Modular Lie Algebras

Dmitriy Rumynin∗

March 11, 2010

How to use these notes

The lecture notes are split into 3 chapters, further split into 30 sections.Each section will be discussed on a separate lecture. There will be a cutoffpoint for the exam: a few of the last lectures will not appear on the exam.

Each section consists of subsections. Regular subsection are examinable.Vista and background subsections are not. Vista subsections contain blue-sky material for further contemplation. Background sections contain mate-rial that should be known to students but is still included to make the notesself-contained. I have no intention to examine background subsections butthey are useful for understanding of the course. If you are missing somebackground, please, ask me and I could add a background subsection toremedy it.

Exercise sections contain exercises that you should attempt: they mayappear on the exam. Proofs of propositions are examinable as soon asthe proof is written or covered in class. Overall, 100% of the exam willbe covered by these lecture notes and in-class proofs. Consult these writtennotes because my planning may not be perfect and some material in a sectioncould be skipped during the lecture. You should still learn it because it couldappear on the exam.

It is worth mentioning that I don’t follow any particular book. Further-more, no book is relevant enough. The lecture notes are not fully ready yet.All the updates will be available on the website. Check whether you havegot the latest version. The final version will be available from the generaloffice.

If you see any errors, misprints, oddities of my English, send me an email.Also write me if you think that some bits require better explanation. If you

∗ c©Dmitriy Rumynin 2010

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want to contribute by writing a proof, an example, a valuable observation,please, do so and send them to me. Just don’t use WORD. Send them astext files with or without LATEX: put all the symbols you want me to latexbetween $ $. All the contributions will be acknowledged and your name willbe covered with eternal (sh)fame.

Introduction

We study modular Lie algebras, that is, Lie algebras over fields of pos-itive characteristic in this course. After careful thinking I have chosen thefollowing three topics to discuss:

• classification of simple Lie algebras,

• restricted Burnside problem,

• irreducible representations of classical simple Lie algebras.

The reasons to choose these three is my perception of what constitutes animportant development in the area. Complete proofs are often long andcomplicated and the emphasis is definitely on understanding the conceptsinstead of proving all the results. The aim is to introduce various conceptsfrom Algebra, Algebraic Geometry and Representation Theory and see howthey are useful for solving mathematical problems.

Contents

1 Basics 4

2 Things that fail in positive characteristic 9

3 Free algebras 15

4 Universal enveloping algebras 18

5 p-th powers 21

6 Uniqueness of restricted structures 24

7 Existence of restricted structures 28

8 Schemes 30

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9 Differential geometry of schemes 33

10 Generalised Witt algebra 35

11 Filtrations 38

12 Witt algebras are generalised Witt algebra 41

13 Differentials on a scheme 44

14 Lie algebras of Cartan type 48

15 Root systems 52

16 Chevalley theorem 56

17 Chevalley reduction 59

18 Simplicity of Chevalley reduction 61

19 Chevalley groups 65

20 Abstract Chevalley groups 67

21 Engel Lie algebras 69

22 Lie algebra associated to a group 72

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1 BasicsWe recall some basic definitions in this lecture. If you know how to asso-

ciate Lie algebra to a Lie group, then all of this should be quite transparent.

1.1 Algebras and multiplications

Recall that an algebra over a field K of characteristic p is a vector spaceA with a K-bilinear multiplication operation µ : A × A → A, which weusually write in the standard shorthand notation a ∗ b or even ab instead ofthe long notation µ(a, b).

To an element a ∈ A one can associate two linear operators La, Ra :A → A where La(b) = ab and Ra(b) = ba. The names of this operatorsare intuitive: La is the left multiplication operator and Ra is the rightmultiplication operator.

These operators are useful for many things. One is that the algebraaxioms can be reformulated in their terms. For instance, we need associativealgebras that are characterised by the associativity identity a(bc) = (ab)cfor all a, b, c ∈ A and the presence of identity element 1 ∈ A such thatL1 = R1 = IdA. The associativity can be reformulated as La ◦ Lb = Labfor all a, b ∈ A in terms of the left multiplication. Equivalently, it saysRc ◦Rb = Rbc for all b, c ∈ A in terms of the right multiplication.

Another important class of algebras we use is commutative algebras.These are associative algebras which satisfy the commutativity identityab = ba for all a, b ∈ A. In terms of multiplication operators, it says thatLa = Ra for all a ∈ A.

Our main protagonist is Lie algebras. They satisfy the anticommuta-tivity and Jacobi identity. Both identities have subtle points which requiresome discussion. We define the anticommutativty as a ∗ a = 0 for all a ∈ A.Notice that this implies that 0 = (a+ b) ∗ (a+ b)− a ∗ a− b ∗ b = a ∗ b+ b ∗ aand, consequently, a ∗ b = −b ∗ a for all a, b ∈ A. This property can bereformulated as La = −Ra for all a ∈ A. Notice that if p 6= 2 then this isequivalent to anticommutativity: a ∗ a = −a ∗ a implies that 2a ∗ a = 0 anda ∗ a = 0.

On the other hand, if p = 2 then 2a ∗ a = 0 always holds for trivialreasons. Moreover, a ∗ b = −b ∗ a is the same as a ∗ b = b ∗ a since 1 =−1. This identity is commutativity. The anticommutativity is a ∗ a = 0.One unintended consequence of this terminology is that anticommutativityimplies commutativity.

The Jacobi identity is (a ∗ b) ∗ c + (b ∗ c) ∗ a+ (c ∗ a) ∗ b = 0. We aregoing to reformulate it twice in the coming two subsections.

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1.2 Endomorphisms and Derivations

The following terminology should be familiar to you from Algebra-II,although we are going to apply it to general algebras. A homomorphismof algebras is a linear map f : A → B such that f(xy) = f(x)f(y). If Acontains identity, we also require f(1A) = 1B . As a consequence the zeromap is a homomorphism of associative algebras if and only if B = 0.

An isomorphism is a bijective homomorphism. An endomorphism is ahomomorphism f : A → A. Finally, an automorphism is an isomorphismf : A→ A.

A new notion, which you may have seen in Lie Algebras is a derivation.A derivation is a linear map d : A→ A such that d(ab) = d(a)b+ ad(b), i.e.it satisfies Leibniz identity.

Proposition 1.1 For an anticommutative algebra A the Jacobi identity isequivalent to the fact that each La is a derivation for each a ∈ A.

Proof: We rewrite Jacobi identity as (a ∗ b) ∗ c+ (c ∗ a) ∗ b = −(b ∗ c) ∗ a.Using anticommutativity we rewrite further as (a∗b)∗c+b∗(a∗c) = a∗(b∗c)that is exactly the fact that La is a derivation. 2

Notice that it is also equivalent to all right multiplications Ra beingderivations.

1.3 Modules and Representations

These notions are usually interchangeable but we make an artificial dis-tinction that we will follow through in the lecture notes. Let A be an algebra(no axioms assumed). We consider vector spaces V equipped with bilinearactions maps A×V → V (denoted by (a, v) 7→ av). Such a vector space canbe a module or a representation if certain axioms are satisfied.

The module axiom is (a ∗ b)v = a(bv) for all a, b ∈ A, v ∈ V . If Acontains 1, we also require 1v = v for all v ∈ V .

The representation axiom is (a ∗ b)v = a(bv) − b(av) for all a, b ∈ A,v ∈ V .

The way we set the distinction up ensures that it only makes sense totalk about modules for associative algebras and representations for Lie alge-bras. The notions of subrepresentation, quotient representation, submoduleand quotient module are standard. Homomorphisms, isomorphisms, directsums and isomorphism theorems work in the same way for modules andrepresentations. Following the convention a representation V is irreducibleif V 6= 0 and 0, V are the only subrepresentations, while a module with thesimilar property is called simple.

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Proposition 1.2 For an anticommutative algebra A the Jacobi identity isequivalent to the fact that A with the multiplication map is a representationof A.

Proof: A is a representation if and only if La is a derivation for each a ∈ A.One does not even need anticommutativity for this as (a∗b)∗c+b∗(a∗c) =a ∗ (b ∗ c) is rewritten as (a ∗ b)c = a(bc) − b(ac). Now use Proposition 1.1.2

1.4 Simple algebras

A vector subspace I of an algebra A is an ideal if it is stable under allmultiplications, i.e., La(I) ⊆ I ⊇ Ra(I) for all a ∈ A. The significance ofideals is that they are kernels of homomorphisms. Besides they allow todefine quotient algebras A/I.

An algebra A is simple if A 6= 0 and 0 and A are the only ideals. Simplealgebras do not have non-trivial quotients. In particular, any homomor-phism from a simple algebra is injective.

It is a popular problem in Algebra to classify simple algebras in a certainclass of algebras. It often leads to interesting mathematics. For instance,in Rings and Modules you have seen Artin-Wedderburn theorem that statesthat simple associative artinian algebras are matrix algebras over divisionrings. In Lie Algebras you have seen the Cartan-Killing classification ofsimple finite dimensional Lie algebras over an algebraically closed field ofcharacteristic zero. Our goal is to understand the following classificationtheorem, whose complete proof is beyond the scope of the present course.

Theorem 1.3 Let K be an algebraically closed of characteristic p ≥ 5. Thena finite dimensional simple Lie algebra over K is either of classical type orof Cartan type or of Melikian type.

1.5 Commutation

If A is an algebra we define a new algebra A[−] as the same vector spacewith the commutator product: [a, b] = a∗b−b∗a. We call it the commutatoralgebra of A. Apriori, it is not clear why the commutator algebra is par-ticularly significant, why the commutator product leads to more interestingmathematics than a ∗ b+ b ∗ a or a ∗ b− 2010b ∗ a. The following propositiongives some explanation.

Proposition 1.4 Let A be an associative algebra. Then A[−] is a Lie alge-bra.

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Proof: The anticommutativity is obvious: [a, a] = a ∗ a − a ∗ a = 0. TheJacobi identity can be done by a straightforward check but we use a slightlymore conceptual method. Let us check that La in A[−] is a derivation of A:La(b)c+bLa(c) = (ab−ba)c+b(ac−ca) = abc−bca = La(bc). It follows thatit is a derivation of A[−]: La([b, c]) = La(bc) − La(cb) = La(b)c − cLa(b) +bLa(c) − La(c)b = [La(b), c] + [b, La(c)]. Now use Proposition 1.1. 2

Notice that A[−] can be a Lie algebra without A being associative (seethe vista subsection below).

If M is an R-module then we denote EndR(M) the set of all R-moduleendomorphisms. It is an algebra under the composition of morphisms. Inparticular, if A is a vector space then EndK(A) is the algebra of linear mapsfrom A to A. It is irrelevant if A is an algebra itself: we still consider alllinear operators.

Proposition 1.5 Let A be an algebra. Derivations form a Lie subalgebrain EndK(A)[−].

Proof: Let c, d be two derivations. It suffices to check that a linear combi-nation αc+βd for some α, β ∈ K and the commutator [c, d] are both deriva-tions. We leave the linear combination to a reader and check the commuta-tor. Pick x, y ∈ A. Then [c, d](xy) = cd(xy) − dc(xy) = c(d(x)y + xd(y)) −d(c(x)y + xc(y)) = c(d(x))y + d(x)c(y) + c(x)d(y) + xc(d(y)) − d(c(x))y −c(x)d(y)−d(x)c(y)−xd(c(y)) = c(d(x))y−d(c(x))y+xc(d(y))−xd(c(y)) =[c, d](x)y + x[c, d](y). 2

We denote this Lie algebra of derivations by Der(A). The geometricnature of this Lie algebra are vector fields under their Lie bracket. To showthis in familiar surroundings, let U be a connected open subset in Rn. Let Abe the algebra of all smooth functions U → R. It is a commutative algebraunder pointwise multiplication and addition. It will require some analysis toshow that every derivation of A is a vector field (and we will avoid this) butthe reverse statement is clear. Indeed, every vector field d =

∑i φi(x)∂/∂xi

can be applied to functions so that d(Ψ(x)) =∑

i φi(x)Ψxi(x) where we use

superscripts to denote partial derivatives. The product rule for d followsfrom the familiar product rule for derivatives. It is instructive to compute[c, d] where c =

∑i θi(x)∂/∂xi:

[c, d] =∑

i

(∑

j

(θjφixj− φjθixj

))∂/∂xi

which can be directly checked by applying to a function Ψ.

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1.6 Vista: affine structures on Lie algebras

An algebra is called Lie-admissible if A[−] is a Lie algebra. It is a wideclass of algebras. We saw that it includes associative algebras but not onlythem. Let us define (a, b, c) = (ab)c − a(bc) for elements a, b, c ∈ A. Thiselement (a, b, c) (or this triple product A × A × A → A) is called an asso-ciator. It measure the failure of associativity, similarly as the commutatormeasure the failure of associativity. An algebra is called left symmetric ofthe associator is left symmetric, i.e., (a, b, c) = (b, a, c) for all a, b, c ∈ A. Itis easy to check that left symmetric algebras are Lie-admissible.

In fact, left symmetric are important in Geometry. An affine structure ona Lie algebra g is a left symmetric multiplication on g whose commutatoris the usual Lie multiplication. Algebraically, affine structure representsg = A[−] for a left symmetric algebra A. Geometrically, it corresponds toleft invariant affine connections on the Lie group. More to follow.

1.7 Exercises

Let A be a finite-dimensional algebra over complex numbers C. Provethat if f : A→ A is a derivation then ef = I+f +f2/2!+ . . . =

∑∞k=0 f

k/k!is an automorphism. Prove that if eαf : A→ A is an automorphism for eachα ∈ C then f is a derivation.

State and prove the first isomorphism theorem for arbitrary algebras.Prove the analogue of Proposition 1.1 for right derivations.Prove that [θ∂/∂x, φ∂/∂y] = θφx∂/∂y − φθy∂/∂x.

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2 Things that fail in positive characteris-

ticHow come so many new Lie algebras slipped into the classification theo-

rem? The reason is that many usual results in characteristic zero suddenlyfail in positive characteristic.

2.1 Complete Reducibility

Define gln = gln(K) = Mn(K)[−], sln its subalgebra of matrices with

zero trace. The standard basis of sl2 is e =

(0 10 0

), f =

(0 01 0

), h =

(1 00 −1

). The products in the standard basis are computed commuting

the matrices: e ∗ f = h, h ∗ f = −2f , h ∗ e = −2e. These three productscompletely determine the Lie algebra structure on sl2.

Proposition 2.1 If p ≥ 3 then sl2 is a simple Lie algebra.

Proof: Pick a nonzero element x and generate an ideal I = (x). It sufficesto show that I = sl2. Pick a non-zero element and multiply it by e twotimes. This gives you e ∈ I and now multiplying by f , you get I = sl2. 2

Let us now consider the polynomial algebra K[x, y]. We define the actionof sl2 on A by formulas

eφ(x, y) = x∂φ

∂y, hφ(x, y) = x

∂φ

∂x− y

∂φ

∂y, fφ(x, y) = y

∂φ

∂x

and then extended by linearity to an arbitrary element.

Proposition 2.2 Under these formulas A is a representation of sl2.

Proof: To conclude that A is a representation we must check that (a∗b)φ =a(bφ)−b(aφ) for all a,b ∈ sl2, φ ∈ A. Because of linearity it is sufficient todo only for the three basic products. We do it for one of them and leave theother two to the reader: h(eφ(x, y))−e(hφ(x, y)) = h(xφy)−e(xφx−yφy) =x(xφy)x − y(xφy)y − x(xφx)y + x(yφy)y = x(φy + xφyx) − xyφyy − x2φxy +x(φy + yφyy) = 2xφy = 2eφ. 2

It is worse pointing out that elements of sl2 act by differentiations andwe get injective homomorphism of Lie algebras sl2 → Der(A). We interpretthese facts geometrically: A are functions on the affine space K2, Der(A) arevector fields on K2, and the representation is a realization of sl2 as vectorfields on K2.

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Let An be the subspace of homogeneous polynomials of degree n, i.e., itis a span of all xn−kyk. As the actions of e, f and h preserve the degree,each An is a subrepresentation of A. We claim that Ap is not completelyreducible.

Let us recall that a representation U is completely reducible if for any sub-representation W ⊆ U there exists another subrepresentation V ⊆ U suchthat U = V ⊕W . A module with the same property is called semisimple.According to Weyl’s complete reducibility, a finite dimensional representa-tion of a finite dimensional simple Lie algebra over an algebraically closedof characteristic zero is completely reducible. We can see now that it failsin characteristic p > 2.

Let us check the claim. Let W be the span of xp and yp. This clearlysubmodule with the trivial action axp = 0 for any a ∈ sl2 because ∂xp/∂x =pxp−1 = 0. Suppose there is a complementary submodule V and pick non-zero φ ∈ V . We can write φ =

∑pk=n αkx

kyp−k with αn 6= 0 and 0 6= n 6= p.Then ep−n(φ) = αnx

p and xp ∈ V , contradiction.

2.2 Lie’s Theorem

Let us recall some standard material from Lie algebras to make thesenotes self-contained. If A is a linear subspace of a Lie algebra g, we defineits commutant [A,A] as the span of all products x ∗ y, x, y ∈ A. Thename comes from thinking of the Lie multiplication as commutation. Theninductively we define the derived series: A(0) = A, A(n+1) = [A(n), A(n)].

Proposition 2.3 If I is an ideal of g then I(n) is an ideal of g for each n.

Proof: We prove it by induction on n. The basis of induction (n = 0)is our assumption. For the induction step, we have to show that [J, J ] isan ideal whenever J is an ideal. It is clear by the product rule for Lx:x(ab) = (xa)b + a(xb) ∈ [J, J ] if x ∈ g, a,b ∈ J . 2

We say that an ideal I is n-soluble if I(n). An ideal is soluble if it isn-soluble for some n. The Lie algebra g is soluble if g is a soluble ideal.

Proposition 2.4 (i) If g is soluble then every subalgebra of g and everyquotient algebra of g is soluble.

(ii) If I is an ideal of g and both I and g/I are soluble then g is soluble.

Proof: The first statement is obvious. For the second statement, observethat the cosets of elements of g(n) are in (g/I)(n). This, if g/I is n-solublethen g(n) ⊆ I. Now if I is m-soluble then g(n+m) = (g(n))(m) ⊆ I(m) = 0. 2

The following proposition immediately follows.

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Proposition 2.5 A sum of two soluble ideals is soluble.

Proof: By the second isomorphism theorem, I + J/J ∼= I/I ∩ J is solubleas the quotient of I. Hence I + J is soluble by Proposition 2.4. 2

This allows to conclude that the radical Rad(g) of a finite-dimensionalLie algebra, defined as the sum of all soluble ideals, is itself soluble.

Lie’s theorem states that over an algebraically closed field of zero char-acteristic, a finite-dimensional irreducible representation of a finite dimen-sional soluble Lie algebra must be one-dimensional. This fails in positivecharacteristic.

Let us consider a two dimensional Lie algebra g with basis x, y. Themultiplication is x ∗ y = y. It is 2-soluble as g(1) = Ky and g(2) = 0. Nowlet us consider a vector space V with a basis ei, i ∈ F where F is a primesubfield with the action of g defined by xei = iei, yei = ei+1 and extendedby bilinearity.

Proposition 2.6 V is an irreducible representation of g of dimension p.

