SYMMETRIC SPACES. PETER HOLMELIN Master s thesis 2005:E3. Centre for Mathematical Sciences Mathematics

Transcription

1 SYMMETRIC SPACES PETER HOLMELIN Master s thesis 2005:E3 Centre for Mathematical Sciences Mathematics CENTRUM SCIENTIARUM MATHEMATICARUM

2

3 Abstract In this text we study the differential geometry of symmetric spaces. We describe how a symmetric space (M, g) can be seen as a homogeneous space G/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 symmetric pairs. Furthermore we investigate the expressions for curvature on a symmetric space. Finally we describe the notion of dual symmetric spaces. To illustrate how well symmetric spaces lend themselves to explicit calculations we calculate the curvature of the real Grassmann manifold and find their dual space. Keywords: homogeneous spaces, symmetric spaces, symmetric pairs, the Killing form, curvature of symmetric spaces, dual symmetric spaces. Throughout this text it has been my intention to give reference to all the sources that have been used.

4

5 Acknowledgements I am very grateful to Sigmundur Gudmundsson for his support and substantial amount of comments. I also wish to thank Martin Svensson his deep knowledge and judicious comments about the material. Peter Holmelin

6 Contents Overview 1 Chapter 1. Homogeneous Spaces 3 1. Group Actions 3 2. Homogeneous Spaces 3 Chapter 2. Symmetric Spaces Definitions The Isometry Group G/K is Diffeomorphic to M 17 Chapter 3. Symmetric Pairs From a Symmetric Space to a Symmetric Pair The Tangent Space of G/K From a Lie Group to a Symmetric Pair From a Symmetric Pair to a Symmetric Space 25 Chapter 4. Curvature of a Symmetric Space The Killing Form The Curvature Formula 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

7

8 Overview History[4] At the end of the nineteenth century, after studying spaces of constant curvature, mathematicians wanted to classify all locally symmetric Riemannian manifolds, i.e. Riemannian manifolds whose curvature tensor is parallel i.e. satisfies R = 0. The whole issue was settled by Elie Cartan in Previously in his thesis he had classified all simple complex Lie algebras and he also classified the simple real Lie algebras. Using this he gave a complete classification of all symmetric spaces which by the way were introduced by himself in Symmetric Spaces A symmetric space is a Riemannian manifold (M, g) such that for every point p M there exist an isometry p of (M, g) called an involution such that (1) p(p) = p and (2) d p = id TpM. 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 Riemanninan manifold can be connected by a geodesic. Therefore the translations along the geodesics makes the isometry group G acting on M transitive. Using the theory of homogeneous spaces one can identify M with G/K where K is the 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 closed subgroup K and an involution, leads to the concept of a symmetric pair which is defined as a Lie group G with a closed subgroup K and an involutive automorphism s on G satisfying (1) (G s ) 0 K G s, 1

9 (2) Ad K is compact where G s is the set of elements left invariant by s. We show that symmetric pairs lead to symmetric spaces. Curvature on a Symmetric Space We show that left invariant vector fields on the isometry group G are mapped to Killing fields in the symmetric space (M, g), which generate Jacobi fields and therefore provide the connection to the curvature tensor. This gives a 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 Lie algebra g of G. Dual spaces have opposite curvature and if one is compact its dual is noncompact and vice versa. Although the path chosen in this paper is more along minimal differential geometry, the standard description is more in the language of Lie groups and Lie algebras. For a reader who wants a more thorough description we refer to [5] Helgason, which is the standard reference on the theory of symmetric spaces. 2

10 CHAPTER 1 Homogeneous Spaces In this chapter we create a manifold structure for the quotient of a Lie group and a closed subgroup. 1. Group Actions Definition 1.1. Let M be a smooth manifold and G be a Lie group with identity element e G. A smooth map : G M M is called a group action on M if (g 1, (g 2, p)) = (g 1 g 2, p) (e, p) = p for all g 1, g 2 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 such that (g, p) = q. The isotropy group K p0 of at p 0 M is given by K p0 = {g G : (g, p 0 ) = p 0 }. 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 = S n be the unit sphere in R n+1 and G = SO(n+1) be the special orthogonal group. Then we have the action : SO(n + 1) S n S n, (A, p) A p For p 0 = (1, 0,..., 0) we get K p0 = SO(1) SO(n). 2. Homogeneous Spaces We need the following lemma to prove that G/K can get a manifold structure. 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 linear complement 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

11 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 diffeomorphism from an open neighborhood U 0 V 0 of (0, 0). We define some sets that will help to show that is one-to-one. The set exp(v 0 ) is an open neighborhood of e in K with the subspace topology so exp(v 0 ) = H K for some open set H in G. So there exists an open set such that U 1 V 1 U 0 V 0 (U 1 V 1 ) K H K = expv 0 If X U 1, Y V 1 and (X, Y ) K then But expx expy = expy for some Y V 0 is a diffeomorphism on U 1 V 1 so X = 0 and Y = Y, thus (U 1 V 1 ) K = exp(v 1 ). Now we will show the injectivity. Let U 2 U 1 be a neighborhood of 0 in m such that exp( U 2 )exp(u 2 ) (U 1 V 1 ) Then exp U2 is injective, since if X, X U 2 and then (exp(x )) = (exp(x )) exp( X )exp(x ) (U 1 V 1 ) K so X = X. The surjectivity on a neighborhood follows since is surjective from U 2 {0} onto (U 2, 0). By definition exp m is continuous. If N is an open subset of U 2 then (exp(n )) = (N, V 1 ) which is open since is a diffeomorphism here and we have the quotient topology. So the inverse is continuous. Thus exp : m G/K is a local homeomorphism at 0 in the quotient topology of G/K. Theorem 1.4. [10] Let G be a Lie group and K a closed subgroup of G, then the quotient space G/K has a unique manifold structure such that (1) the projection : G G/K, g gk is smooth and (2) has smooth local lifts such that for every gk G/K there is a neighborhood U and a map l : U G such that l = id. 4

