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Let $K$ be a finite field. Is there a formula for the number of isomorphism classes of genus $g$ smooth curves over $K$?

In other words does there exists a formula for the number of rational points of the stack $\mathcal{M}_{g}$ over $K$? If so, what is a standard reference for this?

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    $\begingroup$ You should pay attention to the difference between the stack $\mathcal{M}_g$ (whose rational points are curves) and the coarse moduli space $M_g$. More precisely : two non-isomorphic curves over $K$ might become isomorphic over $\bar{K}$ and hence define the same point of $M_g$, and a $K$-point of $M_g$ might not correspond to a curve defined over $K$. $\endgroup$
    – Olivier Benoist
    Commented Dec 9, 2013 at 17:51
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    $\begingroup$ The coarse moduli space $M_g$ is of general type for sufficiently large $g$, so it seems highly unlikely that there would be a simple closed formula for $\#M_g(\mathbb{F}_q)$. $\endgroup$
    – Joe Silverman
    Commented Dec 9, 2013 at 18:12
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    $\begingroup$ Related to Olivier's comment, and especially since you said "moduli stack," you also want to decide whether you want to literally count isomorphism classes or whether you instead want to weight each isomorphism class by $\frac{1}{|\text{Aut}(x)|}$ (the formulas often come out nicer this way). $\endgroup$
    – Qiaochu Yuan
    Commented Dec 9, 2013 at 20:22
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    $\begingroup$ de Jong and Katz have an unpublished note about this: "Counting the number of curves over a finite field", available on de Jong's web page math.columbia.edu/~dejong/papers/curves.dvi. They prove a pretty interesting partial result. $\endgroup$
    – David Zureick-Brown
    Commented Nov 7, 2014 at 20:15

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A formula for the number of isomorphism classes of curves over $\mathbf F_q$ is probably hopeless. As pointed out by Olivier Benoist and Qiaochu Yuan, the much more well behaved number is given by isomorphism classes weighted by their automorphism group, in other words, the groupoid cardinality of the groupoid $\mathcal M_g(\mathbf F_q)$: $$ \# \mathcal M_g(\mathbf F_q) = \sum_{[C]/\cong}\frac 1 {\# \mathrm{Aut}(C)}$$ This is also the number of $\mathbf F_q$-points of the coarse moduli space. It has a cohomological interpretation (Grothendieck-Lefschetz trace formula) that reads $$ \# \mathcal M_g(\mathbf F_q) = \sum_{k}(-1)^k \mathrm{Tr}(\mathrm{Frob}_q \mid H^k_c(\mathcal M_g,\mathbf Q_\ell)).$$ Thus finding a formula for the number of $\mathbf F_q$-points becomes a question about understanding the cohomology of $\mathcal M_g$ as an $\ell$-adic Galois representation.

For small $g$ all the cohomology will be of Tate type which means that $\# \mathcal M_g(\mathbf F_q)$ is a polynomial in $q$, but for large $g$ this will certainly fail to be true and it's then not really clear what it would mean to write down an explicit formula, as suggested by Joe Silverman. Here are the formulas for $g \leq 4$: $$ \# \mathcal M_2(\mathbf F_q) = q^3$$ $$ \# \mathcal M_3(\mathbf F_q) = q^6+q^5+1$$ $$ \# \mathcal M_4(\mathbf F_q) = q^9+q^8+q^7-q^6$$ For the first of these, it's a classical fact that $\mathcal M_2$ has the rational cohomology of a point. Also for $\mathcal M_3$ and $\mathcal M_4$ one actually knows the cohomology and its Galois structure in each degree, not just the alternating sum of cohomology groups: see Looijenga's "Cohomology of $\mathcal M_3$ and $\mathcal M_3^1$" and Tommasi's "Rational cohomology of the moduli space of genus 4 curves". I don't know if $\#\mathcal M_5(\mathbf F_q)$ is in the literature but it's probably something one could figure out. But quite soon writing down an explicit formula is going to get hopeless.

One can also start putting in marked points. The number of $\mathbf F_q$-points of $\mathcal M_{1,n}$ and $\mathcal M_{2,n}$ are in a sense known for all $n$, see Getzler's "Resolving mixed Hodge modules on configuration spaces" and my preprint http://arxiv.org/abs/1310.2508 . Here it's also true that for $n$ large it's not a polynomial in $q$, but instead you find a polynomial expression in $q$ and certain traces of Hecke operators on elliptic and Siegel cusp forms. You might also be interested in papers of (various combinations of) Bergström, Tommasi, Faber, van der Geer...

Addendum, much later: The paper "On the number of curves of genus 2 over a finite field" by Gabriel Cardona determines the actual number of isomorphism classes of genus two curves over a finite field (of odd characteristic). The even characteristic case is treated in a companion paper.

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    $\begingroup$ Theorem 2 of jlms.oxfordjournals.org/content/s1-43/1/57.full.pdf+html plus some elementary computations should give the number of points on $M_{1,n}$ for $n \leq 11$; the Ramaujan $\tau$ function makes an appearance. $\endgroup$
    – David E Speyer
    Commented Dec 9, 2013 at 20:35
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    $\begingroup$ It's surprising to me that the groupoid cardinality is an integer. Is there a good conceptual explanation of this? $\endgroup$
    – Qiaochu Yuan
    Commented Dec 9, 2013 at 20:59
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    $\begingroup$ @DavidSpeyer that's a nice reference and I hadn't seen it before! But maybe it's worth saying that these formulas are a lot easier to obtain using a bit of étale cohomology (which of course Birch didn't have access to); in fact one needs étale cohomology of DM-stacks which wasn't available until Behrend's thesis... $\endgroup$
    – Dan Petersen
    Commented Dec 9, 2013 at 21:08
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    $\begingroup$ @Qiaochu, Olivier: I have no "conceptual" explanation for why the groupoid cardinality is an integer, and why it coincides with the number of points on the coarse space. I always thought of it as a nontrivial consequence of the Grothendieck-Lefschetz trace formula for stacks (together with the fact that the projection to the coarse moduli space is a rational homotopy equivalence/satisfies $R\pi_\ast \mathbf Q_\ell = \mathbf Q_\ell$). Olivier, I think I agree, and that this should follow from Lang's theorem. $\endgroup$
    – Dan Petersen
    Commented Dec 10, 2013 at 7:52
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    $\begingroup$ @OlivierBenoist: I would guess that the automorphism group of the torsor is given as follows: there is a map from $H$ to $Aut(H)$ given by conjugation and so an element of $H^1(k,H)$ gives an element of $H^1(k,Aut(H))$, hence a form of $H$. $\endgroup$
    – naf
    Commented Dec 12, 2013 at 11:11

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