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Choose the Riemann curvature tensor conventionwith positive scalar curvature for a round sphere. The connection preserves the gamma matrices. If $D={\gamma }^{\mu }{\nabla }_{\mu }$, its square isHere ${\nabla }^2$ is the rough Laplacian, including the connection on the derivative index. In the product of two antisymmetric gamma pairs, the fully antisymmetric four-gamma term vanishes by the first Bianchi identity. The two-gamma contractions vanish because the Ricci tensor is symmetric, while the scalar contractions give $-2R$. Thus the Lichnerowicz spinor-square formula isMultiplying $(D+m)\psi =0$ by $-D+m$, with constant $m$, provesIn Riemannian signature $D$ in this positive-Clifford convention is formally anti-self-adjoint; the more usual self-adjoint Dirac operator $iD$ has square $-{\nabla }^2+R/4$. Stating this convention avoids an apparent sign conflict between the two forms of the formula.

Now let $D\psi =0$ on the compact Riemannian spin manifold. Use the positive-definite spinor inner product and integrate the squared equation. Integration by parts has no boundary contribution, givingBoth terms are nonnegative. Smoothness then forces $\nabla \psi =0$ pointwise. Consequently every harmonic spinor is parallel, and $R|\psi {|}^2=0$. A nonzero parallel spinor has constant positive norm on its connected component, so its scalar curvature vanishes there. This is compact harmonic-spinor rigidity under nonnegative scalar curvature.

The full Ricci conclusion follows from integrability, not merely from vanishing scalar curvature. Since a parallel spinor has zero curvature for its spin connection, contract its Ricci identity with ${\gamma }^{\nu }$:The three-gamma term vanishes by the Bianchi identity; the remaining contractions give the displayed Ricci term. For each fixed $\mu$, put $v_{\nu }=R_{\mu \nu }$. Clifford multiplication twice givesPositive definiteness and $\psi \ne 0$ imply $v=0$. Thus the Riemannian parallel-spinor Ricci-flatness result isOn a connected manifold a nontrivial parallel solution is nowhere zero, so this holds everywhere. On a disconnected manifold it holds on every component supporting a nonzero spinor; the stated nowhere-vanishing hypothesis ensures all components are covered. The positive-definite metric tensor is essential to the last inference.