# Q Exercise 4, Hecke Algebras - Daniel Bump

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

It states the following:

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.

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.

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).

• There is no $\tau$ such that $F*\tau =F$. In other words this more general convolution doesn't have an identity. This is obvious once you realize that $F*\tau$ and $F$ belong to different spaces :) The first claim in my answer works around this problem. – Mathematician 42 Sep 14 '18 at 4:05

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?

• Thanks so much! I wrote this function down and got to the point where. I wrote $T(\delta_{1}^{v})(x) = \lambda(x)(v)$, but I was struggling to figure out what the identity I needed for $f(x)$ was. This is perfect! I can certainly go from here and will do so later today. – Adam Higgins Sep 14 '18 at 5:59
• No problem. You still need to do some work, don't forget to show that $\lambda(x)$ defined in this way actually yields a linear map $V\rightarrow W$. I have little experience with the theory, but it looks fun! – Mathematician 42 Sep 14 '18 at 6:42
• I’ll have at it today and I’ll return with an update once done. If you’re interested, the reason I’m studying this is in in preparation for the study of the philosophy of cusp forms, in particular the inductive (parabolic induction) construction of the irreducible representations of the general linear group of finite fields. – Adam Higgins Sep 14 '18 at 6:44
• I added the rest of the details in a comment below! – Adam Higgins Sep 19 '18 at 15:32

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 $T$.

• That seems about right, well done. Enjoy the theory. – Mathematician 42 Sep 20 '18 at 6:37