Mixed SU(3) tensor representation ($V^{\otimes k}\otimes (V^{\ast }{)}^{\otimes l}$)

A mixed tensor has $k$ defining indices and $l$ dual indices. The SU(3) action applies $U$ to every upper index and $U^{-1}$ to every lower index. Young symmetrizers select permutation types, while invariant Kronecker deltas take traces and invariant alternating tensors convert an antisymmetric pair of defining indices into a dual index, or conversely. Repeated fundamental and dual tensor rules decompose it with multiplicities: $(p,q)\otimes (1,0)=(p+1,q)\oplus (p-1,q+1)\oplus (p,q-1)$ and the conjugate rule, omitting negative labels.

For $T^{abc}$, split its first pair into symmetric and antisymmetric parts. The fully symmetric part $S^{abc}$ is a symmetric cubic representation of SU(3). Write $M^{abc}=T^{(ab)c}-S^{abc}$ and $O^a{}_d={ϵ}_{dbc}{M}^{abc}$. Then $O$ is traceless and $M^{abc}=({ϵ}^{bcd}{O}^a{}_d+{ϵ}^{acd}{O}^b{}_d)/3$. The antisymmetric part is $T^{[ab]c}={ϵ}^{abd}{N}^c{}_d$, with $N^c{}_d={ϵ}_{dab}{T}^{abc}/2$. Splitting $N$ into its traceless part and trace gives the second adjoint representation of SU(3) and the singlet. These maps use invariant tensors, so they commute with the group action and explicitly realize the four irreducible summands.

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.