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Is the fundamental group of a connected semisimple Lie group equal to its Schur multiplier?

This is a spin off of A semi simple Lie group whose commutator is not closed? which was getting a bit crowded. The idea is making an analogy between perfect finite groups and connected semisimple Lie groups. With universal perfect central extension of a perfect group being analogous to the universal cover in Lie theory.

Ian Gershon Teixeira
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1 Answers1

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This is true, and can be seen by looking at the Serre spectral sequence in homology for the fibration: $$G\rightarrow EG\rightarrow BG$$

First, lets note that since $\pi_1(G)$ is abelian, its isomorphic to $H_1(G,\mathbb{Z})$, where this is the topological (singular, say) cohomology of $G$, viewing it as a manifold. Then our Schur multiplier is the second group homology of $G$ with coefficients in $\mathbb{Z}$, so this is $H_2(BG,\mathbb{Z})$, where $BG$ is the classifying space for principal $G$ bundles. (Note that I am taking this to be the definition of group homology for a topological group).

So lets analyse this fibration sequence now. Since $G$ is connected, the long exact sequence in homotopy groups implies that $BG$ is simply connected, and we may consider the Serre spectral sequence for integral homology, and no twisted coefficients occur.

So the $E^2_{p,q}$ term is just $H_p(BG,H_q(G,\mathbb{Z}))$, our $E^2$ differential goes from $E^2_{p,q}\rightarrow E^2_{p+2,q-1}$, and since our total space $EG$ is contractible, we know that everything must be killed eventually. This means the only nontrivial differential going out of $E^2_{0,1}$ is this first one on the $E^2$ page, and that the only nontrivial differential going into $E^2_{2,0}$ is this first one on the $E^2$ page, so this differential yields an isomorphism: $$\pi_1(G)\cong H_1(G,\mathbb{Z})\xrightarrow{\text{~}} H_2(BG,\mathbb{Z}).$$

Chris H
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  • Wow this is so cool. Are you using anything other than the fact that $ G $ is a connected topological group? Maybe that $ G $ is a manifold, or at least that $ G $ is path connected? Doesn't seem like you are using anything about semisimplicity. For example does this apply to $ \mathbb{R}^2 $? (Which has nontrivial projective reps (the Heisenberg projective unitary irrep on $ L^2(\mathbb{R}) $) but according to this identification of $ \pi_1(G) $ with the Schur multiplier, $ \mathbb{R}^2 $ has trivial Schur multiplier.) – Ian Gershon Teixeira May 02 '22 at 00:43
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    I think I’m using some (weak) point set assumptions for the fibration stuff, but I reckon the nontrivial question is whether the Schubert multiplier defined in terms of projective representations equals the second homology of the classifying space. Your example shows this is false for the Heisenberg group, and it wouldn’t surprise me if this is where one needs semisimplicity, to ensure the topological and projective representation schur multiplier agree. In general, nilpotent Lie groups are contractible as spaces, so have contractible classifying spaces, so this cohomology sees nothing there. – Chris H May 02 '22 at 00:53
  • Ya that's ok I'm not worried about the point set stuff in my world basically everything is a manifold anyway. So what happened here is that you showed $ \pi_1(G) \cong H_2(BG,\mathbb{Z}) $ for topological homology, but currently its unclear if $ H_2(BG,\mathbb{Z}) $ in topological homology is isomorphic to $ H_2(G,\mathbb{Z}) $ (the Schur multiplier) in group homology? – Ian Gershon Teixeira May 02 '22 at 01:04
  • Yep that sums it up, though I suspect there should be something sensible inbetween topological $H_2(BG,\mathbb{Z})$ which is trivial on nilpotent lie groups, and $H_2(G,\mathbb{Z})$ viewing $G$ as an abstract group, which could be enormous. – Chris H May 02 '22 at 01:12
  • I don't want to come across as mean but should I unaccept your answer then? You seem like a lovely person, and this is definitely some cool math, but if group homology $ H_2(G,Z) $ is not the same as topological homology $ H_2(BG,Z) $ then it seems like what you are saying doesn't actually address the title question? Am I missing something? Is there a way I can rephrase what I asked to make it clearer? – Ian Gershon Teixeira May 02 '22 at 01:30
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    Yea of course, feel free to, I didn't realise before writing that the obvious definition for Schur multiplier for topological groups isn't the one we necessarily care about for lie groups. I'll leave up the answer in case someone can find a proof/reference for topological $H_2$ equals schur multiplier $H_2$ in the semisimple case. – Chris H May 02 '22 at 01:47
  • Ok sounds good. And definitely leave your answer up its very cool math! In the meantime I'll also go back and think more to see if I'm misunderstanding something really basic about some of the definitions here, especially the definition of the Schur multiplier – Ian Gershon Teixeira May 02 '22 at 01:51