1. Curvature of left invariant Riemannian metrics on Lie

groups

Mohamed Boucetta

Cadi-Ayyad Univerity

FSTG Marrakesh

Email: m.boucetta@uca.ma

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CONTENTS

1. Curvature of left invariant Riemannian metrics on Lie groups . . . . . . . . 1

1.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2

1.2 Basic tools on Riemannian manifolds . . . . . . . . . . . . . . . . . . 3

1.2.1 Riemannian metrics and Levi-Civita connection . . . . . . . . 3

1.2.2 Curvature of Riemannian metrics . . . . . . . . . . . . . . . . 9

1.2.3 Ricci curvature, scalar curvature and Killing vector fields . . . 16

1.3 Left invariant Riemannian metrics on Lie groups . . . . . . . . . . . . 18

1.3.1 Definition and basic properties . . . . . . . . . . . . . . . . . . 18

1.3.2 Curvatures on the Lie algebra of a Riemannian Lie group . . . 20

1.3.3 Curvature of bi-invariant Riemannian metrics on Lie groups . 27

1.4 Sectional curvature of Riemannian Lie groups . . . . . . . . . . . . . 30

1.4.1 Riemannian Lie groups with strictly positive sectional curvature 30

1.4.2 Flat Riemannian Lie groups . . . . . . . . . . . . . . . . . . . 30

1.4.3 Riemannian Lie groups with negative sectional curvature . . . 34

1.5 Ricci curvature of Riemannian Lie groups . . . . . . . . . . . . . . . . 36

1.5.1 Riemannian Lie groups with positive Ricci curvature metrics . 37

1.5.2 Riemannian Lie groups with negative Ricci curvature metrics . 37

1.6 Scalar curvature of Riemannian Lie groups . . . . . . . . . . . . . . . 42

1.7 The 3-dimensional case . . . . . . . . . . . . . . . . . . . . . . . . . . 44

1.1 Introduction

This course is based on the survey paper of Milnor [13] which has became a

classic. As its title suggests, this paper deals with the curvature properties of the

large class of so called Riemannian Lie groups, i.e., Lie groups endowed with left

invariant Riemannian metrics. Milnor’s paper gave many known results on the topic,

proved and conjectured many new ones. This triggered a huge interest among the

community of mathematicians and, since, many conjectures in the paper has been

proved and the research on the topic is still going on. It is a well established fact

that there is deep relations between the topology and the geometry of a manifold in

2

3

one side, and the curvature of a given Riemannian metric on this manifold on the

other side. There is a long list of theorems supporting this fact (see for instance

[4]). When the manifold is a Lie group and the metric is left invariant the curvature

is also strongly related to the group’s structure or equivalently to the Lie algebra’s

structure. The main purpose of this course is to illustrate this fact by giving a

detailed and self-contained exposition (based on [13]) of some results on this subject

according to the spirit of this school which is using algebra to study geometric

problems. I will deliberately avoide the use of some strong geometric theorems a

part for Meyer’s Theorem which is used in the proof of Theorem 1.5.2. Although

the content is mainly the same as [13], I will adopt a different approach to establish

basic formulas and to prove some results. For instance, the proof of Theorem 1.4.2

is completely different from the original proof by Milnor. There are some results

which did not appear in Milnor’s paper, namely, Proposition 1.4.3, Theorem 1.6.1

and Theorem 1.5.4 proved in [6].

A part from the introduction, the course is divided into six sections. In the

first one we recall the basic tools in Riemannian geometry. The second one is

devoted to the preliminaries properties of left invariant Riemannian metrics on Lie

groups. Namely, we establish the formulas giving different curvatures at the level

of the associated Lie algebras. We study also the particular case of bi-invariant

Riemannian metrics. In the third section, we study Riemannian Lie groups with

strictly negative, null or strictly positive sectional curvature. The same thing is

done in the fourth and the fifth sections for the Ricci curvature and the scalar

curvature, respectively. In the last section, we give a complete description of the

class of 3-dimensional Riemannian Lie groups.

1.2 Basic tools on Riemannian manifolds

1.2.1 Riemannian metrics and Levi-Civita connection

A Riemannian metric on a smooth manifold M is a map which associate to

any point p ∈M a scalar product 〈 , 〉p on TpM such that, for any local coordinates

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system (x1, . . . , xn) on an open set U , the local functions gij : U −→ R given by

gij = 〈∂xi , ∂xj〉 (1.1)

are smooth for any i, j = 1, . . . , dimM = n. A smooth manifold with a Riemannian

metric is called a Riemannian manifold.

Let (M, 〈 , 〉) be a Riemannian manifold and (x1, . . . , xn) a local coordinates

system. The local expression of 〈 , 〉 is given by

〈 , 〉 =∑i,j

gijdxidxj, (1.2)

where gij = are given by (1.1) and

dxidxj =1

2(dxi ⊗ dxj + dxj ⊗ dxi).

Proposition 1.2.1 Any smooth manifold carries a Riemann metric.

Proof. Choose a locally finite open covering U = {Uα}α∈A of M by chart

domains and a subordinate partition of the unity (fα : M −→ [0, 1])α∈A. For any

α ∈ A define on Uα a Riemannian metric 〈 , 〉α by putting

〈 , 〉α =n∑i=1

(dxi)2.

Now define 〈 , 〉 on M by putting, for any p ∈M and any u, v ∈ TpM ,

〈u, v〉 =∑α∈A

fα〈u, v〉α. (1.3)

One can see easily that 〈 , 〉 is a Riemannian metric on M . �

Let (M, 〈 , 〉) be a Riemannian manifold. For any u ∈ TM , put

‖u‖ =√〈u, u〉.

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For any smooth curve γ : [a, b] −→M , the length of γ is given by

`(γ) =

∫ b

a

‖γ′(t)‖dt.

For any p, q ∈M , put

d(p, q) = infγ∈Cpq

`(γ),

where Cpq is the set of all curve γ joining p to q.

Proposition 1.2.2 (M,d) is a metric space.

Proof. It is obvious that d satisfies:

1. d(p, q) ≥ 0,

2. d(p, q) = d(q, p),

3. d(p, q) ≤ d(p, r) + d(r, q).

To achieve the proof we need to show that if p 6= q then d(p, q) > 0. Fix a point p and

a chart φ : U −→ Rn around p. Then there exists δ > 0 such that φ−1(B(φ(p), δ)) ⊂U where B(φ(p), δ) is the Euclidean ball of center φ(p) with radius δ. There exists

λ > 0 such that for any ξ ∈ Tφ−1(B(φ(p), δ)),

‖ξ‖ ≥ λ

√√√√ n∑i=1

ξ2i ,

where ξ =∑ξi∂xi . Therefore, for q ∈ φ−1(B(φ(p), δ)),

d(p, q) ≥ λ|φ(p)− φ(q)|.

For q ∈M \ φ−1(B(φ(p), δ)), we have obviously

d(p, q) ≥ λδ,

which achieves the proof. �

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A linear connection∇ on a Riemannian manifold (M, 〈 , 〉) is called compatible

with the Riemannian structure if, for any smooth curve γ : I −→ M and for any

t0, t1 ∈ I, the parallel transport τt0,t1 : Tγ(t0)M −→ Tγ(t1)M preserves the scalar

product, i.e.,

〈τt0,t1(u), τt0,t1(v)〉 = 〈u, v〉, u, v ∈ Tγ(t0)M.

Proposition 1.2.3 Let (M, 〈 , 〉) be either a Riemannian manifold and ∇ a linear

connection. The following assertions are equivalent:

1. ∇ is compatible with 〈 , 〉.

2. For any γ : I −→M and any couple of vector field V,W along γ,

d

dt〈V (t),W (t)〉 = 〈DγV (t),W (t)〉+ 〈V (t), DγW (t)〉.

3. For any vector field X, Y, Z on M ,

∇X〈Y, Z〉 := X.〈Y, Z〉 − 〈∇XY, Z〉 − 〈Y,∇XZ〉 = 0.

Proof. 1. =⇒ 2. Fix a smooth curve γ : I −→ M . The fact that ∇ is

compatible with 〈 , 〉 implies that, for any V,W parallel along γ, we have

d

dt〈V (t),W (t)〉 = 0.

Choose a family of vector fields parallel along γ say (E1, . . . , En) such that for any

t ∈ I, (E1(t), . . . , En(t)) is a basis of Tγ(t)M . So if V and W are two vector fields

along γ we have

V =n∑i=1

ViEi and W =n∑i=1

WiEi,

where Vi,Wi ∈ C∞(I,R). Thus

DγV =n∑i=1

V ′iEi and DγW =n∑i=1

W ′iEi,

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Then

d

dt〈V (t),W (t)〉 =

n∑i,j=1

(V ′i (t)Wj(t) + Vi(t)W

′j(t))〈Ei, Ej〉,

and the desired formula holds.

2. =⇒ 3. Fix a point p ∈ M and denote by γ : [0, ε[ the integral curve of X

passing through p. Thus

X.〈Y, Z〉(p) =d

dt |t=0〈Y (γ(t)), Z(γ(t))〉

= 〈(DγY )(γ(0)), Z(p)〉+ 〈Y (p), (DγY )(γ(0))〉

= 〈∇XY (p), Z(p)〉+ 〈Y (p),∇XZ(p)〉,

and the formula follows.

3. =⇒ 1. Exercise. �

Levi-Civita connection The following theorem is a fundamental result in Rie-

mannian geometry.

Theorem 1.2.1 For any Riemannian manifold (M, 〈 , 〉) there exists a unique

linear connection torsion free and compatible with the metric.

This connection is called Levi-Civita connection.

Proof. Since 〈 , 〉 is nondegenerate, to compute ∇XY it suffices to compute

〈∇XY, Z〉. By using repeatedly the fact that ∇ is compatible with the metric and

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torsion free, we get

〈∇XY, Z〉 = X.〈Y, Z〉 − 〈Y,∇XZ〉

= X.〈Y, Z〉 − 〈Y,∇ZX〉 − 〈Y, [X,Z]〉

= X.〈Y, Z〉 − Z.〈Y,X〉+ 〈∇ZY,X〉 − 〈Y, [X,Z]〉

= X.〈Y, Z〉 − Z.〈Y,X〉+ 〈∇YZ,X〉+ 〈[Z, Y ], X〉

−〈Y, [X,Z]〉

= X.〈Y, Z〉 − Z.〈Y,X〉+ Y.〈Z,X〉 − 〈Z,∇YX〉

+〈[Z, Y ], X〉 − 〈Y, [X,Z]〉

= X.〈Y, Z〉 − Z.〈Y,X〉+ Y.〈Z,X〉 − 〈Z,∇XY 〉

+〈Z, [X, Y ]〉+ 〈[Z,X], Y 〉+ 〈[Z, Y ], X〉.

