# 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. – Dylan Thurston 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 $\dim S = d = 2$. – john mangual 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 $d<2$. – Christian Remling May 11 '15 at 22:30
• @ChristianRemling Then the blogger's result is rather pointless! He is measuring $\mathcal{H}^d$ with $d < 2$. – john mangual 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? – Yoav Kallus 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.