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A cycle has zero algebraic intersection with a bounding cycle

Statement

Assume ACω. Let M be a closed oriented n-manifold, Aa⊆M a closed oriented embedded submanifold, and Wb+1⊆M a compact oriented embedded submanifold with boundary, with a+b=n and B:=∂W oriented outward-normal-first. Assume the inclusion of A is transverse to W and to ∂W. Then I(A,B)=0andI2(A,B)=0. The same vanishing holds for the mod 2 number without orientability hypotheses, provided the intersection is transverse. The compactness of W is essential: an extension over a noncompact trace, or an intersection escaping at infinity, need not preserve the count.

Facts & Assumptions

Given: A closed oriented Aa⊆Mn, a compact oriented Wb+1⊆M with boundary B=∂W, a+b=n, and transversality of the inclusion of A to W and to ∂W.

[F1]

Since A is closed in the closed manifold M and W is compact, the inclusion iW:W→M is transverse to A, and its boundary restriction is transverse to A; hence W∩A is a compact embedded submanifold with boundary of W, neat, of dimension b+1−b=1, with ∂(W∩A)=W∩A∩∂W=A∩B (Transverse preimages for maps from manifolds with boundary, Embedded smooth submanifolds with boundary, Transverse embedded submanifolds).

[F2]

Orient S=W∩A by the normal-first quotient Q=TM/TA and kernel-first convention det⁡TW=det⁡TS⊗det⁡Q, and orient ∂S outward-normal-first (Preimage orientation agrees with the local intersection sign, Induced boundary orientation).

[F3]

Local intersection signs compare the written factor order; swapping blocks of dimensions a,b multiplies the sign by (−1)ab (Swapping direct summands scales oriented bases by a sign, The local oriented intersection sign).

[F4]

The signed boundary sum of the compact oriented 1-manifold W∩A vanishes (Oriented boundary counts of a compact oriented 1-manifold cancel), and its boundary has even cardinality (Boundary of a compact 1-manifold has even cardinality); both rest on the classification of compact 1-manifolds, so this corollary inherits ACω through them and steps 1.1-3.1 add no further choice (The Axiom of Countable Choice (ACω)).

[F5]

I(A,B) is the finite sum of the local signs over A∩B, and I2(A,B) is its cardinality modulo two (The oriented intersection number, The mod 2 intersection number).

Proof

technique · direct; count the boundary of the compact oriented $1$-manifold $W\cap A$
1.1F1F2F4F5given

By [F1] the set W∩A is a compact oriented 1-manifold with boundary A∩B, oriented as in [F2]; its boundary is finite by [F4], so A∩B is a finite transverse intersection and both I(A,B) and I2(A,B) are defined.

2.1F2F3F4F5step 1.1algebra

At p∈A∩B, the map TB→Q is an isomorphism. Choose an outward vector r tangent to S; its existence follows from neatness, and let u be a positive determinant of TB. Then (r,u) is positive for TW by the boundary convention. The sign of diB(u) in Q is the local sign ε(B,A), since quotient lifts precede TA. The kernel-first convention therefore assigns the outward r that same sign in TS, so the boundary point sign is ε(B,A)=(−1)abε(A,B). This determinant-element argument also covers b=0. Summing over ∂S=A∩B gives ∑ε∂S=(−1)abI(A,B). The sum vanishes by [F4], hence I(A,B)=0.

3.1F1F4F5step 1.1algebra∎

Independently of orientations, [F4] says that the boundary of the compact 1-manifold W∩A has even cardinality; that boundary is A∩B by [F1], so #(A∩B) is even and I2(A,B)=0 by [F5]. This mod 2 statement assumes no orientability of A, W or M.

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