On knots with trivial Alexander polynomial
A knot \(K\) in \(3\)-space (or, more generally, in an integral homology sphere) is said to have minimal Seifert rank if it admits a Seifert surface whose genus equals the rank of the Seifert form. In this case, the Alexander polynomial of \(K\) is trivial. The authors prove that the converse does not hold: There are knots with trivial Alexander polynomial which do not have minimal Seifert rank. The proof uses the \(2\)-loop part \(Q(K)\) of the rational version of the Kontsevich integral of \(K\), defined by the first author and \textit{A. Kricker} [Geom. Topol. 8, 115--204 (2004; Zbl 1075.57004)], and the results on the effects on \(Q(K)\) of null-moves (which are \(Y\)-clasper surgeries with null-homologous leaves in the knot complement), obtained by the first author and \textit{L. Rozansky} [Topology 43, No. 5, 1183--1210 (2004; Zbl 1052.57011)]. It is also shown that \(Q(K)\) of a knot \(K\) of minimal Seifert rank is completely determined by three finite type invariants of \(K\) of degrees \(3,5,5\). The starting point of the paper is a wrong lemma of \textit{M. H. Freedman} [Invent. Math. 68, 195--226 (1982; Zbl 0504.57016)] contradicting the main result of the paper under review. In the appendix, the authors give a direct argument for the fact that knots with trivial Alexander polynomial are topologically slice, avoiding Freedman's wrong lemma.
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