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LemmaStatement: Literature-sourcedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)
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The boundary intersection pairing is well defined and rationally nondegenerate

Statement

Assume the Axiom of Choice. Let W be a compact connected oriented smooth four-manifold with rational-homology-sphere boundary M. The signed intersection pairing on interior two-cycles descends to a symmetric integer-valued bilinear form on H2(W;Z)/Tor⁡, this group is free of finite rank, and the form is nondegenerate after extending scalars to Q. The form need not be unimodular.

Facts & Assumptions

Given: Full AC and a compact connected oriented smooth four-manifold W with ∂W=M and H∗(M;Q)≅H∗(S3;Q).

[F1]

The long exact sequence of the pair contains H2(M;Q)→H2(W;Q)→H2(W,M;Q)→H1(M;Q). Both end groups vanish by the boundary hypothesis. Long exact sequence of a pair

[F2]

Poincaré–Lefschetz duality identifies H2(W,M;Q) with H2(W;Q). Poincaré–Lefschetz duality

[F3]

Over the field Q, the universal-coefficient theorem identifies H2(W;Q) with Hom⁡Q(H2(W;Q),Q). The universal coefficient theorem for cohomology over a PID

[F4]

Double W along a boundary collar; the resulting closed smooth manifold has finite CW homotopy type under full AC. Folding the second copy into the first makes W a homotopy retract of the double, so H2(W;Z) is a direct summand of a finitely generated group. Collar neighborhood theorem A closed smooth manifold has the homotopy type of a finite CW complex

[F5]

The boundary intersection form is the signed intersection pairing on interior representatives. Intersection lattice of a four-manifold with rational-homology-sphere boundary

[A1]

Full AC is assumed for the finite-CW homotopy-type supplier [F4].

Proof

Given: The data and suppliers [F1]–[F5].

1.1givenF1F5construct

The collar of M lets every absolute two-cycle be moved into int⁡W. The signed transverse intersection number is integer-valued and bilinear. Since H2(M;Q)=0, [F1] shows that H2(W;Q)→H2(W,M;Q) is an isomorphism.

2.1F1F2F3step 1.1algebra

The pairing map over Q is the composite H2(W;Q)→H2(W,M;Q)→PD⁡−1H2(W;Q)→Hom⁡Q(H2(W;Q),Q). Every map in this composite is an isomorphism by step 1.1, [F2], and [F3], so the form is rationally nondegenerate.

3.1F4F5A1step 2.1algebra∎

By [F4] and [A1], H2(W;Z) is finitely generated, so its torsion-free quotient is a finite-rank free abelian group. If x has finite order m, bilinearity gives mQW(x,y)=QW(mx,y)=0 in Z, hence QW(x,y)=0; the same argument applies in the other variable. Thus the form descends to the torsion-free quotient. The four-dimensional cup-product sign is (−1)2⋅2=+1, so the form is symmetric.

Depends on

Used by

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Sources