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A canonical Wess–Zumino model with a cubic superpotential already supplies a microscopic membrane-like object. Choose positive real $\mu ,g$ andThere are two isolated supersymmetric vacua, ${\varphi }_{\pm }=\pm \mu /g$. A planar domain wall has a static scalar profile depending on the transverse coordinate $z=X^3$, with one vacuum at either end. Its energy per unit area isChoosing $\alpha =\arg \Delta W$ gives $T\ge 2|\Delta W|$. The real increasing solution saturates the bound:The tension follows either from integrating $2({\varphi }^{\prime }{)}^2$ or from $W(\mu /g)-W(-\mu /g)=4{\mu }^3/(3g^2)$. This is a supersymmetric domain wall, and the first-order equation also implies the second-order scalar equation. Its thickness is of order $1/\mu$.

The wall breaks transverse translations: changing $z_0$ gives a normalizable scalar zero mode proportional to ${\varphi }^{\prime }(z)$. It also preserves only half the four real supercharges. In the chiral-fermion transformation, $F=-\overline{W^{\prime }}=-{\varphi }^{\prime }$ and the derivative term combine into a rank-half condition on the constant supersymmetry parameter, for example $ϵ=i{\sigma }^3\overline{ϵ}$ in a compatible Weyl convention. The orthogonal two real parameters are broken and generate localized fermion zero modes. One such profile satisfies $({\partial }_z+2g\varphi )f=0$, giving $f\propto {\mathrm{sech}}^2(\mu (z-z_0))$. The scalar and a three-dimensional Majorana fermion therefore form the low-energy wall multiplet. The domain-wall charge in N=1 supersymmetry is a boundary/tensor charge contributing to the extended algebra; the preserved combinations saturate $T=2|\Delta W|$. This avoids falsely applying the unextended particle-algebra argument of Question 3 to a wall with nonzero tension.

Promote the translation collective coordinate to a slowly varying worldvolume field $Y({\xi }^0,{\xi }^1,{\xi }^2)$. At wavelengths long compared with the wall thickness, its leading relativistic geometric action is the Dirac membrane actionIt is invariant under worldvolume reparametrizations and the target Poincaré group. In static gauge $X^i={\xi }^i$, $X^3=Y$, the matrix determinant lemma givesDirectly substituting the leading collective-coordinate profile into the scalar kinetic term gives the same coefficient, since $\int dz({\varphi }^{\prime }{)}^2=T/2$. Target Lorentz symmetry fixes the leading first-derivative nonlinear completion to the invariant area; finite-thickness physics can add higher-curvature terms. The area action is therefore a long-wavelength effective field theory, not an exact replacement for the microscopic wall at arbitrary momenta. Its embedding equation is ${\partial }_i(\sqrt{-h}{h}^{ij}{\partial }_j{X}^m)=0$.

For the four-dimensional supermembrane, extend the embedding to $(X^m(\xi ),{\theta }^{\alpha }(\xi ))$ with a four-real-component Majorana spinor. Pull back the invariant coframe of Question 7:Replacing $\partial X$ by $\Pi$ makes the area term invariant under rigid target supersymmetry, but does not yet give the appropriate fermionic gauge symmetry. Add a Wess-Zumino brane coupling to a superspace three-form potential $C_3$. In one orientation and matching normalization,The supermembrane closed four-form uses the distinct four-dimensional Fierz identity $(C{\gamma }_{mn}{)}_{(\alpha \beta }(C{\gamma }^n{)}_{\gamma \delta )}=0$. Differentiating $G_4$ reduces its coefficient to this symmetric tensor, so $d{G}_4=0$. A local potential therefore exists. Since the curvature is supertranslation invariant, the potential can vary by an exact form; its pullback integral is invariant modulo boundary terms. This is why the three-form $H_3$ of Question 7 is not itself the membrane field strength: integrating a membrane potential requires three-form degree and hence four-form curvature.

