Fractional versions of euclidean space?

This is going to be a somewhat vague question, but I'll be happy if you indulge me.

Euclidean space $\mathbb{R}^n$ is equipped with a lot of nice (algebraic, metric, topological,...) structure and has many nice properties as such, regardless of the dimension $n$ (simple-connectedness, finite-dimensionality, abelianness, having a uniform lattice, being a $\sigma$-finite measure space, and the list goes on...) . It seems interesting to me to try and find geometric/algebraic objects with non-integer dimension while trying to preserve as many properties of classical euclidean space as we can.

Some thoughts: the dimension of a vector space is always an integer so let's forget about vector spaces altogether (but stay in $\mathbb{R}^n$ for some large $n$) and interpret "dimension" as "Hausdorff dimension" (this seems geometrically intuitive to me, though I'm aware that there are many other notions of dimension in $\mathbb{R}^n$ or general metric spaces). One way of thinking of $\mathbb{R}^n$ is as a connected, simply-connected locally compact Hausdorff topological group, so I wonder now whether for every $\alpha > 0$ there is some "canonical" model of a topological group (implicitly assumed to lie in $\mathbb{R}^n$ for some $n$) with these properties and with the additional property that it is $\alpha$-dimensional (in the Hausdorff dimension sense). In the case where $\alpha$ is a positive integer, it is obvious what this model should be.

Solution 1:

A nontrivial metric space of Hausdorff dimension less than 1 cannot be connected, so I'll just ignore that case.

For $\alpha \ge 1$ consider the family of metric spaces $(\mathbb{R}, d_\alpha)$ with $d_\alpha(x, y) = |x-y|^{1/\alpha}$. With $\alpha = 1$ this just gives the usual metric structure.

It is easy to see that the topological structure does not depend on $\alpha$, because the collection of all open balls remains the same; the same balls just get different radii. This means that as a topological field these spaces are all the same.

When it comes to Hausdorff measure the spaces are very different. The largest set with $d_\alpha$-diameter $\varepsilon$ is an interval with length $\varepsilon^\alpha$. Clearly it takes $\Theta(\varepsilon^{-\alpha})$ of these intervals to cover the unit interval. Since smaller intervals don't provide a more efficient cover, this means that $(\mathbb{R}, d_\alpha)$ has Hausdorff dimension $\alpha$. In fact the $\alpha$-dimensional Hausdorff measure is equal to Lebesgue measure. (give or take a constant factor)

If there is a moral to this story, it must be that Hausdorff dimension says very little about topological or algebraic structure.