 Chiral projection by squared supercovariant derivatives

In flat four-dimensional N=1 supersymmetry, let ${\bar{D}}^2={\bar{D}}_{\dot{\alpha }}{\bar{D}}^{\dot{\alpha }}$, contracting the two spinor indices with the antisymmetric epsilon invariant tensor. The supercovariant derivative algebra with left derivatives has $\{{\bar{D}}_{\dot{\alpha }},{\bar{D}}_{\dot{\beta }}\}=0$. There are only two dotted components, so every product of three barred derivatives repeats one component and vanishes. Therefore ${\bar{D}}_{\dot{\alpha }}{\bar{D}}^{2}X=0$ for any superfield $X$: ${\bar{D}}^{2}X$ is a chiral superfield. This map is called chiral projection, although ${\bar{D}}^2$ is not an idempotent projector.

If $\Phi$ is a chiral superfield, holomorphic closure of chiral superfields also makes $W^{\prime }(\Phi )$ chiral. Thus an equation ${\bar{D}}^2{\Phi }^{†}=4W^{\prime }(\Phi )$ is consistent with the chiral constraint: applying any ${\bar{D}}_{\dot{\alpha }}$ gives zero on both sides. Both sides are scalar superfields, and the squared supersymmetric covariant derivative commutes with the supercharges, so the equation is Lorentz invariant and supersymmetry covariant.