Least area Seifert surfaces and periodic knots (Q761662)

The main result is: If a knot \(K\subset S^ 3\) of genus g has a period m then \(m\leq 2g+1\). (Here a period means a topological rotation about an axis which does not meet the knot.) The proof is by constructing a Seifert surface F of (minimal) genus g which is invariant under the cyclic action. This is done by choosing a ''least-area-Seifert surface'' with respect to a suitable Riemannian metric on the knot complement using a result of \textit{M. Freedman}, \textit{J. Hass} and \textit{P. Scott} [Invent. Math. 71, 609-642 (1983; Zbl 0482.53045)]. The rest of the argument is straightforward. It turns out that in fact \(m\leq g\) if the quotient of f under cyclic action is not \(S^ 2\). A further result is: The period of a knot of genus \(g>0\) with trivial Alexander polynomial does not exceed \(g+1\). The proof uses Murasugi's congruence [\textit{K. Murasugi}, Comment. Math. Helv. 46, 162-174 (1971; Zbl 0206.256)].