ExampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passaudited 2026-09-07

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.

A cylinder with oppositely oriented boundary components

Example

For the product orientation on $[0, 1] \times S^{1}$ , the boundary circles ${1} \times S^{1}$ and ${0} \times S^{1}$ receive opposite orientations.

Facts & Assumptions

Given: The interval $[0, 1]$ with positive tangent $\partial_{t}$ , the circle $S^{1}$ with a chosen orientation, and $[0, 1] \times S^{1}$ with the product orientation in that factor order.

[L1]

The product orientation orders the interval tangent before an oriented basis of the circle tangent (Product orientations).

[L2]

A boundary basis is positive exactly when placing an outward vector before it gives the ambient orientation (Induced boundary orientation).

Verification

technique · direct

1.1givenalgebra

Along ${1} \times S^{1}$ , the outward vector is $+ \partial_{t}$ , whereas along ${0} \times S^{1}$ it is $- \partial_{t}$ .

2.1givenL1L2step 1.1algebra∎

If $v$ is a positive tangent vector to $S^{1}$ , then $(\partial_{t}, v)$ is positive by [L1]. Thus [L2] makes $v$ positive on ${1} \times S^{1}$ , while it makes $- v$ positive on ${0} \times S^{1}$ , because $(- \partial_{t}, - v)$ has the same orientation as $(\partial_{t}, v)$ . The two boundary-circle orientations are opposite.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

6 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

Ioan Mărcuț, Manifolds (2017 lecture notes), §§14.5, 15.1 (standard reference, not scraped)
Will Merry, Differential Geometry (2021), Lecture 24 (standard reference, not scraped)