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Let $Y$ be an affine scheme over a field of characteristic zero. Suppose we have a group $G$ acting on $Y$ and that the subset of $Y$ of points with non-trivial stabilizer is in codimension greater or equal than $3$. Then, by a theorem due to Schlessinger $X:=Y/G$ is rigid that is $X$ does not have non-trivial first order deformations.

I know that a quadric cone in $\mathbb{A}^{3}$ admits non-trivial first order deformations.

Does anyone know an example of a $3$-fold with finite quotient singularities and singular in codimension $2$ admitting non-trivial first order deformations?

Is it true the naive statement: "the dimension of the space of deformations is bigger for a bad singularity than for a mild one"?

For instance is it true that non-canonical singularities are not rigid?

Thank you very much.

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    $\begingroup$ For your first question,a stupid example is $\mathbb{A}^1\times Q$, where $Q$ is a quadric cone in $\mathbb{A}^3$. $\endgroup$
    – abx
    Commented Feb 3, 2014 at 18:57
  • $\begingroup$ 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 $2$. $\endgroup$
    – Will Sawin
    Commented Feb 3, 2014 at 19:08
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    $\begingroup$ @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. $\endgroup$
    – Jason Starr
    Commented Feb 3, 2014 at 19:52
  • $\begingroup$ I should have said that locally somewhere it looks like that, not locally everywhere. $\endgroup$
    – Will Sawin
    Commented Feb 3, 2014 at 20:04

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Regarding your last question, the answer is yes, since there are terminal singularities that are not rigid.

For instance, in the recent preprint by Taro Sano On deformations of Fano threefolds with terminal singularities it is shown that any $\mathbb{Q}$-Fano threefold with "ordinary" terminal singularities (the meaning of "ordinary" is explained in the paper) can be deformed to a $\mathbb{Q}$-Fano threefold with only quotient singularities.

Moreover, the Kuranishi space of a $\mathbb{Q}$-Fano 3-fold $X$ is smooth, hence all first-order deformations of $X$ are unobstructed.

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  • $\begingroup$ Thank you for the answer. I look at the paper. You are right there are non-rigid terminal singularities. Do you think the following is true: "Let $X$ be a normal variety with finite quotient canonical singularities in codimension two. Then $X$ is not rigid". Thank you. $\endgroup$
    – Puzzled
    Commented Feb 3, 2014 at 22:00

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