1 Hermitian symmetric spaces: examples and basic properties


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1 Contents 1 Hermitian symmetric spaces: examples and basic properties Almost complex manifolds Hermitian manifolds Symmetric spaces Bergman metric A higherdimensional example: Siegel upper halfspace Isometries of Hermitian symmetric domains Classification of symmetric Hermitian domains Preliminaries: Cartan involutions Hermitian symmetric spaces: examples and basic properties The next goal is to explain classification of Hermitian symmetric domains. We begin with some preliminary explanations of what these are, along with toy examples. 1.1 Almost complex manifolds Definition 1.1. Let M be a smooth real manifold (of class C. We say that it is an almost complex manifold if we are given a smooth tensor field J on M of type (1,1 such that for each vector field X on M holds J(J(X = X. In other words, we ask that to each point p M we associate (in a smooth manner an endomorphism of the real tangent space J p : T p M T p M such that J 2 p = Id. Example 1.2. Let M be a complex manifold of dimension n, i.e. a ringed space (M,O M locally isomorphic to C n with the sheaf of holomorphic functions. Then M may be viewed as a smooth real manifold of dimension 2n. The tangent spaces T p M carry naturally an action of i C, and M is an almost complex manifold. Example 1.3. Let V be a real vector space. To specify an action of C on V means to pick an Ralgebra representation ρ : R[J]/(J End R (V. Since the algebra R[J]/(J is generated by J, everything amounts to picking an endomorphism J : V V satisfying J 2 = Id. Then this gives an action of the whole C by (x + J y v = x v + y J(v. Note that V becomes a Cvector space, so we conclude that dim R V must be an even number, otherwise V does not admit an almost complex structure. Vice versa, if dim R V = 2n, then we can pick a basis (x 1, y 1,..., x n, y n and define an action of C by thinking of these as of coordinates z k = x k + i y k : However, there is nothing canonical about this! J(x k := y k, J(y k := x k. Example 1.4. We see that a real manifold M does not admit an almost complex structure if dim R M is odd. If dim R M is even, then each tangent space T p M carries an almost complex structure J p, but they might not glue to a smooth vector field J on M such that J 2 = Id. For instance, among the spheres S 2n, only S 2 and S 6 admit an almost complex structure (S 2 simply because it may be viewed as the Riemann sphere; for S 6 it is actually not known whether it has a complex structure, it s a famous open problem. See e.g. N. Steenrod, The Topology of Fibre Bundles, 41, and J. P. May, A Concise Course in Algebraic Topology,
2 1.2 Hermitian manifolds Definition 1.5. Let M be a smooth manifold. We say that it is Hermitian if M carries an almost complex structure J. M carries a Riemannian structure, i.e. a smooth tensor field g of type (2,0, such that g (X,Y = g (Y, X for all vector fields X,Y on M, g p : T p M T p M R is a positive definite nondegenerate bilinear form for each p M. The Riemannian structure is invariant under J, i.e. for all vector fields X,Y on M holds g (J X, JY = g (X,Y. Example 1.6. Consider the complex plane C with real coordinates Z = X + i Y. The standard Riemannian metric is given by ( ( ( g p X (p, Y (p = 0, g p X (p, X (p = 1, g p Y (p, Y (p = 1. We recall that usually this is written as d s 2 = d x 2 + d y 2. The action by i is g (i X,i Y = g (X,Y, and C is Hermitian. Example 1.7. Consider the complex upper halfplane H := {z C Im z > 0}. X Y, Y X, and so clearly It carries naturally a complex structure, and real coordinates Z = X + i Y. We consider a Riemannian metric given by d s 2 = (d X 2 +(dy 2, i.e. y 2 ( ( g p X (p, Y (p = 0, g p X (p, X (p = 1 ( y 2, g p Y (p, Y (p = 1 y 2, and we see immediately that g (i X,i Y = g (X,Y, so H is a Hermitian manifold. Example 1.8. Consider P 1 (C = C { } as the Riemann sphere S 2 R 3, which comes with a natural complex and Riemannian structure. z y 0 C x Geometrically, the action of i rotates C counterclockwise, and this satisfies g (i X, i Y = g (X, Y. 2