Proof: To conclude that V is a representation it suffices to follow throughthe following calculation x(yei)−y(xei) = xei+1−yiei = (i+1)ei+1−iei+1 =ei+1 = yei.

To show irreducibility consider a nonzero subrepresentation U ⊆ V andan element 0 6= z ∈ U . Write z =

∑p−1i=0 αiei. Then xz =

∑i iαiei ∈ U

and xkz =∑

i ikαiei ∈ U for each positive k. These conditions can be now

written in the matrix form as

1 1 1 · · · 1 10 1 2 · · · p− 2 p− 102 12 22 · · · (p− 2)2 (p− 1)2

......

... · · · ......

0p−1 1p−1 2p−1 · · · (p− 2)p−1 (p− 1)p−1

α0e0α1e1

...αp−1ep−1

∈ Up

Notice that the column has the coefficients in V while the matrix has thecoefficients in the field K. Hence the product apriori has the coefficients inV and the condition tells us that the coefficients are actually in U . Now thematrix is Vandermonde’s matrix and, in particular, is invertible. Multiplying

by its inverse gives

α0e0α1e1

...αp−1ep−1

∈ Up. We will refer to this consideration

later on as Vandermonde’s trick. As all αiei are in U and one of αi is

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nonzero, one of ei is in U . Applying y p − 1 times we get that all ei are inU . 2

2.3 Cartan’s Criteria

Recall that any finite dimensional algebra A admits Killing form K :A × A → K defined as K(x,y) = Tr(LxLy). Observe that it is a sym-metric bilinear form. In characteristic zero, there are two Cartan’s criteria.Semisimple criterion states that a Lie algebra is semisimple if and only if itsKilling form is non-degenerate, i.e., of maximal possible rank. The solublecriterion states that of K vanishes on g(1) then g is soluble. Both criteriafail in positive characteristic.

Let F be the finite subfield of K of size pn. The Witt algebra W (1, n)is a pn-dimensional vector space with basis ei, i ∈ F and multiplication byei ∗ ej = (i− j)ei+j .

Proposition 2.7 (i) W (1, n) is a Lie algebra.(ii) If p ≥ 3 then W (1, n) is simple.(iii) If p ≥ 5 then the Killing form on W (1, 1) is zero.

Proof: The anticommutativity is obvious because i − j = −(j − i). TheJacobi identity is verified directly (ei ∗ ej) ∗ ek + (ej ∗ ek) ∗ ei + (ek ∗ ei) ∗ej = ((i − j)(i + j − k) + (j − k)(j + k − i) + (k − i)(k + i − j))ei+j+k =(i2 − j2 − ik + jk + j2 − k2 − ji+ ki+ k2 − i2 − kj + ij)ei+j+k = 0.

To check simplicity, let us look at the ideal I generated by a non-zeroelement x =

∑i∈F

αiei =∑

i∈S αiei where S is the subset of all i such thatαi 6= 0. It suffices to show that I = g.

By the ideal property x ∗ e0 =∑

i∈S iαiei ∈ I. Applying Re0 severalmore times, we get Rke0

(x) =∑

i∈S ikαiei ∈ I. Using Vandermonde trick,

ei ∈ I for all i ∈ S.As soon as j 6= 2i for some i ∈ S, it follows that ej = (2i−j)−1ei∗ej−i ∈

I. If p > 2 , we get I = g in two iterations: ei ∈ I implies ej ∈ I for allj 6= 2i, hence at least for one more j 6= i. Then 2i 6= 2j and we concludethat e2i ∈ I.

Let us compute K(ei, ej). Since (ei ∗(ej ∗ek)) = (j−k)(i−j−k)ei+j+k,it is immediate that K(ei, ej) = 0 unless i = −j. Finally, K(ei, e−i) =∑

k∈F(−i − k)(2i − k) =

∑p−1k∈F

(−2i2 − ik + k2). This can computed usingthe usual formula

∑nt=0 t

2 = n(n + 1)(2n + 1)/6 if the characteristic is atleast 5: K(ei, e−i) = −2i2p− ip(p− 1)/2 + p(p+ 1)(2p + 1)/6 = 0. 2

Thus, W (1, 1) is a counterexample to both Cartan criteria. In fact, soare W (1, n), which is the first example of Cartan type Lie algebra.

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2.4 Vista: Semisimple Lie algebras

One usually defines a semisimple Lie algebra as a finite dimensional Liealgebra such that Rad(g) = 0. In characteristic zero, such an algebra is adirect sum of simple Lie algebras. It fails in characteristic p too.

Let A = K[z]/(zp) be the commutative algebra of truncated polynomi-als. In characteristic p this algebra admits the standard derivation ∂(zk) =kzk−1. One needs characteristic p for this to be well-defined: ∂ is always well-defined on K[z] turning K[z] into a representation of the one-dimensional Liealgebra. The ideal (zp) needs to be a subrepresentation. Since ∂(zp) = pzp−1

it happens only in characteristic p. Notice that the product rule for ∂ on Ais inherited from the product rule for ∂ on K[z].

Now extend ∂ to the derivation of the Lie algebra sl2(A): ∂

(f(z) g(z)h(z) −f(z)

)=

(f ′(z) g′(z)h′(z) −f ′(z)

). Finally, we define the Lie algebra g as a semidirect prod-

uct of a one dimensional Lie algebra ans sl2(A).Abstractly speaking, we say that a Lie algebra k acts on a Lie algebra h

if h is a representation of k and for each x ∈ k the action operator a 7→ xa isa derivation of k. The semidirect product is the Lie algebra g = k ⊕ h withthe product

(x, a) ∗ (y, b) = (x ∗ y, xb− ya+ a ∗ b).

Proposition 2.8 The semidirect product is a Lie algebra, h is an ideal, andthe quotient algebra g/his isomorphic to k.

Proof: 2

Hence g = Kd⊕sl2(A) is 3p+1-dimensional Lie algebra. Let us describeits properties.

Proposition 2.9 g admits a unique minimal ideal isomorphic to sl2(K).The commutant g(1) is equal to sl2(A).

Proof: 2

The first statement implies that g is semisimple. The second statementimplies that g is not a direct sum of simple Lie algebras.

2.5 Exercises

Prove that if the Killing of g is nondegenerate then the radical of g iszero.

Compute the Killing form for W (1, 1) in characteristic 3 (note it is non-degenerate there).

13

Show that in characteristic 2 all ei with i 6= 0 span a nontrivial ideal ofW (1, n).

Show that W (1, 1) is solvable in characteristic 2 and that its Killing formhas rank 1.

14

3 Free algebras3.1 Free algebras and varieties

Let X be a set. We define a free “thingy” (X) to be a free set witha binary operation without any axioms. Informally, it consists of all non-associative monomials of elements of X. To define it formally we can userecursion:

(basis) if a ∈ X then (a) ∈ (X),(step) if a, b ∈ X then (ab) ∈ (X).

We will routinely skip all unnecessary brackets. For instance, if X = {a, b},then the recursion basis will give us two elements of X: a and b. Using therecursion step for the first time gives 4 new elements: aa, ab, ba, bb. Using itfor the second time promises to give 36 new elements, but since 4 productsare already there, it gives only 32. We will not attempt to list all of themhere but we will list a few: a(aa), (aa)a, (aa)b, a(ab), etc.

The fact that this defines a set is standard but not-trivial: it is calledRecursion Theorem. Notice the difference with induction. Induction is anaxiom and it is used only to verify a statement. Recursion is a theorem andit is used to construct a set (or a function or a sequence). Have you heardof recursive sequences?

Now the binary operation on (X) is defined by the recursive step: a∗b =(ab) as in the recursion step. Let K(X) be the vector space formally spannedby (X). It consists of formal finite linear combinations

∑t∈(X) αtt, finite

means that αt = 0 except for finitely many elements t. The vector spaceK(X) is an algebra: the multiplication on (X) is extended by bilinearity.We call it the free algebra of X because of the following universal property.

Proposition 3.1 (Universal property of the free algebra) For each algebraA and a function f : X → A there exists a unique algebra homomorphismf : K(X) → A such that f(z) = f(z) for each z ∈ X.

Proof: We start by extending f to a function f : (X) → A preserving themultiplication:

(basis) if a ∈ X define f(a) = f(a),(step) if a, b ∈ X define f(ab) = f(a)f(b).

This is well-defined by recursion and it preserves the multiplication becauseof the recursive step. Any other such extension must also satisfy the recur-sion basis and steps. Hence, the extension is unique.

f is defined on the basis of K(X) and we extend it to f : K(X) → A bylinearity. Uniqueness is obvious. 2

15

We can formulate Proposition 3.1. Think ofX as a subset of K(X). Thenwe have the natural restriction function R : Hom(K(X), A) → Fun(X,A)where Fun(X,A) is just the set of all functions from X to A. Proposition 3.1essentially says that R is a bijection.

The elements of K(X) are nonassociative polynomials. They can beevaluated on any algebra. Formally let X = {x1, . . . xn}, w(x1, . . . xn) ∈K(X), A an algebra, a1, . . . an ∈ A. Consider a function f : X → A definedby f(xi) = ai. It is extended to a homomorphism f : K(X) → A byProposition 3.1. Finally, we define w(a1, . . . an) = f(w).

We say that w(x1, . . . xn) is an identity of A if w(a1, . . . an) = 0 for alla1, . . . an ∈ A. Let us write IdX(A) for the set of all identities of A in K(X).Now we say that an ideal I of K(X) is verbal if w(s1, . . . sn) ∈ I for all w ∈ I,s1, . . . sn ∈ K(X)

Proposition 3.2 The set IdX(A) is a verbal ideal for any algebra A.

Proof: Let us first check that it is a vector subspace. If v,w ∈ IdX(A),α, β ∈ K and f : X → A is an arbitrary function then f(αv + βw) =αf(v) + βf(w) = 0. Hence αv + βw is also an identity of A.

The ideal property is similar. If w ∈ IdX(A), v ∈ K(X) then f(vw) =f(v)f(w) = f(v)0 = 0. Hence vw is an identity. Similarly, wv is an identity.

The verbality is intuitively clear since of w(a1, . . . an) = 0 on A thenw(s1(a1, . . . an), . . . sn(a1, . . . an)) will also be zero. Formally the substitu-tion xi 7→ ai corresponds to a homomorphism f : K(X) → A while the sub-stitution xi 7→ si corresponds to a homomorphism g : K(X) → K(X) andw(s1(a1, . . . an), . . . sn(a1, . . . an)) = f(g(w)). Define h = f ◦ g : X → K(X).Since h(z) = f(g(z)) for each z ∈ X, we conclude that h = f ◦ g andf(g(w)) = h(w) = 0. 2

3.2 Varieties of algebras

A variety of algebras is a class of algebras where a certain set of identitiesis satisfied. We can choose enough variables so that the set of identities isa subset S of K(X). Let (S)v be the verbal ideal generated by S. One canformally define (S)v as the intersection of all verbal ideals containing S. Itfollows from Proposition 3.2 that (S)v consists of the identities in variablesof X satisfied in all the algebras of the variety. We have seen the followingvarieties already.

Ass is given by the identity x(yz) − (xy)z. It consists of all associativealgebras, not necessarily with identity.

16

Com consists of all commutative algebras in Ass. Its verbal ideal is gener-ated by x(yz) − (xy)z and xy − yx.

Lie is the variety of Lie algebras and the identities are xx x(yz) + y(zx) +z(xy).

n− Sol consists of all n-soluble Lie algebras. Its additional identity wnrequires 2n variables. We define it recursively: w1(x1, x2) = x1x2.Thus, 1-soluble Lie algebras are just algebras with zero multiplication.Then wn+1 = wn(x1, . . . x2n) ∗ wn(x2n+1, . . . x2n+1).

Let V be a variety. For each set X, the variety will have a verbal idealIV in K(X) of all identities in variables X that hold in V. We define the freealgebra in the variety V as KV < X >= K(X)/IV .

Proposition 3.3 (Universal property of the free algebra in a variety) Forevery algebra A in a variety V the natural (restriction) function Hom(KV <X >,A) → Fun(X,A) is a bijection.

Proof: Take a function f : X → A and consider its unique extensionhomomorphism f : K(X) → A. It vanishes on IV because elements of IVare identities of A while f(w) are results of substitutions. Hence, f : KV <X >→ A where f(w + IV) = f(w) is well-defined. This proves that therestriction is surjective.

The injectivity of the restriction is equivalent to the uniqueness of thisextension, which follows from the fact that X generates KV < X >. 2

3.3 Vista: basis in free Lie algebra and powers of represen-

tations

Although we have defined the free Lie algebra, it is a pure existencestatement. Fortunately, there are ways to put hands on the free Lie algebra,for instance, by constructing a basis. Contrast this with the free alternativealgebra: this is given by the verbal ideal generated by two identities x(xy)−(xx)y and x(yy)− (xy)y. No basis of the free alternative alternative algebraKAlt(X) is known explicitly if X has at least 4 elements.

The most well-known basis of the free Lie algebra is Hall basis but herewe construct Shirshov basis. More to follow.

3.4 Exercises

Let K(X)n be the space of homogeneous polynomials of degree n. Given|X| = m, compute the dimension of K(X)n.

17

4 Universal enveloping algebras4.1 Free associative algebra and tensor algebra

The free algebra KAss(X) has associative multiplication but no identityelement. This is why it is more conventional to introduce new algebra K <X >= K1 ⊕ KAss(X) with the obvious multiplication: (α1, a) ∗ (β1, b) =(αβ1, αb+βa+ab). This algebra K < X > will be called the free associativealgebra. Its basis is the free monoid (a set with associative binary andidentity) product < X >, generated by X.

Proposition 4.1 (The universal property of K < X >) For every associa-tive algebra A and every function f : X → A there exists a unique homo-morphism of algebras f : K < X >→ A such that f(x) = f(x) for eachx ∈ X.

Proof: The homomorphism of associative algebras takes 1 to 1 and theextension to f : KAss(X) → A exists and is unique by Proposition 3.3).Thus, f(α1, a) = α1A + f(a) is uniquely determined. 2

It is sometimes convenient to reinterpret the free associative algebra astensor algebra. Let V be a vector space. We define the tensor algebraTV as the free associative algebra K < B > where B is a basis of V .Notice that TV depends on B up to a canonical isomorphism. If C isanother basis, the change of basis matrix ci =

∑i αi,jbj gives a function

C → K < B >, ci 7→∑

i αi,jbj , which can be extended to the canonicalisomorphism K < C >→ K < B >.

We are paying a price here for not having defined tensor products ofvector spaces. It is acceptable for a student to use tensor products instead.

Notice that the natural embedding B → K < B > of sets gives a naturallinear embedding V → TV .

Proposition 4.2 (The universal property of TV ) For every associative al-gebra A the natural (restriction) function Hom(TV,A) → Lin(V,A) is abijection.

Proof: Let B be a basis of V . The basis property can be interpreted asthe fact the restriction function Lin(V,A) → Fun(B,V ) is bijective. Noweverything follows from Proposition 4.1. 2

4.2 Universal enveloping algebra

If the vector space V = g is a Lie algebra there is a special subsetHom(g, A) ⊆ Lin(g, A) consisting of Lie algebra homomorphisms. This mo-tivates the definition of a universal enveloping algebra. The universal en-veloping algebra of a Lie algebra g is Ug = Tg/I where I is generated by all

18

ab−ba−a∗b for a, b ∈ g. Notice that ab is the product in Tg while a∗b isthe product in g. Notice also that the natural linear map ω = ωg : g → Ug

defined by ω(a) = a + I is a Lie algebra homomorphism from g to Ug[−]:ω(a)∗ω(b) = (a+I)(b+I)−(b+I)(a+I) = ab−ba+I = a∗b+I = ω(a∗b).

Proposition 4.3 (The universal property of Ug) For every associative al-gebra A and every Lie algebra homomorphism f : g → A[−] there existsa unique homomorphism of associative algebras ϕ : Ug → A such thatϕ(ω(x)) = f(x) for each x ∈ g.

Proof: It follows from Proposition 4.2 and the fact that f(I) = 0 is equiv-alent to f being Lie algebra homomorphism for a linear map f . 2

Corollary 4.4 For a vector space V and a Lie algebra g there is a bi-jection between the set of Ug-module structures on V and the set of Ug-representation structures on V .

Proof: Ug-module structures on V are the same as associative algebra ho-momorphisms Hom(Ug,EndKV ) that are in bijection with Lie algebra ho-momorphisms Hom(Ug,EndKV

[−]) that, in their turn, are g-representationstructures on V . 2

Bijections in Corollary 4.4 leave the notion of a homomorphism of repre-sentations (modules) unchanged. In fact, it can be formulated as an equiv-alence of categories, which we will discuss later on. This seems to makethe notion of a representation of a Lie algebra obsolete. Why don’t we juststudy modules over associative algebras? A point in defense of Lie algebrasis that g is finite-dimensional and easy to get hold of while Ug is usuallyinfinite-dimensional.

The following proposition follows immediately from the Poincare-Birkhoff-Witt Theorem that will not be proved in this lectures.

Proposition 4.5 ωg is injective

4.3 Free Lie algebra

Let L(X) be the Lie subalgebra of KAss(X)[−] generated by

Theorem 4.6 (i) The natural map φ : KLie(X) → KAss(X) gives anisomorphism of Lie algebras between KLie(X) and L(X).

(ii) The natural map ψ : UKLie(X) → K < X > is an isomorphism ofassociative algebras.

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(iii) The following diagram is commutative.

UKLie(X)ψ−−−−→ K < X >

xωKLie(X)

xi

KLie(X)φ−−−−→ L(X)

Proof: Let us first construct both maps. Using the function f : X →KAss(X) and interpreting KAss(X) as a Lie algebra KAss(X)[−] gives a Liealgebra homomorphism φ : KLie(X) → KAss(X)[−] by the universal propertyof the free Lie algebra. Since KLie(X) is generated by X as a Lie algebra,the image of φ is exactly L(X). Hence we have a surjective Lie algebrahomomorphism φ : KLie(X) → L(X).

Now L(X) is a Lie subalgebra in K < X >[−], thus φ gives a Lie algebrahomomorphism KLie(X) → K < X >[−] that can be lifted to a homo-morphism ψ : UKLie(X) → K < X > of associative algebras. Since bothalgebras are generated by X, the homomorphism ψ is surjective.

The commutativity of the diagram follows from the construction becauseψ(ω(z)) = φ(z) for each z ∈ X. Consequently, it should be true for anyelement z ∈ KLie(X) as both ψ◦ω and φ are homomorphisms of Lie algebras.

The inverse of ψ is constructed from the universal property of K < X >:let θ be the extension of function X → UKLie(X) to a homomorphismK < X >→ UKLie(X). The composition θψ is an identity on X, henceidentity on UKLie(X). Similarly, the composition ψθ is an identity on X,hence identity on K < X >.

Finally, ω is injective by Proposition 4.5. Hence, ψω = φ is injective. 2

Now we get a good grasp on the free Lie algebra: it is just L(X) and itsuniversal enveloping algebra is K < X >.

4.4 Vista: Baker-Campbell-Hausdorff formula

4.5 Exercises

State and prove the universal property of the free commutative algebra.Prove the universal enveloping algebra of an abelian Lie algebra is iso-

morphic to the free commutative algebra of any basis.

