Associativity of the Killing form

[Cexy]

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 :)

 

[George Jones]

Expand the last commutator and use \mathrm{Trace} \left( AB \right) = \mathrm{Trace} \left( BA \right) on one of the terms.

 

[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?

 

[George Jones]

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,

[ad(X) , ad(Y)] = ad(X)ad(Y) - ad(Y)ad(X),

where the associative products are defined, since ad(X) and ad(Y) are both linear operators on g considered as a vector space.

 

[Cexy]

It's annoying that it's that simple but I couldn't do it in the exam. :(

Still, I did much more complicated stuff so hopefully the examiners will assume I just had a dim moment - rather than assuming that I'm dim altogether!

 

[Cexy]

I thought I'd post to mention that I had to do this proof (well, variants of it) THREE times in a particle physics exam today, so thanks for clearing it up in my mind - you scored me some marks!

Applied exams are much easier than pure ones.