TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Hamiltonian flows commute iff their Hamiltonians Poisson commute up to locally constant bracket

Statement

The local flows of $X_{F}$ and $X_{G}$ commute wherever both composites are defined if and only if ${F, G}$ is locally constant. In particular, ${F, G} = 0$ is sufficient.

Facts & Assumptions

Given: Smooth functions $F, G$ on a symplectic manifold.

[F1]

$[X_{F}, X_{G}] = - X_{{F, G}}$ . The Hamiltonian vector-field map is a Lie antihomomorphism.

[F2]

Two vector fields have commuting local flows exactly when their Lie bracket vanishes. Two vector fields commute if and only if their local flows commute.

[F3]

The zero field has precisely the locally constant Hamiltonians. Hamiltonians for a fixed vector field differ by a locally constant function.

Proof

technique · direct

1.1

By [F2], the flows commute exactly when $[X_{F}, X_{G}] = 0$ . By [F1], this is equivalent to $X_{{F, G}} = 0$ .

F1F2given

2.1

By [F3], the latter condition holds exactly when ${F, G}$ is locally constant. The zero bracket is one such function.

F3step 1.1∎

Depends on

Used by

Commuting Hamiltonian vector fields integrate to a local ℝⁿ-action Proposition

Dependency tree · two levels

11 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

Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)