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A correlated pair of quarks. Its color, flavor, spin and orbital states are separate degrees of freedom and must collectively satisfy Fermi statistics. In flavor SU(3), the antisymmetric pair space is ${\Lambda }^{2}3\cong \overline{3}$; contraction with the invariant Levi-Civita symbol realizes the isomorphism. The antisymmetric $ud$ flavor pair is an isospin singlet with flavor hypercharge $2/3$. Combining two such flavor-antitriplet pairs with a strange antiquark permits the highest weight of the pentaquark antidecuplet; this flavor construction does not itself establish a bound state.

The zero-isospin, zero-flavor hypercharge state ${\eta }_8=(u\bar{u}+d\bar{d}-2s\bar{s})/\sqrt{6}$ in the pseudoscalar meson octet. It is a pure octet basis state; the physical eta may mix it with the flavor-singlet state.

For flavor symmetry acting on $(u,d,s)$, trace-orthonormal Hermitian diagonal generators are $h_1=\mathrm{diag}(1,-1,0)/\sqrt{2}$ and $h_2=\mathrm{diag}(1,1,-2)/\sqrt{6}$. The isospin and flavor hypercharge operators are $I_3=h_1/\sqrt{2}$ and $Y=\sqrt{2/3}{h}_2$. Their multiples by $i$ lie in the compact SU(3) Lie algebra.

For the light quark flavors, electric charge is $Q=I_3+Y/2$, with flavor hypercharge $Y=B+S$. On $(u,d,s)$ it gives $(2/3,-1/3,-1/3)$, and additivity extends it to the baryon octet and meson octet.

The light-quark $sss$ member of the spin-three-halves baryon decuplet has strangeness $S=-3$, baryon number $B=1$, flavor hypercharge $Y=-2$, and isospin zero. The Gell-Mann--Nishijima formula gives electric charge minus one. Its position at the bottom of the decuplet's triangular weight diagram follows by replacing all three up or down quarks with strange quarks.

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.

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.

When the three light-quark masses are equal, the mass term is proportional to the identity in flavour space, so rotating $q=(u,d,s)$ by SU(3) leaves it unchanged. The gluon coupling is also flavour blind. Nearly equal masses therefore give an approximate flavour symmetry, distinct from the exact colour gauge symmetry. The quark transforms as $3$ and the antiquark as $\overline{3}$.

For mesons, a quark-antiquark tensor is a $3\times 3$ matrix. Its trace and traceless parts are invariant subspaces, givingThe trace is the eta singlet state; the traceless part is the meson octet. For baryons, first separate a pair of quarks into symmetric and antisymmetric tensors:The antisymmetric pair transforms as $\overline{3}$ by contraction with the invariant epsilon tensor. In $6\otimes 3$, complete symmetrization gives ${\mathrm{Sym}}^{3}3$ of dimension $\left(\genfrac{}{}{0ex}{}{5}{3}\right)=10$; the remaining mixed-symmetry subspace has dimension eight. In $\overline{3}\otimes 3$, trace and traceless parts give a singlet and another octet. Thus the tensor cube of the defining SU(3) representation isThe two octets carry the two mixed permutation-symmetry components. Their weight diagrams are identical; the extra copy is a multiplicity, not a new arrangement of weights.

The printed charge relation fixes the half-hypercharge convention, rather than the usual Flavor hypercharge normalization:The quark weights $(I_3,Y)$ are $u:(1/2,1/6)$, $d:(-1/2,1/6)$, $s:(0,-1/3)$; adding three weights gives the baryon weight diagrams in the half-hypercharge convention. The baryon octet has $(p,n)$ at $Y=1/2$, the isotriplet $({\Sigma }^{+},{\Sigma }^0,{\Sigma }^{-})$ and the isosinglet ${\Lambda }^0$ at $Y=0$, and $({\Xi }^0,{\Xi }^{-})$ at $Y=-1/2$. The baryon decuplet has rows $\Delta$ at $Y=1/2$, ${\Sigma }^{\ast }$ at zero, ${\Xi }^{\ast }$ at $-1/2$, and ${\Omega }^{-}$ at $-1$. The three-quark flavour singlet is the antisymmetric $uds$ state at $(0,0)$.

