1
$\begingroup$

In the book `Ergodic Theory and Semisimple groups' (1984 Edition, page 64, Proposition 4.1.8), Zimmer has made this statement "if G is a connected semisimple non-compact Lie group with finite center, then G admits an irreducible representation with a non-relatively compact projective image".

How do we prove this fact?

Improve this question
$\endgroup$
1
  • $\begingroup$ The actual page number is 62, by the way. This statement occurs in a short proof sketch given by Zimmer of a result which he attributes to [Furstenberg 1}; that is a serious 50+ page paper in Ann. of Math. 77 (1963), so it's already challenging to track down the source of the proposition there. Zimmer's book is a concise monograph which relies on a lot of such prior work, so it's nontrivial to read on your own (as I discovered a long time ago). The Lie algebra approach suggested by jef might be useful here. $\endgroup$
    – Jim Humphreys
    Commented Mar 19, 2012 at 0:37

1 Answer 1

Reset to default
3
$\begingroup$

Let us first show that a semisimple Lie group with finite center is non-compact if and only if its Lie algebra contains a copy of $sl(2,R)$.

Indeed, if $G$ is compact, the Killing form is negative definite, and its restriction to a copy of $sl(2,R)$ would be a definite negative invariant form, but such forms do not exist on $sl(2,R)$. Conversely, if $G$ is not compact, then $Lie(G)$ contains "real roots" for suitable Cartan sub-algebras. Any root vector associated to such a real root will be part of a copy of $sl(2,R)$ inside $Lie(G)$.

Now let $V$ be the adjoint representation of $G$. It induces a faithful representation of $Lie(G)$, hence a non-trivial representation of a fixed copy of $sl(2,R)$ inside $Lie(G)$. So it has an irreducible subrepresentation $(\rho,W)$ which induces a non-trivial, hence faithful, representation of $sl(2,R)$. So the Lie algebra of the closure $\rho(G)$ contains a copy of $sl(2,R)$ and in fact, so does the Lie algebra of the closure of the image of $G$ in $PGL(W)$. Therefore the latter closure is non-compact.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Note that in the first paragraph you mean "non-compact" rather than "compact". And in the next paragraph the terminology should be "negative definite". $\endgroup$
    – Jim Humphreys
    Commented Mar 17, 2012 at 22:29
  • $\begingroup$ thanks, I have made corrections. By the way, would you think of a more elementary argument ? $\endgroup$
    – Jef
    Commented Mar 18, 2012 at 20:38
  • 1
    $\begingroup$ But you still haven't changed "definite negative" to "negative definite". This is definitely negative. $\endgroup$
    – David Epstein
    Commented Mar 18, 2012 at 21:44

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .