Timeline for Lagrangian submanifolds in $T^\ast S^n$
Current License: CC BY-SA 3.0
10 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 8, 2014 at 16:46 | answer | added | Nikolaki | timeline score: 3 | |
| Aug 4, 2014 at 12:51 | comment | added | Daniel Pomerleano | @ Acky Right I assume that so that we can use compactly supported Hamiltonian isotopies to possibly deform L to a cotangent fiber. | |
| Aug 3, 2014 at 11:20 | comment | added | Acky | @DanielPomerleano I see, thanks. By "agree with a cotangent fiber" in your assumption of $L$ above, you mean $L$ agrees with a cotangent fiber at infinity? | |
| Aug 3, 2014 at 1:28 | comment | added | Daniel Pomerleano | @ Acky It follows from Koszul duality. Basically, assume M is simply connected. Then we have a fully faithful functor from perfect $C_*(\Omega M)$ modules to homology finite modules over $C^*(M)$. Geometrically that functor can be thought of as taking Hom from the zero section to your given object. | |
| Jul 26, 2014 at 7:40 | comment | added | Acky | @Pomerleano How to show that $L$ is isomorphic to a cotangent fiber? If $L$ is isomorphic to a cotangent fiber, then $L$ has to generate the wrapped Fukaya category. Is this true only for cotangent bundle of spheres or for all cotangent bundles of a closed manifold? | |
| Jul 24, 2014 at 2:33 | comment | added | Daniel Pomerleano | This is not exactly your question, but may be of interest. If we assume our Lagrangian L to be conical at infinity and to agree with a cotangent fiber, then one can prove that in the wrapped Fukaya category, L is isomorphic to a cotangent fiber (shifted by some integer which depends on grading data). This is similar to what happens in recent attacks on the nearby Lagrangian problem. So, at first glance, it seems plausible to ask your question with, say, compactly supported Hamiltonian isotopies. | |
| Jul 22, 2014 at 8:08 | comment | added | Acky | @Pardon Yes, you're correct. | |
| Jul 22, 2014 at 7:57 | history | edited | John Pardon | CC BY-SA 3.0 |
added 24 characters in body
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| Jul 22, 2014 at 7:55 | comment | added | John Pardon | You probably had this in mind when asking, but one should probably add some assumptions about the isotopy being well-behaved near infinity. Otherwise you can just scale $L$ outwards in the fiber direction (this stays Lagrangian) and it will approach a cotangent fiber. | |
| Jul 22, 2014 at 7:46 | history | asked | Acky | CC BY-SA 3.0 |