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For $P=W^{\prime }$ and $S=|P{|}^2-P^{\prime }{\theta }_1{\theta }_2-\overline{P^{\prime }}{\bar{\theta }}_1{\bar{\theta }}_2$, four odd graded derivations preserve $S$: ${\theta }_1{\partial }_z-\bar{P}{\partial }_{{\theta }_2}$, ${\theta }_2{\partial }_z+\bar{P}{\partial }_{{\theta }_1}$ and their barred counterparts. Left Grassmann derivatives fix the signs.

A connection on a vector bundle is a linear operator $\nabla :\Gamma (E)\to {\Omega }^1(M;E)$ satisfying $\nabla (fs)=df\otimes s+f\nabla s$. Choose coordinates $x^i$ and a local frame $e_1,\dots ,e_m$, and define the connection matrix by $\nabla e_b={\sum }_a{A}^a{}_b\otimes e_a$, with $A^a{}_b={\sum }_i{A}^a{}_{bi}d{x}^i$. For a section $s={\sum }_a{s}^a{e}_a$, the coefficient column satisfiesThe symbol $d$ on the right means the ordinary exterior derivative of the coefficient functions.

Extend uniquely to $E$-valued differential forms by the graded Leibniz ruleIn the same local trivialization, an $E$-valued form is a column $\sigma =({\sigma }^a)$ and the covariant exterior derivative isMoving the one-form $A$ past the scalar $r$-form produces exactly the $(-1{)}^r$ in the graded rule. This explains the sign rather than replacing the rule by an ungraded product formula.

The endomorphism bundle has the induced endomorphism bundle connection characterized on sections by ${\nabla }_X^{\mathrm{End}}(T)s={\nabla }_X(Ts)-T{\nabla }_{X}s$. Its coefficient formula is $dT+AT-TA$. On an endomorphism-valued $r$-form $T$, extend as a graded derivation of the endomorphism-valued exterior product, givingExplicitly its $(a,b)$ entry is $d{T}^a{}_b+{\sum }_c{A}^a{}_c\wedge T^c{}_b-(-1{)}^r{\sum }_c{T}^a{}_c\wedge A^c{}_b$. This version is characterized byThe endomorphism-valued version also satisfies $d_A(T\wedge S)=d_{A}T\wedge S+(-1{)}^{r}T\wedge d_{A}S$. Products here combine composition of endomorphisms with wedge products of coefficient forms; matrix factors must retain their order.

These local formulas are independent of the local frame. For $e^{\prime }=eg$, the change of frame of a vector-bundle connection gives $A^{\prime }=g^{-1}Ag+g^{-1}dg$, ${\sigma }^{\prime }=g^{-1}\sigma$ and $T^{\prime }=g^{-1}Tg$. Using $d(g^{-1})=-g^{-1}(dg)g^{-1}$ gives $d_{A^{\prime }}{\sigma }^{\prime }=g^{-1}{d}_A\sigma$ and $d_{A^{\prime }}{T}^{\prime }=g^{-1}(d_{A}T)g$. Thus both extended operators are globally defined.

Here L-superfield means a left-chiral superfield: ${\bar{D}}_{\dot{\alpha }}\Phi =0$, and its complex conjugate is an antichiral superfield, $D_{\alpha }{\Phi }^{\ast }=0$. Interpret the two barred derivatives as the contracted scalar ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$. The key point is chiral projection by squared supercovariant derivatives. Since the two components of $\bar{D}$ anticommute, any product of three barred derivatives repeats a component and is zero. Thusfor every superfield $X$, even when $X$ is antichiral. In particular, ${\bar{D}}^2{\Phi }^{\ast }$ is chiral.

The right side is chiral too. Because $\Phi$ is even, the graded derivation rule gives ${\bar{D}}_{\dot{\alpha }}({\Phi }^2)=({\bar{D}}_{\dot{\alpha }}\Phi )\Phi +\Phi ({\bar{D}}_{\dot{\alpha }}\Phi )=0$. Constant $m$ and $\lambda$ do not contribute derivatives. Consequentlyso applying the chiral constraint to the equation imposes no additional incompatible condition. This is the holomorphic closure of chiral superfields; a generic nonholomorphic right side would not satisfy it.

