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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.

Eliminating the auxiliary field gives $F^{\ast }=-W^{\prime }(\phi )$ and the nonnegative F-term scalar potentialUse $u=\phi -s$ after the critical-point shift, so thatThis notation keeps the original parameters distinct from the shifted mass coefficient $M$. For real $M,g$ with $Mg\ne 0$, the real slice is $V(u)=u^2(M+gu/2{)}^2$. It has zero-energy minima at $u=0$ and $u=-2M/g$, a maximum between them at $u=-M/g$ with height $M^4/(4g^2)$, and grows quartically toward both ends. These statements follow by differentiating: $V^{\prime }(u)=2u(M+gu/2)(M+gu)$.

When $M=0$ it becomes $|g{|}^2{u}^4/4$, with one degenerate zero minimum. When $g=0$, it is $|M{|}^2{u}^2$ after a permissible shift. For the real parameters with negative discriminant, the original real slice can have a positive minimum even though the full complex scalar potential has zero-energy vacua away from that slice. The sketch distinguishes these regimes:

The plots use illustrative normalized coefficients, not values fixed by the exam. A positive minimum on a chosen real slice is not evidence for supersymmetry breaking in the full complex field space.

In the Polonyi model, the Kähler metric is $K_{z\bar{z}}=1$. The relevant Kähler covariant derivative of a superpotential isThe supergravity auxiliary field is $F^z=-e^{|z{|}^2/2}\overline{D_{z}W}$, up to an irrelevant common phase convention. Thus a constant vacuum preserves supersymmetry exactly when this auxiliary field vanishes. For $m=0$, the superpotential and scalar potential vanish identically, and every constant scalar value is a supersymmetric vacuum.

For $m\ne 0$, put $⟨z⟩=x+iy$. Its supersymmetry condition becomesSince $\beta >0$, it requires $y=0$ and $x^2+\beta x+1=0$. Hence the Polonyi supersymmetry branches areAt these points $W\ne 0$, so the supergravity F-term potential is negative, $V=-3e^K|W{|}^2$: these are supersymmetric Anti-de Sitter spacetime vacua, not zero-energy ones. The condition $D_{z}W=0$ also makes them stationary, as follows by differentiating the supergravity F-term potential.

For $0<\beta <2$ and $m\ne 0$, the auxiliary field cannot vanish anywhere, so any vacuum has supersymmetry breaking. For $\beta \ge 2$, a stationary vacuum at any other scalar value still breaks supersymmetry; the parameter condition alone does not determine which vacuum is selected. In particular, a nontrivial zero-energy vacuum cannot preserve supersymmetry: $D_{z}W=0$ and $V=0$ would also imply $W=0$, whereas $W=0$ gives $z=-\beta$ and $D_{z}W=m^2\ne 0$.

Use the minimal four-dimensional Super-Poincaré algebra, with $\{Q_{\alpha },Q_{\beta }\}=0$, $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2{\sigma }_{\alpha \dot{\beta }}^{\nu }{P}_{\nu }$ and commuting translations. Lorentz covariance and closure on the existing supercharges permit onlyThe graded Jacobi identity for $P^{\mu },Q_{\alpha },Q_{\beta }$ therefore givesFor example, with ${\sigma }^0=I$, $\mu =0$ and $\alpha =\beta =1$, the coefficient is a nonzero multiple of $A(P_1-i{P}_2)$. Independence of the translation generators forces $A=0$. Its adjoint gives the barred result. Thus every supercharge commutes with four-momentum:Consequently $[Q_{\alpha },P^2]=0$. If $|u⟩$ has a particular mass-shell condition, the nonzero state $Q_{\alpha }|u⟩$ has the same one. This gives supersymmetric mass degeneracy within an unbroken physical supermultiplet. The qualification matters: after spontaneous supersymmetry breaking, the chosen vacuum is not annihilated by the supercharges, and its one-particle excitations need not constitute degenerate supermultiplets.

