We have seen that if T is a maximal torus of K and ifa = iLie(T ) then a is maximal abelian in p = iLie(K)

and the conjugacy of maximal compact Tori in K impliesthat Ad(K)a = p. Here is a simple proof of that factusing what we already know.

We have shown using Richardson’s Lemma that any twoCartan subalgebras of G are conjugate. Let b be be an-other maximal abelian subalgebra in p. We have seenthat h = ia ⊕ a is a Cartan subalgebra of Lie(G). LetT2 be the metric closure of exp(ib) in K. Then T2 iscontained in a maximal torus T1 of K and we have seenthat

h1 = Lie(T1) + iLie(T1)

is a Cartan subalgebra of Lie(G). There exists g ∈ Gsuch that Ad(g)h = h1. Then g = k expX withX ∈ p.Hence g = exp(Ad(k)X)k. Thus if Y = Ad(k)X then

eadY (Ad(k)h) = h1.

Let g = Lie(G) and let for each λ ∈ R, g(λ) be the λeigenspace for adY . If y ∈ g let yλ be its projection into

the λ—th eigen space. If h ∈ a then ad(Y )2m+1Ad(k)h ∈Lie(K) and ad(Y )2mAd(k)h ∈ p. Now if u ∈ h1 thenu can be written uniquely as uI + uR with uI havingimaginary eigenvalues and uR having real ones. SinceAd(k)h has real eigenvalues we have(

eadYAd(k)h)I

= 0.

But (eadYAd(k)h

)I

=∞∑k=0

ad(Y )2k+1

(2k + 1)!Ad(k)h =

∑λ

sinh(λ) (Ad(k)h)λ .

Thus if (Ad(k)h)λ 6= 0 then sinh(λ) = 0. So if h ∈ athen adY Ad(k)h = 0. Hence

h2 = Ad(k)h

so Ad(k)−1b ⊂a.

Note this also proves the conjugacy of maximal tori inK.

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 theusual U(n) invariant inner product on Cn. We say thatv ∈ Cn is critical if 〈Xv, v〉 = 0 for all X ∈ Lie(G).

We note that the set of critical points is invariant underthe action of K.

Theorem. (Kempf-Ness) Let G,K be as above. Letv ∈ Cn.

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∥∥∥eXv∥∥∥ = ‖v‖

then Xv = 0.

3. If Gv is closed then there exists a critical element inGv.

4. If v is critical then Gv is closed.

Proof. (of 1.,2.,3.) We note that G = K exp(p). IfX ∈ 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

d2

dt2‖exp(tX)v‖2 = 4 〈exp(tX)Xv, exp(tX)Xv〉 .

We now begin the proof. Suppose v is critical. If X ∈ pand k ∈ K we have

‖k exp(tX)v‖2 = ‖exp(tX)v‖2 = α(t).

By the above

α′(0) = 2 〈Xv, v〉 = 0

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 expX as above then

‖gv‖2 = ‖k exp(X)v‖2 = ‖exp(X)v‖2 ≥ ‖v‖2 .

If v ∈ Cn and ‖gv‖ ≥ ‖v‖ for all g ∈ G and if X ∈ pthen

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 as-sume that ‖v‖ = ‖w‖. We define α(t) = ‖exp(tX)v‖2

as above. If Xv 6= 0 then α′′(t) > 0 for all t. Sinceα′(0) = 0 this implies that α(t) > α(0) for t 6= 0 .Butthe hypothesis is α(1) is equal to α(0). This contradic-tion implies that Xv = 0.

We now prove 3. Suppose that Gv is closed . Let m =

infg∈G ‖gv‖. Then since Gv is closed in the S—topologythere exists w ∈ Gv with ‖w‖ = m. 3. now followsfrom 1.

We derive some corollaries to parts 1,2,3 before we prove4.

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 writeg = keX with k ∈ K and X ∈ p. Then we have

‖v‖ = ‖gv‖ =∥∥∥keXv∥∥∥ =

∥∥∥eXv∥∥∥ .Thus 2. implies that Xv = 0 hence gv = kv.

Corollary. If v is critical then Gv = {g ∈ G|gv = v}symmetric subgroup of GL(n,C).

Proof. Let g ∈ Gv with g = keX with k ∈ K andX ∈ p. Then v = keXv implies that

∥∥∥eXv∥∥∥ = ‖v‖thus Xv = 0 by 2 in the Theorem. Thus eX ∈ Gv. Sok ∈ Gv. This implies that g∗ ∈ Gv.

Corollary. If v ∈ V thenGv is closed only ifGv containsa critical element.

Corollary. Let v ∈ V and let m = infg∈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 re-ductive if every regular repesentation of G is completelyreducible.

Theorem. Let G be a linearly reductive affi ne algebraicgroup acting morphically on an affi ne variety X. thenthere exists a linear map RX : O(X) → O(X)G satis-fying:

1. RX1 = 1.

2. If f ∈ O(X)G then RX (fϕ) = fRX(ϕ) for ϕ ∈O(X). (In particular, RXf = f , f ∈ O(X)G.)

3. If S ⊂ O(X) satsifies R∗gS ⊂ S for all g ∈ G thenRXS ⊂ S.

4. If Y ⊂ X isG—invariant andGY ⊂ Y thenRY (ϕ|Y ) =

RX(ϕ)|Y for ϕ ∈ O(X).

Corollary. If X is an affi ne variety with G a linearlyreductive affi ne algebraic group acting morphically on Xand if Y and Z are Z—closed G—invariant subsets thenthere 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 andϕ|Y = 0. Set f = RXϕ Then

f|Y = RX(ϕ)|Y = RY (ϕ|Y ) = RY (1) = 1.

similarly f|Z = 0.

Corollary. If G is linearly reductive acting morphicallyon an affi ne variety X then O(X)G is finitely generated.

Proof. We first prove the result in the special case whenX = Cn and G acts linearly. Let

O(Cn)G+ = {f ∈ O(Cn)G|f(0) = 0}.

Let

I = O(Cn)O(Cn)G+.

Then I is generatef by f1, ..., fr overC with fi ∈ O(Cn)G+.LetA be the subalgebra ofO(Cn)G generated by {1, f1, ..., fr}.We filter by degree. We note that O0(Cn)G = C1 =

A0. Suppose we have shown that A ⊃ Ok(Cn)G. Iff ∈ Ok+1(Cn)G then

f − f(0)1 =∑

ϕifi, ϕi ∈ Ok(Cn).

Now

RCnϕ ∈ RCnOk(Cn) = Ok(Cn)G ⊂ A.

Hence

f = f(0)1 +∑

RCn(ϕi)fi ⊂ A.

But RX(ϕ|X) = RCn(ϕ)|X for ϕ ∈ O(Cn). Hence

O(Cn)G|X = O(X)G

and so {1, f1|X , ..., fr|X

}generate O(X)G.

