# Differential Topology over $\mathbb{Q}$

I have a specific question in mind, but it requires some explanation and context before it can be formally stated. To summarize it in a sentence, this is it:

Are every two rational manifolds of the same dimension diffeomorphic?

Explanation:

It is known that $\mathbb{Q}^n\cong \mathbb{Q}$ (homeomorphic) for every $n\in \mathbb{N}$. It is a corollary to Sierpiński's theorem which states that every countable metric space without isolated points is homeomorphic to $\mathbb{Q}$. (A proof can be found here and a discussion here).

This means that we cannot distinguish topologically between an $n$-rational manifold and an $m$-rational manifold. (Where we define an $n$-rational manifold in the obvious way: A topological space which is locally homeomorphic to $\mathbb{Q}^n$).

However, when we look at two different manifolds from the perspective of differential topology we can distinguish between them:

Let $f:\mathbb{Q}^n \longrightarrow \mathbb{Q}^m$. We can define what it means for such $f$ to be differentiable: (The differential will be a linear transformation $\mathbb{Q}^n \longrightarrow \mathbb{Q}^m$).

The definition (of differentiability) over $\mathbb{R}$ uses the fact it is an ordered field, and the norm structure on $\mathbb{R}^n$. We do not have the standard euclidean norm on $\mathbb{Q}^n$ (we cannot take square roots in $\mathbb{Q}$, and normed spaces are usually defined to be vector spaces over $\mathbb{R}$ or $\mathbb{C}$) but we can use the "rational" $1$-norm instead.

This is only one option which works "intrinsically", i.e does not require using numbers outside $\mathbb{Q}$. We can use other alternatives of course or allow ourselves to go outside of the rational world and measure distances using real numbers.

Now let us say that two rational manifolds $M,N$ are diffeomorphic if there is a differentiable mapping $f:M \longrightarrow N$ whose inverse is also differentiable. (We can of course require stronger conditions like twice differentiabililty etc..).

Note that the chain rule holds (its proof uses only the ordered field structure, not any special propery of $\mathbb{R}$).

Its clear that if $f: \mathbb{Q}^n \longrightarrow \mathbb{Q}^m$ is a diffeomorphism, its differential will be a linear isomorphism (via the chain rule). But this is a contradiction if $n\neq m$.

This shows that the notion of rational-differential-manifold is not trivial (unlike the topological case), and so brings up the following natural questions:

Are every two rational manifolds of the same dimension diffeomorphic?

I guess there are many more questions one could ask. I think it is interesting to find out how much differential topology\geometry can we do over other ordered fields which are not $\mathbb{R}$ ?

(In some sense not much. For example the inverse function theorem does not hold over $\mathbb{Q}$).

• So "differentiable" means differentiable at each point of the domain? So the differential may vary from point to point? And need not depend continuously on the point? – Gerald Edgar May 21 '15 at 14:39
• Also, we need a definition of "rational manifold". Without paracompactness you could have some sort of "long line" not of the same cardinal as the "short line" $\mathbb Q$. – Gerald Edgar May 21 '15 at 14:53
• Yes. I meant only pointwise differentiable. This is the weakest notion of differentiability I could think of. As I said, the chain rule implies that even w.r.t this weak notion of equivalence, manifolds of different dimensions are inequivealent. Of course I see no apparent reason for not choosing to work in a smooth (infinitely differentiable) category or any other degree of differentiability between them. (That is, to use stronger equivalence relations). I will consider any ability to distinguish between two manifolds in every smooth category whatsoever. – Asaf Shachar May 21 '15 at 15:00
• You are probably right about paracompactness. I thought I should add a requirement that the manifold will be second-countable, but was unsure it was necessary in this case. – Asaf Shachar May 21 '15 at 15:04
• For example, an uncountable disjoint union of copies of $\mathbb Q^n$. So the requirement may as well be that the whole space is countable (as Simon assumes). – Gerald Edgar May 21 '15 at 15:08

Claim : any rational manifold is a disjoint union of rational ball.

let's prove it for a countable rational manifold:

assume the point of $X$ are numbered $x_1,\dots,x_n...$.

Pick a neighbourhood of $x_1$ that is diffeomorphic to an open ball in $\mathbb{Q}^n$ and then pick a ball arround $x_i$ in that neighbourhood that is both open and closed (and small enough so that it is also closed in $X$) for example, pick a ball whose radius is not a square root of a rational number. Call it $F_1$.

Because $F$ is open and closed, you can easily check that $X$ is diffeomorphic to $(X-F) \cup F$. Do the same in $X - F$ for the next $x_i$ which is not in $F_1$, you get a clopen ball $F_2$ and keep going...

At then end you obtain a partition of $X$ in clopen ball $F_1, \dots F_n \dots...$ and the canonical map between $X$ and the disjoint union of the $F_i$ is a diffeomorphism. (because it is on all of the $F_i$ which form an open cover of $X$).

To break this argument and allow to recover some behavior from real differential geometry what you need is a uniform structure (or a Riemanian/metric structure for example) on your differential manifold compatible with the manifold structure which will force the existence of a "real completion". Or (almost the same) a notion of admissible cover similar to what we have in analytic rigid geometry.

EDIT : One can improve the argument to show that any to countable manifold of the same positive dimension are diffeomorphic: it suffice to choose all the clopen ball we construct to have comensurable radius in $\mathbb{Q}^n$ (for example raidus $q \pi$) this way any two balls we obtain will be diffeomorphic. If one of the manifold is formed of only a finite number of balls one can always split one of the ball in a infinite number of smaller ball using the same process with ball of radius $R/4^k$ inside the initial ball: a finite family of such balls cannot cover the initial ball by a volume argument.

• Can you please explain the last part in your argument. How can you prove a finite family of such balls cannot cover the initial ball? (what exactly is the volume argument?) – Asaf Shachar May 24 '15 at 16:56
• If a finite union of balls contains all the rational number of the unit ball then the union of the corresponding closed ball contains all of the unit ball, so the volume of the unit ball is smaller than the sum of the volume of the ball involved in the covering. So if you take the volume of the ball in the covering to decrease fast enough so that the sum of there volume is smaller than the volume of the initial ball then any finite familly of those balls cannot form a covering of the (rationals in the) initial ball. – Simon Henry May 24 '15 at 18:03