Timeline for How many Lagrangian submanifolds?
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Sep 5, 2013 at 0:32 | vote | accept | Hwang | ||
| Sep 5, 2013 at 0:32 | comment | added | Hwang | @Oldřich Spáčil. Thank you for the comment. I know Lagrangian intersection is not a topological issue. What I meant by classes was modulo Hamiltonian isotopy, not modulo homologous manifolds, so that self-intersection of nondisplaceable Lagrangian is not zero. | |
| Sep 4, 2013 at 15:51 | comment | added | Oldřich Spáčil | @Hwang Are you aware of displaceability issues of Lagrangians? Take for example the 2-sphere and consider two closed curves, the equator and some other closed curve $C$ which lies fully in the northern hemisphere. Under any Hamiltonian isotopy of the sphere, the image of the equator will intersect itself, but for the curve $C$ you can find an isotopy such that the image of $C$ is disjoint from $C$. Notice that both the equator and $C$ represent the same homology class - I'm trying to point out that intersections of Lagrangians are not a purely topological issue, there is some geometry hidden. | |
| Sep 4, 2013 at 12:27 | comment | added | Hwang | Thank you. I should've thought about it carefully before I post the question. For the second question, my naive hope was to find suitable classes to do intersection theory. It seems to me that people think about Lagrangian intersections, but intersection of two Lagrangians is not Lagrangian. So I was curious if we can consider larger classes modulo Hamiltonian isotopy containing Lagrangians. I have no idea whether this makes sense at all due to my ignorance. | |
| Sep 4, 2013 at 9:48 | history | answered | Robert Bryant | CC BY-SA 3.0 |