Timeline for Two questions about counter example for Torelli theorem for hyperkahler manifolds
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jul 4, 2014 at 11:53 | vote | accept | igor guedz | ||
| Jul 3, 2014 at 20:48 | answer | added | Misha Verbitsky | timeline score: 3 | |
| Jul 3, 2014 at 14:24 | comment | added | igor guedz | yes! thak you, i think you're right, because given the resolution of singularities $Hilb^3(T)\rightarrow Sym^3(T)$ we have that generally the fiber on the singular locus is a $\mathbb{P}^1$, so intersecting with $Ker(\alpha)$ we can say $F$ is bimeromorphic to a $\mathbb{P}^1$-bundle on $T$, which of course has $T$ as Albanese variety! | |
| Jul 3, 2014 at 14:17 | review | First posts | |||
| Jul 3, 2014 at 14:41 | |||||
| Jul 3, 2014 at 14:16 | comment | added | Jason Starr | Regarding (1),it looks to me like $\Sigma$ is the image of an embedding $T\to \overline{K}^2(T)$ sending $p$ to $2(\underline{-p}) + \underline{2p}$. Also, $F$ seems to be bimeromorphic to a projective space bundle over $\Sigma$. Since the Albanese variety is a bimeromorphic invariant, that would imply that the map from $F$ to $\Sigma$ is (equivalent to) the Albanese morphism for $F$. | |
| Jul 3, 2014 at 14:01 | history | asked | igor guedz | CC BY-SA 3.0 |