20

5 p-th powersWe discuss various p-th powers and introduce restricted structures on

Lie algebras. The field K has characteristic p > 0 throughout the lecture.

5.1 In commutative algebra

Proposition 5.1 (Freshman’s dream binomial formula) Let A be an asso-ciative algebra and x, y ∈ A such that xy = yx. Then (x+ y)p = xp + yp.

Proof: Since x and y commute, they satisfy the binomial formula (x+y)p =∑p

n=0

(pn

)xnyp−n. It remains to observe that for any n between 1 and

p − 1 the binomial coefficient

(pn

)= p!/n!(p − n)! vanish since only the

denominator is divisible by p. 2

Corollary 5.2 If T is a matrix with coefficients in K then Tr(T p) = Tr(T )p.

Proof: If λi are eigenvalues of T then λpi are eigenvalues of T p. Hence,Tr(T )p = (

∑i λi)

p =∑

i λpi = Tr(T p). 2

Corollary 5.3 Let A be an associative algebra and x, y ∈ A such that xy =yx. Then (x− y)p−1 =

∑p−1n=0 x

nyp−1−n.

Proof: Let us check this identity in the free com-algebra KCom(X) whereX = {a, b}. There (a − b)p = ap − bp = (a − b)

∑p−1n=0 a

nbp−1−n. Since

KCom(X) is a domain, we can cancel (a−b) proving (a−b)p−1 =∑p−1

n=0 anbp−1−n.

Since x, y commute the function f : X → A, f(a) = x, f(b) = y can beextended to a homomorphism f . Hence, (x − y)p−1 = f((a − b)p−1) =f(

∑p−1n=0 a

nbp−1−n) =∑p−1

n=0 xnyp−1−n. 2

Now look at a Lie algebra g = sln. It is a Lie subalgebra of two differentassociative algebras, U(g) and Mn(K). Let us denote the p-th power inthe former algebra by xp and in the latter algebra by x[p]. Observe that inMn(K), Tr(x[p]) = Tr(x)p as follows from (5.1 ).

5.2 In associative algebra

Let us prove that

Theorem 5.4 If y, z ∈ X then (z + y)p − zp − yp ∈ K < X > belongs toL(X) and contains homogeneous components of degree at least 2.

Proof: We consider polynomials K < X > [T ] in one variable T overK < X >. One can differentiate them using the derivation ∂ : K < X >[T ] → K < X > [T ] where ∂(fT n) = nfT n−1, f ∈ K < X >. One can also

21

evaluate them at α ∈ K using homomorphism evα : K < X > [T ] → K <X >, evα(fT

n) = αnf .Before going further we observe that the associativity that operators

Ls and Rs commute for any s ∈ K < X > [T ]. Hence, (Ls − Rs)p−1 =∑p−1

n=0 LnsR

p−1−ns by Corollary 5.3.

Now we write (zT + y)p = zpT p + yp +∑p−1

n=1 Fn(z, y)Tn for some

Fn(z, y) ∈ K < X >. Differentiating this equality we get∑p−1

n=1 nFn(z, y)Tn−1 =∑p−1

n=0(zT+y)nz(zT+y)p−n−1 =∑p−1

n=0 LnzT+y(R

p−nzT+y(z)) = (LzT+y−RzT+y)

p−1(z) =

[zT + y, . . . [zT + y, z] . . .]. Comparing coefficients at T n−1 we express eachFn(z, y) as a sum of commutators, hence Fn(z, y) ∈ L(X) with homogeneouscomponents of degree 2 and higher.

Finally, using ev1, we get (z+ y)p− zp− yp =∑p−1

n=1 Fn(z, y) ∈ L(X). 2

This polynomial Λp(z, y) = (z+y)p−zp−yp plays a crucial role in whatfollows. Since it is a Lie polynomial it can be evaluated on Lie algebras.

5.3 In Lie algebra

A restricted Lie algebra is a pair (g, γ) where g is a Lie algebra andγ : g → g is a function (often denoted γ(x) = x[p] such that

(i) γ(αx) = αpγ(x),(ii) Lγ(x) = Lpx,(iii) γ(x + y) − γ(x) − γ(y) = Λp(x,y)

for all α ∈ K and x,y ∈ g.The following proposition gives us a tool to produce examples of re-

stricted Lie algebras.

Proposition 5.5 Let g be a Lie subalgebra of A[−] where is an associativealgebra. If xp ∈ g for all x ∈ g then g admits a structure of a restricted Liealgebra.

Proof: The restricted structure is given by the associative p-th power:γ(x) = xp. Axiom (i) is obvious. Axiom (iii) follows from the fact that itis the p-th power in an associative algebra: Λp(z, y) = (z + y)p − zp − yp ∈K < y, z >, hence, in any associative algebra.

To prove axiom (ii), we distinguish Lie multiplication operators by adding∗. Then L∗

z = Lz−Rz, i.e., the Lie multiplication is the difference of two asso-ciative multiplications for any z ∈ A. By associativity Lz and Rz commute,hence we can use Proposition 5.1: (L∗

z)p = (Lz−Rz)p = Lpz−Rpz = Lzp−Rzp .

2

Corollary 5.6 If A is an algebra then Der(A) is a restricted Lie algebra.

22

Proof: Thanks to Proposition 5.5, it suffices to check that ∂p is a deriva-tion whenever ∂ is a derivation. Using induction on n one establishes that

∂n(xy) =∑n

k=0

(nk

)∂k(x)∂(y)n−k. If n = p all the middle terms vanish,

hence ∂p is a derivation. 2

Corollary 5.7 Classical Lie algebras sln, sp2n and son are restricted.

Proof: It suffices to check closeness under p-th power thanks to Proposi-tion 5.5. For sln, it follows from Corollary 5.2. Symplectic or orthogonalmatrices satisfy the condition XJ = −JX where J is the matrix of theform. Clearly, XpJ = −Xp−1JX = . . . = (−1)pJXp = −JXp 2

A homomorphism between Lie algebras is called restricted if it commuteswith γ. An ideal or a subalgebra is restricted if they are closed under γ. Arepresentation V is restricted if x[p]v = x(vx(. . . xv)) where v ∈ V and x ∈ g

appears p times. A direct sum of restricted Lie algebras is the direct sum ofLie algebras with the direct sum of restricted structures of the componentsas the restricted structure.

5.4 Vista: restricted Lie algebroids

5.5 Exercises

Prove by induction on n that ∂n(xy) =∑n

k=0

(nk

)∂k(x)∂(y)n−k.

Prove that the kernel of a restricted homomorphism is a restricted ideal.Formulate and prove the isomorphism theorem for restricted Lie alge-

bras.

23

6 Uniqueness of restricted structuresWe develop more efficient methods for working and constructing re-

stricted structures.

6.1 Uniqueness of restricted structures

Recall the centre of a Lie algebra g consists of all x ∈ g such that x∗y = 0for all y ∈ g. The centre of a Lie algebra is an ideal.

A function f : V →W between vector spaces over the field K of positivecharacteristic p is semilinear if f is an abelian group homomorphism andf(αa) = αpa. The following proposition addresses uniqueness.

Proposition 6.1 If γ and δ are two distinct restricted structures on g thenδ − γ is a semilinear map g → Z(g). On the opposite, if γ is a restrictedstructure and φ : g → Z(g) is a semilinear map then γ + φ is a restrictedstructure too.

Proof: Let φ = δ − γ where γ is a restricted structure. It is clear thatφ(αx) = αpφ(x) if and only if δ(αx) = αpδ(x).

Then φ(x)∗y = (δ(x)−γ(x))∗y = Lpx(y)−Lpx(y) = 0, hence φ(x) ∈ Z(g).In the opposite direction, if φ : g → Z(g) is any function then γ+φ satisfiesaxiom (ii) of the restricted structure by the same argument.

Finally, φ(x+y)−φ(x)−φ(y) = Λp(x,y)−Λp(x,y) = 0. In the oppositedirection, semilinearity of φ implies axioms (iii) for δ. 2

We have proved that the restricted structures form an affine space overthe vector space of all semilinear maps from g to Z(g). In particular, ifZ(g) = 0 then the restricted structure is unique if it exists. To the existencewe turn our attention now.

6.2 Restricted abelian Lie algebras

Let us consider an extreme case Z(g) = g, i.e. the Lie algebra is abelian.Since Λp has no degree 1 terms, the zero map 0(x) = 0 is a restrictedstructure. By Proposition 6.1, any semilinear map γ : g → g is a restrictedstructure and vice versa. We consider the following abelian restricted Liealgebras: a0 is a one-dimensional Lie algebra with a basis x and x[p] = x;

an is an n-dimensional Lie algebra with a basis x1, . . .xn and x[p]k = xk+1,

x[p]n = 0. The following theorem is left without a proof in the course.

Theorem 6.2 Let g be a finite-dimensional abelian restricted Lie algebraover an algebraically closed field K. Then g is a direct sum of an. Themultiplicity of each an is uniquely determined by g.

24

While uniqueness in Theorem 6.2 is relatively straightforward, the exis-tence is similar to existence of Jordan normal forms (proofs will be eitherelementary and pointless or conceptual and abstract). One way you canthink of this is in terms of matrices: on a basis a semilinear map γ is de-fined by a matrix C = (γi,j) so that on the level of coordinate columnsγ(αi) = (γi,j)(α

pj ). The change of basis Q will change C into F (Q)CQ−1

where F (βi,j) = (βpi,j) is the Frobenius homomorphism F : GLn → GLn,which is an isomorphism of groups in this case. Hence, we are just classifyingorbits of this action, very much like in Jordan forms.

One interesting consequences of the theorem is that all nondegeneratematrices form a single orbit: its canonical representative is identity matrixI. The orbit map L : GLn → GLn, L(Q) = F (Q)IQ−1 = F (Q)Q−1 issurjective. This map is called Lang map and its surjectivity is called Lang’stheorem.

The conceptual proof is in terms of the modules over the algebra oftwisted polynomials: A = K[z] but z and scalars do not commute: αz = zαp.The module structure on V comes from linear action of K and the action ofz by the semilinear map. The algebra A is non-commutative left Euclideandomain and the theorem follows from the classification of indecomposablefinite-dimensional modules over A.

6.3 PBW-Theorem and reduced enveloping algebra

After careful consideration, I have decided to include the PBW-theoremwithout a proof. My original plan was to spend two lectures to give a com-plete proof but then I decided against as a partial proof (modulo Grobner-Shirshov’s basis) has featured in Representation Theory last year.

Theorem 6.3 (PBW: Poincare, Birkhoff, Witt) Let X be a basis of a Liealgebra g over any field K. If we equip X with a linear order ≤ (we say thatx < y if x ≤ y and x 6= y) then elements x

k11 x

k22 . . .xkn

n , with ki ∈ {1, 2 . . .}and x1 < x2 < . . . < xn ∈ X form a basis of Ug.

Now we are ready to establish the following technical proposition. Recallthat the centre of an associative algebra A is a subalgebra Z(A) = {a ∈A | ∀b ∈ A ab = ba}.

Proposition 6.4 Let the field be of characteristic p. Let xi (as i runs overlinearly ordered sets) form a basis of g. Suppose for each i we have anelement zi ∈ Z(Ug) such that zi − x

pi ∈ g + K1 ⊆ Ug. Then the elements

zt11 zt22 . . . ztnn xk11 xk22 . . .xkn

n , with ki ∈ {0, 1, 2 . . . , p − 1}, ti ∈ {0, 1, 2 . . . , },and x1 < x2 < . . . < xn ∈ X form a basis of Ug.

25

Proof: We refer to the PBW-basis elements as PBW-terms and to the newbasis elements as new terms. To show that the new terms span Ug it sufficesto express each PBW-term x

s11 x

s22 . . .xsn

n , si ≥ 0 as a linear combination ofthe new terms. We do it by induction on the degree S =

∑i si. If all si

are less than p then the PBW-term is a new term. This gives a basis ofinduction. For the induction step we can assume that si ≥ p for some i. Letyi = x

pi − zi. Then

xs11 x

s22 . . . xsn

n = xs11 . . .xsi−p

i yixsi+1

i+1 . . .xsnn + zix

s11 . . .xsi−p

i xsi+1

i+1 . . .xsnn

The second summand is a new term and the first summand has degreeS + 1 − p < S, i.e. it is a linear combination of PBW-terms of smallerdegrees. Using induction to the terms of the first summand completes theproof.

To show the linear independence we consider Um, the linear span of allPBW-terms of degree less or equal thanm. If a new term zt11 z

t22 . . . ztnn xk11 xk22 . . . xkn

n ,has degree m+ 1 =

∑i(pti + ki) then

zt11 . . . ztnn xk11 . . .xknn = xk11 (xp1 − y1)

t1 . . . xknn (xpn − yn)

tn

and, consequently,

zt11 zt22 . . . ztnn x

k11 x

k22 . . .xkn

n + Um = xpt1+k11 . . .xptn+kn

n + Um .

Note that this gives a bijection f between the new terms and PBW-terms.Consider a non-trivial linear relation on new terms with highest degree m+1.It gives a non-trivial relation on cosets of Um. This is a relation on cosets ofthe corresponding PBW-terms but this is impossible: PBW-terms of degreem+ 1 are linearly independent modulo Um by PBW-theorem. 2

We can interpret this proposition in several ways. Clearly, it implies thatA = K[z1 . . . zn] is a central polynomial subalgebra of Ug. Moreover, Ug isa free module over A with basis xk11 . . .xkn

n , with ki ∈ {0, 1, 2 . . . , p− 1}. Inparticular, if the dimension of g is n then the rank of Ug as an A-moduleis pn. Another interesting consequence is that elements zi generate an idealI which is spanned by all zt11 . . . ztnn ,x

k11 . . . xkn

n with at least one positiveti. Hence the quotient algebra U ′ = Ug/I has a basis xk11 . . .xkn

n ,+I withki ∈ {0, 1, 2 . . . , p− 1}.

The quotient algebra U ′ is called the reduced enveloping algebra and weare going to put some gloss over it slightly later.

26

6.4 Vista: modules over skew polynomial algebra

6.5 Exercises

Prove uniqueness part in Theorem 6.2. (Hint: kernels of semilinear mapsare subspaces)

Prove Theorem 6.2 for one dimensional Lie algebras.Using PBW-theorem, prove that the homomorphism ω : g → Ug is

injective.Using PBW-theorem, prove that the universal enveloping algebra Ug is

an integral domain.

27

7 Existence of restricted structures7.1 Existence of restricted structures

We are ready to prove the following important theorem.

Proposition 7.1 Let g be a Lie algebra with a basis xi. Suppose for eachi there exists an element yi ∈ g such that Lpxi = Lyi

. Then there exists a

unique restricted structure such that x[p]i = yi.

Proof: If γ and δ are two such structure then φ = γ − δ is a semilinearmap such that φ(xi) = yi−yi = 0 for all i. Thus, φ = 0 and the uniquenessis established.

To establish existence, observe that zi = xpi − yi ∈ Z(g). The reason is

that the left Lie multiplication operator is Lzi= Lx

pi−Lyi

== Lpxi−Lyi

= 0.

This allows to construct the reduced enveloping algebra U ′ and g is its Liesubalgebra. It remains to observe that the p-th powers in U ′ behave in theprescribed way: (xi + I)p = x

pi + I = yi + I on the basis. Off the basis, we

need to observe that xp ∈ g if x ∈ g to apply Proposition 5.5. It follows byinduction on the length of a linear combination using the standard formula(αx + βy)p = αpxp + βpyp + Λp(αx, βy). 2

Let us describe the restricted structure onW (1, n): Lpei(ej) = −Rpei

(ej) =

−Rp−1ei ((j − i)ei+j) = −Rp−2

ei ((j − i)je2i+j) = . . . = −∏p−1t=0 (j + ti)ej =

−f(j/i)ipej . The polynomial f(z) =∏p−1t=0 (z + t) has p distinct roots,

all elements of the prime subfield. Elements of the prime subfield satisfyzp − z = 0. Comparing the top degree coefficients, f(z) = zp − z andLpei

(ej) = −(jp− jip−1)ej . If n = 1 then i and j lie in the prime subfield. If

i 6= 0 then ip−1 = 1 and jp − jip−1 = jp − j = 0 and e[p]i = 0. If i = 0 then

jp − jip−1 = jp = j and e[p]0 = e0.

Now if n ≥ 2 then consider i = 0 and j 6= 0. As jp − jip−1 = jp it

forces e[p]0 = jp−1e0. It depends on j and is not well-defined. Thus, W (1, n)

admits no restricted structure in this case.

7.2 Glossing reduced enveloping algebra up

Let (g, γ) be a restricted Lie algebra. It follows that x 7→ xp − x[p] is asemilinear map g → Z(Ug). Its image generates a polynomial subalgebraof the centre of Ug which we call the p-centre from now one and denoteZp(Ug).

Pick a basis xi of g. Now for each χ ∈ g∗ we define zi = xpi−x

[p]i −χ(xi)

p1.Finally, we define reduced enveloping algebra Uχg the algebra U ′ for thisparticular choice of basis. Clearly, it is just the quotient of Ug by the idealgenerated by xp − γ(x) − χ(x)p1 for all x ∈ g.

28

Proposition 7.2 (Universal property of reduced enveloping algebra) For ev-ery associative algebra A and every Lie algebra homomorphism f : g → A[−]

such that f(x)p − f(γ(x)) = χ(x)p1 for all x ∈ g there exists a unique ho-momorphism of associative algebra f : Uχg → A such that f(x) = f(x) forall x ∈ g.

We say that a representation V of g admits a p-character χ ∈ g∗ ifxp(v) − x[p](v) = χ(x)pv for all v ∈ V .

Corollary 7.3 For a vector space V there is a natural bijection betweenstructures of a representation of g with a p-character χ and structures of aUχg-module.

One particular important p-character is zero. Representations with zerop-character are called restricted while the reduced enveloping algebra U0g iscalled the restricted enveloping algebra.

Example. Clearly, Ua0∼= Ua1

∼= K[z], making their representa-tions identical. However, Uχa0 and Uχa1 are drastically different. Uχa0

∼=K[z]/(zp−z−χ(z)p1) and the polynomial zp−z−χ(z)p1 has p distinct rootsbecause (zp− z−χ(z)p1)′ = −1. Hence Uχa0

∼= Kp is a semisimple algebra.Consequently, representations of Uχa0 with any p-character are completelyreducible.

On the other hand, Uχa1∼= K[z]/(zp − χ(z)p1) and the polynomial zp−

χ(z)p1 = (z − χ(z)1)p has one multiple root. Hence Uχa1∼= K[z]/(zp) is a

local algebra. Consequently, there is a single irreducible representation ofUχa1 with any given p-character.

7.3 exercises

Prove Proposition 7.2 and Corollary 7.2.Choose a basis of sln and describe the restricted structure on the basis.

29

8 Schemes8.1 Schemes, points and functions

A scheme X is a commutative algebra A. To be precise we are talkingabout proaffine schemes. To be even more precise, “the category of schemesis the opposite category to the category of commutative algebras”. Thismeans that if a scheme X is an algebra A and Y is an algebra B thenmorphisms from X to Y are algebra homomorphisms from B to A. Toemphasize this distinction between schemes and algebras we say that X isthe spectrum of A and write X = Spec A. We are going to study geometryof schemes.

