg) = X(Y (fg)) − Y (X(fg))

= X(Y (f)g) +X(fY (g)) − Y (X(f)g) − Y (fX(g))

= XY (f)g + Y f(Xg) + fY (Xg) +X(fY )g

− Y X(f)g −Xf(Y g) − fX(Y g) − Y (fX)g

= XY (f)g +X(fY )g − Y X(f)g − Y (fX)g

= [X,Y ](f)g + f [X,Y ](g).

Classical Proof. To see this we need only see that in local coor-dinates so defined [X,Y ] is a smooth first order differential opera-tor. Let f be a smooth function on M and U a neighborhood of p,X(p) =

∑i ηi(p)∂/∂xi and Y (p) =

∑i ζi(p)∂/∂xi. Then


([X,Y ]f)(p) = (∑

i

ηi∂

∂xi

∑

j

ζj∂f

∂xj−

∑

j

ζj∂

∂xj

∑

i

ηi∂f

∂xi)(p)

=∑

i,j

ηi(p)∂ζj∂xi

(p)∂f

∂xj(p) −

∑

i,j

ζj(p)∂ηi∂xj

(p)∂f

∂xi(p)

+∑

i,j

ηi(p)ζj(p)∂2f

∂xi∂xj(p) −

∑

i,j

ζi(p)ηj(p)∂2f

∂xj∂xi(p)

Since f is smooth, the mixed second partials are equal and so thesecond order terms cancel leaving a first-order operator,

∑

i,j

[ηi(p)ζj∂xi

(p)∂f

∂xj(p) − ζj(p)

ηi∂xj

(p)∂f

∂xi(p)].

A curve in M is a smooth map x : R → M , the tangent vector atevery point on the curve is the vector

x′(t) = dtx(1)

where 1 is thought of as a generator for TtR ≃ R. Given a vector fieldand a point p, if we can find a smooth curve through p whose tangentvector at every point coincides with the vector field, we call the curvean integral curve. This amounts to solving a differential equation withan initial condition. If we can only find a local curve then we have alocal solution to our differential equation with initial condition.

We now give the form of the fundamental theorem of ordinary dif-ferential equations which will be of use to us. A proof of this can befound in [37] or [69].

Theorem A.0.17. Let U ⊆ M and V ⊆ Rm be neighborhoods of 0and y0 respectively and v(x, y) be a vector field in M which dependssmoothly on (x, y). For each fixed y ∈ V consider the initial valueproblem f

′(t) = v(f(t), y), f(0) = 0, where f : R → M . Then there is

Appendix A: Vector Fields 401

an ǫ > 0 and a neighborhood V ′ of y such that there is a unique solutionfor t ∈ (−ǫ, ǫ) and y′ ∈ V ′ to the initial value problem. It dependssmoothly on t and y ∈ V ′.

Corollary A.0.18. Let v(x) be a smooth vector field defined in a neigh-borhood U of 0 in M . Consider the initial value problem f

′(t) = v(f(t)),

f(0) = 0. Then there is an ǫ > 0 and a unique smooth solution fort ∈ (−ǫ, ǫ) to the initial value problem.

We recall that a 1-parameter group of diffeomorphisms is a map,φ : R × M → M , where we write φt(p) instead of φ(t, p), such thatφt is a diffeomorphism for each t ∈ R and t 7→ φt is a homomorphismfrom R → Diff(M) and φ0 = I. A similar definition holds for local1-parameter groups of diffeomorphisms. Namely, there is an interval Iabout 0 in R such that for all p ∈ M , φt(φs(p)) = φt+s(p) whenevers, t and s + t ∈ I. Now a local 1-parameter group of diffeomorphismsgives rise to a vector field on M as follows. For each point p0 ∈ Mconsider the smooth curve φt(p0) through p0. Taking its tangent vectorat each point gives a vector field on M . Conversely, given a vector fieldon M and a point p0, there is always a local 1-parameter group of localdiffeomorphisms φt which is the integral curve to this vector field andfor any smooth function f , limt→0(f φt − f) = Xf .

Proof. Let U, x1, . . . , xn be local coordinates around p0 and assume forsimplicity that for i = 1, . . . n, xi(p0) = 0. Let X =

∑i ηi(x1, . . . , xn)

∂∂xi

in U . Consider the following system of ODE, where i = 1, . . . n andf1(t), . . . , fn(t) are the unknown functions,

df i

dt= ηi(f

1(t), . . . , fn(t)).

By the fundamental theorem of ODE, there exists a unique set of func-tions f1(t, x1, . . . , xn), . . . , f

n(t, x1, . . . , xn), defined for |(x1, . . . , xn)| <δ and |t| < ǫ such that for all i, f i(0, x1, . . . , xn) = xi. Let x =(x1, . . . , xn) and φt(x) = (f1(t, x), . . . , fn(t, x)). Clearly, φ0 = I onthis neighborhood. If |x| < δ and |t|, |s| and |t + s| are all less thanǫ, then x and φs(x), (where s is considered fixed), are both in this


neighborhood. Hence the n-tuple of functions, gi(t, x) = f i(t + s, x)are also in the neighborhood and satisfy the same ODE, but with ini-tial conditions, gi(0, x) = f i(s, x). By the uniqueness it follows thatgi(t) = f i(t, φs(x)). Hence φtφs = φt + s on this neighborhood. Thuswe have a local 1-parameter group of local diffeomorphisms which is theintegral curve to our original vector field.

