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Given an irreducible supercuspidal representation $(\pi,V)$ of GL(n), embeds GL(n-1) into GL(n) on the left upper corner. Consider the restriction of $\pi$ to GL(n-1). I want to ask may I find some vector $v\in V$, such that the space generated by translations of $v$ under GL(n-1) is a representation of finite length of GL(n-1)?

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2 Answers 2

Maybe! My nominee is Jacquet-PS-Shalika's "essential vector", that is, the (unique up to scalars) vector fixed by the compact subgroup of $GL(n,k_v)$ consisting of matrices $\pmatrix{A&b\cr c&d}$ with $A\in GL_{n-1}(o_v)$, $b$ $(n-1)$-by-$1$ with entries in $o_v$, and congruence conditions that $c=0\mod \varpi^N$ and $d=1\mod \varpi^N$ for some $N$. (Math. Ann. 256, 1981, 199-214). They prove that there exists an essentially unique such for generic repns (=admitting a Whittaker model), which applies to supercuspidal.

So, at least, any decomposition of the submodule generated by this essential vector would only need spherical repns.

The "correct" local L-factors $L_v(s,\pi\otimes \pi')$ for $GL_n\times GL_{n-1}$ can be viewed as exactly spectral decomposition coefficients of the restriction from $GL_n$ to $GL_{n-1}$, of course, with or without mediation of Whittaker models.

Maybe there are newer results on branching rules in this situation?

Edit: Temporarily forgot the relatively recent Aizenbud, Gourevitch, Sayag result that $GL(n-1)\subset GL(n)$ is a Gelfand pair... also discussed somewhere in MO.

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This is false for n=2. Look at the Kirillov model. (Presumably you mean that all vectors in V are smooth.)

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But for supercuspidal? Matrix coefficients are compactly-supported modulo the center...? Maybe I'm not understanding the implications...? –  paul garrett Jul 25 '11 at 15:43

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