Hecke algebras

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 3 September 2013

Hecke algebras

Let $A\subseteq B$ be semisimple algebras and let $V$ be a representation of $A\text{.}$ Let $$V{\uparrow }_A^B=B{\otimes }_{A}V,$$ be the induced representation of $B$ given by inducing from $A$ to $B\text{.}$ The Hecke algebra $\mathscr{H}(A,B,V)$ is the centralizer of the action of $B$ on $V{\uparrow }_A^B$ in $\text{End}\left(V{\uparrow }_A^B\right),$ $$\mathscr{H}(A,B,V)=\{C\in \text{End}\left(V{\uparrow }_A^B\right)\hspace{0.17em}|\hspace{0.17em}bCv=bCv\hspace{0.17em}\text{for all}\hspace{0.17em}b\in B,v\in V{\uparrow }_A^B\}\text{.}$$

The following results follow from the double centralizer theory.

As $(B,\mathscr{H})$ bimodules $$V{\uparrow }_A^B\cong \underset{\lambda }{⨁}{B}^{\lambda }\otimes H_{\lambda }$$ where $B^{\lambda }$ is an irreducible $B\text{-module,}$ $H_{\lambda }$ is an irreducible $\mathscr{H}$ module, and the sum is over $\lambda$ indexing the irreducible $B\text{-modules}$ appearing in a decomposition of $V{\uparrow }_A^B$ as a $B\text{-module.}$

The trace $\text{tr}:\mathscr{H}\to ℂ$ of the action of $\mathscr{H}$ on $V{\uparrow }_A^B$ is a nondegenerate trace on $\mathscr{H}$ given in terms of the irreducible characters ${\eta }_{\lambda }$ of $\mathscr{H}$ by $$\text{tr}\left(h\right)=\sum _{\lambda }{d}_{\lambda }{\eta }_{\lambda }\left(h\right),$$ for all $h\in \mathscr{H},$ 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 $V{\uparrow }_A^B$ as a $B\text{-module.}$

The center of $\mathscr{H}$ and the center of $B$ coincide in the following sense. The action of a minimal central idempotent $z_{\lambda }\in B$ on $V{\uparrow }_A^B$ gives an endomorphism of $V{\uparrow }_A^B$ 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 $$Ap{\uparrow }_A^B\cong Bp$$ as $B\text{-modules}$ and $$\mathscr{H}(A,B,Ap)\cong pBp$$ as algebras.

		Proof.
	We have that $Ap{\uparrow }_A^B=B{\otimes }_{A}Ap$ and one can check that the map $$\begin{array}{cccc}\varphi :& B{\otimes }_{A}Ap& ⟶& Bp\\ & b\otimes ap& ⟼& bap\end{array}$$ is an isomorphism of $B\text{-modules.}$ Let $C\in \mathscr{H}(A,B,Ap)$ 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 $$Cbp=Cbpp=bpCp=bpcp=bppcp\text{.}$$ 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 $\mathscr{H}(A,B,Ap)\cong pBp\text{.}$ $\square$

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 $$p=\frac{1}{\left|H\right|}\sum _{h\in H}h\text{.}$$ Then $$Ap=ℂ\left[H\right]p$$ is an $H\text{-module}$ and $$Bp=ℂ\left[G\right]p$$ is the $G\text{-module}$ given by inducing the representation $Ap$ to $G\text{.}$ The Hecke algebra $\mathscr{H}(A,B,Ap)$ is also denoted $\mathscr{H}(H,G,1),$ and we have that $$\mathscr{H}(H,G,1)\cong pBp\text{.}$$

Let $w_1,w_2,\dots ,w_k$ be a set of representatives of the double cosets of $H$ in $G$ so that $$G=\bigcup _{w_i}H{w}_{i}H,$$ and the union is disjoint. Define, for each $g\in G,$ $$T_g=T_{H}gH=\frac{1}{\left|H\right|}\sum _{x\in HgH}x\text{.}$$

