Reconstructing a Lie group Banach representation from the Lie algebra rep. on analytic vectors Dear all, 
I have some difficulties with the following assertion in the book of Kirillov.
Let $G$ be a connected Lie group, and T a given (!) representation of G on a Banach space V.
Let $V^\omega$ be the space of analytic vectors. 
Then T induces a representation of $U(\mathcal{G})$ on $V^\omega$, that I denote $T^.$. It is claimed that from $T^.$, we can reconstruct $T$.
If we were in the finite dimensional representations case, I would have no problem. Unfortunately, I don't see how to adapt the proof to the Banach case.
Actually, I don't see how to construct the action of $G$ on $V^\omega$.
By definition, for all $\xi\in V^\omega$, the map $g\mapsto T(g)\xi$ is analytic, so in a neighborhood $U_\xi$ of the neutral element $e$ of $G$, we can reconstruct this map out of $T^.$. That is, we can reconstruct $T()\xi$ on $U_\xi$. But I don't see how to extend this to $G$.
In the finite dimensional case, one doesn't have to take neighborhood depending on $\xi$, since the given $T$ is analytic as map between Lie groups.
Thanks 
 A: I'm not sure which Kirillov book you refer to here (it always helps to give a precise reference), but my understanding is that the basic theorem underlying all of this goes back to Harish-Chandra.   His density theorem is well explained in various books, as well as in his original work.  For instance, see section 5.2 in the 1989 Cambridge Press volume by V.S. Varadarajan An Introduction to Harmonic Analysis on Semisimple Lie Groups.    There are at least mild assumptions imposed on the Lie groups treated here, beyond being connected..   In any case, my limited knowledge comes from hearing various lectures over the years, so I can't say what the optimal reference is.    But keep in mind that some of Kirillov's expository work is informal and might not contain all the details of the standard proof.
A: In general you can't expect to get an action of $G$ on $V^\omega$. Instead, what one has is that if $U$ is a $U(\mathfrak g_{\mathbb C})$-invariant subspace of $V^\omega$ then its closure $\overline{U}$ will be $G$-invariant. This fact makes $V^\omega$ particularly useful---for example, the analogous statement is false for the space $V^\infty$ of smooth vectors. In any case, by applying this to $U=V^\omega$ and using the density theorem $\overline{V^\omega} = V$ mentioned by Jim Humphreys, we recover the $G$-action on all of $V$.
