= Contracted chiral curvature identity
{title2=$\Delta_{AB}\phi^{AB}=0\quad(\phi^{AB}=\phi^{(AB)})$}
For the <Levi-Civita connection>, the action of the curvature commutator on a <bivector> $w^{ab}$ obeys $[\nabla_a,\nabla_b]w^{ab}=0$: both resulting curvature contractions pair the symmetric <Ricci tensor> with an antisymmetric tensor. For a bivector of one chirality, $w^{AA'BB'}=\phi^{AB}\epsilon^{A'B'}$, this is $2\Delta_{AB}\phi^{AB}=0$. In particular it holds for $\phi^{AB}=\alpha^A\alpha^B$; polarization gives $\Delta_{AB}(\alpha^{(A}\beta^{B)})=0$. The polarization step avoids incorrectly calling $\alpha_A\beta_B\epsilon_{A'B'}$ a simple bivector for independent spinors.
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