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Homology splitting for the cyclic cover of zero surgery

Statement

Assume AC. For every smooth oriented knot $K$ and $q \geq 2$ , the first homology of the $q$ -fold cyclic cover $M_{0, q}$ of zero surgery has an isomorphism of abelian groups $H_{1} (M_{0, q}; Z) ≅ H_{1} (Σ_{q} (K); Z) \oplus Z,$ where the free summand is generated by $\tilde{μ}$ , the closed lift of the $q$ -fold meridian loop $μ^{q}$ .

Statement references

The cyclic cover of zero surgery

The meridional branched filling of the knot exterior cover

Meridian and preferred-longitude conventions

The meridian generator and preferred longitude

Existence of a Seifert surface

Total linking as intersection with a Seifert surface

Mayer–Vietoris for the filling decompositions

Classification of connected exterior covers

The lifting criterion for the Seifert-surface inclusion

Charles Livingston and Swatee Naik, Introduction to Knot Concordance, Remark 7.5.3, p. 139

Axiom of Choice

Facts & Assumptions

Given: Full AC, a smooth oriented knot $K$ , the integer $q \geq 2$ , the exterior cover $E_{q}$ , its meridional filling $Σ_{q} (K)$ , and its zero-surgery filling $M_{0, q}$ .

[A1]

Full AC is assumed wherever the cover and orientation conventions use the choice axiom. The Axiom of Choice

[F1]

The zero-surgery cover is obtained by filling the common exterior cover $E_{q}$ along a lift of the preferred longitude. Its filling solid torus has core represented by the closed lift $\tilde{μ}$ of $μ^{q}$ . Cyclic cover of zero surgery on a knot

[F2]

The branched cover $Σ_{q} (K)$ is obtained by filling the same $E_{q}$ along the closed lift of $μ^{q}$ . Cyclic branched cover of a knot

[F3]

The exterior homology is generated by the meridian; the preferred longitude is a degree-one peripheral curve nullhomologous in the exterior. The induced homomorphism $ℓ_{K} : π_{1} (E_{K}) \to Z$ sends $μ$ to $1$ and is the meridian-coordinate map. Homology and preferred longitude of a knot exterior Meridian and preferred longitude

[F4]

Every smooth oriented knot bounds a connected compact smooth oriented Seifert surface. Full AC supplies the countable-choice hypothesis of this supplier. Every smooth oriented knot in $S^{3}$ bounds a Seifert surface The Axiom of Countable Choice ( ${A C}_{ω}$ ) The Axiom of Choice

[F5]

For a one-component knot, the total-linking homomorphism is the meridian-coordinate map $ℓ_{K}$ from [F3]; it equals signed intersection with a Seifert surface. Total linking is intersection with a spanning surface for a link Homology and preferred longitude of a knot exterior

[F6]

Mayer–Vietoris computes the first homology of each filling from the common boundary torus, and the H0 map is injective because all three spaces are connected. Mayer–Vietoris sequence in singular homology

[F7]

The unbranched exterior cover is classified by the kernel of $ℓ_{K} m o d q$ ; a connected, locally path-connected surface whose fundamental-group image lies in this subgroup lifts to the cover, for each prescribed point above its basepoint. Cyclic branched cover of a knot Connected covering spaces are classified by conjugacy classes of fundamental-group subgroups Lifting criterion for maps from path-connected locally path-connected spaces

[L1]

Conway's Lemma 2.31 states the same splitting and gives a Mayer–Vietoris proof. The complete proof in the retrieved §2.6 was read; the explicit meridian retraction below makes the splitting step transparent. Anthony Conway, Algebraic Concordance and Casson–Gordon Invariants, §2.6, Lemma 2.31, pp. 23–24

Proof

1.1A1F3F4F5F7givenconstruct

Choose a Seifert surface for $K$ by [F4] and remove its boundary collar to obtain a properly embedded surface $F_{0}$ in the knot exterior. Its boundary is a degree-one longitude curve, hence represents the preferred longitude by [F3]. Each element of $π_{1} (F_{0})$ has a smooth loop representative. Push that loop slightly to one side of the Seifert surface; the push is homotopic in the exterior and disjoint from the surface, so [F5] gives meridian coordinate zero. Thus $ℓ_{K} m o d q$ vanishes on $π_{1} (F_{0})$ , and [F7] lifts $F_{0}$ to each sheet of $E_{q}$ . Each lifted copy has one boundary component over the preferred longitude, proving that every lifted longitude is nullhomologous in $E_{q}$ .

1.2F2F6algebra

In the branched filling, $\tilde{μ}$ bounds the attached solid torus's meridian disk and a longitude lift maps to its core generator. The same Mayer–Vietoris cokernel calculation [F6] gives $H_{1} (Σ_{q} (K); Z) ≅ H_{1} (E_{q}; Z) / ⟨ [\tilde{μ}] ⟩$ .

2.1F1F6step 1.1algebra

In the zero-surgery filling, the boundary-torus basis consists of the closed meridian lift $\tilde{μ}$ and a longitude lift $\tilde{λ}$ . The latter bounds a meridian disk in the attached solid torus and the former maps to its core generator. The Mayer–Vietoris cokernel in [F6] therefore eliminates the core generator and gives $H_{1} (M_{0, q}; Z) ≅ H_{1} (E_{q}; Z) / ⟨ [\tilde{λ}] ⟩$ . Step 1.1 makes this quotient $H_{1} (E_{q}; Z)$ .

3.1F3F7step 2.1step 1.2algebra∎

The cover subgroup is $\ker (ℓ_{K} m o d q)$ by [F7]. Thus $ℓ_{K} (g)$ is divisible by $q$ for every $g \in π_{1} (E_{q})$ , and $g \mapsto ℓ_{K} (g) / q$ is a homomorphism $π_{1} (E_{q}) \to Z$ . It descends to $r : H_{1} (E_{q}; Z) \to Z$ and satisfies $r ([\tilde{μ}]) = 1$ , since $\tilde{μ}$ projects to $μ^{q}$ . Therefore $Z [\tilde{μ}]$ is a direct summand and $H_{1} (E_{q}) ≅ H_{1} (E_{q}) / ⟨ [\tilde{μ}] ⟩ \oplus Z [\tilde{μ}]$ . Using step 1.2 and then step 2.1 proves the statement.

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Sources

Charles Livingston and Swatee Naik, Introduction to Knot Concordance (standard reference, not scraped)
Anthony Conway, Algebraic Concordance and Casson–Gordon Invariants (standard reference, not scraped)