Theorem. If G ⊂ GL(n,C) is Z—closed and linearly re-ductive then there exists g ∈ GL(n,C) such that gGg−1

is symmetric.

Our proof uses the following variant of Matsushima’s the-orem.

Theorem. Let G be a symmetric subgroup of GL(n,C)

and assume that G acts transitively on an affi ne variety,X. If x ∈ X the group

Gx = {g ∈ G|gx = x}

is conjugate in G to a symmetric subgroup of GL(n,C).

Let X be the categorical quotient of GL(n,C) i.e

SpecmaxO(GL(n,C))G,

relative to theaction of G by right multiplication. Wenote that if g ∈ GL(n,C) then

L∗gO(GL(n,C))G ⊂ O(GL(n,C))G

which gives a morphic action of GL(n,C) on X. Sinceall of the orbits of G are closed we see that GL(n,C)

acts transitively. (The orbits are the cosets gG.)

Let xo be the point in X corresponding to the maximalideal

{f ∈ O(GL(n,C))G|f(I) = 0}

.Then

GL(n,C)xo = G.

Indeed, Suppose gxo = xo then gG = G hence g ∈ G.

Since X is affi ne our variant of Matsushima’s theoremimplies that G is conjugate in GL(n,C) to a symmetricsubgroup.

We will now prove the theorem. Let K = U(n) ∩ GthenG is the Z—closure ofK inGL(n,C). There exists aregular representation (σ, V ) of G, v ∈ V and a bijectivemorphism

φ : X → σ(G)v,

a closed orbit, such that

φ(gx) = σ(g)v.

Thus

Gx = Gv.

We may assume that σ(G) is symmetric by integratingan inner product over K.. Now apply Kempf-Ness tosee that there exists a critical element,in σ(G)v. Wehave seen that if σ(g)v critical then Gσ(g)v is symmetric.Hence

gGvg−1 = gGxg

−1

is symmetric in G. This completes the proof.

We can just use the term reductive.

We note that this implies the following theorem. (ExactlyMatsushima’s theorem.)

Theorem. G is reductive and H ⊂ G is Zarsiki closedthen G/H is affi ne if and only if H is reductive. Further-more, if H is reductive then O(G/H) = O(G)H (thisis relative to the right action).

To prove 4. in the Kempf-Ness theorem we will use an-other Lemma of Richardson.

Lemma. Let v ∈ V and let Gw be the closed orbit inGv then there is k ∈ K such that

Akv ∩Gw 6= ∅.

Proof. We will use the notation of the previous section.We may assume that Gw ⊂ Gv − Gv. Let H be, theCartan subgroup corresponding to T . H is an algebraictorus in G. We note that it is enough to prove that thereexists k ∈ K such that

Hkv ∩Gw 6= ∅.

Indeed, if we have shown this and if T = H ∩ U(n)

then H = TA. Thus

Hkv = TAkv = TAkv

since T is compact. Now Gw is T—invariant and henceAkv ∩Gw 6= ∅.

We will now prove the assertion for H by contradiction.Set Y = Gw.Then by assumption it is Z—closed anddisjoint from Gv. Assume Hkv ∩ Y = ∅ for all k ∈ K.Since Hkv and Y are H invariant and H is symmetricthe separation lemma implies that for each k ∈ K thereexists fk ∈ O(V )H such that fk|Y = 0 and fk|Hkv = 1.Let

Uk = {x ∈ V |fk(x) 6= 0}

Then Uk is open in V and contains kv. This implies(since Kv is compact) that there exist k1, ..., km ∈ Ksuch that

Kv ⊂ Uk1∪ · · · ∪ Ukm.

Let

f(x) =m∑i=1

∣∣∣fki(x)∣∣∣

for x ∈ V . Then f is continuous, H—invariant andf(x) > 0 for all x ∈ Kv. This implies that f attainsa minimum ξ > 0 on the compact set Kv. The H—invariance implies that f(x) ≥ ξ for x ∈ HKv. Since

f(Y ) = 0 and KY = Y we have shown that

∪k∈KkHKv ∩ Y = ∅

. So

KHKv ∩ Y = ∅.

But

KHKv = KHKv = Gv

which is our desired contradiction.

We now prove 4. in the Kempf-Ness theorem. Assumethat v is critical but Gv is not closed.

Then Richardson’s lemma implies that here exists k ∈ Ksuch that the metric closure of Akv contains y with Gythe unique closed orbit in Gv. We may replace v ith kv.Let {aj} be a sequence in A such that

limj→∞

ajv = y.

Let u1, ..., un be a basis of V with respect to whichthe elements of a are diagonal. Since the elements of a

are Hermitian the corresponing weights of a,λi are realvalued. That is

hui = λi(h)ui

for all i = 1, ..., n with λi ∈ a∗. Let aj = expXj withXj ∈ a. We write

v =∑

viui, y =∑

yiui.

We have

limk→∞

eλi(Xk)vi = yi.

Thus of vi 6= 0 then either

(1) limk→∞

eλi(Xk) = 0

or

(2) limk→∞

λi(Xk) = ci

with ci ∈ R. For those i such that (1) is true we mayassume that by taking an appropriate infinite subseqencethat λi(Xk) < 0 for all k. But

0 = 〈Xkv, v〉 =∑

λj(Xk)∣∣∣vj∣∣∣2 .

thus for each k∑j 6=i

λj(Xk)∣∣∣vj∣∣∣2 = −λi(Xk) |vi|2 > 0

and

limk→∞

(−λi(Xk) |vi|2

)= +∞.

This implies that there must be an index j such that

limk→∞

λj(Xk)∣∣∣vj∣∣∣2 = +∞.

This is the desired contradiction if for any i we are incase (1). We may assume that if vj 6= 0 then

limk→∞

λi (Xk) = ci.

Let F = {i|vi 6= 0}. Let PF be the projecton onto thespan of the ui with i ∈ F . Then v = PFv and

limPFXkPF =∑i∈F

ciui = z.

Thus z is in the closure of PFaPF . Since is a subspacehence closed there exists X ∈ a such that z = PFXPF .Now

y = limk→∞

eXkv = limk→∞

eXkPFv = limk→∞

PF eXkPFv =

limk→∞

ePFXkPFv = ePFXPFv = PF eXPFv = eXv.

Thus y ∈ Gv which is a contradiction.

Exercise. Show that the above proof shows directly thatGv is closed if v is critical.

If G is a subgroup of GL(n,R) that is the locus of zerosof polynomials on Mn(R) and is invariant under trans-pose then we call G a symmetric subgroup of GL(n,R).If G is a symmetric subgroup of GL(n,R) then we set

K = O(n) ∩G

and

p = {X ∈ Lie(G)|XT = X}

.Chevalley’s lemma implies

Theorem. (Cartan Decomposition) The map

K × p→ G

k,X 7→ keX

is a homeomorphism. In particular, K is a maximal com-pact subgroup of G.

