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Given an action $\alpha$ of $V$ a Lie group on $B$ a Fréchet space with seminorms $ \{ \| \cdot \|_j \} $, let $B^\infty$ be the space of smooth vectors. Is this dense in $B$? Can I guarantee it is non-empty? Is there any requirements on $G$ or $\alpha$ or $B$? For the case I am (supposed to be) working on right now, we also have that $\alpha$ is isometric for all the seminorms and strongly continuous.

Also, since I am almost illiterate on the subject of Lie groups, I would also be thankful for some easy to read references! Thank you so much.

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3 Answers 3

up vote 9 down vote accepted

The answer is "yes" quite generally, and the following argument can be found in many texts (Knapp, Wallach, Varadarajan, ...): for a test function $f$ on the Lie group $G$, there is the "averaged" action on any quasi-complete, locally convex repn space $\pi,V$ for $G$, namely, $f\cdot v=\int_G f(g)\,\pi(g)(v)\;dg$. These images are smooth vectors. The collection of all such is the Garding subspace. Taking $f$'s to run through an _approximate_identity_ shows density.

(In fact, Dixmier and Malliavin showed that the Garding subspace is the whole space of smooth vectors.)

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+1 for the paper of Dixmier and Malliavin, a pearl! This is (unfortunately) not well-known at all but a really nice statement. –  Stefan Waldmann Mar 14 '12 at 11:42
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If the group $G$ is compact, then your statement is true. More precise, let $A$ be a locally convex complete topological vector space and $G \times A \to A$ be a continuous action of $G$. Let $A_s \subset A$ be the sub vector space generated by finite-dimensional $G$-stable subspaces. Then $A_s$ is dense in $A$. This is a generalization of the Peter-Weyl theorem, compare Bröcker, tom Dieck, Representation of compact Lie groups, Theorem 5.7, p.141.

Vectors in $A_s$ are smooth, because actions of Lie groups on finite-dimensional spaces are smooth. So your question follows from that.

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In full generality, the answer is "no". If you have a unitary irreducible representation of a $p$-adic Lie group on a $p$-adic Banach space, then the locally constant vectors are usually not dense. There may even be no nonzero locally algebraic vector. The locally constant vectors are dense however if you have an $\ell$-adic Lie group acting on a $p$-adic space, with $\ell \neq p$ (this is a theorem of Vigneras).

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Of course, this all depends on what you precisely mean by "smooth". It's a theorem of Schneider and Teitelbaum that if you have a unitary irreducible representation of a p-adic Lie group on a p-adic Banach space, then the locally analytic vectors are dense. –  Laurent Berger Mar 14 '12 at 10:38
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