Satellite links with Brunnian properties
An n-link L is splittable if there exists an \((n+2)\)-disk \(B^{n+2}\) satisfying \(L\cap B^{n+2}\neq \emptyset,L\cap \partial B^{n+2}=\emptyset,\) and \(L\cap(S^{n+2}-B^{n+2})\neq \emptyset.\) Let \({\mathfrak A}\) be the family of those subsets S of \(I=\{1,2,...,m\}\) for which the sublink \(L_ S=\cup_{i\in S}L_ i\) of an n-link \(L=L_ 1\cup L_ 2\cup...\cup L_ m\) does not split. Then we say that L has the Brunnian property of type \({\mathfrak A}'\). For convenience we assume that \(\emptyset,\{i\}\in {\mathfrak A}\) for all \(i\in I\). In this family of subsets \({\mathfrak A}\), the following condition must be satisfied: If \(S,T\in {\mathfrak A}\) and \(S\cap T\neq \emptyset,\) then \(S\cup T\in {\mathfrak A}.\) The main theorem of the paper is that for \(n\geq 1\) and \(m\geq 2\) and for any family of subsets \({\mathfrak A}\) with the above condition, there exists an n-link with m components with the Brunnian property of type \({\mathfrak A}\). For \(n\geq 2\), this theorem was previously obtained by \textit{H. Debrunner} [Math. Z. 85, 154-168 (1964; Zbl 0123.399)], by using a ribbon n-link. The example in this paper is a satellite link, which is defined in a similar way that a satellite knot is defined. To prove that an n-link L is unsplittable, it is shown that its group \(\pi_ 1(S^{n+2}-L)\) is indecomposable with respect to free product.
scientific article; zbMATH DE number 3868244
scientific article