Q Exercise 4, Hecke Algebras - Daniel Bump

Question

I'm struggling with exercise 4 in Bump's Stanford Hecke Algebra notes linked here

It states the following:

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Let $G$ be a finite group and $V,W$ vector spaces. Let $C(G,V)$ denote the spaces of maps from $G$ to $V$, which has the $G$-representation $\rho_{V}$ given by

$$ (\rho_{V}(g)f)(x) = f(xg) $$

Suppose that $T$ is a linear map from $C(G,V)$ to $C(G,W)$ that commutes with the $G$-action, i.e that

$$ T(\rho_{V}(g)f) = \rho_{W}(g)T(f) $$

Prove that there exists a map $\lambda : G \rightarrow \operatorname{Hom}(V,W)$ such that $T(f) = \lambda * f$, where $*$ denotes convolution.

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The notes also has a lemma (Lemma 4) proving the above result in the case that $V$ and $W$ are 1-dimensional, and my hope was that I could adapt that proof to tackle this exercise but it has thus far left me stumped.

In the linear case $\operatorname{Hom}(V,W)$ becomes $\mathbb{C}$ and so the space of maps $f : G \mapsto \mathbb{C}$ is a unital convolution algebra (i.e the group algebra $\mathbb{C}[G]$), with the unit given by the characteristic function of the identity element of $G$, $\delta$ say.

Then $\delta *f = f * \delta$ for all $f \in \mathbb{C}[G]$, and so in particular if such a $\lambda$ existed, then $\lambda = \lambda * \delta = T(\delta)$, and then one checks that $T(\delta)$ satisfies the requirements.

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I would appreciate some assistance \ advice on tackling this exercise. Lemma 4 would suggest that we might want to try and find an element of $C(G,V)$, $\tau$ say, such that $F * \tau = F$ for every $F : G \rightarrow \operatorname{Hom}(V,W)$, and then continue on as in Lemma 4. But I have thus far been unable to construct such a $\tau$ (if one even exists).

Answer 1

Let $v\in V$. Define $\delta_1^v:G\rightarrow V:x\mapsto (\delta_{1,x}|G|)v$. Here $1\in G$ denotes the neutral element and $$\delta_{1,x}=\begin{cases}1 & \mbox{ if } x=1,\\ 0 & \mbox{ else.}\end{cases}$$

We first claim that $(\lambda*\delta_1^v)(x)=\lambda(x)(v)$. Indeed, we find that \begin{eqnarray} (\lambda*\delta_1^v)(x)&=&\frac{1}{|G|}\sum_{g\in G}\lambda(g)\left(\delta_1^v(g^{-1}x)\right)\\ &=& \frac{1}{|G|}\lambda(x)\left(|G|v\right)\\ &=& \lambda(x)(v). \end{eqnarray}

Assume that there exists a $\lambda$ such that $\lambda*f=T(f)$ for all $f$. Then $T(\delta_1^v)(x)=(\lambda*\delta_1^v)(x)=\lambda(x)(v)$ for all $v\in V$ and all $x\in G$. This completely determines $\lambda$. It remains to show that $\lambda$ defined in this way actually works.

Let $f\in C(G,V)$, then $$f=\frac{1}{|G|}\sum_{g\in G} \rho_V(g^{-1})\delta_1^{f(g)}.$$ Indeed, we find that \begin{eqnarray} \frac{1}{|G|}\sum_{g\in G} \rho_V(g^{-1})\delta_1^{f(g)}(x) &=& \frac{1}{|G|}\sum_{g\in G} \delta_1^{f(g)}(xg^{-1})\\ &=& \frac{1}{|G|}\delta_1^{f(x)}(xx^{-1})\\ &=& \frac{1}{|G|}|G|f(x)\\ &=& f(x). \end{eqnarray}

Can you take it from here?

Answer 2

Following the partial response / hint from Mathematician 42 (which shall remain the accepted response), I thought I would fill in the gaps left. I shall use the same notation.

$\textbf{Claim:}$ The map:

\begin{align*} & \lambda : &&G \rightarrow \ \operatorname{Hom}(V,W) \\ & \phantom{} &&x \ \mapsto \ (\ v \mapsto T(\delta_{1}^{v})(x) \ ) \end{align*}

is well defined, and $T(f) = \lambda * f$ for all $f \in C(G,V)$

$\textbf{Proof:}$ That $\lambda(x)$ is an element of $\operatorname{Hom}(V,W)$ for each $x \in G$ follows from the fact that $T$ is a linear map, and that $$ \delta_{1}^{v + au} = \delta_{1}^{v} + a\delta_{1}^{u} \ \ \text{for each} \ v,u \in V, a \in \mathbb{F} $$ Where $\mathbb{F}$ denotes the ground field of $V,W$.

Then since

$$ f(x) = \frac{1}{\left| G \right|} \sum_{g \in G} \rho_{V}(g^{-1})\delta_{1}^{f(g)}(x) $$

it follows that

\begin{align*} T(f)(x) & = \frac{1}{\left| G \right|} \sum_{g \in G} T\left(\rho_{V}(g^{-1})\delta_{1}^{f(g)}\right)(x) \\ & = \frac{1}{\left| G \right|} \sum_{g \in G} \rho_{W}(g^{-1})T(\delta_{1}^{f(g)})(x) \\ & = \frac{1}{\left| G \right|} \sum_{g \in G} \rho_{W}(g^{-1})\lambda(x)f(g) \\ & = \frac{1}{\left| G \right|} \sum_{g \in G} \lambda(xg^{-1})f(g) \\ & = (\lambda * f)(x) \end{align*}

This concludes the proof, and in fact since (as shown by Mathematician 42) the requirement that $\lambda * f = T(f)$ for every $f$, we have shown that in fact our $\lambda$ is uniquely determined by the map $T$.