Let $G$ be a reductive group over some number field $F$ and $\pi$ an irreducible cuspidal automorphic representation of $G(\mathbb{A}_F)$. Let $f$ be a compactly supported function on $G(\mathbb{A}_F)$. Let \begin{equation} K_\pi (x, y) = \sum_{ \varphi } \pi(f) \varphi(x) \overline{\varphi(y)} \end{equation} with $x, y \in G(\mathbb{A}_F)$, and the sum is over an orthonormal basis of $\pi$. Then is it necessarily true that this is a finite sum? If not, why is it convergent?
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1$\begingroup$ Why do you know that it is convergent, if you don't know a proof? $\endgroup$– Stefan Kohl ♦Commented Dec 2, 2013 at 23:29
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Given an irreducible cuspidal automorphic represnetation, then it has character, i.e. the following operator is trace class $$ \pi(f) : v \mapsto \int\limits_{G(A)} f(g) \pi(g) v.$$ Note that $$ tr \pi(f) = \sum\limits <v, \pi(f) v> $$ of course. So this guarantees convergence. Now, if you assume that $f$ is smooth in the sense that it is invariant from the right and the left by an irreudicble representation of a maximal compact subgroup it will indeed be a finite sum for an apropriately chosen basis. These project onto the corresponding $K$-type. Automorphic representations factor into local smooth admissible representations. I am not so sure why your using functions instead of vectors.
