SYMMETRIC SPACES

PETER HOLMELINMaster’s thesis2005:E3

Centre for Mathematical SciencesMathematics

CE

NT

RU

MSC

IEN

TIA

RU

MM

AT

HE

MA

TIC

AR

UM

Abstract

In this text we study the differential geometry of symmetric spaces. Wedescribe how a symmetric space (M , g) can be seen as a homogeneous spaceG/K , the quotient of its isometry group G and a isotropy group K at a point.We study the one-to-one correspondence between symmetric spaces and sym-metric pairs. Furthermore we investigate the expressions for curvature on asymmetric space. Finally we describe the notion of dual symmetric spaces.

To illustrate how well symmetric spaces lend themselves to explicit calcula-tions we calculate the curvature of the real Grassmann manifold and find theirdual space.

Keywords: homogeneous spaces, symmetric spaces, symmetric pairs, theKilling form, curvature of symmetric spaces, dual symmetric spaces.

Throughout this text it has been my intention to give reference to all the sourcesthat have been used.

Acknowledgements

I am very grateful to Sigmundur Gudmundsson for his support and substantialamount of comments. I also wish to thank Martin Svensson his deep knowl-edge and judicious comments about the material.

Peter Holmelin

Contents

Overview 1

Chapter 1. Homogeneous Spaces 31. Group Actions 32. Homogeneous Spaces 3

Chapter 2. Symmetric Spaces 111. Definitions 112. The Isometry Group 143. G/K is Diffeomorphic to M 17

Chapter 3. Symmetric Pairs 191. From a Symmetric Space to a Symmetric Pair 192. The Tangent Space of G/K 203. From a Lie Group to a Symmetric Pair 234. From a Symmetric Pair to a Symmetric Space 25

Chapter 4. Curvature of a Symmetric Space 291. The Killing Form 292. The Curvature Formula 363. The Dual Space 40

Appendix A. The Hopf-Rinow Theorem 45

Appendix B. The Adjoint Representation 47

Appendix C. Inner Products From the Haar Measure 51

Appendix D. Lie Derivatives and Killing Fields 53

Bibliography 59

i

Overview

History[4]At the end of the nineteenth century, after studying spaces of constant cur-

vature, mathematicians wanted to classify all locally symmetric Riemannianmanifolds, i.e. Riemannian manifolds whose curvature tensor is parallel i.e.satisfies ∇R = 0. The whole issue was settled by Elie Cartan in 1932. Previ-ously in his thesis he had classified all simple complex Lie algebras and he alsoclassified the simple real Lie algebras. Using this he gave a complete classifica-tion of all symmetric spaces which by the way were introduced by himself in1926.

Symmetric Spaces

A symmetric space is a Riemannian manifold (M , g) such that for everypoint p ∈ M there exist an isometry �

p of (M , g) called an involution suchthat

(1) �p(p) = p and

(2) d � p = −idTpM .By composing involutions one gets translations along geodesics, which can

be used to extend geodesics to the whole of R i.e. M is geodesically complete.By the Hopf-Rinow theorem any two points in a geodesically complete Rie-manninan manifold can be connected by a geodesic. Therefore the translationsalong the geodesics makes the isometry group G acting on M transitive. Usingthe theory of homogeneous spaces one can identify M with G/K where K isthe isotropy group at a point p ∈ M , i.e K = {k ∈ G : k(p) = p}.

Symmetric Pairs

The description of a symmetric space, in terms of a Lie group G, a closedsubgroup K and an involution � , leads to the concept of a symmetric pairwhich is defined as a Lie group G with a closed subgroup K and an involutiveautomorphism s on G satisfying

(1) (Gs)0 ⊆ K ⊆ Gs,

1

(2) Ad K is compactwhere Gs is the set of elements left invariant by s. We show that symmetricpairs lead to symmetric spaces.

Curvature on a Symmetric Space

We show that left invariant vector fields on the isometry group G aremapped to Killing fields in the symmetric space (M , g), which generate Jacobifields and therefore provide the connection to the curvature tensor. This givesa very simple formula for the curvature tensor on a symmetric space G/K ,namely

R(X , Y )Z = −[[X , Y ],Z ] for X , Y ,Z ∈ p

where p is the linear complement of k which is the Lie algebra of K .Each symmetric space M has a dual space M † which is defined by the

existence of maps(1) A Lie algebra isomorphism � : k→ k† such that

g†( � (X ), � (Y )) = −g(X , Y )(2) A linear isometry ˆ� : p→ p† such that

� ([X , Y ])

= −[ˆ� (X ), ˆ� (Y )]†

where k is the Lie algebra of K and p is a linear complement of k in the Liealgebra g of G. Dual spaces have opposite curvature and if one is compact itsdual is noncompact and vice versa.

Although the path chosen in this paper is more along minimal differentialgeometry, the standard description is more in the language of Lie groups andLie algebras.

For a reader who wants a more thorough description we refer to [5] Hel-gason, which is the standard reference on the theory of symmetric spaces.

2

CHAPTER 1

Homogeneous Spaces

In this chapter we create a manifold structure for the quotient of a Liegroup and a closed subgroup.

1. Group Actions

Definition 1.1. Let M be a smooth manifold and G be a Lie group withidentity element e ∈ G. A smooth map � : G × M → M is called a groupaction on M if

� (g1, � (g2, p)) = � (g1g2, p)

� (e, p) = p

for all g1, g2 ∈ G and p ∈ M . The action � is said to be effective if

� (g, p) = p for all p ∈ M implies that g = e.

The action � is said to be transitive if for all p, q ∈ M there exists a g ∈ G suchthat � (g, p) = q. The isotropy group Kp0 of � at p0 ∈ M is given by

Kp0 = {g ∈ G : � (g, p0) = p0}.If � : G ×M → M is a group action on M , then the group G is said to

act on M and it is customary to write gp for � (g, p).

Example 1.2. Let M = Sn be the unit sphere in Rn+1 and G = SO(n+1)be the special orthogonal group. Then we have the action

� : SO(n + 1)× Sn → Sn, (A, p) 7→ A · pFor p0 = (1, 0, . . . , 0) we get Kp0 = SO(1)× SO(n).

2. Homogeneous Spaces

We need the following lemma to prove that G/K can get a manifold struc-ture.

Lemma 1.3. [10] Let G be a Lie group and let K be a closed subgroup of G.Denote the Lie algebra of G by g and the Lie algebra of K by k. If m is a linearcomplement to k, i.e. g = k ⊕ m and we give G/K the quotient topology, let� : G → G/K be the canonical projection onto the quotient, then the map

� ◦ exp : m→ G/K

3

is a local homeomorphism at 0.

PROOF. [10] Let � be the map

� : g = k⊕m→ G, � (X , Y ) = expX expY

then d � (0,0)(X , Y ) = X + Y , so by the inverse function theorem � is a diffeo-morphism from an open neighborhood U0 × V0 of (0, 0).

We define some sets that will help to show that � ◦ � is one-to-one. Theset exp(V0) is an open neighborhood of e in K with the subspace topology soexp(V0) = H ∩ K for some open set H in G. So there exists an open set

U1 × V1 ⊆ U0 × V0

such that� (U1 × V1) ∩ K ⊆ H ∩ K = expV0

If X ∈ U1, Y ∈ V1 and � (X , Y ) ∈ K then

expX expY = expY ′ for some Y ′ ∈ V0

But � is a diffeomorphism on U1 × V1 so X = 0 and Y = Y ′, thus

� (U1 × V1) ∩ K = exp(V1).

Now we will show the injectivity. Let U2 ⊆ U1 be a neighborhood of 0 inm such that

exp(−U2)exp(U2) ⊆ � (U1 × V1)

Then � ◦ exp|U2is injective, since if X ′,X ′′ ∈ U2 and

� (exp(X ′)) = � (exp(X ′′))

thenexp(−X ′)exp(X ′′) ∈ � (U1 × V1) ∩ K so X ′ = X ′′.

The surjectivity on a neighborhood follows since � is surjective from U2×{0}onto � (U2, 0).

By definition � ◦ exp|m is continuous.If N is an open subset of U2 then � (exp(N )) = � ◦ � (N ,V1) which is open

since � is a diffeomorphism here and we have the quotient topology. So theinverse is continuous. Thus � ◦ exp : m→ G/K is a local homeomorphism at0 in the quotient topology of G/K . �

Theorem 1.4. [10] Let G be a Lie group and K a closed subgroup of G, thenthe quotient space G/K has a unique manifold structure such that

(1) the projection � : G → G/K , g 7→ gK is smooth and(2) � has smooth local lifts such that for every gK ∈ G/K there is a neigh-

borhood U and a map l : U → G such that � ◦ l = id .

4

With this manifold structure a map

f : G/K → N

is smooth if and only iff ◦ � : G → N

is smooth. Furthermore the action

G × G/K → G/K , (g, g ′K ) 7→ gg ′K

is smooth.

PROOF. [10] If one chooses the quotient topology on G/K then we get aHausdorff space by the following. Consider the map

� : G × G → G, � (g1, g2) = g−11 g2

which is continuous, and since K is closed the inverse image � −1(K ) is closed.Now if g1K 6= g2K then g−1

1 g2 6∈ K so there are open sets W1,W2 such that(g1, g2) ∈ W1 ×W2 and W1 × W2 ∩ � −1(K ) = ∅. Now if gK ∈ WiK thenthere is a ki ∈ K such that gki ∈ Wi. So if gK ∈ W1K ∩W2K then gk1 ∈ W1

and gk2 ∈ W2. Therefore (gk1, gk2) ∈ W1 × W2 is mapped to k−11 k2 ∈ K

which contradicts that W1 ×W2 ∩ � −1(K ) = ∅. So G/K is Hausdorff.To get coordinates we need a linear complement m to k. If we use the same

sets as in Lemma 1.3, the maps U2 → G/K ,X 7→ gexp(X )K , for g ∈ G arehomeomorphisms. Then we can choose local coordinates on G/K at gK as theinverses of these maps. We identify a chart with g. Suppose two charts g, g ′

intersect then gexp(X0)K = g ′exp(X ′0)K for X0,X ′0 ∈ U2. Then exp(X0) =g−1g ′exp(X ′0)k0 and exp(X0) ⊆ � (U2 × {0}) ⊆ � (U2 × V1). So if X ′ is closeto X ′0 then there must be an X ∈ U2 and Y ∈ V1 such that g−1g ′exp(X ′)k0 =� (X , Y ) and X depends smoothly on X ′ since � is a diffeomorphism on U2 ×V1 . Which finally gives g ′exp(X ′)K = gexp(X )K , which shows that points inchart g are mapped smoothly to points in chart g ′ and vice versa. So G/K hasa manifold structure.

(1) The projection � is smooth since� (gexp(X )exp(Y )) = gexp(X )K .(2) The lifts l are given by l(gexp(X )K ) = gexp(X ), which are smooth.Uniqueness: If (G/K )′ and (G/K )′′ satisfy the above then � ◦ l ′ : U ′ →

(G/K )′′ is the identity, also � ◦ l ′′ : U ′′ → (G/K )′ is the identity. Thereforeone can make the identity mapping (G/K )′ → (G/K )′′ a diffeomorphism, sothey are identical as manifolds.

If f ◦ � is smooth then f is locally given by f ◦ � ◦ l and is smooth. Thereverse is trivial.

The map (g, g ′K ) 7→ gg ′K is smooth since it is � ◦ (group op. on G) ◦ lwhich is a composition of smooth maps. �

5

The following result tell us important things about the kernel of d � .

Corollary 1.5. [10] Let G be a Lie group, K be a closed subgroup in G, g bethe Lie algebra of G and k be the Lie algebra of K . If m is a linear complement tok in g then

d � e : g→ TeK G/K

is a surjective vector space homomorphism with ker d � e = k.

PROOF. By the proof of Theorem 1.4 the map � ◦exp|m : m→ G/K is lo-cal diffeomorphism, so the differential d( � ◦ exp|m) = d � |m is an isomorphism.Therefore d � e is surjective.

Since g = k⊕ m and d � |k = 0 we get that ker d � = k. �

We now give some examples that illustrates the use of the previous theory.

Example 1.6. Following Example 1.2 let G = SO(n+1),K = SO(n) thenthe quotient SO(n + 1)/SO(n) has manifold structure such that the action

SO(n + 1)×(SO(n + 1)/SO(n)

)→ SO(n + 1)/SO(n)

(A1,A2K ) 7→ A1A2K

is smooth.The Lie algebra g = so(n+1) of SO(n+1) consists of the skew symmetric

matrices. The Lie algebra of k can be identified with the matrices(

0 0T

0 A

)

where A ∈ so(n) and 0 is the n× 1 matrix of zeros. The linear complement mcorresponds to the matrices

(0 vT

−v 0

)

where v is a n × 1 matrix. By Lemma 1.5 the trivial map m → TeK G is anisomorphism.

