Abstract:
We classify smooth Brunnian (i.e., unknotted on both components) embeddings $(S^2\times S^1)\sqcup S^3 \to\mathbb R^6$. Any Brunnian embedding $(S^2\times S^1)\sqcup S^3\to\mathbb R^6$ is isotopic to an explicitly constructed embedding $f_{k,m,n}$ for some integers $k,m,n$ such that $m\equiv n\pmod2$. 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 $S^3\sqcup S^3\to\mathbb R^6$ in our proof. The relation between the embeddings $(S^2\times S^1)\sqcup S^3\to\mathbb R^6$ and $S^3\sqcup S^3\to\mathbb R^6$ is not trivial, however. For example, we show that there exist embeddings $f\colon(S^2\times S^1)\sqcup S^3\to\mathbb R^6$ and $g,g'\colon S^3\sqcup S^3\to\mathbb R^6$ such that the componentwise embedded connected sum $f\#g$ is isotopic to $f\#g'$ but $g$ is not isotopic to $g'$.
Key words and phrases:classification of embeddings, framed cobordism, linked manifolds.