Notation: Let $\mathfrak{g}$ be a Lie algebra, $\operatorname{Aut}(\mathfrak{g})$ denotes the set of automorphisms. $\operatorname{Der}(\mathfrak{g})$ denotes the set of derivations of $\mathfrak{g}$. That is is if $T\in \operatorname{Der}(\mathfrak{g})$ then $T[X,Y]=[T(X),Y]+[X,T(Y)]$.

I’m trying to show that the Lie algebra of $\operatorname{Aut}(\mathfrak{g})$ is given by $\operatorname{Der}(\mathfrak{g})$, I am stuck on a specific step in the proof which I will mark in bold.

Let $T$ be in the Lie algebra of $\operatorname{Aut}(\mathfrak{g})$. Then $\exp(tT)\in \operatorname{Aut}(\mathfrak{g})$. Therefore $\exp(tT)[X,Y]=[\exp(tT)X,\exp(tT)Y]$. Taking the derivative derivative at $t=0$ gives $T[X,Y]$ on the left-hand side but I am not sure how to take the derivative of the right hand side.

Obviously the answer should be $[TX,Y]+[X,TY]$, this is easy to see if $[\,,]$ was a commutator bracket but for general Lie algebras this is not always the case.


1 Answer 1


$${\Bbb d\over {\Bbb dt}}\big[\exp(tT),\exp(tT)Y)\big]=\left[{\Bbb d\over {\Bbb dt}}\exp(tT)X,\exp(tT)Y\right]+\left[\exp(tT)X,{\Bbb d\over{\Bbb dt}}\exp(tT)Y\right]$$

Since the function $b:(X,Y)\rightarrow [X,Y]$ is bilinear, $\Bbb db_{(X,Y)}(U,V)=[U,Y]+[X,V]$ and $t\rightarrow [\exp(tT),\exp(tT)Y)]$ is the composition of $b$ and $t\rightarrow (\exp(tT)X, \exp(tT)Y)$, the result follows from the chain rule.


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