# Proof of an identity for the Killing form involving derivations.

I'm working through Ziller's Lie Groups. Representation Theory and Symmetric Spaces, and in Proposition 1.36, he shows the following identity:

Let $$\mathfrak{g}$$ be a real or complex [finite-dimensional] Lie algebra with Killing form $$B$$... If $$L \in \mathfrak{Der}(\mathfrak{g})$$, then $$B(LX, Y) + B(X, LY) = 0$$.

He's already demonstrated that if $$A \in \operatorname{Aut}(\mathfrak{g})$$, then $$B(AX, AY) = B(X, Y)$$. He says that if $$L$$ is a derivation, then $$e^{tL}$$ is an automorphism of $$\mathfrak{g}$$, so $$B\left( e^{tL} X , e^{tL} Y \right) = B(X, Y)$$. This is where he loses me.

Differentiating at $$t = 0$$ proves [the] claim.

I don't understand how I'd flesh this step out. I think he wants to do something like this.

\begin{align*} B(X, Y) & = B \left( e^{tL} X , e^{tL} Y \right) \\ \Rightarrow \frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} B(X, Y) & = \frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} B \left( e^{tL} X , e^{tL} Y \right) \\ = 0 & = \frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} \operatorname{tr} \left( \operatorname{ad}_{e^{tL} X} \circ \operatorname{ad}_{e^{tL} Y} \right) , \end{align*}

and somehow get to $$\frac{\mathrm{d}}{\mathrm{d}t}|_{t = 0} B \left( e^{tL} X , e^{tL} Y \right) = B(LX, Y) + B(X, LY)$$ or something, but I just don't know what to do with this. I expect it's a fairly straightforward computation, but I just don't have an intuition for where to take this.

• Write $B(e^{tL}X,e^{tL}Y)=B((1+tL+o(t))X,(1+tL+o(t))Y)=B(X,Y)+t(B(LX,Y)+B(X,LY))+o(t)$. Here the Landau notation is meant $X,Y,L$ being fixed, and $t\to 0$. – YCor Mar 24 at 20:04