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Let $G$ be a split semisimple algebraic group and let $C$ be a curve of genus $g$ over $\mathbb F_q$. Assume $g \geq 2$.

The number of $\mathbb F_q$-points of $\# \operatorname{Bun}_G(C)$, where each point is weighted by the inverse of the order of its automorphism group, is $(1+o(1))q^{ (g-1) \dim G}$. This follows from the Tamagawa number 1 theorem, though I think this basic estimate is easier to prove.

Is this still true if we do not divide out by automorphisms? Obviously not. If $G = SL_2$ we can take $L + L^{-1}$ for infinitely many line bundles $L$ to get infinitely many different isomorphism classes of vector bundles with determinant $1$. But it is true if we restrict to stable vector bundles, because the number of automorphisms is bounded.

Let $U$ be an open subset of $\operatorname{Bun}_G(C)$ that is of finite type. For instance, if $G=SL_n$ we can take the set of determinant $1$ vector bundles $V$ such that $V \otimes L^{-1}$ has no sections for a line bundle $L$ of large degree.

Is the number of isomorphism classes of $G$-bundles on $C$ contained in $U(\mathbb F_q)$, not weighted by automorphisms, still $(1+o(1)) q^{(g-1)\dim G}$? Or does the number of points on $\operatorname{Bun}_G$ jump when you add even midly unstable bundles?

I promise I have a really good reason to care about this "evil" count.

I think I can prove this estimate in the case $G=SL_2$. As I already mentioned stable vector bundles are fine. The remainder are split sums $L + L^{-1}$, the number of which with $L$ of degree $n$ is $(1+o(1)) q^g$, and nontrivial extensions of $L^{-1}$ by $L$, which are classified by $H^1(L^2)$, so they are cardinality of them is at most the cardinality of $H^0 ( K L^{-2})$. Summed over all line bundles $L$ of degree $n$, this is at most a constant times the total number of nonzero sections of all line bundles of degree $2g-2-2n$ up to scaling, which is equal to the number of effective divisors of that degree, or $q^{2g-2-2n}$. As $n \geq 0$, either one is less than $q^{3g-3}$.

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    $\begingroup$ Is the pun in the sentence "Or does the number of points on $\operatorname{Bun}_G$ jump" intentional? $\endgroup$ Feb 11, 2016 at 15:29
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    $\begingroup$ @TomDeMedts It is indeed. $\endgroup$
    – Will Sawin
    Feb 12, 2016 at 1:34

1 Answer 1

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I believe that the claim is true, and that this is related to the so-called 'very good' property of the stack $\mathrm{Bun}_G$. Following [Beilinson, Drinfeld, Quantization of Hitchin's integrable system and Hecke eigensheaves], we call a smooth stack $X$ very good if for every $k>0$, the locus of points $$\{x\in X:\dim \mathrm{Aut}(x)= k\}$$ has codimenion that is strictly larger than $k$. Beilinson and Drinfeld show (Proposition 2.1.2) that $\mathrm{Bun}_G$ is very good if $g>1$. Thus, codimension of the locus $$X_k:=\{x\in\mathrm{Bun}_G:\dim \mathrm{Aut}(x)= k\}$$ is at least $k+1$. Keep in mind that this is the codimension in the sense of stacks; its `evil' counterpart is the claim that the dimension of the space corresponding to $X_k$ is less than $\dim(\mathrm{Bun}_G)=(g-1)\dim G$. This implies that $|X_k(\mathbb{F}_q)|=o(q^{(g-1)\dim G})$, which is what you need, right?

Disclaimer. Beilinson and Drinfeld work in characteristic zero. It seems likely that the result is valid in characteristic $p$, but I didn't look at it carefully.

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  • $\begingroup$ Yes, this is precisely what I need. I'm going to go through the argument to make sure it's fine in characteristic $p$... $\endgroup$
    – Will Sawin
    Feb 12, 2016 at 23:23

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