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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.

The chiral spinor curvature operator acts algebraically on a two-dimensional spinor space. We may therefore lower its matrix index using the symplectic form and write ${\Delta }_{AB}{\alpha }_C=X_{ABDC}{\alpha }^D$. Symmetry of ${\Delta }_{AB}$ gives symmetry in the first pair of $X$. Preservation of ${ϵ}_{CD}$ says that its action is symplectic; lowering the endomorphism index of a symplectic infinitesimal transformation makes a symmetric two-index tensor. Thus $X_{ABCD}=X_{(AB)(CD)}$.

The additional Riemann curvature tensor symmetry $R_{abcd}=R_{cdab}$ implies $X_{ABCD}=X_{CDAB}$: project both antisymmetric spacetime index pairs onto the unprimed chiral summand. Equivalently this is the symmetry of the unprimed diagonal block of the curvature operator on differential two-forms. The contracted chiral curvature identity is another way of expressing the same three constraints on the initially nine-component pair-symmetric spinor.

Consequently $X$ is a symmetric bilinear form on the three-dimensional space ${\mathrm{Sym}}^{2}S$, with six independent components. Its total symmetrization ${\Psi }_{ABCD}=X_{(ABCD)}$ has five components. The one-dimensional remainder is spanned byIt is symmetric within and between pairs, and $K_{(ABCD)}=0$. The latter follows from the two-dimensional alternating-form identity, or directly by evaluating its components with ${ϵ}_{01}=1$. Hence the chiral Riemann curvature spinor decomposition isA contraction of a totally symmetric spinor with an symplectic form vanishes. Meanwhile the explicit two-term expression for $K$ gives $K_{ABCD}{ϵ}^{AC}{ϵ}^{BD}=3$. Therefore the scalar in the printed normalization isIn particular, $\Psi =X-\Lambda K$ has zero such trace and is the Weyl curvature spinor. The coefficient is fixed by the factor implicit in the printed symmetrization; substituting a differently normalized scalar-curvature term would change the relation between its scalar parameter and this contraction.