Let us consider the surface $\mathbb{A}^{2}/\mu\_{6}$ 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 $x\_{1}^{3},x\_{2}^{3},x\_{1}x\_{2}$. Therefore we can interpret the surface $\mathbb{A}^{2}/\mu\_{6}$ as
$$S = \{f(x,y,z) = z^{3}-xy = 0\}\subset\mathbb{A}^{3}.$$
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 $Ext^{1}(\Omega\_{S},\mathcal{O}\_{S})\neq 0$ and $S$ admits non-trivial infinitesimal first order deformations. **Is any deformation in $Ext^{1}(\Omega\_{S},\mathcal{O}\_{S})$ locally trivial ?** 



Locally trivial deformations of surfaces with quotient singularities
This is a $A\_2$ surface singularity, in fact it is isomorphic to the quotient $\mathbb{A}^{2}/\mu\_{3}$ 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 explicit expression of its versal deformation is given by $$a+xy-3bz+z^3=0,$$
where $(a, \, b) \in \mathbb{C}^2$.


Taking derivatives, one easily checks that if $a= \pm 2 b \sqrt{b}$ and $(a, b) \neq (0,0)$, the corresponding germ has a ordinary double point at $P\_{\pm}=(0, \, 0, \, \pm \sqrt{b})$ and no further singularities, whereas if $a^2-4b^3 \neq 0$ the germ is smooth.


So we have a $A\_2$-singularity if and only if $(a, \,b)=(0, \, 0)$, which corresponds to the trivial deformation. In other words there are no equisingular deformations apart from the trivial one, hence no locally trivial deformations.


