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 of a function is its differential

Statement

For $f \in C^{\infty} (M) = Ω^{0} (M)$ , $d f (X) = X f$ for every smooth vector field $X$ .

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

For $k = 0$ the bracket sum is empty and the invariant formula gives $(d f) (X) = X f$ .

F1given

2.1

This is exactly the defining action of the ordinary differential on a tangent vector.

step 1.1∎

Depends on

Used by

Cartan's magic formula Theorem
Existence and uniqueness of 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)