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May 10, 2014 at 16:10 vote accept Puzzled
Feb 4, 2014 at 7:00 history edited Francesco Polizzi
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Feb 3, 2014 at 21:46 answer added Francesco Polizzi timeline score: 2
Feb 3, 2014 at 20:04 comment added Will Sawin I should have said that locally somewhere it looks like that, not locally everywhere.
Feb 3, 2014 at 19:52 comment added Jason Starr @WillSawin: "A codimension two finite quotient singularity ... has to locally look like ..." I believe this is incorrect. For instance, let the Klein Viergruppe, $V=(\mathbb{Z}/2\mathbb{Z})^3/\Delta(\mathbb{Z}/2\mathbb{Z})$ act on $\mathbb{A}^3$ by $\overline{e}_1(x_1,x_2,x_3) = (x_1,-x_2,-x_3)$, and similarly for $\overline{e}_2$ and $\overline{e}_3$. The quotient is the hypersurface in $\mathbb{A}^4$ of points $(y_1,y_2,y_3,z)$ such that $y_1y_2y_3-z^2 = 0$, where $y_i = x_i^2$ and $z = x_1x_2x_3$. This germ is not smooth over a two-dimensional germ.
Feb 3, 2014 at 19:08 comment added Will Sawin I think that is a non-stupid example. A codimension two finite quotient singularity in dimension $3$ has to locally look like $\mathbb A^1 \times X$ for $X$ a finite quotient singularity in dimension $2$.
Feb 3, 2014 at 18:57 comment added abx For your first question,a stupid example is $\mathbb{A}^1\times Q$, where $Q$ is a quadric cone in $\mathbb{A}^3$.
Feb 3, 2014 at 18:52 history asked Puzzled CC BY-SA 3.0