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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 $\operatorname{Aut}(G)^{\circ}=\operatorname{Aut}(\mathfrak{g})^{\circ}$?

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

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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.

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

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