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

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A vector field acts as a derivation of smooth functions

Statement

Let $X$ be a smooth vector field on $M$ . Then $f \mapsto X f$ is an $R$ -linear derivation of $C^{\infty} (M)$ :

$X (f g) = f X g + g X f$

for all $f, g \in C^{\infty} (M)$ .

Facts & Assumptions

Given: A smooth vector field $X$ on $M$ and smooth functions $f, g$ on $M$ .

[L1]

Each tangent vector $X_{p} \in T_{p} M$ is a derivation at the point $p$ (Derivations at a point and the tangent space).

[L2]

A smooth vector field has smooth coordinate coefficient functions in every chart (Smoothness of a vector field is equivalent to smooth coordinate components).

Proof

technique · direct

1.1

For each point $p \in M$ , [L1] gives $X_{p} (f g) = f (p) X_{p} (g) + g (p) X_{p} (f)$ . By the definition of the action on functions, this is exactly $(X (f g)) (p) = f (p) (X g) (p) + g (p) (X f) (p)$ .

L1given

1.2

To see that $X f$ is smooth, write locally $X = \sum_{i} X^{i} \partial / \partial x^{i}$ using [L2]. Then $X f = \sum_{i} X^{i} \partial_{i} (f \circ x^{- 1}) \circ x$ , a sum of products of smooth functions.

2.1

Since the equality in step 1.1 holds for every $p$ , one has $X (f g) = f X g + g X f$ as functions on $M$ . The map $f \mapsto X f$ is $R$ -linear for the same pointwise reason.

step 1.1

3.1

Therefore $X$ acts on $C^{\infty} (M)$ as an $R$ -linear derivation.

step 2.1step 1.2∎

Depends on

Used by

Dependency tree · two levels

8 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)