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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?

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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 $V = \tilde{\tilde{V}} = 0$.

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  • $\begingroup$ 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
    Commented 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
    Commented Nov 9, 2011 at 20:04
  • $\begingroup$ In symbols, $\tilde{V} = (V^\ast)^\infty$. $\endgroup$
    – Faisal
    Commented 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
    Commented 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
    Commented Nov 9, 2011 at 20:33
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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.

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