Example 1.7. For x, y ∈ Rn+1 define

f (x, y) = x1y1 −n+1∑

i=2

xiyi

using matrix multiplication we write this as

f (x, y) = xT Qy where Q =

(1 0T

0 −I

)

where 0 is the n× 1 matrix of zeros and I is the n× n identity matrix.

6

Now define

O(1, n) = {A ∈ GLn+1(R) : AT QA = Q}We find the Lie algebra g = o(1, n) of O(1, n) by taking the derivative of acurve at e, in the defining expression for O(1, n). So if X =A(0) where A(t) ∈O(1, n) then

(1.1) 0 =ddt

(A(t)T QA(t)|t=0 = X T Q + QX

if we write X as

X =

(E FG H

)

we get for Equation (1.1)(

ET + E −GT + FF T − G −H t − H

)= 0

So

o(1, n) =

{(0 F T

F H

): F ∈ M(n,1), H ∈ o(n)

}

If A ∈ O(1, n) then det(ATQA) = det(Q) which implies det(A)2 = 1 sodet(A) = ±1. We turn our attention to the subgroup SO(1, n) with determi-nant 1. The hyperboloid

H1,n = {x ∈ Rn+1 : f (x, x) = 1}has two connected components H+

1,n with x1 > 0 and H−1,n with x1 < 0.There are A ∈ SO(1, n) which interchange the two components of H1,n, so werestrict our attention to the subgroup

Lor(1, n) = {A ∈ SO(1, n) : AH+1,n = H+

1,n}If we consider the vector e1 = (1, 0, . . . , 0) which is left unchanged by

matrices of the form(

1 0T

0 B

)where B ∈ SO(n)

and therefore correspond to the isotropy group of e1. ThereforeLor(1, n)/SO(n) has a manifold structure.

We will now show that Lor(1, n) acts transitively on H+1,n. Let u ∈ H+

1,n

and consider the vectors v ∈ Rn+1 such that

0 = uT Qv = u1v1 −n+1∑

i=2

uivi

7

This is an n-dimensional subspace V , since by choosing vi, i = 2 . . . n + 1arbitrarily we get v1

v1 =u · vu1

, this is well defined since u ∈ H+1,n have u1 > 0

It also follows that the vector v satisfies f (v, v) < 0 since

f (v, v) =|u · v|2

u21− |v|2

=|u · v|21 + |u|2 − |v|

2

<|u · v|2|u|2 − |v|2 by Cauchy-Schwartz

≤ |v|2 − |v|2 = 0

So −fV is a positive definite inner product on V and by the Gram-Schmidtprocess we can construct a basis {wi}n+1

i=2 such that

f (wi,wj) = � ij , f (u,wi) = 0

Then the matrix

A =

(u1 w1

1 . . . wn1

u w1 . . . wn

)

is in Lor(1, n) and A(e1) = u. We decompose the Lie algebra of O(1, n) as

k =

{(0 00 A

): A ∈ o(n)

}

and

m =

{(0 AT

A 0

): A ∈ M(n,1)

}

Example 1.8. Let Gk(Rn) denote the set of all k-dimensional subspaces inRn. This space is called the real Grassman manifold. The orthogonal groupO(n) acts transitively on Gk(Rn), since if V is spanned by the k first elementsof the canonical basis e1, . . . , en of Rn let e′1, . . . , e

′n be another basis of Rn. The

matrix A ∈ O(n) with e′1, . . . e′k as the k first columns has the effect

AV = W

where W is the space spanned by e′1, . . . e′k. The isotropy group of V consists

of matrices of the form(B 00 C

)where B ∈ O(k),C ∈ O(n− k)

so we have thatO(n)/(O(k)×O(n− k))

8

is a manifold. Since the Lie algebra of O(n) consists of the skew-symmetricmatrices we get

k =

{(A 00 D

): A ∈ o(k), D ∈ o(n− k)

}

and

m =

{(0 B−BT 0

): B ∈ M(k,n−k)

}

Example 1.9. Let K = R,C or H, then if one considers Kn+1 as a leftK-vector space with the action of

K∗ = {a ∈ K : a 6= 0}on Kn+1

0 = Kn+1 − 0 given by

� : K∗ ×Kn+10 → Kn+1

0 , � z(x) = xz−1

Then set of orbits is denoted by KPn and is called the n-dimensional K-projective space. The orbit is denoted as

[x] = {xz−1 : z ∈ K∗}Let us now restrict ourselves to CPn. Let A ∈ U(n + 1) then

� A[x] = [Ax]

is well defined since A(xz−1) = (Ax)z−1. Now [e1] is stabilized by{(

ei � 00 B

): � ∈ R and B ∈ U(n)

}

So U(n + 1)/(U(1)× U(n)) has a manifold structure. Similarly to the case ofthe sphere the Lie algebra decomposes as

k =

{(ir 00 B

): r ∈ R, B ∈ u(n)

}

and

m =

{(0 A−A∗ 0

): A ∈ M(1,n)

}.

9

CHAPTER 2

Symmetric Spaces

In this section we define the notion of a symmetric space and study someof their properties.

1. Definitions

Here it is assumed that the reader is familiar with the concept of a geodesi-cally complete Riemannian manifold and the Hopf-Rinow theorem. If not,the reader is referred to Appendix A.

Definition 2.1. A Riemannian manifold (M , g) is said to be a symmetricspace if for every point p ∈ M there exists an isometry �

p of (M , g) such that(1) �

p(p) = p, and(2) d � p = −idTpM .

Such an isometry is called an involution at p ∈ M .

Lemma 2.2. [6] Let (M , g) be a symmetric space and let �p : (M , g) →

(M , g) be an involution at p ∈ M. Then �p reverses the geodesics through p, i.e.�

p( � (t)) = � (−t) for all geodesics � ∈ M such that � (0) = p.

PROOF. [6] A geodesic � : I → M is uniquely determined by the initialdata � (0) and ˙� (0). Both the geodesics t 7→ �

p( � (t)) and t 7→ � (−t) take thevalue � (0) and have the tangent − ˙� (0) for t = 0. �

The following lemma entails the core features of a symmetric space.

Lemma 2.3. [6] Let (M , g) be a symmetric space. If � : I → M is a geodesicwith � (0) = p and � ( � ) = q then �

q ◦ � p( � (t)) = � (t + 2 � ). For v ∈ T (t)M,d � q(d � p(v)) ∈ T (t+2 )M is the vector at � (t + 2 � ) obtained by parallel transportof v along � .

PROOF. [6] Let ˜� (t) = � (t + � ) then ˜� is a geodesic with � (0) = q. So byLemma 2.2 it follows that

�q( � p( � (t))) = �

q( � (−t))

= �q(˜� (−t − � ))

= ˜� (t + � )

= � (t + 2 � ).

11

If v ∈ TpM and V is a parallel vector field along � with V (p) = v, then d � p(V )is parallel, since �

p is an isometry. Also

d � q ◦ d � p(V ( � (t))) = V ( � (t + 2 � ))by the above and since d � applied twice cancels direction reversals. �

As a display of the power of Lemma 2.3 we have the following

Corollary 2.4. [6] Every symmetric space (M , g) is geodesically complete andthus any two points p, q ∈ M in the same path component of M can be connectedby a geodesic.

PROOF. [6] By repeatedly composing as in Lemma 2.3 we can get to� (2l � ) until 2l � greater than any real number. This shows that (M , g) isgeodesically complete, so by the Hopf-Rinow theorem any two points can beconnected by a geodesic. �

Definition 2.5. Let (M , g) be a symmetric space and let � : R→ M be ageodesic with p = � (0) and v = ˙� (0). Then for t ∈ R the isometries

�tv : (M , g)→ (M , g)

given by �tv = � (t/2) ◦ � (0)

are called transvections.

It is easily seen, by using Lemma 2.3, that for the particular geodesic � :R→ M in Definition 2.5 the transvection �

tv satisfies

�tv( � (s)) = � (s + t)

Proving things about isometries are facilitated by

Lemma 2.6. [10] Let (M , g) be a connected Riemannian manifold and p ∈M. If � , � ′ : (M , g) → (M , g) are isometries such that � (p) = � ′(p) andd � p = d � ′p then � = � ′.

PROOF. [10] Let B (p) be a normal ball around p ∈ M . Then if � v(t) =Expp(tv) is a geodesic then � ( � v(t)) = � ′( � v(t)) are the same geodesics sincegeodesics are given by initial data. So � = � ′ on an open set but, this set isalso closed, hence it is M . �

Here we have some nice properties of transvections

Proposition 2.7. [10] Let (M , g) be a symmetric space, p ∈ M , v ∈ TpMand define � : R → M by � (t) = Expp(tv) as the unique geodesic with � (0) =

p, ˙� (0) = v. Let � bv be the transvections corresponding to the involution �p, then

(1) �av = � (t+ a

2 ) ◦ � (t) for all t ∈ R12

(2) �av depends only on av

(3) �av( � (t)) = � (t + a)

(4) �av ◦ � bv = �

(a+b)v

(5) �p ◦ � av ◦ � p = �

−av

PROOF. [10] (1) Note that � is the composition of two isometries so it isactually an isometry.For � (s) ∈ M , if � (r) = � (r + t) then � (t) = � (0) and � (s) = ˜� (s − t) so

� (t+ a2 ) ◦ � (t)( � (s)) = �

˜ ( a2 ) ◦ � ˜ (0)(˜� (s − t)) = ˜� (s − t + a)

= � (s + a)

Now we also have � ( a2 ) ◦ � (0)( � (s)) = � (s + a). So the isometries agree at � (s).

If w ∈ T (s)M and V is the parallel vector field along � such that V ( � (s)) = w.Then d( � (t+ a

2 ) ◦ � (t))w is the parallel transport of w along � from � (s) to� (s + a). The same is true for d � avw = d( � ( a

2 ) ◦ � (0))w.So since their initial data agree we have �

av = � (t+ a2 ) ◦ � (t) by Lemma 2.6.

(2) � av = � ( a2 ) ◦ � (0) = �

Expp0( av

2 ) ◦ � p0 , so it depends only on the value of av(3) This follows directly from the proof of 1(4) By (1) � av ◦ � bv = � ( a

2 + b2 ) ◦ � ( b

2 ) ◦ � (0) = � ( a2 + b

2 ) ◦ � (0) = �(a+b)v

(5) � (0) ◦ � ( a2 ) ◦ � (0) ◦ � (0) = � (0) ◦ � ( a

2 ) = �−av �

Here are some concrete examples of involutions and transvections

Example 2.8. At p0 ∈ Rn define the involution�

p0 : x 7→ −(x − p0) + p0

In Rn geodesics are straight lines, so consider the geodesic

� (t) : R→ Rn, t 7→ p0 + tv

where v ∈ Tp0Rn = Rn

Then we get the transvection�

tv(x) = � (t/2) ◦ � (0)(x)

= � (t/2)(−(x − � (0)) + � (0))

= −(−(x − � (0)) + � (0)− � (t/2)) + � (t/2)

= −(−(x − p0) + p0 − (p0 + tv)) + (p0 + tv)

= x + tv.

Now we can verify the translational properties of the transvection�

tv � (s) = (p0 + sv) + tv = p0 + (s + t)v = � (s + t)

13

Example 2.9. If we have a submanifold M in (Rm, g) for some m ∈ Z+

and M = {x ∈ Rm : g(x, x) = r} for example Sn or H+1,n, then we have the

involution at p0

� : M → M

x 7→ 2g(p0, x)

rp0 − x

then�

p0(p0) = 2g(p0, p0)

rp0 − p0 = p0

since Tp0M⊥p0 we get for v ∈ Tp0M

d � p0(v) = 2g(p0, v)

rp0 − v = −v

so we have an involution on the submanifold M .

2. The Isometry Group

Definition 2.10. Let I (M ) be the isometry group of the symmetric space(M , g), and let G be the connected component of I (M ) containing the neutralelement e ∈ G, i.e. G is the identity component of I (M ). Furthermore let Kp0

be the isotropy group of G at p0 ∈ M .

It should be noted that by continuity, all the transvections belong to thegroup G.

Now we can prove the a very important thing concerning the isometrygroup.

Theorem 2.11. [6] Let (M , g) be a symmetric space and G the identity com-ponent of I (M ), then G acts transitively on M

PROOF. [6] By Corollary 2.4 any p, q ∈ M can be connected by a geodesic� . If � (0) = p, ˙� (0) = v and � (s) = q and �

tv is the family of translations along� then q = �

sv(p). So the action is transitive. �Lemma 2.12. [10] Let N be a smooth manifold and let f : N → I (M ) be

a map such that f (n)(p) depends smoothly on (n, p) for n ∈ N and p ∈ M is in aneighborhood of p0 ∈ M. Then the dependence is smooth for all p ∈ M.

