Cubic trace tensor of a Lie algebra representation ($B(X,Y,Z)=\mathrm{Tr}(d(X)d(Y)d(Z))$)

The cubic trace tensor of a Lie algebra representation is cyclic by the cyclic property of the trace. It need not be fully antisymmetric. Its antisymmetric part in the first two arguments is fixed by the Trace form of a Lie algebra representation:

// Target: lie-theory.bigb

Lowering and raising structure constants with the Killing form gives $c_a{}^{mn}{c}_{mn}{}^r=-{\delta }_a{}^r$. Since $c_a{}^{mn}$ is antisymmetric in $m,n$, only the commutator part of the cubic trace tensor of a Lie algebra representation contributes, proving $c_a{}^{mn}{B}_{mnb}=-H_{ab}/2$. In an adapted compact basis with ${\kappa }_{ab}=-{\delta }_{ab}$ and $H_{ab}=-\mu {\delta }_{ab}$, the resulting sign is positive: $+\mu {\delta }_{ab}/2$. Defining the lowered structure constants with a positive Euclidean metric instead would reverse this sign.

// Target: lie-theory.bigb

The trace of a commutator givesIn a basis, cyclicity converts this to $c_{da}{}^{\ell }{B}_{bc\ell }+c_{db}{}^{\ell }{B}_{ca\ell }+c_{dc}{}^{\ell }{B}_{ab\ell }=0$.

// Target: lie-theory.bigb