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I was told that the fundamental group of a connected, compact, semisimple Lie group is finite, with the outline of a possible way to prove this fact. Is there any source however that fleshes this out in detail / are there several ways to prove this fact?


(The result is often known as Weyl's theorem, I think for his take on the proof, Knapp provides a fairly detailed exposition of his perspective.)

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I can think of at least four substantially different proofs off the top of my head and I'm sure there are more, so it is not surprising that Knapp's proof is different from the one that was first suggested to you. So if you want to see a fleshed-out version of the outline that you were first given it might help to include a few details that you remember in your question. – Paul Siegel May 1 '12 at 17:37
While I'm thinking about this, my personal favorite argument is to compute the Ricci curvature of a compact Lie group relative to a bi-invariant Riemannian metric in terms of the Lie bracket and conclude that the Ricci curvature is positive if the group is semi-simple. This implies that the Ricci curvature of the universal cover is also positive and hence the universal cover is compact by the Bonnet-Meyers theorem. – Paul Siegel May 1 '12 at 17:45

There is a quick proof via Lie algebra cohomology: Let $G$ denote your compact, connected, semisimple Lie group, and let $\mathfrak g$ denote its Lie algebra. Then $$ H^1(G;\mathbb R) = H^1(\mathfrak g;\mathbb R) = \text{Hom}_{\mathbb R} (\mathfrak g/[\mathfrak g, \mathfrak g], \mathbb R) = 0, $$ whence $H_1(G;\mathbb Z)$ is finite. But $\pi_1(G)$ is abelian hence is isomorphic to $H_1(G;\mathbb Z)$. QED.

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By the way, this argument is most likely due to Chevalley--Eilenberg. I would give a more precise reference but my erratic internet connection is making this difficult. – Faisal May 1 '12 at 15:18
How easy are the first two equalities? – Igor Rivin May 1 '12 at 23:34
Depends on how you set things up. E.g. let's view $H^\ast(\mathfrak g;\mathbb R)$ as being computed by the complex $(\wedge^q\mathfrak g^\ast,d)$ (I'll omit the formula for $d$...). Then the second equality is trivial: $f\in\mathfrak g^\ast$ is in $H^1$ iff $df=0$ (there are no 1-coboundaries). The formula for $d$ here is $df(X,Y)=f([X,Y])$, whence $f\in H^1 \iff f\in (\mathfrak g/[\mathfrak g,\mathfrak g])^\ast$. – Faisal May 2 '12 at 1:03
The first 'equality' follows from the observation that the complex of left invariant forms computes $H^\ast_{\rm dR}(G)$ (not difficult to prove) together with the fact that said complex can be identified (via evaluation at the identity) with the complex $\wedge^q\mathfrak g^\ast$. You can find all the details in, e.g., Chevalley--Eilenberg. – Faisal May 2 '12 at 1:04
Thanks, I will check it out! Part of my question is which proof is the quickest from nothing... – Igor Rivin May 2 '12 at 1:48

I think you can get a much faster (and maybe easier...) proof using Riemannian geometry, as follows:

First, recall that a semi-simple connected Lie group $G$ is compact if and only if its Killing form $B$ is negative-definite (the proof is easy, see, e.g., Thm 2.28 in these notes). The side we will use ($G$ compact semi-simple $\Rightarrow$ $B$ neg.-def.) actually follows directly from $B(X,X)=tr(ad(X)\cdot ad(X))$ using an orthonormal basis with respect to an auxiliary bi-invariant metric to compute this trace.

Now, the Ricci curvature of any bi-invariant metric on $G$ (that exists because $G$ is compact) can be computed as: $$Ric(X,Y)=-\frac14 B(X,Y),$$ see Remark 2.27 in the same notes. By the observation above, since $G$ is compact and semi-simple, its Killing form $B$ is negative-definite. Hence the above formula gives $Ric>0$. So, by the Bonnet-Myers Theorem, $G$ must have finite fundamental group. Q.E.D.

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Besides Samelson's short 1946 research note linked by Mrc Plm, it's also useful to mention his longer 1952 survey on topology of Lie groups here (see Section 10 and references for various proofs of Weyl's theorem on finiteness of the fundamental group).

By now the whole subject has been treated in numerous textbooks and lecture notes, from a variety of viewpoints. Which approach you take depends a lot on your own background and interests. But the finiteness by itself is too limited a goal, since case-by-case study of the simple compact Lie groups computes each fundamental group in an elegant way relative to the roots and weights of a maximal torus.

P.S. Lucy has found the answer to the question asked about finiteness of the fundamental group (via Knapp's book), though the result is actually Weyl's theorem and not just sometimes called that. As Johannes points out, there is a full treatment in V.7 of the Springer GTM 98 by Brocker and tom Dieck which has the advantage of integrating the topological questions with structure, classification, and representation theory of arbitrary compact connected Lie groups; note their nice summary (7.13). Like other basic theorems, Weyl's theorem has been developed from a variety of directions as indicated in the answers here, though for me the actual computation of the fundamental group for each simple type is an essential part of the picture.

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Another proof is in Bröcker-tom Dieck's book on compact Lie groups, p. 223 ff. It is based on an analysis of root systems.

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See theorem B in Samelson a "Note on Lie groups". He presents two proofs: one via differential forms and one via differential geometry:

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Every connected Liegroup, which has a semisimple Liealgebra with a definite Killing form is compact. The Liealgebra of a compact Liegroup is always the direct sum of an semisimple and abelian Liealgebra, where the killing form of the semisimple part is negative definite.

So we can conclude, that the universal cover of your liegroup is compact and the finiteness of the fundamentalgroup follows immediately. The fundamentalgroup is even always abelian.

The first two things can be found nicely in Serre's "Lie Algebras and Lie Groups" in Theorem 6.2 and 6.3.

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