TITLE: Exotic differentiable structures on R^4?
QUESTION [37 upvotes]: 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?

REPLY [20 votes]: 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.