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Fix the two-component spinor calculus convention ${ϵ}_{01}={ϵ}^{01}=1$, with lowering ${\xi }_A={\xi }^B{ϵ}_{BA}$ and raising ${\xi }^A={ϵ}^{AB}{\xi }_B$. Thus ${ϵ}^{AB}{ϵ}_{AB}=2$; use the same convention for primed indices. The spinor covariant derivative preserves these symplectic forms.

Write $C_{A{A}^{\prime }B{B}^{\prime }}=[{\nabla }_{A{A}^{\prime }},{\nabla }_{B{B}^{\prime }}]$. It is antisymmetric on exchanging the two spacetime index pairs. The tensor-product decomposition of a differential two-form isThis can also be checked directly: symmetrize or antisymmetrize separately in $A,B$ and $A^{\prime },B^{\prime }$. Antisymmetry under simultaneous pair exchange leaves precisely the symmetric–antisymmetric and antisymmetric–symmetric terms. An antisymmetric two-index spinor in a two-dimensional space is a multiple of its symplectic form. Therefore the chiral spinor curvature operators areExchanging $A,B$ in the first expression and relabeling $A^{\prime },B^{\prime }$ shows ${\Delta }_{AB}={\Delta }_{BA}$; the other symmetry follows identically. The projections reconstruct the whole six-component commutator, provingExpanding each projected commutator and relabeling the contracted indices gives the requested derivative expressions:These formulas act tensorially on the field being differentiated, including the covariant derivative's action on the intermediate derivative index. The stated convention removes any ambiguity about signs from contracting alternating spinor forms in the opposite order.

For a torsion-free connection, the covariant Hessian of a scalar is symmetric, so $[{\nabla }_a,{\nabla }_b]f=0$. Moreover, a commutator of two covariant derivatives acts as a derivation on tensor products. Applying the Leibniz rule twice, the terms containing one derivative on each factor cancel between the two orders:Projecting with the parallel symplectic form preserves this identity for the chiral spinor curvature operator. Since its scalar projection obeys ${\Delta }_{AB}f=0$, we obtainThus the chiral spinor curvature operator is algebraic over scalar functions even though its displayed expression uses second derivatives. If torsion tensor is allowed, $[{\nabla }_a,{\nabla }_b]f=-T^c{}_{ab}{\partial }_{c}f$ need not vanish, explaining why metric compatibility alone would not prove the claim.

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.