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The Chern connection decomposes on vector-bundle-valued differential forms as $D=D^{\prime }+D^{\prime \prime }$, where $D^{\prime }={\nabla }_h^{\prime }$ raises holomorphic degree and $D^{\prime \prime }={\nabla }_h^{\prime \prime }={\bar{\partial }}_E$ raises antiholomorphic degree. In a holomorphic local frame, $D^{\prime }=\partial +A\wedge$ and $D^{\prime \prime }=\bar{\partial }$. The type of the curvature form of a connection givesThe Kähler metric and the Hermitian metric define the $L^2$ inner product and the formal adjoints $D^{\prime \ast },D^{\prime \prime \ast }$. The Dolbeault Laplacians areLet $L=\omega \wedge$ be the Lefschetz operator of a Kähler manifold and $\Lambda =L^{\ast }$ its adjoint Lefschetz operator. With the ordinary commutator convention $[A,B]=AB-BA$, the printed Kähler identities giveSubstitution into the two Dolbeault Laplacians, followed by expansion, yieldsThe middle line follows by cancelling the terms with $D^{\prime }\Lambda D^{\prime \prime }$ and $D^{\prime \prime }\Lambda D^{\prime }$; the remaining terms collect the anticommutator of the two differentials. Hence $\boxed{{\Delta }^{\prime \prime }={\Delta }^{\prime }+[i{F}_h,\Lambda ]}$, where $i{F}_h$ denotes its wedge action. This is the Bochner-Kodaira-Nakano identity with the paper's sign convention.

For ungauged chiral superfields, the two-derivative supersymmetric action is a real full-superspace integral plus a holomorphic half-superspace integral:The real Kähler potential $K$ determines the Kähler metric and kinetic terms; the superpotential $W$ determines Yukawa interactions and the F-term scalar potential. For canonical normalization, $K={\sum }_i{\Phi }_i^{†}{\Phi }_i$. Berezin integration is normalized to extract the highest component, $\int d^2\theta {\theta }^2=1$, and the conjugate measure is chosen so the canonical term is $F_i^{\ast }{F}_i$. The D-term and F-term highest components vary by spacetime total derivatives, making the action invariant under supersymmetry. Gauging would replace the canonical bilinear by its gauge-covariant version and add a gauge kinetic F-term; no gauge multiplet is needed for the chiral theory here.

Vary $F_i$ and $F_i^{\ast }$ independently. The auxiliary fields have algebraic, rather than differential, Euler-Lagrange equations:Alternatively complete the square:Substitution therefore yields the on-shell auxiliary fields and nonnegative scalar potentialThe sign in the Lagrangian is minus the F-term scalar potential; changing the metric convention does not change this algebraic result. A supersymmetric vacuum requires all $W_i=0$, namely F-flatness. For a noncanonical positive Kähler metric $K_{i\bar{j}}$, the purely bosonic elimination instead gives $V_F=K^{i\bar{j}}{W}_i\overline{W_j}$; fermionic connection terms must also be included in a complete nonlinear component action.

Use a radial geometric Kähler potential. Write $s=|z_1{|}^2+|z_2{|}^2$ and $y(s)={\phi }^{\prime }(s)$. The coefficient Hermitian matrix of $i\partial \bar{\partial }\phi (s)$ isIts eigenvalues are $y$ in the complex tangential direction and $y+s{y}^{\prime }$ in the complex radial direction. A Kähler metric is positive precisely when both are positive. The standard form has coefficient matrix $\frac{1}{2}I$, so equality of the Riemannian volume forms is the conditionIndeed, the volume form is ${\omega }^2/2$ and its coefficient is proportional to $\det G$. Multiplying the determinant equation by $2s$ gives $(s^2{y}^2{)}^{\prime }=s/2$, so we can takeAn explicit antiderivative isIt is smooth for $s>0$. Its two eigenvalues areThe form is real and closed because it is $i\partial \bar{\partial }$ of a real function, so it is a Kähler metric on the punctured space. Since $a>0$, neither eigenvalue equals $1/2$, and the metric differs from the Euclidean one. The radial and tangential stretches compensate exactly, preserving the volume form:This is a radial Kähler metric with Euclidean volume in complex dimension two.

