Timeline for Do all simple factors of jacobians of curves come from correspondences?
Current License: CC BY-SA 3.0
14 events
| when toggle format | what | by | license | comment | |
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| S Aug 6, 2018 at 3:03 | history | bounty ended | CommunityBot | ||
| S Aug 6, 2018 at 3:03 | history | notice removed | CommunityBot | ||
| Jul 30, 2018 at 21:28 | vote | accept | Maarten Derickx | ||
| Jul 29, 2018 at 14:25 | answer | added | R. van Dobben de Bruyn | timeline score: 11 | |
| S Jul 29, 2018 at 1:34 | history | bounty started | Maarten Derickx | ||
| S Jul 29, 2018 at 1:34 | history | notice added | Maarten Derickx | Draw attention | |
| Oct 21, 2015 at 1:50 | comment | added | t3suji | @Maarten Derickx: No, I don't, but it does not seem particularly hard. Fix base points $e\in E$, $c\in C$, we need to show that for any line bundle $L$ on $E\times C$, there exist $n_1$ and $n_2$ such that the line bundle $L(n_1(\{e\}\times C)+n_2(E\times \{c\}))$ is of the form $O(D)$ for smooth curve $D\subset E\times C$ (which then gives a correspondence between the two curves). However, $O((\{e\}\times C)+(E\times \{c\}))$ is ample, therefore, $L(n(\{e\}\times C)+n(E\times \{c\}))$ is very ample for $n\gg 0$, and the claim follows from Bertini's Theorem. | |
| Oct 20, 2015 at 21:30 | comment | added | Maarten Derickx | @t3suji In my case the Abelian variety that I try to get as image is simple so any curve will generate it. Actually in your argument you might even take the morphism $E=C$ and $J(C) \to J(C)$ a morphism that maps $J(C)$ onto the simple factor. Do you have a reference for the first fact? | |
| Oct 20, 2015 at 21:20 | comment | added | Maarten Derickx | @roysmith: I have no idea wether this is something plausible. I just know one important case, namely that it is true for modular curves if one does it over $\mathbb Q$ instead of over $\mathbb C$ (at least if one does not demand that correspondences are given by irreducible curves). So a related question is: how big is the subring of ${\rm End}_{\mathbb C} J(C)$ generated by correspondences from $C$ to itself? | |
| Oct 20, 2015 at 5:22 | comment | added | t3suji | P.S. A reference for the second fact: Theorem III.10.1 of Milne's jmilne.org/math/CourseNotes/AV.pdf (the section has the telling title `Abelian varieties are quotients of Jacobian varieties :) And the first fact should follow from Bertini's Theorem... | |
| Oct 20, 2015 at 5:14 | comment | added | t3suji | It does seem pretty plausible. First, any morphism $J(E)\to J(C)$ comes from a line bundle on $C\times E$, which in turns comes from a divisor in $C\times E$... Which is kind of a curve (probably singular, but we can resolve this, also probably reducible, which is more annoying). Secondly, for any abelian variety $A$, we can find a curve $E\subset A$ that generates it (again, the obvious construction gives a reducible curve, so some argument would be required?), and then $A$ becomes a quotient of $J(E)$. Combining these two facts we'd get the required claim. | |
| Oct 20, 2015 at 3:30 | comment | added | roy smith | give me a start here. what makes this seem plausible? | |
| Oct 19, 2015 at 23:59 | history | edited | R.P. | CC BY-SA 3.0 |
correction: J(C) -> J(E), plus small stuff
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| Oct 19, 2015 at 22:30 | history | asked | Maarten Derickx | CC BY-SA 3.0 |