The fully symmetric flavour representation is precisely the baryon decuplet, $10={\mathrm{Sym}}^{3}3$. For a spatial ground state, the orbital wavefunction is symmetric. Spin $3/2$ is the symmetric part of three spin-$1/2$ quarks, so a spin-$3/2$ decuplet would have a completely symmetric space-spin-flavour wavefunction. In particular the $uuu$ state with all spins aligned has no antisymmetric factor. This conflicts with the Pauli exclusion principle for the three fermions.

An additional colour degree of freedom resolves the conflict. With three colour labels the normalized colour state isIt is completely antisymmetric and an SU(3) colour singlet. Multiplying it by the symmetric remaining wavefunction gives the required fermionic antisymmetry. Three orthogonal labels are the minimum needed for a nonzero third exterior power: ${\bigwedge }^3{\mathbb{C}}^m=0$ for $m<3$. Thus the decuplet suggests at least three colour states, and three supply the minimal colour-singlet construction used in the quark model. This argument motivates three colours rather than deriving their exact number without further physical input.

The Pauli constraint on three-quark flavour multiplets also explains the absence of a flavour-singlet baryon in the lowest spatial multiplet. After extracting the antisymmetric colour factor, spin-flavour must be symmetric. Singlet flavour is completely antisymmetric, so its spin factor would also have to be completely antisymmetric. Butso three spin-$1/2$ quarks have no such spin state. By contrast, mixed flavour and mixed spin can combine symmetrically to produce the ground-state baryon octet with spin $1/2$. The two abstract flavour-octet copies do not imply two independent ground-state octets after the full Pauli constraint on three-quark flavour multiplets is applied. Excited orbital wavefunctions can permit three-quark flavour singlet; the exclusion is for the spatially symmetric low-energy ground multiplet, not for every baryon at every energy.

A flavour-singlet meson is allowed: a quark and an antiquark are not three identical fermions, and the invariant flavour combination $(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$ can combine with an appropriate spin and colour singlet. The missing three-spin antisymmetric state therefore causes no meson exclusion.

Finally, ${\pi }^{+}p$ has baryon number one, strangeness zero and $I_3=3/2$, so it is a pure $I=3/2$ channel. These quantum numbers match the $uuu$ Delta baryon ${\Delta }^{++}$, the low-energy resonance with spin $3/2$. For $K^{+}p$, the kaon $u\bar{s}$ contributes strangeness plus one, giving total $B=1$, $S=+1$. No ordinary $qqq$ baryon can have positive strangeness, since every $s$ quark contributes minus one. Hence that channel has no matching state in the ordinary low-energy three-quark multiplets. A resonance would require exotic quark content, for example $uuud\bar{s}$, and is not mandated by this spectrum. This is strangeness selection in pion-proton and kaon-proton resonances.

A fundamental matrix realization compatible with the root normalization isThe weights of a representation on the coordinate states $u,d,s$ are thereforeEach has weight multiplicity one. The weight diagram of the defining SU(3) representation is the corresponding equilateral triangle. These are $(H_1,H_2)$ coordinates; the familiar vertical flavor hypercharge coordinate is $Y=2H_2/\sqrt{3}$.

For the adjoint representation of SU(3), the states are elements of the Lie algebra. The two Cartan generators commute with both $H_i$, giving representation weight $(0,0)$ with multiplicity two. Each root generator $E_{\pm }^r$ is an eigenvector of the Cartan adjoint action with representation weight $\pm {\alpha }_r$. ConsequentlyThe six nonzero representation weights form a regular hexagon, with a doubly occupied centre. Its highest weight is ${\alpha }_3={\omega }_1+{\omega }_2$, so its Dynkin labels are $(1,1)$ and its dimension is eight. The triplet and octet panels in the preceding figure give these two weight diagrams explicitly.

