Is there a connection between the construction of $\mathbb{Q}$ and exotic $\mathbb{R}^4$'s

Question

I learned that there exist so-called "exotic" $\mathbb{R}^4$'s. That is, there exist topological spaces which are homeomorphic but not diffeomorphic to $\mathbb{R}^4$. Quite remarkably, it has been proven that $4$ is the only value of $n$ for which there exists an exotic $\mathbb{R}^n$. Moreover, it has also been shown that there are uncountably many exotic $\mathbb{R}^4$'s. https://en.wikipedia.org/wiki/Exotic_R4

I admit that I do not at all understand the existence proof of an exotic $\mathbb{R}^4$, but it seems to be quite non-constructive. It rests upon the existence of a non-trivial smooth 5 dimensional $h$-cobordism which must exist by some other theorem. https://projecteuclid.org/download/pdf_1/euclid.jdg/1214437666

However I had an idea of where these things could be coming from (or at least something interesting if not related to exotic $\mathbb{R}^4$'s).

Recall the standard construction of $\mathbb{Q}$ from $\mathbb{Z}$: on the set $\mathbb{Z}\times(\mathbb{Z}-\{0\})$ we define a relation by $$(a,b)\sim(c,d)\Longleftrightarrow ad=bc$$ Then you show it's an equivalence relation and on the set of equivalence classes, you define addtion, multiplication, show that they're well-defined, etc.

However, recall from elementary set theory the definition of a relation: A relation on $X$ is nothing more than a subset of $X\times X$. So from this perspective, $\mathbb{Q}$ can be identified with a particular subset of $\big(\mathbb{Z}\times(\mathbb{Z}-\{0\})\big)^2\subseteq \mathbb{Z}^4$. But $\mathbb{Z}^4\subseteq \mathbb{R}^4$, so that means that $\mathbb{Q}$ is sitting inside of $\mathbb{R}^4$. Clearly, we already know that $\mathbb{Q}\subseteq \mathbb{R}^4$ by inclusion into any one of the coordinates. However, these are quite different than the $\mathbb{Q}$ coming from the set contruction, which I will call $\tilde{\mathbb{Q}}$ to distingtuish from the "ordinary" $\mathbb{Q}$'s sitting inside $\mathbb{R}^4$.

Of course, $\tilde{\mathbb{Q}}$ has it's natural metric. However, it doesn't coincide with the standard metric as a subset of $\mathbb{R}^4$. So perhaps the completion of $\tilde{\mathbb{Q}}$ with respect to the $\mathbb{R}^4$ metric is what leads to these exotic $\mathbb{R}^4$'s.

Obviously, this is all conjecture but I would appreciate any comments or insight.

Answer 1

I don't think there's any connection between what you describe and the existence or construction of exotic $\mathbb{R}^4$. The reason for the latter is that the $h$-cobordism theorem spectacularly fails in dimension $4$. Above dimension $4$, there's enough "room" to smooth out functions via the Whitney trick; in dimension $4$, there are some structures that just aren't smoothable. (The Whitney trick also fails below dimension $4$, but those are special cases: dimension $1$ is trivial, $2$ is classical, and $3$ is hard but of a completely different nature.) What you're describing is the idea of a general embedding or metric issue, which is different from the very technical issues that arise here.

For the details of this sort of thing, I recommend Scorpan's "Wild World of 4-Manifolds." It's a well-written and fun book, but bear in mind that it unavoidably requires a bit of algebraic and geometric topology to go through.