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The spin-one-half octet of flavor symmetry contains the nucleons, Sigma baryons, Lambda baryon, and Xi baryons. Its weight diagram has six peripheral states and two states at $(I_3,Y)=(0,0)$: ${\Sigma }^0$ and ${\Lambda }^0$, with different total isospin.

The additive flavor symmetry quantum number $Y=B+S$ combines baryon number with strangeness. For the $(u,d,s)$ quark triplet its values are $(1/3,1/3,-2/3)$. The Gell-Mann--Nishijima formula is $Q=I_3+Y/2$. This is distinct from the normalization of electroweak hypercharge.

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.

The gauge-covariant kinetic term for two massless quarks separates into left- and right-handed parts. Independent constant unitary quark flavor rotations preserve these terms, giving classical $U(2{)}_L\times U(2{)}_R$ internal symmetry in addition to local color $SU(3)$. The singlet axial $U(1{)}_A$ has a chiral anomaly, while the nonsinglet chiral symmetry and vector baryon number remain exact quantum symmetries, up to their shared discrete identifications. Equal nonzero quark masses reduce the continuous flavor symmetry to $SU(2{)}_V\times U(1{)}_V$.

The pseudoscalar octet of flavor symmetry contains three pions, four kaons and the Eta octet state. In its weight diagram, the neutral pion and ${\eta }_8$ are separate states at the origin.

A physicist who organized hadrons through flavor symmetry and the quark model.^[ref]

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 $t^A$ be Hermitian color generators with $\mathrm{Tr}(t^A{t}^B)={\delta }^{AB}/2$, and take $D_{\mu }={\partial }_{\mu }-i{g}_s{G}_{\mu }^A{t}^A$. In explicit spinor, color and flavor indices the massless kinetic term isThe full Quantum chromodynamics Lagrangian also has $-G_{\mu \nu }^A{G}^{A\mu \nu }/4$. A local color gauge transformation acts asThus $D_{\mu }q\mapsto V{D}_{\mu }q$ and the kinetic term is gauge-invariant.

Since the gauge coupling is independent of flavor and chirality, the classical internal global transformations are $q_L\mapsto U_L{q}_L$, $q_R\mapsto U_R{q}_R$, with constant flavor matrices $U_L,U_R\in U(2)$. Equivalently these consist of $SU(2{)}_L\times SU(2{)}_R$, a common vector phase, and a singlet axial phase $q\mapsto e^{i\eta {\gamma }_5}q$, modulo shared discrete factors. The vector phase measures quark number, or baryon number after dividing its generator by three. The nonsinglet axial rotations may be written $q\mapsto e^{-i{\theta }_a{\gamma }_5{\tau }_a/2}q$. At the quantum level the singlet axial phase is anomalous; the exact continuous internal symmetry is $SU(2{)}_L\times SU(2{)}_R\times U(1{)}_V$ up to those discrete identifications. The usual spacetime Poincaré transformations also leave the theory invariant; the massless classical theory has dilation symmetry, which is broken by quantum running. These distinctions are massless two-flavor QCD symmetries.

At equal time the canonical relations, including all indices, areNo constraint removes any of the four spinor components of the Dirac field in these equal-time relations.

To evaluate the axial charges, first prove the elementary bilinear identity. For any constant matrix $K$ on the spinor/flavor space,For example $[q_I^{†}{q}_J,q_K]=q_I^{†}\{q_J,q_K\}-\{q_I^{†},q_K\}{q}_J$; the spatial delta function collapses the integral. This proves the charge action directly from the canonical anticommutation relations.

For $K_a={\gamma }_5{\tau }_a/2$, anticommutation of ${\gamma }^0$ and ${\gamma }_5$ then givesThe charge is an even operator, so its commutator obeys the ordinary product rule. ConsequentlySimilarly, using ${\gamma }_5^2=1$ and the Pauli matrices identity $\{{\tau }_a,{\tau }_b\}=2{\delta }_{ab}I$,Composite fields can be defined with a common symmetry-preserving regulator; these nonsinglet identities have no color axial anomaly. This is the nonsinglet axial charge algebra of quark densities.

If every conserved axial charge left the vacuum invariant, the vacuum expectation of each such commutator would vanish. A nonzero quark chiral condensate instead makes $⟨[Q_{5,a},P_b]⟩=-i{\delta }_{ab}⟨S⟩\ne 0$ for $a=b$. It therefore diagnoses spontaneous breaking of the nonsinglet chiral symmetry. In the massless two-flavor limit, the observed hadronic pattern is described by $SU(2{)}_L\times SU(2{)}_R\to SU(2{)}_V$, with three Goldstone boson modes identified with the pions. The vacuum condensate is a dynamical property of QCD, not something derivable from the canonical algebra alone. The light pion multiplet and absence of parity-degenerate light hadrons support this pattern; the singlet axial anomaly prevents interpreting an additional singlet mode as a fourth Goldstone boson.

