# When is compact induction cuspidal?

Let $$G=GL_2(\mathbb{Q}_p)$$, and let $$K$$ be a compact-modulo-center subgroup of $$G$$, $$\rho$$ an irreducible smooth representation of $$K$$.

Question 1: Is $$\mathrm{ind}_K^G \rho$$ cuspidal?

Here cuspidal is meant in the sense that matrix coefficients are compactly supported.

More precisely, I know already from Bushnell-Henniart, The Local Langlands Conjecture for $$GL(2)$$, Thm. 11.4, that $$\mathrm{ind}_K^G \rho$$ is irreducible cuspidal if the following condition holds: $$g\in G$$ intertwines $$\rho$$ if and only if $$g\in K$$. This condition is obviously necessary for irreducibility, but is it also necessary for cuspidality?

Question 2: For which groups $$G$$ other than $$GL(2)$$ does this result hold?

Suppose

(1) $$K$$ is open and contains a finite index subgroup of the center

(2) For every parabolic subgroup $$P$$ of $$G$$ with unipotent radical $$N$$, $$\rho^{ N \cap K}=0$$.

Then $$\rho$$ is cuspidal (i.e. a finite direct sum of super cuspidal representations).

I think the argument is (or finishes on) page 28 of the book Harmonic analysis on reductive $$p$$-adic groups by Harish-Chandra, if you can understand the notation.

But my understanding of the argument is that it is easy to check from (2) and the definition of the Jacquet module that the Jacquet module vanishes. If the Jacquet module vanishes, and the representation is admissible, then it is a finite direct sum of supercuspidals. To check that the induced representation is admissible, we must find its $$K'$$ invariants, which by Frobenius reciprocity reduces to finding the $$g K' g^{-1} \cap K$$ invariants for many all $$g \in K' \backslash G/K$$. Using the (affine) Bruhat decomposition you can check that, for all but finitely many $$g$$'s, $$g K' g^{-1} \cap K$$ contains $$N \cap K$$ for $$N$$ the unipotent radical of a parabolic subgroup.

That these conditions are necessary is straightforward.

A fun (?) exercise is to reprove that certain known constructions of irreducible supercuspidal representations are at least supercuspidal using this method instead of the $$\rho$$-intertwining condition.

• In Induced representations of locally profinite groups. J. Algebra 134 (1990), no. 1, 104–114, Bushnell studies the admissibility of compactly induced representations – Paul Broussous Jun 12 '19 at 6:29
• When the maximal compact subgroup of $K$ is a special maximal compact subgroup of $G$ in Bruhat-Tits's terminology, then one can prove that conditions (1) and (2) are sufficient for the cuspidality of the induced representation. But I think this is not written anywhere. – Paul Broussous Jun 12 '19 at 6:31
• @PaulBroussous Do you think the special condition is essential in any way? it seems to me that the same proof works in each case. – Will Sawin Jun 12 '19 at 14:03
• Savin Harish-Chandra's proof relies on the fact that the maximal compact subgroup is special : he works under the hypotheses of his Theorem 5 page 16. – Paul Broussous Jun 12 '19 at 14:40
• @PaulBroussous do you think that the proof cannot easily be modified to handle the more general case? – Will Sawin Jun 12 '19 at 16:15

The answer to the first question is no. The representation $$V=\operatorname{ind}_K^G\rho$$ is never cuspidal when $$\rho$$ is trivial. In this case V consists of smooth compactly supported functions on K\G while its contragredient consists of smooth functions on K\G. The matrix coefficient corresponding to the indicator functon of K and the constant function is not compactly supported modulo centre.

There is a positive answer to the second question. Bushnell-Henniart state in their first remark after Theorem 11.4 that this result is true assuming unimodularity of the locally profinite group G, with G/K countable for compact open G, such that every irreducible representation is admissible. Thus it holds for every reductive group over a local field, as well as their finite central extensions. A reference for the most difficult part, that irreducible implies admissible for reductive G, is Renard's book: Théorème VI.2.2.

• The trivial representation that you mention is intertwined by all elements of G. Could it be that the induced representation is cuspidal if, say, only finitely many g (modulo K) intertwine \rho? – nikola karabatic Jun 14 '19 at 8:22