arXivDaily arXiv每日学术速递 周一至周五更新
arXiv 1608.06398math.CO
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关于$\mathbb{F}_q^d$中单纯形个数的改进

An improvement on the number of simplices in $\mathbb{F}_q^d$

Thang Pham, Duc Hiep Pham, Anh Vinh Le

摘要:

设$\mathcal{E}$是$\mathbb{F}_q^d$中的一组点。Bennett、Hart、Iosevich、Pakianathan和Rudnev(2016)证明了,如果$|\mathcal{E}|\gg q^{d-\frac{d-1}{k+1}}$,则$\mathcal{E}$决定所有$k$ -单纯形的正比例。本文在$\mathcal{E}$是集合的笛卡儿积的情况下,改进了这个结果.更精确地说,我们证明了如果$\mathcal{E}$是集合与$q^{\frac{kd}{k+1-1/d}}=o(|\mathcal{E}|)$的笛卡儿积,则由$\mathcal{E}$确定的$k$ -单形的同余类的个数至少为$(1-o(1))q^{\binom{k+1}{2}}$,并且在某些情况下我们的结果是尖锐的.

英文摘要:

Let $\mathcal{E}$ be a set of points in $\mathbb{F}_q^d$. Bennett, Hart, Iosevich, Pakianathan, and Rudnev (2016) proved that if $|\mathcal{E}|\gg q^{d-\frac{d-1}{k+1}}$ then $\mathcal{E}$ determines a positive proportion of all $k$-simplices. In this paper, we give an improvement of this result in the case when $\mathcal{E}$ is the Cartesian product of sets. More precisely, we show that if $\mathcal{E}$ is the Cartesian product of sets and $q^{\frac{kd}{k+1-1/d}}=o(|\mathcal{E}|)$, the number of congruence classes of $k$-simplices determined by $\mathcal{E}$ is at least $(1-o(1))q^{\binom{k+1}{2}}$, and in some cases our result is sharp.

机构、总结、中文摘要均由AI分析而来,可能存在错误,请以论文正文为准。

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