Thus

2〈∇XY, Z〉 = X.〈Y, Z〉+ Y.〈X,Z〉 − Z.〈X, Y 〉

+〈[X, Y ], Z〉+ 〈[Z,X], Y 〉+ 〈[Z, Y ], X〉. (1.4)

This formula gives the uniqueness and can be used to define ∇ and to check that ∇satisfies the required properties. �

The formula (1.4) is called Koszul’s formula. Let use it to compute the

Christoffel symbols of Levi-Civita connection. Let (x1, . . . , xn) be a coordinates

system. The Christoffel symbols are given by

∇∂i∂j =n∑l=1

Γlij∂l.

According to (1.4), we have, for any i, j, l = 1 . . . , n,

2〈∇∂i∂j, ∂l〉 = ∂i.gjl + ∂j.gil − ∂l.gij,

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where gij =< ∂i, ∂j >. So

n∑k=1

gklΓkij =

1

2(∂i.gjl + ∂j.gil − ∂l.gij) .

Denote by G = (gij)1≤i,j≤n and its inverse by G−1 = (gij)1≤i,j≤n, we get

Γkij =1

2

n∑l=1

gkl (∂igjl + ∂jgil − ∂lgij) . (1.5)

1.2.2 Curvature of Riemannian metrics

Theorem 1.2.2 Let (M, 〈 , 〉) be a n-dimensional Riemannian manifold and ∇ its

Levi-Civita connection. Then the map

R : X (M)×X (M)×X (M) −→ X (M)

given by

R(X, Y )Z = ∇[X,Y ]Z − (∇X∇YZ −∇Y∇XZ) (1.6)

is a tensor field of type (3, 1). Moreover, R is the unique tensor field satisfying for

any variation (s, t) 7→ Γ(s, t) ∈Mand any vector field along Γ

DSDTY −DTDSY = −R(S, T )Y. (1.7)

Proof. We will show three assertions:

1. R given by (1.6) is a tensor field, i.e., it is C∞(M) 3-linear.

2. R satisfies (1.7).

3. If R′ satisfies (1.7) then R′ = R.

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Let f ∈ C∞(M). we have

−R(fX, Y )Z = ∇fX∇YZ −∇Y∇fXZ −∇[fX,Y ]Z

= f∇X∇YZ −∇Y f∇XZ −∇f [X,Y ]−Y (f)XZ

= f∇X∇YZ − f∇Y∇XZ − Y (f)∇XZ − f∇[X,Y ]Z + Y (f)∇XZ

= −fR(X, Y )Z.

SinceR(X, Y )Z = −R(Y,X)Z, we deduce immediately thatR(X, Y )Z = fR(X, Y )Z.

On the other hand, we have

−R(X, Y )fZ = ∇X∇Y fZ −∇Y∇XfZ −∇[X,Y ]fZ

= ∇X(f∇YZ + Y (f)Z)−∇Y (f∇XZ +X(f)Z)− f∇[X,Y ]Z − [X, Y ](f)Z

= f∇X∇YZ +X(f)∇YZ + Y (f)∇XZ +X(Y (f))Z − f∇Y∇XZ

−Y (f)∇XZ −X(f)∇YZ − Y (X(f))Z − f∇[X,Y ]Z − [X, Y ](f)Z

= −fR(X, Y )Z,

so we have shown the first assertion.

Let (x1, . . . , xn) be a coordinates system and put Γ(s, t) = (x1(s, t), . . . , xn(s, t))

and Y =∑n

i=1 Yi(s, t)∂i. We get

T (s, t) =n∑i=1

∂xi∂t∂i and S(s, t) =

n∑i=1

∂xi∂s

∂i.

We have also

DTY =n∑i=1

∂Yi∂t

∂i +n∑

i,j=1

Yi∂xj∂t∇∂j∂i

DSDTY =n∑i=1

∂2Yi∂s∂t

∂i +n∑

i,j=1

{∂Yi∂t

∂xj∂s

+∂Yi∂s

∂xj∂t

+ Yi∂2xj∂s∂t

}∇∂j∂i

+∑i,j,k

Yi∂xj∂t

∂xk∂s

(∇∂k∇∂j∂i).

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The expression of DTDSY is similar, one needs just to convert the roles of s

and t. So we get

(DSDT −DTDS)Y =∑i,j,k

Yi∂xj∂t

∂xk∂s

(∇∂k∇∂j∂i −∇∂j∇∂k∂i

)=

∑i,j,k

Yi∂xj∂t

∂xk∂s

R(∂k, ∂j)∂i = −R(S, T )Y,

which achieves the proof of the second assertion.

Let R′ be a (3, 1)-tensor field satisfying (1.7). Let p ∈M and φ = (x1, . . . , xn)

a coordinates system around p satisfying φ(p) = 0. For 1 ≤ i < j ≤ n fixed, we

consider the variation Γ given by

Γ(s, t) = φ−1(0, . . . , s, . . . , t, . . . , 0),

s is at the i-place and t at the j-place. We have S = ∂i and T = ∂j. Take Y = ∂k.

Since R′ satisfies (1.7), we get

R′(∂i, ∂j)∂k = −(DSDT −DTDS)Y = R(∂i, ∂j)∂k

thus R′ = R and the proof of the theorem is given. �

Let us express the curvature tensor in a coordinates system (x1, . . . , xn). Since

R is linear in its three entries, there exists locale functions Rlijk such that

R(∂i, ∂j)∂k =n∑l=1

Rlijk∂l.

Let us compute them in function of Christoffel symbols. We have

∇∂j∂k =n∑l=1

Γljk∂l

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which give

∇∂i∇∂j∂k =n∑l=1

∂i(Γljk)∂l +

∑l,m

ΓljkΓmil ∂m.

By gathering these terms, we get

∇∂i∇∂j∂k =n∑l=1

(∂i(Γ

ljk) +

n∑m=1

ΓmjkΓlim

)∂l.

Since [∂i, ∂j] = 0, we get finally

Rlijk = ∂j(Γ

lik)− ∂i(Γljk) +

n∑m=1

(ΓmikΓ

ljm − ΓmjkΓ

lim

). (1.8)

The following proposition summarizes the algebraic properties of the curvature

tensor.

Proposition 1.2.4 Let (M, 〈 , 〉) be a Riemannian manifold. For any p ∈ M and

any x, y, z, w ∈ TpM , we have

1. R(x, y)z = −R(y, x)z (skew-symmetry);

2. R(x, y)z +R(y, z)x+R(z, x)y = 0 (Bianchi’s identity);

3. 〈R(x, y)z, w〉 = −〈R(x, y)w, z〉;

4. 〈R(x, y)z, w〉 = 〈R(z, w)x, y〉.

Proof. The relation 1. and 2. are multi-linear en x, y, z so it suffices to

establish them for ∂i, ∂j, ∂k associated to a local coordinates system (x1, . . . , xn).

The relation 1. is immediate. We have

−R(∂i, ∂j)∂k −R(∂j, ∂k)∂i −R(∂k, ∂i)∂j =(∇∂i∇∂j∂k −∇∂j∇∂i∂k

)+(∇∂j∇∂k∂i −∇∂k∇∂j∂i

)+ (∇∂k∇∂i∂j −∇∂i∇∂k∂j) =

∇∂i

(∇∂j∂k −∇∂k∂j

)−∇∂j (∇∂i∂k −∇∂k∂i) +∇∂k

(∇∂i∂j −∇∂j∂i

)= 0,

since ∇∂i∂j −∇∂j∂i = [∂i, ∂j] = 0.

For the relation 3. we consider a family of vector fields X, Y, Z,W extending

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x, y, z, w. We have, since ∇ preserves the Riemannian metric,

〈∇X∇YZ,W 〉 = X.〈∇YZ,W 〉 − 〈∇YZ,∇XW 〉

= XY.〈Z,W 〉 −X.〈Z,∇YW 〉 − Y.〈Z,∇XW 〉

+〈Z,∇Y∇XW 〉,

〈∇Y∇XZ,W 〉 = Y X.〈Z,W 〉 − Y.〈Z,∇XW 〉 −X.〈Z,∇YW 〉

+〈Z,∇X∇YW 〉,

〈∇[X,Y ]Z,W 〉 = (XY −XY ).〈Z,W 〉 − 〈Z,∇[X,Y ]W 〉.

We deduce by using (1.6) that

〈R(X, Y )Z,W 〉 = −〈Z,R(X, Y )W 〉,

which gives the third formula.

The fourth formula is a purely algebraic consequence of the three first relations. We

will establish it in the following lemma. �

Lemma 1.2.1 Let r : V 4 −→ R be a 4-linear map on a real vector space V satisfying

1. r(x, y, z, w) = −r(y, x, z, w);

2. r(x, y, z, w) + r(y, z, x, w) + r(z, x, y, w) = 0;

3. r(x, y, z, w) = −r(x, y, w, z).

Then r satisfies

r(x, y, z, w) = r(z, w, x, y). (1.9)

Proof. We will get this relation by permuting the second relation in the

following way:

0 = r(x, y, z, w) + r(y, z, x, w) + r(z, x, y, w)

0 = r(y, z, w, x) + r(z, w, y, x) + r(w, y, z, x)

0 = −r(z, w, x, y)− r(w, x, z, y)− r(x, z, w, y)

0 = −r(w, x, y, z)− r(x, y, w, z)− r(y, w, x, z)

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and by using the first and third relation. �

A 4-linear form satisfying the hypothesis of Lemma 1.2.1 is called of curvature

type.

Example 1 Let V be a real vector space endowed with a scalar product 〈 , 〉. Then

the map ρ : V 4 −→ R given by

ρ(x, y, z, w) =

∣∣∣∣∣∣ 〈x,w〉 〈x, z〉〈y, w〉 〈y, z〉

∣∣∣∣∣∣is of curvature type.

Definition 1.2.1 Let (M, 〈 , 〉) be a Riemannian manifold. For any p ∈ M and

for any couple of linearly independent tangent vectors u, v ∈ TpM , put

Q(u, v) =〈R(u, v)u, v〉|u|2|v|2 − 〈u, v〉2

.

The map Q is called sectional curvature of M .

Proposition 1.2.5 The sectional curvature Q(u, v) depends only on the plan spanned

by u, v.

