Geschlecht von Knoten mit zwei Brücken und die Faserbarkeit ihrer Außenräume
An algorithm is given for determining the genus of a knot k with two bridges and for deciding whether such a knot k has a Stallings fibration. One can characterize k by two odd numbers \(\alpha\) and \(\beta \in (- \alpha,\alpha).\) The author produces from such a pair a triple \((n,<n>,<n+1>)\), where n is [\(\alpha\) /\(\beta\) ], \(<n+1>=2(\alpha - n\beta),\) \(<n>=2\beta -<n+1>,\) and introduces a transformation of it into \((n,<n>,| <n+1>-2\lambda <n>|)\) if \(| <n+1>-2\lambda <n>| <<n><<n+1>,\) or into \((n,| <n>-2\lambda <n+1>|,<n+1>)\) if \(| <n>-2\lambda <n+1>| <<n+1><<n>.\) Here \(\lambda\) is uniquely defined. The iteration of this gives a sequence which ends with \(<n,2,0>\) or \(<n,0,2>\). The genus of the corresponding knot is one less in the first case and remains the same in the second one. Also the genus of k corresponding to (n,0,2) is n/2 and that corresponding to (n,2,0) is (n- 1)/2. The knot \(k(n,<n>,<n+1>)\) has a fibration exactly in case \(<n+1>><n>\) and all the \(\lambda\) in the sequence are equal to 1.
scientific article; zbMATH DE number 4214936