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I am curious if the following topological problem is decidable. Let $M,N$ be two closed manifolds. Given a group isomorphism $p: \pi_1(M)\to \pi_1(N)$, is there a homeomorphism $\phi: M\to N$ such that $\phi_*=p$?

EDIT. (Thanks for helpful comments.) There are some relevant results: "Algorithmic aspects of homeomorphism problems" by Nabutovsky and Weinberger. There is a preprint version https://arxiv.org/abs/math/9707232 (click ps, pdf is a mess!), and a published version http://www.ams.org/books/conm/231/. A sort of review of their methods was written by R.I.Soare, "Computability theory and differential geometry" (Bulletin of Symbolic Logic, 2004).

According to the paper, the problem is decidable for simply connected manifolds of dimension at least 5. They also construct a counterexample for a non-simply connected case (Proposition 0.1 in the published paper). However, I would like to point out that this counteraxample does not answer the question. Even if algorithm for solving a problem formulated above exists, it cannot be used unless you have an explicit isomorphism between fundamental groups. In the situation of Proposition 0.1, it leads to no contradiction.

Interestingly, the results of Nabutovsky and Weinberger may be taken as a hint that the answer to the question is positive. At least, this possibility is not excluded for all I know.

EDIT. Actually, the example in Proposition 0.1 does show that the problem is undecidable. (So, I got it wrong. Thanks to Achim Krause for clarifying the details.)

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    $\begingroup$ What do you mean by, "I am curious if the following topological problem is decidable?" Does the word "decidable" mean "does anybody know if the following is always true?" Or does "decidable" mean "is there a Turing machine ...?" $\endgroup$
    – Jason Starr
    Commented Aug 8, 2017 at 12:00
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    $\begingroup$ Anyway it's a bit vague: by closed manifold you probably mean finite triangulation, and you mean the algorithm to halt (with the correct answer) whenever the space it defines is a topological manifold? So the input would be a pair of triangulations and a pair of homomorphisms between fundamental groups, given by images of generators, which composes to identity on both sides? $\endgroup$
    – YCor
    Commented Aug 8, 2017 at 12:03
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    $\begingroup$ Yes, the input may be a pair of triangulations together with an explicit isomorphism between fundamental groups given by appropriate presentations. I do not think it makes any difference except in some pathological cases. (To avoid them, you may assume manifolds to be smooth, which is fine with me.) $\endgroup$
    – Alex Gavrilov
    Commented Aug 8, 2017 at 12:24
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    $\begingroup$ @YCor. I was just trying to get the OP to clarify the question. The answer that was posted (now deleted) seemed to follow the first meaning, whereas the OP has now clarified that the question uses the second meaning. $\endgroup$
    – Jason Starr
    Commented Aug 8, 2017 at 12:32
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    $\begingroup$ @JasonStarr OK indeed I now understand the first interpretation. Actually it meant "decidable" in the sense of set theory, while in the question it's about recursiveness. (Note that the tag "computability-theory" was a hint towards this interpretation) $\endgroup$
    – YCor
    Commented Aug 8, 2017 at 12:46

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This is undecidable. For a construction, see the first page of Nabutovsky and Weinberger, "Algorithmic aspects of homeomorphism problems", Rothenberg Festschrift 1998.

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  • $\begingroup$ There is also the preprint at arxiv.org/abs/math/9707232, but the relevant Proposition 0.1 from the published version doesn't seem to be in there. $\endgroup$
    – Tobias Fritz
    Commented Aug 9, 2017 at 2:27
  • $\begingroup$ Thank you, this is example is really interesting. But I do not think it is an answer to the question. The problem is not about any homeomorphism, it is about a homeomorphism with a given pushforward. Unless I got it wrong, in the construction described by Nabutovsky and Weinberger such a homeomorphism would simply produce an isomorphism between two $\endgroup$
    – Alex Gavrilov
    Commented Aug 9, 2017 at 7:30
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    $\begingroup$ The logic is as follows: Given $g\in G$, you can define $M''(g)$ which due to its construction comes with an isomorphism $\pi_1\simeq G$. If $g=e$, then you can algorithmically find a chain of relations exhibiting $g=e$, and this gives an isotopy from the sphere they surgered on to the trivial sphere, which gives you a homeomorphism $M''(g)\simeq M'$ compatible with the respective isomorphisms of $\pi_1$ to $G$. If $g\neq e$, then they can't be homeomorphic, since they have different $\pi_2$. $\endgroup$
    – Achim Krause
    Commented Aug 9, 2017 at 9:19
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    $\begingroup$ So the problem $g=e$ can be reduced to existence of a homeomorphism of $M''(g)\simeq M'$ which lifts the identity on $G$ under the identification $\pi_1(M''(g)) \simeq G$, $\pi_1(M')\simeq G$. Does this address both of your concerns? $\endgroup$
    – Achim Krause
    Commented Aug 9, 2017 at 9:20
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    $\begingroup$ To be more precise: A proof through a chain of relations that $g=e$ can be used to construct a homeomorphism $M″(g)\simeq M'$ that is the identity on $\pi_1$ under the identifications with $G$, and having such a homeomorphism proves $g=e$. $\endgroup$
    – Achim Krause
    Commented Aug 9, 2017 at 9:22

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