We will regard algebra A as functions on the scheme X = Spec A.Functions can be evaluated at points and the scheme has various notions ofa point. We will just use only one: a point is an algebra homomorphismA → R to another commutative algebra R. If R is fixed we call them R-points and denote them X (R). Notice that a morphism of schemes φ : X →Y gives rise to a function on points φ(R) : X (R) → Y(R) via compositionφ(R)(a) = a ◦ φ.

We have to learn how to compute the value of a function F ∈ A at a pointa ∈ X (R). Well, it is just F (a) = a(F ) as a : A → R is a homomorphismand F is an element of A.

As an example, let us consider A = K[X,Y ]. We can regard the corre-sponding scheme as “an affine plane”. What are its R-points? A homomor-phism a is given by any two elements a(X), a(Y ) ∈ R. Hence, X (R) = R2.

Let us consider “a circle” X = Spec K[X,Y ]/(X2 + Y 2 − 1). Indeed,what are the R-points? To define a homomorphism a(X), a(Y ) ∈ R mustsatisfy a(X)2 + a(Y )2 = 1. Hence, X (R) = {(x, y) ∈ R2 | x2 + y2 = 1}.8.2 Product of schemes

A product of schemes X = Spec A and Y = Spec B is the spectrumof the tensor product of algebras A and B, i.e., X × Y = Spec A ⊗ B.The tensor product of algebras is the standard tensor product of vectorsspaces with the product defined by a⊗ b · a′ ⊗ b′ = aa′ ⊗ bb′. Observe thatX × Y(R) = X (R) × Y(R) for each commutative algebra R.

8.3 Algebraic varieties and local schemes

The circle and the affine plane are very special schemes. Their algebrasare finitely generated domains (with an exception of the circle in character-istic 2, which is no longer a domain). If A is a finitely generated domain,the scheme is called an affine algebraic variety.

We are more interested in local schemes, i.e., spectra of local rings. Recall

30

that a commutative ring A is local if it accepts an ideal I such that everyelement a ∈ A\ I is invertible. One particular scheme of interest is the localaffine line A1,n. Let us emphasize that the characteristic of K is p. Thealgebra A has a basis X(k), 0 ≤ k ≤ pn − 1 and the product

X(k)X(m) =

(m+ kk

)X(m+k).

We think that X(k) = 0, if k ≥ pn.As a scheme, A1,n = (A1,1)n. All we need is to construct an algebra

isomorphism between K[Y1, . . . Yn]/(Ypi ) and A. It is given by φ(Yi) = X(pi).

Hence points A1,n(R) are just collections (r1, . . . rn) ∈ Rn such that rpi = 0for all i. However, we want to distinguish A1,n and (A1,1)n by equippingthem with different PD-structure.

8.4 PD-schemes

Let R be a commutative ring, I it ideal. A PD-structure on I is asequence of functions γn : I → A such that

[i] γ0(x) = 1 for all x ∈ I,[ii] γ1(x) = x for all x ∈ I,[iii] γn(x) ∈ I for all x ∈ I,[iv] γn(ax) = anγn(x) for all x ∈ I, a ∈ A,[v] γn(x+ y) =

∑ni=0 γi(x)γn−i(y) for all x, y ∈ I,

[vi] γn(x)γm(x) =

(n+mn

)γn+m(x) for all x, y ∈ I,

[vii] γn(γm(x)) = An,mγnm(x) where An,m = (nm)!/(m!)nn! ∈ K forall x ∈ I,

The number An,m needs to be integer for the scalar An,m ∈ K to make

sense. It follows from the formula An,m =∏n−1j=1

(jm+m− 1

jm

).

A PD-structure on a scheme X = Spec A is an ideal I of A and a PD-structure on I. Let us describe some basic properties of it. We think thatx0 = 1 for any x ∈ A including zero.

Proposition 8.1 [i] γn(0A) = 0A if n > 0.[ii] n!γn(x) = xn for all x ∈ I and n ≥ 0.

Proof: Statement [i] follows from axiom [iv]. Statement [ii] follows fromaxioms [i], [ii] and [vi] by induction. See lecture notes for details. 2

Statement [ii] has profound consequences in all characteristics. In charac-teristic zero it implies that each ideal has a canonical DP -structure γn(x) =

31

xn/n!. Observe how the axioms and the intuition agrees with this structure.In characteristic p it implies that xp = p!γp(x) = 0 for each x ∈ I. Thus, itis a necessary condition for an ideal I to admit a p-structure.

8.5 Local affine n-space

Let φ : {1, . . . n} → Z≥1 be a function. We introduce the local affinespace An,φ including its PD-structure. It is the spectrum of an algebraAφ whose vector space basis consists of commutative monomials X(a) =

X(a1)1 . . . X

(an)n which we will write in multi-index notation. The multi-index

entries vary in the region determined by φ: 0 ≤ ai < pφ(i) and we will thinkthat X(a) = 0 if one of entries falls outside this region.

The multiplication is defined similarly to the space A1,n above:

X(k)i X

(m)i =

(m+ kk

)X

(m+k)i

while the different variables are multiplied by concatenation (don’t forgetcommutativity). The element X(0,0,...0) is the identity of Aφ. If n = 1,we just write An for Aφ. In the general case, Aφ is isomorphic to thetensor product Aφ(1) ⊗ . . . ⊗ Aφ(n). Hence, on the level of schemes An,φ ∼=∏ni=1 A1,φ(i).Finally, we describe the PD-structure. The ideal I is the unique max-

imal ideal. It is generated by all non-identity monomials X(a). Finally,

γn(X(1)i ) = X

(n)i determines the PD-structure. Indeed,

γn(X(m)i ) = γn(γm(X

(1)i )) = An,mX

(nm)i

from axiom [vii]. Axiom [iv] allows us to extend this to multiples of mono-mials γn(λX

(a)) = λn(n!)sX(na) where n(ai) = (nai) and s+1 is the numberof nonzero elements among ai. Finally, axiom [v] extends γn to any elementof I.

One should ask why this is well-defined. Rephrasing this question, wecan wonder why the repeated use of the axioms on a element x ∈ I willalways produce the same result for γn(x). This will not be addressed in thelecture notes (maybe in the next vista section).

8.6 Vista: divided powers symmetric algebra of a vector

space

8.7 Exercises

Prove that An,m =∏n−1j=1

(jm+m− 1

jm

).

32

9 Differential geometry of schemes9.1 Tangent bundle and vector fields on a scheme

For a scheme X = Spec A we discuss its tangent bundle TX . It is ascheme but it is easier to describe its points than its functions.

As far as points are concerned TX (R) = X (R[z]/(z2)). A homomor-phism φ : A→ R[z]/(z2) can be written as φ(a) = x(a)+τ(a)z using a pair oflinear maps x, τ : A → R. The homomorphism condition φ(ab) = φ(a)φ(b)gets rewritten as x(ab) + τ(ab)z = (x(a) + τ(a)z)(x(b) + τ(b)z) or as a pairof conditions

x(ab) = x(a)x(b), τ(ab) = x(a)τ(b) + x(b)τ(a).

The former condition means x ∈ X (R), i.e., it is a point where the tangentvector is attached. The latter condition is a relative (to x) derivation con-dition. All such elements belong to a vector space Derx(A,R) which we canthink of as the tangent space TxX .

To see that the tangent bundle is a scheme we have to understand dif-ferentials. At this point we judge fudge them up. Associated to an algebraA we consider a new commutative algebra A generated by elements a andda for each a ∈ A. We are going to put three kinds of relations on A. Thefirst kind just says that A → A defined by a 7→ a is a homomorphism ofalgebras. The second kind says that A → A defined by a 7→ da is a linearmap. Finally, we add relations d(ab) = adb+ bda for all a, b ∈ A.

Proposition 9.1 If X = Spec A then we can define TX as Spec A.

Proof: We need to come up with a “natural” bijection between Hom(A,R[z]/(z2))and Hom(A,R). Naturality means that the bijection agrees with homomor-phisms R→ in a sense that the following diagram commutes:

Hom(A,R[z]/(z2)) −−−−→ Hom(A,R)y

y

Hom(A,S[z]/(z2)) −−−−→ Hom(A, S)

We describe the bijections and leave necessary details (well-definedness,inverse bijections, naturality) to the student. An element x+τz ∈ Hom(A,R[z]/(z2))defines φ ∈ Hom(A, R) by φ(a) = x(a) and φ(da) = τ(a). In the op-posite direction φ ∈ Hom(A, R) defines q ∈ Hom(A,R[z]/(z2)) by q(a) =φ(a) + φ(da)z. 2

33

Finally derivations ∂ ∈ Der(A) can be thought of as global vector fieldsbecause they define give sections of tangent sheaf. We leave a proof of thefollowing proposition as an exercise.

Proposition 9.2 If x ∈ X (R), ∂ ∈ Der(A) then x ◦ ∂ ∈ Derx(A,R).

9.2 PD-derivations

We already saw that derivations form a restricted Lie algebra. Now wesay that ∂ is a PD-derivation if ∂(γn(x)) = γn−1(x)∂(x) for all x ∈ I. Thisdefinition is motivated by the following calculation using the chain rule orrepeated product rule: ∂(fn/n!) = ∂(f)fn−1/(n − 1)!. It also show that incharacteristic zero every derivation is a PD-derivation.

It is no longer true in characteristic p even if the PD-structure is trivial.Say A = K[z]/(zp). The PD-structure is γk(f) = fk/k! for k < p andγk(f) = 0 for k ≥ p. Then ∂ = ∂/∂z is not a PD-derivation as it marginallyfails the definition: ∂(γp(x)) = 0 6= γp−1(x)∂(x). Let PD − Der(A, I, γ) bethe set of all PD-derivations.

Proposition 9.3 PD − Der(A, I, γ) is a Lie algebra.

Proof: For the Lie algebra (∂ ∗ δ)(γn(x)) = ∂(δ(γn(x))) − δ(∂(γn(x))) =∂(γn−1(x)δ(x)) − δ(γn−1(x)∂(x)) = γn−1(x)∂(δ(x)) + γn−2(x)∂(x)δ(x) −γn−1(x)δ(∂(x)) − γn−2(x)∂(x)δ(x) = γn−1(x)(∂ ∗ δ)(x). 2

Notice that PD − Der(A) is not restricted, in general. One can show that∂p(γn(x)) = γn−1(x)∂

p(x) + γn−p(x)∂(x)p The best one can conclude fromthis that is that PD-derivations preserving I, i.e ∂(I) ⊆ I form a restrictedLie algebra.

9.3 Exercises

Prove that if A is finitely generated then A is finitely generated.Prove Proposition 9.2.Compute PD − Der(K[z]/(zp), (z), γ).

34

10 Generalised Witt algebra10.1 Definition

We define a derivation ∂i of Aφ by ∂i(X(k)i ) = X

(k−1)i . On a general

monomial ∂i(X(a)) = (X(a−ǫi) where ǫi is the multi-index with the only

nonzero entry 1 in position i. A special derivation is a derivation of Aφ ofthe form

∑ni=1 fi∂i for some fi ∈ Aφ. We define the generalised Witt algebra

as W (n, φ) as the Lie algebra of special derivations of Aφ.We start by stating some basic properties.

Proposition 10.1 (i) W (n, φ) is a free Aφ-module with a basis ∂i, i =1, . . . n.

(ii) W (n, φ) is a Lie subalgebra of Der(Aφ).(iii) W (n, φ) is a restricted Lie algebra if and only if φ(i) = 1 for all i.

Proof: (i) By definition∑n

i=1 fi∂i(Xi) = fi. Hence∑n

i=1 fi∂i = 0 if andonly if all fi are zero.

(ii) It follows from the standard commutations formula f∂i ∗ g∂j =f∂i(g)∂j − g∂j(f)∂i.

(iii) If all φ(i) = 0 then W (n, φ) = Der(Aφ) is restricted. In the oppositedirection, suppose φ(i) ≥ 2 for some i. Observe that L∂i

(F∂j) = ∂i(F )∂j .

Suppose W (n, φ) is restricted. Then Lp∂i= Lx for x = ∂

[p]i =

∑j fj∂j .

Hence, Lp∂i(X

(p)i ∂i) = ∂i while, on the other hand, Lx(X

(p)i ∂i) = (

∑j fj∂j) ∗

X(p)i ∂i = fiX

(p−1)i ∂i −

∑j X

(p)i ∂i(fj)∂j . The resulting equation

∂i = fiX(p−1)i ∂i −

∑

j

X(p)i ∂i(fj)∂j

contains a contradiction as the degrees of the monomials in the right handside are at least p− 1 ≥ 1 while it is zero in the left hand side. 2

It follows that the dimension of W (n, φ) is n · dim(A) = npP

φ(t). Tounderstand the Lie multiplication we write the standard formula

X(n) ∂

∂X∗X(m) ∂

∂X= (X(n)X(m−1) −X(n−1)X(m))

∂

∂X=

(

(n+m− 1

n

)−

(n+m− 1n− 1

))∂

∂X

Let us clear the connection between special and PD-derivations. The spe-cial derivation ∂i is not a PD-derivation as ∂i(γpφ(i)(Xi)) = 0 6= γpφ(i)−1(Xi)∂i(Xi).

Proposition 10.2 Every PD-derivation is special.

35

Proof: Every element δ ∈ PD − Der(A1,n) is determined by n elementsδ(Xi). Indeed, δ(X(a)) =

∑iX

(a−ǫi)δ(Xi). Hence δ =∑

i δ(Xi)∂i is special.2

10.2 Gradings

A graded algebra is an algebra A together with a direct sum decompo-sition A = ⊕n∈ZAn such that An ∗ Am ⊆ An+m. In a graded algebra wedefiance subspaces A≤n = ⊕k≤nAk and A≥n = ⊕k≥nAk. Let us observesome obvious facts. A0 is a subalgebra. If n ≥ 0 then A≥n is a subalgebraand an ideal of A≥0.

A graded Lie algebra is a Lie algebra and a graded algebra. Observe thatin a graded Lie gi is a representation of g0.

The algebra Aφ is a graded commutative algebra. We define the degreeof X(a) as |a| =

∑i |ai|. Now we extend this grading to W (n, φ) by defining

the degree of ∂i to be −1. Let |φ| =∑

t(|φ(t)| − 1).

Proposition 10.3 (i) W (n, φ) = ⊕|φ|−1k=−1W (n, φ) is a graded Lie algebra.

(ii) W (n, φ)0 ∼= gln.(iii) W (n, φ)−1 is the standard representation of gln. In particular, it is

irreducible.(iv) W (n, φ)k = W (n, φ)−1 ∗W (n, φ)k+1 for every k 6= |φ| − 1.(v) W (n, φ)|φ|−1 = W (n, φ)0 ∗W (n, φ)|φ|−1 unless p = 2 and n = 1

Proof: (i) follows from the standard Lie bracket formula..(ii) The isomorphism is given by Xi∂j 7→ Ei,j.(iii) gln has two standard representations coming from the actions on

rows or columns. For the column representation A · v = Av where A is amatrix, v is a column. For the row representation A · v = −AT v where A isa matrix, v is a column. It follows from the standard representation formulathat W (n, φ)−1 is the standard row-representation of gln.

(iv) Since ∂i ∗X(a)∂j = X(a−ǫi)∂j , the statement is obvious as you canincrease one of the indices.

(v) In the top degree |φ|−1 we consider the multi-index a = (φ(1)−, . . . φ(n)−1). The elements X(a)∂i span the top degree space and Xi∂i ∗ X(a)∂j =XiX

(a−ǫi)∂j − δi,jX(a)∂i = (pφ(i) − 1)X(a)∂j − δi,jX

(a)∂i. It is now obviousif p > 2 and we leave the case of p = 2 as an exercise. 2

10.3 Simplicity

Theorem 10.4 If p > 2 or n > 1 then W (n, φ) is a simple Lie algebra.

Proof: Pick a non-zero ideal I � W (n, φ) and a non-zero element x ∈ I.Multiplying by various ∂i, elements of degree -1, we get a nonzero element

36

y ∈ I ∩ g−1.Since W (n, φ)−1 is an irreducible W (n, φ)0-representation, we, multiply-

ing by elements of degree 0, conclude that g−1 ⊆ I.Using part [iv] of proposition 10.3, we get gk ⊆ I for each k < |φ| − 1.

Finally, g|φ|−1| = g0 ∗ g|φ|−1| ⊆ I by part [v] of proposition 10.3. 2

10.4 Exercises

Finish the proof of part (v) of Proposition 10.3 in characteristic 2.Demonstrate a non-trivial ideal of W (1, φ) in characteristic 2.Let p > 2. Describe W (n, φ)1 as a representation of W (n, φ)0. Is it

irreducible?

37

11 Filtrations11.1 Consequences of two descriptions

We intend to establish an isomorphism W (1, φ) ∗ W (1, φ(1)) betweenthe generalised Witt algebra and the Witt algebra. This isomorphism isnon-trivial. Here we discuss four consequences of this isomorphism.

First, W (1, φ) is defined over the prime field F(p) whileW (1, k) is definedover the field F(pk) of cardinality pk. The field of definition is the field wherethe multiplication coefficients µki,j belong. The multiplication coefficients

appear when you write the product on a basis: ei ∗ ej =∑

k µki,jek.

Second, W (1, φ) is graded while there is no obvious grading on W (1, k).To be more precise, the natural grading on W (1, k) (eα ∈ W (1, k)α) is agrading by the additive group of the field F(pk), not by Z.

Third, in characteristic zero there is an analogue of both Lie algebrasand the proof of isomorphism is nearly obvious. One sets it up by φ(ei) =−Xi+1∂. While this map works for some elements in characteristic p it isnot useful, in general. The product ei ∗ ej is never zero unless i = j whileX(a)∂ ∗X(b)∂ = 0 if a+ b > pφ(1).

Finally, these two realisations will lead to two completely different Car-tan decompositions. Recall that a Cartan subalgebra is a soluble subalgebrawhich coincides with its normaliser. The normaliser of a subalgebra l ≤ g

consists of all x ∈ g such that x ∗ l ⊆ l. A Cartan subalgebra l leads toCartan decomposition, which is essentially considering g as a representationof l.

In a simple Lie algebra in characteristic zero, a Cartan subalgebra isabelian and unique up to an automorphism. It is no longer the case in char-acteristic p. For the Witt algebra realisation W (1, k) we see one-dimensionalCartan subalgebra Ke0. From the generalised Witt algebra realisationW (1, φ)we see another Cartan subalgebra ⊕p|nW (1, φ)n. Notice that these coincidefor W (1, 1).

11.2 Filtrations

Gradings on an algebra are very helpful because of the way they structuremultiplication. However, they may be difficult to find. For instance, W (1, k)has no obvious grading as far as we can see. The second best thing after agrading is a filtration.

We fix ε ∈ {1,−1}. A filtration on an algebra A is a collection of vectorsubspaces FnA, n ∈ Z. It has to satisfy two properties: FnA∗FmA ⊆ Fn+mAand FnA ⊆ Fn+εA for all n,m ∈ Z. The filtration is ascending if ε = 1. Itis descending if ε = −1.

38

Every graded algebra has two associated filtrations: ascending FnA =⊕k≤nAk and descending FnA = ⊕k≥nAk. In the opposite direction, froma filtered algebra A we can construct an associated graded algebra grA =⊕nAn/An−ε with multiplication defined by

(a+An−ε)(b+Am−ε) = ab+An+m−ε.

