Let $G$ be a connected semisimple Lie group with finite center. Let $(\pi,V)$ be an admissible representation on a Banach space $V$. Is it true that the following are equivalent?

(a) $\pi$ is irreducible

(b) the $({\mathfrak g},K)$-module of $K$-finite vectors is simple.

I can prove this for unitary representations. If the answer is positive, where do I find a simple proof? Say, a proof that does not use the classification of admissible $({\mathfrak g},K)$-modules?


After some thinking, I realised that my solution for the unitary case also works in general. The difficult direction is (a)$\Rightarrow$(b): Let $K$ be a maximal compact subgroup and let $V_K$ denote the space of $K$-finite vectors. and let $0\ne U\subset V_K$ be a $({\mathfrak g},K)$-submodule. Let $V^\omega$ denote the space of analytic vectors, which is dense in $V$. After projection, the space $V^\omega(\tau)=V^\omega\cap V(\tau)$ is dense in the $K$-isotype $V(\tau)$ for any $\tau\in\widehat K$. Since $V(\tau)$ is finite-dimensional by admissibility, we get that every $K$-finite vector is analytic. Hence for given $u\in U$ and $X\in\mathfrak g$ the curve $\gamma:t\mapsto \pi(\exp(tX))u$ is analytic. It satisfies $\gamma^{(n)}(0)=\pi(X)^nu\in U\subset\overline U$ for every $n\ge 0$. As $\gamma(t)=\sum_{n=0}^\infty\gamma^{(n)}(0)t^n/n!$ for small $t$, it follows that $\pi(g)u\in \overline U$ for all $g$ in some small unit-neighborhood in $G$. By continuity it follows $\pi(g)\overline U\subset \overline U$ for every $g$ in that unit-neighborhood and as this unit-neighborhood generates $G$, we get that $\pi(G)$ preserves $\overline U$. By irreducibility, $\overline U=V$ and so, by finite-dimensionality of the $K$-type $V(\tau)$ we have $V(\tau)=\overline U(\tau)=U(\tau)$, so $U=V_K$.

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