# Discrete Series representations for $SL_{2}$ over $p$-adic field.

I am working on the chamber homology for $SL(2,F)$, and stuck at some basic stuff on D.S. reps of $SL(2,F)$.

Let $I=\left( \begin{array}{cc} \mathcal{O}_{F} & \mathcal{O}_{F} \\ \varpi_{\mathbb{F}}\mathcal{O}_{F} & \mathcal{O}_{F}\\ \end{array} \right)\cap SL(2, F)$. Now, let $w_{0}= \left( \begin{array}{cc} 0 & -1 \\ 1 & 0 \\ \end{array} \right)$ and $w_{1}= \left( \begin{array}{cc} 0 & -\varpi^{-1}_{F} \\ \varpi_{F} & 0 \\ \end{array} \right)$, then $J_{0}=I \cup Iw_{0}I$ and $J_{1}= I \cup Iw_{1}I$ are the two maximal compact subgroups of $SL(2,F)$ where $\varpi_{\mathbb{F}}$ is the uniformizer.

Just wondering if anybody knows how can I induce a cuspidal reps(D.S.) from a charachter belong to $J_{0}$ or/and $J_{1}$?

• 1. The definitions of $J_0$ and $J_1$ look weird: shouldn't $$J_0=SL(2,\mathcal{O}) \textrm{ and } J_1=(w J_0 w^{-1})\cap SL(2,\mathcal{O})?$$ 2. The question is unclear: are you asking whether a d.s. representation of $SL(2,F)$ can be induced from a character of $J_i, i=0,1?$ – Victor Protsak Jun 29 '11 at 15:30
• No mate its as above $J_{0}=\left( \begin{array}{cc} \mathcal{O}_{\mathbb{F}} & \mathcal{O}_{\mathbb{F}} \\ \mathcal{O}_{\mathbb{F}} & \mathcal{O}_{\mathbb{F}} \\ \end{array} \right)\cap SL(2)$, $J_{1}=\left( \begin{array}{cc} \mathcal{O}_{\mathbb{F}}& \varpi_{\mathbb{F}}^{-1}\mathcal{O}_{\mathbb{F}} \\ \varpi_{\mathbb{F}} \mathcal{O}_{\mathbb{F}}& \mathcal{O}_{\mathbb{F}} \\ \end{array} \right)\cap SL(2)$. i.e.$J_{1}=wJ_{0}w^{-1}$. – Dragon Jun 30 '11 at 16:43
• 2- Yes, I am asking if a d.s. representation of $SL(2,F)$ can be induced from a character of $J_{i},i=0,1$ and How? – Dragon Jun 30 '11 at 16:58
• w is not elt of (finite) weyl gp. please remove extraneous tags. – Peter McNamara Jun 30 '11 at 17:35
• If the question is whether discrete-series representations can be induced from 1-dim. characters, then the answer is no. If ‘characters’ allows high dimension, then all so called depth-0 supercuspidals arise this way, as in Paul Garrett's answer. Positive-depth supercuspidals and the Steinberg representation do not arise in this way. – LSpice Jul 1 '11 at 14:15

Inducing a "cuspidal" repn from SL(2,o) produces a finite sum of supercuspidals of SL(2,F). The easiest "cuspidal" repns of SL(2,o) are the ones that factor through SL(2,k), where k is the residue field. The "cuspidal" repns of SL(2,k) can be quasi-explicitly produced via the finite-field version of the Weil/theta pairing, inducing non-trivial characters from a k-not-split $O(2)$ (corresponding to the unique quadratic extension of $k$). Even a simple counting procedure easily shows that induced repns cannot account for all the irreducibles of SL(2,k), so we know that "cuspidal" ones must be there. The Weil/theta correspondence trick happens to produce them.
• The construction described above (together with the similar construction for $J_1$) produces all of the so-called depth zero supercuspidal representations of SL(2, F). There is a nice complete treatment of this example in Joseph Rabinoff's notes: math.stanford.edu/~rabinoff/misc/building.pdf – JGordon Jul 2 '11 at 5:22