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

Sergey Avvakumov

Published 2014-08-18, updated 2015-09-22Version 2

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