Alphabeta Math
LemmaStatement: Literature-sourcedProof: Literature-sourcedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-06
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 negative-gradient trajectory satisfies the energy identity

Statement

If γ:I→M is a negative-gradient trajectory of f for g, then

ddtf(γ(t))=−∥grad⁡gf(γ(t))∥g2(t∈I).

No compactness or completeness is assumed.

Facts & Assumptions

Given: A negative-gradient trajectory γ:I→M of f for g and t∈I.

[F1]

Its velocity is γ˙(t)=−grad⁡gf(γ(t)) (Negative-gradient trajectories of a Morse function).

[F2]

The gradient satisfies dfx(v)=gx(grad⁡gf(x),v) (The Riemannian gradient is the metric dual of the differential).

[F3]

The differential chain rule applies to the smooth maps γ and f (The chain rule for differentials of smooth maps).

Proof

technique · direct
1.1F3given

By [F3], (f∘γ)′(t)=dfγ(t)(γ˙(t)).

2.1F1F2step 1.1

Substituting [F1] into step 1.1 and applying [F2] gives (f∘γ)′(t)=−gγ(t)(grad⁡gf,grad⁡gf).

3.1step 2.1algebra∎

The right-hand side in step 2.1 is −∥grad⁡gf(γ(t))∥g2, proving the identity on I.

Depends on

Used by

Dependency tree · two levels

9 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