9
$\begingroup$

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).

$\endgroup$
2
  • 3
    $\begingroup$ What one would like in general is for the Chen-Ruan "orbifold cohomology" of an orbifold, which in this situation should be $Rep(G)$, to match the ordinary cohomology of a crepant resolution. This is perhaps more of an interpretation than an explanation. $\endgroup$ Nov 7, 2014 at 3:35
  • $\begingroup$ @ksf: Maybe it would help to describe also an exceptional case such as $E_6$. $\endgroup$ Jan 6, 2015 at 17:02

2 Answers 2

7
$\begingroup$

(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.

$\endgroup$
2
  • $\begingroup$ This is suggestive, but it would help to define more explicitly your notation such as $Z.\alpha_i$. (Also, the usual translation for "plus grande racine" is "highest root", relative to the natural "height" defined by a choice of simple roots: the sum of coefficients when $\alpha$ is written as a $\mathbb{Z}$-linear combination.) $\endgroup$ Jan 6, 2015 at 16:59
  • $\begingroup$ Thanks, that definitely helps explain what is "special" about the exceptional curves appearing in the first blowup. It seems like maybe the root lattice explanation could be made natural with some more analysis. Using the negative intersection with the fundamental cycle in combination with a formula from Gonzalez-Sprinberg and Verdier's McKay correspondence paper, there's an interpretation in terms of representations of the group G, which I'll try to post below. $\endgroup$
    – ksf
    May 8, 2015 at 12:05
2
$\begingroup$

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 $-1$.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.