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Suppose $L$ is a Lie Algebra and $x \in L'=\left[L,L\right]$. As a homework problem, I need to show that $\operatorname{tr}(\operatorname{ad} \, x)=0$. I assumed $\dim(L)<\infty$ (not sure if this is necessary) and tried explicitly computing $\operatorname{tr}(\operatorname{ad} \, x)$. I expressed $x$ as a linear combination of commutators, expanded the brackets in terms of structure coefficients, and ended up with a very nasty sum.

There must be a better way to do this. Hints?

Andreas Caranti
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Jackson Walters
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1 Answers1

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There is a much simpler way indeed. As you perceived, it suffices to show that ${\sf tr}({\sf ad}(w))=0$ when $w$ is a commutator $[x,y]$ . The Jacobi identity implies that

$$ {\sf ad}([x,y])=[{\sf ad} (x),{\sf ad} (y)] $$

and now the bracket on the right can be interpreted as $ab-ba$ where $a={\sf ad} (x)$ and $b={\sf ad} (y)$. Now, ${\sf tr}(ab)={\sf tr}(ba)$ is pure linear algebra and does not need anything about Lie algebras.

Ewan Delanoy
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  • Why do we get to assume the Lie bracket is a commutator? Surely there are more general brackets. – Jackson Walters Apr 03 '13 at 13:08
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    @JacksonWalters : I’m not sure I understand your comment, and I’m not sure you understand my proof either. The central point is that the original Lie bracket, which is not a commutator in general, can here be re-interpreted (in its adjoint representation) as a commutator, thanks to the Jacobi identity. – Ewan Delanoy Apr 03 '13 at 15:00