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

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.