PROOF. [10] If we prove the claim for p ∈ M in a normal ball aroundq ∈ M , then we can cover M with intersecting balls from p0 to an arbitrarypoint in M . So if p ∈ B (q) then there exists a v ∈ TqM such that p = Expqvand v depends smoothly on p. Then

f (n)(p) = f (n)(Expqv) = Expf (n)(q)(df (n)(q))v

where the right side depends smoothly on (n, p). �14

Now we have a very important lemma for showing that charts and actionson our manifold are smooth.

Lemma 2.13. [10] Let (M , g) be a symmetric space with involutions �p,

transvections � v and let Kp0 be the isotropy group of the isometries I (M ) acting ata point p0 ∈ M, then

(1) �p(q) depends smoothly on (p, q) ∈ M ×M.

(2) �v(q) depends smoothly on (v, q) ∈ TpM ×M.

(3) �p(q), depends smoothly on (p, q) ∈ M × M, where �

p is � v such that�v(p0) = p.

(4) k(q) depends smoothly on (k, q) ∈ Kp0 ×M.

PROOF. [10] (1) By Lemma 2.12 we only need to show the claim fora point q in a neighborhood of p. Then q = Expp(v) where v dependssmoothly on (p, q). But Expp(tv) goes through p so �

p(q) = �p(Expp(tv))|t=1 =

Expp(−v) which again depends smoothly on (p, q).(2) The claim follows from the definition of � v and 1)(3) Since v in the geodesic Expp0(tv) connecting p0 to p depends smoothly

on p close to p0, then �v depends smoothly on p close to p0. Then by Lemma

2.12 this is true for all p ∈ M .(4) By Lemma 2.12 we show it for q in a neighborhood of p0. Then

q = Expp0(v) so k(q) = k(Expp0(v)) = Expk(p0)(dk)v = Expp0(dk)v whichdepends smoothly on (k, q). �

Note that Kp0 immediately can be defined as a Lie group using Lemma2.6 and the fact that for all f ∈ Kp0 , f (p0) = p0, i.e. an isometry on aconnected manifold is identified by its value at one point and its differential atthe same point. So since df (p0) is an element of the Lie group of isometriesTp0M → Tp0M which is O(dim M ). Since K0 is closed we can identify K0

with a Lie-subgroup of O(dim M ).

Theorem 2.14. [10] Let (M , g) be a symmetric space. Then the isometrygroup I (M ) has the structure of a Lie group such that

(1) The map I (M )×M → M, with (g, p) 7→ g(p) is smooth.(2) If N is a manifold then f : N → I (M ) is smooth if and only if

N ×M → M , (n, q) 7→ f (n)(q)

is smooth.

PROOF. [10] We will define charts in a neighborhood around g0 ∈ I (M ).Let U be a neighborhood of p0 ∈ M . We’ll define a map that will help us withthe charts. Let U × Kp0 → I (M ), (p, k) 7→ g0

�pk, where �

p is defined as inLemma 2.13. This map is injective since we can retrieve both p, v by

g−10 g0

�pkp0 = �

pp0 = p

15

� −1p g−1

0 g0�

pk = k

So we can talk about the inverse image of this map around g0.If we choose local coordinates for M by p 7→ (x1, . . . xm) at p0 and local

coordinates k 7→ (y1, . . . yl ) at k0 in Kp0 , then we get local coordinates aroundg0 in I (M )

g0�

pk 7→ (x1(p), . . . , xm(p), y1(k), . . . , yl (k)).

These coordinates are smooth, since if the image of two such charts intersectwe get g1

�p1k1 = g2

�p2k2 so

p1 = g−12 g2

�p2k2(p0)

dk1(p0) = (d � −1p1

(p1))d(g−11 g2)(p2)(d � p2(p0))(dk2(p0))

these depend smoothly on each other by Lemma 2.13.The topology is Hausdorff since if � , � ′ are two different isometries then

we either have � (p0) 6= � ′(p0) or (d � (p0) 6= (d � (p0)). If the first one holds wecan separate the isometries since M is Hausdorff. If the second hold we musthave that k1 6= k2 so we can separate in O(dim M ).

Now the group operation is smooth since if f , g ∈ I (M ) then f −1g =(g1

�p1k1)−1(g2

�p2k2) = (g3

�p3k3) in local coordinates. So

p3 = �p3k3(p0) = g−1

3 k−11

� −1p1

g−11 g2

�p2k2(p0)

and

dk3 = d � −1p3

dg−13 dk−1

1 d � −1p1

dg−11 dg2d � p2dk2

which is smooth by Lemma 2.13. Next(1) Since g is an isometry then (g, p) 7→ g(p) is smooth(2) Suppose f (n) is smooth. Let n0 ∈ N then f (n) = g0

�pk where (p, k)

depends smoothly on n. So f (n)(q) = g0�

pk(q) depends smoothly on (n, q).Conversely if the assignment (n, q) 7→ f (n)(q) is smooth then

f (n) = g0�

pk

where p = g−10 f (n)(p0) and dk = d � −1

p dg−10 df (n). So (p, k) depends smoothly

on n. �

Example 2.15. If we consider the isometry groups in Examples 1.6-1.9 ofSection 2 then we see that

SO(n + 1) acts smoothly on Sn,Lor(1, n) acts smoothly on H+

1,n,O(n) acts smoothly on Gk(Rn) andU(n + 1) acts smoothly on CPn

16

3. G/K is Diffeomorphic to M

Here we show that the quotient of the isometry group and the isotropygroup can be identified with the symmetric space.

Theorem 2.16. [10] Let (M , g) be a symmetric space and let G = I (M ) acttransitively on M. Let K be the isotropy group of G at p0 ∈ M. Then the map

� : G/K → M with gK 7→ g(p0)

is a diffeomorphism such that

� ◦ � (g) = g ◦ �where � (g ′)gK = g ′(gK ) = g ′gK as in Theorem 1.4.

If the action is not effective one can consider the groups G/N and K /Ninstead, where N is the kernel of the action. So if M is a symmetric space thenI0(M )/K0 is diffeomorphic to M .

PROOF. [10] Clearly � is well defined and bijective since � k(p0) = p0 fork ∈ K . Also

� ◦ � (g ′)gK = � (g ′gK ) = � g′g (p0) = � g′ � g(p0) = � g′( � (gK ))

� is smooth since � ◦ � (g) = � g and � are smooth.We will show that � −1 is smooth. This is done by showing that dim (G/K ) ≥dim M . Then we show d � is injective, so d � is an isomorphism and by theinverse function theorem locally � −1 is smooth. But � −1 exists globally so bysome form of gluing lemma � −1 is smooth.

Now M = ∪∞j=1Uj, where Uj are the images of the charts on G/K . SinceM is locally compact, at least one Uj contains an open subset of M by theBaire category theorem. This open subset on M is in bijective correspondencewith an open set V in G/K since G/K acts smoothly. So using the charts wewould get an onto map Rn → V → Uj → Rm, so we cannot have n < m i.e.dim (G/K ) ≥ dim M .

Thus we only have to show the injectivity which will imply thatdim (G/K ) ≤ dim M . Thus dim (G/K ) = dim M . But � ◦ � (g) = � g ◦ � so� = � g ◦ � ◦ � (g−1). If we differentiate at g ′ we get

(d � )g′K = (d � g ◦ d � )g−1g′K ◦ (d � (g−1))g′K

So if d � is injective at aK then it is injective at g ′K by choosing g such thatg−1g ′K = aK in � = � g ◦ � ◦ � (g−1. Therefore it is enough to show theinjectivity at eK . Let v ∈ TeK (G/K ) with d � (v) = 0. By Corollary 1.5 d �is surjective and has kernel k. So there is a X ∈ g such that d � (X ) = v. Thecurve � (t) = � exp(tX )(p0) satisfies

d � (t)dt

=ddt

(� ◦ � (exp(tX ))

)

17

= d � ◦ d � exp(tX )(X )

= d � ◦ d � ◦ dLexp(tX )(X (e))

= d � ◦ d � (exp(tX )) ◦ d � (X (e))

= d � exp(tX ) ◦ d � ◦ d � (X (e))

= d � exp(tX ) ◦ d � (v)

= 0

so � ∈ K and X ∈ k, i.e. v = 0. �Example 2.17. We have seen in Examples 1.6-1.9 of section 2 groups G

that act transitively on manifolds and their isotropy groups K at certain points.Therefore we have the following diffeomorphisms

SO(n + 1)/SO(n) ∼= Sn

Lor(1, n)/SO(n) ∼= H+1,n

O(n)/O(k)×O(n− k) ∼= Gk(Rn)

U(n + 1)/U(1)× U(n) ∼= CPn

18

CHAPTER 3

Symmetric Pairs

In this chapter we introduce the concept of a symmetric pair and showhow they lead to symmetric spaces.

Definition 3.1. A pair (G,K ) is said to be a Riemannian symmetric pairif G is a Lie group, K a closed subgroup of G, and sp0 an involutive automor-phism on G such that

(1) (Gsp0)0 ⊆ K ⊆ Gsp0

(2) Ad (K ) is a compact subset of GL(g).

Here Gsp0are the elements of G that are left invariant by sp0 , i.e

Gsp0= {g ∈ G : sp0g = g}.

The involutive automorphism sp0 is often uniquely defined, so it is com-mon to only write (G,K ) for a symmetric pair.

1. From a Symmetric Space to a Symmetric Pair

As before let G,Kp0 denote the identity component of the isometry andisotropy groups on M . Then by Theorem 2.16 there is a bijective correspon-dence

G/K ↔ M , gK 7→ g(p0)

We define an involution corresponding to �p0 on G by

sp0 : G → G sp0(g) = �p0 ◦ g ◦ � p0 = �

p0 ◦ g ◦ � −1p0

Then s2p0

= id and sp0(g)p = �p0 ◦ g ◦ � −1

p0(p), so by Lemma 2.13 it depends

smoothly on (g, p). Thus sp0 is an involutive Lie group automorphism of G.Note also that

(dsp0)e : g→ g

is a involutive Lie algebra automorphism.Let Gsp0

= {g ∈ G : sp0g = g}, and let gsp0be its Lie algebra. Then Gsp0

isa Lie group since it is a closed subgroup.

We have the following theorem

19

Theorem 3.2. [10] Let (M , g) be a symmetric space with a fixed point p0, Gbe the identity component of the isometry group and let K be the isotropy group ofG at p0. Then the map G/K → M with K 7→ g(p0) is a bijection. The group Ghas an involutive automorphism sp0 given by

sp0(g) = �p0 ◦ g ◦ � p0

with stabilizer Gsp0such that

(Gsp0)0 ⊆ K ⊆ Gsp0

This says that (G,K ) is a symmetric pair because Ad (K ) is compact, sinceK is closed and bounded and Ad is a homeomorphism.

PROOF. [10] It only remains to prove the last statement.If X ∈ gp0 then exp(tX ) ∈ Gp0 so sp0(exp(tx)) = exp(tX ) therefore

dsp0 = X . Also if dsp0X = X then exp(tX ) = exp(tdsp0X ) = sp0exp(tX ),the implications work in the opposite direction as well, so

gp0 = {X ∈ g : dsp0X = X}.Let k ∈ K then

sp0(k)p0 = �p0 ◦ k ◦ � p0(p0) = p0 = k(p0) and

dsp0(k) = d � p0 ◦ dkp0 ◦ d � p0

= (−idTp0 M ) ◦ dkp0 ◦ (−idTp0 M )

= dkp0.

This means that Lemma 2.6 implies that sp0(k) = k so K ⊆ Gp0 . Furtherif X ∈ gp0 then �

p0(exp(tX ))p0 = �p0(exp(tX )) � p0(p0) = sp0(exp(tX ))p0 =

exp(tX )p0, so exp(tX )p0 is a fixed point of � p0. But if t is small then exp(tX )p0 isclose to p0 and this is the only fixed point so exp(tX )p0 = p0. Thus exp(tX ) ∈K0 = K and X ∈ k. So gp0 ⊆ k⇒ (Gp0)0 ⊆ K .

�

2. The Tangent Space of G/K

Here we will split up the Lie algebra g so that the complement of k can beidentified with the tangent space of M . We know that

k = {X ∈ g : dsp0(X ) = X}and define

p = {X ∈ g : dsp0(X ) = −X}Then since dsp0 is an automorphism k∩ p = {0}. We also have that g = k + pfor all X ∈ g since

X =12

(X + dsp0(X )) +12

(X − dsp0(X ))

20

where the first term is in k and the second is in p. Therefore

g = k⊕ p

Since dsp0 is a Lie algebra automorphism, i.e.

dsp0[X , Y ] = [dsp0X , dsp0Y ]

we have

[k, k] ⊆ k

[p, p] ⊆ k

[k, p] ⊆ p

We now reveal the connection between the isometry group and the sym-metric space.

