I am learning about Lie groups, and I have the following basic question:
Every Lie group $G$ has a (unique) universal covering group $ \bar G $ that is simply connected, and such that the covering map $ p : \bar G \to G $ is a homomorphism. It follows that $\pi _1 ( G ) = \mathrm {ker} (p) \le \bar G $ is a normal discrete subgroup, and hence central.
Using the above one can show that all Lie groups with some fixed Lie algebra $ \mathfrak g $ are quotients of the unique simply connected Lie group $ \bar G$ with $ \mathcal {Lie} (\bar G) = \mathfrak g $, by some central discrete subgroup.
It seems to follow from these remarks that the center $ Z=Z(\bar G) $ is an invariant of $ \mathfrak g $. If $G$ is semisimple this center is discrete abelian group.
So my question is – how can we determine $Z$ given a semi-simple Lie algebra $\mathfrak g$ ?
This seems like a rather basic question but I don't how to answer it. I will be happy to hear an explanation or to have a reference to some standard text.
Best Answer
Well, If I am not mistaken the center of the simply-connected group $G$ is isomorphic to the fundamental group of $Ad G$. That coincides with the abstract fundamental group of the corresponding Lie algebra. It can be done on terms of Stiefel diagrams. See Brocker-Tom Dieck chapter 5 part 7.