Is the fundamental group of a connected semisimple Lie group equal to its Schur multiplier? 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.
 A: 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: $$\pi_1(G)\cong H_1(G,\mathbb{Z})\xrightarrow{\text{~}} H_2(BG,\mathbb{Z}).$$
