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Let $V={\mathbb{C}}^3$ be the defining flavor SU(3) representation. Under $U\in SU(3)$, $q^{\alpha }\mapsto U^{\alpha }{}_{\beta }{q}^{\beta }$. The antiquark transforms in the dual representation:Since $U$ is unitary, its lower-index column of components transforms by $U^{\ast }$, the complex conjugate numerical matrix. Thus $q^{\alpha }{\bar{q}}_{\alpha }$ is invariant. The defining triplet and its conjugate are inequivalent complex representations; conjugation reverses their weights. Write $3=V$ and $\overline{3}=V^{\ast }$, and distinguish a direct sum $\oplus$ from a tensor product $\otimes$.

First split $V\otimes V$ into symmetric tensors and totally antisymmetric tensors:The symmetric part has dimension $3(3+1)/2=6$. Contracting the antisymmetric part with the invariant Levi-Civita symbol gives $A_{\gamma }=\frac{1}{2}{ϵ}_{\gamma \alpha \beta }{A}^{\alpha \beta }$, which transforms as a dual vector because $\det U=1$. Its inverse is $A^{\alpha \beta }={ϵ}^{\alpha \beta \gamma }{A}_{\gamma }$. HenceThe symmetric part is the irreducible highest-weight representation with Dynkin labels $(2,0)$; its six monomials give the standard symmetric-square representation.

For three triplets, the antisymmetric first-pair sector becomes $V^{\ast }\otimes V$. Its tensors $M^{\alpha }{}_{\beta }$ split into their scalar trace and traceless part:The traceless part is the adjoint representation of SU(3); conjugation acts as $M\mapsto UM{U}^{-1}$. It is irreducible because an invariant subspace would be an ideal in the simple complex special linear Lie algebra ${\mathfrak{sl}}_3$.

The symmetric first-pair sector consists of $T^{abc}=T^{bac}$ and has dimension $6\cdot 3=18$. Complete symmetrization maps it onto ${\mathrm{Sym}}^{3}V$, whose dimension is $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$. Its kernel is the mixed-symmetry eight-dimensional sector. More explicitly, the equivariant map $T^{abc}\mapsto M^a{}_e={ϵ}_{ebc}{T}^{abc}$ has trace zero because the first two indices of $T$ are symmetric; it annihilates the completely symmetric part and is not identically zero. Since the target adjoint module is irreducible, the map is onto its eight-dimensional space. Its kernel is consequently exactly the ten-dimensional completely symmetric part. Complete symmetrization supplies an invariant splitting, so this sector is $10\oplus 8$. Combining the sectors gives the tensor cube of the defining SU(3) representationHere the decuplet has Dynkin labels $(3,0)$, and the singlet is the fully antisymmetric contraction ${ϵ}_{abc}{T}^{abc}$. The two octets are distinct copies, not two different irreducible types.

For two conjugate SU(3) triplets with a triplet, begin with $T^a{}_{bc}$ and split its lower pair. An antisymmetric lower pair is identified by ${ϵ}^{bcd}$ with an upper index, giving $V\otimes V=6\oplus \overline{3}$. The symmetric lower-pair sector has dimension $3\cdot 6=18$, and its contraction is the equivariant mapIt is onto $V^{\ast }$, since the invariant embeddingsatisfies $C\iota (v)=v$. Every such tensor therefore splits uniquely as $\iota (C(T))+(T-\iota (C(T)))$. The latter tensor is symmetric in its lower indices and has zero contraction, with dimension $18-3=15$. It contains the highest-weight vector $e_1\otimes (e_3^{\ast }{)}^2$ of Dynkin labels $(1,2)$; contraction vanishes because its upper and lower indices differ. The Weyl dimension formula for an SU(3) module with labels $(p,q)$ is $\dim V_{(p,q)}=(p+1)(q+1)(p+q+2)/2$, which gives $15$ for $(1,2)$. These tensor representations are unitary, so invariant subspaces have invariant orthogonal complements; the irreducible highest-weight summand containing this vector exhausts the fifteen-dimensional kernel. ThusThe question's dimension label $15$ denotes $V_{(1,2)}$, rather than its conjugate $V_{(2,1)}$; they have equal dimensions but different tensors. Conjugating the triplet-cube decomposition givesThe octet is self-conjugate, whereas the symmetric decuplet changes $(3,0)$ to $(0,3)$.

Now consider the flavor diquark $D_{\alpha }\propto {ϵ}_{\alpha \beta \gamma }{q}^{\beta }{q}^{\gamma }$. This notation describes an antisymmetric two-quark flavor state; color and spin factors are separate. In particular $D_3=(ud-du)/\sqrt{2}$ has isospin zero, baryon number $2/3$, strangeness zero and flavor hypercharge $2/3$. The strange antiquark has $I=0$, $B=-1/3$, $S=+1$ and $Y=2/3$. Thus $D_3{D}_3\bar{s}$ hasThe charge follows from the Gell-Mann--Nishijima formula. Each of the three factors has lower flavor index $3$, so their product is the completely symmetric tensor component $A_{333}$. It lies entirely in ${\mathrm{Sym}}^3{V}^{\ast }=\overline{10}$, the pentaquark antidecuplet. Equivalently, the symmetric diquark pair belongs to $\overline{6}$, and $\overline{6}\otimes \overline{3}=\overline{10}\oplus 8$, with this highest-hypercharge state in the antidecuplet.

