Embedding infinite cyclic covers of knot spaces into 3-space

Topology Pub Date : 2006-07-01 DOI:10.1016/j.top.2006.01.005

Boju Jiang , Yi Ni , Shicheng Wang , Qing Zhou

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引用次数: 5

Abstract

We say a knot $k$ in the 3-sphere $S^{3}$ has Property $I E$ if the infinite cyclic cover of the knot exterior embeds into $S^{3}$ . Clearly all fibred knots have Property $I E$ .

There are infinitely many non-fibred knots with Property $I E$ and infinitely many non-fibred knots without property $I E$ . Both kinds of examples are established here for the first time. Indeed we show that if a genus 1 non-fibred knot has Property $I E$ , then its Alexander polynomial $Δ_{k} (t)$ must be either 1 or $2 t^{2} - 5 t + 2$ , and we give two infinite families of non-fibred genus 1 knots with Property $I E$ and having $Δ_{k} (t) = 1$ and $2 t^{2} - 5 t + 2$ respectively.

Hence among genus 1 non-fibred knots, no alternating knot has Property $I E$ , and there is only one knot with Property $I E$ up to ten crossings.

We also give an obstruction to embedding infinite cyclic covers of a compact 3-manifold into any compact 3-manifold.

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将结空间的无限循环覆盖嵌入到三维空间中

如果结外部的无限循环覆盖嵌入到S3中，我们说3球S3中的结k具有属性IE。显然，所有纤维结都具有属性IE。有无限多个具有IE属性的非纤维结和无限多个不具有IE属性的非纤维结。这两种例子在这里都是首次建立。事实上，我们证明了如果一个属1的非纤维结具有性质IE，那么它的亚历山大多项式Δk(t)必须是1或2t2−5t+2，并且我们给出了两个无限族的非纤维属1的结具有性质IE并且分别具有Δk(t)=1和2t2−5t+2。因此，在1属非纤维结中，没有交替结具有IE属性，只有一个结具有10个交叉的IE属性。给出了紧3流形无限循环盖嵌入任意紧3流形的一个障碍。

本文章由计算机程序翻译，如有差异，请以英文原文为准。

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来源期刊

Topology 数学-数学

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1 months