Suppose finite groups and is an irreducible representation of . When is multiplicity-free (c.f. Gelfand pair)? Schur’s lemma tells us that this holds iff is commutative. If is the trivial representation, then the induced representation is essentially , and the endomorphism algebra is the double coset algebra . For general representation , we define the Hecke algebra This is easily seen isomorphic to the endomorphism algebra of and reduced to the double coset algebra if is trivial. Thus we would like to seek a criterion that enables us to show that is commutative. A group is commutative iff is a homomorphism. The key feature is that this is an involution. We have the following criterion (Theorem 1 of this article):
Suppose that there is an involution (i.e. anti-homomorphism) such that for all and (This ensures that induces an involution on ). Then is a Gelfand pair.
The main goal is to show that the Gelfand-Graev representation is multiplicity-free. The irreducible representations of this representation are called generic and include most of the cuspidal representations that are hard to access by means of parabolic induction (those are the representations whose restriction to contains the trivial representation).
The key tool is the Bruhat decomposition, from which we can easily deduce the following variant:
We have the following decomposition where is the set of monomial matrices (i.e. consist of exactly one nonzero entry in each row and each column).
This follows from the fact that where is the diagonal torus. Now if we take where is a nontrivial character, then there is an obvious involution on , which is reflecting about the anti-diagonal. The hard part is to check that for a double coset representative in the Hecke algebra. For details see Theorem 3 of the above article.