The SU(3) Lie algebra is the compact real algebra of traceless skew-Hermitian matrices; its complexification is ${\mathfrak{sl}}_3(\mathbb{C})$, a special linear Lie algebra. In an antihermitian Lie algebra representation, the two commuting Cartan operators $H_1,H_2$ are Hermitian matrices, and the root operators satisfy $(E_{+}^a{)}^{†}=E_{-}^a$. Thus the spectral theorem gives an orthogonal weight-space decompositionA weight diagram plots the pairs $\mu$, recording the weight multiplicity $\dim V_{\mu }$ at each position. One point need not represent only one state. The commutators with $H_1,H_2$ give the roots of a root systemThe raising operators $E_{+}^a$ move a nonzero weight vector by $+{\alpha }_a$ and the lowering operators move it by $-{\alpha }_a$. These six root directions form the A2 root system. Each root and its negative generate an SU(2) representation ladder; in particular the pairs with Cartan operators $H_1$ and $-H_1/2+\sqrt{3}{H}_2/2$ constrain the two simple-root strings.

Choose a linear functional positive on the three positive roots. A maximal weight for this ordering is killed by every raising operator. In an Irreducible Lie algebra representation it generates the entire representation by lowering, and its weight space has dimension one. The two SU(2) representation strings show that its Dynkin labels are nonnegative integers:Conversely, each such pair specifies exactly one finite-dimensional irreducible highest-weight representation. Write it as $R(p,q)$. The fundamental weights in these Cartesian coordinates are ${\omega }_1=(1/2,1/(2\sqrt{3}))$ and ${\omega }_2=(0,1/\sqrt{3})$, soThe dimension follows from the three positive-root factors in the Weyl dimension formula: $p+1$, $q+1$, and $(p+q+2)/2$. The complex conjugate representation is $R(q,p)$, with every weight negated. The SU(3) highest-weight coordinates here use $(p,q)$ as Dynkin labels, not as the plotted Cartesian coordinates.

The weight diagram is invariant under the Weyl group, generated by reflections perpendicular to the two simple roots; its convex hull is the orbit polygon of the highest weight. For $p=q=0$ it is a point, the trivial representation. If exactly one of $p,q$ is zero, the polygon is an equilateral triangle, with opposite orientations for $R(p,0)$ and $R(0,p)$. Its edge has $p$ root steps. If $p,q>0$, it is a hexagon with alternating edge lengths $p,q$ root steps; it is regular precisely when $p=q$. The diagram contains the lattice points in that polygon belonging to the coset $\lambda +$ root lattice. A point in a different weight lattice coset is not a weight merely because it lies inside the polygon.

The shell multiplicities in an SU(3) weight diagram can be stated geometrically. The outer boundary has weight multiplicity one. Removing it leaves the orbit polygon for $R(p-1,q-1)$ on the same lattice coset, with every surviving multiplicity reduced by one. Repeating increases the actual weight multiplicity by one at each inward layer until the smaller label reaches zero. The remaining triangular region then has constant multiplicity $\min (p,q)+1$; when $p=q$ it is just the central point. A triangular weight diagram has multiplicity one everywhere. Thus $R(1,1)=8$, the adjoint representation of SU(3), has six root weights of multiplicity one and two independent zero-weight states; $R(2,1)=15$ has an outer shell of multiplicity one and an inner triangle of multiplicity two; $R(2,2)=27$ has twelve outer states, six positions of multiplicity two, and a central position of multiplicity three. The total is $12+6\cdot 2+3=27$. The dominant weight multiplicity formula for sl3 is an algebraic expression of the same rule.

For a tensor product of Lie algebra representations, the Cartan operators add. Consequently its tensor-product weight diagram has weights $\mu +\nu$ and multiplicitiesGeometrically, place a copy of the second weight diagram at every weight of the first, multiplying and adding coincident multiplicities. Because the representation is antihermitian, invariant subspaces have invariant orthogonal complements. To decompose the product, select a maximal remaining dominant weight, identify its irreducible highest-weight representation, subtract that representation's whole diagram with its weight multiplicities, and repeat. This highest-weight character subtraction must exhaust every weight, not merely the corners. Unlike the Clebsch-Gordan decomposition for SU2, an SU(3) product may contain repeated copies of an irreducible representation.

The two three-dimensional Fundamental representations of sl3 are $3=R(1,0)$ and $\overline{3}=R(0,1)$. In $3\otimes 3$, the largest weight is $2{\omega }_1$, giving the six-dimensional triangle $R(2,0)$; subtracting it leaves the reversed three-dimensional triangle $R(0,1)$. The first summand consists of symmetric tensors and the second of antisymmetric tensors. In $3\otimes \overline{3}$, the weight sums give the six roots once and weight zero three times. Subtracting the octet, whose zero weight has multiplicity two, leaves a singlet. Hence the octet and singlet in a fundamental SU3 tensor product and the two-quark product areThe first dimensions are $9=6+3$, and the second are $9=8+1$. Next $6\otimes 3$ has highest weight $3{\omega }_1$, giving $R(3,0)=10$. Subtraction leaves one $R(1,1)=8$. Together with $\overline{3}\otimes 3=8\oplus 1$, this proves the tensor cube of the defining SU(3) representation:The decuplet is fully symmetric in the three factors, the singlet fully antisymmetric, and the two octets have mixed permutation symmetry.

For the quark model, the relevant approximate flavour symmetry rotates the three light quarks $u,d,s$ in the fundamental $3$. The up quark, down quark, and strange quark basis states have the weight diagram of the defining SU(3) representation. In physical coordinates define isospin $I_3=H_1$ and flavor hypercharge $Y=2H_2/\sqrt{3}$. ThenThe last row follows from the Gell-Mann--Nishijima formula $Q=I_3+Y/2$, in units of the positive elementary electric charge. The antiquarks transform in $\overline{3}$ and have the negatives of these additive quantum numbers. Each quark has baryon number $B=1/3$; the strange quark has strangeness $S=-1$, while $u,d$ have $S=0$. Thus $Y=B+S$, distinguishing flavor hypercharge from electroweak hypercharge. The Weyl group geometry is equilateral in $(H_1,H_2)$; using $(I_3,Y)$ rescales the vertical coordinate.

A conventional meson has valence content $q\bar{q}$, so its flavour symmetry gives a meson octet and a singlet. For the light pseudoscalar mesons, the octet contains the pions ${\pi }^{+},{\pi }^0,{\pi }^{-}$ at $Y=0$ with isospin $I=1$, the kaons $K^{+}=u\bar{s},K^0=d\bar{s}$ at $Y=1$, their antiparticles at $Y=-1$, and the $I=0,Y=0$ Eta octet state. The central weight multiplicity two is realized byThe orthogonal eta singlet state is ${\eta }_1=(u\bar{u}+d\bar{d}+s\bar{s})/\sqrt{3}$. The eta and eta prime mesons are mixtures of ${\eta }_8,{\eta }_1$, rather than exact octet and singlet mass eigenstates; this gives the pseudoscalar meson nonet. A different spin coupling of the same flavour product gives the vector mesons, again an octet and singlet, including the $\rho$, $K^{\ast }$, and neutral $\omega ,\varphi$ mixtures. The lowest orbital states have $J^P=0^{-}$ and $1^{-}$ respectively.

