Center of a simply-connected simple compact Lie group and McKay correspondence

Question

Let $G$ be a simply-connected simple compact Lie group. Its center $Z(G)$ is a finite abelian group, say $Z(G) = \mathbb Z/k\mathbb Z$ for $G=SU(k)$.

I find the following interpretation of $Z(G)$ in terms of McKay correspondence:

Let $\Gamma$ be a finite subgroup of $SU(2)$. Let $\{ \rho_i\}_{i\in I}$ be the set of (isomorphism classes of) irreducible representations of $\Gamma$. Let $Q$ be the $2$-dimensional representation of $\Gamma$ given by the inclusion $\Gamma\subset SU(2)$. Consider the tensor product decomposition $\rho_i\otimes Q = \bigoplus \rho_j^{\oplus a_{ij}}$. McKay correspondence says $2\delta_{ij} - a_{ij}$ is an affine Cartan matrix of type ADE.

Let $J$ be the subset of $I$ consisting of $1$-dimensional irreducible representations. It consists of $i$ such that the coefficient of $\alpha_i$ in the imaginary root $\delta$ is $1$. (Sometimes they are called special vertices.) We put $J$ an abelian group structure by the tensor product as representations of $\Gamma$. Then $J$ is isomorphic to $Z(G)$.

This observation can be checked by case-by-case analysis. It is easy for type $A_{k-1}$. The group $\Gamma$ is $\mathbb Z/k\mathbb Z$, $J$ is the whole $I$, and is isomorphic to $\Gamma$ itself as a group. For type $D_n$, $J$ consists of $4$ extremal vertices. One needs to know the dual representation of $\rho_j$ for $j\in J$ in order to determine the group $J$. It is given in Gozalez-Sprinberg Verdier paper, and the answer is: $J$ is either $\mathbb Z/2\mathbb Z\oplus \mathbb Z/2\mathbb Z$ or $\mathbb Z/4\mathbb Z$ according to $n$ is even or odd. This coincides with the center of $Spin(2n)$. Exceptional cases can be determined in the same way.

My questions are

1. Did anybody find this observation before ?

2. Is there a conceptual explanation of this observation ? In particular, is there a proof without case-by-case check ?

Answer 1

I believe the right reference is Borel-de Siebenthal.

A finite-dimensional proof is as follows. The space of conjugacy classes $G/\sim$ can be identified with $T/W = (Lie(T)/\Lambda)/W = Lie(T)/(\Lambda \rtimes W) = Lie(T)/{\widehat W}$, where $\Lambda = ker(\exp:Lie(T)\to T)$, and the final equality uses $G$ simply connected. Then this last space is the Weyl alcove $\Delta$.

There's an obvious action of $Z(G)$ on $G/\sim$, and the corresponding action on $\Delta$ is by isometries. In particular, it takes corners to corners. Some of those corners are in the orbit of the identity. So far we have an injection $Z(G) \to$ {corners of $\Delta$}, which of course then bijects to vertices of the affine Dynkin diagram. Lastly, we have to figure out the image.

Given a point $t \in T$ lying over some point $p$ in the relative interior of a $k$-face of $\Delta$, the centralizer $C_G(t)$ has central rank $k$, semisimple rank $rank(G)-k$. If $p$ is a corner, then $C_G(t)$ is semisimple, and the neighborhood of $p$ in $\Delta$ is the Weyl chamber of $C_G(t)$. So the root lattice of $C_G(t)$ is finite index in that of $G$, and this index is the coefficient of the corresponding simple root in the affine root.

The center is mapping to those vertices for which $C_G(t) = G$, i.e. this coefficient is therefore $1$.

Answer 2

Dave Morrison give me the following answer.

The center $Z(G)$ is known to be isomorphic to $P/Q$, where $P$ is the weight lattice and $Q$ is the root lattice.

On the other hand, let $X$ be the minimal resolution of $\mathbb C^2/\Gamma$. This is the place where McKay correspondence is realized by geometry. The configuration of the exceptional set is the finite Dynkin diagram, thus $Q$ is $H_2(X,\mathbb Z)$ with the intersection matrix. The weight lattice $P$ is isomorphic to $H^2(X,\mathbb Z)$ and the inclusion $Q\to P$ is realized by the intersection pairing, in other words, $Q\cong H^2_c(X,\mathbb Z)$ and $Q\to P$ is the natural homomorphism $H^2_c$ to $H^2$. Thus the quotient $P/Q$ is the cohomology of the `boundary' $H^2(S^3/\Gamma,\mathbb Z)$.

McKay correspondence a la Gonzalez-Spinberg Verdier gives a homomorphism from the representation ring $R(\Gamma)$ to $P$ by the first Chern classes of tautological bundles. The first Chern class is computed by the connecting homomorphism $H^1(S^3/\Gamma, U(1))\to H^2(S^3/\Gamma,\mathbb Z)$, which is an isomorphism in our case. Thus $H^2(S^3/\Gamma,\mathbb Z)$ is given by 1-dimensional characters of $\Gamma$.

Answer 3

This is not quite the answer, but we can use the representation theory of affine Lie algebras at level one to get a very similar observation, which is written e.g. di Francesco-Mathieu-Sénéchal.

1. Using the general theorem, unitary integrable representations at level k is given by the weights $w>0$ such that $w.\delta\le k$. In particular, at level 1, the set of unitary integrable representations can naturally be identified with $J$. An algebra structure on $\mathbb{C} J$ can be put by the fusion rule. Turns out it comes from a group structure on $J$.

2. At level 1, you can use the free-boson construction; then it's clear that the set of unitary integrable representation is given by $P/Q=Z$, and the fusion rule is given by the group structure of $Z$.

Comparing 1. and 2., we see $Z=J$, including the group structure. Now the question is to relate the affine algebra representations and the McKay correspondence, but I'm sure Hiraku can do that for me.