MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

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

Suppose $V_1$ and $V_2$ are two $(g,K)$ modules of some reductive group $G$ with maximal compact $K$. Let $P$ be the minimal parabolic of $G$, $U$ its unipotent part, and $u$ its Lie algebra. Suppose the quotients $V_1/uV_1$ and $V_2/uV_2$ are isomorphic as modules for the Levi component of $P$, then what else do we need to know to conclude that $V_1$ and $V_2$ are isomorphic as $G$ modules?

edit:Thanks for Kevin and Emerton's comments and sorry for the confusion about the base field. Here I'm assuming REAL reductive group.

share|cite|improve this question
up vote 4 down vote accepted

This is a comment, mostly on terminology that I think caused some confusion, not an answer, but I don't have enough "influence" to post this as a comment.

For real reductive groups, the Jacquet functor $V\mapsto J(V)$ (Jacquet-Casselman functor, Jacquet module, etc) is defined differently from the nonarchimedean case: it is $\varinjlim (V/n^iV)^*$, which is dual to the n-adic completion of V. Thus J(V) is a p-finite g-module (n is the nilradical, p is the parabolic Lie subalgebra) and the target category of the functor J is parabolic category O for g. On the one hand, this provides more structure: we get a g-module and, for example, the infinitesimal character of V can be read off the infinitesimal character of J(V) (Casselman-Osborne); more generally, V and J(V) have the same annihilator in U(g). On the other hand, J(V) is morally a highest weight module, not a Harish-Chandra module, so some information is lost. The 0th n-homology that this problem is asking about, V/nV, is just the the top layer of J(V), so even more information is lost. By the way, unlike the p-adic Jacquet functor, n-homology is not exact (there may be higher homology), but V/nV is always non-zero (short proof was given by Beilinson and Bernstein).

I assume that the modules V1 and V2 in the formulation were presumed to be simple?

share|cite|improve this answer
By the way, can we test irreducibility from J(V)? – user1832 May 24 '10 at 13:54
I don't think so: Jacquet module of a simple module may have multiplicities. – Victor Protsak May 25 '10 at 4:50

EDIT: I assumed the OP was asking about reductive groups over non-arch local fields. Emerton raises the possibility that the question is actually about groups over R or C, and he's probably right. So the answer below is probably irrelevant.

Do correct me if I'm wrong; I'm not an expert. But I thought that if $V$ was any supercuspidal representation of, say, $GL_2(\mathbf{Q}_p)$, then $V/uV=0$. So in fact you know very little about $V$ if you only know $V/uV$. Can't $V$ basically be recovered from $V/uV$ in the $GL_2$ case when it's principal series or Steinberg, and in the supercuspidal case you have nothing? Actually, even in the Steinberg case you might have trouble distinguishing $V$ from a 1-dimensional representation...

share|cite|improve this answer
This is very vague (but then again your question is also vague)---perhaps for GL_n on the Galois side, the Jacquet module sees something like the Frobenius action on the inertia-invariant vectors, so to see the rest of $V$ (assuming $V$ is irreducible) you'll need to know the inertia action, which somehow boils down to seeing the type of $V$. So perhaps you need to know "enough about the action of $K$". – Kevin Buzzard Apr 27 '10 at 8:43
Kevin, This question is about Harish-Chandra modules, i.e. real groups, not p-adic groups. There are no supercuspidals ... . – Emerton Apr 27 '10 at 12:21
Hah! He didn't say what base field he was over! – Kevin Buzzard Apr 27 '10 at 12:37
But the question does say $(g,K)$-module ... . – Emerton Apr 27 '10 at 13:14

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.