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Let $G$ be a compact topological group. By the Peter-Weyl theorem, the complex Hilbert space $L^2(G)$ is the Hilbert space direct sum of the spaces of matrix coefficients of all the irreducible representations of $G$. When $G$ is commutative, these irreducible representations are one-dimensional, and the Peter-Weyl theorem just says that the unitary characters of $G$ form an orthonormal basis of $L^2(G)$.

For example, when $G = \mathbb R/\mathbb Z$, the functions $f_n(x) = e^{2\pi i nx}$ are an orthonormal basis of $L^2(G)$. Thus if $f \in L^2(G)$, then there are unique complex numbers $c_n$, with $\sum\limits_n |c_n|^2 < \infty$, which gives $f$ its Fourier expansion:

$$f = \sum\limits_{n\in \mathbb Z} c_n f_n$$ where the sum on the right hand side converges in the $L^2$-norm to $f$. It is a much deeper theorem that when $f$ is continuous, the right hand side also converges pointwise to $f$ almost everywhere.

Is the analogue of this deep theorem known for other compact groups? That is, suppose we take an orthonormal basis $f_i : i \in I$ of $L^2(G)$ via matrix coefficients of irreducible representations as in the Peter-Weyl theorem, so that any $f \in L^2(G)$ has a "Fourier expansion" for uniquely determined $c_i \in \mathbb C$,

$$f = \sum\limits_{i \in I} c_i f_i$$ so the right hand side converges to $f$ in the $L^2$-norm. Suppose that $f$ is continuous. Then, do we know that the right hand side converges pointwise to $f$ almost everywhere?

If this is not known in general, is it known for, say, $G = \mathbb A_k/k$ for $k$ a global field?

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  • $\begingroup$ How are you performing your summation for a general $I$? That is: your infinite sum must be interpreted as some kind of limit, but how do you wish to do it? This is a problem even if $G={\bf T}^2$, if I recall correctly $\endgroup$
    – Yemon Choi
    Commented Jan 2, 2019 at 4:48
  • $\begingroup$ A hint at some of the potential problems with making your question more precise can be found in the remarks on the Wikipedia page for Carleson's theorem: en.wikipedia.org/wiki/Carleson%27s_theorem $\endgroup$
    – Yemon Choi
    Commented Jan 2, 2019 at 4:49
  • $\begingroup$ I see. Maybe the statement of my question is unclear for general $G$. $\endgroup$
    – D_S
    Commented Jan 2, 2019 at 4:53
  • $\begingroup$ D_S, I was gently trying to point out that it's not even clear fo the particular case of ${\bf T}^2$. However, perhaps for the kinds of groups you ask about in your final line, one can say more? I am not familiar enough with Fourier analysis on those kinds of groups to hazard a guess $\endgroup$
    – Yemon Choi
    Commented Jan 2, 2019 at 4:56
  • $\begingroup$ For $k$ a global field (which is countable), and $\mathbb A$ the adele group of $k$, the quotient group $\mathbb A/k$ is compact, and its character group identifies with $k$. So in the Fourier expansion, I would want the infinite sum over $k$ to be taken in any order, since there is no predetermined order on $k$. $\endgroup$
    – D_S
    Commented Jan 2, 2019 at 5:02

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