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For canonical Kähler potential, set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$. The supplied supergravity F-term potential is $|\lambda {|}^2{e}^{{\kappa }^2(a+b+c)}[ab+ac+bc+3{\kappa }^{2}abc+{\kappa }^{4}abc(a+b+c)]$. Adding $g(b-c-\zeta {)}^2$ with $g>0$, $\lambda \ne 0$ and $\zeta >0$ gives a supersymmetric vacuum at $a=c=0$, $b=\zeta$. The branch $b=c=0$ has positive energy $g{\zeta }^2$, D-term breaking and a flat direction of a scalar potential in ${\varphi }_0$. Its charged squared masses are $|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}\mp 2g\zeta$, so it is a local minimum valley only when both are positive. For negative $\zeta$, exchange the charged fields. These are branches rather than two isolated minima; a conventional constant Fayet–Iliopoulos term additionally needs a consistent supergravity gauge completion.

Canonical gauge interactions preserve the tree-level supertrace mass sum rule at a vacuum with vanishing D-terms. Put $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. The extra real-scalar squared-mass trace is $2C$, the extra Weyl spinor squared-mass trace from gaugino mixing is $4C$, and the gauge boson squared-mass trace is $2C$. Their spin-weighted supertrace is $2C-2(4C)+3(2C)=0$. The scalar contribution comes from differentiating $\frac{1}{2}{\sum }_a{D}_a^2$; the fermionic one counts both blocks of the symmetric mixing matrix with entries $\sqrt{2}{g}_a{T}_{a}v$; the vector one follows from the covariant scalar kinetic term. Any zero-mass Goldstone bosons in the unreduced scalar Hessian do not change its trace.

Adding a holomorphic function $f$ and its conjugate does not change the Kähler metric. In global supersymmetry, these terms integrate to zero in the D-term action. In Planck-unit supergravity, the simultaneous transformation $W\mapsto e^{-f}W$ leaves the scalar potential invariant; shifting $K$ while holding $W$ fixed generally does not.

With canonical tree-level global supersymmetry, neutral F-term breaking and no D-term mass shifts, the tree-level supertrace mass sum rule applies in conserved electric and color charge blocks. The corresponding scalar squared-mass average equals the fermion squared-mass average, so not every scalar partner can lie above that average. This obstructs making every squark heavy through direct canonical visible-sector breaking while retaining light quarks. Effective soft supersymmetry breaking, noncanonical interactions, radiative effects or supergravity invalidate the assumptions of that argument.

Work with the local Wilsonian effective action, retaining the elementary chiral superfield and using a nonzero infrared cutoff. Its superpotential is holomorphic in both the field and chiral spurions representing the couplings. A convenient pair of spurion symmetries assigns chargesBoth couplings and $\Phi$ transform nontrivially under each displayed symmetry. Each superpotential monomial has ordinary charge zero and R-charge two, so the integral $\int d^2\theta W$ is invariant. This R assignment differs from the conventional $(1,0,-1)$ assignment for $(\Phi ,m,g)$ by twice the ordinary charge, and describes the same two-dimensional charge space. These are formal spurion symmetries; fixing the couplings to numerical values need not leave them as actual symmetries of the fields alone.

For $m\ne 0$, $m{\Phi }^2$ has the required charges and mass dimension three, while $z=g\Phi /m$ has zero charge under both symmetries and is dimensionless. Thus the allowed holomorphic form isOne can see its generality directly. A holomorphic monomial ${\Phi }^u{m}^v{g}^w$ must obey $u-2v-3w=0$ and $3u-4v-7w=2$. Solving gives $u=w+2$ and $v=1-w$, hence precisely $m{\Phi }^2(g\Phi /m{)}^w$. A field-independent constant is irrelevant to this global supersymmetric action and can be dropped.

The symmetries alone leave the function arbitrary. To determine its perturbative form, use regularity as well as the two limits. A regular weak-coupling expansion at $g=0$ hasThe $g\to 0$ limit is a free massive theory, so $c_0=1/2$. Since the elementary field has not been integrated out, the Wilsonian superpotential must remain regular term by term as $m\to 0$ with the infrared cutoff fixed. Every term with $r\ge 2$ would introduce a pole in $m$ and is excluded. Negative powers of $g$ were already excluded by perturbative regularity at $g=0$. Consequently $F=c_0+c_{1}z$.

