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I’m taking an introductory course on finite group representations/character theory and just read about this:

Given a representation $\rho:G\rightarrow V$ of a group $G$ with associated isotypic decomposition $\rho=\rho_1^{a_1}\oplus \ldots \oplus \rho_k^{a_k}$, or alternatively $V \cong V_1^{\oplus a_1}\oplus \ldots \oplus V_k^{\oplus a_k}$, then it can be shown that $a_i=(\chi_V,\chi_{V_i})$. Furthermore, we also have the identity $$(\chi_V,\chi_V) =\sum_{i=1}^k a_i^2$$ and it can be shown that $V$ is an irreductible representation iff $(\chi_V,\chi_V)=1$.

This identity reminded me of Parseval’s identity from Fourier analysis (which I only know from undergraduate courses): the sum of squares of the Fourier coefficients of a square-integrable function $f$ is equal to the integral of the square of the function, $$ \Vert f\Vert _{L^{2}(-\pi ,\pi )}^{2}=\int _{-\pi }^{\pi }|f(x)|^{2}\,dx=2\pi \sum _{n=-\infty }^{\infty }|c_{n}|^{2}$$

where $c_{n}$ are the Fourier coefficients of $f$. I was wondering if the two identities were related, having read something about Fourier series having something to do with representation theory when $G=S^1$ is acting on itself (on this Math Exchange answer).

If that’s the case, I was curious to know what’s exactly the connection in this particular case, and if the representation’s properties (such as being irreductible) translate to something meaningful in Fourier analysis. References would be welcome too.

dahemar
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  • Isn't it the same identity (that is, these are particular forms about a general fact). – markvs Nov 11 '21 at 21:33
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    @markvs Hope then that it isn’t an obvious question, but what would this general fact be? – dahemar Nov 11 '21 at 21:57
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    @dahemar If my memory is ok then the link is that you have two vector spaces both equipped with an inner product. The identities look alike because they are both the result of a generalisation of pythagoras' theorem. Perhaps a more knowledgeable user than I would be able to give you a more detailed explanation an post it as an answer. – Digitallis Nov 12 '21 at 17:40

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