The real coordinate functions are real and imaginary parts of the holomorphic coordinates $z_j$. Part (a), together with the real coefficients of the Laplace-Beltrami operator, therefore shows that every $x_l$ is a harmonic function. For the given divergence-of-gradient convention, applying the operator to $x_l$ givesThese are the harmonic coordinate identities. Expanding the operator on an arbitrary smooth function now givesThe bracketed coefficients vanish. Hence there is no first-derivative term in these coordinates:The reason is that the real components of holomorphic coordinates are harmonic coordinates for a Kähler metric; a general real coordinate change need not retain this simplification.

Choose a constant positive Hermitian form on $V$. Its underlying flat Kähler metric is translation-invariant and descends through the Euclidean lattice to the compact complex torus $X$. The Dolbeault Hodge decomposition gives a unique Dolbeault Laplacian-harmonic representative of every Dolbeault cohomology class. By the Kähler Laplacian identity, these are also Hodge Laplacian-harmonic.

Choose a constant coframe on $V$. A lifted harmonic form is a sum of constant wedge basis forms with smooth periodic coefficients. For the flat metric, the Hodge Laplacian acts coefficientwise by the scalar constant-coefficient Laplacian. Each coefficient is a harmonic function on a compact flat real torus, hence is constant: integration by parts gives $\int |Da{|}^2=0$. Conversely every constant-coefficient form is harmonic. Thus the harmonic representatives are exactly the translation-invariant differential forms of type (p, q).

The constant $(1,0)$ covectors form $V^{\ast }={\mathrm{Hom}}_{\mathbb{C}}(V,\mathbb{C})$; the constant $(0,1)$ covectors form the space of antilinear maps $\overline{V^{\ast }}={\mathrm{Hom}}_{\overline{\mathbb{C}}}(V,\mathbb{C})$. Sending a constant wedge form to its cohomology class givesThe map is intrinsic and does not depend on the auxiliary flat metric: that metric proves every class has exactly one constant representative, while the inclusion of constant forms is defined directly by the quotient. The Dolbeault cohomology of a complex torus consists of constant forms, and in particularFor indices outside $0,\dots ,g$, both the exterior-power expression and the cohomology group vanish.

First assume global four-dimensional N=1 supersymmetry, canonical positive Kähler potential, and tree level. Write $W_i={\partial }_{i}W$ and $m_{ij}=W_{ij}$ at the stationary vacuum. The chiral-superfield fermion mass matrix is the symmetric matrix $m$, and the sum of squared Weyl spinor masses is $\mathrm{tr}(m^{†}m)$. From the F-term scalar potential $V={\sum }_k|W_k{|}^2$, its mixed and holomorphic Hessian matrix blocks areFor canonically normalized real and imaginary parts of the scalar fields, the real scalar mass matrix has trace twice the mixed trace. Its holomorphic blocks can split the two real scalar masses but do not change their sum. HenceA Weyl spinor has two spin states, so this proves the tree-level supertrace mass sum rule,The paper defines its supertrace with $(-1{)}^{2j+1}$, the negative of the conventional boson-minus-fermion supertrace used here. The asserted zero is the same in either convention. The result requires the stated canonical tree-level hypotheses; a noncanonical Kähler metric, supergravity, or radiative corrections can change the sum rule.

If vector multiplets are also present, the gauge contributions to the F-term supertrace cancel rather than being omitted. At $D_a=0$, define $C={\sum }_a{g}_a^2\Vert T_{a}v{\Vert }^2$ for the scalar expectation vector $v$. Differentiating $\frac{1}{2}{\sum }_a{D}_a^2$ adds $2C$ to the real-scalar trace. The symmetric gaugino-matter mass matrix has off-diagonal entries $\sqrt{2}{g}_a{T}_{a}v$, giving an extra $4C$ in its fermion squared-mass trace. The covariant scalar kinetic term gives vector squared-mass trace $2C$. Thus their contribution is $2C-2(4C)+3(2C)=0$. The gauge-theory result uses all spin states, including the vector weight three; it is not obtained by dropping massive gauge partners.

