Question

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 $J_{1}$?

Answer 1

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.

This kind of discussion already appeared a long time ago, I think in Jacquet's 1970 Montecatini lectures. In more recent times, work of Kutzko et al classifies supercuspidals of GL(n).