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Earlier this year it was asked on MO, "Are there only countably many compact topological manifolds?" Thanks to Cheeger and Kister, the answer is yes. On the other hand, Manolescu recently debunked the triangulation conjecture. A natural follow-up question asks if there is some other way to enumerate topological n-manifolds, in the sense of creating a Turing machine that will eventually output an example from every homeomorphism class of topological manifolds, given enough time.

Of course, for $n \leq 3$, TOP = PL, so I'm really interested in the cases $n\geq 4$. It's entirely possible that the answer still depends on $n$, so you can interpret the question with either $n$ fixed or variable.

If the answer is no, is it known how hard the problem of enumerating manifolds is? Is it harder than the halting problem?

Edit in response to comments below: I do not mean to jump the gun. To even have a hope that the answer to the question is yes, one would have to have some finitely computable description of topological manifolds. As BjørnKjos-Hanssen indicates in comments, this might take the form of some sequence of approximations. If a direct answer to my question seems out of reach, I would be happy with an answer explaining what is and isn't known. (I also removed the madness about reference to Turing degrees above.)

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Eric S.
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    Did you look at the proof of Cheeger and Kister? I thought that it amounts to an enumeration with possible repetitions. – Andreas Thom Oct 23 '15 at 09:04
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    This is not immediately clear to me, especially considering their proof is by contradiction. – Eric S. Oct 23 '15 at 09:14
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    Turing machines output numbers or strings, not manifolds, so you first need to label your manifolds somehow to make this precise. The answer will of course depend on how you do this (say you give the compact topological manifolds the labels $2,4,6,8,\ldots$; then I could sell you a Turing machine that lists them). – Christian Remling Oct 23 '15 at 14:53
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    @Christian, the whole point is I don't know how to create an algorithm that eventually labels all $n$-manifolds. If I did, I would already have the answer to this question. But yes, the output of this Turing machine is supposed to be a string which somehow encodes a manifold. For example, if I asked about PL $4$-manifolds, the Turing machine could just output some binary description of all admissible triangulations of 4-manifolds. – Eric S. Oct 23 '15 at 18:18
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    @Eric S. : I believe that Christian Remling's point is that before you even think about computability, a more fundamental question is whether there is even a way to describe a compact topological manifold using a finite number of bits. And the question of what constitutes a reasonable description is a question that you have to answer before anyone else can answer the computability question, or else there is nothing to stop someone from saying that the number 1 is a description of some compact manifold, the number 2 is a description of some other compact manifold, etc. – Timothy Chow Oct 23 '15 at 22:02
  • Yes, I think that (Timothy's comment) sums it up. Or, in the context of the halting problem you refer to: Is there a TM that decides if TM #n, run on input $0$, say, stops? That of course depends on how you number TM's. It makes no sense whatsoever to ask the question without first clarifying how one wants TM's numbered (there are or course agreed on admissible numberings in this case that are standard, but I wouldn't know of a standard numbering of manifolds). – Christian Remling Oct 23 '15 at 22:35
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    @ChristianRemling I think the question is okay if you interpret it more loosely. Even if it can't be triangulated, we can still ask for a sequence of better and better approximations (as measured by $\epsilon=1/k>0$) of an embedding of our manifold in $\mathbb R^n$ for a suitable $n$. And we can make a double sequence of approximations $A_{n,\epsilon}$ of the embedding $A_n$ of the $n$th manifold. – Bjørn Kjos-Hanssen Oct 23 '15 at 22:41
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    @BjørnKjos-Hanssen: The OP asks about a Turing degree, among other things, so I really think we must insist on a set of integers that we are asked to investigate. – Christian Remling Oct 23 '15 at 22:48
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    @ChristianRemling Yes, and I mean, say, the set of integers $2^n3^k5^f$ where $n$ and $k$ are as in my other comment, and $f$ is a number encoding the finite approximation (in the sense of a bitmap with low resolution, I guess) to the embedded manifold. – Bjørn Kjos-Hanssen Oct 23 '15 at 22:57
  • @ChristianRemling, to avoid confusion, I have removed the question about Turing degrees, and clarified that I am in part asking for an answer to the issues you bring up. – Eric S. Oct 24 '15 at 00:55
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    I suppose an in-between possibility exists: You could possibly be able to enumerate all topological manifolds (in a matter similar to @BjørnKjos-Hanssen's suggestion, or as the index of a description of a computable topological space), but your enumeration contains homeomorphic duplicates of each manifold type, and moreover, it is impossible to find an enumeration containing only one representative of each homeomorphism class. – Jason Rute Oct 24 '15 at 02:32
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    I guess another question is whether every topological manifold is even homeomorphic to (1) a computable subset of $\mathbb{R}^n$ (for some $n$), (2) a computable metric space, or (3) a computable topological space. – Jason Rute Oct 24 '15 at 02:40
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    Via Wikipedia (https://en.wikipedia.org/wiki/Topological_manifold): "The full classification of $n$-manifolds for $n$ greater than three is known to be impossible; it is at least as hard as the word problem in group theory, which is known to be algorithmically undecidable. In fact, there is no algorithm for deciding whether a given manifold is simply connected. There is, however, a classification of simply connected manifolds of dimension $\geq 5$." I don't know the citation, what they mean by classification, or if the result applies to the class of compact manifolds. – Jason Rute Oct 24 '15 at 02:48
  • Thanks, @BjørnKjos-Hanssen! @Jason Rute, I interpret the classification issue as a harder question, since it is known to be unsolvable, even for PL manifolds. One might also ask about the Turing degree in this case, but it is probably halting complete. The questions you ask about computability above are along the lines of what I am asking. – Eric S. Oct 24 '15 at 04:35
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    The classification problem is much harder than the problem in the question, which is just to enumerate all manifolds. For example, the question of whether a finitely presented group is trivial is undecidable, but the problem "enumerate all finitely presented groups" has an essentially trivial answer (list all finite sets of words in all finite alphabets of the form $(x_1,\cdots,x_n)$).