Ascending filtrations are more common. Given generators xi of an alge-bra A, we define F−1A = 0 and F0A as either 0 or K1, the latter in case ofassociative or commutative algebras. For a positive n we define FnA as thelinear span of all products (any order of brackets) of k generators, for allk ≤ n. This filtration is particularly useful for the universal enveloping alge-bra Ug. Its associated graded algebra is the symmetric algebra Sg which canbe defined as the universal enveloping algebra of g with zero multiplication.This fact follows from the PBW-theorem and we leave it as exercise.

Let us explain a method of obtaining a descending filtration on a Liealgebra. Let g be a Lie algebra, l its subalgebra. We define vector subspacesFng recursively:

F−1g = g, F0g = l, Fn+1g = {x ∈ Fng | x ∗ g ⊆ Fng}.For example, if l is an ideal then all positive Fng are equal to l.

Proposition 11.1 Subspaces Fng form a descending filtration and ∩Fng isan ideal of g.

Proof: The descending property is clear. The multiplicative property of thefiltration is observe by induction. As an induction basis, F−1g∗Fng ⊆ Fn−1g

be the definition of Fng and F0g∗F0g ⊆ F0g because F0g = l is a subalgebra.Now we assume that we have proved Fig ∗ Fjg ⊆ Fi+jg for i + j ≤ k andconsider the case of i + j = k. Using Jacobi’s identity, (Fig ∗ Fjg) ∗ g =(Fig ∗ g) ∗ Fjg +Fig ∗ (Fjg ∗ g) = Fi−1g ∗ Fjg +Fig ∗ Fj−1g ⊆ Fi+j−1g. Thisimplies that Fig ∗ Fjg ⊆ Fi+jg.

Finally, g ∗ ∩jFjg ⊆ ∩jFj−1g ⊆ ∩jFjg. 2

11.3 Filtration on Witt algebra

To conclude we exhibit a useful subalgebra of W (1, k) to construct afiltration on.

Proposition 11.2 l = {∑i αiei ∈W (1, k) | ∑i αi = 0} is a Lie a subalge-

bra of W (1, k).

Proof: Let∑

i αi = 0 =∑

i βi. Then (∑

i αiei) ∗ (∑

i βiei) =∑

i,j αiβj(i−j)ei+j belongs to l because

∑i,j αiβj(i−j) =

∑i αii

∑j βj−

∑j βjj

∑i αi =

0. 2

39

11.4 Exercises

Prove that the normaliser of a subalgebra (in a Lie algebra) is a subal-gebra.

Prove that the normaliser of a subalgebra a (in a Lie algebra) is theunique maximal subalgebra that contains a as an ideal.

Prove that Ke0 and ⊕p|nW (1, φ)n are both Cartan subalgebras.What is the dimension of ⊕p|nW (1, φ)n? When is it non-abelian?Let ei be a basis of a Lie algebra g. Prove that gr(Ug) is isomorphic to

the (commutative) polynomial algebra K[ei] in variables ei.Let g = ⊕kgk be a graded Lie algebra. We consider the standard filtra-

tion FnU given by g as generators of Ug. Choosing a basis of homogeneouselements of g and using PBW, we can see that Ug is graded. We denote Units n-th graded piece. Let F ′

nU be the span of all FiU ∩Uj for all 2i+ j ≤ n.Prove that this is an ascending filtration1. Prove that if ei is a homoge-neous basis of g then gr(Ug) is isomorphic to the (commutative) polynomialalgebra K[ei] in variables ei.

1called Kazhdan filtration

40

12 Witt algebras are generalised Witt al-

gebra12.1 Graded algebras of certain filtered Lie algebras

Theorem 12.1 Let g be a simple Lie algebra of dimension pk with a subal-gebra of codimension 1. Using filtration of Proposition 11.1, the associatedgrade algebra gr(g) is isomorphic to W (1, φ) where φ(1) = k.

Proof: We start with two general observations. First, ∩nFng = 0 since g

is simple.Second, the codimension of Fn+1g in Fng is 1 unless both spaces are

zero. If the codimension is zero then Fng = Fn+1g = Fn+2g = . . . andFng = ∩kFkg is an ideal. Hence, Fng = Fn+1g = 0. To prove that thecodimension is always less than 2, suppose the contrary. Pick the smallestn when it is at least 2. For each k = −1, 0, . . . n− 1 choose ak ∈ Fkg so thatak = ak+Fk+1g forms a basis of Fkg/Fk+1g. When we reach n we choose atleast two elements b1,b2, . . . ∈ Fng so that bj = bj+Fn+1g forms a basis ofFng/Fn+1g. In the graded algebra a−1 ∗bj = λjan−1, which is equivalent toa−1 ∗bj ∈ λjan−1 +Fng in g itself. Notice that λj 6= 0 because λj = 0 wouldimply bj ∈ Fn+1g and bj = 0 contradicting the basis property. Finally,a−1 ∗ (λibj − λjbi) = 0 implying as just above that λibj − λjbi ∈ Fn+1g,contradiction.

Since the codimension is always 1 unless things degenerate into zero,we can choose choose a−1 ∈ g so that a−1 = a−1 + F0g forms a basis ofF−1g/F0g. Also for the top degree N = |φ| − 1 we choose nonzero aN ∈Fng. With this choices, we define all the elements in the middle recursively:an = a−1 ∗ an+1. Notice that an 6= 0 because if an ∈ Fn+1g then Fn+1g isan ideal that contradicts simplicity of g.

When we reach the bottom degree the element is already predefinedthere. Nevertheless, we can conclude that λa−1 + x0 = a−1 ∗ a0 for somex0 ∈ F0g. Without loss of generality λ = 1 because we can replace aN byλ−1aN . In the associated graded algebra, ai ∗ aj = αi,jai+j and we cancompute the coefficients αi,j recursively:

α−1,j = 1 = −αj,−1, αi,j = αi−1,j + αi,j−1.

The second formula follows from αi,jai+j−1 = a−1 ∗ αi,jai+j = (a−1 ∗ ai) ∗aj + ai ∗ (a−1 ∗ aj) = ai−1 ∗ aj + ai ∗ aj−1. By induction, it follows that

αi,j =

(i+ j + 1i+ 1

)−

(i+ j + 1

i

)

41

and ak 7→ X(k+1)∂ is an isomorphism since X(n)∂ ∗X(m)∂ = (X(n)X(m−1)−X(n−1)X(m))∂ = (

(n+m− 1

n

)−

(n+m− 1n− 1

))X(n+m−1)∂. 2

We have used the dimension assumption only at the end. In fact, it is notrequired. One can use the properties of binomial coefficients to fiddle theargument at the end to show a stronger statement that any finite dimensionalsimple Lie algebra with subalgebra of codimension 1 and the associatedfiltration has gr(g) isomorphic to W (1, φ) for some φ.

A more interesting question is whether we can change a−1 to force b0 =0. This would lead to an even stronger statement that g ∼= W (1, φ). Whilethe answer to this question is positive we are in position to tackle it here. Itinvolves subtleties of the deformation theory2 of W (1, φ). In characteristiczero, a simple Lie algebra is rigid, i.e., all its deformations are isomorphicto itself. W (1, φ) is no longer rigid but its non-trivial deformations are nolonger simple.

12.2 Filtration properties of the Witt algebra

We are ready to describe the filtration on g = W (1, k) described inProposition 11.1,

Proposition 12.2 If n ≥ 0 then Fng = {∑i αiei ∈W (1, k) |∀k ∈ {0, . . . , n} ∑i αii

k =0}.Proof: We go by induction on n. The basis of the induction is the definitionof l.

Suppose we have proved it for n−1. Consider an element x =∑

i αiei ∈Fn−1g. By induction assumption,

∑i αii

k = 0 for k < n. Now x ∗ et =∑i αi(i − t)ei+t. Clearly, x ∈ Fng if and only if x ∗ et ∈ Fn−1g for all t if

and only if∑

i αi(i − t)(i + t)k = 0 for all k < n and t. Now observe that∑i αi(i− t)(i+ t)n−1 =

∑i αii

n while∑

i αi(i− t)(i+ t)k = 0 for k ≤ n− 2by induction assumption. 2

12.3 Main theorem

Let F(pk) be the field of pk elements.

Lemma 12.3 If 0 < m < pk− 1 then∑

α∈F(pk) αm = 0. If m = pk− 1 then∑

α∈F(pk) αm = −1.

Proof: Let d be the greatest common divisor of m and pk − 1. Writem = dm′. Since the multiplicative group of F(pk) is the cyclic group3 of order

2never mind, it is not examinable3In general, any finite subgroup A of the multiplicative group of any field is finite as

you have learnt in Algebra-II. This follows from the unique factorisation property of the

42

pk−1, a 7→ am′

is a permutation of the field and∑

α∈F(pk) αm =

∑α∈F(pk) α

d.

Now let pk − 1 = dq. The map a 7→ ad is an endomorphism of themultiplicative group whose image consists of all the elements with ordersdividing q. Calling this image A we get

∑α∈F(pk) α

d = d∑

α∈A α.

Observe that A is a cyclic group of order q. If m = pk − 1 then q = 1,d = pk − 1 and we get the second statement. If 0 < m < pk − 1 then the qelements of A are the roots of zq − 1. Hence, zq − 1 =

∏α∈A(z − α) and,

consequently,∑

α∈A α = 0 as the coefficients at zq−1 is zero. 2

Theorem 12.4 Let n = 1 and φ(1) = k. Then W (1, φ) ∼= W (1, k).

Proof: (partial, k = 1 only) Lemma 12.3 and Proposition 12.2 give us anon-zero element in the top of the filtration: apk−2 =

∑i ei. It remains

to choose a−1 and I can do this painlessly only if k = 1. Let a−1 = −e1.Observe that by induction

apk−2−t = a−1 ∗ apk−1−t =∑

i

(−1)t(1 − i)(0 − i)(−1 − i) . . . (2 − t− i)ei+t

for non-negative grades. If k = 1 each consecutive element has one less termas zeros will appear in the product which will give a−1 ∗ a0 = (−1)p−1(p −1)!e1 = a−1 because (p− 1)! = −1 by Wilson’s theorem4 2

12.4 Exercises

Prove that∏α∈F(pk) α = −1.

polynomials as every z − α, α ∈ A divides zm − 1 where m is the exponent of A. Thus,the exponent A must be equal to the order of A that characterises cyclic groups.

4Observe that the product of all elements in an abelian group is equal to the productof elements of order 2 as the rest of the elements can be paired up with their inverses.Now the multiplicative group of F(p) is cyclic and has only one element −1 of order 2.

43

13 Differentials on a schemeWe already know how to do vector fields on a scheme and we are about

to learn how to do differentials.

13.1 Calculus

We would like to consider a module M over a commutative algebra Aas a geometric object over the scheme Spec (A). The correct geometricnotion is the one of quasicoherent sheaf but we would like it have morestructure. We define a calculus on a scheme Spec (A) as a pair (M,dM )where M is an A-module M and dM : A → M is a linear map such thatdM (ab) = adM b+ bdMa.

As a primary example, let us consider the universal calculus. As an A-module, C(A) is generated by symbols da for all a ∈ A. There are two typesof relations:

d(αa+βb) = αda+βdb and d(ab) = ad(b)+bd(a) for all a, b ∈ A,α, β ∈ K.

The first type of relations makes d linear and the second type enforces theproduct rule. Since these relations hold in any calculus, we can immedi-ately observe the universal property of the universal calculus. For any othercalculus (M,dM ) there exists a unique homomorphism φ : C(A) → M ofA-modules such that φ(da) = dM (a) for each a ∈ A. This property justifieswriting da rather than dC(A)a while working with universal calculus.

For example:, consider A = K[z]/[z2]. If p 6= 2, zdz = d(z2)/2 = 0 andC(A) is one-dimensional vector space with a basis dz. If p = 2, zdz 6= 0as it does not formally follows from the relations and the universal calculusis a two-dimensional vector space, a free A-module with a basis dz. It issignificant as the freeness of the universal calculus is a sign of smoothness.We will not discuss it here but Spec A, indeed, becomes very special incharacteristic 2 as this becomes a group scheme.

We will now give a less formal view of the universal calculus. The mul-tiplication µ : A× A → A is a bilinear map and as such it uniquely definesa linear map from the tensor product µ : A⊗A→ A, which we can denoteby the same letter. Notice that A ⊗ A is a commutative algebra (undera ⊗ b · a′ ⊗ b′ = aa′ ⊗ bb′) and µ is a an algebra homomorphism. Hence,its kernel I is an ideal. It has two distinct A-modules structures givenby multiplication in the left and right factors: a ⋆ (b ⊗ c) = ab ⊗ c anda ∗ (b⊗ c) = b⊗ ac.

We define M = I/I2 with one of these structures and dM : A → I/I2

by dM (f) = 1 ⊗ f − f ⊗ 1 + I2.

44

Proposition 13.1 Both A-module structures on I/I2 are the same. Themap dM defines a calculus structure on M , isomorphic to the universalcalculus C(A).

Proof: Let us check the first statement: a ∗ x − a ⋆ x = (1 ⊗ a − a ⊗ 1)xbelongs to I2 if x ∈ I. Hence a ∗ (x+ I2) = a ⋆ (x+ I2) if x ∈ I.

Now we define linear maps φ : M → C(A), φ(∑

i ai⊗ bi+ I2) =∑

i aidbiand ψ : C(A) → M , ψ(

∑i aidbi) =

∑i(ai ⊗ bi − aibi ⊗ 1) + I2. The first

task is to establish that these are well-defined.To show that φ is well-defined we need to prove that φ(x) = 0 for each x ∈

I2. Such x is a linear combination of (∑

i ai⊗bi)(∑

j cj⊗dj) with∑

i aibi =∑j cjdj = 0. Hence φ(x) is a linear combination of

∑i,j aicjd(bidj) =∑

i,j aibicjd(dj) +∑

i,j aicjdjd(bi) = 0.To show that ψ is well-defined it is necessary and sufficient to check the

axiom of calculus on M . Indeed, ψ defines a map on the free module withgenerators da. Well-definedness is equivalent to vanishing on the submoduleof relations. The first relation is just linearity and it is obvious. The secondrelation is equivalent to the product rule for dM and needs verification:dM (ab) = 1⊗ ab− ab⊗ 1+ I2 = a · (1⊗ b− b⊗ 1)+ b ∗ (1⊗ a− a⊗ 1)+ I2 =adM (b) + bdM (a).

Thus, we know that M is a calculus and it remains to establish that φand ψ are inverse bijections: ψ(φ(

∑i ai⊗bi+I2)) = ψ(

∑i aidbi) =

∑i(ai⊗

bi−aibi⊗1)+I2 =∑

i ai⊗bi+I2 since∑

i aibi = 0. In the opposite direction,φ(ψ(

∑i aidbi)) = φ(

∑i(ai⊗bi−aibi⊗1)+I2) =

∑i(aidM (bi)−aibidM (1)) =∑

i(aidM (bi) since dM (1) = 0. 2

Again in the example of A = K[z]/[z2], I is two dimensional with a basis1⊗ z− z⊗ 1 and z⊗ z. If p 6= 2 then I2 contain z⊗ z = −(1⊗ z− z⊗ 1)2/2and I/I2 is one dimensional with basis 1⊗ z − z ⊗ 1 + I2 = dM (z). On theother hand, if p = 2 then I2 = 0 and M = I is two-dimensional.

In analysis you may have got used to the fact that differentials are linearfunctionals on tangent spaces. The same is true here. Each ω = gdf ∈ C(A)defines a linear map ωx = g(x)dxf : TxX → R where x is a point of Spec (A)over R. As a tangent vector τ ∈ TxX is a derivation τ : A → R, we candefine ωx(τ) = g(x)τ(f). We leave it to the reader to check that this iswell-defined.

13.2 Differential calculus

We would like to see how the universal calculus behaves with respectto derivations. We consider the Lie algebra g = Der(A) and its naturalrepresentation on A. It is instructive to work with the second realization ofthe universal calculus in the next proposition.

45

Proposition 13.2 The universal calculus C(A) is a representation of Der(A).The differential d : A → C(A) is a homomorphism of representations. Theproduct rule ∂(aω) = ∂(a)ω + a∂(ω) holds for all ∂ ∈ Der(A), a ∈ A,ω ∈ C(A).

Proof: A tensor product of representations is a representation. This definesa representation structure on A⊗A via ∂(a⊗b) = ∂(a)⊗b+a⊗∂(b). To seethat C(A) is a representation it suffices to establish that both I and I2 aresubrepresentations. If x =

∑i ai ⊗ bi ∈ I then ∂(x) =

∑i(∂(ai) ⊗ bi + ai ⊗

∂(bi)). Observe that ∂(x) ∈ I because µ(∂(x)) =∑

i(∂(ai)bi + ai∂(bi)) =∂(

∑i aibi) = 0. Finally, because of the product rule ∂(I2) ⊆ I∂(I) ⊆ I2.

Since ∂(1) = 0, the second statement is straightforward: ∂(da) = ∂(1⊗a− a⊗ 1 + I2) = 1 ⊗ ∂(a) − ∂(a) ⊗ 1 + I2 = d(∂a).

Finally, ∂(a · (∑i bi ⊗ ci + I2)) = ∂(∑

i abi⊗ ci + I2) =∑

i(∂(a)bi ⊗ ci +a∂(bi)⊗ ci+abi⊗∂(ci))+ I2 = ∂(a) · (∑i bi⊗ ci+ I2)+a ·∂(

∑i bi⊗ ci+ I2).

2

Such situation should be really called a differential calculus, i.e., a dif-ferential calculus is a triple (M,d, g) where (M,d) is a calculus, g is a Liesubalgebra of Der(A) and the statements of Proposition 13.2 hold. Whilewe use this terminology, it is definitely non-standard. Thus, we will notexamine the usage of this terminology.

13.3 PD-calculus

This short section is of little relevance because the special derivationsare not PD-derivations. They come quite close, so this section comes quiteclose to defining the necessary differential calculus. Hence, we think it isinstructive to spell out the necessary structures.

A calculus (M,dM ) on a PD-scheme Spec (A, I, γn) is called a PD-calculus if dM (γn(x)) = γn−1(x)dMx for all x ∈ I and n.

One can turn a calculus (M,dM ) into a PD-calculus by imposing theserelation, i.e. let N be the A-submodule of M generated by all dM (γn(x))−γn−1(x)dMx for all x ∈ I and n. We the new calculus M = M/N withd

fM(x) = dM (x) +N . Clearly, it is a PD-calculus.

The universal PD-calculus C(A) is the quotient of C(A). It satisfies theuniversal property among all PD-calculi.

As an example, consider C(A) for A = K[z]/[z2] in characteristic 2. ThePD-condition forces 0 = d(γ2(z)) = zd(z) making C(A) one-dimensionalvector space with a basis d(z), similarly to the universal calculus in oddcharacteristic.

We leave the following proposition as an exercise and we will not use it

46

later.

Proposition 13.3 If A is an algebra with a PD-structure then (C(A),d,PD − Der(A)is a differential calculus.

13.4 Exercises

Prove that in a calculus dM (α1) = 0 for any α ∈ K.Prove that, given a calculus on an algebra A, the set of all x ∈ A such

that dM (x) = 0 form a subalgebra (constants of the calculus) of A0. Provethat dM is a homomorphism of A0-modules.

