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 $γ : I \to M$ is a negative-gradient trajectory of $f$ for $g$ , then

$\frac{d}{d t} f (γ (t)) = - ∥ {grad}_{g} f (γ (t)) ∥_{g}^{2} (t \in I) .$

No compactness or completeness is assumed.

Facts & Assumptions

Given: A negative-gradient trajectory $γ : I \to M$ of $f$ for $g$ and $t \in I$ .

[F1]

Its velocity is $\dot{γ} (t) = - {grad}_{g} f (γ (t))$ (Negative-gradient trajectories of a Morse function).

[F2]

The gradient satisfies $d f_{x} (v) = g_{x} ({grad}_{g} f (x), v)$ (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

1.1

By [F3], $(f \circ γ)^{'} (t) = d f_{γ (t)} (\dot{γ} (t))$ .

F3given

2.1

Substituting [F1] into step 1.1 and applying [F2] gives $(f \circ γ)^{'} (t) = - g_{γ (t)} ({grad}_{g} f, {grad}_{g} f)$ .

F1F2step 1.1

3.1

The right-hand side in step 2.1 is $- ∥ {grad}_{g} f (γ (t)) ∥_{g}^{2}$ , proving the identity on $I$ .

step 2.1algebra∎

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

Ralph L. Cohen, Bundles, Manifolds, and Homotopy, Lemma 13.1 (standard reference, not scraped)