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 nontrivial infinitesimal first order deformations. Is any deformation in $Ext^{1}(\Omega_{S},\mathcal{O}_{S})$ locally trivial ?

$\begingroup$ If you consider the local to global spectral sequence for $Ext$, it follows that the locally trivial deformations are classified by an $H^1$, so these will always vanish on affine schemes. (This is with the natural meaning of locally trivial  below someone has suggested that perhaps you mean equisingular.) $\endgroup$ – Tom Graber Feb 10 '14 at 3:58
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 $$a+xy3bz+z^3=0,$$ 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 $a^24b^3 \neq 0$ 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.

$\begingroup$ It seems that your argument implies that there are not deformations at all i.e. that $S$ is rigid. Is this true? $\endgroup$ – A. M. Flat_Line Feb 9 '14 at 16:50

$\begingroup$ Not at all: look at the versal deformation given by Francesco, it depends on two parameters. And the general deformation is smooth. $\endgroup$ – abx Feb 9 '14 at 16:57

$\begingroup$ Let me interpret the answer in another way. Consider the exact sequence $0\mapsto H^{1}(S,T_{S})\rightarrow Ext^{1}(\Omega_{S},\mathcal{O}_{S})\rightarrow H^{0}(S,\mathcal{E}xt^{1}(\Omega_{S},\mathcal{O}_{S}))$. Then $H^{1}(S,T_{S}) = 0$ and $H^{0}(S,\mathcal{E}xt^{1}(\Omega_{S},\mathcal{O}_{S}))\cong Ext^{1}(\Omega_{S},\mathcal{O}_{S})\cong k[z]/(z^{2})$. Is this correct? $\endgroup$ – A. M. Flat_Line Feb 9 '14 at 16:58

$\begingroup$ Could you give me an example of a locally trivial deformation of a surface singularity? Thank you. $\endgroup$ – A. M. Flat_Line Feb 9 '14 at 17:08

$\begingroup$ By "locally trivial" you mean equisingular? In this case, some cones do the job, since they have nontrivial, equisingular deformations. See link.springer.com/article/10.1007%2FBF01303329 $\endgroup$ – Francesco Polizzi Feb 9 '14 at 17:29