3 1.3 Symmetric spaces Definition 1.9. Let M be a connected complex manifold. Then its underlying smooth manifold M has a canonical almost complex structure. Assume that M also carries a Riemannian structure which is compatible with the almost complex structure, in the sense that M is a Hermitian manifold. We say that M is a Hermitian symmetric space if for each point p M there exists a symmetry s p : M M that satisfies the following properties: 1 s p is an involution, i.e. s 2 p = Id, 2 s p is holomorphic with respect to the complex structure on M, 3 s p is an isometry with respect to the specified Riemannian structure on M, i.e. g (s p (X, s p (Y = g (X,Y, 4 s p has p as its isolated fixed point. Example As we observed, the complex plane C is naturally a Hermitian manifold. To see that it is also symmetric, note that the group of plane translations acts transitively on C, and these actions are holomorphic isometries. Hence we may check the conditions 1 4 for one point, for example 0. Then a symmetry is given by z z: it is an involution, a holomorphic isometry, and p is its only fixed point. Similarly, we have Hermitian symmetric spaces C/Λ, where Λ = ω 1 Z ω 2 Z is a lattice. Example Recall that the group acts on H by Möbius transformations ( a b SL 2 (R := {γ = M c d 2 (R detγ = ad bc = 1} γ z := az + b cz + d. The maps z γ z are holomorphic, and they are also isometries. Indeed, the Riemannian metric d s 2 = (d X 2 +(dy 2 may be written in coordinates z = x + i y and z = x i y as Using the formula d(γ z = detγ d z = (cz+d 2 d s 2 = 4 d z d z (z z 2. d z (cz+d 2, we see that d(γ zd(γ z 4 (γ z γ z 2 = 4 d z d z (z z 2. ( y x/ y The action is transitive, since x + i y = 0 1/ i for any x + i y H. Hence we again may consider one y particular point, e.g. i H. Now z 1/z is an involutive holomorphic isometry, having i as the only fixed point. We conclude that H is a Hermitian symmetric space. Example Consider P 1 (C with the natural Riemannian structure on the Riemann sphere C { } R 3. Any rotation of the sphere is a holomorphic isometry, and the group of rotations acts transitively. Rotation by π along the Z axis (that is, z z on C and is an involution, leaving fixed the poles 0 and. So P 1 (C is a Hermitian symmetric space. y 2 3
4 1.4 Bergman metric Let Ω C n be a bounded domain, i.e. a nonempty bounded open connected subset. It turns out that there is a canonical way to assign a Riemannian metric to Ω, so that every biholomorphic map Ω Ω is an isometry with respect to it. We sketch the construction following S.G. Krantz, Function Theory of Several Complex Variables, referred as [Krantz] below. As always, for functions f, g : Ω C we have an inner product f, g := f (z g (zd z, Ω which is correctly defined on the space of squareintegrable functions L 2 (Ω := {f : Ω C f, f < }. We have the subspace H(Ω L 2 (Ω of holomorphic square integrable functions, and in fact it is a closed subspace. This means that H(Ω is a Hilbert space, just as L 2 (Ω namely, it is complete with respect to the associated norm f := f, f. Definitiontheorem There exists a unique map K : Ω Ω C, such that 1 for fixed w Ω the function z K (z, w lies in H(Ω, 2 K (z, w = K (w, z, 3 it satisfies the reproducing property for all f H(Ω. f (z = K (z, w f (wd w Ω We call K (z, w the Bergman kernel 1. The construction is roughly the following. Let (φ k k N be an orthonormal basis of H(Ω. We set K (z, w := k φ k (zφ k (w. This converges, and in fact satisfies 1 [Krantz, Proposition 1.4.7]. It is clear that 2 is satisfied as well. Finally, we check 3: f = f,φ k φ k = K (, w f (wd w. k Ω The uniqueness of K (z, w, and hence the fact that it does not depend on the choice of an orthonormal basis of H(Ω, is due to 2 and 3. Indeed, suppose K (z, w is another Bergman kernel. Then K (z, w = K (w, z = K (z, tk (w, td t = K (w, tk (z, td t = K (z, w = K (z, w. Ω Ω Example As one can imagine, explicitly calculating the Bergman kernel is difficult. It is possible in some easy cases, e.g. for the unit open disk D := {z C z < 1}. The Bergman kernel for D is given by K (z, w = 1 1 π (1 z w 2 see the calculation in [Krantz, Theorem ]. 1 STEFAN BERGMAN ( , a Polishborn American mathematician 4
