桥式三等分诱导的立方体图形

IF 0.6 3区 数学 Q3 MATHEMATICS
Jeffrey Meier, Abigail Thompson, Alexander Zupan
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引用次数: 0

摘要

$4$球中的每个内嵌曲面 $\mathcal{K}$ 都有一个桥三段论,即把 $(S^4, \mathcal{K})$ 分解成三个简单的部分。在这种情况下,曲面 $\mathcal{K}$ 是由一个内嵌的 1 复数决定的,这个 1 复数被称为桥式三等分的 1$骨架。作为一个抽象图,1-skeleton 是一个立方图 $\Gamma$,它继承了自然的泰特着色(Tait coloring),即 $\Gamma$ 边集的三色着色,使得每个顶点都与三种颜色的边相连。在本文中,我们逆转了这种关联:我们证明了每个泰特色立方图都与对应于无结曲面的桥式三剖面的 1 骨架同构。当曲面不可定向时,我们证明这种嵌入存在于所有可能的法欧拉数中。作为推论,每个打结曲面的三平面图都可以通过交叉变化和内部莱德米斯特移动转换为非打结曲面的三平面图。用于证明主定理的工具包括两个新的桥式三等分运算,即交叉求和和管式运算,这两个运算可能会引起独立的兴趣。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
Cubic graphs induced by bridge trisections
Every embedded surface $\mathcal{K}$ in the $4$-sphere admits a bridge trisection, a decomposition of $(S^4, \mathcal{K})$ into three simple pieces. In this case, the surface $\mathcal{K}$ is determined by an embedded 1‑complex, called the $1$-skeleton of the bridge trisection. As an abstract graph, the 1‑skeleton is a cubic graph $\Gamma$ that inherits a natural Tait coloring, a 3‑coloring of the edge set of $\Gamma$ such that each vertex is incident to edges of all three colors. In this paper, we reverse this association: We prove that every Tait-colored cubic graph is isomorphic to the 1‑skeleton of a bridge trisection corresponding to an unknotted surface. When the surface is nonorientable, we show that such an embedding exists for every possible normal Euler number. As a corollary, every tri-plane diagram for a knotted surface can be converted to a tri-plane diagram for an unknotted surface via crossing changes and interior Reidemeister moves. Tools used to prove the main theorem include two new operations on bridge trisections, crosscap summation and tubing, which may be of independent interest.
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来源期刊
CiteScore
1.40
自引率
0.00%
发文量
9
审稿时长
6.0 months
期刊介绍: Dedicated to publication of complete and important papers of original research in all areas of mathematics. Expository papers and research announcements of exceptional interest are also occasionally published. High standards are applied in evaluating submissions; the entire editorial board must approve the acceptance of any paper.
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