At $m=0$, the only surviving allowed interaction is proportional to $g{\Phi }^3$. Its coefficient is first order in $g$, and matching the tree three-field vertex gives $c_1=1/3$. No closed-loop three-field diagram can be made from a single cubic vertex; extra interaction vertices would produce higher coupling order. Equivalently free superspace propagators join chiral to antichiral fields, so a closed loop would require additional vertices or conjugate couplings, which the holomorphic charge constraints exclude for this term. This fixes the coefficient without assuming the desired conclusion. ThereforeThe regularity condition is essential: it concerns the local Wilsonian superpotential with the field retained, not a nonlocal one-particle-irreducible functional at zero infrared cutoff or a superpotential obtained by eliminating a massive field. Such different operations can generate inverse masses or infrared singularities without contradicting this proof.

The Kähler potential is a real D-term, not a holomorphic F-term, and is not protected by this argument. For example its quadratic part becomes $Z(g,\bar{g},m,\bar{m};\mu ){\Phi }^{†}\Phi$, with higher real operators also allowed. Terms involving $|g{|}^2$ are compatible with the symmetries and produce wave-function renormalization. After the canonical rescaling ${\Phi }_c=Z^{1/2}\Phi$, the parameters in the superpotential expressed using canonical fields are $m_c=m/Z$ and $g_c=g/Z^{3/2}$, so physical couplings can run despite the holomorphic non-renormalization statement.

Holomorphy is crucial. Without it, neutral factors such as $1+c|g{|}^2$ could multiply either superpotential monomial while respecting both spurion symmetries. Excluding conjugate couplings, together with regularity and tree matching, is what turns the charge argument into the non-renormalization theorem.

Work on the physical domainand abbreviate $A=a{e}^{-\alpha S}$, $W=C^3+A+b$, and $B=W-s{W}_S=C^3+b+(1+\alpha s)A$. There are no vector multiplets specified, hence no additional gauge D-term. In Planck units the supergravity F-term potential and upper-index supergravity auxiliary fields areThe overall auxiliary-field sign is a convention; its vanishing and norm are invariant statements.

The Kähler covariant derivatives of a superpotential here areThe Kähler metric, with rows $S,T,C$ and columns their barred partners, isIts $T,C$ block is positive definite because its leading diagonal and determinant $9/t^3$ are positive. In the inverse-tensor convention $K_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, the nonzero upper components areIn particular the conjugations in the off-diagonal components matter.

To verify the no-scale supergravity cancellation explicitly, temporarily write $u=D_{T}W=-3W/t$ and $v=D_{C}W=W_C+3\bar{C}W/t$. Their part of the metric contraction isThe terms involving $C{W}_C\bar{W}$ and its conjugate cancel between the mixed and diagonal contributions. The remaining $3|W{|}^2$ cancels the universal negative term. Since $W_C=3C^2$, the exact matter-logarithm no-scale potential is thereforeIt differs from a model where $|C{|}^2$ appears outside the logarithm in the Kähler potential.

Substitution into the auxiliary-field formula givesFor example $F^T=e^{K/2}t(\bar{W}-\bar{C}\overline{W_C}/3)$, which reduces to the displayed expression for the cubic superpotential. To test unbroken supersymmetry at any finite point, $F^C=0$ first forces $C=0$. Then $F^T=0$ forces $A+b=0$, whereas $F^S=0$ forces $(1+\alpha s)A+b=0$. Subtracting gives $\alpha sA=0$. Because $s>0$, $\alpha >0$ and $e^{-\alpha S}\ne 0$ at a finite point, all auxiliaries can vanish only if $a=b=0$. Thus every nontrivial choice of $(a,b)$ breaks supersymmetry at finite field values.

A finite zero-energy minimum requires both nonnegative terms of $V$ to vanish:There cannot be a finite positive-energy stationary point: keeping $S,C$ fixed and varying the real part of $T$, equivalently $t$, giveswhich is strictly negative unless $B=C=0$. Hence these zero-energy solutions are all the possible finite minima.

