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Material on Algebraic and Lie Groups 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 C1 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 nf2 2 j 6 g 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 ]. f g 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. 2 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 C1. ! 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 . f 2 j 2 2 g 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 kp k ≤ k k k k 1 Xm eX = m! m=0 X then this series converges absolutely and uniformly in compacta. In particular the implies that X eX ! defines a C1 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 1 Xm log(I X) = m m=1 X converges absolutely and uniformly on compacta. This says that if > 0 is so small that if X < 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). 2 2 Proof. We have sX tY sX+tY + st [X,Y ]+O(( t + s )3) e e = e 2 j j j j . 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 areO(( t + s )3). j j j j (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 ! 1 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 ]. 2 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 2 2 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 2 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 K 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, , 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 : C(K) C is a measure if therek existsk a constant2 j suchj that ! (f) C f , f C(K). j j ≤ k k 2 We say that is positive if (f) 0 if f(K) [0, ). We say that it is normalized if it is positive and (1)≥ = 1 (1 the constant 1 function taking the 1 value 1). Finally, if k K we define Lkf(x) = f(k x) for k, x K and 2 2 f C(K). Then Lk : C(K) C)(K). Left invariant means that Lk = . One2 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 : K GL(V ) be a continuous homomorphism. Then there exists a Hermitian inner! product, (..., ...), on V such that ((k)v, (k)w) = (v, w) for v, w V and k K. 2 2 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) = K (k (k) z, (k) w ). Then one checks that this form is Hermitian, and positive! h definite since thei measure is positive. We also note that if u K then K 2 1 1 ((u)z, (u)w) = (k (k) (u)z, (k) (u)w ) = K ! 1 1 1 1 1 1 k(k (u k) z, (u k) w ) = K (k (k) z, ( 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 continuous2 function, \ ; f, on U such that f K = 1 and f Ku = 0. The Stone-Weierstrass theorem j j 1 implies that there is C[xij] such that f(x) (x) < 4 for x U. Thus (ku) < 1 and (k)2> 3 for all k Kj . Let (X)j = (k 2 (kX)). j j 4 j j 4 2 K ! Then is a polynomial on Mn(C) (expand in monomials and note that we are just integrating coeffi cients). We note that (ku) 1 and (k) 3 j j ≤ 4 j j ≥ 4 for all k K. On the other hand, if Y Mn(C) then then the function 2 2 X (XY ) is a polynomial on Mn(C) thus in (GL(n, C)) and it takes the constant! value (Y ) on K. Thus since G is theO Z-closure of K we see that (gY ) = (Y ) for all g G hence for all g U. But then (x) = (I) for all u U. This is a contradiction2 since (I)2 3 and (u) = 0. 2 j j ≥ 4 j j 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 2 2 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 fact2 every row of g G is an element of the 2n 1 sphere so the group is topologically a closed2 subset of a product of n spheres).
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