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Re: Unimodular Lie groups
Posted:
Feb 16, 2006 5:45 PM


In article <dt2n6g$e6a$1@dizzy.math.ohiostate.edu>, Christopher J. Henrich <chenrich@monmouth.com> wrote:
>In article <dt2ff5$co5$1@dizzy.math.ohiostate.edu>, John Baez ><baez@math.removethis.ucr.andthis.edu> wrote:
>> Right now it seems to me that all semisimple real Lie groups >> are unimodular...
>> Am I hallucinating?
>Now that somebody has bred phosphorescent green pigs, that becomes a >very difficult question.
Only if you're seeing phosphorescent green pigs, not unimodular semisimple groups.
>But you're right that semisimple Lie groups are unimodular, as is that >semidirect product.
Great!
Here's another argument, thanks to Yves de Cornulier.
Just for fun, I'll fill in enough details in his argument so that anyone who knows about Lie groups and Haar measure has a chance of understanding it.
Every Lie group has a leftinvariant Borel measure that's unique up to an overall scale factor, called Haar measure.
When we do "right translation by g" to this measure, we clearly get another leftinvariant measure, but not necessarily the same one. The only way it can change is by getting multiplied by some number f(g). This defines the "modular function" f: G > R. Clearly this is a homomorphism from G to the multiplicative group of nonzero real numbers.
The kernel of f is a normal Lie subgroup of G. So, if G is semisimple, it has to be everything or nothing... and since dim(G) > dim(R) for such groups, it has to be everything. So in this case f = 1 and the leftinvariant and rightinvariant Haar measures agree! In this situation we say G is "unimodular".
So, semisimple Lie groups are unimodular. That includes most of the SO(p,q)'s.



