2 Lie groups and algebraic groups.

2.1 Basic Definitions. In this subsection we will introduce the class of groups to be studied. We first recall that a Lie group is a group that is also a differentiable manifold 1 and multiplication (x, y xy) and inverse (x x− ) are C∞ maps. An algebraic group is a group7−→ that is also an algebraic7→ variety such that multi- plication and inverse are morphisms. Before we can introduce our main characters we first consider GL(n, C) as an affi ne algebraic group. Here Mn(C) denotes the space of n n matrices × and GL(n, C) = g Mn(C) det(g) =) . Now Mn(C) is given the structure n{2 ∈ | 6 } of affi ne space C with the coordinates xij for X = [xij] . This implies that GL(n, C) is Z-open and as a variety is isomorphic with the affi ne variety 1 Mn(C) det . This implies that (GL(n, C)) = C[xij, det− ]. { } O Lemma 1 If G is an algebraic group over an algebraically closed field, F , then every point in G is smooth.

Proof. Let Lg : G G be given by Lgx = gx. Then Lg is an isomorphism → 1 1 of G as an algebraic variety (Lg− = Lg− ). Since isomorphisms preserve the set of smooth points we see that if x G is smooth so is every element of Gx = G. ∈

Proposition 2 If G is an algebraic group over an algebraically closed field F then the Z-connected components

Proof. Theorem 18 in section 1.2.6 implies that every element of G is con- tained in a unique irreducible component.

Theorem 3 A closed subgroup of GL(n, C) is a Lie group.

This theorem is a special case of the fact that a closed subgroup of a Lie group is a Lie group. We should also explain what “is” means in these contexts. The result needed is

Theorem 4 Let G and H be Lie groups then a continuous homomorphism f : G H is C∞. →

1 This implies that there is only one Lie group structure associated with the structure of G as a topological group. If G is a closed subgroup of GL(n, C) then we define the Lie algebra of G to be tX Lie(G) = X Mn(C) e G for all t R . { ∈ | ∈ ∈ } Some explanations are in order. First we define X,Y = trXY ∗ and X = h i k k X,X We use this to define the metric topology on Mn(C). We note that XYh i X Y so if we set pk k ≤ k k k k ∞ Xm eX = m! m=0 X then this series converges absolutely and uniformly in compacta. In particular the implies that X eX → defines a C∞ map of Mn(C) to GL(n, C) in fact real (even complex) analytic. We also note that if X < 1 then the series k k ∞ Xm log(I X) = m − m=1 X converges absolutely and uniformly on compacta. This says that if δ > 0 is so small that if X < δ then I eX < 1 we have log(eX ) = log(I (I eX )) = X. k k − − −

Proposition 5 Let G be a closed subgroup of GL(n, C) then Lie(G) is an R- subspace of Mn(C)such that if X,Y Lie(G), XY YX = [X,Y ] Lie(G). ∈ − ∈ Proof. We have

sX tY sX+tY + st [X,Y ]+O(( t + s )3) e e = e 2 | | | | .

To to sketch a proof this we take s and t so small that I esX etY < 1. Expand the series log(I (I esX etY )) ignoring terms that are−O(( t + s )3). − − | | | | (c.f. [GW], ???). For details. We now prove the Lemma. We note that for fixed t

t X t Y m t(X+Y )+O( 1 ) (e m e m ) = e m .

2 Taking the limit as m shows that Lie(G) is a subspace. Also for fixed t → ∞ t X t Y t X t Y m2 t2[X,Y ]+O( 1 ) (e m e m e− m e− m ) = e m . Thus et2[X,Y ] G. Take inverses to get negative multiples of [X,Y ]. ∈ Theorem 6 The Lie algebra of a linear algebraic group,G GL(n, C), is ⊂ a complex subspace of Mn(C) . Furthermore Lie(G) = TI (G)in the sense of algebraic geometry. 1 X 1 Ad(g)X G acts on Lie(G) via Ad(g)X = gXg− , We note that ge g− = e so 1 gLie(G)g− = Lie(G) for g G. We also note that if X,Y Mn(C) then ∈ ∈ d tX tX tX e Y e− = Ad(e )[X,Y ]. dt Then is d Ad(etX ) = Ad(etX )ad(X) dt with ad(X)Y = [X,Y ]. This implies that as an endomorphism of Mn(C) Ad(etX ) = etad(X). If X g the ad(X)g g so these formulas are true for ∈ ⊂ any closed subgroup of GL(n, C).

