Hecke algebra generalities

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

Last updated: 27 November 2014

Hecke algebra generalities

Let $G$ be a group and let $B$ be a subgroup of $G\text{.}$ Let $W$ be a set of representatives of the double cosets of $B$ in $G$ so that $$G=\bigcup _{w\in W}BwB,$$ where the union is disjoint. For each $w\in W$ define $$\text{ind}\left(w\right)=\text{Card}(BwB/B)=\text{\# of left cosets of}\hspace{0.17em}B\hspace{0.17em}\text{in}\hspace{0.17em}BwB$$ and assume that $$\text{ind}\left(w\right)<\infty ,\text{for all}\hspace{0.17em}w\in W\text{.}$$

Let $ℳ$ be the collection of unions of left cosets of $B\text{.}$ For each $A\in ℳ$ define $$\mu \left(A\right)=\text{Card}(A/B)=\text{\# of left cosets of}\hspace{0.17em}B\hspace{0.17em}\text{in}\hspace{0.17em}A\text{.}$$

$ℳ$ is a $\sigma \text{-algebra}$ on $G$ and $\mu$ is a measure on $G$ with respect to $ℳ\text{.}$

Define $H_{ℂ}(G,B)$ to be the set of complex valued $B\text{-biinvariant}$ functions on $G$ with $\mu \text{-finite}$ support, i.e. a function $f:G\to ℂ$ is in $H_{ℂ}(G,B)$ if

(a)	$f\left(b_{1}g{b}_2\right)=f\left(g\right),$ for all $b_1,b_2\in B,$ $g\in G,$ and
(b)	$\mu \left(\text{supp}\left(f\right)\right)<\infty \text{.}$

The convolution product $$(f_1*f_2)\left(h\right)={\int }_G{f}_1\left(hg\right)f_2\left(g^{-1}\right)d\mu \left(g\right)$$ makes $H_{ℂ}(G,B)$ into an associative algebra.

(a)	The characteristic functions $I_{BwB},$ $BwB\in B\backslash G/B,$ form a basis of $H_{ℂ}(G,B)\text{.}$
(b)	The structure constants ${\mu }_{u,v}^w$ defined by $$I_{BuB}*I_{BvB}=\sum _w{\mu }_{u,v}^w{I}_{BwB}$$ are given by $${\mu }_{w,v}^u=\text{Card}((BvB\cap uB{w}^{-1}B)/B)\text{.}$$
(c)	We have ${\mu }_{w,v}^u=0$ unless $BuB\subseteq \left(BwB\right)\left(BvB\right)\text{.}$

		Proof.
	(b) Let $u,v,w\in G\text{.}$ Then $$\begin{array}{rcl}{\displaystyle (I_{BuB}*I_{BvB})\left(w\right)}& =& {\displaystyle {\int }_G{I}_{BuB}\left(wg\right)I_{BvB}\left(g^{-1}\right)d\mu \left(g\right)}\\ & =& {\displaystyle {\int }_{B{v}^{-1}B\cap w^{-1}BuB}d\mu \left(g\right)}\\ & =& {\displaystyle \mu (BuB\cap wB{v}^{-1}B)}\\ & =& {\displaystyle \text{Card}\left((BuB\cap wB{v}^{-1}B)\text{.}B\right)}\\ & =& {\displaystyle \text{\# left cosets of}\hspace{0.17em}B\hspace{0.17em}\text{in}\hspace{0.17em}BuB\cap wB{v}^{-1}B\text{.}}\end{array}$$ (c) It follows from the formula for ${\mu }_{u,v}^w$ in part (b) that if ${\mu }_{u,v}^w\ne 0$ then there exist $b_1,b_2,b_3\in B$ such that $b_{1}u{b}_2=w{b}_3{v}^{-1}\text{.}$ Thus $w=b_{1}u{b}_{2}v{b}_3^{-1}\text{.}$ It follows that $BwB\subseteq \left(BuB\right)\left(BvB\right)\text{.}$ $\square$

(a)	The map $$\begin{array}{cccc}\text{ind}:& H_{ℂ}(G,B)& ⟶& ℂ\\ & f& ⟼& {\int }_{G}f\left(g\right)d\mu \left(g\right)\end{array}$$ is an algebra homomorphism.
(b)	For each $w\in W,$ $\text{ind}\left(I_{BwB}\right)=\text{ind}\left(w\right)=\text{Card}(BwB/B)\text{.}$

