Suppose $X^n$ and $M^{n-2}$ are manifolds, and $f_1,f_2 : M \to X$ to two homotopic embeddings of $M$ into $X$. We can then embed $M$ into both boundary components in $X \times I$ using $f_1$ and $f_2$, respectively. I am wondering if there will always be some submanifold $N^{n-1} \subset X \times I$ with boundary equal to these two embeddings of $M$ into $X \times I$.

If we choose a homotopy between $f_1$ and $f_2$, say $H$, then we can look at the track of the homotopy $M \times I \to X \times I$, but this will not be an embedded $N$. This then shows that the two different images of $M$ are homologous in $X \times I$ - is there some sort of Steenrod realization thing that allows we to conclude that there is such a desired $N$? Maybe I need assumptions on $n$...

For what its worth, I'm really interested in the case where everything is connected and oriented.

  • $\begingroup$ I think this is called concordance (but this is outside of my area). See en.wikipedia.org/wiki/Link_concordance (and the remark on higher dimensions) $\endgroup$ – Thomas Rot Aug 21 '19 at 12:15
  • $\begingroup$ With concordance you want $N = M \times I$ - this can not always be done, for example the trefoil and unknot are homotopic in $S^3$ but not concordant in $S^3 \times I$. So I want to let $N$ be more general. $\endgroup$ – user101010 Aug 21 '19 at 12:23
  • $\begingroup$ @user101010 what you describe is the higher dimensions entry in that Wikipedia article. $\endgroup$ – Fernando Muro Aug 21 '19 at 12:57

You are talking about the notion of L-equivalence, studied by Thom in his seminal paper

Thom, René, Quelques propriétés globales des variétés différentiables, Comment. Math. Helv. 28, 17-86 (1954). ZBL0057.15502.

(Nowadays some people, myself included, might call this cobordism of embeddings.)

The discussion starts with the definition on page 71. At the bottom of page 74 you find the statement that (assuming everything is oriented) in codimensions $1$ or $2$, two submanifolds are $L$-equivalent if and only if they are homologous. This is slightly stronger than what you ask (since homotopic implies homologous, but not the other way round).

In general it should be possible to find embeddings which are homotopic but not $L$-equivalent. Note that $L$-equivalence classes of codimension $k$ embeddings in $X$ are in one-to one correspondence with homotopy classes of maps from $X$ to the Thom space $MSO(k)$ (in the oriented case, or $MO(k)$ forgetting orientations) so finding such examples involves studying the unstable homotopy of Thom spaces.

Edit: I must have learnt this from Thom's paper, as I can't think of a canonical reference in English. This is the basic Pontrjagin-Thom construction, so it should be in many textbooks. An expository paper (in English) is here.

It should be noted that the statement that two codimension $2$ oriented embeddings in an oriented $n$-manifold $X$ are cobordant if and only if they are homologous is down to the (somewhat miraculous, in my opinion) fact that $MSO(2)\simeq \mathbb{C}P^\infty$ is an Eilenberg-Mac Lane space $K(\mathbb{Z},2)$. Then there are isomorphisms $$ [X,MSO(2)]\cong [X,K(\mathbb{Z},2)]\cong H^2(X;\mathbb{Z})\cong H_{n-2}(X;\mathbb{Z}), $$ the last being Poincaré duality.

  • $\begingroup$ Thank you very much for the answer :-)- I have been directed to this Thom reference before - as a human question, is there a nice English (and possibly more modern but still readable for the uninitiated) reference that people like for this stuff? $\endgroup$ – user101010 Aug 21 '19 at 14:21
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    $\begingroup$ Thom's paper is already translated into English. See worldscientific.com/worldscibooks/10.1142/6379 $\endgroup$ – Min Hoon Kim Aug 22 '19 at 16:02

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