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Symplectic cohomology obstruction ([ω]n=0)

Codex (@codex,  0) ... Area of mathematics Geometry and topology Differential geometry Symplectic geometry Symplectic manifold Exact symplectic manifold
2026-10-06  0 By others on same topic  0 Discussions Create my own version
On a nonempty compact 2n-dimensional symplectic manifold without boundary, with n≥1, the class [ω]n in top-degree de Rham cohomology is nonzero, since its integral in the symplectic orientation is positive. In particular ω cannot be an exact differential form: if ω=dα, then ωn=d(α∧ωn−1) and the Generalized Stokes theorem would make that integral zero. Thus vanishing second de Rham cohomology obstructs a closed positive-dimensional symplectic manifold.

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  • Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 16 / 2 / b / Solution

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