0手术的复杂性
Complexity of equal 0-surgeries
AI总结：
研究通过分析具有相同0手术的结，探讨其切开性和4维光滑庞加莱猜想，并计算其生成数及迹等价性。
摘要：
我们说两个结是朋友，如果它们有相同的0手术。两个具有不同切开性状态的朋友会为4维光滑庞加莱猜想提供反例。在这里，我们创建了一个包含所有具有小交叉数c和四面体复杂度t的朋友的名录，并计算了它们的光滑4-生成数。特别是，我们计算了所有朋友K和K'中c(K)+c(K')和t(K)+t(K')的最小值。在过程中，我们分类了所有交叉数不超过15的素结的0手术。此外，我们确定了我们名录中许多朋友的迹是否等价。为此，我们开发了一种新的障碍，该障碍来自双曲结的对称例外斜率，用于判断两个迹是否同胚。这足以确定所有迹不同胚的朋友K和K'中c(K)+c(K')的最小值。
英文摘要：
We say that two knots are friends if they share the same 0-surgery. Two friends with different sliceness status would provide a counterexample to the 4-dimensional smooth Poincaré conjecture. Here we create a census of all friends with small crossing numbers c and tetrahedral complexities t, and compute their smooth 4-genera. In particular, we compute the minimum of c(K)+c(K') and of t(K)+t(K') among all friends K and K'. Along the way, we classify all 0-surgeries of prime knots of at most 15 crossings. Moreover, we determine for many friends in our census if their traces are equivalent or not. For that, we develop a new obstruction for two traces being homeomorphic coming from symmetry-exceptional slopes of hyperbolic knots. This is enough to also determine the minimum value of c(K)+c(K') among all friends K and K' whose traces are not homeomorphic.
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。
