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 $X(G)=\operatorname{Hom}(G,\operatorname{GL}(1))$ 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 $G=\operatorname{GL}(n,F)$ 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 $R:=\operatorname{End}_G(\operatorname{Ind}_{G^0}^G(\pi_i))$ 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 $\operatorname{GL}(1,\mathbf{C})$'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.
 A: $\DeclareMathOperator\Res{Res}$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 $\operatorname{GL}(1,C)$'s.
A: 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.  
A: A lot of time has passed, so perhaps this question is moot for you [I mean, I saw you give the Plücker lectures last week on something very different],
or perhaps you now know the answer.  But since the question was of interest to me for a long time, and since I didn't see the answer until a few years after it was published, here it is for the record.
Multiplicity can occur.  Henniart and Vignéras give an example of a group $G$, a maximal compact subgroup $\tilde K$, and a depth-zero cuspidal representation (really, a character) $\chi$ of $\tilde K$ such that the Hecke algebra $H(G,\chi)$ is noncommutative.  This implies the existence of higher multiplicities.  The example is somewhat hidden by the title of the paper: A Satake isomorphism for representations modulo $p$ of reductive groups over local fields, J. reine angew. Math., 2015.
See §4.4, and in particular the remark there: the example can be made to work when the coefficient field is complex.
A: 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.
