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The Adjoint representation of a Lie algebra isThe Killing form is the symmetric invariant bilinear form

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Invariance of the Killing form givesIf $x\in {\mathfrak{g}}^{\perp }$, the right-hand side vanishes for every $y$, so $[a,x]\in {\mathfrak{g}}^{\perp }$. Thus ${\mathfrak{g}}^{\perp }$ is an ideal.

Let $\mathfrak{h}=\mathrm{ad}({\mathfrak{g}}^{\perp })\subseteq \mathfrak{gl}(\mathfrak{g})$. For $x\in [{\mathfrak{g}}^{\perp },{\mathfrak{g}}^{\perp }]$ and $y\in {\mathfrak{g}}^{\perp }$,The Cartan solvability criterion makes $\mathfrak{h}$ solvable. The kernel of the adjoint map on ${\mathfrak{g}}^{\perp }$ lies in its center and is abelian, so ${\mathfrak{g}}^{\perp }$ is itself solvable. This proves the Solvability of the radical of the Killing form.

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The Killing form is $\kappa (X,Y)=\mathrm{Tr}({\mathrm{ad}}_X{\mathrm{ad}}_Y)$. From part (a),where the middle equality uses cyclicity of the trace.

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Each structure constant of a Lie algebra satisfies $f_{ab}{}^d=-f_{ba}{}^d$, so $f_{abc}=f_{ab}{}^d{\kappa }_{dc}$ is antisymmetric in $a,b$. Invariance of the Killing form givesThis cyclic symmetry together with antisymmetry in the first pair implies antisymmetry under every transposition. Thus $f_{abc}$ is totally antisymmetric.

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For the trace trilinear form of a Lie algebra representation, the trace of a commutator vanishes:Taking $X=T_b$, $Y=T_a$, $Z=T_c$, $W=T_d$ and expanding each Lie bracket givesSince $f_a{}^{bc}$ is antisymmetric in $b,c$,Raising indices with the inverse Killing form gives $f_a{}^{bc}{f}_{bc}{}^e={\delta }_a{}^e$ in the stated normalization. Using $H(T_e,T_d)=-\mu {\delta }_{ed}$ proves

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