发表机构
Colorado State University; Sandia National Laboratories（科罗拉多州立大学; 桑迪亚国家实验室）
机构由 AI 辅助整理，请以论文原文为准。AI 中文总结
本文精确计算了单位球面上六个独立均匀随机点形成的闭六边形为三叶结的概率，通过凸包类型分类和球极投影转化为Sylvester型问题，得到三叶结概率为1/(16π²)。
AI 中文摘要
我们确定了将单位球面上均匀分布的六个独立点按循环方式连接得到的闭六边形多边形为纽结的精确概率。唯一可能的非平凡纽结是三叶结。几乎必然地，这六个点的凸包具有两种单纯组合类型之一：组合正则八面体和组合非正则八面体。我们证明了，对于组合正则八面体类型的凸包的六个顶点上的直线完全图，恰好存在一个无向三叶结哈密顿圈。另一方面，连接非正则八面体顶点的任何圈都不能产生三叶结。然后，我们利用球极投影将正则凸包类型的概率转化为平面beta-prime概率测度$d\mu(x,y)=\frac{dx dy}{\pi\\ (1+x^2+y^2)^2}$的Sylvester型问题。应用Stokes定理和Blaschke-Petkantschin公式，我们计算了随机三角形的期望平方$\mu$-含量。由此得到正则八面体概率为$\frac{15}{4\pi^2}$，三叶结概率为$\frac{1}{16\pi^2}$。
英文摘要
We determine the exact probability that the closed hexagonal polygon obtained by cyclically joining six independent points uniformly distributed on the unit sphere is knotted. The only possible nontrivial knot is a trefoil. Almost surely, the convex hull of the six points has one of two simplicial combinatorial types: a combinatorially regular octahedron and a combinatorially non-regular octahedron. We show that the straight-line complete graph on the six vertices of the hull of the combinatorially regular octahedral type admits exactly one unoriented trefoil Hamiltonian cycle. On the other hand, no cycle connecting vertices of the irregular octahedron can produce a trefoil. We then use stereographic projection to transform the probability of the regular hull type to a Sylvester-type problem for the planar beta-prime probability measure $dμ(x,y)=\frac{dx dy}{π (1+x^2+y^2)^2}$. Applying Stokes' theorem and the Blaschke-Petkantschin formula, we compute the expected squared $μ$-content of a random triangle. This yields a regular octahedral probability of $\frac{15}{4π^2}$ and a trefoil probability of $\frac{1}{16π^2}$.
Comments12 pages, 6 figures
