## Plancherel’s Formula

Let $G$ be a finite group and let $f, h : G \rightarrow \mathbf{C}$ be functions. On page 13 of the book, Diaconis proves the following `Plancherel’ formula

$\sum_{s \in G} f(s^{-1})h(s) = \frac{1}{|G|} \sum_{i=1}^k d_i \mathrm{Tr} \bigl( \hat{f}(\rho_i)\hat{h}(\rho_i) \bigr)$

where $\rho_1, \ldots, \rho_k$ are the irreducible representations of $G$ and $d_i = \mathrm{dim}\ \rho_i$.

First of all I will give a fairly short proof of the formula using the group algebra $\mathbf{C}G$. Define

$x = \bigl( \sum_{u \in G} f(u) u \bigr) \bigl( \sum_{s \in G} h(s) s \bigr) \in \mathbf{C}G.$

On the representation $\rho_i$, the element $x$ acts as $\hat{f}(\rho_i) \hat{h}(\rho_i)$. As a representation of $G$, the group algebra affords the regular representation of $G$. Hence, by Corollary 1 on page 12, $\mathbf{C}{G}$ has exactly $d_i$ irreducible summands on which $G$ acts according to $\rho_i$. Therefore, the trace of $x$ acting on $\mathbf{C}G$ is

$\sum_{i=1}^k d_i \mathrm{Tr} \bigl( \hat{f}(\rho_i)\hat{h}(\rho_i) \bigr)$

On the other hand, the character of the regular representation was found in Proposition 5 to be $0$ on non-identity elements, and $|G|$ on the identity. Hence, by linearity, the trace of $x$ acting on the regular representation is

$|G| \sum_{s \in G} f(s^{-1}) h(s).$

This completes the proof of the Plancherel formula. One might object to this name, on the grounds that the bilinear forms that appear on either side are not inner products. To get round this, we note that if $g$ is defined by $g(s) = f(s^{-1})$ then

$\hat{g}(\rho) = \sum_{s \in G} g(s) \rho(s) = \sum_{s \in G} f(s^{-1}) \rho(s) = \sum_{s \in G} f(s) \rho(s)^\star.$

Hence the original formula implies that

$\sum_{s \in G} g(s)h(s) = \frac{1}{|G|} \sum_{i=1}^k d_i \mathrm{Tr} \bigl( \hat{g}(\rho_i)^\star \hat{h}(\rho_i) \bigr)$

where we now assume that the representations $\rho_i$ are unitary. (The above formula appears in Remark 3 on page 14.) The form on matrices defined by $(A,B) = \mathrm{Tr} (A^\star B) = \sum_{jk} \bar{A}_{jk}B_{jk}$ is an inner product, while on the left-hand side we have an inner product provided we restrict to real valued functions. This restriction seems a little surprising, but does appear to be necessary.