Representations of algebraic groups and their Lie algebras Jens Carsten Jantzen Lecture I Set-up. Let K be an algebraically closed field. By convention all our algebraic groups will be linear algebraic groups over K. For such a group G let K[G] denote the algebra of regular functions on G. For the background on algebraic groups I refer to the books Linear Algebraic Groups by J. E. Humphreys, T. A. Springer, and A. Borel. Modules. Let G be an algebraic group. A G{module is by definition a module V over the group algebra KG of G as an abstract group such that V satisfies two additional conditions: • V is locally finite: each element is contained in a finite dimensional KG{submodule. ∗ • For each v 2 V and ' 2 V the matrix coefficient c';v given by c';v(g) = '(g v) for all g 2 G is a regular function: c';v 2 K[G]. For example, any finite dmensional vector space V is a module under the natural action for the general linear group GL(V ) as well as for any closed subgroup of GL(V ). For any algebraic group G the algebra of regular functions K[G] is a G{module under the (left) regular representation given by (gf)(g0) = f(g−1g0). It is the easy to chech that the category of all G{modules is closed under: • submodules • factor modules • direct sums • tensor products • symmetric and exterior powers. If a G{module V is finite dimensional, then also its dual space V ∗ is a G{module. Frobenius twist. Assume char(K) = p > 0. In this case we have another method to construct new G{modules from known ones. If V is a vector space over K we denote by V (1) the vector space over K that coincides with V as an additive group and where the scalar multiplication is given by p a • v = p a v for all a 2 K, v 2 V where the left hand side is the new multiplication and the right hand side the old one. (Since K is algebraically closed, the Frobenius morphism a 7! ap is a bijection K ! K, so any element in K has a unique p{th root.) 1 If V is a G{module, then V (1) is a G{module using the given action of any g 2 G on the additive group V (1) = V . The action is again K{linear. A finite dimensional G{ stable subspace of V is also a finite dimensional G{stable subspace of V (1) (of the same dimension); so also V (1) is locally finite. For any ' 2 V ∗ the map '(1): V (1) ! K; v 7! '(v)p is K{linear. One checks that (V (1))∗ = f '(1) j ' 2 V ∗ g and gets finally for all g 2 G (1) p p c'(1);v(g) = ' (gv) = '(gv) = c';v(g) p hence c'(1);v = c';v 2 K[G]. One defines higher Frobenius twists inductively: Set V (r+1) = (V (r))(1). Exercise: Consider K[G] as a G{module under the (left) regular representation. Show that f 7! f p is a homomorphism K[G](1) ! K[G] of G{modules. Induction. Let H be a closed subgroup of an algebraic group G. Any G{module V is by restriction an H{module since the restriction to H of any function in K[G] belongs to K[H]. This restriction of modules is clearly a functor from G{modules to H{modules. G We can define an induction functor indH in the opposite direction, from H{modules to G{modules. For any H{module M set G −1 indH M = f f: G ! M regular j f(gh) = h f(g) for all g 2 G; h 2 H g: Here a function f: G ! M is called regular if there exists m1; m2; : : : ; mr 2 M and Pr f1; f2; : : : ; fr 2 K[G] such that f(g) = i=1 fi(g) mi for all g 2 G. All regular functions G G ! M form a vector space over K and indH M is then a subspace. We get a G{action on this subspace by setting (gf)(g0) = f(g−1g0). In order to show that this KG{module is a G{module one uses the fact that K[G] is a G{module. G A trivial example for an induced module is indf1g = K[G]. We'll soon see more interesting examples for G = SL2(K). G The functor indH is right adjoint to the restriction functor from G{modules to H{modules. This means that we have functorial isomorphisms G ∼ HomG(V; indH M) −! HomH (V; M) G Here any G{homomorphism ': V ! indH M is sent to ': V ! M with '(v) = '(v) (1). G Conversely any H{homomorphism : V ! M goes to e: V ! indH M such that e(v)(g) = −1 (g v). 