By John R. Stallings

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3 There is a natural isomorphism d Hom of n -modules, n for every left n⊗ M M -module M. 2. 10]. 4 There is a natural isomorphism indn M for every finite dimensional Proof The functor indn Hence it is isomorphic to Hom indn d M -module M. -mod → n -mod is right adjoint to resn . n ? by uniqueness of adjoint functors. 3 (with swapped and d replaced by d−1 ). 8 Intertwining elements We will need certain elements of i < n, define i n which go back Cherednik. 22) means that for every w ∈ Sn we obtain a well-defined element w ∈ n , namely, w = i1 im where w = si1 sim is any reduced expression for w.

If is such an orbit we set = a for any a ∈ . Now let M be a finite dimensional n -module and ∈ F n /∼. We let M denote the generalized eigenspace of M over Z n that corresponds to the central character , that is M = v∈M z− z k v = 0 for all z ∈ Z n and k 0 First results on 38 n -modules Observe this is an n -submodule of M. Now, for any a ∈ F n with a ∈ , · · · L an via the central character . 1. decomposes as M= M ∈F n /∼ as an n -module. 2 shows that every such central character does arise in some finite dimensional n -module.

To see that L a Frobenius reciprocity and the fact just proved that L an is irreducible. (ii) The fact that all composition factors of resn L an are isomorphic · · · L a r follows by formal characters and (i). To see that to L a 1 n n ⊗ L of res L an soc res L a is irreducible, note that the submodule · · · L a l . This module is irreducible, and so it is isomorphic to L a 1 -submodule is contained in the socle. Conversely, let M be an irreducible n of L a . 1 as in the proof of (i), we see that M must ⊗ L.