LemmaStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-05

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Integral manifolds are locally contained in plaques

Statement

Let $D$ be an integrable rank- $k$ distribution, let $φ = (x, y) : U \to R^{k} \times R^{n - k}$ be a flat chart for $D$ , and let $i : N \to M$ be a connected integral manifold of $D$ . Then each connected component of $i^{- 1} (U)$ is mapped by $i$ into a single plaque of $U$ .

Facts & Assumptions

Given: A flat chart $φ = (x, y)$ and a connected integral manifold $i : N \to M$ .

[A1]

Let $C$ be a connected component of $i^{- 1} (U)$ .

Proof

technique · direct

1.1

The transverse coordinate map $y \circ i : C \to R^{n - k}$ has zero [given] differential. Indeed, the tangent image of $i$ is $D$ , and in a flat chart the distribution is exactly the kernel of $d y$ .

given

1.2

A smooth map with zero differential is locally constant, hence constant on [given] each connected component of its domain. Therefore $y \circ i$ is constant on $C$ .

given

1.3

The image $i (C)$ is therefore contained in the slice with that fixed [given] transverse coordinate, namely in a single plaque of the flat chart.

given∎

Depends on

Used by

Dependency tree · two levels

20 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

Will J. Merry, Differential Geometry (standard reference, not scraped)