Alphabeta Math
LemmaStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 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.

Normalized gradient crosses a compact regular band in controlled time

Statement

Assume ACω. Let f:MR be smooth on a boundaryless manifold, a<b, and let K=f1([a,b]) be compact with df0 on K. For any Riemannian metric there is a compactly supported smooth field Y agreeing with gradf/gradf2 near K. Its complete flow Φ satisfies f(Φt(x))=f(x)+t for xK and af(x)tbf(x). Thus every intervening level is reached in exactly its value difference.

Facts & Assumptions

[F1]

Closed sublevel and level set of a smooth function: Let f:MR be smooth on a boundaryless smooth n-manifold. Write Ma=f1((,a]), Ma=f1({a}), and f1([a,b]) for the closed band. Both endpoints are included. A regular value may have empty fiber. The smooth-manifold convention is def-smooth-manifold.

[F2]

The Riemannian gradient is the metric dual of the differential: Let g be a Riemannian metric on a smooth manifold M and let f:MR be smooth. The Riemannian gradient of f is the smooth vector field gradgf characterized by gx((gradgf)x,v)=dfx(v)for every xM and vTxM. Pointwise, it is the inverse metric-dual of dfx. In a local frame with metric matrix (gij) and inverse (gij), it is gradgf=i,jgijfxjxi; the displayed coefficients are smooth, so this pointwise definition is a smooth vector field.

[F3]

Assuming countable choice, every smooth manifold admits a Riemannian metric: Assume ACω. Every smooth manifold admits a Riemannian metric.

[F4]

A manifold bump for a compact set inside an open set: Let M be a smooth manifold, let KM be compact, and let WM be open with KW. Then there exists a smooth function ρ:M[0,1] that equals 1 on an open neighbourhood of K and satisfies supp(ρ)W.

[F5]

Compactly supported smooth vector fields are complete: Every compactly supported smooth vector field on a smooth manifold is complete.

Proof

Given: The objects and hypotheses in the statement.

1.1

Use the closed-band convention and choose a metric. If K=, the zero field suffices and all trajectory assertions are vacuous. Otherwise the metric exists under the stated choice axiom.

F1F3given
1.2

The open set {df0} contains K. Cover K by finitely many coordinate neighborhoods with compact closures in this open set; their union W is relatively compact. Choose ρ=1 near K with support in W.

F4
2.1

On W set Y=ρgradf/gradf2, and set it to zero outside W. The support condition makes this smooth with compact support. Since df(gradf)=gradf2, df(Y)=1 near K.

F2step 1.2algebra
3.1

The field is complete. On every trajectory segment contained in K, differentiation gives d(fΦt)/dt=1. Starting at an endpoint the same identity holds on its open neighborhood, so the trajectory enters the band in the required time direction. A first exit before the claimed level would have value strictly between a and b, contradicting continuity. Integrating gives the identity through both endpoints; strict unit speed gives the asserted hitting time.

F5step 2.1algebra

Depends on

Used by

Dependency tree · two levels

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Sources