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

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Overlapping plaques through a point have compatible germs

Statement

Let $P_{1}$ and $P_{2}$ be plaques from flat charts of the same integrable distribution, and suppose $p \in P_{1} \cap P_{2}$ . Then there is a neighborhood $W$ of $p$ in $M$ such that

$P_{1} \cap W = P_{2} \cap W .$

Facts & Assumptions

Given: Two plaques $P_{1}$ and $P_{2}$ through the same point $p$ .

[A1]

Each plaque is itself a local integral manifold of the distribution.

Proof

technique · direct

1.1

Apply the previous lemma to the connected integral manifold $P_{1}$ inside a [given] flat chart producing $P_{2}$ . Near $p$ , the set $P_{1}$ must lie in the plaque of that chart through $p$ , namely in $P_{2}$ .

given

1.2

Reversing the roles of $P_{1}$ and $P_{2}$ gives the opposite inclusion on [given] possibly smaller neighborhoods. Intersecting those neighborhoods yields an open set $W$ with $P_{1} \cap W = P_{2} \cap W$ .

given

2.1

Thus plaques through the same point determine the same germ. [given] ∎

given

Depends on

Used by

Existence and uniqueness of maximal connected integral manifolds Theorem

Dependency tree · two levels

14 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)