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

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A cohomology class obstructs a global symplectomorphism

Counterexample

Let $ω$ be the standard positive area form on $S^{2}$ . The symplectic manifolds $(S^{2}, ω)$ and $(S^{2}, 2 ω)$ are not symplectomorphic.

Facts & Assumptions

Given: The oriented sphere and its positive area form.

[F1]

A symplectomorphism $f$ must satisfy $f^{*} ω_{1} = ω_{0}$ . Symplectomorphisms, local symplectomorphisms, and symplectic embeddings.

[F2]

On a compact connected oriented surface, integration identifies top de Rham cohomology with $R$ . Integration is an isomorphism on top compactly supported de Rham cohomology.

Verification

technique · direct

1.1

Both forms are closed and positive, hence symplectic. By [F2], their cohomology classes are distinct because their integrals are $A > 0$ and $2 A$ .

F2givenalgebra

2.1

If $f^{*} (2 ω) = ω$ , then $f$ is orientation preserving and change of variables gives $A = \int_{S^{2}} f^{*} (2 ω) = 2 A$ , impossible. Thus [F1] fails for every diffeomorphism, exhibiting the global cohomology obstruction.

F1F2step 1.1∎

Depends on

Used by

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Dependency tree · two levels

15 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)