My question has survived. Therefore I try another one. Consider some elementary operations on closed compact 3-manifold $M \subset R^4$. These elementary operations are e.g. $0$-surgery or $1$-surgery such that corresponding disk of dimension $1$ or $2$ is nicely embedded in $R^4$ (not intersecting manifold except the boundary). This way we can for given manifold $M$ already embedded in $R^4$ construct new one $M'$ which will also be embedded in $R^4$ by its construction. Let's assume that we are working in smooth category.

I am still digesting the comment of Ryan Budney on my other question. This comment was. "Compute the torsion subgroup of $H_1$, and check that for each prime power $p^k$ the subgroup $\mathbb Z_p^k$ occurs an even number of times in the prime-power direct-sum factorization of the torsion subgroup."

I appreciate if someone can comment on it in terms of fundamental group. What can we say about finite order elements in fundamental group of 3-manifold embedded in $R^4$ ? I have understood that out of spherical manifolds only few embed in $R^4$.

Going back to my original question. Can such "step-by-step" method produce all closed 3-sub-manifolds of $R^4$ ? If "surgery" step is not enough, what other step could be considered in order to achieve this result ?

The same question can be asked for compact manifold with boundary. In such case we should also admit non-orientable ones.

My motivation for this question is to catalog manifolds by it's topological complexity in popular language called "number of holes". The "holes" can be one or two-dimensional. Connected problem is how to imagine 3-manifold. Embedding in $R^4$ seems to be more accessible then in $R^5$. Definition of manifold by surgery on certain knot or link is good on one hand, because it is fairly easy to draw a know. On the other hand it is not easy (for me ) to see what impact surgery has on fundamental group of the manifold.

The other aspect is analogy. We have already surfaces embedded in $R^3$ (orientable) and in $R^4$ (non-orientable). It should somehow be possible to find analogy between (known) surfaces and (less known) 3-manifolds.

  • $\begingroup$ You can construct all 3-manifolds in $\mathbb R^4$ via such surgery constructions. Roughly the proof is to put a standard linear height function on $\mathbb R^4$ which is a Morse function on the 3-manifold (any 3-manifold in $\mathbb R^4$). This gives you the corresponding surgery description of the 3-manifold. This is theoretically nice but if you want to reverse the process it is difficult, i.e. given a 3-manifold determine if it has a surgery description corresponding to an embedding. There are a few simple cases of this construction that appear in my preprint, ref in the prev. thread. $\endgroup$ – Ryan Budney Jun 6 '18 at 16:04
  • $\begingroup$ Thank you for this response. I will take a look into "11 tetrahedron census" again. I should probably install SnapPea and Regina software. On wikipedia I read that 3-manifolds can be classified algorithmically (Matveev 2003) ! $\endgroup$ – Marek Mitros Jun 7 '18 at 7:05
  • $\begingroup$ @RyanBudney If we consider 4-dimensional manifold $N$ with boundary $M$ in $R^4$. Do we know what are possible retracts of $N$ ? Is it always 2-complex ? I saw this mathoverflow.net/questions/32787/… question which you asked 8 years ago. $\endgroup$ – Marek Mitros Jun 7 '18 at 9:59
  • $\begingroup$ One argument which come to my mind is consider the Morse function $f$ from your comment. Cut in function regular point $f^{-1}(p)$ of $N$ is 3-manifold with boundary in $R^3$ which can be retracted to 1-complex. $\endgroup$ – Marek Mitros Jun 7 '18 at 12:31
  • $\begingroup$ Retracts of compact 4-dimensional submanifolds of $\mathbb R^4$ are generally 3-complexes. $\endgroup$ – Ryan Budney Jun 7 '18 at 17:43

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