Timeline for If it quacks like an abelian variety over a finite field
Current License: CC BY-SA 4.0
16 events
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| Sep 2, 2020 at 21:18 | comment | added | Will Sawin | Perhaps if the integral $\ell$-adic cohomology ring looks like the cohomology of an abelian variety for all $\ell$, including crystalline cohomology if $\ell=p$? This could suffice to show that the Albanese map has degree $1$ (by pulling back cohomology classes in $H^1$ from the Albanese and then cupping them to degree $2n$ to calculate the degree) and that it is finite (because the class of any curve that is contracted by the map would have zero cup product with any $2n-1$ classes in degree $1$) which would imply it is an isomorphism. | |
| Sep 2, 2020 at 21:06 | history | edited | Nguyen | CC BY-SA 4.0 |
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| Aug 27, 2020 at 22:17 | comment | added | paul garrett | @DavidESpeyer, a funny business! Thanks for the illuminating points, even if not exactly annihilating the original question. | |
| Aug 27, 2020 at 20:24 | history | became hot network question | |||
| Aug 27, 2020 at 19:41 | comment | added | David E Speyer | Okay, I read further in the Catanese et al paper, and I retract my warning. According to Theorem 2.3, if $X$ is a compact complex manifold which (1) has the integer cohomology ring of a torus and (2) is bimeromorphic to a compact Kahler manifold, then $X$ is a complex torus. In particular, if (1) holds and $X$ is algebraic then Chow's lemma tells us it is birational to a projective variety, so this should show it is an abelian variety. That doesn't prove anything in char p, but it means that I no longer consider this a reason for caution. | |
| Aug 27, 2020 at 18:16 | comment | added | David E Speyer | Sommese, Quaternion Manifolds (1975) link.springer.com/article/10.1007/BF01357140 constructed a complex 3-fold which is diffeomorphic to $(S^1)^6$ but not a complex torus. So there is grounds for caution. I found a readable description of the construction in the first pages of projecteuclid.org/euclid.kjm/1291041217 | |
| Aug 27, 2020 at 17:57 | answer | added | Jef | timeline score: 8 | |
| Aug 27, 2020 at 17:04 | history | edited | Nguyen |
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| Aug 27, 2020 at 15:20 | comment | added | Hacon | I don't know about finite fields, but over $\mathbb C$ by arxiv.org/abs/math/9903184 and arxiv.org/abs/math/0011042, if $h^0(2K_X)=1$ and $h^0(\Omega ^1_X)=\dim X$, then $X$ is birational to an abelian variety. If $k$ is an algebraically closed field of char $p>0$, then related results are contained in arxiv.org/abs/1703.06631. | |
| Aug 27, 2020 at 14:59 | history | edited | Nguyen |
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| Aug 27, 2020 at 12:55 | comment | added | Nguyen | @JoeSilverman the classification of surfaces in positive characteristic is kind of confusing to me but I tried. In higher dimensions I don't know anything | |
| Aug 27, 2020 at 12:52 | comment | added | Joe Silverman | Have you considered the dimension 2 case? Clearly a necessary condition is that the Kodaira dimension be 0. There are just a few such classes of varieties, so you can check to see whether their discrete invariants suffice to pick out the abelian surfaces. | |
| Aug 27, 2020 at 12:50 | comment | added | Joe Silverman | @Wojowu There is actually a much easier proof that every genus 1 curve over a finite field has a point, and it generalizes to the statement that if $F$ Is a finite field and $A/F$ is a variety such that $A/\overline{F}$ is an abelian variety, then $A(F)\ne\emptyset$, so $A/F$ is itself already an abelian variety. (I have a recollection this may be due to Lang, but I could be wrong.) | |
| Aug 27, 2020 at 12:48 | comment | added | Wojowu | Ah yeah, right, my bad. I messed up my Hasse-Weil bound :P | |
| Aug 27, 2020 at 12:25 | review | First posts | |||
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| Aug 27, 2020 at 12:22 | history | asked | Nguyen | CC BY-SA 4.0 |