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The Cabibbo-Kobayashi-Maskawa matrix $V=U_{uL}^{†}{U}_{dL}$ compares the left-handed unitary matrices that diagonalize the up-type and down-type Yukawa matrices. Its rows label up quarks, charm quarks and top quarks; its columns label down quarks, strange quarks and bottom quarks. In the weak charged current, the coefficient of ${\bar{u}}_{Li}{\gamma }^{\mu }{d}_{Lj}{W}_{\mu }^{+}$ is $V_{ij}$. A single CKM matrix therefore connects quark masses, flavour-changing decay amplitudes, and CP violation.

Write $s_{ij}=\sin {\theta }_{ij}$ and $c_{ij}=\cos {\theta }_{ij}$. An exactly unitary matrix with three mixing angles and one CP-violating phase isThis convention agrees with Makoto Kobayashi's account of three-generation quark mixing.^[ref]

The Christenson-Cronin-Fitch-Turlay kaon CP violation experiment established the phenomenon that required an explanation. Makoto Kobayashi and Toshihide Maskawa showed that two-generation quark mixing cannot supply it, whereas adding a third quark generation permits an irreducible weak phase. Their 1973 paper also considered additional fields as alternative mechanisms; the three-generation conclusion concerns CP violation from the minimal unitary quark mixing mechanism.^[ref]

A common weak phase makes CP violation across mesons testable through the BABAR time-dependent CP violation experiment. It does not explain the cosmic baryon asymmetry.^[ref]

A baryon containing one or more strange quarks and no heavier quark. The light hyperon multiplets include lambda, sigma, xi and omega baryons. Strong and electromagnetic interactions conserve strangeness, whereas a weak interaction can change it; a photon in the final state does not by itself make a hyperon decay electromagnetic.

A neutral strange B meson has valence content $\bar{b}s$ in the conventional $B_s^0$ assignment; its antiparticle has $b\bar{s}$. In its charged-current spectator amplitude for $J/\psi \varphi$, $\bar{b}\to \bar{c}{W}^{+}$ and $W^{+}\to c\bar{s}$, while the original strange quark is carried into the final strange pair. This is one Standard-Model contribution, not a complete treatment of mixing, loop amplitudes or hadronization.

The 1964 report exposed Brookhaven's 80-inch hydrogen bubble chamber to a mass-separated $5.0\mathrm{GeV}/c$ negative-kaon beam. A photographed decay chain, track momenta and bubble density supported the particle identification. Two converted photons helped reconstruct a neutral pion. Alternative assignments were tested kinematically. The authors reported an Omega baryon candidate with $Q=-1$, $S=-3$ and mass $1686\pm 12\mathrm{MeV}/c^2$, reserving a detailed systematic-error discussion for further events.^[ref]

This is an inference from several linked observations. A charged particle can leave a visible bubble-chamber track, while a neutral segment must be reconstructed from its production and decay products. Agreement between possible decay channels, electric charge, momentum conservation and the reconstructed mass is stronger evidence than a suggestive track shape alone. The event tests a predicted hadron pattern; it does not photograph the three strange quarks individually.

For example, an independently constructed two-body kinematic check usesIn a parent rest frame with equal and opposite daughter momenta, the momentum sum vanishes, giving $M{c}^2=E_1+E_2$. In units $c=1$, stipulated daughter masses $m_1=1$, $m_2=1/4$ and momentum magnitude $p=3/4$ give $E_1=5/4$, $E_2=\sqrt{10}/4$ and $M=(5+\sqrt{10})/4\simeq 2.04057$. A common Lorentz boost changes the daughter energies and momenta while preserving this reconstructed invariant mass. These chosen units and numbers are an illustration, not a reanalysis of the 1964 event.

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.

Use the printed hypercharge convention; reversing every hypercharge relative to another common convention does not change the gauge invariants. Write $U,D,E,N$ for the singlet chiral superfields labelled with a superscript $R$ in the paper, and define the weak-doublet contraction $A\cdot B={ϵ}_{ab}{A}^a{B}^b$. The most general homogeneous cubic superpotential with right-handed-neutrino superfields separates as $W_3=W_{B,L}+W_{\mathrm{viol}}$, whereandRepeated generation and color indices are summed. Because the chiral superfields commute while the gauge contractions are antisymmetric,The factors $1/2$ and $1/6$ remove the corresponding overcounting.

