LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-30

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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.

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Tangent-bundle chart transitions are smooth with smooth inverses

Statement

If $(U, x)$ and $(V, y)$ are smooth charts on $M$ , then the transition map $\tilde{y} \circ {\tilde{x}}^{- 1}$ on $\tilde{x} (π^{- 1} (U \cap V))$ is smooth, and so is its inverse.

Facts & Assumptions

Given: Smooth charts $(U, x)$ and $(V, y)$ with nonempty overlap.

[F1]

The induced tangent-bundle chart records the base coordinate together with the coefficients in the coordinate tangent basis (The induced tangent bundle chart).

[L1]

Tangent bases transform by the Jacobian of the coordinate change (Change-of-coordinate formula for tangent bases).

[L2]

Matrix inversion preserves $C^{k}$ regularity on the general linear group (Matrix inversion preserves $C^{k}$ regularity where the determinant is nonzero).

Proof

technique · direct

1.1

If $v = \sum_{i} v^{i} \partial_{x^{i}} ∣_{p}$ , then [L1] gives $v = \sum_{j} w^{j} \partial_{y^{j}} ∣_{p}$ with $w = J (a) v$ , where $a : = x (p)$ and $J (a) = D (y \circ x^{- 1}) (a)$ . Hence $\tilde{y} \circ {\tilde{x}}^{- 1} (a, v) = (y \circ x^{- 1} (a), J (a) v)$ .

F1L1given

2.1

The base part $y \circ x^{- 1}$ is smooth, the matrix-valued map $a \mapsto J (a)$ is smooth, and matrix-vector multiplication is polynomial in the entries; therefore the transition map is smooth.

step 1.1

3.1

Reversing the roles of $x$ and $y$ gives the inverse transition, whose fiber matrix is $J (a)^{- 1}$ . The smoothness of this inverse matrix field follows from [L2], so the inverse transition is smooth.

L1L2step 1.1∎

Depends on

Used by

Assuming countable choice, the tangent bundle has a canonical smooth 2n-manifold structure Theorem

Dependency tree · two levels

11 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 (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)