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I am working on the Fourier transform over finite non-abelian groups, specifically following Diaconis. He defines it as follows (p.7):

Let $P$ be a probability on a finite group $G$. The Fourier transform of $P$ at the representation $\rho$ is the matrix \begin{equation*} \widehat{P}(\rho) = \sum_{s\in G}P(s)\rho(s) \end{equation*}

My understanding is that the Fourier transform is a ring isomorphism between the set of functions $L(G) = \{f: G \rightarrow \mathbb{C}\}$ (with the operations of pointwise addition and convolution) and some other set (with pointwise addition and pointwise multiplication) via the convolution theorem. What is this other set?

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    $\begingroup$ the ring isomorphism is explained in en.wikipedia.org/wiki/… $\endgroup$ Commented Jun 21, 2019 at 20:48
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    $\begingroup$ @CarloBeenakker I've seen that characterization but it's still unclear to me. $\endgroup$ Commented Jun 22, 2019 at 13:48

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The transform, which is a ring isomorphism, $$ P \mapsto \widehat{P} = \sum_{g\in G} P(g)g $$ goes from your $L(G)$ to the group algebra ${\mathbb C}G$. The convolution corresponds to the multiplication in the group algebra: http://mathworld.wolfram.com/GroupAlgebra.html

If you apply an irreducible representation $\rho$ to the righthand side, this corresponds to the projection to the matrix direct summand in the Wedderburn decomposition: $$ {\mathbb C}G = \oplus_{\mu\in Irr(G)} M_{d(\mu)} \rightarrow M_{d(\rho)} $$ where $d(\rho)$ is the degree of your representation. Thus, you end up in the matrix algebra, while the convolution corresponds to the matrix multiplication. It is no longer an isomorphism, just a homomorphism.

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