Relating deformations of a scheme to deformations of its singular locus

Question

Let $X$ be a normal scheme with quotient singularities and $Y\subset X$ its singular locus. The first order deformations of $X$ are parametrized by $\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X})$. This sheaf is supported on $Y$.

I would like to relate $\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X})$ and $\mathcal{E}xt^{1}(\Omega_{Y},\mathcal{O}_{Y})$ (here I am considering first order deformations of $Y$ as an abstract scheme not its embedded deformations as a subscheme of $X$). To do this I am thinking to consider the conormal exact sequence $$\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2}\rightarrow \Omega_{X|Y}\rightarrow\Omega_{Y}\mapsto 0$$ Applying $\mathcal{H}om(-,\mathcal{O}_{Y})$ we get $$0\mapsto \mathcal{H}om(\Omega_{Y},\mathcal{O}_{Y})\rightarrow \mathcal{H}om(\Omega_{X|Y},\mathcal{O}_{Y})\rightarrow \mathcal{H}om(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y})\rightarrow \mathcal{E}xt^{1}(\Omega_{Y},\mathcal{O}_{Y})\rightarrow \mathcal{E}xt^{1}(\Omega_{X|Y},\mathcal{O}_{Y})\rightarrow \mathcal{E}xt^{1}(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y})\rightarrow ....$$ Is there any reason why one should have $\mathcal{E}xt^{1}(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y}) = 0$ or $\mathcal{H}om(\mathcal{I}_{Y}/\mathcal{I}_{Y}^{2},\mathcal{O}_{Y}) = 0$ ?

Answer 1

Here is a partial answer.

Let $Y$ be a smooth variety over a field $k$ of characteristic zero. Let $G$ be a finite group acting on $Y$, and let $X = Y/G$ be the quotient. Assume that the set of points where the isotropy is not trivial is in codimension greater or equal than three, that is the singular locus of $X$ is in codimension greater or equal than three. Then $Ext^1(\omega_X,\mathcal{O}_X) = 0$, that is $X$ is rigid.

One can find this in:

• M. Schlessinger, "Rigidity of quotient singularities", Inventiones mathematicae, 1971, Volume 14, Issue 1, pp. 17-26.

This could fail is $X$ is singular in codimension two. For instance, we may consider the a singular point of type $\frac{1}{6}(2,4)$. Then, étale locally, in a neighborhood of $p$ the surface $X$ is isomorphic to $\mathbb{A}^{2}/\mu_{6}$ where the action is given by $$ \begin{array}{ccc} \mu_{6}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1},x_{2}) & \longmapsto & (\epsilon^{2}x_{1},\epsilon^{4}x_{2}) \end{array} $$ The invariant polynomials with respect to this action are clearly $x_{1}^{3},x_{2}^{3},x_{1}x_{2}$. Therefore, étale locally, in a neighborhood of $p$ the surface $X$ is isomorphic to an étale neighborhood of singularity $$S = \{f(x,y,z) = z^{3}-xy = 0\}\subset\mathbb{A}^{3}.$$ Now, we have $$Ext^{1}(\Omega_{S},\mathcal{O}_{S})\cong K[x,y,z]/(f,\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}) = K[x,y,z]/(z^{3}-xy,-y,-x,3z^{2})\cong K[z]/(z^{2}).$$

Therefore $X$ is not rigid.