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

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De rham homotopy formula for a smooth homotopy

Statement

If $F : M \times [0, 1] \to N$ is smooth up to the endpoints and $F_{t} (x) = F (x, t)$ , then $F_{1}^{*} - F_{0}^{*} = d (K F^{*}) + (K F^{*}) d$ .

Facts & Assumptions

Given: A smooth homotopy $F$ and a smooth form $ω$ on $N$ .

[F1]

De rham homotopy formula on a product: For endpoint inclusions $i_{t} : M \to M \times [0, 1]$ , $i_{1}^{*} - i_{0}^{*} = d K + K d$ on smooth forms of every degree.

[F2]

The exterior derivative commutes with pullback: For every smooth map $F : M \to N$ and every form $ω$ on $N$ , $d (F^{*} ω) = F^{*} (d ω) .$

Proof

technique · direct

1.1

Apply the product identity to the smooth form $F^{*} ω$ : $i_{1}^{*} F^{*} ω - i_{0}^{*} F^{*} ω = d K F^{*} ω + K d F^{*} ω$ . Evaluation on tangent tuples shows $i_{t}^{*} F^{*} ω = F_{t}^{*} ω$ , including functions.

F1given

2.1

Naturality gives $d F^{*} ω = F^{*} d ω$ . Substitution yields $F_{1}^{*} ω - F_{0}^{*} ω = d (K F^{*} ω) + K F^{*} d ω$ , the required identity for every degree; zero terms require no separate extension.

F2step 1.1∎

Source locator

Lee, Introduction to Smooth Manifolds, 2nd ed., Lemma 17.9 and Proposition 17.10, pp.444–445; the proof here computes the product differential directly.

Depends on

Used by

Dependency tree · two levels

9 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

Nigel Hitchin, Differentiable Manifolds (2014) (standard reference, not scraped)
John M. Lee, Introduction to Smooth Manifolds, second edition (standard reference, not scraped)