Let v be a finite place of a number field F. Let $\pi_{v}$ be an irreducible tempered representation of $ GL_{n}(F_v)$. Is it true that we can find some irreducible cuspidal automorphic representation $\pi$ of $GL_{n}(\mathbb{A_{F}})$ with $v$-component isomorphic to $\pi_{v}$ ?
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$\begingroup$ See also this question and its answers: mathoverflow.net/questions/43962/… $\endgroup$– EmertonCommented Nov 23, 2011 at 21:24
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2 Answers
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This is not true, because the set of equivalence classes of irreducible tempered representation of $Gl_n(F_v)$ has the power of the continuum (think of the principal series for example) while the number of automorphic representations for $GL_n$ over $F$ is countable (if we fix the central character), since the separable Hilbert space $L^2_\chi(Gl_n({\bf{A}})/Gl_n(F))$ contains the Hilbert direct sum of all the cuspidal automorphic representations.
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As Joel points out, this is not possible. Even in the case of a discrete series rep'n, one can't control the value of the central character on the non-compact part of $F_v^*$. What one can do (in some generality; for precise references recent work of Sug Woo Shin might be relevant) is show that the set of tempered representations which are local components of automorphic forms are "large" in the space of all tempered rep'ns, where "large" can be taken to mean something like "equidistributed according to Plancherel measure", or also "Zariski dense in the Spec of the Bernstein centre". (There may well be caveats to both these statements, though, so I would look at Shin's work and whatever references are contained therein.)