The required additional local fermionic gauge invariance is kappa symmetry. Define the worldvolume gamma matrices ${\gamma }_i={\Pi }_i^m{\gamma }_m$ andWith mostly-plus target signature, a local orthonormal worldvolume frame gives ${\Gamma }_{\kappa }={\gamma }_0{\gamma }_1{\gamma }_2$, so ${\Gamma }_{\kappa }^2=1$ and $\mathrm{tr}{\Gamma }_{\kappa }=0$. Thus $(1\pm {\Gamma }_{\kappa })/2$ are rank-two kappa symmetry projectors. The embedding variation gives $\delta {\Pi }_i^m=2i\delta \bar{\theta }{\gamma }^m{\partial }_i\theta$. The area variation is consequently proportional to $-2iT\sqrt{-h}\delta \bar{\theta }{\gamma }^i{\partial }_i\theta$. Contracting the displayed $G_4$ with the fermionic variation and using ${ϵ}^{ijk}{\gamma }_{ij}=2\sqrt{-h}{\Gamma }_{\kappa }{\gamma }^k$ gives the opposite Wess-Zumino contribution with an extra ${\Gamma }_{\kappa }$. Their sum is proportional toFor the boxed kappa variation, $\delta \bar{\theta }=\bar{\kappa }(1+{\Gamma }_{\kappa })$, so it vanishes by $(1+{\Gamma }_{\kappa })(1-{\Gamma }_{\kappa })=0$. Reversing membrane orientation reverses the Wess-Zumino sign and the matching projector. The cancellation fixes the relative charge magnitude to the tension; rigid supersymmetry alone would not fix it.

Worldvolume reparametrizations remove three of the four embedding scalars, leaving one transverse mode. Kappa symmetry removes half the four real embedding spinor components, leaving a two-component three-dimensional Majorana field, whose first-order equation has one on-shell fermionic polarization. The physical worldvolume modes are one scalar and one Majorana fermion, matching the wall zero modes. Without kappa symmetry the fermionic degrees would be doubled. A planar configuration preserves the target supersymmetries that can be compensated by the kappa projector, reproducing the wall's half-supersymmetry condition. This is the four-dimensional effective domain-wall interpretation of a supermembrane, rather than the eleven-dimensional M2-brane particle-content problem.

Here a supersymmetry-preserving combination means a real combination of Hermitian Majorana spinor supercharges, in a positive-norm unitary representation of the ordinary Super-Poincaré algebra. This reality qualification is essential. Let $|\psi ⟩$ be normalized and in the charge domains, and define the real symmetric matrixFor real $u$, the Hermitian operator $q_u=u^a{Q}_a$ satisfiesTherefore $M$ is a positive semidefinite matrix, andThe reverse implication uses positivity: diagonalize $M$, so a zero quadratic form has zero component in every positive-eigenvalue direction. This converts the problem into nullity of real N=1 supercharges, not just a condition on the determinant of an operator-valued expression.

In the equivalent Weyl spinor form, $\{q_{\alpha },q_{\beta }^{†}\}=2{\sigma }_{\alpha \dot{\beta }}^m{P}_m$. After a fixed invertible change of real charge basis, $M$ is the realification of the Hermitian two-by-two matrixup to the harmless spatial-sign convention. A complex Hermitian matrix has the same eigenvalues in its realification, each repeated twice. The Pauli matrix multiplication law gives $(\mathbf{p}\cdot \mathbf{\sigma }{)}^2=|\mathbf{p}{|}^{2}I$, so these eigenvalues are $E+|\mathbf{p}|$ and $E-|\mathbf{p}|$, each twice. Positivity forces $E\ge |\mathbf{p}|$. If one nonzero real $u$ annihilates the state, $M$ has a kernel: either $E=|\mathbf{p}|>0$, with exactly two zero eigenvalues, or $E=|\mathbf{p}|=0$, with four. ThusFor a momentum eigenstate these are the null-momentum and zero-momentum cases; the expectation-matrix proof also covers states not assumed to have definite momentum. There are no massive states preserving a real charge in this unextended algebra.

If “linear combination” instead allows arbitrary complex coefficients, the unrestricted claim is false. In a massive fermionic Fock space with two lowering operators $a_1,a_2$ and a spin-half Clifford vacuum spectator, takeThen $a_1\psi =0$. The three vectors $a_2\psi$, $a_2^{†}\psi$ and $a_1^{†}\psi$ are linearly independent, by their occupation numbers and spectator spin labels. The complex annihilator is therefore only one-dimensional. Such a lowering operator is not a Hermitian generator of preserved real supersymmetry. The boxed conclusion applies to the physical real-charge interpretation; it also assumes no domain-wall charge in N=1 supersymmetry.

