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?
Best Answer
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
insteadget open manifolds that are homeomorphic to $\mathbb{R}^4$ (because they are contractibleI'm not sure if other conditions are neededand 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.