Given an irreducible supercuspidal representation $(\pi,V)$ of GL(n), embeds GL(n1) into GL(n) on the left upper corner. Consider the restriction of $\pi$ to GL(n1). I want to ask may I find some vector $v\in V$, such that the space generated by translations of $v$ under GL(n1) is a representation of finite length of GL(n1)?
Maybe! My nominee is JacquetPSShalika'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 $A\in GL_{n1}(o_v)$, $b$ $(n1)$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, 199214). 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 Lfactors $L_v(s,\pi\otimes \pi')$ for $GL_n\times GL_{n1}$ can be viewed as exactly spectral decomposition coefficients of the restriction from $GL_n$ to $GL_{n1}$, 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 $GL(n1)\subset GL(n)$ is a Gelfand pair... also discussed somewhere in MO. 


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