12 With this manifold structure a map is smooth if and only if is smooth. Furthermore the action is smooth. f : G/K N f : G N G G/K G/K, (g, g K ) gg K PROOF. [10] If one chooses the quotient topology on G/K then we get a Hausdorff space by the following. Consider the map : G G G, (g 1, g 2 ) = g 1 1 g 2 which is continuous, and since K is closed the inverse image 1 (K ) is closed. Now if g 1 K g 2 K then g1 1 g 2 K so there are open sets W 1, W 2 such that (g 1, g 2 ) W 1 W 2 and W 1 W 2 1 (K ) =. Now if gk W i K then there is a k i K such that gk i W i. So if gk W 1 K W 2 K then gk 1 W 1 and gk 2 W 2. Therefore (gk 1, gk 2 ) W 1 W 2 is mapped to k1 1 k 2 K which contradicts that W 1 W 2 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 U 2 G/K, X gexp(x )K, for g G are homeomorphisms. Then we can choose local coordinates on G/K at gk as the inverses of these maps. We identify a chart with g. Suppose two charts g, g intersect then gexp(x 0 )K = g exp(x 0)K for X 0, X 0 U 2. Then exp(x 0 ) = g 1 g exp(x 0 )k 0 and exp(x 0 ) (U 2 {0}) (U 2 V 1 ). So if X is close to X 0 then there must be an X U 2 and Y V 1 such that g 1 g exp(x )k 0 = (X, Y ) and X depends smoothly on X since is a diffeomorphism on U 2 V 1. Which finally gives g exp(x )K = gexp(x )K, which shows that points in chart g are mapped smoothly to points in chart g and vice versa. So G/K has a 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. Therefore one can make the identity mapping (G/K ) (G/K ) a diffeomorphism, so they are identical as manifolds. If f is smooth then f is locally given by f l and is smooth. The reverse is trivial. The map (g, g K ) gg K is smooth since it is (group op. on G) l which is a composition of smooth maps. 5

13 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 be the Lie algebra of G and k be the Lie algebra of K. If m is a linear complement to k in g then d e : g T ek 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 local 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) then the 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) (A 1, A 2 K ) A 1 A 2 K 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 0 T 0 A where A so(n) and 0 is the n 1 matrix of zeros. The linear complement m corresponds to the matrices ( ) 0 v T v 0 where v is a n 1 matrix. By Lemma 1.5 the trivial map m T ek G is an isomorphism. Example 1.7. For x, y R n+1 define n+1 f (x, y) = x 1 y 1 x i y i using matrix multiplication we write this as ( ) 1 0 f (x, y) = x T T Qy where Q = 0 I where 0 is the n 1 matrix of zeros and I is the n n identity matrix. 6 i=2

14 Now define O(1, n) = {A GL n+1 (R) : A T QA = Q} We find the Lie algebra g = o(1, n) of O(1, n) by taking the derivative of a curve at e, in the defining expression for O(1, n). So if X =Ȧ(0) where A(t) O(1, n) then (1.1) 0 = d dt (A(t)T QA(t) t=0 = X T Q + QX if we write X as ( ) E F X = G H we get for Equation (1.1) ( E T + E G T + F F T G H t H So {( 0 F T o(1, n) = F H ) ) = 0 } : F M (n,1), H o(n) If A O(1, n) then det(a T QA) = det(q) which implies det(a) 2 = 1 so det(a) = ±1. We turn our attention to the subgroup SO(1, n) with determinant 1. The hyperboloid H 1,n = {x R n+1 : f (x, x) = 1} has two connected components H + 1,n with x 1 > 0 and H 1,n with x 1 < 0. There are A SO(1, n) which interchange the two components of H 1,n, so we restrict our attention to the subgroup Lor(1, n) = {A SO(1, n) : AH + 1,n = H+ 1,n } If we consider the vector e 1 = (1, 0,..., 0) which is left unchanged by matrices of the form ( ) 1 0 T where B SO(n) 0 B and therefore correspond to the isotropy group of e 1. Therefore Lor(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 R n+1 such that n+1 0 = u T Qv = u 1 v 1 u i v i 7 i=2