Before go to the conjugacy theorem let me show how wecan move many of our results from reductive algebraicgroups to real symmetric groups. If G ⊂ GL(n,R) is

symmetric define θ(g) =(g−1

)−T. Let GC be the Z—

closure of G in GL(n,C).

Lemma. GC is a symmetric subgroup of GL(n,C).

Proof. It is clear that GC is closed under transpose.Let σ be the automorphism (not regular!) of GL(n,C)

defined by complex conjugation of matrix entries. Thenσ|G = I. Thus if f ∈ O(Mn(C) is such that f|G = 0

then f(σ(g)) = 0 for all g ∈ G. Hence GC is invariantunder complex conjugation hence under adjoint.

Theorem. If a ⊂ p is a subspace that is maximal subjectto the condition that [a, a] = 0 then Ad(K)a = p.

Proof. We note that if ZC is the center of GC then

σ(ZC) = ZC.

Thus if Z is the center of G then Z = G∩ZC. We cantherefore replace G and GC by their commutator groups.If KC = GθC then (

Ad(KC)|pC, pC)

is a Vinberg pair furthermore the complexification, aC,of a in pC is a Cartan subspace. If X ∈ p then X isdiagonalizable. Thus there is k ∈ KC such that

Ad(k)X ∈ aC.

We write

Ad(k)X = H1 + iH2

with H1, H2 ∈ a. Now

Ad(k)X = kXk−1

has real eigenvalues. Hence H2 = 0. Writing k = eiY u

with Y ∈ Lie(K) and u ∈ K we have

eiadYAd(u)X = H1.

Applying σ we have

e−iadYAd(u)X = H1.

Thus e2iadYAd(u)X = Ad(u)X.Chevalley’s lemma im-plies that [Y,Ad(u)X] = 0. Hence Ad(u)X ∈ a.

This implies that we should call a a Cartan subspace ofp and if we write A = exp a then we have

Theorem.

G = KAK.

Let 〈..., ...〉 denote the standard inner product on Rn.Then v ∈ Rn will be said to be critical if 〈Xv, v〉 = 0

for all X ∈ p (hence for all X ∈ Lie(G)).

Theorem. Let G,K be as above. Let v ∈ Rn.

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∥∥∥eXv∥∥∥ =

‖v‖then Xv = 0. If w ∈ Gv is such that ‖v‖ =

‖w‖then w ∈ Kv.

3. If Gv is closed then there exists a critical element inGv.

Exercise. Prove this theorem. (Hint. Exacty the sameas the proof over C.)

Theorem. Let G be a symmetric subgroup of GL(n,R)

and let K1 be a compact subgroup of G then there exists

X ∈ p = {X ∈ Lie(G)|XT = X}

such that

eXK1e−X ⊂ K = O(n) ∩G.

The proof will use a collection of observations. Let

pn = {X ∈Mn(R)|XT = X}

and

Pn = {g ∈Mn(R)|gT = g}

and g is positive definite}. If A ∈ Pn we set

fA(X) = tr(AeX +A−1e−X)

for X ∈ pn.

1. Let A ∈ Pn. If C ∈ R and C > 0 then the set

L(A,C) = {X ∈ pn|fA(X) ≤ C}

is compact.

To prove this assertion consider A be symmetric and pos-itive definite. Then A =

∑λ∈S λPλ with S = Spec(A)

the spectrum of A and Pλ is the orthogonal projectiononto the λ eigenspace of A. Then

fA(X) = tr(AeX +A−1e−X) =∑λ∈S

(λtr(PλeXPλ) + λ−1tr(Pλe

−XPλ)).

Let a = min(S∪{λ−1|λ ∈ S}. Then since tr(PλeZPλ) >

0 for Z symmetric and λ ∈ S we have

fA(X) ≥ a∑λ∈S

tr(Pλ(eX+e−X)Pλ) = atr(eX+e−X).

Let X =∑µ∈Spec(X) µQµ be its spectral decomposi-

tion. Then we have

fA(X) ≥ a∑

µ∈Spec(X)

(eµ+e−µ)trQµ = 2a∑µ,k

µ2k

(2k)!trQµ ≥ 2a+atrX2.

Thus L(A,C) ⊂ {X ∈ pn|trX2 ≤ Ca }. Since L(A,C)

is closed this implies 1.

We also note that we have shown

2. If a = minSpec(A)∪Spec(A−1) then fA(X) ≥ 2a.

3. Let

uo =∫K1

kTkdk.

then uo ∈ Pn and kTuok = uo for all k ∈ K1.

This follows since the K1 invariant inner product

(v, w) =∫K1

〈kv, w〉 dk

is given by 〈uov, w〉.

Note that 1. implies that for all C > 0 the set

L(uo, C) ∩ p

is compact. Let

c = infX∈p

fuo(X) > 0

(by 2.). Set C = c. Then the compactness of L(uo, C)∩p implies that the infimum is attained. Let Xo ∈ p besuch that

fuo(Xo) = c.

4.

e−Xo2 K1e

Xo2 ⊂ K.

To prove this we note that Chevalley’s lemma implies that

Pn ∩G = ep.

Also the Cartan decomposition implies that

Pn ∩G = {ggT |g ∈ G}

Let k ∈ K1. Then there exist g ∈ G and Z ∈ p suchthat eXo = ggT and

keXokT = geZgT .

Indeed, take g = eXo2 and since

g−1keXokT (g−1)T ∈ Pn ∩G

we can write g−1keXokT (g−1)T = eZ with Z ∈ p. Wenote that

tr(uogetZgT+u−1

o (getZgT )−1) = tr(AetZ+A−1e−tZ)

with A = gTuog. We set

α(t) = tr(AetZ +A−1e−tZ)

. We also note that

α(0) = tr(uoggT+u−1

o (ggT )−1) = tr(uoeXo+u−1

o e−Xo) = fuo(Xo).

the value at t = 0 is a minimum for α by the definitionof Xo. Thus

α′(0) = 0

and if Z 6= 0 then we would have

α′′(t) = tr(AZ2etZ +A−1Z2e−tZ) =

tr(A12Z2etZA

12) + tr(A−

12Z2etZA−

12) > 0

for all t so α(t) would be strictly increasing. But 3. andthe choice of g with eXo = ggT and

keXokT = geZgT

imply that α(0) = α(1) this implies that Z = 0 and so

keXokT = eXo.

Now write eXo = eXo2 e

Xo2 multiplying the equation by

e−Xo2 on the left and right yields

e−Xo2 ke

Xo2 e

Xo2 kT e−

Xo2 = I

which is the assertion of 3.