The MSSM tree-level sfermion mass constraint explains the phenomenological difficulty with direct visible-sector breaking. If F-term breaking is neutral under an unbroken electric and color gauge symmetry, and no D-term shifts are present, the same trace argument applies within each conserved-charge fermion block. Its corresponding scalar partners cannot all have squared masses above the mean of the light fermion squared masses. Thus canonical tree-level visible-sector supersymmetry breaking alone cannot make every squark and other scalar partner heavy while leaving the observed fermions light. The usual effective soft supersymmetry breaking terms arise after communicating breaking from a separate sector; integrating out that sector, noncanonical interactions and radiative effects evade the hypotheses. The global trace identity alone would not identify a particular light squark without the conserved-charge block argument.

For the specified O'Raifeartaigh model, phases may be chosen so that $\lambda ,\mu ,M$ are real and positive. Its superpotential derivatives areand its F-term scalar potential isFor fixed $x,z$, choose $y=-2\lambda xz/M$ to minimize the final square. Since $|z^2-{\mu }^2{|}^2\ge (|z{|}^2-{\mu }^2{)}^2$,Thus $M^2>2{\lambda }^2{\mu }^2$ makes $z=y=0$ a global minimum with arbitrary $x$. The hierarchy $M\gg \mu$ ensures this for fixed perturbative $\lambda$, rather than for arbitrarily large $\lambda$. The origin is one member of this classically flat family, with $V_0={\lambda }^2{\mu }^4>0$ and $F_X=\lambda {\mu }^2$. The complex field $x$ is a pseudomodulus.

At the origin the chiral-superfield fermion mass matrix, in the $({\psi }_X,{\psi }_Y,{\psi }_Z)$ basis, isIts physical fermion masses are $0,M,M$; the two massive Weyl spinors form one massive four-component fermion. The massless ${\psi }_X$ is the goldstino. More generally, stationarity gives $W_{ij}\overline{W_j}=0$, so the nonzero auxiliary field direction is a null vector of $m$, proving the goldstino zero mode from vacuum stationarity.

Writing $z=(z_R+i{z}_I)/\sqrt{2}$ and similarly for $x,y$, the quadratic scalar potential isThe mass spectrum of the quadratic-cubic O'Raifeartaigh model is thereforeThe scalar squared-mass sum is $4M^2$, while twice the fermion squared-mass sum is also $4M^2$, so the supertrace is zero, as required. The two massless real $x$ modes are tree-level pseudomoduli, which can be lifted by a quantum effective potential; their tree-level masslessness is not the exact symmetry protection enjoyed by the goldstino.

Use left Grassmann derivatives for the Grassmann variables. A chiral superfield satisfies ${\bar{\mathcal{D}}}_{\dot{\alpha }}\Phi =0$. In the derivative convention printed in the PDF, differentiating $\theta {\sigma }^{\mu }\bar{\theta }$ with respect to a barred coordinate brings a minus sign from moving the odd derivative through $\theta$. Consequently ${\bar{\mathcal{D}}}_{\dot{\alpha }}{y}^{\mu }=0$, so the general solution depends only on $y$ and $\theta$:Here $\varphi$ is a complex scalar field, $\psi$ a Weyl spinor and $F$ a complex auxiliary field. This is a finite Grassmann algebra expansion, not an assumption that the spacetime fields are constant.

For a completely convention-independent way to organize the ordinary-coordinate expansion, let $v^{\mu }=\theta {\sigma }^{\mu }\bar{\theta }$. Translation by $iv$ givesAll higher terms vanish because there are only two unbarred and two barred odd coordinates. For example take ${\sigma }^{\mu }=(I,\mathbf{\sigma })$, signature $(+---)$, ${\theta }^2={\theta }^{\alpha }{\theta }_{\alpha }$ and ${\bar{\theta }}^2={\bar{\theta }}_{\dot{\alpha }}{\bar{\theta }}^{\dot{\alpha }}$, with both lower epsilon tensors having ${ϵ}_{12}=1$. Then $v^{\mu }{v}^{\nu }=\frac{1}{2}{\theta }^2{\bar{\theta }}^2{\eta }^{\mu \nu }$ and the chiral-superfield component expansion readsThe frequently written positive quarter coefficient corresponds to the opposite convention for $\square$ or the spinor contractions; the preceding shift formula fixes the sign without ambiguity. The local TeX's barred derivative is damaged: the original PDF has ${\partial }_{\mu }$, not a barred-coordinate factor at its end.

