Simplified SFT moduli spaces for Legendrian links

Pub Date : 2023-09-28 DOI:10.4310/jsg.2023.v21.n2.a2
Russell Avdek
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引用次数: 1

Abstract

We study moduli spaces $\mathcal{M}$ of holomorphic maps $U$ to $\mathbb{R}^4$ with boundaries on the Lagrangian cylinder over a Legendrian link $\Lambda \subset (\mathbb{R}^3, \xi_{std})$. We allow our domains, $\dot{\Sigma}$ , to have non-trivial topology in which case $\mathcal{M}$ is the zero locus of an obstruction function $\mathcal{O}$, sending a moduli space of holomorphic maps in $\mathbb{C}$ to $H^1 (\dot{\Sigma})$. In general, $\mathcal{O}^{-1} (0)$ is not combinatorially computable. However after a Legendrian isotopy $\Lambda$ can be made left-right-simple, implying that any $U$ 1) of index $1$ is a disk with one or two positive punctures for which $\pi_\mathbb{C} \circ U$ is an embedding. 2) of index $2$ is either a disk or an annulus with $\pi_\mathbb{C} \circ U$ simply covered and without interior critical points. Therefore any SFT invariant of $\Lambda$ is combinatorially computable using only disks with $\leq 2$ positive punctures.
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Legendrian链路的简化SFT模空间
我们学习模空间 $\mathcal{M}$ 全纯映射的 $U$ 到 $\mathbb{R}^4$ 拉格朗日柱体上的边界 $\Lambda \subset (\mathbb{R}^3, \xi_{std})$. 我们允许我们的域, $\dot{\Sigma}$ ,在这种情况下具有非平凡拓扑 $\mathcal{M}$ 零点轨迹是一个阻碍函数吗 $\mathcal{O}$的全纯映射的模空间 $\mathbb{C}$ 到 $H^1 (\dot{\Sigma})$. 一般来说, $\mathcal{O}^{-1} (0)$ 不是组合可计算的。然而,经过一个传奇的同位素 $\Lambda$ 可以使左右简单,暗示任何 $U$ 1)指数 $1$ 椎间盘有一个或两个阳性穿刺是什么原因 $\pi_\mathbb{C} \circ U$ 是一种嵌入。2)指数 $2$ 是圆盘还是环 $\pi_\mathbb{C} \circ U$ 简单覆盖,没有内部临界点。的任意SFT不变量 $\Lambda$ 仅使用带有的磁盘是否可组合计算 $\leq 2$ 阳性穿刺。
本文章由计算机程序翻译,如有差异,请以英文原文为准。
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