Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 59 / 1 / d

First use a genuine simple bivector of the specified chirality, with spinor part ${\varphi }^{AB}={\alpha }^A{\alpha }^B$. It is a exterior product: choose independent primed spinors ${\rho }^{A^{\prime }},{\sigma }^{A^{\prime }}$ and take $u^{A{A}^{\prime }}={\alpha }^A{\rho }^{A^{\prime }}$, $v^{B{B}^{\prime }}={\alpha }^B{\sigma }^{B^{\prime }}$. Their antisymmetric product is proportional to $w^{A{A}^{\prime }B{B}^{\prime }}={\alpha }^A{\alpha }^B{ϵ}^{A^{\prime }{B}^{\prime }}$.

For any antisymmetric $w^{ab}$, the Riemann curvature tensor acts on both tensor indices. In the printed curvature convention its complete contraction isbecause the Ricci tensor is symmetric. Substituting the one-chirality bivector and the chiral spinor curvature operator decomposition, the primed-operator term vanishes on contraction with the symmetric ${\alpha }^A{\alpha }^B$. The other term is $2{\Delta }_{AB}({\alpha }^A{\alpha }^B)$. Hence ${\Delta }_{AB}({\alpha }^A{\alpha }^B)=0$.

Repeat with $\alpha +\beta$ and subtract the identities for $\alpha$ and $\beta$. Linearity leaves twice the mixed symmetric product, giving the contracted chiral curvature identityThe polarization works for spinor fields as well as pointwise spinors: the chiral spinor curvature operator is algebraic by the preceding part.

The displayed source expression ${\alpha }_A{\beta }_B{ϵ}_{A^{\prime }{B}^{\prime }}$ needs this clarification. For independent $\alpha ,\beta$ it is not antisymmetric in the spacetime indices. Its antisymmetric part is ${\alpha }_{(A}{\beta }_{B)}{ϵ}_{A^{\prime }{B}^{\prime }}$, which is generally not simple. The proof above uses a genuinely simple bivector first and then polarization, so it establishes the requested conclusion without relying on the incorrect simplicity assertion.