Let φ be a diffeomorphism of M and dφ its differential. For a vectorfield X on M , φ∗X will denote the vector field induced by the actionof Diff(M) on X (M) mentioned above. If the 1-parameter group gen-erated by X is φt, then the smooth vector field φ∗X also generates a1-parameter group. It is

φ φt φ−1.

Proof. Now φ φt (φ)−1 is clearly a 1-parameter group of diffeomor-phisms, so let Y be its vector field. We must show Y = φ∗X. Let p ∈Mand q = φ−1(p). Since φt induces X, the vector Xq ∈ Tq is tangent tothe curve φt(q) at t = 0. Therefore (φ∗X)p = φ∗(Xq) ∈ Tp is tangent toφ φt(q)) = φ φt φ−1(p).

Corollary A.0.19. Let φ be a diffeomorphism of M . A vector field isφ fixed (i.e. φ∗X = X) if and only if φ commutes with all φt in Diff(M)as t varies.

Appendix B

The Kronecker

Approximation Theorem

Let n be a fixed integer n ≥ 1 and consider n-tuples α1, . . . , αn,where αi ∈ R. We shall say that α1, . . . , αn is generic if whenever∑n

i=1 kiαi ∈ Z for ki ∈ Z, then all ki = 0.Here is an example of a generic set. Let θ be a transcendental real

number and consider the powers, αi = θi. Then for any positive integern, θ1 . . . , θn is generic. For if k1θ

1 + . . . knθn = k, where the ki and

k are integers then since Z ⊆ Q, this is polynomial relation of degreebetween 1 and n which θ satisfies. This is a contradiction.

Proposition B.0.20. The set α1, . . . , αn is generic if and only if1, α1, . . . , αn is linearly independent over Q.

Proof. Suppose 1, α1, . . . , αn is linearly independent over Q. Let∑ni=1 kiαi = k, where k ∈ Z. We may assume k 6= 0. For if k = 0

then since the subset α1, . . . , αn is linearly independent over Q andki ∈ Q for each i we get ki = 0. On the other hand if k 6= 0 we divideand get

∑ni=1

ki

k αi = 1. So 1 is a Q-linear combination of αi’s. Thiscontradicts our hypothesis regarding linear independence.

Conversely suppose α1, . . . , αn is generic and q1 + q1α1 + . . . +qnαn = 0, where q and all the qi ∈ Q. If q = 0, then clearing denomi-nators gives a relation k1α1 + . . . + knαn = 0, where ki ∈ Z. Since the

403

404 Appendix B: The Kronecker Approximation Theorem

αi are generic and 0 ∈ Z we get each ki = 0. Hence also each qi = 0.Thus 1, α1, . . . , αn is linearly independent over Q. On the other handif q 6= 0, by dividing by q we get, 1 + s1α1 + . . . + snαn = 0, wheresi ∈ Q. Again clear denominators and get k + k1α1 + . . . + knαn = 0,where k and ki ∈ Z. Since k1α1 + . . .+ knαn = −k and the original αiis generic, each ki = 0. Therefore each si is also 0 and thus 1 = 0, acontradiction.

Here is another way to “find” generic sets. We consider R to be avector space over Q. Let B be a basis for this vector space. Then anyfinite subset of this basis gives a generic set after removing 1.

Before proving Kronecker’s approximation theorem we define thecharacter group G of a locally compact abelian group G. Here

G = Hom(G,T)

consists of continous homomorphisms and is equipped with the compact-open topology and pointwise multiplication. G is a locally compactabelian topological group.

Proposition B.0.21. Let G and H be locally compact abelian groups(written additively) and β : G × H → T be a nondegenerate, jointlycontinuous bilinear function. Consider the induced map ωG : G →H given by ωG(g)(h) = β(g, h). Then ωG is a continuous injectivehomomorphism with dense range. Similarly, ωH : H → G given byωH(h)(g) = β(g, h) is also a continuous injective homomorphism withdense range.

Proof. By symmetry we need only consider the case of ωG. ClearlyωG : G → H is a continuous homomorphism. If ωG(g) = 0 then forall h ∈ H, β(g, h) = 0. Hence g = 0 so ωG is injective. To prove that

ωG(G) is a dense subgroup of H we show that its annihilator inH is

trivial. Identifying H with its second dualH, its annihilator consists

of all h ∈ H so that β(g, h) = 0 for all g ∈ G. By nondegeneracy (thistime on the other side) the annihilator of ωG(G) is trivial. Hence ωG(G)is dense in H (see [70]).

Appendix B: The Kronecker Approximation Theorem 405

We now come to the Kronecker theorem itself. What it saysis that one can simultaneously approximate (x1, . . . , xn)mod(1) byk(α1, . . . , αn). If we denote by π : R → T the canonical projec-tion with Kerπ = Z, the Kronecker theorem says that any point,(π(x1), . . . , π(xn)) on the n-torus, Tn, can be approximated to any re-quired degree of accuracy by integer multiples of (π(α1), . . . , π(αn)).

Of course a fortiori any point on the torus can be approximatedto any degree of accuracy by real multiples of (π(α1), . . . π(αn)). Theimage under π of such a line (namely the real multiples of (α1, . . . , αn))is called the winding line on the torus. So winding lines and generic setsalways exist.

Theorem B.0.22. Let α1, . . . , αn be a generic set, x1, . . . , xn ∈ Rand ǫ > 0. Then there exists a k ∈ Z and ki ∈ Z such that |kαi − xi −ki| < ǫ.

Proof. Consider the bilinear form β : Z × Zn → T given byβ(k, (k1, . . . kn)) = π(k

∑ni=1 kiαi). Then β is additive in each vari-

able separately and of course is jointly continuous since here the groupsare discrete. The statement is equivalent to saying that image of themap ωG : Z→ Zn ≃ Tn is dense.

