PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6-luna)

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Positive scalar curvature annihilates untwisted spinor harmonic sections

Statement

Assume full AC. Let $M$ be a compact boundaryless even-dimensional Spin Riemannian manifold, and suppose its scalar curvature satisfies ${scal}_{g} (x) > 0$ for every $x \in M$ . Then every smooth untwisted spinor $ψ$ satisfying $D_{S} ψ = 0$ is zero.

Facts & Assumptions

Given: The closed Spin manifold, its untwisted spinor bundle and spin connection, and a smooth harmonic spinor $ψ$ .

[F1]

For the untwisted spinor module, the relative curvature vanishes, so the Lichnerowicz identity is $D_{S}^{2} = \nabla^{*} \nabla + {scal}_{g} / 4$ (Lichnerowicz formula for a geometric Dirac operator, Spinor and geometric Clifford bundles and connections are well defined).

[F2]

On a compact boundaryless Riemannian manifold, integration by parts gives $\int_{M} ⟨ \nabla^{*} \nabla ψ, ψ ⟩ = \int_{M} ∣ \nabla ψ ∣^{2}$ (Riemannian divergence theorem).

[F3]

The identity function on a nonempty compact subset of $R$ attains its minimum (Extreme value theorem: a continuous real function on a nonempty compact subset of $R$ attains a greatest and a least value); the scalar-curvature image of compact $M$ is compact because scalar curvature is continuous.

[A1]

The actual spinor module source is Complex spinor modules and complete reducibility for Euclidean Spin. Full AC is inherited through the connection and integration suppliers (The Axiom of Choice).

Proof

Given: $ψ \in C^{\infty} (S)$ and $D_{S} ψ = 0$ .

1.1A1F1F2given

Apply [F1], take the Hermitian inner product with $ψ$ , and integrate. Since $D_{S} ψ = 0$ , [F2] gives $0 = \int_{M} ∣ \nabla ψ ∣^{2} d {vol}_{g} + \frac{1}{4} \int_{M} {scal}_{g} ∣ ψ ∣^{2} d {vol}_{g}$ . Both integrands are nonnegative.

2.1A1F3step 1.1givenalgebra∎

If $M$ is nonempty, let $K = {scal}_{g} (M) \subseteq R$ . For any open cover of $K$ , its preimages under the continuous scalar-curvature map cover compact $M$ ; a finite subcover of $M$ maps to a finite subcover of $K$ . Thus $K$ is nonempty and compact. Apply [F3] to the identity function on $K$ to obtain $c = \min K$ . Since every value in $K$ is strictly positive and $c \in K$ , one has $c > 0$ . Step 1.1 then implies $0 \geq (c / 4) ∥ ψ ∥_{L^{2}}^{2}$ , so $ψ = 0$ almost everywhere and, by smoothness, everywhere. If $M$ is empty there is only the zero section. No twist is present; for a twisted connection the curvature term need not vanish.

Depends on

Used by

The round two-sphere has zero untwisted Spin Dirac kernel Example

Dependency tree · two levels

66 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Liviu I. Nicolaescu, Notes on the Atiyah–Singer Index Theorem, revision 15 November 2013 (standard reference, not scraped)
Konstantin Wernli, Lecture Notes on Spin Geometry (2019) (standard reference, not scraped)