I read that for a compact Hausdorff group $G$, the set $\hat{G}$ of isomorphism classes of irreducible unitary representations of $G$ is countable. Why is this the case?

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    $\begingroup$ I don't believe this. What about an uncountable direct product of finite groups? $\endgroup$ Jun 27, 2020 at 10:53

1 Answer 1


Perhaps you want to use the following fact: if $G$ is a locally compact topological group that is second-countable, then $L^2(G)$ is a separable Hilbert space, and in fact this holds for all $L^p$, $p\geq 1$ as explained in this MO answer of Dmitri Pavlov. In fact second-countability is necessary as well according to his answer.

Now if $G$ is in addition compact,the Peter-Weyl theorem says that the matrix coefficients of finite-dimensional unitary irreducible representations are an orthonormal basis of $L^2(G)$, and that every irreducible unitary representation is finite-dimensional. Now your claim follows from the fact. It would seem in the comments there's a counterexample when $G$ is not second-countable.


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