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.

  • $\begingroup$ For which $\mathfrak{g}$ do you know the answer already? $\endgroup$ – Qiaochu Yuan Feb 22 '13 at 21:15
  • $\begingroup$ I know the answer for some cases where I understand the corresponding Lie group well enough - for $\mathfrak g = \mathfrak {sl_2}$ the Lie group is $SL_2 (\mathbb R)$, which has center $ { \pm I } $ and fundamental group $\mathbb{Z}$. Hence the universal covering group has center $\mathbb Z$. Similarly, with $ \mathfrak g = \mathfrak {sl_n} $ for $ n > 2 $ the group $SL_2 (\mathbb R)$ is simply connected (this follows from the polar decomposition $ SL_n = SO_nP $ ). Hence the center of the universal cover is either $\mathbb{Z}/2 \mathbb{Z}$ or $\mathbb Z$ depedning on the parity of $n$. $\endgroup$ – the_lar Feb 23 '13 at 18:56
  • $\begingroup$ @Yuan- this example shows that somewhat interesting is happening. I suppose that the other cases can be analyzed similarly, but I am wondering what is the general theory behind this. $\endgroup$ – the_lar Feb 23 '13 at 18:59
  • $\begingroup$ I just realized I made a mistake in my first comment above - for $\mathfrak{sl_n}$ the fundamental group of the universal cover is either $\mathbb Z / 2 \mathbb Z$ or $\{e\}$. $\endgroup$ – the_lar Feb 24 '13 at 7:00

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.


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