LemmaStatement: AI-adaptedProof: AI-generatedPipeline-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 term eliminates the coupled Laplace kernel on round S4

Statement

Let $E \to S^{4}$ be a finite-rank Hermitian vector bundle with a smooth unitary connection $\nabla$ , where $S^{4}$ is the unit round sphere with no boundary. Let $c > 0$ be a real constant. If a smooth section $s \in Γ (E)$ satisfies $(\nabla^{*} \nabla + c I) s = 0$ , where $\nabla^{*}$ is the formal adjoint for the round volume and Hermitian metric, then $s = 0$ .

Statement references

Donaldson, §5, printed p. 8

Facts & Assumptions

Given: The compact boundaryless round sphere, the Hermitian bundle and unitary connection, a constant $c > 0$ , and a smooth section $s$ satisfying the stated equation.

[F1]

The round $S^{4}$ and its metric are those fixed in the ADHM/twistor definitions. Zero-FI regular ADHM and framed twistor data

[F2]

The formal adjoint satisfies $\int_{S^{4}} ⟨ \nabla^{*} \nabla s, s ⟩ d v o l = \int_{S^{4}} ∣ \nabla s ∣^{2} d v o l$ on a compact manifold without boundary, by its defining integration-by-parts identity.

Proof

1.1givenF1F2algebra

Pair $(\nabla^{*} \nabla + c I) s = 0$ with $s$ using the Hermitian metric and integrate over the compact boundaryless $S^{4}$ in [F1]. By [F2], the result is $0 = \int_{S^{4}} ∣ \nabla s ∣^{2} d v o l + c \int_{S^{4}} ∣ s ∣^{2} d v o l$ . Both terms are nonnegative and $c > 0$ , so $\int_{S^{4}} ∣ s ∣^{2} d v o l = 0$ .

2.1givenstep 1.1algebra∎

The pointwise function $∣ s ∣^{2}$ is continuous and nonnegative. If it were positive at some point, it would be bounded below by a positive constant on a nonempty open neighborhood, which has positive Riemannian volume, contradicting the zero integral in step 1.1. Therefore $s = 0$ everywhere.

Positive scalar term eliminates the coupled Laplace kernel on round S4

Statement

Statement references

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources