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My question is rather philosophical : without using advanced tools as Perlman-Thurston's geometrisation, how can we get convinced that the class of closed oriented $3$-manifolds is large and that simple invariants as Betti number are not even close to classify ?

For example i would start with :

  1. If $S_g$ is the closed oriented surface of genus $g$, the family $S_g \times S^1$ gives an infinite number of non pairwise homeomorphic $3$-manifolds.

  2. Mapping tori of fiber $S_g$ gives as much as non-diffeomorphic $3$-manifolds as conjugacy classes in the mapping class group of $S_g$ which can be shown to be large using the symplectic representation for instance.

I think that I would like also say that Heegaard splittings give rise to a lot of different $3$-manifolds which are essentially different, but I don't know any way to do this.

So if you know a nice construction which would help understanding the combinatorial complexity of three manifolds, please share it :)

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    $\begingroup$ The class is not that large, it's countable, and you can easily produce countable (non-exhaustive) families of non-homeomorphic 3-manifolds, even non-homotopy equivalent. $\endgroup$
    – Fernando Muro
    Commented Oct 6, 2015 at 13:52
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    $\begingroup$ @FernandoMuro: I would read “a lot” here as referring to richness and complexity, not cardinality — in the same way as there are “a lot more” finite groups than finite fields, say, even though of course in terms of cardinality there are precisely $\aleph_0$-many of each. $\endgroup$
    – Peter LeFanu Lumsdaine
    Commented Oct 6, 2015 at 14:36
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    $\begingroup$ @PeterLeFanuLumsdaine you're right, I didn't intend to minimize the relevance of the question, but you maybe agree that the question is a little bit diffuse, and that the background provided doesn't clearly imply that facts, like the one I mentioned, are known to Selim. To me, honestly, the class of manifolds is not so large but very rich, and this is what I wanted to reflect in my small contribution. $\endgroup$
    – Fernando Muro
    Commented Oct 6, 2015 at 14:44
  • $\begingroup$ Not sure if you are aware of this (2. kind of suggests the contrary) but mapping tori with different fibers can still be diffeomorphic. You can find restrictions on the genus for this though. $\endgroup$
    – Daniel Valenzuela
    Commented Oct 6, 2015 at 15:05
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    $\begingroup$ @SelimG: Even more is true: if a closed orientable $3$-manifold has first betti number at least $2$ and fibers over the circle in one way, then it fibers over the circle in infinitely many distinct ways. The keyword to search for is "Thurston norm". $\endgroup$
    – Andy Putman
    Commented Oct 6, 2015 at 21:03

6 Answers 6

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Here are two examples suggesting the complexity of the world of $3$-manifolds.

The first is the classical result that any $3$-manifold can be obtained by integral surgery on a link in $S^3$. If you believe that knots and links form a complex Universe, than this result should suggest that $3$-manifolds cannot be much simpler.

The next example comes from the striking work of Dunfield and Thurston on random $3$-manifolds. You can get such things by picking random elements in the mapping class group, where randomness is generated by a random walk on this group. This has lead to the discovery of strange $3$-manifolds. For more recent work on this topic see also this paper of Lubotzky, Maher and Wu.

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    $\begingroup$ Dear Liviu, I'd be interested to know a bit more about your first comment. How do you catch the complexity of 3-manifolds from the complexity of links ? Why (just for the sake of the argument) this operation wouldn't lead to only a finite number of manifolds ? In other word why essentially different links give rise to two different manifolds ? $\endgroup$
    – Selim G
    Commented Oct 6, 2015 at 14:08
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    $\begingroup$ The surgery construction is surjective but not injective. Kirby calculus explains when two surgery presentations yield the same manifold. However, the simplest example of $0$ surgery on a knot shows that you get infinitely many examples that are differentiated by the Alexander polynomial of the knot. $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 6, 2015 at 15:15
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    $\begingroup$ @LiviuNicolaescu Is there a simple reason the Alexander polynomial is an invariant of the 3 manifold given by 0 surgery on a knot? Coincidentally it was recently shown (by Yasui) that there exists knots with the same 0-surgery which aren't even concordant. $\endgroup$
    – PVAL
    Commented Oct 6, 2015 at 23:50
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    $\begingroup$ @PVAL The Reidemeister torsion of the $3$-manifold obtained by $0$-surgery is determined by the Alexander polynomial of the knpt. See the notes below for more details. www3.nd.edu/~lnicolae/Torsion.pdf $\endgroup$
    – Liviu Nicolaescu
    Commented Oct 7, 2015 at 10:10
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Maybe by looking first at homology $3$-spheres and in particular to Brieskorn manifolds $M(p,q,r)$: the link of the singular point $(0,0,0)$ of the hypersurface $$z^p_1+z_2^q+z^r_3=0$$ with integers $p,q,r\geq 2$. Milnor studied these manifolds in his beautiful paper "On the 3-dimensional Brieskorn manifolds M(p,q,r)". in: Knots, groups, and 3-manifolds (Papers dedicated to the memory of R. H. Fox), pp. 175–225. Ann. of Math. Studies, No. 84, Princeton Univ. Press, Princeton, N. J., 1975.

