What is the Killing form of $\mathfrak{gl}_{m}$? 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.
 A: I'm currently doing my homework and came here to check the result. It looks OK so I'll post my solution (basically the same way but without playing with $\delta_{ij}$ which in my opinion greatly simplifies the solution).
I rename $m$ as $n$, and I use the notation $\operatorname{ad}_A$ for $\operatorname{ad}\left(A\right)$. The $\left(i,j\right)$-th entries of the matrices $A$ and $B$ will be called $a_{ij}$ and $b_{ij}$, respectively. We let $e_{ij}$ be the $n\times n$-matrix with a $1$ in cell $\left(i,j\right)$ and a $0$ in all the other cells.
Fix $\left(i,j\right)\in\left\{1,2,\ldots,n\right\}^2$. Then,
$$\operatorname{ad}_A e_{ij} = \sum_{k=1}^n (a_{ki}e_{kj}-a_{jk}e_{ik})$$
so
$$
\begin{align*}
\operatorname{ad}_B \operatorname{ad}_A e_{ij}
&= \operatorname{ad}_B(\sum_{k=1}^n (a_{ki}e_{kj}-a_{jk}e_{ik}))
=
\sum_{k=1}^n (a_{ki}\operatorname{ad}_Be_{kj}-a_{jk}\operatorname{ad}_Be_{ik}) \\
&=
\sum_{k=1}^n (a_{ki}\sum_{l=1}^n (b_{lk}e_{lj}-b_{jl}e_{kl})-a_{jk}\sum_{l=1}^n (b_{li}e_{lk}-b_{kl}e_{il})) \\
&= 
\sum_{k=1}^n \sum_{l=1}^n(a_{ki}b_{lk}e_{lj}-a_{ki}b_{jl}e_{kl}-a_{jk}b_{li}e_{lk}+a_{jk}b_{kl}e_{il}) \\
&= 
\sum_{k=1}^n \sum_{l=1}^n a_{ki}b_{lk}e_{lj} - \sum_{k=1}^n \sum_{l=1}^n a_{ki}b_{jl}e_{kl} - \sum_{k=1}^n \sum_{l=1}^n a_{jk}b_{li}e_{lk} + \sum_{k=1}^n \sum_{l=1}^n a_{jk}b_{kl}e_{il} .
\end{align*}
$$
Now we want to find the coefficient with which $e_{ij}$ appears in this sum. (Later, we shall sum these coefficients over all $\left(i,j\right)\in\left\{1,2,\ldots,n\right\}^2$, and thus obtain the trace of $\operatorname{ad}_B \operatorname{ad}_A$.)
For a single pair $\left(i,j\right)$, we find the coefficient of $e_{ij}$ in the above sum by finding its coefficient in each of the four summands:
considering the first summand we are interested only at such entries that $l = i$, obtaining:
$$\sum_{k=1}^na_{ki}b_{ik}$$
from the second only at such that $k=i$ and $l=j$:
$$-a_{ii}b_{jj}$$
from the third similarly we are taking only coefficients for $k=j$, $l=i$:
$$-a_{jj}b_{ii}$$
and from the fourth at $l=j$:
$$\sum_{k=1}^na_{jk}b_{kj}.$$
Now after summing all these coefficients up over all pairs $(i, j)$ we are obtaining the expected result:
$$
\begin{align*}
\operatorname{tr}(\operatorname{ad}_B \operatorname{ad}_A)
&= \sum_{i=1}^n\sum_{j=1}^n(\sum_{k=1}^na_{ki}b_{ik} -a_{ii}b_{jj} -a_{jj}b_{ii} + \sum_{k=1}^na_{jk}b_{kj}) \\
&= 
\sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^na_{ki}b_{ik} + \sum_{i=1}^n\sum_{j=1}^n\sum_{k=1}^na_{jk}b_{kj}- \sum_{i=1}^n\sum_{j=1}^na_{ii}b_{jj} -\sum_{i=1}^n\sum_{j=1}^n a_{jj}b_{ii} \\
&=
n\sum_{i=1}^n\sum_{k=1}^na_{ki}b_{ik} + n\sum_{j=1}^n\sum_{k=1}^na_{jk}b_{kj} - 2\operatorname{tr}(A)\operatorname{tr}(B) \\
&= 2n \operatorname{tr}(AB) - 2\operatorname{tr}(A)\operatorname{tr}(B).
\end{align*}
$$