Write $X=x+\sqrt{2}\theta G+{\theta }^2{F}_X$ in chiral coordinates. Multiplication gives the three component constraintsOn the regular branch where the ordinary commuting part of $F_X$ is nonzero, the nilpotent chiral superfield is thereforeThe goldstino $G$ and auxiliary field $F_X$ are independent; the scalar is a composite, not another independent degree of freedom. Its first two constraints follow because any product of three identical two-component odd spinor entries vanishes. It is essential not to divide by $F_X$ on a branch with zero commuting part: for example $X=0$ satisfies the nilpotency condition without breaking supersymmetry.

For a regular local action with no superspace derivatives in $K$ or $W$, nilpotency truncates the most general superpotential and Kähler potential toThe last inequality makes the kinetic Kähler metric healthy. In global four-dimensional N=1 supersymmetry, the constant and holomorphic pieces of $K$ integrate to zero, and $w_0$ does not affect the action. On the bosonic background $G=0$, the constraint sets $x=0$ and the auxiliary terms are $\kappa F_X{\bar{F}}_X+f{F}_X+\bar{f}{\bar{F}}_X$. Eliminating them givesA nonzero $f$ is necessary for this regular constrained branch to be an on-shell vacuum. It gives nonzero $F_X$ and supersymmetry breaking, with $G$ the goldstino. Nilpotency alone is not a proof that every possible branch breaks supersymmetry. With $f=0$ the above composite parametrization is singular on shell, rather than evidence for a regular supersymmetric vacuum containing an independent Goldstino.

If these data are instead embedded in supergravity, the constant and linear terms cannot simply be discarded independently: a Kähler transformation also transforms $W$. Before such a transformation, at $x=0$ the supergravity F-term potential isThus nonzero $F^X$ still breaks supersymmetry, but the vacuum energy need not be positive. This distinguishes the global answer from the local theory explicitly introduced in the next question.

Now write $Y=y+\sqrt{2}\theta \chi +{\theta }^2{F}_Y$. The condition $XY=0$ becomesUsing the same nonzero-$F_X$ branch gives the chiral superfield constrained by a nilpotent superfieldThe spinor $\chi$ and the auxiliary $F_Y$ remain independent; the scalar is removed. Substituting this expression verifies the other two product constraints using two-component spinor identities. It does not impose $Y^2=0$. It does imply cubic nilpotency from a mixed chiral constraint: $Y^3=0$. Indeed $(G\chi {)}^2=-\frac{1}{2}(GG)(\chi \chi )$, soThese are exactly the three components needed for $Y^3=0$. The generic nonzero expression for $y^2$ shows why imposing quadratic nilpotency on $Y$ would lose allowed interactions.

The analytic chiral monomials consequently reduce to $1,X,Y,Y^2$. The most general superpotential isFor the Kähler potential, put $U=(X,Y,Y^2{)}^T$. Every allowed mixed monomial is included inThis includes $X\bar{X}$, $X\bar{Y}$, $X{\bar{Y}}^2$, $Y\bar{Y}$, $Y{\bar{Y}}^2$, $Y^2{\bar{Y}}^2$ and the necessary conjugates; products containing $XY$ or $\bar{X}\bar{Y}$ vanish. In a global theory the holomorphic $\ell$ terms again integrate to zero; in supergravity they can be changed by a Kähler transformation. Positivity requires the kinetic metric on the independent multiplets to be positive definite at the chosen background, not necessarily the entire coefficient matrix on the redundant composite list $U$. The most general claim here concerns regular analytic two-derivative actions; singular functions of the constrained fields or higher-derivative operators are outside it.

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.

The Dolbeault theorem identifies $H^2(M,{\mathcal{O}}_M)$ with $H_{\bar{\partial }}^{0,2}(M)$. It vanishes by hypothesis, and complex conjugation in the Hodge decomposition theorem for compact Kähler manifolds makes $H^{2,0}$ vanish too. Thus every real degree-two cohomology class has type $(1,1)$.

Let $\omega$ be a Kähler form. Choose a rational cohomology class $\eta$ sufficiently close to $[\omega ]$. More concretely, using harmonic differential forms as representatives for a fixed Kähler metric, the harmonic form representing $\eta$ is close to $\omega$ in every smooth norm: harmonic differential forms constitute a finite-dimensional vector space. It is a real closed $(1,1)$ form, and is positive if sufficiently close to $\omega$, by compactness. This is the openness of the Kähler cone. Multiplying by a positive integer gives an integral Kähler class $\kappa =N\eta$.

