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"it is an open question if the known compact manifolds in 3-D are complete."

This is a quote from Eric Weisstein's CRC Concise Encyclopedia of Mathematics, Second Edition. 2010, p.480. (Google Books link)

Is this still the case, post-Perelman? What are the known compact manifolds in $\mathbb{R}^3$?

I ask this as someone (obviously!) naive in these areas. Thanks for educating me!

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    $\begingroup$ just a feeling, Thurston's geometrization conjecture is true; so we can describe at least one "geometry" for each compact manifold. That may or may not count as "knowing" en.wikipedia.org/wiki/Geometrization_conjecture $\endgroup$ – Will Jagy Nov 3 '13 at 1:34
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    $\begingroup$ That's about it. We feel pretty happy about saying we know the compact orientable surfaces, because a single number says which one you have, despite there being other equivalence relations (conformal structure) that are not decided by that. Jeff Cheeger once made a relevant comment in an undergraduate class I was taking, "How equal do two things have to be before they are the same thing?" $\endgroup$ – Will Jagy Nov 3 '13 at 1:58
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    $\begingroup$ I'm finding this question rather ambiguous; do you mean 3-dimensional compact manifolds or manifolds embedded in $\mathbb{R}^3$? (The latter are easy: they're precisely the orientable 2- and 1-manifolds) Is completeness being used in the normal English or the technical sense (from my reading either understanding is sensible but false). $\endgroup$ – Ben Webster Nov 3 '13 at 2:04
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    $\begingroup$ I did read the link, it was still confusing. In part, the issue is that I think that sentence is simply wrong, so I think I was casting around for an alternate interpretation that made more sense to me. $\endgroup$ – Ben Webster Nov 3 '13 at 3:08
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    $\begingroup$ The quote in the Weisstein book makes no sense to me. It's not clear to me what the notation $N(D)$ means, or what he means by "complete". Is $N(D)$ meant to be the number of prime orientable manifolds, up to some equivalence? $\endgroup$ – Ryan Budney Nov 3 '13 at 11:37
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Actually, it's very easy to write down a complete list of orientable 3-manifolds (and thus all, since every nonorienatble 3-manifold has its orientation cover), using a tool called Heegaard splitting. This list has the following structure: break it up into $\mathbb{N}$ many sublists corresponding to different genera of 2-dimensional manifolds. Having fixed $g$, look at all mapping classes from a surface with that genus to itself (these form a finitely generated group), and then consider the 3-manifold obtained by gluing two handlebodies (filled genus $g$ surfaces) using that map to identify the boundaries.

The proof that this is a complete list of 3-manifolds is quite easy; it can be explained in a few minutes to a smart undergrad. The hard part isn't listing all 3-manifolds; it's that the list I gave above (and various other lists one can produce with similar techniques, such as surgery) is massively redundant. We even know the rules (analogues of Reidemeister moves) that tell you all the redundancies, but just as Reidemeister moves don't help that much with classifying knots, this doesn't mean you can take two Heegaard splittings and efficiently test if they give you the same 3-manifold.

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ADDED: What I find unsatisfactory about my answer below is that I do not say who did what. I hope someone would post an answer that does this. I also have to say that although this classification of Thurston is (inevitably) more elaborate than for 2-manifolds, I find it quite beautiful.

Let me venture an answer, even though there are way, way more qualified people than me who will answer it correctly. I just want to see how close my understanding is to the correct answer. I will speak only about orientable 3-manifolds. I have no idea what happens for non-orientable ones.

First, if there are any 2-spheres in a compact 3-manifold that do not bound balls, then they can be made disjoint. By cutting along these spheres and plugging each hole with a ball, one gets a finite set of compact manifolds that are called "prime".

Next, one looks for incompressible torii (which I believe means that its fundamental group injects into the fundamental group of the manifold). These can also be made disjoint and the manifold chopped along these torii. At this point you don't try to fill in the toroidal hole.

Now everything reduces to a manifold without any incompressible spheres or torii, except it might have toroidal ends. There are three possible geometric structures for such manifolds:

  • Spherical geometry
  • Hyperbolic geometry
  • Seifert fibrations (flat, solvable, nilpotent geometries)

What's notable about Seifert fibrations, as well as the 3-sphere, is that they are fibered by circles and admit a family of "collapsing" Riemann structures in the sense of Cheeger-Gromov-Fukaya. In other words, a family of Riemannian metrics where the sectional curvature remains bounded but the injectivity radius goes to $0$ everywhere. Thurston's conjecture implies that the only non-collapsible geometry is the hyperbolic geometry.