We prove that β is nondegenerate. That is if β(k, (k1, . . . kn)) = 0 forall k then (k1, . . . kn) = 0 and if β(k, (k1, . . . kn)) = 0 for all (k1, . . . kn)then k = 0.

If β(k, (k1, . . . kn)) = 0 for all k, then (k1, . . . kn) = 0. The hypothesishere means just that π(k

∑ni=1 kiαi) = 0, or k

∑ni=1 kiαi is an integer.

Choose any k 6= 0. Then∑n

i=1 kkiαi is an integer, because of ourhypothesis regarding the α’s we conclude all kki = 0 therefore ki = 0.On the other hand, suppose β(k, (k1, . . . kn)) = 0 for all (k1, . . . kn), thenwe show k = 0. Hence we have k

∑ni=1 kiαi is an integer for all choices

of (k1, . . . kn). Arguing as before suppose k 6= 0. Choose ki not all zero.This gives kk0 = as the αi is a generic set therefore k = 0.

Hence by Proposition B.0.21 we get an injective homomorphism ωG :Z→ Zn = Tn with dense range. Thus the cyclic subgroup ω(Z) in densein Tn.

406 Appendix B: The Kronecker Approximation Theorem

Exercise B.0.23. (1) Show that in R2 a line is winding if and onlyif it has irrational slope.

(2) Find the generic sets when n = 1. What does this say about densesubgroups of T?

Appendix C

Properly discontinuous

actions

Let Γ × X → X be a (continuous) group action of a locally compactgroup Γ on a locally compact spaceX. We shall say the action is properlydiscontinuous if given a compact set C of X there is a finite subset FCof Γ so that C ∩ (

⋃γ∈C\FC

γC) is empty. In particular, for each pointx ∈ X, the orbit, Γx, has no accumulation point. In particular, Γ mustbe discrete. Also clearly the isotopy group Γx of each point x ∈ X isfinite.

We now look at the converse in the case of an isometric action.

Proposition C.0.24. Let (X, d) be a metric space on which Γ actsisometrically. Suppose each orbit, Γx, has no accumulation points andeach isotopy group Γx is finite. Then Γ acts properly discontinuously.

Proof. If not, there is some compact set C ⊆ X so that C ∩ γ ·C is nonempty for infinitely many γ ∈ Γ. Thus there is a sequence γi of distinctelements of Γ with γi(ci) ∈ C, where ci ∈ C. By compactness thereis a convergent subsequence which we relabel γi(ci) → c ∈ C. Againpassing to a subsequence, using compactness of C and relabeling we findci → c′, c′ ∈ C. Now

d(γic, c′) ≤ d(γic, γici) + d(γici, c

′),

407

408 Appendix C: Properly discontinuous actions

Since Γ acts isometrically d(γic, γici) = d(c, ci) which therefore tends tozero. Also d(γici, c

′) tends to zero. Hence γic→ c′. Since Γc is finite, foreach i there are only finitely many j with γic = γjc. Hence by choosinga subsequence there is a sequence γic→ c′ where the terms are distinct.This contradicts the second condition and proves the result.

However, being properly discontinuous is stronger than being dis-crete. For example consider the action of Z on Tn where n ≥ 2. Thisaction is one in which a discrete group acts by isometries on a (com-pact) metric space. If we have an irrational flow , then every orbit isdense by Kronecker’s approximation theorem. Therefore this action isnot properly discontinuous. Now consider a rational flow. Since its anaction on a metric space by isometries we have only to check the orbitsare discrete and the isotropy groups are finite. In this case both theseconditions are satisfied so the action is properly discontinuous,

We require the following lemma. Here the group, Homeo(X), thehomeomorphisms of X takes the topology of uniform convergence oncompacta which we call the compact open topology.

Lemma C.0.25. Let Γ ×X → X be a continuous group action where(X, d) is a compact metric space and the countable discrete group, Γ,acts isometrically. Then the image of Γ ∈ Homeo(X) is also discrete.

Proof. Denote the map γ 7→ Φ(γ) by Φ, where Φ(γ)(x) = γ · x, x ∈ X.Then for each γ ∈ Γ, Φ(γ) is a homeomorphism, in fact an isometry,of X. Notice that Φ(γ)(X) = X. For if it were smaller, then applyingΦ(γ−1) would yield a contradiction. Also Φ is evidently a continuoushomomorphism Γ → Homeo(X). To complete the proof we need toshow this map is open. Since Γ is countable discrete the open mappingtheorem will do this if we know the image is locally compact. Now inthe compact open topology a neighborhood of I in the image is givenby N(C, ǫ), together with the inverses, where C is compact and ǫ > 0.However, since X is compact we can always take a smaller neighborhoodN0 = N(X, ǫ) of I. These are the homeomorphisms (actually isometries)h such that d(h(x), x) < ǫ for all x ∈ X. The condition of beginan isometry automatically shows any such N , in fact all of Isom(X),

Appendix C: Properly discontinuous actions 409

is equicontinuous. Evidently N0 is pointwise bounded. Hence by theAscoli theorem N0 has compact closure so Φ(Γ) is locally compact. Theopen mapping theorem says Φ is open and therefore Φ(Γ) is discrete.

Let G be a connected semisimple Lie group of non-compact type,X = G/K the associated symmetric space. Then G is the connectedcomponent of the isometry group of X. Let Γ be a torsion free discretecocompact subgroup of G. Then Γ, the fundamental group of S = X/Γacts on S and S is a smooth connected manifold locally isometric withX so S is also metric and Γ acts by isometries. The cocompactness ofΓ implies S is compact.