Proof. It is a consequence of the following lemma. �

Lemma 1.2.2 Let r, ρ : V 4 −→ R be two 4-linear forms of curvature type. If

P = span{x, y} = span{x′, y′} and Q = span{z, w} = span{z′, w′} where P and Q

are two plans in V . Then

r(x, y, z, w)

ρ(x, y, z, w)=r(x′, y′, z′, w′)

ρ(x′, y′, z′, w′).

Proof. Exercise. �

The curvature tensor gives the sectional curvatures. The convert is also true.

Proposition 1.2.6 The curvature tensor is entirely determined by the sectional

curvatures.

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Proof. The formula which gives R in function of K is given in the following

lemma. �

Lemma 1.2.3 Let r : V 4 −→ R be of curvature type and let s given by s(x, y) =

r(x, y, y, x). Then

24r(x, y, z, w) = s(x+ w, y + w)− s(x− w, y + z)

−s(x+ w, y − z)− s(x− w, y − z)

−s(y + w, x+ z)− s(y − w, x+ z)

+s(y + w, x− z)− s(y − w, x− z).

Proof. We find first an intermediary relation:

r(x, y + z, y + z, w)− r(x, y − z, y − z, w)− r(y, x+ z, x+ z, w) + r(y, x− z, x− z, w)

= 2r(x, y, z, w) + 2r(x, z, y, w)− 2r(y, x, z, w)− 2r(y, z, x, w)

= 4r(x, y, z, w) + 2r(x, z, y, w) + 2r(z, y, x, w)

= 6r(x, y, z, w) + 2r(y, x, z, w) + 2r(x, z, y, w) + 2r(z, y, x, w)

= 6r(x, y, z, w).

Therefore we use

r(u+ w, v, v, u+ w)− r(u− w, v, v, u− w) = 2r(u, v, v, w) + 2r(w, v, v, u)

= 4r(u, v, v, w).

�

A Riemannian manifold has a positive (resp. negative) sectional curvature if

for any p ∈M and for any plan P ⊂ TpM , Q(P ) ≥ 0 (resp. Q(P ) ≤ 0).

A Riemannian manifold has a strictly positive (resp. strictly negative) sectional

curvature if for any p ∈M and for any plan P ⊂ TpM , Q(P ) > 0 (resp. Q(P ) < 0).

A Riemannian manifold has a constant sectional curvature κ if for any p ∈ M and

for any plan P ⊂ TpM , Q(P ) = κ.

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Proposition 1.2.7 (M, 〈 , 〉) is of constant sectional curvature κ if and only if the

curvature tensor is given by

R(x, y)z = κ(〈x, z〉y − 〈y, z〉x).

Proof. One can see easily that if the curvature tensor is given by the formula

above then the sectional is constant. For the converse, put

r(x, y, z, w) = 〈R(x, y)z, w〉 − κ(〈x, z〉〈y, w〉 − 〈y, z〉〈x,w〉).

Then r is of curvature type and for any x, y, r(x, y, x, y) =constante= 0. Then by

lemma 1.2.3 r = 0 and the lemma follows. �

1.2.3 Ricci curvature, scalar curvature and Killing vector fields

Let (M, 〈 , 〉) be a Riemannian metric and R its curvature tensor. The Ricci

curvature is the trace of R, namely, if (e1, . . . , en) is an orthonormal basis of TpM ,

we have for any u, v ∈ TpM ,

ric(u, v) = tr(x 7→ R(u, x)v)

=n∑i=1

〈R(u, ei)v, ei〉

=n∑i=1

〈R(v, ei)u, ei〉 (see Proposition 1.2.4)

= ric(v, u). (1.10)

We can see the Ricci curvature as an endomorphism Ricp : TpM −→ TpM via the

formula

ric(u, v) = 〈Ricp(u), v〉 = 〈Ricp(v), u〉, ∀u, v ∈ TpM. (1.11)

From (1.10), we have

Ricp(u) =n∑i=1

R(u, ei)ei. (1.12)

We call Ric Ricci operator. It is a symmetric field of endomorphism and hence has

real eigenvalues in each point p ∈ M say λ1(p) ≤ . . . ≤ λn(p). The metric has

17

positive (resp. strictly positive) Ricci curvature if, for any i = 1, . . . , n, λi(p) ≥ 0

(resp. λi(p) > 0). In analogue way, we can define metric of negative Ricci curvature

and strictly negative curvature. The metric is called Einstein if ric = λ〈 , 〉 where

λ is a constant.

The scalar curvature of (M, 〈 , 〉) is the C∞ function s : M −→ R given by

s(p) = trRicp =n∑i=1

λi(p) =n∑i=1

ric(ei, ei). (1.13)

A Killing vector field of (M, 〈 , 〉) is a vector field X such that its local flow

is consisting of isometries of 〈 , 〉. This is equivalent to the Lie derivative of 〈 , 〉along X vanishes, i.e.,

LX(〈 , 〉)(Y, Z) = X.〈Y, Z〉 − 〈[X, Y ], Z〉 − 〈[X,Z], Y 〉 = 0, (1.14)

for any Y, Z ∈ X (M). It is obvious that the space of Killing vector fields Kill(〈 , 〉)is a real Lie algebra for the Lie bracket. It is a non obvious fact (see [11]) that,

actually when the metric is geodesically complete Kill(〈 , 〉) is a finite dimensional

Lie algebra.

We end this section by stating Meyer’s Theorem. For the proof, one can see

[12] pp. 201.

Theorem 1.2.3 Suppose M is a complete, connected Riemannian n-manifold whose

Ricci tensor satisfies the following inequality for any u ∈ TM :

ric(u, u) ≥ n− 1

ρ2|u|2.

Then M is compact, with finite dimensional fundamental group, and diameter at

most πρ.

18

1.3 Left invariant Riemannian metrics on Lie groups

1.3.1 Definition and basic properties

Let G be a n-dimensional Lie group, X `(G) the Lie algebra of left invariant

vector fields on G and g = TeG where e stands for the neutral element of G. For

any u ∈ g, we denote by u` ∈ X `(G) so that

u`(a) = TeLa(u), (1.15)

where TeLa : g −→ TaG is the tangent application at e for the left multiplication by

a, La : G −→ G, b 7→ ab. It is obvious that the map u 7→ u` is an isomorphism of

real vector spaces between g and X `(G). Thus g carries a Lie bracket [ , ]g given by

[u, v]g = [u`, v`](e), ∀u, v ∈ g.

A left invariant Riemannian metric on G is a Riemannian metric h on G such that

for any a ∈ G, La : G −→ G is an isometry of h. This means that, for any a, b ∈ Gand any u, v ∈ TbG,

h (TeLa(u), TeLa(v)) = h(u, v).

We call Riemannian Lie group a Lie group with a left invariant Riemannian metric.

Proposition 1.3.1 Let h be a Riemannian metric on G. Then the following asser-

tions are equivalent.

(i) The metric h is left invariant.

(ii) For any couple of left invariant vector fields X, Y , h(X, Y ) is a constant func-

tion.

Proof. If a, b ∈ G and u, v ∈ TbG. There exists an unique couple of left invariant

vector fields Y, Z such that Y (b) = u and Z(b) = v. Then

h (TeLa(u), TeLa(v)) = h (TeLa(Y (b)), TeLa(Z(b))) = h(Y (ab), Z(ab)).

This shows that (i) is equivalent to (ii) and completes the proof. �

19

Proposition 1.3.2 Let h be a Riemannian metric on a connected Lie group G.

Then the following assertions are equivalent.

(i) The metric h is left invariant.

(ii) The Lie algebra of right vector fields X r(G) is a Lie subalgebra of the Lie

algebra of Killing vector fields Kill(h).

Proof. If X is a right invariant vector field then it is complete and its flow

φX is given, for any t ∈ R, φXt = Lexp(tX(e)). This shows that (i) implies (ii).

On the other hand, let X be a right invariant vector field and Y and Z a couple of

left invariant vector field. Since [X, Y ] = [X,Z] = 0, we get

LXh(Y, Z) = X.h(Y, Z).

This shows that (ii) implies (i) of Proposition 1.3.1 and hence (i). �

LetG be a Lie group. We denote byM`(G) the set of left invariant Riemannian

metrics on G andM(g) the set of definite positive inner products on g. According to

Proposition 1.3.1, the map I :M`(G) −→M(g), h 7→ h(e) is a bijection. Moreover,

we have:

Proposition 1.3.3 Let (G, h) be a Riemannian Lie group. Then the tensor curva-

ture R, the sectional curvature Q, the Ricci curvature ric and the scalar curvature s

of (G, h) are left invariant. Thus for any left invariant vector field X, Y, Z and for

any a ∈ G,

R(X(a), Y (a))Z(a) = TeLa(R(X(e), Y (e))Z(e)), ric(X(a), Y (a)) = ric(X(e), Y (e)).

Moreover, s is a constant and Q(X(a), Y (a)) = Q(X(e), Y (e)).

Proof. This an immediate consequence of the fact that the metric is left

invariant. �

This proposition shows that the sign of the sectional curvature, Ricci curvature

or scalar curvature of a Riemannian Lie group is entirely determined by the values

of these curvatures on the neutral element. We will devote the next section to give

20

useful formulas of the different curvatures on the Lie algebra of a Riemannian Lie

group.

To end this section, we point out that a left invariant Riemannian metric on

a Lie group is geodesically complete.

1.3.2 Curvatures on the Lie algebra of a Riemannian Lie group

Let (G, h) be a Riemannian Lie algebra, (g, [ , ]g) its Lie algebra and 〈 , 〉 =

h(e). For any u ∈ g, we denote by adu : g −→ g the endomorphism given by

aduv = [u, v]g and for any endomorphism F : g −→ g, we denote by F ∗ : g −→ g

the adjoint of F with respect to 〈 , 〉 given by

〈F (u), v〉 = 〈u, F ∗(v)〉, ∀u, v ∈ g.

Through this paper, we denote by Z(g) the centre of g and D(g) its derived ideal.

We denote also by B : g× g −→ g the Killing form given by

B(u, v) = tr(adu ◦ adv).

We call a Lie group G unimodular if for any u ∈ g, tr(adu) = 0.1

We have shown in Proposition 1.3.2 that any right invariant vector field is a

Killing vector field for (G, h), however, not any left invariant vector field is a Killing

vector field. So we consider

L(g) = {u ∈ g, u` ∈ Kill(h)}.

It is clear that L(g) is subalgebra of g, and from (1.14) we can deduce that

L(g) = {u ∈ g, adu + ad∗u = 0}. (1.16)

Note that the centre Z(g) of g is contained in L(g).

The Levi-Civita product on g is the product L : g × g −→ g, (u, v) 7→ Luv

1One can see [13] for a precise definition of unimodular Lie groups.

