Take the 2-minute tour ×  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 anal