# Show that the Lie algebra of $\operatorname{Aut}(\mathfrak{g})$ is given by $\operatorname{Der}(\mathfrak{g})$.

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.

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