Qusisymmetric dimension distortion of Ahlfors regular subsets of a metric space

TL;DR

This paper investigates how quasisymmetric maps affect the Hausdorff dimension of Ahlfors regular sets in metric spaces, establishing almost-everywhere dimension preservation and providing sharp bounds and counterexamples.

Contribution

It proves that quasisymmetric maps preserve Hausdorff dimension for almost every Ahlfors regular set and extends results to Carnot groups and planar cases, including sharp bounds and counterexamples.

Findings

Dimension is non-increasing under quasisymmetric maps for almost every set.

In Loewner spaces, Hausdorff dimension is preserved for almost every set.

Constructed examples show sharpness of bounds and existence of sets with no rectifiable sub-arcs.

Abstract

We show that if $f:X\to Y$ is a quasisymmetric mapping between Ahlfors regular spaces, then $\dim_H f(E)\leq\dim_H E$ for "almost every" bounded Ahlfors regular set $E\subseteq X$. If additionally, $X$ and $Y$ are Loewner spaces then $\dim_H f(E)=\dim_H E$ for "almost every" Ahlfors regular set $E\subset X$. The precise statements of these results are given in terms of Fuglede's modulus of measures. As a corollary of these general theorems we show that if $f$ is a quasiconformal map of $\mathbb{R}^N$, $N\geq 2$, then for Lebesgue a.e. $y\in\mathbb{R}^N$ we have $\dim_H f(y+E) = \dim_H E$. A similar result holds for Carnot groups as well. For planar quasiconformal maps, our general estimates imply that if $E \subset \mathbb{R}$ is Ahlfors $d$-regular, $d<1$, then some component of $f(E \times \mathbb{R})$ has dimension at most $2/(d+1)$, and we construct examples to show this bound is sharp. In addition, we show there is a $1$-dimensional set $S\subseteq \mathbb R$ and planar quasiconformal map $f$ such that $f(\mathbb{R} \times S)$ contains no rectifiable sub-arcs. These results generalize work of Balogh, Monti and Tyson \cite{Tyson:frequency} and answer questions posed in \cite{Tyson:frequency} and \cite{AimPL}.