The Slice–Ribbon conjecture for knots

Other statements of this same problem, merged from separate entries. Each is equivalent to the statement above — proving any one settles them all.

1. The slice-ribbon conjecture for knots

   A knot in $S^3$ is ribbon if it bounds a ribbon disk, and slice if it bounds a properly embedded disk in the 4-ball $D^4$. Slice-ribbon conjecture. A knot in $S^3$ is ribbon if and only if it is slice. The conjecture is a central problem in knot theory; the supplied text gives no resolution.

   source: Alessio Carrega, “Shadows and quantum invariants”, arXiv:1610.04728 (2016).

2. The slice-ribbon conjecture for knots

   Let $K\subseteq S^3$ be a knot. A knot is ribbon if it bounds a smoothly immersed disc in $S^3$ with only ribbon singularities, while it is smoothly slice if it bounds a smoothly embedded disc in $B^4$. Slice-ribbon conjecture. Every smoothly slice knot is ribbon. This is a longstanding open problem asking whether every smooth slice knot admits a ribbon presentation.

   source: Arunima Ray, “Slice knots and knot concordance”, arXiv:2311.12168 (2023).

3. The Slice–Ribbon Conjecture for knots

   A knot is a smooth embedding of $S^1$ in $S^3$. It is slice if it bounds a smoothly embedded disk in $B^4$, and ribbon if it bounds a ribbon disk, meaning a disk admitting a Morse handle decomposition with only $0$- and $1$-handles.

   Slice–Ribbon Conjecture. Every slice knot is ribbon.

   The conjecture asks whether every smooth slice disk can be replaced by a ribbon disk for the same knot. It remains open.

   source: Melissa Zhang, “Notes on Khovanov homology”, arXiv:2501.03115 (2025).