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I need some philosophical explanation for JSJ decomposition theorem. It says that closed orientable irreducible 3-manifold can be cut along set of incompressible tori onto pieces which are:

  • atoroidal or Seifert-fibered
  • hyperbolic or Seifert-fibered
  • hyperbolic or spherical or Seifert-fibered with infinite fundamental group.

The three bullets are just different wording of the the same theorem.

My question is what is the method for producing all irreducible closed 3-manifolds ? We start with the ones having no incompressible torus in JSJ theorem. Let's call them family zero manifolds which are closed hyperbolic, spherical or Seifert-fibered with infinite $\pi_1$. In Friedl, Introduction to 3 manifolds, I read that Seifert-fibered manifolds are finitely covered by $S^1$-bundle over surface. The hyperbolic closed manifolds are kind of mystery for me yet. In the same place I read that closed hyperbolic manifold is finitely covered by surface bundle over circle (famous virtually fibered conjecture). Still it does not help me to understand hyperbolic manifolds but this would be the topic for another question.

Next we go family one i.e. irreducible closed manifold which has one uncompressible torus. When we cut along this torus we obtain either one or two pieces. How can we describe these pieces ? How can we prepare hyperbolic or Seifert-fibered manifold having one or two tori as boundary ? The good candidate seems to be knot complement in sphere $S^3$. How about circle complement in family zero member ? Say, we have such pieces in hand, what are possible ways of gluing the boundary tori ? The automorphism group of torus up to homotopy is $SL_2(\mathbb Z)$ - it is named mapping class group - MCG in Friedl introduction. Thus we have a lot of freedom in gluing these pieces. How can we control this ? Our goal is to enlist all manifolds in family one.

Next we continue with family two i.e. irreducible closed manifolds having two incompressible tori in JSJ decomposition. In this family we again define all possible pieces and all possible gluings. And so on...

Can someone explain me whether this procedure works and we can name all irreducible closed 3-manifolds by such method ?

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  • $\begingroup$ Are you only considering orientable 3-manifolds? I would encourage this as a starting point. $\endgroup$
    – Neil Hoffman
    Commented Sep 19, 2018 at 16:53
  • $\begingroup$ Ok, I added word "orientable" to my question. These are also described in Friedl 's introduction. $\endgroup$
    – user21230
    Commented Sep 19, 2018 at 18:28
  • $\begingroup$ Have you read Bonahon's manuscript? Geometric structures on 3-manifolds, in: Handbook of geometric topology, 93-164, North-Holland, Amsterdam, 2002. I think it explains geometrization very well. $\endgroup$
    – Ryan Budney
    Commented Sep 19, 2018 at 18:39
  • $\begingroup$ @RyanBudney I printed it and I will read it. What is the answer to my question ? If you were to list all prime orientable closed 3-manifolds - how would you do it ? Say, that I would like to construct all such 3-manifolds. What should I do ? $\endgroup$
    – user21230
    Commented Sep 20, 2018 at 8:35
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    $\begingroup$ You have many questions. Regarding listing 3-manifolds, I doubt there's a great way to do it. For hyperbolic 3-manifolds there's fairly reasonable ways to go about it, mostly based on volume. I suppose if I was taking the task seriously I'd do what we do in Regina: list the isomorphism signatures of all the minimal triangulations. In principle that could be an enormous amount of work, though. $\endgroup$
    – Ryan Budney
    Commented Sep 20, 2018 at 14:41

1 Answer 1

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Nobody knows how to classify the family zero manifolds. The Seifert ones are pretty well-known and were classified by Seifert himself: each such may be represented as a string like $$(S, (p_1,q_1), \ldots, (p_k,q_k)$$ where $S$ is a surface and $(p_i,q_i)$ are coprime numbers, and there is a theorem that says very clearly when two different strings represent the same manifold. However, on the hyperbolic side there is yet no classification at all. In principle there are algorithms that "produce all the hyperbolic 3-manifolds" without repetition, but they are not useful in practice. There are of course various important theorems around, the first one to cite should probably be Thurston's hyperbolic Dehn filling. There is also a very concrete fantastic tool, SnapPy, to investigate these manifolds. But we do not have a global satisfying classification.

On the other hand, the higher level manifolds are not a problem. Once you know all the level zero ones, every irreducible manifold is almost uniquely described as a graph with a level zero manifold at each vertex and a $2\times 2$ integer matrix at each edge (to be precise, every level zero manifold should be equipped with a homology basis at each torus boundary). The "almost" here stands for the fact that one should take into account the self-diffeomorphisms of the level zero manifolds: in the Seifert cases these are pretty understood and may have infinite order, in the hyperbolic case they are finite and form precisely the isometry group of the manifold.

Edit : By "level zero" I mean every compact manifold, possibly with boundary made of tori, that is either Seifert or hyperbolic or a torus (semi-)fibration. The third class contains only closed manifolds.

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    $\begingroup$ However I do not understand one point. You say about graph "with a level zero manifold at each vertex". But level zero manifold is closed and piece in JSJ decomposition has torus boundary. How do we obtain manifold with torus boundary from closed one ? Do we remove neighborhood of some trivial circle or knot ? If trivial circle then it is just one result. If knot then again we have infinite number of possibilities. $\endgroup$
    – user21230
    Commented Sep 20, 2018 at 13:07
  • $\begingroup$ Well, you can regard the Virtual Fibring theorem as a topological classification of finite-volume hyperbolic 3-manifolds. $\endgroup$
    – HJRW
    Commented Sep 21, 2018 at 4:10
  • $\begingroup$ @HJRW Is it known for given - say - torus bundle over circle - how many closed hyperbolic manifolds it covers ? $\endgroup$
    – user21230
    Commented Sep 21, 2018 at 7:14
  • $\begingroup$ It depends what you mean by 'known'. The problem is decidable in theory, but I doubt there's a simple formula. $\endgroup$
    – HJRW
    Commented Sep 21, 2018 at 13:14
  • $\begingroup$ I wonder whether random irreducible 3-manifold is atoroidal or not. I am still learning... I assume that by hyperbolic Dehn filling we obtain closed hyperbolic manifold. Since most of links in $S^3$ are hyperbolic and most of surgeries on given link satisfy hyperbolic Dehn filling criteria then I conclude that most of the closed manifolds are hyperbolic. At this point I do not know whether closed hyperbolic manifold must be irreducible. I found your book on the Internet, so I will try to read it. $\endgroup$
    – user21230
    Commented Sep 24, 2018 at 10:55

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