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

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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.

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Coordinate formula for the differential

Statement

Let $F : M \to N$ be smooth, let $(U, x)$ and $(V, y)$ be smooth charts with $p \in U$ and $F (p) \in V$ , and let $\tilde{F} : = y \circ F \circ x^{- 1}$ . Then $d F_{p} (\partial_{x^{i}} ∣_{p}) = \sum_{j = 1}^{m} \frac{\partial {\tilde{F}}^{j}}{\partial x^{i}} (x (p)) \partial_{y^{j}} ∣_{F (p)} .$

Facts & Assumptions

Given: A smooth map $F : M \to N$ , charts $(U, x)$ and $(V, y)$ , and the coordinate representative $\tilde{F}$ .

[F1]

The differential acts on a derivation by precomposing target germs with $F$ (The differential of a smooth map).

[L1]

Coordinate derivations form bases of the tangent spaces (Coordinate derivations form a basis of the tangent space).

Proof

technique · direct

1.1L1given

By [L1], it is enough to compute the values of $d F_{p} (\partial_{x^{i}} ∣_{p})$ on the coordinate germs $[y^{j}]$ at $F (p)$ .

2.1F1step 1.1

By [F1], one has $d F_{p} (\partial_{x^{i}} ∣_{p}) ([y^{j}]) = \partial_{x^{i}} ∣_{p} ([y^{j} \circ F]) = \partial {\tilde{F}}^{j} / \partial x^{i}$ at $x (p)$ .

3.1L1step 2.1∎

The displayed linear combination of the $\partial_{y^{j}} ∣_{F (p)}$ has exactly the same values on all coordinate germs $[y^{j}]$ , so it equals $d F_{p} (\partial_{x^{i}} ∣_{p})$ .

Coordinate formula for the differential

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources