Universal cover of a compact simple Lie Group 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.    
 A: 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.
