= Chiral projection by squared supercovariant derivatives
{title2=$\bar D_{\dot\alpha}\bar D^2X=0$}
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 $\bar D_{\dot\alpha}\bar D^2X=0$ for any <superfield> $X$: $\bar D^2X$ 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 $W'(\Phi)$ chiral. Thus an equation $\bar D^2\Phi^\dagger=4W'(\Phi)$ 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.
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