Consider A = K[X] and a finite-dimensional A-module M . Show thatfor each m ∈ M the assignment dM (X) = m can be uniquely extended toa calculus. Prove that constants depend only on the generalised minimalpolynomial µX,m(Z) of X (as it acts on M) and describe the constantsexplicitly.

Prove that if A is a finitely generated algebra then C(A) is finitely gen-erated A-module.

Prove Proposition 13.3.

47

14 Lie algebras of Cartan typeWe define what it means for a Lie algebra to be of Cartan type and state

their classification.

14.1 Differential forms

Given a commutative algebra A and an A-module M , we construct itsexternal powers ∧nAM as an A-module in two steps. As a first step, thetensor power T nAM is the quotient of the tensor power vector space T n

KM =

M⊗M⊗ . . .M by the vector subspace V spanned by tensors x1⊗ . . . xk−1⊗(axk⊗xk+1−xk⊗axk+1)⊗xk+2⊗. . . xn for all a ∈ A, xi ∈M . Clearly, T nAMhas a canonical A-module structure as the A-action on each factor producesthe same result. At the second step ∧nAM is the quotient of T nAM by theA-submodule, generated by all u⊗x⊗ v⊗x⊗w+V where x ∈M , u, v andw are elements of various tensor products. Notice that if the characteristicsis not 2, the latter submodule is generated by all x1⊗ . . . xk−1⊗(xk⊗xk+1+xk ⊗ xk+1) ⊗ xk+2 ⊗ . . . xn + V .

We denote x1 ∧ . . . ∧ xn = x1 ⊗ . . . ⊗ xn +W where W is the kernel ofthe natural linear map T n

KM → ∧nAM .

In the case of calculus (M,dM ) the elements of ∧nAM should be referredas n-forms. You may have seen them in differential geometry and analysis.

Proposition 14.1 If (M,dM , g) is a differential calculus then each ∧nAMis an g-module.

Proof: By routine check that ∂ω ∈W when ω ∈W . Consult your lecturenotes. 2

14.2 Special differential calculus

We consider the PD-algebra Aφ where φ : {1, 2 . . . n} → Z≥1. It is arepresentation of the Witt algebra g = W (n, φ). We construct a specialdifferential calculus to go along with it. As a module M = ⊕n

j=1AφdXj isa free Aφ-module with a basis dXj . The differential is given by the usualformula dM (f) =

∑j ∂j(f)dXj using the basis of the special derivations.

Finally, the action of W (n, φ) is given by the standard formula: ∂(fdXj) =∂(f)dXj + fdM(∂(Xj)).

Proposition 14.2 With the maps defined above, (M,dM ,W (n, φ)) is a dif-ferential calculus.

Proof: This routine check is left as an exercise. The following formula isuseful: g∂i(fdXj) = g∂i(f)dXj + δi,j

∑k f∂k(g)dXk . 2

48

14.3 Contact, symplectic and volume forms and certain graded

Lie algebras

In differential geometry, various forms play various important roles andusually have names as certain “structures” on manifolds. In particular, apointwise non-vanishing n-form ω on an n-dimensional manifold X is a vol-ume form. Indeed, it locally records the volume and allows one to integratefunctions by

∫X f(X) =

∫X fω. Here we are more in special n-forms on Aφ

and by analogy with the the differential geometry we call

ωS = dX1 ∧ dXk . . . ∧ dXn ∈ ∧nAφM

the volume form. It may be worse pointing out that in differential geometryevery volume form looks like this in a certain choice of local coordinates. Itis an easy (but non-examinable) exercise.

A pointwise non-vanishing 1-form ω on an n-dimensional manifold X is acontact form if it satisfies a certain “integrability condition”, which amountsto the vector fields in the kernel of this form comprising a Lie subalgebrainside all vector fields. You can observe this form and appreciate its nameby riding an bicycle (non-examinable too). Observe that while the rotationof the front wheel is unrestricted, its velocity lies the direction of the wheelat any given moment. This says that the velocity (tangent vector) lies in thekernel of the contact structure. Contact structures play significant role incontrol theory and thermodynamics but I doubt that this will be discussedin undergraduate courses on these topics. Contact structures exist only onodd dimensional spaces, so we define

ωK = dXn +

t∑

j=1

(XjdXj+t −Xj+tdXj) ∈M

where n = 2t + 1. Again in differential geometry every contact form lookslike this in a certain choice of local coordinates. It is a non-trivial resultcalled Darboux’s Theorem.

Finally, a non-degenerate closed 2-form ω on an n-dimensional manifoldX is a symplectic form. They are very similar to contact forms. Closenessplays role of “the integrability condition”. Non-degeneracy (the local skew-symmetric matrix has rank n) replaces non-vanishing. Symplectic formsfirst appear in Hamiltonian approach to classical mechanics but becameubiquitous throughout modern mathematics. Symplectic structures existonly on even dimensional spaces, so we define

ωH =

t∑

j=1

dXj ∧ dXj+t ∈ ∧2AφM

49

where n = 2t. Again in differential geometry every symplectic form looks likethis in a certain choice of local coordinates, which is again called Darboux’sTheorem.

Using these forms we define a series of new Lie algebras, which are ofCartan type, although we have not defined what it means. Cartan hasstudied these Lie algebras in differential geometry where they remain simplebut become infinite-dimensional:

S(n, φ) = {∂ ∈W (n, φ) | ∂(ωS) = 0},

H(n, φ) = {∂ ∈W (n, φ) | ∂(ωH) = 0},K(n, φ) = {∂ ∈W (n, φ) | ∂(ωK) ∈ AφωK},CS(n, φ) = {∂ ∈W (n, φ) | ∂(ωS) ∈ KωS},CH(n, φ) = {∂ ∈W (n, φ) | ∂(ωH) ∈ KωH}.

While it is easy to observe that all five are Lie subalgebras of the Wittalgebra, the following theorem is more involved. While we have all thenecessary tools to prove it, we do not have time, so it is given without aproof.

Theorem 14.3 Algebras S(n, φ), H(n, φ), K(n, φ), CS(n, φ), CH(n, φ)are graded Lie algebras. Of p ≥ 3 then Lie algebras S(n, φ), H(n, φ), K(n, φ)are simple.

Maybe, it is worse pointing out that the grading on K(n, φ) is slightlyunusual. The remaining 4 algebras inherit their grading for W (n, φ) butin K(n, φ) we need to assign degree 2 to Xn and −2 to ∂n. See also theexercises below.

The Lie algebra S(n, φ) is called the special Lie algebra, H(n, φ) thehamiltonian Lie algebra and K(n, φ) the contact Lie algebra.

14.4 Lie algebras of Cartan type

We have constructed several Lie algebra of Cartan type but there aremore. The following definition is rather technical but it has been adoptedthroughout the classification. A finite-dimensional simple Lie algebra g isa Lie algebra of Cartan type if it admits a descending filtration Fng suchthat the associated graded Lie algebra gr(g) is a transitive graded Lie sub-algebra of one of the algebras W (n, φ), S(n, φ), H(n, φ), K(n, φ), CS(n, φ),CH(n, φ). transitivity can be formulated as for each x ∈ gr(g)k with k ≥ 0the condition x ∗ gr(g)−1 = 0 implies x = 0.

50

The new algebras that appear are no longer graded. They appear becausethere are more volume and symplectic forms on Aφ. Amazingly, Darboux’sTheorem holds for contact forms but fails for symplectic. We consider forms

ω1S = eX1ωS = (

∑

k

X(k)1 )dX1 ∧ dXk . . . ∧ dXn ∈ ∧nAφ

M,

ω1S = (1 −X(1,1,...1))ωS = (1 −

∏

k

Xk)dX1 ∧ dXk . . . ∧ dXn ∈ ∧nAφM,

ω1H = d∧2M (eX1

t∑

j=1

XjdXj+t) = eX1ωH + . . . ∈ ∧2AφM,

ω2H = ωH +

∑

i,j

αi,jXiXjdXi ∧ dXj + . . . ∈ ∧2AφM.

In the definition of ω1H we use d∧2M (fdXi) = dM (f)∧dXi =

∑j ∂j(f)dXj∧

dXi. In the definition of ω2H we use a nonsingular skew-symmetric matrix.

It is know which matrices define equivalent forms but the relation is toocumbersome to attempt to state it here. The forms give rise to new specialand hamiltonian Lie algebras

Si(n, φ) = {∂ ∈W (n, φ) | ∂(ωiS) = 0},

H i(n, φ) = {∂ ∈W (n, φ) | ∂(ωiH) = 0},The following theorem is the classification of Lie algebras of Cartan type.The proof is too long even to contemplate to explain it here.

Theorem 14.4 Let g be a finite-dimensional simple Lie algebra of Cartantype over an algebraically closed field of characteristic p ≥ 5. Then g isisomorphic to one of the Lie algebras W (n, φ), S(n, φ), H(n, φ), K(n, φ),Si(n, φ), H i(n, φ) where i ∈ {1, 2}.

14.5 Exercises

Prove Proposition 14.2.Prove that S(n, φ) is an ideal of CS(n, φ).Prove that H(n, φ) is an ideal of CH(n, φ).We define divergence Div : W (n, φ) → Aφ as a linear map defined by the

formula Div(∑

j fj∂j) =∑

j ∂j(fj). Prove that S(n, φ) consists of vectorfields with zero divergence.

51

15 Root systemsWe recall some important information about semisimple Lie algebras

that you should have covered in Lie Algebras. So we just state the resultswithout proofs.

15.1 Roots

A finite-dimensional semisimple Lie algebra g over C admits a Cartansubalgebra h which is abelian. This leads to the Cartan decomposition

g = h ⊕⊕

α∈R

gα where gα = {x ∈ g | ∀h ∈ hh ∗ x = α(h)x} .

Elements of R are called roots and R is a subset of h∗. The subspaces gαare called root subspaces and all of them are one dimensional. They satisfygα ∗gβ ⊆ gα+β with equality if α+β is a root, so the Cartan decompositionis a grading by the free abelian group Λ generated by all α ∈ R insidethe group h∗. The group Λ is called the root lattice but its elements arenot called roots!! You may call them weights, although there may be moreweights than elements of the root lattice. There is some rationale in callingzero a root and g0 = h a zero root subspace but there are also good reasonsnot to do it. So we won’t.

Now R is a root system and have all the structures discusses in Liealgebras. One particularly useful feature is an ability to choose a basis α1,α2, . . .αn, elements of each are called simple roots. A basis of the rootsystem must be a basis of the vector space h∗ but needs to satisfy a furtherproperty: each root must be a linear combination of simple roots with eitherintegral non-negative or integral non-positive coefficients.

The choice of such a basis is far from unique but they are all isomorphic,in a sense that one can move one basis into another basis by an element ofthe Weyl group, which will be discussed in the next subsection.

15.2 Coroots

The crucial bit of structure is the Killing form on g. It is a non-degeneratesymmetric bilinear form on g by Cartan criterion. Taking into account thatK(gα, gβ) = 0 unless α + β = 0. This forces the restriction of the Killingform to h being a non-degenerate symmetric bilinear form as well. Thisform provides a canonical linear isomorphism h → h∗ via x 7→ K(x, ).We find this isomorphism particularly useful in the opposite direction, sowe write α∗ ∈ h for the image of an element α ∈ h. One useful thing thisisomorphism does is transferring the Killing form from h to h∗. We denote

52

the transferred form by α • β. Observe that α • β = K(α∗, β∗) by thedefinition of the transferred form.

As you know from Algebra-I, bilinear forms are rather dull on complexvector spaces. The useful trick is to turn to real vector spaces, namely let h∗

R

be the real subspace of h spanned by roots. From the properties of the rootsystem it is clear that simple roots form a basis of h∗

R. It is a slightly more

subtle observation that the form restricted to h∗R

is positive definite. Thus,h∗

Ris naturally a Euclidean space and we can usual Euclidean things there

such as talking about lengths, angles, etc. In particular, ||α|| =√α • α for

each coroot α.Now we are ready to define the coroots. For a root α ∈ R, we define its

coroot asα∨ = 2α∗/||α||2 ∈ h .

Thus, a coroot is an element of the Lie algebra g! One can play the rootgames with the coroots. Coroots form a root system R∨. Coroots α∨

i ofsimple roots form a basis the (co)root system R∨ and fully deserve thename of simple coroots. If R is simple then R∨ is usually of the same typewith an exceptions of types Bn and Cn, n ≥ 3. These two are swapped, i.e.,the root system of type Bn has the coroot system of type Cn and vice versa.

One useful thing a coroot does is it defines a reflection Sα(x) = x −α∨(x)α. We can equally write α∨(x) or x(α∨) as we are ambivalent whichof the vector spaces h, h∗ consist of “elements” and which one consists of“functionals on elements”. I hope you do not need reminding that the canon-ical map V → (V ∗)∗ is an isomorphism for finite-dimensional vector spaces.One can define reflections Sα as either complex linear transformations of h

or real linear transformations of hR or lattice automorphisms of h, dependingon particular aims. The group W generated by them is the same in eachcase. It is called the Weyl group. For instance, to see that W is finite itis instructive to look at hR. One observes that W is a discrete subgroup ofthe orthogonal group, and as the orthogonal group is compact, W must befinite.

Another useful thing derived from coroots is Cartan matrix

(Ci,j) = α∨j (αi) .

It is a matrix with integral coefficients with diagonal elements Ci,i equalto 2. Off-diagonal elements are either 0, −1, −2 or −3 depending on theangle between αi and αj as well as their lengths. One interpretation ofthe Cartan matrix is that its columns are coordinate expressions of simpleroots in the basis of fundamental weights. Recall that a fundamental weight

53

ωi ∈ h∗ is an element of the dual basis to the basis of simple coroots, thatis, ωi(α

∨j ) = δi,j .

15.3 Cartan automorphism

There exists a unique Lie algebra automorphism κ : g → g satisfying thefollowing properties:

(i) κ2 = Ig,(ii) κh = −Ih,(iii) κ(gα) = g−α.

The easiest way to see this is via Serre’s theorem. The latter defines g

via generators hi, ei, fi and Serre’s relations. One can κ on generators byκ(hi) = −hi, κ(ei) = −fi, κ(fi) = −ei. This defines an automorphism of thefree Lie algebra with the same generators. It remains to check that the idealgenerated by relations is stable under κ. We leave details as an exercise.

15.4 Series of roots

You have seen in Lie Algebras that, given a root α, the intersectionCα ∩ R is equal to {α,−α}. This property ensures that the root system isreduced.

We need a slightly more subtle property which was probably covered butwe will sketch a proof to improve self-consistency of these notes. Given tworoots α, β the intersection β + Zα ∩R is called the α-series via β.

Proposition 15.1 Let α, β ∈ R such that α 6= ±β. The following state-ments hold.

(i) β+Zα∩R = {β−rα, β−(r−1)α, . . . , β+qα} for some r, q ∈ {0, 1, 2, 3}.(ii) r − q = α∨(β).

(iii) If α+ β ∈ R then r + 1 = q ||α+β||2

||β||2.

Proof: Everything happens inside the span of α and β. This span’s inter-section with R is a root system of rank 2. It suffices to go over all pairs ofroots in the rank 2 systems A1 × A1, A2, B2, G2. Actually, one does nothave to go over all pairs but over each possible angle between roots. Thepossible angles are π/2, π/3, 2π/3, π/4, 3π/4, π/6, 5π/6 and going throughall the cases is not long.

Refer to in-class lecture notes where I went through 3 of the angles. 2

15.5 Lie algebras of classical type

Let K be a field of characteristic p ≥ 5. A simple finite-dimensional Liealgebra g over K is a Lie algebra of classical type if it satisfies the following4 conditions:

54

(i) there exists an abelian Cartan subalgebra h,(ii) there exists a Cartan decomposition g = h⊕⊕α∈h∗gα where gα = {x ∈

g | ∀h ∈ hh ∗ x = α(h)x} with respect to h,(iii) if gα 6= 0 then gα ∗ g−α is one-dimensional,(iv) if α 6= 0 6= β and gα 6= 0 6= gβ then there exists k in the prime subfield

of K such that gα+kβ = 0.

Over an algebraically close field, such Lie algebras follow Cartan-Killingclassification. We do not intend to prove this but we would like to constructthese Lie algebras.

15.6 Exercises

Prove that the angle between two simple roots is obtuse, i.e. greater orequal to π/2.

Verify that the columns of Cartan matrix are coordinate expressions ofsimple roots in the basis of fundamental weights.

Verify that κ is well-defined by writing images of Serre’s relations underκ.

Argue why κ is unique.Complete the proof of Proposition 15.1 by going through all possible

angles.

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16 Chevalley theorem16.1 Representations of sl2

In proposition 2.2, we constructed n+1 dimensional representation An ofLie algebra sl2(K) over any field. Over the field of complex numbers these arethe only non-isomorphic representations of sl2(C). Furthermore, by Weyl’sComplete Reducibility Theorem, every finite-dimensional representation ofsl2(C) is completely reducible. This means that every finite-dimensionalrepresentation of sl2(C) is isomorphic to a direct sum of representations An.This has also been done in Lie Algebras and we will use this fact soon.

16.2 Chevalley basis

We would like to pick a very special basis of a simple finite-dimensionalLie algebra g of C. In the Cartan we use simple coroots α∨

i . In each rootspace we choose a non-zero eα ∈ gα.

Let us see what we can say about the products in this basis. By thedefinition of a root space α∨

i ∗ eβ = α∨i (β)eβ . Since the Cartan subalgebra

is abelian, α∨i ∗ α∨

j = 0. It remains to see what happens with root vectors.First, eα ∗ e−α ∈ h but it is not immediately obvious what this element is.Second, if α 6= −β but α+β is not a root then eα ∗ eβ = 0. Finally, if α+βis a root then eα ∗ eβ = cα,βeα+β but the values of the constants cα,β ∈ C

are not immediately clear.The point of the following fact (Chevalley Theorem) is that one can do

much better job determining the remaining products.

Theorem 16.1 It is possible to choose eα ∈ gα so that the following twoproperties hold:

(i) eα ∗ e−α = α∨ for each root α,(ii) cα,β = ±(r + 1) for each pair of roots such that α + β ∈ R where

β + Zα ∩R = {β − rα, β − (r − 1)α, . . . , β + qα}.Proof: Let us choose eα ∈ gα for each positive root α. For the negativeroots we define eα = −κ(e−α) using Cartan’s automorphism κ. We willreadjust this basis once during the course of the proof to arrive at a basiswith desired properties.

Let us first compute eα ∗ e−α ∈ h. For arbitrary x ∈ h, we computeKilling form K(x, eα ∗ e−α) = K(x ∗ eα, e−α) = α(x)K(eα, e−α) =. On theother hand, K(x, α∨) = 2K(x, α∗)/||α||2 = 2α(x)/||α||2 . Since the form isnon-degenerate, we conclude that

eα ∗ e−α =1

2||α||2K(eα, e−α)α∨ .

56

Notice that K(eα, e−α) 6= 0 because the Killing form is non-degenerate andK(gα, gβ) = 0 unless α+ β = 0.

Now we define

eα =

√2

||α||√K(eα, e−α)

eα

for all the roots and these elements satisfy property (i). It suffices to verifyproperty (ii).

