Determining knots and links by cyclic branched coverings
A cyclic covering \(M\) of \(S^3\) branched along a link \(L\) is ``strongly cyclic'', if each meridian of \(L\) is projected onto a generator of the cyclic group. Let \(M\) be a hyperbolic 3-manifold and let the covering group act by isometries (\(L\) is then \(2\pi/n\)-hyperbolic). The main result states: If \(M\) and \(M'\) are two such manifolds and \(L\) and \(L'\) the respective links, then the homeomorphy of \(M\) and \(M'\) implies the equivalence of \(L\) and \(L'\), if \(n\geq 2\) does not divide the order of the orientation-preserving symmetry group \(\text{Sym}_+(L)\) of \(L\). The proof uses a simple argument concerning \(p\)-groups. The result is applied to maximally symmetric \(D_6\)-manifolds (\(D_6=\) dihedral group of order 12) which turn out to be the 3-fold cyclic coverings of 2-bridge links.
scientific article; zbMATH DE number 1038318