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

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.

Change-of-coordinate formula for tangent bases

Statement

Let $(U, x)$ and $(V, y)$ be smooth charts on $M$ with $p \in U \cap V$ . Then $\partial_{x^{i}} ∣_{p} = \sum_{j = 1}^{n} \frac{\partial (y^{j} \circ x^{- 1})}{\partial x^{i}} (x (p)) \partial_{y^{j}} ∣_{p}$ for each $i$ .

Facts & Assumptions

Given: Two smooth charts $(U, x)$ and $(V, y)$ containing $p$ .

[L1]

The coordinate derivations form a basis of the tangent space (Coordinate derivations form a basis of the tangent space).

Proof

technique · direct

1.1L1given

Both sides are derivations by [L1], so it is enough to compare their values on the coordinate germs $[y^{k}]$ , which form a separating family in the chart $y$ .

2.1L1step 1.1

Applying the left-hand side to $[y^{k}]$ gives $\partial (y^{k} \circ x^{- 1}) / \partial x^{i}$ at $x (p)$ by definition. Applying the right-hand side to $[y^{k}]$ gives $\sum_{j} (\partial (y^{j} \circ x^{- 1}) / \partial x^{i}) (x (p)) \partial_{y^{j}} ∣_{p} ([y^{k}])$ , and only the $j = k$ term survives because $\partial_{y^{j}} ∣_{p} ([y^{k}]) = δ_{j k}$ . So the two sides agree on every coordinate germ $[y^{k}]$ .

3.1step 2.1∎

Therefore the two derivations agree on a basis of germs and hence are equal.

Depends on

Coordinate derivations form a basis of the tangent space

Used by

A coordinate tuple is not an intrinsic tangent vector Counterexample
Hermitian metric and L² pairing on a compact Riemann surface Definition
Tangent basis change between Cartesian and polar coordinates Example
Curve velocity coordinates depend on the chart False statement
Cotangent coordinate changes use the inverse transpose Jacobian Lemma
Tangent-bundle chart transitions are smooth with smooth inverses Lemma
Smoothness of a vector field is equivalent to smooth coordinate components Proposition
Tensor transition laws define a smooth vector bundle Theorem

Dependency tree · two levels

5 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)