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

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The commutator of vector-field derivations is again a derivation

Statement

Let $X$ and $Y$ be smooth vector fields on $M$ . Then the commutator $[X, Y] : C^{\infty} (M) \to C^{\infty} (M)$ defined by $[X, Y] f = X (Y f) - Y (X f)$ is an $R$ -linear derivation.

Facts & Assumptions

Given: Smooth vector fields $X$ and $Y$ on $M$ and smooth functions $f, g$ .

[L1]

Each smooth vector field acts on $C^{\infty} (M)$ as a derivation (A vector field acts as a derivation of smooth functions).

Proof

technique · direct

1.1

By [L1], both $X$ and $Y$ are $R$ -linear derivations, so their commutator is automatically $R$ -linear. It remains to prove the Leibniz rule.

L1given

1.2

Expand $X (Y (f g))$ using [L1] twice: $X (Y (f g)) = X (f Y g + g Y f) = X f Y g + f X (Y g) + X g Y f + g X (Y f) .$

L1given

1.3

Similarly, $Y (X (f g)) = Y f X g + f Y (X g) + Y g X f + g Y (X f) .$

L1given

2.1

Subtracting step 1.3 from step 1.2 cancels the mixed first-order products, leaving $[X, Y] (f g) = f [X, Y] g + g [X, Y] f .$

step 1.2step 1.3algebra

3.1

Therefore $[X, Y]$ is an $R$ -linear derivation of $C^{\infty} (M)$ .

step 1.1step 2.1∎

Depends on

Used by

Coordinate formula for the Lie bracket Proposition
Related vector fields have related Lie brackets Proposition
Smooth vector fields form a Lie algebra under the Lie bracket Theorem

Dependency tree · two levels

5 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

Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)