Timeline for Proving that a generic variety with ample canonical bundle has no automorphisms
Current License: CC BY-SA 3.0
6 events
| when toggle format | what | by | license | comment | |
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| May 13, 2013 at 13:09 | comment | added | Francesco Polizzi | I have added some further details | |
| May 13, 2013 at 13:08 | history | edited | Francesco Polizzi | CC BY-SA 3.0 |
added 611 characters in body; added 6 characters in body
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| May 12, 2013 at 23:41 | vote | accept | Jonathan | ||
| May 12, 2013 at 21:34 | comment | added | Francesco Polizzi | @Jonathan: 1. If you take the branch locus in a very ample linear system, by Bertini theorem the general element of the system will be smooth, hence the general cover will be smooth. 2. By the universal property of the Albanese map, the cover $\alpha \colon X \to A$ factors through the Albanese morphism $X \to \textrm{Alb}(X)$. Since $\deg \alpha =2$ and $X$ is of general type, the isogeny $\textrm{Alb}(X) \to A$ must be an isomorphism. 3. $X$ has non trivial deformations coming from the deformations of $A$ (and from those of the branch locus). | |
| May 12, 2013 at 10:19 | comment | added | Jonathan | I like this counterexample a lot. I do have some questions. 1. Is it really possible that $X$ is nonsingular. I thought $X$ would have cyclic quotient singularities above the singular locus of the branch locus. Or maybe this is where you use "degree two"? 2. This is not so important, but how do you show that your double cover really equals the albanese map. Is it not possibly necessary to compose with an isogeny of A to really get the albanese map? 3. Can you give rigid examples of your $X$, and examples where $X$ has infinitely many deformations? Thanks again! | |
| May 12, 2013 at 8:37 | history | answered | Francesco Polizzi | CC BY-SA 3.0 |