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

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Smooth vector fields form a Lie algebra under the Lie bracket

Statement

The space $X (M)$ of smooth vector fields on $M$ , together with the Lie bracket, is a Lie algebra over $R$ .

Facts & Assumptions

Given: Smooth vector fields $X, Y, Z$ on $M$ .

[L1]

The commutator of two vector-field derivations is again a derivation (The commutator of vector-field derivations is again a derivation).

[L2]

Every derivation of $C^{\infty} (M)$ comes from a unique smooth vector field (Derivations of smooth functions are exactly smooth vector fields).

Proof

technique · direct

1.1

By [L1] and [L2], the commutator $[X, Y]$ is again a smooth vector field, so the bracket closes on $X (M)$ .

L1L2given

1.2

Bilinearity and antisymmetry follow from the corresponding identities for commutators of $R$ -linear endomorphisms of $C^{\infty} (M)$ : $[a X + b Y, Z] = a [X, Z] + b [Y, Z], [X, Y] = - [Y, X] .$

givenalgebra

1.3

The operator commutator satisfies the Jacobi identity $[X, [Y, Z]] + [Y, [Z, X]] + [Z, [X, Y]] = 0$ on $C^{\infty} (M)$ , again by direct expansion in the endomorphism algebra.

givenalgebra

2.1

Steps 1.1-1.3 are exactly the Lie-algebra axioms, so smooth vector fields form a Lie algebra under the Lie bracket.

step 1.1step 1.2step 1.3∎

Depends on

Used by

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Sources

John M. Lee, Introduction to Smooth Manifolds, 2nd ed. (standard reference, not scraped)
Will J. Merry, Differential Geometry (standard reference, not scraped)
Nigel Hitchin, Differentiable Manifolds (standard reference, not scraped)