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If $G$ is a finite group acting on a finite set $X$, we have Burnside's formula that counts the number of orbits $|X/G|$ as: $$ |X/G| = \frac1{|G|} \sum_{g\in G} |X^g|, $$ with $X^g$ being the set of elements in $X$ fixed by $g$.

Now, consider the finite set $X$ of $\mathbb F_p$-points of $\mathbb P^1_{\mathbb F_p}$, the one dimensional projective space over the finite field $\mathbb F_p$, with $p$ elements, and the finite set $Y:=X\times X$ (Cartesian product). The 2-element group $G:=\{e,g\}$ acts by permuting the factors on $Y$ ($G$ is the symmetric group on 2 elements, and $e$ its identity), and it is well-known that (it follows, for example, from the Zeta function computation in https://math.stackexchange.com/q/799101): $$ Y/G = Sym^2(\mathbb P^1_{\mathbb F_p}) = \mathbb P^2_{\mathbb F_p}. $$

As such, we have $|Y/G|=|\mathbb P^2_{\mathbb F_p}|=p^2+p+1$.

On the other hand, one would say that $|Y^e|=|Y|=|X|^2=(p+1)^2$, and $|Y^g|=|X|=p+1$, since the elements fixed under the permutation are in the diagonal of $Y=X\times X$, isomorphic to $X$, so Burnside's formula gives, apparently, the wrong answer: $$ |Y/G|=\frac12((p+1)^2+(p+1))=\frac{p^2+3p+2}{2}.$$

Maybe I am using the wrong notion of $\mathbb F_p$-points of a variety (/scheme?), but I believe there should be a simple explanation of why the computations do not match. Any help is appreciated.

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    $\begingroup$ The quotient map might not be surjective on $\mathbf{F}_p$-points. I didn't look carefully at this example, but for another action of a group of order $2$: the homomorphism $\mathrm{SL}_2\to\mathrm{PGL}_2$ is surjective, but not surjective on $F$-points when $F$ is a finite field of odd cardinal. $\endgroup$
    – YCor
    Sep 22 at 11:03
  • $\begingroup$ Of course, +2 instead of +1, that was a typo, thanks! And I also edited the notation for over $\mathbb F_p$ $\endgroup$ Sep 22 at 12:10
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    $\begingroup$ The usual identification of symmetric powers of the line with projective space goes via encoding polynomials as their roots, which seems to mess with the point counting equalities you gave. Someone more knowledgeable can hopefully see what’s going on here. $\endgroup$
    – Chris H
    Sep 22 at 12:12

1 Answer 1

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The issue is that $X(\mathbb{F}_p)/G$ is not the same thing as $(X/G)(\mathbb{F}_p)$. A simpler example is to take $p$ odd, $X = \mathbb{A}^1$ and let $S_2$ act by $\pm 1$. There are $\tfrac{p+1}{2}$ orbits, but the quotient space is $\mathbb{A}^1$ with the quotient map $x \mapsto x^2$. The quotient space has $p$ $\mathbb{F}_p$-points, but only $\tfrac{p+1}{2}$ of them (the squares) are in the image of $X(\mathbb{F}_p)$.

Points of $(X/G)(\mathbb{F}_p)$ index Frobenius stable $G$-orbits, so the other $\tfrac{p-1}{2}$ points correspond to the orbits of the form $\{ \pm x \}$ where $x^p = -x$ (other than the point $x=0$).

If you want to count $(X/G)(\mathbb{F}_p)$, there is a combined Burnside/Lefschetz formula. Choose $\ell$ relatively prime to $p$ and $|G|$, then a formula of Grothendieck tells us that $H^j(X/G, \mathbb{Q}_{\ell}) \cong H^j(X, \mathbb{Q}_{\ell})^G$, and this isomorphism is Frobenius equivariant. So the trace of Frobenius on $H^j(X/G, \mathbb{Q}_{\ell})$ is the same as the trace of Frobenius restricted to the subspace $H^j(X, \mathbb{Q}_{\ell})^G$. Now, the linear operator $\tfrac{1}{|G|} \sum_{g \in G} g$ on $H^j(X, \mathbb{Q}_{\ell})$ is an idempotent whose image is $H^j(X, \mathbb{Q}_{\ell})^G$. So the trace of Frobenius restricted to $H^j(X, \mathbb{Q}_{\ell})^G$ is the same as the trace of $\tfrac{1}{|G|} \sum_{g \in G} \text{Frob} \circ g$. So we get $$\#((X/G)(\mathbb{F}_p)) = \tfrac{1}{|G|} \sum_{g \in G} \sum_j (-1)^j \text{Tr}{\big(}\text{Frob} \circ g : H^j(X, \mathbb{Q}_{\ell}) \longrightarrow H^j(X, \mathbb{Q}_{\ell}){\big)}.$$

I am not sure that there is a Burnside/Lefschetz style formula for $\#X(\mathbb{F}_p)/G$. Here is a troubling example: Take $p$ odd, $X = \mathbb{P}^1$ and let $S_2$ act by $[x:y] \mapsto [cy : x]$. Then $X(\mathbb{F}_p)$ has $p+1$ points. If $c$ is a quadratic residue, then there are two fixed points for $S_2$, namely $[\pm \sqrt{c}:1]$, otherwise there are none. So there are either $\tfrac{p+1}{2}$ or $\tfrac{p-1}{2}$ orbits for $S_2$ on $X(\mathbb{F}_p)$ depending on whether or not $c$ is a quadratic residue. I find it hard to imagine a Burnside/Lefschetz style formula which could take this information into account.


It might help to say that there is a purely topological version of this question. Let $G$ be a finite group and let $X$ be a compact topological space with an action of $G$ and also with an endomorphism $\phi$ that commutes with the $G$-action. (More generally, we could imagine that there is an automorphism $\sigma$ of $G$ with $\phi \circ g = \sigma(g) \circ \phi$.)

Then the two questions are how to count $(X/G)^{\phi}$ and how to count $(X^{\phi})/G$. These aren't the same thing: If there is a $G$-orbit which $\phi$ permutes nontrivially, then it will contribute to the first count but not the second.

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