Let $\mathfrak{g}$ be a finite dimensional Lie algebra. The Killing form $K:\mathfrak{g}\times\mathfrak{g}\rightarrow\mathbb{C}$ is given by $$K(x,y) = tr(ad_xad_y)$$ I have two questions about the Killing form:
- How can it be computed? And more generally, how can $ad_x$ be expressed for a general $x\in\mathfrak{g}$? In particular, if $\mathfrak{g}\subset\mathfrak{gl}(V)$ for some vector space $V$ (and it is always the case, by Ado's theorem) then $ad_x$ can be expressed as a matrix. How can this matrix be determined?
- I don't really see why $K([x,y],z)\neq 0$. Indeed we have $$K([x,y],z) = tr(ad_{[x,y]}ad_z) = tr(ad_xad_yad_z) - tr(ad_yad_xad_z)$$ and since we can commute matrices in the trace, this should give zero. Could someone help me see what I'm overlooking?
I'm not sure what you mean by your first question. If you have a lie algebra, with basis $ \{ e_1, \cdots, e_n \} $ with structure constants $ C^{i}_{jk} $, then it is easy to write the matrix $ ad_{e_j} $: $ C^{i}_{jk} $ with $ j $ fixed. You should also keep in mind that, in the case the lie algebra is embedded in $ \mathfrak{g}\subset\mathfrak{gl}(V)$, the dimension of the matrix $ ad_x $ will be different than the dimension of $ V $. Maybe you should try with the case $ g = \mathfrak{gl}(2,\mathbb{R}) $ and use the basis of four matrices each with 1 in one spot and 0 everywhere else.
In the second question, the $ \text{tr} $ is not invariant under arbitrary permutations, only cyclic permutations.