15 This is an n-dimensional subspace V, since by choosing v i, i = 2... n + 1 arbitrarily we get v 1 v 1 = u v u 1, this is well defined since u H + 1,n have u 1 > 0 It also follows that the vector v satisfies f (v, v) < 0 since f (v, v) = = < u v 2 v 2 u 2 1 u v u 2 v 2 u v 2 v 2 by Cauchy-Schwartz u 2 v 2 v 2 = 0 So f V is a positive definite inner product on V and by the Gram-Schmidt process we can construct a basis {w i } n+1 i=2 such that f (w i, w j ) = ij, f (u, w i ) = 0 Then the matrix ( ) u1 w1 A = 1... w1 n u w 1... w n is in Lor(1, n) and A(e 1 ) = u. We decompose the Lie algebra of O(1, n) as {( ) } 0 0 k = : A o(n) 0 A and {( 0 A T m = A 0 ) } : A M (n,1) Example 1.8. Let G k (R n ) denote the set of all k-dimensional subspaces in R n. This space is called the real Grassman manifold. The orthogonal group O(n) acts transitively on G k (R n ), since if V is spanned by the k first elements of the canonical basis e 1,..., e n of R n 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 0 where B O(k), C O(n k) 0 C so we have that O(n)/(O(k) O(n k)) 8

16 is a manifold. Since the Lie algebra of O(n) consists of the skew-symmetric matrices we get {( ) } A 0 k = : A o(k), D o(n k) 0 D and m = {( 0 B B T 0 ) } : B M (k,n k) Example 1.9. Let K = R, C or H, then if one considers K n+1 as a left K-vector space with the action of K = {a K : a 0} on K n+1 0 = K n+1 0 given by : K K n+1 0 K n+1 0, z(x) = xz 1 Then set of orbits is denoted by KP n 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 CP n. Let A U(n + 1) then A[x] = [Ax] is well defined since A(xz 1 ) = (Ax)z 1. Now [e 1 ] is stabilized by {( e i ) } 0 : R and B U(n) 0 B So U(n + 1)/(U(1) U(n)) has a manifold structure. Similarly to the case of the sphere the Lie algebra decomposes as {( ) } ir 0 k = : r R, B u(n) 0 B and m = {( 0 A A 0 ) : A M (1,n) }. 9

17

18 CHAPTER 2 Symmetric Spaces In this section we define the notion of a symmetric space and study some of their properties. 1. Definitions Here it is assumed that the reader is familiar with the concept of a geodesically 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 symmetric space if for every point p M there exists an isometry p of (M, g) such that (1) p(p) = p, and (2) d p = id TpM. 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 initial data (0) and (0). Both the geodesics t p ( (t)) and t ( t) take the value (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 geodesic with (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 transport of v along. PROOF. [6] Let (t) = (t + ) then is a geodesic with (0) = q. So by Lemma 2.2 it follows that q( p ( (t))) = q ( ( t)) = q ( ( t )) = (t + ) = (t + 2 ). 11

19 If v T p M 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 and thus any two points p, q M in the same path component of M can be connected by a geodesic. PROOF. [6] By repeatedly composing as in Lemma 2.3 we can get to (2 l ) until 2 l greater than any real number. This shows that (M, g) is geodesically complete, so by the Hopf-Rinow theorem any two points can be connected by a geodesic. Definition 2.5. Let (M, g) be a symmetric space and let : R M be a geodesic with p = (0) and v = (0). Then for t R the isometries given by are called transvections. tv : (M, g) (M, g) tv = (t/2) (0) 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) and d p = d p then =. PROOF. [10] Let B (p) be a normal ball around p M. Then if v(t) = Exp p (tv) is a geodesic then ( v(t)) = ( v(t)) are the same geodesics since geodesics are given by initial data. So = on an open set but, this set is also 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 T p M and define : R M by (t) = Exp p (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 R 12

20 (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 is actually an isometry. For (s) M, if (r) = (r + t) then (t) = (0) and (s) = (s t) so (t+ a 2 ) (t)( (s)) = ( a 2 ) (0)( (s t)) = (s t + a) = (s + a) Now we also have ( a 2 ) (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+ a 2 ) (t) by Lemma 2.6. (2) av = ( a 2 ) (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 + b 2 ) ( b 2 ) (0) = ( a 2 + b 2 ) (0) = (a+b)v (5) (0) ( a 2 ) (0) (0) = (0) ( a 2 ) = av Here are some concrete examples of involutions and transvections Example 2.8. At p 0 R n define the involution p 0 : x (x p 0 ) + p 0 In R n geodesics are straight lines, so consider the geodesic (t) : R R n, t p 0 + tv where v T p0 R n = R n Then we get the transvection tv(x) = (t/2) (0)(x) = (t/2)( (x (0)) + (0)) = ( (x (0)) + (0) (t/2)) + (t/2) = ( (x p 0 ) + p 0 (p 0 + tv)) + (p 0 + tv) = x + tv. Now we can verify the translational properties of the transvection tv (s) = (p 0 + sv) + tv = p 0 + (s + t)v = (s + t) 13

21 Example 2.9. If we have a submanifold M in (R m, g) for some m Z + and M = {x R m : g(x, x) = r} for example S n or H + 1,n, then we have the involution at p 0 : M M then x 2 g(p 0, x) p 0 x r p 0 (p 0 ) = 2 g(p 0, p 0 ) p 0 p 0 = p 0 r since T p0 M p 0 we get for v T p0 M d p0 (v) = 2 g(p 0, v) p 0 v = v r so we have an involution on the submanifold M. 2. The Isometry Group Definition 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 neutral element e G, i.e. G is the identity component of I(M). Furthermore let K p0 be the isotropy group of G at p 0 M. It should be noted that by continuity, all the transvections belong to the group G. Now we can prove the a very important thing concerning the isometry group. Theorem [6] Let (M, g) be a symmetric space and G the identity component 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 [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 a neighborhood of p 0 M. Then the dependence is smooth for all p M. PROOF. [10] If we prove the claim for p M in a normal ball around q M, then we can cover M with intersecting balls from p 0 to an arbitrary point in M. So if p B (q) then there exists a v T q M such that p = Exp q v and v depends smoothly on p. Then f (n)(p) = f (n)(exp q v) = Exp f (n)(q) (df (n)(q))v where the right side depends smoothly on (n, p). 14

