# 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

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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 $V$.
It may be that you're right, I admit that this part in this book (actually some other as well) is written so loosely that I didn't try to double check everything. Having said this, to quote it, it's written "The representation of G in $V^\omega$ obtained in this way can be extended to $V$" (p. 155). Now for my question, it's modest I think : given a $U(\mathcal(G))$ -module, why can we extend it to a $G$-module ? Perhaps it's a trick that I'm missing. –  Amin Feb 14 '13 at 19:33