The \((2,1)\)-cable of the figure-eight knot is not smoothly slice
The paper under review shows that a potential counterexample to the slice-ribbon conjecture is in fact not such. The \textit{slice-ribbon conjecture} posits that every slice knot is a ribbon knot. \textit{Ribbon knots} are knots that bound discs having only \textit{ribbon singularities}. This type of singularity can be removed in \(4\)-dimensions, so that a ribbon knot bounds a smooth disc in the \(4\)-ball whose boundary is the \(3\)-sphere containing the knot. This says precisely that every ribbon knot is a \textit{slice knot}.\N\NThe \((2,1)\)-cable of the figure-eight knot was known not to be ribbon by work of \textit{K. Miyazaki} [Trans. Am. Math. Soc. 341, No. 1, 1--44 (1994; Zbl 0816.57007)]. Miyazaki observed more generally that a whole family of cables (i.e. \((2n,1)\)-cables) of knots with extra properties, enjoyed in particular by the figure-eight knot, (that is, fibred, negatively amphicheiral, with irreducible Alexander polynomial) cannot be ribbon, although various invariants cannot exclude them from being slice.\N\NIf a knot is slice, its branched double cover must bound a \({\mathbb Z}/2{\mathbb Z}\)-homology ball, which is the branched double cover of the slice disc in the \(4\)-ball. This fact can thus be used as a criterion to test whether a knot is not slice. A stronger criterion is obtained if one takes into account the existence of a covering involution. This observation is the core of the proof of the result. \N\NBesides proving that the \((2,1)\)-cable of the figure-eight knot has infinite order in the smooth concordance group, the authors also provide a family of composite knots with the same property.
scientific article; zbMATH DE number 7935201
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