Let $(V,\pi)$ be an irreducible, admissible, supercuspidal representation of $G = \operatorname{GL}_n(F)$ for $F$ a $p$-adic field. Let $B = TU$ be the usual Borel subgroup, maximal torus, and unipotent radical of $G$. Let $f_{v^{\ast},v}(g) = \langle v^{\ast}, \pi(g)v \rangle$ be a matrix coefficient for $v \in V$ and $v^{\ast} \in V^{\ast \infty}$. Then $f = f_{v^{\ast},v}$ is locally constant and compactly supported modulo the center $Z$ of $G$. If $\chi$ is a generic character of $U$, the integral

$$\int\limits_U f(ug)\chi(u^{-1}) du $$

can be shown to converge absolutely for every $g \in G$. Also, a change of variables shows that for fixed $v^{\ast}$,

$$v \mapsto \int\limits_U f_{v^{\ast},v}(u)\chi(u^{-1})du \tag{1} $$

defines a linear functional $\lambda: V \rightarrow \mathbb{C}$ which satisfies $\lambda(\pi(u_1)v) = \chi(u_1)\lambda(v)$ for all $u_1 \in U$. However, this linear functional might be the zero functional.

In general, $\pi$ is called generic if there exists a nonzero linear functional satisfying the property of the previous paragraph for $\chi$. If $\pi$ is generic for $\chi$, it is also generic for every other generic character.

1 . Is every irreducible, admissible supercuspidal representation of $G$ generic?

2 . If $\pi$ is generic, does there exist a smooth linear functional $v^{\ast}$ such that the map (1) is not the zero map? In other words, if there is a nonzero Whittaker functional, can it always be defined by an integral?


The answer to both questions is yes.

  1. All irreducible supercuspidal representations of ${\rm GL}(N,F)$ are generic. See e.g. I. M. Gelfand and D. A. Kajdan, Representations of the group ${\rm GL}(n,K)$ where $K$ is a local field, Lie groups and their representations.

  2. All ingredients to prove 2. are in

Paskunas, Vytautas; Stevens, Shaun On the realization of maximal simple types and epsilon factors of pairs. Amer. J. Math. 130 (2008), no. 5, 1211–1261.

First it is known by Bushnell and Kutzko that any irreducible supercuspidal representation of $G$ is of the form $\pi ={\rm ind}_J^G \Lambda$ (compactly induced representation), where $(J,\Lambda )$ is a maximal simple type in the sense of B. and K. Paskunas and Stevens prove that one can arrange the data so that $Hom_{U\cap J} (\Lambda , \chi ) \not= 0$ (1). It is a classical fact that if $c$ is a coefficient of $\Lambda$, then viewed as a fonction on $G$ (extend by $0$ off $J$), $c$ is a coefficient of $\pi$ (of the form $f_{v* ,v}$) (cf. e.g. Carayol Ann. Scient. ENS). Now writing condition (1) as the non-vanishing of an integral, you get exactly what you want.

Tell me if you want some more detail.

  • $\begingroup$ Thank you. A related question: the integral (1) can be defined for any quasisplit group. So does my question 2 have the same answer in general? $\endgroup$ – D_S Feb 4 '18 at 15:41
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    $\begingroup$ @D_S By the same argument the answer will be positive as soon as one knows that the supercuspidal representation is compactly induced. This is known in a very wilde range of cases. $\endgroup$ – Paul Broussous Feb 4 '18 at 17:44
  • $\begingroup$ Is this a result that is expected to be true of all supercuspidal representations? $\endgroup$ – D_S Feb 5 '18 at 4:31
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    $\begingroup$ It should be mentioned that it is known by Kim - Supercuspidal representations: an exhaustion theorem (MSN) that all supercuspidal representations of $p$-adic groups are compactly induced if "$p$ is big". If you'll forgive the joke, small $p$ is a big problem, and that's part of the significance of the work of Stevens et al.; but already a lot of interesting cases are covered by Ju-Lee's impressive result. $\endgroup$ – LSpice Feb 7 '18 at 19:03
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    $\begingroup$ However, it cannot be true that "all supercuspidals are generic" follows in general from "all supercuspidals are compactly induced", since the latter is often true and the former is rarely true. $\endgroup$ – LSpice Feb 7 '18 at 19:04

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