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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.

Keep the PDF's hypercharge signs, which are the negatives of the frequently used convention. Write $U,D,E,N$ for the displayed $U^R,D^R,E^R,N^R$ chiral superfields. Their gauge group representations are conjugate right-handed-matter representations; their Weyl spinor components are left-chiral charge conjugates. Thus assign baryon number $B(Q)=\frac{1}{3}$, $B(U)=B(D)=-\frac{1}{3}$, and lepton number $L(L)=1$, $L(E)=L(N)=-1$, with both Higgs fields neutral under $B,L$. The corresponding electromagnetic convention is $Q_{\mathrm{em}}=T_3-Y$.

Let $X\cdot Y={ϵ}_{ab}{X}^a{Y}^b$ contract the two weak doublets, and use the color Levi-Civita symbol for three color antitriplets. The most general homogeneous cubic superpotential with right-handed-neutrino superfields separates as $W_3=W_{B,L}+W_{/B\text{or}/L}$, whereandRepeated generation and color indices are summed. The four Yukawa coupling matrices are arbitrary. The weak and color contractions imply ${\lambda }_{ijk}=-{\lambda }_{jik}$ and ${\lambda }_{ijk}^{\prime \prime }=-{\lambda }_{ikj}^{\prime \prime }$, while ${\rho }_{ijk}$ is completely symmetric. The $\lambda ,{\lambda }^{\prime }$ terms have $\Delta L=1$; ${\lambda }^{\prime \prime }$ has $\Delta B=-1$; $\kappa$ and $\rho$ have respectively $\Delta L=-1,-3$. Their conjugate interactions have the opposite changes. All four Yukawa terms preserve both numbers; all five remaining types violate at least one. In particular, the neutral $N$ fields printed in the PDF require both $N{H}_1{H}_2$ and $NNN$ terms.

For completeness of this list, a cubic color gauge-invariant can contain a $Q$–$U$ or $Q$–$D$ pair and one colorless doublet, or three color antitriplets. Vanishing hypercharge gives $Q{H}_{2}U$, $Q{H}_{1}D$, $LQD$ and $UDD$. Three $Q$ doublets have no weak singlet. Among colorless terms, two doublets and a singlet give $LLE$, $L{H}_{1}E$, $L{H}_{2}N$ and $H_1{H}_{2}N$; the apparent $H_1{H}_{1}E$ contraction vanishes since the identical bosonic superfields commute. Three singlets yield only $NNN$. This also explains why no omitted cubic monomial is available. The word cubic excludes the linear $N$ term and the quadratic $H_1{H}_2$, $L_i{H}_2$, $N_i{N}_j$ terms; these are separate possible terms in the full renormalizable superpotential.

To demonstrate squark-mediated proton decay, take ${\lambda }_{11k}^{\prime }$ and ${\lambda }_{11k}^{\prime \prime }$ with $k=2$ or $3$. Identical down-flavor indices would make ${\lambda }_{111}^{\prime \prime }$ vanish. Both interactions couple to the same scalar ${\tilde{D}}_k$ of mass $M$. Suppressing overall convention-dependent vertex signs, their relevant two-component fermion bilinears areThe color index on $A$ is contracted with that on ${\tilde{D}}_k$; Lorentz spinor indices within each bilinear are contracted too. Together with $-M^2|{\tilde{D}}_k{|}^2$, its algebraic equation at momenta small compared with $M$ gives ${\tilde{D}}_k=-A^{†}/M^2$. Substitution gives $A^{†}A/M^2$, whose cross term is the four-fermion matching for squark-mediated proton decayIts electron term contains three quark fields of flavors $uud$ and a lepton, and changes baryon number and lepton number together. The Hermitian-conjugate terms describe both orientations of the process. The hadronic matrix element of the three-quark proton-decay operator between a proton and a neutral pion has the right quantum numbers and is not forbidden by a remaining symmetry. Hence the operator permits $p\to e^{+}{\pi }^0$: the lepton field creates the positron and the three-quark matrix element connects the proton to the pion. This supplies the mechanism rather than merely naming the two violating interactions.

Each fermion field has mass dimension $\frac{3}{2}$, so this effective operator has dimension six and coefficient $C_6={\lambda }^{\prime }{\lambda }^{\prime \prime }{}^{\ast }/M^2$. By dimensional analysis its contribution to the decay rate has the formHere $c_h$ collects phase space, running and the dimensionless hadronic factor; a dimensional estimate takes it of order unity rather than pretending to calculate a strong-interaction matrix element. Assuming this exchange dominates and there is no tuned cancellation, the supplied lifetime gives the proton-lifetime bound on a product of R-parity-violating couplingsThe last estimate uses $m_p\simeq 1\mathrm{GeV}$ and ${\tau }_{\min }=2.4\times 10^{64}{\mathrm{GeV}}^{-1}$. Keeping $m_p=0.938\mathrm{GeV}$ instead changes $6.5$ to about $7.6$. A numerical limit independent of the squark mass or hadronic factor is not determined by the supplied data.

