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Let $X$ be an affine variety defined over $\mathbb{Z}$ and let $G$ be an algebraic group defined over $\mathbb{Z}$. Let $q$ be a power of a prime number. We write $\mathbb{F}_q$ for the field with $q$ elements and we define $$n_q:= \text{number of } G(\mathbb{F}_q) \text{ orbits in } X(\mathbb{F}_q).$$

Question 1: Under what condition $n_q$ has a closed formula which is a polynomial in $q$?

Question 2: In case of a positive answer for Question 1, do the coefficients of this polynomial relate to any known invariants of $X$ and $G$?

Question 3: given a polynomial with rational coefficients, is there any way to detect if it arises in the above situation?

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    $\begingroup$ If what you were after was the groupoid cardinality of the quotient stack then there is Behrend's trace formula: en.wikipedia.org/wiki/Behrend%27s_trace_formula $\endgroup$
    – Qiaochu Yuan
    Oct 30, 2020 at 1:24
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    $\begingroup$ An example: If X is affine n space and G = S_n, then the number of orbits is polynomial in q for q larger than n. So perhaps you only want to require that $n_q$ is polynomial for large enough q? $\endgroup$
    – Asvin
    Oct 30, 2020 at 1:47
  • $\begingroup$ Is there a condition for the number of points of a variety to be polynomial in q (the case when G is trivial)? By Weil, this is saying that the Frobenius acts trivially on the cohomology but does this imply that X is rational? If we restrict attention to X over a finite field, then there are examples like supersingular elliptic curves but how about over Z? $\endgroup$
    – Asvin
    Oct 30, 2020 at 1:57
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    $\begingroup$ Umm, that's true but if you allow a piece wise polynomial definition, then every function is polynomial... $\endgroup$
    – Asvin
    Oct 30, 2020 at 2:32
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    $\begingroup$ @Asvin I think Ehud means that the number is given by $\binom{q+n-1}{n}$ both for small and large values of $q$! $\endgroup$
    – Dan Petersen
    Oct 30, 2020 at 9:08

1 Answer 1

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If the group $G$ is connected then there is a nice cohomological expression for the quantity $\#X(\mathbf F_q)/\#G(\mathbf F_q)$. Indeed in this case Lang's theorem implies that this equals the number of $\mathbf F_q$-points of the quotient stack $[X/G]$ (counted, as always, weighted by their stabilizer group) which can be interpreted cohomologically by the Grothendieck--Lefschetz trace formula, which was generalized to stacks by Behrend, as alluded to by Qiaochu. There is a very large literature about schemes and stacks with the property that their number of $\mathbf F_q$-points is given by a polynomial in $q$. Unless there is some "unexpected cancellation" in the cohomology, this corresponds to all cohomology groups being of Tate type.

If $G$ is disconnected then the quantity $\#[X/G](\mathbf F_q)$ is still well-behaved, but it generally not equal to $\#X(\mathbf F_q)/\#G(\mathbf F_q)$, since an $\mathbf F_q$-point of $[X/G]$ no longer has any reason to lift to an $\mathbf F_q$-point of $X$.

I doubt that you will find any nice expression or general result concerning the actual number of orbits, beyond what you get by combining the Grothendieck--Lefschetz trace formula and Burnside's lemma.

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