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The professor of an introduction to general relativity made a remark that confused me. It is not merely that I find it unintuitive, I also find it hard to wrap my head around what it means:

In dimensions $1$ to $3$, a topological manifold can be given only one differential structure (and finitely many in $5-7$) [I think what he means here is that we can give it an infinite amount of differentially-compatible atlasses, but they are all diffeomorphic], but in dimension $4$, we can give it an uncountable number of non-diffeomorphic differential structures.

First of all, I am not 100% sure if I understand what it means. For example is it correct to say that to give a topological manifold a particular differential structure is to choose a smooth Atlas $A$ or any other one that has smooth transition maps to $A$'s charts?

To test my understanding, please tell me if the following re-formulation is correct:

Take any ${1,2,3}$-dimensional differentiable manifold $M$. Take the set $X$ of all manifolds that are homeomorphic to $M$. Then all elements of $X$ are also diffeomorphic to eachother.

Take any $4$-dimensional differentiable manifold $M$. Take the set $Y$ of all manifolds that are homeomorphic to $M$. Then there are an uncountable number of subsets $U_{\alpha \in R}$ of $Y$ such that for all $\alpha \in R$, all elements of $U_{\alpha}$ are diffeomorphic to eachother, but for every $\alpha, \beta \in R, ($with $\alpha \neq \beta)$, no element of $U_{\alpha}$ is diffeomorphic to an element of $U_{\beta}$.

Then the questions this post is all about:

  • Why is there only one particular differential structure for ${1,2,3}$ dimensional manifolds (proof-wise perhaps, but mostly intuitively)? I feel like I can't understand the $4$ dimensional weirdness, until I intuitively understand the ${1,2,3}$ case.
  • What should we even imagine when we say that a $4$-dimensional manifold has multiple differential structures? We can't easily picture $4$ dimensional space, but is there some intuitive way to show what this would mean?

Additional remarks

$(1)$. Additional analysis/question: If my above re-formulation is correct, then the theorem does not imply that for any particular differentiable $4$-manifold (i.e. any particular set of points in $R^{d>=4}$ forming a $4$-sub-manifold) there are infinite possible differential structures. For example, if we take any particular differentiable $4$-manifold embedded in $R^{d>=4}$, then that manifold will only have a unique differentiable structure, correct?

Community
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user56834
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    It's never been 'intuitive' (whatever that means) to me. I did study the proof once. Your reformulation is in essence OK. As a set theory student, I have to object to "the set of all manifolds", there is no such thing ,it's a proper class. – Henno Brandsma Feb 05 '17 at 06:23
  • Could you elaborate on your point about set vs class, or give me the name of a theorem/concept so that I can study it myself? (Though this is not a major point within thia question, I believe?) – user56834 Feb 05 '17 at 06:25
  • It's not really relevant, but it's better to avoid such formulations. You are trying to be exact. I propose below a reformulation that does not use this idea of set of manifolds. One should see it as a (hard!) theorem instead. – Henno Brandsma Feb 05 '17 at 06:32
  • E.g. see https://en.wikipedia.org/wiki/Class_(set_theory) as a start. Using classes as if they are sets led to Russell's paradox, showing one has to be careful – Henno Brandsma Feb 05 '17 at 06:36
  • It is not true that a 1-manifold has a unique smooth structure. – Mariano Suárez-Álvarez Feb 05 '17 at 07:22

3 Answers3

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The $1$-dimensional case is trivial, and the $2$-dimensional case is classical (but harder than one might expect). The case of dimension $3$ was proved by Moise in the 50s. Higher dimensions are different: The first distinct smooth structures on the same manifold were presented by Milnor for $S^7$. Furthermore, any compact, PL manifold in dimension $n\not = 4$ has only finitely many distinct smooth structures.