A scalar superfield transformation is a pullback under a supertranslation. Since that transformation mixes $x$ and $\theta$, an odd-coordinate derivative acquires an extra spacetime-derivative term under the coordinate Jacobian. It therefore does not transform as a spinor superfield merely by differentiating the scalar transformation law. This is partial derivatives and superfield covariance.

For explicit signs, split the Majorana spinor coordinates into ${\theta }^{\alpha },{\bar{\theta }}^{\dot{\alpha }}$ and use left Grassmann derivatives. Choose the differential generatorsFor example, $\{{\partial }_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=i{\sigma }_{\alpha \dot{\beta }}^m{\partial }_m$, which is nonzero. Commuting the even supersymmetry variation through ${\partial }_{\alpha }$ consequently gives the unwanted translation term.

Replace the ordinary odd-coordinate derivatives by the supersymmetric covariant derivativesThe graded product rule cancels the mixed coordinate/derivative terms, giving $\{D,Q\}=\{D,\bar{Q}\}=\{\bar{D},Q\}=\{\bar{D},\bar{Q}\}=0$. Thus $D_{\alpha }V$ and ${\bar{D}}_{\dot{\alpha }}V$ transform as spinor superfields. Their own mixed algebra is $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}=-2i{\sigma }_{\alpha \dot{\beta }}^m{\partial }_m$, with equal-chirality anticommutators zero. This covariant differentiation is also what makes a chiral superfield constraint ${\bar{D}}_{\dot{\alpha }}Z=0$ supersymmetry invariant. Negating every barred derivative is another convention, but changes the displayed mixed-algebra sign.

In the ordinary local two-derivative matter class, the general ungauged supersymmetric action for a chiral superfield isThe Kähler potential $K$ is real, and the superpotential $W$ is a holomorphic function. Adding $f(Z)+\bar{f}(\bar{Z})$ to $K$ changes the action only by a boundary term: full superspace integration of a chiral or antichiral function vanishes modulo that term. This is a Kähler transformation. A positive Kähler metric $G=K_{z\bar{z}}$ is required for a nonsingular positive-norm kinetic term, not just for derivative counting.

To see the differential order, the bosonic component Lagrangian in mostly-plus signature containsThe auxiliary field equation is algebraic: $F=-G^{-1}\overline{W_z}$ on a purely bosonic configuration. It gives the scalar potential $U=G^{-1}|W_z{|}^2$. The scalar equations have at most two spacetime derivatives, the fermion equations from the sigma-model kinetic term have one, and the remaining interaction terms do not raise this order. The fermionic corrections to the auxiliary equation are still algebraic.

For four-dimensional renormalizability, assign engineering dimensions $[Z]=1$, $[d^4\theta ]=2$, and $[d^2\theta ]=1$. Hence, up to a Kähler transformation and a constant field normalization, the restrictions areA mixed higher-degree term in $K$ or a term of degree above three in $W$ introduces a negative-mass-dimension coupling. The constant $w_0$ has no local effect in global supersymmetry, as in the constant superpotential in global supersymmetry.

The boxed general-action statement uses the conventional two-derivative matter interpretation. It is not a classification of all derivative-dependent effective superfield actions by the order of their bosonic equations alone. For example, a scalar Lagrangian quartic in first derivatives still yields a second-order Euler–Lagrange equation, and supersymmetric derivative-dependent interactions can contain such terms. Their auxiliary and fermionic equations require separate checking. This distinction is explicit in the original construction of supersymmetric derivative-dependent scalar actions. Such actions are outside the ordinary $K+W$ class used for the renormalizable answer here.

