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?
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3 Answers
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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, ...).
answered Jul 1, 2012 at 19:19
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$\begingroup$ So in the last statement, the reductive group is defined over a global field a priori? $\endgroup$ Commented Jul 2, 2012 at 16:44
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$\begingroup$ No, in all statements G is simply assumed to be defined over a local field. $\endgroup$ Commented Jul 3, 2012 at 5:23
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$\begingroup$ How does one define then large enough? $\endgroup$ Commented Jul 3, 2012 at 8:20
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1$\begingroup$ Roughly speaking large enough is so that all the field extensions involved are tamely ramified. $\endgroup$– manderCommented Jul 3, 2012 at 13:25
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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.
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$\begingroup$ This answer really belongs as a comment on Broussous's answer. $\endgroup$– MartyCommented Jul 2, 2012 at 3:11
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$\begingroup$ That would be great indeed ! $\endgroup$ Commented Jul 3, 2012 at 5:21
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Not known. See the work of J. Kim (JAMS) or S. Stevens (Inventiones) for some general results.