# Automorphism group of a lie algebra as a lie subgroup of $GL(\frak g)$

Let $G$ be a lie group with lie algebra $\frak{g}$. Let $Aut(\frak g)$ be the automorphism group of $\frak{g}$.

Its clear to me that $Aut(\frak{g})$ $\subset GL(\frak{g})$ since any automorphism of $\frak{g}$ is also a linear transformation. Now i have several questions:

1. How do i prove that $Aut(\frak{g})$ is actually a lie subgroup of $GL(\frak{g})$? I've tried playing with actions to realize $Aut(\frak g)$ as an orbit or an isotropy subgroup but nothing worked out...

2. Say $H \subset G$ is an abstract subgroup (not necessarily lie). Are the following statements equivalent?

$$\{T_e H \text{ is closed under the lie bracket of } \frak{g} \}$$ $$\{H \text{is a lie subgroup of } G\}$$

1. How does the lie algebra of $Aut(\frak g)$ look like? When does $Aut(\frak g)$ $= G$? what does $Aut(\frak g)$ tell me about $G$?

1. You are almost there: The group $Aut(\mathfrak{g})$ of automorphisms of a Lie algebra $\mathfrak{g}$ is closed in the group $End(\mathfrak{g})^×$ of vector space automorphisms, hence it is a Lie group.
2. No, only one direction is true in general: If $H$ is a Lie subgroup of $G$, then $\mathfrak{h}\simeq T_eH\subset T_eG\simeq \mathfrak{g}$ is a Lie subalgebra. Conversely we only know that if $\mathfrak{h}\subset \mathfrak{g}$ is a Lie subalgebra, there exists a unique connected Lie subgroup $H \subset G$ with Lie algebra $\mathfrak{h}$.
3. The group $Aut(\mathfrak{g})$ is a Lie subgroup of $GL(\mathfrak{g})$ ( and not of $G$). Its Lie algebra is given by $Der(\mathfrak{g})$, the algebra of derivations. It is an important invariant for $\mathfrak{g}$ resp. $G$. It can be "small", i.e., $Der(\mathfrak{g})=Inn(\mathfrak{g})$, for example for semisimple Lie groups, or "big", i.e., $Der(\mathfrak{g})=End(\mathfrak{g})$, for example for abelian Lie groups. Analoguous results hold for the automorphism group. We have the exponential mapping $\exp : \mathfrak{g}\rightarrow G$.
• Thanks! Problem is I'm not sure how to prove that the subset $Aut(\frak g)$ is closed... – Saal Hardali Jul 17 '15 at 8:43
• See the proof of Proposition $7.1$ here, which shows that $Aut(A)$ is a closed subgroup of $GL(A)$ and hence is a Lie group for any finite-dimensional non-associative algebra $A$. – Dietrich Burde Jul 17 '15 at 9:46
• Thanks, I can't believe how close I came. I did the exact same thing but still had doubts so I continued to search for a different proof. Say, is there something deep that makes $Der(\frak G)$ isomorphic to the algebra of derivations on the algebra of germs of smooth functions at the identity (I.e. tangent space)? – Saal Hardali Jul 17 '15 at 9:53