21

given by

Luv = (∇u`v`)(e),

where ∇ is the Levi-Civita connection associated to (G, h). By using (1.4), we get

for any u, v, w ∈ g

〈Luv, w〉 =1

2{〈[u, v]g, w〉+ 〈[w, u]g, v〉+ 〈[w, v]g, u〉} . (1.17)

For any u ∈ g, we denote by Ru, Ju, adu : g −→ g the endomorphisms of g given

by Ruv = Lvu and Juv = ad∗vu. It is easy to check that Ju is a skew-symmetric

endomorphism and Ju = 0 if and only if u ∈ D(g)⊥.

Proposition 1.3.4 We have the following formulas:

(i) For any u, v ∈ g, Luv − Lvu = [u, v]g, i.e., Lu − Ru = adu.

(ii) For any u, v, w ∈ g, 〈Luv, w〉+ 〈v,Luw〉 = 0. This means that , for any u ∈ g,

Lu is skew-symmetric, i.e., L∗u = −Lu.

(iii) For any u ∈ g, Lu = 12(adu − ad∗u)− 1

2Ju.

(iv) For any u ∈ g, Ru = −12(adu + ad∗u)− 1

2Ju.

Proof. The assertions (iii) and (iv) follow easily from (1.17), (i) and (ii)

follow from (iii) and (iv). �

Remark 1 One can deduce from Proposition 1.3.4 (i) that

L(g) = {u ∈ g, Ru + R∗u = 0}.

We denote by K : g× g −→ End(g) the curvature of (G, h) at e. We have, for

any u, v ∈ g,

K(u, v) = L[u,v]g − [Lu,Lv].

According to Proposition 1.2.4, K satisfies:

1. K(u, v) = −K(v, u) (skew-symmetry);

22

2. K(u, v)w + K(v, w)u+ K(w, u)v = 0 (Bianchi’s identity);

3. 〈K(u, v)w, z〉 = −〈K(u, v)z, w〉;

4. 〈K(u, v)z, w〉 = 〈K(z, w)u, v〉.

Proposition 1.3.5 For any u, v ∈ g,

〈K(u, v)u, v〉 = −3

4|aduv|2 +

1

4|ad∗uv + ad∗vu|2 −

1

2〈aduv, ad∗uv − ad∗vu〉 − 〈ad∗uu, ad∗vv〉.

If the sectional curvature is negative (resp. strictly negative) then L(g) ⊂ D(g)⊥

(resp. L(g) = {0}).

Proof. The formula follows from a direct computation using (1.17) and (i) of

Proposition 1.3.4. Indeed,

〈K(u, v)u, v〉 = 〈L[u,v]gu, v〉 − 〈LuLvu, v〉+ 〈LvLuu, v〉

=1

2〈[[u, v]g, u]g, v〉+

1

2〈[v, [u, v]g]g, u〉+

1

2〈[v, u]g, [u, v]g〉

+〈Lvu,Luv〉 − 〈Luu,Lvv〉

=1

2〈[[u, v]g, u]g, v〉+

1

2〈[v, [u, v]g]g, u〉+

1

2〈[v, u]g, [u, v]g〉

+〈[v, u]g,Luv〉+ 〈Luv,Luv〉 − 〈Luu,Lvv〉

=1

2〈[[u, v]g, u]g, v〉+

1

2〈[v, [u, v]g]g, u〉+

1

2〈[v, u]g, [u, v]g〉

−1

2|[u, v]g|2 +

1

2〈[[v, u]g, u]g, v〉+

1

2〈[[v, u]g, v]g, u〉

+〈Luv,Luv〉 − 〈Luu,Lvv〉

= |Luv|2 − 〈Luu,Lvv〉 − 〈adv ◦ advu, u〉 − |[u, v]g|2.

The desired formula follows from the relation (see Proposition 1.3.4 (iii))

Luv =1

2(aduv − ad∗uv − ad∗vu).

23

Indeed, we have

〈K(u, v)u, v〉 = |Luv|2 − 〈Luu,Lvv〉 − 〈adv ◦ advu, u〉 − |[u, v]g|2

=1

4|aduv − ad∗uv − ad∗vu|2 − 〈ad∗uu, ad∗vv〉 − 〈advu, ad∗vu〉 − |aduv|2

=1

4|aduv|2 +

1

4|ad∗uv|2 +

1

4|ad∗vu|2 −

1

2〈aduv, ad∗uv〉 −

1

2〈aduv, ad∗vu〉

+1

2〈ad∗uv, ad∗vu〉 − 〈ad∗uu, ad∗vv〉 − 〈advu, ad∗vu〉 − |aduv|2

= −3

4|aduv|2 +

1

4|ad∗uv|2 +

1

4|ad∗vu|2 −

1

2〈aduv, ad∗uv〉+

1

2〈aduv, ad∗vu〉

+1

2〈ad∗uv, ad∗vu〉 − 〈ad∗uu, ad∗vv〉

= −3

4|aduv|2 +

1

4|ad∗uv + ad∗vu|2 −

1

2〈aduv, ad∗uv − ad∗vu〉 − 〈ad∗uu, ad∗vv〉.

Let u ∈ L(g). By using adu + ad∗u = 0 and the formula we have just established, we

get for any v ∈ g,

〈K(u, v)u, v〉 =1

4|ad∗vu|2 ≥ 0.

If the sectional curvature is negative then ad∗vu = 0 and hence u ∈ D(g)⊥.

If the sectional curvature is strictly negative then u = 0. �

Example 2 Let G be a non abelian n-dimensional Lie group such that its Lie al-

gebras satisfies the following properties: there exists a linear form ` : g −→ R such

that, for any u, v ∈ g,

[u, v]g = `(u)v − `(v)u.

Let h be a left invariant metric on G and 〈 , 〉 its value on e. Let compute the curva-

ture of h at e by using Proposition 1.3.5. Let u, v be two vector such that 〈u, v〉 = 0,

|u| = |v| = 1. We complete to have an orthonormal basis (u, v, e1, . . . , en−2). A

direct computation shows:

ad∗uv = `(u)v, ad∗vu = `(v)u, ad∗uu = −`(v)v−n−2∑i=1

`(ei)ei, ad∗vv = −`(u)u−n−2∑i=1

`(ei)ei.

24

By using Proposition 1.3.5, we get

Q(u, v) = −`(u)2 − `(v)2 −n−2∑i=1

`(ei)2 = −|`|2 < 0.

So the sectional curvature is a strictly negative constant.

Recall that the Ricci curvature at e is defined by ric(u, v) = tr (w −→ K(u,w)v).

In order to give an useful formula of ric we introduce the mean curvature vector on

g which is the vector given by

〈H, u〉 = tr(adu), ∀u ∈ g. (1.18)

It follows thatG is unimodular if and only ifH = 0. Since tr(ad[u,v]g) = tr([adu, adv]) =

0 then H ∈ D(g)⊥.

Proposition 1.3.6 1. If τ(u, v) : g −→ g is the endomorphism given by τ(u, v)w =

K(u,w)v then

τ(u, v) = −Rv ◦ Ru + Ru.v − [Lu,Rv],

where u.v = Luv.

2. For any u, v ∈ g,

ric(u, v) = −1

2B(u, v)− 1

2tr(adu◦ad∗v)−

1

4tr(Ju◦Jv)−

1

2(〈adHu, v〉+〈adHv, u〉).

Proof.

1. For any u, v, w ∈ g,

τ(u, v)w = K(u,w)v

= L[u,w]gv − Lu ◦ Lwv + Lw ◦ Luv

= Rv ◦ aduw − Lu ◦ Rvw + Ru.vw

= Rv ◦ Luw − Rv ◦ Ruw − Lu ◦ Rvw + Ru.vw (Proposition 1.3.4 (i))

= −Rv ◦ Ruw + Ru.vw + [Rv,Lu]w.

25

This establishes the first formula.

2. From the first formula, we get

ric(u, v) = tr(τ(u, v)) = −tr (Rv ◦ Ru) + tr(Ru.v).

Choose an orthonormal basis (ei)ni=1 of g. We have

tr(Ru.v) =n∑i=1

〈Ru.vei, ei〉

=n∑i=1

〈Lei ◦ Luv, ei〉

(1.17)= −

n∑i=1

〈[Luv, ei]g, ei〉

= −tr(adu.v)

(1.18)= −〈H,Luv〉

(1.17)= −1

2〈adHu, v〉 −

1

2〈adHv, u〉 −

1

2tr(ad[u,v]g)

= −1

2〈adHu, v〉 −

1

2〈adHv, u〉.

So

ric(u, v) = −tr (Rv ◦ Ru)−1

2〈adHu, v〉 −

1

2〈adHv, u〉. (1.19)

To achieve the proof, from Proposition 1.3.4 (iv), we have

Ru = −1

2(adu + ad∗u)−

1

2Ju and Rv = −1

2(adv + ad∗v)−

1

2Jv.

By replacing in (1.19) and using the fact that, since adu+ad∗u is symmetric and

Jv skew-symmetric then tr ((adu + ad∗u) ◦ Jv) = 0, we get the desired formula.

�

Corollary 1.3.1 Let (G, h) be a Riemannian nilpotent Lie group. Then its Ricci

curvature at e is given by

ric(u, v) = −1

2tr(adu ◦ ad∗v)−

1

4tr(Ju ◦ Jv),

26

for any u, v ∈ g.

Proof. It is a consequence of Proposition 1.3.6 and the fact that the Lie

algebra of G is nilpotent and hence, for any u, v ∈ g, tr(adu) = tr(adu ◦ adv) = 0, in

particular H = 0. �

Define now the auto-adjoint endomorphisms Ric, B, J1 and J2 by

〈Ric u, v〉 = ric(u, v),

〈Bu, v〉 = B(u, v),

〈J1u, v〉 = tr(adu ◦ ad∗v),

〈J2u, v〉 = −tr(Ju ◦ Jv).

From Proposition 1.3.6, we get

Ric = −1

2

(B + J1

)+

1

4J2 −

1

2(adH + ad∗H). (1.20)

Proposition 1.3.7 With the notations above we have:

(i) J1, J2 and B + J1 are positive skew-symmetric endomorphisms and hence

their eigenvalues are real positive.

(ii) kerJ1 = Z(g), kerJ2 = D(g)⊥ and ker(B + J1) = L(g).

(iii) For any orthonormal basis (e1, . . . , en) of g,

J1 =n∑i=1

ad∗ei ◦ adei and J2 =n∑i=1

adei ◦ ad∗ei ,

in particular trJ1 = trJ2 ≥ 0 and trJ1 = 0 if and only if g is abelian.

Proof.

