What can we say about connected groups with the same Lie algebra?

Question

Each (finite-dimensional) Lie algebra has exactly one simply connected Lie group associated to it (up to isomorphism). What can we say about all other connected groups with the same Lie algebra ?

Thank you in advance

Answer 1

Every connected Lie group $G$ has a universal cover $\widetilde{G}$ which is a simply connected Lie group, and the corresponding covering map $\widetilde{G} \to G$ is a morphism of Lie groups. It fits into a short exact sequence

$$1 \to \pi_1(G) \to \widetilde{G} \to G \to 1$$

meaning that $G$ is a quotient of $\widetilde{G}$ by the discrete normal abelian subgroup $\pi_1(G)$. It is a further exercise from here to show that $\pi_1(G)$ is in fact central, so the conclusion is that, fixing $\widetilde{G}$, the possible choices of $G$ are parameterized by the discrete central subgroups of $\widetilde{G}$.

Example. Suppose $\widetilde{G} = \mathbb{R}^n$. Up to change of coordinates, the discrete subgroups all look like $\mathbb{Z}^k$ for some $k \le n$, and the quotients look like $T^k \times \mathbb{R}^{n-k}$.

It sometimes happens that the center of $\widetilde{G}$ is discrete (this is not the case above, but e.g. is the case whenever $\mathfrak{g}$ is semisimple). In this case things simplify yet further: the possible choices of $G$ are parameterized by the subgroups of $Z(\widetilde{G})$.

Example. Suppose $\widetilde{G} = SU(2) \times SU(2) \cong Spin(4)$. This group has center $\mathbb{Z}_2 \times \mathbb{Z}_2$, which has five subgroups. The corresponding quotients are $SU(2) \times SU(2)$ itself, $SU(2) \times SO(3), SO(3) \times SU(2)$, $SO(4)$, and $SO(3) \times SO(3)$.