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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.

flag	asked Sep 30 2011 at 10:25 strygwyr 145●4

Try $X=\mathbb{C}$ and $G=\{1,-1\}$ acting multiplicatively – Michael Sep 30 2011 at 10:49

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Michael, your example is smooth, but $\mathbb{C}^2/\pm 1$ is not. – Donu Arapura Sep 30 2011 at 11:38

@Donu: you're absolutely right! My mistake was in thinking of $\mathbb{C}$ as $\mathbb{R}^2$. – Michael Sep 30 2011 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 at 17:19

ulrich · Accepted Answer · 2011-09-30 12:39:48Z UTC

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.

A question about quotient singularity

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