Spontaneous supersymmetry breaking preserves the action and conserved supercharges, but the vacuum fails to be invariant. This differs from explicit breaking, which changes the symmetry of the action itself. In global four-dimensional N=1 supersymmetry, summing diagonal entries of the Super-Poincaré algebra writes the Hamiltonian as a sum of positive charge anticommutators. Its expectation is a sum of squared norms. Consequently a zero-energy vacuum is supersymmetric, whereas an existing vacuum with broken global supersymmetry has positive energy density. The energy zero is fixed by the algebra. Neither this positivity statement nor the particle-multiplet mass relations should be transferred without qualification to supergravity.

For supersymmetric quantum mechanics, let $H=\{q,q^{†}\}/2$, $q^2=0$, and let fermion parity anticommute with both charges. On a positive-energy eigenspace, $(q+q^{†})/\sqrt{2E}$ is an odd involution: its square is one, it preserves energy, and it reverses parity. Thus every positive-energy bosonic state has a fermionic partner and conversely. Their contributions cancel in the Witten indexFor a discrete spectrum with a well-defined trace, this proves independence of $\beta$. Continuous changes of parameters also preserve the index if the trace stays well defined and no state escapes to infinity or crosses an unregulated continuum threshold. A nonzero index guarantees an unbroken supersymmetric ground state; a zero index is inconclusive. For example, de Rham supersymmetric quantum mechanics on a circle has a bosonic constant zero-form and a fermionic constant one-form, both normalizable zero modes, so it is unbroken with index zero.

A more explicit one-dimensional family illustrates how normalizability determines the answer. DefinePositivity of these Hamiltonian operators shows that a zero mode must solve $A{\psi }_{-}=0$ or $A^{†}{\psi }_{+}=0$. Thus the only candidates areFor $w(x)=x$, the Gaussian minus-sector wavefunction is normalizable and its reciprocal is not, giving an unbroken model with $I=1$ when the minus sector is assigned even parity. For $w(x)=x^2+a$, $a>0$, the exponent $x^3/3+ax$ tends to opposite infinities at the two ends, so neither candidate is normalizable. Both partner potentials $[w^2\mp w^{\prime }]/2$ grow quartically; their spectra are discrete and their lowest energies cannot be zero. The lowest states are therefore paired at positive energy, $I=0$, and supersymmetry is spontaneously broken. This is supersymmetric factorization and zero-mode normalizability. In quantum mechanics the paired ground states illustrate the charge algebra; a relativistic massless-particle conclusion requires field-theoretic hypotheses.

In relativistic global field theory, Goldstone's theorem has a fermionic version. If a conserved supercurrent charge does not annihilate the vacuum, its Ward identity has a nonzero vacuum contact term. Fourier transforming, the divergence of the current correlator cannot tend to zero smoothly at zero momentum. Its spectral representation therefore contains a massless fermionic pole: the Goldstino, with spin one-half. This statement assumes a Poincaré-invariant vacuum and the usual local unitary field-theory conditions. It does not require a bosonic Goldstone particle for the fermionic symmetry. In supergravity the Goldstino can instead supply the longitudinal polarization of a massive gravitino.

For canonical chiral superfields ${\Phi }_i$, elimination of their auxiliary fields givesA supersymmetric vacuum solves every $W_i=0$. At a stationary broken vacuum,The same matrix $W_{ij}$ is the chiral-fermion mass matrix, so the nonzero vector $\overline{W_j}$ is a zero mode. This derives the Goldstino zero mode from vacuum stationarity. The transformation $\delta {\psi }_i=\sqrt{2}ϵF_i+\cdots$ independently identifies that direction through its inhomogeneous shift.

For one canonical chiral field, $W=fZ$, $f\ne 0$, has $U=|f{|}^2$ everywhere, a massless fermion and a classically flat scalar direction. It is flat F-term breaking with a linear superpotential, rather than an isolated vacuum. In contrast, a nonconstant polynomial $W^{\prime }$ of positive degree has a complex root and hence a supersymmetric vacuum; in particular a generic quadratic or cubic single-field superpotential cannot remove all such roots. This argument assumes canonical nonsingular kinetic terms and does not claim that every runaway or nonpolynomial model has a vacuum.

