2
$\begingroup$

Let $\pi$ be an admissible representation of a locally compact totally disconnected group. I have a technical problem about the proof of

$\pi$ is irreducible if and only if its contragredient is so

given in 2.15(c) of the '76 article of Bernstein and Zelevinsky. There $\pi$ is assumed to have a nontrivial proper subrepresentation $E_1$, and it is asserted that the orthogonal complement of $E_1$ be a nontrivial proper subrepresentation of the contragredient, whence the result. What I cannot figure out is the nontriviality of this orthogonal complement. We simply have to find a nonzero smooth functional which vanishes on $E_1$; this shall follow from $E_1\neq E$ (as properness of the orthogonal complement follows from $E_1\neq 0$), but how?

Improve this question
$\endgroup$
4

2 Answers 2

Reset to default
2
$\begingroup$

This follows from two facts:

  1. The complement $E_1^\perp$ of $E_1$ in $\tilde{E}$ is isomorphic to the contragredient of $E/E_1$.

  2. If $V$ is admissible and nonzero then $\tilde{V}$ is nonzero (and admissible). For if $\tilde{V}=0$ then $V = \tilde{\tilde{V}} = 0$.

Improve this answer
$\endgroup$
5
  • $\begingroup$ If I understand the question, the worry is about algebraic (=smooth) vectors-- here $x\in E$ is smooth if the stabilizer of $x$ for the $\pi$ action is an open subgroup of $G$. Could you say some words about this?? $\endgroup$
    – Matthew Daws
    Nov 9, 2011 at 19:57
  • $\begingroup$ The contragredient is by definition the "smooth dual" representation (i.e. the subrep of the dual rep on the subspace of smooth vectors). $\endgroup$
    – Faisal
    Nov 9, 2011 at 20:04
  • $\begingroup$ In symbols, $\tilde{V} = (V^\ast)^\infty$. $\endgroup$
    – Faisal
    Nov 9, 2011 at 20:14
  • $\begingroup$ @Faisel: Sure! But I think the original question wanted to know why there is an algebraic vector in the perp of $E_1$. That is, your answer seemed a bit brief, given what the original question asked (so I think maybe it won't be easy to understand). However, on a close reading of the original question, I see that $\pi$ is assumed "admissible". I personally don't understand what implications this has, but it seems to imply lots of powerful things; so maybe one needs to fully understand this definition...?? $\endgroup$
    – Matthew Daws
    Nov 9, 2011 at 20:21
  • $\begingroup$ As far as this question is concerned, admissibility is important because it gives us the equivalence $\tilde{\tilde{\pi}}=\pi$ (which I used in (2)). This equivalence doesn't hold for general inadmissible $\pi$. $\endgroup$
    – Faisal
    Nov 9, 2011 at 20:33
0
$\begingroup$

If you believe that the monomorphism $\pi \to \widetilde{\widetilde{\pi}}$ is an isomorphism for admissible $\pi$, then you can see that $\pi \mapsto \widetilde{\pi}$ is a contravariant autoequivalence on the abelian category of admissible (smooth) representations. Thus it carries simple objects to simple objects.

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.