(i) This is a consequence of the following relations

〈J1u, u〉 = tr(adu ◦ ad∗u) ≥ 0, 〈J2u, u〉 = tr(Ju ◦ J∗u) ≥ 0,

27

and

〈(B + J1)u, u〉 =1

2tr((adu + ad∗u)

2) ≥ 0.

(ii) This is a consequence of the relations above and the fact that Ju = 0 if and

only if u ∈ D(g)⊥.

(iii) We have

〈J1u, v〉 = tr(ad∗u ◦ adv)

=n∑i=1

〈ad∗u ◦ advei, ei〉

=n∑i=1

〈[v, ei], [u, ei]g〉

=n∑i=1

〈ad∗ei ◦ adeiu, v〉.

This shows the first relation, the second one can be obtained in the same way.

�

From (1.20) and the relation trJ1 = trJ2, we deduce that the scalar curvature

of 〈 , 〉 is given by

s = −1

4

(2tr(B) + trJ1

)− |H|2. (1.21)

Since for a nilpotent Lie algebra B = 0 and H = 0 we deduce:

Proposition 1.3.8 Let G be a Riemannian nilpotent non abelian Lie group. Then

its scalar curvature is given by s = −14trJ1 < 0.

1.3.3 Curvature of bi-invariant Riemannian metrics on Lie groups

A Riemannian metric on a Lie group is bi-invariant if it is both left and right

invariant.

Proposition 1.3.9 Let G be a Lie group and h is a left invariant Riemannian

metric on G. If h is also right invariant then L(g) = g, i.e., for any u ∈ g,

adu + ad∗u = 0. If G is connected the converse is also true.

28

Proof. We have seen in Proposition 1.3.1 that if h is left invariant then

X r(G) ⊂ Kill(h). In a similar way if h is right invariant then X `(G) ⊂ Kill(h)

which is equivalent to L(g) = g. As in Proposition 1.3.1 the converse is true if G is

connected. �

Proposition 1.3.10 Let (G, h) be a Lie group with a bi-invariant Riemannian met-

ric. Then, for any u, v ∈ g,

Lu =1

2adu, K(u, v) =

1

4ad[u,v], 〈K(u, v)u, v〉 =

1

4|[u, v]|2, ric(u, v) = −1

4B(u, v).

In particular, the sectional curvature and the Ricci curvature of h are positive and

ric(u, u) = 0 if and only if u ∈ Z(g).

Proof. It is a an immediate consequence of the fact that ad∗u = −adu, Ju =

adu, H = 0 and Propositions 1.3.4 and 1.3.6. Moreover, for any u ∈ g, ric(u, u) =

14tr(adu ◦ ad∗u) ≥ 0 and ric(u, u) = 0 if and only if adu = 0. �

Theorem 1.3.1 (i) A connected Lie group carries a bi-invariant Riemannian

metric if and only if it is isomorphic to the cartesian product of a compact Lie

group and an abelian Lie group.

(ii) If the Lie algebra of a compact Lie group G is simple then there is up to mul-

tiplication by a strictly positive constant an unique bi-invariant Riemannian

metric and this metric is Einstein.

Proof.

(i) Let (g, [ , ]g, 〈 , 〉) be the Lie algebra of a Riemannian Lie group (G, h) endowed

with a bi-invariant Riemannian metric. From the relation

〈[u, v]g, w〉+ 〈v, [u,w]g〉,

we can deduce easily that g splits orthogonally

g = Z(g)⊕D(g).

29

The Killing form of D(g) is given by

B(u, v) = tr(adu ◦ adv) = −tr(adu ◦ ad∗v),

from any u, v ∈ D(g). This formula shows that B is definite negative and

hence D(g) is semi-simple compact (see for instance [8]). So the universal

covering G of G splits as the cartesian product of a compact Lie H group and

an abelian group Rm and G = G/Γ where Γ is a discrete normal subgroup of

G. Projecting Γ into Rm, let V be the vector space spanned by its image and

let V ⊥ be the orthogonal complement in Rm. Then G is the cartesian product

of the compact Lie group (H × V )/Γ and the vector group V ⊥.

Conversely, any left invariant metric on an abelian Lie group is also bi-invariant

and let G be a compact Lie group. Choose an arbitrary metric µ on g and

define 〈 , 〉 on g by

〈u, v〉 =

∫G

µ(Adgu,Adgv)dλ,

where λ is a Haar measure on G. Then the left invariant Riemannian metric

on G associated to 〈 , 〉 is clearly also right invariant.

(ii) Suppose that the Lie algebra of G is simple, pick two bi-invariant Riemannian

metrics h1 and h2 on G and denote by 〈 , 〉1, 〈 , 〉2 their value at e. For any

u, v ∈ g, the relation

〈u, v〉2 = 〈Au, v〉1 = 〈u,Av〉1

defines a symmetric (with respect both 〈 , 〉1 and 〈 , 〉2) isomorphism of g.

Since h1 and h2 are bi-invariant and according to Proposition 1.3.9, we have

30

for any u, v, w ∈ g,

〈[u, v]g, w〉2 = 〈[u, v]g, Aw〉1

= −〈[u,Aw]g, v〉1

= −〈[u,Aw]g, A−1v〉2

= 〈Aw, [u,A−1v]g〉2

= 〈w,A[u,A−1v]g〉2.

This relation shows that, for any u ∈ g, adu ◦A = A ◦ adu. This implies that,

for any real eigenvalue λ ∈ g of A, the associated eigenspace is an ideal of g.

Now g is simple and A is diagonalizable over R so we must have A = λIdg and

hence 〈 , 〉2 = λ〈 , 〉1 and finally h2 = λh1.

The Ricci curvature of a bi-invariant metric 〈 , 〉 on g satisfies, according to

Proposition 1.3.10, ric = −14B where B is the Killing form of g. Since the

Killing form is bi-invariant argument similar to what above gives B = µ〈 , 〉which shows that 〈 , 〉 is Einstein. �

Corollary 1.3.2 Every compact Lie group admits a left invariant (and in fact bi-

invariant) metric so that all sectional curvatures are positive.

1.4 Sectional curvature of Riemannian Lie groups

1.4.1 Riemannian Lie groups with strictly positive sectional curvature

The following theorem have been proved in [14], we give it without proof since

it needs the introduction of tools which are beyond this course.

Theorem 1.4.1 The 3-sphere group SU(2), consisting of 2× 2 unitary matrices of

determinant 1, is the only simply connected Lie group which admits a left invariant

metric of strictly positive sectional curvature.

1.4.2 Flat Riemannian Lie groups

Let (G, h) be a Riemannian Lie group and (g, [ , ]g, 〈 , 〉) its Lie algebra

endowed with 〈 , 〉 = h(e). One can see easily that the orthogonal of the derived

31

ideal of g is given by

D(g)⊥ = {u ∈ g,Ru = R∗u}. (1.22)

Put

N`(g) = {u ∈ g,Lu = 0} and Nr(g) = {u ∈ g,Ru = 0} .

We have obviously

Nr(g) = (gg)⊥, (1.23)

where gg = span{Luv, u, v ∈ g}.

Proposition 1.4.1 If the curvature of h vanishes then H = 0 and

L(g) = D(g)⊥ = {u ∈ g, Ru = 0}.

Proof. By using (1.17) one can see easily that, for any u ∈ L(g) ∪D(g)⊥, Luu = 0

and deduce from Proposition 1.3.6 (i) that

[Ru,Lu] = R2u.

From this relation, one can deduce by induction that for any k ∈ N∗

[Rku,Lu] = kRk+1

u ,

and hence tr(Rku) = 0 for any k ≥ 2 which implies that Ru is nilpotent. A skew-

symmetric or a symmetric endomorphism is nilpotent iff it vanishes so

L(g) = D(g)⊥ = {u ∈ g, Ru = 0}.

On the other hand, for any u, v ∈ g, tr(ad[u,v]) = 0, thus H ∈ D(g)⊥. Now, for any

u ∈ D(g)⊥, Ru = 0 and hence

tr(adu) = tr(Ru) = 〈H, u〉 = 0,

which implies H ∈ D(g) and hence H = 0. �

32

A Riemannian Lie group (G, h) is flat if its curvature at e vanishes identically.

This is equivalent to the fact that g endowed with the Levi-Civita product is a left

symmetric algebra, i.e., for any u, v, w ∈ g,

ass(u, v, w) = ass(v, u, w),

where ass(u, v, w) = (u.v).w − u.(v.w) and u.v = Luv. Left symmetric algebras (or

under other names like pre-algebras, quasi-associative algebras, Vinberg algebras)

constitute an important topic in both algebra and geometry. For a left invariant

connection on a Lie group its flatness is equivalent to to left symmetry of the induced

product on the Lie algebra. The following theorem was proved first in [13] using

some strong geometric theorems. We give here a more precise statement and an

algebraic proof based on an important result about left symmetric Lie algebras,

namely the fact proved in [9] that for a left symmetric algebra g, g 6= D(g). The

proof we will give appeared first in [1].

Theorem 1.4.2 A Riemannian Lie group Lie group (G, h) is flat if and only L(g)

and D(g) are abelian, D(g)⊥ = L(g) and

g = L(g)⊕D(g).

Moreover, in this case the dimension of D(g) is even and the Levi-Civita product is

given by

La =

ada if a ∈ L(g),

0 if a ∈ D(g).(1.24)

Proof

Suppose that G is a Riemannian flat Lie group and g its Lie algebra. Let

(D(g)k)k∈N denote the commutator series of g defined recursively by

D(g)0 = g, D(g)1 = D(g) and D(g)k+1 = [D(g)k,D(g)k].

33

We deduce from Proposition 1.4.1 that

L(g) = D(g)⊥ = Nr(g) = (gg)⊥. (1.25)

These relations imply that L(g) is abelian, D(g) = gg and hence D(g) is a two-

sided ideal of the Levi-Civita product so D(g) endowed with the restricted metric

is a Riemannian flat Lie algebra and, by induction, for any k ∈ N, D(g)k is a

Riemannian flat Lie algebra. On the other hand, it is known that a non null left

symmetric algebra cannot be equal to its derived ideal (see [9] pp.31). So g must be

solvable and hence D(g) is nilpotent. If D(g) is non abelian then the splitting

D(g) = L(D(g))⊕D2(g)

is non trivial. But the center of D(g) is contained in L(D(g)) and it intersects

non trivially D2(g) (D(g) is nilpotent) so D(g) must be abelian. This achieves the

direct part of the theorem. The equation (1.24) is easy to establish and the converse

follows immediately from this equation.

Suppose that G is flat. Hence L(g) is abelian, D(g) is abelian and

g = L(g)⊕D(g).

