Fibered knots with the same $0$-surgery and the slice-ribbon conjecture

Abstract

Abstract. Akbulut and Kirby conjectured that two knots with the same 0-surgery are concordant. In this paper, we prove that if the slice-ribbon conjecture is true, then the modified Akbulut-Kirby's conjecture is false. We also give a fibered potential counterexample to the slice-ribbon conjecture.

Non-slice linear combinations of iterated torus knots

“…In particular, if the slice-ribbon conjecture holds, then Rudolph's conjecture holds. In fact, Baker [2] and Abe-Tagami [1] recently noticed that the slice-ribbon conjecture implies a statement stronger than Rudolph's conjecture:…”

“…Conjecture 2 (Abe-Tagami [1] and Baker [2]). The set of prime fibered strongly quasi-positive knots is linearly independent in the smooth knot concordance group C.…”

On rational sliceness of Miyazaki's fibered, −amphicheiral knots

“…Remark It is interesting to compare Theorem with the result of Abe and Tagami [, Lemma 3.1] based on the work of Baker that any non‐trivial linear combination of tight, prime fibered knots is not ribbon.…”

“…In [, Theorem 5.1], it is proved that a fibered knot is homotopy ribbon if and only if the monodromy of its fiber extends over the handlebody bounded by the fiber. Hinging on this theorem, several non‐homotopy ribbon knots have been constructed (for example, see ). Most of these examples are algebraically slice, and detecting their non‐sliceness is an interesting problem.…”

Knot traces and concordance

“…Let denote the ( smooth ) 4 ‐genus of , the minimal genus of any smoothly embedded surface that bounds in . We observe that the annulus twisting construction as used in , and produces pairs of knots with diffeomorphic 0‐traces which each have 4‐genus either 0 or 1. By Corollary , any such pair of knots must have the same 4‐genus.…”