This was going to be a comment to Differentiable structures on R^3, but I thought it would be better asked as a separate question.

So, it's mentioned in the previous question that $\mathbb{R}^4$ has uncountably many (smooth) differentiable structures. This is a claim I've certainly heard before, and I have looked a little bit at the construction of exotic $\mathbb{R}^4$s, but it's something that I really can't say I have an intuitive understanding of.

It seems reasonable enough to me that a generic manifold can have more than one differentiable structure, just from the definition; and is in fact, a little surprising to me that manifolds have only one differentiable structure for dimension $d \le 3$.

But it's very odd to me that $\mathbb{R}^d$ has exactly one differentiable structure, unless $d=4$, when it has way too many!

Naively, I would have thought that, since $\mathbb{R}^4 = \mathbb{R}^2 \times \mathbb{R}^2$, and $\mathbb{R}^2$ has only one differentiable structure, not much can happen. Although, we know $\text{Diff}(M\times N)$ cannot generically be reasonably decomposed in terms of $\text{Diff}(M)$ and $\text{Diff}(N)$ in general, I would not have expected there to be obstructions for this to happen in this case.

I would have also thought, that since $\mathbb{R}^5$ has only one differentiable structure, and $\mathbb{R}^4$ is a submanifold of $\mathbb{R}^5$, and $\mathbb{R}^3$ is a submanifold of $\mathbb{R}^4$ with only one differentiable structure, this would be fairly restrictive on the differentiable structures $\mathbb{R}^4$ can have.

Although it seems that this only restricts the "inherited" differentiable structure to be a unique one, it still seems odd to me that the there are "non-inherited" structures in $d=4$ from $d=5$, and somehow all of these non-inhereted structures are identical on the submanifold $\mathbb{R}^3$!

Anyway, can anyone provide a intuitively sensible explanation of why $\mathbb{R}^4$ is so screwed up compared to every other dimension? Usually I would associate multiple differentiable structures with something topologically "wrong" with the manifold. Is something topologically "wrong" with $\mathbb{R}^4$ compared to every other dimension? Or is this a geometric problem somehow?


I once heard Witten say that topology in 5 and higher dimensions "linearizes". What he meant by that is that the geometric topology of manifolds reduces to algebraic topology. Beginning with the Whitney trick to cancel intersections of submanifolds in dimension $d \ge 5$, you then get the h-cobordism theorem, the solution to the Poincare conjecture, and surgery theory. As a result, any manifold in high dimensions that is algebraically close enough to $\mathbb{R}^d$ is homeomorphic or diffeomorphic to $\mathbb{R}^d$.

By the work of Freedman and others using Casson handles, there is a version of or alternative to the Whitney trick in $d=4$ dimensions, but only in the continuous category and not in the smooth category. Otherwise geometric topology does not "linearize" in Witten's sense. But in $d \le 3$ dimensions, the dimension is too low for the smooth category to separate from the continuous category, at least for the question of classification of manifolds.

What you have in 3 dimensions is examples such as the Whitehead manifold, which is contractible but not homeomorphic to $\mathbb{R}^3$. In 4 dimensions you instead get open manifolds that are homeomorphic to $\mathbb{R}^4$ (because they are contractible I'm not sure if other conditions are needed and simply connected at infinity), but not diffeomorphic to $\mathbb{R}^4$. You have to be on the threshold between low dimensions and high dimensions to have the phenomenon. I would say that these exotic $\mathbb{R}^4$s don't really look that much like standard $\mathbb{R}^4$, they just happen to be homeomorphic. The homeomorphism has fractal features, and so does the Whitehead manifold.

Meanwhile 2 dimensions is too low to have non-standard contractible manifolds. In the smooth category, the Riemann uniformization theorem proves that smooth 2-manifolds are very predictable, or you can get the same result in the PL category with a direct combinatorial attack on planar graphs. And as mentioned, smooth, PL, and topological manifolds don't separate in this dimension.

