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The Lie group $G := SO(n, 1)$ acts on the hyperbolic space $\mathbb{H}^n$ by isometries. In particular, we have a representation of $G$ on $F := C^\infty(\mathbb{H}^n)$. I assume that $F$ is endowed with its standard Fréchet structure.

I am interested in classifying the finite dimensional representations that appear in $F$.

Let me denote by $\mathcal{H}_k$ the space of homogeneous (wave) harmonic polynomials on $\mathbb{R}^{n, 1}$ (Minkowski space). Each $\mathcal{H}_k$ is an irreducible finite dimensional representation of $G$. Let me identify $\mathcal{H}_k$ with the subspace of $F$ it induces by restriction on $\mathbb{H}^n$. The direct sum $\bigoplus_{k \in \mathbb{N}} \mathcal{H}_k$ is isomorphic as a $G$-space to the space of regular functions (in the sense of algebraic geometry) of $\mathbb{H}^n$. As a consequence, a simple argument using the Stone-Weierstrass theorem shows that $$ F = \overline{\bigoplus_{k \in \mathbb{N}} \mathcal{H}_k}. $$ How can one conclude from this that the $\mathcal{H}_k$ are the only finite dimensional irreducible subrepresentations of $F$?

In the compact setting this would be a simple consequence of the Peter-Weyl theorem but in the non-compact case I don't know how to do.

Thanks in advance for your answer!

-- Original question:

Assume F is a Fréchet space which is a representation of a simple Lie group G. Let V be a finite dimensional subrepresentation of G in F. Does there exist a closed subspace W of F which is stable for the action of G and such that $V \oplus W = F$?

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    $\begingroup$ I am leaving this as a comment for now since I don't have time to check details, but I think that if G is any locally compact group acting on $L^\infty(G)$, then the one-dimensional copy of the trivial representation is complemented as a submodule in the sense you describe if and only if $L^\infty(G)$ has a (left or right) invariant mean, i.e. G would have to be amenable. (I'll come back to this later when I have more time, and either delete the comment if I've made a silly mistake, or turn it into a proper answer.) $\endgroup$
    – Yemon Choi
    Commented Feb 17, 2016 at 13:58
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    $\begingroup$ Not without some further conditions. E.g., certain of the smooth principal series (not necessarily unitary or unitarizable) of $G=SL_2(\mathbb R)$ have the finite-dimensional irreducibles of $G$ as subrepns, but these are not complemented. There are intertwinings that have these finite-dimensional subs as kernels, yes, and the images are (anti-/) holomorphic discrete series, but there's no direct sum decomposition. (Yes, these repns are Frechet.) $\endgroup$ Commented Feb 17, 2016 at 15:39
  • $\begingroup$ Thanks a lot guys! I finally found a way out of my problem without using the question I was mentioning :). $\endgroup$ Commented Feb 17, 2016 at 17:53
  • $\begingroup$ Argh... nope... Let me state the original problem. $\endgroup$ Commented Feb 18, 2016 at 7:55

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