Timeline for What does the Néron model of the dual abelian variety parametrize?
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8 events
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| Jun 2, 2023 at 16:56 | comment | added | Hans | Thanks a lot for your comments and thoughts! I will have a look at these references. | |
| Jun 2, 2023 at 9:44 | comment | added | Jason Starr | "semi-normal" --> "semi-factorial" | |
| Jun 1, 2023 at 20:18 | comment | added | Jason Starr | I read more about this in a survey article of Cedric PEPIN: math.univ-paris13.fr/~cpepin/neron-picard-final.pdf There is "almost" a functorial description. There is a "semi-normal", proper $R$-model $\overline{A}$ of $A_K$, and the N'eron model $A^\dagger$ is the unique "group smoothening" of the relative $\text{Pic}^0$ of $\overline{A}/R$. As noted by user @Jef, when the groups of connected components of the closed fibers of $A$ and $A^\dagger$ are nontrivial, the Poincar'e bundle almost never extends. So the group smoothening is nontrivial. | |
| Jun 1, 2023 at 19:01 | comment | added | Jason Starr | I doubt that there is a good answer to your question in general. If we knew an elementary functorial property of the N'eron model, then it would be much easier to construct (e.g., by checking Artin's algebraization axioms). The fact that the current constructions of the N'eron model are not so easy (e.g., using the auxiliary invariant differential forms) suggests that there is no good answer. On the other hand, if you find a good answer, you can use that to try to give a simple construction of N'eron models! | |
| Jun 1, 2023 at 18:27 | comment | added | Jef | Also this question seems very related: mathoverflow.net/questions/114337/… | |
| Jun 1, 2023 at 18:23 | comment | added | Jef | Of course $X^t(R) = A^t(K)$ just corresponds to degree zero line bundles on $A$, but there is a more subtle relationship too. The universal line bundle on $A \times A^t$ (Poincare bundle) gives rise to a so-called biextension of $A\times A^t$. Grothendieck showed that it extends to a biextension on $X^{0}\times (X^t)^{0}$. For (rather intricate) details, SGA7, Expose IX. | |
| Jun 1, 2023 at 18:13 | comment | added | Jef | This is not an answer, but a special case that's already quite interesting. If $A$ is the Jacobian of a smooth projective curve $C/K$ with minimal regular model $\mathcal{C}/R$, then $A = A^t$ and the identity component of $X_k$ is isomorphic to $\text{Pic}^0_{C_k/k}$, when the gcd of the multiplicities of $C_k$ is one; this is a theorem of Raynaud and explains your elliptic curve example. | |
| Jun 1, 2023 at 13:55 | history | asked | Hans | CC BY-SA 4.0 |