Proposition C.0.26. The action of Γ on a compact locally symmetricspace S is properly discontinuous.

Proof. If not, there is a point s ∈ S and an infinite number of distinctγi so that Φ(γi)(s) converges to something in S. By Lemma C.0.25Φ(Γ) is a discrete subgroup of Homeo(S). Now the set Φ(Γ1) of the γiis equicontinuous since all of Isom(S) acts equicontinuously. Let t ∈ Sbe fixed. Then Γ1(t) ⊆ N(γis), d(s, t))

− which is compact since S is.Hence Γ1 is uniformly bounded. Since it is also equicontinuous Γ1 hascompact closure. On the other hand Φ(Γ) is discrete. Therefore Γ1 isfinite, a contradiction. This means Γ acts properly discontinuously.

410 Appendix C: Properly discontinuous actions

Appendix D

The Analyticity of Smooth

Lie Groups

Here we sketch the proof of the analyticity of a connected smooth Liegroup G. In the complex case this is just a fact of complex analysis sohere we focus on the real case, although the proof that follows worksequally well in the case of complex Lie groups.

If left and right translations are analytic, to prove the claim it issufficient to prove that multiplication and inversion are analytic in aneighborhood of 1 in G. For suppose we were at a neighborhood of(p, q). Let x1 = p−1x and y1 = q−1y. Then xy−1 = px1y

−11 q−1 =

LpRq−1x1y−11 . If the function (x, y) 7→ xy−1 is analytic at the origin

and, as above, left and right translations are analytic on G, then as acomposition of analytic functions (x, y) 7→ xy−1 is analytic at (p, q).

Now we prove the analyticity in a neighborhood of the 1. Sincethis is a local question and any Lie group is locally isomorphic to alinear Lie group, as mentioned in Section 1.7, we may assume G islinear. Let U be a canonical neighborhood of 1 in G. We identify Uwith an open ball B about 0 in g using Exp which is analytic. SinceExp(x) 7→ Exp(−x) is evidently analytic, i.e. x 7→ −x being linear, itis sufficient to prove multiplication is analytic on U . We can considereach u = (u1, . . . , un) ∈ B, where n = dimG. Let z = xy, where x andy ∈ U . Then for each i, zi = fi(x1, . . . , xn, y1, . . . , yn), fi ∈ C∞(U ×U).

411

412 Appendix D: The Analyticity of Smooth Lie Groups

Now ∂fi

∂yj= δij at (x, y) = (1, 1). However, at y = 1 with x varying ∂fi

∂yj

is a function of x: vij(x) = vij(x1, . . . , xn). If b = (b1, . . . , bn) ∈ B, the1-parameter group Exp(tb) satisfies the system of differential equations,

dxidt

= Σni=1bivij(x1(t), . . . , xn(t)), xi(0) = 0.

Since Exp(tb) is the unique solution, this system of equations is nothingmore than the matrix differential equation dx

dt = bExp(tb), x(0) = I.Thus the matrix, (vij(x)) = Exp(x) and since x 7→ Exp(x) is analyticso are the vij .

Now the product functions zi = fi(x, y) satisfy a system of partialdifferential equations:

Σjvij(z)∂zj∂xk

(x) = vik(x), i, k = 1, . . . , n,

called the fundamental differential equations of the group, G which de-termine the z’s if the v’s are known and certain integrability conditionsare satisfied. These link the v’s and their derivatives to the structureconstants of g. Since these conditions are necessary and sufficient andG is a smooth Lie group, the vij certainly satisfy these integrabilityconditions. The only question remaining is whether the zi are analytic.But since we know the v’s are analytic, so are the z’s. This follows fromthe Frobenius theorem (see [66] Theorem 211.9).

Finally we prove that left and right translations are analytic. Mul-tiplication is analytic in a neighborhood U of 1 in G. Hence so is lefttranslation Lg on U when g ∈ U . Therefore, because of the way we putthe manifold structure on G, such Lg’s are analytic on all of G. Nowlet g ∈ G be arbitrary. Then g = g1 . . . gn, where each gi ∈ U . HenceLg = Lg1 . . . Lgn , a composition of analytic functions and therefore eachLg is analytic. Similarly each Rg is analytic.

Bibliography

[1] Adams, J.F. Lectures on Lie groups, W. A. Benjamin Inc., NewYork-Amsterdam 1969.

[2] Auslander L. Lecture notes on nil-theta functions, CBMS Reg.Conf. Series Math. 34, Amer. Math. Soc., Providence, 1977.

[3] Ballmann W., Gromov M. and Schroeder V. Manifolds of nonpos-itive curvature, Progress in Mathematics 61, Birkhuser BostonInc., Boston, MA, 1985.

[4] Barbano P. Automorphisms and quasiconformal mappings ofHeisenberg type groups, J. of Lie Theory, 8 (1998), 255-277.

[5] Borel A. Density properties for certain subgroups of semi-simplegroups without compact components, Ann. of Math. (2) 72, 1960179–188.

[6] Borel A. and Harish-Chandra Arithmetic subgroups of algebraicgroups, Annals of Math. (2) 75 (1962), 485-535.

[7] Borel A. Compact Clifford-Klein forms of symmetric spaces,Topology (2) 1963, 111–122.

[8] Borel A. Introduction aux groupes arithmetiques, Publications del’Institut de Mathematique de l’Universite de Strasbourg, XV, Ac-tualites Scientifiques et Industrielles, No. 1341, Hermann, Paris1969, 125 pp.

