On the conformal dimension of product measures

Abstract

Given a compact set $E\subset {\mathbb{R}}^{d-1}$, $d\ge 1$, 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 $f:K_E\to (X,d)$ is a quasisymmetric homeomorphism, then $${\dim }_{\mathrm{H}}f(K_E)\ge {\dim }_{\mathrm{H}}{K}_E.$$ We prove that the measure-theoretic analogue of the result is not true. For any $d\ge 2$ and $0\le s<d-1$, 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 $F:K_E\to {\mathbb{R}}^d$ such that ${\dim }_{\mathrm{H}}{F}_{♯}\nu <\epsilon$.

arXiv comment27 pages. v2: incorporated minor referee comments. To appear in Proc. LMS