This is a follow-up to my question here.*
I'm interested in knowing about (finite-dimensional) Lie group extensions, where $G$ is simple and $H$ is connected (and hopefully abelian): $$1\to H\to G'\to G \to 1$$
Specifically, I want to know if it's possible for this sequence to split topologically, but not as groups. This is probably equivalent, at least when $H$ is abelian, to the non-vanishing of the continuous group cohomology $H^2(G, H)$ for some action of $G$ on $H$.
We have a corresponding exact sequence of Lie algebras: $$0\to \mathfrak{h}\to\mathfrak{g}'\to\mathfrak{g}\to 0$$
By Whitehead's second lemma, this sequence always splits. I'm aware that if $G,H$ are simply connected, this splitting lifts to the groups. But is it possible that this lifting is obstructed, while $G'\to G$ still ends up being a trivial fibration of topological spaces?
* In the previous question, the Heisenberg group came up as an example of an extension that splits topologically but not as groups. But the reducibility of the base leads to undesirable properties (let's say it has a "partial splitting") so I'm asking a sharper question.
Best Answer
Yes, indeed, if $H$ is abelian then the Lie group central extensions of $G$ by $H$ are classified by the 2ned continuous cohomology $H^2_c(G; H)$. A good survey can be found in
Stasheff, James D., Continuous cohomology of groups and classifying spaces, Bull. Am. Math. Soc. 84, 513-530 (1978). ZBL0399.55009.
In particular, the isomorphism $$ H^2_c(G; {\mathbb R})\cong H^2(G_u/K; {\mathbb R}) $$ appearing on page 520 of the survey proves that $$ H^2_c(SL(2, {\mathbb R}); {\mathbb R})\cong H^2(S^2; {\mathbb R})\cong {\mathbb R}.$$ In particular, one obtains a nontrivial central extension of $SL(2, {\mathbb R})$ by ${\mathbb R}$ (which splits topologically, see the top of the page 516 of the survey).