# P-adic representations corresponding to the same cuspidal pair

Let $$G(F)$$ be a reductive $$p$$-adic group. A result of Bernstein says that we can correspond each smooth irreducible representation to a “cuspidal pair” where it is embedded, and at most finitely many smooth irreducible representations correspond to the same pair.

My question is: Can we distinguish between them in a natural way? Specifically, I was thinking that maybe the following is true: For any two irreducible representations $$V,V’$$ corresponding to the same cuspidal pair, there exists some compact open $$K$$ such that $$V$$ has $$K$$-fixed points while $$V’$$ doesn’t, or the converse. Or at least that there exists some compact open K such that the dimensions $$dimV^K$$ and $$dimV’^K$$ are different.

Does someone know a proof of the above statements somewhere? Or any result like that?

Edit: I actually realized I may have a proof of that but does it already exist?

• By the way, the language "$p$-adic representation" can be misleading; there are plenty of people who consider actions on $p$-adic vector spaces, rather than (as you seem to mean) actions of $p$-adic groups on complex vector spaces. (I prefer the explicit "representations of $p$-adic groups" for exactly this reason.) – LSpice Dec 6 '18 at 19:16
• My impression is that, if you allow yourself too much flexibility with the choice of $K$, the fixed-point dimensions are fairly uncontrollable, and aren't likely to be very useful. I think a distinction based on the existence or non- of non-$0$ fixed vectors is likely to be more useful (see, for example, the Moy–Prasad definition of the depth of a representation). – LSpice Dec 6 '18 at 20:40
• I also believe that but the dimensions of the fixed point spaces is an intermediate result which is enough for the purpose I want it. I do believe it should be enough to know just the triviality or non-triviality of them though. Do you know of any result in this case? – Ioannis Zolas Dec 6 '18 at 21:11
• I do not. I defer to @PL.'s answer. – LSpice Dec 6 '18 at 21:30

## 1 Answer

I don't think anything quite so simple should work in general, but something close to this should (conjecturally).

Consider the cuspidal pair in GL(2) consisting of the trivial representation of the Borel subgroup. Parabolically induce this and you get a length two indecomposable representation. Its unique sub is the trivial rep, and the quotient is the Steinberg. It's easy to check that while these both have vectors fixed by the Iwahori subgroup, only the trivial rep has vectors fixed by GL(2,O). The same kind of thing holds for GL(n), with the various representations containing the cuspidal pair (Borel,trivial) being picked out by which subgroups intermediate between the Iwahori and GL(n,O) they have fixed vectors under.

More generally, in GL(n) a cuspidal pair determines a semisimple type in the sense of Bushnell--Kutzko. This is a representation of a compact open subgroup J of G which can be taken to be contained in GL(n,O). Induce it up to GL(n,O) and you'll get a decomposable representation in general. Given a component of this induced representation, the property of containing this component upon restriction to GL(n,O) will single out a representation containing your cuspidal pair.

The same most likely holds for general groups, but this relies on lots of conjectures about being able to perform analogous constructions to the above. In general groups, there is also some uncertainty about choices which must be made.

• A cuspidal representation is (usually, and conjecturally always) induced; and can't one produce a cover of the inducing datum as a type on the big group? (I'm not up on the state of the art on types.) – LSpice Dec 6 '18 at 19:26
• Under reasonable assumptions (p coprime to order of Weyl group), or for various special cases of cuspidal supports (e.g. corresponding to maximal proper parabolics in classicals), yes. Almost certainly this all should work because conjecturally there is an "inertial Langlands correspondence" relating the components of reps of maximal compacts induced from semisimple types to reps of inertia + monodromy operators. Monodromy operator in an L-parameter tells you which component of parabolic induction you're working with on p-adic side, and the choice of irr component should be doing the same... – PL. Dec 6 '18 at 19:32
• ...But I think to make sense of anything like this requires an awful lot of work, doing the analogue of the Bernstein--Zelevinsky classification. And then what if your semisimple type is contained in maximal parahorics of different types (say both hyperspecial and not)? Then maybe you need to work at the right parahoric to see the picture. I think this is probably very hard in general! – PL. Dec 6 '18 at 19:34
• For a fixed cuspidal pair then I wouldn't be surprised if it suffices to look at dimensions of fixed points under any compact open. But there are a huge number of strange-looking compact opens, and so that's kind of a useless thing to ask really. If you really want to pursue this then it might be worth looking at the Schneider--Stuhler paper on coefficient systems, where they show that a representation is determined by the homology of the coefficient system on the building given by vectors invariant under various pro-p subgroups. But I'm not sure how hard it will be to spot any connection. – PL. Dec 6 '18 at 22:44
• Right. But a type lives on J, which isn't maximal compact. Induce it up to a maximal compact K. You'll get a sum of reps which aren't types, but are "typical reps". Each of these is precisely picking out a component of the representation parabolically induced from your cuspidal pair. As I've said previously, this is the correct way to think about it. I don't think that merely looking at dimensions will be sufficient unless you're willing to look at ridiculous subgroups, at which point you're just looking at the semisimple type in disguise. – PL. Dec 8 '18 at 20:36