Finally, matter parity is $P_M=(-1{)}^{3(B-L)}$: all matter chiral superfields, including $N$, are odd, and the Higgs chiral superfields are even. R-parity of a component field is $R_p=P_M(-1{)}^{2s}$. In a superpotential, the spin factor is irrelevant to the allowed monomials: each of $Q{H}_{2}U,Q{H}_{1}D,L{H}_{1}E,L{H}_{2}N$ has two odd matter factors and is even. Each of $LLE,LQD,UDD,NNN$ has three odd matter factors, and $N{H}_1{H}_2$ has one, so all are odd and forbidden. Higgs vacuum expectation values are even; the allowed Yukawa interactions therefore still generate ordinary fermion masses.

R-parity forbids all the baryon/lepton-number-violating cubic terms above, not every possible violating operator. In particular, $\frac{1}{2}{M}_{ij}{N}_i{N}_j$ is even under matter parity but changes lepton number by two: matter parity allows Majorana neutrino masses. Thus the literal unrestricted claim would be false for these very fields; its valid cubic interpretation gives precisely the requested exclusion while preserving the Yukawa coupling mass terms.

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.

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 baryon asymmetry is an excess of net baryon number over zero, conveniently measured by ${\eta }_B=(n_B-n_{\bar{B}})/n_{\gamma }$. Ordinary matter contains abundant baryons, whereas no comparable primordial antibaryon population is observed. Large nearby domains of matter and antimatter would produce radiation through particle-antiparticle annihilation at their boundaries. Antiparticles produced in energetic collisions are not evidence for a cosmological antimatter domain. The small nonzero baryon-to-photon ratio inferred from Big Bang nucleosynthesis is independently tested by baryon-density consistency between nucleosynthesis and the microwave background.

Starting with zero net baryon number, the Sakharov conditions require baryon-number violation, C and CP violation, and departure from thermal equilibrium. Conservation of baryon number would keep the initial value zero. C symmetry or CP symmetry would pair baryon production with equal antibaryon production; CP rate cancellation in baryogenesis explains why C violation alone does not suffice. In thermal equilibrium with zero imposed charge chemical potentials, CPT symmetry forbids the required net baryon number. An asymmetric rate in one reaction must not be mistaken for net production after all reverse reactions are included.

A simple example is heavy-boson decay baryogenesis. Take equally many heavy $X$ bosons and their antiparticles, with decay channelsThe symbols in parentheses are branching fractions. The two $X$ channels carry baryon numbers $2/3$ and $-1/3$, so both cannot conserve one assigned baryon number for $X$. Their mean baryon number is $r-1/3$, while the mean for $\bar{X}$ is $-\bar{r}+1/3$. Consequently,where $ϵ$ is the asymmetry per decaying particle. CP violation permits $r\ne \bar{r}$ without violating the equality of total decay widths required by CPT symmetry. Microscopically a direct CP asymmetry requires interference between decay amplitudes with both relative weak phases and absorptive phases; specifying complex couplings alone would not establish unequal partial decay widths.

For radiation domination, a sufficient decay condition is ${\Gamma }_X\lesssim H(T=M_X)=1.66\sqrt{g_{\ast }}{M}_X^2/m_P$. The heavy population can then survive as the equilibrium abundance drops, before decaying while inverse production is suppressed. Its actual abundance differs from its thermal equilibrium value. If subsequent baryon-number washout is negligible and cosmological entropy conservation holds, an initial total abundance $Y_X=(n_X+n_{\bar{X}})/s$ producesHere $s$ is the entropy density. Residual baryon-number washout multiplies this expression by an efficiency factor below one. The example realizes all three Sakharov conditions; it illustrates a mechanism, rather than identifying the actual cosmological source of the baryon asymmetry.

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.

R-parity is the multiplicative ${\mathbb{Z}}_2$ quantum numberwhere $B,L,s$ are baryon number, lepton number and spin. Ordinary Standard Model particles have $R_p=+1$ and their superpartners have $R_p=-1$. Gauge invariance alone allows MSSM superpotential terms $LL{E}^c$, $LQ{D}^c$, $U^c{D}^c{D}^c$ and $L{H}_u$. Simultaneous baryon- and lepton-number violation can induce rapid proton decay. Exact R-parity forbids these renormalizable terms, since the associated matter parity is odd; it does not forbid every higher-dimensional proton-decay operator, for example $QQQL$.

Two consequences are superpartner production in pairs from ordinary-particle initial states and stability of the lightest supersymmetric particle: a single parity-odd particle cannot decay only into parity-even particles. Cascades therefore end in the lightest supersymmetric particle; if it is neutral and weakly interacting, it produces missing-energy signatures.

