Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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?

share|improve this question
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. –  Jim Humphreys Mar 19 '12 at 0:37

1 Answer 1

up vote 2 down vote accepted

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.

share|improve this answer
Note that in the first paragraph you mean "non-compact" rather than "compact". And in the next paragraph the terminology should be "negative definite". –  Jim Humphreys Mar 17 '12 at 22:29
thanks, I have made corrections. By the way, would you think of a more elementary argument ? –  Jef Mar 18 '12 at 20:38
But you still haven't changed "definite negative" to "negative definite". This is definitely negative. –  David Epstein Mar 18 '12 at 21:44

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.