It cannot be in an octet. The octet has the weights of a traceless $q^{\alpha }{\bar{q}}_{\beta }$ tensor, whose hypercharges are $Y_{\alpha }-Y_{\beta }$. For $(u,d,s)$ the quark values are $1/3,1/3,-2/3$, so these differences are only $0,+1,-1$; there is no $Y=2$ weight. The octet's top row instead has $Y=1$ and $I=1/2$. This is a flavor-representation assignment for a possible pentaquark, not a proof that the configuration binds or that its full color-spin-orbital wavefunction satisfies all Fermi statistics constraints.

Treat the quarks as the fundamental triplet of approximate flavour symmetry. The flavour product follows by splitting the first two quarks into symmetric and antisymmetric pieces:ThereforeThe dimension check is $27=10+8+8+1$. The baryon decuplet is the completely symmetric flavour sector, the three-quark flavour singlet is completely antisymmetric, and the two copies of the baryon octet carry mixed permutation symmetry. The two octet copies are a multiplicity space for permutations of the three quark slots, not automatically two distinct ground-state baryon octets.

For the weight diagrams use isospin projection $I_3$ and flavour hypercharge $Y$. The quark weights areWeights add in a tensor product. In a three-quark composition, $Y=1-n_s$ and $I_3=(n_u-n_d)/2$. The baryon decuplet has rows $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$. Its upper-right weight is $uuu$, the ${\Delta }^{++}$, while its bottom weight is $sss$, the ${\Omega }^{-}$. In the baryon octet, the upper weights are $uud$ and $udd$, giving the proton and neutron. The origin has two independent states with content $uds$: the $I=1$ ${\Sigma }^0$ and the $I=0$ $\Lambda$. Their equal weights do not make them the same state. The three-quark flavour singlet has only $(I_3,Y)=(0,0)$ and content $uds$, with normalized flavour wavefunctionThis singlet is a different representation from the octet $\Lambda$, despite the same quark content and weight.

The Pauli exclusion principle requires the full three-quark wavefunction to change sign under exchange of any two quarks, including their spatial, spin, flavour and colour labels. A three-quark colour singlet has the antisymmetric colour factor ${ϵ}_{abc}/\sqrt{6}$. Consequently the remaining spatial-spin-flavour factor must be symmetric. The flavour representation alone is not the full exchange wavefunction.

For the lowest orbital state, the spatial wavefunction is symmetric. Completely symmetric decuplet flavour then requires the symmetric spin-$3/2$ wavefunction; this includes states such as $uuu$ with aligned spins and does not violate Pauli because their colours are antisymmetrized. Mixed octet flavour combines with mixed spin-$1/2$ wavefunctions to give a symmetric spin-flavour factor. More explicitly, the two-dimensional permutation representation of mixed symmetry tensored with itself contains the trivial representation, which selects the physical symmetric combination. These are the familiar ground-state spin assignments.

Antisymmetric singlet flavour in a symmetric orbital state would instead require a completely antisymmetric three-quark spin state. But ${\bigwedge }^3{\mathbb{C}}^2=0$, so three spin-$1/2$ quarks have no such state. There is no flavour-singlet three-quark ground-state S-wave baryon. A singlet is allowed with orbital excitation: mixed spatial and mixed spin symmetry can combine antisymmetrically, and their product with the antisymmetric flavour sector is symmetric. For example an $L=1$, spin-$1/2$ configuration can give negative-parity total spins $1/2$ or $3/2$. This Pauli constraint on three-quark flavour multiplets distinguishes a permitted representation in the flavour tensor product from its possible orbital-spin realization.

For a pentaquark, choose each of two quark pairs in $\overline{3}$. Their symmetric flavour combination lies in $\overline{6}$, and combining with the antiquark givesThus $3^{\otimes 4}\otimes \overline{3}$ contains a pentaquark antidecuplet. This identifies a flavour sector; it does not by itself prove binding or fix the spin and orbital structure needed for overall fermion antisymmetry.

The antidecuplet is the conjugate of the symmetric decuplet, so its rows are $(Y,I)=(2,0),(1,1/2),(0,1),(-1,3/2)$. A three-quark state only has $Y=1,0,-1,-2$, and at $Y=-1$ it contains two strange quarks and one light quark, allowing only $I_3=\pm 1/2$. Hence the exotic weights are preciselyPossible minimal contents are $uudd\bar{s}$, $ddss\bar{u}$, and $uuss\bar{d}$. By the Gell-Mann--Nishijima formula, their charges are respectively $+1,-2,+1$. Every other antidecuplet weight is also a weight of some three-quark composition, although its total isospin representation can differ.

Exotic weights do not imply a weak-decay lifetime. The strong interaction can conserve all the quantum numbers in baryon-plus-meson channels, for exampleThese are quark rearrangements into a three-quark baryon and a quark-antiquark meson, not decay into a single three-quark state. Accordingly they are generically short-lived strong resonances if these channels are kinematically open. Flavour representation theory alone gives no masses or widths; a state below all strong thresholds, or one with dynamically suppressed couplings, can be longer-lived. The quantum numbers provide no general protection against the displayed strong decays.