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Let $X$ be a normal projective surface with just two singular points $x_1,x_2\in X$, where $X$ has rational quotient singularities.

Assume that both the singularities in $x_1$ and in $x_2$ admit a non-trivial first order infinitesimal deformation.

Can we conclude that $X$ itself admit a non-trivial first order infinitesimal deformation?

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  • $\begingroup$ If I understand the cotangent complex correctly, this is computed by the spectral sequence for the hypercohomology of the tangent complex, but there is a differential from $H^0 R^2$ to $H^2 R^0$ that might be problematic. $\endgroup$ – Will Sawin Jul 7 '15 at 15:31
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You need to know that the natural map $$Ext^{1}(\Omega_{X},\mathcal{O}_{X})\rightarrow H^{0}(X,\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X}))$$ is surjective. For instance, this is true when the deformations of $X$ are unobstructed that is $H^{2}(X,T_X)=0$.

The vector space of locally trivial first order infinitesimal deformations of $X$ is $H^{1}(X,T_{X})$, where $T_{X} = \mathcal{H}om(\Omega_{X},\mathcal{O}_{X})$. While the vector space of first order infinitesimal deformations of $X$ is $Ext^{1}(\Omega_{X},\mathcal{O}_{X})$. In general, these two spaces are linked by the following exact sequence $$0\mapsto H^{1}(X,T_{X})\rightarrow Ext^{1}(\Omega_{X},\mathcal{O}_{X})\rightarrow H^{0}(X,\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X}))\rightarrow H^{2}(X,T_{X})$$ The sheaf $\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X})$ is supported on the singular locus of $X$.

Now, let $S$ be an étale neighborhood of $x_1$. Since your singularity admits non-trivial first order infinitesimal deformations we have $Ext(\Omega_S,\mathcal{O}_S)\neq 0$. Since $\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X})$ is supported on $x_1$ and $x_2$ we get that $H^{0}(X, \mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X}))\neq 0$.

Now, assume that the map $$Ext^{1}(\Omega_{X},\mathcal{O}_{X})\rightarrow H^{0}(X,\mathcal{E}xt^{1}(\Omega_{X},\mathcal{O}_{X}))$$ in the above exact sequence is surjective. This is the case when the deformations of $X$ are unobstructed that is $H^{2}(X,T_X)=0$. Clearly, this yields that $Ext^{1}(\Omega_{X},\mathcal{O}_{X})\neq 0$ and you are done.

As an example you can consider the weighted projective plane $\mathbb{P}(1,2,3)$. The singularity of type $\frac{1}{3}(1,2)$ admits non-trivial first order infinitesimal deformations. To see this just observe that is isomorphic to $\mathbb{A}^{2}/\mu_{3}$ where the action is given by $$ \begin{array}{ccc} \mu_{3}\times\mathbb{A}^{2} & \longrightarrow & \mathbb{A}^{2}\\ (\epsilon,x_{1},x_{2}) & \longmapsto & (\epsilon x_{1},\epsilon^{2}x_{2}) \end{array} $$ 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 the singularity the surface is isomorphic to an étale neighborhood of the vertex of the cone $$S = \{f(x,y,z) = z^{3}-xy = 0\}\subset\mathbb{A}^{3}$$ Now, we have $Ext^{1}(\Omega_{S},\mathcal{O}_{S})\cong K[x,y,z]/(f,\frac{\partial f}{\partial x},\frac{\partial f}{\partial y},\frac{\partial f}{\partial z}) = K[x,y,z]/(z^{3}-xy,-y,-x,3z^{2})\cong K[z]/(z^{2})$.

We have the following exact sequence $$0\mapsto \mathcal{O}_{\mathbb{P}(1,2,3)}\rightarrow\mathcal{O}_{\mathbb{P}(1,2,3)}(1)\oplus\mathcal{O}_{\mathbb{P}(1,2,3)}(2)\oplus\mathcal{O}_{\mathbb{P}(1,2,3)}(3)\rightarrow T_{\mathbb{P}(1,2,3)}\mapsto 0$$ Taking cohomology we get $h^{0}(T_{\mathbb{P}(1,2,3)}) = 5$ and $h^{1}(T_{\mathbb{P}(1,2,3)}) = h^{2}(T_{\mathbb{P}(1,2,3)}) = 0$. We conclude that $H^{i}(\mathbb{P}(1,2,3),T_{\mathbb{P}(1,2,3)}) = 0$ for $i\geq 1$.

Then, $Ext^{1}(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})\cong H^{0}(\mathbb{P}(1,2,3),\mathcal{E}xt^{1}(\Omega_{\mathbb{P}(1,2,3)},\mathcal{O}_{\mathbb{P}(1,2,3)})) \neq 0$, that is $\mathbb{P}(1,2,3)$ is not rigid.

For an example of infinitesimally rigid surfaces whose singularities admit non-trivial deformations you may consider Beauville surfaces. Roughly speaking, these are quotients $(C_1\times C_2)/G$ where $C_1, C_2$ are smooth curves of genus at least two, and $G$ is a finite group. Some of these surfaces have non rigid singularities, on the other hand Beauville surfaces are infinitesimally rigid. For this you may take a look at:

Ingrid Bauer, Shelly Garion, Alina Vdovina, Beauville Surfaces and Groups, Springer, 2015.

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    $\begingroup$ It would be more interesting (and a better answer to the question) to give an example where the map on your second line is not surjective. $\endgroup$ – abx Jul 7 '15 at 16:27

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