# What is the lie algebra of the group of automorphisms of a lie algebra?

Let $\mathfrak{g}$ be a Lie algebra. Then its automorphism group $Aut(\mathfrak{g})$ is a Lie group, and hence we may take its Lie algebra $Lie(Aut(\mathfrak{g}))$.

I'd like to say that this is equal to the Lie algebra of derivations $Der(\mathfrak{g})$. Is this true? Where can I find a reference?

• Yes, it is true, see here. – Dietrich Burde Jun 1 '18 at 8:25

This is a standard fact from the theory of Lie algebras. There are several references, say over $\Bbb{R}$ and $\Bbb{C}$, e.g., Proposition 1.25 in the book The Structure of Complex Lie Groups.
It is in fact true that $$\text{Lie}(\text{Aut}(L))=\text{Der}(L)$$, as you say. There are a lot of questions on Stack Exchange asking about this, and how to prove it. But no answers seem to give a proof.
And it actually is not. Let $$G=\{g\in \text{GL}(L)| g[x,y]=[gx,gy], x,y\in L\}$$. Then, in order to get the tangent space of $$G$$ at $$1$$, i.e. the Lie algebra of $$G$$, we just have to differentiate the equation $$g[x,y]=[gx,gy]$$ and plug in $$g=1$$. If we differentiate at $$g$$ in the direction of $$\delta$$, we get $$\delta([x,y])=[gx,\delta(y)]+[\delta(x),gy]$$, and so plugging in $$g=1$$ gives us $$\delta([x,y])=[x,\delta(y)]+[\delta(x),y]$$. Thus, $$\delta$$ is in the tangent space of $$1$$ if and only if $$\delta$$ is a derivation on $$L$$. (And the Lie products are the same, because they are both given by the commutator $$[\delta_1,\delta_2]=\delta_1\delta_2-\delta_2\delta_1$$).