PropositionStatement: Literature-sourcedProof: AI-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-04

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.

Smoothness of a vector field is equivalent to smooth coordinate components

Statement

Let $(U, x)$ be a smooth chart on an $n$ -manifold $M$ , and let $X$ be a vector field on $U$ . Then $X$ is smooth if and only if there exist smooth functions $X^{1}, \dots, X^{n} : U \to R$ such that

$X = \sum_{i = 1}^{n} X^{i} \frac{\partial}{\partial x^{i}}$

on $U$ .

Facts & Assumptions

Given: A chart $(U, x)$ and a vector field $X$ on $U$ .

[L1]

Smoothness of a section of a smooth vector bundle is equivalent to smoothness of its local frame coefficients (Smoothness of a section is equivalent to smooth local components).

[L2]

On an overlap of charts, tangent bases transform by the Jacobian matrix of the coordinate change (Change-of-coordinate formula for tangent bases).

Proof

technique · direct

1.1L1given

In the induced tangent-bundle chart over $U$ , the coordinate fields $\partial / \partial x^{1}, \dots, \partial / \partial x^{n}$ form a local frame of $T M ∣_{U}$ , so [L1] says that $X$ is smooth exactly when it can be written with smooth coefficient functions $X^{1}, \dots, X^{n}$ in that frame.

2.1L2step 1.1

If one changes charts, [L2] expresses the new coefficients as linear combinations of the old ones with smooth Jacobian entries. Hence the criterion from step 1.1 is independent of the chosen chart.

3.1step 1.1step 2.1∎

Therefore a vector field is smooth exactly when its coordinate components in a chart are smooth.

Smoothness of a vector field is equivalent to smooth coordinate components

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources