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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.

The Kronecker delta is the identity map on the defining representation. Its transformation is $A^{\alpha }{}_{\gamma }{\delta }^{\gamma }{}_{\delta }(A^{-1}{)}^{\delta }{}_{\beta }={\delta }^{\alpha }{}_{\beta }$. For the alternating Levi-Civita symbol, the determinant identity givesThe lower-index alternating tensor transforms with three inverse matrices and acquires $\det (A^{-1})=1$ instead. Thus ${\delta }^{\alpha }{}_{\beta }$, ${ϵ}^{\alpha \beta \gamma }$ and ${ϵ}_{\alpha \beta \gamma }$ are invariant tensors of SU(3). These identities use unimodularity; unitarity identifies the dual representation with the complex-conjugate defining representation.

For a mixed SU(3) tensor representation $T^a{}_b$, the matrix trace is invariant. HencegivesThe trace is the singlet. The traceless matrix transforms by conjugation, so it is the adjoint representation of SU(3). It is irreducible: an invariant complex subspace is invariant under commutators with the complexified algebra ${\mathfrak{sl}}_3$, and would be an ideal in that simple Lie algebra. Only the zero and full traceless-matrix spaces are possible.

For the cubic tensor, define its symmetric and antisymmetric first-pair parts byand let $S^{abc}=T^{(abc)}$ be its completely symmetric part. This has $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$ independent components and transforms as the symmetric cubic representation of SU(3). Its irreducibility can be seen in the homogeneous cubic-polynomial realization: the simultaneous highest-weight conditions $z_1{\partial }_{z_2}P=0$ and $z_2{\partial }_{z_3}P=0$ leave only $P\propto z_1^3$. A finite-dimensional unitary representation is completely reducible, and each irreducible summand has a highest-weight vector, so this single highest-weight line implies a single irreducible summand. Lowering generates all ten monomials.

Remove this fully symmetric part from $T_{+}$, putting $M^{abc}=T_{+}^{abc}-S^{abc}$. DefineIts trace vanishes because $M$ is symmetric in its first two indices. Invariance of $ϵ$ makes $O$ transform as one upper and one lower index. The inverse map isIndeed contraction with ${ϵ}_{ebc}$ gives $(2O^a{}_e+O^a{}_e-{\delta }^a{}_e\mathrm{tr}O)/3=O^a{}_e$. The inverse has the required first-pair symmetry and zero completely symmetric part. Thus this eight-dimensional remainder is another adjoint representation of SU(3).

For the antisymmetric part, defineNow decompose $N^c{}_d=U^c{}_d+{\delta }^c{}_d\mathrm{tr}N/3$, with $U$ traceless. The first term supplies a second independent octet; the trace supplies a singlet proportional to ${ϵ}^{abc}$. With $q={ϵ}_{abc}{T}^{abc}/6=\mathrm{tr}N/3$, the explicit cubic SU3 tensor projections reconstruct the entire tensor:All projections commute with the SU(3) action because they use permutations and invariant tensors. Their inverse maps and the dimension count $10+8+8+1=27$ establishThe two octets are equivalent as group representations but distinct subspaces in this decomposition; one comes from the first-pair symmetric sector and the other from the first-pair antisymmetric sector. General permutations can mix these equivalent copies.

In the quark model, the up quark, down quark and strange quark form the defining triplet of approximate flavor symmetry. Three-quark baryons therefore have precisely these possible flavor representations. The fully symmetric decuplet corresponds to the observed spin-$3/2$ baryon decuplet: the four Delta baryons, three ${\Sigma }^{\ast }$ states, two ${\Xi }^{\ast }$ states and the Omega baryon. Their strangeness rows have flavor hypercharge and isospin $(Y,I)=(1,3/2),(0,1),(-1,1/2),(-2,0)$; $uuu$ and $sss$ occupy the extreme ${\Delta }^{++}$ and ${\Omega }^{-}$ weights.

The observed spin-$1/2$ baryon octet contains the nucleons, three Sigma baryons, the Lambda baryon and two Xi baryons. In its $Y=0$ row, ${\Sigma }^0$ and $\Lambda$ share the same third-component/hypercharge weight but have different total isospin. Approximate flavor symmetry explains multiplet organization, not exact equality of all their masses: the strange-quark mass and electromagnetic interactions break it.

