Contracted chiral curvature identity (${\Delta }_{AB}{\varphi }^{AB}=0({\varphi }^{AB}={\varphi }^{(AB)})$)

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, $w^{A{A}^{\prime }B{B}^{\prime }}={\varphi }^{AB}{ϵ}^{A^{\prime }{B}^{\prime }}$, this is $2{\Delta }_{AB}{\varphi }^{AB}=0$. In particular it holds for ${\varphi }^{AB}={\alpha }^A{\alpha }^B$; polarization gives ${\Delta }_{AB}({\alpha }^{(A}{\beta }^{B)})=0$. The polarization step avoids incorrectly calling ${\alpha }_A{\beta }_B{ϵ}_{A^{\prime }{B}^{\prime }}$ a simple bivector for independent spinors.