# Equivariant projective modules and skew group algebras

This is a question related to the two dimensional McKay correspondence. Let $$R = \mathbb{C}[x,y]$$, and $$G$$ a finite group acting on $$R$$. Recall that a $$G$$-equivariant $$R$$-module is an $$R$$-module with a $$G$$ action, such that the multiplication map $$R \otimes M \to M$$ is $$G$$-equivariant, i.e., $$g *(rm) = (g\cdot r) (g *m)$$.

It is well known that the category of such modules, $$R\text{-Mod}_G$$, is equivalent to the category of modules over the (noncommutative, unital) skew group algebra $$R \# G$$, which is the algebra on the underlying vector space $$R \otimes_{\mathbb{C}} \mathbb{C}[G]$$, and multiplication defined by $$(r\otimes g)(r' \otimes g') = r(g\cdot r')\otimes gg'$$.

I would like to know what the projective objects are in the latter category. I know that Quillen-Suslin implies that finitely generated projective $$R$$-modules are free, and so I suspect that all the finitely generated projective equivariant $$R$$-modules are also free. However I would like to know a proof of this in the setting of the skew group ring. I am aware that the group ring $$\mathbb{C}[G]$$ is semisimple, so one would hope that $$R \# G$$ also has a similar decomposition, but I can only see this as vector spaces, rather then algebras. Moreover I'm still not sure how I would use this to get projective objects in $$R\# G$$-Mod.

• Almost immediately (sigh...) after posting this I found this question which is very related. I would still like to know why the decomposition of $R \# G$ holds as algebras, and I think that the question about projectives is still unclear, as certainly $R\# G$ can't be semisimple (there are $R\# G$ modules which are not projective, right?) – DKS Mar 22 at 18:03

## 2 Answers

A first observation is that the centre $$Z=Z(R\#G)$$ is isomorphic to the fixed point ring $$R^G$$. Next, we can define a ring homomorphism $$R\# G \to \mathrm{End}_Z(R), \quad r\otimes g \mapsto (t\mapsto rg(t)).$$ Auslander showed that this is actually an isomorphism.

We therefore have an equivalence between the category $$\mathrm{proj}(R\# G)$$ of finitely generated projective $$R\# G$$-modules, and the category $$\mathrm{add}_Z(R)$$ of $$Z$$-module direct summands of $$R^n$$ for $$n\geq1$$.

We also have an equivalence between $$\mathrm{proj}(R\# G)$$ and the semisimple category $$\mathrm{rep}_{\mathbb C}\,G=\mathrm{mod}\,\mathbb C[G]$$.

This sends a $$G$$-representation $$V$$ to $$R\otimes_{\mathbb C}V$$, with its natural $$R\#G$$-module structure. This will be projective since there exists some $$W$$ such that $$V\oplus W$$ is a free $$\mathbb C[G]$$-module. Alternatively it is extension of scalars along the inclusion $$\mathbb C[G]\subset R\#G$$.

Conversely it sends a projective $$R\#G$$-module $$P$$ to the quotient $$P/(xP+yP)$$. Alternatively this is extension of scalars along the quotient map $$R\#G\to\mathbb C[G]$$ which sends $$x,y$$ to zero.

This latter equivalence shows that the isomorphism classes of indecomposable projective $$R\#G$$-modules are in bijection with the isomorphism classes of irreducible $$G$$-representations.

Update: The paper by Auslander is 'On the purity of the branch locus' Amer J Math 84 (1962), and note that he takes $$R=[[x,y]]$$ with $$G$$ acting linearly.

A good reference is chapter 5 of 'Cohen-Macaulay Representations' by Leuschke and Wiegand.

Note that if we pass to the quotient fields, then this is just Galois theory. Let $$L=\mathbb C(x,y)$$ and $$K=L^G$$. Then $$L/K$$ is a Galois extension with Galois group $$G$$, and so $$L\#G\cong \mathrm{End}_K(L)$$.

• Thank you for your excellent answer. Do you happen to have a reference for the result of Auslander that you quoted? – DKS Apr 2 at 12:00

You need the action to be faithful to have the centre be the fixed point ring. The endomorphism ring and the skew group ring are isomorphic when the group is a small subgroup of GL.