Let $\Gamma$ be a finite subgroup of $SL_2({\mathbb C})$, and $Q$ the set of its irreducible representations. McKay makes $Q$ into a directed graph by having $V \to W$ if $W \leq V \otimes {\mathbb C}^2$, where the latter comes from the natural action of $\Gamma$ on ${\mathbb C}^2$. (But since ${\mathbb C}^2 \otimes {\mathbb C}^2$ has an $SL_2$-invariant hence $\Gamma$-invariant vector, the directed graph is actually undirected: each edge comes with its reverse.) In this way we get a graph with a vector space at each edge.

McKay observes that the graphs so arising are exactly the simply-laced affine Dynkin diagrams, with the trivial rep as the affine node. (In particular, the extra symmetry of the affine diagram comes here from $(\Gamma / \Gamma')^*$, which is therefore identifiable with $Z(G)$, for $G$ the corresponding simply-connected Lie group. I wonder if there's some larger correspondence there... but that's not my question.)

If we toss that node, and orient the edges (i.e. throw out half), we can look at the "roots", or indecomposable representations, of the resulting quiver. McKay observes further (in effect) that the largest such quiver representation has the same-dimensional vector spaces as in the first construction. But now, since it's a quiver representation, there are maps between the spaces. So my question:

In McKay's construction, the vertices of a Dynkin diagram are labeled by nontrivial irreps $\{V\}$ of $\Gamma$. Given an orientation on the diagram and an edge $V \to W$, is there a natural linear map $V \to W$, such that the result is the largest indecomposable quiver representation?

Obviously these maps aren't $\Gamma$-equivariant. The natural map is $V \otimes {\mathbb C}^2 \to W$, so maybe these other maps correspond to choosing a vector, or a list of vectors, in ${\mathbb C}^2$. So a more specific version of the question:

If $\vec x$ is a generic vector in ${\mathbb C}^2$, e.g. with no $\Gamma$-stabilizer, do the resulting composite maps $V \cong V \otimes \vec x \hookrightarrow V \otimes {\mathbb C}^2 \twoheadrightarrow W$ give the largest indecomposable? If so, what if $\vec x$ isn't generic?

Feel free to add tags; I couldn't think of anything other than rt.representation-theory.

  • $\begingroup$ This is true (and easy) in the type A case (just avoid the coordinate lines). I think it's also pretty easy in type D (you have avoid all the reflecting lines). Of course, in type E, it's less clear; probably it works, but I wouldn't be that shocked if it failed. $\endgroup$
    – Ben Webster
    Oct 12, 2012 at 20:51
  • $\begingroup$ There's an equivalence of derived categories (G-equivariant coherent sheaves on C^2) = (coherent sheaves on a minimal resolution of the Duval singularity). It sends a nontrivial irrep of G (supported at the origin in C^2) to the structure sheaf of a component of the exceptional fiber. I wonder which sheaf on C^2 corresponds to a skyscraper on a node in the exceptional fiber. Whatever it is it has maps to and from the irrep. $\endgroup$ Oct 12, 2012 at 21:32
  • $\begingroup$ The structure sheaf of the primary component, or the irreducible component? I'm pretty sure the exceptional fiber isn't reduced for D,E. $\endgroup$ Oct 12, 2012 at 23:22
  • 1
    $\begingroup$ I was a little off: a nontrivial irrep matches to O(-1) on a reduced P^1. The trivial irrep matches to the structure sheaf of the scheme-theoretic exceptional fiber, shifted by 1. More stuff like this available here arxiv.org/abs/math/9812016 $\endgroup$ Oct 13, 2012 at 3:39

1 Answer 1


Part of what makes this question more interesting outside type $A$ is that the highest root can't be projective or injective. A test case to consider is $D_4$, say with all three arrows pointing to the central vertex. A representation of $D_4$ with dimension vector (1,1,1,2) will be indecomposable provided you don't choose any zero maps or have any two of the maps be multiples.

In your setting, this translates into saying that you shouldn't choose the zero vector at any of the three vertices, and they shouldn't map to the same line in the 2-dimensional vector space over the central vertex either. It's not quite clear to me how to express this as a condition on the choice of vectors at the three vertices, but independent generic choices would work.

It's a general fact about quiver representations that if the dimensions come from a real root, then if you choose the maps generically, you will get the indecomposable representation. So one way to view the question is whether choosing a vector $v$ for each vertex is sufficiently generic.

For $D_n$ with $n>4$, I find it plausible that someone a bit more comfortable with representation theory of finite groups than I am, could convince themselves that this works fine as well. The situation for $E$ seems more complicated.

The following is not an answer to the question, but if it's unfamiliar to you, it might be of interest. This setup is called the "algebraic McKay correspondence", and is due to Auslander.

Let $S=\mathbb C[x,y]$. Let $R=S^G$. As an $R$-module, $S$ is a direct sum of the maximal Cohen-Macaulay $R$-modules. The maximal CM $R$-modules can also be constructed from the irreps of $\Gamma$ as follows: for $V$ an irrep, take $(V\otimes_R S)^G$.

The endomorphism ring of $S$ as an $R$-module is the preprojective algebra of the corresponding affine type. (If we throw away the node for $R$, we get the finite type.)

From this point of view, the arrows of the McKay quiver come with maps for free. In the $A_n$ case, they are just multiplication by $x$ and $y$, so the preprojective relation is just the fact that $xy=yx$.

However, I don't see how to get back to finite-dimensional $\mathbb C$-vector spaces in a natural way now.


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.