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

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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

Change-of-coordinate formula for tangent bases

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources