# I have a conjecture/ possibly already a theorem

1. Dec 20, 2008

### Jim Kata

I am not very well read so this may already exist as a theorem. If not, try to prove it, or disprove it.

Let G be a compact group over the reals, then the maximally compact subgroup of the complexification of G is just G over the reals.

That is the maximally compact subgroup of $$G_\mathbb{C}$$ is just $$G\left( \mathbb{R} \right)$$

Here's a simple example:

SU(2) is the maximal compact subgroup of $$Sl\left( {2,\mathbb{C}} \right)$$

And $$SU(2)_\mathbb{C} \cong Sl\left( {2,\mathbb{C}} \right)$$

2. Dec 22, 2008

### mathwonk

definitions would be nice. or does maximal compact mean just that? no larger compact subgroup exists? or no larger proper compact subgroup?

3. Dec 22, 2008

### mathwonk

a few minutes web search reveals, without even knowing wjhat these things mean, that any compact group is a maximal compact of its chevalley complexification.