Is there a space of smooth structures on a manifold, analogous of the moduli space of complex structures on a manifold? If so, what is the natural topology?

Also, I am a bit confused about why smooth structures usually form a discrete space; for example there are finitely many smooth structures on spheres. I think one possible exception is that $\mathbb{R}^4$ admits a continuum of smooth structures but I think this is with some unusual topology. Intuitively this is because if we deform a smooth structure, we are deforming "smoothly" so we can undo this deformation using a diffeomorphism and hence smooth structures that are deformation equivalent are diffeomorphic. We can similarly deform complex structures using diffeomorphisms but then nearby complex structures are not connected by biholomorphism but by diffeomorphism and hence are different in the moduli space. It would be great if someone could make this argument rigorous.

As a separate but related question: is it true that any compact topological space admits at most finitely many smooth structures? (up to homeomorphism of the topological space). This is true for spheres. This fails for $\mathbb{R}^4$, which is non-compact.

  • $\begingroup$ You express confusion about why "smooth structures usually form a discrete space". Can you supply some citations to support the assertion that they do usually form a discrete space? $\endgroup$
    – Lee Mosher
    Feb 1, 2017 at 15:32
  • $\begingroup$ @LeeMosher I edited my question and provided exotic spheres as an example. $\endgroup$
    – user39598
    Feb 2, 2017 at 0:29
  • 1
    $\begingroup$ The correct statement about smooth structures on $S^n$ is not that there are finitely many of them. Instead, there are finitely many smooth structures on $S^n$ up to the action of the homeomorphism group of $S^n$. $\endgroup$
    – Lee Mosher
    Feb 2, 2017 at 0:32
  • $\begingroup$ @LeeMosher What's a good candidate for the space of smooth structures, anyway? The concordance space? $\endgroup$
    – user98602
    Feb 2, 2017 at 0:52
  • $\begingroup$ I don't have any particularly good idea on that question, mostly I've been trying to find out what the OP is really asking. As an unmotivated guess, starting with the compact open topology on $\text{Homeo}(M)$, with respect to the natural left action of $\text{Homeo}(M)$ on the set of smooth structures one could take the quotient topology on the set of left cosets of the stabilizer of one point in the orbit. $\endgroup$
    – Lee Mosher
    Feb 2, 2017 at 1:02

1 Answer 1


Let me answer your last question.

Theorem. (Kirby, Siebenmann) Let $M^n$ be a closed $n$-dimensional topological manifold, where $n\ge 5$. Then the set of isomorphism classes of smooth structures on $M$ is finite.

You can extract this from their Classification theorem, page 155 of Essay IV of their book "Foundational essays on topological manifolds, smoothings and triangulations", vol. 88 of Annals of Mathematics Studies, Princeton University Press, 1977.

The basic reason of finiteness is that (according to their classification theorem) the isotopy classes of smooth structures on $M$ are in bijective correspondence with vertical homotopy classes of sections of a certain bundle $E\to M$ and the homotopy groups of the fiber of this bundle are all finite. (The latter is because of finiteness of the group of smooth structures on $S^n$ with fixed $n$, which was proven by Keraire and Milnor.)

In dimensions $\le 3$ every topological manifold (compact or not) has unique (up to isotopy) smooth structure.

What happens in dimension 4 is anybody's guess. There are examples of closed 4-manifolds supporting infinitely many nondiffeomorphic smooth structures (R. Friedman and J. Morgan, On the diffeomorphism types of certain algebraic surfaces, I and II, J. Diff. Geom. 27 (1988), 297-398). It is conceivable that this is the case for all closed 4-manifolds. It is known (again Kirby and Siebenmann) that in dimension 4 PL category is isomorphic to DIFF category (every PL manifold admits a smooth and the latter is unique). From this you can easily see that every closed 4-manifold has at most countably many smooth structures.

Edit 1. A direct proof of the fact that there are only countably many diffeomorphism classes of smooth compact manifolds is a corollary of

S. Peters, Cheeger's finiteness theorem for diffeomorphism classes of Riemannian manifolds, Journal für die reine und angewandte Mathematik 349 (1984) p. 77-82.

Namely, he gives a self-contained differential-geometric proof of Cheeger's theorem that given $n$, $D$, $V$ and $K$, there are only finitely many diffeomorphism classes of Riemannian $n$-manifolds of volume $\ge V$, diameter $\le D$ and sectional curvature in the interval $[-K, K]$. (Cheeger's original proof used results of Kirby and Siebenmann.) Now, take $D$ and $K$ to be natural numbers and $V$ be of the form $1/N$, where $N$ is a natural number. As I said in my comments, the proof is quite painful and you need to know some basic Riemannian geometry (say, the first 5 chapters of do Carmo's "Riemannian Geometry") to appreciate the proof. Of course, it is still much-much easier than to read Kirby and Siebenmann. If you really decide to understand his proof, you can do it in less than two months (starting with the definition of a smooth manifold). In contrast, you probably will never get to the point of understanding any proofs in Kirby-Siebenmann.

Edit 2. Here is a possible topology on the space of (isomorphism classes of ) smooth structures on an $m$-dimensional compact manifold $M$, which is inspired by the proof of Cheeger's theorem. Fix a finite smooth atlas for a smooth structure $s$ on $M$. This atlas determines (and is determined by) a collection of its transition maps, which are diffeomorphisms between open bounded subsets of $R^m$, $f_{ij}: U_{ij}\to V_{ij}$. Then you declare an open $\epsilon$-neighborhood of $s$ to consist of those smooth structures $s'$ on $M$ which admit a finite atlas with the connection of transition maps $f'_{ij}: U'_{ij}\to V'_{ij}$ such that:

  1. The domains $U_{ij}, U'_{ij}$ are within $\epsilon$-Hausdorff distance from each other.

Set $U''_{ij}:= U_{ij}\cap U'_{ij}$.

  1. The $C^1$-uniform distance between the maps $f_{ij}|U''_{ij}, f'_{ij}|U''_{ij}$ is $<\epsilon$.

One needs to check that this defines a basis of topology (this seems OK). I think this topology will be discrete because a smooth map between closed manifolds (sufficiently) $C^1$-close to a diffeomorphism is a diffeomorphism. However, I do not want to do either one of these things.

  • $\begingroup$ Thank you for your answer. Is there more an intuitive reason that there are at most countably many smooth structures for closed manifolds in any dimensions other than the "vertical homotopy classes of sections of the bundle" argument? Is it that any PL manifold admits at most countable many smooth structure and any topological manifold admits at most countably many PL structures? Can these statements be proven in an elementary way? $\endgroup$
    – user39598
    Feb 4, 2017 at 23:28
  • $\begingroup$ @user39598 there is an elementary but rather painful argument using Riemannian Geometry. I do not know if it qualifies as intuitive. Hard math becomes intuitive only once you know it. $\endgroup$ Feb 4, 2017 at 23:31
  • $\begingroup$ Can you provide a reference for this elementary argument? $\endgroup$
    – user39598
    Feb 4, 2017 at 23:57
  • $\begingroup$ and it would be great if you could explain rigorously why this is false for complex structures; there is a continuous moduli of complex structures on surfaces of higher genus. $\endgroup$
    – user39598
    Feb 5, 2017 at 2:26
  • $\begingroup$ @user39598: The key is that it is much easier to construct diffeomorphisms then holomorphic maps. If you want to see formal proofs, check any book covering Teichmuller spaces, for instance, Abikoff's "Real analytic theory of Teichmuller spaces". $\endgroup$ Feb 5, 2017 at 3:21

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