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Let $G$ be a compact (assumed Hausdorff) group and $\hat{G}$ be the set of all characters of irreducible, finite-dimensional representations of $G$. It might occur that $\hat{G}$ is uncountable. It seems that, as a corollary of the Peter-Weyl theorem, $\hat{G}$ is an orthonormal basis for the Hilbert space of square-integrable class functions $L^2(G)^G$ (with respect to the Haar measure of $G$).

If $\hat{G}$ was countable and $f \in L^2(G)^G$, I'd have a Fourier expansion $f=\sum_{\hat{G}} \langle f, \chi \rangle \chi$. I'd like to do the same with uncountable $\hat{G}$. So I'd have to build a measure space $(\Omega, \Sigma, \mu)$ such that $\hat{G}$ is measurable and $f= \int_{\hat{G}} \langle f, \chi \rangle \chi d\mu$. Is there a natural choice for this measure? How could I make such Fourier expansion?

P.S.: I'm interested in the case where $G$ is not abelian.

José Siqueira
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  • While this is not exactly an answer to my question, I've found out it is not necessary in many reasonable situations. A result found in Pontryagin's book (Theorem 22) states that if $G$ is compact and second-countable, then any orthogonal system of functions on $G$ is at most countable. In particular, you can have at most countably many characters of finite-dimensional irreducible representations. – José Siqueira Apr 03 '14 at 11:57

1 Answers1

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Countability is completely irrelevant. For any compact group $G$, $\hat{G}$ is an orthonormal basis for $L^2(G)^G$, whether $\hat{G}$ is countable or not. Thus if $f\in L^2(G)^G$, we get a Fourier expansion $f=\sum_{\hat{G}}\langle f,\chi\rangle \chi$ (note that it is then also automatic that for any fixed $f$, the inner product $\langle f,\chi\rangle$ is nonzero for only countably many $\chi$, since otherwise this sum cannot converge). If you like, you can take your measure space to be $\hat{G}$ with counting measure.

Eric Wofsey
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