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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.

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.