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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 Seifert-matrix map in the infinite cyclic cover

Statement

Assume the Axiom of Choice. Let K be a smooth oriented knot in the oriented standard three-sphere, and let F be a connected compact smooth oriented embedded surface with boundary orientation equal to that of K. Let g≥0 satisfy χ(F)=1−2g. Choose a small product tubular neighborhood N(K) adapted to the inward conormal of F, put E=S3∖int⁡N(K) and S=F∩E, and cut E open along a product bicollar of S to obtain the compact cut-open complement Y. Orient the normal line n of F by requiring (n,u,v) to be positive in S3 whenever (u,v) is a positive tangent basis of F; name the face on the +n side S+ and the face on the −n side S−. Let j:S↪F be the collar-removal inclusion and ι±:S→Y the two face inclusions; the inclusion induces an isomorphism on H1, and set i±=(ι±)∗∘j∗−1. Form E~F from copies Yn by gluing S−(Yn) to S+(Yn+1), and let t:Yn→Yn+1 be the deck shift. Because the boundary orientation of F agrees with K, this shift is the positive-meridian shift. Let a1,…,a2g be an ordered integral basis of H1(F;Z) and let Vij=lk⁡(ai+,aj), where ai+ is the positive normal push-off. In the linking-dual basis given by [γ]↦([a]↦lk⁡(γ,a)) under H1(Y;Z)≅Hom⁡(H1(F;Z),Z), the map i−−ti+ is represented over Λ=Z[t±1] by V−tVT. Consequently H1(E~F;Z) is the cokernel of V−tVT.

Facts & Assumptions

Given: Full AC, the oriented knot/surface pair, an integral basis of H1(F;Z), its Seifert matrix V, the cut-open complement Y, and the gluing convention S−(Yn)=S+(Yn+1).

[A1]

Full AC is the choice-function axiom and supplies the Alexander-duality and countable-choice interfaces. The Axiom of Choice, AC implies DC implies countable choice

[F1]

The pairing convention is Vij=lk⁡(ai+,aj) with positive normal fixed by the ambient orientation. Seifert pairing and matrix of an oriented knot

[F2]

Linking is symmetric and unchanged by moving both components through a disjoint product collar; the pairing is defined on homology classes. Well-definedness of the Seifert pairing

[F3]

Linking identifies H1(Y) with the integral dual of H1(F), the stack has the positive meridian deck shift, and Mayer–Vietoris presents its homology by i−−ti+. The cut-open cover presentation of the Alexander module

[F4]

The deck shift defines the Λ-module structure with t acting by Yn↦Yn+1. The Laurent polynomial ring as the principal localisation of Z[t] at t

Proof

1.1A1F1F2F3F4givenconstruct

Matrices of the two inclusions. Use the dual coordinate of γ∈H1(Y;Z) at ai given by lk⁡(γ,ai) from [F3]. The jth source basis vector under i− is the negative push aj−. Its ith dual coordinate is lk⁡(aj−,ai). By symmetry [F2], this equals lk⁡(ai,aj−). Translating both curves by a small positive normal amount preserves their disjointness and changes the pair to (ai+,aj), so this coordinate is Vij by [F1]. Thus the jth column of i− is the jth column of V. For i+ the ith dual coordinate of the image of aj is lk⁡(aj+,ai)=Vji by [F1], so the matrix of i+ is VT.

2.1F3F4step 1.1givenalgebra∎

The deck variable multiplies the i+ term by t under the seam convention in [F3]. Therefore the coordinate matrix of i−−ti+ is V−tVT by step 1.1. The exact Mayer–Vietoris sequence in [F3] identifies its cokernel with H1(E~F;Z), proving both claims.

Remarks

The coordinate square records the map and its matrix. The vertical arrows send a surface class to its chosen basis coordinates and a cut-complement class to its linking evaluations on that basis; step 1.1 proves commutativity entry by entry.

diagram failed to render:
\begin{tikzpicture}[>=stealth,node distance=1.6cm]
  \node (h1) {$H_1(F)\otimes\Lambda$};
  \node[right=3.2cm of h1] (h2) {$H_1(Y)\otimes\Lambda$};
  \node[below=1.2cm of h1] (z1) {$\Lambda^{2g}$};
  \node[below=1.2cm of h2] (z2) {$\Lambda^{2g}$};
  \draw[->] (h1) -- node[above] {$i_- -t i_+$} (h2);
  \draw[->] (h1) -- node[left] {basis coordinates} (z1);
  \draw[->] (h2) -- node[right] {linking dual} (z2);
  \draw[->] (z1) -- node[below] {$V-tV^T$} (z2);
\end{tikzpicture}

Depends on

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Sources