22 Now we have a very important lemma for showing that charts and actions on our manifold are smooth. Lemma [10] Let (M, g) be a symmetric space with involutions p, transvections v and let K p0 be the isotropy group of the isometries I(M) acting at a point p 0 M, then (1) p(q) depends smoothly on (p, q) M M. (2) v(q) depends smoothly on (v, q) T p M M. (3) p(q), depends smoothly on (p, q) M M, where p is v such that v(p 0 ) = p. (4) k(q) depends smoothly on (k, q) K p0 M. PROOF. [10] (1) By Lemma 2.12 we only need to show the claim for a point q in a neighborhood of p. Then q = Exp p (v) where v depends smoothly on (p, q). But Exp p (tv) goes through p so p (q) = p (Exp p (tv)) t=1 = Exp p ( 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 Exp p0 (tv) connecting p 0 to p depends smoothly on p close to p 0, then v depends smoothly on p close to p 0. 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 p 0. Then q = Exp p0 (v) so k(q) = k(exp p0 (v)) = Exp k(p0 )(dk)v = Exp p0 (dk)v which depends smoothly on (k, q). Note that K p0 immediately can be defined as a Lie group using Lemma 2.6 and the fact that for all f K p0, f (p 0 ) = p 0, i.e. an isometry on a connected manifold is identified by its value at one point and its differential at the same point. So since df (p 0 ) is an element of the Lie group of isometries T p0 M T p0 M which is O(dim M). Since K 0 is closed we can identify K 0 with a Lie-subgroup of O(dim M). Theorem [10] Let (M, g) be a symmetric space. Then the isometry group I(M) has the structure of a Lie group such that (1) The map I(M) M M, with (g, p) g(p) is smooth. (2) If N is a manifold then f : N I(M) is smooth if and only if is smooth. N M M, (n, q) f (n)(q) PROOF. [10] We will define charts in a neighborhood around g 0 I(M). Let U be a neighborhood of p 0 M. We ll define a map that will help us with the charts. Let U K p0 I(M), (p, k) g 0 pk, where p is defined as in Lemma This map is injective since we can retrieve both p, v by g0 1 g 0 pkp 0 = pp 0 = p 15

23 1 p g0 1 g 0 pk = k So we can talk about the inverse image of this map around g 0. If we choose local coordinates for M by p (x 1,... x m ) at p 0 and local coordinates k (y 1,... y l ) at k 0 in K p0, then we get local coordinates around g 0 in I(M) g 0 pk (x 1 (p),..., x m (p), y 1 (k),..., y l (k)). These coordinates are smooth, since if the image of two such charts intersect we get g 1 p 1 k 1 = g 2 p 2 k 2 so p 1 = g2 1 g 2 p 2 k 2 (p 0 ) dk 1 (p 0 ) = (d 1 p 1 (p 1 ))d(g1 1 g 2 )(p 2 )(d p 2 (p 0 ))(dk 2 (p 0 )) these depend smoothly on each other by Lemma The topology is Hausdorff since if, are two different isometries then we either have (p 0 ) (p 0 ) or (d (p 0 ) (d (p 0 )). If the first one holds we can separate the isometries since M is Hausdorff. If the second hold we must have that k 1 k 2 so we can separate in O(dim M). Now the group operation is smooth since if f, g I(M) then f 1 g = (g 1 p 1 k 1 ) 1 (g 2 p 2 k 2 ) = (g 3 p 3 k 3 ) in local coordinates. So and p 3 = p 3 k 3 (p 0 ) = g 1 dk 3 = d 1 p 3 dg 1 3 k1 1 1 p 1 g 1 1 g 2 3 dk1 1 1 d p 1 dg 1 p 2 k 2 (p 0 ) 1 dg 2 d p 2 dk 2 which is smooth by Lemma Next (1) Since g is an isometry then (g, p) g(p) is smooth (2) Suppose f (n) is smooth. Let n 0 N then f (n) = g 0 pk where (p, k) depends smoothly on n. So f (n)(q) = g 0 pk(q) depends smoothly on (n, q). Conversely if the assignment (n, q) f (n)(q) is smooth then f (n) = g 0 where p = g0 1 f (n)(p 0 ) and dk = d 1 p on n. dg 1 pk 0 df (n). So (p, k) depends smoothly Example If we consider the isometry groups in Examples of Section 2 then we see that SO(n + 1) acts smoothly on S n, Lor(1, n) acts smoothly on H + 1,n, O(n) acts smoothly on G k (R n ) and U(n + 1) acts smoothly on CP n 16