Also, concerning your question about Cartesian products: Obviously the famous results imply that there is a fibration of standard $\mathbb{R}^5$ by exotic $\mathbb{R}^4$. The Whitehead manifold cross $\mathbb{R}$ is also homeomorphic to $\mathbb{R}^4$. (I don't know if it's diffeomorphic.) These fibrations are also fractal or have fractal features.

  • 1
    $\begingroup$ I would like to add that while the failure of the Whitney trick opens up for the possibility that something new can happen in 4 dimensions (and Freedman et al closed that possibility in the topological case) it seems to be the special properties of differential geometry that gives an actual realisation of that possibility. In the case of Donaldson's work it seems to be the fact that 2-forms are of degree half the dimension of the manifold. $\endgroup$ May 17 '10 at 7:43
  • 2
    $\begingroup$ Greg, from your discussion one may (mistakenly) conclude that any open contractible 4-manifold is homeomorphic to $\mathbb R^4$; in fact any homology 3-sphere bounds a topological contractible manifold. $\endgroup$ May 17 '10 at 12:29
  • $\begingroup$ Igor: Thanks much for that correction. If I had thought it through better, I might have noticed; it's even there on the Wikipedia page for the Whitehead manifold. $\endgroup$ May 17 '10 at 13:35

This started as a comment on Greg's post, but my comments are getting too long...

The additional topological condition you need for an open contractible 4-manifold to be homeomorphic to $\mathbb{R}^4$ is that it be "simply connected at infinity" (this is definitely necessary, and I would bet that it is sufficient, though I am definitely not an expert at $4$-manifold topology). This excludes interiors of contractible manifolds w/ boundary a homology 3-sphere. By the way, the result Igor quoted is a deep theorem of Friedman.

Here's something else that will blow your mind. There actually exist exotic $\mathbb{R}^4$'s $U$ that embed as open submanifolds of $\mathbb{R}^4$ (the so-called "small" exotic $\mathbb{R}^4$'s). We thus get that $U \times \mathbb{R}$ embeds as an open submanifold of $\mathbb{R}^5$. By Stallings's theorem, $U \times \mathbb{R}$ is thus diffeomorphic to $\mathbb{R}^5$ even though $U$ is only homeomorphic to $\mathbb{R}^4$.

This is actually a very common type of thing -- often bizarre examples become nice when you "stabilize them" by, say, crossing with $\mathbb{R}$ or some other related operation.

  • 5
    $\begingroup$ Speaking of stabilization, I think it is true that the product of any contractible 3-manifold with $\mathbb R$ is homeomorphic to $\mathbb R^4$; is it diffeomorphic to $\mathbb R^4$? $\endgroup$ May 17 '10 at 13:18
  • 6
    $\begingroup$ Yes (the following might not be a geodesic to the proof, but it works <grin>). Let U be a smooth contractible 3-manifold. In "Cartesian products of contractible open manifolds", McMillan proved (assuming the Poincare conjecture, which I suppose we now all believe) that UxR is PL isomorphic to R^4. You can now quote Corollary 6.6 of "Obstructions to the smoothing of piecewise-differentiable homeomorphisms" by Munkres, which says that any smooth manifold which is PL isomorphic to R^n is diffeomorphic to R^n. $\endgroup$ May 17 '10 at 23:19
  • $\begingroup$ Thanks, Andy! I missed that McMillan works in PL category. $\endgroup$ May 18 '10 at 12:59
  • $\begingroup$ The third paragraph is definitely surprising (to me, at least). This might be a dumb question, but is the embedding U -> R^4 a smooth embedding or just a topological one? $\endgroup$ May 20 '10 at 6:42
  • 1
    $\begingroup$ Just a note for future visitors - as Andy correctly bet, simply connected at infinity is sufficient. That is, Corollary 1.2 of Freedman's The Topology of Four-Dimensional Manifolds says that any topological 4-manifold which has trivial $\pi_1$ and $H_2$, and is simply connected at infinity, is homeomorphic to $\mathbb{R}^4$. $\endgroup$
    – Aru Ray
    Apr 10 '18 at 9:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.