If $\mathfrak g_1=\mathfrak g_2$ and they are simple, what can we say about the corresponding Lie groups?

Question

If there are two matrix Lie groups $G_1$ and $G_2$ and $\mathfrak{g}_1$ and $\mathfrak{g}_2$ be the corresponding Lie algebras. If it is known that $\mathfrak{g}_1=\mathfrak{g}_2$ and additionally that $\mathfrak{g_1}$ and $\mathfrak{g_2}$ are simple lie algebras then what can we comment about $G_1$ and $G_2$ ? I don't have much intuition as to what happens to the lie groups when their corresponding lie algebras are the same.

Answer 1

Recall the following important fact:

Theorem. Let $G$ be a Lie group with Lie algebra $\mathfrak{g}$. For each subalgebra $\mathfrak{h}$ in $\mathfrak{g}$ there is a unique connected Lie subgroup $H$ of $G$ with Lie algebra $\mathfrak{h}$.

(However, $H$ is in general not necessarily a closed subgroup of $G$.) Any way, what is important here is uniqueness. If you have two connected matrix Lie groups in $\mathrm{GL}(n,\mathbb{R})$, say $H_1,H_2\subseteq\mathrm{GL}(n,\mathbb{R})$, such that $\mathfrak{h}_1=\mathfrak{h}_2\subseteq\mathrm{Mat}(n,\mathbb{R})$, then $H_1=H_2$. If they are not necessarily connected, then you can still say that their identity components are equal: $(H_1)_0=(H_2)_{0}$.

However, if $H_1\subseteq\mathrm{GL}(n_1,\mathbb{R})$ and $H_2\subseteq\mathrm{GL}(n_2,\mathbb{R})$ for $n_1\neq n_2$, then you cannot necessarily conclude that $H_1=H_2$.

In general, if $\mathfrak{h}_1\cong\mathfrak{h}_2$ and $H_1,H_2$ are connected, all you can say is that their universal cover are equal: $\tilde{H}_1=\tilde{H}_2$.

Answer 2

Well, the third Lie theorem states that for each Lie algebra $\mathfrak{g}$, there exist a unique connected and simply connected Lie group $G$ such that the Lie algebra of $G$ is isomorphic to $\mathfrak{g}$.

Hence, if your groups are connected and simply connected, they shall be equal. Otherwise, anything could happen. For example, the Lie groups $SO(n)$ and $O(n)$ have the same Lie algebra, consisting of skew symmetric matrices. Note that in this case, $SO(n)$ is the connected component of $O(n)$ at the identity.