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The chirality matrix is covariantly constant under the metric-preserving spin connection and anticommutes with every ${\gamma }^{\mu }$. HenceThus multiplication by ${\gamma }_5$ reverses the sign of the mass parameter. At $m=0$ it maps solutions into solutions of the same Dirac equation.

Use the spinor curvature identityHere the rough Laplacian includes the affine connection on the index of the first derivative. Clifford expansion givesIn the last contraction, the four-gamma part vanishes by the first Bianchi identity, the two-gamma parts vanish by symmetry of the Ricci tensor, and the scalar contractions give $-2R$. Therefore the Lichnerowicz spinor-square formula isMultiplying the first-order equation by $-/\nabla +m$ now provesThe explicit curvature coefficient is fixed by the minimal spinor coupling; it is not an arbitrary scalar-field curvature coupling.

For the general metric-preserving connection, define $T^{\rho }{}_{\mu \nu }={\Gamma }_{\mu \nu }^{\rho }-{\Gamma }_{\nu \mu }^{\rho }$ and let $R^T$ be its curvature. Distinguish the spin connection acting on the spinor from the full second derivative acting also on its derivative index. Their antisymmetric relation isThus the torsion correction to the Dirac square gives the exact squared field equationGeneral torsion removes the curvature symmetries used in the preceding simplification. The curvature contraction may now contain nonscalar Clifford terms, in addition to the displayed first-order torsion term; simply replacing $R$ by a torsionful scalar in the torsion-free answer is not valid.

Equivalently, write the connection as Levi-Civita plus contorsion tensor $C_{\mu ab}$. Then $/{\nabla }_T=/{\nabla }_0+Q$, where $Q={\gamma }^{\mu }{C}_{\mu ab}{\gamma }^{ab}/4$, and the square contains $\{/{\nabla }_0,Q\}+Q^2$. This exhibits the extra couplings through derivatives of contorsion, first-order spinor derivatives and quadratic contorsion. Metric compatibility still preserves ${\gamma }_5$, so the mass-sign result remains valid. Spin therefore couples directly to curvature through its connection and the universal $R/4$ term, and to torsion through additional spin-dependent terms. The scalar reduction in the torsion-free square does not mean the full spinor transport is insensitive to the remaining Riemann curvature.