Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedPipeline-generatedverified 2026-10-22 (gpt-6.1-sol)
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Blackboard self-linking equals diagram self-writhe

Statement

Assume the Axiom of Countable Choice. Let D be a regular oriented diagram of a knot K with blackboard framing bD. The self-linking number of the framed knot (K,bD) equals the diagram writhe: sl⁡(K,bD)=wr⁡(D). More generally, let D be a regular oriented diagram of an oriented link L=K1∪⋯∪Kr, and let bD,i be the blackboard framing of Ki induced by its component subdiagram. Then sl⁡(Ki,bD,i)=wr⁡i(D)(1≤i≤r). Total writhe of a link still includes the signs of mixed crossings, while the sum of component self-linking numbers uses only self-crossings.

Facts & Assumptions

Given: The Axiom of Countable Choice and a regular oriented diagram D of a knot K, or, for the component claim, of a link L=K1∪⋯∪Kr.

[A1]

The Axiom of Countable Choice, written ACω (The Axiom of Countable Choice (ACω)).

[F1]

The self-linking number of a framed knot is the linking number with a sufficiently small embedded push-off along its framing section (Small framed push-offs define a well-defined self-linking number, Framed knots, self-linking, and the Seifert framing).

[F2]

For a two-component diagram the algorithmic linking number is one half of the signed sum of its mixed crossing signs (Compute oriented linking number from a diagram, The diagram linking-number scan terminates and returns the stated integer).

[F3]

The blackboard push-off has exactly two mixed crossings of the original sign for each original self-crossing and no other mixed crossings (A blackboard push-off doubles each crossing into mixed crossings).

Proof

1.1A1F3givenalgebra

By [F3], the mixed crossing list for K∪KbD has two crossings of sign sgn⁡(c) for each crossing c of D and no others. Therefore its signed mixed sum is 2∑csgn⁡(c)=2wr⁡(D).

2.1A1F1F2step 1.1construct∎

By [F2], the linking number of K with its blackboard push-off is half this mixed sum; by [F1] it is the self-linking number. Thus sl⁡(K,bD)=wr⁡(D). For a link diagram D and each component Ki, apply the same calculation to its regular component subdiagram with induced blackboard framing bD,i; crossings with other components are not mixed crossings of that component and its push-off, so precisely its self-crossings contribute.

Depends on

Used by

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Sources