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Closed defining forms have vanishing Godbillon-Vey class

Statement

Assume Countable Choice ACω. Let F be a transversely oriented codimension-one foliation of a smooth manifold M defined by a closed nowhere-vanishing 1-form ω with TF=ker⁡ω (so that ker⁡ω is integrable by Closed constant-rank one-forms define integrable hyperplane fields). Then GV(F)=0 in HdR3(M;R).

Facts & Assumptions

Given: A transversely oriented codimension-one foliation F of a smooth manifold M defined by a closed nowhere-vanishing one-form ω with TF=ker⁡ω, and the standing countable choice assumption.

[F1]

For a defining form ω and a one-form η with dω=η∧ω, the Godbillon-Vey class is the de Rham class GV(F)=[η∧dη]. (The Godbillon-Vey class of a codimension-one foliation).

[F2]

For a transversely oriented codimension-one foliation with nowhere-vanishing defining form ω there is a smooth one-form η with dω=η∧ω. (Frobenius divisibility: d omega equals eta wedge omega).

Proof

technique · direct
1.1F2given

For the closed defining form ω the choice η=0 satisfies dω=0=0∧ω, so it is one of the forms whose existence the divisibility lemma [F2] guarantees.

2.1F1step 1.1∎

The Godbillon-Vey form of this choice is η∧dη=0∧0=0, so the class defined in [F1] is the class of the zero form, namely GV(F)=0 in HdR3(M;R); this applies in particular to fibre foliations of bundles over S1 defined by pullbacks of volume forms on the circle, and no choice principle is used.

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