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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 unitary representation of $G$ is trivial.

How to prove it?

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  • $\begingroup$ One way to prove this is to use c-forms (due to Vogan and his collaborators). Here's a rough sketch: Given a finite dimensional irrep V of G, view it as a rep of the complexified Lie algebra g_C. Then g_C has a compact real form u, and there exists a u-invariant inner (-,-) product on V. Now twist (-,-) by the Cartan involution to get a new form <-,-> = (-,\theta(-)), so called c-form. One can check that this is g-invariant (g = Lie(G)). Since V is irred, <-,-> is the unique G-inv form on V up to scaling. If you write this form down and analyze the k- and p-action, .... $\endgroup$
    – Qixian Zhao
    Mar 18 at 22:03
  • $\begingroup$ ... you will see that <-,-> is definite iff p acts trivially on V. Since k=[p,p], this would imply that g = k+p acts trivially on V, i.e. V is a trivial rep. By connectedness of G, G also acts trivially. $\endgroup$
    – Qixian Zhao
    Mar 18 at 22:03

1 Answer 1

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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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    $\begingroup$ Why is the map $\mathfrak{g}\rightarrow \mathfrak{su}(n)$ surjective? $\endgroup$
    – Jason DeVito - on hiatus
    Jul 13, 2014 at 18:26
  • $\begingroup$ It's not necessarily. I was being dumb. I'll fix it in a bit. $\endgroup$
    – Siddharth Venkatesh
    Jul 14, 2014 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, 2015 at 17:35
  • $\begingroup$ Why does the image of OP's unitary representation sit inside $SU(n)$ [as opposed to $U(n)$]? $\endgroup$
    – Qmechanic
    May 26, 2019 at 13:53

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