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Let $\mathfrak{g}$ be a Lie algebra over $\mathbb{K}=\mathbb{C}$ or $\mathbb{R}$ associated with the Lie group $G$. A smooth linear map $\delta:\mathfrak{g}\to\mathfrak{g}$ is a derivation of $\mathfrak{g}$ if$$\delta([X,Y])=[\delta(X),Y]+[X,\delta(Y)]$$ for all $X,Y\in\mathfrak{g}$. The automorphism group $\text{Aut}(G)$ is the set of all group isomorphism of $G$, which is itself a Lie group therefore has a Lie algebra $\mathfrak{Aut}(G)$ associated with it. Now consider the vector space of all derivations of $\mathfrak{g}$, denoted $\text{der}(\mathfrak{g})$, I am looking for a clear proof on the equality$$\text{der}(\mathfrak{g})=\mathfrak{Aut}(G)$$

It would be extremely helpful if one could explain it to me, thanks.

Edit: at this point I have realised that indeed $\mathfrak{Aut}(G)\cong\mathfrak{Aut}(\mathfrak{g})$ by Lie's first and second theorem, it is still not clear even from the answers provided in other posts how to make the connection to $\text{der}(\mathfrak{g})$.

Rescy_
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  • At first I thought https://math.stackexchange.com/q/1335428/96384, https://math.stackexchange.com/q/2803956/96384, https://math.stackexchange.com/q/3271971/96384, https://math.stackexchange.com/q/3917157/96384 were duplicates, but they talk about what in your notation would be $\mathfrak{Aut}(\mathfrak g)$ as opposed to $\mathfrak{Aut}(G)$. It's quite possible these two are the same though. – Torsten Schoeneberg Nov 25 '22 at 17:24
  • Yes its a typo, fixed now. thanks for pointing out. Ive looked at these answers but unfortunately could not understand most of what they are saying – Rescy_ Nov 25 '22 at 17:35
  • Sure they are not literally the same. I meant there might be an isomorphism. – Torsten Schoeneberg Nov 25 '22 at 18:26
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    For $G$ simply connected, $\operatorname{Aut}(G) \cong \operatorname{Aut}(\mathfrak{g})$ (so they have the same Lie algebra). Otherwise I think we get an inclusion $\operatorname{Aut}(G) \hookrightarrow \operatorname{Aut}(\mathfrak{g})$ but I don't know if we drop dimension here and thus get a smaller Lie algebra or not. – Callum Nov 25 '22 at 23:26

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