Let us consider constants cα,β in this basis. As the basis still satis-fies κ(eα) = −e−α, applying κ to eα ∗ eβ = cα,βeα+β gives e−α ∗ e−β =−cα,βe−α−β . This implies that c−α,−β = −cα,β. Now we compute the prod-uct (eα ∗ eβ) ∗ (e−α ∗ e−β) = cα,βc−α,−βeα+β ∗ e−α−β = −c2α,β(α + β)∨ =

−c2α,β/||α + β||2(α∗ + β∗).Let us recompute this product using Jacobi’s identity first: (eα ∗ eβ) ∗

(e−α∗e−β) = eα∗(eβ∗(e−α∗e−β))+(eα∗(e−α∗e−β))∗eβ = −eα∗(eβ∗(e−β∗e−α))− eβ ∗ (eα ∗ (e−α ∗ e−β)). These two products can be computed usingrepresentations of sl2. The elements eα, e−α and α∨ span an sl2-subalgebrain g. We consider g as a representation of this sl2. Subspaces gγ , as γruns through the α-series via −β, span an irreducible subrepresentation ofdimension r+ q+ 1 forcing it to be isomorphic to Ar+q. Since Lr+1

eα(e−β) =

Lr+1e−α

(e−β) = 0, an isomorphism must send e−β into a multiple of xqyr. Itfollows that eα ∗ (e−α ∗e−β) = q(r+1)e−beta and the whole product is equalto −q′(r′ + 1)α∗ − q(r + 1)β∨. It follows that

c2α,β||α + β||2 β

∗ = q(r + 1)β∨ =q(r + 1)

||β||2 β∗

and, finally, by part (iii) of Proposition 15.1

c2α,β = q(r + 1)||α + β||2||β||2 = (r + 1)2 .

2

Any basis satisfying conditions of Theorem 16.1 will be called Chevalleybasis. This basis depends on a choice of Cartan subalgebra. Besides, wehave not addressed the issue of the signs. This means that the basis is notunique even if the choice of Cartan subalgebra is fixed.

16.3 G2

As an example, we would like to write a part (positive Borel subalgebra)of multiplication table in Chevalley basis for Lie algebra of type G2. The

Cartan matrix is

(2 −3

−1 2

). The 12 roots are vertices of David’s start.

57

Let α = α1 be the long simple root, β = α2 the short simple root. Simplecoroots are

α∨ =2

3α∗, β∨ = 2β∗ .

Now we can fill the multiplication table (up to signs) of the positive Borelsubalgebra of Lie algebra of type G2:

eα eβ eα+β eα+2β eα+3β e2α+3β

α∨ 2eα −eβ eα+β 0 −eα+3β e2α+3β

β∨ −3eα 2eβ −eα+β eα+2β 3eα+3β 0eα 0 ±eα+β 0 0 ±eα+3β 0eβ ±eα+β 0 ±2eα+2β ±3eα+3β 0 0

eα+β 0 ±2eα+2β 0 ±3e2α+3β 0 0eα+2β 0 ±3eα+3β ±3e2α+3β 0 0 0eα+3β ±e2α+3β 0 0 0 0 0

e2α+3β 0 0 0 0 0 0

.

To fill the rest of the table one needs the remaining coroots:

(α+ β)∨ = 2(α∗ + β∗) = 3α∨ + β∨, (α+ 2β)∨ = 2(α∗ + 2β∗) = 3α∨ + 2β∨,

(α+3β)∨ =2

3(α∗+3β∗) = α∨+β∨, (2α+3β)∨ =

2

3(2α∗+3β∗) = 2α∨+β∨ .

16.4 Exercises

In a Lie algebra of type A2 with a fixed choice of Cartan subalgebradescribe all possible Chevalley bases.

Fill the rest of the table for G2.Fill the full multiplication (up to signs) in type B2.

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17 Chevalley reduction17.1 Reduced Lie algebra

Let gC be a simple finite-dimensional Lie algebra over C. The abeliansubgroup of gC generated by the Chevalley basis is a Lie ring. By Theo-rem 16.1, it is closed under the products and the only axioms of a Lie algebrait fails are the ones related to vectors space structure. We call this ring gZ.As an abelian group it is free with the Chevalley basis as a basis.

Using this ring we define the Lie algebra gK = gZ ⊗Z K over any fieldK. As a vector space, it has the Chevalley basis as a basis. It has thesame multiplication as the one in Chevalley basis but the coefficients areconsidered in K.

We call gK the Chevalley reduction of gC, which is a sensible name overthe fields of positive characteristic. In zero characteristic a better name maybe the split form of gC.

In the rest of the lecture we consider some examples but now we wouldlike to describe the restricted structure on g.

Proposition 17.1 Let p ≥ 5. Then a restricted structure on gK is deter-

mined by eα[p] = 0 and α

[p]i = αi.

Proof: The operator Lpα sends Lβ to Lβ+pα. By Proposition 15.1, β + 4αis not a root. Hence, Lβ+pα = 0 and Lpα = 0 = L0. On the other hand,Lα∨ sends an element x of Chevalley basis into nx where n is an integermodulo p, i.e. an element of the prime subfield. Hence, it satisfies np = nand Lpα∨ = Lα∨ . We are done by Proposition 7.1. 2

To describe all restricted structures we need to know the centre of gK,which we will compute in the next lecture.

17.2 sl2 in characteristic 2

The Chevalley basis is the standard basis e, f , h with the productse∗f = h, h∗f = −2f , h∗e = 2e. The resulting Lie algebra gK is isomorphicto sl2(K). It is simple unless the characteristic is 2. In characteristic 2, it isa nilpotent Lie algebra: g(1) = Kh = Z(g).

There is a second integral form on gC. One takes r = h/2 insteadof h. In this basis the products are e ∗ f = 2r, r ∗ f = −f , r ∗ e = e.The resulting Lie algebra gK is isomorphic to gK if the characteristic isnot two. In characteristic 2, it is a 2-soluble non-nilpotent Lie algebra:g(1) = Ke + Kf and g(2) = 0 but there are non-zero products of arbitrarylong length: r ∗ (r . . . (r ∗ e)) = e. One can think of gK as the projectivespecial linear algebra psl2(K).

59

17.3 sln in characteristic p

The standard choice of Cartan subalgebra is diagonal matrices h. Theroot spaces are spanned by elementary matrices ei,j with i 6= j. If h ∈ h withentries hi on the diagonal then h ∗ ei,j = (hi − hj)ei,j, so the correspondingroot αi,j ∈ h∗ is defined by αi,j(h) = hi−hj . The standard choice of simpleroots is αi = αi,i+1, i = 1, . . . n− 1. These root vectors give Chevalley basisand the corresponding simple coroots are α∨

i = ei,i − ei+1,i+1.If p does not divide n+1, the reduced Lie algebra gK is simple. If p does

divide n+1, the reduced Lie algebra gK admits a central element∑

i iα∨i . If

n > 2 the quotient by the centre is simple and should be denoted psln(K).

17.4 G2 in characteristic 2

All Lie algebras experience some collapse of the root structure in char-acteristic 2. For each root α ∈ R we consider α ∈ h∗

K. These functionals

determine Cartan decomposition of gK. Since α = −α, the root spaces areat least 2-dimensional in characteristic 2.

For G2, there is a further collapse: α = −α = α+ 2β = −α− 2β,α+ β = −α− β = α+ 3β = −α− 3β and β = −β = 2α+ 3β = −2α− 3β.Thus, there are three 4-dimensional root spaces. Amazingly enough, the Liealgebra gK of type G2 is still simple in characteristic 2.

17.5 G2 in characteristic 3

There is a mild collapse of the root structure: α = α+ 3β = −2α− 3βand −α = −α− 3β = 2α+ 3β. So long root vectors join together to formtwo 3-dimensional root spaces. Short roots stay distinct, so short root vec-tors give six more 1-dimensional root spaces.

A similar accident happens with long coroots (coroots as short roots). Asa result the short root vectors generate a 7-dimensional ideal in gK. Checkdetails yourself...

17.6 Exercises

Describe the Cartan decomposition of the Lie algebra of type B2 incharacteristic 2. Is the Lie algebra simple?

Describe the dimensions of root spaces of psl3(K) in characteristic 3 andof psl4(K) in characteristic 2.

Consider gK of type G2 in characteristic 3. Prove that its 7-dimensionalideal I is isomorphic to psl3(K) as Lie algebras. Prove that the quotientalgebra gK/I is isomorphic to psl3(K) as Lie algebras.

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18 Simplicity of Chevalley reductionWe prove simplicity of the reduced algebras gK. Throughout the lecture

gK is the reduction of a simple finite-dimensional Lie algebra gC to the fieldK of characteristic p > 0.

18.1 Centres of reduced algebras

First, we would like to compute the centre of gK. The centre is anideal, thus, the presence of non-zero centre is incompatible with simplicity.Besides, the centre makes the restricted structures non-unique.

Proposition 18.1 The dimension of the centre of gK is equal to either 2(if p = 2 and g is of type D2n), or 1 (if p = 3 and g is of type E6, or p = 2and g is of type Bn or Cn or D2n+1 or E6, or p|(n+ 1) and g is of type An)or 0 other wise.

Proof: This is done in several steps and you should look in your lecturenotes for further details of each step.

(Step 1) We observe that any central element x ∈ Z(gK) must lie in theCartan. Otherwise, x = aeα + . . . where a 6= 0 and the dots denotesome linear combination of the remaining elements of the Chevalleybasis. From “Lie Algebras”, it is possible to choose a basis of the rootsystem so that α is one of simple roots. Let β be another simple rootconnected to α by an edge. Notice that such β does not exist if g is oftype A1 but one can use β = −α in this case.

As there is an edge α+ β is a root and β − α is not a root as bothare simple. Hence, x ∗ eβ = ±aeα+β + . . ., contradicting centrality ofx. In case of A1 one gets x ∗ e−α = ±aα∨ + . . ., arriving at the sameconclusion.

(Step 2) Hence we are confined to considering linear combinations of simplecoroots x =

∑i aiα

∨i . Such x commutes with coroots. For root vectors

x∗eβ = β(x)eβ , so x is central if and only if αi(x) =∑

j αi(α∨j )aj = 0

for each simple root αi. This says that the matrix product (Ci,j)(aj) =0, so the dimension of the centre is equal to the corank (nullity) of theCartan matrix (considered as the matrix of K).

(Step 3) To get an idea of coranks it would useful to compute determinantsof Cartan matrices (over Z). Let xn = det(Xn) be the determinant ofthe Cartan matrix of type Xn, By abuse of notation Xn is a type andthe Cartan matrix and the Dynkin diagram!

We derive a formula for computing determinants explicitly. Let Γbe a Dynkin diagram. Pick a vertex that is connected to a single other

61

vertex and the arrow is single. Let Γ1 be the Dynkin diagram obtainedby removing the former vertex, Γ1 the Dynkin diagram obtained byremoving both vertices. Expanding first the row, then the column ofDynkin diagram gives the recursive formula:

det(Γ) = 2det(Γ1) − det(Γ2) .

(Step 4) Before we do the recursions, we precompute for ranks 1 and 2:a1 = 2, a2 = 2 · 2 − (−1) · (−1) = 3, b2 = 2 · 2 − (−1) · (−2) = 2,g2 = 2 · 2 − (−1) · (−3) = 1. In type An we remove the side vertex toderive that an = 2an−1 − an−2 = 2n− (n− 1) = n+ 1. In type Bn wealso remove the side vertex to derive that b3 = 2b2 − a1 = 2 · 2− 2 = 2and bn = 2bn−1 − bn−2 = 2 · 2 − 2 = 2. Cartan matrices in types Bnand Cn are related by transposition, so cn = bn. In type Dn we removeone the two (or three in type D4) end vertices connected to the onlytrivalent vertex to derive that dn = 2an−1−a1·an−3 = 2n−2(n−2) = 4.In type En we remove the end vertex connected to the only trivalentvertex to derive that en = 2an−1−a2 ·an−4 = 2n−3(n−3) = 9−n. Inparticular, e8 = 1, e7 = 2, e6 = 3. Finally, f4 = 2b3−a2 = 2 ·2−3 = 1.

If xn = 1 this means that the corank is always zero and the Liealgebra of type Xn has no centre in any characteristic. If xn = p, aprime number then the corank is zero except characteristic p whereit is 1. The Lie algebra of type Xn has one dimensional centre incharacteristic p and no centre, otherwise. It remains to sort compositenumber an = n+ 1 and dn = 4.

(Step 5) In type An we observe that the elementary transformations overZ are also elementary transformations in any characteristic p. Thus,we can use reduction of Smith’s normal over Z to determine the rankin characteristic p. Using the following chain of transformations recur-sively

... −1 0...

. . . 2 −1 0

. . . −1 2 1 − k

. . . 0 −1 k

;

... −1 0...

. . . 2 −1 −k

. . . −1 2 k + 1

. . . 0 −1 0

;

... −1 0...

. . . 2 −k 0

. . . −1 k + 1 0

. . . 0 0 1

leads as to conclude that the Smith’s normal of Cartan matrix of typeAn has n + 1 and n ones on the diagonal. Hence, the centre will beone-dimensional whenever p divides n+ 1.

(Step 6) In type Dn we have to understand what happens in characteristic2 as the Smith’s normal form could have either (4, 1, . . .) or (2, 2, 1, . . .)

62

on the diagonal. Consider the Cartan matrix (Cj,i) reduced to charac-teristic 2 with the last two columns corresponding to the two verticesconnected to the trivalent vertex. This makes the last two rows andthe last two columns equal. Hence, the rank of (Cj,i) is the same asthe rank of its (n− 1)× (n− 1)-minor obtained deleting n-th row andcolumn. The minor is Cartan’s matrix of type An−1 which, by step 5,has rank n− 2 if n is even and n− 1 if n is odd.

2

18.2 Root structure collapse

Proposition 18.2 Let p ≥ 5 and (n + 1) not divisible by p if g is of typeAn. For any root α 6= β ∈ R the root space gα ⊆ gK is one dimensional.

Proof: It suffices to show that for any two distinct root α 6= β ∈ R theirrestrictions α, β ∈ h∗ are distinct. Because of the restriction on p, g has nocentre and simple roots form a basis of the dual space h∗. Because of this,the proposition becomes a “local” question, i.e., a question about the rank2 root system (Zα+ Zβ)∩R. If Λ is the root lattice, we would like to knowwhether it is possible that α − β ∈ pΛ. Going over all roots in the rank 2systems A1 ×A1, A2, B2, G2 shows that it is impossible. 2

Notice that the columns of Cartan matrix are coordinates of simple rootsin the basis of fundamental weights. It follows that the cokernel of Cartan’smatrix, which we essentially computed, is the quotient Φ/Λ of the weightlattice by the root lattice. This group has further significance: it is isomor-phic to the centre of the corresponding simply-connected group Gsc and thefundamental group of the corresponding adjoint group Gad.

18.3 Simplicity

Theorem 18.3 Let p ≥ 5 and (n+ 1) not divisible by p if g is of type An.Then the reduced Lie algebra gK is simple.

Proof: By Proposition 18.1, the centre of gK is zero. By Proposition 18.2,all root spaces gα in the Cartan decomposition of gK are one-dimensional.

We consider a nonzero ideal I � gK. Let Iα = I ∩ gα. First, we observethat I = ⊕αIα. Pick arbitrary x =

∑α xα ∈ I where xα ∈ gα. It suffices

to prove that all xα ∈ I. We use Vandermonde trick. Pick hi ∈ h so thatαj(hi) = δj,i. In characteristic zero, these are fundamental weights but wehave to argue why these exist in characteristic p: conditions on p ensurethat simple roots αi ∈ h∗ form a basis, hence hi is just the dual basis. Nowhi ∗ x =

∑α α(hi)xα ∈ I. Repeating this k times,

Lkhi(x) =

∑

α

α(hi)kxα ∈ I .

63

By Vandermonde trick, ∑

α, hi(α)=m

xα ∈ I

for all m. Repeating this for all i we conclude that all xα are in I andI = ⊕αIα.

If I0 6= 0, pick h ∈ I0. Since the centre is zero, there exists a root αsuch that α(h) 6= 0 ∈ K. Hence, eα = α(h)−1h ∗ eα ∈ I and Iα 6= 0.Even if I0 = 0, there exist α with Iα 6= 0 since I 6= 0. By proposition 15.1,whenever α+β is a root and eα∗eβ = cα,βeα+β ∈ gC, the coefficient satisfies1 ≤ |cα,β | ≤ 4 < p, so it is non-zero in K. As the root system is irreducible,we conclude that all eα are in I. Finally, all α∨ = eα ∗ e−α are in I too andI = gK. 2

18.4 Exercises

Write the Cartan matrix of type An in characteristic 2 and prove directlywhat its rank is.

Write a non-zero central element of gK explicitly when gK is of type F4

and the characteristic of K is 2.Write a non-zero central element of gK explicitly when gK is of type E6

and the characteristic of K is 3.

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19 Chevalley groupsWe associate a group, called Chevalley group, to the reduced Lie algebra

gK. If the field K is finite, this is a finite simple group.

19.1 Exponentials

Let gC be a simple finite-dimensional Lie algebra, gZ its Lie subringspanned over Z by Chevalley basis. We would like to extend scalars to thepolynomial ring Z[t] by considering gZ[t] = gZ ⊗Z Z[t]. In an elementarylanguage, gZ[t] is a free abelian group with basis tnx, n ∈ Z≥0, x an elementof Chevalley basis. The product is Z-bilinear and defined on the basis bytnx ∗ tmx = tn+m(x ∗ y).

In fact, the product gZ[t] is Z[t]-bilinear turning gZ[t] into a Lie algebraover the ring Z[t]. We would like to define exponential

Xα(t) = eteα =

∞∑

n

tn

n!Lneα

to be a Z[t]-linear map from gZ[t] to itself. The most urgent question why isit well-defined?

Proposition 19.1 For each root Xα(t) a well-defined Z[t]-linear Lie algebraautomorphism of gZ[t].

Proof: To see that Xα(t) is well-defined, we will derive explicit formulasfor its action on Chevalley basis. First of all, since 2α is not a root

Xα(t)(eα) = eα, Xα(t)(β∨) = β∨−β∨(α)teα, Xα(t)(e−α) = e−α+tα∨−t2eα .

If β 6= ±α is another root then we need the α-series via β: β + Zα ∩ R ={β − rα, β − (r − 1)α, . . . , β + qα}. Since β + 4α is not a root,

Xα(t)(eβ) = eβ±(r+1)teα+β±(r + 1)(r + 2)

2t2e2α+β±

(r + 1)(r + 2)(r + 3)

6t3e3α+β .

Clearly, Xα(t) is a Z[t]-linear map and its inverse is Xα(t). It is anautomorphism because Leα is a derivation, similarly to exercises to Lecture1. 2

19.2 Specialized exponentials and Chevalley groups

Now consider the reduced algebra gK. For each root α and each s ∈ K

we get a Lie algebra automorphism Xα(s) of gK by specialising Xα(t). In ahigh-tech language, we choose a ring homomorphism Z[t] → K by sendingt to s and define Xα(s) = Xα(t) ⊗ I, an automorphism of gZ[t] ⊗Z[t] K =

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(gZ ⊗Z Z[t]) ⊗Z[t] K ∼= gZ ⊗Z (Z[t] ⊗Z[t] K) ∼= gK, with the two naturalisomorphisms. In an elementary language, we define Xα(s) by formulas inProposition 19.1 with substituting s for t.

