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.

  • $\begingroup$ Try $X=\mathbb{C}$ and $G=\{1,-1\}$ acting multiplicatively $\endgroup$ – Michael Bächtold Sep 30 '11 at 10:49
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    $\begingroup$ Michael, your example is smooth, but $\mathbb{C}^2/\pm 1$ is not. $\endgroup$ – Donu Arapura Sep 30 '11 at 11:38
  • $\begingroup$ @Donu: you're absolutely right! My mistake was in thinking of $\mathbb{C}$ as $\mathbb{R}^2$. $\endgroup$ – Michael Bächtold Sep 30 '11 at 17:49
  • $\begingroup$ @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? $\endgroup$ – 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.

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    $\begingroup$ 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. $\endgroup$ – Jason Starr Sep 30 '11 at 12:54

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