For the <trace trilinear form of a Lie algebra representation>, the trace of a commutator vanishes:
$$
B([X,Y],Z,W)+B(Y,[X,Z],W)+B(Y,Z,[X,W])=0.
$$
Taking $X=T_b$, $Y=T_a$, $Z=T_c$, $W=T_d$ and expanding each <Lie bracket> gives
$$
f^e{}_{ba}B_{cde}+f^e{}_{bc}B_{dae}+f^e{}_{bd}B_{ace}=0.
$$
Since $f_a{}^{bc}$ is antisymmetric in $b,c$,
$$
\begin{aligned}
f_a{}^{bc}B_{bcd}
&=\frac12f_a{}^{bc}\operatorname{Tr}([d(T_b),d(T_c)]d(T_d))\\
&=\frac12f_a{}^{bc}f_{bc}{}^eH(T_e,T_d).
\end{aligned}
$$
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
$$
f_a{}^{bc}B_{bcd}=-\frac\mu2\delta_{ad}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 302 / 3 / f / Solution

{title2=$B(X,Y,Z)$}

For a finite-dimensional representation $d$, the trilinear form $B(X,Y,Z)=\operatorname{Tr}(d(X)d(Y)d(Z))$ is invariant under simultaneous adjoint action. This follows by writing the sum of its three infinitesimal variations as the trace of a commutator.


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