This is what I understand. Let $G$ be a simply connected Lie group. Given an isomorphism $\phi : G \to G$, we get an isomorphism $d\phi : \frak{g} \to \frak{g}$, and this induces an abstract group homomorphism (by the chain rule) $\operatorname{Aut}(G) \to \operatorname{Aut}(\frak{g})$. Since $G$ is simply connected, for any homomorphism $\psi : \frak{g} \to \frak{g}$ there exists a unique homomorphism $\phi : G \to G$ such that $\psi = d\phi$. This fact allows us to identify the abstract groups $Aut(G)$ and $Aut(\frak{g})$.
Here is where my understanding is very shaky. Now, $\frak{g}$ is just a finite dimensional vector space (endowed with a Lie bracket, ofc. Fix a basis so that $\frak{g} = \mathbb{R}^n$. Now, $\operatorname{Aut}(\frak{g})$ can be identified with a subgroup of matrices in $\operatorname{GL}(n, \mathbb{R})$. (Namely, the matrices $A$ such that $A[e_i, e_j] = [Ae_i, Ae_j]$ for the chosen basis.) As such, $\operatorname{Aut}(\frak{g})$ inherits a differential structure, and we just slap that onto $\operatorname{Aut}(G)$ through the above identification. Functoriality has to be checked, ofc.
Is my understanding correct?
Best Answer
The fact that the automorphism group $Aut({\mathfrak g})$ of a Lie algebra ${\mathfrak g}$ is a Lie group is not immediate. However, it follows from Cartan's Theorem that every closed subgroup of a Lie group is a Lie subgroup. In your case, $Aut({\mathfrak g})$ is a subgroup of $GL({\mathfrak g})$ given by the condition that you have mentioned. This condition clearly defines a closed subset of $GL({\mathfrak g})$ from which it follows that $Aut({\mathfrak g})$ is a Lie subgroup of $GL({\mathfrak g})$.