Representations of algebraic groups and their Lie algebras Jens Carsten Jantzen Lecture I

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1 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 V and ϕ V the matrix coefficient c ϕ,v given by c ϕ,v (g) = ϕ(g v) for all g G is a regular function: c ϕ,v 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) (g ) = f(g 1 g ). 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 a v = p a v for all a K, v 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 a p is a bijection K K, so any element in K has a unique p th root.) 1

2 If V is a G module, then V (1) is a G module using the given action of any g 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 ϕ V the map ϕ (1) : V (1) K, v ϕ(v) p is K linear. One checks that (V (1) ) = { ϕ (1) ϕ V } and gets finally for all g G hence c ϕ (1),v = c p ϕ,v K[G]. c ϕ (1),v(g) = ϕ (1) (gv) = ϕ(gv) p = c ϕ,v (g) p 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 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. We can define an induction functor ind G H in the opposite direction, from H modules to G modules. For any H module M set ind G H M = { f: G M regular f(gh) = h 1 f(g) for all g G, h H }. Here a function f: G M is called regular if there exists m 1, m 2,..., m r M and f 1, f 2,..., f r K[G] such that f(g) = r i=1 f i(g) m i for all g G. All regular functions G M form a vector space over K and ind G H M is then a subspace. We get a G action on this subspace by setting (gf)(g ) = f(g 1 g ). In order to show that this KG module is a G module one uses the fact that K[G] is a G module. A trivial example for an induced module is ind G {1} = K[G]. We ll soon see more interesting examples for G = SL 2 (K). The functor ind G H is right adjoint to the restriction functor from G modules to H modules. This means that we have functorial isomorphisms Hom G (V, ind G H M) Hom H (V, M) Here any G homomorphism ϕ: V ind G H M is sent to ϕ: V M with ϕ(v) = ϕ(v) (1). Conversely any H homomorphism ψ: V M goes to ψ: V ind G H M such that ψ(v) (g) = ψ(g 1 v). Example: SL 2 (K). Consider G = SL 2 (K) acting on its natural module V = K 2 with its standard basis e 1, e 2. Then G acts also on V and on the symmetric algebra S(V ) that we can identify with the algebra K[X 1, X 2 ] of polynomial functions on V. Here X 1, X 2 is the basis of V dual to e 1, e 2. 2

3 The action of G on S(V ) is given by (gf) (v) = f(g 1 v). Each symmetric power S n (V ) is a submodule. We choose as basis all v i = ( 1) i X1X i 2 n i, 0 i n, and get (e.g.) for all a K 1 a v 0 1 i = i j=0 i a i j v j j and 1 0 v a 1 i = n j=i n i a j i v j (1) n j and for all t K t 0 0 t 1 v i = t n 2i v i. (2) Consider the closed subgroups in G U = { 1 0 t 0 a K } and B = { a 1 a t 1 a K, t K }. Let us first observe that we have an isomorphism of G modules S(V ) where K is the trivial U module such that ind G U K ind G U K = { f K[G] f(gu) = f(g) for all g G, u U }. Well, any ψ S(V ) defines ψ: G K setting ψ(g) = ψ(ge 2 ). One checks that ψ ind G U K and that ψ ψ is a morphism of G modules. On the other hand, any f ind G U K defines f: V \ {0} K: For any v V, v 0 there exists g G with v = ge 2 ; we then set f(v) = f(g). This is well defined as ge 2 = g e 2 implies g gu, hence f(g) = f(g ) since f ind G U K. One checks that f is a regular function on V \ {0}. Algebraic geometry tells you that f has a (unique) extension to a regular function on V. So there exists ψ S(V ) with f(v) = ψ(v) for all v 0, hence with f(g) = ψ(ge 2 ) for all g G, i.e., with f = ψ. For each n Z let K n denote K considered as a B module such that t 0 a t 1 v = t n v for all a, t, v K, t 0. Since any u U acts trivially on any K n, it is clear that ind G B K n ind G U K. The inverse image of ind G B K n under the isomorphism S(V ) ind G U K from above consists of all ψ S(V ) with ψ(tv) = t n ψ(v) for all t K and v V. This shows that we get isomorphisms { ind G B K n S n (V ) if n 0, 0 if n < 0. Simple modules for SL 2 (K). Consider again G = SL 2 (K). If char K = 0, then one can show that all S n (V ) with n N are simple G modules. [Using (2) above one checks that 3

4 each non-zero submodule contains some v i. Then (1) shows that the submodule contains all v j. Here one uses that all binomial coefficients are non-zero in K.] Furthermore, each simple G module is isomorphic to S n (V ) for a unique n. If char K = p > 0, then the same argument shows that all S n (V ) with 0 n < p are simple G modules. But one can quickly see that S p (V ) is not simple: The map ϕ: V S p (V ), f f p is semi-linear (i.e., satisfies ϕ(f + f ) = ϕ(f) + ϕ(f ) and ϕ(af) = a p ϕ(f) for all a K) and G equivariant. So its image is a G submodule of S p (V ) that is non-zero and not the whole module. Going back to the definition of a Frobenius twist one checks that ϕ is an isomorphism of G modules (V ) (1) KG X p 2 onto the submodule of S p (V ) generated by X p 2. For general n the characteristic 0 argument shows first that each non-zero submodule of S n (V ) contains some v i. Then the first formula in (1) implies that the submodule contains v 0 = X n 2. This shows that the submodule L(n) := KG v 0 = KG X n 2 of S n (V ) generated by X n 2 is simple, in fact, it is the only simple submodule of S n (V ). It then turns out that each simple G module is isomorphic to L(n) for exactly one n. One can describe L(n) alternatively as follows. Write n in p adic expansion: Then the map n = n 0 + n 1 p + n 2 p n r p r with 0 n i < p for all i. is an isomorphism of G modules f 0 f 1 f 2 f r f 0 f p 1 f p fr pr S n 0 (V ) S n 1 (V ) (1) S n 2 (V ) (2) S n r (V ) (r) L(n). Notations. Let G be a connected reductive group over K. Choose a maximal torus T in G. We want to generalise the SL 2 example. This involves some notation: X = Hom alg.gr (T, K ), the character group of T X = Hom alg.gr (K, T ), the cocharacter group of T,, the pairing X X Z such that λ(ϕ(t)) = t λ,ϕ for all t K Φ X, the root system of G with respect to T Φ = {α α Φ} X the corresponding dual roots Φ + a chosen system of positive roots 4

