Simplified SFT moduli spaces for Legendrian links

IF 0.6 3区 数学 Q3 MATHEMATICS
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.
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。
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来源期刊
CiteScore
1.30
自引率
0.00%
发文量
0
审稿时长
>12 weeks
期刊介绍: Publishes high quality papers on all aspects of symplectic geometry, with its deep roots in mathematics, going back to Huygens’ study of optics and to the Hamilton Jacobi formulation of mechanics. Nearly all branches of mathematics are treated, including many parts of dynamical systems, representation theory, combinatorics, packing problems, algebraic geometry, and differential topology.
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