413

414 Biliography

[9] Borel A. Linear algebraic groups, Second edition Graduate Textsin Mathematics 126, Springer-Verlag, New York, 1991.

[10] Borel A. Semisimple Lie Groups and Riemannian SymmetricSpaces, Hindustan Book Agency, 1998.

[11] Brocker T. and tom Dieck T. Representations of compact Liegroups. Translated from the German manuscript. Correctedreprint of the 1985 translation.Graduate Texts in Mathematics,98, Springer-Verlag, New York 1995.

[12] Cartan E. Groupes simples clos et ouverts et geometrie Rieman-nienne , Journal de Math. Pures et Appliques 8 (1929), 1-33.

[13] Chabauty C. Limites d’ensembles et geometrie des nombres, BullSoc. Math. de France 78 (1950), 143-151.

[14] Cheeger J. and Ebin D. Comparison Theorems in RiemannianGeometry, North Holland, Amsterdam, 1975.

[15] Chevalley C. The Theory of Lie Groups, Princeton UniversityPress, Princeton, 1946.

[16] Corwin L. and Moskowitz M. A note on the exponential map of areal or p-adic Lie group. J. Pure Appl. Algebra 96 (1994), no. 2,113-132.

[17] Djokovic D. and Thang N. On the exponential map of almost sim-ple real algebraic groups, J. of Lie Theory 5 (1996), 275-291.

[18] Dubrovin B.A., Fomenko A.T. and Novikov S.P. Moderngeometry—methods and applications. Part II. The geometry andtopology of manifolds, Graduate Texts in Mathematics, 104.Springer-Verlag, New York, 1985.

[19] Faraut J. Analyse harmonique sur les espaces hyperboliques Topicsin modern harmonic analysis, Vol. I, II (Turin/Milan, 1982), 445–473, Ist. Naz. Alta Mat. Francesco Severi, Rome.

Biliography 415

[20] Farkas H. and Kra I. Riemann surfaces Graduate Texts in Math-ematics, 71. Springer-Verlag, New York-Berlin, 1980.

[21] Furstenberg H. A Poisson formula for semi-simple Lie groups,Ann. of Math. (2) 77 (1963) 335–386.

[22] Furstenberg H. A note on Borel’s density theorem, Proc. AMS 55

(1976), 209-212.

[23] Gallot S., Hulin D. and Lafontaine J. Riemannian geometry, Sec-ond Edition, Springer-Verlag, Berlin 1990.

[24] Garland H. and Goto M. Lattices and the adjoint group of a Liegroup, Trans. Amer. Math. Soc. 124 1966 450–460.

[25] Gelfand, I., Graev I. and Pyatetskii-Shapiro I. Representation the-ory and automorphic functions, Translated from the Russian byK. A. Hirsch. Reprint of the 1969 edition. Generalized Functions,6. Academic Press, Inc., Boston, MA, 1990.

[26] Glushkov V.M. The structure of locally compact groups andHilbert’s fifth problem, AMS Translations 15 (1960) 55-93.

[27] Grosser S. and Moskowitz M. Representation theory of centraltopological groups, Trans. Amer. Math. Soc. 129 (1967) 361–390.

[28] Grosser S. and Moskowitz M. Harmonic analysis on central topo-logical groups, Trans. Amer. Math. Soc. 156 (1971) 419–454.

[29] Greenleaf F., Moskowitz M. and Rothschild L. Compactness ofcertain homogeneous spaces of finite volume Amer. J. Math., 97

no.1, (1975), 248-259.

[30] Guillemin V. and Pollack A. Differential topology, Prentice-HallInc., Englewood Cliffs, N.J., 1974.

[31] Hausner M. and Schwarz J.T. Lie groups; Lie algebras, Gordonand Breach Science Publishers, New York-London-Paris 1968.

416 Biliography

[32] Helgason S. Differential Geometry, Lie Groups and SymmetricSpaces, Pure and Applied Mathematics 80, Academic Press, NewYork, 1978.

[33] Hochschild G.P. The Structure of Lie Groups, Holden Day Inc.,San Francisco-London-Amsterdam, 1965.

[34] Hochschild G.P. and Mostow G. Representations and representa-tive functions of Lie groups, Ann. of Math (2) 66 1957, 495–542.

[35] Hunt G. A Theorem of E. Cartan, Proc. Amer. Math. Soc. 7

(1956), 307–308.

[36] Kaplansky I. Infinite abelian groups, University of Michigan Press,Ann Arbor 1954.

[37] Kolmogorov A.N. and Fomin S.V. Elements of the Theory of Func-tions and Functional Analysis, Translated from the 1st (1954)Russian ed. by Leo F. Boron, Rochester, N.Y., Graylock Press.

[38] Mahler K. On Lattice Points in n-dimensional Star Bodies I, Ex-istence theorems, Proc. Roy. Soc. London Ser. A, 187 (1946) 151–187.

[39] Malcev A. On a class of homogeneous spaces, Amer. Math. Soc.Translation 1951, (1951) no. 39.

[40] Massey W.S. A basic course in algebraic topology Graduate Textsin Mathematics 127, Springer-Verlag, New York, 1991.

[41] Margulis G. Discrete Subgroups of Semisimple Lie Groups,Springer-Verlag, Ergebnisse der Mathematik 3. Folge Bd 17.Berlin Heidelberg New York, 1990.

[42] Milnor J. Morse theory, Based on lecture notes by M. Spivak andR. Wells. Annals of Mathematics Studies, No. 51,Princeton Uni-versity Press, Princeton N.J. 1963.

