Timeline for A functor on Abelian varieties corresponding to this operation on Weil numbers
Current License: CC BY-SA 4.0
25 events
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| S Nov 5, 2019 at 22:03 | history | bounty ended | CommunityBot | ||
| S Nov 5, 2019 at 22:03 | history | notice removed | CommunityBot | ||
| Nov 4, 2019 at 7:34 | comment | added | Vivek Shende | @WillSawin well, the map could be something like, fixed locus in some space (canonically associated to the variety) of some automorphism, which then could have varying dimensions. | |
| Nov 4, 2019 at 6:18 | comment | added | Will Sawin | If you fix a Newton strata where the $k$ is the gcd of the coordinates of the vertices with $d$, then this map takes abelian varieties of genus $g$ to abelian varieties of genus $gd/k$. From the formula for the dimension of the central leaf of the Newton stratum (which is relevant because our varieties are defined up to isogeny and the central leaf is the whole stratum up to isogeny), we can see that its dimension is multiplied by $d/k^2$ by this operation. It is unlikely that we will have a canonical map, except possibly if $k^2=d$. | |
| Nov 4, 2019 at 6:14 | comment | added | Will Sawin | @VivekShende The subtlety we were discussing earlier is that sometimes the list of eigenvalues will involve each member of the conjugacy class repeated a certain number of times. This is done to make the Newton polygon of the characteristic polynomial with respect to the $q$-adic valuation integral. | |
| Nov 4, 2019 at 5:20 | comment | added | Asvin | Exactly the conjugacy class. It has to be a conjugacy class since the Frobenius eigenvalues satisfy a monic polynomial equation over $\mathbb Z$. | |
| Nov 4, 2019 at 5:15 | comment | added | Vivek Shende | Thanks! Does the Honda-Tate assertion mean: exactly the list of eigenvalues on H^1, or among the list of eigenvalues? | |
| Nov 4, 2019 at 5:12 | comment | added | Asvin | @VivekShende I added some explanations, let me know if anything is confusing. | |
| Nov 4, 2019 at 5:12 | history | edited | Asvin | CC BY-SA 4.0 |
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| Nov 4, 2019 at 4:55 | comment | added | Vivek Shende | it would be great if this question was readable to a let us say hypothetical person who was an algebraic geometer but didn't know what Weil numbers or Honda-Serre-Tate were. | |
| Oct 29, 2019 at 18:50 | comment | added | Dror Speiser | Already for the simplest case of elliptic curves over a prime field and odd $d$, the situation seems pretty complicated: for the supersingular curves you can just take extension-of-scalars to $\mathbb{F}_{q^d}$, while for the ordinary curves the corresponding abelian varieties are of dimension $d$. | |
| Oct 29, 2019 at 4:29 | history | edited | Asvin |
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| S Oct 28, 2019 at 20:48 | history | bounty started | Asvin | ||
| S Oct 28, 2019 at 20:48 | history | notice added | Asvin | Draw attention | |
| Oct 28, 2019 at 20:47 | history | edited | Asvin | CC BY-SA 4.0 |
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| Oct 24, 2019 at 1:37 | comment | added | Will Sawin | Over $\mathbb F_{q^d}$, it's the $q^d$-adic valuation, so the integrality does change. | |
| Oct 24, 2019 at 0:41 | comment | added | Asvin | Right, but since I am only multiplying by a power of $q$, the integrality shouldn't change. The first condition is positivity in integers if I am reading it right, but that shouldn't change either? | |
| Oct 24, 2019 at 0:40 | comment | added | Will Sawin | The formula depends on the $q$-adic valuation being an integer, not being nonzero. | |
| Oct 24, 2019 at 0:38 | comment | added | Asvin | pg 64 math.mit.edu/~poonen/papers/curves.pdf | |
| Oct 24, 2019 at 0:38 | comment | added | Will Sawin | Where are you getting that? The formula here math.stanford.edu/~conrad/vigregroup/vigre04/hondatate.pdf seems more complicated than that. | |
| Oct 24, 2019 at 0:34 | comment | added | Asvin | @WillSawin Actually, given that the minimal dimension depends only on the q-valuation of (some products of) the roots being positive over $\overline{\mathbb Q_p}$, I believe the dimension shouldn't change after all. | |
| Oct 24, 2019 at 0:07 | comment | added | Asvin | I guess the question still makes sense modulo the dimension | |
| Oct 23, 2019 at 23:15 | comment | added | Asvin | Oh that's right! | |
| Oct 23, 2019 at 23:06 | comment | added | Will Sawin | This does not produce an abelian variety of the same dimension as $A$. The dimension in Honda-Serre-Tate (for simple abelian varieties) is not simply the degree of the polynomial, as you can see when $q$ is a perfect square and $\alpha =1$. | |
| Oct 23, 2019 at 21:28 | history | asked | Asvin | CC BY-SA 4.0 |