Consider the Killing form of the Lie algebra $\mathfrak{gl}_{m}$. Then $\{e_{ij}\}$ is a basis for this Lie algebra where $e_{ij}$ is a matrix with 1 in the $i$th row, $j$th column and 0 everywhere else. Then every $A = (a_{ij}) \in \mathfrak{gl}_{m}$ can be written as $A = \sum_{i, j = 1}^{m}a_{ij}e_{ij}$. Since $[e_{ij}, e_{kl}] = \delta_{jk}e_{il}-\delta_{li}e_{kj}$, we have $ad(a)e_{ij} = \sum_{kl}(a_{ki}\delta_{lj} -a_{jl}\delta_{ki})e_{kl}$. From this we have $$ad(a)ad(b)e_{ij} = \sum_{k, r, s}(a_{rk}b_{ki}\delta_{sj} + a_{ks}b_{jk}\delta_{ri} - a_{ri}b_{jk}\delta_{sk} - a_{js}b_{ki}\delta_{rk})e_{rs}.$$ From this how can I see that the Killing form is $2m tr(ab) - 2tr(a)tr(b)$. I know that the Killing form is just the trace of the above centered equation, but I seem to be getting confused with all the indices.
Computation of the Killing form of $\mathfrak{gl}_{m}$.
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lie-algebras
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0You can look up Fulton and Harris, it might be easier... – 2012-04-23
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1You need to take the coefficient of $e_{ij}$ on the right hand side (i. e., the term for $r=i$ and $s=j$) and then sum over all $i,j$. You get $\sum\limits_{i,j}\sum\limits_{k} \left( a_{ik}b_{ki} + a_{kj}b_{jk} - a_{ii}b_{jj}\delta_{jk} - a_{jj}b_{ki}\delta_{ik}\right)$. Now, split this sum into four small ones, and the way to proceed should be clear... (Note: If you sum over $k$ a term of the type $u_k \delta_{ik}$, then you get exactly $u_i$.) – 2012-04-23