Skip to main content

All Questions

Filter by
Sorted by
Tagged with
4 votes
1 answer
363 views

The numbers of isomorphism classes of abelian variety over finite fields

It is known that there are only finitely many isomorphism classes of abelian variety over a finite field. I am curious about the exact number of these isomorphism classes. Explicitly, fix $g$, let $\...
Yuan Yang's user avatar
  • 547
8 votes
2 answers
703 views

Isomorphic endomorphism algebras implies isogenous (for abelian varieties over finite fields)?

$\newcommand{\F}{\mathbb{F}} \newcommand{\End}{\mathrm{End}} \newcommand{\Q}{\mathbb{Q}} \newcommand{\Z}{\mathbb{Z}}$ I would like to know if the following is true: Proposition A : Let $A_1, A_2$ ...
Watson's user avatar
  • 1,742
5 votes
0 answers
459 views

A functor on Abelian varieties corresponding to this operation on Weil numbers

Let $A/\mathbb F_q$ be an abelian variety over a finite field with Weil numbers $q^{1/2}\alpha_1,\dots,q^{1/2}\alpha_n$. Consider the numbers $q^{d/2}\alpha_1,\dots,q^{d/2}\alpha_n$. These are still ...
Asvin's user avatar
  • 7,746
14 votes
3 answers
978 views

Zeta function of Abelian variety over finite field

Let $A/\mathbf{F}_q$ be an Abelian variety of dimension $g$. Suppose one knows $|A(\mathbf{F}_{q^n})|$ for all $1 \leq n \leq g$. Does one know then $\zeta(A,s)$ (equivalently, $|A(\mathbf{F}_{q^n})|$ ...
user avatar
15 votes
0 answers
427 views

Counting abelian varieties over finite fields in a given isogeny class

Let $f(x) \in \mathbb Z[x]$ be a monic polynomial of degree $g$ with all roots having absolute value $\sqrt{q}$. How many principally polarized abelian varieties over $\mathbb F_q$ have $f(x)$ as the ...
Will Sawin's user avatar
  • 148k
9 votes
1 answer
633 views

Motives over finite field not generated by hyperelliptic curves

So the question is that, over a finite field, does there exist an abelian variety $A$ for which there does not exist a generically one-to-one morphism from a hyperelliptic curve $C$ to $A$. p.s. A ...
Cheng-Chiang Tsai's user avatar