The case $n = 4$ is very different from the lower- and higher-dimensional cases, and the intuition there probably doesn't apply. The glib one-line answer is that in low-dimensions, geometry dominates; in high-dimensions, the $h$-cobordism theorem and its extensions dominate, and the subject becomes surgery theory. The problem with dimension $4$ is that the Whitney trick fails spectacularly. There are nice $4$-manifolds that have no smooth structure (i.e., a manifold $X$ not homeomorphic to any smooth manifold $Y$), and there are nice $4$-manifolds that have multiple smooth structures. For example, there exist uncountably many manifolds that are homeomorphic to $\mathbb{R}^4$, but no two of which are diffeomorphic. The $E_8$-manifold is compact and simply connected, but it can't be given a smooth structure. The details of which manifolds have a smooth structure are complicated, but the existence of a PL structure for a compact manifold $X$ is detected by the Kirby-Siebenmann class $\kappa\in H^4(X, \mathbb{Z}_2)$. In particular, if $X$ has dimension $<4$, then this class vanishes. (Of course, that obscures exactly where the class comes from; it's a bit like reading the punchline without the joke.)

You asked for the intuition behind that, and the best answer I can come up with is that the naive intuition that one can take a improve a reasonable homeomorphism $X \to Y$ to a "nearby" smooth map fails completely in higher dimensions. Along similar lines, there are characteristic classes attached to manifolds that are invariant under diffeomorphisms but not under arbitrary homeomorphisms, and so the two categories of structures are distinguishable. Dimension $4$ is just particularly weird. Low-dimensional topology has a very different flavor from high-dimensional topology, and dimension $4$ is very different even from dimension $3$.

anomaly
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  • "The claim the top box in your post is incorrect." Could you specify what claim you are referring to that is incorrect? – user56834 Feb 05 '17 at 07:14
  • You use certain concepts I am not very familiar with (cobordism, Whitney trick,..). Perhaps you could clarify at least somewhat by responding to my additional remark (1)? Also, could you clarify/elaborate on this point: "the naive intuition that one can take a improve a reasonable homeomorphism to a "nearby" smooth map fails completely in higher dimensions". By nearby smooth map I assume you mean e.g. changing the RELU function to a smooth RELU function. Why would this not work in 4 dimensions? – user56834 Feb 05 '17 at 07:23
  • @Programmer2134: I made that remark before the edit to the OP (and have since edited my own response). The box said that the smooth structure was unique in dimensions $5$ through $7$, which is incorrect. It is finite, though; if I remember correctly, the proof is to show that you can take the manifold to be smooth except at a single point, then use a bit of complicated algebraic topology to show that the number of distinct smooth structures on the sphere is finite in every dimension. – anomaly Feb 05 '17 at 07:59
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    @Programmer2134: Unfortunately, this subject gets very technical fast; codifying exactly where and how the intuition fails involves a lot of complicated geometric constructions and algebraic topology machinery. The seminal result for the high-dimensional case is the $h$-cobordism theorem, which states in rough terms that cobordism can be promoted up to a homeomorphism, PL-homeomorphism, diffeomorphism, etc. under very mild restrictions. It only applies above dimension $4$ (or $5$, depending on the statement of the theorem). – anomaly Feb 05 '17 at 08:05
  • @Programmer2134: In dimension $1$ through $3$, things are relatively familiar for other reasons: $1$ and $2$ are almost trivial, and the geometry in dimension $3$ is well-understood. That leaves dimension $4$ as the outlier, and it is in fact pathological. As for why homeomorphism does't imply diffeomorphism in dimension $4$, the short answer is that it doesn't in general; $4$ is just the first dimension where it fails. For example, there exist manifolds in dimension $4$ that have no smooth structure at all. Their construction is nontrivial, but take a look at Donaldson's theorem. – anomaly Feb 05 '17 at 08:09
  • For your remark (1): If you mean a topological embedding, then no: Any compact $C^0$ manifold $X$ can be embedded in $\mathbb{R}^N$ for some $N$ (take a finite open cover by coordinate patches), but not all of them admit a smooth structure. Similarly, there's a nice description of the various exotic smooth structures on $S^7$ in terms of an explicit variety in $\mathbb{C}^n$ for small $n$. If you mean a smooth embedding, then $X$ has to inherit its smooth structure from $\mathbb{R}^n$. – anomaly Feb 05 '17 at 08:16
  • On your first criticism, you are right, I've added "differentiable" in the remark. On your second criticism: when you say that there are multiple smooth structures on $S^7$, don't you mean that there are multiple actual differentiable $7$-sub-manifolds embedded in $R^{d>=7}$ (each one defined by a set of points in $R^d$), which are homeomorphic to eachother but not diffeomorphic? In other words, you say they are the same "manifold" with a different smooth structure, but in fact they are different individual sub-manifolds in $R^d$ which happen to be homeomorphic but not diffeomorphic. – user56834 Feb 05 '17 at 08:29
  • My remark then conjectured: if you take any particular instance of a sub-manifold in $R^d$, it only has one differential structure. Does it make sense what I'm trying to say? Is my remark then consistent with your criticism? And is it then correct? – user56834 Feb 05 '17 at 08:33
  • Why do you want to consider only submanifolds of $\mathbb{R}^n$? By a different smooth structure on $S^7$, I mean a smooth manifold that is homeomorphic to but not diffeomorphic to the usual $S^7$. Any smooth manifold embeds in $\mathbb{R}^N$ by Whitney's theorem, but the embedding isn't canonical or easily constructed, and homeo-/diffeomorphisms aren't required to extend over the ambient space. – anomaly Feb 05 '17 at 09:41
  • I wanted to consider only submanifolds, just in order to clarify what the theorem does and doesn't say. If any particular set of points in $R^d$ that forms a 4-submanifold could have multiple differential structures, that would surprise me a lot more than if it is merely the case that two different such sub manifolds which are nevertheless homeomorphic have two different differential structures (which it seems to me is what the theorem is saying) – user56834 Feb 05 '17 at 09:48
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I would personally put it as follows:

