相对线性扩展比的重叠构造
An Overlap Construction for Relative Linear Extension Ratios
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中文总结 AI 辅助
本文通过重叠构造方法，证明了在d≥(1+ε)c条件下，相对线性扩展比的最小元素数目的上界，并去除了陈和帕克结果中的因子3。
中文摘要 AI 辅助
Chan和Pak引入了相对线性扩展比$\rho(P,x)=e(P)/e(P - x)$，其中$e(P)$是有限偏序集$P$的线性扩展数，并设$\nu(c,d)$是实现$\rho(P,x)=d/c$的偏序集的最少元素数。他们证明了对于$d\geq3c$，$\nu(c,d)\leq d/c + O(\log d\log\log d)$，并询问该假设$d\geq3c$是否可放宽到$d\geq(1 + \varepsilon)c$或去除。我们证明了这个问题的固定间隙形式：对于每个固定的$\varepsilon>0$，当$d\geq(1 + \varepsilon)c$时，$\nu(c,d)\leq\frac{d}{c}+O_{\varepsilon}(\log d\log\log d)$，并且一旦$d\geq2c$，隐含常数是绝对的。新要素是单元素重叠构造：若$x$在$P$中是极小元，$y$在$Q$中是极小元，则存在偏序集$R$，$|R| = |P| + |Q| - 1$及元素$z$使得$\rho(R,z)=\rho(P,x)+\rho(Q,y)-1$。结合Chan和Pak的连分数构造以及Rukavishnikova对部分商和的尾部界，去除了他们结果中的因子3。我们还表明固定间隙假设对于此构造本质上是最优的。在$1 < d/c < 2$范围内，设$h = d - c$，该构造能证明的大小界至少为$\lfloor c/h\rfloor$，所以只有当$h$至少为$c/(\log c\log\log c)$阶时，该方法才达到所述误差项。去除该假设的剩余障碍是部分商和的短区间问题，我们对此进行了描述。论证的演绎部分已用Lean证明助手进行了检查。
英文摘要
Chan and Pak introduced the relative linear extension ratio $ρ(P,x)=e(P)/e(P-x)$, where $e(P)$ is the number of linear extensions of a finite poset $P$, and let $ν(c,d)$ be the least number of elements of a poset that realizes $ρ(P,x)=d/c$. They proved that $ν(c,d)\le d/c+O(\log d\log\log d)$ for $d\ge 3c$, and asked whether the hypothesis $d\ge 3c$ can be relaxed to $d\ge(1+\varepsilon)c$ or removed. We prove the fixed-gap form of this question: for every fixed $\varepsilon>0$, $ν(c,d)\le \frac{d}{c}+O_{\varepsilon}(\log d\log\log d)$ whenever $d\ge(1+\varepsilon)c$, and the implied constant is absolute once $d\ge 2c$. The new ingredient is a one-element overlap construction: if $x$ is minimal in $P$ and $y$ is minimal in $Q$, then there is a poset $R$ with $|R|=|P|+|Q|-1$ and an element $z$ such that $ρ(R,z)=ρ(P,x)+ρ(Q,y)-1$. Together with the continued-fraction construction of Chan and Pak and Rukavishnikova's tail bound for sums of partial quotients, this removes the factor $3$ in their range. We also show that the fixed-gap hypothesis is essentially optimal for this construction. In the range $1 < d/c < 2$, with $h=d-c$, the size bound the construction can certify is at least $\lfloor c/h\rfloor$, so the method reaches the stated error term only when $h$ is at least of order $c/(\log c\log\log c)$. The remaining obstruction to removing the hypothesis is a short-interval problem for sums of partial quotients, which we describe. The deductive part of the argument has been checked with the Lean proof assistant.
