Timeline for Translates of abelian subvarieties
Current License: CC BY-SA 4.0
13 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 1, 2020 at 15:37 | vote | accept | user unknown | ||
| Apr 1, 2020 at 15:30 | answer | added | Ariyan Javanpeykar | timeline score: 2 | |
| Apr 1, 2020 at 14:39 | comment | added | user unknown | @AriyanJavanpeykar: That's very very helpful! Thanks again! Yes please make it into an answer:) | |
| Apr 1, 2020 at 8:22 | comment | added | Ariyan Javanpeykar | Does this answer all your questions? If so, then I can make the above comments into an answer. | |
| Apr 1, 2020 at 8:21 | comment | added | Ariyan Javanpeykar | ...This means that images of rational maps of abelian varieties are images of morphisms of abelian varieties. But these images are translates of subabelian varieties (by closed points!) of $A $ because morphisms of abelian varieties are homomorphisms up to translation. | |
| Apr 1, 2020 at 8:21 | comment | added | Ariyan Javanpeykar | Finally, I claim that Sp(X) equals the "groupless-exceptional locus" as defined in Definition 12.4. of arxiv.org/abs/2002.11981 . To prove this, you have to combine some of the statements in arxiv.org/abs/1909.12187 with the following fact: If $B$ is an abelian variety and $B\dashrightarrow X$ is a rational map, then $B\dashrightarrow X$ is a morphism (i.e., everywhere defined). (This is because $B$ is smooth and $X$ has no rational curves.) ..con'd below | |
| Apr 1, 2020 at 8:17 | comment | added | Ariyan Javanpeykar | If you want to avoid using the fact that Sp(X) is closed, you will have to think a bit harder. I will look into it once I have some more time if you really want to know if it is possible to argue without using that Sp(X) is closed....con't below... | |
| Apr 1, 2020 at 8:16 | comment | added | Ariyan Javanpeykar | The fact that $Sp(X)$ is closed in $X$ was proven first by Kawamata; see Thm. 4 in Y. Kawamata, On Bloch’s conjecture, Invent. Math. 57 (1980), 97-100. Ueno subsequently proved that $Sp(X)\neq X$ if and only if $X$ is of general type. Similar statements are true for closed subvarieties of semi-abelian varieties; see Abramovich numdam.org/item/CM_1994__90_1_37_0 ...cont'd below... | |
| Apr 1, 2020 at 3:48 | comment | added | user unknown | @AriyanJavanpeykar: Also, if you have a nontrivial map from a group variety to $X$, would it factor through $Sp(X)$? | |
| Mar 31, 2020 at 23:48 | comment | added | user unknown | @AriyanJavanpeykar: That's really helpful! Thanks a lot! So I assume if we want an argument as Proposition 3.7, we would need to know $Sp(X_L)$ is closed？Do you think there is a way to do it without assuming closedness of the set (so $Sp(X)_L$ just means the inverse image of $Sp(X)$ in $X_L$)? | |
| Mar 31, 2020 at 17:09 | comment | added | Ariyan Javanpeykar | ...the positive-dimensional translates of abelian subvarieties by closed points of $A_L$ contained in $X_L$. Thus, the Mordell exceptional locus behaves well with respect to field extensions. You can prove this using spreading out and specialization arguments; see also Proposition 3.7 in arxiv.org/abs/1909.12187 | |
| Mar 31, 2020 at 17:09 | comment | added | Ariyan Javanpeykar | Usually, when one speaks of "translate of an abelian subvariety" it really is in the sense "translate of an abelian subvariety by a closed point". To answer your question: Let $X\subset A$ be a closed subvariety over an abelian variety $A$ over $k$. Assume $k$ is algebraically closed (of characteristic zero?). Let $Sp(X)$ be the union of positive-dimensional translates (by closed points of $A$) of abelian subvarieties contained in $X$. Let $L/k$ be an extension of algebraically closed fields. Then $Sp(X)_L = Sp(X_L)$, where $Sp(X_L)$ is the union of ... | |
| Mar 31, 2020 at 15:04 | history | asked | user unknown | CC BY-SA 4.0 |