Timeline for Elliptic curves on abelian surface
Current License: CC BY-SA 3.0
12 events
when toggle format | what | by | license | comment | |
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Nov 4, 2011 at 12:41 | comment | added | user5117 | Qfwfq: insert the word "abelian" before the word "subvarieties" in the quote. | |
Nov 4, 2011 at 12:01 | comment | added | Qfwfq | @rita: "An abelian variety has at most countably many subvarieties" - I'm missing something: what about the translates of a fixed subvariety? | |
Nov 4, 2011 at 11:56 | answer | added | Qfwfq | timeline score: 1 | |
Nov 4, 2011 at 11:47 | comment | added | Qfwfq | See also: mathoverflow.net/questions/21439/… | |
S Nov 4, 2011 at 7:50 | vote | accept | fds | ||
Nov 3, 2011 at 21:07 | comment | added | Qing Liu | @Rita: the same arguments hold over uncountable fields. | |
Nov 3, 2011 at 20:49 | comment | added | rita | @Qing Liu: your last statement is true at least over the complex numbers. An abelian variety has at most countably many subvarieties, hence by Baire's theorem the union of all the abelian subvarieties cannot be all the space. | |
Nov 3, 2011 at 20:20 | comment | added | Qing Liu | So as explained by Sandor, if $Y$ is a simple abelian surface, the anwser is no. If $Y$ contains an elliptic curve $E$, then by any point $y\in Y$, it passes a curve $y+E$ of genus $1$. If we want a real elliptic curve (abelian subvariety) passing through a general $P$, then it is easy to see that $Y$ is isogeneous to the square of an elliptic curve. I think that even when $Y=E^2$, the answer is no for a general $P$. | |
Nov 3, 2011 at 17:06 | vote | accept | fds | ||
S Nov 4, 2011 at 7:50 | |||||
Nov 3, 2011 at 17:04 | answer | added | Sándor Kovács | timeline score: 20 | |
Nov 3, 2011 at 16:15 | answer | added | Simon Rose | timeline score: 12 | |
Nov 3, 2011 at 16:00 | history | asked | fds | CC BY-SA 3.0 |