Brunnian links are determined by their complements
Abstract
If L1 and L2 are two Brunnian links with all pairwise linking numbers 0, then we show that L1 and L2 are equivalent if and only if they have homeomorphic complements. In particular, this holds for all Brunnian links with at least three components. If L1 is a Brunnian link with all pairwise linking numbers 0, and the complement of L2 is homeomorphic to the complement of L1, then we show that L2 may be obtained from L1 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.
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