Biliography 417

[43] Moore C.C. Decomposition of unitary representations defined bydiscrete subgroups of nilpotent groups, Ann. of Math. (2) 82 1965,146–182.

[44] Moore C.C. Ergodicitiy of flows on homogeneous spaces., Amer.J. Math. (88) 1966 154–178.

[45] Mosak R.D. and Moskowitz M. Zariski density in Lie groups, Is-rael J. Math. 52 (1985), no. 1-2, 1–14.

[46] Mosak R.D. and Moskowitz M. Analytic density in Lie groups,Israel J. Math. 58 (1987), no. 1, 1–9.

[47] Mosak R.D. and Moskowitz M. Stabilizers of lattices in Lie groupsJ. Lie Theory 4 (1994), no. 1, 1–16.

[48] Moskowitz M. A remark on faithful representations Atti della Ac-cademia Nationale dei Lincei, ser. 8., 52 (1972), 829-831.

[49] Moskowitz M. Faithful representations and a local property of Liegroups Math. Zeitschrift, 143 (1975), 193-198.

[50] Moskowitz M. Some Remarks on Automorphisms of Bounded Dis-placement and Bounded Cocycles, Monatshefte fur Math., 85

(1978), 323-336.

[51] Moskowitz M. On the density theorems of Borel and Furstenberg,Ark. Mat. 16 (1978), no. 1, 11–27.

[52] Moskowitz M. On the surjectivity of the exponential map in certainLie groups, Annali di Matematica Pura ed Applicata, Serie IV-Tomo CLXVI (1994), 129-143.

[53] Moskowitz M. Correction and addenda to: “On the surjectivityof the exponential map for certain Lie groups” [Ann. Mat. PuraAppl. (4) 166 (1994), 129–143, Ann. Mat. Pura Appl., (4) 173

(1997), 351–358

[54] Moskowitz M., Complete reducibility and Zariski density in linearLie groups Math. Z. 232 (1999), no. 2, 357–365.

418 Biliography

[55] Moskowitz M. A course in complex analysis in one variable, WorldScientific Publishing Co., Inc., River Edge, NJ, 2002

[56] Moskowitz M. and Wustner M. Exponentiality of certain real solv-able Lie groups, Canad. Math. Bull. 41 (1998), no. 3, 368–373.

[57] Mostow G.D. Self-adjoint Groups, Annals of Math.(2) 62 (1955),44-55.

[58] Mostow G.D. Equivariant embeddings in Euclidean space, Ann. ofMath. (2) 65 (1957), 432–446.

[59] Mostow G.D. Cohomology of topological groups and solvable man-ifolds Ann. of Math., 73, 20-48 (1961).

[60] Mostow, G.D. Homogeneous spaces with finite invariant measure,Ann. of Math. (2) 75 (1962) 17–37.

[61] Mostow G.D. Strong rigidity of locally symmetric spaces, Annals ofMathematics Studies, No. 78. Princeton University Press, Prince-ton, N.J.; University of Tokyo Press, Tokyo, 1973.

[62] Mostow G.D. Discrete subgroups of Lie groups, Advances in Math.16 (1975), 112–123.

[63] Mostow G.D. Discrete subgroups of Lie groups, The mathematicalheritage of Elie Cartan (Lyon, 1984). Asterisque (1985), NumeroHors Serie, 289–309.

[64] Mostow G.D. and Tamagawa T. On the compactness of arithmeti-cally defined homogeneous spaces, Ann. of Math. (2) 76 (1962),446–463.

[65] Nachbin L. The Haar integral D. Van Nostrand Co. Inc., Prince-ton, N.J.-Toronto-London 1965.

[66] Narasimhan R. Analysis on Real and Complex Manifolds, Ad-vanced Studies in Pure Mathematics, Masson & Cie-Paris 1973.

Biliography 419

[67] Myers S. and Steenrod N. The group of isometries of a riemannianmanifold, Ann. of Math., 40 (1939), 400-416.

[68] Palais R. Imbedding of compact, differentiable transformationgroups in orthogonal representations, J. Math. Mech. 6 (1957),673–678.

[69] Perko, L. Differential equations and dynamical systems, Secondedition. Texts in Applied Mathematics, 7. Springer-Verlag, NewYork, 1996.

[70] Pontryagin L.S. Topological groups, Translated from the secondRussian edition by Arlen Brown, Gordon and Breach Science Pub-lishers Inc., New York-London-Paris 1966.

[71] Raghunathan M.S. Discrete subgroups of Lie groups, Ergebnisseder Mathematik und ihrer Grenzgebiete, Band 68, Springer-Verlag, New York-Heidelberg, 1972.

[72] Rothman R. The von Neumann kernel and minimally almost peri-odic groups,Trans. Amer. Math. Soc. 259 (1980), no. 2, 401–421.

[73] Royden H.L., Real analysis, Third edition, Macmillan PublishingCompany, New York, 1988.

[74] Rudin, W. Principles of mathematical analysis, International Se-ries in Pure and Applied Mathematics, McGraw-Hill Book Co.,New York-Auckland-Dusseldorf 1976.

[75] Siegel C.L. Uber Gitterpunkte in convenxen Korpern und eindamit zusammenhangendes Extremalproblem, Acta Mathematicavol 65 (1935), 307-323.

[76] Steenrod N. The Topology of Fibre Bundles, Princeton Mathe-matical Series vol. 14, Princeton University Press, Princeton, N.J. 1951.

