# A question about quotient singularity

If a finite group G acts on a smooth variety X over complex number field and the fixed locus of G is smooth subvariety of codimension 1, will the resulting quotient variety be smooth? What will happen if the fixed locus has lager codimension ? thanks.

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Try $X=\mathbb{C}$ and $G=\{1,-1\}$ acting multiplicatively –  Michael Bächtold Sep 30 '11 at 10:49
Michael, your example is smooth, but $\mathbb{C}^2/\pm 1$ is not. –  Donu Arapura Sep 30 '11 at 11:38
@Donu: you're absolutely right! My mistake was in thinking of $\mathbb{C}$ as $\mathbb{R}^2$. –  Michael Bächtold Sep 30 '11 at 17:49
@Donu and Michael, about the example $\mathbb C^2/ \pm$: It looks to me like $G$ is generated by a pseudo reflection, so by the answer below the quotient should be smooth. And indeed the ring of invariants is $\mathbb C[x,y^2]$, which is a polynomial ring. Am I missing something? –  Drew Jan 11 '13 at 17:19

In general, for a finite group $G$ acting faithfully on a smooth variety $X$, whether or not the quotient is smooth is determined by the Chevalley-Shephard-Todd theorem:
For $x \in X$, let $G_x\subset G$ be the stabilizer of $x$. Then a necessary and sufficient condtion for the quotient to be smooth is that each $G_x$ should be generated by pseudoreflections i.e. elements which fix pointwise a codimension $1$ subvariety of $X$ containing $x$.
In particular, one cannot just look at the fixed locus of $G$ to determine whether the quotient is smooth. It could well be empty but the quotient could still be singular.
Just to repeat what Ulrich said, it is not enough to look at the fixed locus, i.e., the set of points fixed by every element of $G$. You have to look at the points with nontrivial stabilizer, i.e., the set of points fixed by at least one non-identity element. –  Jason Starr Sep 30 '11 at 12:54