It is a theorem of Auslander that if $G< GL(2,\mathbb C)$ is a finite subgroup without pseudo-reflections, then the Auslander-Reiten quiver of $K[x,y]^G$ coincides with the McKay quiver of $G$ with the canonical representation. I would like to know what is known for arbitrary finite subgroups of $GL(2,\mathbb C)$, even those which contain pseudo-reflections. Do we still get this correspondence? If no, why not? Is there some specific cases where this is still true? Thank you.


This is not true.

First up, you need to take the completion of $k[x,y]^G$ before the statement even makes sense; for an AR sequence to exist you need your endomorphism rings to be local. Also, you might want a Krull-Schmidt category too, and this for these reasons Auslander studies $k[[x,y]]^G$.

Now in the case of pseudoreflections, being the quiver of the quotient stack, roughly speaking the McKay quiver remembers the action of G, as well as the lines of reflection. On the other hand, the category of CM modules of $k[[x,y]]^G$ only sees the singularity, not how it is created, which in the case of pseudoreflections is much "smaller" a singularity than you might expect. Thus, based on this intuition, we should expect the AR quiver to be smaller than the McKay quiver in general.

Indeed this is the case. Consider the example $\frac{1}{4}(1,2)$, then the invariants are $x^4$, $x^2y$ and $y^2$, which if we label as $A$, $B$ and $C$, satisfy the equation $AC=B^2$. Thus the AR quiver of the category of CM modules is (by Auslander) the McKay quiver for the group $\frac{1}{2}(1,1)$ which has only two vertices. On the other hand, the McKay quiver of $\frac{1}{4}(1,2)$ has four vertices.

The general statement is that the AR quiver of $k[[x,y]]^G$ is the same as the McKay quiver for the group $G/N$, where $N$ is the subgroup of $G$ generated by pseudoreflections.

  • $\begingroup$ Thank you very much for this complete answer. I think it makes sense when thinking about the result of Chevalley: If a group $G$ acting on $k[[x,y]]$ is generated by reflections, then there exist $G$-invariants $u,v$ such that $k[[x,y]]^G\cong k[[u,v]]$. Thus, in this case the Auslander-Reiten quiver as only one vertex, while the McKay quiver of $G$ is more complicated in general. $\endgroup$ – user42547 Nov 15 '13 at 18:47

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