Theorem 3.3. [10] Let (M , g) be a symmetric space, G be the identity com-ponent of I (M ), K the isotropy group of G at p0 ∈ M, g the Lie algebra ofG, k the Lie algebra of K and p a linear complement of k in g. As usual let� : G → M , g 7→ � g(p0) then

d � |k(p0) = 0

d � |p(p0) ∼= Tp0M

If X ∈ p then exp(X ) = �d � X (p0) and � exp(X )(p0) = Expp0(d � X (p0)).

PROOF. [10] d � |k(p0) = 0 since � (p0) is constant on K . Now let v ∈Tp0M then the map t 7→ �

tv is a smooth group homomorphism R → G.There is a unique X ∈ g such that � tv = exp(tX ). For this X we have

exp(tdsp0X ) = sp0(exp(tX )) = �p0 ◦ � tv ◦ � p0 = �

−tv.

This means that dsp0(x)) = −X so X ∈ p. Also

d � X (p0) =ddt

� exp(tX )|t=0(p0)

=ddt

exp(tX )(p0)|t=0

=ddt

�tv(p0)|t=0

=ddt

Expp0(tv)|t=0 = v

This shows that d � (p0) : p→ Tp0M is a surjective map.Since the coordinates of G were the coordinates for K and Tp0 ,

dim G = dim K + dim M

21

Also since g = k + p, we get dim M=dim p. Therefore d � (p0) : p → Tp0Mmust be injective as well.

Now the last formulas of the claim follow immediately

expX = �v = �

d � X (p0)

and� expX (p0) = exp(X )p0 = �

d � X (p0)p0 = Expp0(d � X (p0))

�

So the map � : G/K → M in Theorem 2.16 defined by gK 7→ g(p) is asubmersion. We apply this theorem in some of the symmetric spaces we know.

Example 3.4. By Theorem 3.3 for G = SO(3), K = SO(2) acting on S2

at

p0 =

100

∈ S2

we have for

X =

0 1 0−1 0 00 0 0

∈ p

� exptX p0 =

cos t sin t 0− sin t cos t 0

0 0 1

100

=

cos t− sin t

0

is the geodesic in S2 starting at p0 in direction

d � X p0 =

0 1 0−1 0 00 0 0

100

=

0−10

To get a geodesic going in an arbitrary direction we apply any element of K tothe above geodesic which gives

1 0 00 cos � sin �0 − sin � cos �

cos t− sin t

0

Example 3.5. By Theorem 3.3 for G = Lor(1, 2), K = SO(2) acting onH+

1,2 at

p0 =

100

∈ H+

1,2

22

we have for

X =

0 1 01 0 00 0 0

∈ p

� exptX p0 =

cosh t sinh t 0sinh t cosh t 0

0 0 1

100

=

cosh tsinh t

0

is the geodesic in H+1,2 starting at p0 in direction

d � X p0 =

0 1 01 0 00 0 0

100

=

010

To get a geodesic going in an arbitrary direction we apply any element of K tothe above geodesic which gives

1 0 00 cos � sin �0 − sin � cos �

cosh tsinh t

0

3. From a Lie Group to a Symmetric Pair

Now we are ready for the theorem that tell us if a Lie group and a closedsubgroup can be made into a symmetric pair.

Lemma 3.6. Let G be a Lie group and K a closed subgroup of G. Let k ∈ K .If we denote by In g(g ′) = gg ′g−1. Then

� (k) ◦ � = � ◦ In k

and by differentiating at e

d � (k)eK ◦ d � e = d � e ◦ Ad k.

PROOF. [10] Let g ∈ G and k ∈ K then � (k)(gK ) = kgK = kgk−1K =In (k)gK = � (In (k)g), i.e. � (k) ◦ � = � ◦ In k. �

Theorem 3.7. [10] Let G be a Lie group and let K be a closed subgroup, withidentity components G0,K0. Form the quotient manifold G/K . Let � denote theaction � (g ′)gK = g ′gK . Let J ∈ Z (K ) ⊆ K be such that � (J )eK = eK andsuppose

(1) d � (J )eK = −idG/K

Then for every compact subset K ′ in K where

K0 ⊆ K ′ ⊆ K ∩ G0

(G0,K ′) is a Riemannian symmetric pair with

seK (g) = JgJ−1 = In(J )g

23

The assumption (1) is equivalent to

(2) X + Ad J (X ) ∈ k for all X ∈ g.

condition (2) holds if k has a linear complement m in g, i.e. g = k⊕m, on whichAd J = −id

PROOF. [10] By Lemma 3.6 � (k) ◦ � = � ◦ In k and

d � (k)eK ◦ d � e = d � e ◦ Ad k

If we put k = J we get −d � = d � ◦ Ad J . So d � (X + Ad (J )X ) = 0 and thusX + Ad (J )X ∈ k for all X ∈ g. Thus (1) implies (2). By working backwardswe get −d � e = d � ◦ Ad J = d � (J )eK ◦ d � e so d � (J )eK = −idG/K so also (2)implies (1).

Now we prove that we have a symmetric pair. Since J ∈ Z (K ),

In J|K = idK and Ad J|k = idk.

If X ∈ m where m is the linear complement of k in g. Then

X + Ad (X ) = Y

where Y ∈ k but Y depends linearly on X so we write Y = T (X ) whereT : m→ k is a linear map. Then

(Ad J )2(X ) = Ad J (−X + T (X )) = −(−X + T (X )) + T (X ) = X .

Now since (Ad J|k)2 = idk we have Ad (J )2 = idg. So if X ∈ g then

In (J )2(exp(x)) = exp(dIn (J )2X ) = exp(Ad (J )2X ) = exp(X ).

So In is an involutive automorphism on G0.Now we’ll show ((G0)In J )0 ⊆ K ′ ⊆ (G0)In J . First ((G0)In J )0 = (GIn J )0

since (GIn J )0 = (G0)In J . Also (G0)In J = G0 ∩ GIn J . So

K ′ ⊆ (G0)In J

since K ′ ⊆ G0,K ′ ⊆ GIn J because J ∈ Z (K ) and K ′ ⊆ K . Secondly ifgIn J ⊆ k′ = k then (GIn J )0 ⊆ K ′. So let X ∈ gIn J then

X = dIn (J )X = Ad J (X )

and

d � (X ) = d � (Ad (J )X ) = d � (J )d � (X ) = −d � (X ).

Thus d(X ) = 0 and so X ∈ k.Finally Ad K ′ is compact since Ad is a homeomorphism. �

24

Example 3.8. Let G = O(n + 1) and K = O(n) in Theorem 3.7 with

J =

1−1

. . .−1

with J ∈ Z (K ). If we form G/K = O(n+1)/O(n) and since for X ∈ o(n+1)

X =

(A B−BT D

), Ad J

(A B−BT D

)=

(A −B

BT D

)

we have

X + Ad J (X ) = 2

(A 00T D

)∈ k

so J satisfies 2) in Theorem 3.7 so we have the symmetric pair

(SO(n + 1), SO(n)) with s = In J

Example 3.9. Let G = O(n) and K = O(k)×O(n− k) in Theorem 3.7with

J =

(Ik 00T −In−k

)

with J ∈ Z (K ). If we form G/K = O(n)/O(k) × O(n − k) and since forX ∈ o(n + 1)

X =

(A B−BT D

), Ad J

(A B−BT D

)=

(A −B

BT D

)

we have

X + Ad J (X ) = 2

(A 00T D

)∈ k

so J satisfies 2) in Theorem 3.7 so we have the symmetric pair(SO(n), S(O(k)×O(n− k))

)with s = In J

4. From a Symmetric Pair to a Symmetric Space

We will soon prove a theorem which shows that symmetric pairs inducesymmetric spaces. With this it is possible to show that several familiar groupquotients produce symmetric spaces.

First we need to produce some nice metrics for the intended symmetricspace.

Lemma 3.10. [10] Let G be a Lie group and K be a closed subgroup of G. IfAd K is compact then there exists a G-invariant inner product on p ∼= TeK G/K ,such that the action of G on G/K is an isometry. Here p is the linear complementof k in g, i.e. g = k⊕ p.

25

PROOF. [10] If we use Lemma C.2 with H = Ad K , V = p and � =id , we get an Ad K -invariant inner product {·, ·} on p. We transfer this toTeK G/K by d � (eK ) by Corollary 1.5 and get an Ad K -invariant inner product〈·, ·〉eK on TeK G/K . Denote the action � of G on G/K by

� (g ′)gK = g ′(gK ) = g ′gK .

Define(·, ·)gK = 〈d � (g−1) ·, d � (g−1) ·〉eK

This is well defined since if aK = bK then a = bk so we get

(·, ·)aK = 〈d � (a−1) ·, d � (a−1) ·〉eK= 〈d � (k−1b−1) ·, d � (k−1b−1) ·〉eK= 〈d � (k−1)d � (b−1) ·, d � (k−1)d � (b−1) ·〉eK= 〈Ad (k−1)d � (b−1) ·,Ad (k−1)d � (b−1) ·〉eK by Lemma 3.6

= 〈d � (b−1) ·, d � (b−1) ·〉eK since 〈·, ·〉 is Ad (K )-invariant

= (·, ·)bK

�Note that the submersion of Theorem 3.3 becomes a isometric submer-

sion.Finally we have the long awaited theorem.

Theorem 3.11. [10] Let (G,K ) be a symmetric pair with involutive auto-morphism s. Denote the action of G on G/K by � (g ′)gK = g ′gK . Then there is aG-invariant metric g on M = G/K which makes (M , g) a symmetric space withinvolution �

eK such that�

eK ◦ � = � ◦ s i.e. � eK ( � (g)eK ) = s(g)eK

For X ∈ p ≡ {x ∈ g : ds(X ) = −X} we have

� (exp(X )) = �d � (X )eK and � (exp(X ))eK = ExpeK (d � (X ))

PROOF. For a symmetric pair Ad K is compact so by Lemma 3.10 we canpick a G-invariant inner product on the tangent space of G/K , as metric forG/K . Define the involution by

�eK ( � (g)eK ) = � (s(g))eK

this is well defined since�

eK ( � (gk)eK ) = � (s(g)) � (s(k))eK = � (s(g))eK = �eK ( � (g)eK ).

It is smooth since � ◦ � = � ◦ s which is smooth.Further

� 2eK ( � (g)eK ) = �

eK ( � (s(g))eK )

26

= � (s(s(g)))eK

= � (s2(g))eK

= � (e)eK = eK

so � 2eK = e so it is involutive. Therefore � −1

eK is also smooth, so �eK is a

diffeomorphism. To show that it is an isometry we note that by differentiating

(d � eK )eK ◦ (d � ) = (d � )e ◦ (ds)e

so on p we have d � eK ◦ d � = −d � . But at p0 = eK� : g → geK � (g)eK = geK

so since d � |p is an isomorphism so d � |p is also that. Therefore

(d � eK )eK = −id .

So since dpi|p is an isomorphism by Corollary 1.5 d � |p is also that. Moreover

� (s(g)) � eK ( � (g ′)eK ) = � (s(g)) � (s(g ′))eK

= � (s(gg ′))eK

= �ek( � (gg ′)eK )

= �eK ( � (g) � (g ′)eK )

So

(3.1) � (s(g)) ◦ � eK = �eK � (g)

Let u, v ∈ TgK M = T � (g)eK M then

〈d � eK (u), d � eK (v)〉 � eK ( � (g))eK

= 〈d � eK (u), d � eK (v)〉 � (s(g))eK

= 〈d � (s(g)−1)d � eK (u), d � (s(g)−1)d � eK (v)〉eK by G-invariance

= 〈d � eK ◦ d � (g−1)u, d � eK ◦ d � (g−1)v〉eK by eq (3.1)

= 〈−d � (g−1)u,−d � (g−1)v〉eK since d � ek = −id

= 〈−u,−v〉gK by G-invariance

= 〈u, v〉gK

So �ek is an isometry and

(d � eK )eK = id�eK (eK ) = seK ( � (e)eK )

= � (s(e))eK

= eK .

So we have shown that (M , g, � eK ) is a symmetric space.The last claims follow from Theorem 3.3. �

27

Example 3.12. By applying Theorem 3.11 to Examples 3.8-3.9 we get thefollowing symmetric spaces from the corresponding symmetric pairs

(SO(n + 1), SO(n)

)→ SO(n + 1)/SO(n) = Sn

(SO(n), S(O(k)×O(n− k))

)→ SO(n)/S(O(k)×O(n− k))

= Gk(Rn).

In a similar fashion one can get the symmetric spaces

Lor(1, n)/SO(n) = H+1,n

SU(n + 1)/SU(1)× SU(n) = CPn

28

CHAPTER 4

Curvature of a Symmetric Space

The aim of this chapter is to show how to transfer the curvature calcula-tions from the symmetric space to the Lie group which is “behind” the sym-metric space. So we need something like an inner product on the Lie algebra.