For the O'Raifeartaigh model, write $X={\varphi }_1$, $Y={\varphi }_2$, $Z={\varphi }_3$. Field phases allow $g,M,m$ to be positive. The canonical Kähler potential and elimination of the auxiliary fields give the F-term scalar potentialA supersymmetric vacuum would require $W_Y=MZ=0$, hence $Z=0$, while $W_X=g(Z^2-m^2)=-g{m}^2$ could not vanish. Thus F-flatness is impossible. More precisely, with $Z=u+iv$,The stated strict inequality makes every nonconstant term nonnegative. The entire classical vacuum family isHere $F_X=-\overline{W_X}=g{m}^2$, while $F_Y=F_Z=0$. The Weyl spinor ${\psi }_X$ is the massless goldstino, and the complex scalar field $X$ is a pseudomodulus with two zero tree-level squared masses. The hierarchy $M\gg m$ separates the massive fields from the breaking scale, but it does not itself select $x=0$ along this flat direction.

At the representative vacuum $x=0$, the chiral-superfield fermion mass matrix isIts physical masses are its singular values: $0,M,M$. The two massive Weyl spinors can be combined into one massive Dirac spinor. For $Y=(y_R+i{y}_I)/\sqrt{2}$, $Z=(z_R+i{z}_I)/\sqrt{2}$, the real scalar squared masses areIn particular, there is no tachyon under the stated inequality. The splitting of the $Z$ masses displays supersymmetry breaking directly.

For completeness, the mass spectrum can be given throughout the classical vacuum family, rather than silently fixing the pseudomodulus. A phase rotation makes $x$ real and nonnegative without changing physical masses. The massive fermion squared masses areThe real and imaginary scalar sectors have two-by-two scalar squared-mass matriceswhose eigenvalues areBoth determinants are $M^2(M^2+2ϵg^2{m}^2)>0$ and their traces are positive, so stability holds at every finite $x$. Quantum corrections can lift a pseudomodulus; these formulas describe the classical, tree-level spectrum requested here.

Define the physical mass supertrace byEach real scalar contributes once and each Weyl spinor twice with a minus sign. At $x=0$ this gives $4M^2-2(2M^2)=0$. At general $x$, ${\sum }_{ϵ,\pm }{b}_{ϵ,\pm }=4M^2+8g^2|x{|}^2=2(f_{+}+f_{-})$. Hence the tree-level mass supertrace vanishes despite broken supersymmetry:This is the tree-level supertrace mass sum rule, not a claim that the individual masses are equal.

For direct breaking in the Minimal supersymmetric Standard Model, the difficulty is both field content and the tree-level spectrum. A linear gauge-invariant superpotential term needs a gauge-singlet chiral superfield, which the minimal model does not contain. Moreover, with canonical kinetic terms and purely neutral F-term breaking, without D-term mass shifts, the MSSM tree-level sfermion mass constraint applies separately to conserved charge sectors. For one electron pair it gives $m_{{\tilde{e}}_1}^2+m_{{\tilde{e}}_2}^2=2m_e^2$: both scalar partners cannot be heavy. This is an illustrative charge-sector consequence under those assumptions, not an unrestricted inference from the total supertrace alone. A separate hidden supersymmetry-breaking sector communicating effective soft supersymmetry breaking through loops or suppressed operators avoids the direct canonical tree-level obstruction.

In the conventions $\{Q_{\alpha },Q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^{\mu }{P}_{\mu }$, the trace over the two spinor components of the Super-Poincaré algebra givesHence, on a normalized state in the domain of the supercharges,This proves energy positivity in global supersymmetry. The zero of energy is fixed by the algebra; adding an arbitrary constant to $H$ while keeping that algebra unchanged is not allowed.

For a finite-dimensional supermultiplet at fixed positive energy and four-momentum, choose a nonzero component $q$ with $\{q,q^{†}\}=cI$, $c>0$, and $q^2=0$. The Hermitian odd operatorIt is therefore an invertible map from the even fermion parity subspace to the odd subspace and conversely. This proves boson-fermion degeneracy in a supermultiplet, $n_B=n_F$, for every positive-energy physical particle supermultiplet. Equivalently, the supertrace of the supercharge anticommutator vanishes because odd operators interchange these two subspaces. A zero-energy vacuum is an exception to the literal claim about every possible representation: it can be a one-dimensional bosonic singlet with all supercharges acting trivially.

