# Uncountable differential structures on $4$-manifolds?

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?

• 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 5 '17 at 6:23
• 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 5 '17 at 6:25
• 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 5 '17 at 6:32
• E.g. see 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 5 '17 at 6:36
• It is not true that a 1-manifold has a unique smooth structure. – Mariano Suárez-Álvarez Feb 5 '17 at 7:22

## 2 Answers

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's 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 $3$.

• "The claim the top box in your post is incorrect." Could you specify what claim you are referring to that is incorrect? – user56834 Feb 5 '17 at 7:14
• 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 5 '17 at 7:23
• @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 5 '17 at 7:59
• @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 5 '17 at 8:05
• @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 5 '17 at 8:09

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.

• 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 5 '17 at 6:42
• @Programmer2134 yes, of course, I added that. It was implicit, or we could not have talked about diffeomorphisms at all. – Henno Brandsma Feb 5 '17 at 6:46
• 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 5 '17 at 9:44
• @DanielRobert-Nicoud what happens above 4, exactly? Finitely many options in some dimensions right? – Henno Brandsma Feb 5 '17 at 9:45
• @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 5 '17 at 9:57