24 3. G/K is Diffeomorphic to M Here we show that the quotient of the isometry group and the isotropy group can be identified with the symmetric space. Theorem [10] Let (M, g) be a symmetric space and let G = I(M) act transitively on M. Let K be the isotropy group of G at p 0 M. Then the map is a diffeomorphism such that : G/K M with gk g(p 0 ) (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 /N instead, where N is the kernel of the action. So if M is a symmetric space then I 0 (M)/K 0 is diffeomorphic to M. PROOF. [10] Clearly k K. Also is well defined and bijective since k(p 0 ) = p 0 for (g )gk = (g gk ) = g g(p 0 ) = g g(p 0 ) = 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 the inverse function theorem locally 1 is smooth. But 1 exists globally so by some form of gluing lemma 1 is smooth. Now M = j=1 U j, where U j are the images of the charts on G/K. Since M is locally compact, at least one Ū j contains an open subset of M by the Baire category theorem. This open subset on M is in bijective correspondence with an open set V in G/K since G/K acts smoothly. So using the charts we would get an onto map R n V U j R m, so we cannot have n < m i.e. dim (G/K ) dim M. Thus we only have to show the injectivity which will imply that dim (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 1 g K (d (g 1 )) g K So if d is injective at ak then it is injective at g K by choosing g such that g 1 g K = ak in = g (g 1. Therefore it is enough to show the injectivity at ek. Let v T ek (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. The curve (t) = exp(tx )(p 0 ) satisfies d (t) = d ( ) (exp(tx )) dt dt 17

25 = d d exp(tx )(X ) = d d dl exp(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 We have seen in Examples 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) = S n Lor(1, n)/so(n) = H + 1,n O(n)/O(k) O(n k) = Gk (R n ) U(n + 1)/U(1) U(n) = CP n 18

26 CHAPTER 3 Symmetric Pairs In this chapter we introduce the concept of a symmetric pair and show how they lead to symmetric spaces. Definition 3.1. A pair (G, K ) is said to be a Riemannian symmetric pair if G is a Lie group, K a closed subgroup of G, and s p0 an involutive automorphism on G such that (1) (G sp0 ) 0 K G sp0 (2) Ad(K ) is a compact subset of GL(g). Here G sp0 are the elements of G that are left invariant by s p0, i.e G sp0 = {g G : s p0 g = g}. The involutive automorphism s p0 is often uniquely defined, so it is common to only write (G, K ) for a symmetric pair. 1. From a Symmetric Space to a Symmetric Pair As before let G, K p0 denote the identity component of the isometry and isotropy groups on M. Then by Theorem 2.16 there is a bijective correspondence G/K M, gk g(p 0 ) We define an involution corresponding to p0 on G by s p0 : G G s p0 (g) = p0 g p0 = p0 g 1 p 0 Then sp 2 0 = id and s p0 (g)p = p0 g 1 p 0 (p), so by Lemma 2.13 it depends smoothly on (g, p). Thus s p0 is an involutive Lie group automorphism of G. Note also that (ds p0 ) e : g g is a involutive Lie algebra automorphism. Let G sp0 = {g G : s p0 g = g}, and let g sp0 be its Lie algebra. Then G sp0 is a Lie group since it is a closed subgroup. We have the following theorem 19

27 Theorem 3.2. [10] Let (M, g) be a symmetric space with a fixed point p 0, G be the identity component of the isometry group and let K be the isotropy group of G at p 0. Then the map G/K M with K g(p 0 ) is a bijection. The group G has an involutive automorphism s p0 given by with stabilizer G sp0 such that s p0 (g) = p0 g p0 (G sp0 ) 0 K G sp0 This says that (G, K ) is a symmetric pair because Ad(K ) is compact, since K is closed and bounded and Ad is a homeomorphism. PROOF. [10] It only remains to prove the last statement. If X g p0 then exp(tx ) G p0 so s p0 (exp(tx)) = exp(tx ) therefore ds p0 = X. Also if ds p0 X = X then exp(tx ) = exp(tds p0 X ) = s p0 exp(tx ), the implications work in the opposite direction as well, so Let k K then g p0 = {X g : ds p0 X = X }. s p0 (k)p 0 = p0 k p0 (p 0 ) = p 0 = k(p 0 ) and ds p0 (k) = d p0 dk p0 d p0 = ( id Tp0 M) dk p0 ( id Tp0 M) = dk p0. This means that Lemma 2.6 implies that s p0 (k) = k so K G p0. Further if X g p0 then p0 (exp(tx ))p 0 = p0 (exp(tx )) p0 (p 0 ) = s p0 (exp(tx ))p 0 = exp(tx )p 0, so exp(tx )p 0 is a fixed point of p 0. But if t is small then exp(tx )p 0 is close to p 0 and this is the only fixed point so exp(tx )p 0 = p 0. Thus exp(tx ) K 0 = K and X k. So g p0 k (G p0 ) 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 be identified with the tangent space of M. We know that k = {X g : ds p0 (X ) = X } and define p = {X g : ds p0 (X ) = X } Then since ds p0 is an automorphism k p = {0}. We also have that g = k + p for all X g since X = 1 2 (X + ds p 0 (X )) (X ds p 0 (X )) 20

