Classical simple Lie algebras and their Killing forms

On $\mathrm{End}(k^n)$, ${\mathrm{ad}}_X=L_X-R_X$. The trace identities $\mathrm{tr}(L_X{L}_Y)=n\mathrm{tr}(XY),\mathrm{tr}(R_X{R}_Y)=n\mathrm{tr}(XY),\mathrm{tr}(L_X{R}_Y)=\mathrm{tr}(X)\mathrm{tr}(Y)$follow by applying the maps to the matrix units $E_{ij}$. Hence${\mathrm{tr}}_{{\mathfrak{g}\mathfrak{l}}_n}({\mathrm{ad}}_X{\mathrm{ad}}_Y)=2n\mathrm{tr}(XY)-2\mathrm{tr}(X)\mathrm{tr}(Y)\mathrm{.}$ For traceless $X,Y$, the central line in ${\mathfrak{g}\mathfrak{l}}_n=kI\oplus {\mathfrak{s}\mathfrak{l}}_n$ contributes zero to the adjoint trace, proving the first formula.

Use the standard split Cartan and root matrices, all defined over $k$. For a Cartan element write its coordinates as $(t_1,\dots ,t_r)$. The roots of ${\mathfrak{s}\mathfrak{o}}_{2r}$ are $\pm {\epsilon }_i\pm {\epsilon }_j$; those of ${\mathfrak{s}\mathfrak{o}}_{2r+1}$ add $\pm {\epsilon }_i$. Summing $\alpha (H)\alpha (H^{\mathrm{\prime }})$ over the roots gives respectively $4(r-1){\sum }_i{t}_i{s}_i$ and $(4r-2){\sum }_i{t}_i{s}_i$. Since the defining matrix trace is $2{\sum }_i{t}_i{s}_i$, both results equal $(n-2)\mathrm{tr}(H{H}^{\mathrm{\prime }})$.

The roots of ${\mathfrak{s}\mathfrak{p}}_{2r}$ are $\pm {\epsilon }_i\pm {\epsilon }_j$ and $\pm 2{\epsilon }_i$. Their sum gives $4(r+1){\sum }_i{t}_i{s}_i$; dividing by the same defining trace $2{\sum }_i{t}_i{s}_i$ yields $2(r+1)\mathrm{tr}(H{H}^{\mathrm{\prime }})$. Invariance pairs a root space only with its opposite root space, and direct multiplication of the standard root matrices gives the same nonzero scalar there. Thus the identities on the Cartan and opposite-root pairs determine the displayed orthogonal and symplectic forms on the whole algebra.

The trace pairing is nondegenerate on each displayed matrix algebra: diagonal Cartan coordinates pair coordinatewise, while every root matrix pairs nontrivially with its opposite. The scalar multipliers $2n$, $n-2$, and $2(n+1)$ are nonzero in characteristic zero, so all three Killing forms are nondegenerate.

The excluded orthogonal ranks explain the endpoints: ${\mathfrak{s}\mathfrak{o}}_1=0$, ${\mathfrak{s}\mathfrak{o}}_2$ is one-dimensional abelian and has zero Killing form, and ${\mathfrak{s}\mathfrak{o}}_4\cong {\mathfrak{s}\mathfrak{l}}_2\oplus {\mathfrak{s}\mathfrak{l}}_2$ is semisimple but not simple (its formula is still $2\mathrm{tr}(XY)$). Thus none is silently included among the simple orthogonal cases. All calculations are finite and choice-free.