		Proof.
	This is an easy calculation. $\square$

Example 1: Let $G$ be a locally compact topological group and let $B$ be a compact open subgroup. Let $dg$ be a Haar measure on $G$ normalized so that ${\int }_{B}dg=1\text{.}$ Then the pair $(G,B)$ satisfies the condition in () and $d\mu \left(g\right)=dg$ where $\mu$ is the measure defined in (). The Hecke algebra $H_{ℂ}(G,B)$ is a subalgebra of the convolution algebra $C_c\left(G\right)$ of continuous functions on $G$ with compact support.

Example 2: Let $G$ be a finite group. Then the discrete topology on $G$ makes $G$ into a locally compact group and with this topology any subgroup $B$ is compact and open. This is a particularly nice special case of example 1.

(a)	The Haar measure on $G,$ normalized so that ${\int }_{B}dg=1$ is given explicitly by $${\int }_{G}f\left(g\right)dg=\frac{1}{\mid B\mid }\sum _{g\in G}f\left(g\right),$$ for a function $f:G\to ℂ\text{.}$
(b)	The map $$\begin{array}{cccc}\mathrm{\Phi }:& C_c\left(G\right)& ⟶& ℂ\left[G\right]\\ & f& ⟼& \frac{1}{\mid B\mid }\sum _{g\in G}f\left(g\right)g\end{array}$$ is an isomorphism of algebras.
(c)	The Hecke algebra $H(G,B)$ is a subalgebra of $C_c\left(G\right)$ and restriction of the isomorphism $\mathrm{\Phi }$ to $H(G,B)$ gives an isomorphism $$\mathrm{\Phi }:H(G,B)⟶eℂ\left[G\right]e,\text{where}$$ $$e=\frac{1}{\mid B\mid }\sum _{x\in B}x\text{.}$$
(d)	$\mathrm{\Phi }\left(I_{BwB}\right)=\frac{1}{\mid B\mid }\sum _{x\in BwB}x\text{.}$

		Proof.
	$\square$

The module $1_B^G$

Define $$\begin{array}{rcl}{\displaystyle C_{\mu }(G/B)}& =& {\displaystyle \{f\in C_{\mu }\left(G\right)\hspace{0.17em}|\hspace{0.17em}f\left(gb\right)=f\left(g\right)\hspace{0.17em}\text{for all}\hspace{0.17em}b\in B\}}\\ {\displaystyle H_{ℂ}(G,B)=C_{\mu }(B\backslash G/B)}& =& {\displaystyle \{f\in C_{\mu }\left(G\right)\hspace{0.17em}|\hspace{0.17em}f\left(b_{1}g{b}_2\right)=f\left(g\right)\hspace{0.17em}\text{for all}\hspace{0.17em}{b}_1,b_2\in B\}}\end{array}$$

(a)	The vector space $C_{\mu }\left(G\right)$ with product given by convolution $$(f_1*f_2)\left(h\right)={\int }_G{f}_1\left(hg\right)f_2\left(g^{-1}\right)d\mu \left(g\right)$$ is an associative algebra over $ℂ,$ $$\begin{array}{rcl}{\displaystyle C_{\mu }(B\backslash G/B)}& =& {\displaystyle I_B*C_{\mu }\left(G\right)*I_B}\\ {\displaystyle C_{\mu }(G/B)}& =& {\displaystyle C_{\mu }\left(G\right)*I_B\text{.}}\end{array}$$
(b)	$C_{\mu }(B\backslash G/B)$ is a subalgebra of the $ℂ\text{-algebra}$ $C_{\mu }\left(G\right)$ with identity $I_B\text{.}$
(c)	The vector space $C_{\mu }(G/B)$ is a left $C_{\mu }\left(G\right)$ module and a right $C_{\mu }(B\backslash G/B)$ module where the action of $C_{\mu }\left(G\right)$ is by convolution on the left and the action of $C_{\mu }(B\backslash G/B)$ is by convolution on the right.