If L(g) = {0} then g = D(g) = {0} and the result follows trivially. Suppose now

that L(g) 6= {0}. Let (s1, ..., sp) be a basis of L(g). The restriction of ads1 to D(g)

is a skew-symmetric endomorphism, thus its kernel K1 is of even codimension in

D(g). Now, ads2 commutes with ads1 and K1 is invariant by ads2 . By using the

same argument as above, we deduce that K2 = K1 ∩ ker ads2 is of even codimension

in K1. Finally K2 is of even codimension in D(g). Thus, by induction, we show that

Kp = D(g) ∩ (∩pi=1 ker adsi)

is an even codimensional subspace of D(g). Now from its definition Kp is contained

in the center of g which is contained in L(g) and then Kp = {0} and the second

part of the theorem follows. �

34

1.4.3 Riemannian Lie groups with negative sectional curvature

The study of Riemannian Lie groups with negative sectional curvature is based

on the following result established by Heintze in [7].

Proposition 1.4.2 A connected homogeneous manifold of non-positive curvature

can be represented as a connected solvable Lie group with a left invariant metric.

From this proposition we deduce, in particular, that a Riemannian Lie group

with negative sectional curvature must be solvable. Based on this remark Heintze

gave a classification of Lie groups which admits a left invariant metric of strictly

negative sectional curvature. The main result of Heintze is the following theorem.

Theorem 1.4.3 Let G be a solvable Lie group. Then the following conditions are

equivalent:

(i) G admits a left invariant Riemannian metric of strictly negative curvature.

(ii) dimD(g) = dim g − 1, and there exists A ∈ g such that the eigenvalues of

adA|D(g) have positive real part.

We cannot give the proof of this result, it constitutes the object of [7]. However, we

can give now a key proposition in [7].

Proposition 1.4.3 Let G be a solvable Riemannian Lie group with strictly negative

sectional curvature. Then the following conditions hold:

(i) dimD(g) = dim g− 1.

(ii) there exists a unit vector u in the orthogonal of D(g) such that D : D(g) −→D(g) is positive definite, where D is the symmetric part of adu|D(g) : D(g) −→D(g).

(iii) If S is the skew-symmetric part of adu|D(g) : D(g) −→ D(g), then also D2 +

DS − SD : D(g) −→ D(g) is positive definite.

Proof.

35

(i) Let u ∈ D(g)⊥ and consider (adu)|u⊥ : u⊥ −→ D(g). We will prove that

(adu)|u⊥ is injective. Let v a vector in the kernel of (adu)|u⊥ . From proposition

1.3.5, we get

〈K(u, v)u, v〉 =1

4|ad∗uv|2.

Since Q < 0 this implies that v = 0, (adu)|u⊥ is injective and hence dimD(g) =

dim g− 1.

(ii) Let u be an unit vector in D(g)⊥. Since g is solvable, D(g) is nilpotent and

hence there exists z 6= 0 in the center of D(g). From Proposition 1.3.5, we get

for any v ∈ D(g),

〈K(z, v)z, v〉 =1

4|ad∗zv + ad∗vz|2 − 〈ad∗zz, ad∗vv〉

=1

4|ad∗zv + ad∗vz|2 − 〈z, [z, u]g〉〈v, [v, u]g〉,

since from g = D(g)⊕ Ru, we get

ad∗zz = 〈z, [z, u]g〉u and ad∗vv = 〈v, [v, u]g〉u+ q(v),

where q(v) ∈ D(g). Let D be the symmetric part of adu|D(g). Then

〈K(z, v)z, v〉 =1

4|ad∗zv + ad∗vz|2 − 〈Dz, z〉〈Dv, v〉.

Suppose that v ∈ kerD. From the relation above and the fact that Q < 0 we

must have v = λz. If λ 6= 0 then [u, z]g = 0 which implies that z is in the

center of g. But we have seen in Proposition 1.3.5 that if Q < 0 the center of

g is trivial so z = 0 which is impossible. Thus D is an isomorphism. From

the expression of 〈K(z, v)z, v〉 given above and the fact that Q < 0 we deduce

that D is either definite positive or definite negative, so by replacing u by −uwe get the assertion.

(iii) Fix u ∈ g as in (ii). We have, for any v ∈ D(g), [u, v] = Dv + Sv. From

36

Proposition 1.3.5 and since ad∗uu = ad∗vu = 0, we get for any v ∈ D(g) \ {0},

〈K(u, v)u, v〉 = −3

4|aduv|2 +

1

4|ad∗uv|2 −

1

2〈aduv, ad∗uv〉

= −3

4|Dv + Sv|2 +

1

4|Dv − Sv|2 − 1

2〈Dv + Sv,Dv − Sv〉

= −〈Dv,Dv〉 − 2〈Dv, Sv〉

= −〈(D2 − (SD −DS))v, v〉.

This achieves the proof. �

The Riemannian Lie groups with Q ≤ 0 have been classified by Azencott and

Wilson [3]. Since the statements are complicated we will content ourselves with the

following qualitative result.

Theorem 1.4.4 If a connected Lie group G has a left invariant metric with all

sectional curvatures Q ≤ 0, the it is solvable. If G is unimodular, then any such

metric with Q ≤ 0 must actually be flat (Q ≡ 0).

Remark 2 To our knowledge, unlike the negative case, there is no systematic study

of Riemannian Lie group with positive sectional curvature.

1.5 Ricci curvature of Riemannian Lie groups

We start this section by this general result.

Theorem 1.5.1 Suppose that the Lie algebra of G is nilpotent but not abelian.

Then for any left invariant metric on G the scalar curvature is strictly negative and

there exists a direction of strictly negative Ricci curvature and a direction of strictly

positive Ricci curvature.

Proof. From (1.21) and Proposition 1.3.7, we have s = −14trJ1 < 0.

Since s < 0 there exists v ∈ g such that ric(v, v) < 0. On the other hand, since g

is nilpotent there exists a non null vector u ∈ D(g)∩Z(g). From Proposition 1.3.6,

we get ric(u, u) = −14tr(J2

u). Since Ju is skew-adjoint tr(J2u) ≤ 0 and tr(J2

u) = 0 if

and only if Ju = 0. This is equivalent by Proposition 1.3.7 to u ∈ D(g)⊥ which is

impossible. �

37

1.5.1 Riemannian Lie groups with positive Ricci curvature metrics

Proposition 1.5.1 If a connected Lie group admits a left invariant metric with all

Ricci curvatures positive then it must be unimodular.

Proof. Since H ∈ D(g)⊥, JH = 0 and then from Proposition 1.3.6 we get

ric(H,H) = −1

4tr((adH + ad∗H)2). (1.26)

Since tr(adH) = |H|2 we get adH + ad∗H = 0 iff H = 0. Thus if H 6= 0 then

ric(H,H) < 0. �

Theorem 1.5.2 A connected Lie group admits a left invariant metric with all Ricci

curvatures strictly positive if and only if it is compact with finite fundamental group.

Proof. The direct sense follows from Myers’s Theorem (see Theorem 1.2.3).

In the other direction, if G is compact then we can choose a bi-invariant metric on

G which Ricci curvature, according to Proposition 1.3.10, is given by

ric(u, u) =1

4tr(adu ◦ ad∗u) ≥ 0,

and ric(u, u) = 0 if and only if u ∈ Z(g). Now the hypothesis that G has a finite

fundamental group implies that its universal covering is also compact and hence

Z(g) = {0}. So all Ricci curvatures are strictly positive. �

A proof of the following theorem can be found in [2]. It is based on a geometric

theorem. It can be interesting to find an algebraic proof of this result.

Theorem 1.5.3 A left invariant Riemannian metric on a Lie group is Ricci flat if

and only if it is flat.

1.5.2 Riemannian Lie groups with negative Ricci curvature metrics

It remains open the classification of Lie groups admitting left invariant metrics

with all Ricci curvatures < 0 or ≤ 0. In [6] we have the following result.

Theorem 1.5.4 A solvable unimodular Lie group G with a left invariant metric

has all Ricci curvatures ≤ 0 if and only if its derived Lie algebra D(g) is abelian,

38

D(g)⊥ is a commutative subalgebra and the linear transformation adb is normal for

every b ∈ D(g)⊥.

To prove this theorem, we need the following lemma.

Lemma 1.5.1 Let (g, 〈 , 〉) be an unimodular Lie algebra such that g splits orthog-

onally as

g = u⊕ b,

where u is an abelian ideal and b is an abelian subalgebra. Then

(i) Given an orthonormal basis b1, . . . , br of b and u ∈ u, we have

Ric(u) =1

2[Li, L

∗i ](u),

where Li is the restriction of adbi to u.

(ii) The self-adjoint transformation Ric ≡ 0 on u if and only if, for any b ∈ b, the

restriction of adb to u is a normal transformation.

Proof.

(i) Since G is solvable, the Cartan’s criteria of solvability (see [10]) implies that

for any u ∈ D(g) and any v ∈ g, B(u, v) = 0. So, by (1.20), for any u ∈ D(g)

and any v ∈ g,

ric(u, v) = −1

2〈J1u, v〉+

1

4〈J2u, v〉.

Choose an orthonormal basis (e1, . . . , ep) an orthonormal basis of D(g). From

Proposition 1.3.7, we have

J1u =

p∑i=1

ad∗ei ◦ adeiu+

p∑i=1

ad∗bi ◦ adbiu

=

p∑i=1

ad∗bi ◦ adbiu,

J2u =

p∑i=1

adei ◦ ad∗eiu+

p∑i=1

adbi ◦ ad∗biu.

39

Now

ad∗eiu =r∑j=1

〈u, [ei, bj]g〉bj,

ad∗biu =

p∑j=1

〈u, [bi, ej]g〉ej.

Thus

Ric(u) = −1

2

p∑i=1

ad∗bi ◦ adbiu+1

4

∑i,j

〈u, [ei, bj]g〉[ei, bj]g +1

4

∑i,j

〈u, [bi, ej]g〉[bi, ej]g

= −1

2

p∑i=1

ad∗bi ◦ adbiu+1

2

∑i,j

〈u, [ei, bj]g〉[ei, bj]g

= −1

2

p∑i=1

ad∗bi ◦ adbiu+1

2

∑j

[bj,∑i

〈u, [bj, ei]g〉ei]g

= −1

2

p∑i=1

ad∗bi ◦ adbiu+1

2

∑j

[bj,∑i

〈ad∗bju, ei〉ei]g

=1

2

∑j

[adbi , ad∗bi ](u).