This completes the proof.

Exercises.Let Go be an open subgroup of a symmetricsubgroup of GL(n,R) and let π : G → Go be a finitecovering group of Go. This is up to isomorphism theclass of real groups studied in and is exactly the classof groups that are called real reductive in my book RealReductive Groups 1. The point of the next two exercisesis to prove the conjugacy of maximal compact subgroupsof these groups.

1.Show that Go ∩O(n) is a maximal compact subgroupof Go and K = π−1(Go∩O(n)) is a maximal compactsubgroup of G.

2.Prove that ifK1 is a compact subgroup of G then thereexists g ∈ G such that gK1g

−1 ⊂ K.

Let G be a symmetric subgroup of GL(n,R) and let Kand p be as above.

Theorem. Let v ∈ V , if 0 ∈ Gv then there is anelement u ∈ Kv and h ∈ a such that

limt→+∞

exp(th)u = 0.

Proof. We first note that the compactness of K impliesthat Gv = K(AKv). So the hypothesis implies that

0 ∈ (AKv).

Thus there exist sequences hj with hj ∈ a and kj ∈ Ksuch that

limj→∞

exp(hj)kjv = 0.

Since K is compact we may assume

limj→∞

kj = k ∈ K.

On Mn(R) we use the Hilbert-Schmidt inner product

(tr(XY T ) = 〈X,Y 〉)

. We write hj = tjuj with uj ∈ a,tj real, tj > 0 and∥∥∥uj∥∥∥ = 1. We may assume that

limj→∞

uj = u ∈ a.

We write V for Rn. If λ ∈ a∗ then define

Vλ = {x ∈ V |hx = λ(h)v, h ∈ a}

.Set

Σ = {λ ∈ a∗|Vλ 6= 0}.

If x ∈ V then we write

x =∑λ∈Σ

xλ

with

xλ ∈ Vλ.

Then, since

〈Vλ, Vµ〉 = 0, λ 6= µ,

we have

limj→∞

∑λ

e2tjλ(uj)∥∥∥(kjv)λ

∥∥∥2= 0.

Thus for every λ ∈ Σ we have

limj→∞

e2tjλ(uj)∥∥∥(kjv)λ

∥∥∥2= 0.

For all λ ∈ Σ with λ(u) > 0 there exists N such thatλ(uj) > 0 for j ≥ N . Thus

etjλ(uj)∥∥∥(kjv)λ

∥∥∥2 ≥∥∥∥(kjv)λ

∥∥∥2

for j > N and all λ with λ(u) > 0. Hence we musthave

(kv)λ = 0 if λ(u) > 0.

Suppose that

λ(u) = 0

and

(kv)λ 6= 0.

If for an infinite number of j we have

λ(uj) ≥ 0

we will again run into the contradiction

(kv)λ = 0.

We therefore see that we may assume that if

λ(u) = 0 and (kv)λ 6= 0 then λ(uj) < 0

all j. We choose N to be so large that if

λ(u) < 0

then

λ(u+ uN) < 0.

We note that by our choice of subsequence above

λ(uN) < 0 if λ(u) = 0 and (kv)λ 6= 0

so in that case we we also have

λ(u+ uN) < 0.

Hence

λ(u+ uN) < 0

for all λ such that

(kv)λ 6= 0.

Take

h = u+ uN .

Then

ethkv → 0 as t→ +∞.

Let G be a linearly reductive subgroup of GL(n,C) andlet (σ, V ) be a regular representation of G. The HilbertMumford theorem states

Theorem. A necessary and suffi cient condition for v ∈V to be in the null cone is that there exist an algebraicgroup homomorphism

ϕ : C× → G

such that

limz→0

σ(ϕ(z))v = 0.

Wemay assume thatG is a symmetric subgroup ofGL(n,C)

and that V = Cn. We take a maximal compact torus,

T , in K and H to be the Z—closure of T in G. ThenH = T exp(iLie(T )). As usual we set

a = iLie(T )

and

A = exp(a).

We note that H is isomorphic with (C×)m and any al-gebraic homomorphism of C× to (C×)m is given by

z → (zq1, zq2, ..., zqm).

Thus if we take as a basis of Lie(H) the elements e1, ..., emwith

exp(∑

zjej) = (ez1, ez2, ..., ezm)

then since

t = (ez1, ez2, ..., ezm) ∈ Tthen

z ∈ iR

hence

ej ∈ a.

We also see that an algebraic homomorphism φ : C× →H is given by

φ(ez) = exp(z(∑

kjej))

with kj ∈ Z. Now let

h =∑

hiei

be chosen as in the Theorem above. Let wi ∈ Q be suchthat ∑

|wi − hi| < δ

for some small δ (to be determined). Taking δ suffi cientlysmall λ(

∑wiei) < 0 for all λ ∈ Σ (the weights of a)

such that υλ 6= 0. Now let p be a positive integer suchthat pwi is an integer for each i. Set X = −p∑wiei.Then

φ(ez) = exp(zX)

defines an algebraic homomorphism from C× toH. Since

limz→0

φ(z)u = 0

we have completed the proof the theorem.

Exercises.

1. Let G be a connected, linearly reductive, algebraicgroup over C and assume that (ρ, V ) is a regular, finitedimensional representation of G such that V 6= 0 andO(V )G 6= C1. Then the null cone of V is non-zero.

2. Let G and V be as in 1. and let H be a maximalalgebraic torus of G (i.e. a Cartan subgroup). Show thatif X is the null-cone of H then GX is the nullcone of G.

3. Let G = SL(2,C)× SL(2,C) acting on V = C2 ⊗C2 by the tensor product action. Show that the nullconeis the set of all (v ⊗ w|v, w ∈ C2}.

4. Let G = SL(2,C)×SL(2,C)×SL(2,C) acting onV = C2⊗C2⊗C2 by the tensor product action. Showthat the null cone is the Z—closure of Gw with

w = e1 ⊗ e1 ⊗ e2 + e1 ⊗ e2 ⊗ e1 + e2 ⊗ e1 ⊗ e1.

We will need the theory of Borel and Parabolic subgroupsof a reductive group to give a descriprion fo null conesfor reductive group actions.

We now give a more concrete discription of the metrictopology on V//G = SpecmaxO(V )G.

Let G be a linearly reductive subgroup of GL(n,C). Wemay assume (and do) assume that G is a symmetric sub-group of GL(n,C) and we take K to be G ∩ U(n).Let (σ, V ) be a regular representation of G with a K—invariant inner product 〈..., ...〉. We will use module no-tation (i.e. drop the σ if it is understood). Let

Crit(V ) = {v ∈ V | 〈Xv, v〉 = 0, X ∈: Lie(G)}.

