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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.

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2 Answers 2

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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.

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    $\begingroup$ It seems like it might be a good idea to combine your two answers, rather than to make two separate answers. $\endgroup$
    – LSpice
    Apr 26, 2021 at 15:11
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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.

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

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