PropositionStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Tangent space of a free proper quotient

Statement

For a smooth free proper action and $x \in M$ , the quotient differential is surjective and

$\ker (d q_{x}) = T_{x} (G \cdot x) .$

Consequently it induces a canonical linear isomorphism

$T_{x} M / T_{x} (G \cdot x) \overset{≅}{\to} T_{[x]} (M / G) .$

Facts & Assumptions

Given: A smooth free proper action of $G$ on $M$ , its quotient map $q : M \to M / G$ , and $x \in M$ .

[F1]

The quotient map is a smooth surjective submersion. Free proper action quotient manifold.

[F2]

A slice gives product coordinates $G \times S \to G \cdot S$ around $x$ . Local slice for a free proper action.

[F3]

A linear map vanishing on a subspace factors uniquely through the vector space quotient. A module homomorphism vanishing on $N$ factors uniquely through $M / N$ .

Proof

technique · compute in slice-product coordinates

1.1

Choose the slice $S$ through $x$ from [F2]. Under the diffeomorphism $A : G \times S \to G \cdot S$ , the quotient map is the projection $(g, s) \mapsto s$ , followed by the slice chart $S ≅ q (G \cdot S)$ . Its differential at $(e, x)$ is therefore the projection $T_{e} G \oplus T_{x} S \to T_{x} S$ .

F1F2

2.1

The kernel of that projection is $T_{e} G \oplus 0$ . Its image under $d A_{(e, x)}$ is exactly the tangent space to the orbit map $g \mapsto g \cdot x$ , namely $T_{x} (G \cdot x)$ . Thus $\ker (d q_{x}) = T_{x} (G \cdot x)$ , and the same coordinate projection shows that $d q_{x}$ is surjective.

step 1.1F2

3.1

By step 2.1, $d q_{x}$ vanishes exactly on $T_{x} (G \cdot x)$ , so [F3] gives an injective induced linear map from $T_{x} M / T_{x} (G \cdot x)$ to $T_{[x]} (M / G)$ . It is surjective because $d q_{x}$ is, and hence is an isomorphism. The formula holds also for a zero-dimensional group and uses no choice principle.

F3step 2.1∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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)