That is, the set of critical points in V . We note thatCrit(V ) is K—invariant. We endow it with the subspacetopology in V . Let Crit(V )/K be the K— orbit spaceendowed with the quotient S—topology.

The Kempf-Ness theorem implies that if v ∈ V is suchthat Gv is closed then

Gv ∩ Crit(V ) = Kw

for some w ∈ Gv and if w ∈ Crit(V ) then Gw isclosed. This establishes a bijection

Φ : V//G→ Crit(V )/K.

Theorem. The map Φ is a homeomorphism in the S—topology.

Clearly Crit(V ) has the structure of an algebraic va-riety over R. Furthermore, the real polynomial invari-ants of K acting on Crit(V ) give an algebraic structureon Crit(V )/K (since the algebra is finitely generated).However, if we choose a set of generators for the invari-ants and use them to define an imbedding of Crit(V )

into Rm (m the number of generators) then the imagewill not necessarily be the entire set that is given as thelocus of points that satisfy the relations among the invari-ants. That is, this procedure just makes Crit(V )/K intoa subset with interior in a real algebraic variety (in fact aset satisfying algebraic inequalities, i.e a quasi-algebraicset). Whereas V//G is a full affi ne variety over C. If welook at it as an algebraic variety over R then it is the fullvariety. It is therefore not reasonable to think of V//Gas being a real variety isomorphic with the variety definedby Crit(V )/K as above.

Example: Let G = SO(2,C) acting on C2 by the matrixaction. If z1, z2 are the standard coordinates on C2 then

O(C2)G = C[z21 + z2

2].

Thus

C2//G ∼=,C

hence, it is isomorphic with R2 as a real variety. Nowwriting

zj = xj + iyj.

Then the critical set is

Crit = {(x1, y1, x2, y2)|x1y2 − x2y1 = 0}.

The invariants of R4 under the action of

K = SO(2,C) ∩ U(2) = SO(2)

are generated by

x21 + x2

2, y21 + y2

2, x1y1 + x2y2

If we define the mapΨ : Crit/K → R3 byΨ(x1, x2, y1, y2) =

(x1y1+x2y2, x21+x2

2, y21 +y2

2) then this defines a home-omorphism if Crit//K into a real algebraic variety. One

can check that if t1, t2, t3 are the standard coordinateson R3 then the corresponding algebraic variety is the cone

{(t1, t2, t3)|t21 − t2t3 = 0}.

However, the image is exactly the points on the varietywith t2, t3 ≥ 0.One can therefore describe the set as

{((√t2t3, t2, t3)|t2 ≥ 0, t3 ≥ 0}∪{((−

√t2t3, t2, t3)|t2 ≥ 0, t3 ≥ 0}.

It is easy to see that this set is homeomorphic with R2

but it looks like half of an elliptical cone. This indicatesthat one cannot expect the theorem to be much strongerthan it is.

Let

f1, ..., fm

be homogeneous generators of

O(V )G

of positive degree. If v ∈ V then we set

Ψ(v) = (f1(v), ..., fm(v)).

Ψ(gv) = Ψ(v) for g ∈ G, v ∈ V .We have seen thatthe image of Ψ is Z—closed and is an affi ne variety, X,isomorphic with V//G. Then by construction

Ψ∗O(X) = O(V )G

The Kempf-Ness theorem implies that

Ψ(Crit(V )) = Ψ(V ).

Since

Ψ(gv) = Ψ(v), g ∈ G,

this map descends to a map

F : Crit(V )/K → X

continuous in the quotient topology. Since the map is abijection, to prove that it is a homeomorphism in the met-ric topology onto X we need only show that it is closed.Thus, we need to show that if zj ∈ Crit(V ) and iflimj→∞Ψ(zj) = u then u = Ψ(w) with w ∈ Crit(V ).If an infinite subsequence of {zj} were bounded then asubsequence of {zj} converges and since Ψ is continuous

our result would follow. We may thus assume that eachzj 6= 0 and

limj→∞

∥∥∥zj∥∥∥ =∞.

We derive a contradiction which will complete the proofthat F is a closed mapping. Consider the sequence

uj =zj∥∥∥zj∥∥∥.

Then replacing the sequence by a subsequence we mayassume that

limj→∞

uj = uo ∈ Crit(V ).

Let deg fj = dj. We note that dj > 0 (by assumption).We have

fj(ui) =fj(zi)

‖zi‖dj.

Thus

limi→∞

fj(ui) = limi→∞

fj(zi)

‖zi‖dj

which is 0 since the fi are the coordinates of Ψ and wehave assumed that limj→∞Ψ(zj) = u. Applying thisargument to all j we find that if u 6= 0 then Ψ(uo) = 0

(that is, uo is in the null cone). But since u0 is criticalthe Kempf-Ness Theorem would then imply that u0 is0.This is impossible.

Let H be an algebraic torus and let (ρ, V ) be a regularrepresentation of H. We may assume that

H =(C×

)mand that V is Cn with the elements of ρ(H) diagonal.Thus there is an n×m integral matrix

A = [aij]

such that if Ai = [ai1, ..., aim] and if zJ = zj11 · · · z

jmm

for J ∈ Zm then we have

ρ(z)ii = zAi, i =, ..., n

and all other intries are 0. If

ϕ : C× → H

is a regular homomorphism then there exists b ∈ Zm

such that

ϕ(z) = (zb1, ..., zbm)

hence

ρ(ϕ(z))ii = z(Ab)i.

We write ϕ as ϕb. The Hilbert-Mumford theorem impliesthat v = (v1, ..., vn) is in the nullcone for H then thereexists b ∈ Zm such that if vi 6= 0 then (Ab)i > 0. Wecan define for each b ∈ Zm,

Vb = {v ∈ V |vi = 0 if (Ab)i ≤ 0}.Each of these is a vector subspace of V so it is Z—closedand irreducible. We order Zm by

b ≤ c if Vb ⊂ Vc.Then the set of irreducible components of NH the null-cone is the union of the sets

{Vb|b ∈ Zm, Vb maximal}.Finally, if

J ⊂ {1, 2, ..., n}

then we set

V J = {v|vi = 0, i /∈ J}.Set Λ(H) equal to the set of such J such that there existsb ∈ Zm such that

V J = Vb.

Finally set Λmax(H) equal to the minimal elements ofΛ(H) relative to inclusion.

Lemma. NH =⋃J∈Λmax(H) V

J is the decompositioninto irreducible components.

Exercise. Describe the null cone for the algebraic subtoruscorresponding to 1 −2

−2 34 −3

.

Let G ⊂ GL(n,C) be a Z-closed reductive subgroupand let H be a maximal algebraic torus (that is a Cartansubgroup). Set NG equal to the null cone of G actingon Cn.

Lemma. We have NG = GNH .

