Tensor-product Casimir trace identity (${\sum }_{i}d(R_i)C_2(R_i)=d(R_1)d(R_2)[C_2(R_1)+C_2(R_2)]$)

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 $\mathrm{tr}{t}_a=\mathrm{tr}{s}_a=0$. Decomposition into irreducible representations, with multiplicities counted, gives the displayed identity. Without tracelessness an extra term $2{\sum }_a\mathrm{tr}{t}_a\mathrm{tr}{s}_a$ 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.