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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.

Use the canonical Kähler potential and normalize superspace integration by $\int d^2\theta {\theta }^2=1$. The superspace Lagrangian density isThe full superspace integration produces a D-term, while the chiral superfield integral produces an F-term. Work in chiral coordinates for the chiral-superfield component expansion, so the omitted spacetime-derivative terms do not enter the extraction of $F$.

Set $\delta \Phi =\sqrt{2}\theta \psi +{\theta }^{2}F$. A Taylor expansion of the superpotential givesHigher terms vanish because there are only two Grassmann variables ${\theta }^{\alpha }$. With $\theta \psi ={\theta }^{\alpha }{\psi }_{\alpha }$ and $\psi \psi ={\psi }^{\alpha }{\psi }_{\alpha }$, their anticommutation gives $(\theta \psi {)}^2=-\frac{1}{2}{\theta }^2\psi \psi$. ThusThe canonical Kähler potential contributes $F^{\ast }F$. Consequently the part of the Lagrangian density containing the auxiliary field isThe fermion terms coming from the superpotential areThe Weyl spinor kinetic term and the complex scalar field kinetic term are canonical because $K={\Phi }^{†}\Phi$.

Treat $F$ and $F^{\ast }$ as independent variables in their algebraic Euler-Lagrange equations. They giveEquivalently,After eliminating the auxiliary field, the Lagrangian density contains $-V$, so the F-term scalar potential isIt is nonnegative for arbitrary complex $m,g,\phi$, since it is a modulus squared. A supersymmetric vacuum satisfies F-flatness, so its zeros are:If $m=0$ and $g\ne 0$ these coincide; if $g=0$ and $m\ne 0$ only the origin remains; if $m=g=0$ every constant $\phi$ is a supersymmetric vacuum.

To compare masses, expand around a supersymmetric vacuum $v$ satisfying $W^{\prime }(v)=0$, writing $\phi =v+\eta$. With $M=W^{\prime \prime }(v)=m+2gv$,Writing $\eta =(A+iB)/\sqrt{2}$ gives the scalar field mass term $-\frac{1}{2}|M{|}^2(A^2+B^2)$; the Weyl spinor has physical mass $|M|$. Therefore the supersymmetric mass degeneracy isIn particular, at $v=0$ the boson and fermion masses are both $|m|$. At the second supersymmetric vacuum $v=-m/g$ the sign of $M$ reverses, but the physical mass is again $|m|$. Equality here concerns fluctuations about a supersymmetric vacuum, not arbitrary backgrounds with $W^{\prime }(v)\ne 0$.

In the convention $V\supset \lambda |\phi {|}^4$ and ${\mathcal{L}}_{\mathrm{Yukawa}}=-y\phi \psi \psi +h.c.$, the same superpotential coefficient gives $\lambda =|g{|}^2$ and $y=g$. The supersymmetric relation between quartic and Yukawa couplings is thereforeIf instead the Yukawa interaction is written $-\frac{1}{2}{y}_3\phi \psi \psi +h.c.$, then $y_3=2g$ and the same physical relation is $\lambda =|y_3{|}^2/4$. The numerical factor depends on the definition of the Yukawa coupling.

The perturbative non-renormalization theorem applies to the local holomorphic superpotential in the Wilsonian effective action, using regularization in quantum field theory that preserves supersymmetry. There are two complementary ways to see why it applies here.

First, in supergraph calculations in perturbative quantum field theory, the algebra of supersymmetric covariant derivatives puts any candidate local correction to the superpotential into a full $d^4\theta$ superspace integration. It therefore corrects a D-term, such as the Kähler potential, rather than a local chiral $d^2\theta$ F-term. Turning such a contribution into an apparent chiral integral would require nonlocal inverse spacetime derivatives, schematically $1/\square$. These are excluded from the local Wilsonian effective action with a fixed nonzero momentum cutoff. This gives the diagrammatic non-renormalization theorem.

Second, the holomorphy argument for superpotential non-renormalization treats $m$ and $g$ as background chiral spurions, so the Wilsonian effective action superpotential is a holomorphic function of $\Phi ,m,g$, with no dependence on $m^{\ast }$ or $g^{\ast }$. Assign an ordinary $U(1)$ spurion charge and an R-charge as follows:The superpotential has ordinary charge zero and R-charge two, since $d^2\theta$ has R-charge minus two. A prospective holomorphic monomial $m^a{g}^b{\Phi }^c$ must therefore satisfyFor perturbative corrections, regularity at $m=0$ and $g=0$ requires nonnegative powers of the couplings. Regularity in $m$ is justified by keeping a nonzero Wilsonian momentum cutoff and retaining $\Phi$ as a light field. Hence only $c=2$ and $c=3$ survive: the existing $m{\Phi }^2$ and $g{\Phi }^3$ structures. The quadratic coefficient is fixed by the free theory at $g=0$, where there are no interaction loops. The cubic coefficient is fixed at first order in $g$, where a connected three-field interaction has only its tree-level Feynman diagram; loop corrections would require further powers of the interaction couplings, which the holomorphy argument for superpotential non-renormalization charge constraints forbid. ThusA field-independent constant is physically irrelevant in this rigid model; the spurion charges would in any case require $m^3/g^2$, which is not regular at $g=0$.

