0

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?

Jotabeta
  • 1,253

1 Answers1

2

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 $\rho=\operatorname{ad}$.

Dietrich Burde
  • 130,978
  • Maybe I am a little bit confused, but it is not true that the Lie bracket is $[X,Y]=XY-YX$ in general, or it is? I thought it was only true for matrices Lie algebras – Jotabeta Jan 21 '21 at 16:08
  • Yes, but the image of $\rho$ consists of matrices, or linear maps. Indeed, ${\rm ad}(\mathfrak{g})$ is a linear Lie algebra, consisting of matrices. And there indeed $[X,Y]=XY-YX$. – Dietrich Burde Jan 21 '21 at 16:09