A renormalizable three-field O'Raifeartaigh model gives a less degenerate illustration:The auxiliary equations require $W_2=m{\varphi }_1=0$, then $W_1=hX{\varphi }_1+m{\varphi }_2=0$, leaving $W_X=f\ne 0$. They cannot vanish simultaneously. For real positive $f,h,m$ with $m^2>fh$, the potential obeysTherefore its global tree-level minimum has ${\varphi }_1={\varphi }_2=0$, arbitrary $X$, and energy $f^2$. The massless ${\psi }_X$ is the Goldstino, while $X$ is a pseudomodulus. At $X=0$, the scalar mass-squared spectrum is $0,0,m^2,m^2,m^2+fh,m^2-fh$, and the fermion masses are $0,m,m$. This exhibits both the stability condition and the loss of ordinary supermultiplet mass degeneracy in a broken vacuum. Quantum corrections can lift the pseudomodulus and must be analysed rather than assumed away; the noncompact classical valley also prevents an unqualified finite-volume Witten index trace argument. The index, the vacuum equations and the Goldstino theorem consequently answer complementary questions.

Ordinary electromagnetism uses a one-form gauge potential, its gauge transformation and its curvature. On superspace the corresponding object is a superspace gauge connection, with both vector and spinor components. Write ${\nabla }_A=D_A+i{\mathcal{A}}_A$. Its graded commutator contains the flat-frame torsion plus supercurvature. The conventional constraints set the two undotted-spinor curvatures, the two dotted-spinor curvatures and the conventional mixed spinor curvature to zero; the mixed derivative algebra then determines the vector connection. The remaining independent gauge curvature is packaged into a chiral field-strength superfield. A real vector superfield $V=V^{†}$ is a prepotential solving these constraints, rather than four unrelated ordinary gauge potentials.

Use the left-derivative conventions of Question 5, with $D^2=D^{\alpha }{D}_{\alpha }$ and ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. An Abelian supergauge transformation has a chiral parameter $\Lambda$, and one consistent normalization isThe parameter is chiral so that the transformed matter is still a chiral superfield; its complex components remove more of the prepotential than an ordinary real gauge parameter. In Wess-Zumino gauge, one convenient expansion isIts components are the gauge potential $A_m$, the gaugino $\lambda$ and the real auxiliary field $D$. Wess-Zumino gauge removes the extra scalar and spinor components of a general real superfield. Supersymmetry transformations usually leave that gauge slice and require a compensating supergauge transformation. For a residual chiral parameter with real lowest component $a(x)$, the expansion gives $A_m^{\prime }=A_m+2{\partial }_{m}a$ and ${\varphi }^{\prime }=e^{-2iqa}\varphi$; thus $\alpha =2a$ is the ordinary Maxwell gauge parameter.

The Abelian field strength is the Abelian field-strength chiral projectionThree barred derivatives vanish because there are only two anticommuting dotted components, so ${\bar{D}}_{\dot{\beta }}{W}_{\alpha }=0$. Gauge invariance follows directly: $D_{\alpha }\bar{\Lambda }=0$ by antichirality, while the mixed derivative algebra gives ${\bar{D}}^2{D}_{\alpha }\Lambda =0$ for chiral $\Lambda$. In chiral coordinates its component structure iswhere ${\sigma }^{mn}=({\sigma }^m{\bar{\sigma }}^n-{\sigma }^n{\bar{\sigma }}^m)/4$ fixes the tensor normalization. Overall gaugino phase choices do not change the gauge-invariant field strength or its Bianchi constraint.

Reality of $V$ and the derivative algebra imply the superspace Yang-Mills Bianchi identityIndeed, the two sides are the same differential operator on $V$, because $D^{\alpha }{\bar{D}}^2{D}_{\alpha }={\bar{D}}_{\dot{\alpha }}{D}^2{\bar{D}}^{\dot{\alpha }}$. This is the superspace counterpart of $dF=0$, and its components contain ${\partial }_{[m}{F}_{np]}=0$. It is a defining curvature constraint, not the Maxwell equations. Varying the pure gauge action additionally sets the common real superfield $D^{\alpha }{W}_{\alpha }$ to zero; matter supplies a current instead.

