2
$\begingroup$

In this archiv paper which is continuation of following:

Borodzik, Maciej; Némethi, András; Ranicki, Andrew, Morse theory for manifolds with boundary, Algebr. Geom. Topol. 16, No. 2, 971-1023 (2016). ZBL1342.57018.

authors prove that manifold $M$ codimension 2 bounds so called Seifert surface of codimension 1 when manifold $M$ normal bundle is trivial. Consider now 3-manifold $M$ embedded in 5-dimensional Euclidean space having trivial normal bundle. Let $N_1$ be 4-manifold which boundary is $M$ according to above work. I would like to consider $M$ which is not knotted. For the moment let's consider following definition. $M$ embedded in 5-space is not knotted when exist closed 4-manifold $N$ embracing $M$ and $M$ separates $N$. This "definition" is wrong - see comments below from Danny Rubermann.

This is just work definition, because for giving embedded $M$ I would like to construct such $N$.

My next step is to see whether I can make $N$ 1-connected. If yes then let's look at second homotopy or homology of $N$. I saw somewhere statement that in such case $\pi_2=H_2$ has no torsion - I cannot find it now here. Now we look at Mayer-Vietoris to see what possibilities for $N, N_1, N_2$ are for given $H_1M$. The plan is to find simplest possible $N$. If $N$ is sphere then it means that $M$ is embeddable in 4-space. Otherwise - if we cannot make $N$ sphere then $M$ is not embeddable in 4-space.

Does this plan make sense ?

$\endgroup$
7
  • $\begingroup$ Perhaps you could explain what you mean by "embracing". Since N has trivial normal bundle (equivalent to being orientable, which you get for free) there's a section of the normal bundle and so a copy of $N_1\times I$ in $R^5$. Then $M\subset N = \partial N_1 \times I$. ($N$ is the double of $N_1$.) Does this $N$ "embrace" $M$? $\endgroup$ Commented Oct 17, 2018 at 13:24
  • $\begingroup$ @DannyRuberman Sorry, I explained now, that $N$ must be closed. $\endgroup$
    – user21230
    Commented Oct 17, 2018 at 14:02
  • $\begingroup$ “Does this plan make sense?” is not the kind of precise and specific question that MO is intended for. I have voted to close. $\endgroup$ Commented Oct 17, 2018 at 15:24
  • 1
    $\begingroup$ @MarekMitros The N I proposed is closed. $\endgroup$ Commented Oct 18, 2018 at 0:08
  • 1
    $\begingroup$ @MarekMitros Yes, I did mean $N = \partial(N_1 \times I) = N_1 \times \{0\} \cup \partial N_1 \times I \cup N_1 \times \{1\}$ which is the double of $N_1$ along $M$. So $M = \partial N_1 \times \{1/2\}$ does in fact separate $N$. $\endgroup$ Commented Oct 18, 2018 at 12:59

1 Answer 1

7
$\begingroup$

Since you've asked for an opinion, my answer is also an opinion. I would say that the embedding in $R^5$ is not going to be helpful. You are seeking to take $M = \partial N_i$ and `improve' the $N_i$ to make their union into a 4-sphere. This is a very reasonable approach; maybe you can do surgery on the $N_i$ to give them reasonable homotopy or homology properties to get their union to look like a sphere. However, what you have proposed is to do all of that in an ambient fashion (ie lying in 5-space). It seems to me that insisting that all of those surgeries be ambient is not going to be helpful, and may even make your life more difficult.

By the way, the approach of `improving' bounding manifolds in some fashion to get an embedding in a sphere is classical in higher dimensions. To see how this works one dimension up (embedding 4-manifolds in 5-space) you can look at some early papers of Tim Cochran (Embedding 4-manifolds in S5. Topology 23 (1984), no. 3, 257–269, 4-manifolds which embed in R6 but not in R5, and Seifert manifolds for fibered knots. Invent. Math. 77 (1984), no. 1, 173–184.) One thing that you will observe is the importance of the bounding manifolds being spin, which is relevant for embedding 3-manifolds too.

$\endgroup$
2
  • $\begingroup$ You can call it opinion. For me you have given hints and further reading. It is much, much better then closing my question, as it sometimes happen here...thanks a lot ! $\endgroup$
    – user21230
    Commented Oct 17, 2018 at 14:13
  • $\begingroup$ On the other hand if my surgeries were not ambient then I do not know whether result is embeddable in $R^5$...OK, I can do surgeries outside $M$, then I do not care.... It is nice to hear that someone already thought about embedding of 4-manifolds in $R^5$ and $R^6$. I wonder how many of them are 1-connected. Of course I have many other questions. For example when $M$ is embeddable in $S^4$ what are possible different embeddings in terms of homology of $N_1$ and $N_2$ (components onto which $M$ cuts the sphere) ? Are they as many as different decompositions of $H_1M$ onto direct summands ? $\endgroup$
    – user21230
    Commented Oct 17, 2018 at 14:31

You must log in to answer this question.