Hecke algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia

Last update: 3 September 2013

Hecke algebras

Let AB be semisimple algebras and let V be a representation of A. Let VAB= BAV, be the induced representation of B given by inducing from A to B. The Hecke algebra (A,B,V) is the centralizer of the action of B on VAB in End(VAB), (A,B,V)= { CEnd(VAB) |bCv=bCv for allbB,vV AB } .

The following results follow from the double centralizer theory.

As (B,) bimodules VABλ BλHλ where Bλ is an irreducible B-module, Hλ is an irreducible module, and the sum is over λ indexing the irreducible B-modules appearing in a decomposition of VAB as a B-module.

The trace tr: of the action of on VAB is a nondegenerate trace on given in terms of the irreducible characters ηλ of by tr(h)= λdλ ηλ(h), for all h, where the dλ denotes the dimension of the irreducible B-module indexed by λ. As above the sum is over λ indexing the irreducible B-modules appearing in a decomposition of VAB as a B-module.

The center of and the center of B coincide in the following sense. The action of a minimal central idempotent zλB on VAB gives an endomorphism of VAB which is a minimal central idempotent of H.

Idempotent representations

Let p be an idempotent of A. Ap is an A-module with action of A by left multiplication. Then ApABBp as B-modules and (A,B,Ap) pBp as algebras.


We have that ApAB=BAAp and one can check that the map ϕ: BAAp Bp bap bap is an isomorphism of B-modules.

Let C(A,B,Ap) so that C is an operator on Bp, and let cB be such that Cp=cp. Then for each bB we have Cbp=Cbpp=bpCp= bpcp=bppcp. So C is equivalent to right multiplication on Bp by pcp. Conversely, for every cB, one has that right multiplication by pcp on Bp commutes with left multiplication by elements of B. This shows that (A,B,Ap)pBp.

Finite groups

Let G be a finite group and let H be a subgroup of G. Let A=[H] be the group algebra of H and let B=[G] be the group algebra of G. Let pA be the idempotent p=1|H| hHh. Then Ap=[H]p is an H-module and Bp=[G]p is the G-module given by inducing the representation Ap to G. The Hecke algebra (A,B,Ap) is also denoted (H,G,1), and we have that (H,G,1) pBp.

Let w1,w2,,wk be a set of representatives of the double cosets of H in G so that G=wiHwiH, and the union is disjoint. Define, for each gG, Tg=THgH= 1|H| xHgHx.

The ring of functions 𝒞 on G constant on double cosets is the set of functions f:G such that for each gG, f(h1gh2)= f(g), for all h1,h2H. The multiplication is by convolution, for functions f1,f2 the product f1*f2 is given by (f1*f2)(g)= tGf1(t) f2(t-1g). Let gG and let fg denote the function fg(g)= { 1|H| ifgHgH; 0 otherwise. Let w1,w2,,wk be a set of representatives of the double cosets of H in G. The functions fwi form a basis of 𝒞.

a) h1,h2H h1gh2= |HgHg-1| xHgHx.
b) |HgH||H|= |H||HgHg-1| .
c) Tg=Tg if g and g are in the same double coset of H in G.
d) Tg=1|H| xHgHx.
e) Tg=|HgH||H| pgp.
f) Tg= 1|H||HgHg-1| h1,h2H h1gh2.
g) The elements Twi form a basis of the Hecke algebra .
h) is isomorphic to the ring of functions constant on double cosets of H where the multiplication is given by convolution.


Let xHgH; suppose that x=b1gb2 with b1,b2H. Given h1,h2H, h1gh2=x if and only if b1-1h1gh2b2-1=g. Conversely if c1gc2=1 with c1,c2H then b1c1gc2b2=x. This shows that |{h1gh2=x}| =|{h1gh2=g}|. Now h1gh2=g if and only if h1=gh2g-1. So |{h1gh2=g}|= |HgHg-1|.

The number of terms on the left side of a) is |H|2. The number of terms on the right side of a) is |HgHg-1||HgH|. b) follows.

c) is clear from the definition since HgH=HgH. d) is just the definition of Tg. e) follows from a) and b).