A conventional baryon has three valence quarks, so the tensor-cube decomposition supplies the flavour possibilities. The observed lowest baryon octet has spin $1/2$ and positive parity. Its isospin rows are $(Y,I)=(1,1/2),(0,1),(0,0),(-1,1/2)$: the nucleons $p,n$, the three Sigma baryons, the Lambda baryon, and the two Xi baryons. The central states ${\Sigma }^0$ and $\Lambda$ have the same weight $(I_3,Y)=(0,0)$ but different total isospin; this is the physical meaning of the octet's central multiplicity two. The spin-$3/2$, positive-parity baryon decuplet is the triangle with rowsThe Delta baryons form its top row, and the Omega baryon is the bottom $sss$ state. All ten weights have multiplicity one. Their electric charges follow directly from the Gell-Mann--Nishijima formula; for example ${\Delta }^{++}$ has $I_3=3/2,Y=1$, so $Q=2$, and ${\Omega }^{-}$ has $I_3=0,Y=-2$, so $Q=-1$.

The two algebraic octets do not mean that two independent ground-state octets must occur. A three-quark colour singlet has a completely antisymmetric colour wavefunction. For a symmetric spatial ground state, the Pauli exclusion principle therefore requires a symmetric combined spin-flavour wavefunction. The symmetric flavour decuplet pairs with symmetric spin $3/2$; mixed flavour octets pair with mixed spin $1/2$ in one symmetric combination. Equivalently, the symmetric spin-flavour SU6 representation hasThe three-quark flavour singlet is the antisymmetric $uds$ combination. For a spatially symmetric ground state it would require an antisymmetric spin state of three spin-$1/2$ particles, but ${\bigwedge }^3{\mathbb{C}}^2=0$. The Pauli constraint on three-quark flavour multiplets thus excludes it from this ground-state sector; orbital excitation permits other multiplets.

Finally, this flavour symmetry is approximate: the unequal light-quark masses, especially the larger strange-quark mass, and electromagnetic interactions split masses within the multiplets and permit octet-singlet mixing. The weight diagrams classify flavour states, rather than asserting exact equality of their masses. Colour SU(3) is a separate gauge action on each flavour and supplies the colour-singlet constraint; it is not the approximate flavour action used in the eightfold way. The light conventional mesons organize into flavour octets and singlets, while the spatial ground-state three-quark baryons form one spin-$1/2$ octet and one spin-$3/2$ decuplet.

Start with a nonzero highest-weight vector $v_0=|n⟩$ and set $v_r=E_{-}^r{v}_0$. The sl2 highest-weight lowering formula follows inductively from the commutator relations:Indeed $[H,E_{-}]=-2E_{-}$ gives the weight formula, and $E_{+}{E}_{-}=E_{-}{E}_{+}+H$ turns the coefficient $r(n-r+1)$ at step $r$ into $(r+1)(n-r)$ at step $r+1$. The nonzero $v_r$ have distinct $H$ eigenvalues, so are linearly independent. Finite dimensionality therefore forces a last nonzero vector $v_N$. Applying $E_{+}$ to $v_{N+1}=0$ gives $(N+1)(n-N)v_N=0$, hence $N=n$. This also proves that no earlier vector can vanish. Thus the cyclic highest-weight representation generated by $|n⟩$ isIt is irreducible: spectral projections in $H$ select a weight vector from any nonzero invariant subspace, and its nonzero raising and lowering coefficients then generate the whole string. This gives the classification of finite-dimensional sl2 representations used for each root direction below.

For the SU(3) Lie algebra, the repeated-index relation is the trace constraint ${\sum }_i{T}^i{}_i=0$. The displayed generators are a basis of its complexification, not nine independent generators or three individually vanishing diagonal elements. Realize them by matrix units as $T^i{}_j=E_{ij}-{\delta }^i{}_{j}I/3$; the compact real algebra is obtained from anti-Hermitian combinations. Thenand $E_{1+}=E_{12}$, $E_{1-}=E_{21}$, $E_{2+}=E_{23}$, $E_{2-}=E_{32}$. Substituting into the given commutators givesThus both root pairs obey the required sl2 Lie algebra relations. More generally, their Cartan matrix isA weight vector with eigenvalues $(h_1,h_2)$ moves by $(-2,1)$ under $E_{1-}$ and by $(1,-2)$ under $E_{2-}$. The third negative-root operator is $[E_{2-},E_{1-}]=E_{31}$ and moves it by $(-1,-1)$. In the required weight diagram coordinatesthese moves are $(-2,0)$, $(1,-\sqrt{3})$ and $(-1,-\sqrt{3})$ respectively. Successive lowerings from a highest-weight vector, with linear dependencies removed, construct the following bases.