[77] Tits J. Free subgroups in linear groups, J. Algebra 20 (1972), 250–270.

420 Biliography

[78] Varadarajan V.S. Lie groups, Lie algebras, and their representa-tions, Prentice-Hall Series in Modern Analysis. Prentice-Hall, Inc.,Englewood Cliffs, N.J. 1974.

[79] Wang H.C. On the deformations of lattices in a Lie group, Amer.J.Math., 89 189–212 (1963).

[80] Warner F.W. Foundations of differentiable manifolds and Liegroups, Corrected reprint of the 1971 edition, Graduate Texts inMathematics 94, Springer-Verlag, New York-Berlin 1983.

[81] Whitney H. Elementary structure of real algebraic varieties, Ann.of Math. 66 (1957), 545–556.

[82] Zariski O. and Samuel P. Commutative algebra, Vol. 1. With thecooperation of I. S. Cohen. Corrected reprinting of the 1958 edi-tion. Graduate Texts in Mathematics, No. 28. Springer-Verlag,New York-Heidelberg-Berlin, 1975.

Index

∗, 269

1-parameter subgroup, 32

A∗, xiii

At, xiii

B(G), 373

Bτ , 343

Bθ, 343

C(G)G, 246

DX(Y ), 279

G-equivalent, 17

G#, 389

G0, xiii

GR, 44

H, 264

K, 268

L1(G), 224

L2(G), 224

Mn(C), 3

Mn(R), 3

Mn(k), xiii

NG, 218

P , 264, 268

R(G), 236

R(ρ), 238

Tk, 228

Vλ, 313

X(T ), 175, 177

ZG, 218

AdG, xiii, 55

Ad, xiii

AdG(H), xiii

Aut(G), 16

Aut(g), 29, 321

Der(g), 59

Exp, 33

Homeo(X), 408

ℑ(H), 134

ℑ, xiii

Ind(H ↑ G,σ), 250

O(n,C), 3

O(n,R), 3

R-points, 44

Rad(G), 390

ℜ, xiii

SL(n,C), 3

SL(n,R), 3

SO(n,C), 3

SO(n,R), 3

SO(p, q), 4

Sp(n,C), 4

Sp(n,R), 4

SU(n,C), 4

Spec(T ), xiii

U(n,C), 4

421

422 Index

ad g, xiii, 29, 59

ad, xiii

ad-nilpotent, 193

H, 264

P, 264

R(G), 225

X (G), 244

χ(M), 398

χρ, 243

exp, 36

gl(V ), 26

gl(n, k), 26

gλ,X , 315

gλ, 315

k, 268

o(n, k), 28

p, 267

gk, 135

gk, 133

nk(V ), 135

s(V ), 135

z(g), 130

sl(n, k), 127

so(n, k), 28

u(n), 28

zg(X), 139

Inn(g), 321

T, 2

Z(p), 10

sp(n,R), 45

rad(g), 136

reg(ρ, g, V ), 317

ρ(X), 29

ρX , 29

ρg, 7

ax+ b-Lie algebra, 27, 137

ax+ b-group, 96

l.s.k, xiii

l.s.C(Ω), 235

p-adic integers, 10

311

1-parameter group of diffeomor-phisms, 401

2-step nilpotent, 134, 371

n(V ), 135

min(ρ, g, V ), 317

nil(g), 135

σ-compact group, 18

action, 15

simply transitive, 15

transitive, 15

adjoint, 27

adjoint algebra, 29

adjoint group, xiii

adjoint representation, 29, 55

Ado’s theorem, 75, 186

affine group, 9

algebra of invariants, 122

algebraic group, 43

algebraic hull, 389

amenable, 394

analytically dense, 392

approximate identity, 235

arithmetic, 393

automorphism, 29

automorphism group, 7

Baire’s category theorem, 19

Index 423

Baker-Campbell-Hausdorffformula, 40, 74

BCH formula, 74

binomial theorem, 150

block triangular form, 157

Bochner linearization theorem, 119

Borel Density Theorem, 377

bounded part, 373

canonical coordinates of the 2ndkind, 83

Cartan criteria, 162

Cartan decomposition, 267, 270

Cartan involution, 343

Cartan relations, 28, 273

Cartan subalgebra, 316

Cartan’s fixed point theorem, 273,293

Cartan’s solvability criterion, 162

Cartan, Elie, 261

Cartier, 115

Casimir element, 173

Casimir index, 173

Casimir operator, 173

Cayley-Hamilton theorem, 196

center, 127, 153

central extension, 136

central function, 246

central groups, 223

central ideal, 130

centralizer, 139

character, 243

character group, 404

Chevalley’s Theorem, 125

class function, 246

cocompact, 106

cofinite volume, 106commutative operators, 154compact real form, 337

complete reducibility, 173completely reducible, 186complexification, 147

conjugation, 338conjugation relative to the real

form, 338covering map, 10

covering space, 11

density theorem, 379derivation, 59

inner, 59

derived series, 135derived subalgebra, 130diagonalizable, 158

direct sum, 129distribution, 45

integrable, 45involutive, 45smooth, 45

Engel’s theorem, 152

equivariantly equivalent, 17Erlanger Program, 262essentially algebraic group, 44

essentially algebraic subgroup, 293exponential, 213exponential map, 36

exponential submanifolds, 267external direct sum, 129

faithful representation, 186

424 Index

fiber preserving map, 11

field extension, 147

finite generation of an algebra, 123

first isomorphism theorem, 52

first-order differential operator,398

flag manifolds, 23

flag of ideals, 156

Fourier transform, 240

Frobenius reciprocity theorem, 254

Frobenius theorem, 46

fundamental differential equationsof the group, 412

fundamental domain, 374, 375

fundamental theorem of invarianttheory, 124

Furstenberg, 379

general linear group, 3

Grassmann Space, 22

group

affine, 9

complex Lie, 6

real Lie, 6

topological, 1

