## Hecke algebras

Last update: 3 September 2013

## Hecke algebras

Let $A\subseteq B$ be semisimple algebras and let $V$ be a representation of $A\text{.}$ Let $V↑AB= B⊗AV,$ be the induced representation of $B$ given by inducing from $A$ to $B\text{.}$ The Hecke algebra $ℋ\left(A,B,V\right)$ is the centralizer of the action of $B$ on $V{↑}_{A}^{B}$ in $\text{End}\left(V{↑}_{A}^{B}\right),$ $ℋ(A,B,V)= { C∈End(V↑AB) | bCv=bCv for all b∈B,v∈V ↑AB } .$

The following results follow from the double centralizer theory.

As $\left(B,ℋ\right)$ bimodules $V↑AB≅⨁λ Bλ⊗Hλ$ where ${B}^{\lambda }$ is an irreducible $B\text{-module,}$ ${H}_{\lambda }$ is an irreducible $ℋ$ module, and the sum is over $\lambda$ indexing the irreducible $B\text{-modules}$ appearing in a decomposition of $V{↑}_{A}^{B}$ as a $B\text{-module.}$

The trace $\text{tr}:ℋ\to ℂ$ of the action of $ℋ$ on $V{↑}_{A}^{B}$ is a nondegenerate trace on $ℋ$ given in terms of the irreducible characters ${\eta }_{\lambda }$ of $ℋ$ by $tr(h)= ∑λdλ ηλ(h),$ for all $h\in ℋ,$ where the ${d}_{\lambda }$ denotes the dimension of the irreducible $B\text{-module}$ indexed by $\lambda \text{.}$ As above the sum is over $\lambda$ indexing the irreducible $B\text{-modules}$ appearing in a decomposition of $V{↑}_{A}^{B}$ as a $B\text{-module.}$

The center of $ℋ$ and the center of $B$ coincide in the following sense. The action of a minimal central idempotent ${z}_{\lambda }\in B$ on $V{↑}_{A}^{B}$ gives an endomorphism of $V{↑}_{A}^{B}$ which is a minimal central idempotent of $H\text{.}$

Idempotent representations

Let $p$ be an idempotent of $A\text{.}$ $Ap$ is an $A\text{-module}$ with action of $A$ by left multiplication. Then $Ap↑AB≅Bp$ as $B\text{-modules}$ and $ℋ(A,B,Ap)≅ pBp$ as algebras.

 Proof. We have that $Ap{↑}_{A}^{B}=B{\otimes }_{A}Ap$ and one can check that the map $ϕ: B⊗AAp ⟶ Bp b⊗ap ⟼ bap$ is an isomorphism of $B\text{-modules.}$ Let $C\in ℋ\left(A,B,Ap\right)$ so that $C$ is an operator on $Bp,$ and let $c\in B$ be such that $Cp=cp\text{.}$ Then for each $b\in B$ we have $Cbp=Cbpp=bpCp= bpcp=bppcp.$ So $C$ is equivalent to right multiplication on $Bp$ by $pcp\text{.}$ Conversely, for every $c\in B,$ one has that right multiplication by $pcp$ on $Bp$ commutes with left multiplication by elements of $B\text{.}$ This shows that $ℋ\left(A,B,Ap\right)\cong pBp\text{.}$ $\square$

Finite groups

Let $G$ be a finite group and let $H$ be a subgroup of $G\text{.}$ Let $A=ℂ\left[H\right]$ be the group algebra of $H$ and let $B=ℂ\left[G\right]$ be the group algebra of $G\text{.}$ Let $p\in A$ be the idempotent $p=1|H| ∑h∈Hh.$ Then $Ap=ℂ[H]p$ is an $H\text{-module}$ and $Bp=ℂ[G]p$ is the $G\text{-module}$ given by inducing the representation $Ap$ to $G\text{.}$ The Hecke algebra $ℋ\left(A,B,Ap\right)$ is also denoted $ℋ\left(H,G,1\right),$ and we have that $ℋ(H,G,1)≅ pBp.$

Let ${w}_{1},{w}_{2},\dots ,{w}_{k}$ 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 $g\in G,$ $Tg=THgH= 1|H| ∑x∈HgHx.$