For a vacuum $|\Omega ⟩$, the sum-of-squares formula gives $E_0=0$ exactly when all supercharges and their adjoints annihilate it. If global supersymmetry is spontaneously broken and a vacuum exists, at least one such norm is nonzero, so the vacuum energy is positive. In infinite volume this is the statement about the vacuum energy density, understood with a finite-volume regulator. Explicit supersymmetry breaking need not preserve the algebra used in the proof. The conclusion also does not extend to the supergravity F-term potential, whose negative $-3e^K|W{|}^2$ term permits, for example, the zero-energy broken vacua of no-scale supergravity.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.

Use left Grassmann derivatives for the Grassmann variables. A chiral superfield satisfies ${\bar{\mathcal{D}}}_{\dot{\alpha }}\Phi =0$. In the derivative convention printed in the PDF, differentiating $\theta {\sigma }^{\mu }\bar{\theta }$ with respect to a barred coordinate brings a minus sign from moving the odd derivative through $\theta$. Consequently ${\bar{\mathcal{D}}}_{\dot{\alpha }}{y}^{\mu }=0$, so the general solution depends only on $y$ and $\theta$:Here $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor and $F$ a complex auxiliary field. This is a finite Grassmann algebra expansion, not an assumption that the spacetime fields are constant.

For a completely convention-independent way to organize the ordinary-coordinate expansion, let $v^{\mu }=\theta {\sigma }^{\mu }\bar{\theta }$. Translation by $iv$ givesAll higher terms vanish because there are only two unbarred and two barred odd coordinates. For example take ${\sigma }^{\mu }=(I,\mathbf{\sigma })$, signature $(+---)$, ${\theta }^2={\theta }^{\alpha }{\theta }_{\alpha }$ and ${\bar{\theta }}^2={\bar{\theta }}_{\dot{\alpha }}{\bar{\theta }}^{\dot{\alpha }}$, with both lower epsilon tensors having ${ϵ}_{12}=1$. Then $v^{\mu }{v}^{\nu }=\frac{1}{2}{\theta }^2{\bar{\theta }}^2{\eta }^{\mu \nu }$ and the chiral-superfield component expansion readsThe frequently written positive quarter coefficient corresponds to the opposite convention for $\square$ or the spinor contractions; the preceding shift formula fixes the sign without ambiguity. The local TeX's barred derivative is damaged: the original PDF has ${\partial }_{\mu }$, not a barred-coordinate factor at its end.

Write $X=x+\sqrt{2}\theta G+{\theta }^2{F}_X$ in chiral coordinates. Multiplication gives the three component constraintsOn the regular branch where the ordinary commuting part of $F_X$ is nonzero, the nilpotent chiral superfield is thereforeThe goldstino $G$ and auxiliary field $F_X$ are independent; the scalar is a composite, not another independent degree of freedom. Its first two constraints follow because any product of three identical two-component odd spinor entries vanishes. It is essential not to divide by $F_X$ on a branch with zero commuting part: for example $X=0$ satisfies the nilpotency condition without breaking supersymmetry.

For a regular local action with no superspace derivatives in $K$ or $W$, nilpotency truncates the most general superpotential and Kähler potential toThe last inequality makes the kinetic Kähler metric healthy. In global four-dimensional N=1 supersymmetry, the constant and holomorphic pieces of $K$ integrate to zero, and $w_0$ does not affect the action. On the bosonic background $G=0$, the constraint sets $x=0$ and the auxiliary terms are $\kappa F_X{\bar{F}}_X+f{F}_X+\bar{f}{\bar{F}}_X$. Eliminating them givesA nonzero $f$ is necessary for this regular constrained branch to be an on-shell vacuum. It gives nonzero $F_X$ and supersymmetry breaking, with $G$ the goldstino. Nilpotency alone is not a proof that every possible branch breaks supersymmetry. With $f=0$ the above composite parametrization is singular on shell, rather than evidence for a regular supersymmetric vacuum containing an independent Goldstino.

