The following exercise comes from Diaconis' book Group Representations in Probability and Statistics. Here $G$ is a finite group, and $U$ is the uniform distribution on $G$, i.e., $U(s) = 1/|G|$ for all $s\in G$.

Exercise 5. Let $P$ be a probability on $G$. Define $\overline P(s) = P(s^{-1})$. Show that $U = P\ast \overline P$ if and only if $P$ is uniform.

A probability $P\colon G\to\Bbb R_{\ge 0}$ is a function satisfying $\sum_{s\in G}P(s) = 1$. Note that if $P$ and $Q$ are probabilities, then by definition, $$ P\ast Q(s) = \sum_{t\in G}P(st^{-1})Q(t). $$ If $P = U$, then it's not hard to show $U = P\ast \overline P$. My question is about the other direction. I have tried using the Fourier inversion formula: $$ P(s) = \sum_id_i\operatorname{Tr}(\rho_i(s^{-1})\hat P(\rho_i)), $$ where the $\rho_i$ are the irreducible representations of $G$ and $d_i$ is the degree of $\rho_i$. By definition, $\hat P(\rho_i) = \sum_{s\in G} P(s)\rho_i(s)$. However, this didn't yield anything besides $P(s) = 1/|G| \cdot P(s)\cdot |G|$.

I did manage to show that $\sum_{s\in G}P(s)^2 = 1/|G|$ by looking at the regular representation of $G$, which is certainly a property that the uniform distribution on $G$ has, but I haven't been able to make further progress. Also relevant here is the Plancherel formula: $$ \sum_{s\in G}f(s^{-1})h(s) = \frac{1}{|G|}\sum_id_i\operatorname{Tr}(\hat{f}(\rho_i)\cdot\hat h(\rho_i)), $$ where $f,h$ are any two functions $G\to\Bbb C$ and the sum on the right is over all irreducible representations of $G$. Another relevant equation is $$ \widehat{P\ast Q}(\rho) = \hat P(\rho)\cdot\hat Q(\rho). $$ Any hints or suggestions are welcome in figuring this out.


It actually follows immediately from $$\sum_{s\in G}P(s)^2 = \frac{1}{|G|}$$ and Cauchy-Schwarz that $P$ is uniform. Using the inner product $$\langle P,Q\rangle=\sum_{s\in G} P(s)\overline{Q(s)}$$ on $\mathbb{C}^G$, we have $\langle U,U\rangle=\frac{1}{|G|}$ and also $\langle P,U\rangle=\frac{1}{|G|}$ since $P$ is a probability. The equation $\sum_{s\in G}P(s)^2 = \frac{1}{|G|}$ says that also $\langle P,P\rangle=\frac{1}{|G|}$. But by Cauchy-Schwarz, $$|\langle P,U\rangle|^2\leq \langle P,P\rangle\langle U,U\rangle$$ with equality iff $P$ is a scalar multiple of $U$. Since both sides are $\frac{1}{|G|^2}$, equality does hold and so $P$ is a scalar multiple of $U$, and hence $P=U$.

  • $\begingroup$ This was a very nice idea, using the Cauchy-Schwarz inequality like that. Thanks very much. $\endgroup$ – Alex Ortiz Jun 10 '18 at 5:59

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