Proof. If v ∈ NG then there exists ϕ : C× → G aregular homomorphism such that

limz→0

ϕ(z)v = 0.

ϕ(C×) is Z—closed connected, algebraic and reductivethus it is in an algebraic torus. Hence it is contained ina conjugate of H. That is there exists g ∈ G such thatgv ∈ NH .

Maintaining the notation we see that

NG =⋃

J∈Λmax(H)

GV J .

We will show that each of sets GV J is Zariski closed.

For this we must develop some GIT for reductive groupactions on projective varieties.

Let V be a finite dimensional vector space over C. Recallthat P(V ) is the set of one dimensional subspaces of V .If v ∈ V − {0} then we write [v] for the space Cv.

Choosing a basis of V we may assume that V = Cn+1

and we write Pn for P(Cn+1). We set

Pni = {[v]|vi 6= 0}.Then we have a bijection

Ti : Cn → Pnigiven by

(z1, ..., zn) 7−→ [z1, ..., zi−1, 1, zi, ..., zn].

We take the structure sheaf to be the sheaf of functionsOPn on Pn with

OPn|Pni =(T−1i

)∗OCn.This makes Pn into an algebraic variety. We note thatthe open subvarieties Pni are isomorphic with Cn. Wenote that the topology of Pn is completely determined(both the Zariski and Metric topology) by taking Pni tobe open and homeomorphic with with Cn.

Lemma. Let X be a topological space and let U be anopen covering of X then Y ⊂ X is closed if and only ifY ∩ U is relatively closed in U for all U ∈ U .

Proof. We prove the suffi ciency the necessity is obvi-ous. If U ∈ U then the complement in U of Y ∩ U is(X − Y ) ∩ U . Thus (X − Y ) ∩ U is open and hence

X − Y =⋃U∈U (X − Y ) ∩ U

which is open. Thus Y is closed.

A variety isomorphic with a Zariski closed subset X ofPn is called a projective variety. Its structure sheaf isOPn|X . A variety isomorphic with a Zariski open subsetof a projective variety is called a quasi-projective variety.

Theorem. A quasi-projective variety is projective if andonly if it is compact in the metric topology.

Proof. A projective variety is homeomorphic with a Z—closed subset of Pn for some n, hence it is closed in themetric topology of Pn thus to prove the suffi ciency it isenough to prove that Pn is compact. We note that themap

π : Cn+1 − {0} → Pnπ(z) = [z]

is regular and surjective. If z ∈ Cn+1−{0} then we canwrite z as ‖z‖w with w ∈ S2n+1 = {z ∈ Cn+1| ‖z‖ =

1}. Thus π(S2n+1) = Pn and π is continuous in themetric topology we see that Pn is compact.

Assume that X is compact and Z—open and non-emptyin a projective variety Y . Then the metric closure of Xis the same as its Z—closure in Y . But the Z—closure isprojective and the metric closure is X.

Theorem. If X is an irreducible affi ne variety that iscompact in the metric topology then it is a point.

Proof. We may assume that X ⊂ Cn for some n as aZ—closed subset. Let xi be the i—th standard coordinate.Then xi : X → C is a regular function with imagein C having Z—interior in its closure. Since C is onedimensional and the Z—closure of xi(X) is irreducibleit is either a point or C. Since X is compact xi(X) isclosed in the metric topology thus since C is not compactxi(X) is a point.

Corollary. If X is an affi ne variety that is compact in themetric topology then X is finite.

Proof. X = X1 ∪ ... ∪Xm with Xi Z—closed and irre-ducible. Thus since X is compact in the metric topologythe Xi are all compact hence points.

Examples. 1. A flag in Cn is a sequence

V = V1 ⊂ V2 ⊂ ... ⊂ Vn

with

dimVi = i.

The standard flag is

Vo = Ce1 ⊂ Ce1 + Ce2 ⊂ ... ⊂ Ce1 + ...+ Cei ⊂ ...

GL(n,C) acts on a flag by

gV = gV1 ⊂ gV2 ⊂ ... ⊂ Vn.

If

V = V1 ⊂ V2 ⊂ ... ⊂ Vn

is a flag then choose a compatible basis

{v1, ..., vn}of Cn as follows: v1 a basis of V1, v1, v2 a basis ofV2, v1, ..., vi a basis of Vi, ...If g ∈ GL(n,C) is suchthat

gei = vi

then gVo = V. If gVo = Vo then g ∈ Bn the uppertriangular matices in GL(n,C). Let Fn be the set of allflags in Cn. We have a map

GL(n,C)/Bn → Fndefined by

gBn 7−→ gVo.

We make Fn into a quasi-projctive variety by using thestructure onGL(n,C)/Bn.We note that U(n) acts tran-sitively on Fn since for each V we can choose an ortho-normal compatible basis.

Thus

Fn = U(n)Vo.

So Fn is compact hence projective.

2. Let Gk,n be the set of all k dimensional subspaces ofCn. Then we note that if V is aK—dimenisonal subspaceof Cn then

V = {v ∈ Cn|v ∧ (∧kV ) = 0}.

Thus we can map Gk,n into P(∧kCn) by V 7−→ ∧kV .The image is the set of all decomposable elements ofP(∧kCn) (i.e elements of the form v1 ∧ v2 ∧ · · · ∧ vk ).Let G = GL(n,C) act on Gk,n then using orthonormalbases it is easy to see that U(n) acts transitively. TakingV0 = Span{e1, ..., ek} then

GVo =

{[A B0 C

]|A ∈ GL(k,C), B ∈ GL(n− k,C)

}.

The image ofG/GVo is a realization as a quasiporjectiovevariety. But the image is compact this realizes Gk,n as aprojective variety.

Exercise. Find a set of homogeneous quadratic polyno-mials on ∧kCn that define the decomposable elements.

Now for some theory of solvable groups.

Theorem. If G is a connected affi ne algebraic groupthen [G,G] is Z—closed in G and connected.

Proof. Let

Φk : ×2kG→ G

be defined by

Φk(g1, h1, ..., gk, hk) = g1h1g−11 h−1

1 · · · gkhkg−1k h−1

k .

Then Φk is regular and thus Φk(×2kG) has interior inits closure. We set Uk = Φk(×2kG) then we have

dimUk ≤ dimUk+1.

Thus for dimension reasons there exists ko such that

dimUko = dimUko+j

for all j ≥ 0. Since ×2kG is irreducible, the varieties Ukare irreducible. Thus

Uko = Uko+j

for all j ≥ 0. We note that

UkUl = Uk+l.

Thus

UkoUko ⊂ UkoUko ⊂ Uko+ko = Uko.

Indeed, if f is a regular function on G and

f(UkoUko) = 0

then f(uUko) = 0 all u ∈ Uko. Thus f(uUko) = 0 forall u ∈ Uko thus

UkoUko ⊂ U2ko,

Using the same argument we see that if f(U2ko) = 0

then by the above f(Ukou) = 0 so f(Ukou) = 0. Thus

UkoUko ⊂ Uko+ko = Uko.

