ExampleConstruction: AI-generatedVerification: 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.

Local transforms versus global injectivity

Example

Let $U_{N}$ and $U_{S}$ be the two stereographic charts of $S^{4}$ , and let $ξ \in H^{1} (Z, F \otimes O_{Z} (- 2))$ be a global class under the coupled theorem's hypotheses. On each chart compute the fibre-harmonic section from a global closed Dolbeault representative. On $U_{N} \cap U_{S}$ the two local sections agree because they are the harmonic projection on the same fibre and are unchanged by replacing the representative by a Dolbeault-exact form. If the glued section is zero, the global class is zero by injectivity. Thus a local contour expression alone establishes neither compatibility on the overlap nor injectivity of the global class map.

Example references

Positive incidence geometry for the coupled Penrose transform

Dolbeault groups of the negative twistor-fibre twists

Injective fibre-harmonic edge for coupled Penrose cohomology

Rawnsley, §§2–3, pp. 46–54

Wells, §3, pp. 488–494

Facts & Assumptions

Given: The two stereographic charts, the global twistor fibration, a global analytic cohomology class, and its global Dolbeault representative.

[A1]

The theorem uses the local analytic sheaf/Dolbeault resolution comparison; its holomorphic-bundle and full AC hypotheses are inherited here. Coupled Penrose vanishing on the round four-sphere

[F1]

Each base point has one fibre $L_{x} ≅ C P^{1}$ in the fixed global incidence fibration, and the normalized harmonic generator is intrinsic on that fibre. Positive incidence geometry for the coupled Penrose transform, Dolbeault groups of the negative twistor-fibre twists

[F2]

The fibre-harmonic coefficient is independent of representative and the resulting global map is injective: its zero class has horizontal remainder in $O (- 1)^{\oplus 2}$ , which has no holomorphic sections. Injective fibre-harmonic edge for coupled Penrose cohomology

Verification

Given: The global class and exact construction in [F1]–[F2].

1.1F1F2algebra

For $x \in U_{N} \cap U_{S}$ , both chart computations restrict the same global closed form to the same fibre $L_{x}$ . They use the same fibre Dolbeault operator and normalized harmonic projection, so both give the same value $s (x)$ . If the global representative changes by ${\overset{ˉ}{\partial}}_{F} β$ , its vertical restriction changes by ${\overset{ˉ}{\partial}}_{V} β$ ; harmonic forms are orthogonal to exact forms, so the harmonic projection is unchanged. Hence the chart sections agree and are independent of the representative, and they glue to a global section on $S^{4}$ .

2.1A1F2step 1.1algebra∎

If the glued section is zero, the decomposition in [F2] writes the global representative as a horizontal $(0, 1)$ -form plus a global Dolbeault-exact term. Closedness makes the horizontal form fibrewise holomorphic with coefficient bundle $O (- 1)^{\oplus 2} \otimes E_{x}$ . Its holomorphic-section space is zero by the fibre calculation in [F2], so the horizontal form vanishes; the original class is exact and is therefore zero. Thus the global map is injective. A formula on one chart supplies neither the overlap equality in step 1.1 nor this zero-kernel argument.

Local transforms versus global injectivity

Example

Example references

Facts & Assumptions

Verification

Depends on

Dependency tree · 2 levels

Sources