2.1.1 Some remarks about compact groups.

Let K GL(n, C) be a compact subgroup. Then K is closed and hence a Lie ⊂ group. We will denote by µK the left invariant, positive normalized measure on K(Haar measure). We recall what this means. We define a (complex) measure on K to be a continuous linear map, µ, of C(K) (continuous func- tions from K to C) to C where C(K) is endowed with the uniform topology. That is, we set f = maxx K f(x) and a linear map µ : C(K) C is a measure if therek existsk a constant∈ | such| that → µ(f) C f , f C(K). | | ≤ k k ∈ We say that µ is positive if µ(f) 0 if f(K) [0, ). We say that it is normalized if it is positive and µ(1)≥ = 1 (1 the constant⊂ ∞ function taking the 1 value 1). Finally, if k K we define Lkf(x) = f(k− x) for k, x K and ∈ ∈ f C(K). Then Lk : C(K) C)(K). Left invariant means that µ Lk = µ. One∈ proves that a normalized→ invariant measure on K is unique and◦ that it is given by integration of a differential form. For our purposes we will only need the following property (and its proof) that follows from the fact that it is given by the integration against a differential form.

3 Theorem 7 Let V be a finite dimensional vector space over C. Let σ : K GL(V ) be a continuous homomorphism. Then there exists a Hermitian inner→ product, (..., ...), on V such that (σ(k)v, σ(k)w) = (v, w) for v, w V and k K. ∈ ∈ Proof. Let ..., ... be an inner product on V (e.g. choose a basis and use h i 1 1 the standard inner product). Define (z, w) = µK (k σ(k)− z, σ(k)− w ). Then one checks that this form is Hermitian, and positive→ h definite since thei measure µ is positive. We also note that if u K then K ∈ 1 1 (σ(u)z, σ(u)w) = µ (k σ(k)− σ(u)z, σ(k)− σ(u)w ) = K → 1 1 1 1 1 1 µk(k σ(u− k)− z, σ(u− k)− w ) = µK (k σ(k)− z, σ( k)− w ). → → Let W be the orthogonal complement of V with respect to (..., ...) and let P be the projection onto V corresponding to the decomposition Cm = V W. ⊕

Theorem 8 Let K be a compact subgroup of GL(n, C) and let G be the Zariski closure of K in GL(n, C) then K is a maximal compact subgroup of G. Proof. Let U be a compact subgroup of G containing K. Suppose that K = U we show that this leads to a contradiction. Then if u U K then Ku6 K = . Since both sets are compact there exists a continuous∈ − function, ∩ ∅ f, on U such that f K = 1 and f Ku = 0. The Stone-Weierstrass theorem | | 1 implies that there is φ C[xij] such that f(x) φ(x) < 4 for x U. Thus φ(ku) < 1 and φ(k)∈> 3 for all k K| . Let− γ(X)| = µ (k ∈ φ(kX)). | | 4 | | 4 ∈ K → Then γ is a polynomial on Mn(C) (expand in monomials and note that we are just integrating coeffi cients). We note that γ(ku) 1 and γ(k) 3 | | ≤ 4 | | ≥ 4 for all k K. On the other hand, if Y Mn(C) then then the function ∈ ∈ X γ(XY ) is a polynomial on Mn(C) thus in (GL(n, C)) and it takes the constant→ value γ(Y ) on K. Thus since G is theO Z-closure of K we see that γ(gY ) = γ(Y ) for all g G hence for all g U. But then γ(x) = γ(I) for all u U. This is a contradiction∈ since γ(I)∈ 3 and γ(u) = 0. ∈ | | ≥ 4 | | 2.2 Symmetric Subgroups. 2.2.1 Definition.