28 where the first term is in k and the second is in p. Therefore g = k p Since ds p0 is a Lie algebra automorphism, i.e. we have ds p0 [X, Y ] = [ds p0 X, ds p0 Y ] [k, k] k [p, p] k [k, p] p We now reveal the connection between the isometry group and the symmetric space. Theorem 3.3. [10] Let (M, g) be a symmetric space, G be the identity component of I(M), K the isotropy group of G at p 0 M, g the Lie algebra of G, k the Lie algebra of K and p a linear complement of k in g. As usual let : G M, g g(p 0 ) then d k(p 0 ) = 0 d p(p 0 ) = Tp0 M If X p then exp(x ) = d X (p 0 ) and exp(x )(p 0 ) = Exp p0 (d X (p 0 )). PROOF. [10] d k(p 0 ) = 0 since (p 0 ) is constant on K. Now let v T p0 M then the map t 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(tds p0 X ) = s p0 (exp(tx )) = p0 tv p0 = tv. This means that ds p0 (x)) = X so X p. Also d X (p 0 ) = d dt exp(tx ) t=0 (p 0 ) = d dt exp(tx )(p 0) t=0 = d dt tv(p 0 ) t=0 = d dt Exp p 0 (tv) t=0 = v This shows that d (p 0 ) : p T p0 M is a surjective map. Since the coordinates of G were the coordinates for K and T p0, dim G = dim K + dim M 21

29 Also since g = k + p, we get dim M=dim p. Therefore d (p 0 ) : p T p0 M must be injective as well. Now the last formulas of the claim follow immediately and expx = v = d X (p 0 ) expx (p 0 ) = exp(x )p 0 = d X (p 0 )p 0 = Exp p0 (d X (p 0 )) So the map : G/K M in Theorem 2.16 defined by gk g(p) is a submersion. We apply this theorem in some of the symmetric spaces we know. at Example 3.4. By Theorem 3.3 for G = SO(3), K = SO(2) acting on S 2 we have for exptx p 0 = X = p 0 = S cos t sin t 0 sin t cos t p is the geodesic in S 2 starting at p 0 in direction d X p 0 = = = cos t sin t 0 To get a geodesic going in an arbitrary direction we apply any element of K to the above geodesic which gives cos sin cos t sin t 0 sin cos 0 Example 3.5. By Theorem 3.3 for G = Lor(1, 2), K = SO(2) acting on H 1,2 + at p 0 = H + 1,2

30 we have for exptx p 0 = X = cosh t sinh t 0 sinh t cosh t p is the geodesic in H 1,2 + starting at p 0 in direction d X p 0 = = = cosh t sinh t 0 To get a geodesic going in an arbitrary direction we apply any element of K to the above geodesic which gives cos sin cosh t sinh t 0 sin cos 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 closed subgroup 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 1 K = 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, with identity components G 0, K 0. Form the quotient manifold G/K. Let denote the action (g )gk = g gk. Let J Z(K ) K be such that (J)eK = ek and suppose (1) d (J) ek = id G/K Then for every compact subset K in K where K 0 K K G 0 (G 0, K ) is a Riemannian symmetric pair with s ek (g) = JgJ 1 = In(J)g 23

31 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 which Ad 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 thus X + Ad (J)X k for all X g. Thus (1) implies (2). By working backwards we get d e = d Ad J = d (J) ek d e so d (J) ek = id G/K so also (2) implies (1). Now we prove that we have a symmetric pair. Since J Z(K ), In J K = id K and Ad J k = id k. 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 ) where T : 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 = id k we have Ad (J) 2 = id g. So if X g then In (J) 2 (exp(x)) = exp(din (J) 2 X ) = exp(ad (J) 2 X ) = exp(x ). So In is an involutive automorphism on G 0. Now we ll show ((G 0 ) In J ) 0 K (G 0 ) In J. First ((G 0 ) In J ) 0 = (G In J ) 0 since (G In J ) 0 = (G 0 ) In J. Also (G 0 ) In J = G 0 G In J. So K (G 0 ) In J since K G 0, K G In J because J Z(K ) and K K. Secondly if g In J k = k then (G In J ) 0 K. So let X g In 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

32 Example 3.8. Let G = O(n + 1) and K = O(n) in Theorem 3.7 with 1 1 J =... 1 with J Z(K ). If we form G/K = O(n+1)/O(n) and since for X o(n+1) ( ) ( ) ( ) A B A B A B X = B T, Ad J D B T = D B T D we have ( ) A 0 X + Ad J(X ) = 2 0 T k D 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 3.7 with ( ) Ik 0 J = 0 T I n k with J Z(K ). If we form G/K = O(n)/O(k) O(n k) and since for X o(n + 1) ( ) ( ) ( ) A B A B A B X = B T, Ad J D B T = D B T D we have ( ) A 0 X + Ad J(X ) = 2 0 T k D 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 induce symmetric spaces. With this it is possible to show that several familiar group quotients produce symmetric spaces. First we need to produce some nice metrics for the intended symmetric space. Lemma [10] Let G be a Lie group and K be a closed subgroup of G. If Ad K is compact then there exists a G-invariant inner product on p = T ek G/K, such that the action of G on G/K is an isometry. Here p is the linear complement of k in g, i.e. g = k p. 25

33 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 to T ek G/K by d (ek ) by Corollary 1.5 and get an Ad K -invariant inner product, ek on T ek 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 1 b 1 ), d (k 1 b 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 submersion. Finally we have the long awaited theorem. Theorem [10] Let (G, K ) be a symmetric pair with involutive automorphism s. Denote the action of G on G/K by (g )gk = g gk. Then there is a G-invariant metric g on M = G/K which makes (M, g) a symmetric space with involution 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 = Exp ek (d (X )) PROOF. For a symmetric pair Ad K is compact so by Lemma 3.10 we can pick a G-invariant inner product on the tangent space of G/K, as metric for G/K. Define the involution by this is well defined since It is smooth since Further ek ( (g)ek ) = (s(g))ek ek ( (gk)ek ) = (s(g)) (s(k))ek = (s(g))ek = ek ( (g)ek ). = s which is smooth. 2 ek ( (g)ek ) = ek ( (s(g))ek ) 26

34 = (s(s(g)))ek = (s 2 (g))ek = (e)ek = ek 2 1 so ek = e so it is involutive. Therefore 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 p 0 = 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 So (s(g)) ek ( (g )ek ) = (s(g)) (s(g ))ek = (s(gg ))ek = ek ( (gg )ek ) = ek ( (g) (g )ek ) (3.1) (s(g)) ek = ek (g) Let u, v T gk 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 So = u, v gk ek is an isometry and (d ek ) ek = id ek (ek ) = s ek ( (e)ek ) = (s(e))ek = ek. So we have shown that (M, g, ek ) is a symmetric space. The last claims follow from Theorem

35 Example By applying Theorem 3.11 to Examples we get the following symmetric spaces from the corresponding symmetric pairs ( SO(n + 1), SO(n) ) SO(n + 1)/SO(n) = S n ( SO(n), S(O(k) O(n k)) ) SO(n)/S(O(k) O(n k)) = G k (R n ). 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) = CP n 28

36 CHAPTER 4 Curvature of a Symmetric Space The aim of this chapter is to show how to transfer the curvature calculations from the symmetric space to the Lie group which is behind the symmetric 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 the isometry group G on the symmetric space. Definition 4.1. Let g be a Lie algebra and let be an involutive automorphism on g with fixed point set k, which is a Lie subalgebra or g. If Int k(see Appendix B) with Lie algebra ad (k) is compact, then (g, ) is called an orthogonal symmetric algebra. The orthogonal symmetric algebra is called effective if z(g) k = {0}. Example 4.2. Let g = R, = id R, k = {0} then we have an orthogonal symmetric algebra which corresponds to and also G = R, K = {0}, s = id R G = T = R/Z, K = {0, 1/2}, s = id T Definition 4.3. Let g be a Lie algebra, then the Killing form B of g over a field F is the bilinear form B : g g F, (X, Y ) tr(ad X ad Y ). The Lie group G and its Lie algebra g are called semisimple if B is nondegenerate, i.e. if B(X, Y ) = 0 for all Y g then X = 0. Example 4.4. To calculate 1 the trace of linear maps from a vector space to 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 1 To save space we use the convention to sum over repeated indices without explicit summation expressions, i.e. x i y i = P n i=1 x iy i. 29

37 becomes especially easy if we have an inner product,, so that we have an orthonormal basis {e i } i I where I is some finite set. Then the trace is given by trace = e i, (e i ) since has the matrix representation ij = e i, (e j ). We do an explicit calculation for so(n) and display some others afterwards. For so(n) we chose the basis to be 1 2 (e ij e ji ) for 1 i < j n, where e ij is the matrix with a 1 at position (i, j) and zeros elsewhere. trace (ad X ad Y ) = f k, [X, [Y, f k ]] by Lemma B.4 where f k are the basis elements. This gives trace (ad X ad Y ) = 1 2 eij e ji, [X, [Y, e ij e ji ]] (4.1) = 1 2 eij, [X, [Y, e ij ]] 1 2 eij, [X, [Y, e ji ]] 1 2 eji, [X, [Y, e ij ]] eji, [X, [Y, e ji ]] 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 (A T B) and the basis elements are orthogonal for this choice. The first expression in Equation (4.1) is 1 { (e ij ) T X Y e ij (e ij ) T X e ij Y (e ij ) T Y e ij X 2 +(e ij ) T } e ij Y X Now (e ij ) T = e ji so we get = 1 { e ji X Y e ij 2 } +e ji e ij Y X e ji X e ij Y e ji Y e ij X Also e ij X = i X j and X e ij = X i j, e ij e kl = i jk l so we get = 1 { j X i Y j i j X i i Y j j Y i i X j + j ii j Y X 2 = 1 { } jj X i Y i X ii Y jj Y ii X jj + ii Y j X j 2 } 30

38 The fourth term of Equation (4.1) gives = 1 { ii X j Y j X jj Y ii Y jj X ii + 2 jj Y i X i } adding these gives (4.2) 1 2 ( jj X i Y i + Y jj X ii ( ii Y j X j + ii X j Y j) X jj Y ii jj Y i X i) Since so(n) is skew symmetric the two middle terms in Equation (4.2) vanish and summing the other term over 1 i < j n gives 1 (ntrace (XY ) + ntrace (YX )) = ntrace (XY ) 2 The second term in Equation (4.1) becomes 1 { (e ij ) T X Y e ji (e ij ) T X e ji Y (e ij ) T Y e ji X 2 +(e ij ) T } e ji Y X = 1 { e ji X Y e ji e ji X e ji Y e ji Y e ji X 2 } +e ji e ji Y X = 1 { j X i Y i j j X i j Y i j Y i j X i + j ij i Y X 2 = 1 { } 0 X ij Y ij Y ij X ij + 0 since i j we get the zeros 2 = trace (XY ) after summing over i, j the last equality follows since X ij = X ji for X so(n). The third term of equation (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 }

