I'm trying to prove the Lie algebra of $Aut(\mathfrak{g})$ is $Der(\mathfrak{g})$, the set of all derivations over $\mathfrak{g}$. Where $\mathfrak{g}$ is an arbitrary finite lie algebra.

Due to $\mathfrak{g}$ is finite Lie algebra, I am using that $Lie(Aut(\mathfrak{g}))=\{X \in M(n, \mathbb{C}) / e^{tX}\in Aut(\mathfrak{g})\}$.

How can I prove if $\delta \in Der(\mathfrak{g})$ then $e^{t\delta}([X,Y])=[e^{t\delta}(X),e^{t\delta}(Y)]$?

And if $\delta \in Lie(Aut(\mathfrak{g}))$ then $\delta \in Der(\mathfrak{g})$?


1 Answer 1


Let $\varphi(t)$ be a smooth path of automorphisms with $\varphi(0)=\mathrm{id}$. Then for $X,Y\in\mathfrak g$ you've got:

$$0=\frac d{dt}[X,Y]\lvert_{t=0}= \frac d{dt}[\varphi(t) X,\varphi(t)Y]\lvert_{t=0}=[\varphi'(0)X, Y]+[X,\varphi'(0)Y]$$ and $\varphi'(0)$ is a derivation. Hence the elements of Lie algebra of the automorphism group are derivations.

And the other direction? Let $\delta$ be a derivation, then:

$$[\delta^n X,\delta^m Y]= (-1)^m[\delta^{n+m}X,Y]$$


$$[\sum_n\frac{(t\delta)^n}{n!}X, \sum_m \frac{(t\delta)^m}{m!}Y]=\sum_{n,m}\frac{(-1)^mt^{n+m}}{n!\,m!}[\delta^{n+m}X,Y]$$ do a substitution of the sums, let $k=n+m$ and multiply by $\frac{k!}{k!}$ and $1^{k-m}$. Then the sum becomes:

$$\sum_k\frac{t^k}{k!}\sum_{m≤k}\frac{(-1)^m 1^{k-m}k!}{m!\,(k-m)!} [\delta^{k}X,Y] = \sum_k \frac{t^k}{k!}\sum_{m≤k}(1+(-1))^k[\delta^kX,Y]$$

Now only the $k=0$ term survives, so you are left with $[X,Y]$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .