ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-12

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.

Gauge transformation of a connection one form

Example

On an open set with a real line-bundle frame $e$ and connection form $ω$ , change frame to $e^{'} = e \exp (f)$ for a smooth real function $f$ . Then $ω^{'} = ω + d f$ . Thus zero coefficients can become nonzero without changing the connection.

Facts & Assumptions

Given: The supplied frame, connection and smooth $f$ on its domain.

[F1]

For $e^{'} = e A$ , $ω^{'} = A^{- 1} ω A + A^{- 1} d A$ (Connection one form transformation law).

Verification

1.1F1given

Since $e^{f} > 0$ , $e^{'}$ is a frame everywhere. Scalar coefficients commute, and $d (e^{f}) = e^{f} d f$ , so [F1] gives $ω^{'} = ω + d f$ . This changes coordinates of the same derivative, not the intrinsic connection.

2.1step 1.1∎

On the trivial line over $R$ with $ω = 0$ and $f = x$ , the new form is $d x$ . In particular $\nabla_{\partial_{x}} e^{'} = e^{'}$ whereas $\nabla_{\partial_{x}} e = 0$ . The old constant section has new coefficient $e^{- x}$ , and its new covariant derivative is $d (e^{- x}) + d x e^{- x} = 0$ , confirming agreement on an actual section. Constant $f$ gives $d f = 0$ ; the frame never vanishes even when $f = 0$ .

Depends on

Connection one form transformation law

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 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

Ved Datar, Lectures on Riemannian Geometry (standard reference, not scraped)