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 $\mathbb{Q}$).

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