We will take as the main examples over C the Z-closed subgroups, G, of GL(n, C) that have the additional property: if g G then g∗ G (here, as ∈ ∈ 4 usual, [gij]∗ = [gji]). We will call such a group symmetric. Let U(n) denote the group of all g GL(n, C) such that gg∗ = I. Then U(n) is a compact Lie group (in fact∈ every row of g G is an element of the 2n 1 sphere so the group is topologically a closed∈ subset of a product of n spheres).− Examples. 1. Obviously. GL(n, C) is a symmetric subgroup of itself. 2. SL(n, C) = g GL(n, C) det g = 1 . This group is a hypersurface { ∈ | } in Mn(C). T T 3. O(n, C) = g GL(n, C) gg = 1 (here [gij] = [gji]). Thus the { ∈ | } equations defining O(n, C) are

xikxkj = δij. k X 4. SO(n, C) = g O(n, C) det g = 1 . { ∈ | } T 5. Sp(n, C) = Sp2m(C) = g GL(2n, C) gJg = J . Here { ∈ | } 0 I J = I 0  −  with I the n n identity matrix. × 6. (Most general example) Let K U(n) be a closed (hence compact sub- ⊂ group). Let G be the Z-closure of K in GL(n, C). If f (GL(n, C))vanishes ∈ O on K then f(X∗) (the overbar denotes complex conjugation) is also in (GL(n, C)) O and vanishes on K. Thus if g G then g∗ is in G. ∈ Exercise. Prove that Sp(n, C) is symmetric. Also, show that det g = 1 if g Sp(n, C). ∈ 2.2.2 Some decompositions of symmetric groups.

We define the exponential map to be exp : Mn(C) GL(n, C) given by exp(X) = eX . →

Theorem 9 If G is a symmetric subgroup of GL(n, C) and if K = G U(n) then ∩ Lie(G) = Lie(K) + iLie(K). Furthermore, the map K iLie(K) G × →

5 given by k, Z keZ → defines a homeomorphism. We note that the map is actually a diffeomorphism but we will not need this slightly harder fact. This theorem is a special case of a more general result that we will explain after deriving a few consequences.

Corollary 10 If G is a symmetric subgroup of GL(n, C) then G is the Z- closure of K = G U(n). ∩ tX Proof. Let φ (GL(n, C)) be such that φ K = 0. Then φ(ke ) = 0 for ∈ O | k K,X Lie(K) and t R. But then by looking at this function in t as ∈ ∈ ∈ the restriction of a holomorphic function on C we see that the above equation is also true for t C. Thus φ(keZ ) = 0 for k K and Z iLie(K).Hence G is contained in the∈ Z-closure of K. Since G is∈ Z-closed G∈is the Z-closure.

Corollary 11 If G is a symmetric subgroup of GL(n, C) then G is irre- ducible in the Z-topology if and only if G is connected in the S-topology. We note that this is a special case of fact that an irreducible algebraic variety over C is connected in the S-toplogy. We will now describe the generalization. Let G be a subgroup of GL(n, R) that is given as the locus of zeros of a set of polynomials on Mn(R) (thought 2 of as Rn ). We will say that G is a symmetric real group if whenever g ∈ G, gT G. Let O(n) be the compact Lie group O(n, C) U(n) and set K = G∈ O(n). ∩ Assume∩ that we have such a group. Let g = Lie(G) and let k = Lie(K). Define X,Y =trXY T is real valued on, positive definite and invariant under h i the action of Ad(K). Set p = k⊥ g. Then Ad(k)p p for k K. We note that if X p then ∩ ⊂ ∈ ∈ ad(X)Y,Z = tr[X,Y ]ZT = tr(XYZT YXZT ) = h i − tr(YZT X YXZT ) = tr(Y [X,Z]T ) = Y, ad(X)Z . − h i This implies that if X p then adX is self adjoint relative to ..., ... .It is therefore diagonalizable∈ as an endomorphism of g. Let a be a subspaceh i of p that is maximal subject to the condition that [X,Y ] = 0 for X,Y a. Such a subspace is called a Cartan subspace. ∈ The following are basic theorems of E. Cartan.

6 Theorem 12 Let G, K, p be as above then the map K p G given by k, X keX is a homeomorphism. In particular, G is connected× → if and only if K 7−→is connected.

We will use the following Lemma of Chevalley.

