Given a finitely presented group $G$, how many necessary conditions do people know for $G$ to be isomorphic to the fundamental group of some closed connected 3manifold? (e.g. residually finite)

1$\begingroup$ How do you tell whether a finitely presented group is residually finite? It is hard enough to tell if it is trivial... $\endgroup$– Igor RivinOct 22, 2012 at 2:49

2$\begingroup$ @Li: There are way too many restrictions: Coherence, virtual cohomological dimension $\le 3$, every infinite amenable subgroup is polyciclic; either virtually $PD(3)$ or (virtually) splits as a free product; either contains $Z^2$ or is hyperbolic; either finite or has virtually infinite abelianization, etc. You may want to narrow down the scope of the algebraic properties you are interested in. As Igor said, most of these properties are hard (or impossible) to verify algorithmically. $\endgroup$– MishaOct 22, 2012 at 3:00

1$\begingroup$ I know it is difficult to use them in practice. But I just want to know some such kind of conditions. $\endgroup$– Li YuOct 22, 2012 at 3:10

1$\begingroup$ The word problem is solved for 3manifold groups, so whatever conditions you'd have to check, they'd be noncomputable in practice. $\endgroup$– Ryan BudneyOct 22, 2012 at 3:13

1$\begingroup$ It seems to me that this question is asking for a list, and hence should be community wiki. $\endgroup$– HJRWOct 22, 2012 at 9:38
3 Answers
Apologies for the shameless selfpromotion, but as you ask for necessary conditions, you seem to want a list of theorems of the form 'If G is a 3manifold group then G has property P'.
Aschenbrenner, Friedl and I have given what I think is a pretty exhaustive list in our survey paper here.


6$\begingroup$ Why not this paper? arxiv.org/abs/1210.2101v1 $\endgroup$– Ian AgolOct 22, 2012 at 11:50

$\begingroup$ Well, I thought the question was asking for an exhaustive list, so the survey paper seemed like the 'right' answer. But if you actually want to recognise whether or not something's a 3manifold group then, yes, that paper is the one to look at! $\endgroup$– HJRWOct 22, 2012 at 13:21

$\begingroup$ Regarding the main theorem of GMW, that recognizing presentations of 3manifold groups is reducible to the word problem; doesn't this follow from general theorems about c.e. sets? The word problem is equivalent to the membership problem for a universal c.e. set. Deciding whether a presentation is of a 3manifold group is equivalent to the membership problem for a particular c.e. set, because one can list all 3manifolds (say, as simplicial complexes) and hence the presentations of their groups. Hence the latter problem is reducible to the word problem, or am I missing something? $\endgroup$ Oct 25, 2012 at 22:47

1$\begingroup$ @John Stillwell: See Remark 2.6 of our paper, in which we make (I believe) exactly the point that you do in your comment(s) above. We mean something different by `recursive modulo the word problem'. See Definition 2.4 (and proceeding definitions) for a precise statement of what we mean. $\endgroup$ Oct 26, 2012 at 2:30
since Henry started the shameless selfpromotion, let me also do so... Given any group $\pi$ one can study the corresponding Alexander polynomial $\Delta_\pi$ which lies in the group ring of $H:=H_1(\pi;\Bbb{Z})/\mbox{torsion}$. If $\pi$ is the fundamental group of a closed 3manifold, then the Alexander polynomial $\Delta_{\pi}$ is symmetric and the onevariable specializations have even degree. (see F, Kim, Kitayama: Poincaré duality and degrees of twisted Alexander polynomials)
The symmetry holds also if $\pi$ is a 3dimensional Poincare duality group, but I am not sure whether the degree condition holds in that case.
The advantage is that this condition can be checked easily, and by checking it for finite index subgroups one gets even more necessary conditions. I would guess that in practice this is a very effective way for weeding out non 3manifold groups.
At least it allowed me to make the right bet on Ryan's example...
As a demonstration of difficulty, a counterquestion: Is this a 3manifold fundamental group, and if so, which one?
$$\langle a, b  a^2b^{1}a^{2}ba^{1}ba^{2}b^{1}a^2b, a^{1}b^{1}a^2ba^{4}ba^2b^{1}a^{1}b^{1}a^2ba^{2}ba^2b^{1} \rangle$$

1$\begingroup$ The program Heegaard (written by Berge) says the presentation is realizable. So yes, it is a threemanifold group. Also, the homology is Z. If pressed I would take the diagram that his program produces, convert that to a triangulation, and feed the result to SnapPy. $\endgroup$– Sam NeadOct 26, 2012 at 4:54

1$\begingroup$ If I haven't made a mistake: I used a variety of tools, ending with snappy, to obtain a triangulation of this threemanifold M. I then used regina to find fundamental normal tori. Cutting proves that M is obtained by gluing X, the figure eight knot complement, to Y, the Seifert fibered space with base orbifold D(2,2) (aka the orientation Ibundle over the Klein bottle). Unfortunately, regina doesn't track gluing data when we cut along a normal torus. So we can't recover the gluing data that way. $\endgroup$– Sam NeadJan 10, 2016 at 10:15

$\begingroup$ Thinking about homology considerations tells us that the longitude of X is glued to the longitude of Y  that reduces the set of possible gluings to a "line" of Dehn twists. I think that your "challenge" could now be answered computationally if we had a facility to glue manifolds with torus boundary. Does that exist? $\endgroup$– Sam NeadJan 10, 2016 at 10:49