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

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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

1.1

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.

L1given

2.1

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.

L2step 1.1

3.1

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

step 1.1step 2.1∎

Depends on

Used by

FALSE: every pointwise assignment of a tangent vector is a smooth vector field False statement
A vector field acts as a derivation of smooth functions Proposition
A vector field tangent to an embedded submanifold restricts to a vector field on it Proposition
Coordinate formula for the Lie bracket Proposition
Derivations of smooth functions are exactly smooth vector fields Theorem
Local existence, uniqueness, and smooth dependence for manifold integral curves Theorem

Dependency tree · two levels

10 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, 2nd ed. (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)