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Let $M$ be a manifold. Must the number of non-diffeomorphic smooth structures on $M$ be either countable or continuum? Could it be something in between when the continuum hypothesis fails?

Edit:

  1. By manifold I mean a connected Hausdorff second-countable locally Euclidean space (so without boundary). I'm not sure how dropping Hausdorffness or allowing boundary would change the problem, but let's keep things simple.

  2. Although I'm asking about second countable manifolds, it seems the problem remains nontrivial without this assumption. As pointed out in the comments, the long line has $2^{\aleph_1}$ non-diffeomorphic smooth structures, but $2^{\aleph_1}$ is not stricly between countable and continuum...

  3. WLOG we may focus on open (non-compact) manifolds, since the number of smooth structures on a closed manifold is always countable, in fact always finite except in dimension $4$.

  4. There are examples showing the number can be finite, countably infinite, or continuum.

    • Finite: any closed manifold of dimension other than $4$.

    • Countably infinite: for example $S^2\times S^2$ (I couldn't find an open example). Ryan Budney mentions in a comment here the conjecture that every smoothable closed 4-manifold has countably infinitely many smooth structures.

    • Continuum: $\mathbb{R}^4$

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    $\begingroup$ The general strategy for such questions is to show that there is a natural topology on the set of relevant objects - in this case, smooth structures - which yields a (Borel subset of a) Polish space and with respect to which the relevant equivalence relation (in this case, diffeomorphism) is coanalytic or better. If you can do that, then Silver's dichotomy tells you that you do indeed have either $\le\aleph_0$ or exactly $2^{\aleph_0}$ many objects of the relevant type. I believe this works in this case, but my differential geometry isn't good enough to be certain. $\endgroup$
    – Noah Schweber
    Commented Jul 1 at 4:06
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    $\begingroup$ Out of curiosity, what is an example of such an $M$ admitting countably many smooth structures? $\endgroup$
    – Alessandro Codenotti
    Commented Jul 1 at 8:27
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    $\begingroup$ The long line supports $2^{\aleph_1}$ many pairwise non-diffeomorphic differential structures. So at least its consistent that the statement fails for nonmetrisable manifolds. $\endgroup$
    – Tyrone
    Commented Jul 1 at 11:38
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    $\begingroup$ Naively this equivalence relation is analytic, because it posits the existence of a diffeomorphism. I don’t think there is a simple trick to proving this conjecture. In the 4D case, it will probably requiring actually overcoming the longstanding problem of finding nontrivial invariants for smooth structures. $\endgroup$
    – Elliot Glazer
    Commented Jul 2 at 4:10
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    $\begingroup$ Hjorth and Kechris wrote a paper about the classification of Riemann surfaces, which might be relevant. I once heard from Solecki that descriptive-set-theoretic classification for the class of manifolds has not been well-studied. $\endgroup$
    – Hanul Jeon
    Commented Jul 2 at 6:24

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