The holomorphic exponential sequence contains the cohomology segmentChoose an integral lift of $\kappa$. Exactness gives a holomorphic line bundle $\mathcal{L}$ having that First Chern class. We must ensure that it has a positive metric, rather than merely a positive representative of its class.

Choose any Hermitian metric $h_0$ and let ${\theta }_0$ be its normalized curvature form of a Hermitian holomorphic line bundle. If $\theta$ is the positive form representing $\kappa$, then $\theta -{\theta }_0$ is an exact real $(1,1)$ form. The ddbar lemma gives a real smooth function $u$ withReplace $h_0$ by $h=e^{-u}{h}_0$. Its normalized curvature is $\theta$, so $\mathcal{L}$ is a positive holomorphic line bundle. The Kodaira embedding theorem now applies: sufficiently many sections of a sufficiently high tensor power define the desired embedding. Therefore

For the formal adjoints $d^{\ast }$, ${\partial }^{\ast }$ and ${\bar{\partial }}^{\ast }$ defined by the Kähler metric,andThese are respectively the Hodge Laplacian and the two Dolbeault Laplacians.

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 $V_F={\sum }_i|W_i{|}^2$.

There is a conflict in the printed question: with dynamical $X$ and the displayed superpotential, $F_X=\tilde{\varphi }\varphi =M$ forces $M=0$, so the $N=1$ theory has no vacuum branch parametrized by nonzero $M$. The natural intended calculation is the Kähler quotient of the $N=1$ D-flat SQED matter fields before imposing that extra F-term.

On this quotient, $|\varphi |=|\tilde{\varphi }|$ and the gauge-invariant coordinate is $M=\tilde{\varphi }\varphi$. Hence $|\varphi {|}^2=|\tilde{\varphi }{|}^2=|M|$, and the canonical Kähler potential descends toThe associated Kähler metric isIt is smooth for $M\ne 0$ and has a conical singularity at $M=0$, locally $\mathbb{C}/{\mathbb{Z}}_2$. Physically, the charged fields become massless and the U(1) gauge symmetry is restored there, so integrating out the vector multiplet to obtain a sigma-model metric ceases to be valid.

For $s=|z_1{|}^2+|z_2{|}^2>0$ and $a>0$, a geometric Kähler potential with derivative ${\phi }^{\prime }(s)=\sqrt{s^2+a^2}/(2s)$ defines a Kähler metric on ${\mathbb{C}}^2\setminus \{0\}$. Its tangential and radial eigenvalues are $\sqrt{s^2+a^2}/(2s)$ and $s/(2\sqrt{s^2+a^2})$. Their product is $1/4$, giving the same Riemannian volume form as ${\omega }_0=(i/2)\sum d{z}_j\wedge d{\bar{z}}_j$, although the metric is different. This construction makes preservation of volume compatible with anisotropic stretching.

For a Hermitian metric on a complex manifold, the Chern connection on its holomorphic tangent bundle corresponds to a real metric-compatible connection preserving the almost complex structure. Its mixed torsion vanishes, and in holomorphic coordinates its remaining coefficients are ${\Gamma }_{ij}^k=h^{k\bar{l}}{\partial }_i{h}_{j\bar{l}}$. Symmetry in $i,j$ is equivalent to $\partial \omega =0$; its conjugate gives $d\omega =0$. Thus zero torsion is exactly the Kähler metric condition, and then uniqueness identifies this connection with the Levi-Civita connection.

For canonical positive kinetic terms in global supersymmetry, $V={\sum }_k|W_k{|}^2$ has mixed Hessian matrix $V_{i\bar{j}}={\sum }_k{W}_{ki}\overline{W_{kj}}$ and holomorphic block $V_{ij}={\sum }_k\overline{W_k}{W}_{kij}$. The $2n$ real scalar squared-mass sum is $2\mathrm{tr}(m^{†}m)$ for the chiral-superfield fermion mass matrix $m=W_{ij}$. Each Weyl spinor contributes two states, giving the same fermion weighted sum. Thus their boson-minus-fermion supertrace vanishes even with F-term breaking. The holomorphic Hessian splits real scalar masses without changing their sum. Noncanonical Kähler metrics, supergravity, and radiative corrections require different formulas.