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

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Every ribbon knot is smoothly slice

Statement

Assume the Axiom of Countable Choice. Every smooth oriented knot $K \subset S^{3}$ that bounds a smooth immersed disk in $S^{3}$ with only ribbon double arcs bounds a smooth properly embedded disk in $B^{4}$ with boundary $K$ . In particular, its smooth slice genus is zero.

Statement references

Ribbon immersion and ribbon-knot conventions

A ribbon immersion resolves rel boundary to a smooth proper disk

Smooth slice genus and its ACω convention

Ray, Slice Knots and Knot Concordance, §1.1

Owens, Knots and 4-Manifolds, Lecture 1

Livingston, A Survey of Classical Knot Concordance, §2.1

Facts & Assumptions

Given: ACω, a smooth oriented knot $K \subset S^{3}$ , and a ribbon disk for $K$ .

[F1]

A ribbon disk is a smooth immersion of the oriented disk with embedded boundary $K$ and only finitely many ribbon double arcs, whose preimages are one proper arc and one interior arc. (Ribbon disks and ribbon knots)

[F2]

Such an immersion has a smooth proper embedded resolution in $B^{4}$ fixed on its boundary. (Local resolution of ribbon singularities)

[F3]

The smooth slice genus is the minimum genus of a connected compact smooth oriented properly embedded surface in $B^{4}$ with boundary $K$ ; the disk has genus zero and all genera are nonnegative. (Smooth and topological slice genus)

[A1]

ACω is the sequence-choice axiom assumed in [F3]. (The Axiom of Countable Choice ( ${A C}_{ω}$ ))

Proof

1.1givenF1F2construct

Choose a ribbon immersion $f : D^{2} \to S^{3}$ for $K$ as in [F1]. Applying [F2] gives a smooth proper embedding $\tilde{f} : D^{2} \to B^{4}$ that agrees with $f$ on $\partial D^{2}$ ; hence its boundary is exactly the oriented knot $K$ .

2.1givenF1F2F3A1step 1.1algebra∎

The image of $\tilde{f}$ is a connected orientable disk, so its Euler characteristic is $1$ and its genus is $0$ . By [F3], this disk is an admissible smooth spanning surface in the set defining $g_{4}^{s m} (K)$ ; since that set contains $0$ and consists of nonnegative integers, $g_{4}^{s m} (K) = 0$ . Thus $K$ is smoothly slice. The argument from [F1] to [F2] itself is choice-free; ACω is present only for the library's slice-genus convention [F3].

Remarks

This proves ribbon $\Rightarrow$ smooth sliceness. It makes no assertion that every smooth slice knot is ribbon.

Depends on

Used by

Under full AC, the right-handed trefoil is not ribbon Counterexample
The square knot is a nontrivial ribbon knot Example

Dependency tree · two levels

24 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

Arunima Ray, Slice Knots and Knot Concordance, Winter Braids Lecture Notes 8 (standard reference, not scraped)
Brendan Owens, Knots and 4-Manifolds, Winter Braids IX lecture notes (standard reference, not scraped)
Charles Livingston, A Survey of Classical Knot Concordance, Handbook of Knot Theory, Chapter 7 (standard reference, not scraped)