# When representations of reductive Lie group in a Banach space and in its Garding space have the same length?

Let $$G$$ be a real reductive Lie group (e.g. $$G=\operatorname{GL}(n,\mathbb{R})$$). Let $$\rho$$ be a continuous representation of $$G$$ in a Banach space $$V$$. Let $$V^\infty\subset V$$ be the subspace of smooth vectors equipped with the Garding topology. Let $$\rho^\infty$$ be the natural representation of $$G$$ in $$V^\infty$$.

Under what precise technical conditions the representations $$\rho$$ and $$\rho^\infty$$ have the same length?

As far as I understand this situation is rather typical in the theory.

A reference would be very helpful.

I do not know if you need K-admissibility. You can check if it fails in the following example: write the usual Iwasawa for $$\operatorname{SL}_2(R)=KAN$$, $$MAN$$ usual parabolic. Fix $$T$$ a continuous linear operator in a Banach space without closed invariant subspaces (Enflo's example). Now, it is known that $$\operatorname{Ind}_{MAN}^{\operatorname{SL}_2(\mathbb R)}(1\otimes e^{T\cdot}\otimes 1)$$ is irreducible and it is not admissible. Let me call to your attention Section 11.8 in Wallach Real reductive groups II.

• It seems like it might be a good idea to combine your two answers, rather than to make two separate answers. Apr 26, 2021 at 15:11

If you consider $$V^\infty$$ with its Fréchet topology. and irreducibility means no closed invariant subspaces except for the trivial ones. Then $$V$$ is irreducible iff $$V^\infty$$ is irreducible. In the book Warner, Harmonic Analysis on semisimple Lie groups, Chapter 4, you will find more about this.

• Does it require some admissibility of $V$?
– asv
Apr 24, 2021 at 17:00
• @makt's comment is addressed in another answer. Apr 26, 2021 at 15:14