Let $M$ be a differential manifold $M$ of dimension $\dim(M) < 4$ (this comes with a particular atlas), if we have $N$ a differential manifold of the same dimension and $M$ is homeomorphic to $N$ (so topologically the same), the $M$ and $N$ are even diffeomorphic.

Also, there are uncountably many $M_\alpha$, $\alpha \in A$, some uncountable index set, such that every $M_\alpha$ is a 4-dimensional differential manifold (including atlas etc.) such that for $\alpha \neq \beta$, we have $M_\alpha$ is homeomorphic to $M_\beta$ but $M_\alpha$ is not diffeomorphic to $M_\beta$.

For technical info, also see wikipedia, as usual.

Henno Brandsma
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  • Shouldn't you also require $M$ to be differentiable? Otherwise it would imply that the regular sphere is diffeomorphic to the regular cube, since it is homeomorphic to it? – user56834 Feb 05 '17 at 06:42
  • @Programmer2134 yes, of course, I added that. It was implicit, or we could not have talked about diffeomorphisms at all. – Henno Brandsma Feb 05 '17 at 06:46
  • The first paragraph is only true for $\dim(M)<4$ (or at least it has exceptions in higher dimensions; I am not completely familiar with this kind of phenomenology). – Daniel Robert-Nicoud Feb 05 '17 at 09:44
  • @DanielRobert-Nicoud what happens above 4, exactly? Finitely many options in some dimensions right? – Henno Brandsma Feb 05 '17 at 09:45
  • @DanielRobert-Nicoud I find it surprising that the number of diffeomorphism classes of the manifolds $M_\alpha \times \mathbb{R}$ collapse so much in number. – Henno Brandsma Feb 05 '17 at 09:57
  • @HennoBrandsma On compact topological manifolds of dimension greater than $4$, there are at finitely many differentiable structures (up to diffeomorphisms). See https://en.wikipedia.org/wiki/Differential_structure#Differential_structures_on_topological_manifolds – Daniel Robert-Nicoud Feb 05 '17 at 10:30
  • @HennoBrandsma As for non-compact, I would be surprised if you couldn't get uncountably many simply by considering something like $M\times\mathbb{R}^4$... But then, it's already surprising that you have uncountably many differentiable structures on $\mathbb{R}^4$, so who knows? – Daniel Robert-Nicoud Feb 05 '17 at 10:31
  • @HennoBrandsma: For a concrete example, there are uncountably many spaces $X$ that are exotic $\mathbb{R}^4$ (i.e., smooth manifolds homeomorphic but diffeomorhic to the usual $\mathbb{R}^4$) such that all $X\times \mathbb{R}$ are diffeomorphic to the usual $\mathbb{R}^5$. See Akbulut corks, for example. The intuition is that the exotic $\mathbb{R}^4$ are based on the failure of the Whitney trick in dimension $4$, but the extra dimension in $X\times \mathbb{R}$ allows you to "smooth out" the obstructions with the extra dimension. – anomaly Feb 06 '17 at 00:24
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Let me clarify some of the questions in the post:

