Timeline for Fibrations with isomorphic fibers, but not Zariski locally trivial
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Feb 25, 2014 at 17:56 | comment | added | Alex Degtyarev | @Brenin Yes, constant $j$-invariant. No, this is not birational to product. If the fiber has nontrivial automorphism, the monodromy $\pi_1(\text{base})\to\operatorname{Aut}(\text{fiber})$ may be nontrivial. I don't see how this can be undone birationally. | |
| Feb 25, 2014 at 15:09 | comment | added | Brenin | @AlexDegtyarev: Just to be sure: is "isotrivial" a synonym of "having isomorphic fibers", isn't it? This also means, for the family, to be birational to the product family (I guess). | |
| Feb 23, 2014 at 21:22 | comment | added | Alex Degtyarev | The key word is isotrivial. They are easily constructed as quotients of $(\text{disk})\times(\text{elliptic curve})$. | |
| Feb 23, 2014 at 21:21 | comment | added | Damian Rössler | The fibres of an elliptic surface are not isomorphic to each other in general. | |
| Feb 23, 2014 at 17:50 | history | answered | Alex Degtyarev | CC BY-SA 3.0 |