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Read Chapter 4 of Thurston's notes http://library.msri.org/books/gt3m/. He produces infinitely many closed hyperbolic 3-manifold of different volumes, and hence non-homeomorphic, just by doing Dehn surgery on the figure 8 knot in $S^3$. This does not used advanced geometrisation theorems, just some basic (but clever) hyperbolic geometry, combined with the Mostow rigidity theorem.

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I'm not sure if this is what you have in mind by "without using advanced tools" or "a nice construction which would help understanding the combinatorial complexity of three manifolds," but how about identifying pairs of faces of different polyhedra? I guess this is what convinced Poincaré...

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  • $\begingroup$ To me it is not obvious that I build this way a lot of different manifolds. I glue faces of a polyhedron and then get by Poincaré's theorem a presentation of the fundamental group of my manifold. But how can I quickly deduces that, for instance, this method builds infinitely many different manifolds ? Even if so, are these differentiated by their first Betti number ? Building this way only one homology sphere is (a little bit) painful and requires non-obvious computation on the fundamental group of the resulting manifold. $\endgroup$
    – Selim G
    Commented Oct 9, 2015 at 8:18
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This does depend (as others have said) on what you mean by "advanced tools", but what about looking at Seifert-Fibred manifolds? These are ones that are locally products $S^1 \times D^2 / C_n$ for some cyclic group $C_n$. Alternatively, you can view them as appropriate notions of a bundle over an orbifold surface $S$.

It's worth noting that, as I recall, a large variety of three-manifolds can be Seifert-fibred. But even ignoring that, you get the fact that you can construct them easily from an orbifold surface (quite a concrete object) together with some extra discrete data.

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  • $\begingroup$ That is for sure another interesting construction. But the question I am asking is how can I convince myself that these constructions lead to non homeomorphic manifolds ? An how much of the combinatorial complexity of 3-manifolds can I catch this way ? $\endgroup$
    – Selim G
    Commented Oct 9, 2015 at 8:12
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    $\begingroup$ I can't remember the exact details off the top of my head, but there are strong conditions (that are quite explicit, as I recall) for when two of these manifolds will be homeomorphic. Some information can be found here: en.wikipedia.org/wiki/Seifert_fiber_space $\endgroup$
    – Simon Rose
    Commented Oct 9, 2015 at 8:14
  • $\begingroup$ As for how much combinatorial complexity... Again, my recollection is a bit shaky, and maybe this uses a bit too much fancy machinery, but of the 8 three-dimensional geometries, six can be Seifert-fibred. So that's a good sign that many of the local geometries can be captured with this construction. $\endgroup$
    – Simon Rose
    Commented Oct 9, 2015 at 8:15
  • $\begingroup$ I get the fact that the geometrisation conjecture combined with the recent achievement on hyperbolic 3-manifolds gives a good picture of the exact complexity of the world 3-manifolds. But without this machinery, how good can be an approximation of this complexity ? $\endgroup$
    – Selim G
    Commented Oct 9, 2015 at 8:23
  • $\begingroup$ You don't need advanced tools to distinguish them. There's a nice elementary account of their classification in Peter Scott's article 'The geometry of 3-manifolds'. $\endgroup$
    – HJRW
    Commented Mar 2, 2016 at 12:07
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I would explain it like this:

Start with a framed link in $S^3$ and do Dehn-surgery along it to get a new $3$-manifold. In fact one can get every compact oriented $3$-manifold in this way as mentioned earlier.

To distinguish this $3$-manifolds one can use the fundamental group. Actually the fundamental group can distinguish almost all $3$-manifolds. It is easy to give a presentation of the fundamental group out of a surgery diagram:

First use the Wirtinger presentation of the link exterior and then add a new realtion (of the form $p\mu+q\lambda=1$) for every surgery.

Of course its difficult to distinguish two such presentations in general. But to convince someone that there are many different should not be hard. For example one can do the following:

  • make this groups abelian and use the classification of abelian groups.
  • compare the orders of the groups (this is how Poincaré proved the Poincaré sphere to be not $S^3$).
  • show that some of this groups are non-abelian (for example by finding surjective group homomorphisms in non-abelian groups) others are abelian.
  • compare orders of the elements.
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