## Is the real Jacquet module of a Harish-Chandra module still a Harish-Chandra module?

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Casselman defined the real Jacquet module for a Harish-Chandra module, if we view the Jacquet module as a module corresponding to the Levi subgroup, the question is is it still a Harish-Chandra module? In particular is it still admissible?

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## 1 Answer

As I understand it, the Jacquet module for $(\mathfrak g, K)$-modules is defined so as to again be a $\mathfrak g$-module, and in fact it is a Harish-Chandra module, not for $({\mathfrak g},K)$, but rather for $(\mathfrak g,N)$ (where $N$ is the unipotent radical of the parabolic with respect to which we compute the Jacquet module). (I am probably assuming that the original $(\mathfrak g, K)$-module has an infinitesimal character here.)

I am using the definitions of this paper, in particular the discussion of section 2. This in turn refers to Ch. 4 of Wallach's book. So probably this latter reference will cover things in detail.

Added: I may have misunderstood the question (due in part to a confusion on my part about definitions; see the comments below), but perhaps the following remark is helpful:

If one takes the Jacquet module (say in the sense of the above referenced paper, which is also the sense of Wallach), say for a Borel, then it is a category {\mathcal O}-like object: it is a direct sum of weight spaces for a maximal Cartan in ${\mathfrak g},$ and any given weight appears only finitely many times. (See e.g. Lemma 2.3 and Prop. 2.4 in the above referenced paper; no doubt this is also in Wallach in some form; actually these results are for the geometric Jacquet functor of that paper rather than for Wallach's Jacquet module, but I think they should apply just as well to Wallach's.

Maybe they also apply with Casselman's definition; if so, doesn't this give the desired admissibility?

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Thank, Emerton. The definition in Wallach's book is different from Casselman's, in some sense it's the dual of Casselman's. In Casselman's definition, it becomes a (g,P)-module, and in particular a module for the corresponding Levi. – unknown (google) Feb 10 2010 at 18:17
Sorry, I didn't know that.  Maybe ignore the above, then! – Emerton Feb 10 2010 at 18:25
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