The ring of functions $𝒞$ on $G$ constant on double cosets is the set of functions $f:G\to ℂ$ such that for each $g\in G,$ $$f\left(h_{1}g{h}_2\right)=f\left(g\right),$$ 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 $$(f_1*f_2)\left(g\right)=\sum _{t\in G}{f}_1\left(t\right)f_2\left(t^{-1}g\right)\text{.}$$ Let $g\in G$ and let $f_g$ denote the function $$f_g\left(g\prime \right)=\{\begin{array}{cc}\frac{1}{\left|H\right|}& \text{if}\hspace{0.17em}g\prime \in HgH\text{;}\\ 0& \text{otherwise.}\end{array}$$ 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)	$$\sum _{h_1,h_2\in H}{h}_{1}g{h}_2=|H\cap gH{g}^{-1}|\sum _{x\in HgH}x\text{.}$$
b)	$$\frac{\left|HgH\right|}{\left|H\right|}=\frac{\left|H\right|}{|H\cap gH{g}^{-1}|}\text{.}$$
c)	$T_g=T_{g\prime }$ if $g$ and $g\prime$ are in the same double coset of $H$ in $G\text{.}$
d)	$$T_g=\frac{1}{\left|H\right|}\sum _{x\in HgH}x\text{.}$$
e)	$$T_g=\frac{\left|HgH\right|}{\left|H\right|}pgp\text{.}$$
f)	$$T_g=\frac{1}{\left|H\right||H\cap gH{g}^{-1}|}\sum _{h_1,h_2\in H}{h}_{1}g{h}_2\text{.}$$
g)	The elements $T_{w_i}$ form a basis of the Hecke algebra $\mathscr{H}\text{.}$
h)	$\mathscr{H}$ 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 $$\left|\{h_{1}g{h}_2=x\}\right|=\left|\{h_{1}g{h}_2=g\}\right|\text{.}$$ Now $h_{1}g{h}_2=g$ if and only if $h_1=g{h}_2{g}^{-1}\text{.}$ So $$\left|\{h_{1}g{h}_2=g\}\right|=|H\cap gH{g}^{-1}|\text{.}$$ The number of terms on the left side of a) is ${\left|H\right|}^2\text{.}$ The number of terms on the right side of a) is $|H\cap gH{g}^{-1}|\left|HgH\right|\text{.}$ 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 $\mathscr{H}\text{.}$ This proves g). The map $\varphi :\mathscr{H}\to 𝒞$ given by $\varphi \left(T_g\right)=f_g$ is an isomorphism. $\square$

Let $g,t\in G\text{.}$ $$T_g{T}_t=\sum _{w_k}{c}_{gt}^k{T}_{w_k},$$ where $$c_{gt}^k=\left|\right|$$

		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 $\mathscr{H}$ on $Bp$ is given by $\text{tr}\left(T_g\right)=\frac{\left|G\right|}{\left|H\right|}{\delta }_{T_g{T}_1}\text{.}$
b)	Let $\overrightarrow{t}$ be the linear functional on $\mathscr{H}$ given by taking the coefficient of the identity, i.e., $$\overrightarrow{t}\left(T_g\right)={\delta }_{T_g{T}_1}\text{.}$$ Then $\overrightarrow{t}$ is a nondegenerate trace on $\mathscr{H}\text{.}$
c)	The dual basis to the $T_{w_k}$ with respect to the trace $\overrightarrow{t}$ is $\left|\right|T_{w_k^{-1}}\text{.}$

		Proof.
	Let ${\tau }_i$ be a set of representatives for the left cosets ${\tau }_{i}H,$ of $H$ in $G\text{.}$ Then $$\text{tr}\left(T_g\right)=\sum _{{\tau }_i}{\tau }_{i}p{T}_g{|}_{{\tau }_{i}p}=\sum _{{\tau }_i}\frac{\left|HgH\right|}{\left|H\right|}{\tau }_{i}pgp{|}_{{\tau }_{i}p}=\sum _{{\tau }_{i}p}\frac{\left|H\right|}{|H\cap gH{g}^{-1}|}pgp{|}_p=|G/H|\frac{\left|H\right|}{|H\cap gH{g}^{-1}|}pgp{|}_p=\{\begin{array}{cc}0,& \text{if}\hspace{0.17em}g\notin H\text{;}\\ \frac{\left|G\right|}{\left|H\right|}& \text{if}\hspace{0.17em}g\in H\text{.}\end{array}$$ The fact that $\overrightarrow{t}$ is a trace follows from a) as $\overrightarrow{t}=\left|H\right|/\left|G\right|\text{tr}\text{.}$ $$\overrightarrow{t}\left(T_v{T}_w\right)=\overrightarrow{t}\left(\sum _{w_{\lambda }}{c}_{vw}^k{T}_{w_k}\right)=c_{vw}^1=\left|\right|=\{\begin{array}{cc}\left|\right|,& \text{if}\hspace{0.17em}v\in H{w}^{-1}H\text{;}\\ 0,& \text{otherwise.}\end{array}$$ $\square$

