ExampleConstruction: Literature-sourcedVerification: AI-generatedPipeline-generatedprecheck passverified 2026-10-22 (gpt-6.1-sol)

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Worked cases of torus knots

Example

Assume the full Axiom of Choice. For $T (2, 3)$ , the exterior group is $⟨ x, y ∣ x^{2} = y^{3} ⟩$ , a central fiber is $h = x^{2} = y^{3}$ , a positive meridian is $μ = y^{2} x^{- 1}$ , and a preferred longitude is $λ = h μ^{- 6}$ . Its Alexander polynomial is $Δ (t) ≐ t^{2} - t + 1$ and its genus is $1$ . For $T (1, 5)$ , the group presentation reduces to $Z$ , the knot is the unknot, the Alexander polynomial is $1$ , and the genus is $0$ .

Example references

The group, central fiber, and peripheral-word formulas

The Alexander polynomial and genus formulas

The unit-slope unknot result

Cattabriga and Francaviglia, §§5.1.1 and 5.1.6–5.2.4

Calegari, Example 1.2 and §4.3

Facts & Assumptions

Given: Full AC, the coprime torus knots $T (2, 3)$ and $T (1, 5)$ , and the formulas of the A page.

[A1]

Full AC supplies the countable-choice hypotheses used by the knot and genus statements (The Axiom of Choice).

[L1]

Full AC implies Countable Choice (AC implies DC implies countable choice, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[F1]

Torus-knot exterior group and peripheral system gives $π_{1} (E_{T (p, q)}) = ⟨ x, y ∣ x^{p} = y^{q} ⟩$ , $h = x^{p} = y^{q}$ , and $μ = y^{α} x^{β}$ , $λ = h μ^{- p q}$ when $p α + q β = 1$ .

[F2]

Torus-knot Alexander polynomial and genus formulas gives $Δ_{T (p, q)} ≐ (t^{∣ p q ∣} - 1) (t - 1) / ((t^{∣ p ∣} - 1) (t^{∣ q ∣} - 1))$ and $g (T (p, q)) = (∣ p ∣ - 1) (∣ q ∣ - 1) / 2$ .

[F3]

Unit-slope torus knots are unknots says $T (p, q)$ is the unknot when $∣ p ∣ = 1$ or $∣ q ∣ = 1$ .

Verification

1.1A1F1givenalgebra

For $(p, q) = (2, 3)$ the group relation is $x^{2} = y^{3}$ , and this common element $h$ is central. The equation $2 α + 3 β = 1$ has solution $(α, β) = (2, - 1)$ , so [F1] gives $μ = y^{2} x^{- 1}$ and $λ = h μ^{- 6}$ .

1.2F2step 1.1algebra

Substituting $(2, 3)$ into [F2] gives $Δ (t) ≐ \frac{(t^{6} - 1) (t - 1)}{(t^{2} - 1) (t^{3} - 1)} = \frac{t^{3} + 1}{t + 1} = t^{2} - t + 1,$ and $g = (2 - 1) (3 - 1) / 2 = 1$ .

1.3F1givenalgebra

For $(p, q) = (1, 5)$ , the relation $x = y^{5}$ eliminates $x$ and leaves the infinite cyclic group generated by $y$ . The meridian equation has solution $(α, β) = (1, 0)$ , so [F1] gives $μ = y$ and $λ = h μ^{- 5} = 1$ in the group.

2.1A1L1F2F3step 1.3algebra∎

Full AC [A1] supplies the countable-choice hypothesis of [F3] by [L1], so [F3] identifies $T (1, 5)$ with the unknot. In [F2] the quotient cancels to $1$ and the genus expression is $0$ , giving the stated Alexander polynomial and genus.

Depends on

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Sources

Alessia Cattabriga and Stefano Francaviglia, Knot Theory and Applications: Lecture Notes (standard reference, not scraped)
Danny Calegari, Introduction to Knot Theory (standard reference, not scraped)