MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $G$ be a reductive group over a local non-archimedean field $F$.

Can every irreducible supercuspidal representation of $G(F)$ be realized as the induction from an open subgroup, which is compact modulo the center?

share|cite|improve this question
up vote 8 down vote accepted

It is known for GL(N) and SL(N) (Bushnell and Kutzko), for classical groups when the residue characteristic is not $2$ and when no quaternionic algebra is involved (Stevens), for GL(N) of a division algebra (Stevens and Sécherre), for a general reductive group when the residue characteristic of $F$ is large enough (Kim, Yu, ...).

share|cite|improve this answer
So in the last statement, the reductive group is defined over a global field a priori? – Marc Palm Jul 2 '12 at 16:44
No, in all statements G is simply assumed to be defined over a local field. – Paul Broussous Jul 3 '12 at 5:23
How does one define then large enough? – Marc Palm Jul 3 '12 at 8:20
Roughly speaking large enough is so that all the field extensions involved are tamely ramified. – mander Jul 3 '12 at 13:25

It is perhaps worth noting that all the proofs Broussous mentions are along the lines of: Construct some types that induce irr scs and then show that all irr scs are thus obtained. In other words they are explict constructions. But that is stronger than the original question asked. One might hope that there would be a general proof that all irr scs are induced for reductive p-adic groups that doesn't produce the types.

share|cite|improve this answer
This answer really belongs as a comment on Broussous's answer. – Marty Jul 2 '12 at 3:11
That would be great indeed ! – Paul Broussous Jul 3 '12 at 5:21

Not known. See the work of J. Kim (JAMS) or S. Stevens (Inventiones) for some general results.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.