Hence Uko is closed under multiplication so

Uko = [G,G].

But Uko contains an open dense subset of Uko which iscontained in [G,G]. This implies that [G,G] is open in

[G,G] hence it is closed. Since [G,G] is the image of×koG under Φko it is connected.

If G is a group then we define D1(G) = [G,G] andDk+1(G) = [G,Dk(G)]. Then G is said to be solvableif Dk(G) = {e} for some k.

Lemma. If S is a solvable subgroup of GL(n,C) thenits Z—closure is solvable.

Proof. Since S is solvable, there exists an r such thatDr(S) = {I}. For each s define φs to be the map

φs : GL(n,C)2s → GL(n,C)

given by

g1, h1, g2, h2, ..., gs, hs 7−→ (g1h1g−11 h−1

1 ) · · · (gshsg−1s h−1

s ).

This map is regular and if U ⊂ GL(n,C) is a subgroupthen

∞⋃s=1

φs(U2s) = D1(U).

This implies that (here over—bar is Z—closure) D1(U) ⊂D1(U). SinceD1(U) is closed this implies thatD1(U) =

D1(U). Thus if Dk(U) = Dk(U) then

Dk+1(U) = D1(Dk(U)) =

D1(Dk(U)) = D1(Dk(U)) = Dk+1(U).

Since {I} = Dr(S) we have

{I} = Dr(S) = Dr(S).

Exercise. Show that a subgroup of a solvable group issolvable.

Theorem. If S is a connected solvable affi ne algebraicgroup acting algebraically on a projective variety then Shas a fixed point.

Proof. We will first prove the result in the case when Sis commutative. Let Y be the projective variety on which

S is acting. Then S has a closed orbit Sy. Hence Sy iscompact. The group

Sy = {s ∈ S|sy = y}

is normal and closed. Thus S/Sy is connected and affi neand it is compact in the standard topology. so it is apoint.

We now prove the full result by induction on dimS. IfdimS is 1 then S is abelian so the result is true in thiscase. Assume for

1 ≤ dimS < k.

Let S be of dimension k and acting on Y a projec-tive variety. Then D1(G) is connected, solvable anddimD1(S) < dimS.Thus

Z = {y ∈ Y |D1(S)y = y}

is non-empty and Z—closed with S acting on it throughS/D1(S). Now S/D1(S) is a connected and abelianaffi ne algebraic group so the first part of the proof impliesthat it has a fixed point.

Corollary. If S is a connected solvable affi ne algebraicgroup and if (ρ, V ) is a regular representation of G thenthere exists a basis of V such that the matrices ρ(s) areupper triangular.

Proof. Let F be the flag variety of V . The fixed pointtheorem implies that S has a fixed point Vo in F . Henceif {v1, ..., vn} is a basis adapted to Vo then the matricesof the elements of ρ(S) are upper triangular.

An affi ne algebraic group, U , is said to be unipotent if foreach regular representation (ρ, V ) the set ρ(U) consistof unipotent elements (i.e. I − ρ(u) is nilpotent).

Lemma. If G ⊂ GL(n,C) is a Z—closed subgroup andif g ∈ G is unipotent then

g = eX

with X ∈ Lie(G) a nilptent element.

Proof. Set

log g =∑k≥0

(−1)k+1(g − 1)k+1

k + 1.

then

elog g = g.

X = log g

is nilpotent. If f is a polynimal on Mn(C) such thatf(G) = 0 then f(etX) is a polynomial in t vanishing atevery integer. Hence X ∈ Lie(G).

Lemma. 1,A unipotent group, U , is connected and solv-able.

2. Lie(U) consists of nilpotent elements.

3. If u ⊂Mn(C) is a Lie algebra consisting of nilpotentelements then exp(u) is a unipotent group.

Proof. We may assume that U is Z—closed in GL(n,C).Let X ∈ Lie(U) then the matrices etX are unipotentfor all t. Thus X is nilpotent. Since

0 =

(etX − I

t

)n= Xn(I +

∑k≥1

tkXk

(k + 1)!)n

for all t. The preceding Lemma implies that exp : Lie(U)→U is an isomophism of varieties. Thus U is connected.

We prove that U solvable and 3. by induction on dimU

(resp u). If dimU = 1 then U is abelian hence solv-able. If dim u = 1 then clearly eu is a group. Assume forall subgroups of lower dimension (subalgebras of lowerdimension). Then if u1 is a proper Lie subalgebra ofLie(U) of maximal dimension then U1 = exp u1 is aproper closed subgroup. This implies that it is solvable.Hence it leaves invariant a flag in Lie(U)/ exp u1. Inparticular there is z ∈ Lie(U) such that z /∈ u1 and[u1, z] = 0. Thus Cz + u1 is a subalgebra of Lie(U).The maximality implies that Cz + u1 = Lie(U) henceu1 is an ideal containing Lie[U,U ]. This implies that

dim[U,U ] < dimU. We now complete the proof forLie algebras consinting of nilpotent elements. Let u1

be a maximal proper subalgebra. Then by the inductionhypthesis

U1 = eu1

is a uniptent subgroup. Hence Ad(U) leaves invariant aflag in u/u1. Hence as above there exists z ∈ u suchthen [z, u1] ⊂ u1 and hence

eCzU1

is a subgroup with Lie algebra u. We leave it as an exer-cise using the Campbell-Hausdorff formula to see that

etzeX = eZ

with Z ∈ u.

Let G be an affi ne algebraic group and let S be the setof Z-closed connected solvable subgroups ordered by in-clusion.

Theorem. There exists a maximal element in S and all ofthe maximal elements are conjugate. Furthermore, G/Sis projective for each such S.

Proof. Let S be a Z—closed connected solvable subgroupof maximal maximal dimension. If S ⊂ S1 a connectedsolvable subgroup then since dimS1 = dimS we musthave S = S1.

We have shown that there exists an injective regular rep-resentation (ρ, V ) of G and an element v ∈ V − {0}such that S is the stabilzer in G of the line [v]. Let Xbe the set of all flags in V

U : U1 ⊂ U2 ⊂ ...

such that U1 = [v]. We assert that X is Z-closed. Let

Φ : F(V)→ P(V )

be given by

U1 ⊂ U2 ⊂ ... 7−→ U1

then Φ is a morphism and X = Φ−1[v].

If V ∈ X and if g ∈ G is such that gV = V theng[v] = [v] hence g ∈ S. Clearly S acts on X and henceit has a fixed point inX, Vo. NowGVo gives a realizationofG/S as a Z—open subset of the Zarisky closure ofGVo,Y . We assert that GVo = Y . Let GU be a closed orbitin Y . Then the identity component of GU must havedimension at most dimS. Thus

dimGU ≥ dimGVohence

GU = GVo.