39 Lemma 4.5. [6] The Killing form B of g is symmetric. Also B is invariant under automorphisms of g. In particular B((Ad g)x, (Ad g)y ) = B(X, Y ) for all X, Y g and g G also B((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, 1 Y ] = [X, 1 Y ] = ad X 1 Y 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) = d dt B(Ad (exp(tx ))Y, Z) t=0 = d dt 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 compact. By the proof of Lemma 3.10 there is an Ad G 0 -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 matrix representation of ad X is skew symmetric i.e. (ad X ) ij = a ij = a ji. Thus B(X, X ) = tr(ad X ad X ) = a ij a ji = i,j a 2 ij 0 32

40 Now a ij = e i, (ad X )e j so if i,j a2 ij = 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. Let X ij = 1 2 (e ij e ji ) be an arbitrary basis element in p of so(n) then B(X ij, X ij ) = (n 2) trace (e ij e ji )(e ij e ji ) 2 = (n 2) trace 2e ij e ji 2 = (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 is compact 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 automorphisms of g, g be the Lie algebra of Aut g and let Int g be the identity component of 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 DY i.e [D, ad X ] = ad DX so [ g, ad g] ad g. The bi-linear form C(F 1, F 2 ) = tr(f 1 F 2 ) is nondegenerate on ad g by the assumptions 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. By construction g ad g + a and (ad g) a = {0}. Now 2 see Example 4.21 for the definition tr([f 1, F 2 ]F 3 ) = tr(f 1 F 2 F 3 F 2 F 1 F 3 ) = tr(f 1 F 2 F 3 F 1 F 3 F 2 ) 33

41 so for A a, D g and X g = tr(f 1 [F 2, F 3 ]) 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 [10] Let B be the Killing form on the Lie algebra g then then following 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 can assume that G is connected otherwise we consider G 0. Then Int g = Ad G since 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 for X 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 orthogonal matrices i.e exp(t ad X ) for X g is an orthogonal matrix so ad X is a skew symmetric i.e. a ij = a ji so for X g B(X, X ) = tr(ad X ) 2 = a ij a ji = i,j a 2 ij 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 is semisimple, 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, then g 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 represented by orthogonal matrices, i.e. Int g O(dim Int g ). Also by Lemma 4.9 Int g = (Aut g) 0 so Int g is closed and thus compact. So g is the Lie algebra of a compact Lie group. It turns out that the Killing form decomposes the orhogonal symmetric algebra of a symmetric pair into a very convenient form. 34

42 Theorem [6] Let (g, ) be an effective orthogonal symmetric algebra of a symmetric pair (G, K ). Then we can decompose g as g = k p 1... p m where p i p j, i j with respect to B. We can also find an Ad K -invariant inner product on g given by g(, ) = B k + 1 B p B pm 1 m We also have [p i, p j ] = 0 for i j. PROOF. [6] By the proof of Lemma 3.10 we get a Ad K invariant inner product (, ) on p. Define B(X, Y ) X, Y k g(x, Y ) = (X, Y ) X, Y p 0 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 an orthonormal basis of eigenvectors {X j } of T such that T (X j ) = jx j. Note that all j 0 since B is nondegenerate. Also since B is symmetric, the eigenspaces corresponding to different eigenvalues are orthogonal. In fact since {X k } are orthonormal with respect to g(, ) we have that k = B(X k, X k ). So p = p 1... p m The expression for g(, ) follows by Equation (4.3). For the last claim let Y i p i, Y j p j then but also B([Y i, Y j ], [Y i, Y j ]) = B(Y i, [Y j, [Y i, Y j ]]) = ig(y i, [Y j, [Y i, Y j ]]) B([Y i, Y j ], [Y i, Y j ]) = B([Y j, Y i ], [Y i, Y j ]) = B(Y j, [Y i, [Y i, Y j ]]) = jg(y j, [Y i, [Y i, Y j ]]) = jg(y i, [Y j, [Y i, Y j ]]) where the last equality follows from Theorem 4.13 and the symmetries of the curvature tensor Rjkl i. So if i j then i j so B([Y i, Y j ], [Y i, Y j ]) = 0 but g is semisimple hence [Y i, Y j ] = 0. Thus [p i, p j ] = 0 for i j. 35

43 With the metric g in Theorem 4.11 the symmetric pair (G, K ) makes the quotient 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 from G to the symmetric space M = G/K. By Theorem 3.3 we get the vector fields on M by parallel translation and the map d pp 0 T p M. The images X (p 0 ) = d X (p 0 ) are Killing fields (see Appendix D) i.e. their local flows are isometries. We start with some properties of these Killing fields. Lemma [6] Let G be the isometry group on the symmetric space (M, g), K G be the isotropy group at p 0 M. If g is the Lie algebra of G with the standard decomposition g = k p, then X (p 0 ) = 0 for all X k v X (p 0 ) = 0 for all X p, v T p M PROOF. [6] Let X k then X (p 0 ) = d X (p 0 ) = d dt (exp(tx )p 0) t=0 = d dt (p 0) t=0 = 0. Next let : t M be a curve such that (0) = v T p M then so X (p 0 ) = d dt v X (p 0 ) = t s = s t tx (p 0 )(p 0 ) t=0 tx (p 0 )( (s)) s=t=0 by Theorem3.3 tx (p 0 )( (s)) since s=t=0 [ s, ] = 0 t = d v s=t=0 t but by Lemma 2.3 v is a parallel = 0 transport of v along (t), so Since we can parallel translate all vector fields on M to the origin it is enough to be able to do the curvature calculations for vector fields at the origin. 36