The ring of functions $𝒞$ on $G$ constant on double cosets is the set of functions $f:G\to ℂ$ such that for each $g\in G,$ $f(h1gh2)= f(g),$ for all ${h}_{1},{h}_{2}\in H\text{.}$ The multiplication is by convolution, for functions ${f}_{1},{f}_{2}$ the product ${f}_{1}*{f}_{2}$ is given by $(f1*f2)(g)= ∑t∈Gf1(t) f2(t-1g).$ Let $g\in G$ and let ${f}_{g}$ denote the function $fg(g′)= { 1|H| if g′∈HgH; 0 otherwise.$ Let ${w}_{1},{w}_{2},\dots ,{w}_{k}$ be a set of representatives of the double cosets of $H$ in $G\text{.}$ The functions ${f}_{{w}_{i}}$ form a basis of $𝒞\text{.}$

 a) $∑h1,h2∈H h1gh2= |H∩gHg-1| ∑x∈HgHx.$ b) $|HgH||H|= |H||H∩gHg-1| .$ c) ${T}_{g}={T}_{g\prime }$ if $g$ and $g\prime$ are in the same double coset of $H$ in $G\text{.}$ d) $Tg=1|H| ∑x∈HgHx.$ e) $Tg=|HgH||H| pgp.$ f) $Tg= 1|H||H∩gHg-1| ∑h1,h2∈H h1gh2.$ g) The elements ${T}_{{w}_{i}}$ form a basis of the Hecke algebra $ℋ\text{.}$ h) $ℋ$ is isomorphic to the ring of functions constant on double cosets of $H$ where the multiplication is given by convolution.

 Proof. Let $x\in HgH\text{;}$ suppose that $x={b}_{1}g{b}_{2}$ with ${b}_{1},{b}_{2}\in H\text{.}$ Given ${h}_{1},{h}_{2}\in H,$ ${h}_{1}g{h}_{2}=x$ if and only if ${b}_{1}^{-1}{h}_{1}g{h}_{2}{b}_{2}^{-1}=g\text{.}$ Conversely if ${c}_{1}g{c}_{2}=1$ with ${c}_{1},{c}_{2}\in H$ then ${b}_{1}{c}_{1}g{c}_{2}{b}_{2}=x\text{.}$ This shows that $|{h1gh2=x}| =|{h1gh2=g}|.$ Now ${h}_{1}g{h}_{2}=g$ if and only if ${h}_{1}=g{h}_{2}{g}^{-1}\text{.}$ So $|{h1gh2=g}|= |H∩gHg-1|.$ The number of terms on the left side of a) is ${|H|}^{2}\text{.}$ The number of terms on the right side of a) is $|H\cap gH{g}^{-1}||HgH|\text{.}$ b) follows. c) is clear from the definition since $HgH=Hg\prime H\text{.}$ d) is just the definition of ${T}_{g}\text{.}$ e) follows from a) and b). The elements ${T}_{{w}_{i}}$ are linearly independent as they are linearly independent in $ℂG\text{.}$ It follows from e) and () that these elements span $ℋ\text{.}$ This proves g). The map $\varphi :ℋ\to 𝒞$ given by $\varphi \left({T}_{g}\right)={f}_{g}$ is an isomorphism. $\square$

Let $g,t\in G\text{.}$ $TgTt= ∑wkcgtk Twk,$ where $cgtk=||$

 Proof. $\square$

A basis of the representation $Bp$ is given by the elements ${g}_{i}p$ where the ${g}_{i}$ are a set of coset representatives of $G/H\text{.}$

 a) The trace of the action of $ℋ$ on $Bp$ is given by $\text{tr}\left({T}_{g}\right)=\frac{|G|}{|H|}{\delta }_{{T}_{g}{T}_{1}}\text{.}$ b) Let $\stackrel{\to }{t}$ be the linear functional on $ℋ$ given by taking the coefficient of the identity, i.e., $t→(Tg)= δTgT1.$ Then $\stackrel{\to }{t}$ is a nondegenerate trace on $ℋ\text{.}$ c) The dual basis to the ${T}_{{w}_{k}}$ with respect to the trace $\stackrel{\to }{t}$ is $||{T}_{{w}_{k}^{-1}}\text{.}$

 Proof. Let ${\tau }_{i}$ be a set of representatives for the left cosets ${\tau }_{i}H,$ of $H$ in $G\text{.}$ Then $tr(Tg) = ∑τiτip Tg|τip = ∑τi |HgH||H| τipgp|τip = ∑τip |H||H∩gHg-1| pgp|p = |G/H| |H||H∩gHg-1| pgp|p = { 0, if g∉H; |G||H| if g∈H.$ The fact that $\stackrel{\to }{t}$ is a trace follows from a) as $\stackrel{\to }{t}=|H|/|G|\text{tr}\text{.}$ $t→(TvTw) = t→(∑wλcvwkTwk) = cvw1 = || = { ||, if v∈Hw-1H; 0, otherwise.$ $\square$

