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Re: Unimodular Lie groups
Posted:
Feb 16, 2006 5:45 PM
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In article <dt2n6g$e6a$1@dizzy.math.ohio-state.edu>,
Christopher J. Henrich <chenrich@monmouth.com> wrote:
>In article <dt2ff5$co5$1@dizzy.math.ohio-state.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
>semi-direct 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 left-invariant 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 left-invariant 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
left-invariant and right-invariant 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.
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