1. The Killing Form

The Killing form will provide us with a metric on the Lie algebra g of theisometry group G on the symmetric space.

Definition 4.1. Let g be a Lie algebra and let � be an involutive automor-phism on g with fixed point set k, which is a Lie subalgebra or g. If Int k(seeAppendix B) with Lie algebra ad (k) is compact, then (g, � ) is called an orthog-onal symmetric algebra. The orthogonal symmetric algebra is called effective ifz(g) ∩ k = {0}.

Example 4.2. Let

g = R, � = −idR, k = {0}then we have an orthogonal symmetric algebra which corresponds to

G = R,K = {0}, s = −idR

and alsoG = T = R/Z,K = {0, 1/2}, s = −idT

Definition 4.3. Let g be a Lie algebra, then the Killing form B of g over afield F is the bilinear form

B : g× g→ F, (X , Y ) 7→ tr(ad X ◦ ad Y ).

The Lie group G and its Lie algebra g are called semisimple if B is nondegener-ate, i.e. if B(X , Y ) = 0 for all Y ∈ g then X = 0.

Example 4.4. To calculate1 the trace of linear maps from a vector spaceto itself we calculate the trace of the corresponding matrix representations.Then we will have an expression that can be used for an arbitrary vector. It

1To save space we use the convention to sum over repeated indices without explicit summation expressions, i.e.xiyi =

Pni=1 xiyi .

29

becomes especially easy if we have an inner product 〈·, ·〉, so that we have anorthonormal basis {ei}i∈I where I is some finite set. Then the trace is given by

trace � = 〈ei, � (ei)〉since � has the matrix representation � ij = 〈ei, � (ej)〉. We do an explicitcalculation for so(n) and display some others afterwards.

For so(n) we chose the basis to be 1√2(eij − eji) for 1 ≤ i < j ≤ n, where

eij is the matrix with a 1 at position (i, j) and zeros elsewhere.

trace (ad X ◦ ad Y ) = 〈fk, [X , [Y , fk]]〉 by Lemma B.4

where fk are the basis elements. This gives

trace (ad X ◦ ad Y ) =12〈eij − eji, [X , [Y , eij − eji]]〉

=12〈eij, [X , [Y , eij]]〉 − 1

2〈eij, [X , [Y , eji]]〉(4.1)

−12〈eji, [X , [Y , eij]]〉+

12〈eji, [X , [Y , eji]]〉

we only need to calculate the first two on the right hand side of Equation (4.1)since the last two are the first with ij switched. The inner product we use is

〈A,B〉 = trace (AT B)

and the basis elements are orthogonal for this choice. The first expression inEquation (4.1) is

12

{(eij)T��� X � Y � eij � −(eij)T��� X � eij� Y � − (eij)T��� Y � eij� X �

+(eij)T��� eij� Y � X �}

Now (eij)T��� = eji��� so we get

=12

{eji��� X � Y � eij � −eji��� X � eij� Y � − eji��� Y � eij� X �

+eji��� eij� Y � X �}

Also eij��� X � = � i� Xj and X ��� eij� = X � i � j , eij��� ekl� = � i� � jk � l so we get

=12

{� j� Xi Y i � j� − � j� Xi �� i Yj

� − � j� Yi �� i Xj� + � j� � ii � j Y � X �

}

=12

{� jjXi Y i − XiiYjj − YiiXjj + � iiYj X j

}

30

The fourth term of Equation (4.1) gives

=12

{� iiXj Y j − XjjYii − YjjXii + � jjYi X i

}

adding these gives

12

( � jjXi Y i + � iiXj Y j)− XjjYii −(4.2)

YjjXii +12

( � iiYj X j + � jjYi X i)Since so(n) is skew symmetric the two middle terms in Equation (4.2) vanishand summing the other term over 1 ≤ i < j ≤ n gives

12

(ntrace (XY ) + ntrace (YX )) = ntrace (XY )

The second term in Equation (4.1) becomes

−12

{(eij)T��� X � Y � eji � − (eij)T��� X � eji� Y � − (eij)T��� Y � eji� X �

+(eij)T��� eji� Y � X �}

= −12

{eji��� X � Y � eji � − eji��� X � eji� Y � − eji��� Y � eji� X �

+eji��� eji� Y � X �}

= −12

{� j� Xi Y j � i� − � j� Xi �� j Yi

� − � j� Yi �� j Xi� + � j� � ij � i Y � X �

}

= −12

{0− XijYij − YijXij + 0

}since i 6= j we get the zeros

= −trace (XY ) after summing over i, j

the last equality follows since Xij = −Xji for X ∈ so(n). The third term ofequation (4.1) similarly gives

−trace (XY )

So adding it all up gives the expression for the Killing form on so(n) as

B(X , Y ) = (n− 2)trace (XY )

Some other explicit expressions of Killing forms are[1]

B(X , Y ) = (n− 2)trace (XY ) on o(n)B(X , Y ) = 2ntrace (XY ) on su(n)B(X , Y ) = 2ntrace (XY )− 2trace X trace Y on u(u)

The Killing form has several nice symmetry properties.

31

Lemma 4.5. [6] The Killing form B of g is symmetric. Also B is invariantunder automorphisms of g. In particular

B((Ad g)X , (Ad g)Y ) = B(X , Y ) for all X , Y ∈ g and g ∈ G

alsoB((ad X )Y ,Z ) + B(Y , (ad X )Z ) = 0 for all X , Y ,Z ∈ g

PROOF. [6] The symmetry follows from trAB = trBA.If � is an automorphism of g then

(ad � X )Y = [ � X , Y ] by Lemma B.4

= [ � X , ��� −1Y ]

= � [X , � −1Y ]

= � ◦ ad X ◦ � −1Y

So

tr(ad � X ◦ ad � Y ) = tr( � ad X � −1 ◦ � ad Y � −1) = tr(ad X ◦ ad Y )

by the cyclic property of the trace.Now if � = Ad exp(tX ) then

B((ad X )Y ,Z ) =ddt

B(Ad (exp(tX ))Y ,Z )|t=0

=ddt

B(Y ,Ad (exp(−tX ))Z )|t=0

= B(Y , (ad X )Z )

�

Theorem 4.6. [10] Let (g, � ) be an effective orthogonal symmetric algebra,then the Killing form B is negative definite on k.

PROOF. [10] Since (g, � ) is an orthogonal symmetric algebra Int g is com-pact. By the proof of Lemma 3.10 there is an Ad G0-invariant inner product〈·, ·〉 on g so

〈Ad (exp(tX ))Y ,Z〉 = 〈Y ,Ad (exp(−tX ))Z〉therefore by differentiating at t = 0 we get

〈(ad X )Y ,Z〉 = −〈Y , (ad X )Z〉.If we choose an orthonormal basis with respect to 〈·, ·〉 for g, then the matrixrepresentation of ad X is skew symmetric i.e. (ad X )ij = aij = −aji . Thus

B(X ,X ) = tr(ad X ◦ ad X ) = aijaji =∑

i,j

−a2ij ≤ 0

32

Now aij = 〈ei, (ad X )ej〉 so if∑

i,j −a2ij = 0 then ad X = 0 so X ∈ z(g) but

k ∩ z(g) = {0} so X = 0. Thus B|k is negative definite. �The sign of the Killing form on p leads to the following classifications,

Definition 4.7. An orthogonal symmetric algebra (g, � ) is said to be of(1) compact type if B is negative definite on p,(2) noncompact type if B is positive definite on p,(3) Euclidean type if B is identically zero on p.

Example 4.8. We make show that so(n) is a compact Lie algebra. LetX ij = 1√

2(eij − eji) be an arbitrary basis element in p of so(n) then

B(X ij,X ij) =(n− 2)

2trace (eij − eji)(eij − eji)

= − (n− 2)2

trace 2eijeji

= −(n− 2).

Hence it is negative definite.It is also true that su(n), u(n) and o(n) are compact, while o(p, q)2 and

u(p, q) are noncompact Lie algebras.

We now want to prove that the Lie algebra of a compact Lie group iscompact i.e. agrees with the above classifications. To do so we need a lemma.

Lemma 4.9. [10] Let g be a Lie algebra, Aut g be the Lie algebra automor-phisms of g, ∂g be the Lie algebra of Aut g and let Int g be the identity componentof Aut g. If g is semisimple then

ad g = ∂g and therefore Int g = (Aut g)0

For the reader unfamiliar with this terminology we refer to Appendix B.

PROOF. [10] Let D ∈ ∂g then (D ◦ ad X )Y = ad DX (Y ) + ad X ◦ DYi.e

[D, ad X ] = ad DXso [∂g, ad g] ⊆ ad g.

The bi-linear form C (F1, F2) = tr(F1F2) is nondegenerate on ad g by theassumptions of the lemma. Let a be the orthogonal complement of ad g in∂g with respect to C . We will show that a = 0, then the claims follows. Byconstruction ∂g ⊆ ad g + a and (ad g) ∩ a = {0}. Now

tr([F1, F2]F3) = tr(F1F2F3 − F2F1F3)

= tr(F1F2F3 − F1F3F2)

2see Example 4.21 for the definition

33

= tr(F1[F2, F3])

so for A ∈ a,D ∈ ∂g and X ∈ g

tr([A,D]ad X ) = tr(A[D, ad X ]) = tr(Aad DX ) = 0

so [A,D] ∈ a, thus a is a Lie ideal.Let A ∈ a,X ∈ g then ad AX = [A, ad X ] ⊆ a ∩ ad g so ad AX = 0

therefore AX = 0 since C is semisimple on ad g, so A = 0 and thus a ={0}. �

Theorem 4.10. [10] Let B be the Killing form on the Lie algebra g then thenfollowing are equivalent

(1) g is the Lie algebra of a compact Lie group,(2) g = z(g)⊕ g′ where B is negative definite on the ideal g′.

PROOF. [10] Let G be a compact Lie group with Lie algebra g. We canassume that G is connected otherwise we consider G0. Then Int g = Ad Gsince G is connected and so Int g is compact.

By Lemma C.2 we can pick an Int g-invariant inner product 〈·, ·〉 on g.Let g′ be the orthogonal complement of z(g) with respect to 〈·, ·〉. Then forX ∈ g, Y ∈ g′,Z ∈ z(g) we have

〈[X , Y ],Z〉 = 〈Y , [Z ,X ]〉 = 〈Y , 0〉 = 0

so [g, g′] ⊆ g′, and g′ is an ideal.With an orthonormal basis of g w.r.t. 〈·, ·〉, the elements of Int g are or-

thogonal matrices i.e exp(t ad X ) for X ∈ g is an orthogonal matrix so ad X isa skew symmetric i.e. aij = −aji so for X ∈ g′

B(X ,X ) = tr(ad X )2 = aijaji =∑

i,j

−a2ij ≤ 0

which is zero if and only if ad X = 0 is X ∈ z(g) i.e. X = 0. Thus g =z(g)⊕ g′ with B|g′×g′ negative definite.

Conversely if g = z(g)⊕ g′ with B negative definite on g′. Note that g′ issemisimple, since if B(X , Y ) = 0 for all Y ∈ g′ then B(X ,X ) = 0 so X = 0.

We will show that g′ is the Lie algebra of a compact Lie group H, theng′ ⊕ z(g) correspond to H × T dim z(g), where T n is the n dimensional torus,which is compact.

Since B is Ad G invariant, B is invariant under Int g′, so Int g′ is repre-sented by orthogonal matrices, i.e. Int g′ ⊆ O(dim Int g′). Also by Lemma4.9 Int g′ = (Aut g)0 so Int g′ is closed and thus compact. So g′ is the Liealgebra of a compact Lie group. �

It turns out that the Killing form decomposes the orhogonal symmetricalgebra of a symmetric pair into a very convenient form.

34

Theorem 4.11. [6] Let (g, � ) be an effective orthogonal symmetric algebra ofa symmetric pair (G,K ). Then we can decompose g as

g = k⊕ p1 ⊕ . . .⊕ pm

where pi⊥pj, i 6= j with respect to B. We can also find an Ad K -invariant innerproduct on g given by

g(·, ·) = −B|k +1�1B|p1 + . . .+

1�m

B|pm

We also have [pi, pj] = 0 for i 6= j.

PROOF. [6] By the proof of Lemma 3.10 we get a Ad K invariant innerproduct (·, ·) on p. Define

g(X , Y ) =

−B(X , Y ) X , Y ∈ k

(X , Y ) X , Y ∈ p0 X ∈ k, Y ∈ p or vice versa.

