Timeline for Homology classes represented by $J$-holomorphic curves
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 8, 2013 at 1:07 | comment | added | JHM | @Sam: I was not thinking clearly. Of course, if $\Sigma$ is closed and $\omega$ is exact, then $\Sigma$ has (by Stokes theorem) zero area (w.r.t. the metric $\omega(J\cdot, \cdot)$). | |
| Apr 8, 2013 at 0:12 | comment | added | Nathaniel Bottman | For the record, as long as $\Sigma$ is closed and $(M,J,\omega)$ is almost-Kaehler the reason that $u_*[\Sigma]$ can't be torsion is the energy identity. | |
| Apr 7, 2013 at 22:53 | comment | added | Mark Gross | See mathoverflow.net/questions/110615/difference-of-curve-classes/… | |
| Apr 7, 2013 at 22:44 | history | edited | Hwang | CC BY-SA 3.0 |
deleted 1 characters in body
|
| Apr 7, 2013 at 22:39 | history | edited | Hwang | CC BY-SA 3.0 |
aa; added 1 characters in body; added 4 characters in body
|
| Apr 7, 2013 at 21:07 | comment | added | Sam Lisi | @J.Martel : If the form on $M$ is exact and $\Sigma$ is closed, then the curve is constant. Perhaps you are thinking of punctured curves? @Hwang: This is probably a stupid comment, but what do you mean by the free part? I thought the splitting wasn't natural. | |
| Apr 7, 2013 at 20:19 | comment | added | JHM | If the symplectic form on $M$ is exact, then any J-holomorphic curve is not necessarily a nontrivial $\mathbb{R}$-cycle. So i don't think your last sentence is correct. | |
| Apr 7, 2013 at 15:46 | history | asked | Hwang | CC BY-SA 3.0 |