CounterexampleConstruction: AI-generatedVerification: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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The unknot has a zero-order Seifert matrix

Statement refuted

Every Seifert matrix of an oriented knot has positive order, meaning that its number of rows and columns is at least one.

Statement references

A Seifert matrix has order equal to the rank of the first homology of the chosen surface

Facts & Assumptions

Given: Full AC, the oriented round unknot, its oriented spanning disk, and the positive-push Seifert-matrix definition.

[A1]

Full AC supplies the countable-choice interface used by the surface definition and normal-form supplier. The Axiom of Choice, AC implies DC implies countable choice

[L1]

The standard countable-choice principle obtained from full AC is equivalent to the surface normal-form supplier's countable-choice formulation for foliated chart pairs. The Axiom of Countable Choice ( ${A C}_{ω}$ ), The countable-choice principle used in the foliation pair, Countable choice is equivalent to nonempty countable products

[F1]

A Seifert surface is connected, compact, smooth, oriented, and has boundary the oriented knot; for a one-boundary-component surface its genus satisfies $χ (F) = 1 - 2 g$ . Oriented Seifert surfaces in the three-sphere

[F2]

A genus-zero compact surface with one boundary component has $H_{1} (F; Z) ≅ Z^{0} = 0$ . Finite surface normal forms, Jordan disks, and torsion control

[F3]

The Seifert matrix is indexed by an ordered integral basis of $H_{1} (F; Z)$ and therefore has order equal to the number of basis elements. Seifert pairing and matrix of an oriented knot

Counterexample

technique · direct

1.1A1L1F1givenconstruct

Take the round unknot $K \subset S^{3}$ with its standard oriented flat disk $F$ in a coordinate ball. The disk is connected, compact, and smooth, and its Euler characteristic is $1$ , so [F1] gives genus $g = 0$ .

2.1F2F3step 1.1givenalgebra∎

By [F2], $H_{1} (F; Z) = 0$ and its integral basis is empty. The matrix in [F3] has one row and one column for each basis element, hence is $0 \times 0$ . Its order is zero, contradicting the refuted positive-order claim.

Sources

The genus-zero surface has free first homology of rank zero

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

43 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, Chapter 8.A (standard reference, not scraped)
Alessia Cattabriga and Stefano Francaviglia, Knot Theory and Applications: Lecture Notes, Section 4.3 and 6.2.3 (standard reference, not scraped)