Chiral-coordinate solution and supersymmetry covariance

The left derivatives with respect to distinct odd generators anticommute by [F2]; each ${\mathrm{\partial }}_{\mu }$ commutes with them and with every other ${\mathrm{\partial }}_{\nu }$ because the coefficient functions are smooth. Expanding the displayed formulas for $D$ and $\bar{D}$, the only nonzero mixed terms in $\{D_{\alpha },{\bar{D}}_{\dot{\beta }}\}$ 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 $\{i{\sigma }_{\alpha \dot{\gamma }}^{\mu }{\bar{\theta }}^{\dot{\gamma }}{\mathrm{\partial }}_{\mu },-{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\beta }}}^L\}=-i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }$. Their sum is the stated mixed anticommutator. In each same-type anticommutator, the derivative-only terms vanish by [F2], the terms with one coordinate and one derivative cancel or are zero because the differentiated coordinate is of the other type, and the products of two identical-type odd coordinates vanish after symmetrization. Thus both same-type anticommutators are zero.

With $y=x+i\theta \sigma \bar{\theta }$, the left derivative gives ${\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L{y}^{\mu }=-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }$: the minus sign comes from differentiating the rightmost odd factor past the left odd ${\theta }^{\beta }$. The finite Taylor definition and ordinary algebra therefore give, on functions expressed in $(y,\theta ,\bar{\theta })$, ${\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L{\mathrm{\mid }}_x={\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L{\mathrm{\mid }}_y-i{\theta }^{\beta }{\sigma }_{\beta \dot{\alpha }}^{\mu }{\mathrm{\partial }}_{y^{\mu }}$ and ${\mathrm{\partial }}_{x^{\mu }}={\mathrm{\partial }}_{y^{\mu }}$. Substitution into $\bar{D}$ cancels the two $\theta \sigma {\mathrm{\partial }}_y$ terms, leaving ${\bar{D}}_{\dot{\alpha }}=-{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\alpha }}}^L{\mathrm{\mid }}_y$.

Direct expansion of $Q_{\alpha }$ against ${\bar{D}}_{\dot{\beta }}$ gives two opposite terms: $\{{\mathrm{\partial }}_{{\theta }^{\alpha }}^L,-i{\theta }^{\gamma }{\sigma }_{\gamma \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }\}=-i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }$ and $\{-i{\sigma }_{\alpha \dot{\gamma }}^{\mu }{\bar{\theta }}^{\dot{\gamma }}{\mathrm{\partial }}_{\mu },-{\mathrm{\partial }}_{{\bar{\theta }}^{\dot{\beta }}}^L\}=+i{\sigma }_{\alpha \dot{\beta }}^{\mu }{\mathrm{\partial }}_{\mu }$, so $\{Q_{\alpha },{\bar{D}}_{\dot{\beta }}\}=0$. All other terms vanish by the same independent-generator and odd-square identities. The two barred operators ${\bar{Q}}_{\dot{\alpha }}$ and ${\bar{D}}_{\dot{\beta }}$ also anticommute: their barred derivative parts anticommute, their $\theta {\mathrm{\partial }}_x$ parts anticommute, and each barred derivative anticommutes with each $\theta$ factor because it differentiates a different generator. Therefore ${\bar{D}}_{\dot{\beta }}({\delta }_{ϵ,\overline{ϵ}}\mathrm{\Phi })=0$ whenever ${\bar{D}}_{\dot{\beta }}\mathrm{\Phi }=0$, accounting for the odd parameter crossing the odd derivative.

Expand a function in the unique ordered basis of ${\mathrm{\Lambda }}_4$ with coefficients in $C^{\mathrm{\infty }}({\mathbb{R}}^{1,3},\mathcal{A})$. In $(y,\theta ,\bar{\theta })$ coordinates, ${\bar{D}}_{\dot{\alpha }}\mathrm{\Phi }=0$ for both dotted indices means both left derivatives in the barred generators vanish. For a fixed generator, distinct basis monomials containing it map under its derivative to distinct basis monomials not containing it, with coefficients $+1$ or $-1$; hence its derivative has zero kernel exactly on the span of monomials not containing it. Applying this to both barred generators shows $\mathrm{\Phi }$ is independent of $\bar{\theta }$. The remaining ordered basis is $1,{\theta }^1,{\theta }^2,{\theta }^1{\theta }^2$, so its coefficients are unique and give the four displayed component functions and parities.

The transformed field is smooth because it is a finite linear combination of spacetime derivatives of smooth coefficient functions, and it is chiral by step 2.2. Applying the unique basis expansion in step 3.1 to this transformed field yields one and only one tuple of four smooth components. This is the component transformation defined in the chiral-superfield definition; no field equation or quantum operator-domain assertion is involved.