Timeline for Unique symplectic form in an adapted complex structure
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
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| Mar 28, 2016 at 16:30 | history | edited | Ben McKay | CC BY-SA 3.0 |
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| Oct 14, 2012 at 10:15 | vote | accept | Joan | ||
| Oct 9, 2012 at 11:00 | answer | added | Brendan Foreman | timeline score: 1 | |
| Oct 8, 2012 at 18:34 | comment | added | Eugene Lerman | For lagrangian tori in Kaehler toric manfolds yes and yes. See Abreu's paper I pointed to you yesterday: arxiv.org/abs/math/0004122 Another good place is Burns and Guillemin, Potential functions and actions of tori on Kaehler manifolds (arxiv.org/abs/math/0302350). | |
| Oct 8, 2012 at 15:22 | comment | added | Joan | is there any reference ? | |
| Oct 8, 2012 at 15:21 | comment | added | Joan | does this set of metrics have a manifold structure or something like that ? | |
| Oct 8, 2012 at 11:33 | comment | added | Eugene Lerman | Well, have you looked at the reference I gave you? There the real torus is Lagrangian. More concretely, take $M=S^1$. Then $T^*M$ is $\mathbb{C} -\{0\}$. There are many different Kaehler structures on $T^*M$. You can embed it in $\mathbb{C}$ with a flat metric. Or you can embed it in $\mathbb{C}P^1$ with the Fubini-Study metric. Basically toric manifolds provide plenty of examples. | |
| Oct 8, 2012 at 7:06 | comment | added | Joan | But I am asking not only the forms compatible with $J$ but also that fixes $M$ as a Lagrangian manifold! | |
| Oct 7, 2012 at 11:47 | comment | added | Eugene Lerman | If you start with a Riemannian manifold $(M, g_0)$ then $T^*M = TM$ acquires a natural Riemannian metric $g$ from $g_0$; this is what I thought you meant. Since this is not the case, are you asking: "I have a complex manifold $(N, J)$. What are all symplectic forms compatible with $J$." ? If this is the case, I believe there are lots of these forms. See arxiv.org/abs/math/0004122 which applies to your question when $M$ is a torus (so $T^*M = (C^\times)^n$). | |
| Oct 7, 2012 at 4:24 | comment | added | Joan | But I didn't fix $g$. $g$ comes with the symplectic form. | |
| Oct 6, 2012 at 21:41 | comment | added | Eugene Lerman | On a Kaehler manifold, two out of three structures --- $J$, $g$, $\omega$ --- determine the third. Since you are fixing $g$ and $J$, there doesn't seem to be any choice for $\omega$. | |
| Oct 6, 2012 at 6:07 | history | asked | Joan | CC BY-SA 3.0 |