Last update: 3 September 2013
Let be semisimple algebras and let be a representation of Let be the induced representation of given by inducing from to The Hecke algebra is the centralizer of the action of on in
The following results follow from the double centralizer theory.
As bimodules where is an irreducible is an irreducible module, and the sum is over indexing the irreducible appearing in a decomposition of as a
The trace of the action of on is a nondegenerate trace on given in terms of the irreducible characters of by for all where the denotes the dimension of the irreducible indexed by As above the sum is over indexing the irreducible appearing in a decomposition of as a
The center of and the center of coincide in the following sense. The action of a minimal central idempotent on gives an endomorphism of which is a minimal central idempotent of
Let be an idempotent of is an with action of by left multiplication. Then as and as algebras.
We have that and one can check that the map is an isomorphism of
Let so that is an operator on and let be such that Then for each we have So is equivalent to right multiplication on by Conversely, for every one has that right multiplication by on commutes with left multiplication by elements of This shows that
Let be a finite group and let be a subgroup of Let be the group algebra of and let be the group algebra of Let be the idempotent Then is an and is the given by inducing the representation to The Hecke algebra is also denoted and we have that
Let be a set of representatives of the double cosets of in so that and the union is disjoint. Define, for each
The ring of functions on constant on double cosets is the set of functions such that for each for all The multiplication is by convolution, for functions the product is given by Let and let denote the function Let be a set of representatives of the double cosets of in The functions form a basis of
|c)||if and are in the same double coset of in|
|g)||The elements form a basis of the Hecke algebra|
|h)||is isomorphic to the ring of functions constant on double cosets of where the multiplication is given by convolution.|
Let suppose that with Given if and only if Conversely if with then This shows that Now if and only if So
The number of terms on the left side of a) is The number of terms on the right side of a) is b) follows.
c) is clear from the definition since d) is just the definition of e) follows from a) and b).
The elements are linearly independent as they are linearly independent in It follows from e) and () that these elements span This proves g).
The map given by is an isomorphism.
A basis of the representation is given by the elements where the are a set of coset representatives of
|a)||The trace of the action of on is given by|
|b)||Let be the linear functional on given by taking the coefficient of the identity, i.e., Then is a nondegenerate trace on|
|c)||The dual basis to the with respect to the trace is|
Let be a set of representatives for the left cosets of in Then The fact that is a trace follows from a) as
The dimensions of the irreducible representations of appearing in are given by
From Theorem 3.9 in Dissertation Chapter 1
A Tits system is a quadruple where is a group, is a subgroup, is a subgroup, is a set of elements of such that
|(T2)||is normal in consists of elements of order 2 in which generate|
|(T3)||For each and each|
The following terminology is standard. is a Borel subgroup of the subgroup is the torus of and is the Weyl group of the Tits system. For each the Schubert cell associated to is the double coset Since the elements of generate each element can be written as a word where If this product is of minimal length then is a reduced word for and the length is the length of the element
Axiom (T3) implies that for every and We always have that
Let and let Then
The proof is by induction on The statement is trivial if Assume Then
Bruhat decomposition. One has the following double coset decomposition of where the union is a disjoint union.
Let If then since if then then Proposition () shows that It is clear the contains and Since is closed under the group operations and contains and axiom (T1) implies that
It remains to show that the union is disjoint, i.e., that if and then This is by induction on the length of We can assume that
Let such that Then So and But if it were possible to have then we would have Thus
This is a copy of lectures notes for 18.318 Topics in Combinatorics, MIT Spring 1992, Prof. G.-C. Rota, given by Arun Ram.