Then g(·, ·) is positive definite and Ad K invariant.Now for a fixed X ∈ p consider the functional f (Y ) = B(X , Y ) for all

Y ∈ p. Then the Riesz-theorem implies

(4.3) B(X , Y ) = g(Y ,T (X ))

Since B is symmetric T is self adjoint with respect to g(·, ·), so we can find anorthonormal basis of eigenvectors {Xj} of T such that T (Xj) =

�jXj . Note that

all�

j 6= 0 since B is nondegenerate. Also since B is symmetric, the eigenspacescorresponding to different eigenvalues are orthogonal. In fact since {Xk} areorthonormal with respect to g(·, ·) we have that

�k = B(Xk,Xk). So

p = p1 ⊕ . . .⊕ pm

The expression for g(·, ·) follows by Equation (4.3).For the last claim let Yi ∈ pi, Yj ∈ pj then

B([Yi, Yj], [Yi, Yj]) = B(Yi, [Yj, [Yi, Yj]]) =�

ig(Yi, [Yj, [Yi, Yj]])

but also

B([Yi, Yj], [Yi, Yj]) = −B([Yj, Yi], [Yi, Yj])

= −B(Yj, [Yi, [Yi, Yj]])

= − �jg(Yj, [Yi, [Yi, Yj]])

=�

jg(Yi, [Yj, [Yi, Yj]])

where the last equality follows from Theorem 4.13 and the symmetries of thecurvature tensor Ri

jkl . So if i 6= j then�

i 6=�

j so B([Yi, Yj], [Yi, Yj]) = 0 but g

is semisimple hence [Yi, Yj] = 0. Thus [pi, pj] = 0 for i 6= j. �35

With the metric g in Theorem 4.11 the symmetric pair (G,K ) makes thequotient G/K into a nice Riemannian manifold (G/K , g).

2. The Curvature Formula

To do calculations on the symmetric space we turn our attention fromG to the symmetric space M = G/K . By Theorem 3.3 we get the vectorfields on M by parallel translation and the map d � pp0 → TpM . The imagesX ∗(p0) = d � X (p0) are Killing fields (see Appendix D) i.e. their local flows areisometries.

We start with some properties of these Killing fields.

Lemma 4.12. [6] Let G be the isometry group on the symmetric space (M , g),K ⊆ G be the isotropy group at p0 ∈ M. If g is the Lie algebra of G with thestandard decomposition g = k⊕ p, then

X ∗(p0) = 0 for all X ∈ k

∇vX ∗(p0) = 0 for all X ∈ p, v ∈ TpM

PROOF. [6] Let X ∈ k then

X ∗(p0) = d � X (p0) =ddt

(exp(tX )p0)|t=0 =ddt

(p0)|t=0 = 0.

Next let � : t → M be a curve such that ˙� (0) = v ∈ TpM then

X ∗(p0) =ddt

�tX ∗(p0)(p0)|t=0 by Theorem3.3

so

∇vX ∗(p0) = ∇∂∂s

∂∂t�

tX ∗(p0)( � (s))|s=t=0

= ∇∂∂t

∂∂s�

tX ∗(p0)( � (s))|s=t=0

since[ ∂∂s,∂

∂t

]= 0

= ∇∂∂t

d � v|s=t=0

but by Lemma 2.3 �v is a parallel

transport of v along � (t), so

= 0

�

Since we can parallel translate all vector fields on M to the origin it isenough to be able to do the curvature calculations for vector fields at the origin.

36

Theorem 4.13. [6] Let (M , g) be a symmetric space. With the identification

p ∼= TpM X 7→ d � X (p)

in Theorem 3.3, the curvature tensor of M satisfies

R(X ∗, Y ∗)Z ∗ = −[[X ∗, Y ∗]Z ∗](p)

for all X ∗, Y ∗,Z ∗ ∈ d � p(p) = TpM

PROOF. [6] We denote the image d � p(p) by p∗. Let X ∗ ∈ p∗, Y ∗ ∈ p∗.Then X ∗(p) = d � X (p). The geodesic c(t) = ExpptY

∗(p) satisfies Y ∗(c(t)) =

c(t), since by theorem 3.3

Y ∗(c(t)) =dds

exp(sY (c(t)))c(t)|s=0

=dds

�sY (c(t))∗(c(t))|s=0

=dds

c(t + s)|s=0 = c(t)

By Lemma D.10 the Killing field X ∗ is a Jacobi field along c so we havethe Jacobi equation

∇Y ∗∇Y ∗X∗ + R(X ∗, Y ∗)Y ∗ = 0.

But p∗ is a subspace so for Y ∗,Z ∗ ∈ p∗ we have Y ∗ + Z ∗ ∈ p∗. Inserting thisinto the above equation we obtain

0 = ∇(Y ∗ + Z ∗)∇(Y ∗ + Z ∗)X∗ + R(X ∗, Y ∗ + Z ∗)(Y ∗ + Z ∗)

= ∇Y ∗∇Y ∗X∗ +∇Y ∗∇Z ∗X

∗ +∇Z ∗∇Y ∗X∗ +∇Z ∗∇Z ∗X

∗ +

R(X ∗, Y ∗)Y ∗ + R(X ∗,Z ∗)Y ∗ + R(X ∗, Y ∗)Z ∗ + R(X ∗,Z ∗)Z ∗

= ∇Y ∗∇Z ∗X∗ +∇Z ∗∇Y ∗X

∗ + R(X ∗,Z ∗)Y ∗ + R(X ∗, Y ∗)Z ∗

(But R(X ∗,Z ∗)Y ∗ = −R(Z ∗,X ∗)Y ∗)

= ∇Y ∗∇Z ∗X∗ +∇Z ∗∇Y ∗X

∗ − R(Z ∗,X ∗)Y ∗ + R(X ∗, Y ∗)Z ∗

(By the Bianchy identity

−R(Z ∗,X ∗)Y ∗ = R(X ∗, Y ∗)Z ∗ + R(Y ∗,Z ∗)X ∗)

= ∇Y ∗∇Z ∗X∗ +∇Z ∗∇Y ∗X

∗ + 2R(X ∗, Y ∗)Z ∗ + R(Y ∗,Z ∗)X ∗

(Now R(Y ∗,Z ∗)X ∗ = ∇Y ∗∇Z ∗X∗ −∇Z ∗∇Y ∗X

∗ −∇[Y ∗,Z ∗]X∗

But for Y ∗,Z ∗ ∈ p∗ [Y ∗,Z ∗] ∈ k∗ since

[Y ∗,Z ∗] = −[Y ,Z ]∗by Appendix D so [Y ∗,Z ∗](p) = 0

Thus R(Y ∗,Z ∗)X ∗ = ∇Y ∗∇Z ∗X∗ −∇Z ∗∇Y ∗X

∗)

= 2∇Y ∗∇Z ∗X∗ + 2R(X ∗, Y ∗)Z ∗.

37

Hence

(4.4) ∇Y ∗∇Z ∗X∗ + R(X ∗, Y ∗)Z ∗ = 0.

Therefore starting with the Bianchy identity

R(X ∗, Y ∗)Z ∗(p) = −R(Y ∗,Z ∗)X ∗(p) + R(X ∗,Z ∗)Y ∗(p)

= ∇Z ∗∇X ∗Y∗(p)−∇Z ∗∇Y ∗X

∗(p)

by eq. (4.4)

= ∇Z ∗[X ∗, Y ∗](p) since ∇ is torison free

= ∇[X ∗, Y ∗]Z ∗(p)− [[X ∗, Y ∗],Z ∗](p)

again since ∇ is torision free

= −[[X ∗, Y ∗],Z ∗](p) since [X ∗, Y ∗](p) = 0

�

For M = G/K we give M the Riemann structure induced by Theorem3.11. There are a few details about Lie algebra operations in p∗ versus p, forour purposes it is the fact that [X ∗, Y ∗] = −[X , Y ]∗, see Appendix D.

Corollary 4.14. [6] Let (G/K , g) be a symmetric space. Then the sectionalcurvature of X ∗, Y ∗ ∈ TpG/K is

K (X ∗, Y ∗)(p) =−g([[X , Y ], Y ]∗,X ∗)

g(X ∗,X ∗)g(Y ∗, Y ∗)− g(X ∗, Y ∗)2

PROOF. This follows from the definition of the sectional curvature andthat [X ∗, Y ∗] = −[X , Y ]∗. �

Now consider the curvature tensor R(X ∗, Y ∗)Z ∗ = −[[X ∗, Y ∗],Z ∗] whichcan be rewritten as R(X ∗, Y ∗)Z ∗ = −[[X , Y ],Z ]∗. Since we know thatR(X ∗, Y ∗)Z ∗ ∈ TpM we formally denote its corresponding vector in p byR(X , Y )Z therefore we formally have

R(X , Y )Z = −[[X , Y ],Z ] for X , Y ,Z ∈ p

Since g(·, ·) on TM derives from 〈·, ·〉 on p we can always write

−g([[X , Y ], Y ]∗,X ∗)g(X ∗,X ∗)g(Y ∗, Y ∗)− g(X ∗, Y ∗)2

=−〈[[X , Y ], Y ],X )

〈X ,X 〉〈Y , Y 〉 − 〈X , Y 〉2

in p. So in terms of this Corollary 4.14 becomes

K (X , Y ) =−〈[[X , Y ], Y ],X )

〈X ,X 〉〈Y , Y 〉 − 〈X , Y 〉2 for X , Y ∈ p

38

Example 4.15. By Corollary 4.14 and an Ad g invariant metric g the for-mula for the sectional curvature can be expressed as

K (X ∗, Y ∗) =g([X , Y ]∗, [X , Y ]∗)

g(X ∗,X ∗)g(Y ∗, Y ∗)− g(X ∗, Y ∗)2

If we define a metric in the same way as in Lemma 3.10 we can write theformula in p as

K (X , Y ) =〈[X , Y ], [X , Y ]〉

〈X ,X 〉〈Y , Y 〉 − 〈X , Y 〉2where we use the same notation for the metrics. If we work with the real Grass-mann manifolds we can choose a basis eij− eji . Before we start the calculationswe produce some intermediate results.

[eij, ekl ] � = � i� � jk � l − � k� � li � j = � jkeil − � liekj

this gives

[eij − eji, ekj − elk] = [eij, ekl ]− [eij, elk]− [eji, ekl ] + [eji, elk]

= � jkeil − � liekj − � jl eik + � kielj − � ikejl + � ljeki

+ � il ejk − � jkeil − � kjeli

= � jk(eil − eli) + � il (ejk − ekj) + � ik(elj − ejl )

+ � jl (eki − eik)

Now if we choose the metric

〈X , Y 〉 = −12

trace (XY )

which is proportional to the Killing form on so(n). Therefore

〈eij − eji, ekl − elk〉 = −12

trace ( � jkeil − � jl eik − � ikejl + � il ejk)

= −12

( � jk � il − � jl � ik − � ik � jl + � il � jk)

= − � jk � il + � ik � jl but i < j, k < l so

= � ik � jl

So it is indeed an orthogonal basis. Therefore

g([eij − ejk, ekl − elk], [eij − ejk, ekl − elk]

)= � jk + � il + � ik + � jl

Note that only none or one of these can be satisfied, so

K (eij − eji, ekl − elk) =

1 if (i = k, j 6= l), (i = l , j 6= k),, (i 6= k, j = l) or (i 6= l , j = k)

0 else

39

So the sphere Sn has constant sectional curvature 1, while the real Grassman-nian manifolds have sectional curvature 0 or 1.

3. The Dual Space

Definition 4.16. Let (G/K , g, � ) be a symmetric space such that the innerproduct g on g is both Ad G and d � invariant. Such that we can decomposethe Lie algebra as

g = k⊕ p

The symmetric space is then called a normal symmetric space.

Note that when g is semisimple the Killing form B makes

(G/K ,B, � )

a normal symmetric space.The concept of dual spaces makes it possible to pair up symmetric spaces

to that if one knows properties of one in the pair, one also knows the sameproperties of the dual and vice versa.

Definition 4.17. Two normal symmetric spaces M = G/K and M † =G†/K † are said to be dual provided

(1) There is a Lie algebra isomorphism � : k→ k† such that

g†( � (X ), � (Y )) = −g(X , Y )

(2) A linear isometry ˆ� : p→ p† such that

[ˆ� (X ), ˆ� (Y )]† = − � [X , Y ],

where all the corresponding entities in M † have been denoted with an †.Remark 4.18. Because � is a Lie algebra isomorphism we have

[ � (X ), � (Y )]† = [X , Y ]

and since ˆ� is an isometry

g†(ˆ� (X ), ˆ� (Y )) = g(X , Y ).

The above isometry ˆ� induce an isometry

TeK M → TeK †M†

via the following commutative diagram

pˆ�−−−→ p†

� (p)

y � (p†)

yTeK M −−−→ TeK †M †

For compact spaces the following choice of metric is common

40

Definition 4.19. Let G/K be a homogeneous space with G compact andg semisimple then the standard homogeneous Riemannian metric on G/K ischosen as the negative Killing form i.e. g = −B.

As promised duality provides important information about the dual spaces.

