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

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A two-meridian figure-eight presentation and its preferred longitude word

Statement

Assume full AC and use the standard oriented four-crossing figure-eight code of the explicit oriented Wirtinger-code lemma, with its positive based under-arc meridians $x_{1}, x_{2}, x_{3}, x_{4}$ and compatible diagram access paths. Put $u = x_{1}$ , $v = x_{2}$ , and $w = u v^{- 1} u^{- 1} v$ . Then $π_{1} (E_{4_{1}}) = ⟨ u, v ∣ v w = w u ⟩,$ with $u$ a positive meridian. In this presentation $X_{3} = u v u^{- 1}, X_{4} = v^{- 1} u v u^{- 1} v = w^{- 1} v$ are the old $x_{3}, x_{4}$ . The preferred longitude based compatibly with $u$ is represented by $λ = X_{4} u^{- 1} v X_{3}^{- 1} .$ In particular its trace in any representation is the trace of this explicit word. The formula does not identify a zero-homology centralizing word as a longitude without the diagram-parallel calculation.

Statement references

Actual oriented four-crossing code

Actual based blackboard-parallel and self-writhe correction

Facts & Assumptions

Given: The stated oriented figure-eight code, compatible positive based diagram meridians, and the selected initial strand $x_{1}$ .

[A1]

The Axiom of Choice is retained for the named knot-exterior branch; all new operations are finite.

[F1]

Explicit unknot and figure-eight oriented Wirtinger codes gives $R_{1} : x_{1} x_{2} = x_{3} x_{1}$ , $R_{2} : x_{1} x_{4} = x_{4} x_{2}$ , $R_{3} : x_{3} x_{2} = x_{2} x_{4}$ , and $R_{4} : x_{3} x_{4} = x_{1} x_{3}$ . Its full proof shows $R_{4}$ follows from the first three. Its actual oriented under-arc cycle is $x_{1} \to x_{2}$ at $R_{2}$ , $x_{2} \to x_{3}$ at $R_{1}$ , $x_{3} \to x_{4}$ at $R_{3}$ , $x_{4} \to x_{1}$ at $R_{4}$ ; the local coordinate charts specify the source and target sides and positive crossing signs.

[F2]

Diagram word for a preferred peripheral longitude proves that a positive component traversal contributes the over-meridian with exponent $+ 1$ for source-to-target underpasses and exponent $- 1$ for target-to-source. Their ordered product is the based blackboard parallel; multiplying by the initial meridian to minus the self-writhe gives its preferred zero-self-linking longitude.

Proof

1.1F1givenconstructalgebra

In [F1] eliminate $x_{3}$ using $R_{1}$ to get $X_{3} = u v u^{- 1}$ , then eliminate $x_{4}$ using $R_{3}$ to get $X_{4} = v^{- 1} X_{3} v = v^{- 1} u v u^{- 1} v$ . These are eliminations of explicitly defined generators, reversible by substitution, so impose exactly the remaining relation $R_{2}$ ; $R_{4}$ adds none by [F1]. With $w = u v^{- 1} u^{- 1} v$ the expression is $X_{4} = w^{- 1} v$ . Hence $R_{2}$ reads $u w^{- 1} v = w^{- 1} v^{2}$ , which after right cancellation of $v$ is $u w^{- 1} = w^{- 1} v$ , equivalently $v w = w u$ . This proves the asserted presentation with the original positive meridian $u = x_{1}$ .

1.2F1F2givenconstructalgebra

Start the oriented traversal at $x_{1}$ . The four successive underpasses in [F1] give the table $(R_{2}, x_{4}, + 1), (R_{1}, x_{1}, - 1), (R_{3}, x_{2}, + 1), (R_{4}, x_{3}, - 1),$ listing crossing, over-meridian and source/target exponent. Indeed $R_{2}, R_{3}$ go from the lower source side to the upper target side in those actual coordinate charts, while $R_{1}, R_{4}$ go from the upper target side to the lower source side. The determinants of their ordered over/under tangents have these same signs: each over-tangent points right and the under-tangent points respectively upwards or downwards. Thus the self-writhe is $1 - 1 + 1 - 1 = 0$ . This checks both the traversal order and the sign instead of deriving a longitude solely from homological exponent zero.

2.1A1F2step 1.1step 1.2constructalgebra∎

By [F2], the based blackboard parallel for that actual code is $x_{4} x_{1}^{- 1} x_{2} x_{3}^{- 1}$ . The self-writhe correction in step 1.2 is zero, so this very word is the preferred longitude. Substitution from step 1.1 gives $X_{4} u^{- 1} v X_{3}^{- 1}$ . The positive component traversal has degree one in its peripheral tube by [F2], and the zero-writhe correction gives zero self-linking; this is the actual preferred peripheral representative with the chosen initial access path. Evaluating it in any representation proves the trace assertion. No other longitude convention or external two-bridge peripheral formula is assumed.

Depends on

Used by

Figure-eight A-polynomial calculation Example
Figure-eight character curve and longitude trace Lemma

Dependency tree · two levels

17 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

Petersen, A-polynomials of a family of two-bridge knots (standard reference, not scraped)