# 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.

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tangentially related: mathoverflow.net/questions/60375/… – Grigory M Oct 21 '11 at 11:37

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)
I guess I was somewhat vague in my requirements. I would expect (among other things) that the Hausdorff dimension of the group (its given metric, actually) would be identical to the growth rate of the group. This isn't satisfied by your family of examples unless $\alpha = 1$. So this would be one relation between the algebraic structure and the metric structure which we would like to have. – Mark Oct 23 '11 at 19:17