TITLE: Resolving ADE singularities by blowing up
QUESTION [8 upvotes]: Let's say we have a finite subgroup $\Gamma \subseteq SL(2,\mathbb{C})$ and consider the quotient variety $\mathbb{C}^2/\Gamma$, which will have one of the well-known ADE or du Val surface singularities and can be embedded into $\mathbb{C}^3$ as a hypersurface with a singular point at the origin.  These singularities have crepant resolutions where the exceptional fiber is a union of $\mathbb{P}^1$s connected according to an ADE Dynkin diagram.  They can be obtained by repeatedly blowing up at singular points, starting with blowing up at the origin in $\mathbb{C}^3$.
When doing this it seems that after the first blowup, the exceptional curve or curves you get correspond to the natural representation of $\Gamma$ on $\mathbb{C}^2$ that is given by its inclusion into $SL(2,\mathbb{C})$.  (This is under the McKay correspondence, which gives a one-to-one correspondence between nodes in the Dynkin diagram and nontrivial irreducible representations of $\Gamma$ over $\mathbb{C}$.)  For instance, if you start with an $A_n$ singularity, defined by $x^2+y^2+z^{n+1}=0$ in $\mathbb{C}^3$, and blow up at the origin, the exceptional divisor will be $X^2+Y^2=0$ in $\mathbb{P}^2$ which is two $\mathbb{P}^1$s meeting at a point.  These two $\mathbb{P}^1$s correspond to the outermost nodes in the $A_n$ Dynkin diagram which are dual representations of $\Gamma$, the cyclic group of order $n+1$ in this case, that you get from its natural action on $\mathbb{C}^2$.  (The node or nodes for each $\Gamma$ that I'm describing are also the node(s) that you would connect to the additional node on an extended Dynkin diagram.)  Although I haven't checked recently I believe this also works for all the other ADE groups.   
My question is, is there some kind of "natural" explanation for this?  I was hoping this might be explained by one of the geometric interpretations of the McKay correspondence out there, like the Bridgeland-King-Reid "Mukai implies McKay" paper (which I don't thoroughly understand).

REPLY [2 votes]: We can use inkspot's key observation that the curves appearing in the first blowup are exactly the ones having negative intersection with the fundamental cycle.  Let $[\rho_i]$ be the exceptional curve corresponding to $\rho_i \in Irr(\Gamma)$ where $Irr(\Gamma)$ is the set of nontrivial irreducible representations of $\Gamma$.  Also let $c$ be the natural representation of $\Gamma$ on $\mathbb{C}^2$.  If we then further accept that:

The fundamental cycle is given by the formula $$\sum_{\rho_i \in Irr(\Gamma)} \deg(\rho_i) [\rho_i].$$
Let $C$ be the square matrix of dimension $|Irr(\Gamma)|$ defined by $C = (a_{ij})$, where $a_{ij}$ is the multiplicity of $\rho_i$ in $c \otimes \rho_j$.  For any $\rho_i, \rho_j \in Irr(\Gamma)$, not necessarily distinct, the intersection $[\rho_i].[\rho_j]$ is given by the $ij$ entry of the matrix $C-2I$.  ($2I-C$ is also the Cartan matrix of the appropriate root system.)

These facts are mentioned in Gonzalez-Sprinberg and Verdier's paper "Construction geometrique de la correspondance de McKay", Theorem 2.2, and they also have a natural explanation of fact 1.  In Prop. 1.4 they prove the formula
$$2-c = \sum_{\rho_i \in Irr(\Gamma)} (\deg \rho_i)(c-2)\rho_i$$ in the representation ring $R(\Gamma)$.  Combining the facts 1 and 2, we can see that after expanding everything out in the RHS, the coefficient on $\rho_i$ will be the intersection of $\rho_i$ with the fundamental cycle, for any $\rho_i \in Irr(\Gamma)$.  So the LHS says exactly that $[\rho_i]$ will have zero intersection with the fundamental cycle, unless $\rho_i$ appears in $c$, in which case it has intersection $-1$.