 Lichnerowicz spinor-square formula

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 $-{\nabla }^2+R/4$.