Symmetric traceless SU(3) tensor representation ($d(p,q)=\frac{1}{2}(p+1)(q+1)(p+q+2)$)

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.

The degree-three monomials $z_1^a{z}_2^b{z}_3^c$, with $a+b+c=3$, form a basis of this ten-dimensional highest-weight representation. Its highest monomial is $z_1^3$, and $E_{21}=z_2{\partial }_{z_1}$ and $E_{32}=z_3{\partial }_{z_2}$ generate every basis monomial. Its Dynkin labels are $(3,0)$, and the weight of a monomial is $(a-b,b-c)$. In coordinates $(h_1,(h_1+2h_2)/\sqrt{3})$, the ten weights form four horizontal rows of lengths $4,3,2,1$, all with weight multiplicity one. The baryon decuplet realizes this representation of approximate flavor symmetry.

// Target: special-relativity.bigb

For $d(p,q)=(p+1)(q+1)(p+q+2)/2$, the exact largest constituent dimension of a mixed SU(3) tensor representation isEvery constituent highest weight lies below $k{\omega }_1+l{\omega }_2$, so its labels have the form $(p,q)=(k-2a+b,l+a-2b)$ for nonnegative integers $a,b$. If both are positive, reducing each by their minimum increases both $p,q$ and strictly increases the dimension. The resulting boundary labels occur: antisymmetrize disjoint upper pairs into lower indices, or lower pairs into upper indices, then take the symmetric traceless component. This proves the maximum formula. For $(k,l)=(5,0)$, the fully symmetric dimension is 21 but $(3,1)$ has dimension 24. For $k=l$, the fully symmetric traceless dimension $(k+1{)}^3$ is the maximum.