For Dynkin labels $(1,0)$, use the defining action of SU(3) on ${\mathbb{C}}^3$. Its highest vector is $e_1$; lowering gives $e_2=E_{21}{e}_1$ and $e_3=E_{32}{e}_2$. There are no further independent vectors. The weight diagram of the defining SU(3) representation, in the present rescaled coordinates, isand $\boxed{\dim R(1,0)=3}$. Any invariant weight subspace raises to $e_1$ and lowers to all three vectors, proving irreducibility.

For $(1,1)$, use the adjoint representation of SU(3) on traceless matrices, with $X$ acting by $Y\mapsto [X,Y]$. The highest vector is $E_{13}$: both $[E_{12},E_{13}]$ and $[E_{23},E_{13}]$ vanish, while $[H_i,E_{13}]=E_{13}$. Lowering gives $[E_{21},E_{13}]=E_{23}$ and $[E_{32},E_{13}]=-E_{12}$. Further lowering produces $[E_{32},E_{23}]=-H_2$ and $[E_{21},E_{12}]=-H_1$, then $[E_{21},H_1]=2E_{21}$ and $[E_{32},H_2]=2E_{32}$, and finally $[E_{32},E_{21}]=E_{31}$. Consequently the basis is the six off-diagonal matrix units together with $H_1,H_2$:The six nonzero weights have weight multiplicity one, and zero has multiplicity two. Thus $\boxed{\dim R(1,1)=6+2=8}$. Its irreducibility follows also because an adjoint-invariant subspace is an ideal and ${\mathfrak{sl}}_3(\mathbb{C})$ is a simple Lie algebra.

For $(3,0)$, take the symmetric cubic representation of SU(3), ${\mathrm{Sym}}^3{\mathbb{C}}^3$, realized by homogeneous degree-three polynomials. Hereso $z_1^3$ is a highest-weight vector with eigenvalues $(3,0)$. For $a+b+c=3$,which is nonzero and constructs every monomial. These monomials form a basis and have distinct weights $(a-b,b-c)$. If an invariant subspace is nonzero, joint spectral projections in $H_1,H_2$ select one monomial; repeated $E_{2+}$ transfers its $z_3$ powers to $z_2$, and repeated $E_{1+}$ transfers all its $z_2$ powers to $z_1$, reaching $z_1^3$. Lowering then generates the whole basis, proving irreducibility without merely assuming the dimension formula. The plotted weights arewith multiplicity one at every point. Hence $\boxed{\dim R(3,0)=\left(\genfrac{}{}{0ex}{}{5}{2}\right)=10}$.

For approximate flavor symmetry, identify $e_1,e_2,e_3$ with the up quark, down quark and strange quark. The defining representation is their flavour triplet. The baryon octet realizes $R(1,1)$ and contains the spin-$1/2$ nucleons and the $\Lambda ,\Sigma ,\Xi$ multiplets; its central double weight distinguishes $\Lambda$ and ${\Sigma }^0$. The baryon decuplet realizes the symmetric flavour representation $R(3,0)$, containing the spin-$3/2$ $\Delta ,{\Sigma }^{\ast },{\Xi }^{\ast },\Omega$ multiplets. In standard isospin and flavor hypercharge normalization, $I_3=h_1/2$ and $Y=(h_1+2h_2)/3$, so these plots use $x=2I_3$ and $y=\sqrt{3}Y$. Three-quark spin and flavour combine subject to the Pauli constraint on three-quark flavour multiplets and the antisymmetric three-quark colour singlet. The adjoint flavour representation also describes a meson octet. A distinct, exact colour SU(3) gauge symmetry places quarks in its defining representation and gluons in its adjoint octet; colour and flavour here are different physical group actions.

