Timeline for Correspondences and Albanese
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Jun 28 at 12:14 | comment | added | Ariyan Javanpeykar | If $X\to Y$ is a morphism of smooth projective schemes over $k$, then you get $Alb(X) \to Alb(Y)$ and $Alb(Y)\to Alb(X)$ as naf explains. So, if you have a correspondence $Z\to X, Z\to Y$, then you get a triangle of morphisms, hence a morphism from $Alb(X)$ to $Alb(Y)$. | |
| Jun 28 at 1:37 | comment | added | naf | Yes, one does get an induced morphism on Albanese varieties. It might be easier to see this using the description of the Albanese variety as the dual of the Picard variety: by basic intersection theory one has a map (in the opposite direction) on Picard groups over any extension of $k$. This is enough to get the morphism. | |
| Jun 27 at 17:26 | history | edited | curious math guy | CC BY-SA 4.0 |
R.van Dobben de Bruyn made me realize to phrase question more carefully
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| Jun 21 at 18:12 | comment | added | curious math guy | @R.vanDobbendeBruyn I do mean to ask whether a correspondence from $X$ to $Y$ induces a morphism between the Albanese. Doesn't a correspondence induce a Galois-equivariant morphism of the $H^1$ which then (by Tate) induces a morphism in the isogeny category between the Albanese? (although this seems to work for elliptic curves and Albanese). Does this make sense? | |
| Jun 21 at 16:36 | comment | added | R. van Dobben de Bruyn | Sorry, why does a correspondence give a morphism in the curve case? If $X$ and $Y$ are both elliptic curves, there are many correspondences that are not morphisms. Or do you mean whether a correspondence from $X$ to $Y$ induces a correspondence from $\operatorname{Alb}(X)$ to $\operatorname{Alb}(Y)$? | |
| Jun 21 at 16:18 | history | edited | LSpice | CC BY-SA 4.0 |
Typo
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| Jun 21 at 15:58 | history | asked | curious math guy | CC BY-SA 4.0 |