Why can you deform singularities in two dimensions but not in higher dimensions?

I've been trying to read this paper to understand deformations of surface quotient singularities. I'm particularly interested in when one can deform certain cyclic quotient singularities into other cyclic quotient singularities.

I've heard that in higher dimensions, isolated quotient singularities are rigid. More precisely, if $X$ is an isolated quotient singularity by a finite group and $\dim X \geq 3$ then the only deformation of $X$ is the trivial one. What is some explanation or intuitive reason for why you can deform quotient surface singularities but you can't deform higher dimensional quotient singularities?

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This is a classical result of Schlessinger (Inventiones math. 14 (1971), 17-26). It is a nontrivial result and it is not easy to give an intuition of why it holds. –  abx Jan 30 '14 at 10:08

The rigidity of quotient singularities in dimension greater or equal than $3$ was established by Schlessinger in his paper Rigidity of quotient singularities, Invent. Math. 14 (1971). Roughly speaking, he proved that if $(X, \,x)$ is a local scheme with an isolated singularity at $x$ and $\dim X \geq 3$, then deforming $X$ is equivalent to deform the punctured spectrum $U=X-x$.

More precisely, the key result in the paper is the following

Proposition. Let $X$ be a geometric local scheme such that $X$ has an isolated singularity at his closed point $x$. If $\textrm{depth}_xX \geq 3$ then there is an isomorphism $$T_X^1 \cong H^1(U, \, \Theta_X),$$ where $U:=X - x$. More precisely, under these assumptions the formal deformation theories of $X$ and its punctured spectrum $U$ are equivalent.

Now, assume that $X=Y/G,$ where $Y$ is smooth and $G$ is a finite group of automorphisms acting with an isolated fixed point $y \in Y$. Let $p \colon Y \to X$ the quotient map and $x=p(y)$. Then $\textrm{depth}_xX \geq \textrm{depth}_yY$ and, if $\textrm{depth}_yY \geq 2$, one has $\Theta_Y=(p_*\Theta_X)^G.$

From this, if $\textrm{depth}_yY \geq 3$, it is not too difficult to show that $H^1(U, \, \Theta_X)=0$, hence using the Proposition one obtains $T_X^1=0,$ that is $X$ is rigid.

You can read Schlessinger's paper for further details.

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