# Cartan-Killing metric and Lie-groups

We defined the Cartan-Killing metric of a Lie-Group $$G$$ as $$g_{ab}\equiv C_{acd}C_{bdc},$$ where $$C_{abc}$$ are the structure constants of the Lie-algebra $$\mathfrak{g}$$. According to my professor it is possible to show that $$\operatorname{tr}(A_aA_b)=g_{ab},$$ where $$A_i$$ denote the generators in the adjoint representation.

I'm honestly having a really hard time on trying to show that. As far as I understand we have $$A_a=\left.\frac{\partial \tau_A(g)}{\partial \alpha_a}\right|_{g=e},$$ where $$\tau_A$$ is the adjoint representation and $$e$$ the unit-element of $$G$$. I know that one can write $$\|A_a\|_{bc}=C_{abc}$$. This would then imply that $$g_{ab}=C_{acd}C_{bdc}= \sum_{c,d=1}^3\|A_a\|_{cd}\|A_{b}\|_{dc}.$$ The problem is that I don't really see what this has to do with the trace of the two matrices $$A_a$$ and $$A_b$$.

• Hint: What is your definition of the structure constants? Also, forget about the group, think only about the Lie algebra. – Moishe Kohan Feb 2 '19 at 15:11
• @MoisheCohen We defined the structure constants over $[A_a, A_b]= C_{abc}A_c$. Thanks for the hint, will think about it... – Sito Feb 2 '19 at 15:15
• @MoisheCohen Alright, I thought about it now for some time, but I just can't figure out what the trick is supposed to be... To make use of the definition of the structure constants I would need to multiply $g_{ab}$ with $A_c$ and $A_d$, which doesn't seem to lead anywhere... Could you maybe help a little more.. – Sito Feb 2 '19 at 18:51

If $$e_1,...,e_N$$ is a basis of the Lie algebra $${\mathfrak g}$$, then $$[e_i,e_j]=\sum_{l} c^l_{ij} e_l.$$ Now, $$ad(e_i)\circ ad(e_j)(x)= [e_i, [e_j, x]]$$ for $$x\in {\mathfrak g}$$. If $$x=e_k$$ then $$[e_i, [e_j, e_k]] = [e_i, \sum_{l} c^l_{jk} e_l]= \sum_{l,m} c^l_{jk} c^m_{il} e_m.$$ Thus, the linear map $$ad(e_i)\circ ad(e_j)$$ sends $$e_k\mapsto \sum_{l,m} c^l_{jk} c^m_{il} e_m$$ Computing the trace of this map means setting $$k=m$$ and taking the sum over $$k=1,...N$$: $$tr(ad(e_i)\circ ad(e_j))= \sum_{l,k} c^l_{jk} c^k_{il}.$$ That's your professor's formula.