Unitary representations of non-compact Lie Groups

Question

I have often read the following statement:

Let $G$ be a connected, simple, non-compact Lie Group of dimension $n \geq 2$. Let $ρ: G \to U(H)$ be a unitary representation of $G$ on the Hilbert Space $H$. If $H$ is finite-dimensional, then $ρ$ is trivial.

Unfortunately, I have never found a complete proof. Does anyone know where I can find it or provide some hints?

• Moishe Kohan: Consider for instance Lie algebras and signatures of Killing forms of compact and noncompact simple Lie groups. Or unipotent elements...
• Luca: I see that the Killing form of the Lie algebra $\mathfrak g$ of $G$ must be non-degenerate but not negative-definite (so it has at least one positive eigenvalue) since $G$ is non-compact, while the Killing form on $\mathfrak{su}(n)$ is negative-definite, but I can't see how to proceed from here. Also I can't understand the second hint.
• Callum: @Luca Consider for the second hint that $\operatorname{U}(H)$ (or $\operatorname{SU}(H)$ where the image must lie in fact since $G$ is simple) has no unipotent elements except the trivial one, while any noncompact simple group must be generated by its unipotent elements (easy proof of the latter is to show the subgroup they generate is normal)
• Luca: @Callum Actually, if $\omega$ is a $n$-th root of unity and $k \in \mathbb Z,ドル the matrix $\omega^k I$ lies in $\mathrm{SU}(n)$ and is unipotent, should your argument work nonetheless? Also, I see that the subgroup generated by the unipotent elements in $G$ is normal, but I can't say why couldn't it just be $\{e\}$.
• Callum: @Luca, ah, good point, this is what I get for trying to port over Lie algebra concepts to Lie groups without proper consideration. What I said is true of nilpotent elements in the Lie algebra, so it should provide only a discrete set of unipotents in the group (I hope), which should then be amenable to our argument, or we could just do this over the Lie algebras for ease. The subgroup contains the unipotent elements themselves so must be bigger than $\{e\}$ unlesss $e$ is the only unipotent element. But you can easily construct a nontrivial unipotent element using (restricted) root spaces.