Here is a completeness check, not merely a list of possible terms. A color-singlet cubic can have no colored fields, a fundamental–antifundamental pair, or three fundamentals or three antifundamentals. A colored pair must be $QU$ or $QD$: the remaining weak doublet and hypercharge select respectively $H_2$ or $H_1,L$. Three antifundamentals allow only $UDD$ at zero hypercharge; three $Q$ doublets have no weak singlet. With no colored fields, weak invariance requires zero or two doublets. Two positive-hypercharge doublets require $E$, giving $L{H}_{1}E$ and $LLE$, while $H_1\cdot H_1=0$. Opposite-hypercharge doublets require $N$, giving $L{H}_{2}N$ and $H_1{H}_{2}N$. With no doublets, only $NNN$ has zero hypercharge. These are exactly the nine structures above.

For baryon number and lepton number, assignand zero charges to the Higgs fields. This is the conjugate-singlet convention, so the neutrino Yukawa coupling conserves lepton number. The first four terms conserve both charges. The $LLE$ and $LQD$ terms carry $L=1$, the $UDD$ term carries $B=-1$, the $N{H}_1{H}_2$ term carries $L=-1$, and $NNN$ carries $L=-3$. The gauge-singlet neutrino interactions must therefore be included in the violating sector.

If “cubic” instead means a polynomial of degree at most three, there are also the lower-degree gauge invariantsThe Higgs mass term conserves both charges; $t_i{N}_i$, $M_{ij}{N}_i{N}_j$ and ${\mu }_i{L}_i\cdot H_2$ violate lepton number. These additional terms distinguish the full renormalizable superpotential from its homogeneous cubic part.

To obtain squark-mediated proton decay, combine a lepton number-violating $LQD$ vertex with a baryon number-violating $UDD$ vertex. A nonzero example is ${\lambda }_{112}^{\prime }$ with ${\lambda }_{112}^{\prime \prime }$, exchanging the singlet strange quark squark of mass $M_{\tilde{s}}$. The antisymmetry forbids ${\lambda }_{111}^{\prime \prime }$, so using three first-generation singlets would not be a valid example. The relevant component vertices contain $e_L{u}_L{\tilde{s}}^c$ and $u^c{d}^c{\tilde{s}}^c$, with the required color contraction. Connecting one vertex to the complex conjugate of the other through the massive scalar propagator produces schematicallyParentheses denote Lorentz scalar two-spinor contractions, dotted for the right-handed pair and undotted for the left-handed pair. This dimension-six proton-decay operator contains the three quarks $uud$ and an electron field that can create a Positron. Its hadronic matrix element $⟨{\pi }^0|{ϵ}_{abc}(u_R^a{d}_R^b)u_L^c|p⟩$ gives the channel $p\to e^{+}{\pi }^0$. The internal strange flavor need not appear in the final state. The two interactions together violate both continuous charges, as required for this channel.

The coefficient of the dimension-six effective operator has mass dimension $-2$. Thus dimensional analysis giveswhere $c_h$ is a dimensionless hadronic and phase-space factor. Its detailed value cannot be fixed by dimensional reasoning. Applying the historical lower bound supplied in the paper gives the proton-lifetime bound on a product of R-parity-violating couplingsThe numerical coefficient is only the accuracy of the requested dimensional estimate; the squark mass was not fixed by the question.

Finally, R-parity is $R_p=(-1{)}^{3(B-L)+2s}$. At the superfield level use matter parity $P_M=(-1{)}^{3(B-L)}$: $Q,U,D,L,E,N$ are odd and $H_1,H_2$ are even. Each conserving Yukawa coupling contains two matter chiral superfields and one Higgs field, so it is even. Every violating homogeneous cubic above is odd: $LLE,LQD,UDD,NNN$ contain three matter fields, while $N{H}_1{H}_2$ contains one. R-parity forbids all the listed cubic violations and preserves the Yukawa terms that generate fermion masses. The Higgs bilinear is also even, whereas the lepton bilinear $L{H}_2$ and the singlet linear term are odd.

The final assertion must not be read as forbidding every possible violation of continuous baryon number or lepton number. In particular, matter parity allows Majorana neutrino masses: $N_i{N}_j$ is even but carries $L=-2$, a concrete counterexample once quadratic terms are included. Higher-dimensional operators such as $QQQL$ and $(L{H}_2{)}^2$ are likewise even and violate the continuous charges. The verified statement here concerns the homogeneous cubic sector, including the neutrino singlet terms.