Before Perelman, proving the Thurston conjecture had been reduced to the following: Show that an orientable prime atoroidal 3-manifold, either compact or with toroidal ends that is not Seifert fibered or the 3-sphere has a hyerbolic structure.

Thurston had already proved that if these assumptions hold and the manifold is Haken (it has an incompressible surface of higher genus), then it is hyperbolic. Curt McMullen gave a beautiful proof of this via a proof of the theta conjecture.

ADDED: Note that this does not really answer the question. I don't believe that there is a complete classification of hyperbolic 3-manifolds (as there is for 2-manifolds). If that's right, then we don't really know what all 3-manifolds are.

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  • $\begingroup$ Thanks, Deane, your added note may highlight the core issue here. $\endgroup$ – Joseph O'Rourke Nov 3 '13 at 2:15
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    $\begingroup$ Technically there is a complete classification of 3-manifolds post-Perelman. In the Hyperbolic case, you know that the manifolds are determined by their fundamental groups (by Mostow Rigidity), and the isomorphism problem for torsion-free hyperbolic groups is solvable thanks to a theorem of Sela. $\endgroup$ – Autumn Kent Nov 3 '13 at 2:26
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    $\begingroup$ A classification is a list without redunancies. I think in this sense there is a classification of oriented compact 3-manifolds, which uses that (a) prime and JSJ decomposition is canonical. (b) lists of geeometric 3-manifolds (where for the hyperbolic case with toral boundary one has to quote the isomorphism problem for total relatively hyperbolic groups of Damani-Groves. See ldtopology.wordpress.com/2010/01/26/… for some detail. $\endgroup$ – Igor Belegradek Nov 3 '13 at 12:08
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    $\begingroup$ Igor, in the case of manifolds with boundary, one can use the algorithm for recognition of Haken manifolds instead. $\endgroup$ – Misha Nov 3 '13 at 16:05
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I like to think that compact 3-manifolds will not be known until we know "the list" of compact, oriented, hyperbolic 3-manifolds, in the way that we know "the list" of compact, oriented, hyperbolic surfaces: the genus 2 surface, the genus 3 surface, the genus 4 surface, ...

We don't know "the list" yet, as the other answers indicate. But the potential for ordering hyperbolic 3-manifolds by volume was established by Jorgensen who proved that the set of volumes of finite volume hyperbolic 3-manifolds is a well-ordered set with order type $\omega^\omega$ (the same order type as polynomials with natural number coefficients), and each volume occurs for only finitely many manifolds. And with that in mind, we know the first entry: the Weeks manifold, which is the unique lowest volume closed hyperbolic 3-manifold, as proved by Gabai, Meyerhoff, and Milley. I'm not up on the very latest developments of this technology, but I think that the next few lowest volumes have also been completely listed, using the MOM technology of the same three authors.

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    $\begingroup$ Even so, I guess that we will never understand the set of 3-manifolds as well as we understand the set of 2-manifolds. It's a bit like prime numbers: we know how to list them, we know how to recognize them, but we still don't know whether the twin prime or strong Goldbach conjectures are true. So knowing seems to me even wider than expressed in the various answers. $\endgroup$ – Benoît Kloeckner Nov 3 '13 at 10:02
  • $\begingroup$ I think the existence of a list is sketched at ldtopology.wordpress.com/2010/01/26/… $\endgroup$ – Igor Belegradek Nov 3 '13 at 12:11
  • $\begingroup$ @Igor: that link does not emphasize the point I am trying to make here, that specific theorems of hyperbolic 3-manifolds make a natural well-ordered list possible. $\endgroup$ – Lee Mosher Nov 3 '13 at 14:14
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    $\begingroup$ @Igor: The list of surfaces that I gave is described by the identify function on the natural numbers $\ge 2$: one such surface for each natural number. That is a highly concrete way of "knowing". The problem with listing 3-manifolds by their groups is that it depends on computationally difficult methods embedded in, say, Sela's proof, which pale in comparison to the identity function as a way of "knowing". The volume listing based on Jorgensen's theorem seems to me to have the potential to realize "knowing" much more intimately, based on Thurston's tools of hyperbolic 3-manifolds. $\endgroup$ – Lee Mosher Nov 3 '13 at 15:06
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    $\begingroup$ It hasn't stopped folks from plugging on. My understanding, after a little more digging, is that the 10 lowest volume one-cusped hyperbolic manifolds are now known. It might be interesting to know if all of the manifolds of volume less than the least limit volume can be explicitly listed with what we know nowadays. $\endgroup$ – Lee Mosher Nov 4 '13 at 0:48

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