Timeline for Hodge structure of abelian surfaces
Current License: CC BY-SA 3.0
8 events
| when toggle format | what | by | license | comment | |
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| Jan 13, 2014 at 12:26 | comment | added | abx | It is true -- it is just an elementary fact about cohomology of surfaces. There might be an isometry, but it is certainly not $p^*$. | |
| Jan 13, 2014 at 11:47 | comment | added | Li Yutong | This is certainly not true - abelian variety and its dual are derived equivalent and hence when $G=K(L)$ we should have the isometry by Orlov & Mukai... | |
| Jan 12, 2014 at 13:05 | comment | added | abx | $p^*$ is not an isometry unless $G$ is trivial. I suggest that you begin by a good book on the subject, for instance Barth et al. | |
| Jan 12, 2014 at 10:25 | history | edited | Li Yutong |
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| Jan 12, 2014 at 10:04 | history | edited | Li Yutong | CC BY-SA 3.0 |
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| Jan 12, 2014 at 9:40 | comment | added | Li Yutong | (1)$\phi = p^*$ (2) $G$ may not be abelian (3) I don't know what is the role of the polarization in my problem, but that's the (only) information I know about the abelian surface. | |
| Jan 12, 2014 at 6:54 | comment | added | abx | What you ask is not clear at all. Is $\phi=p^*$, where $p:A\rightarrow A/G$ is the quotient map? Is $G$ abelian? What is the role of the polarization in your question? | |
| Jan 12, 2014 at 0:36 | history | asked | Li Yutong | CC BY-SA 3.0 |