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Let $(J,\pi)$ be a cuspidal type in $SL(2,F)$, $F$ is a non-Arch. local field and let $I$ be the Iwahori subgroup of $SL(2,F)$.

Is there any possibility that $J\subset I$ or even a subgroup?

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up vote 5 down vote accepted

Here is a novel, explicit answer to your question :

Benedict Gross and Mark Reeder have recently discovered a family of supercuspidal representations, called "simple supercuspidal representations", of simply connected, split, almost simple, reductive groups. These representations are examples of what you are looking for.

Here's an example of a simple supercuspidal representation, and the general case is similar. Consider $G(\mathbb{Q}_2) = SL(2,\mathbb{Q}_2)$. Let $I$ be an Iwahori subgroup of $SL(2,\mathbb{Q}_2)$, and let $I^+$ be the pro-unipotent radical of $I$. For example, we can take

$$I = \begin{bmatrix} \mathfrak{o} & \mathfrak{o}\\\ \mathfrak{p} & \mathfrak{o} \end{bmatrix}$$

$$I^+ = \begin{bmatrix} 1 + \mathfrak{p} & \mathfrak{o}\\\ \mathfrak{p} & 1 + \mathfrak{p} \end{bmatrix}$$

Let $\chi$ be an "affine generic character" of $I^+$. Since we are dealing with $SL(2,\mathbb{Q}_2)$, this just means the following : Let $\eta$ be the character of $\mathbb{Z}_2$ given by $$\eta : \mathbb{Z}_2 \rightarrow \mathbb{C}^*$$ $$2 \mathbb{Z}_2 \mapsto 1$$ $$1 + 2 \mathbb{Z}_2 \mapsto -1$$ Then define the character

$$\chi : I^+ \rightarrow \mathbb{C}^*$$ $$\begin{bmatrix} a & b\\\ 2c & d \end{bmatrix} \mapsto \eta(b) \eta(c)$$

Then $Ind_{I^+}^{SL(2,\mathbb{Q}_2)} \chi$ is a supercuspidal representation (which Gross and Reeder call a "simple supercuspidal representation", since it is so "simple" to define). The general situation is similar : If $G$ is simply connected, split, almost simple, $F$ is a $p$-adic field, $I \subset G(F)$ is an Iwahori subgroup, $I^+ \subset I$ is the pro-unipotent radical, and $\chi : Z I^+ \rightarrow \mathbb{C}^*$ is an "affine generic character" (where $Z$ is the center of $G(F)$), then $Ind_{Z I^+}^{G(F)} \chi$ is a supercuspidal representation (called a "simple supercuspidal representation"). For more details, check out Section 9 of

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In your example with ${\mathbb Q}_2$, I think you need either to extend your character to the center $\pm I_2$ first, or replace ${\mathbb Q}_2$ by ${\mathbb F}_2 ((X))$. – Paul Broussous Jul 6 '11 at 19:08
In this case (unless I'm mistaken) I don't think I need to, since $-1 \in 1 + \mathfrak{p}$, since $-1 = 1 + 2 + 2^2 + 2^3 + 2^4 + ...$ Thus, the center of $SL(2, \mathbb{Q}_2)$ is already in $I^+$, so $Z I^+ = I^+$. – Moshe Adrian Jul 6 '11 at 19:46
Sorry, you're totally right ! – Paul Broussous Jul 7 '11 at 8:05

Yes this is indeed possible. You may read the proof of theorem (2.1.) of

Bushnell and Kutzko, The admissible dual of ${\rm SL}(N)$ - I Ann. Sci. École Norm. Sup. (4) 26 (1993), no. 2,

where the supercuspidal representation are described as induced representations from a subgroup wich can always be taken of the form $U({\mathfrak A })\cap {\rm SL}(N)$, where $\mathfrak A$ is a principal order in ${\rm M}(N,F)$.

But of course, you may replace the group $J$ by a maximal compact subgroup by inducing the type to a maximal compact subgroup $K$ containing $J$.

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While possible it only can happen if the quadratic extension associated to the type is ramified.

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The Iwahori is a type for the trivial representation on the Levi for $SL_2(F)$. Moreover all the types for $SL_2(F)$ of non supercuspidal bernstein components are subgroups of the Iwahori. There is a paper by Kutzko where he describes all the types for $SL_2(F)$.

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