a = ilie(t ) then a is maximal abelian in p = ilie(k)

Transcription

1 We have seen that if T is a maximal torus of K and if a = ilie(t ) then a is maximal abelian in p = ilie(k) and the conjugacy of maximal compact Tori in K implies that Ad(K)a = p. Here is a simple proof of that fact using what we already know. We have shown using Richardson s Lemma that any two Cartan subalgebras of G are conjugate. Let b be be another maximal abelian subalgebra in p. We have seen that h = ia a is a Cartan subalgebra of Lie(G). Let T 2 be the metric closure of exp(ib) in K. Then T 2 is contained in a maximal torus T 1 of K and we have seen that h 1 = Lie(T 1 ) + ilie(t 1 ) is a Cartan subalgebra of Lie(G). There exists g G such that Ad(g)h = h 1. Then g = k exp X with X p. Hence g = exp(ad(k)x)k. Thus if Y = Ad(k)X then e ady (Ad(k)h) = h 1. Let g = Lie(G) and let for each λ R, g(λ) be the λ eigenspace for ady. If y g let y λ be its projection into

2 the λ th eigen space. If h a then ad(y ) 2m+1 Ad(k)h Lie(K) and ad(y ) 2m Ad(k)h p. Now if u h 1 then u can be written uniquely as u I + u R with u I having imaginary eigenvalues and u R having real ones. Since Ad(k)h has real eigenvalues we have ( e ady Ad(k)h ) I = 0. But ( e ady Ad(k)h ) I = λ k=0 ad(y ) 2k+1 sinh(λ) (Ad(k)h) λ. (2k + 1)! Ad(k)h = Thus if (Ad(k)h) λ 0 then sinh(λ) = 0. So if h a then ady Ad(k)h = 0. Hence so Ad(k) 1 b a. h 2 = Ad(k)h Note this also proves the conjugacy of maximal tori in K.

3 This implies that if A = exp a then since exp p = k K exp kak 1 KAK we have G = KAK. Let G be a connected symmetric subgroup of GL(n, C) and let, as usual, K = U(n) G. Let...,... be the usual U(n) invariant inner product on C n. We say that v C n is critical if Xv, v = 0 for all X Lie(G). We note that the set of critical points is invariant under the action of K. Theorem. (Kempf-Ness) Let G, K be as above. Let v C n. 1. v is critical if and only if gv v for all g G. 2. If v is critical and X p is such that then Xv = 0. e X v = v

4 3. If Gv is closed then there exists a critical element in Gv. 4. If v is critical then Gv is closed. Proof. (of 1.,2.,3.) We note that G = K exp(p). If X p then d dt exp(tx)v 2 = d dt exp(tx)v, exp(tx)v = exp(tx)xv, exp(tx)v + exp(tx)v, exp(tx)xv = 2 exp(tx)xv, exp(tx)v. since X = X. Arguing in the same way we have d 2 dt 2 exp(tx)v 2 = 4 exp(tx)xv, exp(tx)xv. We now begin the proof. Suppose v is critical. If X p and k K we have By the above k exp(tx)v 2 = exp(tx)v 2 = α(t). α (0) = 2 Xv, v = 0

5 and α (t) = 4 exp(tx)xv, exp(tx)xv 0. This implies that t = 0 is a minimum for α. In particular, if g = k exp X as above then gv 2 = k exp(x)v 2 = exp(x)v 2 v 2. If v C n and gv v for all g G and if X p then 0 = d dt exp(tx)v 2 t=0 = 2 Xv, v. Since Lie(G) = ip + p we have proved 1. in the Theorem. We will now prove 2. Suppose w = k exp(x)v and assume that v = w. We define α(t) = exp(tx)v 2 as above. If Xv 0 then α (t) > 0 for all t. Since α (0) = 0 this implies that α(t) > α(0) for t 0.But the hypothesis is α(1) is equal to α(0). This contradiction implies that Xv = 0.

6 We now prove 3. Suppose that Gv is closed. Let m = inf g G gv. Then since Gv is closed in the S topology there exists w Gv with w = m. 3. now follows from 1. We derive some corollaries to parts 1,2,3 before we prove 4. Corollary. If v, w are critical then w Gv implies w Kv. Proof. If w = gv with g G then w v by 1. But also v = hw with h G so v w. Now write g = ke X with k K and X p. Then we have v = gv = ke X v = e X v. Thus 2. implies that Xv = 0 hence gv = kv. Corollary. If v is critical then G v = {g G gv = v} symmetric subgroup of GL(n, C).

7 Proof. Let g G v with g = ke X with k K and X p. Then v = ke X v implies that e X v = v thus Xv = 0 by 2 in the Theorem. Thus e X G v. So k G v. This implies that g G v. Corollary. If v V then Gv is closed only if Gv contains a critical element. Corollary. Let v V and let m = inf g G gv. Then {u Gv u = v } = Kz (i.e. a single K orbit). An affi ne algebraic group, G, is said to be linearly reductive if every regular repesentation of G is completely reducible. Theorem. Let G be a linearly reductive affi ne algebraic group acting morphically on an affi ne variety X. then there exists a linear map R X : O(X) O(X) G satisfying: 1. R X 1 = 1.

8 2. If f O(X) G then R X (fϕ) = fr X (ϕ) for ϕ O(X). (In particular, R X f = f, f O(X) G.) 3. If S O(X) satsifies R gs S for all g G then R X S S. 4. If Y X is G invariant and GY Y then R Y (ϕ Y ) = R X (ϕ) Y for ϕ O(X). Corollary. If X is an affi ne variety with G a linearly reductive affi ne algebraic group acting morphically on X and if Y and Z are Z closed G invariant subsets then there exists f O(X) G such that f Y = 1 and f Z = 0. Proof. Let ϕ O(X) be such that ϕ Y = 1 and ϕ Y = 0. Set f = R X ϕ Then f Y = R X (ϕ) Y = R Y (ϕ Y ) = R Y (1) = 1. similarly f Z = 0. Corollary. If G is linearly reductive acting morphically on an affi ne variety X then O(X) G is finitely generated.

9 Proof. We first prove the result in the special case when X = C n and G acts linearly. Let Let O(C n ) G + = {f O(C n ) G f(0) = 0}. I = O(C n )O(C n ) G +. Then I is generatef by f 1,..., f r over C with f i O(C n ) G +. Let A be the subalgebra of O(C n ) G generated by {1, f 1,..., f r }. We filter by degree. We note that O 0 (C n ) G = C1 = A 0. Suppose we have shown that A O k (C n ) G. If f O k+1 (C n ) G then Now Hence f f(0)1 = ϕ i f i, ϕ i O k (C n ). R C nϕ R C no k (C n ) = O k (C n ) G A. f = f(0)1 + R C n(ϕ i )f i A. But R X (ϕ X ) = R C n(ϕ) X for ϕ O(C n ). Hence O(C n ) G X = O(X)G

10 and so { 1, f1 X,..., f r X } generate O(X) G.