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Let $G$ be a real reductive Lie group, $P$ its parabolic subgroup with Levi decomposition $P=MN$, let $\mathfrak{n}$ be the nilpotent Lie algebra of $N$. Suppose given a smooth representation $(\pi,V)$ of $G$ with $V$ some Frechet space. We can form both $V/(\mathfrak{n}V)$ and $V/(N.V),$ where
$$ N.V=\operatorname{span}\{n.v-v|v\in V, n\in N\} $$

(here we take the closure of $\mathfrak{n}V$ and $N.V$ in the two quotients). Now question is there any difference between these two quotients? or equivalently, what's the relation between the closure of $\mathfrak{n}V$ and that of $N.V$? On one direction we have $\mathfrak{n}V$ is contained in $N.V$, so the question amounts to ask whether there is some $V$, so that the latter contains the former properly.

A related phenomenon is that, when considering the minimal parabolic subgroup $B=TU$, if $\chi$ is a generic character of $U$, $\eta$ the derivative of $\chi$, so $\eta$ is a nondegenerate complex linear form on the Lie algebra $\mathfrak{u}$ of $U$. There are two versions of Whittaker functional in literature, defined either in terms of pair $(U,\chi)$ or $\mathfrak{u},\eta$. And I'm wondering if they are equivalent.

Edit (Victor Protsak): definition of N.V has been made explicit.

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  • $\begingroup$ By $N.V$, do you mean the space generated by $n\cdot v-v$ (a la the p-adic Jacquet module)? $\endgroup$
    – B R
    Mar 29, 2011 at 17:22
  • $\begingroup$ yes, I should be careful. $\endgroup$
    – user1832
    Mar 29, 2011 at 17:25
  • $\begingroup$ If the Jacquet module is finite-dimensional then it follows from the Lie correspondence that the two spaces are the same; this is certainly the case if $V$ is irreducible. Is this sufficient for your purposes or are you interested in a more general situation? $\endgroup$
    – Victor Protsak
    Mar 29, 2011 at 19:41
  • $\begingroup$ Thanks Victor. May I know that's the Lie correspondence you mentioned? To my knowledge, the Jacquet module is finite dimensional in case of maximal nilpotent Lie subalgebra (i.e.,\mathfrak{u}), even if $V$ is not irreducible. But does this still hold for general nilpotent $\mathfrak{n}$? $\endgroup$
    – user1832
    Mar 29, 2011 at 21:29

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Maybe I'm misreading the set-up, but the answer to the first question seems to be obviously no. Since the group $N$ contains the identity element, the quotient by the closure of $N.V$ must be zero. On the other hand, when $V$ is finite dimensional and irreducible for $SL_2(\mathbb{R})$, the subspace $\mathfrak{n}.V$ is the sum of all but the lowest weight space and the quotient is nonzero.

ADDED: Assuming your interest is in the real groups and their infinite dimensional representations (with some added structure), as opposed to $p$-adic analogues, there are a lot of papers to consult. Being a non-specialist in all of this, the only source that comes to mind is a long paper by Kostant, On Whittaker vectors and representation theory, lnvent. Math. 48, 101- 184 (1978). That might be relevant to your comparison between the group and Lie algebra situations.

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  • $\begingroup$ Thanks for your post, Jim. It should be $N.V-V$ in stead of $N.V$. When $V$ is a nice smooth representation, the subrepresentation theorem says that the quotient of $V$ by $\mathfrak{n}V$ is never zero. $\endgroup$
    – user1832
    Mar 29, 2011 at 17:30

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