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Homology of a cyclic branched cover as Alexander-module coinvariants

Statement

Assume the Axiom of Choice. Let $K$ be a smooth oriented knot, let $X_{K} = S^{3} ∖ K$ , and let $q \geq 2$ . Let $A_{K} = H_{1} ({\tilde{X}}_{K}; Z)$ be the one-variable Alexander module with the positive-meridian deck action $t$ . Then, as abelian groups,

H_{1} (Σ_{q} (K); Z) ≅ A_{K} / (t^{q} - 1) A_{K} .

Statement references

The meridian-mod- $q$ exterior cover and filling that define $Σ_{q} (K)$

The infinite cyclic cover and positive-meridian deck action

The compact exterior is a deformation retract of the open complement

First homology is the fundamental group abelianization

Homology exactness for the filling

Facts & Assumptions

Given: The Axiom of Choice, a smooth oriented knot $K$ , its chosen exterior $E_{K}$ , positive meridian $μ$ , and an integer $q \geq 2$ .

[A1]

Full AC is the choice-function axiom. The Axiom of Choice

[F1]

The connected cover $E_{q} \to E_{K}$ is classified by the positive-meridian homomorphism modulo $q$ ; its filling kills the closed lift of $μ^{q}$ . Cyclic branched cover of a knot

[F2]

The inclusion $E_{K} ↪ X_{K}$ is a strong deformation retraction equivalence and carries the positive meridian to the positive meridian. Tubular exterior and boundary torus

[F3]

The module $A_{K}$ is $H_{1}$ of the connected infinite cyclic cover of $X_{K}$ classified by total linking, with $t$ acting by the positive-meridian deck transformation. The one-variable Alexander module of an oriented link

[F4]

For a path-connected space, the first Hurewicz map identifies $H_{1}$ with the abelianization of its fundamental group. The first Hurewicz map is abelianization

[F5]

Connected covers of a path-connected, locally path-connected, semilocally simply connected base are classified by subgroups of its fundamental group. Connected covering spaces are classified by conjugacy classes of fundamental-group subgroups

[F6]

Mayer–Vietoris is exact for an open cover, and a deformation retraction induces an isomorphism on singular homology. Mayer–Vietoris sequence in singular homology, Singular homology is invariant under deformation retracts

Proof

Given: The data in the Statement and Facts & Assumptions. Choose a basepoint on $T_{K}$ and let $G = π_{1} (E_{K})$ , with $s \in G$ the based positive meridian loop. Write $ϕ : G \to Z$ for the homomorphism with $ϕ (s) = 1$ , and put $N = \ker ϕ$ .

1.1A1F2F3F4given

By [F2], inclusion of the compact exterior into the open complement induces an isomorphism on fundamental groups and identifies the positive meridian classes. The total-linking homomorphism in [F3] therefore corresponds to $ϕ$ ; by [F4], $A_{K}$ is isomorphic to $N^{a b}$ with the action induced by conjugation by a positive meridian, up to replacing that action by its inverse according to the left-versus-right lift convention.

2.1F1F5step 1.1construct

Put $G_{q} = \ker (ϕ m o d q)$ . For $g \in G_{q}$ , write $ϕ (g) = q k$ and set $n = g s^{- q k} \in N$ . Then $g = n s^{q k}$ , and the expression is unique because $ϕ (s^{q k}) = q k$ . Thus $G_{q}$ is the split extension of $N$ by the infinite cyclic group generated by $s^{q}$ , acting on $N$ by conjugation. By [F5] and [F1], $G_{q}$ is the fundamental group of $E_{q}$ .

3.1F3F4step 2.1algebra

Abelianizing this split extension kills the commutators in $N$ and the relations $s^{q} n s^{- q} = n$ for $n \in N$ . By [F3], those relations impose $(t^{q} - 1) A_{K} = 0$ on $N^{a b}$ ; if the inverse deck convention is used, $t^{- q} - 1 = - t^{- q} (t^{q} - 1)$ generates the same submodule. The homomorphism $G_{q} \to Z$ , $g \mapsto ϕ (g) / q$ , splits by $1 \mapsto s^{q}$ , so $H_{1} (E_{q}; Z) ≅ A_{K} / (t^{q} - 1) A_{K} \oplus Z [s^{q}]$ .

4.1F1F6step 3.1construct∎

The boundary of $E_{q}$ is a torus cover with basis the lift $\tilde{μ}$ of $μ^{q}$ and a lifted longitude. Thicken $E_{q}$ and the filling solid torus by collars to form an open cover. By [F6], its Mayer–Vietoris map presents $H_{1} (Σ_{q} (K))$ as $H_{1} (E_{q})$ modulo $[\tilde{μ}]$ : the longitude lift maps to the core generator of the attached solid torus and eliminates that generator, while the filling meridian maps to zero there. The class $[\tilde{μ}]$ is $[s^{q}]$ , the generator of the displayed free summand in step 3.1. Quotienting by it gives the stated isomorphism.

Remarks

The quotient by $t^{q} - 1$ is unchanged if the positive deck action is represented by its inverse, since $t^{- q} - 1$ differs from $t^{q} - 1$ by the unit $- t^{- q}$ .

The group extension and filling relation are shown below. The proof uses the element calculation in steps 2.1–4.1, not the diagram.

Depends on

Used by

Root-product test for cyclic branched-cover homology Lemma

Dependency tree · two levels

51 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Dale Rolfsen, Knots and Links (standard reference, not scraped)