Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-10
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.

Riemannian divergence theorem

Statement

Assume countable choice. On an oriented Riemannian manifold with boundary, n1, a smooth compactly supported vector field satisfies M(divgX)volg=Mg(X,ν)volg, with outward unit normal ν and outward-normal-first boundary orientation.

Facts & Assumptions

Given: The stated oriented manifold, vector field, and countable choice.

[F1]

Coordinate formula for riemannian divergence: In coordinates, divgX=(detG)1/2i=1ni((detG)1/2Xi).

[F2]

The riemannian volume form is the unique positive unit top form: The Riemannian volume form is the unique positive unit section of nTM for the specified orientation and normalized exterior metric.

[F3]

Pullback of a riemannian metric is riemannian exactly for immersions: Fh is Riemannian if and only if F is an immersion. In general it is positive semidefinite, with radical kerdFp at p.

[F4]

The general Stokes theorem: Assume ACω. Let M be an oriented smooth n-manifold with boundary, n1, and let ηΩcn1(M). With j:MM and the outward-normal-first orientation, Mdη=Mjη. An empty boundary contributes zero; in dimension one its integral is a finite signed sum of point values.

[F6]

Induced boundary orientation: For an oriented manifold with boundary, orient TpM by the outward-normal-first rule: an outward vector first, followed by a positive boundary determinant, is a positive determinant of TpM.

[F7]

The boundary tangent space is the boundary-tangent hyperplane: For pM of an n1 dimensional manifold and the inclusion i:MM, the differential dip identifies TpM with the hyperplane of boundary-tangent vectors in TpM.

[F8]

The gradient is characterized by inner products: The gradient is the unique smooth vector field Y satisfying g(Y,X)=Xf for every smooth vector field X.

[F9]

Riemannian metrics induce metrics on dual tensor and exterior bundles: A Riemannian metric induces smooth metrics on dual, tensor and exterior bundles. On decomposable covectors, α1αk,β1βk=det(αi,βj); increasing orthonormal wedge monomials have norm one.

[F10]

The Axiom of Countable Choice (ACω): The Axiom of Countable Choice, written ACω, is the following statement. > For every family (Xn)nN of nonempty sets indexed by > N there is a function f with domain N such that > f(n)Xn for every nN. Equivalently, in the vocabulary of def-choice-function: every at most countable family of nonempty sets (def-countable) has a choice function.

Proof

technique · direct
1.1

In a boundary chart let u0 be its inward boundary coordinate. The boundary tangent space is kerdu. The gradient identity gives g(gradu,v)=du(v), so the nonzero gradient is perpendicular to that hyperplane. Therefore ν=gradu/gradu is smooth, unit and outward, since du(ν)=gradu<0. The orthogonal complement is a line and exactly one of its two unit vectors is outward; hence these local definitions agree. The boundary inclusion is an immersion, so the induced metric is Riemannian.

F3F7F8given
2.1

Write X=g(X,ν)ν+XT, with XT tangent to the boundary. The term volg(XT,v1,,vn1) vanishes because these n vectors lie in a hyperplane. On a positive orthonormal boundary basis the form j(ινvolg) is positive and unit by the outward-first convention and the normalized exterior pairing. Uniqueness of the boundary volume form gives j(ιXvolg)=g(X,ν)volg. For n=1, this identity uses the induced signed zero-form: its value is volg(ν)=ε, so the same scalar identity holds.

F2F6F9step 1.1
3.1

The form η=ιXvolg is smooth and supported in the compact support of X. In coordinates dη=ii(detGXi)dx1dxn=(divgX)volg. Stokes under countable choice therefore gives M(divgX)volg=Mjη. Substitute step 2.1 to obtain the claimed formula. Empty boundary contributes zero, a zero field gives zero on both sides, and compact M permits every smooth field. In dimension one the oriented boundary integral is the finite signed endpoint sum.

F1F4F10step 2.1

Source locator

Lee, Propositions 15.32–15.33, pp.390–391, Lemma 16.30 and Theorem 16.32, pp.423–424; boundary and choice hypotheses are checked explicitly.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

36 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