The two octet copies in the flavor tensor do not require two identical ground-state octets. A three-quark colour singlet is antisymmetric in color; for a symmetric ground-state orbital wavefunction, the Pauli exclusion principle requires a symmetric combined spin-flavor state. The mixed flavor octet tensors combine with the mixed spin-$1/2$ tensors to supply that symmetry. The symmetric flavor decuplet combines with symmetric spin $3/2$. Together they form the symmetric spin-flavour SU6 representation, with $10\times 4+8\times 2=56$ spin-flavor states. A completely antisymmetric flavor singlet cannot appear in this spatial ground state because three spin-$1/2$ indices have no fully antisymmetric spin state: ${\bigwedge }^3{\mathbb{C}}^2=0$. Three-quark flavour singlet configurations instead require a suitable excited spatial/spin symmetry. Flavor SU(3) here is distinct from the exact color gauge symmetry; treating them as the same group action would misidentify the physical multiplets.

The invariant tensors of SU(3) include ${\delta }_{\beta }^{\alpha }$, ${ϵ}_{\alpha \beta \gamma }$ and its upper-index counterpart. Consequently an antisymmetric pair of upper fundamental indices can be converted into one lower antifundamental index, and conversely; a completely antisymmetric triple is a determinant singlet. In the Young-symmetrizer tensor decomposition, complete columns of three can therefore be removed. A remaining diagram has rows of lengths $k+l,l$: its two-box columns become $l$ dual indices and its remaining boxes give $k$ defining indices. The irreducible highest-weight component is represented by a mixed SU(3) tensor representation symmetric in all upper indices, symmetric in all lower indices, and with every upper-lower contraction removed. Such contractions produce the lower-weight components. The resulting symmetric traceless SU(3) tensor representation has Dynkin labels $(k,l)$; this is an equivalent canonical realization, not a claim that an arbitrary tensor already has those symmetries.

There are $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)$ symmetric upper-index components and $\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$ symmetric lower-index components. All contractions are equivalent under the separate symmetries. Their image is the separately symmetric tensor space with $(k-1,l-1)$ indices. To subtract its dimension legitimately, we must show that contraction is onto.

Represent such tensors by polynomials of bidegree $(k,l)$ in three variables $x$ and three dual variables $y$. Up to a nonzero normalization, contraction isIf $P$ has bidegree $(k-1,l-1)$, differentiation givesChoose $a_0=1/(k+l+1)$ and $a_j=-a_{j-1}/[(j+1)(k+l+1-j)]$. Thenhas bidegree $(k,l)$ and $\Delta Q=P$: consecutive terms cancel, and the last further derivative vanishes. This constructs a preimage of every contraction target. If $k=0$ or $l=0$, there is no contraction. Therefore the number of trace-free components isThe highest-weight vector $x_1^k{y}_3^l$ lies in this kernel. The Young-symmetry construction identifies the traceless highest-weight component with the irreducible representation of labels $(k,l)$; equivalently its Weyl dimension formula agrees with the independently calculated kernel dimension above.

For traceless matrices $T,S$, the tensor square of the SU(3) adjoint representation decomposes asThe component maps make the decomposition concrete. The singlet is $\mathrm{tr}(TS)$. The two octets are the antisymmetric commutator $[T,S]$ and symmetric traceless anticommutator $\{T,S\}-\frac{2}{3}\mathrm{tr}(TS)I$. Symmetrizing the two upper and two lower indices in $T_{\beta }^{\alpha }{S}_{\delta }^{\gamma }$ and removing contractions gives $R(2,2)$, the 27. Its highest component is already visible in $E_{13}\otimes E_{13}$.

For the 10, contract the two lower indices with ${ϵ}^{\beta \delta \rho }$ and symmetrize the three upper indices $\alpha ,\gamma ,\rho$; for its conjugate, contract the upper pair with ${ϵ}_{\alpha \gamma \rho }$ and symmetrize the three lower indices. These maps are antisymmetric under exchange of $T,S$. They are nonzero: $T=E_{12},S=E_{13}$ produces a nonzero all-upper-1 component, and $T=E_{21},S=E_{31}$ produces the conjugate all-lower-1 component. Thus they reach the irreducible $R(3,0)$ and $R(0,3)$, each of dimension ten. The octet maps are also nonzero, for example on a pair of noncommuting matrix units and on equal nonzero traceless diagonal matrices. These equivariant maps supply distinct irreducible constituents, with the two octets separated by exchange symmetry. Their dimensions exhaust the product:

