Let $G$ and $G'$ be compact connected Lie groups (which are not necessarily simply connected) with Lie algebra $\mathfrak{g}$ and $\mathfrak{h}$. Suppose that the two Lie algebras are isomorphic, under which conditions I can deduce that the Lie groups are also isomorphic?
If two Lie algebras are isomorphic, under which conditions will their Lie groups also be isomorphic?

3$\begingroup$ My guess would be that isomorphism of $\pi_0$ and $\pi_1$ is sufficient and necessary. $\endgroup$– Vít TučekDec 9, 2023 at 11:23

6$\begingroup$ In general, if $G$ and $G^{\prime}$ are simplyconnected (not necessarily compact), then $\mathfrak{g}\cong\mathfrak{g}^{\prime}$ implies $G\cong G^{\prime}$. Without simply connetedness, this is in general not true: $\mathfrak{su}(2)\cong\mathfrak{so}(3)$ but $\mathrm{SO}(3)$ is not isomorphic to $\mathrm{SU}(2)$. Simply connectedness is just a sufficient condition though. $\endgroup$– G. BlaicknerDec 9, 2023 at 14:02

1$\begingroup$ But there exist some cases where the groups are isomorphic without being simply connected for example $SO(3) \times SO(3)$ is isomorphic to $SO(4)/ \lbrace +I, I \rbrace$. $\endgroup$– user32415Dec 9, 2023 at 15:22

4$\begingroup$ In general, if $G$ and $G'$ are adjoint (i.e., have trivial center), then $\mathfrak g\simeq\mathfrak g'$ implies $G\simeq G'$ as well. Your example of $SO(3)\times SO(3)$ and $SO(4)/\{\pm1\}$ also falls under this. $\endgroup$– Kenta SuzukiDec 9, 2023 at 18:25

1$\begingroup$ @VítTuček Isomorphism of $\pi_0$ and $\pi_1$ is NOT sufficient. Consider $\mathbb Z_2\times S^1$ versus $\mathbb Z_2\ltimes S^1$. $\endgroup$– André HenriquesDec 10, 2023 at 23:10
1 Answer
A Lie algebra $\frak g$ determines a simply connected Lie group. Let's call it $G$. It is the only simply connected Lie group with that Lie algebra. More precisely, if $H$ is a simply connected Lie group whose Lie algebra is isomorphic to $\frak g$ then there is a unique isomorphism $H\cong G$ inducing this isomorphism.
If $H$ is a connected but not necessarily simply connected Lie group whose Lie algebra is isomorphic to $\frak g$ then there is a unique smooth homomorphism $G\to H$ that induces this isomorphism of Lie algebras. Its kernel, say $D$, is a discrete normal subgroup of $G$, and $H$ is isomorphic to $G/D$, and $D$ is canonically isomorphic to $\pi_1(H)$. (By the way, a discrete normal subgroup of a connected group $G$ is necessarily contained in the center.) So, for a given Lie algebra $\frak g$, all of the possible connected Lie groups that have that Lie algebra correspond in a very rigid way to all of the discrete normal (= central) subgroups of the corresponding simply connected group $G$. For two connected Lie groups $H$ to be isomorphic (by an isomorphism inducing a given isomorphism of Lie algebras) you need their fundamental groups to be not just isomorphic, but also to be both embedded in the simply connected group in the same way. For example, $SO(4)$ and $SO(3)\times SU(2)$ are two nonisomorphic connected Lie groups both having fundamental group of order $2$ and both having the same Lie algebra as the simply connected group $SU(2)\times SU(2)$.

$\begingroup$ Dropped an S above. Should be SU(2) not U(2) $\endgroup$ Dec 9, 2023 at 19:23

$\begingroup$ SO(4), SO(3)xU(2), and U(2)xU(2) have three different Lie algebras with three different dimensions. $\endgroup$ Dec 9, 2023 at 19:38

$\begingroup$ @Daniel Sebald. Thank you. Of course I was thinking of $SU(2)$ but wrote $U(2)$ in my haste. I've corrected it. $\endgroup$ Dec 9, 2023 at 19:42

$\begingroup$ You may want to include a reference to Adams or something. $\endgroup$ Dec 12, 2023 at 11:35