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

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Symplectic vector fields modulo Hamiltonian vector fields are first de Rham cohomology

Statement

There is a natural vector-space isomorphism

$X_{s y m p} (M) / X_{h a m} (M) ≅ H_{d R}^{1} (M), [X] ⟼ [ι_{X} ω] .$

Facts & Assumptions

Given: A symplectic manifold $(M, ω)$ .

[F1]

Symplectic fields correspond under $ω^{♭}$ to closed one-forms. A vector field is symplectic iff $ι_{X} ω$ is closed.

[F2]

Hamiltonian fields correspond under the same map to exact one-forms. Hamiltonian vector field and Hamiltonian function.

Proof

technique · direct

1.1

The linear bundle isomorphism $ω^{♭}$ gives a linear bijection between all vector fields and all one-forms. By [F1] it restricts to a bijection from symplectic fields to closed one-forms.

F1given

2.1

By [F2], the inverse image of the exact one-forms is precisely the Hamiltonian fields. Passing to quotients in step 1.1 therefore gives $Z^{1} / d C^{\infty} = H_{d R}^{1} (M)$ and the displayed natural isomorphism.

F2step 1.1∎

Depends on

Used by

A symplectic non-Hamiltonian vector field on the two-torus Example
Every symplectic vector field has a global Hamiltonian function False statement

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

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