False statementConstruction: AI-adaptedVerification: 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.

False: all exact forms integrate to zero everywhere

Statement

False assertion: every exact smooth top form has zero total integral whenever that integral exists, on every manifold.

Facts & Assumptions

[F1]

The general Stokes theorem: Assume ${A C}_{ω}$ . Let $M$ be an oriented smooth $n$ -manifold with boundary, $n \geq 1$ , and let $η \in Ω_{c}^{n - 1} (M)$ . With $j : \partial M ↪ M$ and the outward-normal-first orientation, $\int_{M} d η = \int_{\partial M} j^{*} η .$ An empty boundary contributes zero; in dimension one its integral is a finite signed sum of point values.

[F2]

Stokes agrees with the fundamental theorem of calculus: For $a < b$ , orient $[a, b]$ increasingly. Every smooth $f$ on this interval satisfies $\int_{[a, b]} d f = f (b) - f (a),$ where the boundary point signs are $- 1$ at $a$ and $+ 1$ at $b$ . This agrees with the Riemann fundamental theorem of calculus.

Refutation

Given: The proposed assertion; use the data constructed below.

1.1

On the increasingly oriented compact interval $[0, 1]$ , the exact form $d t = d (t)$ has integral $1 - 0 = 1$ . Stokes includes its nonzero boundary contribution. This alone refutes the assertion.

F1F2

2.1

The primitive support condition also matters without boundary. Let $b (t) = e^{- 1 / (1 - t^{2})}$ for $∣ t ∣ < 1$ , zero otherwise; its derivatives vanish at the cutoff endpoints. Put $c = \int_{- 1}^{1} b > 0$ and $G (t) = c^{- 1} \int_{- 1}^{t} b (s) d s$ , extending $b$ by zero. Then $G$ is smooth, equals zero for $t \leq - 1$ and one for $t \geq 1$ . The exact form $d G = c^{- 1} b (t) d t$ has compact support and integral one by the interval FTC, whereas $G$ has noncompact support. Hence this example also fails the compact-primitive hypothesis of Stokes on the line.

F2step 1.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

8 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

Lee Theorem 16.11 and Corollary 16.13 hypotheses, pp.411–414 (standard reference, not scraped)