For equal nonzero masses, add ${\mathcal{L}}_m=-m\bar{q}q$. It is invariant under local color gauge transformations and common vector flavor rotations, but not under independent left/right rotations. The exact continuous flavor symmetry becomes $SU(2{)}_V\times U(1{)}_V$. The kinetic term and canonical anticommutators are unchanged, and the two proved equal-time commutators still hold. The charges, however, are no longer conserved:The first equation follows from the massive Dirac equations, or the last from $H_m=m\int S$ and $\dot{Q}=i[H,Q]$. This is the flavor-nonsinglet axial Ward identity with equal quark masses. The pions become pseudo-Goldstone bosons with $m_{\pi }^2=O(m)$, rather than exact massless modes. To leading order, in the convention $S=\bar{u}u+\bar{d}d$, $f_{\pi }^2{m}_{\pi }^2=-m⟨S⟩+O(m^2)$. The condensate persists for small masses, but the symmetry is now explicitly broken, so the exact-symmetry Goldstone argument is replaced by an approximate one. The singlet anomaly and gauge symmetry are unaffected.

The Adjoint double cover from SU(2) to SO(3) is obtained by writing a traceless Hermitian matrix as $X=x_i{\sigma }_i$, where the ${\sigma }_i$ are the Pauli matrices, and defining $AX{A}^{†}=(R(A)x{)}_i{\sigma }_i$. Explicitly,Conjugation preserves $\frac{1}{2}\mathrm{tr}{X}^2=|x{|}^2$, so $R(A)$ is orthogonal; connectedness of SU(2) gives $\det R(A)=1$. Composition of conjugations makes $R$ a group homomorphism. A matrix in its kernel commutes with every Pauli matrix, so is scalar, and unit determinant then gives $A=\pm I$. Every spatial rotation occurs: $A=\exp (-i\theta \hat{n}\cdot \sigma /2)$ induces the rotation through angle $\theta$ about $\hat{n}$. ThusSince SU(2) is the simply connected three-sphere, this is the universal double cover; the two Lie algebras are isomorphic, although the groups are not.

Define the conjugate fundamental spinor of SU(2) by ${\bar{\eta }}_{\alpha }=({\eta }^{\alpha }{)}^{\ast }$. Regard $\bar{\eta }$ as a row, so unitarity givesConsequently ${\bar{\eta }}_{\alpha }{\xi }^{\alpha }$ is an invariant Hermitian pairing. The basic invariant tensors are the Kronecker delta ${\delta }^{\alpha }{}_{\beta }$ and the alternating tensors ${ϵ}_{\alpha \beta }$ and ${ϵ}^{\alpha \beta }$. Choose ${ϵ}_{12}=1$, ${ϵ}^{12}=-1$, so that ${ϵ}^{\alpha \gamma }{ϵ}_{\gamma \beta }={\delta }^{\alpha }{}_{\beta }$. The unit-determinant identity $A^TϵA=ϵ$ makes $[\eta ,\xi ]={ϵ}_{\alpha \beta }{\eta }^{\alpha }{\xi }^{\beta }$ invariant; its inverse gives the corresponding upper-index invariant. These contractions generate the tensor invariants of the defining representation. In particular ${\tilde{\eta }}^{\alpha }={ϵ}^{\alpha \beta }{\bar{\eta }}_{\beta }$ transforms by $A$, showing that the conjugate representation is equivalent to the defining one. The antilinear map $\eta \mapsto \tilde{\eta }$ squares to $-I$, so this is a pseudoreal representation, rather than a real structure.

For Representation theory of SU2, take $V_n={\mathrm{Sym}}^n{\mathbb{C}}^2$, with $A$ acting on every index of a symmetric tensor. Its independent components are determined by the number $r$ of indices equal to $2$, so there are $n+1$ of them. Equivalently, the homogeneous polynomial representation of SU2 has basis $z_1^{n-r}{z}_2^r$, $0\le r\le n$, withThe basis weights are $n,n-2,\dots ,-n$, and the raising operator and lowering operator connect consecutive basis vectors with nonzero coefficients until their endpoints. Any nonzero invariant subspace contains a weight vector, by applying polynomial spectral projections in $H$. Raising it reaches $z_1^n$, and lowering then obtains the entire basis. Thus $V_n$ is an irreducible representation, of spin $j=n/2$. The classification of finite-dimensional representations of SU2 gives every finite-dimensional complex irreducible representation in this way. The central element $-I$ acts on a rank-$n$ tensor as $(-1{)}^n$, so descent of an SU(2) representation to SO(3) gives

