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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.

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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?

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  • $\begingroup$ By the way, can we test irreducibility from J(V)? $\endgroup$
    – user1832
    Commented May 24, 2010 at 13:54
  • $\begingroup$ I don't think so: Jacquet module of a simple module may have multiplicities. $\endgroup$
    – Victor Protsak
    Commented May 25, 2010 at 4:50
  • $\begingroup$ Do Jacquet modules in the real case also kill off the discrete series (as they do cuspidals in the non-Archimedean case)? If so, then connecting irreducibility of a representation and its Jacquet module can fail "the other way": the Jacquet module of a reducible module may be irreducible. $\endgroup$
    – LSpice
    Commented Nov 26, 2018 at 16:07
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    $\begingroup$ No, at least not generically. For example, consider a holomorphic discrete series representation $V=V_k, k\geq 2$ for $G={\rm SL}(2,\Bbb {R})$. Choose the standard basis $\{e,h,f\}$ for $\frak {g}=\frak {sl}_2$. Then $V$ is a lowest weight module with lowest weight $k$, the subspace $e^{i}V$ is spanned by the vectors of weight at least $k+2i$, and so the Jacquet module with respect to the nilradical $\frak{n}$ of the standard (upper) Borel, as defined above, is the contragredient $V^*$. $\endgroup$
    – Victor Protsak
    Commented Nov 27, 2018 at 7:12
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    $\begingroup$ However, an analogous computation with $f$ shows that the Jacquet module with respect to the nilradical of the lower Borel is $0.$ The subtlety here is that, unlike in the $p$-adic case, the Jacquet functor depends on the choice of the parabolic: only the action of $\frak{k}$ exponentiates to the group, and so the isomorphism class of the Jacquet module is determined by the $K$-orbit of $\frak{n}$. The Beilinson-Bernstein result mentioned in my answer asserts that for an open dense orbit of $K$ on the flag variety the corresponding $\frak{n}$-homology and Jacquet module are non-zero. $\endgroup$
    – Victor Protsak
    Commented Nov 27, 2018 at 7:19
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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...

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  • $\begingroup$ 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 $K$". $\endgroup$
    – Kevin Buzzard
    Commented Apr 27, 2010 at 8:43
  • $\begingroup$ Kevin, This question is about Harish-Chandra modules, i.e. real groups, not p-adic groups. There are no supercuspidals ... . $\endgroup$
    – Emerton
    Commented Apr 27, 2010 at 12:21
  • $\begingroup$ Hah! He didn't say what base field he was over! $\endgroup$
    – Kevin Buzzard
    Commented Apr 27, 2010 at 12:37
  • $\begingroup$ But the question does say $(g,K)$-module ... . $\endgroup$
    – Emerton
    Commented Apr 27, 2010 at 13:14

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