most recent 30 from http://mathoverflow.net 2013-06-18T22:00:04Z http://mathoverflow.net/feeds/question/95686 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/95686/source-on-the-proof-that-codimension-2-is-sufficient-for-knottings Oscar Guajardo 2012-05-01T19:36:11Z 2012-05-01T20:09:35Z

<p>Hi all. </p> <p>I'm not even sure that this is a theorem, but a while ago I heard a topologist friend of mine (whom I haven't been able to reach) saying that given a continuous embedding $f:M^{n}\to N^{r}$ ($M$ and $N$ being topological manifolds), a sufficient condition for $f(M)$ to be (possibly) knotted in $N$ is that $r=2n$. Does anyone know if this is even true? Also, if it is true then the following question seems natural: will the embedding still be knotted if $r>2n$. Thanks a lot.</p>

http://mathoverflow.net/questions/95686/source-on-the-proof-that-codimension-2-is-sufficient-for-knottings/95690#95690 Daniele Zuddas 2012-05-01T20:01:29Z 2012-05-01T20:09:35Z

<p>This question is badly posed. If you have a not simply connected manifold (of any dimension $> 3$) then two connected closed curves are isotopic iff they are homotopic, hence you have as many isotopy classes of curves as the conjugacy classes in the fundamental group. Also, following your comment, in codimension two (which means $r = n + 2$) there are a lot of "knotted" submanifolds: classical knots in $S^3$, 2-knots in $S^4$ and in general knotted spheres of dimension $r-2$ in $S^r$. In any manifold there are knotted codimension two submanifolds (i.e. pairs of non-isotopic embeddings of the same manifold in the ambient space). And this satisfies $r &lt; 2 (r - 2)$ if $r > 4$.</p>