= Tensor-product Casimir trace identity
{title2=$\sum_i d(R_i)C_2(R_i)=d(R_1)d(R_2)[C_2(R_1)+C_2(R_2)]$}
For traceless generators, the <tensor product of Lie algebra representations> has generators $t_a\otimes I+I\otimes s_a$. The <trace> of their summed squares is the sum of the two factor Casimir <traces>; the cross term vanishes because $\operatorname{tr}t_a=\operatorname{tr}s_a=0$. Decomposition into <irreducible representations>, with multiplicities counted, gives the displayed identity. Without tracelessness an extra term $2\sum_a\operatorname{tr}t_a\operatorname{tr}s_a$ is present. A one-dimensional abelian representation of charge one tensored with itself has Casimir four, not the sum two; semisimple <Lie algebras> avoid this problem because every generator is a sum of <commutators>.
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