5 Proposition Let Ω C n be a bounded domain and let K (z, w be the corresponding Bergman kernel. Then K (z, z > 0 for all z Ω. Proof. We have from the construction of the Bergman kernel K (z, z = k φ k (zφ k (z = k φ k (z 2, and there are k such that φ k (z 0, since there are nonzero functions in H(Ω. Definitiontheorem Let Ω C n be a bounded domain. Then there exists a canonical Hermitian metric on Ω given by This is called the Bergman metric on Ω. g = g i j d z i d z j, g i j (z = 2 z i z j logk (z, z. (Checking that g i j (z is positive definite for all z Ω requires some thought; see [Krantz, Chapter 1, Exercise 39]. Here is a truly remarkable property of the Bergman metric: Theorem Let Ω 1,Ω 2 C n be two bounded domains, and let f : Ω 1 Ω 2 be a biholomorphic map. Then f is an isometry with respect to the Bergman metric. (This is [Krantz, Proposition ]. Example For the open unit disk D we have K (z, z = 1 π, and a little calculation (e.g. in real coordinates ( (1 z ( 2 2 x + i y = z, keeping in mind that z = 1 2 x i y and z = 1 2 x + i y gives the Bergman metric g (z = 2 z z logk (z, z = 2 (1 z In fact, this is the socalled Poincaré metric. We recall that H and D are two different models of the hyperbolic plane, and the upper half plane H corresponds to the open disk D via the Cayley transform H D, z i z i + z i 1 z z. 1 + z 5
6 i 0 A little calculation shows that the canonical Bergman metric on D corresponds to the metric on H that we considered above (up to scaling the latter by an appropriate constant. Indeed, let z, z denote the coordinates on H, and let w, w be the coordinates on D. ( i z d w = d i + z ( i + z d w = d i z = 2i d z (i + z 2, = 2i d z (i z 2, 2d w d w (1 w w 2 = 2 2i d z (i + z 2 2i d z (i z 2 ( 1 i z i + z i + z 2 = i z 8 d z d z ((i + z(i z (i z(i + z 2 = 8 d z d z (2i (z z 2 = 2 d z d z (z z 2. Remark Any Hermitian symmetric domain D may be embedded in some C n as a bounded symmetric domain Ω. This means that D has a unique Hermitian metric that maps to the Bergman metric on Ω. 1.5 A higherdimensional example: Siegel upper halfspace We have the transitive action of SL 2 (R on the upper halfplane H by Möbius transformations. We see that the stabilizer of i H identifies with ( a b SO 2 (R = { a,b R, a 2 + b 2 = 1}, b a which is a maximal compact subgroup, and we have H = SL 2 (R/SO 2 (R. 6
7 This has a higherdimensional generalization, the Siegel upper halfspace, consisting of symmetric complex n n matrices with positive definite imaginary part: H n := {Z = X + i Y X,Y M n (R, X = X, Y = Y, Y > 0} (in particular, H 1 = H is the upper halfplane. We may see this as an open subset of C n (n+1/2 by sending a matrix (z i j to the point (z i j j i C n (n+1/2, so there is a natural complex structure. We have the symplectic group ( A B Sp 2n (R := {γ = C D M 2n (R γ ( 0 In γ = I n 0 ( 0 In } I n 0 (in particular, SL 2 (R = Sp 2 (R, and Sp 2n (R acts transitively on H n by Möbius transformations We have a subgroup of unitary matrices γ Z = (A Z + B(C Z + D 1. U n = {Z M n (C Z Z = Z Z = I n }, where Z = Z denotes the conjugate transpose (in particular, SO 2 (R U 1. This may be identified with a subgroup of Sp 2n (R: ( X Y X + i Y, Y X and this is a maximal compact subgroup of Sp 2n (R. In fact, Now consider the set of matrices H n = Sp 2n (R/U n. D n := {Z M n (C Z = Z and I n Z Z > 0}, which may be identified with a bounded domain in C n (n+1/2 (in particular, for n = 1 it is just the open unit disk, so we know that for D n there is a canonical Hermitian metric the Bergman metric. There is a Cayley transform, identifying H n and D n : H n D n, i (I n Z (I n + Z 1 Z. Z (i I n Z (i I n + Z 1 This allows us to conclude that H n also carries an invariant Hermitian metric. Further, it is a Hermitian symmetric space ( (and hence D n. For this observe that the point i I n H n is the only fixed point of the involution 0 In Sp I n 0 2n (R. 1.6 Isometries of Hermitian symmetric domains Theorem Let (M, g be a symmetric space, with group of orientation preserving isometries Isom(M, g. Denote by Isom(M, g + the connected component of the identity (in the analytic topology. For a point p M denote by K p Isom(M, g + be the subgroup leaving p fixed. 7
