= Solution
The <chirality matrix> is covariantly constant under the metric-preserving <spin connection> and anticommutes with every $\gamma^\mu$. Hence
$$
\not\nabla(\gamma_5\psi)=-\gamma_5\not\nabla\psi=m\gamma_5\psi,
\qquad
\boxed{(\not\nabla-m)\gamma_5\psi=0.}
$$
Thus 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 identity>
$$
[\nabla_\mu,\nabla_\nu]\psi=\frac14R_{\mu\nu ab}\gamma^{ab}\psi.
$$
Here the <rough Laplacian> includes the affine connection on the index of the first derivative. Clifford expansion gives
$$
\not\nabla^2\psi
=\nabla^2\psi+\frac12\gamma^{\mu\nu}[\nabla_\mu,\nabla_\nu]\psi
=\nabla^2\psi+\frac18\gamma^{\mu\nu}R_{\mu\nu ab}\gamma^{ab}\psi.
$$
In 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> is
$$
\not\nabla^2=\nabla^2-\frac R4.
$$
Multiplying the first-order equation by $-\not\nabla+m$ now proves
$$
\boxed{\left(-\nabla^2+\frac R4+m^2\right)\psi=0.}
$$
The 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 $T^\rho{}_{\mu\nu}=\Gamma^\rho_{\mu\nu}-\Gamma^\rho_{\nu\mu}$ 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 is
$$
\nabla^T_\mu\nabla^T_\nu\psi-\nabla^T_\nu\nabla^T_\mu\psi
=\frac14R^T_{\mu\nu ab}\gamma^{ab}\psi-T^\rho{}_{\mu\nu}\nabla^T_\rho\psi.
$$
Thus the <torsion correction to the Dirac square> gives the exact squared field equation
$$
\boxed{\left[-(\nabla^T)^2+\frac12\gamma^{\mu\nu}T^\rho{}_{\mu\nu}\nabla^T_\rho
-\frac18\gamma^{\mu\nu}R^T_{\mu\nu ab}\gamma^{ab}+m^2\right]\psi=0.}
$$
General 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 $\not\nabla_T=\not\nabla_0+Q$, where $Q=\gamma^\mu C_{\mu ab}\gamma^{ab}/4$, and the square contains $\{\not\nabla_0,Q\}+Q^2$. 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.
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