Embedding problem for 3-manifolds attacked via 4-manifolds

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 ?

• 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" $M$? – Danny Ruberman Oct 17 '18 at 13:24
• @DannyRuberman Sorry, I explained now, that $N$ must be closed. – user21230 Oct 17 '18 at 14:02
• “Does this plan make sense?” is not the kind of precise and specific question that MO is intended for. I have voted to close. – Andy Putman Oct 17 '18 at 15:24
• @MarekMitros The N I proposed is closed. – Danny Ruberman Oct 18 '18 at 0:08
• @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 $N$. – Danny Ruberman Oct 18 '18 at 12:59

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.
• 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 ? – user21230 Oct 17 '18 at 14:31