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The no-scale supergravity identity cancels the negative gravitino contribution, leaving $V=e^{|C{|}^2}|3C^2+\bar{C}(C^3+B){|}^2/(T+\bar{T}{)}^3$. Zero-energy minima occur at $C=0$ and, for $B\ne 0$, three phase-related nonzero solutions with $|B|=|C|(3+|C{|}^2)$. Their $T$ supergravity auxiliary field is nonzero if $B\ne 0$, while both real components of $T$ are flat on each vacuum branch. For $B=0$, the $C=0$ branch is instead supersymmetric and has zero gravitino mass from a superpotential.

If a no-scale supergravity scalar potential has the displayed form with $y>0$ and nonnegative $A,B$ independent of the volume coordinate $y$, every point with $A+B>0$ has a direction of strictly decreasing potential. Thus a finite stationary vacuum requires $A=B=0$. If that common zero set is empty, the potential has infimum zero along $y\to \infty$ but no finite minimum. If the common zero set is nonempty, varying $y$ on it is a flat direction of a scalar potential. This distinguishes an attained zero-energy vacuum from a decompactification runaway.

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.

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.

Set $x=T+\bar{T}>0$, the domain in which the Kähler potential is defined, and use Planck units. The nonzero entries of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areDenote $W=C^3+B$ and $S=3C^2+\bar{C}W$. The supergravity F-term potential isHere the $T$ contribution is exactly $3|W{|}^2$, cancelling the negative term. This is the no-scale supergravity identity, givingThe cancellation is important: replacing the covariant derivatives by ordinary derivatives would miss both the matter dependence and the possibility of zero vacuum energy with supersymmetry breaking.

In the convention $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$, the chiral supergravity auxiliary fields areAn overall auxiliary phase or sign convention does not change their vanishing conditions. These are upper-index fields, including the inverse Kähler metric, rather than just the covariant derivatives.

Since $V$ is nonnegative and vanishes at $C=0$, its minimum energy isThere are additional minima that should not be discarded by examining only real $C$. For $C=\rho e^{i\alpha }\ne 0$, the condition $S=0$ becomesIf $B\ne 0$, the magnitude equationhas exactly one positive root: its left side is strictly increasing from zero to infinity. There are three phases, differing by $2\pi /3$, determined by $e^{3i\alpha }=-B/|B|$. Thus for nonzero $B$ the finite zero-energy matter vacua consist of $C=0$ and these three nonzero phase-related values, each with arbitrary $T$ in the allowed half-plane. At a nonzero matter vacuum,At $C=0$, $W=B$.

For $B\ne 0$, every such minimum has $F^C=0$ but $F^T\ne 0$, so supersymmetry is spontaneously broken despite zero vacuum energy. Indeed no finite point anywhere is fully supersymmetric: $D_{T}W=0$ would force $W=0$, after which $D_{C}W=3C^2=0$ forces $C=0$ and then $B=0$. This proves breaking for the specified nonzero-constant case without relying just on the positivity of the potential.

For the allowed exceptional value (B=0), the requested breaking statement is false. Then $S=C^2(3+|C{|}^2)$ vanishes only at $C=0$, where $W=F^T=F^C=0$. Every finite allowed $T$ gives an unbroken supersymmetric vacuum with zero energy. This is an explicit counterexample to unqualified breaking for an arbitrary complex $B$.

Both real components of $T$ are flat directions of a scalar potential along every minimum: the imaginary component never appears in $V$, and when $S=0$ the entire $x$ dependence is multiplied by zero. The matter vacua are isolated in the $C$ plane; there is no continuous matter vacuum direction. At $B=0$ the potential begins quartically in $C$, so zero quadratic mass is not a flat vacuum manifold. Away from $S=0$, the real $T$ direction instead has $V\propto x^{-3}$ and a decompactification-type runaway towards zero as $x\to \infty$, not a further finite positive-energy minimum.

The gravitino mass from a superpotential isAt the $C=0$ branch it is $|B|/x^{3/2}$; at each nonzero branch it is $3\rho e^{{\rho }^2/2}/x^{3/2}$. Its value is not fixed because the $T$ modulus is flat. For $B=0$ it vanishes at the supersymmetric minimum. For nonzero $B$, the super-Higgs mechanism absorbs the Goldstino associated with $F^T$ into the massive gravitino. The zero-energy relation is explicitly satisfied:which balances the auxiliary contribution against the gravitational negative term.

Work in the supplied Planck units and write $t=T+\bar{T}>0$. The nonzero components of the Kähler metric and its inverse areThe Kähler covariant derivatives of a superpotential areIn the supergravity F-term potential, $K^{T\bar{T}}|D_{T}W{|}^2=3|W{|}^2$ cancels its negative universal term. This is the no-scale supergravity identity. The exact answer is thereforeIn particular $C=0$ is a global minimum for every $T$ in the domain, with zero vacuum energy.

Use the standard convention for a supergravity auxiliary field, $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. In this conventionAt the $C=0$ minimum these becomeThus supersymmetry breaking occurs if $B\ne 0$, despite the vanishing vacuum energy. Indeed $K_{T\bar{T}}|F^T{|}^2=3|B{|}^2/t^3$ is precisely canceled by the negative gravitino contribution. For $B=0$ the asserted breaking does not occur: both supergravity auxiliary fields vanish at $C=0$, so this vacuum is supersymmetric.

For completeness, the full set of finite zero-energy minima includes additional branches. For $C=r{e}^{i\theta }\ne 0$, the condition $D_{C}W=0$ reduces, after factoring $r{e}^{-i\theta }$, toFor every $B\ne 0$ there is a unique positive solution of $r(3+r^2)=|B|$, since the left side increases strictly from zero to infinity. There are three associated phases,These also give $V=0$, and on each branch $W=3B/(3+r^2)\ne 0$. Consequently $F^C=0$ but $F^T\ne 0$, so these are also supersymmetry breaking no-scale vacua with a cubic matter superpotential. A supersymmetric finite vacuum would require $D_{T}W=0$, hence $W=0$, and then $D_{C}W=3C^2=0$; this is possible only at $C=0$ with $B=0$.

The scalar potential is independent of $\mathrm{Im}T$ everywhere. On every zero-energy vacuum branch it is also independent of $\mathrm{Re}T$, so both real components of $T$ are flat directions of a scalar potential at the vacuum. Away from $D_{C}W=0$, the real direction is not flat: ${\partial }_{t}V=-3V/t$, giving a runaway towards $t\to \infty$ rather than a finite positive-energy stationary point. The zero-energy branches have no continuous flat direction of a scalar potential in $C$; their allowed $C$ values are isolated. This is why the vacuum does not select a unique numerical gravitino mass: the modulus $t$ is unfixed.

The gravitino mass from a superpotential is $m_{3/2}=e^{K/2}|W|$. On the conventional $C=0$ branch,On a nonzero-$C$ branch it is insteadWhen $B=0$, the only finite zero-energy matter solution is $C=0$ and $m_{3/2}=0$, with unbroken supersymmetry. These results exhibit the distinction between vanishing vacuum energy and vanishing supersymmetry breaking auxiliary fields in a no-scale supergravity model.

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.

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.