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TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Nondegeneracy is equivalent to a nonvanishing top wedge

Statement

Let M have dimension 2n and let ω be a smooth two-form. Then ω is pointwise nondegenerate if and only if the top-degree form ωn is nowhere zero.

Facts & Assumptions

Given: A smooth 2n-manifold M and ωΩ2(M).

[F1]

Wedge products of differential forms are defined pointwise. The wedge product of differential forms.

[F3]

Pointwise nondegeneracy is the linear clause in the definition of a symplectic form. Symplectic form and symplectic manifold.

Proof

technique · direct
1.1

Fix pM. If ωp is nondegenerate, [F2] supplies a basis e1,,en,f1,,fn with ωp=ieifi. Hence ωpn=n!e1f1enfn0.

F1F2
1.2

Conversely, if 0v lies in the radical of ωp, then the graded contraction rule gives ιv(ωpn)=n(ιvωp)ωpn1=0. A nonzero top covector has nonzero contraction by every nonzero vector: extend v to a basis and evaluate on the remaining basis vectors. Therefore ωpn=0.

F1algebra
2.1

Steps 1.1--1.2 prove the equivalence at every p, which is exactly [F3]. For n=0, ω0=1 and the zero tangent space is nondegenerate, so the same conclusion holds. Closedness is irrelevant to this pointwise equivalence.

F3step 1.1step 1.2

Depends on

Used by

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