Brunnian links are determined by their complements

Abstract

If $L_1$ and $L_2$ are two Brunnian links with all pairwise linking numbers $0$, then we show that $L_1$ and $L_2$ are equivalent if and only if they have homeomorphic complements. In particular, this holds for all Brunnian links with at least three components. If $L_1$ is a Brunnian link with all pairwise linking numbers $0$, and the complement of $L_2$ is homeomorphic to the complement of $L_1$, then we show that $L_2$ may be obtained from $L_1$ by a sequence of twists around unknotted components. Finally, we show that for any positive integer $n$, an algorithm for detecting an $n$–component unlink leads immediately to an algorithm for detecting an unlink of any number of components. This algorithmic generalization is conceptually simple, but probably computationally impractical.