# Relation between $\operatorname{Aut}(G)$ and $\operatorname{Aut}(\mathfrak g)$

Let $$G$$ be a connected Lie group with Lie algebra $$\mathfrak{g}$$. We know that when $$G$$ is simply connected, $$\operatorname{Aut}(G)=\operatorname{Aut}(\mathfrak{g})$$ (this should follow from the fact that we can lift a Lie algebra homomorphism to a Lie group homomorphism whose differential at $$1$$ is the Lie algebra homomorphism).

Now remove the simple connectedness hypothesis and replace it with semi-simplicity, does it hold that $$\operatorname{Aut}(G)^{\circ}=\operatorname{Aut}(\mathfrak{g})^{\circ}$$?

• @Rócherz If you’re going to edit after 9 years, please do it right. Commented Dec 18, 2023 at 20:40

We have a map $$\phi: G \rightarrow \operatorname{Aut} \frak{g}$$ given by the derivative (at $$e$$) of conjugation. Now, the Lie algebra of $$\operatorname{Aut} \frak{g}$$ is the space of derivations of $$\frak{g}$$. So we have $$\phi_\ast :\mathfrak{g} \rightarrow \operatorname{Der} \frak{g}.$$

The image of $$\phi _\ast$$ is the space $$\operatorname{ad}(\mathfrak{g}):=\{\operatorname{ad}(E): E\in \frak{g}\}.$$

In Chapter 1, Section 15 of Knapp’s Lie Groups Beyond an Introduction it’s proved that for semisimple $$\frak{g}$$, $$\operatorname{Der} (\mathfrak{g})= \operatorname{ad} (\mathfrak{g})$$, so that $$\phi _\ast$$ is surjective. Therefore, every automorphism in the identity component of $$\operatorname{Aut} \frak{g}$$ lifts to an inner automorphism of $$G$$. Of course, the derivative of an automorphism of $$G$$ determines the automorphism (for $$G$$ connected), so this lift is unique.

All the assertions I make are proved in chapter 1 of Knapp’s book.

• @vap What does $^\circ$ mean? Commented Nov 16, 2014 at 0:41
• Connected component of the identity.
– vap
Commented Nov 16, 2014 at 11:41
• I should have accepted this way long ago, I'm sorry.
– vap
Commented Sep 29, 2016 at 14:34
• Lol, no worries Commented Sep 29, 2016 at 19:47