Lichnerowicz spinor-square formula ($/{\nabla }^2={\nabla }^2-R/4$)

With $\{{\gamma }^{\mu },{\gamma }^{\nu }\}=2g^{\mu \nu }$ and spin curvature $[{\nabla }_{\mu },{\nabla }_{\nu }]=R_{\mu \nu ab}{\gamma }^{ab}/4$, the Dirac equation operator squares to the rough Laplacian minus $R/4$. In the curvature contraction the four-gamma term vanishes by the first Bianchi identity, the two-gamma terms vanish by Ricci symmetry, and the scalar term is $-2R$. In Riemannian signature $i/\nabla$ is the self-adjoint convention and its square is $-{\nabla }^2+R/4$.

The four-gamma product decomposes into fully antisymmetric, two-gamma and scalar parts. The first part vanishes on contraction by the first Bianchi identity, and the two-gamma parts vanish because they contract symmetric Ricci curvature against antisymmetric gamma matrices. The scalar contractions are $-g^{ac}{g}^{bd}{R}_{abcd}+g^{ad}{g}^{bc}{R}_{abcd}=-2R$. Here $R=g^{ac}{g}^{bd}{R}_{abcd}$ fixes the curvature-slot convention.

// Target: second-fundamental-form.bigb

For a metric-compatible connection with torsion tensor $T^{\rho }{}_{\mu \nu }={\Gamma }_{\mu \nu }^{\rho }-{\Gamma }_{\nu \mu }^{\rho }$, expand the square using Clifford multiplication and the full affine second derivative. Its antisymmetric part is spin curvature minus $T^{\rho }{}_{\mu \nu }{\nabla }_{\rho }$. This gives the displayed formula. The curvature need not have the torsion-free pair symmetry or first Bianchi identity, so its Clifford contraction need not reduce to a scalar $-R/4$. General torsion produces additional first-order and nonscalar curvature couplings.

On a compact boundaryless Riemannian spin manifold with $R\ge 0$, integrating the Lichnerowicz spinor-square formula against a harmonic spinor gives $\int (|\nabla \psi {|}^2+R|\psi {|}^2/4)=0$. Positivity forces the spinor to be parallel. Its norm is constant on each connected component, so a nontrivial solution is nowhere zero there. Strictly positive scalar curvature at any point on such a component excludes nonzero harmonic spinors.

A parallel spinor has zero spin curvature. Contracting the spinorial Ricci identity with a gamma matrix gives $R_{\mu \nu }{\gamma }^{\nu }\psi =0$. For each fixed $\mu$, squaring Clifford multiplication by that Ricci row gives $|R_{\mu \cdot }{|}^2\psi =0$. Riemannian positivity and a nonzero spinor force every row to vanish. This implication is local and does not itself require compactness; the compact hypothesis is used to obtain parallelism from harmonicity.