For $\mathfrak{sl}_3(\mathbb C)$, the <Killing form of the special linear Lie algebra> is
$$
\kappa(X,Y)=6\operatorname{tr}(XY).
$$
For $H_1=E_{11}-E_{22}$ and $H_2=E_{22}-E_{33}$, its matrix on the Cartan part is
$$
\begin{pmatrix}12&-6\\-6&12\end{pmatrix},
$$
whose determinant is $108$. Each $E_{ij}$ pairs only with $E_{ji}$, with value $6$. In the stated ordering, the remaining block is
$$
\begin{pmatrix}0&6I_3\\6I_3&0\end{pmatrix},
$$
whose determinant is $-6^6$. Therefore
$$
\boxed{\det\kappa=-108\cdot6^6=-5\,038\,848}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 102 / 1 / f / Solution

{c}
{title2=$\kappa(X,Y)=2n\operatorname{tr}(XY)$}

For $\mathfrak{sl}_n(\mathbb C)$,
$$
\kappa(X,Y)=2n\operatorname{tr}(XY).
$$


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 Incoming links: Killing form of the special linear Lie algebra