The Supersymmetric Yang-Mills action and minimal matter coupling areThe chiral phase factors cancel in $\bar{\Phi }{e}^{2qV}\Phi$. The component matter derivative is $({\partial }_m+iq{A}_m)\varphi$, which transforms covariantly under the residual gauge transformation just derived. The same superspace term generates the associated matter–gaugino Yukawa coupling and $qD|\varphi {|}^2$ auxiliary coupling, so minimal coupling is not only replacing an ordinary derivative in the scalar action. The gauge action contains $-F_{mn}{F}^{mn}/(4g^2)$, the gaugino kinetic term and $D^2/(2g^2)$. Rescaling the vector multiplet to canonical kinetic normalization gives the physical coupling $qg$. The holomorphic superpotential must be gauge invariant; a single field of nonzero continuous Abelian charge admits no nonconstant polynomial superpotential without additional fields. For several fields, invariance requires ${\sum }_i{q}_i{\Phi }_i{\mathcal{W}}_i=0$. A Fayet–Iliopoulos term is additionally possible for an Abelian factor, shifting the algebraic $D$ equation.

For a non-Abelian gauge group, $V$ and $\Lambda$ are Lie-algebra-valued, and multiplication cannot be reordered. For matter in a specified representation, setThe same matter quadratic form remains invariant. Its curvature isAt first order in an Abelian $V$, this reduces to $-{\bar{D}}^2{D}_{\alpha }V/4$, fixing the factor $1/8$. In the chiral gauge frame, $\bar{\nabla }=\bar{D}$ and ${\nabla }_{\alpha }=e^{-2V}{D}_{\alpha }{e}^{2V}$ as an operator. The gauge-covariant Bianchi identity is ${\nabla }^{\alpha }{W}_{\alpha }={\bar{\nabla }}_{\dot{\alpha }}{\tilde{\bar{W}}}^{\dot{\alpha }}$, where $\tilde{\bar{W}}=e^{-2V}\bar{W}{e}^{2V}$ is the conjugate curvature expressed in that same frame. It follows from the graded Jacobi identity of the covariant derivatives. Taking a trace of $W^{\alpha }{W}_{\alpha }$ makes the chiral gauge action invariant. Its bosonic curvature becomes the ordinary Yang-Mills curvature, and the gaugino transforms in the Adjoint representation. A constant Fayet–Iliopoulos coefficient cannot be inserted for a simple non-Abelian group, since it would need an invariant linear functional annihilating every Lie bracket. These constructions connect the constrained superspace connection, its prepotential and its component gauge multiplet.

TeV-scale softly broken supersymmetry addresses the technical hierarchy problem: equal bosonic and fermionic couplings cancel quadratic ultraviolet contributions to the Higgs mass. Soft mass splittings leave corrections of order $\delta m_H^2\sim g^2{M}_{\mathrm{soft}}^2\log (\Lambda /M_{\mathrm{soft}})/(16{\pi }^2)$, so a weak-scale soft spectrum protects the weak scale against a much higher cutoff. Unbroken four-dimensional extended supersymmetry is incompatible with the observed complex chiral gauge spectrum: charged hypermultiplets contain conjugate representations, while vector multiplets have real adjoint matter. It must therefore be broken to at most $\mathcal{N}=1$ before reproducing low-energy Standard Model chirality.

In a supersymmetric Higgs mechanism, an internal gauge symmetry breaks while supersymmetry can remain unbroken. A massless vector multiplet absorbs a chiral multiplet and becomes a massive multiplet with spins $(1,\frac{1}{2},\frac{1}{2},0)$: the gauge boson acquires its longitudinal particle polarization from the Goldstone mode. In the super-Higgs mechanism, local supersymmetry itself breaks: the spin-$3/2$ gravitino absorbs the spin-$1/2$ goldstino, gaining two longitudinal polarizations and becoming massive. With only global supersymmetry breaking, the goldstino instead remains a physical massless fermion.

