## Structure of cuspidal Bernstein components---do non-commutative endomorphism rings ever really show up?

Let $F$ be a finite extension of $\mathbf{Q}_p$ with integers $\mathscr{O}$, let $\mathbb{G}$ be a connected reductive group over $F$ and let $G=\mathbb{G}(F)$ be its $F$-points. Let $X(G)=Hom(G,GL(1))$ denote the group of algebraic characters of $G$ and let $G^0$ denote the intersection of $\chi^{-1}(\mathscr{O}^\times)$ as $\chi$ runs over $X(G)$. For example, if $G=GL(n,F)$ then $G^0$ is the elements of $G$ with determinant equal to a unit. Informally, $G^0$ is generated by the derived subgroup of $G$ and the maximal compact subgroup of the centre of $G$.

Let $\pi$ be an irreducible supercuspidal representation of $G$ (over the complexes, as usual) and consider the restriction of $\pi$ to $G^0$. If $Z$ is the centre of $G$ then $G^0Z$ is a normal subgroup of finite index in $G$, and $G/G^0Z$ is abelian. Hence the restriction of $\pi$ to $G^0$ can be written as a finite direct sum of irreduibles $\pi_i$.

My question is: is there an example known where the isomorphism class of some $\pi_i$ occurs with multiplicity greater than one in $\pi$? That is---is any supercuspidal $\pi$ always "multiplicity-free" as a representation of $G^0$?

Why do I want to know this?

Those who know about Bernstein components will know the motivation behind this question. When analysing the cuspidal Bernstein component of the category of smooth representations of $G$ associated to $\pi$, life can be very easy (Schur orthogonality relations etc) if all matrix coefficients of $\pi$ are actually compactly-supported. But a general connected reductive group may have a non-trivial centre, whose $F$-points are typically non-compact, which means compactly-supported matrix coefficients are very rare. The point about $G^0$ is that it has compact centre so these problems go away, but the representation theory of $G^0$ is very close to that of $G$. However, when moving from the study of the representation theory of $G^0$ to $G$ one has to induce back up; a typical ring that shows up in this procedure is $R:=End_G(Ind_{G^0}^G(\pi_i))$ for a $\pi_i$ as in the question. If $\pi_i$ shows up (up to isomorphism) with multiplicity greater than one then this ring will be non-commutative (in fact this is an iff). The Bernstein component corresponding to $\pi$ is isomorphic to the category of right $R$-modules, and if $R$ is commutative then it seems to me that it's always isomorphic to the algebraic functions on a product of $GL(1,\mathbf{C})$'s so it's a really "easy" ring, but the literature seems to leave open the possibility that non-commutative $R$s can occur.

What do I know?

If $G/G^0$ is isomorphic to the integers then multiplicities greater than one cannot occur. In particular if $G=GL_n$ then multiplicities greater than one cannot occur. If $G$ is a product of $GL_n$'s then the same arguments apply and multiplicities greater than one cannot occur. If $\pi$ admits a Whittaker model then $\pi$ is multiplicity-free as a representation of $G_{der}$ and hence as a representation of $G^0$, and so again multiplicites cannot occur. I learnt this from Remark 1.6.1.3 of "The Bernstein decomposition and the Bernstein centre" by Alan Roche. Beyond this I know nothing.

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 Do we expect that non-commutative R's can occur? (ie Kevin, did your real life conversation you mentioned below indicate in which direction this question might be settled?) – Peter McNamara Mar 17 2011 at 7:45

For classical groups, V. Heiermann has proved that $Res^G_{G^0}(\pi)$ has no multiplicity, see :

• Opérateurs d'entrelacement et algèbres de Hecke avec paramètres d'un groupe réductif p-adique - le cas des groupes classiques - Selecta Mathematica.

In general, the endomorphism algebra $R$ is a free module of rank $m^2$ over its center Z, where $m$ is the multiplicity of an irreducible in $Res^G_{G^0}(\pi)$. See p. 181 in my book "Représentations des groupes réductifs $p$-adiques", SMF, cours spécialisés 17. I suspect that $R$ and $Z$ could be Morita equivalent.

$Z$ is always indeed isomorphic to the ring of algebraic functions on a product of GL(1,C)'s.

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 Thanks for your answer. I tried to get hold of your book recently (we had a study group on the Bernstein Centre in London) but it wasn't in any of the libraries I tried :-/ I should redouble my efforts! – Kevin Buzzard Apr 5 2011 at 20:24

First, for generalities related to the question, I'd recommend the article of Bushnell and Henniart, " Generalized Whittaker Models and the Bernstein Center," in Amer. J. of Math., Vol. 125, No. 3, Jun. 2003, pp.513-547. Starting in Section 8, you can find a nice and explicit survey and expansion on the results of Bernstein. Some of the "what do I know" facts you mention can be found in this paper of Bushnell-Henniart (see Example 8.6).

I started to write a complete answer here and accidentally hit the "Post" button. It seems more difficult than I thought to cook up an example. I'll try to think more about it, and I'll post at a later time/date.

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Aaahhhh! I clicked on post accidentally! How do I undo that... Now I really have to finish this answer!! – Marty Mar 15 2011 at 18:34
I'll try to finish the answer later! – Marty Mar 15 2011 at 19:07
Marty: it might not be so easy. I was in Oberwohlfach with Bushnell and Henniart last week and I asked them both. Colin seemed to think that this question might have been resolved relatively recently, but I can't remember if he gave a precise reference. I wondered whether someone else could fill me in on any details---what I'm saying is that an answer might well be of the form "see this recent paper" rather than "here's an example I just cooked up". Of course I could be wrong! PS you can just delete any answer you post here as far as I know---and then undelete it later if you feel like it! – Kevin Buzzard Mar 15 2011 at 20:44
I think you're right -- it's not so easy, and I'm not aware of a reference. This sort of question seems related to general questions along the line of "how do Bernstein components behave under isogeny?" I'll let you know if I think of anything... – Marty Mar 16 2011 at 0:21

I don't have an answer, and can't seem to figure out how to post as a comment above. So sorry for writing this in the space reserved for answers.

Anyway what I wanted to say is, the paper of Adler and (Dipendra) Prasad "On Certain Multiplicity One Theorems", in Israel Journal of Mathematics says that restrictions from GSp to Sp and GO to O of irreducible admissible representations decompose with multiplicity one.

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 Sorry that is probably useless : for G = GSp, G/G^0 is isomorphic to integers via valuation of similitude, so that is covered by what you said. For GO, translating into a result about GSO may not be obvious. – anādimadhyānta Mar 18 2011 at 12:45 [it might be the case that a new user can't leave comments---this might be explanation.] – Kevin Buzzard Mar 21 2011 at 19:11 Sorry if you know this already, but it seems to be expected that the restriction of an irreducible admissible representation of a p-adic reductive group to its derived group is of multiplicity one : fields.utoronto.ca/programs/scientific/06-07/… Actually I haven't registered in mathoverflow, which should be why, I now suspect, I cannot leave comments. – anādimadhyānta Mar 23 2011 at 15:07