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

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Symplectic manifolds have a canonical orientation and volume form

Statement

If $(M^{2 n}, ω)$ is symplectic, then $μ_{ω} = ω^{n} / n!$ is a nowhere-zero volume form. Its positive ray gives the canonical symplectic orientation of $M$ .

Facts & Assumptions

Given: A symplectic $2 n$ -manifold $(M, ω)$ .

[F1]

Nondegeneracy makes $ω^{n}$ nowhere zero. Nondegeneracy is equivalent to a nonvanishing top wedge.

[F2]

An orientation is a smooth choice of ray in the determinant line. Oriented smooth manifolds and oriented charts.

Proof

technique · direct

1.1

By [F1], $ω^{n}$ is a smooth nowhere-zero top form; division by the positive number $n!$ preserves that property. Thus $μ_{ω}$ is a volume form.

F1algebra

2.1

A nonzero top covector $(μ_{ω})_{p} \in \det T_{p}^{*} M$ selects the ray of tangent determinants $v \in \det T_{p} M$ on which $(μ_{ω})_{p} (v) > 0$ . This ray varies smoothly and therefore defines an orientation by [F2]. For $n = 0$ , $μ_{ω} = 1$ selects the positive sign at each point, so the boundary case is included.

F2step 1.1algebra∎

Depends on

Used by

Liouville volume preservation Theorem

Dependency tree · two levels

8 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

Eckhard Meinrenken, Symplectic Geometry (standard reference, not scraped)