Because $a,b$ were allowed to be arbitrary, existence must be checked. If both are nonzero, put $x=\alpha s>0$ and $r=|b/a|$. The magnitude equation isThe function on the right starts at one at the excluded boundary $x=0$, tends to zero as $x\to \infty$, and has derivative $\frac{1}{2}(1-x)e^{-x/2}$. Its maximum is $2e^{-1/2}$ at $x=1$. Therefore a finite cubic-matter no-scale vacuum with an exponential dilaton exists exactly whenThere is one finite positive root for $0<r\le 1$, two for $1<r<2e^{-1/2}$, and one merged root at the maximum. The complex phase equation fixes $\mathrm{Im}S$ modulo $2\pi /\alpha$. If exactly one of $a,b$ vanishes, or the ratio exceeds the bound, no finite minimum exists; the potential instead has infimum zero along the runaway $t\to \infty$. A zero infimum at an infinite-field boundary is not an attained finite vacuum.

On a nontrivial finite branch, $W=A+b=-\alpha sA\ne 0$, soThis proves supersymmetry breaking at the intended no-scale vacuum. Its zero energy is consistent with the negative gravitational term: $K_{T\bar{T}}|F^T{|}^2=3e^K|W{|}^2$, which cancels $3e^K|W{|}^2$ in $V$. It is not a counterexample to the global-supersymmetry positivity argument in question 3.

Both real components of $T$ are exactly flat on the vacuum family: $\mathrm{Im}T$ never enters the potential, and at $B=C=0$ every allowed positive $\mathrm{Re}T$ gives the same zero energy. Generically both components of $S$ are fixed. Although the matter field has no quadratic mass there, it is not an exact flat direction: for $S$ fixed at its root, $B=C^3$ and the leading positive potential is $3|C{|}^4/(s{t}^2)$. At the special merged root $x=1$, the real-$S$ quadratic curvature also vanishes, but higher orders lift it; this does not add a continuous flat vacuum direction. If $a=b=0$, by contrast, $C=0$ is a supersymmetric zero-energy family with both complex $S$ and $T$ flat and all auxiliaries zero. This exception prevents an unconditional breaking assertion for arbitrary constants.

Finally the gravitino mass from a superpotential isAt a nontrivial finite minimum it becomesIts value varies along the flat volume modulus; the classical potential does not select a unique supersymmetry-breaking or gravitino-mass scale. On the trivial supersymmetric branch $a=b=C=0$, it is zero.

At the usual two-derivative order, the two-derivative global supersymmetric gauge action isHere ${\mathcal{W}}_{\alpha }^a$ is the chiral field-strength superfield, distinguished from the superpotential $W$. The Kähler potential $K$ is a real gauge-invariant function, defined up to a Kähler transformation; $K_{i\bar{j}}$ supplies the scalar kinetic metric. The superpotential is a gauge-invariant holomorphic function of the chiral superfields. The symmetric gauge kinetic function $f_{ab}$ is holomorphic and transforms so that its contraction with the field strengths is gauge invariant. Its real part supplies the gauge and gaugino kinetic coefficients and must be positive on physical configurations; its imaginary part supplies the topological angle. The final constants are Fayet–Iliopoulos terms, normalized so that they contribute ${\xi }_a{D}^a$ when $V^a$ has highest component ${\theta }^2{\bar{\theta }}^2{D}^a/2$.

For the simple non-Abelian gauge group specified in the question, $\xi =0$. Invariance of a proposed ${\xi }_a{D}^a$ requires ${\xi }_a{f}^a{}_{bc}=0$. A simple Lie algebra equals its commutator algebra, so no nonzero invariant covector exists. This proves that Fayet–Iliopoulos terms require an Abelian gauge factor. The last term above describes the customary Abelian extension, not an additional arbitrary coupling of the simple group.

For a renormalizable theory, after choosing matter bases the functions reduce toOnly gauge-invariant monomials are included. A nonconstant $f$ would produce interactions of dimension greater than four. Here the gauge coupling is absorbed into the connection in $e^{2V}$, while $\mathrm{Re}Y=1/g_h^2$ in the holomorphic normalization. A constant $c$ has no dynamical effect in global supersymmetry. The original arbitrary functions remain useful for a nonrenormalizable two-derivative effective theory, but the following argument restricts to the requested renormalizable setting.

To analyze perturbative corrections, use a local Wilsonian effective action with a nonzero infrared cutoff and keep the same elementary fields. Introduce nondynamical background chiral superfields $X,Y$ into its F-term part:One may additionally promote each polynomial coefficient to its own chiral spurion to track flavor charges. Treating the couplings as fields ensures that a local chiral integral depends holomorphically on $\Phi ,X,Y$: antichiral dependence would violate the chirality constraint unless accompanied by derivatives or rewritten as a D-term. This is the key use of holomorphy, rather than an assumption that all effective-action terms are holomorphic.

