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Restriction of a foliation transverse to the boundary

Statement

Let W be a smooth (n+1)-manifold with boundary and let F be a codimension-one regular foliation of W transverse to ∂W, i.e. Tp(∂W)+TpF=TpW for every p∈∂W. Let ω be a nowhere-vanishing smooth defining 1-form for F. Then: (i) Tp(∂W)∩TpF has dimension n−1 for every p∈∂W, and these subspaces form a codimension-one regular foliation F∣∂W of the smooth n-manifold ∂W; (ii) the restriction ω∣∂W is nowhere vanishing and defines F∣∂W, with ω∣∂W∧d(ω∣∂W)=0; (iii) every leaf of F∣∂W is a connected component of L∩∂W for a leaf L of F, with the intersection taken in the intrinsic leaf topology; (iv) if F is transversely oriented by ω, then F∣∂W is transversely oriented by ω∣∂W; (v) if η is a 1-form on W with dω=η∧ω, then the pullback of η to ∂W satisfies d(ω∣∂W)=(η∣∂W)∧(ω∣∂W).

Facts & Assumptions

Given: A smooth (n+1)-manifold W with boundary, a codimension-one regular foliation F of W transverse to ∂W, and a nowhere-vanishing smooth defining one-form ω for F.

[F1]

For a smooth map of manifolds, pullback sends smooth forms to smooth forms, is functorial, and satisfies F∗(α∧β)=F∗α∧F∗β (Pullback of forms is smooth functorial and preserves wedges, and for boundary manifolds The de Rham complex and pullback extend to manifolds with boundary).

[F2]

For every smooth map F and every form ω on the target, d(F∗ω)=F∗(dω) (The exterior derivative commutes with pullback, and for boundary manifolds The de Rham complex and pullback extend to manifolds with boundary).

[F3]

For a nowhere-zero one-form α, the hyperplane distribution ker⁡α is integrable if and only if α∧dα=0 (The codimension-one Frobenius criterion).

[F4]

For p∈∂M of a manifold with boundary and the inclusion i:∂M↪M, the differential dip identifies Tp∂M with the hyperplane of boundary-tangent vectors in TpM (The boundary tangent space is the boundary-tangent hyperplane).

[F5]

On a manifold, an integrable rank-k distribution defines a regular foliation atlas whose leaves are its maximal connected integral manifolds (Regular foliations and integrable distributions correspond).

Proof

technique · direct
1.1givenF4

Write i:∂W↪W for the inclusion and fix p∈∂W. By [F4] the subspace Tp∂W sits inside TpW as a hyperplane, and the transversality hypothesis reads Tp∂W+TpF=TpW; since ω defines F, moreover TpF=ker⁡ωp.

2.1step 1.1

If ωp vanished on all of Tp∂W, then Tp∂W⊆ker⁡ωp=TpF, so the sum Tp∂W+TpF=TpF would have dimension n instead of n+1; hence ω∣∂W is nowhere vanishing, the restricted one-form has ker⁡(ω∣∂W)p=Tp∂W∩TpF as its kernel, and the dimension formula for two hyperplanes with sum TpW gives dim⁡(Tp∂W∩TpF)=n−1, which is the dimension count of (i) and the kernel description of (ii).

3.1F1F2F3F5step 2.1

Pulling back ω∧dω=0 along i with [F1] and [F2] gives (ω∣∂W)∧d(ω∣∂W)=0; since ω∣∂W is nowhere vanishing by step 2.1, [F3] makes its kernel an integrable hyperplane distribution on the smooth n-manifold ∂W, and [F5] turns that distribution into a codimension-one regular foliation F∣∂W defined by ω∣∂W, completing (i) and (ii).

3.2step 2.1

A defining form that orients F transversely restricts to the nowhere vanishing form ω∣∂W of step 2.1, whose kernel is the restricted distribution, so the restricted foliation is transversely oriented by ω∣∂W; this is (iv).

4.1step 2.1step 3.1

Fix a leaf L′ of F∣∂W and p∈L′. As a connected manifold tangent to TF and contained in ∂W, the leaf L′ lies in a leaf L of F and, being connected, in the intrinsic component C of L∩∂W containing p. Conversely, at a point q of C a boundary chart with ∂W={t=0} and a foliation chart for F present L locally as a level set {y=y0}, and because TqL=TqF and Tq∂W are transverse the functions y and t have independent differentials at q; hence L∩∂W is near q an integral manifold of ker⁡(ω∣∂W) of dimension n−1, that is, a plaque of the restricted foliation, and C is covered by such plaques. The set of points of C lying in the leaf L′ is then both open and closed in C and nonempty, so it equals C; therefore L′=C, which is (iii).

5.1F1F2step 3.1step 4.1∎

Finally, pulling back the identity dω=η∧ω along i and applying [F1] and [F2] gives d(ω∣∂W)=i∗(dω)=(i∗η)∧(i∗ω)=(η∣∂W)∧(ω∣∂W), which is (v); together with steps 2.1, 3.1, 3.2 and 4.1 this proves all five assertions.

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