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Let $f_1,f_2,f_3$ be the dual basis of $W^{\ast }$, and use $L_1+L_2+L_3=0$. The elementary tensors $e_i\otimes e_j\otimes f_k$ have weight $L_i+L_j-L_k$. Grouping them gives every weight space explicitly:The counts are $6+3\cdot 2+3\cdot 5=27$, so this table exhausts the tensor product. The diagrams below use equilateral coordinates $L_1=(1,0)$, $L_2=(-1/2,\sqrt{3}/2)$, $L_3=(-1/2,-\sqrt{3}/2)$; numbers at points are weight multiplicities.

To decompose the module, first split the first two factors into their symmetric and alternating parts. Put $t_{ij}=(e_i\otimes e_j+e_j\otimes e_i)/2$, including $t_{ii}=e_i\otimes e_i$. The equivariant contractionis onto, since $C(t_{ii}\otimes f_i)=2e_i$. Its kernel $K$ has dimension fifteen and contains the highest-weight vector $t_{11}\otimes f_3$ of weight $2L_1-L_3$.

Here is an explicit weight basis of the trace-free symmetric-square dual tensor module, and thus the requested fifteen-vector basis:Every displayed vector contracts to zero. The first two families occupy distinct one-dimensional weights; at each weight $L_i$ the two vectors in the last family are independent because their respective $t_{ij}\otimes f_j$ terms are distinct. Thus all fifteen are independent and span $K$.

For completeness, $K$ is irreducible, not just an invariant kernel of the right dimension. On elementary tensors the raising action isAt the six weights $2L_i-L_k$, both $E_{12}$ and $E_{23}$ kill the vector only for $(i,k)=(1,3)$. At each weight $-2L_k$, at least one of those operators is nonzero. At weight $L_1$, write a vector as $a(t_{12}\otimes f_2-t_{11}\otimes f_1/2)+b(t_{13}\otimes f_3-t_{11}\otimes f_1/2)$. The two raising equations are $3a+b=0$ and $-a+b=0$, forcing zero. On the two-dimensional weight-$L_2$ space $E_{12}$ is injective, and on the weight-$L_3$ space $E_{23}$ is injective, as substitution in the same action formula shows. Therefore the common raising kernel in $K$ is the single line $\mathbb{C}(t_{11}\otimes f_3)$. By Weyl complete reducibility theorem, every irreducible summand supplies a highest-weight line, so $K$ has only one summand and is irreducible.

The symmetric-factor contraction splits off a defining copy $W$; explicitly $e_i\mapsto {\sum }_j{t}_{ij}\otimes f_j$ is equivariant and its contraction is $4e_i$. For the alternating factor, the invariant volume form identifies ${\Lambda }^{2}W\cong W^{\ast }$, sending $e_1\wedge e_2$ to $f_3$, $e_2\wedge e_3$ to $f_1$ and $e_3\wedge e_1$ to $f_2$. ThereforeThe symmetric dual module is irreducible of dimension six and highest weight $-2L_3$; its symmetric-monomial basis has weights $-2L_i$ and $-L_i-L_j=L_k$, all of multiplicity one. In the original tensor product its highest-weight vector isThe raising action above annihilates it, and its weight is $L_1+L_2-L_3=-2L_3$.

Altogether the Triple tensor decomposition for the defining sl3 representation isThe fifteen-dimensional module has multiplicity one at each $2L_i-L_k$ and $-2L_i$, and multiplicity two at each $L_i$. The six-dimensional module has multiplicity one at $-2L_i$ and $L_i$. Each defining module has the three weights $L_i$, once each. These add to the original table of multiplicities.

The two defining summands are not intrinsically distinguished: other diagonal copies inside their isotypic component have the same three-point weight diagram. Thus the displayed diagrams account for every irreducible submodule type, as well as showing both copies in this decomposition.

Use the full definition of the special unitary group: $A^{†}A=I$ and $\det A=1$. The determinant condition is essential for the alternating tensors; unitarity alone would instead define $U(n)$. Assume $n\ge 2$ when discussing proper invariant subspaces.

Let $V={\mathbb{C}}^n$ be the fundamental representation. A $(j,k)$ tensor belongs to $V^{\otimes j}\otimes (V^{\ast }{)}^{\otimes k}$. Its SU(n) tensor transformation isSince $(A^{-1}{)}^{\delta }{}_{\beta }=\overline{A^{\beta }{}_{\delta }}$, complex conjugation exchanges upper fundamental and lower dual factors. Reordering the two index groups shows that $\overline{T}$ is a $(k,j)$ tensor.

