6
$\begingroup$

Let $G$ be an unramified reductive group over $Q_p$. I want to prove that the group $G(Q_p)$ has a supercuspidal representation (complex coefficients).

I have been looking in many parts of the literature, and it seems that many people are convinced that it is true; however up to now I never saw it stated explicitly.

By the works of L. Morris I reduce to showing that a reductive group $M$ over $F_p$ has a cuspidal representation (L. Morris, level zero G-types, p 140).

So then, I should prove that $M(F_p)$ has a cuspidal representation. The article of Deligne–Lusztig provides such a representation when given a minisotropic torus $T$ in $M$ and a character $\chi$ of $T(F_p)$ which is in general position.

Let me recall that "character in general position" means that the rational Weil group acts freely on the character.

So now comes my doubt and question. Is it true that such a pair $(T, \chi)$ can always be found for all reductive groups $M$ over $F_p$ ?

I am "afraid" of "small" groups that have tori with "large" Weyl groups.

The supercuspidal representations that come from the above construction are of "level 0".

In the book of Carter (Finite Groups of Lie Type) I found a result pointing in this direction. Lemma 8.4.2 p. 281 (with an easy proof) shows that for $T$ given and $q$ sufficiently large, the torus $T(F_q)$ has a character in general position.

Improve this question
$\endgroup$
5
  • $\begingroup$ I forgot the little $p$-adic theory I knew, but in the real case, the $\textit{existence}$ of the discrete series ("cuspidals"), while considered a high point of Harish-Chandra theory, was proved using characters, without explicitly constructing them. What did the great man have to say about the $p$-adic case? $\endgroup$
    – Victor Protsak
    Jul 1, 2010 at 20:33
  • $\begingroup$ In the real case it is not true in general. I think that for example, $SL_3(R)$ does not have any discrete series representations. If I understand things correctly, the main result of HC is that a connected semi simple real lie group has discrete series representations if and only if it has the same rank as a maximal compact subgroup. $\endgroup$
    – mnr
    Jul 2, 2010 at 9:07
  • $\begingroup$ Well, for that matter no real group has supercuspidal representations, either. My point, which you didn't seem to grasp, was that it wasn't necessary to $\textit{construct}$ discrete series in order to prove their existence, and Harish-Chandra's "cuspidal forms philosophy" in the $p$-adic case was an extension of his work in the real case. $\endgroup$
    – Victor Protsak
    Jul 2, 2010 at 20:13
  • $\begingroup$ The answer is "yes", such a pair (T, chi) can be found easily, by computing with the root systems. For all classical groups I have done this computation, so at least for those groups I am certain that cuspidal representations exist. For the exceptional groups there is another method to obtain find cuspidal representations. $\endgroup$
    – mnr
    Feb 18, 2011 at 9:50
  • 1
    $\begingroup$ There is a fairly general argument. Let $F$ be a non-archimedean, local field. The compact induction of a representation $\rho$ of an open, compact modulo center subgroup $K$ in the $F$-points of a $F$-reductive group $G$ is a supercuspidal (not necessarily irreducible) representation if $1 \not\subset Res_{N \cap K} \rho$ for every unipotent $F$-subgroup of $G$. E.g., you seem to want to prove that every reductive group over a finite field has a cuspidal representation, which is certainly sufficient, but not necessary. $\endgroup$
    – Marc Palm
    Jul 1, 2012 at 13:34

1 Answer 1

Reset to default
2
$\begingroup$

Various special constructions make the point for particular groups, which tends to convince one about the general case. E.g., for $G=SL(2,\mathbb Q)$, a well-known (Shalika, Jacquet, c. 1970) idea is that inducing "supercuspidals" from $K=SL(2,\mathbb Z_p)$ to $G$ is a finite direct sum of supercuspidals. The "level zero" cuspidals on $SL(2,\mathbb Z)$ factor through $SL(2,\mathbb Z/p)$. The irreducibles of the latter finite group are ascertained in various ways, and one finds that a significant fraction are "cuspidal", so compose to "supercuspidal" on $K$, and (with proof) induce to finite direct sums of supercuspidals on $G$.

The critical point is that induced repns (from Iwahori subgroups) do not exhaust the irreds of $K$. For $SL(2)$ this is easily proven by a counting argument, and/or by a theta-correspondence argument. For general groups, some of these choices persist.

Improve this answer
$\endgroup$
2
  • $\begingroup$ I think that $\mathrm{SL}(2, \mathbb Z)$ is supposed to be $\mathrm{SL}(2, \mathbb Z_p)$, and I'm almost certain that $\mathrm{SL}(2, \mathbb Q)$ is supposed to be $\mathrm{SL}(2, \mathbb Q_p)$. $\endgroup$
    – LSpice
    Apr 2, 2018 at 1:06
  • $\begingroup$ @LSpice, yikes, yes, indeed! $\endgroup$
    – paul garrett
    Apr 2, 2018 at 12:28

Your Answer

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

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