8 For example, (M, g isometries orientationpreserving p K p H PGL 2 (R PSL 2 (R Z/2 PSL 2 (R i SO 2 (R/{±1} P 1 (C O 3 (R SO 3 (R 0 U 1 = SO 2 (R Then K p is compact, and we have an isomorphism of smooth manifolds Isom(M, g + /K p M, γ K p γ(p; in particular, the action of Isom(M, g + on M is transitive. For a proof we refer to [Helgason, Differential geometry, Lie Groups, and Symmetric Spaces, II.4.3]. Theorem Let (M, g be a Hermitian symmetric space. Consider the following groups: Isom(M, g = isometries of the underlying real manifold M. Hol(M = automorphisms M M as a complex manifold. Isom(M, g = Isom(M, g Hol(M = holomorphic isometries. Then in fact in particular Isom(M, g + = Hol(M + = Isom(M, g +, Hol(M + /K p M. this is claimed in Milne s notes. Theorem Let (M, g be a Hermitian symmetric domain. Consider the real Lie group H := Hol(M + and its Lie algebra h := Lie(H. There is a unique connected algebraic subgroup G of GL(h such that inside GL(h G(R + = Hol(M +. Moreover, G(R + = G(R Hol(M. Example For the upper halfplane H every isometry is either holomorphic (and orientationpreserving, or differs from a holomorphic map by the orientation change z z 1, which is antiholomorphic. We have G = PGL 2, and PGL 2 (R + = PSL 2 (R. Theorem Let M be a Hermitian symmetric domain. For each point p M there exists a unique homomorphism u p : U 1 Hol(M (where U 1 := {z C z = 1} is the circle group, such that u p (z fixes p and acts on T p M as multiplication by z. (A proof of this is sketched in Milne s notes 8
9 Example Consider M = H. Then Hol(M PSL 2 (R := SL 2 (R/{±1}. We want morphisms that fix i, and we keep in mind that Stab SL2 (R(i = SO 2 (R. In fact, U 1 may be viewed as SO 2 (R via h : U 1 SL 2 (R, ( a b z = a + i b. b a h(z fixes the point i, and we compute that it acts on T i H as multiplication by z 2 : d d X f (( a b b a X X =i Since we want action by multiplication by z, we define = d ( ax + b d X f bx + a X =i = d ax + b d d X bx + a X =i d X f (X = a2 + b 2 (a bi 2 d d X f (X X =i = (a + bi 2 d d X f (X. X =i X =i u : U 1 PSL 2 (R := SL 2 (R/{±1}, z h( z (mod ± 1. The choice of z does not change the class modulo ±1, and so u is welldefined and u(z acts on T i H by multiplication by z. 2 Classification of symmetric Hermitian domains 2.1 Preliminaries: Cartan involutions Definitiontheorem 2.1. Let G be a connected reductive algebraic group over R. Then there exists an automorphism θ : G G (as an algebraic group over R with θ 2 = Id, such that the group G (θ := {g G(C g = θ(g } is compact (by g we denote complex conjugation. Such θ is called a Cartan involution, and it is unique up to conjugation by an element of G(R. Example 2.2. Let V be a finite dimensional real vector space. Consider the group GL(V. A choice of some basis defines an isomorphism GL(V GL n, and on GL n we have the usual matrix transpose X X. The map θ : X (X 1 is a Cartan involution: GL (θ n (R = {Z GL n(c Z = (Z 1 } = U n is compact. A different choice of basis of GL(V differs from θ by conjugation by an element of GL n (R. Example 2.3. If G is a reductive algebraic group over R and G GL(V is a faithful representation over R, then there exists some basis for V for which G is stable under transpose X X (exactly because G is assumed to be reductive!. The restriction of X (X 1 to G is a Cartan involution. In fact, all Cartan involutions arise this way. 9
10 ( a b Example 2.4. For instance, take SL 2 GL 2. For a matrix M = c d take a Cartan involution ( ( a b d c θ : c d b a ( 0 1 Note that it is the same as conjugation by : 1 0 ( ( a b c d ( = 1 0 the inverse is given by. ( d c b a. 1 det M ( d b c a, so we Now we see that SL (θ 2 = { ( a b b a a 2 + b 2 = 1} = SU 2, which is a closed bounded subset in C 2, hence a compact subgroup. 10
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