The Kähler potential can still receive wave-function renormalization. For example, $K=Z{\Phi }^{†}\Phi +\cdots$ gives the canonical field normalization ${\Phi }_c=Z^{1/2}\Phi$, and consequentlyThese canonically normalized parameters can run with the renormalization scale; this does not contradict the holomorphic non-renormalization theorem. When massless modes are included in the full quantum effective action, nonlocal terms associated with infrared divergences can also imitate an F-term. Distinguishing that action from the local Wilsonian effective action is part of the theorem's convention.

Put $\theta ={\theta }^{+}$, $\bar{\theta }={\bar{\theta }}^{+}$ and $y^{+}=x^{+}+i\theta \bar{\theta }$. The general two-dimensional N=(0,2) supersymmetry chiral superfield isIn ordinary coordinates this is $\Phi =\varphi +\sqrt{2}\theta {\psi }_{+}+i\theta \bar{\theta }{\partial }_{+}\varphi$. A supersymmetric kinetic action iswhose component form, up to light-cone conventions and total derivatives, is

A Fermi superfield satisfying ${\bar{D}}_{+}{\Lambda }_{-}=f(\Phi )$ has expansionHere $G$ is a complex bosonic auxiliary field. With the normalization in the question, the component action isup to equivalent sign conventions for the fermions.

The chiral integral in $S_3$ is supersymmetric only if its integrand is chiral. Applying ${\bar{D}}_{+}$ gives the necessary and sufficient conditionwith every $J^a$ holomorphic and with gauge charges chosen so that each product ${\Lambda }_{-a}{J}^a$ is gauge invariant. Its component expansion couples $G_a$ linearly to $J^a$ and supplies the corresponding Yukawa term ${\lambda }_{-a}{\psi }_{+}^i{\partial }_i{J}^a$.

Eliminating each $G_a$ by its algebraic field equation gives the nonnegative scalar potential. In the normalization displayed above and in the question,If the conventional definitions ${\bar{D}}_{+}{\Lambda }_{-}^a=\sqrt{2}{E}_a$ and $S_J=-\int d\theta {\Lambda }_{-a}{J}^a/\sqrt{2}+h.c.$ are used instead, the same result is written $V={\sum }_a(|E_a{|}^2+|J^a{|}^2)$; the two forms differ only by the normalization of $f$ and $J$.

The Kähler potential $K(\Phi ,{\Phi }^{†})$ is a real function integrated over all four fermionic coordinates, $\int d^4\theta K$. Its complex Hessian gives the scalar Kähler metric and therefore the kinetic terms. The superpotential $W(\Phi )$ is holomorphic and is integrated over chiral superspace, $\int d^2\theta W+h.c.$; its derivatives determine Yukawa couplings and the F-term scalar potential.

For the canonical Kähler potential $K={\Phi }^{†}\Phi$, extracting the ${\theta }^2{\bar{\theta }}^2$ component of the supplied chiral-superfield component expansion and integrating by parts gives the bosonic actionin the mostly-plus convention. Extracting the ${\theta }^2$ component of a holomorphic function givesso its bosonic term is $F{W}^{\prime }(\varphi )$, with the Hermitian conjugate understood in a real action. For several fields, eliminating each algebraic auxiliary field by $F_i^{\ast }=-W_i$ produces $V_F={\sum }_i|W_i{|}^2$.

A chiral superfield obeysExamples in the Minimal supersymmetric Standard Model include the quark, lepton and Higgs chiral superfields. Such a multiplet contains a complex scalar $\varphi$, a two-component Weyl spinor $\psi$, and a complex auxiliary field $F$; only the scalar and fermion propagate on shell.

The chiral coordinate $y^{\mu }=x^{\mu }+i\theta {\sigma }^{\mu }\bar{\theta }$ is annihilated in the required combination by ${\bar{D}}_{\dot{\alpha }}$, so the solution has the simple formTaylor-expand each component about $x$. Nilpotence truncates the series, and the Grassmann identitygives the chiral-superfield component expansionTherefore

In the chiral coordinate $y^{\mu }=x^{\mu }+i\bar{\theta }{\bar{\sigma }}^{\mu }\theta$, a chiral superfield isHere $\varphi$ is a complex scalar, ${\psi }_{\alpha }$ is a two-component Weyl spinor, and $F$ is a complex auxiliary field. Off shell, $\varphi$ and $F$ contribute two real bosonic components each, while $\psi$ has two complex Grassmann components. Thus there are four real bosonic and four real fermionic degrees of freedom.

Eliminating the auxiliary field givesExplicitly,The two supersymmetric vacua for $g\ne 0$ are $\varphi =0$ and $\varphi =-m/g$.