		Proof.
	Let us only show that if $f_1\in C_c\left(G\right)$ and $f_2\in C_c(G/K)$ then $f_1*f_2\in C_c(G/K)\text{.}$ The other facts are proved similarly. Let $f_1\in C_c\left(G\right)$ and $f_2\in C_c(G/K)\text{.}$ Then, if $h\in G$ and $k\in K,$ $$(f_1*f_2)\left(hk\right)={\int }_G{f}_1\left(hkg\right)f_2\left(g^{-1}\right)dg\text{.}$$ Putting $p=kg$ we have $$(f_1*f_2)\left(hk\right)={\int }_G{f}_1\left(hp\right)f_2\left(p^{-1}k\right)dp={\int }_G{f}_1\left(hp\right)f_2\left(p^{-1}\right)dp=(f_1*f_2)\left(h\right)\text{.}$$ It remains to show that $f_1*f_2$ is $\mu \text{-finite.}$ Let $P\in ℳ$ such that $\text{supp}\left(f_2\right)\subseteq P$ and let $Q\in ℳ$ such that $\text{supp}\left(f_1\right)\subseteq Q\text{.}$ Then $f_1\left(hp\right)f_2\left(p^{-1}\right)\ne 0$ only if $p\in P$ and $hp\in Q,$ i.e. only if $h\in PQ,$ which is $\mu \text{-finite.}$ Thus $$(f_1*f_2)\left(h\right)={\int }_G{f}_1\left(hp\right)f_2\left(p^{-1}\right)dp\ne 0\text{only if}\hspace{0.17em}h\in PQ\text{.}$$ Let us show that if $f_1\in C_{\mu }\left(G\right)$ and $f_2\in C_{\mu }(G/B)$ then $f_1*f_2\in C_{\mu }(G/B)\text{.}$ Let $f_1\in C_{\mu }\left(G\right)$ and $f_2\in C_{\mu }(G/B)\text{.}$ Then, if $h\in G$ and $b\in B,$ $$(f_1*f_2)\left(hb\right)={\int }_G{f}_1\left(hbg\right)f_2\left(g^{-1}\right)dg\text{.}$$ Putting $p=bg$ we have $$\left(f_1{f}_2\right)\left(hb\right)={\int }_G{f}_1\left(hp\right)f_2\left(p^{-1}b\right)dp={\int }_G{f}_1\left(hp\right)f_2\left(p^{-1}\right)dp=(f_1*f_2)\left(h\right)\text{.}$$ Let $f\in C_{\mu }(G/B)\text{.}$ Then $$(f*I_B)\left(h\right)={\int }_{G}f\left(hg\right)I_B\left(g^{-1}\right)dg={\int }_{B}f\left(hg\right)dg={\int }_{B}f\left(h\right)dg=f\left(h\right)\text{.}$$ So $f=f*I_B\in C_c\left(G\right)*I_B\text{.}$ $\square$

For each $f\in C_{\mu }\left(G\right)$ define $$\begin{array}{ccccc}{L}_f:& C_{\mu }(G/B)& ⟶& C_c(G/B)& \text{by}\\ & \phi & ⟼& f*\phi \text{.}\end{array}$$ For each $\psi \in C_{\mu }(B\backslash G/B)$ define $$\begin{array}{ccccc}{R}_{\psi }:& C_c(G/K)& ⟶& C_c(G/K)& \text{by}\\ & \phi & ⟼& \phi *\psi \text{.}\end{array}$$

Define $${\text{End}}_{C_{\mu }\left(G\right)}\left(C_{\mu }(G/B)\right)=\{\text{linear maps}\hspace{0.17em}T:C_c(G/K)\to C_c(G/K)\hspace{0.17em}|\hspace{0.17em}{TL}_f=L_{f}T\hspace{0.17em}\text{for all}\hspace{0.17em}f\in C_{\mu }\left(G\right)\}\text{.}$$ The map $$\begin{array}{cccc}\mathrm{\Phi }:& C_{\mu }(B\backslash G/B)& ⟶& {\text{End}}_{C_{\mu }\left(G\right)}\left(C_{\mu }(G/B)\right)\\ & \psi & ⟼& R_{\psi }\end{array}$$ is an anti-isomorphism of algebras.

