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?

  • $\begingroup$ Sorry, I was thinking too much about finite-dimensional representations. Delete my answer. $\endgroup$ – jmc Jan 12 '15 at 7:48
  • $\begingroup$ Have you checked out the work by Toshiyuki Kobayashi on discrete decomposability? $\endgroup$ – Vít Tuček Jan 12 '15 at 9:02
  • $\begingroup$ See Marty's answer to this question. $\endgroup$ – Kimball Jan 12 '15 at 9:39

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.

  • $\begingroup$ Thank you for your enlightening example. It helped me very much! $\endgroup$ – Monty Jan 12 '15 at 15:49

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