Give $\theta$ R-charge one, the matter fields charge zero, $X$ charge two and $Y$ charge zero. Then the chiral measure has charge minus two, $X{W}_{\mathrm{tree}}$ has charge two, and ${\mathcal{W}}_{\alpha }$ has charge one. These assignments define formal spurionic symmetries even if fixed numerical couplings do not enjoy the corresponding physical symmetry. An anomalous transformation is accompanied by the appropriate shift of the gauge-coupling spurion; nonperturbative dependence can record that anomaly.

For the superpotential non-renormalization argument, perturbation theory is insensitive to continuous shifts of the topological angle, which shifts the imaginary part of $Y$. A holomorphic coefficient invariant under every such shift is independent of $Y$ by the Cauchy-Riemann equations. This proves perturbative gauge-coupling independence of the Wilsonian superpotential. At fixed infrared cutoff the perturbative expansion is regular near zero superpotential couplings. R-charge two therefore forces an allowed correction to be linear in $X$, rather than involving $X^2$, $\bar{X}$ or inverse powers of $X$. Flavor charges impose the corresponding monomial selection rules.

The remaining coefficient can be evaluated at zero gauge coupling. Free chiral propagators join a chiral and an antichiral endpoint. A single holomorphic superpotential vertex cannot close a loop; adding antichiral vertices violates holomorphy, while adding holomorphic vertices violates the linear spurion degree. Consequently the coefficient is its tree value. This proves, rather than merely quotes, the perturbative resultThe fixed-cutoff regularity matters: integrating out a whole massive multiplet can introduce inverse masses at tree level, and infrared-singular one-particle-irreducible expressions are not this local Wilsonian superpotential.

For the gauge kinetic function, R-charge zero excludes regular positive powers of $X$ in its perturbative coefficient. Flavor spurions cannot repair this without forbidden antichiral or singular dependence. A topological-angle shift changes the tree gauge term only by its topological contribution, so the correction $f-Y$ is invariant under the shift. Holomorphy then makes it independent of $Y$. An $L$-loop correction to the inverse gauge coupling carries the weak-coupling behavior $g_h^{2L-2}$: only one loop has the required coupling-independent behavior. Hence the one-loop exactness of the holomorphic gauge kinetic function givesup to a one-loop matching constant. The coefficient follows from the usual one-loop field contributions: gauge bosons and an adjoint Weyl gaugino contribute $(11/3-2/3)C_2=3C_2$, while a chiral Weyl fermion and complex scalar contribute $-(2/3+1/3)T(R)=-T(R)$. Holomorphic logarithmic thresholds appear when massive fields are eliminated. This statement is about the holomorphic Wilsonian coupling, not a canonically normalized physical gauge coupling, whose running need not be one-loop exact.

The Kähler potential is integrated over all of superspace. It can depend on $X^{†}X$, on both $Y$ and $Y^{†}$, and on both matter chiralities. These neutral combinations allow ordinary loop corrections, such as a Yukawa-dependent correction to ${\Phi }^{†}\Phi$, and no analogous argument forbids higher loops. ThusIts wave-function renormalization changes physical superpotential couplings after canonical normalization even though holomorphic vertex coefficients do not change. For a single field with kinetic coefficient $Z$, for instance, $m_c=m/Z$ and $y_c=y/Z^{3/2}$.

Finally consider perturbative renormalization of a Fayet–Iliopoulos term in an Abelian extension. A term $\int d^4\theta H(X,X^{†},Y,Y^{†})V$ must be invariant under $V\mapsto V+\Lambda +{\Lambda }^{†}$ for arbitrary chiral $\Lambda$. Projection of the variation requires ${\bar{D}}^{2}H=D^{2}H=0$. An arbitrary algebraic function of background chiral spurions and their conjugates fails this requirement unless it is constant. Therefore a general coupling-dependent local FI counterterm is forbidden. The one-loop $D$ tadpole from charged scalars can still giveIt is independent of interaction couplings in the connection normalization used above and proportional to the trace of the Abelian charge generator. Higher-loop diagrams require interactions and cannot produce the permitted coupling-independent additive term. Thus a traceless Abelian charge generator has no additive perturbative FI renormalization in this exactly supersymmetric setting; otherwise a one-loop, regulator-dependent tadpole is possible. Vector-field normalization can change the reported canonical parameter, and soft breaking falls outside this proof. For the specified simple group, gauge invariance forbids $\xi$ before and after corrections.

