Knot commensurability and the Berge conjecture (Q422834)

This paper is about knot commensurability. Two knots in the 3-sphere are called commensurable if their complements are commensurable, i.e.~have homeomorphic finite sheeted covers. \textit{A. W. Reid} [J. Lond. Math. Soc., II. Ser. 43, No. 1, 171--184 (1991; Zbl 0847.57013)] has shown that the figure eight knot is the only arithmetic knot in \(S^3\), which implies that it is the unique knot in its commensurability class.NEWLINENEWLINEReid and Walsh have conjectured that when a knot \(K\) is hyperbolic, there are at most three distinct knots in the commensurability class of \(K\). In the paper under review, this conjecture is proved for so-called flexible knots, a condition which is expected to be generic. Another result states that (1) knots without hidden symmetries which are commensurable are cyclically commensurable, and (2) a cyclic commensurability class contains at most three hyperbolic knot complements.NEWLINENEWLINEThere are other results concerning fibred knots and periodic knots.