  1. Take any $4$-dimensional differentiable manifold $M$. Take the set $Y$ of all manifolds that are homeomorphic to $M$. Then there are an uncountable number of subsets $U_{\alpha \in R}$ of $Y$ such that for all $\alpha \in R$, all elements of $U_{\alpha}$ are diffeomorphic to each other, but for every $\alpha, \beta \in R, ($with $\alpha \neq \beta)$, no element of $U_{\alpha}$ is diffeomorphic to an element of $U_{\beta}$.

Actually, you do not need this $Y$. The point is that every 4-dimensional (nonempty) smooth manifold $M$ contains an open subset diffeomorphic to $R^4$. The latter contains continuum of open subsets $U_\alpha$ which are all homeomorphic to $R^4$ but are all pairwise non-diffeomorphic.

  1. A compact 4-dimensional manifold admits at most countably many pairwise non-diffeomorphic smooth structures. Furthermore, there are examples of compact 4-manifolds with countably infinitely many non-diffeomorphic smooth structures.

The following are unknown:

a. Does every smooth compact 4-manifold admits at least two non-diffeomorphic smooth structures. The most famous case is that of $S^4$.

b. Does every smooth compact 4-manifold admits infinitely many pairwise non-diffeomorphic smooth structures?

c. Does every noncompact 4-manifold admits continuum of pairwise non-diffeomorphic smooth structures?

  1. For some reason (probably since this feels "easier") the question concludes with a discussion of submanifolds in $R^d$. This makes no difference whatsoever. Of course, every smooth submanifold of $R^d$ has a canonical smooth structure, namely, the one coming from the assumption that it is a submanifold in $R^d$. However, canonical does not mean unique.

For example, if we take any particular differentiable 4-manifold embedded in $^$, then that manifold will only have a unique differentiable structure, correct?

No, it will have a canonical smooth structure, but this smooth structure need not be unique. This already happens when $d=4$.

  1. What should we even imagine when we say that a $4$-dimensional manifold has multiple differential structures? We can't easily picture $4$ dimensional space, but is there some intuitive way to show what this would mean?

This is actually a wrong way to approach the issue. Nobody constructs exotic smooth structures, say, on $R^4$, by describing an atlas of charts on the 4-dimensional Euclidean space which have smooth transition maps. Instead, one builds some 4-dimensional manifold $M_\alpha$ which is manifestly smooth since it is given by gluing some "standard" building blocks in a differentiable fashion. The gluing is done in such a way that one can show that the manifold is homeomorphic but not diffeomorphic to the standard $R^4$, by applying some hard topological results. In other words, you "know" what $M_\alpha$ is, since you built it with your bare hands. Then somebody comes by and says: "By the way, what you constructed is homeomorphic but not diffeomorphic to the standard $R^4$.''

Moishe Kohan
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