An octet operator between baryon octet states defines an intertwining operator $8_{\mathrm{operator}}\otimes 8_{\mathrm{state}}\to 8_{\mathrm{state}}$. Its covariance follows from the commutator transformation law in the question. By Schur lemma, the number of independent reduced matrix elements is the multiplicity of the output octet in the product: there are exactly two independent coefficients. To display them, use a Hermitian octet matrix basis $B_s={\lambda }_s$, $t_i={\lambda }_i/2$, normalized by $\mathrm{tr}({\lambda }_r{\lambda }_s)=2{\delta }_{rs}$. The two allowed actions areWith $[{\lambda }_i,{\lambda }_s]=2i{f}_{isk}{\lambda }_k$ and $\{{\lambda }_i,{\lambda }_s\}=\frac{4}{3}{\delta }_{is}I+2d_{isk}{\lambda }_k$, this gives the two reduced octet matrix elements:up to the common normalization absorbed into $F,D$. All component dependence is fixed by the invariant symmetric and antisymmetric tensors. For a Hermitian operator and this Hermitian basis, $F,D$ may be chosen real; neither their values nor a relation between them follows from symmetry alone.

Let $V={\mathbb{C}}^3$ be the defining SU(3) representation. A mixed SU(3) tensor representation belongs to $V^{\otimes k}\otimes (V^{\ast }{)}^{\otimes l}$, with transformation lawThe unrestricted tensor has dimension $3^{k+l}$. The upper and lower counts are independent; setting them equal would lose part of the printed generality.

For an irreducible tensor, impose the appropriate Young symmetrizer on each set of indices and subtract all upper-lower contractions. Antisymmetric pairs can be converted using the invariant alternating tensor, since ${\bigwedge }^{2}3\cong \overline{3}$ and ${\bigwedge }^2\overline{3}\cong 3$. A full alternating triple is a singlet because $\det U=1$. Traces use the invariant Kronecker delta and yield lower-rank tensors. Repeating these operations gives invariant irreducible subspaces, with multiplicities retained.

An explicit algorithm for the complete decomposition starts at $(p,q)=(0,0)$, tensors with $3$ exactly $k$ times, and then with $\overline{3}$ exactly $l$ times. At each step applyomitting terms with negative labels and adding multiplicities of coincident terms. The first is the sl3 highest-weight tensor rule: a Young diagram of row lengths $(p+q,q,0)$ gains a box in one of its three rows. The third-row term loses a determinant column of height three, leaving labels $(p,q-1)$. The second rule follows by conjugation, which exchanges $p$ and $q$. Thus the procedure determines all constituents of the general tensor, not just one selected symmetry type.

The separately symmetric, traceless component has labels $(k,l)$. Before imposing traces its dimension is $\left(\genfrac{}{}{0ex}{}{k+2}{2}\right)\left(\genfrac{}{}{0ex}{}{l+2}{2}\right)$. The contraction map has target the separately symmetric tensors of degrees $(k-1,l-1)$ and is onto: in polynomial coordinates it is ${\sum }_i{\partial }_{x_i}{\partial }_{y_i}$, the adjoint of multiplication by ${\sum }_i{x}_i{y}_i$ in the monomial-factorial inner product. Multiplication is injective, so the adjoint is surjective. Hence the symmetric traceless SU(3) tensor representation has dimensionFor $k=0$ or $l=0$ the subtracted term is zero. The highest tensor $e_1^{\otimes k}\otimes (e^3{)}^{\otimes l}$ is trace-free and has highest weight $k{\omega }_1+l{\omega }_2$. The three positive-root factors in the Weyl dimension formula are $k+1$, $l+1$ and $(k+l+2)/2$, giving exactly the kernel dimension above. Thus the whole symmetric traceless kernel is that irreducible constituent.

There is a genuine distinction between this highest-weight constituent and the literal largest-dimensional constituent. For example, $3^{\otimes 5}$ contains $(3,1)$, obtained by antisymmetrizing one pair, with dimension $24$, whereas the fully symmetric $(5,0)$ has dimension $21$. Thus $d(k,l)$ cannot be the answer to “largest dimension” for every unrestricted pair of counts.