(ii) This is a consequence of the following property. ”Let L1, . . . , Lr be a family

of commuting endomorphisms on Rn such that∑

i[Li, L∗i ] = 0, where L∗i is

the adjoint of Li with respect to an inner definite positive product 〈 , 〉 on

Rn. Then L1, . . . , Lr are normal transformation”. We will prove this property

by induction on n. The property is obviously true for n = 1. Since the Li

commute, the associated transformation Li on Cn have a common eigenvector

x with eigenvalues λ1, . . . , λr. The new family Li − λiI satisfies

r∑i=1

[Li − λiI, (Li − λiI)∗] = 0.

(Here we are considering Cn with the Hermitian product associated to 〈 , 〉).

40

Then

0 =r∑i=1

〈[Li − λiI, (Li − λiI)∗]x, x〉

=r∑i=1

|(Li − λiI)∗|2.

This shows that x is a common eigenvector of L∗i . Therefore, we may re-

strict each Li to the the orthogonal complement of x and use the induction

hypothesis to conclude. �

Proof of Theorem 1.5.4. Since G is solvable, the Cartan’s criteria of solv-

ability (see [10]) implies that for any u ∈ D(g) and any v ∈ g, B(u, v) = 0. Moreover,

D(g) is nilpotent and hence its center, say Z0, is non trivial with Z0∩ [D(g),D(g)] 6={0}. Put

g = Z0 ⊕ h⊕D(g)⊥,

and choose an orthonormal basis (z1, . . . , zp, e1, . . . , eq, f1, . . . , fr) respecting this

splitting. From (1.20) we get, for any z ∈ Z0,

ric(z, z) = −1

2〈J1z, z〉+

1

4〈J2z, z〉.

By using the Jacobi identity, one can see easily that adfi preserves Z, we get for

any i = 1, . . . , r,

[fi, z]g =

p∑j=1

〈[fi, z]g, zj〉zj.

From Proposition 1.3.7, we have

〈J1z, z〉 =r∑i=1

〈[fi, z]g, [fi, z]g〉,

=∑i,j

〈[fi, z]g, zj〉2,

〈J2z, z〉 =

p∑i=1

〈ad∗ziz, ad∗ziz〉+

p∑i=1

〈ad∗eiz, ad∗eiz〉+

p∑i=1

〈ad∗fiz, ad∗fiz〉.

41

Now

ad∗ziz =r∑j=1

〈[zi, fj], z〉fj,

ad∗eiz =

q∑j=1

〈[ei, ej], z〉ej +r∑j=1

〈[ei, fj], z〉fj,

ad∗fiz =

p∑j=1

〈[fi, zj], z〉zj +

q∑j=1

〈[fi, ej], z〉ej +r∑j=1

〈[fi, fj], z〉fj.

So we get

r(z, z) = −1

2

∑i,j

〈[fi, z]g, zj〉2+1

2

∑i,j

〈[zi, fj ], z〉2+1

2

∑i,j

〈[fi, ej ], z〉2+1

4

∑i,j

〈[ei, ej ], z〉2+1

4

∑i,j

〈[fi, fj ], z〉2.

Now since ric ≤ 0, we get

0 ≥∑k

r(zk, zk) =1

2

∑i,j,k

〈[fi, ej], zk〉2 +1

4

∑i,j,k

〈[ei, ej], zk〉2 +1

4

∑i,j,k

〈[fi, fj], zk〉2.

This shows that, for any i, j, k,

〈[fi, ej], zk〉 = 〈[ei, ej], zk〉 = 〈[fi, fj], zk〉 = 0.

This shows that Z0 is orthogonal to [D(g),D(g)] = 0, so Z0 = D(g). We deduce

also that D(g)⊥ is also abelian. We have shown also that for any u, v ∈ D(g),

ric(u, v) = 〈Ricu, v〉 = 0. From Lemma 1.5.1 (i) we have also that Ric preserves

D(g) so for any u ∈ D(g), Ricu = 0. Hence from Lemma (ii), we can achieve the

direct sense of the theorem.

Conversely, if g = D(g) ⊕ D(g)⊥, where D(g) and D(g)⊥ are abelian and adb is

normal for any b ∈ D(g)⊥, we get from Lemma 1.5.1 that for any u ∈ D(g) and

b ∈ D(g)⊥,

ric(u+ b, v + b) = ric(b, b) = −1

2tr((adb + ad∗b)

2) ≤ 0.

This achieve the proof of the theorem. �

42

We can deduce from Theorem 1.5.4 and its proof the following corollary.

Corollary 1.5.1 No unimodular solvable Lie group carries a left invariant Rieman-

nian metric with ric < 0.

Moreover, we can combine the proof of Theorem 1.5.4, Proposition 1.5.1 and

Theorem 1.4.2 to get the following result which is a particular case of Theorem 1.5.3.

Proposition 1.5.2 A left invariant Riemannian metric on a solvable Lie group is

Ricci flat if and only if it is flat.

1.6 Scalar curvature of Riemannian Lie groups

Theorem 1.6.1 Let (g, 〈 , 〉) be a solvable Lie algebra. Then

2tr(B) + tr(J1) ≥ 0

and the equality holds if and only if 〈 , 〉 is flat.

Proof. Note first that by Cartan’s criteria for solvability (See [10] pp. 50) we have

Bu = 0 for any u ∈ D(g). We consider an orthonormal basis B = {e1, . . . , ep, f1, . . . , fq}of g where {e1, . . . , ep} is a basis of D(g) and {f1, . . . , fq} is a basis of D(g)⊥. We

have, according to Proposition 1.3.7,

tr(J1) =

p∑i=1

tr(adei ◦ ad∗ei

)+

q∑i=1

tr(adfi ◦ ad∗fi

),

2tr(B) = 2

q∑i=1

tr (adfi ◦ adfi) .

For i = 1, . . . , p and j = 1, . . . , q, we denote by Mi and Nj the matrix of adei and

adfj in B respectively. We have

Mi =

Ai Xi

0 0

and Nj =

Bj Yj

0 0

,

43

where Mi and Nj are p-square matrix and Xi and Yj are (p, q)-matrix. We have

tr(J1) =

p∑i=1

(tr(AiA

ti

)+ tr

(XiX

ti

))+

q∑j=1

(tr(BjB

tj

)+ tr

(YjY

tj

)),

2tr(B) = 2

q∑i=1

tr(B2i

).

The key formula isp∑i=1

tr(XiX

ti

)=

q∑j=1

tr(BjBtj)

which implies that

2tr(B) + tr(J1) =

p∑i=1

tr(AiA

ti

)+

q∑j=1

tr(YjY

tj

)+

q∑j=1

tr(Bj +Btj)

2.

This formula shows that 2tr(B) + tr(J1) ≥ 0 and equality occurs when Ai = 0,

Yj = 0 and Bj + Btj = 0 for i = 1, . . . , p and j = 1, . . . , q. This means that D(g)

and D(g)⊥ are both abelian and adu is skew-adjoint for any u ∈ D(g)⊥. According

to Theorem 1.4.2, this is equivalent to that 〈 , 〉 is flat. �

An important consequence of this theorem is a new proof of the following

result du to Heintze and Jensen.

Theorem 1.6.2 Let G be a Riemannian non flat solvable Lie group. Then

s = −|H|2 − 1

4

(2tr(B) + tr(J1)

)< 0.

Theorem 1.6.3 If the Lie algebra of G is non abelian then G possesses a left in-

variant metric of strictly negative scalar curvature.

Proof. First suppose that there exists linearly independent vectors x, y, z in g

such that [x, y] = z and choose a basis b1, . . . , bn such that b1 = x, b2 = y and b3 = z.

For any real number ε > 0, consider an auxiliary basis e1, . . . , en defined by ei = aibi

with (a1, . . . , an) = (ε, ε, ε2, . . . , ε2). Define a left invariant metric by requiring that

(e1, . . . , en) should be orthonormal. Let gε denote the Lie algebra provided with this

44

particular metric and this particular orthonormal basis. Sitting [ei, ej]g =∑αkijek,

the structure constants are clearly function of ε. Indeed if [bi, bj]g =∑Ckijbk,

[ei, ej]g = aiaj[bi, bj] =∑

Ckijaiajbk =

∑ aiajak

Ckijek,

thus, for any 1 ≤ i, j, k ≤ n,

αkij =aiajak

.

From this relation and the definition of the ai, one can see that each αkij tends to a

well define limit. Thus we obtain a limit algebra g0 with a prescribed metric and

prescribed orthonormal basis. One can see easily that the bracket on g0 is given by

[e1, e2]0g = e3

with [ei, ej]0g = 0 otherwise. Thus g0 is nilpotent and hence, from Proposition 1.3.8,

its scalar curvature s(g0) < 0. It follows by continuity that s(gε) < 0 whenever ε is

sufficiently close to 0.

On the other hand, if x, y and [x, y]g are always linearly independent, then g

is isomorphic to the special example in 2, hence g has strictly negative curvature for

any choice of metric. �

Theorem 1.6.4 If the universal covering of G is not homeomorphic to Euclidean

(or if G contains a compact non-commutative subgroup) then G admits a left invari-

ant metric of strictly positive scalar curvature.

Proof. See [13]. �

1.7 The 3-dimensional case

In this section, we study the curvatures of left invariant Riemannian metric on

3-dimensional Lie groups. The main tool to be used is the Euclidean cross product.

If V is a three dimensional vector space endowed with a positive definite metric

and a fixed orientation, then there exists a bilinear map V × V −→ V called cross

product satisfying:

45

1. for any u, v ∈ V , u × v = −v × u is orthogonal to u and v and |u × v| =√〈u, u〉〈v, v〉 − 〈u, v〉2,

2. if u, v are linearly independent then (u, v, u× v) is a positively oriented basis

of V .

Let G be a connected 3-dimensional Lie group with a left invariant Riemannian

metric. Choose an orientation of g such that the cross product is definite. The

following lemma plays a crucial role in the the study of the curvature of G.

Lemma 1.7.1 There exists a unique endomorphism L : g −→ g such that, for any

u, v ∈ g,

[u, v]g = L(u× v).

Moreover, G is unimodular if and only if L∗ = L.

Proof. Choose an oriented orthonormal basis (e1, e2, , e3) of g and put

L(e1) = [e2, e3]g, L(e2) = [e3, e1]g and L(e3) = [e1, e2]g.

It is easy now to check that L satisfies the desired property. Moreover, if L(ei) =

a1ie1 + a2ie2 + a3ie3 then

tr(ade1) = a23 − a32, tr(ade2) = −a13 + a31 and tr(ade3) = a12 − a21.

