Linking topological spheres

TL;DR

This paper constructs a specific embedding of a circle in five-dimensional space demonstrating that no 3-sphere can be topologically linked with it, revealing new insights into high-dimensional linking phenomena.

Contribution

It provides an explicit example of a topological embedding in -dimensional space where the third homotopy group of the complement vanishes, showing no linking with 3-spheres.

Findings

Existence of a circle embedding in -space with trivial -homotopy of the complement

No 3-sphere can be linked with this embedding

Advances understanding of high-dimensional linking and embedding properties

Abstract

There is a topological embedding $\iota:\mathbb{S}^1\to\mathbb{R}^5$ such that $\pi_3(\mathbb{R}^5\setminus\iota(\mathbb{S}^1))=0$. Therefore, no $3$-sphere can be linked with $\iota(\mathbb{S}^1)$.

Full text

P.H. was supported by NSF grant DMS-1800457.

Dedicated to MathOverflow

If $\mathbb{S}^{k}$ is smoothly embedded in $\mathbb{R}^{n}$, $n>k+1$, then there is an embedding of $\mathbb{S}^{n-k-1}$ so that the two spheres are linked with the liking number equal $1$. Now, if $\iota:\mathbb{S}^{k}\to\mathbb{R}^{n}$ is an arbitrary topological embedding, then by the Alexander duality $H_{n-k-1}(\mathbb{R}^{n}\setminus\iota(\mathbb{S}^{k}))=\mathbb{Z}$, so the homology of the complement is the same for all embeddings. Therefore, one might expect that $\iota(\mathbb{S}^{k})$ is linked with some ($n-k-1$)-dimensional sphere embedded into the complement of $\iota(\mathbb{S}^{k})$, so that this embedding of $\mathbb{S}^{n-k-1}$ gives a generator of $H_{n-k-1}(\mathbb{R}^{n}\setminus\iota(\mathbb{S}^{k}))=\mathbb{Z}$. This is the case when $k=n-2$. Indeed, then $n-k-1=1$ and the existence of a linked sphere follows from the fact that that $H_{1}$ is the abelianization of $\pi_{1}$, so a generator of $H_{1}=\mathbb{Z}$ is given by a mapping from a circle; this mapping can be approximated by an embedding leading to an embedded $\mathbb{S}^{1}$ that is linked with $\iota(\mathbb{S}^{k})$ with the linking number $1$. However, the case when $n-k-1>1$ is more subtle.

The purpose of this short note is to construct a somewhat surprising example: there is a topological embedding $\iota_{1}:\mathbb{S}^{1}\to\mathbb{R}^{5}$ such that for every embedding $\iota_{2}:\mathbb{S}^{3}\to\mathbb{R}^{5}\setminus\iota_{1}(\mathbb{S}^{1})$, the spheres $\iota_{1}(\mathbb{S}^{1})$ and $\iota_{2}(\mathbb{S}^{3})$ are unlinked. More precisely $\pi_{3}(\mathbb{R}^{5}\setminus\iota_{1}(\mathbb{S}^{1}))=0$ so the embedding of $\mathbb{S}^{3}$ in the complement of $\iota_{1}(\mathbb{S}^{1})$ is always homotopic to a constant map.

The result is a simple consequence of the celebrated Cannon-Edwards theorem [3, 4, 5] according to which the double suspension of the integer homology sphere is a topological sphere. This deep and counterintuitive result is often used to construct various counterexamples that are similar in the spirit to the one presented below.

Since the linking number of highly non-smooth topological spheres has recently been used in geometric analysis (cf. [6, 8, 7, 12]), the author hopes that this example will be of some interest.

Using a one point compactification of $\mathbb{R}^{5}$ it suffices to prove the following result.

Acknowledgements. The author would like to express his deepest gratitude to the Mathoverflow community; without their help this short note would have never been written. In particular he would like to thank Ian Agol, Jason DeBlois, Neil Hoffman, John Klein, Thilo Kuessner, Andrew Putman, Daniel Ruberman, see [10, 11]. The author is also grateful to the referee for the valuable comments.