---
tag: EQB39X4HGB
kind: conjecture
name: Slice-ribbon conjecture
msc: [57K10, 57K40]
uses: [QYEAZTYGPT, P68G28E4AD]
about: [P68G28E4AD]
sources:
  - url: https://en.wikipedia.org/wiki/Slice_knot
history:
  posed_by: "Ralph Fox"
  first_stated: "1962"
---

## Statement

A tame [[knot|QYEAZTYGPT]] $K \subset S^3$ is a *ribbon knot* if it bounds an immersed disk $D$ in $S^3$ whose only self-intersections are ribbon singularities.
A ribbon singularity is an arc of double points with two preimages in $D$.
One preimage is an arc interior to $D$, and the other is an arc meeting $\partial D$ in its two endpoints.
The conjecture states that every smoothly [[slice knot|P68G28E4AD]] is a ribbon knot.

## Remarks

The converse holds: pushing the interior of a ribbon disk into $B^4$ removes the singularities, so every ribbon knot is slice.
Equivalently, $K$ is ribbon if and only if it bounds a smoothly embedded disk in $B^4$ on which the radial distance function has no local maxima.
Lisca proved the conjecture for two-bridge knots.
The conjecture is open in general.

## Explanations

### L1 human

A ribbon knot bounds a disk in ordinary space that passes through itself only in a mild way.
The disk looks like a ribbon threaded through a slit in itself.
Such a disk can be pushed into four dimensions so that it no longer meets itself, so ribbon knots are slice.
The conjecture says that this is the only way a knot can be slice.
It would turn a four-dimensional question into a three-dimensional one.