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.

The commuting Hermitian operators $H_1,H_2$ admit a weight-space decomposition. The six root vectors shift their simultaneous eigenvalues byor their negatives. Put $h_i=2{\alpha }_i\cdot H$, $e_i=\sqrt{2}{E}_{+}^i$ and $f_i=\sqrt{2}{E}_{-}^i$. The supplied relations giveThese are three sl2 triples, coming from the complexifications of the three SU(2) Lie algebra root-pair subalgebras. They act through homomorphisms into $\mathrm{End}(V)$. For a nontrivial representation, simplicity of the SU(3) Lie algebra makes its kernel zero, so these images are actual ${\mathfrak{sl}}_2(\mathbb{C})$ subalgebras; a trivial representation gives zero actions. The real compact generators can be taken as $i{h}_i$, $i(E_{+}^i+E_{-}^i)$ and $E_{+}^i-E_{-}^i$.

Decomposing under any one of these subalgebras gives finite weight strings in its root direction. The two simple-root highest weights must be nonnegative integers. In the SU(3) highest-weight coordinates used here, this condition isHere $(a,b)$ are the Dynkin labels; $(p,q)$ are the eigenvalues of $H_1,H_2$. An irreducible representation is determined by these labels and has dimension $(a+1)(b+1)(a+b+2)/2$. An arbitrary finite-dimensional anti-Hermitian representation is a direct sum of such representations: the orthogonal complement of an invariant subspace remains invariant.

The weight diagram has Weyl group symmetry, namely reflections and rotations by $120$ degrees. Its convex boundary is a hexagon with alternating side lengths $a,b$, degenerating to a triangle when one label vanishes. Generic representations need not have $60$-degree symmetry. Boundary weight multiplicities are one. Moving inward through hexagonal layers increases multiplicity by one, up to $\min (a,b)+1$; triangular inner layers retain that maximum. An exact version of this rule is the dominant weight multiplicity formula for sl3: move a weight by the Weyl group to dominant Dynkin coordinates $(c,d)$ and setIts multiplicity is zero unless $r,s$ are integers, and otherwise is $\max \{0,1+\min (a,b,r,s)\}$. These rules construct the diagram while counting coincident weights correctly.

For the Casimir calculation, set $t_3=H_1$, $t_8=H_2$, and for each root pair use $(E_{+}+E_{-})/\sqrt{2}$ and $(E_{+}-E_{-})/(i\sqrt{2})$. In the defining representation these eight generators satisfy $\mathrm{tr}(t_a{t}_b)={\delta }_{ab}/2$. Their summed squares are precisely the question's $X$. The invariant trace form makes their structure constants of a Lie algebra antisymmetric, sothe antisymmetric coefficient contracts a symmetric anticommutator. Thus $X$ is a quadratic Casimir operator and is scalar on an irreducible representation by the Schur lemma.

Let $v$ be a highest-weight vector, annihilated by all $E_{+}^i$. On $v$, each $E_{-}^i{E}_{+}^i$ vanishes and each $E_{+}^i{E}_{-}^i$ reduces to the root commutator. Summing those three commutators gives $H_1+\sqrt{3}{H}_2$. Henceand therefore the SU(3) quadratic Casimir eigenvalue isFor a reducible representation the same calculation applies separately to its irreducible blocks.

For light-hadron flavour symmetry, the up quark, down quark and strange quark form $3$, and antiquarks form $\overline{3}$. Conventional mesons have $3\otimes \overline{3}=1\oplus 8$. Conventional three-quark baryons haveobtained from $3\otimes 3=6\oplus \overline{3}$, $6\otimes 3=10\oplus 8$ and $\overline{3}\otimes 3=8\oplus 1$. Their Dynkin labels are $(0,0)$, $(1,1)$ and $(3,0)$, respectively. Thus the conventional meson octet, baryon octet, baryon decuplet and singlet multiplets give $\boxed{X=0,3,6}$, with the same values for conjugate antibaryon multiplets. More generally, the zero flavour triality of a light-quark hadron imposes $a+2b\equiv 0(\mathrm{mod}3)$ on multiplets constructed entirely from light quarks, antiquarks and gluons. Their Casimir values are given by the same formula for these labels. Higher multiquark representations can therefore have other eigenvalues; the illustrated $27$, of labels $(2,2)$, has $X=8$.

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.