## Associativity of the Killing form

This came up in an exam on Lie algebras that I had today, and it's been bugging me. How do you prove that

$$B([X,Y],Z)=B(X,[Y,Z])$$?

The best I've managed is writing

$$B([X,Y],Z)=\mathrm{Tr}(\mathrm{ad}([X,Y])\mathrm{ad}(Z))=\mathrm{Trace}([\mathrm{ad}(X),\mathrm{ad}(Y)]\mathrm{ad}(Z))$$

but I have no idea where to go from there. Hints and/or a complete proof are both appreciated :)
 Mentor Expand the last commutator and use $\mathrm{Trace} \left( AB \right) = \mathrm{Trace} \left( BA \right)$ on one of the terms.
 Sure, that's how I would do it if the Lie bracket was defined as a commutator, but it's not in a general Lie algebra. Oh, hold on... since $$\mathrm{ad}:g\to gl(g)$$ does that mean that I can expand the product [ad(X),ad(Y)] as a commutator?

Mentor

## Associativity of the Killing form

 Quote by Cexy Sure, that's how I would do it if the Lie bracket was defined as a commutator, but it's not in a general Lie algebra. Oh, hold on... since $$\mathrm{ad}:g\to gl(g)$$ does that mean that I can expand the product [ad(X),ad(Y)] as a commutator?
Yes,