Timeline for What are the compact Lagrangian submanifolds of a twisted cotangent bundle?
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11 events
| when toggle format | what | by | license | comment | |
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| Dec 10, 2017 at 19:43 | vote | accept | MBIS | ||
| Dec 9, 2017 at 0:21 | answer | added | Nikolaki | timeline score: 4 | |
| Nov 28, 2017 at 13:23 | comment | added | MBIS | @YHBKJ Ok, still sounds interesting. It is definitely a different kind of answer than what I was expecting :). Thanks. | |
| Nov 28, 2017 at 13:05 | comment | added | YHBKJ | @MadsR.Bisgaard It's only a prediction, which tells you which kind of monotone Lagrangians are not allowed, but does not say anything about existence existence. Moreover, it's usually not easy to classify $A_\infty$-modules if the endomorphism algebra of a cotangent fiber is not formal and trivially graded. | |
| Nov 28, 2017 at 9:23 | comment | added | MBIS | @YHBKJ thanks! This sounds interesting. I will definitely have a look. | |
| Nov 24, 2017 at 21:34 | comment | added | YHBKJ | After turning on the magnetic potential, it should still be true that a cotangent fiber generates the wrapped Fukaya category $\mathcal{W}(T^\ast N,\omega_\sigma)$, see the introduction part of the following paper: arxiv.org/pdf/1201.5880.pdf. Because of this, the classification of monotone Lagrangians is equivalent to the problem of classifying proper $A_\infty$-modules over the endomorphism algebra of a cotangent fiber. | |
| Nov 24, 2017 at 16:14 | comment | added | MBIS | @ThomasRot Thanks. I agree that as long as $\sigma$ is exact it is quite easy to construct closed Lagrangians. One way to look at this is that $\omega_{\sigma}$ is an exact perturbation of $\omega$, so it should be symplectomorphic to it by Moser's argument. The question I am interested in is the case when $\sigma$ is not exact. In this case it doesn't seem so easy to me. | |
| Nov 24, 2017 at 12:58 | comment | added | Thomas Rot | I would start with considering the graphs of one forms. For $\sigma=0$ the graph of a one-form is Lagrangian iff it is closed. If $\sigma$ is exact, you can see that the condition that the graph is lagrangian is exactly the condition that the one-form is a primitive of $\sigma$. These Lagrangians are of course all diffeomorphic to the base. | |
| Nov 24, 2017 at 10:13 | history | edited | MBIS | CC BY-SA 3.0 |
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| Nov 24, 2017 at 8:45 | history | edited | MBIS | CC BY-SA 3.0 |
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| Nov 23, 2017 at 19:45 | history | asked | MBIS | CC BY-SA 3.0 |