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

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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An essential disk in a knot exterior has longitude boundary

Statement

Let $E$ be the compact exterior of a smooth knot with supplied product boundary torus $T = S^{1} \times S^{1}$ , fibre meridian $μ$ , and unique degree-one zero-homology longitude class $[λ]$ from the product data. Let $D \subset E$ be a properly embedded piecewise-linear disk, meaning an embedded copy of the closed disk with $D \cap T = \partial D$ . If $\partial D$ is not null-homotopic in $T$ , then $[\partial D] = [λ]$ or $- [λ]$ in $H_{1} (T; Z)$ , and its mod-two intersection number with $μ$ is $1$ .

Statement references

Choice-free homology and longitude computation for a supplied knot tube — the kernel of boundary-to-exterior homology is generated by the degree-one longitude, and meridian/longitude form a basis.

Essential simple curves on a torus are primitive — a non-null-homotopic simple curve on the torus represents a primitive class.

Facts & Assumptions

Given: $E, T, μ, λ$ and the properly embedded disk $D$ in the statement.

[F1]

Choice-free homology and longitude computation for a supplied knot tube: $H_{1} (T; Z)$ has basis $([μ], [λ])$ , the class $[λ]$ generates the kernel of $i_{*} : H_{1} (T) \to H_{1} (E)$ , and product projection sends $[μ]$ to $0$ and $[λ]$ to $1$ .

[F2]

Essential simple curves on a torus are primitive: every non-null-homotopic embedded PL simple closed curve on a product torus has a nonzero primitive homology class.

Proof

1.1givenF1construct

The boundary $\partial D$ is an embedded PL simple closed curve in $T$ . Since it bounds the disk $D$ in $E$ , its image under inclusion $i : T ↪ E$ is zero. By [F1], the kernel is $Z [λ]$ , so $[\partial D] = k [λ]$ for an integer $k$ .

2.1givenF1F2step 1.1algebra

The hypothesis and [F2] say $[\partial D]$ is nonzero and primitive. Since $[λ]$ is a basis vector by [F1], the multiple $k [λ]$ is primitive only if $k = 1$ or $k = - 1$ . Hence $[\partial D] = \pm [λ]$ .

3.1F1step 2.1algebraconstruct

Put the boundary curve in PL transverse position to the fibre $μ = p^{- 1} (s_{0})$ , where $p : T \to S^{1}$ is product projection. Its intersection points are exactly the preimages of $s_{0}$ under $p ∣_{\partial D}$ . A lift of this PL circle map to $R$ changes by its degree, which is the product-projection coefficient of $[\partial D]$ , namely $+ 1$ or $- 1$ by step 2.1. For a regular value, the signed crossing count of the lift through the integer translates of that value equals this degree; reducing modulo $2$ removes the signs, so the number of preimages is odd. Thus $\partial D$ has mod-two intersection $1$ with $μ$ .

4.1step 2.1step 3.1algebra∎

Steps 2.1 and 3.1 give the longitude class and the asserted parity.

Depends on

Used by

Dependency tree · two levels

10 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

Joel Hass, Jeffrey C. Lagarias, and Nicholas Pippenger, The Computational Complexity of Knot and Link Problems, Journal of the ACM 46 (1999) (standard reference, not scraped)
Jeff Erickson, Normal Surfaces and Knots, Computational Topology lecture notes (2009) (standard reference, not scraped)