    I think the answer to the question should be "yes," which should be some combination of the statements "a compact M is determined by finitely many gluing maps," (cont...)

     – Peter Samuelson Oct 24 '15 at 19:49
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    "the space of gluing maps between two balls in $\mathbb R^n$ has a dense countable subset," and "if the finitely many gluing maps needed to describe $M$ are approximated well enough, the two resulting manifolds are homeomorphic." But I don't know enough topology to make this precise off the top of my head. (edit: we'd also need each point in the countable dense set in the previous comment to be described by a finite amount of data.) – Peter Samuelson Oct 24 '15 at 19:50
  • @PeterSamuelson, I don't see any immediate reason something like that couldn't be made to work, but then again I don't know enough to be sure that it would. One might try to formulate this approach in terms of some notion of "computable pseudogroups." – Eric S. Oct 24 '15 at 23:06
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    @EricS. I'm not sure either - it could be the case that this strategy works for PL manifolds but doesn't work for topological ones for a subtle (at least for me) reason. – Peter Samuelson Oct 24 '15 at 23:15
  • I am probably burning my hands here, as this is very far from what I usually think about. Are (smooth?) manifolds not more or less the same thing as connected components of zero sets of rational polynomials, which can be enumerated: http://mathoverflow.net/questions/71415/manifolds-and-polynomials – Thomas Rot Nov 23 '15 at 16:36
  • That's an interesting link, Thomas, but it only seems to apply to smooth manifolds, not general topological manifolds. – Eric S. Nov 23 '15 at 16:45
  • How does a Turing machine output a topological manifold if it's not triangulable? – YCor May 31 '23 at 16:48

2 Answers2

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In a note of Freedman and Zuddas, they show that this is true for dimensions $\geq 4$.

In the "Background" section of the paper, they describe the solution in the higher dimensional case using surgery theory, but without any references.

Then they proceed to describe the 4-dimensional case. Here they use the fact that the complement of a point in a 4-manifold is smoothable. Hence one can describe a triangulation of a finite part of the complement of a point, together with a certificate of a 3-sphere tamely embedded.

It's frustrating that they don't give any references for the higher-dimensional case.

Ian Agol
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  • Yes, assuming the higher dimensional case works as they describe, this does seem to answer my question. Also, I haven't been on MO much lately, and hadn't updated my profile since moving to Illinois. It would be a pretty long walk to Station Q now. – Eric S. Jul 16 '20 at 17:04
  • Maybe someone else on MO knows some references. – Eric S. Jul 16 '20 at 17:06
  • Ah, I forgot you had moved. In any case, I also had forgotten that Mike had asked this question on MO a couple of years ago (someone should have pointed out the relation to your question at the time).

    https://mathoverflow.net/q/292525/1345

    It appears that the paper resulted from that posting (Zuddas gave an answer).

    So someone on MO knows the answer, but Mike uses the site intermittently (and usually creates a new profile each time, which the moderators then have to merge). https://mathoverflow.net/users/58457/michael-freedman https://mathoverflow.net/users/122296/michael-freedman

     – Ian Agol Jul 16 '20 at 17:16
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At the risk of getting on everyone's nerves (and special apologies to the OP), I would still maintain that the question in this form is too vague.

If I understand the suggestions in the comments correctly, they amount to something like this: interpret the output of a Turing machine as a sequence of points (possibly empty or finite) $x_0,x_1,x_2,\ldots\in\mathbb Q^{2n+1}$. Now we can ask: for what $e$ will TM number $e$ output a sequence of points whose closure in $\mathbb R^{2n+1}$ is an (embedded) manifold of the desired type? (This is certainly not exactly what Bjorn suggested, but it feels close enough.)

However, if we formalize like this, then part of the problem immediately disappears because any such set is non-recursive by Rice's theorem and this doesn't feel very satisfying because it had nothing to do with manifolds. (Admittedly, we could still ask about other properties of my set of $e$'s.)

Christian Remling
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    It seems to me that the whole point of the question is that the OP doesn't presume to know what sort of formalisation might work. Sure, that makes the question vague, but it's still a reasonable mathematical question. Often in mathematics, we know that some definition doesn't go the job we want, and wonder if there's another one which does. – HJRW Oct 24 '15 at 15:57
  • If you think about the classification of 2-manifolds (spheres and Klein bottles with varying numbers of handles), it is definitely computable. – Bjørn Kjos-Hanssen Oct 24 '15 at 16:49
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    HJRW sums the situation up well. If I were to propose a specific model in my question, I run the risk of getting answers about the infeasibility of that model. What I really want is either positive evidence, or a broader discussion of why no feasible model or definition is likely to exist. – Eric S. Oct 24 '15 at 18:10