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

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Hamiltonian vector fields are symplectic and symplectic fields are locally Hamiltonian

Statement

Every Hamiltonian vector field is symplectic. Conversely, every symplectic vector field is Hamiltonian on a sufficiently small neighbourhood of each point.

Facts & Assumptions

Given: A vector field $X$ on a symplectic manifold.

[F1]

$X$ is symplectic exactly when $ι_{X} ω$ is closed. A vector field is symplectic iff $ι_{X} ω$ is closed.

[F2]

On a star-shaped open subset of Euclidean space, every closed $C^{1}$ coefficient field is the gradient of a potential. Poincare's lemma on a star-shaped domain: every closed C1 field is exact.

Proof

technique · direct

1.1

If $X = X_{H}$ , then $ι_{X} ω = d H$ is exact and therefore closed; [F1] makes $X$ symplectic.

F1given

2.1

If $X$ is symplectic, [F1] makes $ι_{X} ω$ closed. Around any point restrict to a coordinate ball that is star-shaped in coordinates. The coefficient vector of this smooth one-form satisfies the symmetric-partial equations for a closed field, so [F2] supplies $H$ there with $d H = ι_{X} ω$ . Thus $X = X_{H}$ locally.

F1F2givenalgebra∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

9 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)