Alphabeta Math
CounterexampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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A higher cyclic branched cover can have infinite first homology

Statement refuted

Every cyclic branched cover of a smooth oriented knot has finite first homology.

Facts & Assumptions

Given: Full AC, the right-handed trefoil matrix VT=(−101−1), and the primitive sixth root ζ6=eiπ/3.

[A1]

Full AC is the choice-function axiom required by the cyclic-cover and Alexander-polynomial theorems. The Axiom of Choice

[F1]

The Seifert determinant det⁡(V−tVT) represents the Alexander polynomial up to a unit ±tk. The Seifert determinant is the knot Alexander polynomial, The Alexander polynomial from the zeroth elementary ideal

[F2]

The cyclic-cover order theorem states that a zero at any nontrivial q-th root makes the first homology infinite. Cyclic-cover homology order from the Alexander polynomial

[F3]

The Alexander normalization gives ΔK(1)=±1, and the double-cover theorem gives finiteness when ∣ΔK(−1)∣ is this nonzero odd integer. The square Seifert presentation generates the knot Alexander polynomial, Knot determinant is the order of double-cover homology

Counterexample

technique · direct
1.1F1givenalgebra

Direct calculation gives det⁡(VT−tVTT)=det⁡(t−1−t1t−1)=t2−t+1. By [F1], this represents the trefoil Alexander polynomial up to ±tk. Since ζ62−ζ6+1=0, the Alexander polynomial vanishes at this nontrivial sixth root.

2.1A1F2F3step 1.1given∎

Applying [F2] with q=6 shows that H1(Σ6(T);Z) is infinite, contradicting the refuted universal claim. For the boundary comparison, [F3] gives ΔT(−1) odd and nonzero, so H1(Σ2(T);Z) is finite. Thus the failure occurs at a higher cyclic cover even though the double-cover case is always finite. The two supplier theorems use the Axiom of Choice in [A1].

Depends on

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Sources