Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-13
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.

Straight lines as Euclidean geodesics

Example

On Euclidean Rn with its Levi–Civita connection, every affinely parametrized geodesic on an interval I has the form γ(t)=p+tv(t∈I) for fixed p,v∈Rn, and every such curve is a geodesic. Here v=0 gives a constant geodesic; a nonconstant curve traces a straight line. If n=0, the only curves are constant.

Facts & Assumptions

Given: The Euclidean metric gij=δij in Cartesian coordinates and an interval I of affine parameter values.

[F1]

Christoffel formula for the levi civita connection gives Γkij=12gkℓ(∂igjℓ+∂jgiℓ−∂ℓgij).

[F2]

Coordinate geodesic equation says that γ is a geodesic exactly when x¨k+Γkijx˙ix˙j=0 in every coordinate.

[F3]

A function continuous on an interval I whose derivative vanishes at every interior point of I is constant on I; consequently two such functions with the same derivative differ by a constant says that a continuous function on an interval whose interior derivative is zero is constant, including when the interval has endpoints.

Verification

1.1F1given

Every gij=δij is constant, so all its partial derivatives vanish. Formula [F1] therefore gives Γkij=0 for every index.

2.1F2F3step 1.1

By [F2] and step 1.1, the geodesic equation is x¨k=0 for each k. Applying [F3] first to x˙k gives a constant vk; applying it to xk(t)−tvk gives a constant pk. Thus γ(t)=p+tv throughout the interval, not merely near one parameter value. For an included endpoint the equality extends by continuity.

3.1F2step 1.1step 2.1∎

Conversely, xk(t)=pk+tvk has x¨k=0, so [F2] and step 1.1 make it a geodesic. If v=0, it is constant; if v≠0, its image lies on the straight line p+Rv. In dimension zero there are no coordinate equations and the unique curve is constant. The argument makes no choice beyond the given curve's own coordinates.

Depends on

Used by

Dependency tree · two levels

17 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