The dimensions $t_{\lambda }$ of the irreducible representations of $B$ appearing in $Bp$ are given by $$t_{\lambda }=\frac{d_{\lambda }\left|\right|}{{\sum }_{w_k}\left|\right|{\chi }^{\lambda }\left(T_{w_k}\right){\chi }^{\lambda }\left(T_{w_k^{-1}}\right)}\text{.}$$

		Proof.
	From Theorem 3.9 in Dissertation Chapter 1 $$t_{\lambda }=\frac{d_{\lambda }}{{\sum }_{w_k}{\chi }^{\lambda }\left(T_{w_k}\right){\chi }^{\lambda }\left(T_{w_k^{-1}}\right)}\text{.}$$ $\square$

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 $W=N/(N\cap N),$ 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 $w\in W$ $$sBw\subseteq BswB\cap BwB\text{;}$$
(T4)	For each $s\in S$ $$sBs\not\subset B\text{.}$$

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 $$C\left(w\right)=BwB\text{.}$$ Since the elements of $S$ generate $W$ each element $w\in W$ can be written as a word $$s_{i_1}{s}_{i_2}\cdots s_{i_p}=w,$$ 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 $w\in W$ $$C\left(s\right)C\left(w\right)=\{\begin{array}{cc}C\left(sw\right)\cup C\left(w\right),& \text{if}\hspace{0.17em}C\left(w\right)\subset C\left(sw\right)\text{;}\\ C\left(sw\right),& \text{if}\hspace{0.17em}C\left(w\right)\not\subset C\left(sw\right)\text{.}\end{array}$$ We always have that $$C\left(sw\right)\subseteq C\left(w\right)\text{.}$$

Let $s_1,s_2,\dots ,s_q\in S,$ and let $w\in W\text{.}$ Then $$C\left(s_1{s}_2\cdots s_q\right)C\left(w\right)\subseteq \bigcup _{\{i_1,i_2,\dots ,i_p\}\subset [1,q]}C\left(s_{i_1}\cdots s_{i_p}w\right)\text{.}$$

		Proof.
	The proof is by induction on $q\text{.}$ The statement is trivial if $q=0\text{.}$ Assume $q>0\text{.}$ Then $$\begin{array}{ccc}C\left(s_1{s}_2\cdots s_q\right)C\left(w\right)& \subseteq & C\left(s_1\right)C\left(s_2\cdots s_q\right)C\left(w\right)\\ & \subseteq & C\left(s_1\right)\bigcup _{\{j_1,j_2,\cdots ,j_p\}\subseteq [2,q]}C\left(s_{j_1}{s}_{j_2}\cdots c_{j_p}w\right)\\ & \subseteq & \bigcup _{\{j_1,\cdots ,j_p\}}C\left(s_{j_1}\cdots s_{j_p}w\right)\cup C\left(s_1{s}_{j_1}\cdots s_{j_p}w\right)\text{.}\end{array}$$ $\square$

Bruhat decomposition. One has the following double coset decomposition of $G\text{.}$ $$G=\bigcup _{w\in W}BwB,$$ 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 $N\subseteq N/(B\cap N)·B=WB\subseteq BWB\text{.}$ 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 $$\begin{array}{c}\ell \left(sw\right)\le \ell \left(sw\prime \right)\\ \ell \left(sw\right)\le \ell \left(w\prime \right)\end{array}$$ 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 $$C\left(sw\right)\subseteq C\left(s\right)C\left(w\right)=C\left(s\right)C\left(w\prime \right)\subseteq C\left(sw\prime \right)\cup C\left(w\prime \right)\text{.}$$ 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.

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