Theorem 4.20. [9] Let (M , g) and (M †, g†) be dual spaces. Then (M , g and(M †, g†) have opposite sectional curvature i.e.

K †M†(ˆ� X , ˆ� Y ) = −KM (X , Y ) for all X , Y ∈ p

PROOF.

KM†(ˆ� X , ˆ� Y ) =g†([ˆ� X , ˆ� Y ], [ˆ� X , ˆ� Y ])

g(ˆ� X , ˆ� X )g(ˆ� Y , � Y )− g(ˆ� X , ˆ� Y )2

=g†(− � [X , Y ],− � [X , Y ])

g(X ,X )g(Y , Y )− g(X , Y )2

=−g([X , Y ], [X , Y ])

g(X ,X )g(Y , Y )− g(X , Y )2

= −KM (X , Y )

�

To illustrate duality we find the the dual space of the real Grassmannianmanifold.

Example 4.21. If we now generalize Example 1.7 by considering

(4.5) Q =

(Ip 0T

0 −Iq

)

Then we define

O(p, q) = {A ∈ GLp+q(R) : AT QA = Q}Then O(p, q) represents orthogonal matrices with respect to the inner productdefined by Equation (4.5). We find the Lie algebra g = o(p, q) of O(p, q)by taking the derivative of a curve at the origin in the defining expression forO(p, q). Then as in Example 1.7

o(p, q) =

{(A B

BT C

): A ∈ o(p), B ∈ M(p,q), C ∈ o(q)

}

If we let O(p, q) act on the real Grassmann manifold Gp(Rp+q) as in Example1.8, then the isotropy group consists of matrices of the form

(B 00 C

)where B ∈ O(p),C ∈ O(q)

41

so we have thatO(p, q)/(O(p)×O(q))

is a manifold. Since the Lie algebra of O(k) consists of the skew symmetricmatrices we get

k =

{(A 00 D

): A ∈ o(p), D ∈ o(q)

}

and

m =

{(0 B

BT 0

): B ∈ M(p,q)

}

Example 4.22. Let G = O(p, q) and K = O(p)×O(q) be as in Theorem3.7 with

J =

(Ip 00T −Iq

)

with J ∈ Z (K ). If we form G/K = O(p, q)/O(p)×O(q) and observe that forX ∈ o(p, q)

X =

(A B

BT D

), Ad J

(A B

BT D

)=

(A −B−BT D

)

we have

X + Ad J (X ) = 2

(A 00T D

)∈ k

so J satisifies 2) in Theorem 3.7 so we have the symmetric pair(SO0(p, q), S(O(p)×O(q))

)with s = In J

Example 4.23. Let (G,K ) be the symmetric pair

(G,K ) = (SO(p + q), (SO(p)× SO(q))

with involutive automorphism s given by Example 3.9

� (A) = In S(A) = SAS−1 where S =

(Ip 00 −Ip

)

on the manifold G/K and the Ad G-invariant metric

B(X , Y ) = −12

trace (XY )

at the origin. Then by extending the inner product B to all of M as in Theorem3.11 � becomes an isometry, so

M = SO(p + q)/(SO(p)× SO(q))

becomes a normal symmetric space. Note that B is positive definite on p.

42

On the symmetric pair (G†,K †) = (SO0(p, q), SO(p) × SO(q)) we pickthe same involutive automorphism s as in Example 4.22. And on the manifoldG†/K † we take the metric

B†(X , Y ) =12

trace (XY )

at the origin, which again is Ad G† invariant and � is an isometry after extend-ing the metric. So we have the normal symmetric space

M † = SO0(p, q)/(SO(p)× SO(q))

Again B† is positive definite on p†.Let

� : k→ k† = idk

which certainly is a Lie algebra isomorphism such that

B(X , Y ) = −12

trace XY = −(−12

trace XY ) = −B†(X , Y )

and

ˆ� : p→ p†,

(0 x−xT 0

)7→(

0 xxT 0

)

which is an isometry such that

[ˆ� (X ), ˆ� (Y )]† =

(0 x

xT 0

)(0 yyT 0

)−(

0 yyT 0

)(0 x

xT 0

)

=

(xyT 00 xT y

)−(

yxT 00 yT x

)

= −(

0 x−xT 0

)(0 y−yT 0

)+

(0 x−xT 0

)(0 y−yT 0

)

= −[X , Y ]

= − � ([X , Y ])

Therefore M ,M † are dual.

43

Example 4.24. Further examples of dual pairs of symmetric spaces can befound in the following table,

Noncompact compact Dimension

SL(n,R)/SO(n) SU(n)/SO(n) 12 (n− 1)(n + 2)

SU(p, q)/S(U(p)× U(q)) SU(p + q)/S(U(p)× U(q)) 2pqSO0(p, q)/SO(p)× SO(q) SO(p + q)/SO(p)× SO(q) pqSp(n,R)/U(n) Sp(n)/U(n) n(n + 1)Sp(p, q)/Sp(p)× Sp(q) Sp(p + q)/Sp(p) + Sp(q) 4pq

see also [5].

44

APPENDIX A

The Hopf-Rinow Theorem

Definition A.1. A Riemannian manifold (M , g) is said to be geodesicallycomplete if for all p ∈ M the exponential map Expp is defined on all of TpM ,i.e. all geodesics � ∈ M can be defined for all t ∈ R.

Definition A.2. A normal ball B (p) at p ∈ M on a Riemanninan mani-fold (M , g) is the image of an open ball B (p) in the tangent space TpM of theexponential map Expp.

Definition A.3. Let p, q ∈ M then define the distance d (p, q) as the infi-mum of the lengths of all curves on M from p to q.

Remark A.4. It can be shown that the distance function is continuous andthat geodesics locally minimize the distance between two points, also that if acurve is a minimum then it must be a geodesic.

Theorem A.5. [3] Hopf-RinowLet (M , g) be a Riamannian manifold if (M , g) is geodesically complete then anytwo points p, q ∈ M can be connected by a geodesic.

PROOF. [3] If M is geodesically complete then the exponential map Expp

is defined on all of TpM .Let p, q ∈ M and r = d (p, q). Let B (p) be a normal ball in M and

let S (p) be the boundary of B (p). Let x0 be a point where the continuousfunction d (q, x), x ∈ S (p) takes on its infimum(It does so since S (p) iscompact). Then x0 = Expp � v where v ∈ TpM and |v| = 1. Let � be thegeodesic given by � (s) = Exppsv. Now let

A = {s ∈ [0, r] : d ( � (s), q) = r − s}.We will show that r ∈ A and � (r) = q.

Now A is not empty since 0 ∈ A and A is closed by continuity of thedistance. We are going to show that if s0 ∈ A and s0 < r then there exists a� > 0 such that s0 + � ∈ A. This will imply that supA = r and r ∈ A since Ais closed. Thus d ( � (r), q) = r − r = 0 so � (r) = q.

Let B ′( � (s0)) be a normal ball at � (s0) and S ′ be the boundary of B ′( � (s0)).Let x′0 be the point where d (x, q), x ∈ S ′ takes on its minimum. So

d ( � (s0), q) = � ′ + min{d (x, q) : x ∈ S ′} = � ′ + d (x′0, q)

45

But d ( � (s0), q) = r − s0 so r − s0 = � ′ + d (x′0, q). Now if x′0 = � (s0 + � ′) weget

r − s0 = � ′ + d ( � (s0 + � ′, q) so s0 + � ′ ∈ A.We only have to show that x′0 = � (s0 + � ′). But by the triangle inequality

d (p, x′0) ≥ d (p, q)− d (q, x′0) = r − (r − s0 − � ′) = s0 + � ′which is no surprise. But more importantly the broken curve joining p and x ′0by going along � from p to � (s0) and along the exponential ray from � (s0) tox′0 in B ′( � (s0)), has length s0 + � ′ so d (p, x′0) = s0 + � ′. By minimality of thedistance that broken curve must be a geodesic. But at it is starting out as � itmust remain � , so x′0 = � (s0 + � ′). �

46

APPENDIX B

The Adjoint Representation

Definition B.1. Let G be a Lie group. For each h ∈ G we define the innerautomorphism of G by

In h : G → G g 7→ hgh−1

Denote by GL(g) the group of vector space automorphisms of g.

Definition B.2. Let G be a Lie group, then the adjoint representation of Gis given by

Ad : G → GL(g) h 7→ deIn(h)

The map In g is a Lie group automorphism so for fixed g ∈ G Ad g is aLie algebra automorphism, thus it satisfies

Ad g([X , Y ]) = [Ad g(X ),Ad g(Y )] for all X , Y ∈ g

andexp(Ad g(X )) = In g(exp(X ))

But we can also view Ad as a map Ad : G → GL(g). Then

Ad g1 ◦ Ad g2 = Ad g1g2

so it is a Lie group homomorphism. Therefore it induces a Lie algebra auto-morphism.

Definition B.3. Let g be a Lie algebra, then the adjoint representation of gis defined as

ad : g→ End(g), X 7→ (deAd )(X )

where End(g) is the space of linear maps g→ g

So we havead [X , Y ] = [ad X , ad Y ]

Lemma B.4. [10] Let g be the Lie algebra of the Lie group G then

(adX )Y = [X , Y ]

for X , Y ∈ g.

47

PROOF. For g ∈ G, Y ∈ g we have

Ad gYe = d(In g)(Ye) = d(Rg−1Lg )(Ye) = dRg−1Yg

Let exp(tX ) be an integral curve of X then

ad X (Y ) = limt→0

1t

(Ad exp(tX )(Ye)− Ad idG(Ye))

= limt→0

1t

(dRexp(−tX )YexptX − Ye)

= limt→0

1t

(dRexp(−tX )YRexp(tX )e − Ye)

= LX Y = [X , Y ]

since RexptX is the flow of X in G by Lemma D.15. �

Proposition B.5. [6] Let g be a Lie algebra, then for all X ∈ g

exp(ad X ) = Ad exp(X )

PROOF. [6]

ddt

exp(tad X )|t=0 = ad X

= (dAd )X

= =ddt

Ad exp(tX )|t=0.

So the claim follows by the uniqueness of ordinary differential equations withinitial data. �

Let Aut g be the set of Lie algebra automorphisms on g. Then Aut g is aclosed subgroup of the Lie group GL(g). Let Int g be the identity componentof Aut g, with Lie algebra ad g. By exponentiating ad g we get the entire Int gwhich is contained in Ad g ⊆ Aut g.

Definition B.6. Let ∂g be the Lie algebra of Aut g. Then

Aut g ⊆ GL(g) and ∂g ⊆ End(g).

Definition B.7. And endomorphism D ∈ End(g) is called a derivation if

D[X , Y ] = [DX , Y ] + [X ,DY ]

Theorem B.8. [10] Let g be a Lie algebra and ∂g be the Lie algebra of Aut gthen ∂g is the set of derivations on g.

PROOF. [10] Let D ∈ ∂g then for t ∈ R, exp(tD) ∈ Aut g so

exp(tD)[X , Y ] = [exp(tD)X , exp(tD)Y ].

48

Therefore

D[X , Y ] =ddt

exp(tD)[X , Y ]|t=0

=ddt

[exp(tD)X , exp(tD)Y ]|t=0

= [DX , Y ] + [X ,DY ],

so D is a derivation.Conversely let D be a derivation. Then

D[X , Y ] = [DX , Y ] + [X ,DY ]

so inductively

Dk[X , Y ] =

k∑

i=0

(ki

)[DiX ,Dk−iY ].

So we get

exp(tD)[X , Y ] =∞∑

l=0

t l Dl

l ![X , Y ]

=

∞∑

l=0

t l

l !

l∑

i=0

l !(l − i)!i!

[DiX ,Dl−iY ]

=∞∑

l=0

l∑

i=0

[ t iDiXi!

,t l−iDl−iY

(l − i)!

]

if we switch summation order we get

=∞∑

i=0

∞∑

l=i

[ t iDiXi!

,t l−iDl−iY

(l − i)!

]let k = l − i

=

∞∑

i=0

∞∑

k=0

[ t iDiXi!

,tkDkY

k!

]

= [exp(tD)X , exp(tD)Y ].

The above manipulations of the summations are legitimate since exp(tD) con-verges absolutely so can replace upper limit∞ with N and do manipulations.Thus exp(tD) ∈ Aut g so D ∈ ∂g. �

Theorem B.9. [10] Let G be a Lie group and let Z (G) be the center of G. IfG is connected then Z (G) = kerAd g.

PROOF. [10] Obviously Z (G) ⊆ kerAd g. Now suppose Ad g = id then

exp(X ) = gexp(X )g−1 by exponentiating,

so g ∈ Z (G0) = Z (G) since G is connected. �49

We define z(g) = {X ∈ g : ad X = 0}Theorem B.10. Let G be a Lie group with Lie algebra g. If G is connected

then z(g) is the Lie algebra of Z (G).

