# Space of derivations of a Lie algebra of a Lie Group is the Lie algebra of the Group of Automorphism on that Lie Group

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 $$\text{der}(\mathfrak{g})$$.

• At first I thought math.stackexchange.com/q/1335428/96384, math.stackexchange.com/q/2803956/96384, math.stackexchange.com/q/3271971/96384, 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. Commented Nov 25, 2022 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 Commented Nov 25, 2022 at 17:35
• Sure they are not literally the same. I meant there might be an isomorphism. Commented Nov 25, 2022 at 18:26
• 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. Commented Nov 25, 2022 at 23:26