These are perturbative statements. Instantons or other nonperturbative effects may produce symmetry-allowed holomorphic terms, for example dependence on $e^{-8{\pi }^{2}Y}$, so neither the superpotential result nor the one-loop gauge result should be misread as forbidding all nonperturbative corrections.

Assume global four-dimensional N=1 supersymmetry with canonical kinetic terms and no explicit gaugino mass. Write $W_i={\partial }_{i}W$ and $W_{ij}={\partial }_i{\partial }_{j}W$. To match the displayed stationarity equation, use $F^i=\overline{W_i}$; this differs by a minus sign from the eliminated auxiliary field in the component convention of Solution 2. The real D-terms are $D^a=-g^a({\bar{\varphi }}_i(T^a{)}^i{}_j{\varphi }^j+{\xi }^a)$, with Hermitian gauge generators. A constant Fayet–Iliopoulos term is permitted only for an appropriate Abelian factor. The scalar potential isDifferentiate with respect to ${\varphi }^i$, holding $\bar{\varphi }$ independent:Thus stationary point conditions gives the first requested identity. Gauge invariance of the superpotential gives $W_i(T^a{)}^i{}_j{\varphi }^j=0$. Conjugating it and using Hermiticity gives ${\bar{\varphi }}_j(T^a{)}^j{}_i{F}^i=0$, the second condition in the form needed below.

At the vacuum, define $H_{ij}=⟨W_{ij}⟩$ and $G_{ia}=g^a⟨{\bar{\varphi }}_j⟩(T^a{)}^j{}_i$. The two conditions combine asHere the upper equation is stationarity and the lower equation is conjugated gauge invariance. The chiral-superfield fermion mass matrix contributes $-\frac{1}{2}{H}_{ij}{\psi }^i{\psi }^j$, and the gauge Yukawa interaction contributes the mixed chiral-fermion/gaugino bilinear. With canonical Weyl fermion and gaugino kinetic terms, and choosing the gaugino phase to put the mixed bilinear in real-sign convention, these terms areThe factor $\sqrt{2}$ is important: the matrix directly acting on the printed $(F,D)$ is the same mass bilinear with rescaled gauginos, rather than literally the canonical matrix. In detail, for $S=\mathrm{diag}(I,\sqrt{2}I)$,Since supersymmetry breaking makes this vector nonzero, there is at least one zero mass eigenvalue. A normalized massless fermion isWith the consistent gaugino phase, the inhomogeneous vacuum supersymmetry transformation of $(\psi ,\lambda )$ is proportional to the same vector $(F,D/\sqrt{2})$. Therefore $\eta$ shifts by a nonzero constant times the transformation parameter: it is the Goldstino, the Goldstone fermion of broken global supersymmetry. The argument guarantees at least one Goldstino zero mode, not the absence of other massless fermions.

For the second calculation, first take only canonical chiral superfields and F-term breaking. At a stationary vacuum $v$, let $h=\varphi -v$, $m_{ij}=W_{ij}(v)$ andThese follow by differentiating $V={\sum }_k|W_k{|}^2$ twice. The quadratic scalar potential isThe doubled scalar mass matrix has blocksTransforming to the $2N$ canonically normalized real components of $h$ preserves its trace, so the sum of real scalar mass squares is $\mathrm{tr}{M}_s^2=2\mathrm{tr}A$. The Weyl fermion mass squares are the eigenvalues of $m^{†}m=A$ and their sum is $\mathrm{tr}A$. Each Weyl mass eigenstate has two physical spin states. The printed supertrace has scalar sign $-1$ and fermion weight $+2$, henceThe frequently used boson-minus-fermion definition has the opposite overall sign and gives the same zero. The mixing block $B$ can split scalar masses but cannot change this trace. This proves the tree-level supertrace mass sum rule; it uses canonical kinetic terms and a renormalizable global theory, rather than a general supergravity or noncanonical effective theory.

