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Prime-power cyclic branched covers of knots have finite first homology

Statement

Assume AC. Let K be a smooth oriented knot and let q=pk for a prime p and k≥1. Then H1(Σq(K);Z) is finite, so Σq(K) is a rational homology sphere and carries a nonsingular torsion linking form.

Facts & Assumptions

Given: Full AC, a smooth oriented knot K, and q=pk with p prime and k≥1.

[A1]

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

[F1]

For any connected Seifert surface with positive-push matrix V, the group H1(Σq(K);Z) is finite exactly when D(ζqj)≠0 for all 1≤j<q, where D(t)=det⁡(V−tVT). Root-product test for cyclic branched-cover homology

[F2]

D(1)=det⁡(V−VT)=±1, since V−VT is the integral intersection matrix of the Seifert surface. The skew part of a Seifert matrix is the surface intersection form

[F3]

For each s≥1, the minimal polynomial over Q of a primitive ps-th root is the monic cyclotomic polynomial Φps(t), which has integer coefficients and Φps(1)=p. Φn is irreducible in Q[t] for every n≥1

[F4]

If Y is closed, connected, oriented and has finite first homology, its bounding-chain torsion linking pairing is nonsingular. The torsion linking pairing of a rational homology sphere is nonsingular

[F5]

For a closed oriented 3-manifold with finite first homology, the homology universal coefficient theorem and Poincaré duality give rational homology of S3. The universal coefficient theorem for homology over a PID Poincaré duality for oriented topological manifolds

Proof

1.1F3givenalgebraA1

Suppose a nontrivial q-th root of unity ω were a zero of D. Its order is ps for some 1≤s≤k. By [F3], the minimal polynomial Φps divides D in Q[t]. Since Φps is monic and both polynomials have integer coefficients, polynomial long division gives an integer-coefficient quotient with zero remainder. Thus p=Φps(1) divides D(1).

2.1F1F2step 1.1algebra

This contradicts D(1)=±1 from [F2]. Therefore D has no zero at any nontrivial q-th root of unity. The criterion [F1] gives finiteness of H1(Σq(K);Z).

3.1F1F4F5step 2.1algebra∎

By [F5], the finite first homology from step 2.1 makes Σq(K) a rational homology sphere. Its torsion linking form is nonsingular by [F4], which proves the final assertion.

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