# Can the image of a disk have nontrivial Hausdorff measure for $1 < d < 2$?

I am reading a blog which talks about a $C^1$ diffeomorphism $f: \mathbb{D}\{ x^2 + y^2 < 1\} \to \mathbb{R}^2$ and estimates the Hausdorff dimension of its image $\mathcal{H}_\sqrt{2}^d (f(\mathbb{D}))$. In the blog it is shown:

$$\mathcal{H}_\sqrt{2}^d (f(\mathbb{D})) \leq 170 \pi \cdot \max \{ K, L\}^{2-d}\cdot L^{d-1}$$

My question is how can the diffeomorphic image of a disk have Hausdorff dimension between 1 and 2?

To remind myself, I read up on Hausdorff measure. The Hausdorff measure of $\mathbb{D}$ is zero in dimensions bigger than 2 and infinite in dimensions less than 2.

So how can the diffeomorphic image of a disk have $\mathcal{H}_\sqrt{2}^d (f(\mathbb{D}))$ which is neither $0$ nor $\infty$?

Related What is the Point of a Horseshoe Map ? and some more discussion where the authors announce finding Hausdorff dimension of the (un)stable set to be less than 2. Fractal Geometry of non-Uniformly Hyperbolic Horseshoes

$\sqrt{2}$ seems to have to do with the diameter of the unit square, so these Hausdorff measures are counting squares. These result is only meaningful with dervatives $||Df(p)|| < K$ and $|\det Df(p)| < L$ are both large. Therefore, $f$ must exhibit quite a bit of distortion, even thought it is $C^1$.

This measure rewards you for stretching. A very long thin rectangle with unit Euclidean area must be covered by (very inefficiently) by a lot of unit squares.

• Several comments: 1. A $C^1$ diffeomorphism is not a diffeomorphism. $C^1$ implies that only the first derivatives are defined, while plain "diffeomorphism" is usually interepreted to be $C^\infty$. 2. Hausdorff dimension is defined as a limit as balls get smaller and smaller. From the context in that blog post, $\mathcal{H}^d_{\sqrt 2}$ is a non-limiting value. The first equation in that lemma is the definition of the Hausdorff measure. It's some finite value, regardless of which sets you are looking at. May 11 '15 at 21:54
• @DylanThurston I am trying to understand why the infimimum goes down with $\delta$. If we allow sets of size $\delta/3$ instead of $\delta$ we should need to cover with $3^{\dim S}$ as many sets, width $3^d$ measure. The measure should increase by factor of $3^{\dim S - d}$. I can't tell if that's increasing or decreasing. If the image of the neighborhood is a point then at least locally $\dim S = d = 2$. May 11 '15 at 22:26
• A diffeo (or just homeo)morphism maps the open disk to an open set again, so the image trivially has infinite $d$ dimensional Hausdorff measure for all $d<2$. May 11 '15 at 22:30
• @ChristianRemling Then the blogger's result is rather pointless! He is measuring $\mathcal{H}^d$ with $d < 2$. May 11 '15 at 22:32
• $H_\sqrt{2}^d (f(B))$ is not the Hausdorff measure or the Hausdorff dimension, as you write in your question. The blog author clearly defines it as $\inf_{\mathrm{diam}\,U_i \le \sqrt{2},f(B)\subseteq\bigcup_i U_i} \sum_i (\mathrm{diam}\,U_i)^d$. If $\sqrt{2}$ is replaced with $r$ and the limit $r\to0$ is taken, then you get the Hausdorff measure. Now are you asking the question in your title or are you asking for this clarification? May 12 '15 at 4:24

The question has been answered in the comments, but just for the record: The homeomorphic image of a disc (even embedded in $\mathbb{R}^n$ for $n>2$, or indeed in any metric space) cannot have Hausdorff dimension less than two, since the topological dimension is a lower bound for the Hausdorff dimension. The diffeomorphic image of a disc will always have Hausdorff dimension two.