A weight of a representation of $SU(3)$ is a joint eigenvalue of two commuting generators spanning a Cartan subalgebra. Equivalently it describes the character by which the diagonal maximal torus acts on a weight vector. Use the Hermitian generatorsTheir multiples by $i$ are in the SU(3) Lie algebra. A vector with weight $(a,b)$ acquires the phase $e^{i(a\theta +b\phi )}$ under $\exp (i\theta I_3+i\phi Y)$. The conventional generator $H_8$ is related by $Y=2H_8/\sqrt{3}$; using $H_8$ instead just rescales the vertical weight coordinate.

In the defining group representation, the coordinate vectors $e_u,e_d,e_s$ are joint eigenvectors. The weight diagram of the defining SU(3) representation therefore hasComplex conjugation reverses all torus phases, so the weights of $\overline{3}$ are $-{\mu }_u,-{\mu }_d,-{\mu }_s$, each with multiplicity one. The conjugation bar is essential: the tensor product here is $3\otimes \overline{3}$.

Weights add in a tensor product of group representations. Thus the nine weights of this product are ${\mu }_i-{\mu }_j$. There are three zero weights from $i=j$, and the remaining six areTo turn this weight calculation into a direct-sum decomposition, identify the tensor product with $\mathrm{End}({\mathbb{C}}^3)$ by $v\otimes \overline{w}\mapsto v{w}^{†}$. The group acts by $M\mapsto UM{U}^{-1}$. This gives two invariant spaces,The scalar line has weight zero and is the trivial group representation $1$. The traceless space has the six nonzero weights just computed, represented by the off-diagonal matrix units $E_{ij}$, and two zero weights, represented by traceless diagonal matrices. Its dimension is eight and it is the complex Adjoint representation $8$. ThereforeThis proves the octet and singlet in a fundamental SU3 tensor product with the correct central weight multiplicities.

The singlet is irreducible because it is one-dimensional. For the octet, an invariant complex subspace of ${\mathfrak{sl}}_3$ is stable under commutators with the complexified Lie algebra. Commuting Cartan generators project it into weight spaces. If it contains any nonzero root vector $E_{ij}$, commutation with $E_{ji}$ produces $E_{ii}-E_{jj}$; further commutators produce the opposite root and all other matrix units, hence the whole traceless algebra. If it contains only a nonzero diagonal traceless matrix $D$, two diagonal entries differ, so $[E_{ij},D]=(D_{jj}-D_{ii})E_{ij}$ supplies a root vector and reduces to the preceding case. There is no nonzero proper invariant subspace. The octet is irreducible, not a sum of six one-dimensional weight spaces and two singlets: the nondiagonal generators connect those spaces.

In the quark model, take $u,d,s$ as the defining flavour triplet and their antiquarks as the conjugate triplet. A colour-singlet quark-antiquark state with relative orbital angular momentum $L=0$ and total spin $S=0$ has $J=0$ and parity $(-1{)}^{L+1}=-1$. Thus this flavour decomposition classifies a pseudoscalar meson nonet: a meson octet and a singlet. The use of approximate flavor symmetry is important; the strange quark is heavier, so these states need not have equal masses.

In the $(I_3,Y)$ diagram, $Y$ is flavor hypercharge, not electroweak hypercharge. Mesons have baryon number zero, so $Y$ equals their strangeness, and electric charge is $Q=I_3+Y/2$. The outer six weight states areThe pions form an isospin triplet, completed at the centre by ${\pi }^0=(u\overline{u}-d\overline{d})/\sqrt{2}$. The four kaons occupy the two hypercharge-one and two hypercharge-minus-one weights. In particular electrically neutral kaons are not at the centre of the weight diagram.