# Is the quotient of resolution the same as resolution of the quotient?

Suppose $X$ is a singular variety over a field $k$, which admits an action by a finite group $G$. Suppose the quotient $X/G$ is also a variety over $k$. If $Y_G$ is a resolution of $X/G$, does there exist a variety $Y$ which is a resolution of $X$ such that the action on $X$ could be extended to an action on $Y$ and

$$Y/G \simeq Y_G$$

Put it in another way, is the resolution of the quotient "the same thing" as the quotient of (equivariant) resolution?

• No. It can happen that $X$ is smooth, but $X/G$ is singular. E.g., $X= \mathbb P^1\times \mathbb P^1$ and $G =\mathbb{Z}/2\mathbb{Z}$ acts by permuting the factors. Then $X/G$ is singular. Clearly, in this case, no desingularization of the quotient is the "same thing" as the quotient $X/G$ of the (trivial) resolution. But maybe you meant to consider a free action of $G$ on $X$? Commented Dec 13, 2017 at 21:08
• The quotient of $\mathbb{P}^1\times \mathbb{P}^1$ by the action of $\mathbb{Z}/2$ permuting the factors is $\operatorname{Sym}^2\mathbb{P}^1\cong \mathbb{P}^2$. It is smooth.
– abx
Commented Dec 13, 2017 at 21:12
• @AriyanJavanpeykar Thank you. Indeed, I am interested in the case when the action is free. What happens in this case? Commented Dec 13, 2017 at 21:21
• @abx Yes, that's correct. I meant to write $\mathbb{P}^2\times \mathbb{P}^2$. Commented Dec 13, 2017 at 23:39
• If the action is free, then $X\to X/G$ is étale, so why not take $Y=X\times_{X/G} Y_G$? Commented Dec 14, 2017 at 8:41

As I said in the comments, the answer is no in general. Indeed, let $X = \mathbb{P}^2\times \mathbb{P}^2$ and let $G=\mathbb{Z}/2\mathbb{Z}$ act by permuting the factors. Then $X$ is smooth and $X/G$ is singular. If $(X/G)'\to X/G$ is a resolution of singularities, then $(X/G)' \not \cong X/G$ (clearly).
But, if $G$ acts freely on $X$ (in the scheme-theoretic sense), then $X\to X/G$ is a $G$-torsor (for the etale topology). Let $Y_G \to X/G$ be a desingularization. Then, the pull-back $Y:= X\times_{X/G} Y_G$ of $X\to X/G$ along $Y_G \to X/G$ is a $G$-torsor over $Y_G$ and maps $G$-equivariantly to $X$. This means that the answer to your question is positive when $G$ acts freely on $X$.
• Thank you very much. A further question, suppose $G$ acts on $X$ freely, and the singular loci of $X$ is of the form $\cup_{g \in G} gZ$ while $g_1Z \cap g_2 Z=\emptyset,g_1 \neq g_2$, where $Z$ is a subvariety of $X$. If the singularity $Z$ could be resolved, could we get a $G$-equivariant resolution of $X$? Commented Dec 14, 2017 at 10:58