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.Comment: Published by Algebraic and Geometric Topology at http://www.maths.warwick.ac.uk/agt/AGTVol1/agt-1-7.abs.htm