2 Example: SL2(K). Consider G = SL2(K) acting on its natural module V = K with its ∗ ∗ standard basis e1; e2. Then G acts also on V and on the symmetric algebra S(V ) that we can identify with the algebra K[X1;X2] of polynomial functions on V . Here X1;X2 is ∗ the basis of V dual to e1; e2. 2 The action of G on S(V ∗) is given by (gf)(v) = f(g−1v). Each symmetric power Sn(V ∗) i i n−i is a submodule. We choose as basis all vi = (−1) X1X2 , 0 ≤ i ≤ n, and get (e.g.) for all a 2 K i n 1 a X i 1 0 X n − i v = ai−jv and v = aj−iv (1) 0 1 i j j a 1 i n − j j j=0 j=i and for all t 2 K× t 0 v = tn−2i v : (2) 0 t−1 i i Consider the closed subgroups in G 1 0 t 0 U = f j a 2 K g and B = f j a 2 K; t 2 K× g: a 1 a t−1 Let us first observe that we have an isomorphism of G{modules ∗ ∼ G S(V ) −! indU K where K is the trivial U{module such that G indU K = f f 2 K[G] j f(gu) = f(g) for all g 2 G, u 2 U g: ∗ G Well, any 2 S(V ) defines b: G ! K setting b(g) = (ge2). One checks that b 2 indU K G and that 7! b is a morphism of G{modules. On the other hand, any f 2 indU K defines f: V n f0g ! K: For any v 2 V , v 6= 0 there exists g 2 G with v = ge2; we then set 0 0 0 f(v) = f(g). This is well defined as ge2 = g e2 implies g 2 gU, hence f(g) = f(g ) since G f 2 indU K. One checks that f is a regular function on V n f0g. Algebraic geometry tells you that f has a (unique) extension to a regular function on V . So there exists 2 S(V ∗) with f(v) = (v) for all v 6= 0, hence with f(g) = (ge2) for all g 2 G, i.e., with f = b. For each n 2 Z let Kn denote K considered as a B{module such that t 0 v = tn v for all a; t; v 2 K, t 6= 0: a t−1 G G Since any u 2 U acts trivially on any Kn, it is clear that indB Kn ⊂ indU K. The inverse G ∗ G image of indB Kn under the isomorphism S(V ) ' indU K from above consists of all 2 S(V ∗) with (tv) = tn (v) for all t 2 K× and v 2 V . This shows that we get isomorphisms n n ∗ indG K −!∼ S (V ) if n ≥ 0, B n 0 if n < 0. Simple modules for SL2(K). Consider again G = SL2(K). If char K = 0, then one can show that all Sn(V ∗) with n 2 N are simple G{modules. [Using (2) above one checks that 3 each non-zero submodule contains some vi. Then (1) shows that the submodule contains all vj. Here one uses that all binomial coefficients are non-zero in K.] Furthermore, each simple G{module is isomorphic to Sn(V ∗) for a unique n. If char K = p > 0, then the same argument shows that all Sn(V ∗) with 0 ≤ n < p are simple G{modules. But one can quickly see that Sp(V ∗) is not simple: The map ': V ∗ −! Sp(V ∗); f 7! f p is semi-linear (i.e., satisfies '(f + f 0) = '(f) + '(f 0) and '(af) = ap'(f) for all a 2 K) and G{equivariant. So its image is a G{submodule of Sp(V ∗) that is non-zero and not the whole module. Going back to the definition of a Frobenius twist one checks that ' is an isomorphism of G{modules ∗ (1) ∼ p (V ) −! KGX2 p ∗ p onto the submodule of S (V ) generated by X2 . For general n the characteristic 0 argument shows first that each non-zero submodule n ∗ of S (V ) contains some vi. Then the first formula in (1) implies that the submodule n contains v0 = X2 . This shows that the submodule n L(n) := KG v0 = KGX2 n ∗ n n ∗ of S (V ) generated by X2 is simple, in fact, it is the only simple submodule of S (V ).