Timeline for Hodge decomposition of a symplectic form.
Current License: CC BY-SA 3.0
17 events
| when toggle format | what | by | license | comment | |
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| S Dec 18, 2011 at 8:00 | vote | accept | Mirjana | ||
| S Dec 18, 2011 at 8:00 | vote | accept | Mirjana | ||
| S Dec 18, 2011 at 8:00 | |||||
| Dec 18, 2011 at 7:56 | vote | accept | Mirjana | ||
| S Dec 18, 2011 at 8:00 | |||||
| Dec 18, 2011 at 7:56 | vote | accept | Mirjana | ||
| Dec 18, 2011 at 7:56 | |||||
| Dec 18, 2011 at 7:56 | vote | accept | Mirjana | ||
| Dec 18, 2011 at 7:56 | |||||
| Dec 18, 2011 at 7:49 | vote | accept | Mirjana | ||
| Dec 18, 2011 at 7:56 | |||||
| Dec 8, 2011 at 2:43 | history | edited | Allen Knutson | CC BY-SA 3.0 |
corrected English
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| Dec 7, 2011 at 16:06 | comment | added | Tim Perutz | You're right - they do use that term - but all the same, their notation makes clear that they mean the type decomposition. On a Kaehler manifold, the Hodge and type decompositions are compatible (the harmonic forms decompose into types) - but that is not the setting of this proposition. | |
| Dec 7, 2011 at 15:48 | comment | added | Mirjana | @Tim In Proposition 4 it is explicitly written that it is the Hodge decomposition... | |
| Dec 7, 2011 at 14:00 | comment | added | Tim Perutz | Mirjana: Ah, so I misunderstood too. But what metric do you want to use for the Hodge theory? Note that Prop. 4 of the paper of Alekseevsky et al is about the type decomposition w.r.t. $J$, not the Hodge decomposition. [You've been leaving comments, but please edit the question to make it clearer.] | |
| Dec 7, 2011 at 5:52 | comment | added | Mirjana | @Tim, "how you know that such a complex manifold actually has a Hodge decomposition" I thought that every symplectic form can be written as a direct sum of a closed, coclosed and harmonic form? | |
| Dec 6, 2011 at 15:30 | answer | added | Spiro Karigiannis | timeline score: 1 | |
| Dec 6, 2011 at 15:23 | comment | added | Tim Perutz | Mirjana, I guess you are referring to the notion of "special symplectic manifold" introduced by Alekseevsky et al as a variant of the more popular "special Kaehler" geometry. It would be good to edit your question to include a full definition and references, explaining how you know that such a complex manifold actually has a Hodge decomposition. The need for this level of detail is illustrated by the fact that Spiro, an expert on manifolds with special holonomy, does not understand what you are asking. | |
| Dec 6, 2011 at 14:20 | history | edited | Spiro Karigiannis |
edited tags
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| Dec 6, 2011 at 13:28 | answer | added | yael fregier | timeline score: 0 | |
| Dec 6, 2011 at 12:24 | comment | added | Spiro Karigiannis | What exactly are you asking? What does a special symplectic manifold mean? Do you mean a Kaehler manifold? The standard Hodge decomposition does not make sense unless you have a Riemannian metric. Do you mean some other, less common, Hodge decomposition? If you mean the usual Hodge decomposition on a Kaehler manifold, then the symplectic (Kaehler) form is already harmonic, so it does not decompose further. | |
| Dec 6, 2011 at 10:52 | history | asked | Mirjana | CC BY-SA 3.0 |