If these data are instead embedded in supergravity, the constant and linear terms cannot simply be discarded independently: a Kähler transformation also transforms $W$. Before such a transformation, at $x=0$ the supergravity F-term potential isThus nonzero $F^X$ still breaks supersymmetry, but the vacuum energy need not be positive. This distinguishes the global answer from the local theory explicitly introduced in the next question.

Now write $Y=y+\sqrt{2}\theta \chi +{\theta }^2{F}_Y$. The condition $XY=0$ becomesUsing the same nonzero-$F_X$ branch gives the chiral superfield constrained by a nilpotent superfieldThe spinor $\chi$ and the auxiliary $F_Y$ remain independent; the scalar is removed. Substituting this expression verifies the other two product constraints using two-component spinor identities. It does not impose $Y^2=0$. It does imply cubic nilpotency from a mixed chiral constraint: $Y^3=0$. Indeed $(G\chi {)}^2=-\frac{1}{2}(GG)(\chi \chi )$, soThese are exactly the three components needed for $Y^3=0$. The generic nonzero expression for $y^2$ shows why imposing quadratic nilpotency on $Y$ would lose allowed interactions.

The analytic chiral monomials consequently reduce to $1,X,Y,Y^2$. The most general superpotential isFor the Kähler potential, put $U=(X,Y,Y^2{)}^T$. Every allowed mixed monomial is included inThis includes $X\bar{X}$, $X\bar{Y}$, $X{\bar{Y}}^2$, $Y\bar{Y}$, $Y{\bar{Y}}^2$, $Y^2{\bar{Y}}^2$ and the necessary conjugates; products containing $XY$ or $\bar{X}\bar{Y}$ vanish. In a global theory the holomorphic $\ell$ terms again integrate to zero; in supergravity they can be changed by a Kähler transformation. Positivity requires the kinetic metric on the independent multiplets to be positive definite at the chosen background, not necessarily the entire coefficient matrix on the redundant composite list $U$. The most general claim here concerns regular analytic two-derivative actions; singular functions of the constrained fields or higher-derivative operators are outside it.

The exponent in the original PDF is $e^{-\alpha S}$ with real $\alpha >0$, not the $e^{-aS}$ produced by the local TeX. Use that authoritative expression. PutThe domain conditions ensure a real Kähler potential and positive Kähler metric. There are no gauge multiplets specified, so the scalar potential is the supergravity F-term potentialThe Kähler covariant derivative of a superpotential and metric entries areState the inverse with its indices explicitly, since transposing the off-diagonal complex entries would change the answer:Here ${\sum }_j{K}_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$, so the displayed upper-index array is the transpose of the ordinary matrix inverse of the displayed lower-index array.

Substitution shows that the cross terms involving $W$ cancel and that the $T,C$ sector givesThe first term cancels the universal negative term. Thus the no-scale supergravity potential isThis positivity is an algebraic cancellation, not an assumption that the individual Kähler derivatives vanish.

Choose the supergravity auxiliary field convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. Direct multiplication givesA conventional common phase or overall sign on the auxiliary fields changes none of the vanishing conditions. A supersymmetric configuration requires every $F^i$ to vanish. First $F^C=0$ implies $C=0$, then $F^T=0$ implies $w=0$, and $F^S=0$ implies $w_S=0$. Since $w_S=-\alpha a{e}^{-\alpha S}$, at a finite point of the physical domain this is possible only when $a=0$, followed by $b=0$. Consequently If $(a,b)\ne (0,0)$, every finite configuration has a nonzero auxiliary field, so any finite vacuum breaks supersymmetry. The printed request cannot hold for arbitrary parameters without this exception: $a=b=0$, $C=0$ is a zero-energy supersymmetric family with both $S$ and $T$ unfixed.

