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A gauge generator is the representation matrix of a basis element of the Lie algebra of a continuous gauge group on a specified set of fields. For Hermitian generators $T^a$, an infinitesimal transformation is $\delta \varphi =i{\alpha }^a{T}^a\varphi$, and $[T^a,T^b]=i{f}^{ab}{}_c{T}^c$. Gauge invariance of a superpotential consequently requires $W_i(T^a{)}^i{}_j{\varphi }^j=0$ for each generator.

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

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

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

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

Use an unfluxed toroidal zero-mode dimensional reduction, with no projection removing fields. A ten-dimensional gauge field has $10-2=8$ physical polarizations per gauge generator. Its Majorana-Weyl spinor partner has sixteen real components before the massless equation, and eight propagating fermionic states. Thus ten-dimensional super Yang-Mills theory has sixteen real supercharges and eight bosonic plus eight fermionic states per generator.

Split the spacetime index as $M=(\mu ,m)$, with $\mu =0,\dots ,3$ and six internal directions. The zero modes of $A_M$ become $A_{\mu }$ and six real adjoint scalar fields $X_m=A_m$. The spinor representation decomposes under $\mathrm{Spin}(1,3)\times \mathrm{Spin}(6)$ into four four-dimensional Weyl spinors; the Majorana condition supplies their conjugates rather than independent additional spinors. All sixteen supercharges survive, giving four-dimensional N=4 super Yang-Mills theory. The supermultiplet is therefore one gauge vector, four Weyl gauginos and six real scalars, all in the Adjoint representation, with $2+6=8$ bosonic and $4\times 2=8$ fermionic states. The internal $\mathrm{Spin}(6)\simeq SU(4)$ acts as R-symmetry, with the scalars in its six-dimensional representation and the Weyl gauginos in its four-dimensional representation.

Since the internal derivatives vanish, the internal Yang-Mills field strength is $F_{mn}=-i{g}_4[X_m,X_n]$ in a Hermitian-generator convention. The internal part of the gauge kinetic term becomes the scalar potentialHere the invariant trace is normalized by $\mathrm{tr}(T^a{T}^b)={\delta }^{ab}$ and $[T^a,T^b]=i{f}^{abc}{T}^c$; the coupling and scalar kinetic terms use the same normalization. The sum includes both orders of each pair. For Hermitian $X_m$, the commutator is anti-Hermitian, so equivalently $V=-\frac{g_4^2}{4}{\sum }_{m,n}\mathrm{tr}[X_m,X_n{]}^2$. It is nonnegative and vanishes exactly when all six adjoint scalars commute. This gives the classical vacuum moduli space of a supersymmetric gauge theory.

Interpret $A_{MNP}$ as a massless three-form gauge field, with transformation $A_3\mapsto A_3+d{\Lambda }_2$ and kinetic term built from $F_4=d{A}_3$. The massless p-form gauge field count is an on-shell count, rather than the raw $\left(\genfrac{}{}{0ex}{}{D}{3}\right)$ tensor components. For a null plane wave, gauge transformations remove the components along the null gauge direction, while the field equation removes the complementary longitudinal components. The independent polarization is consequently an antisymmetric tensor on the $D-2$ transverse directions. ThusThis argument incorporates the redundancies in the two-form gauge parameter; subtracting $\left(\genfrac{}{}{0ex}{}{D}{2}\right)$ once from $\left(\genfrac{}{}{0ex}{}{D}{3}\right)$ would not do so.

Let $n=D-4$ and split indices into four-dimensional and internal ones. The toroidal reduction of a p-form gauge field gives one $A_{\mu \nu \rho }$, $n$ two-forms $A_{\mu \nu i}$, $\left(\genfrac{}{}{0ex}{}{n}{2}\right)$ vectors $A_{\mu ij}$ and $\left(\genfrac{}{}{0ex}{}{n}{3}\right)$ scalars $A_{ijk}$. A massless three-form in four dimensions has no local propagating state; each massless two-form has one, each vector two and each scalar one. The two-form state can equivalently be represented by a scalar using massless p-form duality: its three-form field strength satisfies $H_3=\ast da$. ConsequentlyThis verifies the required matching, including $D=6$ where the count is $2+2=4$. The nondynamical four-dimensional three-form can carry a constant flux parameter, but that is not an additional propagating degree of freedom.

Eleven-dimensional supergravity contains the metric $G_{MN}$, a three-form $A_{MNP}$ and a Majorana gravitino ${\Psi }_M$. The massless graviton is a symmetric traceless tensor of the nine-dimensional transverse rotation group, giving $9\times 10/2-1=44$ states. The three-form gives $\left(\genfrac{}{}{0ex}{}{9}{3}\right)=84$. The gravitino is a gamma-traceless transverse vector-spinor: a spinor representation of $\mathrm{Spin}(9)$ has sixteen real components, so it gives $9\times 16-16=128$ states. Hence the eleven-dimensional balance is $44+84=128$ bosonic states and $128$ fermionic states.

Reduction on a circle yields type IIA supergravity, with thirty-two real supercharges and opposite ten-dimensional chiralities. The metric decomposes into the ten-dimensional metric, the Kaluza-Klein vector $C_{\mu }=G_{\mu 11}$ and the radius scalar, or dilaton. Their counts are $35+8+1=44$. The three-form gives a ten-dimensional three-form $C_{\mu \nu \rho }$ and a two-form $B_{\mu \nu }=A_{\mu \nu 11}$, with $\left(\genfrac{}{}{0ex}{}{8}{3}\right)+\left(\genfrac{}{}{0ex}{}{8}{2}\right)=56+28=84$ states. The eleven-dimensional Majorana spinor decomposes into two opposite-chirality Majorana-Weyl spinors. The external gravitino components give two gravitini, each with $8\times 8-8=56$ states; the internal component supplies two spin-one-half dilatini, each with eight states. ThereforeThe two-form is in the NS--NS sector and the one-form and three-form are the Ramond--Ramond potentials. This reduction is nonchiral IIA rather than IIB.

Reduction on a seven-torus yields four-dimensional N=8 supergravity, again preserving thirty-two real supercharges. From $G_{MN}$ one gets one graviton, seven vectors $G_{\mu i}$ and $7\times 8/2=28$ scalars $G_{ij}$, giving $2+7\times 2+28=44$ states. From $A_{MNP}$ one gets a nonpropagating four-dimensional three-form, seven two-forms, $\left(\genfrac{}{}{0ex}{}{7}{2}\right)=21$ vectors and $\left(\genfrac{}{}{0ex}{}{7}{3}\right)=35$ scalars, giving $0+7+21\times 2+35=84$ states. Dualizing the seven two-forms to scalars gives the familiar field contentThe gravitino reduces to eight four-dimensional Majorana gravitini and $7\times 8=56$ spin-one-half Majorana fermions, after the external gravitini are separated from their gamma-trace mixing with the internal components. Their count is $8\times 2+56\times 2=128$. Together these fields form the single maximal supergravity multiplet.