 Symmetric traceless SU(3) tensor representation

Symmetrize the $p$ upper and $q$ lower indices separately, and remove every upper-lower trace. The resulting representation has highest weight $p{\omega }_1+q{\omega }_2$. Its dimension isThe subtraction is valid because contraction is onto: in the polynomial realization its adjoint, for the monomial-factorial inner product, is multiplication by the nonzero polynomial ${\sum }_i{x}_i{y}_i$, an injective map. The resulting kernel contains the highest-weight constituent and has precisely its Weyl dimension formula dimension, hence is irreducible. This highest-weight constituent is not always the largest-dimensional constituent of the full mixed tensor.