PROOF. Let X ∈ g then if X is belongs to the Lie algebra of Z (G) thenexp(tX ) ∈ Z (G) so

Ad exp(tX ) = idGL(g).

This is equivalent to that

exp(ad tX ) = idGL(g)

which happens if and only if ad X = 0 so X ∈ z(g). �

50

APPENDIX C

Inner Products From the Haar Measure

Fact C.1. Let G be a compact topological group and let C (G) be the set ofcontinuous real valued functions on G. Then there is a unique map

I : C (G)→ R, I (f ) 7→∫

Gf (g)dg

such that(1) I (e) = 1,(2) I (f ) ≥ 0 for f ≥ 0,(3) I (

�f + � g) =

�I (f ) + � I (g) for

�, � ∈ R,

(4) I is G − invariant.The function I is called the Haar-integral.

Lemma C.2. [1] Let H be a Lie group, V be a vector space and � : H →Aut(V ) be a representation of the group H on V . If H is compact there exists anH invariant inner product on V .

PROOF. Since H is compact, we can define the Haar-integral

I (f ) =

∫

Hf (h)dh

The Haar-integral is H -invariant i.e.

I (f ) = I (f ◦ Rg ) = I (Lg ◦ f ).

So if 〈·, ·〉 is an inner product on V then we define the H invariant innerproduct on V

(u, v) =

∫

H〈 � (h)u, � (h)v〉dh.

�

51

APPENDIX D

Lie Derivatives and Killing Fields

We state the following result from the theory of ordinary differential equa-tions

Lemma D.1. [6] Let X be a vector field on the manifold M. Then for everypoint p ∈ M there exists an open neighborhood U of p and an open open intervalI of R such that 0 ∈ I , with the property that for all q ∈ U there is a curvecq : I → M satisfying

cq(t) = X (cq(t)), cq(0) = q.

Then map (t, q)→ cq(t) from I × U to M is smooth.

Definition D.2. Let X , c be as in Lemma D.1, then the map (t, q) 7→ cq(t)is called the local flow of the vector field X. The curve cq is called the integralcurve of X through q.

For a fixed t and varying q we write � t(q) = cq(t). Note that

(D.1) � t ◦ � s(q) = � t+s(q) if s, t, s + t ∈ Iq

sinceddt

� s+t(q)|t=0 = X ( � s(q)) =ddt

� t( � s(q))|t=0

and they have the same initial data for t = 0 at q.It can be shown that if � t is defined on an open subset U of M , it maps U

diffeomorphically onto its image.

Definition D.3. A family ( � t)t∈I of diffeomorphisms from M to M satis-fying equation (D.1) is called a local 1-parameter group of diffeomorphisms.

Definition D.4. Let � : M → N be a diffeomorphism between differen-tiable manifolds and let X be a vector field on M. Then we define the followingvector field on N

( � ∗X )(p) = d � (X ( � −1(p))) for p ∈ N

The following lemma follows from the definition of the differential map.

Lemma D.5. [6] Let N be a differentiable manifold. For any differentiablefunction f : N → R we have

( � ∗X )(f )(p) = X (f ◦ � )( � −1(p)).

53

Now consider the special case when N = M so � −t = � −1t .

Let f be a real valued function defined in the codomain of � t , then thepull-back of f via � t is defined by

� ∗t f ≡ f ◦ � t

For a vector field X = ak∂∂xk on the codomain of � t

� ∗t X (p) ≡ ( � −t)∗X (p) = ak∂ � i−t

∂xk

∂

∂yievaluated at � t(p)

where x,y are local coordinates in the codomain and domain and

� i−t = (y ◦ � −t)

i.

For a 1-form � = � idyi in the codomain of � t we define the pull-back by

� ∗t ( � )(p) = � i( � t(p))∂ � i

t

∂xkdxk

The action of � ∗t on higher order tensors is obtained by its action on the atomictensors.

Definition D.6. Let X be a vector field on the differentiable manifold Mwith local 1-parameter group � t of local diffeomorphisms and S a tensor fieldon M . Then the Lie-derivative of S in the direction of X is defined by

LX S =ddt

( � ∗t S)|t=0

Definition D.7. Let X be a vector field and S be a tensor of type (0, r)then the inner product � X S is the contraction

( � X S)(X1,X2, . . . ,Xr−1)(p) ≡ C1,1(X ⊗ S)(X1,X2, . . . ,Xr−1)(p)

= S(X ,X1, . . . ,Xr−1)

where C1,1 is the contraction operator.

Theorem D.8. [6] Let M be a differenitiable manifold and X a vector fieldon M.

(1) If f : M → R be a differentiable function, then

LX (f ) = df (X ) = X (f )

(2) If Y be a vector field on M, then

LX Y = [X , Y ]

(3) Let � = � jdxj be a 1-form on M. If X = X i ∂∂xi , then

LX � = d( � X � ) + � x(d � ).

For higher order tensors one simply uses the product rule for differentiations on theatomic tensors.

54

PROOF. [6] 1)

LX (f ) =ddt

� ∗t f|t=0 =ddt

f ◦ � t|t=0 =∂f∂xi

X i = X (f )

2) Let Y = Y i ∂∂xi

LX Y =ddt

� ∗t(

Y i ∂

∂xi

)|t=0

=ddt

( � −t)∗(

Y i ∂

∂xi

)|t=0

=ddt

(Y i( � t)

∂ � j−t

∂xi

∂

∂xj

)|t=0

=∂Y i

∂xkX k � j

i∂

∂xj+ Y i

(− ∂X j

∂xi

) ∂∂xj

since

{∂ � j−t

∂xi |t=0=∂(xj ◦ � −t)

∂xi |t=0=∂xj( � 0)∂xi

= � ji

}and

{ ddt

� −t|t=0 = −X}

=(

X k ∂Y j

∂xk− Y k∂X j

∂xk

) ∂∂xj

= [X , Y ]

3)

LX � =ddt

( � ∗t � )|t=0 =ddt

(� j( � t)

∂ � jt

∂xkdxk)|t=0

=∂ � j

∂xiX i � j

kdxk + � j∂X j

∂xkdxk as in the previous

=(∂ � j

∂xiX i + � i

∂X i

∂xj

)dxj.

Now if we expand

d( � X � ) + � X (d � ) = d( � jX j) + � X (∂ � j

∂xidxi ∧ dxj)

=∂ � j

∂xiX jdxi +

∂X i

∂xj� idxj +

∂ � j

∂xiX idxj

−∂ � j

∂xiX jdxi

=∂X i

∂xj� idxj +

∂ � j

∂xiX idxj

55

So indeedLX � = d( � X � ) + � X (d � ).

�Definition D.9. Let (M , g) be a Riemannian manifold with a metric g =

gijdxi⊗dxj then a vector field X on M is called a Killing field or an infinitesimalisometry if

LX (g) = 0.

Lemma D.10. [6] A vector field X on a Riemannian manifold (M , g) is aKilling field if and only if the local 1-parameter group generated by X consists oflocal isometries.

PROOF. [6] By definition

ddt

( � ∗t g)|t=0 = 0

holds at every point of M . Hence � ∗t g = g for all t ∈ I , so the diffeomor-phisms � t are isometries.

Conversely if � t are isometries then

g(( � t)∗(X ), ( � t)∗(Y )) = g(X , Y )

soddt

� ∗t g = 0.

�Lemma D.11. [8] Let M be a differentiable manifold. Two derivations

D1,D2 agree if they agree on vector fields and functions on M.

PROOF. [8] Since the product rule applies for derivations on tensors, weonly have to show the claim for one forms, since a general tensor can be writtenas a tensor product of vector fields, functions and one forms.

Let � = � idxi be an arbitrary one form on M and let V = V i ∂∂xi be an

arbitrary vector field on M . � (V ) is a function on M so if D is a derivation onM then

D( � (V )) = D(C 11 � ⊗ V ) = C 1

1 (D � ⊗ V + � ⊗ DV ) = (D � )(V ) + � (DV )

where C 11 is the contraction operating on � and V . This defines the derivation

on one forms if one knows how the derivation operates on vector fields andfunctions. �

Corollary D.12. [8] Let M be a differentiable manifold. Then the Lie de-rivative satisfies

L[U ,V ] = [LU , LV ]for arbitrary vector fields U ,V on M.

56

PROOF. [8] Since both [LU , LV ] and L[U ,V ] are derivations, Lemma D.11implies that it is enough that they agree on functions and vector fields on M .

Let f be a function on M then

[LU , LV ]f = LU LV f − LV LU f

= U ◦ V (f )− V ◦ U (f )

= [U ,V ](f )

= L[U ,V ]f .

Similarly if W is a vector field on M then

[LU , LV ]W = LU LV W − LV LU W

= [U , [V ,W ]]− [V , [U ,W ]]

= [[U ,V ],W ]

= L[U ,V ]W ,

where we have used the Jacobi identity and part 2 of Theorem D.8. �Lemma D.13. [6] The Killing fields of a Riemannian manifold (M , g) is a

Lie-algebra.

PROOF. [8],[6] Since vector fields on a differentiable manifold constitute aLie-algebra, we only have to show that if X , Y are Killing fields then L[X ,Y ]g =0. But by Corollary D.12 the equation L[X ,Y ] = [LX , LY ] and the fact thatX , Y are Killing fields LX g = 0, LY g = 0 so L[X ,Y ]g = 0. �

Proposition D.14. [6] Every Killing field X on a Riemannian manifold(M , g) is a Jacobi field along any geodesic � ∈ M.

PROOF. [6] By Lemma D.10 each Killing field X generates a local 1-parameter group of isometries. Isometries map geodesics to geodesics, so Xgenerates a variation of geodesics c(t, s) = � t( � (s)), where � is the local 1-parameter group of X . Since every variation of geodesics generates a Jacobifield by X = ∂

∂t c(t, s)|t=0, the claim follows. �Lemma D.15. [9] Let G be a Lie group with Lie algebra g. Let � be the

one-parameter subgroup of X ∈ g then the flow of X is R � t

PROOF. [9] Here Lg is the left translation, i.e it has nothing to do with theLie derivative. If g ∈ G then Lg ◦ � is the integral curve of dLgX = X startingat g, so Lg � t = g � t = R � t g. Thus R � t is the flow of X. �

Proposition D.16. [9] Let G be an isometry group that acts on a Riemann-ian manifold (M , g) by � p(g) = gp. Denote the Lie algebra of G by g. LetX , Y ∈ g and define the Killing field

X ∗ = d � p(X ) =ddt

exp(tX )p

57

and similarly for Y ∗ then

[X ∗, Y ∗] = −[X , Y ]∗

PROOF. [9] Again Lg is the left translation. Let � t be the local flow of X ∗.Then for Y ∈ G we have

� −t ◦ exp(Y ) ◦ � t(p) = ( � pR � t L � −t )exp(Y )

so by the left-invariance of Y we have

d � −t(Y ∗� t (p)) = (d � pdR � t dL � −t )Ye = d � pdR � t Y � −t = d � pdR � t YR −t e.

Furthermore

[X ∗, Y ∗] = LX ∗Y ∗

= limt→0

1t

(dR � −t Y∗� t− Y ∗p )

= limt→0

1t

(d � pdR � t YR −t e − d � pYe)

= d � p limt→0

1t

(dR � t YR −t e − Ye)

= d � pL−X Y

= [−X , Y ]∗

The second last equality follows from the formula that if � t is the flow of Xthen

LX Y =ddt

� −tY � t .

By Lemma D.15 if � t is the integral curve of X then R � t is the flow of X inG. �

58

Bibliography

[1] A. Arvanitoyeorgos, An Introduction to Lie Groups and the Geometry of Homogeneous Spaces, Student MathematicalLibrary vol. 22, AMS (2003).

[2] A. Baker, Matrix Groups, Springer Undergraduate Mathematics Series, Springer (2002).[3] M. P. DoCarmo, Riemannian Geometry, Birkhauser (1992).[4] S. Hassani, Mathematical Physics, A Modern Introduction to Its Foundations, Springer (2002)[5] S. Helgason, Differential Geometry, Lie Groups and Symmetric Spaces, Academic Press (1978).[6] J. Jost, Riemannian Geometry and Geometric Analysis, Universitext, Springer (2002).[7] S. Kobayashi and K. Nomizu, Foundations of Differential Geometry vol. II, John Wiley & Sons (1963).[8] M. Kriele, Spacetime, Foundations of General Relativity and Differential Geometry, Lecture Notes in Physics -

Monographs, Springer (1999).[9] B. O’Neill, Semi Riemannian Geometry, Academic Press (1983).[10] P. Sjogren, Riemannska Symmetriska Rum, Matematiska Institutionen, Chalmers Tekniska Hogskola och

Goteborgs Universitet (1987).

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Master’s Theses in Mathematical Sciences 2005:E3ISSN 1404-6342

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