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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 $\mathcal{M_g}:Sch_{\mathbb{F}_p}\rightarrow\mathcal{Sets}$ be the functor of such that $\mathcal{M}(X)=\{\text{isomorphism classes of abelian varieties of dim g over $X$}\}$.

What is #$\mathcal{M_g}(\mathbb{F}_{p^n})$?

If such a functor is represented by an algebraic variety, then these numbers are well studied by Weil conjecture. But unfortunately $\mathcal{M_g}$ is only represented by a stack.

Is there any pattern between these numbers? Can someone calculate some explicit examples?

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    $\begingroup$ arxiv.org/abs/1511.02212 $\endgroup$
    – alpoge
    Oct 10, 2021 at 8:21
  • $\begingroup$ @alpoge Thanks Mr.Alpoge. I roughly went through the whole passage, it contains a lot of estimates and asymptotics. But I want something more algebraic and precise, like if we build the zeta function like what we do in Weil conjecture, is it rational?&what can we say about zeros and poles? Maybe I missed something in the passage? $\endgroup$
    – Yuan Yang
    Oct 10, 2021 at 9:35

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I did a little search work on this problem, and it seems that I found the following article. \

The Lefschetz trace formula for algebraic stacks \

The result is that: If $\mathcal{X}$ is an algebraic stack, the we will have a same Lefschetz trace formula for $\mathcal{X}$, but it would not always be a finite sum. But if it is Deligne-Mumford, then it is known that it is a finite sum, so we do have the rationality of its zeta-function.

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    $\begingroup$ Since we do not fix polarizations, the stack is far from being DM or even algebraic. $\endgroup$
    – Piotr Achinger
    Oct 11, 2021 at 14:19
  • $\begingroup$ @PiotrAchinger Ahhhh, is that so,, I really don’t know $\endgroup$
    – Yuan Yang
    Oct 11, 2021 at 14:20
  • $\begingroup$ @PiotrAchinger On Wikipedia, it says that "The moduli stack of elliptic curves is a smooth separated Deligne–Mumford stack of finite type over $spec(\mathbb{Z})$". So in dim 1, it is actually Deligne-Mumford? $\endgroup$
    – Yuan Yang
    Oct 25, 2021 at 14:28
  • $\begingroup$ correct. The zero section gives an ample divisor on an elliptic scheme, and so they are automatically polarized. The issue starts in dimension two. $\endgroup$
    – Piotr Achinger
    Oct 25, 2021 at 19:43

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