关于海森堡群上高斯圆问题的类比
数论
2019-12-16 v1
摘要
我们考虑估计误差项E q ( x ) = ∣ Z 2 q + 1 ∩ δ x B ∣ − vol ( B ) x 2 q + 2 \mathcal{E}_{q}(x)=\big|\mathbb{Z}^{2q+1}\cap\delta_{x}\mathcal{B}\big|-\textit{vol}\big(\mathcal{B}\big)x^{2q+2} E q ( x ) = Z 2 q + 1 ∩ δ x B − vol ( B ) x 2 q + 2 E q ( x ) \mathcal{E}_{q}(x) E q ( x ) q ≥ 3 q\geq3 q ≥ 3 ∣ E q ( x ) ∣ ≪ x 2 q − 2 / 3 |\mathcal{E}_{q}(x)|\ll x^{2q-2/3} ∣ E q ( x ) ∣ ≪ x 2 q − 2/3 E q ( x ) \mathcal{E}_{q}(x) E q ( x ) x 2 q − 1 x^{2q-1} x 2 q − 1 Ω \Omega Ω E q ( x ) = Ω ( x 2 q − 1 ( log x ) 1 / 4 ( log log x ) 1 / 8 ) \mathcal{E}_{q}(x)=\Omega\big(x^{2q-1}\big(\log{x}\big)^{1/4}\big(\log{\log{x}}\big)^{1/8}\big) E q ( x ) = Ω ( x 2 q − 1 ( log x ) 1/4 ( log log x ) 1/8 ) q ≥ 3 q\geq3 q ≥ 3 κ q = sup { α > 0 : ∣ E q ( x ) ∣ ≪ x 2 q + 2 − α } ≥ 8 3 \kappa_{q}=\sup\big\{\alpha>0:\big|\mathcal{E}_{q}(x)\big|\ll x^{2q+2-\alpha}\big\}\geq\frac{8}{3} κ q = sup { α > 0 : E q ( x ) ≪ x 2 q + 2 − α } ≥ 3 8 κ q = 3 \kappa_{q}=3 κ q = 3
引用
@article{arxiv.1912.06263,
title = {On an Analogue Of the Gauss Circle Problem For the Heisenberg Groups},
author = {Yoav A. Gath},
journal= {arXiv preprint arXiv:1912.06263},
year = {2019}
}
备注
49 pages. Comments are welcome