On the best possible exponent for the error term in the lattice point counting problem on the first Heisenberg group

TL;DR

This paper proves that the error term in counting lattice points on the first Heisenberg group with a specific norm cannot be improved beyond an exponent of 1/2, confirming the optimality of previous bounds.

Contribution

The authors establish the sharp exponent for the error term in the lattice point counting problem on the first Heisenberg group with the Cygan-Korányi norm, resolving a longstanding question.

Findings

Error term exponent is exactly 1/2, matching the known upper bound.

The result confirms the optimality of previous estimates.

The lattice point counting problem is fully resolved for this case.

Abstract

We use classical methods from analytic number theory to resolve the lattice point counting problem on the first Heisenberg group, in the case where the gauge function is taken to be the Cygan-Kor$\acute{a}$nyi Heisenberg-norm $\mathcal{N}_{4,1}(z,w)=(|z|^{4}+w^{2})^{1/4}$. In this case, our main theorem establishes the estimate $\mathcal{E}(x)=\Omega_{\pm}(x^{\frac{1}{2}})$, where $\mathcal{E}(x)=\mathcal{S}(x)-\frac{\pi^{2}}{2}x$ is the error term arising in the lattice point counting problem, $\mathcal{S}(x)$ is given by $$\mathcal{S}(x)=\sum_{0\leq m^2+n^2<\, x}r_2(m)$$ and $r_2(m)=|\{a,b\in\mathbb{Z}:\,a^{2}+b^{2}=m\}|$ is the familiar sum of squares function. As a corollary, we deduce that the exponent $\frac{1}{2}$ in the upper-bound $\left|\mathcal{E}(x)\right|\ll x^{\frac{1}{2}}\log{x}$ obtained by Garg, Nevo & Taylor can not be improved and is thus best possible, thereby resolving the lattice point counting problem for the case in hand.