With the stipulated vanishing vacuum expectation value of $C$, define $A(S)=b+(1+\alpha s)a{e}^{-\alpha S}$. The potential reduces to $|A{|}^2/(s{t}^3)$. Its derivative along $t$ is $-3V/t$, so a finite stationary point must have $A=0$. Such a point is a global minimum of the full nonnegative potential, since both squares vanish at $C=0$. The finite zero-energy vacuum for a single exponential superpotential therefore satisfiesFor nonzero $a,b$, write $S=\sigma +i\chi$ and $x=\alpha \sigma >0$. The modulus and phase conditions areThe function $(1+2x)e^{-x}$ increases up to $x=1/2$ and then decreases to zero; its maximum is $2e^{-1/2}$. Hence a finite minimum exists exactly whenThere is one positive solution for $0<\rho \le 1$, two for $1<\rho <2e^{-1/2}$, and one coalesced solution at the upper bound. The formal $x=0$ solution at $\rho =1$ is outside $s>0$. At each allowed $x$, the phase fixes $\chi$ modulo $2\pi /\alpha$. Equivalently $x=-\frac{1}{2}-W_k(-\rho /(2\sqrt{e}))$, using the real branches of the Lambert W function and retaining only $x>0$.

At any of these nontrivial minima, $w=-\alpha sa{e}^{-\alpha S}\ne 0$. The auxiliary fields becomeThis is a supersymmetry breaking Minkowski minimum in which the nonzero auxiliary field belongs to $T$. Both real components of $T$ are exact flat directions of a scalar potential at the minimum. They change the metric and auxiliary-field magnitudes but not the zero potential. Generically the two real components of $S$ are fixed. At the coalesced solution $x=1/2$, the radial quadratic restoring term vanishes, but the leading restoring term is quartic; this is not an additional exact flat direction of a scalar potential. The matter field $C$ likewise has a positive leading quartic potential, not an exact flat direction of a scalar potential, despite its zero quadratic mass here.

The exceptional parameter cases must also be stated. If $a=0,b\ne 0$ or $a\ne 0,b=0$, no finite zero-energy minimum with $C=0$ exists. The same holds when $|b/a|>2e^{-1/2}$. Because the $t$ derivative is nonzero at any positive-energy point on $C=0$, none is a finite minimum there; the energy approaches zero along the runaway $t\to \infty$. If $a=b=0$, all physical $S,T$ at $C=0$ give supersymmetric zero-energy minima. Thus an unconditional finite minimum or unconditional breaking would be a false claim for the printed arbitrary parameters.

For the homogeneous model, assume $\Gamma >0$ is twice differentiable, of degree one in the real moduli ${\tau }_i$, and that its Hessian matrix for $K$ is invertible on the sector considered. Differentiate $K=-3\log \Gamma$:The two Euler theorem for homogeneous functions identities in the question implyMultiplying by the inverse gives $K_{ij}^{-1}{\Gamma }_j/\Gamma ={\tau }_i/3$, so contracting once more yieldsThis is the no-scale identity from degree-one homogeneity. It does not assert that every degree-one function gives a positive or invertible metric: for example $\Gamma ={\tau }_1+{\tau }_2$ has a rank-one Kähler Hessian, and its inverse is undefined. The inverse hypothesis is necessary.

For complex moduli with ${\tau }_i=T_i+{\bar{T}}_i$, the same derivatives are the mixed Kähler metric; choosing real parts with a factor of two only introduces factors that cancel in the contraction. If $W$ is independent of these moduli, $D_{i}W=K_{i}W$, and their contribution is $3e^K|W{|}^2$. It cancels the universal $-3e^K|W{|}^2$, leaving no scalar potential from this isolated no-scale sector. Other chiral sectors can still contribute positive terms. For this conclusion in a larger theory, the displayed metric must be the appropriate decoupled no-scale block, or the full inverse metric must itself satisfy the corresponding identity; arbitrary mixed additions to $K$ do not inherit the cancellation automatically.