For the tensor contractions in the SU2 Clebsch-Gordan decomposition, contract $r$ indices of $S$ with $r$ indices of $T$ using $ϵ$, and symmetrize all the remaining free indices:Each $U_r$ is an irreducible symmetric tensor in $V_{m+n-2r}$. To prove these are precisely the constituents, use two sets of formal variables $z,w$. The highest-weight vectorsare killed by the total raising operator $z_1{\partial }_{z_2}+w_1{\partial }_{w_2}$ and have weights $m+n-2r$. The determinant factor is invariant; lowering the remaining factor gives an irreducible string of dimension $m+n-2r+1$. Distinct highest weights give inequivalent summands. The unitary representation on this tensor product makes invariant complements available, and the dimension checkexhausts the space. Hence the contractions above are its irreducible projections, up to normalization, and the Clebsch-Gordan decomposition for SU2 is

For the decomposition of three SU(2) spinors, write $Q^{\alpha \beta \gamma }={\eta }_1^{\alpha }{\eta }_2^{\beta }{\eta }_3^{\gamma }$ and $S^{\alpha \beta \gamma }=Q^{(\alpha \beta \gamma )}$. The contractionstransform as defining spinors. Alternation in three indices in dimension two vanishes, giving $d_1-d_2+d_3=0$; there are exactly two independent contractions. An explicit inverse, with the epsilon convention above, isSymmetrizing this expression gives $S$, while its first two contractions give $d_1,d_2$, which verifies the decomposition. Thus

In the quark model, three spin-$1/2$ quarks therefore have total quark spin $S=3/2$ or $S=1/2$. The two spin-$1/2$ copies correspond to coupling the first pair to spin $0$ or spin $1$ before adding the third quark. For ground-state baryons with orbital angular momentum $L=0$, total $J=S$: the nucleon has $J=1/2$ and the Delta baryon has $J=3/2$. These two coupling copies alone do not assert two unrestricted physical baryon families. The antisymmetric three-quark colour singlet, the spatial wavefunction and flavor symmetry must also satisfy the Pauli constraint on three-quark flavour multiplets; orbital excitations can give other total angular momenta.

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.

An orthonormal basis of the three central flavour combinations isThe Eta octet state is the second zero weight of the octet; the eta singlet state is the separate invariant scalar. Both have isospin zero, whereas the neutral pion has isospin one. All have $I_3=Y=Q=0$, so the location of a weight alone does not determine either isospin or irreducible multiplet. Their neutral flavour-diagonal $L=S=0$ states have charge conjugation $C=(-1{)}^{L+S}=+1$ and hence $J^{PC}=0^{-+}$.

The physical eta and eta prime mesons are mixtures of the octet and singlet combinations, conventionally described at this level byFlavour breaking and the singlet axial anomaly affect their masses and mixing. The neutral pion is much lighter, about $135\mathrm{MeV}$, and decays predominantly to two photons. The eta has mass about $548\mathrm{MeV}$ and important two-photon and three-pion decay modes; the eta prime has mass about $958\mathrm{MeV}$ and important decays to $\eta \pi \pi$. The singlet axial anomaly explains why the eta prime is not an additional light Goldstone boson merely because the flavour tensor product contains a singlet. The centre contains two octet directions and one singlet direction; physical eta mixing combines the latter two, leaving the isospin-one neutral pion separate to a good approximation.

Use flavor hypercharge $Y=B+S$, where $B$ is baryon number and $S$ is strangeness. The isospin coordinate is $I_3$. The baryon octet has coordinatesHere the nucleons have $S=0$, the Sigma baryons and Lambda baryon have $S=-1$, and the Xi baryons have $S=-2$. The central Sigma baryon belongs to an isospin triplet while the central Lambda baryon is an isospin singlet; equal coordinates do not identify the states.

The pseudoscalar meson octet isThe pions have $S=0$, the upper kaons have $S=+1$, and their lower antiparticles have $S=-1$. The two central states are the neutral pion and the Eta octet state. This is the octet basis of flavor symmetry; the physical eta can also mix with the flavor-singlet state.

For the flavor SU(3) Cartan generators, choose the Hermitian physics convention and the inner product $⟨A,B⟩=\mathrm{tr}(AB)$. Then an orthonormal diagonal basis isStrictly, $i{h}_1,i{h}_2$ are elements of the anti-Hermitian SU(3) Lie algebra; $h_1,h_2$ are the corresponding Hermitian observables. In the quark basis $(u,d,s)$, the up and down quarks form an isospin doublet and the strange quark is a singlet. Their $I_3$ values are $(1/2,-1/2,0)$. Each quark has baryon number $1/3$, and their strangeness values are $(0,0,-1)$. It follows thatThese flavor hypercharge conventions differ from the electroweak hypercharge convention. If instead the inner product is $2\mathrm{tr}(AB)$, the orthonormal basis is $t_3=h_1/\sqrt{2}$, $t_8=h_2/\sqrt{2}$, and the same operators are $I_3=t_3$, $Y=2t_8/\sqrt{3}$.