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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.

Use Planck units and restrict to the physical Kähler metric domainPut $A=a{e}^{-\alpha S}$ and $w=A+b$, so that $W=C^3+w$. There are no gauged interactions specified, hence no additional D-term scalar potential. The supergravity F-term potential is $V=e^K(K^{i\bar{j}}{D}_{i}W\overline{D_{j}W}-3|W{|}^2)$, with $e^K=1/(x{y}^3)$. Its Kähler covariant derivatives of a superpotential areThe nonzero metric entries areThe determinant of this $T,C$ block is $9/y^3>0$, so it is positive definite. With the inverse-index convention $K_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, the raised components areKeeping these complex-conjugate off-diagonal entries in their correct index positions is essential.

To exhibit the no-scale supergravity cancellation, write the $T,C$ norm asHere $D_{C}W+\bar{C}{D}_{T}W=3C^2$. The $3|W{|}^2$ term cancels the negative term in the supergravity F-term potential. The result is the nonnegative scalar potentialThis is the matter-logarithm no-scale potential, specialized to the exponential superpotential.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Substitution into the full inverse Kähler metric givesAn overall auxiliary-field phase is conventional. The $C$ dependence of $F^T$ cancels between the two metric terms. If all three auxiliary fields vanished at a finite physical point, $F^C=0$ would force $C=0$, $F^T=0$ would force $A+b=0$, and $F^S=0$ would then force $\alpha xA=0$. Thus $a=b=0$ is necessary. For any nontrivial choice of $a,b$, supersymmetry is broken at every finite physical point. The allowed exception $a=b=0$, $C=0$ is supersymmetric; the statement cannot hold for all arbitrary parameters without this qualification.

The no-scale volume runaway criterion also determines whether the minimum is attained. For fixed $S,C$, differentiation with respect to $y$ givesIt is negative unless both numerators vanish. Therefore a finite stationary vacuum must satisfyEvery such point is a global minimum with $V_{\min }=0$. For $a,b\ne 0$, setting $r=|b/a|$ and $z=\alpha x>0$ reduces existence toThe derivative of the magnitude function is $(1-z)e^{-z/2}/2$. It rises from $1$ at the excluded boundary $z=0$ to $2e^{-1/2}$ at $z=1$, then falls to zero. Consequently a finite zero-energy minimum exists exactly whenor in the separate trivial family $a=b=0$. There is one positive root for $0<r\le 1$, two for $1<r<2e^{-1/2}$, and one double root at equality. If only one of $a,b$ vanishes, or $r$ exceeds this maximum, there is no finite minimum: the infimum is zero as $\mathrm{Re}T\to \infty$. Calling this runaway a finite zero-energy vacuum would be incorrect.

At a nontrivial zero-energy minimum, $W=A+b=-\alpha xA\ne 0$, and the auxiliary expectation values reduce toThis explicitly verifies supersymmetry breaking with zero vacuum energy in no-scale supergravity. Both real components of $T$ are exact flat directions of a scalar potential on the minimum locus. Away from that locus, $\mathrm{Im}T$ is still absent from the potential, but $\mathrm{Re}T$ is not flat. At a minimum, $C$ is lifted at quartic order: at fixed $S,T$, $V=3|C{|}^4/[x(T+\bar{T}{)}^2]+O(|C{|}^6)$. A zero quadratic mass does not make $C$ an exact flat direction. The quartic stabilization at a double no-scale root likewise gives a vanishing quadratic mass, but no exact flat direction, for $\mathrm{Re}S$ at $z=1$. For $a=b=0$, the whole $C=0$ family has both $S$ and $T$ flat and supersymmetry unbroken.

Finally, the gravitino mass from a superpotential isAt a nontrivial zero-energy minimum this becomesIt depends on the unfixed volume modulus, so the model does not predict one numerical mass. For $a=b=0$, $C=0$, the gravitino is massless.