Let $F$ be local field of characteristic zero and $\pi$ be a irreducible admissible representation of $GL\_n(F)$.


Let us consider its restriction to $GL\_{n-1}(F)$. Then I want to know whether $\pi|\_{GL\_{n-1}(F)}$ is completely reducible, namely, $\pi|\_{GL\_{n-1}(F)}=\oplus\_{i\in A}m\_i \cdot \tau\_i$ where $\tau\_i$ is an irreducible representation of $GL\_{n-1}(F)$ and $m\_i$'s are non-negative integers.


Is it true? If so, from which theorem and how does it follow?



Is the restriction of a representation semisimple?
The anwser to your question is "no in general" since you already have a counter-example in the case $n=2$.


Take for $\pi$ an irreducible supercuspidal representation of ${\rm GL}(2,F)$. Then its restriction to ${\rm GL}(1,F)\simeq F^\times$ is known by the Kirillov model. This is the space $S(F^\times )$ of locally constant functions with compact support on $F^\times$, where the action is given by $\pi (t)f(x) = f(tx)$, $t\in F^\times$. If the restriction of $\pi$ were completely reducible then $S(F^\times )$ would have an invariant subspace spanned by a non-zero function $f\_0\in S(F^\times )$. So for some abelian character $\chi\_0$ of $F^\times$, one would have:


$$
f\_0 (x) =\chi\_0 (x) f\_0 (1)\ .
$$
But such a function $f\_0$ cannnot have compact support; a contradiction. 


