Off-shell $N=1$ supersymmetry of the Wess–Zumino sigma-model action

For any even smooth superspace function $X$, substitute the operators from [F1] into $\delta X$. Since the odd parameters are constant, the odd Leibniz rule gives ${ϵ}^{\alpha }{\mathrm{\partial }}_{{\theta }^{\alpha }}^{L}X=-{\mathrm{\partial }}_{{\theta }^{\alpha }}^L({ϵ}^{\alpha }X)$ and $-{\overline{ϵ}}^{\dot{\alpha }}{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^{L}X={\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L({\overline{ϵ}}^{\dot{\alpha }}X)$. The remaining terms are ordinary spacetime derivatives, so $\delta X=-{\mathrm{\partial }}_{{\theta }^{\alpha }}^L({ϵ}^{\alpha }X)+{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L({\overline{ϵ}}^{\dot{\alpha }}X)+{\mathrm{\partial }}_{\mu }(-i{ϵ}^{\alpha }{\sigma }_{\alpha \dot{\beta }}^{\mu }{\bar{\theta }}^{\dot{\beta }}X+i{\overline{ϵ}}^{\dot{\alpha }}{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }X)$.

For a chiral superfield $X(y,\theta )$, use $y=x+i\theta \sigma \bar{\theta }$ and the left-derivative rules in [F1]–[F3]. The chain rule gives $Q_{\alpha }={\mathrm{\partial }}_{{\theta }^{\alpha }}^L{\mathrm{\mid }}_y$ and ${\bar{Q}}_{\dot{\alpha }}=-{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L{\mathrm{\mid }}_y+2i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\mathrm{\partial }}_{y^{\mu }}$. Since $X$ is independent of $\bar{\theta }$, $\delta X={ϵ}^{\alpha }{\mathrm{\partial }}_{{\theta }^{\alpha }}^{L}X+2i{\overline{ϵ}}^{\dot{\alpha }}{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\mathrm{\partial }}_{\mu }X$.

Expand the supercharge anticommutators from [F1]. The only nonzero terms in $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}$ are $\{{\mathrm{\partial }}_{{\theta }^{\alpha }}^L,i{\theta }^{\gamma }{\sigma }_{\gamma \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }\}=i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }$ and $\{-i{\sigma }_{\alpha \dot{\gamma }}^{\mu }{\bar{\theta }}^{\dot{\gamma }}{\mathrm{\partial }}_{\mu },-{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\beta }}}^L\}=i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }$. Terms with two odd coordinates cancel under symmetrization, and derivatives in distinct generators anticommute by [F3]. Hence $\{Q_{\alpha },{\bar{Q}}_{\dot{\beta }}\}=2i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }$ and both same-chirality anticommutators are zero.

Apply step 1.1 to the even D-term integrand $X=K(\bar{\mathrm{\Phi }},\mathrm{\Phi })$, whose finite nilpotent Taylor expansion is smooth. Each odd-coordinate derivative has zero top coefficient in the ordered Grassmann basis from [F3], so its Berezin integral vanishes. The spacetime-divergence term integrates to zero because the field-dependent part of $K$ is compactly supported and its constant exterior value has zero supersymmetry variation. The fixed physical factor $-1\mathrm{/}4$ multiplies zero, so the D-term action is invariant.

The first term in step 1.2 integrates to zero under $d^2\theta$ by the same coefficient-extraction calculation as step 2.1. The second is the spacetime divergence ${\mathrm{\partial }}_{\mu }(2i{\overline{ϵ}}^{\dot{\alpha }}{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }X)$; compact support removes its boundary term. For the finite Taylor extension in the postulate, holomorphicity ${\mathrm{\partial }}_{\bar{i}}W=0$ and the odd derivation rule give the chain identity ${\bar{D}}_{\dot{\alpha }}W(\mathrm{\Phi })={\sum }_i{W}_i(\mathrm{\Phi }){\bar{D}}_{\dot{\alpha }}{\mathrm{\Phi }}^i=0$, because each ${\mathrm{\Phi }}^i$ is chiral; hence $W(\mathrm{\Phi })$ is chiral. Complex conjugation gives the corresponding antichirality of its conjugate. The fixed physical factor $1\mathrm{/}2$ and its conjugate preserve this zero variation, so the F-term action is invariant.

Work in ${\mathcal{G}}_{ϵ}$ for one variation, and adjoin a second independent exterior parameter copy for commutators of two variations. The operators extend by the graded derivative rule and annihilate the coefficient algebra, exactly as in [F1]. By [F2], the transformations preserve the chiral constraint, so the transformed field has the unique component expansion in [F1]. Coefficient extraction is linear; therefore the operator identities in step 1.3 induce the same translation algebra on every full formal component, including the independent even auxiliary coefficient ${\mathcal{F}}^i$. Products of odd parameters and odd component maps lie in the total even subspace $({\mathcal{G}}_{ϵ}{)}_{\bar{0}}$, while fermionic variations lie in its total odd subspace, and derivatives preserve coefficientwise compact support; thus this argument does not restrict the auxiliary coefficient to $\mathbb{C}$ or assert closure on the bosonic slice. No Euler–Lagrange equation was used, proving off-shell closure.