thus have a natural perfect pairing

$$\langle\,,\,\rangle:\M\times\M^\vee\rightarrow\R.$$

For a finite set $A$, denote by $\C[A]$ the $\C$-vector space of all functions $A\rightarrow \C$. It is endowed with the inner product

$$(f,g)_A:=\sum_{x\in A}f(x)\overline{g(x)},$$ for all $f,g\in\C[A]$.

Let $X$ be a finite non-empty set. Consider the correspondence

$$\xymatrix{&X\times\M\times \M^\vee\ar[dl]_{p_{12}}\ar[dr]^{p_{13}}&\\
X\times \M&&X\times\M^\vee}$$ and define the Fourier transform $\mathcal{F}:\C[X\times\M]\rightarrow\C[X\times\M^\vee]$ by

$$\mathcal{F}(f):=(p_{13})_!\left((p_{12})^*(f)\otimes (p_{23})^*(\varpi\langle,\rangle)\right),$$ for all $f\in\C[X\times \M]$. Namely

$$\mathcal{F}(f)(x,v)=\sum_{u\in\M}\varpi(\langle u,v\rangle)f(x,u),$$ for all $(x,v)\in X\times\M^\vee$.

We have the following standard proposition.

\(1) The rescaled Fourier transform $|\M|^{-\frac{1}{2}}\,\mathcal{F}$ is an isometry with respect to $\left(\,,\,\right)_{X\times \M}$ and $\left(\,,\,\right)_{X\times \M^\vee}$.

\(2) For $f,g\in\C[X\times \M]$, we have

$$\calF(f*g)=\calF(f)\,\calF(g),$$ where the convolution $f*g$ of $f$ by $g$ is the function $z\mapsto
(f*g)(z)=\sum_{x+y=z}f(x)g(y)$. \[propFour\]

Indeed

$$\begin{aligned}
\left(\mathcal{F}(f),\mathcal{F}(g)\right)_{X\times\M^\vee}&=\sum_{(x,v)\in X\times\M^\vee}\left(\sum_{u\in\M}\varpi(\langle u,v\rangle)f(x,u)\right)\left(\sum_{w\in\M}\varpi(\langle -w,v\rangle)\overline{g(x,w)}\right)\\
&=\sum_{x,u,w}f(x,u)\overline{g(x,w})\sum_v\varpi(\langle u-w,v\rangle).\end{aligned}$$

We need to see that for any $u\in\M$, we have

$$\sum_{v\in\M^\vee}\varpi\big(\langle u,v\rangle\big)=\begin{cases}|\M|&\text{ if } u=0,\\0&\text{ otherwise}.\end{cases}$$ The map $\M^\vee\rightarrow\C^\times$, $v\mapsto\varpi(\langle u,v\rangle)$ is an additive character. Since the kernel of $\varpi$ does not contain non-zero ideals of $\R$, this map is non-trivial as long as there exists an element $v$ of $\M^\vee$ such that $\langle u,v\rangle\neq 0$. Since the pairing $\langle\,,\,\rangle$ is non-degenerate such an element $v$ exists when $u\neq 0$.

We note that if a finite group $\G$ acts 