# Can a simply-connected Lie group have a non-trivial central extension by a discrete group?

(1) Suppose a (finite-dimensional, real) Lie group $$G$$ is connected and simply-connected. Can $$G$$ have a non-trivial central extension by a discrete group, i.e. can there exist a Lie group $$\tilde G$$ such that $$G \simeq \tilde G/Z$$ where $$Z$$ is a discrete central subgroup of $$\tilde G$$, and where $$\tilde G$$ is not just a direct product $$G\times Z$$?

(2) Related question: Suppose $$G$$ is a Lie group that is connected, simply-connected and semisimple. Can it have any non-trivial central extensions at all?

• (1) no: indeed then $\tilde{G}=Z\times\tilde{G}_0$ and $\tilde{G}_0\to G$ is a topological isomorphism. (2) is a bit imprecise. If you also mean in the category of Lie groups the answer is no too (use Levi factors). – YCor Feb 24 at 7:20

(1) Such an extension can not exist if you want $$\widetilde{G}$$ to be connected, because then $$G=\widetilde{G}/Z$$ for discrete $$Z$$ implies $$Z\subset\pi_1G$$.
To see this, take $$z\in Z$$, a basepoint $$\tilde{x}_0\in\widetilde{G}$$ and a curve connecting $$\tilde{x}_0$$ to $$z\cdot\tilde{x}_0$$. Its image is a loop in $$G$$, which is not 0-homotopic. Indeed, if it was, then by the homotopy lifting property, the curve in $$\widetilde{G}$$ would be homotopic to a point by a homotopy which fixes $$Z\cdot \tilde{x}_0$$ as a set. Because this set is discrete, it would have to be fixed pointwise. So you get a 0-homotopy of a curve that fixes both endpoints, which is absurd.
If $$\widetilde{G}\not=G\times Z$$, then at least some connected component of $$\widetilde{G}$$ is preserved by some nontrivial subgroup of $$Z$$ and you can apply the above argument to that component.
(2) If $$Z$$ is not discrete, then Lie group extensions correspond to Lie algebra extensions, which are classified by $$H^2(\mathfrak{g},\mathfrak{z})$$. The latter is zero for semisimple Lie algebras, by the second Whitehead Lemma.
• (1) $\pi_0$ applied to $1\to Z\to\tilde G\to G\to1$ gives an isomorphism $Z\to\pi_0\tilde G$, whence a morphism $\tilde G\to Z$ which splits the sequence. – Francois Ziegler Feb 24 at 7:20