Timeline for Exterior derivative on almost complex manifolds
Current License: CC BY-SA 2.5
9 events
| when toggle format | what | by | license | comment | |
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| Nov 23, 2010 at 16:16 | comment | added | solbap | hmm yeah it seems I've just reversed the orientation of $\mathbb{C}^n$. I guess I was just thinking that the identity map $(\mathbb{C}^n,i)→(\mathbb{R}^{2n},J)$ doesn't satisfy $i∘D(id)=D(id)∘J$,so this doesn′t give a holomorphic chart for $\mathbb{R}^{2n}$,but I guess $\overline{\mathbb{C}^n}$ does. | |
| Nov 23, 2010 at 15:12 | answer | added | Spiro Karigiannis | timeline score: 10 | |
| Nov 23, 2010 at 15:01 | comment | added | Spiro Karigiannis | @solbap: Your example doesn't work. You just replaced the original J by its negative, which is still an integrable complex structure. Francesco and Eric gave the correct reason. | |
| Nov 23, 2010 at 14:14 | comment | added | solbap | For an explicit example of $d \ne \partial + \bar \partial$ consider $\mathbb{C}^n$ as $\mathbb{R}^{2n}$ with coordinates $x_i, y_i$. The tan. sp. has basis $\partial/\partial x_i, \partial/\partial y_i$. The usual complex structure of $\mathbb{C}^n$ uses the almost complex structure $i(\partial/\partial x_i) = \partial/\partial y_i$ (this determines $i$ on the other basis vectors using $i^2 = -1$. If instead you used another complex structure $J(\partial/\partial x_i) = -\partial/\partial y_i$ and then proceeded to use $J$ to define $(p,q)$ forms then $d \ne \partial + \bar \partial$ | |
| Nov 23, 2010 at 13:44 | vote | accept | miramo | ||
| Nov 23, 2010 at 13:40 | vote | accept | miramo | ||
| Nov 23, 2010 at 13:44 | |||||
| Nov 23, 2010 at 13:28 | comment | added | Francesco Polizzi | In your computation of $d \omega$ you are assuming that $M$ is complex, since you are using holomorphic coordinates $z_i$. For a general almost complex manifold it makes no sense to write $\partial /\partial z$ and $\partial / \partial \bar{z}$, just because no holomorphic coordinates are available. There is just a complex structure on the tangent space, but to write $f(z)$ you need such a structure to be integrable. | |
| Nov 23, 2010 at 13:27 | answer | added | Eric Zaslow | timeline score: 15 | |
| Nov 23, 2010 at 11:25 | history | asked | miramo | CC BY-SA 2.5 |