PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-06

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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The exterior derivative is local

Statement

If $ω = η$ on an open set $U$ , then $d ω = d η$ on $U$ .

Facts & Assumptions

Given: The manifolds, forms, vector fields, maps, and coordinates explicitly named in the statement.

[F1]

The preceding result states that For $ω \in Ω^{k} (M)$ , define the candidate $d ω$ on smooth vector fields by $d ω (X_{0}, \dots, X_{k}) = \sum_{i} (- 1)^{i} X_{i} ω (X_{0}, {\hat{X}}_{i}, \dots, X_{k}) + \sum_{i < j} (- 1)^{i + j} ω ([X_{i}, X_{j}], X_{0}, {\hat{X}}_{i}, {\hat{X}}_{j}, \dots, X_{k}) .$ The next lemma proves that this candidate is a form. (The exterior derivative by the invariant vector-field formula).

Proof

technique · direct

1.1

At a point of $U$ , extend the prescribed tangent vectors by vector fields on $U$ ; the invariant formula uses only the values of the form and these fields in $U$ .

F1given

2.1

Applying the same formula to the equal restrictions of $ω$ and $η$ gives equal values at every point of $U$ .

step 1.1∎

Depends on

Used by

The exterior derivative commutes with restriction Proposition
The exterior derivative does not enlarge support Proposition
The exterior derivative squares to zero Theorem
The local coordinate formula for the exterior derivative Theorem

Dependency tree · two levels

7 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

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)