So G is unimodular if and only if the matrix (aij) is symmetric or L∗ = L. �

Let us specialize to the unimodular case. If L is self-adjoint there exists an

oriented orthonormal basis (e1, e2, e3) such that L(ei) = λiei with λ1 ≤ λ2 ≤ λ3.

Thus

[e1, e2]g = λ3e3, [e2, e3]g = λ1e1 and [e3, e1]g = λ2e2. (1.27)

The three eigenvalues λ1, λ2, λ3 are well defined up to order. However, the construc-

tion is based on a choice of orientation. If we reverse the orientation, L will changes

sign and λ1, λ2, λ3 also.

46

Let compute the Ricci and scalar curvature. We will use the formula

ric(u, v) = −1

2B(u, v)− 1

2tr(adu ◦ ad∗v)−

1

4tr(Ju ◦ Jv)−

1

2(〈adHu, v〉+ 〈adHv, u〉).

established in Proposition 1.3.6. Since G is unimodular, H = 0. Now, by identifying

an endomorphism with its matrix in the basis (e1, e2, e3), we get

ade1 =

0 0 0

0 0 −λ20 λ3 0

, ade2 =

0 0 λ1

0 0 0

−λ3 0 0

, ade3 =

0 −λ1 0

λ2 0 0

0 0 0

,

ad∗e1 =

0 0 0

0 0 λ3

0 −λ2 0

, ad∗e2 =

0 0 −λ30 0 0

λ1 0 0

, ad∗e3 =

0 λ2 0

−λ1 0 0

0 0 0

.

From the definition of Ju, we have

Juv = 〈u, [v, e1]g〉e1 + 〈u, [v, e2]g〉e2 + 〈u, [v, e3]g〉e3,

and by a direct computation, we get

Je1 =

0 0 0

0 0 −λ10 λ1 0

, Je2 =

0 0 λ2

0 0 0

−λ2 0 0

, Je3 =

0 −λ3 0

λ3 0 0

0 0 0

.

By using the expression of the Ricci curvature given above, we get

(ric(ei, ej)) =

λ2λ3 + 1

2(λ21 − λ22 − λ23) 0 0

0 λ3λ1 + 12(λ22 − λ21 − λ23) 0

0 0 λ1λ2 + 12(λ23 − λ21 − λ23)

.

47

Put µi = 12(λ1 + λ2 + λ3)− λi. Then

(ric(ei, ej)) = 2

µ2µ3 0 0

0 µ3µ1 0

0 0 µ1µ2

, (1.28)

and the scalar curvature is given by

s = 2(µ1µ2 + µ1µ3 + µ2µ3).

To compute the sectional curvature, we can use the following formula which holds

in any 3-dimensional manifold (see [5] pp. 48):

Q(u, v) =1

2|u× v|2s− ric(u× v, u× v).

From what above we can deduce easily the following result.

Proposition 1.7.1 In the 3-dimensional unimodular case, the determinant 8(µ1µ2µ3)2

of ric is always ≥ 0. If this determinant is zero, then at least two of the principal

Ricci curvatures must be zero.

Let us give now a precise description of the possible group according to the signs

of λ1, λ2, λ3. By changing signs if necessary, we can assume that at most one of the

constant structure λ1, λ2, λ3 is ≤ 0. So get six distinct cases we tabulate as follows.

Signs of λ1, λ2, λ3 Associated Lie group Description

+,+,+ SU(2) or SO(3) compact, simple

+,+,− SL(2,R) or O(1, 2) noncompact, simple

+,+,0 E(2) solvable

+,−,0 E(1, 1) solvable

+,0,0 Heisenberg group nilpotent

0,0,0 R3 commutative

It is not difficult that these six possibilities do really give rise to nonisomorphic

Lie algebras. They can be distinguished for example by computing the signature of

48

Killing form. We have

1. SU(2) is the group of 2× 2 unitary matrices of determinant 1; homeomorphic

to the unit 3-sphere.

2. SO(3) is the rotation group of R3, isomorphic to SU(2)/{I3,−I3}.

3. SL(2,R) is the group of 2× 2 real matrices of determinant 1.

4. O(1, 2) is Lorentz group consisting of linear transformations the quadratic from

t2 − x2 − y2. Its identity component is isomorphic to SL(2,R)/{I3,−I3}.

5. E(2) is the group of rigid motion of the Euclidean 2-space.

6. E(1, 1) is the group of rigid motion of the Minkowski 2-space. This group is a

semi-direct product of subgroup isomorphic to R2 and to R, where each τ ∈ R

acts on R2 by the matrix

eτ 0

0 e−τ

.

7. Finally, the Heisenberg group can be described as the group of all 3 × 3 real

matrices of the form

1 x z

0 1 y

0 0 1

.

Let study now how the curvature can be altered by a change of metric.

Proposition 1.7.2 Depending on the choice of left invariant metric, the Ricci cur-

vature for SU(2) can have signature either (+,+,+) or (+, 0, 0) or (+,−,−) and

the scalar curvature can be either positive, negative or zero.

Proof. This follows from (1.28). �

As in the commutative case all the left invariant metrics all flat, in the nilpotent

case, the curvature properties are independent from the metric.

Proposition 1.7.3 For any metric on the Heisenberg group, the Ricci form has

signature (+,−,−) and the scalar curvature s is strictly negative. Furthermore, we

have

|ric(e1, e1)| = |ric(e2, e2)| = |ric(e3, e3)| = |s|.

49

Proof. We have λ1 > 0, λ2 = λ3 = 0. So µ1 = −µ2 = −µ3 = −12λ1. So

ric(e1, e1) = −ric(e2, e2) = −ric(e3, e3) = −s =1

2λ21.

The simple group SL(2,R) and the solvable group E(1, 1) are difficult to dis-

tinguish by curvature properties.

Proposition 1.7.4 Let G be either SL(2,R) or E(1, 1). Then depending on the

choice of left invariant metric the signature of the Ricci form can be either (+,−,−)

or (0, 0,−). However, the scalar curvature s must always be strictly negative.

Proof. If λ1 = 0 while λ2, λ3 have opposite sign, then s = −12(λ2 − λ3)2 < 0.

If the λi are all non zero say λ1 < 0 < λ2, λ3, then

∂s

∂λ1= −λ1 + λ2 + λ3 ≤ 0

shows that s is strictly monotone as a function of λ1 (keeping λ2, λ3 fixed) for λ1 ≤ 0.

Therefore

s(λ1, λ2, λ3) < s(0, λ2, λ3) = −1

2(λ2 − λ3)2 < 0.

Further details will be left to the reader. �

Proposition 1.7.5 The Euclidean group E(2) is non-commutative, but admits a

flat left invariant metric. Every nonflat left invariant metric has Ricci form of

signature (+,−,−) with scalar curvature s < 0.

Proof. Left to the reader. �

Now let return to the nonunimodular case. The possible algebras can be

described as follows.

Lemma 1.7.2 If the connected 3-dimensional G is nonunimodular then its Lie al-

gebra has a basis (e1, e2, e3) so that

[e1, e2]g = αe2 + βe3, [e1, e3]g = γe2 + δe3 and [e2, e3]g = 0,

50

and so that the matrix A =

α β

γ δ

has trace α + δ = 2. If we exclude the

case where A = I2 (compare Example 2), the determinant D = αδ − βγ provides a

complete isomorphism invariant for this Lie algebra.

Proof. We consider a 3-dimensional Lie algebra g which is not unimodular and

define u = {u ∈ g, tr(adu) = 0}. Since g is nonunimodular then u is 2-dimensional

unimodular ideal and hence it is commutative. Choose e1 ∈ g such that tr(ade1) = 0.

Since u is commutative, the linear transformation

L(u) = [e1, u]g

from u to itself, with trace 2, is independent of the particular choice of e1.

If L maps each vector to a multiple of itself, then we are in the special case of

Example 2 (and in fact L must be the identity). otherwise, the determinant D of

L provides a complete isomorphism invariant. For choosing e2 so that the vector e2

and L(e2) = e3 are linearly independent, the conditions tr(L) = 2 and det(L) = D

imply

L(e2) = e3 and L(e3) = −De2 + 2e3.

Thus the bracket product operation is uniquely determined. �

Curvature properties can be described as follows. Consider a Lie group as in

Lemma 1.7.2.

Theorem 1.7.1 If the determinant D ≤ 0 then every left invariant metric has Ricci

form of signature (+,−,−). But if D ≥ 0 the signature (0,−,−) is also possible,

and if D > 0 the signature (−,−,−) is also possible. In fact, for D > 0 there exists

a left invariant metric of strictly negative sectional curvature and for D > 1 there

exists a left invariant metric of constant negative sectional curvature. In all cases

the scalar curvature is strictly negative.

Proof. See [13]. �

51

References

[1] M. Ait Haddou, M. Boucetta, H. Lebzioui, Left-invariant Lorentzian flat

metrics on Lie groups, Journal of Lie Theory 22 (2012), No. 1, 269-289.

[2] D.V. Alekseevsky, B.N. Kimelfeld, The structure of homogeneous Riemannian

spaces with zero Ricci curvature, Functional Anal. Appl. 9 (2), 5-11 ( 1975).

[3] R. Azencott and E. Wilson, Homogeneous manifolds with negative curvature,

Part I, Trans. Amer. Math. Soc. 215 (1976), 323-362.

[4] M. Berger, A Panoramic View of Riemannian Geometry, Springer (2002).

[5] A. L. Besse, Einstein manifolds, A Series of Modern Surveys in Mathematics,

Springer (1987).

[6] I. Dotti, Ricci Curvature of Left Invariant Metrics on Solvable Unimodular

Lie Groups, Math. Z. 180,257-263(1982).

[7] E. Heintze, On homogeneous manifolds of negative curvature, Math. Ann. 211

(1974), 23-34.

[8] S. Helgason, Differential geometry, Lie groups and symmetric spaces,

Academic press Inc (1978).

[9] J. Helmstetter, Radical d’une algebre symetrique a gauche, Annales de

l’Institut Fourier 29 (1979) 17–35.

[10] A. Knapp, Lie groups beyond an introduction, Progress in Mathematics

(2002).

[11] Shoshichi Kobayashi, Transformations groups in differential geometry,

Classics in Mathematics Springer-Verlag Berlin Heidelberg New York (1995).

[12] John M. Lee, Riemannian manifold, An introduction to curvature, Graduate

Texts in Mathematics (1997).

[13] J. Milnor, Curvatures of left invariant metrics on Lie Groups, Advances in

Mathematics 21 (1976), 293–329.

52

[14] N. Wallach, Compact homogeneous Riemannian manifolds with strictly

positive curvature, Ann. Math. 96 (1972), 277-29.