The maximal dimension in mixed SU(3) tensor powers can nevertheless be determined exactly. Every constituent highest weight differs from $k{\omega }_1+l{\omega }_2$ by $a{\alpha }_1+b{\alpha }_2$ with nonnegative integers $a,b$, where ${\alpha }_1=2{\omega }_1-{\omega }_2$ and ${\alpha }_2=-{\omega }_1+2{\omega }_2$. Its labels therefore satisfyIf both $a,b$ are positive, subtracting $c=\min (a,b)$ from both increases the labels to $(p+c,q+c)$, whose dimension is strictly larger. The resulting boundary labels occur in the tensor: use $a$ disjoint antisymmetric upper pairs or $b$ disjoint antisymmetric lower pairs, and then take the separately symmetric traceless component. Consequently the exact answer is the finite formulaThis supplies a practical closed finite computation for arbitrary $k,l$. To reduce it to at most four evaluations, put $K=k+1$, $L=l+1$ and $r=1+\sqrt{3}$. The first continuous maximum is at $a_{\ast }=(K-rL)/(2+r)$, clipped to $[0,k/2]$; test its two neighbouring allowed integers. The second is similarly at $b_{\ast }=(L-rK)/(2+r)$, clipped to $[0,l/2]$. Indeed differentiating the first cubic gives a derivative with the sign of $(K-2a{)}^2-2(K-2a)(L+a)-2(L+a{)}^2$, which changes sign once at that root. For equal upper and lower counts, both maxima are at zero and the familiar result is $(k+1{)}^3$.

For three quarks, the unconstrained flavour product isThe brackets indicate permutation symmetry: symmetric, mixed, and alternating. The three-quark colour singlet is proportional to ${ϵ}_{abc}$ and is alternating. The fermion wavefunction must be alternating overall. In the usual spatially symmetric ground-state S-wave, the spin-flavour part must therefore be symmetric.

The three-spin-one-half space has a symmetric spin-$3/2$ sector and mixed-symmetry spin-$1/2$ sectors; it has no fully alternating sector because ${\bigwedge }^3{\mathbb{C}}^2=0$. Symmetric flavour can pair with symmetric spin, giving the baryon decuplet of spin $3/2$. The flavour-octet multiplicity space and the spin-$1/2$ multiplicity space both transform as the two-dimensional standard representation of the symmetric group. Their product contains exactly one symmetric combination: its characters at the identity, a transposition and a three-cycle are $4,0,1$, so the symmetric multiplicity is $(4+3\cdot 0+2\cdot 1)/6=1$. This gives one baryon octet of spin $1/2$. Alternating singlet flavour would need alternating spin to make the spin-flavour part symmetric, and that sector does not exist. Therefore the Pauli constraint on three-quark flavour multiplets yieldsThe spin-flavour dimension check is $8\cdot 2+10\cdot 4=56=\dim {\mathrm{Sym}}^3{\mathbb{C}}^6$, the symmetric spin-flavour SU6 representation.

The spatial-symmetry qualification matters if the wording is interpreted as total $L=0$ alone. Rotationally scalar excited orbital functions can have mixed permutation symmetry. For example, the two independent differences among $r_{12}^2,r_{23}^2,r_{31}^2$, multiplied by a symmetric radial factor, span such a space while remaining rotational scalars. Combining this standard permutation representation with the spin-$1/2$ standard representation gives an alternating spin-orbital combination; multiplying by alternating singlet flavour makes it symmetric, and the colour factor restores overall fermionic antisymmetry. Thus total $L=0$ without a symmetric ground-state orbital assumption would not by itself exclude a singlet.

For a quark-antiquark pair, colour singletness selects the contraction ${\delta }^a{}_b$, and the independent flavour product givesQuark and antiquark are distinguishable, so there is no three-identical-quark exclusion of either flavour multiplet or either spin. Both the octet and singlet admit spin singlets and spin triplets. Their baryon number is zero. The flavour octet has rows $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$, while the singlet has $(Y,I)=(0,0)$. The octet's zero-isospin state is $(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$, and the singlet is $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. Each row contains all $I_3=-I,\dots ,I$, and electric charge is $Q=I_3+Y/2$.

For ground-state mesons with $L=0$, opposite intrinsic quark-antiquark parities give $P=-1$. The neutral self-conjugate states have charge conjugation $C=(-1{)}^{L+S}$, so both flavour multiplets occur as $0^{-+}$ pseudoscalar mesons and $1^{--}$ vector mesons, forming a flavour nonet in each channel. Charged and open-flavour states have no individual charge-conjugation eigenvalue. The meson request does not restrict orbital excitation: in general $J=|L-S|,\dots ,L+S$, $P=(-1{)}^{L+1}$, and the same formula for $C$ applies where defined. Thus the flavour representations stay $1\oplus 8$, with these additional orbital-spin quantum numbers.