# On the conformal dimension of product measures

## TL;DR

This paper shows that, unlike sets minimal for conformal dimension, certain measure-theoretic analogues of these sets can have arbitrarily small conformal dimension, challenging previous assumptions about measure and conformal invariance.

## Contribution

The authors construct examples of sets with positive Hausdorff measure whose associated measures have zero conformal dimension, revealing limitations of conformal dimension invariance under quasisymmetric maps.

## Key findings

- Existence of sets with finite positive Hausdorff measure but zero conformal dimension.
- Construction of quasisymmetric embeddings reducing measure dimension arbitrarily close to zero.
- Counterexample to the measure-theoretic analogue of a known conformal dimension minimality theorem.

## Abstract

Given a compact set $E \subset \mathbb{R}^{d - 1}$, $d \geq 1$, write $K_{E} := [0,1] \times E \subset \mathbb{R}^{d}$. A theorem of C. Bishop and J. Tyson states that any set of the form $K_{E}$ is minimal for conformal dimension: if $(X,d)$ is a metric space and $f \colon K_{E} \to (X,d)$ is a quasisymmetric homeomorphism, then $$\dim_{\mathrm{H}} f(K_{E}) \geq \dim_{\mathrm{H}} K_{E}.$$ We prove that the measure-theoretic analogue of the result is not true. For any $d \geq 2$ and $0 \leq s < d - 1$, there exist compact sets $E \subset \mathbb{R}^{d - 1}$ with $0 < \mathcal{H}^{s}(E) < \infty$ such that the conformal dimension of $\nu$, the restriction of the $(1 + s)$-dimensional Hausdorff measure on $K_{E}$, is zero. More precisely, for any $\epsilon > 0$, there exists a quasisymmetric embedding $F \colon K_{E} \to \mathbb{R}^{d}$ such that $\dim_{\mathrm{H}} F_{\sharp}\nu < \epsilon$.