In the MSSM, trying to make all squarks and sleptons heavy through direct canonical tree-level F-term breaking encounters this constraint. In an unbroken electric-charge and color sector, the scalar mass-squared trace is twice the trace for its ordinary fermionic partners. More explicitly, take a unit vector $u$ minimizing $u^{†}Au$. For $h=e^{i\alpha }u$, choose $\alpha$ so that $\mathrm{Re}(e^{2i\alpha }{u}^{T}Bu)\le 0$. The curvature along this real scalar direction is then at most $u^{†}Au$, so at least one scalar mass square is no larger than the smallest fermion mass square in that sector. For one isolated pair the scalar masses are $m_f^2\pm |B|$. Thus light quarks can force an unacceptably light squark, or a tachyon, under these assumptions. A hidden supersymmetry-breaking sector with mediation produces effective soft supersymmetry breaking; radiative effects, noncanonical kinetic terms, or supergravity can change the assumptions and allow a viable spectrum. The trace over an arbitrary enlarged theory alone would not locate the light state in the visible sector; the MSSM conclusion uses the conserved-charge block structure.

Write $S={\Phi }_0$, $u={\Phi }_{+}$, $v={\Phi }_{-}$, $M=M_p$, $P=uv-\zeta$, $x=|S{|}^2/M^2$ and $q=|u{|}^2+|v{|}^2$. Assume $g\ne 0$ and, initially, $\zeta \ne 0$. The supergravity F-term potential isThe canonical Kähler potential gives the identity Kähler metric. The three Kähler covariant derivatives of a superpotential areSubstituting, including the negative term rather than treating the potential as just a sum of squares, gives the exact answerHere $(1+x{)}^2-3x=1-x+x^2$ explains the first coefficient. It is strictly positive for real $x$, since $1-x+x^2=(x-1/2{)}^2+3/4$. Thus this model's scalar potential is nonnegative everywhere.

In the stated limit $q/M^2\ll 1$, the exponential can be replaced by $e^x[1+O(q/M^2)]$. The other displayed terms retain the effects of the Kähler covariant derivatives of a superpotential. Small waterfall fields alone do not justify discarding $\zeta /M^2$ in those derivatives, or expanding in $x$. If also $|\zeta |/M^2\ll 1$ and $x\ll 1$, the leading result reduces to the global F-term scalar potentialNo gauge sector or gauge coupling is supplied, so there is no specified D-term contribution to add.

The supersymmetric vacuum manifold of a bilinear hybrid superpotential isIndeed $W=0$ and all three Kähler covariant derivatives of a superpotential vanish there, so all supergravity auxiliary fields vanish and supersymmetry is unbroken. The nonnegative scalar potential makes this a global minimum. Conversely zero potential requires $uv=\zeta$, and then $|S{|}^{2}q=0$; for $\zeta \ne 0$ this forces $S=0$. With just the supplied fields this is a complex vacuum manifold before imposing any gauge quotient: the product is fixed but the ratio of $u$ and $v$ is not. More precisely, it is an F-flat vacuum manifold whether or not a gauge symmetry is specified. If one additionally gauges opposite charges, D-flatness typically imposes $|u|=|v|$, giving $|u|=|v|=\sqrt{|\zeta |}$ up to phases and gauge equivalence; that is an extra assumption, not a consequence of $K$ and $W$ alone.

The other familiar branch of F-term hybrid inflation has $u=v=0$. Its nonzero derivative $D_{S}W=-g\zeta (1+x)$ establishes supersymmetry breaking, and its exact scalar potential isThe canonical-Kähler correction to an F-term hybrid valley begins asIn particular the inflaton quadratic mass cancels at this order; the first lift of the global flat valley is quartic in $|S|$.

To identify when this branch is a transverse minimum, rephase the waterfall fields so that $\zeta$ is real positive. To quadratic order in $u,v$, the global scalar potential isThe normalized combinations $(u\pm \bar{v})/\sqrt{2}$ diagonalize this quadratic form, withThus $|S{|}^2>|\zeta |$ gives a valley of transverse minima, while below the threshold the negative eigenvalue produces the waterfall instability in F-term hybrid inflation.

The exact supergravity correction to this stability test is also available without taking $x$ small. Put $b=|\zeta |/M^2$. Expansion of the exact supergravity F-term potential about $u=v=0$ gives equal diagonal coefficients $|g{|}^2{M}^2{e}^x[x+b^2(1+x^2)]$ and off-diagonal magnitude $|g{|}^2{M}^2{e}^{x}b(1+x+x^2)$. HenceFor the usual sub-Planckian symmetry-breaking scale $0<b\ll 1$, the lower eigenvalue is $|g{|}^2{M}^2{e}^x(1-b)[x-b(1+x^2)]$. It is positive for the familiar sub-Planckian valley $b\ll x\ll 1$, and changes sign near $x=b$.