		Proof.
	Surjectivity: Let $T$ be as in the statement and let $\psi =T{I}_K\text{.}$ Then $$\psi =T{I}_K=T(I_K*I_K)=I_K*\left(T{I}_K\right)\in C_c(K\backslash G/K)$$ and, if $\phi \in C_c(G/K),$ then $$T_{\phi }=T(\phi *I_K)=\phi *T{I}_K=\phi *\psi =R_{\psi }\phi \text{.}$$ Injectivity: Suppose that $\psi \in C_{\mu }(B\backslash G/B)$ and $R_{\psi }=0\text{.}$ Then $$0=R_{\psi }\left(I_B\right)=I_B*\psi =\mathrm{\Psi }\text{.}$$ so $\mathrm{\Psi }=0\text{.}$ Thus $R$ is injective. anti-Homomorphism $$R_{{\psi }_1}{R}_{{\psi }_2}\left(f\right)=R_{{\psi }_1}(f*{\psi }_2)=f*{\psi }_2*{\psi }_1=R_{{\psi }_2*{\psi }_1}\left(f\right)\text{.}$$ $\square$

The trace and the bilinear form

Define a function $$\begin{array}{ccccc}\tau :& C_c(K\backslash G/K)& ⟶& ℂ& \text{by}\\ & \psi & ⟼& \psi \left(1\right)\end{array}$$ and define a bilinear map $$⟨,⟩:C_{\mu }(B\backslash G/B)⟶ℂ$$ by $$⟨{\psi }_1,{\psi }_2⟩=\tau ({\psi }_1*{\psi }_2)=({\psi }_1*{\psi }_2)\left(1\right)\text{.}$$

If the group $G$ is unimodular then

(a)	$\tau \left(I_{KpK}\right)=I_{KpK}\left(1\right)=\{\begin{array}{ll}1,& \text{if}\hspace{0.17em}p\in K,\\ 0,& \text{otherwise.}\end{array}$
(b)	$\tau (I_{KpK}*I_{KqK})=\{\begin{array}{ll}\text{ind}\left(p\right),& \text{if}\hspace{0.17em}p=q^{-1},\\ 0,& \text{if}\hspace{0.17em}p\ne q^{-1}\text{.}\end{array}$
(c)	$\tau ({\psi }_1*{\psi }_2)=\tau ({\psi }_2*{\psi }_1)$ for all ${\psi }_1,{\psi }_2\in C_c\left(G\right)\text{.}$
(d)	If $G/K$ is finite then $\tau \left(\psi \right)=\frac{1}{\text{Card}(G/K)}\text{Tr}\left(R_{\psi }\right)\text{.}$
(e)	The bilinear form $⟨,⟩$ is symmetric and nondegenerate and the dual basis of the basis of $C_c(K\backslash G/K)$ given by $${\left\{I_{KpK}\right\}}_{p\in K\backslash G/K}\text{is the basis}{\left\{\frac{I_{K{p}^{-1}K}}{\text{ind}\left(p\right)}\right\}}_{p\in K\backslash G/K}\text{.}$$

		Proof.
	(c) is always true if the group $G$ is unimodular (or the measure $\mu$ is both left and right invariant). (2) We have $$\begin{array}{rcl}{\displaystyle (I_{pK}*I_{KqK})\left(p\right)}& =& {\displaystyle {\int }_G{I}_{pK}\left(pq\right)I_{KqK}\left(g^{-1}\right)dg}\\ & =& {\displaystyle \text{\# of left cosets of}\hspace{0.17em}K\hspace{0.17em}\text{in}\hspace{0.17em}K\cap K{q}^{-1}K}\\ & =& {\displaystyle \{\begin{array}{ll}1,& \text{if}\hspace{0.17em}q=1,\\ 0,& \text{otherwise.}\end{array}}\end{array}$$ Thus $$\text{Tr}\left(R_{I_{KqK}}\right)=\{\begin{array}{ll}\text{Card}(G/K),& \text{if}\hspace{0.17em}q=1,\\ 0,& \text{otherwise.}\end{array}$$ It follows that $\text{Tr}\left(R_{\psi }\right)=\text{Card}(G/K)\tau \left(\psi \right)$ for all $\psi \in C_c(K\backslash G/K)\text{.}$ $\square$

It is interesting to note that if $G/K$ is finite then $$\text{Card}(G/K)=\sum _{w\in K\backslash G/K}\text{ind}\left(w\right)\text{.}$$

It is a consequence of the trace property of $\tau$ that $\text{ind}\left(p\right)=\text{ind}\left(p^{-1}\right)$ for all $p\in G\text{.}$

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