The classification of certain linked $3$-manifolds in $6$-space

Abstract

We work entirely in the smooth category. An embedding $f:(S^2\times S^1)\bigsqcup S^3\to {\mathbb{R}}^6$ is {\it Brunnian}, if the restriction of $f$ to each component is isotopic to the standard embedding. For each triple of integers $k,m,n$ such that $m\equiv n(\mathrm{mod}2)$, we explicitly construct a Brunnian embedding $f_{k,m,n}:(S^2\times S^1)\bigsqcup S^3\to {\mathbb{R}}^6$ such that the following theorem holds. Theorem: Any Brunnian embedding $f:(S^2\times S^1)\bigsqcup S^3\to {\mathbb{R}}^6$ is isotopic to $f_{k,m,n}$ for some integers $k,m,n$ such that $m\equiv n(\mathrm{mod}2)$. Two embeddings $f_{k,m,n}$ and $f_{k^{\prime },m^{\prime },n^{\prime }}$ are isotopic if and only if $k=k^{\prime }$, $m\equiv m^{\prime }(\mathrm{mod}2k)$ and $n\equiv n^{\prime }(\mathrm{mod}2k)$. We use Haefliger's classification of embeddings $S^3\bigsqcup S^3\to {\mathbb{R}}^6$ in our proof. The following corollary shows that the relation between the embeddings $(S^2\times S^1)\bigsqcup S^3\to {\mathbb{R}}^6$ and $S^3\bigsqcup S^3\to {\mathbb{R}}^6$ is not trivial. Corollary: There exist embeddings $f:(S^2\times S^1)\bigsqcup S^3\to {\mathbb{R}}^6$ and $g,g^{\prime }:S^3\bigsqcup S^3\to {\mathbb{R}}^6$ such that the componentwise embedded connected sum $f\#g$ is isotopic to $f\#g^{\prime }$ but $g$ is not isotopic to $g^{\prime }$.