The dimensions ${t}_{\lambda }$ of the irreducible representations of $B$ appearing in $Bp$ are given by $tλ= dλ|| ∑wk|| χλ(Twk) χλ(Twk-1) .$

 Proof. From Theorem 3.9 in Dissertation Chapter 1 $tλ= dλ ∑wkχλ (Twk) χλ(Twk-1) .$ $\square$

## Tits systems

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

 (T1) $B$ and $N$ generate $G\text{;}$ (T2) $B\cap N$ is normal in $N,$ $S$ consists of elements of order 2 in $W$ which generate $W\text{;}$ (T3) For each $s\in S$ and each $w\in W$ $sBw⊆BswB∩ BwB;$ (T4) For each $s\in S$ $sBs⊄B.$

The following terminology is standard. $B$ is a Borel subgroup of $G,$ the subgroup $B\cap N=T$ is the torus $T\subseteq B$ of $G,$ and $W$ is the Weyl group of the Tits system. For each $w\in W,$ the Schubert cell associated to $w$ is the double coset $C(w)=BwB.$ Since the elements of $S$ generate $W$ each element $w\in W$ can be written as a word $si1si2⋯sip =w,$ where ${s}_{{i}_{1}},\dots {s}_{{i}_{p}}\in W\text{.}$ If this product is of minimal length then ${s}_{{i}_{1}}\cdots {s}_{{i}_{p}}$ is a reduced word for $w$ and the length $p$ is the length $\ell \left(w\right)$ of the element $w\text{.}$

Axiom (T3) implies that for every $s\in S$ and $w\in W$ $C(s)C(w)= { C(sw)∪C(w), if C(w)⊂ C(sw); C(sw), if C(w)⊄C(sw).$ We always have that $C(sw)⊆C(w).$

Let ${s}_{1},{s}_{2},\dots ,{s}_{q}\in S,$ and let $w\in W\text{.}$ Then $C(s1s2⋯sq) C(w)⊆ ⋃{i1,i2,…,ip}⊂[1,q] C(si1⋯sipw).$

 Proof. The proof is by induction on $q\text{.}$ The statement is trivial if $q=0\text{.}$ Assume $q>0\text{.}$ Then $C(s1s2⋯sq) C(w) ⊆ C(s1) C(s2⋯sq) C(w) ⊆ C(s1) ⋃{j1,j2,⋯,jp}⊆[2,q] C(sj1sj2⋯cjpw) ⊆ ⋃{j1,⋯,jp} C(sj1⋯sjpw) ∪C(s1sj1⋯sjpw).$ $\square$

Bruhat decomposition. One has the following double coset decomposition of $G\text{.}$ $G=⋃w∈WBwB,$ where the union is a disjoint union.

 Proof. Let $BWB={\cup }_{w\in W}BwB\text{.}$ If $x\in BWB$ then ${x}^{-1}\in BWB$ since if $x\in BwB$ then ${x}^{-1}\in B{w}^{-1}B\text{.}$ $x,y\in BWB$ then Proposition () shows that $xy\in BWB\text{.}$ It is clear the $BWB$ contains $B$ and $N\subseteq N/\left(B\cap N\right)·B=WB\subseteq BWB\text{.}$ Since $BWB$ is closed under the group operations and contains $B$ and $N$ axiom (T1) implies that $BWB=G\text{.}$ It remains to show that the union is disjoint, i.e., that if $w,w\prime \in W$ and $w\ne w\prime$ then $C\left(w\right)\ne C\left(w\prime \right)\text{.}$ This is by induction on the length $\ell \left(w\right)$ of $w\text{.}$ We can assume that $q=\ell \left(w\right)\le \ell \left(w\prime \right)\text{.}$ Let $s\in S$ such that $\ell \left(sw\right)\le \ell \left(w\right)\text{.}$ Then $ℓ(sw)≤ ℓ(sw′) ℓ(sw)≤ ℓ(w′)$ So $C\left(sw\right)\ne C\left(w\prime \right)$ and $C\left(sw\right)\ne C\left(sw\prime \right)\text{.}$ But if it were possible to have $C\left(w\right)=C\left(w\prime \right)$ then we would have $C(sw)⊆C(s) C(w)=C(s)C (w′)⊆C(sw′) ∪C(w′).$ Thus $C\left(w\right)\ne C\left(w\prime \right)\text{.}$ $\square$

## 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.