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For a Twisted Dirac operator on a closed even-dimensional spin manifold, the theorem identifies its analytic Fredholm index with the integral of the top-degree part of the A-hat form times the Chern character. This is one geometric specialization of the general theorem for elliptic differential operators, which expresses the analytic index in terms of the topological class of the principal symbol. The heat proof pairs all positive eigenvalues by the odd Dirac operator, leaving only the index in the supertrace; Getzler rescaling computes its local limit from curvature.

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.

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.

The quantized fermions obey a Clifford algebra, $\{{\hat{\psi }}^a,{\hat{\psi }}^b\}=g^{ab}$. When $M$ is a spin manifold, they act on the spinor bundle andthe Hilbert space of square-integrable spinor fields. The supercharge becomes the Dirac operator, the Hamiltonian is one half of its square, and because $M$ is even dimensional, $(-1{)}^F$ is the spinor chirality operator.

A chosen spin structure on a spin manifold defines the complex spinor bundle associated to the spin representation. In dimension $2m$ its vector bundle rank is $2^m$. Its chirality matrix is $\Gamma =i^{m}c(e^1)\cdots c(e^{2m})$ in the negative Clifford multiplication convention. It satisfies ${\Gamma }^2=1$ and anticommutes with each $c(e^i)$, giving $S=S^{+}\oplus S^{-}$ and an odd Dirac operator. In odd dimension there is no such canonical chiral splitting of the irreducible spinor bundle.

On a spin manifold, tensor the spinor bundle $S$ with a Hermitian vector bundle $W$ carrying a unitary connection. Composing the induced tensor-product connection on a vector bundle with Clifford multiplication defines the twisted Dirac operator. In even dimension it maps $S^{+}\otimes W$ to $S^{-}\otimes W$ and conversely. On a closed manifold its chiral restriction $D_W^{+}:H^1(S^{+}\otimes W)\to L^2(S^{-}\otimes W)$ is a Fredholm operator.