2
$\begingroup$

Let $G$ be a reductive group over a nonarchimedean local field $F$. Let $\pi$ be an irreducible, cuspidal representation of $G$, with contragredient $\tilde{\pi}$. Then $\tilde{\pi}$ is cuspidal.

A character of $G$ is unramified if it is trivial on all compact subgroups of $G$. The group of unramified characters of $G$ acts on the set of isomorphism classes of irreducible, cuspidal $G$-representations. Under this action, is $\tilde{\pi} \sim \pi$?

Note: I asked this question on math.stackexchange but it has not been answered.

$\endgroup$
  • 2
    $\begingroup$ In addition to @PaulBroussous's nice answer, another easy way to make this false is if ($\pi$ is not self dual and) $G$ has no (non-trivial) characters, as happens, for example, for $G = \mathrm{SL}_2(F)$ (at least if $\mathrm{char}(F) \ne 2$). $\endgroup$ – LSpice Mar 23 '18 at 13:49
5
$\begingroup$

This is already false for $G={\rm GL}(1,F)$. In that case a cuspidal irreducible representation is a smooth character $\chi$ of $F$. The contragredient is $\chi^{-1}$. We have $\chi \sim \chi^{-1}$ iff $\chi^2$ is unramified. There are easy counter-examples.

Let us give another counter-example in higher rank. Take $G={\rm PGL}(n,F)$ and $\pi$ compactly induced from an irreducible smooth representation $\lambda$ of ${\rm PGL}(n,O_F )$, $O_F$ the ring of integers of $F$. Assume that $\lambda$ is lifted from a cuspidal representation $\sigma$ of ${\rm PGL}(n,k_F )$, where $k_F$ is the residue field of $F$ (so $\pi$ is of level $0$). Then $\lambda$ is a type for the Bernstein block of $\pi$ in the sense of Bushnell and Kutzko's type theory. If $\pi\sim {\tilde \pi}$ then $\lambda$ and $\tilde \lambda$ are types for the same Bernstein block. By a standard result of type theory for ${\rm PGL}(n,F)$, $\lambda$ and $\tilde \lambda$, whence $\sigma$ anb $\tilde \sigma$, are isomorphic. By Green's parametrization, $\sigma$ is attached to a regular character $\chi$ of $l^\times$, trivial on $k_F^\times$, where $l/k_F$ is a degree $n$ extension. Then $\sigma\sim {\tilde \sigma}$ is equivalent to $\chi$ and $\chi^{-1}$ belonging to the same ${\rm Gal}(l/k_F )$-orbit. It is then an easy exercise to build up a counter-example.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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