# Which closed 3-manifolds can be embedded in $R^4$?

I wonder which closed orientable 3-manifolds can be embedded in $\mathbb R^4$ and which in $\mathbb R^5$. Is there a way to determine whether given closed 3-manifold, obtained, say by Dehn surgery on knot, can be embedded into $R^4$ ?

Is the answer known for spherical 3-manifolds (finite fundamental group) ?

I am mainly interested in topological properties of manifolds. The known answers for PL or smooth embedding are of value as well for me. I appreciate any kind of answer. I am not interested in such peculiarities as exotic $R^4$.

• They all embed in $R^5$ (C.T.C. Wall, BAMS 1956), but the question of embedding in $R^4$ seems a hard one. See openproblemgarden.org/op/… – BS. Feb 29 '16 at 14:12
• related question: mathoverflow.net/q/219313/1345 see also arxiv.org/abs/0810.2346 – Ian Agol Feb 29 '16 at 14:48
• Do you mean smooth or PL embeddings, or just topological embeddings? – Daniele Zuddas Feb 29 '16 at 15:11
• This latter question is maybe a question to be asked on math.stackexchange, rather than in a comment here. – Marco Golla Feb 29 '16 at 17:26
• Note that every closed orientable 3-manifold immerses into $\mathbb{R}^4$. This follows from Hirsch-Smale. So there are no "easy" obstructions (e.g. characteristic classes) to embedding. – Qiaochu Yuan Feb 29 '16 at 18:40

There is an analogy to surfaces in a sense. For 3-manifolds that fibre over surfaces there is a complete answer. For a variety of Seifert-fibred manifolds there are complete answers -- but not all. For example, Seifert-fibred homology spheres are still problematic. The preprint that Ian linked to in his comments has much more results of this kind in it.

At present, in summary:

1) We likely do not have a complete set of invariants that obstruct embedding into $\mathbb R^4$.

2) We appear to be far from knowing all the "natural" constructions of embeddings of 3-manifolds into $\mathbb R^4$ for the manifolds that are known to embed.

It is quite possible there are elements of formal logic obstructing both 1 and 2.

For example, if a compact boundaryless connected 3-manifold embeds in $S^4$ it separates it into two components. It is possible that one or even both of these components has a fundamental group with an unsolvable word problem. This would restrict the kinds of techniques one could use for creating obstructions in (1).

edit: I see Agol and Freedman's paper on this topic as connected to this last concern. 2-manifolds in $S^3$ have the Fox re-embedding theorem. So you could hope for some nice re-embedding theorems for $3$-manifolds in $S^4$. You shouldn't expect too nice a re-embedding theorem in $S^4$, since the tool that makes Fox's theorem work is Dehn's lemma, and the analogies to Dehn's lemma in 4-manifold theory are generally not true.

• Thank you very much for the answers ! I am reading your paper now suggested by Ian Agol. At my present knowledge it is a mistery why Poincare homology sphere has tame topological embedding in $S^4$ but it does not have smooth embedding in $S^4$. – Marek Mitros Mar 3 '16 at 14:06
• @MarekMitros: you are not alone. In principle there is a semi-constructive embedding but it is not easy to visualize. This is a result of Mike Freedman's. The embedding construction involves a type of infinite handle adjunctions. So the fact that you have a tubular neighbourhood is not obvious at all. – Ryan Budney Mar 4 '16 at 22:37

Results on the subject seem spotty, but the last of them seems to be:

MR3271270 Reviewed
Donald, Andrew(4-GLAS-SMS)
Embedding Seifert manifolds in S4. (English summary)
Trans. Amer. Math. Soc. 367 (2015), no. 1, 559–595.

• @ScottMorrison I tried the citation thing, it generates the citation, but seems to fail to insert it - am I supposed to cut and paste? – Igor Rivin Feb 29 '16 at 14:07
• Scott Morrison can't be notified this way since he did not participate in this thread. However, I'll alert him to your comment. – Todd Trimble Mar 1 '16 at 11:34
• After you search, click on your chosen result, then click the "Insert citation" button down in the bottom right of the screen. – Scott Morrison Mar 1 '16 at 11:50
• @ScottMorrison I could not find such a button, probably a browser issue... – Igor Rivin Mar 1 '16 at 14:27
• Let's sort this out via email. – Scott Morrison Mar 1 '16 at 22:44