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The following is an exercise appearing page 148 in Knapp's book, representation theory of semisimple groups.

Let $G$ be a connected linear non-compact Lie group with simple Lie algebra $\mathfrak g$. Let $\mathfrak{g}=\mathfrak{k}\oplus \mathfrak{p}$ be the Cartan decomposition of $\mathfrak{g}$. We have $[\mathfrak{p},\mathfrak{p}]=\mathfrak{k}$ since $\mathfrak{g}$ is simple.

Prove that any finite dimensional uniatry representation of $G$ is trivial.

How to prove it ?

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$\def\su{\mathfrak{su}} \def\g{\mathfrak{g}}$

Suppose there exists a non-trivial $n$-dimensional unitary representation of $G$ i.e. a homomorphism of Lie groups from $G$ to $SU(n)$. Then, since $\su(n)$ is a simple Lie algebra, the induced map from $\g \rightarrow \su(n) $ is an isomorphism. Hence, the map from $G$ to $SU(n)$ has discrete kernel and is surjective (as $SU(n)$ is connected).This implies that $G$ is a covering group of $SU(n)$.

However, note that $SU(n)$ is a compact group, and the fundamental group of compact groups is finite. Hence, $G$ is a fiber product of compact $SU(n)$ and a finite (and hence compact) group. Hence, $G$ is compact, by a variation of Tychonoff's theorem of products of compact spaces. This is a contradiction. Hence, $G$ does not have a nontrivial finite dimensional unitary representation.

EDIT: I made a pretty big mistake in the above paragraph. The map from $\g \rightarrow \su(n)$ is only injective, not surjective. Hence, we only get a covering map from $G$ onto a connected subgroup of $SU(n)$. This shows that the existence of a unitary representation of a connected, noncompact, "simple" Lie group $G$ is equivalent to the existence of a noncompact, connected, "simple" subgroup of $SU(n)$ for some $n.$ I can't think of an explanation for why such a subgroup does not exist.

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  • 2
    $\begingroup$ Why is the map $\mathfrak{g}\rightarrow \mathfrak{su}(n)$ surjective? $\endgroup$ – Jason DeVito Jul 13 '14 at 18:26
  • $\begingroup$ It's not necessarily. I was being dumb. I'll fix it in a bit. $\endgroup$ – Siddharth Venkatesh Jul 14 '14 at 7:39
  • $\begingroup$ If the center of $G$ is finite, then your same argument should yield a finite covering map $G\rightarrow SU(n)$ with image a subgroup $H$ of $SU(n)$. Finite covering maps are closed, and so $H$ should be closed, hence compact. We then have an isomorphism $G/\operatorname{Ker}(\rho )\rightarrow H$, $\rho :G\rightarrow SU(n)$ the representation. As the quotient of a noncompact group $G$ by a finite group $\operatorname{Ker}(\rho )$ cannot be compact, this is a contradiction. $\endgroup$ – Jonathan Gleason Sep 21 '15 at 17:35

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