Question

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?

Answer 1

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

Answer 2

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.