arXiv 2101.007482021-01-04math.CO

$\mathbb{F}_q^d$上距离图的任意长圈

Cycles of arbitrary length in distance graphs on $\mathbb{F}_q^d$

Alex Iosevich, Gail Jardine, Brian McDonald

摘要：

对于$E \subset {\Bbb F}_q^d$，$d \ge 2$，其中${\Bbb F}_q$是具有$q$元素的有限域，我们考虑距离图${\mathcal G}^{dist}_t(E)$，$t \not=0$，其中顶点是$E$的元素，如果$||x-y|| \equiv {(x_1-y_1)}^2+\dots+{(x_d-y_d)}^2=t$，则两个顶点$x$，$y$由一条边连接.我们证明，如果$|E| \ge C_k q^{\frac{d+2}{2}}$，那么${\mathcal G}^{dist}_t(E)$包含一个统计正确的周期长度$k$的数量。我们还将考虑点积图${\mathcal G}^{prod}_t(E)$，$t \not=0$，其中顶点是$E$的元素，并且如果$x \cdot y \equiv x_1y_1+\dots+x_dy_d=t$，则两个顶点$x$，$y$由边连接。在这种情况下，我们得到了类似的结果，使用更复杂的方法所必需的事实，即功能$x \cdot y$不是平移不变的。对于足够长的周期，指数$\frac{d+2}{2}$得到改善。

英文摘要：

For $E \subset {\Bbb F}_q^d$, $d \ge 2$, where ${\Bbb F}_q$ is the finite field with $q$ elements, we consider the distance graph ${\mathcal G}^{dist}_t(E)$, $t \not=0$, where the vertices are the elements of $E$, and two vertices $x$, $y$ are connected by an edge if $||x-y|| \equiv {(x_1-y_1)}^2+\dots+{(x_d-y_d)}^2=t$. We prove that if $|E| \ge C_k q^{\frac{d+2}{2}}$, then ${\mathcal G}^{dist}_t(E)$ contains a statistically correct number of cycles of length $k$. We are also going to consider the dot-product graph ${\mathcal G}^{prod}_t(E)$, $t \not=0$, where the vertices are the elements of $E$, and two vertices $x$, $y$ are connected by an edge if $x \cdot y \equiv x_1y_1+\dots+x_dy_d=t$. We obtain similar results in this case using more sophisticated methods necessitated by the fact that the function $x \cdot y$ is not translation invariant. The exponent $\frac{d+2}{2}$ is improved for sufficiently long cycles.

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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英文摘要：

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