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Let $G$ be a finite group acting on $\mathbb A^n_{\mathbb C}$. Let $Y$ be a dense open whose complement is of codimension at least two.

Assume $Y$ is $G$-stable, the action of $G$ is free on $Y$, and that $Y/G$ is a quasi-affine scheme.

Is the action of $G$ on $\mathbb A^n_{\mathbb C}$ free?

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    $\begingroup$ What about $\mathbb{Z}/2 \mathbb{Z}$ acting on $\mathbb{A}^2$ by $$(x, \, y) \mapsto (-x, -y)?$$ $\endgroup$
    – Francesco Polizzi
    Commented Oct 18, 2016 at 8:11
  • $\begingroup$ @FrancescoPolizzi Perfect. I should have realized that myself. Could I ask you to post it as an answer? By the way, just out of curiosity, what is $Y/G$ explicitly? It is a smooth quasi-affine variety with Picard group $\mathbb Z/2\mathbb Z$, as far as I can see. Is there a more explicit way of describing $Y/G$ (as an explicit open in some affine variety)? $\endgroup$
    – Miele
    Commented Oct 18, 2016 at 12:04
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    $\begingroup$ In this example $\mathbb{A}^2/G$ is the quadratic cone in $\mathbb{A}^3$ defined by $v^2-uw=0$ (just put $u=x^2$, $v=xy$, $w=y^2$). Thus $Y/G$ is this quadratic cone minus its vertex. $\endgroup$
    – abx
    Commented Oct 18, 2016 at 12:20
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    $\begingroup$ I have posted my comment above as an answer. $\endgroup$
    – Francesco Polizzi
    Commented Oct 18, 2016 at 13:05

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The answer in general is no, as shown by the following example.

Take the group $G=\mathbb{Z}/2 \mathbb{Z}$, acting on $\mathbb{A}^2$ as $$(x, \, y) \mapsto (-x, \, -y).$$ This action is free outside the unique fixed point $p=(0, \, 0)$, so $Y= \mathbb{A}^2-\{p\}$ is $G$-stable.

Furthermore, we have $$\mathbb{A}^2/G = \textrm{Spec} \, \mathbb{C}[x, \, y]^G=\textrm{Spec} \, \mathbb{C}[x^2, \, xy, \, y^2] = \textrm{Spec} \, \mathbb{C}[u, \, v, \, w]/(v^2-uw).$$ So $\mathbb{A}^2/G$ is an affine variety isomorphic to a quadric cone in $\mathbb{A}^3$, whereas $Y/G$ is this cone minus its vertex.

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  • $\begingroup$ $Y/G$ is the total space of $\mathcal O_{\mathbb P^1}(-2)$ minus its zero section, a $\mathbb C^*$-bundle over $\mathbb P^1$ without a section. $\endgroup$
    – pgraf
    Commented Oct 18, 2016 at 14:09
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    $\begingroup$ Well, an affine quadric cone in $\mathbb{A}^3$ minus its vertex is a $\mathbb{C}^*$-bundle over $\mathbb{P}^1$, or am I missing something? $\endgroup$
    – Francesco Polizzi
    Commented Oct 18, 2016 at 14:15
  • $\begingroup$ You are right. I was making that remark only since the OP asked about an explicit description of $Y/G$. $\endgroup$
    – pgraf
    Commented Oct 20, 2016 at 13:44
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Just for your information: the action of a non-trivial finite group on a complex affine space is never free. More precisely, any $p$-Sylow subgroup will have a fixed point. This can be either proved by Smith theory (Borel) or by an arguments involving reduction mod $\ell\ne p$ (Shafarevitch).

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