transformation, 15

group action, 15

group homomorphism, 4

Haar measure, 89

Hadamard manifold, 273, 290

Heisenberg Lie algebra, 132

Hilbert basis theorem, 123

Hilbert’s 14th problem, 121

Hilbert’s fifth problem, 7

Hilbert-Schmidt inner product,278

Hilberts 14th problem, 124homogeneous space, 296hyperboloid, 303

ideal, 127

characteristic, 131nilpotent, 133solvable, 135

identity component, 7index of nilpotence, 133index of solvability, 135

induced representations, 250inner derivations, 59integral curve, 400

integral distribution, 45integral manifold, 45intertwining operator, 140, 224

invariantvector, 151

invariant form, 145, 203invariant measure, 89, 102invariant set, 17

invariant vector, 151involution, 343irrational flow, 408

isometric, 407Iwasawa decomposition, 343

J-M condition, 193Jacobi identity, 25

Jordan decomposition, 158

Kazdan, 366kernel function, 229

Index 425

Killing form, 146

Klein’s Erlanger Program, 16

Klein, Felix, 16

Kronecker’s approximation theo-rem, 404

lattice, 106

log, 372

Lebesgue measure, 95

left invariant, 31

left invariant subspace, 239

left translation, 16

Levi decomposition, 185

Levi’s splitting theorem, 180

Lie algebra, 25, 31

ax+ b, 27, 137

Heisenberg, 198

abelian, 26

affine, 137

compact type, 203

complete, 139

Heisenberg, 132

linear, 26

nilpotent, 133

reductive, 188

semisimple, 138, 163

simple, 138

solvable, 135

Lie algebra representation, 29

equivalent, 140

Lie bracket, 25

Lie group

compact, 202

exponential, 213, 304

Lie homomorphism, 6, 29

Lie subgroup, 6

Lie’s theorem, 153, 157light cone, 300linear actions, 16

linearly reductive, 189log lattice, 372Lorentz group, 274

Lorentz model, 277, 303

Malcev uniqueness theorem, 183Margulis, 84, 366Margulis Lemma, 85

Margulis lemma, 86maximal abelian subalgebra, 294maximal compact subgroup, 270

maximal torus, 207minimally almost periodic groups,

386modular function, 100

monodromy principle, 50Morozov’s lemma, 193

nilradical, 135niltriangular, 135

norm, 228normalizer, 138

operatorCasimir, 173

compact, 228nilpotent, 150

self adjoint, 229semisimple, 158skew symmetric, 28

symmetric, 28operators

426 Index

finite rank, 228

orbit, 17orbit map, 17orthogonal group, 3

Plancherel theorem, 240polar decomposition, 264

polar decomposition theorem, 266Pontrjagin, 16

properly discontinuous, 407

quasi generator, 208

radical, 136, 390

rank, 211, 294, 317rational form, 148real form, 328, 337

real points, 44regular elements, 317regular measure, 89

representation, 7, 29adjoint, 29, 55admissible, 382

completely reducible, 141, 224equivalent, 140, 224faithful, 7, 29

irreducible, 140, 224Lie algebra, 29reducible, 140

strongly admissible, 382unitary, 224

representation space, 7representative functions, 236restricted root space, 348

root, 153, 315root space, 315

root string, 326root vector, 154, 315

Schur orthogonality relations, 226Schur’s lemma, 141self-adjoint subgroup, 269semi direct sum, 131semi-invariant, 153semidirect product, 9, 97semidirect products, 9semisimple Lie algebra, 163semisimple operator, 158Siegel generalized upper half space,

263simply connected, 50small subgroups, 58spherical harmonics, 256stabilizer, 18Stiefel manifolds, 23structure constants, 26subalgebra, 26symmetric space, 288symplectic form, 4symplectic group, 4system of differential equations,

412

theoremBochner, 119Cartan’s fixed point, 273Cayley-Hamilton, 196Chevalley normalization, 332Engel, 152first isomorphism, 5, 52, 128Frobenius, 46Harish Chandra, 365

Index 427

Jacobson-Morozov, 193

Kronecker’s approximation,404

Lagrange Interpolation, 160Levi’s splitting, 180

Lie’s, 153, 157Mahler’s compactness crite-

rion, 363Malcev uniqueness, 183

Mostow’s rigidity, 366Mostow-Tamagawa, 365open mapping, 20

Peter-Weyl, 237second isomorphism, 52, 130Serre isomorphism, 332

spectral, 230third isomorphism, 53, 130

Weyl’s finiteness, 118totally geodesic, 288transformation group, 15

triangular form, 154two-fold transitively, 297two-fold transitivity, 296

two-point homogeneous space, 297

uniform lattice, 106uniform subgroup, 106

unimodular groups, 96unipotent, 371unitary group, 3

universal cover, 11

vector field, 397

weight, 153, 313weight space, 313

weight vector, 153weight vectors, 313Weyl group, 218Weyl’s finiteness theorem, 118Weyl’s theorem, 173, 175, 341Whitehead’s lemma, 179

Zariski dense, 377Zassenhaus, 84

Documents

Basic Lie Theory · their Lie algebras, subalgebras and ideals, the functorial relationship determined by the exponential map, the topology of the classical groups, the Iwasawa decomposition