Lemma 13 Let φ be a polynomial on Mn(C) and let X Mn(C) be such mX ∈ that X∗ = X. Then if φ(e ) = 0 for all m N = 0, 1, 2, ... then ∈ { } φ(etX ) = 0 for all t R. ∈ 1 Proof. There exists u U(n) such that uXu− is diagonal with real diagonal ∈ 1 entries ai, i = 1, ..., n. Replacing φ by φ Ad(u)− we may assume that X is diagonal with the indicated diagonal◦ entries. Now observing that the tX only monomials in the xij that are non-zero on e are of the form γm = m1 m2 mn tX tA tX x11 x22 xnn and γm(e ) = e with A = i miai. Thus if φ(e ) is not identically··· 0 then it must be of the form P r tX tAj φ(e ) = cje j=1 X for some cj C and the Aj are of the form mijai with mij N. We group ∈ ∈ the terms so that A1 > ... > Ar and we may assume c1 = 0. We show that this leads to a contradiction. In fact, P 6

r tA1 tX t(A1 Aj ) e− φ(e ) = c1 + cje− − . j=2 X

mX mA1 mX But φ(e ) = 0 for m N. Taking the limit of e− φ(e ) as m in ∈ → ∞ N yields the contradiction c1 = 0. We will now prove the theorem. We note that the polar decomposition of an element of GL(n, R) as kp T with k O(n) and p Mn(R) with p = p and all eigenvalues real and ∈ ∈ strictly positive let Pn be the set of such matrices. This decomposition is T unique. We also note that if pn = X Mn(R) X = X then Pn is an open { ∈ | } subset and exp : pn Pn is a homeomorphism. → Let φ1, ..., φm be polynomials defining G GL(n, R). Then if g G we X ⊂ T X T X 2X∈ have g = ke with k O(n) and X pn. Now g g = e k ke = e . Thus since G is invariant under∈ transposition∈ we have e2X G. Thus e2mX G ∈ ∈ 7 2mX for all m Z. Thus implies that φi(e ) = 0 for all i and m Z. The ∈ 2tX ∈ Lemma above implies that φi(e ) = 0 For all t R. Thus X Lie(G). Thus X p. This implies that k K. But then ∈G = K exp(p).∈ That the map in the∈ statement is a homeomorphism∈ follows from the assertion for the polar decomposition.

Theorem 14 Let G, K, p be as above. Then if a1 and a2 are Cartan sub- spaces of p then there exists k K such that Ad(k)a1 = a2. ∈ For the proof we need the following

Lemma 15 If a is a Cartan subspace of p then there exists H a such that a = X p [H,X] = 0 . ∈ { ∈ | }

Proof. If λ a∗define gλ = X g [h, X] = λ(h)X, h a . Since a consists ∈ { ∈ | ∈ } of elements that commute and so [ad(hi), ad(h2)] = 0 for h1, h2 a we see that the elements of ad(a) simultaneously diagonalize and we therefore∈ see that g = g0 λ=0gλ. Let Λ(g, a) denote the set of all λ a∗ with λ = 0 ⊕ ⊕ 6 ∈ 6 and gλ = 0. Then Λ(g, a) is finite so there exists H a such that λ(H) = 0 6 ∈ 6 if λ Λ(g, a). Fix such an H. If X g and [H,X] = 0 then X g0. By ∈ ∈ ∈ definition of Cartan subspace g0 p = a. This proves the Lemma. ∩ We will now prove the theorem. The argument we will give is due to Hunt. Let a1 and a2 be Cartan subspaces of p. Let Hi ai be as in the ∈ Lemma. Since K is compact the function f(k) = Ad(k)H1,H2 achieves a h i minimum ko. Thus if X k then ∈ d tX Ad(e )Ad(ko)H1,H2 = 0. dtt=0

Using ??? this implies that

0 = [X, Ad(ko)H1],H2 = X, [Ad(ko)H1,H2] . h i h i If x, y p then [x, y] k. This implies that since X is an arbitrary element of ∈ ∈ k we must have [Ad(ko)H1,H2] = 0. This implies that H2 Ad(ko)a1. Since ∈ a1 is abelian this implies that a2 Ad(ko)a1. Maximality implies equality. ⊂ Corollary 16 Let G, K be as above and let a be a Cartan subspace of p. Set A = exp(a). Then G = KAK.

8 Proof. If X p then since [X,X] = 0 it is contained in a Cartan subspace. The preceding∈ theorem now implies that

p = Ad(K)a.