The printed assertion of two full minima requires qualification. The broken branch is a transverse inflationary valley, not generally a second stationary minimum of the complete supergravity potential. In factIts only stationary point in $S$ is the origin. For $0<|\zeta |<M^2$ that point has a negative waterfall eigenvalue $|g{|}^2(|\zeta {|}^2/M^2-|\zeta |)$. A direct counterexample is the path $S=0$, $u=v=a$ with real small $a$, after rephasing $\zeta >0$:The negative quadratic coefficient proves it is not a minimum. No loop corrections or additional stabilizing interactions were specified that could create the requested second stationary vacuum. The mathematically supported interpretation is a SUSY-breaking transverse valley and a zero-energy supersymmetric vacuum minimum manifold. The limit on $u,v$ alone does not fix the parameter range $|\zeta |/M^2$; the familiar waterfall conclusion uses the sub-Planckian range just stated.

In the corresponding global theory the exponential prefactor, Kähler corrections to $D_{i}W$ and universal $-3|W{|}^2/M^2$ term are absent. Its $u=v=0$ valley has exactly constant tree-level energy $|g\zeta {|}^2$ for all $S$, whereas canonical supergravity lifts it by the quartic correction above. The supersymmetric vacuum $S=0$, $uv=\zeta$ and its zero energy persist in both descriptions. If $\zeta =0$, the origin and the $S$ axis instead have vanishing F-terms; the broken-valley interpretation does not apply.

Use the scalar components of the three chiral superfields, and set $a=|{\varphi }_0{|}^2$, $b=|{\varphi }_{+}{|}^2$, $c=|{\varphi }_{-}{|}^2$ and $\Sigma =a+b+c$. We assume the physical sign $g>0$ and a nonzero coupling $\lambda$. The coefficient $g$ below is exactly the coefficient in the supplied D-term scalar potential; it should not be silently replaced by a differently normalized squared gauge coupling. The canonical Kähler potential has identity Kähler metric, and the three Kähler covariant derivatives of a superpotential areSubstitution in the supergravity F-term potential givesIn particular, the subtraction cancels only half of the six ${\kappa }^{2}abc$ terms contributed by the three squares. The full scalar potential isIts nonnegativity follows term by term from $a,b,c\ge 0$ and $g>0$. This is a special property of this cubic superpotential, not of a general supergravity F-term potential.

The D-term vacuum branches of a cubic superpotential must be described as branches: there are not two isolated field-space minima. First suppose $\zeta >0$. Vanishing of $V_F$ requires $ab=ac=bc=0$, and vanishing of the D-term scalar potential requires $b-c=\zeta$. Together these conditions give the vacuum expectation valuesThe phase of ${\varphi }_{+}$ lies on a gauge orbit, and one may choose it positive and real. At this global minimum, all $D_{i}W$ and the real auxiliary field of the vector multiplet vanish. It is a supersymmetric vacuum, even though the internal gauge symmetry is spontaneously broken. For $\zeta <0$, interchange ${\varphi }_{+}$ and ${\varphi }_{-}$: the nonzero vacuum expectation value has magnitude $\sqrt{-\zeta }$. For $\zeta =0$, the supersymmetric vacuum is the entire family ${\varphi }_{+}={\varphi }_{-}=0$, with ${\varphi }_0$ arbitrary.

The other branch has ${\varphi }_{+}={\varphi }_{-}=0$ and arbitrary ${\varphi }_0$. Here $W=D_{i}W=0$ but the real auxiliary field is nonzero for $\zeta \ne 0$, andThus this branch has pure D-term supersymmetry breaking and a flat direction of a scalar potential in the neutral scalar. It is not stable at every neutral-field value. At a fixed $a=|{\varphi }_0{|}^2$, expansion in the charged scalars yieldsHence the two charged squared masses and the stability condition areAbove this threshold, both charged directions have positive mass squared. Continuity makes them uniformly positive in a sufficiently small neighborhood of the chosen neutral-field value, so the stationary branch is a genuine, non-strict local minimum valley. Its neutral direction stays flat. Below the threshold, a charged mass squared is negative and the branch is a saddle point. At equality it is a boundary of the stable valley, not a local minimum in the full field space: changing $a$ slightly downward and switching on the unstable charged field lowers $V$. For example, for $\zeta >0$, put $a=a_c-ϵ$, $b={ϵ}^2$, $c=0$. Since $M(a)=|\lambda {|}^{2}a{e}^{{\kappa }^{2}a}$ has $M^{\prime }(a_c)>0$, one finds $V-g{\zeta }^2=-M^{\prime }(a_c){ϵ}^3+O({ϵ}^4)<0$.

