How can one prove the Killing form given by $B(X,Y)=Tr(ad(X) \circ ad(Y))$, where $X,Y \in \mathfrak{g}$ and $Tr$ is the trace, is $ad$-invariant. I mean, it verifies:
$B(ad(X)(Y),Z)+B(Y,ad(X)(Z))=0$, with $ad$ the adjoint representation of Lie algebras and $ad(X)(Y)=[X,Y]$,
Applying two times the Jacobi identity to $[Z,[X,[Y,W]]]$, I got:
$ad(Z) \circ ad(X) \circ ad(Y)=ad(ad(Z)X) \circ ad(Y)+ad(X) \circ ad(ad(Z)Y)+ad(X) \circ ad(Y) \circ ad(Z)$
Then I tried to apply the trace, but I do not achieve what I want. Some ideas? Is there an easier way to do it?
For any representation $\rho\colon \mathfrak{g}\rightarrow \mathfrak{gl}(V)$ we have \begin{align*} B_{\rho}([x,y],z) & = {\rm tr} (\rho([x,y])\rho(z)) \\ & = {\rm tr}(\rho(x)\rho(y)\rho(z))-{\rm tr} (\rho(y)\rho(x)\rho(z)) \\ & = {\rm tr}(\rho(y)\rho(z)\rho(x))-{\rm tr} (\rho(y)\rho(x)\rho(z)) \\ & = {\rm tr} (\rho(y) \rho ([z,x])) \\ & = - B_{\rho}(y,[x,z]). \end{align*} So the Killing form is $\rho$-invariant. Now take $\rho=\operatorname{ad}$.