Is a cyclic subspace of a compact unitary representation finite dimensional? Let $K$ be a compact Lie group and let $\rho_k: H \rightarrow H$ be a (strongly continuous) unitary representation of K on a Hilbert space H. Why does the orbit, $\rho(K)v$ ,of any $v\in H$ generate a finite dimensional subspace of H?  Is it because such a subspace is irreducible? Or is it because a compact subset (the orbit of v) of the unit sphere in H is contained in a finite dimensional space?
Or maybe it's not true?
 A: It is false. For instance, let $G$ be any second countable compact group, and let $\lambda$ be the regular representation of $G$ on $L^2(G)$ by left translation:
$$[\lambda(x)f](y) = f(x^{-1}y) \qquad (f \in L^2(G);\, x,y \in G).$$
Then $\lambda$ is cyclic. In fact, a representation of $G$ is cyclic if and only if it is unitarily equivalent to a subrepresentation of $\lambda$. See this article by Greenleaf and Moskowitz. 
For example, let $\mathbb{T}$ be the group of unimodular complex numbers under multiplication. Let $f \in L^2(\mathbb{T})$ be any function with non-vanishing Fourier coefficients. That is, $\hat{f}(k) \neq 0$ for all $k \in \mathbb{Z}$. Then $f$ is a cyclic vector for the regular representation of $\mathbb{T}$. This is a consequence of Wiener's classification of translation invariant spaces in terms of the support of the Fourier transform. Namely, to a subset $E \subset \mathbb{Z}$ we associate the space
$$ V_E = \{ g \in L^2(\mathbb{T}) : \hat{g}(k) = 0 \text{ for $k \notin E$}\}. $$
Then $E \mapsto V_E$ is a bijection between subsets of $\mathbb{Z}$ and (closed) translation invariant subspaces of $L^2(\mathbb{T})$. Since the support of $\hat{f}$ is all of $\mathbb{Z}$, the only invariant subspace that can contain $f$ is $V_\mathbb{Z} = L^2(\mathbb{T})$ itself.
By the way, cyclic representations need not be irreducible. You could use the example above to see this, but there are much simpler examples using finite groups (for instance, $\mathbb{Z}/2\mathbb{Z}$).