For global four-dimensional $\mathcal{N}=1$ supersymmetry, useLet $\Pi =(-1{)}^F$ be fermion parity. On a finite one-particle supermultiplet at fixed positive-energy momentum, the supercharges act within the multiplet and $\Pi Q_{\alpha }=-Q_{\alpha }\Pi$. Cyclicity of the finite-dimensional trace givesThus the trace with $\Pi$ of every displayed diagonal anticommutator vanishes. Summing the two components, ${\sum }_{\alpha }\{Q_{\alpha },Q_{\alpha }^{†}\}=4E1$, soSince $E>0$, this proves $N_B=N_F$, for both massive and massless one-particle multiplets. The restriction matters: a zero-momentum supersymmetric vacuum singlet, annihilated by every supercharge, need not contain a fermionic partner. The algebra does not imply equal counts for such a trivial zero-energy representation or for an arbitrary unregulated infinite-dimensional trace.

The same Super-Poincaré algebra gives the operator identityFor a normalized physical state $|v⟩$ in a positive-norm Hilbert space,An unbroken supersymmetric vacuum must be annihilated by every supercharge and its adjoint, hence has zero energy. Conversely, zero expectation of this positive sum forces each norm separately to vanish. Therefore a vacuum with strictly positive energy necessarily breaks supersymmetry. For an infinite-volume field theory the corresponding statement is about vacuum energy density, obtained with an appropriate finite-volume regulator. The zero of energy is fixed by the algebra; it cannot be arbitrarily shifted while preserving this identity. This is the global-supersymmetry result and does not assert positivity of the general local-supergravity scalar potential.

For the unheaded extended-algebra request, fix the central-charge normalization explicitly:The central charges are scalar numbers on an irreducible massive representation. A unitary skew-diagonalization of an antisymmetric matrix puts $Z$ into two-by-two blocks $\left(\begin{array}{cc}0& z_r\\ -z_r& 0\end{array}\right)$ and, for odd $\mathcal{N}$, a leftover zero entry. Since this change of supercharge basis is unitary, it preserves the mixed anticommutator. Go to the rest frame $P^{\mu }=(M,0,0,0)$.

For one block set $q_{\alpha }=Q_{\alpha }^1$ and $r_{\alpha }={ϵ}_{\alpha \gamma }(Q_{\gamma }^2{)}^{†}$, with ${ϵ}_{12}=1$. Direct use of the algebra givesWrite $z=|z|e^{i\phi }$ and form the combinations suggested by the hint:ThenEach diagonal anticommutator has nonnegative expectation because it is the sum of squared norms of an operator and its adjoint. Therefore $M-|z_r|\ge 0$ for every block, provingIf the algebra is instead written with $\{Q,Q\}=ϵZ$ without the factor two, the same proof gives $M\ge \frac{1}{2}{\max }_r|Z_r|$. This is the normalization used by some definitions of the BPS bound in supersymmetry; the physical bound is unchanged by the naming of the central charge.

A massive BPS state saturates at least one block bound. At saturation $\{B_{\alpha },B_{\alpha }^{†}\}=0$, so both $B_{\alpha }$ and $B_{\alpha }^{†}$ annihilate every state in its multiplet. The two complex inactive oscillators are four preserved real supersymmetries per saturated block. If $r$ blocks saturate, the preserved fraction is $r/\mathcal{N}$ and only $2\mathcal{N}-2r$ complex fermionic oscillators remain active. For a spin-$j$ Clifford vacuum,compared with $(2j+1)2^{2\mathcal{N}}$ for a long massive multiplet. Thus BPS states have an exact mass-charge relation, preserve part of the supersymmetry, and belong to shortened representations. A physical CPT-invariant spectrum may additionally require the conjugate-charge multiplet. The short representation protects the saturated mass relation while it exists; it does not prohibit a decay at a threshold where central-charge phases align and the sum of daughter masses equals the bound.

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.

The Polyakov action has independent worldsheet metric ${\gamma }_{ab}$ and embedding fields $X^{\mu }$. Varying the metric before imposing a gauge givesWorldsheet diffeomorphisms and Weyl transformations allow conformal gauge ${\gamma }_{ab}=e^{2\omega }{\eta }_{ab}$ locally. The Weyl factor then cancels out of the classical action, leavingThese are free two-dimensional scalar fields, but the surviving metric equations requireEquivalently, ${\dot{X}}^2+X^{\prime 2}=0$ and $\dot{X}\cdot X^{\prime }=0$. Gauge fixing does not license dropping these Virasoro constraints.