Let S1 be another maximal element in S. Then S1 musthave a fixed point in Y , gVo. This implies that g−1S1g

fixes Vo and hence is contained in S.

A maximal element of S is called a Borel subgroup.

Theorem. If S is a solvable, affi ne algebraic group thenthe subset of unipotent elements, U , forms a Z-closed,normal subgroup.

Proof. We may assume that S ⊂ GL(n,C) as a Z—closed subgroup. If g is unipotent then g ∈ So the iden-tity component of S. Since So has a fixed point in Fnwe may assume that So is contained in Bn the groupof upper triangular elements of GL(n,C). Let Un bethe subgroup of Bn consiteng of the elements with oneson the main diagonal. Then U = So ∩ Un which is aZ—closed subgroup. Since any conjugate of a unipotentelement is unipotent the result follows.

If G is an affi ne algebraic group then we define the unipo-tent radical to be the union of all Z—closed unipotentnormal subgroups of G.

Exercise. The unipotent radical is a Z—closed, unipotent,normal subgroup.

Lemma. Let G be an affi ne algebraic group with unipo-tent radical U . Then U acts trivially on any irreducible,regular, representation of G.

Proof. Let (ρ, V ) be an irreducible, regular, non-zero,representation. Since U is solvable and connected, thereis a basis of V such that ρ(U) consists of upper triangularmatrices with ones on the main diagonal. This impliesthat V U 6= 0. Since U is normal V U is G—invariant.

Theorem. IfG is Z—closed subgroup ofGL(n,C) actingcompletely reducibly on Cn then G is linearly reductive.

Theorem. Let G be an affi ne algebraic group then Gis linearly reductive if and only if its unipotent radical istrivial.

Proof. Suppose that G is not reductive. Then thereexists a regular representation, (ρ, V ), of G that is notcompletely reducible. We may assume that this represen-tation is faithful. Let

V1 % V2 % ... % Vm % Vm+1 = {0}be a composition series for the repesentation. Set Zi =

Vi/Vi+1 then we have a representation, µ,of G on

Z = Z1 ⊕ Z2 ⊕ ...⊕ Zm.

If g ∈ kerµ then

(µ(g)− I)Vi ⊂ Vi+1

for all i. This implies kerµ is a normal subgroup consist-ing of unipotent elements. Hence trivial unipotent radicalimplies linearly reductive.

If G ⊂ GL(n,C) is linearly reductive. Then

Cn = V1 ⊕ V2 ⊕ ...⊕ Vm

with Vi irreducible. The unipotent radical of G acts triv-ially on each of the Vi and hence on Cn.

Theorem. Let H be a Z—closed subgroup of an affi nealgebraic group G over C. Then the following are equiv-alent:

Theorem 1 1. H contains a Borel subgroup.

2. G/H is compact in the S—topology.

3. G/H is projective.

Under any of these conditions we will call H a parabolicsubgroup of G.

Proof. We have seen that a quasi-projective variety iscompact in the S—topology if and only if it is projective.Thus 2 and 3 are equivalent. If G/H is projective thenthe Borel fixed point theorem implies that if B is a Borelsubgroup of G then B has a fixed point in G/H. Thisimplies that B is conjugate to a subgroup of H. Sincea conjugate of a Borel subgroup is a Borel subgroup 3.implies 1. If H contains a Borel subgroup then we havea surjective regular map f : G/B → G/H. Since G/Bis projective it is compact and f is continuous in the S—topology we see that if B ⊂ H then G/H is compact inthe S—topology. Thus 1. implies 2.

Exercise. Is there a similar theorem if we only assumethat H is S—closed in G?

Let G be a reductive subgroup of GL(n,C) and let Hbe a Cartan subgroup of G. Set h = Lie(H) and letΦ be the root system of g = Lie(G) with respect toH. Let hR be the span of the coroots. If h ∈ hR thenα(h) ∈ R for all α ∈ Φ. There exists h ∈ hR such thatα(h) 6= 0 if α ∈ Φ.

Proposition. Let

b = h⊕ ⊕

α∈Φ,α(h)>0

gα

.Then there exists a Borel subgroup B in G such that bis Lie(B).

Proof. We note that b1 = [b, b] =⊕α∈Φ,α(h)>0 gα.

Let a1 = minα∈Φ,α(h)>0α(h) then

b2 = [b1, b1] =⊕

α∈Φ,α(h)>a1

gα.

If we define a2 = minα∈Φ,α(h)>a1α(h) then

b3 = [b2, b2] =⊕

α∈Φ,α(h)>a2

gα.

Thus procesure will eventually lead to 0. Thus if H is theconnected Lie subgroup of G then G is solvable. The Z—closure, H1, of H is solvable and the identity componentof H1 contains B. We assert that

Lie(H1) = Lie(H).

Indeed if Lie(H1) % b then there must be

gα ⊂ Lie(H1), α(h) < 0.

Hence Lie(H1) contains a TDS. This is a contradiction.Hence H1 = B. The same argument shows that B ismaximal connected solvable.

Theorem. LetG be as above. LetB be a Borel subgroupof G and let K be a maximal compact subgroup of G.Then G = KB.

Proof. First consider B constucted from H as abovesuch that H ∩K is a maximal compact torus, T , of K.We can assume that G ⊂ GL(n,C) and the elements ofB are upper trangular. We note

K ∩B = T.

The elements of

n =⊕

α∈Φ,α(h)>0

gα

are nilpotent thus

U = exp n

is unipotent. Thus

B = HU.

Let

a = iLie(T ), A = exp a.

Then we assert that the map

K ×A× U → G

k, a, u 7−→ kau

is bijective. If

kau = k1a1u1.

Thus

k−11 k ∈ AU.

But this implies that k−11 k has real posoitive eigenvalues.

Hence k = k1. similarly a = a1. Thus the map isinjective. Note if X ∈ Lie(K), h ∈ Lie(A) and u ∈Lie(U) then

d

dtt=0ketXaethuetY = kXau+ kahu+ kauY.

If this expression is 0 then

X = −aha−1 − auY (au)−1

which is upper trianagular with real eigenvalues. ThusX = 0. Thus

h = −uY u−1

since Y is nilpotent and h is semisimple h = 0. Thisimplies that the set KAU is open that the image is openin the metric topology. But AU is closed in G. Hencethe image is open and closed. Since K intersects everyconnected component of G. G = KAN.

Suppose B1 is another Borel subgroup. Then there existsg ∈ G with gBg−1 = B1. Thus writing

g = kb, k ∈ K

and b ∈ B then

kBk−1 = B1.

Hence

k−1KB1k = KB = G.