Resolving ADE singularities by blowing up

Question

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).

Answer 1

(Maybe this is an answer, maybe it merely moves the lump around under the carpet.) M. Artin showed ["On rational singularities of surfaces"] that for any rational surface singularity (in particular, for du Val singularities) the fundamental cycle $Z$ (by definition, the smallest non-zero effective divisor $Y$ supported on the exceptional locus of the minimal resolution $X$ such that $Y.C\le 0$ for all exceptional curves $C$ on $X$) is defined, as a scheme, by the maximal ideal of the original singularity. Therefore $X$ factors through the first blow-up and the exceptional curves that appear there are exactly those whose strict transforms on $X$ have strictly negative intersection with $Z$.

Now think of $Z$ merely as an element of the appropriate irreducible root lattice, bearing in mind that in this context the roots $\alpha$ have square (= self-intersection) $\alpha.\alpha$ equal to $-2$. Then $Z$ is characterized by the property that $Z$ is positive linear combination of the simple roots $\alpha_i$, the intersection number $Z.\alpha_i\le 0$ for each $i$ and $Z$ is minimal with respect to these conditions. Separate consideration of each of these (I don't know how to make this step "natural") shows that $Z$ equals the highest root ("plus grande racine"). Therefore the irreducible curve(s) that appear in the first blow-up correspond to the simple root(s) $\alpha_j$ such that $Z.\alpha_j<0$. In turn, further inspection of each Dynkin diagram shows that such $\alpha_j$ correspond to the node(s) that "you would connect to the additional node on an extended Dynkin diagram", exactly as you say.

Answer 2

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:

1. The fundamental cycle is given by the formula $$\sum_{\rho_i \in Irr(\Gamma)} \deg(\rho_i) [\rho_i].$$

2. 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 $-1$.