The Chevalley group G(g,K) is defined to be the subgroup of GL(gK)generated by all Xα(s) for all roots α and s ∈ K. It is unfortunate that wehad to make several choice to arrive at this group. How justifiable is thenotation G(g,K)? Don’t we have to add all the choices we have made to thelist of arguments?

Proposition 19.2 A different set of choices leads to isomorphic reducedalgebra gK and Chevalley group G(g,K).

Proof: We have made three choices in the process of construction:

(i) Cartan subalgebra h,(ii) simple roots α1, . . .αn,(iii) root vectors eα in the Chevalley basis.

Choice (i) is up to conjugation. Any two Cartan subalgebras h, h′ of gC

can be moved one to another by non-unique element of the adjoint group(from Lie algebras). Two different choices (ii) (of simple roots) are relatedby a canonical element of the Weyl group. Choice (iii) of root vectors is justup to sign (one can replace some eα with −eα) as we saw in the proof ofTheorem 16.1.

In fact choices (ii) and (iii) do not change the Lie ring gZ but may changeXα(s) into Xα(−s) but both elements are among generators of the Chevalleygroup anyway. Hence, the Lie algebra gK and the Chevalley group G(g,K)are the same if one makes different choices (ii) and (iii).

Different choice in (i) will lead to isomorphic gK and G(g,K) but thereis no canonical isomorphism. 2

19.3 Exercises

Prove that Xα(s)Xα(s′) = Xα(s+ s′).

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20 Abstract Chevalley groups20.1 Definition

Let (g, γ) be a restricted Lie algebra over a field K of characteristic p,V its restricted representation. For all purposes, it is sufficient to considerfinite-dimensional V only.

The restricted representation gives a homomorphism of restricted Liealgebras ρ : g → gl(V ). This means that ρ(γ(x)) = ρ(x)p, i.e, gl(V ) carriesits canonical restricted structure.

Let N(g) be the set of all x ∈ g such that γ(x) = 0. We call this set thep-nilpotent cone of g. For each x ∈ N(g) we can define an exponential

eρ(x) =

p−1∑

k=0

1

k!ρ(x)k ∈ gl(V ) .

Moreover, eρ(x) ∈ GL(V ) because (eρ(x))−1 = eρ(−x). We define an abstractChevalley group G(g, V ) as the subgroup of GL(V ) generated by all expo-nentials eρ(x) for all x ∈ N(g). If one consider adjoint representation then,barring accidents in small characteristic, say p ≥ 5, one arrives at the usualChevalley group. Only one inclusion is obvious but we will not prove theopposite inclusion here.

Proposition 20.1 If ρ is a restricted representation of a restricted Lie al-gebra g then

ρ(eLx(y)) = eρ(x)ρ(y)e−ρ(x)

for all x ∈ N(g), y ∈ g.

Proof: First, observe by induction that for each k = 1, 2, . . . p− 1

ρ(1

k!Lkx(y)) =

k∑

j=0

(−1)j

(k − j)!j!ρ(x)k−jρ(y)ρ(x)j .

For k = 1 it is just the definition of a representation: ρ(Lx(y)) = ρ(x ∗y) = ρ(x)ρ(y) − ρ(y)ρ(x). Going from k to k + 1, ρ( 1

(k+1)!Lk+1x (y)) =

1k+1(ρ(x)ρ( 1

k!Lkx(y))−ρ( 1

k!Lkx(y)ρ(x)) =

∑kj=0

(−1)j

k+1 ( 1(k−j)!j!ρ(x)k−j+1ρ(y)ρ(x)j−

1(k−j)!j!ρ(x)k−jρ(y)ρ(x)j+1 =

∑k+1i=0

(−1)i

(k+1)(k−i)!(i−1)! (1i+ 1k+1−i)ρ(x)k+1−iρ(y)ρ(x)i =

∑k+1i=0

(−1)i

(k+1−i)!i!ρ(x)k+1−iρ(y)ρ(x)i.

Finally, ρ(eLx(y)) =∑p−1

k=0 ρ(1k!L

kx(y)) =

∑p−1i,j=0

(−1)j

i!j! ρ(x)iρ(y)ρ(x)j =

eρ(x)ρ(y)e−ρ(x). 2

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20.2 Examples

Let us consider g = sl2(K) and its natural representation ρ : g → gl2(K).For each x ∈ N(sl2) we have x2 = 0 in gl2 which implies that

eρ(x) = I + x , in particular ete =

(1 t0 1

), etf =

(1 0t 1

).

Since these matrices generate SL2(K) (see the exercise), we can use Propo-sition 20.1 to derive a surjective homomorphism φ : SL2(K) → G(g,K).Indeed, we define φ(eρ(x)) = eLx on generators. Whenever, there is a rela-tion in SL2(K), say

eρ(x1)eρ(x2)) . . . eρ(xn) = I ,

using Proposition 20.1, we conclude that

ρ(eLx1 eLx . . . eLxn (y)) = ρ(y)

for all y. Since ρ is injective it follows that that

eLx1 eLx . . . eLxn = I

and the homomorphism ρ is well-defined. It remains to figure out the ker-nel of φ. Again using Proposition 20.1, φ(A)(y) = AyA−1 and the kernelconsists of scalar matrices. It follows that G(g,K) is isomorphic to PSL2(K).

I believe Witt was originally interested in finding new finite simple groupsusing other simple Lie algebras. It was a disappointment. For instance,G(W (1, 1),W (1, 1)) is soluble.

20.3 Exercises

Prove that matrices

(1 t0 1

)and

(1 0t 1

)for all t ∈ K generate

SL2(K).

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21 Engel Lie algebras21.1 Engel Theorem

Let us consider the following two elements of the free Lie algebra:

N2(x, y) = E1(x, y) = x ∗ y ,Nn+1(x1, . . . xn, y) = x1 ∗ Nn(x2, . . . xn, y) = Lx1Lx2 . . . Lxn(y) ,

En(x, y) = x ∗ En−1(x, y) = Lnx(y)

A Lie algebra is called n-Engel if it satisfies En. It is called Engel if it isn-Engel for some n. A Lie algebra is called n-nilpotent if it satisfies Nn. Itis nilpotent if it is n-nilpotent for some n.

Theorem 21.1 A finitely generated Engel Lie algebra is nilpotent.

For finite-dimensional Lie algebras it is classical Engel’s theorem. It isusually formulated for ad-nilpotent Lie algebra, i.e. such Lie algebras whereeach Lx is nilpotent. But clearly then Lnx = 0 where n is the dimension ofthe Lie algebra, i.e., it satisfies En.

Let us observe what happens for small n. Since N2(x, y) = E1(x, y) =x ∗ y, 1-Engel Lie algebras are the same as 2-nilpotent ones.

Let us look at the free 2-Engel algebra g = L(X)/(N2)w. For everyx,y ∈ g, we can see that 0 = L2

x+y−L2x−L2

y = LxLy +LyLx. On the otherhand, (LxLy−2LyLx)(z) = x∗(y∗z)−2y∗(x∗z) = (x∗y)∗z−y∗(x∗z) =(LzLy + LyLz)(x) = 0. We conclude that 3LyLx = and g is 3-nilpotentunless p = 3. If p = 3 then further analysis along the same line reveals thatg (and consequently any 2-Engel Lie algebra) is 4-nilpotent.

So far we have never used the fact that the Lie algebra is finitely gener-ated. It starts playing a role in 3-Engel algebras. Let g = L(X)/(N3)w bethe free 3-Engel algebra. If p 6= 5 then it is 7-nilpotent. However, if p = 5then it is (2|X|+1)-nilpotent and finite generation is crucial: if X is infinite,g is no longer nilpotent. nevertheless, it is known that in characteristic zero,any Engel Lie algebra is nilpotent.

21.2 Linearized Engel identity

The identity En(x,y) has degree n in x. In particular, it is not linear ifn > 1. However, one can polarise it to derive a linear identity. Consider thefollowing polynomial in the free Lie algebra:

LEn(x1,x2, . . .xn,y) =∑

σ∈Sn

Nn(xσ(1), . . .xσ(n),y)

The identity LEn is called linearised n-Engel identity.

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Proposition 21.2 An n-Engel Lie algebra g satisfies LEn. In the oppositedirection, if g is a Lie algebra satisfying LEn and the ground field has eithercharacteristic 0 or p > n then g is n-Engel.

Proof: The second statement is obvious: LEn(x,x, . . . x,y) = n!En(x,y)and the condition on characteristic allows us to divide by n! in the field K.

For the first statement, rewrite En(a1x1 +a2x2 +anxn,y) = Q0 +a1Q1 +a1a2Q2 + . . . + a1a2 . . . anQn in the free Lie algebra K[a1, . . . an](x1, . . . xn)over the polynomial ring K[a1, . . . an]. The Lie polynomial a1a2 . . . asQs isa combination of all the terms divisible by a1, a2 . . . as but not by as+1.Plugging concrete values from K for all as gives us an identity of an n-EngelLie algebra over K. Plug a1 = 0. Hence, Q0 is an identity. Now plug a1 = 1and a2 = 0. Hence, both Q0 + Q1 and Q1 are identities. Repeating thisargument n times, we derive that all Qi are identities of an n-Engel Liealgebra.

Notice that Qi may still depend on aj , j > i. These identities Qi are(partial) linearisations (or polarisations) of En. Observe, that Qn = LEn,which is called the full linearisation. 2

21.3 Properties of group commutators

A commutator of two elements in a group is [x, y] = x−1y−1xy. Wedenote xy = y−1xy and x−y = y−1x−1y. We will write x∗ and x−∗ for xy

and x−y if we are to lazy to specify y. If H and F are subgroups then [H,F ]is the subgroup generated by all commutators.

Using commutators a group starts looking a bit like Lie algebra. Identi-ties in the next proposition look like anti-commutativity, distributivity andJacobi identity.

Proposition 21.3 The following commutator identities hold in any group:

(i) [x, y]−1 = [y, x] ,(ii) [x, yz] = [x, z][x, y]z ,(iii) [[x, y]zx][[z, x]yz ][[y, z]xy] = 1 ,(iv) [[x, y−1]z]y[[y, z−1]x]z[[z, x−1]y]x = 1 .

Proof: We check only identity (iii) (call The Hall identity). We keep brack-ets around what is left of the commutators: [[x, y]zx][[z, x]yz ][[y, z]xy ] =[y, x]z−x[x, y]zx[x, z]y−z [z, x]yz[z, y]x−y [y, z]xy =(y−1x−1y)z−1(y−1xy)x−1z(z−1x)y−1(x−1zx)z−1y(y−1z)x−1(z−1yz)y−1xy =(y−1x−1y)z−1(y−1xy)y−1(x−1zx)x−1(z−1yz)y−1xy =(y−1x−1y)z−1(y−1x)(x−1)(yz)y−1xy = (y−1x−1y)y−1xy = 1. 2

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21.4 Exercises

Prove all the identities in Proposition 21.3.Linearise the elasticity identity (xy)x = x(yx) in the free algebra. Show

that if the characteristic is not two, the elastic algebras are the same aslinearised elastic algebras.

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22 Lie algebra associated to a group22.1 Lie algebra associated to a strongly central series in a

group

Let G be a group, G = G1 �G2 � . . . Gn . . . a descending chain of normalsubgroups, i.e., each Gn is normal in G. It is called a central series ifGn−1/Gn is in the centre of G/Gn for each n. It is called a strongly centralseries if [Gn, Gm] ⊆ Gn+m for all n and m. Observe that each stronglycentral series is central since the centrality is equivalent to [Gn, G] ⊆ Gn+1

for all n. The opposite is false: not every central series is strongly central.My lack of group theory background doesn’t let me give you an example ofthe top of my head but, in general, there is no reason for the upper centralseries of a nilpotent group to be strongly central. On the other hand, lowercentral series of G is strongly central. We discuss a slight generalisation ofthis in the next section.

To each central series we can associate a (graded) Lie ring. As an abeliangroup, L(G,Gn) = ⊕∞

n=1Gn−1/Gn. The Lie product on this group is definedusing group commutator [x, y] = x−1y−1xy via

xGn+1 ∗ yGm+1 = [x, y]Gn+m+1 , x ∈ Gn , y ∈ Gm .

Theorem 22.1 If Gn is a strongly central series then L(G,Gn) is a Liering. If there exists a prime p such that each Gn−1/Gn is an elementaryabelian p-group then L(G,Gn) is a Lie algebra over the field of p elements.

Proof: Since each Gn/Gn+1 is an abelian group then their direct sumL(G,Gn) is also an abelian group.

Let us see that the product is well-defined. If x ∈ Gn, y ∈ Gm then theircommutator [x, y] belongs to Gn+m. The product needs to be independent ofthe representatives of the cosets. Pick z ∈ Gm+1, so that yGm+1 = yzGm+1.Then [x, yz]Gn+m+1 = [x, z][x, y]zGn+m+1 = [x, y]zGn+m+1 since [x, z] ∈Gn+m+1 and Gn+m+1 is normal. Finally, [x, y]zGn+m+1 = [x, y]Gn+m+1

because [x, y]Gn+m+1 is in the centre of G/Gn+m+1, hence commutes withzGn+m+1.

Let us see that the product is bilinear. Pick x ∈ Gn, y ∈ Gm, z ∈ Gk.If m 6= k then xGn+1 ∗ (yGm+1 − zGk+1) = (xGn+1 ∗ yGm+1) − (xGn+1 ∗zGk+1) by definition of the product. If m = k then the left hand side is[x, yz−1]Gn+m+1 = [x, z−1][x, y]z

−1Gn+m+1 = [x, y]Gn+m+1[x, z

−1]Gn+m+1

equal to the right hand side. Here again, we used the fact that [x, y]Gn+m+1

is central in G/Gn+m+1. Similarly, (xGn+1 − yGm+1) ∗ zGk+1 = (xGn+1 ∗zGk+1) − (yGm+1 ∗ zGk+1).

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Let us see observe anticommutativity. A general element e is a Z-linear combinations of various cosets xGn. For the cosets xGn+1 ∗ xGn+1 =[x, x]G2n+1 = 0 and xGn+1 ∗ yGm+1 = [x, y]Gn+m+1 = [y, x]−1Gn+m+1 =yGm+1 ∗ xGn+1. It follows that e ∗ e = 0.

Finally, the Jacobi identity can be checked on homogeneous elements(cosets). From the Hall identity, [[x, y−1]z]yGn+m+k+2+[[y, z−1]x]zGn+m+k+2+[[z, x−1]y]xGn+m+k+2 = 0. Using the centrality trick, [[x, y−1]z]yGn+m+k+2 =[[x, y−1]z]Gn+m+k+2 = −[[x, y]z]Gn+m+k+2. 2

22.2 Lower central m-series

Let m ∈ Z≥0. We define Gn recursively. The definition depends onm, which is not reflected in the notation but we find alternatives such asGn,m or Gn(m) too cumbersome. We set G1 = G. Recursively, Gn+1 is thesubgroup of G generated by all [x, y] and ym for all x ∈ G, y ∈ Gn.

Proposition 22.2 For each m, the series Gn is strongly normal.

Proof: First observe inductively that Gn is normal. Clearly, G = G1 isnormal. Let Gn be normal. Since [x, y]z = [xz, yz] and (ym)z = (yz)m forall x, z ∈ G, y ∈ Gn, the next subgroup Gn+1 is normal too.

Now we prove that [Gn, Gk] ⊆ Gn+k by induction on minimum of n andk. By the definition of Gn+1 it contains [Gn, G]. Now we assume that n ≥ kand that we have proved [Gn, Gk−1] ⊆ Gn+k−1. Now Gk is generated bytwo types of elements and it is sufficient to check that [x, [y, z]] and [x, ym]belong to Gn+k for all x ∈ Gn, y ∈ Gk−1, z ∈ G.

For the first element, [x, [y, z]] = [[x∗, y]z∗][[z, x∗]y∗], thanks to part (iii)of Proposition 21.3. Since all the subgroup are normal, x∗ ∈ Gn, y∗ ∈ Gk−1.Then [[x∗, y]z∗] ∈ [Gn+k−1, G1] ⊆ Gn+k and [[z, x∗]y∗] ∈ [Gn+1, Gk−1] ⊆Gn+k by the induction assumptions.

For the second element, [x, ym] = [x, y][x, y]∗ . . . [x, y]∗ with m commu-tators on the right, thanks to part (ii) of Proposition 21.3. Since [x, y] ∈Gn+k−1 and Gn+k−1/Gn+k is central in G/Gn+k we conclude that on thelevel of cosets [x, y]∗Gn+k = [x, y]Gn+k. Hence, [x, ym]Gn+k = ([x, y]Gn+k)

m =[x, y]mGn+k = Gn+k since Gn+k contains m-th powers. Hence [x, ym] ∈Gn+k. 2

We write Lm(G) = L(G,Gn) where Gn is the lower central m-series.Some values of m are more useful then the others. L1(G) is always zeroand quite useless. L0(G) is the Lie ring of the group G but may not bea Lie algebra. Lp(G) is a Lie algebra over the field of p elements and isparticularly useful.

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Let us consider three concrete examples. As the first example, G =D2n =< a, b|an = b2 = 1 , ab = a−1 > is the dihedral group of order 2nwhere n = 2sm for oddm. Working out the central 0-series, the commutatorsubgroup is G2 =< a2 >, and, in general, Gt =< a2t−1

>. Hence, startingfrom s + 1, Gs+1 = Gs+2 = . . . =< am >. The elements b = bG2, ai =a2i−1

Gi+1, 1 ≤ i ≤ s form a basis L0(D2n) over the field of two elements.It follows that L0(D2n) = L2(D2n). The products of the basis elements areb ∗ ai = ai+1 and b ∗ as = ai ∗ aj = 0.

As the second example, we consider one of two non-abelian groups oforder p3, p is prime, the extraspecial group C2

p ⋋Cp. The normal subgroupis a vector space over Fp. The generator of the second group acts via matrix(

1 10 1

)that has order p in characteristic p. In generators and relations,

G =< x, y, z|xp = yp = zp = 1, yz = yx, xy = yx, xz = x > Its 0-series isG2 =< x > and G3 = {1}. The elements y = yG2, z = zG2 and x = xG3

form a basis of L0(G) = Lp(G) over Fp. The Lie structure is given byy ∗ z = x, x ∗ z = x ∗y = 0. Thus, L0(G) is the Heisenberg Lie algebra overFp.

As the final example, we consider the modular group G = Cp2 ⋋ Cp. In

generators and relations, G =< y, z|yp2 = zp = 1, yz = yp+1 >. Notice that(p+1)p ≡ 1 modulo p2. Again G2 =< yp >, G3 = {1} and y = yG2, z = zG2

and x = ypG3 form a basis of L0(G) = Lp(G) over Fp. The Lie structure isthe same as for the extraspecial group: y ∗ z = x, x ∗ z = x ∗ y = 0.

22.3 Exercises

Prove that each strongly central series is central.Prove that the intersection of finitely many subgroups of finite index has

finite index too.Use this to prove that a subgroup of finite index contains a normal

subgroup of finite index.Prove that every finite p-group is nilpotent.

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