TITLE: Unitary representations of non-compact Lie groups
QUESTION [5 upvotes]: This question is somewhat of a continuation of this question that I had asked earlier -
Representations of a non-compact group are labeled by its maximal compact subgroup?

I want to know when or is it always true that an unitary representation of a non-compact Lie group is infinite dimensional? If yes then why? If no then  kindly give examples. 
Also when can one be sure that an infinite dimensional representation of some non-compact Lie group is labeled by a single representation of its maximal compact subgroup ?

REPLY [3 votes]: In addition to the existing answer, let me give a proof of the fact that any smooth, finite-dimensional unitary representation $\rho: G \to GL_n(\mathbb C)$ of a non-compact, simple Lie group is always trivial. 
We will crucially make use of the highly non-trivial fact that a finite-dimensional representation of a semisimple Lie group has closed image in $GL_n(\mathbb C)$, see for example Corollary 14.5.7 in Hilgert, J., Neeb, K.-H.: Structure and Geometry of Lie Groups. Springer, New York, (2013). This is false in general, as Stephen has pointed out in the comments below. 
Thus, let $\rho: G \to GL_n(\mathbb C)$ be a unitary representation, i.e, we may assume that $\rho(G) \subseteq U(n)$. Since $G$ is simple and $\ker(\rho) \subseteq G$ is a normal subgroup, either $\ker(\rho) = \{1\}$ or $\ker(\rho) = G$. In the second case, $\rho$ is the trivial map and we are done. 
We want to show that the first case $\ker(\rho) = \{1\}$ cannot hold, i.e, $\rho$ cannot be injective. For this, note first that $\rho(G) \subseteq U(n)$ is a closed subgroup by the first paragraph. This implies by Lie's Theorem that $\rho(G)$, equipped with the subspace topology of $G$, is a Lie group.
Since a group admits, up to diffeomorphism, at most one Lie group topology, this implies that that the injective map $\rho: G \to U(n)$ is a diffeomorphism onto its image, so that in particular, the inverse map $\rho^{-1}: \rho(G) \to G$ is continuous as well. However, $\rho(G) \subseteq U(n)$ is compact as a closed subspace of the compact group $U(n)$, so $G = \rho^{-1}(\rho(G))$ would be compact as well, a contradiction.
Edit: In my original answer, I was unaware of the fact that a faithful representation $\rho: G \to GL_n(\mathbb C)$ might not have closed image for general non-compact Lie groups $G$, which Stephen kindly reminded me of. Valter Moretti, over at Physics.stackexchange, provided me with the appropriate reference that the result holds whenever $G$ is semisimple.