On singular integrals with non-negative kernels in the Heisenberg group

TL;DR

This paper characterizes uniform 1-rectifiability in the Heisenberg group using the $L^2$-boundedness of certain singular integrals and explores related geometric measure theory questions.

Contribution

It establishes a new characterization of uniform 1-rectifiability via singular integral boundedness and provides counterexamples addressing open questions.

Findings

Boundedness of $K_4$-associated SIO implies $E$ lies in a 1-Ahlfors regular curve.

Existence of 1-Ahlfors regular curves where $K_eta$ operators are unbounded for $eta eq 4$.

Existence of a 1-Ahlfors regular, purely 1-unrectifiable set with bounded singular integral.

Abstract

In this paper we revisit nonnegative kernels in the first Heisenberg group $\He$, and in particular we further study the family $$K_\alpha(x,y,z)= \frac{|z|^{\alpha/2}}{\|(x,y,z)\|_{H}^{\alpha+1}}, \quad \alpha>0,$$ which was introduced in \cite{CL}. We first show that if $E \subset \He$ is a $1$-Ahlfors regular set and the SIO associated with the kernel $K_4$ is $L^2(E)$-bounded, then $E$ is contained in a $1$-Ahlfors regular curve. Combined with the converse implication which was obtained by F\"assler and Orponen in \cite{FO1dim}, our result provides a characterization of uniform $1$-rectifiability in the Heisenberg group via the $L^2$-boundedness of a singular integral. We also give a negative answer to a question of F\"assler and Orponen from \cite{FO1dim} by showing that for any $\alpha \in (0,2)$ there exists a $1$-Ahlfors regular curve $E_a$ such that the operators associated with the kernels $K_\alpha$ are not bounded in $L^2(E_\alpha)$. We finally show that there exists a $1$-Ahlfors regular and purely $1$-unrectifiable set $E$ such that the singular integral associated with $|x| \|(x,y,z)\|^{-2}$ is $L^2(E)$ -bounded.