5 B and B + the Borel subgroups of G containing T that correspond to Φ + and Φ + respectively U and U + the unipotent radicals of these Borel subgroups. So we have semi-direct decompositions B = T U and B + = T U + as algebraic groups. Each λ X can be extended to a character λ: B K such that λ(tu) = λ(t) for all t T and u U. Denote by K λ the corresponding one dimensional B module. The general theory of connected soluble algebraic groups yields: Proposition: Each simple B module is isomorphic to K λ for a unique λ X. A corresponding statement holds for B + instead of B. Simple G modules. Any G module is by definition locally finite, i.e., a union of finite dimensional submodules. Therefore any simple G module V is finite dimensional. Considered as a B module it then has a composition series. This implies in particular that there is a surjective homomorphism of B modules from V onto a simple B module, hence by the proposition above onto some K λ with λ X. The basic property of induction implies then Hom G (V, ind G B K λ ) Hom B (V, K λ ) 0. Using the simplicity of V we get now: Claim 1: Each simple G module is isomorphic to a simple submodule of some ind G B K λ with λ X. In the case of SL 2 (K) this implies that any simple module is (as claimed above) isomorphic to some L(n). The next step towards a classification of the simple modules requires more subtle arguments that I cannot discuss here. They yield: Claim 2: ind G B K λ 0 λ dominant where the set X + of dominant characters is defined by X + = { λ X λ, α 0 for all α Φ + }. As a third step we want to see that each ind G B K λ with λ dominant contains a unique simple submodule. This is done by looking at fixed points for the subgroup U +. For any non-zero U + module M the space of fixed points M U + = { x M u x = x for all u U + } is non-zero as U + is connected and unipotent. The definition of ind G B K λ and of the G action on this space shows for any f (ind G B K λ ) U + that f(u 1 tu 2 ) = λ(t) 1 f(1) for all u 1 U +, t T, u 2 U. 5

6 So the restriction of f to the subset U + T U of G is determined by f(1). As U + T U is dense in G, it follows that f itself is determined by f(1). This implies that (ind G B K λ ) U + = K f λ where f λ K[G] is the unique regular function on G with f λ (u 1 tu 2 ) = λ(t) 1 for all u 1 U +, t T, u 2 U. Now every non-zero G submodule M of ind G B K λ satisfies M U + 0, hence f λ M and KG f λ M. It follows that L(λ) := KG f λ is the unique simple submodule of ind G B K λ. Theorem: Each simple G module is isomorphic to L(λ) for a unique λ X +. In order to get uniqueness one notes that L(λ) U + = Kf λ and that T acts on this subspace via λ. In case char K = 0 one can show that L(λ) = ind G B K λ for all λ X +. The examples above show that this fails in prime characteristic already for G = SL 2 (K). Steinberg s tensor product theorem. In order to simplify notation assume now that G is semi-simple and simply connected. Our choice of positive roots Φ + determines a system of simple roots. The corresponding dual roots form under our assumption a basis for the free abelian group X and X has then a dual basis consisting of fundamental weight. For any integer m > 0 the set X m = { λ X 0 λ, α < m for all simple roots α } X + consists of those characters whose coefficients with repect to the basis of fundamental weights belong to the interval from 0 to m 1. Suppose now that char K = p > 0. Any λ X + has a p adic expansion λ = λ 0 + pλ 1 + p 2 λ p r λ r with all λ i X p. Now Steinberg s tensor product theorem says (generalising the result for SL 2 ) that we have an isomorphism L(λ 0 ) L(λ 1 ) (1) L(λ 2 ) (2) L(λ r ) (r) L(λ). Note that each L(λ i ) ind G B K λi K[G] is a submodule of K[G]; similarly for L(λ). The isomorphism is then simply given by f 0 f 1 f 2 f r f 0 f p 1 f p fr pr Application to finite groups of Lie type. Assume again that G is semi-simple and simply connected. Suppose that char K = p > 0 and that G is defined over a finite subfield F q of K. (So q = p n for some n > 0.) The group G(F q ) of points of G over F q is then a finite group of Lie type. We can describe G(F q ) as the fixed points of a suitable Frobenius map on G. 6

7 In this situation a theorem due to Curtis for q = p and to Steinberg in general says: Theorem: Each L(λ) with λ X q is still simple when considered as a KG(F q ) module. Any simple KG(F q ) module is isomorphic to L(λ) for a unique λ X q. Note that the Ree and Suzuki groups are finite groups of Lie type, which are not of the form G(F q ). However, a modification of this theorem yields also a classification of their irreducible representations over an algebraically closed field of their defining characteristic. References. For a more detailed survey of the material in Lectures I and II see my article Modular representations of reductive groups in the proceedings of a conference in Beijing published under the title Group Theory, Beijing 1984 as volume 1185 of the Springer Lecture Notes in Mathematics. 7

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