A continuous R-symmetry rotates $\theta$ and the supercharges. With $R(\theta )=1$, the superpotential has R-charge two and the fermion in a chiral superfield has one less charge than its scalar. R-parity is only a discrete restriction, represented by combining matter parity with $\theta \mapsto -\theta$, and need not preserve a full continuous R-symmetry. For example, a gaugino Majorana mass term violates an unbroken continuous R-symmetry but preserves R-parity. Thus proton protection by parity and continuous R-charge selection rules are distinct requirements.

Use natural units $\hslash =c=k_B=1$ and the unreduced Planck mass $M_P=G^{-1/2}$. Neutrino decoupling occurs when the weak interaction rate falls below the Hubble parameter. The weak-decoupling temperature estimate equates the given rate with the radiation-era expansion rate:Here $g_{\ast }$ counts the effective number of relativistic energy degrees of freedom, and the Fermi constant is $G_F\simeq 10^{-5}{\mathrm{GeV}}^{-2}$. The omitted reaction-rate coefficients make this an order-of-magnitude estimate: decoupling occurs at a temperature of order one MeV.

The charge-conserving channel is $n+{\nu }_e\leftrightarrow p+e^{-}$, together with $p+{\bar{\nu }}_e\leftrightarrow n+e^{+}$. The positron sign in the printed neutron-capture channel violates electric charge conservation. Chemical equilibrium requires ${\mu }_n+{\mu }_{{\nu }_e}={\mu }_p+{\mu }_e$. Negligible lepton chemical potentials therefore give ${\mu }_n\simeq {\mu }_p$. Since neutrons and protons are nonrelativistic particles, their Maxwell-Boltzmann distribution givesBoth nucleons have two spin states, and their masses agree closely in the prefactor. Thus neutron-proton chemical equilibrium impliesThe heavier neutron is suppressed by a Boltzmann factor.

For the deuterium equilibrium abundance, retain the paper's convention $B=m_D-m_n-m_p<0$; the positive nuclear binding energy is ${\epsilon }_D=-B\simeq 2.22\mathrm{MeV}$. Chemical equilibrium of $n+p\leftrightarrow D+\gamma$ gives ${\mu }_D={\mu }_n+{\mu }_p$ because the equilibrium photon chemical potential vanishes. Using the nonrelativistic Maxwell--Boltzmann number density for each species and dividing givesHere $g_D=3$ and $g_n=g_p=2$. With the baryon-to-photon ratio $\eta =n_B/n_{\gamma }$, the photon number density $n_{\gamma }=2\zeta (3)T^3/{\pi }^2$, and $m_D\simeq 2m_N$, $m_n\simeq m_p\simeq m_N$, this becomesIn this convention $X_D=n_D/n_B$ counts nuclei, so its contribution to the fraction of baryons in deuterium is $2X_D$.

The deuterium bottleneck explains why helium-4 formation waits until $T$ is far below ${\epsilon }_D$. At higher temperature, there are vastly more photons than baryons; even the dilute energetic tail can destroy newly formed deuterium. A crude estimate balances ${\eta }^{-1}{e}^{-{\epsilon }_D/T}$ against one, giving $T\sim {\epsilon }_D/\log ({\eta }^{-1})\sim 0.1\mathrm{MeV}$ for $\eta$ of order $10^{-9}$. The power-law factors in the deuterium equilibrium abundance and the nuclear reaction rates refine this estimate. Once stable deuterium accumulates, further reactions can efficiently place nearly all surviving neutrons into helium-4.

Write $r_N=n_n/n_p$ at the onset of nuclear burning. Each helium-4 nucleus requires two neutrons and contains four baryons. Neutron-limited helium synthesis therefore gives the primordial helium mass fractionFor a realistic rough cosmological weak freeze-out estimate, $r_f$ is about $1/6$ and free-neutron decay after freeze-out reduces it to about $1/7$, giving roughly one quarter of the baryonic mass in helium-4. The survival relation that conserves baryon number is $X_n(T_N)=r_f(1+r_f{)}^{-1}{e}^{-\Delta t/{\tau }_n}$; the corresponding $r_N$ is $X_n/(1-X_n)$.

The rough neutrino decoupling rate by itself does not determine that one-quarter value. Substituting $T_d=1.8\mathrm{MeV}$ literally into $e^{-Q/T_d}$ would give $r\simeq 0.49$ and $Y_4\simeq 0.65$ before neutron decay. Neutron-proton chemical equilibrium persists to a lower temperature, approximately $0.7$--$0.8\mathrm{MeV}$ once the actual conversion rates are included. The rate scaling fixes the MeV scale; accurate abundances need the conversion coefficients and the delay through the deuterium bottleneck. This sequence is described in David Tong's discussion of nucleosynthesis.