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Suppose we have a compact group $G$ with continuous $f$ on $G$ that is also $G$-finite.

I am told that then, out of all the irreducible representations $\pi$ of $G$ we must have $\pi(f)\neq 0$ only for finitely many. Here the representation is thought of as a transformation on the space of functions.

Why is this true? I feel like it is very short, and I was just missing something quick. Can someone enlighten me on this one.

Thanks again!!

(Also, by $G$-finite I mean that the vector space of right and left translates is finite.)

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possible duplicate of Compact Group $G$ with a $G$-finite function. –  Zev Chonoles May 11 '11 at 23:30
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@Zev: Accident! I deleted the other one so it should be ok. –  Math Student May 11 '11 at 23:40
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1 Answer 1

I assume by $\pi(f)$ you mean that $\pi(f)(v) = \int_G f(g) \pi(g) (v) dg$. Let $E$ be the vector space generated by the left translates of $f$ (it is also a representation of $G$), and $V$ the vector space of the representation $\pi$.

Then check that $W = \left\{ \sum_i \pi(f_i)(v_i) \ |\ f_i \in E,\ v_i \in V \right\}$ is a subrepresentation of $(\pi,V)$, so it is either zero or the whole of $V$. Show that $W$ is (as a representation of $G$) a quotient of $E$.

This shows that $\pi(f) \neq 0$ only if $\pi$ is one of the subrepresentations of $E$ (which is finite dimensional).

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