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The Godbillon-Vey form eta wedge d eta is closed

Statement

Assume Countable Choice ACω. Let F be a transversely oriented codimension-one foliation with defining form ω and dω=η∧ω. Then dη∧ω=0; the 2-form dη is divisible by ω, i.e. dη=ζ∧ω for a smooth 1-form ζ; dη∧dη=0; and d(η∧dη)=0. Consequently η∧dη is a closed 3-form on M.

Facts & Assumptions

Given: A transversely oriented codimension-one foliation F of a smooth manifold M with defining one-form ω and a one-form η satisfying dω=η∧ω, and the standing countable choice assumption.

[F1]

Under ACω, if ω is nowhere vanishing and θ∧ω=0 for a smooth two-form θ, then θ=β∧ω for a smooth one-form β. (Divisibility by a nowhere-vanishing one-form).

[F2]

For homogeneous smooth forms α,β one has d(α∧β)=dα∧β+(−1)deg⁡αα∧dβ. (The exterior derivative is a graded derivation).

[F3]

For every differential form ω, d(dω)=0. (The exterior derivative squares to zero).

[F4]

The wedge product is associative and graded-commutative, so ζ∧ζ=0 for a one-form ζ and ω∧ω=0. (The wedge product is associative and graded commutative).

Proof

technique · direct
1.1F2F3F4given

Differentiating dω=η∧ω with the Leibniz rule [F2] and d2=0 [F3] gives 0=dη∧ω−η∧dω=dη∧ω−η∧η∧ω, and η∧η=0 by graded commutativity [F4], so dη∧ω=0.

2.1F1step 1.1

Since ω is nowhere vanishing and the two-form dη satisfies dη∧ω=0, the divisibility lemma [F1] gives a smooth one-form ζ with dη=ζ∧ω.

3.1F4step 2.1

Then dη∧dη=(ζ∧ω)∧(ζ∧ω)=−ζ∧ζ∧ω∧ω=0 by associativity and graded commutativity with ζ∧ζ=0 and ω∧ω=0 [F4].

4.1F2F3step 3.1∎

Finally d(η∧dη)=dη∧dη−η∧d(dη)=0 by the graded Leibniz rule [F2] and d2=0 [F3], so η∧dη is a closed three-form. The standing ACω assumption licenses the global divisibility result in step 2.1; the remaining calculations are formal exterior-algebra identities.

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