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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.

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.

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.

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.

For a single chiral superfield, the supergravity F-term potential, with $\kappa =m_p^{-1}$ restored, isThe canonical Kähler potential has $K_{z\bar{z}}=K^{z\bar{z}}=1$, so the Kähler covariant derivative of a superpotential isConsequently the Polonyi model has the scalar potentialThe negative term is essential; retaining only the square of the supergravity auxiliary field would give the global F-term scalar potential rather than the supergravity F-term potential.

Set $x=\kappa z=a+iy$ and $b=\kappa \beta =2-\sqrt{3}$. In this notation $V=|\mu {|}^2{m}_p^2{e}^{|x{|}^2}U$, whereAt $a_0=\sqrt{3}-1$, $y=0$, one has $a_0+b=1$ and $1+a_0(a_0+b)=\sqrt{3}$, so $U=0$. To prove that this is a local minimum in both real directions, rather than checking only the real axis, put $u=a-a_0$. Direct expansion gives the stable zero-energy Polonyi vacuum identityEvery coefficient outside the square is strictly positive, and the exponential factor is positive. Thus $V\ge 0$ everywhere and equality requires $u=y=0$, provided $\mu \ne 0$. This proves the stronger conclusiona unique global minimum of this Polonyi model. If $\mu =0$, the scalar potential vanishes identically and this uniqueness and supersymmetry breaking conclusion do not hold.

For an independent local stability check, the Hessian matrix of $U$ at this global minimum is diagonal:Since $U$ and its first derivatives vanish there, differentiating the exponential introduces no additional Hessian matrix terms at the global minimum. For canonically normalized real fluctuations $z=z_0+(s+it)/\sqrt{2}$, the squared scalar masses are thereforewhich are both positive.

Although the vacuum energy is zero, the supergravity auxiliary field is not. In the convention $F^z=-e^{{\kappa }^{2}K/2}\overline{D_{z}W}$, its vacuum expectation value obeysThe matter fermion transformation contains a term proportional to $F^zϵ$, so no nonzero constant supersymmetry parameter leaves this vacuum invariant. Hence supersymmetry is spontaneously broken even though the vacuum energy vanishes: the positive auxiliary field contribution cancels the negative gravitational term. The matter fermion supplies the goldstino, which is absorbed by the gravitino in the super-Higgs mechanism.

Using the gravitino mass from a superpotential, and $z_0+\beta =m_p$, givesAn auxiliary field $F^z$ has mass dimension two. Defining the supersymmetry breaking scale by ${\Lambda }_{\mathrm{SUSY}}=\sqrt{|F^z|}$, one obtainsThus the gravitino mass is small compared with the supersymmetry breaking scale when that scale is sub-Planckian, as in the usual hierarchy $|\mu |\ll m_p$. This smallness is gravitational suppression; it is not a conclusion valid for arbitrary choices of the mass parameter.

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 $D_{z}W=m^2\ne 0$.

For the stable branch found above, $t=x+\beta =1$ and $s=1$. Hence the vacuum expectation value isin the stated Planck units. To verify that it is a vacuum rather than merely a zero-energy stationary point, evaluate the Hessian matrix. At either zero-energy stationary branch,Because $H$ and its first derivatives vanish there, the Hessian matrix of $V$ is just $|m{|}^4{e}^{x^2}$ times this Hessian matrix. For $s=1$, both eigenvalues are positive. This proves a strict local minimum in both real scalar directions. For $s=-1$, $H_{xx}<0$, so the alternative $\beta =2+\sqrt{3}$, $⟨z⟩=-\sqrt{3}-1$ is a saddle point and is excluded from the stable zero-energy Polonyi vacuum.