Both sides have zero Grassmann parity and are Lorentz scalars: the two dotted derivative indices are contracted with the invariant epsilon tensor, and $\Phi$ is a scalar superfield. Moreover, each covariant derivative anticommutes with every supercharge, so its squared contraction commutes with the supercharges. Hence the equation transforms covariantly under supersymmetry. It is the chiral field equation of motion of the Wess–Zumino model with superpotential $\mathcal{W}(\Phi )=m{\Phi }^2/2+\lambda {\Phi }^3/3$, for which ${\mathcal{W}}^{\prime }=m\Phi +\lambda {\Phi }^2$, using the normalization in the PDF. With mass dimensions $[\Phi ]=1$, $[\bar{D}]=1/2$, $[m]=1$ and $[\lambda ]=0$, each side has dimension two. No component field equations of motion are needed for this consistency argument.

Factor the odd parameter on the left, $\delta =ϵs$. The resulting left-acting BRST differential obeys the graded Leibniz ruleFor the odd Grassmann field $c$, the bracket in the transformation is a graded commutator: $[c,c{]}_{\mathrm{gr}}=2c^2$, not the identically zero ordinary commutator of a matrix with itself. Thus $sc=i{c}^2$, while $s{A}_{\mu }=D_{\mu }c$, $s\bar{c}=h$ and $sh=0$.

On the ghost,On the gauge field, variation of the connection and the adjoint covariant derivative givesHere $D_{\mu }$ is even and therefore obeys the ordinary product rule. Also $s^2\bar{c}=sh=0$ and $s^{2}h=0$, without using any field equation; this is off-shell nilpotence supplied by the Nakanishi-Lautrup field.

Applying the graded Leibniz rule twice cancels the two cross terms:The square is consequently an even graded derivation. Since it vanishes on every generator, it vanishes inductively on every polynomial in the fields. Hence $s^2\mathcal{O}=0$ for every such operator. This genuine result is stronger than the automatic vanishing obtained by merely setting ${ϵ}^2=0$; two independent transformation parameters also give a vanishing commutator.

Set $P(z)=W^{\prime }(z)$ and $\overline{P}(\bar{z})=\overline{W^{\prime }(z)}$, treating $z,\bar{z}$ as independent coordinates for Wirtinger derivatives. Use left Grassmann derivatives, so ${\partial }_{{\theta }_2}({\theta }_1{\theta }_2)=-{\theta }_1$. The given odd vector field acts by $V_{1}z={\theta }_1$, $V_1{\theta }_2=-\overline{P}$, and zero on ${\theta }_1,\bar{z},{\overline{\theta }}_i$. Consequently its square vanishes on every coordinate. The square of an odd graded derivation is an even graded derivation, so the nilpotent operator property holds on every function:

For the action in this zero-dimensional supersymmetric field theory, ${\partial }_{z}S=P^{\prime }\overline{P}-P^{\prime \prime }{\theta }_1{\theta }_2$ and ${\partial }_{{\theta }_2}S=P^{\prime }{\theta }_1$. The cubic term vanishes after multiplication by ${\theta }_1$, and the other two terms cancel:Three further independent odd symmetries areFor $V_2$, use ${\partial }_{{\theta }_1}S=-P^{\prime }{\theta }_2$; the same cancellation proves invariance. The barred calculations exchange barred and unbarred variables and coefficients. Each graded derivation is also a nilpotent operator, by its coordinate action. Their different odd coefficients and coordinate derivatives make them linearly independent. These are odd symmetries of a zero-dimensional polynomial model.

If an odd graded derivation $Q$ preserves the action and integration measure, then $⟨QX⟩=0$ whenever boundary contributions and anomalies vanish. This applies to finite-dimensional zero-dimensional supersymmetric field theories as well as suitably regulated field integrals.

An $\mathbb{R}$-linear degree-one graded derivation of the exterior algebra of smooth differential forms that equals the differential on functions and squares to zero is unique. A smooth bump function and the graded Leibniz rule first establish locality. Then $d(d{x}^i)=0$, and applying the graded Leibniz rule to $\alpha ={\sum }_I{a}_{I}d{x}^I$ gives the displayed coordinate formula. The chain rule makes that formula coordinate-independent.

A zero-dimensional model is a finite-dimensional integral with bosonic and Grassmann variables, rather than a field depending on spacetime. Odd graded derivations preserve its action and integration measure. Such models expose supersymmetric Ward identities and supersymmetric localization without spacetime dynamics.