{ "id": "1408.3918", "version": "v2", "published": "2014-08-18T08:04:39.000Z", "updated": "2015-09-22T09:24:42.000Z", "title": "The classification of certain linked $3$-manifolds in $6$-space", "authors": [ "Sergey Avvakumov" ], "categories": [ "math.GT" ], "abstract": "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 $g'$.", "revisions": [ { "version": "v1", "updated": "2014-08-18T08:04:39.000Z", "comment": null, "journal": null, "doi": null }, { "version": "v2", "updated": "2015-09-22T09:24:42.000Z" } ], "analyses": { "keywords": [ "smooth category", "brunnian embedding", "theorem holds", "haefligers classification", "restriction" ], "note": { "typesetting": "TeX", "pages": 0, "language": "en", "license": "arXiv", "status": "editable", "adsabs": "2014arXiv1408.3918A" } } }