1 Thus exp(p) = k K kAk− . Thus G = K exp(p) KAK. ∪ ∈ ⊂ 2.2.3 Compact and Algebraic Torii.

We note that GL(1,F ) = [x] x F × We will therefore think of F ×as the { | ∈ } n algebraic group GL(1,F ). An algebraic group isomorphic with (F ×) will be called an n-dimensional algebraic torus. Thus the group of diagonal matrices in GL(n, F ) is an n dimensional algebraic torus.

Examples. 1. Consider the subgroup, H, of diagonal matrices in SL(n, F ). We can write such a matrix as

z1 0 0 0 ··· 0 z2 0 0  . . ···. . .  z = ......    0 0 zn 0   0 0 ··· 0 1   z1z2 zn   ··· ···  n   Thus the map (F ×) H with (z1, ..., zn) z (as displayed) defines an isomorphism. So H is→ an n-dimensional algebraic7−→ torus. 2. Consider the group, B, of all matrices of the form

a b a2 + b2 = 1 . b a |  −   Then if i is a choice of √ 1 then the map − a b a + ib b a →  −  is a group isomorphism of B onto C×, the one dimensional algebraic torus.

1 We note that S = z C z = 1 is a compact subgroup of C×.A compact n-torus is a Lie{ group∈ || isomorphic| } with (S1)n (the n-fold product

9 group). We will write T n for the group (S1)n. We can realize T n as the group of diagonal elements of U(n). It is easily seen that the Zariski closure of this realization in GL(n, C) is an algerbraic n-torus. Conversely, if H is an algebraic n-torus taken to be a Z-closed subgroup of GL(m, C) for some m then the subgroup, T , of H corresponding to T n under the isomormphism n with (C×) is a compact subgroup of GL(m, C) we leave it to the reader to check that H is the Z-closure of T . One can also prove that a compact connected commutative Lie group is a compact torus.

Now let G be a symmetric subgroup of GL(n, C). Let K = G U(n). Let T K be a maximal compact torus in K. Then iLie(T ) iLie∩ (K) = p. Let⊂a be a maximal abelian subspace of p containing iLie(⊂T ). Then ia is an abelian subalgebra of Lie(K). Let T1 be the S-closure of exp(ia) then T1 is the closure of a connected set so T1 is a connected compact abelian group. Since T is a maximal such group we see that T1 = T . This implies a =iLie(T ).

We now put all of these observations together.

Theorem 17 Let G be a symmetric subgroup of GL(n, C) and let K = G U(n). Then ∩ 1. A maximal compact torus of K is of the form exp(ia) with a a Cartan subspace of p. 2. All maximal compact tori are conjugate in K. 3. If T is a maximal compact torus in K and a = iLie(T ) then T exp(a) is a maximal algebraic torus in G. 4. If a is a Cartan subspace of p then exp(ia) is a maximal compact torus in K.

We also have

Theorem 18 Let G and K be as in the previous theorem. Let a be a Cartan subspace of p = iLie(K). Let H be the unique maximal algebraic torus containing exp(ia) then H = exp(a + ia) = T exp(a) with T = exp(ia) a maximal compact torus in K, Furthermore, G = KHK.

Proof. The first assertions are a direct consequence of the previous result. The last follows since HK = AK in the notation of Corollary 16.

10 2.2.4 General reductive algebraic groups over C. We will say that an affi ne algebraic group, G, is linearly reductive if when- ever σ : G GL(V ) is an algebraic group homomorphism with V a finite → dimensional vector space over C whenever W is a subspace invariant under σ(G) there exists V1 a subspace invariant under C such that V = W V1. There is a more general notion of reductive over any algebraically closed⊕ field which is equivalent to linearly reductive in characteristic 0. The following theorems are true. We will not need them in these lectures.

Theorem 19 Let G be an affi ne algebraic, reductive group then G is iso- morphic with a symmetric subgroup of GL(n, C) for some n.

Theorem 20 Let G be a linearly reductive affi ne algebraic group that all maximal compact subgroups (in the S-topology) are conjugate.

Theorem 21 If K is a compact Lie group then there exists n and a closed subgroup H in U(n) such that K is Lie group isomorphic with H.

11