There are no further local-minimum branches under these generic assumptions. If $b+c>0$, the expression for $V_F$ is strictly increasing in $a$, so a stationary configuration with a charged condensate must have $a=0$. At $a=0$, an interior critical point with $b,c>0$ is impossible, becauseOn the boundary $c=0$, the relevant scalar potential is $g(b-\zeta {)}^2$; on $b=0$ it is $g(c+\zeta {)}^2$. Their nonnegative-domain minima give exactly the supersymmetric vacuum described above, or the origin of the charged-field branch. Thus, for nonzero $\zeta$, the two types are the zero-energy supersymmetric gauge orbit and the positive-energy D-breaking minimum valley above the threshold. If $\zeta =0$, the two types coalesce into the flat supersymmetric vacuum; the literal two-minimum assertion needs this parameter qualification.

In the global supersymmetry limit $\kappa \to 0$, the F-term scalar potential becomes $|\lambda {|}^2(ab+ac+bc)$ and the D-term scalar potential retains its form. The same two types of branches occur, with the threshold $|\lambda {|}^{2}a>2g|\zeta |$. Supergravity instead introduces the exponential and additional positive terms. The goldstino of global supersymmetry breaking is a physical fermion; in local supersymmetry it participates in the super-Higgs mechanism. A positive vacuum energy also gravitates in supergravity. In particular, the positive-energy branch here has $W=0$ and hence vanishing gravitino mass from a superpotential; it is not the tuned zero-energy broken vacuum of Question 3.

There is a further gauge-consistency qualification. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and requires a corresponding nontrivial transformation of the superpotential. The apparent charges $0,+1,-1$ make the supplied cubic superpotential neutral, so a nonzero independent constant $\zeta$ does not by itself specify a consistent conventional supergravity action. The computed branches are those of the supplied scalar potential; a full embedding needs compatible charges or additional fields. This restriction is explained in Van Proeyen's derivation of the superpotential gauge-covariance condition.

More generally, the supergravity moment-map constraint on D-term breaking follows directly from gauge covariance. If $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and the moment map is $\mathcal{P}=i(k^i{K}_i-r)$, then at a point with $W\ne 0$,Consequently all chiral auxiliary fields vanishing implies the D-term also vanishes there. This relation has no analogue forcing that implication in an arbitrary global supersymmetry model. Its $W\ne 0$ hypothesis matters: the formal D-breaking branch above has $W=0$, so it does not contradict the identity. These conventional gauge-completion restrictions are additional to the elementary minimization calculation.

The chiral superfield condition is $\boxed{{\bar{D}}_{\dot{\alpha }}\Phi =0}$ for both dotted components. Use left Grassmann derivatives with the displayed supersymmetric covariant derivatives. The odd product rule gives $\partial (\theta {\sigma }^{\mu }\bar{\theta })/\partial {\bar{\theta }}^{\dot{\alpha }}=-(\theta {\sigma }^{\mu }{)}_{\dot{\alpha }}$, soThe chirality constraint therefore says that $\Phi$ is independent of $\bar{\theta }$ at fixed $y$. The Grassmann algebra has only two independent $\theta$ components, so every cubic product of them vanishes. Its general even scalar superfield consequently has a scalar coefficient, a term linear in $\theta$ with a Weyl spinor coefficient, and one quadratic term. Choosing the conventional component normalization gives the chiral-superfield component expansionHere $\varphi$ is a complex scalar field, $\psi$ an odd Weyl spinor and $F_{\mathrm{aux}}$ a complex nonpropagating auxiliary field. Conversely this expansion is annihilated by $\bar{D}$ because $y$ and $\theta$ are. At fixed ordinary $x$, the same result is the terminating translation $\Phi =e^{i\theta {\sigma }^{\mu }\bar{\theta }{\partial }_{\mu }}[\varphi (x)+\sqrt{2}\theta \psi (x)+(\theta \theta )F_{\mathrm{aux}}(x)]$; this retains the printed contraction convention without changing the order in the barred square.