5
$\begingroup$

Can we talk about a canonical space of dimension $\pi$? Is there anything like $\mathbb R^\pi$?

Have anyone met any fractal of dimension $\pi$?

$\endgroup$
  • 7
    $\begingroup$ There's more than one definition of dimension, which are you interested in? (Some are restricted to the natural numbers) $\endgroup$ – Ben Millwood Sep 18 '12 at 13:29
  • $\begingroup$ Your ordinary every-day dimension is going to be a cardinal number, but then again it might be one of the numerous dimensions Ben M mentions, which I know nothing about. $\endgroup$ – rschwieb Sep 18 '12 at 13:47
  • $\begingroup$ I believe he's talking about Hausdorff Dimension, in which case the Hausdorff Dimension Theorem says such spaces exist. $\endgroup$ – JSchlather Sep 18 '12 at 15:05
  • $\begingroup$ I'm interested in it for every suitable meaning of 'dimension'. $\endgroup$ – Berci Sep 18 '12 at 15:54
  • 1
    $\begingroup$ @rschwieb, I agree that $\mathbb R^\alpha$ looks like a vector space. I was guessing hausdorff dimension based on his question concerning a fractal of dimension $\pi$. I don't know too much about spaces with fractional Hausdorff dimension. But it seems like there isn't a canonical space of dimension $\pi$ or a way to make sense of $\mathbb R^\pi$. But there are subsets of euclidean space with Hausdorff dimension $\pi$. $\endgroup$ – JSchlather Sep 18 '12 at 17:35
5
$\begingroup$

The answer to the question as stated is "No, there is no canonical space of dimension $\pi$" (where I am assuming from context that dimension is meant to be something like the Hausdorff dimension, which makes sense in light of the reference to fractals). However, it is possible to construct such spaces in ways that have been well-studied. Two quick examples come to mind:

Snowflaking

Suppose that $(X,d)$ is a metric space. For any $\alpha\in(0,1)$, we may define a new metric space $(X,d^{\alpha})$ by setting $$ d^{\alpha}(x,y) := d(x,y)^{\alpha} $$ (this process is called "snoflaking" a space). It can be shown that if $(X,d)$ is Ahlfors regular of dimension $D$, then $(X,d^{\alpha})$ will be Ahlfors regular of dimension $D/\alpha$. Moreover the Hausdorff dimension of an Ahlfors regular space coincides with the regularity dimension. The basic thory is outlined nicely in Chapter 2 of David and Semmes' book:

David, Guy; Semmes, Stephen, Fractured fractals and broken dreams. Self-similar geometry through metric and measure, Oxford Lecture Series in Mathematics and its Applications. 7. Oxford: Clarendon Press. ix, 212 p. (1997). ZBL0887.54001.

Let $(X,d) = (\mathbb{R}, d)$, where $d$ is the usual Euclidean metric. Note that $\mathbb{R}$ is Ahlfors regular of dimension 1. Now, set $\alpha = \frac{1}{\pi}$. Then the snowflaked version of $\mathbb{R}$ has dimension $$ \dim(\mathbb{R},d^{1/\pi}) = \frac{\dim(\mathbb{R},d)}{1/\pi} = \frac{1}{1/\pi} = \pi,$$ as desired.

Product Spaces

In general, the Hausdorff dimension is not stable under products. That is, we don't generally have the (possibly desirable) result that $$ \dim(X \times Y) = \dim(X) + \dim(Y) $$ where $X\times Y$ is the Cartesian product of $X$ and $Y$. In fact, the left-hand side is generally larger. However, if $X$ and $Y$ are "nice enough", then we will get equality. Here is an example with spaces that are "nice enough":

First, we are going to build a "Cantor set" of dimension $\pi - 3$:

For $\alpha \in (0,\frac{1}{2})$, let $K_{\alpha}$ be the Cantor set obtained as the attractor of a self-similar IFS consisting of two maps $\mathbb{R}\to\mathbb{R}$, each with contraction ratio $\alpha$, so that the system satisfies the open set condition for $(0,1)$ (the above can be read as "blah blah blah technical details that ensure we get what we want"). Then the Hausdorff dimension $D_{\alpha}$ of $K_{\alpha}$ will be the solution to the Moran equation, which in this case gives $$ D_{\alpha} = \frac{\log(2)}{\log(\alpha)}. $$ Since we want a set of dimension $\pi - 3$, let this be $D_{\alpha}$, and solve for $\alpha$ to get $$ \pi - 3 = \frac{\log(2)}{\log(\alpha)} \implies \alpha = 2^{(\pi-3)^{-1}}.$$ Therefore we have $$ \dim\left( K_{2^{(\pi-3)^{-1}}} \right) = \pi - 3.$$

Cantor sets and closed boxes are "nice enough" that the product formula holds, from which it follows that $$ \dim\left( K_{2^{(\pi-3)^{-1}}} \times [0,1]^3 \right) = \dim\left( K_{2^{(\pi-3)^{-1}}} \right) + \dim([0,1]^3) = (\pi-3) + 3 = \pi. $$

Thus the product of a Cantor set of dimension $\pi-3$ and a cube gives another example of (non-canonical, but somewhat constructive) set with dimensin $\pi$. For details about product spaces, a good place to start would be Chapter 7 of Falconer's book:

Falconer, Kenneth, Fractal geometry: mathematical foundations and applications, Chichester: John Wiley & Sons (ISBN 0-471-96777-7/pbk). xxii, 288 p. (1997). ZBL0871.28009.


If one is interested in other notions of dimension, then the answer continues to be "No, there is no canonical example of a space of [insert notion of dimension here]-dimension $\pi$, but there [are | are not] examples of such spaces." Note that there are notions of dimension such that spaces of dimension $\pi$ are impossible (the usual definition of topological dimension, for example, is integer valued).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.