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**Brunnian links are determined by their complements.**
*(English)*
Zbl 0970.57002

C. McA. Gordon and J. Luecke proved in [J. Am. Math. Soc. 2, No. 2, 371-415 (1989; Zbl 0678.57005)] that knots in the three-dimensional sphere \(S^{3}\) are determined by their complement. On the other hand there exist non-equivalent links with homeomorphic complements. A simple way to construct such links is by twisting along unknotted components.

The main result of this paper is that two Brunnian links with all pairwise linking numbers zero are equivalent if and only if they have homeomorphic complements. Here, an \(n\)-component link is called Brunnian if every \((n-1)\)-component sublink is trivial. Moreover, the authors prove the following: Let \(L_{1}\) be a Brunnian link with all pairwise linking numbers zero and let \(L_{2}\) be any link whose complement is homeomorphic to that of \(L_{1}\). Then \(L_{2}\) is obtained from \(L_{1}\) by successive twists around unknotted components.

The main result of this paper is that two Brunnian links with all pairwise linking numbers zero are equivalent if and only if they have homeomorphic complements. Here, an \(n\)-component link is called Brunnian if every \((n-1)\)-component sublink is trivial. Moreover, the authors prove the following: Let \(L_{1}\) be a Brunnian link with all pairwise linking numbers zero and let \(L_{2}\) be any link whose complement is homeomorphic to that of \(L_{1}\). Then \(L_{2}\) is obtained from \(L_{1}\) by successive twists around unknotted components.

Reviewer: Michael Heusener (Aubière)

### MSC:

57M25 | Knots and links in the \(3\)-sphere (MSC2010) |

### Citations:

Zbl 0678.57005
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\textit{B. Mangum} and \textit{T. Stanford}, Algebr. Geom. Topol. 1, 143--152 (2001; Zbl 0970.57002)

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