Equivariant projective modules and skew group algebras Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing $mathbb C G$ modules are projectiveCondition for a ring on projective and free modules problemWhat is the relation between graded modules and finitely generated modulesFinite projective dimension and vanishing of ext on f.g modulesProving that free modules are flat (without appealing projective modules)Graded projective modules and vector bundles on projective varietiesFinitely generated projective modules over a simple algebraic ring extension of a polynomial ringProjective objects in BGG category $mathcalO$ are projective $U(mathfrakg)$-modules?projective resolution of finitely generated modulesFinitely generated projective modules over polynomial rings with integral coefficientsProjective injective modules over algebras.

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Equivariant projective modules and skew group algebras



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)Showing $mathbb C G$ modules are projectiveCondition for a ring on projective and free modules problemWhat is the relation between graded modules and finitely generated modulesFinite projective dimension and vanishing of ext on f.g modulesProving that free modules are flat (without appealing projective modules)Graded projective modules and vector bundles on projective varietiesFinitely generated projective modules over a simple algebraic ring extension of a polynomial ringProjective objects in BGG category $mathcalO$ are projective $U(mathfrakg)$-modules?projective resolution of finitely generated modulesFinitely generated projective modules over polynomial rings with integral coefficientsProjective injective modules over algebras.










1












$begingroup$


This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, 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., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, 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 $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



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 $mathbbC[G]$ 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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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?)
    $endgroup$
    – DKS
    Mar 22 at 18:03















1












$begingroup$


This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, 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., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, 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 $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



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 $mathbbC[G]$ 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.










share|cite|improve this question











$endgroup$











  • $begingroup$
    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?)
    $endgroup$
    – DKS
    Mar 22 at 18:03













1












1








1


1



$begingroup$


This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, 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., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, 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 $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



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 $mathbbC[G]$ 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.










share|cite|improve this question











$endgroup$




This is a question related to the two dimensional McKay correspondence. Let $R = mathbbC[x,y]$, 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., $g *(rm) = (gcdot r) (g *m)$.



It is well known that the category of such modules, $Rtext-Mod_G$, 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 $R otimes_mathbbC mathbbC[G]$, and multiplication defined by $(rotimes g)(r' otimes g') = r(gcdot r')otimes gg'$.



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 $mathbbC[G]$ 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.







abstract-algebra algebraic-geometry representation-theory homological-algebra






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Mar 22 at 18:08







DKS

















asked Mar 22 at 17:53









DKSDKS

633414




633414











  • $begingroup$
    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?)
    $endgroup$
    – DKS
    Mar 22 at 18:03
















  • $begingroup$
    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?)
    $endgroup$
    – DKS
    Mar 22 at 18:03















$begingroup$
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?)
$endgroup$
– DKS
Mar 22 at 18:03




$begingroup$
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?)
$endgroup$
– DKS
Mar 22 at 18:03










1 Answer
1






active

oldest

votes


















4





+50







$begingroup$

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
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ 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 $R#Gtomathbb C[G]$ 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 $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$












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












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1 Answer
1






active

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votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4





+50







$begingroup$

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
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ 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 $R#Gtomathbb C[G]$ 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 $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$












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
















4





+50







$begingroup$

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
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ 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 $R#Gtomathbb C[G]$ 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 $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$












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














4





+50







4





+50



4




+50



$begingroup$

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
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ 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 $R#Gtomathbb C[G]$ 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 $L#Gcong mathrmEnd_K(L)$.






share|cite|improve this answer











$endgroup$



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
$$ R# G to mathrmEnd_Z(R), quad rotimes g mapsto (tmapsto rg(t)). $$
Auslander showed that this is actually an isomorphism.



We therefore have an equivalence between the category $mathrmproj(R# G)$ of finitely generated projective $R# G$-modules, and the category $mathrmadd_Z(R)$ of $Z$-module direct summands of $R^n$ for $ngeq1$.



We also have an equivalence between $mathrmproj(R# G)$ and the semisimple category $mathrmrep_mathbb C,G=mathrmmod,mathbb C[G]$.



This sends a $G$-representation $V$ to $Rotimes_mathbb CV$, with its natural $R#G$-module structure. This will be projective since there exists some $W$ such that $Voplus W$ 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 $R#Gtomathbb C[G]$ 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 $L#Gcong mathrmEnd_K(L)$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Apr 2 at 14:34

























answered Apr 2 at 8:23









Andrew HuberyAndrew Hubery

58625




58625











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

















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
















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





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


















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