The elements Twi are linearly independent as they are linearly independent in G. It follows from e) and () that these elements span . This proves g).

The map ϕ:𝒞 given by ϕ(Tg)=fg is an isomorphism.

Let g,tG. TgTt= wkcgtk Twk, where cgtk=||


A basis of the representation Bp is given by the elements gip where the gi are a set of coset representatives of G/H.

a) The trace of the action of on Bp is given by tr(Tg)=|G||H|δTgT1.
b) Let t be the linear functional on given by taking the coefficient of the identity, i.e., t(Tg)= δTgT1. Then t is a nondegenerate trace on .
c) The dual basis to the Twk with respect to the trace t is ||Twk-1.


Let τi be a set of representatives for the left cosets τiH, of H in G. Then tr(Tg) = τiτip Tg|τip = τi |HgH||H| τipgp|τip = τip |H||HgHg-1| pgp|p = |G/H| |H||HgHg-1| pgp|p = { 0, ifgH; |G||H| ifgH. The fact that t is a trace follows from a) as t=|H|/|G|tr. t(TvTw) = t(wλcvwkTwk) = cvw1 = || = { ||, ifvHw-1H; 0, otherwise.

The dimensions tλ of the irreducible representations of B appearing in Bp are given by tλ= dλ|| wk|| χλ(Twk) χλ(Twk-1) .


From Theorem 3.9 in Dissertation Chapter 1 tλ= dλ wkχλ (Twk) χλ(Twk-1) .

Tits systems

A Tits system is a quadruple (G,B,N,S) where G is a group, B is a subgroup, N is a subgroup, S is a set of elements of W=N/(NN), such that

(T1) B and N generate G;
(T2) BN is normal in N, S consists of elements of order 2 in W which generate W;
(T3) For each sS and each wW sBwBswB BwB;
(T4) For each sS sBsB.

The following terminology is standard. B is a Borel subgroup of G, the subgroup BN=T is the torus TB of G, and W is the Weyl group of the Tits system. For each wW, the Schubert cell associated to w is the double coset C(w)=BwB. Since the elements of S generate W each element wW can be written as a word si1si2sip =w, where si1,sipW. If this product is of minimal length then si1sip is a reduced word for w and the length p is the length (w) of the element w.

Axiom (T3) implies that for every sS and wW C(s)C(w)= { C(sw)C(w), ifC(w) C(sw); C(sw), ifC(w)C(sw). We always have that C(sw)C(w).

Let s1,s2,,sqS, and let wW. Then C(s1s2sq) C(w) {i1,i2,,ip}[1,q] C(si1sipw).


The proof is by induction on q. The statement is trivial if q=0. Assume q>0. Then C(s1s2sq) C(w) C(s1) C(s2sq) C(w) C(s1) {j1,j2,,jp}[2,q] C(sj1sj2cjpw) {j1,,jp} C(sj1sjpw) C(s1sj1sjpw).

Bruhat decomposition. One has the following double coset decomposition of G. G=wWBwB, where the union is a disjoint union.


Let BWB=wWBwB. If xBWB then x-1BWB since if xBwB then x-1Bw-1B. x,yBWB then Proposition () shows that xyBWB. It is clear the BWB contains B and NN/(BN)·B=WBBWB. Since BWB is closed under the group operations and contains B and N axiom (T1) implies that BWB=G.

It remains to show that the union is disjoint, i.e., that if w,wW and ww then C(w)C(w). This is by induction on the length (w) of w. We can assume that q=(w)(w).

Let sS such that (sw)(w). Then (sw) (sw) (sw) (w) So C(sw)C(w) and C(sw)C(sw). But if it were possible to have C(w)=C(w) then we would have C(sw)C(s) C(w)=C(s)C (w)C(sw) C(w). Thus C(w)C(w).

Notes and References

This is a copy of lectures notes for 18.318 Topics in Combinatorics, MIT Spring 1992, Prof. G.-C. Rota, given by Arun Ram.

page history