7
$\begingroup$

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.

$\endgroup$
4
$\begingroup$

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 $P=U$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.