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Nearby Lagrangian fibers and Whitney sphere links

Published online by Cambridge University Press:  20 February 2018

Tobias Ekholm
Affiliation:
Department of Mathematics, Uppsala University, Box 480, 751 06 Uppsala, Sweden Institut Mittag-Leffler, Aurav. 17, 182 60 Djursholm, Sweden email tobias.ekholm@math.uu.se
Ivan Smith
Affiliation:
Centre for Mathematical Sciences, University of Cambridge, Wilberforce Road, CB3 0WB, UK email is200@cam.ac.uk
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Abstract

Let $n>3$ , and let $L$ be a Lagrangian embedding of $\mathbb{R}^{n}$ into the cotangent bundle $T^{\ast }\mathbb{R}^{n}$ of $\mathbb{R}^{n}$ that agrees with the cotangent fiber $T_{x}^{\ast }\mathbb{R}^{n}$ over a point $x\neq 0$ outside a compact set. Assume that $L$ is disjoint from the cotangent fiber at the origin. The projection of $L$ to the base extends to a map of the $n$ -sphere $S^{n}$ into $\mathbb{R}^{n}\setminus \{0\}$ . We show that this map is homotopically trivial, answering a question of Eliashberg. We give a number of generalizations of this result, including homotopical constraints on embedded Lagrangian disks in the complement of another Lagrangian submanifold, and on two-component links of immersed Lagrangian spheres with one double point in $T^{\ast }\mathbb{R}^{n}$ , under suitable dimension and Maslov index hypotheses. The proofs combine techniques from Ekholm and Smith [Exact Lagrangian immersions with a single double point, J. Amer. Math. Soc. 29 (2016), 1–59] and Ekholm and Smith [Exact Lagrangian immersions with one double point revisited, Math. Ann. 358 (2014), 195–240] with symplectic field theory.

Information

Type
Research Article
Copyright
© The Authors 2018 
Figure 0

Figure 1. The fronts of: $L$ (top), $\unicode[STIX]{x1D6E4}(L)$ (middle) and $C(L)$ (bottom).

Figure 1

Figure 2. Top: the evaluation point $1$ is far from the glued in holomorphic disk $v$, as in Lemma 3.9. Middle: the glued in holomorphic disk passes the evaluation point $1$, as in Lemma 3.10. Bottom: the passage as viewed from the holomorphic disk, where the evaluation point sweeps the whole boundary.