CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-6-luna)verified 2026-10-22 (gpt-6.1-sol)

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Diagram crossing count can differ from unknotting number

Statement refuted

Under the Axiom of Countable Choice, the proposed universal statement is that for every smooth oriented knot $K \subset S^{3}$ and every regular oriented diagram $D$ of a representative of $K$ (a projection with finitely many transverse double points, no other multiple points, oriented strands, and an over/under choice at each double point), the number $c (D)$ of double points equals the unknotting number $u (K)$ .

Statement references

The separate definitions of diagram crossing count and unknotting number

Reidemeister-I kink creation and deletion

The countable-choice hypothesis used to define the unknotting number

Facts & Assumptions

Given: AC $_{ω}$ , the standard crossing-free unknot diagram, and a positive integer $k$ .

[A1]

AC $_{ω}$ is the countable-choice axiom under which $u (K)$ is defined. The Axiom of Countable Choice ( ${A C}_{ω}$ )

[F1]

The standard unknot has bridge number one and unknotting number zero by their definitions. Bridge number and unknotting number of an oriented knot

[F2]

Each oriented Reidemeister-I creation adds exactly one transverse double point, and each oriented Reidemeister move preserves the represented knot type by an ambient isotopy. Oriented Reidemeister moves, Each oriented Reidemeister move is realized by an ambient isotopy

Counterexample

Let $U$ be the standard unknot and start from its crossing-free regular diagram. Perform $k$ oriented Reidemeister-I kink creations in pairwise disjoint disks along the diagram, obtaining a regular diagram $D_{k}$ of the same unknot. Then $c (D_{k}) = k$ , whereas $u (U) = 0$ and $b (U) = 1$ . For every $k \geq 1$ , this gives $c (D_{k}) \neq u (U)$ and refutes the proposed universal equality.

1.1F2giveninduction

By [F2], each of the $k$ moves preserves the represented knot type and adds exactly one crossing, while disjoint supports ensure that no move changes crossings created in another disk. Induction on the finite number of moves gives $c (D_{k}) = k$ and shows that $D_{k}$ still represents $U$ .

2.1F1step 1.1algebra∎

By [F1], $u (U) = 0$ and $b (U) = 1$ . Since $k \geq 1$ , step 1.1 yields $c (D_{k}) = k \neq 0 = u (U)$ , proving the counterexample.

Diagram

This local schematic shows the one-crossing kink created by an R1 move. The horizontal segment is the underpassing branch, the diagonal segment is the overpassing branch, and the outer curved portion returns the strand to itself. Repeating the move in disjoint disks produces the family $D_{k}$ ; the count proof uses only the R1 definition, not the drawing.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

18 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

Ozsváth, Stipsicz, and Szabó, Grid Homology for Knots and Links (standard reference, not scraped)
Danny Calegari, Introduction to Knot Theory (standard reference, not scraped)