The invariant tensors follow directly. The Kronecker delta transforms to $A^{\alpha }{}_{\gamma }(A^{-1}{)}^{\gamma }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating tensors, the determinant expansion givesThe lower tensor is preserved in the same way using $\det A^{-1}=1$. Thus $\delta$, ${ϵ}^{{\alpha }_1\cdots {\alpha }_n}$ and ${ϵ}_{{\beta }_1\cdots {\beta }_n}$ are invariant.

When $j\ge 2$, the permutation of two upper slots commutes with the tensor product representation. Its $+1$ and $-1$ eigenspaces are nonzero proper invariant subspaces. The same argument uses two lower slots when $k\ge 2$. The remaining rank-two case $(j,k)=(1,1)$ splits into scalar multiples of $\delta$ and traceless second-rank tensors, since tensor contraction is equivariant. This covers every case except the fundamental, dual fundamental and scalar cases specified in the question. Those three are irreducible: the special unitary group acts transitively on unit vectors in its fundamental space, so a nonzero invariant subspace must be the full space, and the dual has the same property.

A symmetric tensor of upper rank $j$ has one component for each multiplicity vector $(r_1,\dots ,r_n)$ with nonnegative entries and $\sum r_i=j$. Equivalently it is a homogeneous degree-$j$ polynomial in $n$ variables. Counting these monomials givesFor a totally antisymmetric tensor, any repeated index gives zero and the remaining components are indexed by increasing $j$-element subsets. HenceThese are the dimensions of the symmetric power and exterior power, respectively.

For SU(4), the two-index antisymmetric representation has complex dimension six. To establish a six-dimensional real representation, one must exhibit a real structure; simply counting six complex components would give twelve real components. Normalize ${ϵ}^{1234}=1$ and define the antilinear mapThe invariant Hermitian form identifies the conjugate indices with lower indices, and the invariant volume tensor then returns an upper pair. Thus $\mathcal{J}$ commutes with SU(4). The identitygives ${\mathcal{J}}^2=1$. Its fixed space hasThree arbitrary complex entries specify all the others, giving exactly six real degrees of freedom. The real structure of the SU4 exterior square therefore provesThe ordinary Hermitian norm restricts to an invariant positive real inner product on this fixed space.

Irreducible representations classify states because an exact symmetry commutes with the Hamiltonian, preserving energy eigenspaces. Irreducible multiplets are the smallest sets closed under all symmetry transformations; their labels and Casimir operators provide quantum numbers. Approximate flavour symmetry gives approximate multiplets and mass relations, with splittings caused by its breaking. Spin and internal quantum numbers are distinct labels.

For the light-quark flavour symmetry in Quantum chromodynamics, the quark flavours form $3$. Their Triple tensor decomposition for the defining sl3 representation isThis algebraic decomposition alone does not determine the allowed ground-state baryons. The three-quark colour singlet is antisymmetric. The no-orbital-excitation ground state has a symmetric spatial wavefunction, so the Pauli exclusion principle requires the combined spin-flavour wavefunction to be symmetric. The three spin-one-half factors haveHere the four-dimensional spin representation has spin $3/2$, while each doublet has spin $1/2$. Symmetric decuplet flavour pairs with symmetric spin $3/2$. The two flavour-octet copies and two spin-doublet copies carry the two-dimensional standard permutation multiplicity space; its tensor square contains one symmetric singlet. Their invariant pairing therefore supplies one octet with spin $1/2$, not two independent ground-state octets. Antisymmetric flavour would need a completely antisymmetric three-quark spin state, which does not exist.

Equivalently the symmetric spin-flavour SU6 representation hasThus the Pauli constraint on three-quark flavour multiplets gives the concise spectrum classificationThese are the baryon octet and baryon decuplet. The three-quark flavour singlet requires a different spatial permutation symmetry and belongs to excited states, outside the symmetric ground-state approximation used here. If low-lying spatially excited states are also included while their orbital labels are merely suppressed, singlets can occur; the full flavour tensor cube has types $1,8,10$. The absence of the singlet from the unexcited ground multiplet is a permutation-symmetry constraint, not a prohibition by $SU(3)$ alone. Light-quark mass differences make the flavour symmetry approximate rather than exact.

Let $V=V({\omega }_1)$, so $V^{\ast }=V({\omega }_2)$. First,The highest-weight tensor-product rule givesandThe dimensions $15+3+6+3=27$ check the decomposition. Therefore the Triple tensor decomposition for the defining sl3 representation isThus one may take