I am looking for a reference that provides a full proof of the fact that any compact simple Lie Group admits a compact universal cover.

In particular, it would be desirable to have a proof that does not use methods from Riemannian Geometry (e.g. the Bonnet-Myers theorem on curvature), as I understood that there is also a proof using the fundamental group / first homology group or the like.

I am asking this question as it is related to a post I made earlier: Infinite-dimensionality of unitary representations of non-compact simple Lie Groups

In order to finish the proof of this statement, I would require the above assertion.

Any help would be greatly appreciated.


We simply use the theorem that a Lie group with finite center (which of course naturally includes simple groups) is compact if and only if its Killing form is negative definite.

A universal cover naturally exists, by the standard procedure for building a universal cover by defining a new group with elements comprising pairs of the original group's elements together with homotopy classes of paths joining them to the group identity and then one defines a new group operation with original group product acting on the group element in the pair and path composition acting on the homotopy classes.

So now we deploy the theorem I mentioned to both the group and the universal cover: they both have the same Lie algebra, thus the same Killing form. We can prove that theorem a number of ways.

For the "if" direction, that the group is compact if its Killing form is negative definite, the group's adjoint representation image must be a subgroup of $O\left(\mathbf{g},\,-B\right)$ (where $\mathbf{g}$ is the group's Lie algebra and $B$ the Killing form), since the Killing form is Ad-invariant. (here $O\left(\mathbf{g},\,-B\right)$ means the orthogonal group, with the notion of orthogonality defined by the Killing form, which, by its negative definiteness is an inner product). Thus the original group must be a finite cover (owing to the finite centre) of this compact beast.

There is a tricky step to this one in that one must prove that the image of the original group under the adjoint representation is closed in $O\left(\mathbf{g},\,-B\right)$ to make the whole thing work, but it can be done. Once we have proved closedness, we can then appeal to the known compactness of $O\left(\mathbf{g},\,-B\right)$ to infer compactness of our subgroup. The proof of closedness proves that the image of Ad is an embedding and rules out noncompact subgroups of the compact $O\left(\mathbf{g},\,-B\right)$ (like an irrational slope line on a torus, for example).

I believe this one fits your requirements of minimizing the use of Riemannian geometry.

A better known proof uses Myer's theorem; see:

Helgason, S. (1978). Differential geometry, Lie groups and Symmetric Spaces. American Mathematical Society. Chapter 2, Section 6, proposition 6.6.

In the "only if" direction, one uses the compactness to build an Ad-invariant average over the group $G$ of an arbitrary pair of elements $x$ and $y$ the Lie algebra $B\left(x,y\right)=\int_{\gamma\in G} \left<{\rm Ad}_\gamma x, {\rm Ad}_\gamma y\right> d\mu(G)$, with $\mu$ the Haar measure. Some simple gymnastics on this integral then shows it is negative definite.

  • $\begingroup$ Thanks for the elaborate answer! This argument was also sketched by my thesis advisor. I have a few further questions and remarks: $\endgroup$
    – Thomas Bakx
    Mar 7 '18 at 23:21
  • $\begingroup$ if I am not mistaken, we can deduce compactness of the Ad-image by looking at the Lie Algebra of both groups in some sense, is this correct? Secondly, it seems that you are saying that the finiteness of the cover then automatically implies the compactness of the original group? I hope to hear back from you. Unfortunately I do not have enough reputation to upvote your answer. $\endgroup$
    – Thomas Bakx
    Mar 7 '18 at 23:27
  • $\begingroup$ @ThomasBakx Also, feel free to contact me in chat / email (rod dot vance at protonmail.ch) I also have my own proof of closedness of the Ad image written out in full, although it has never been reviewed. It would be great to discuss it in detail with you and your thesis advisor if it helps your research. $\endgroup$ Mar 7 '18 at 23:37
  • $\begingroup$ @ThomasBakx Ha. That's pretty funny. I've just looked at your other question, and you have exactly the same doubts as i did about these arguments (the need to prove the Ad image is an embedding) some time ago, which is why I came up with the closedness proof I'm talking about. This MO thread may be relevant: mathoverflow.net/q/182159/14510 $\endgroup$ Mar 7 '18 at 23:44
  • $\begingroup$ @ThomasBakx "finiteness of the cover then automatically implies the compactness of the original group, although that would seem to be a line of argument" This doesn't work. Counterexample: $U(1)\cong SO(2)$ : Universal cover is $\tilde{U(1)}=(\mathbb{R},\,+)$. The argument above is simply deducing compactness from the Killing form: both groups must fulfill the negative definiteness criterion, because they both have the same Killing form. $\endgroup$ Mar 7 '18 at 23:56

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.