After eliminating the chiral auxiliary fields in Planck-unit supergravity, one convention gives $F^i=-e^{K/2}{K}^{i\bar{j}}\overline{D_{j}W}$. The upper-index inverse satisfies $K_{i\bar{j}}{K}^{k\bar{j}}={\delta }_i^k$. Nonzero auxiliary expectation values diagnose supersymmetry breaking; their common phase convention does not affect the vanishing condition.

Let $k^i$ generate a holomorphic gauge transformation, with $k^i{\partial }_{i}W=-{\kappa }^{2}rW$ and moment map $\mathcal{P}=i(k^i{K}_i-r)$. Substitution of the Kähler covariant derivative of a superpotential gives the displayed identity. Consequently, a finite point with $W\ne 0$ and all $D_{i}W=0$ also has $\mathcal{P}=0$. Pure D-term breaking with vanishing chiral auxiliary fields can therefore evade this argument only where $W=0$ or outside its hypotheses. In conventional two-derivative matter-coupled supergravity, a constant Fayet–Iliopoulos term gauges an R-symmetry and constrains the transformation of $W$; a neutral superpotential and an independently assigned nonzero constant shift are not generally consistent. The gauge-covariance restriction is derived in Van Proeyen's discussion of Fayet–Iliopoulos terms and R-symmetry.

A gauge connection on superspace has both vector and spinor components. Subtracting the flat-frame torsion from the graded commutator of its covariant derivatives defines its supercurvature. The conventional four-dimensional $\mathcal{N}=1$ spinor-curvature constraints leave one real vector superfield as a prepotential. Its chiral field-strength superfield contains the ordinary gauge curvature, gaugino and real auxiliary field.

In two-dimensional N=(1,1) superspace, one convention for the supersymmetric Liouville equation is $i{D}_{-}{D}_{+}\Phi =e^{\Phi }$. With the ordered expansion $\Phi =\varphi +i{\theta }^{-}{\psi }_{+}+i{\theta }^{+}{\psi }_{-}+i{\theta }^{-}{\theta }^{+}F$ and left Grassmann derivatives, its left side isWriting $N=\Phi -\varphi$, Grassmann parity gives $N^2=2{\theta }^{-}{\theta }^{+}{\psi }_{+}{\psi }_{-}$ and $N^3=0$. HenceComparing coefficients eliminates the auxiliary field as $F=e^{\varphi }$ and gives ${\partial }_{-}{\psi }_{-}=e^{\varphi }{\psi }_{+}$, ${\partial }_{+}{\psi }_{+}=-e^{\varphi }{\psi }_{-}$, ${\partial }_{-}{\partial }_{+}\varphi =-e^{2\varphi }+i{e}^{\varphi }{\psi }_{+}{\psi }_{-}$. The fermion bilinear sign depends on the stated ordering and derivative convention.

In a canonical Wess–Zumino model with cubic superpotential $g{\Phi }^3/3$, eliminating the auxiliary field gives $V\supset |g{|}^2|\phi {|}^4$ and $\mathcal{L}\supset -g\phi \psi \psi +h.c.$. Thus the coefficient of the quartic scalar field interaction is the squared modulus of this convention for the Yukawa coupling.

A supersymmetry transformation is the infinitesimal action generated by an odd supercharge, with constant anticommuting spinor parameter in a global theory. In superspace it acts as $\delta \mathcal{S}=(ϵQ+\overline{ϵ}\bar{Q})\mathcal{S}$, with possible conventional overall factors of $i$. On the fermion of a chiral superfield, the transformation includes $\delta \psi =\sqrt{2}ϵF_{\mathrm{aux}}+i\sqrt{2}{\sigma }^{\mu }\overline{ϵ}{\partial }_{\mu }\varphi$. A nonzero vacuum auxiliary field therefore gives an inhomogeneous fermion shift, identifying the Goldstino direction when global supersymmetry is spontaneously broken.

For canonical Kähler potential and $W=m{\Phi }^2/2+g{\Phi }^3/3$, eliminating the auxiliary field gives $V=|m\varphi +g{\varphi }^2{|}^2$. The cubic interaction Lagrangian is $-m^{\ast }g{\varphi }^{\ast }{\varphi }^2-m{g}^{\ast }\varphi ({\varphi }^{\ast }{)}^2$. Differentiating with respect to the three external fields gives the all-incoming Feynman rules $V_{\varphi \varphi {\varphi }^{\ast }}=-2i{m}^{\ast }g$ and $V_{{\varphi }^{\ast }{\varphi }^{\ast }\varphi }=-2im{g}^{\ast }$; the factor two counts identical scalar legs. For real $m,g$ and $\varphi =(A+iB)/\sqrt{2}$, the cubic Lagrangian is $-mgA(A^2+B^2)/\sqrt{2}$, giving $V_{AAA}=-3\sqrt{2}img$ and $V_{ABB}=-\sqrt{2}img$.