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Knot determinant is the order of double-cover homology

Statement

Assume the Axiom of Choice. Let $K$ be a smooth oriented knot with Alexander polynomial $Δ_{K} (t)$ normalized up to multiplication by $\pm t^{k}$ . Then $H_{1} (Σ_{2} (K); Z)$ is finite if and only if $Δ_{K} (- 1) \neq 0$ , and in that case

∣ H_{1} (Σ_{2} (K); Z) ∣ = ∣ Δ_{K} (- 1) ∣ .

Moreover, $Δ_{K} (- 1)$ is odd, so the first homology of every double branched cover is finite.

Statement references

The double-cover homology presentation by $V + V^{T}$

The determinant criterion for a square integer presentation

The Seifert determinant represents the Alexander polynomial

The integer Laurent-polynomial normalization $Δ_{K} (1) = \pm 1$

The definition and genus convention for a Seifert surface

Rolfsen, Knots and Links, Chapter 8.D

Existence of a connected compact oriented Seifert surface for every knot

The compact surface normal form and free fundamental group

The positive-push Seifert matrix convention

First homology is the abelianization of the fundamental group

Facts & Assumptions

Given: Full AC and a smooth oriented knot $K$ .

[A1]

Full AC is the choice-function axiom required by the cyclic branched-cover definition and Seifert determinant presentation. The Axiom of Choice

[F1]

For a positive-push Seifert matrix $V$ , $H_{1} (Σ_{2} (K); Z) ≅ coker (V + V^{T})$ . Homology of the double branched cover from a Seifert matrix

[F2]

The cokernel of a square integer matrix is finite exactly when its determinant is nonzero, and then its order is the absolute determinant. Finiteness and order of the cokernel of a square integer matrix

[F3]

$\det (V - t V^{T})$ represents $Δ_{K} (t)$ up to multiplication by $\pm t^{k}$ . The Seifert determinant is the knot Alexander polynomial

[F4]

$Δ_{K} (t)$ is an integer Laurent polynomial and $Δ_{K} (1) = \pm 1$ . The Alexander polynomial from the zeroth elementary ideal, The square Seifert presentation generates the knot Alexander polynomial

[F5]

Full AC supplies the countable-choice hypotheses used by the surface-existence and surface-normal-form results: apply its choice function to the range of any sequence of nonempty sets and reuse the selected element at each index. Thus the sequence assertions in The Axiom of Countable Choice ( ${A C}_{ω}$ ) and The countable-choice principle used in the foliation pair hold. The Axiom of Choice

[F6]

Every smooth oriented knot bounds a connected compact smooth oriented embedded surface with its given boundary, which is a Seifert surface under Oriented Seifert surfaces in the three-sphere. Every smooth oriented knot in $S^{3}$ bounds a Seifert surface

[F7]

For a connected compact oriented surface with one boundary component and genus $g$ , the fundamental group is free of rank $2 g$ . Finite surface normal forms, Jordan disks, and torsion control

[F8]

A connected smooth surface $F$ is locally path-connected, so its path components are open and it is path-connected. The first Hurewicz map identifies $H_{1} (F; Z)$ with the abelianization of $π_{1} (F)$ ; for the genus- $g$ surface in [F7], this gives $H_{1} (F; Z) ≅ Z^{2 g}$ and an ordered integral basis. The first Hurewicz map is abelianization

Proof

Given: The data in the Statement and Facts & Assumptions. By [F5]--[F8], choose a connected compact smooth oriented Seifert surface $F$ with oriented boundary $K$ and an ordered integral basis of $H_{1} (F; Z)$ . Let $V$ be its positive-push Seifert matrix as defined in Seifert pairing and matrix of an oriented knot, and put $D (t) = \det (V - t V^{T})$ . [A1, F5, F6, F7, F8, given, choose]

1.1A1F1F2givenalgebra

By [F1], the double-cover homology is presented by the square integer matrix $V + V^{T} = V - (- 1) V^{T}$ . Its determinant is $D (- 1)$ . By [F2], the homology is finite exactly when $D (- 1) \neq 0$ , and then its order is $∣ D (- 1) ∣$ . The hypothesis needed by [F1] is the Axiom of Choice in [A1].

2.1F3step 1.1algebra

By [F3], $D (- 1) = \pm (- 1)^{k} Δ_{K} (- 1)$ for some integer $k$ , so $∣ D (- 1) ∣ = ∣ Δ_{K} (- 1) ∣$ and the zero tests agree. This proves the finiteness criterion and order formula.

3.1F4step 1.1step 2.1algebra∎

Write $Δ_{K} (t) = \sum_{r = m}^{n} c_{r} t^{r}$ with integers $c_{r}$ and finitely many terms. Since $(- 1)^{r} \equiv 1 (m o d 2)$ for every integer $r$ , including negative $r$ , [F4] gives $Δ_{K} (- 1) \equiv \sum_{r} c_{r} = Δ_{K} (1) \equiv 1 (m o d 2)$ . Thus $Δ_{K} (- 1)$ is odd and nonzero; step 1.1 then gives finiteness for every knot.

Depends on

Used by

A higher cyclic branched cover can have infinite first homology Counterexample

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Sources

Dale Rolfsen, Knots and Links (standard reference, not scraped)
Charles Livingston, Seifert Forms and Concordance (standard reference, not scraped)