Group Action on a Scheme
$begingroup$
Let $X$ be scheme and $G subset Aut(X)$ be a subgroup of automorphism group of $X$. By definition $G$ acts espectially on local sections $mathcal{O}_X(U)$ for open $U$ and one can therefore define the subsheaf $mathcal{O}_X^G$ of $G$-invariants on $X$.
My question is if and how there exist a canonical way to extend the action of $G$ to all coherent sheaves $mathcal{F}$ over $X$ or resp $mathcal{O}_X$ modules?
Background: How to define the sub sheaves of invariants $mathcal{F}^G$ for arbitrary $mathcal{F}$?
If yes, does $G$ induce a functor $^G$ on the category of sheaves over $X$? When is it exact?
algebraic-geometry sheaf-theory group-actions coherent-sheaves
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
$endgroup$
add a comment |
$begingroup$
Let $X$ be scheme and $G subset Aut(X)$ be a subgroup of automorphism group of $X$. By definition $G$ acts espectially on local sections $mathcal{O}_X(U)$ for open $U$ and one can therefore define the subsheaf $mathcal{O}_X^G$ of $G$-invariants on $X$.
My question is if and how there exist a canonical way to extend the action of $G$ to all coherent sheaves $mathcal{F}$ over $X$ or resp $mathcal{O}_X$ modules?
Background: How to define the sub sheaves of invariants $mathcal{F}^G$ for arbitrary $mathcal{F}$?
If yes, does $G$ induce a functor $^G$ on the category of sheaves over $X$? When is it exact?
algebraic-geometry sheaf-theory group-actions coherent-sheaves
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
$endgroup$
add a comment |
$begingroup$
Let $X$ be scheme and $G subset Aut(X)$ be a subgroup of automorphism group of $X$. By definition $G$ acts espectially on local sections $mathcal{O}_X(U)$ for open $U$ and one can therefore define the subsheaf $mathcal{O}_X^G$ of $G$-invariants on $X$.
My question is if and how there exist a canonical way to extend the action of $G$ to all coherent sheaves $mathcal{F}$ over $X$ or resp $mathcal{O}_X$ modules?
Background: How to define the sub sheaves of invariants $mathcal{F}^G$ for arbitrary $mathcal{F}$?
If yes, does $G$ induce a functor $^G$ on the category of sheaves over $X$? When is it exact?
algebraic-geometry sheaf-theory group-actions coherent-sheaves
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
$endgroup$
Let $X$ be scheme and $G subset Aut(X)$ be a subgroup of automorphism group of $X$. By definition $G$ acts espectially on local sections $mathcal{O}_X(U)$ for open $U$ and one can therefore define the subsheaf $mathcal{O}_X^G$ of $G$-invariants on $X$.
My question is if and how there exist a canonical way to extend the action of $G$ to all coherent sheaves $mathcal{F}$ over $X$ or resp $mathcal{O}_X$ modules?
Background: How to define the sub sheaves of invariants $mathcal{F}^G$ for arbitrary $mathcal{F}$?
If yes, does $G$ induce a functor $^G$ on the category of sheaves over $X$? When is it exact?
algebraic-geometry sheaf-theory group-actions coherent-sheaves
algebraic-geometry sheaf-theory group-actions coherent-sheaves
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
asked Dec 27 '18 at 1:28
KarlPeterKarlPeter
4521315
4521315
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Ok, this question makes more sense to me than your other (earlier) question.
$G$ doesn't act on local sections. $G$ isn't guaranteed to preserve open sets, (i.e. $gUne U$ necessarily) so how could it possibly act on $newcommandcalO{mathcal{O}}calO_X(U)$ for all open $U$?
Instead, if $U$ is $G$-invariant, $G$ descends to an action on $calO_X(U)$ in the obvious way. Note, however that $G$ acts by ring homomorphisms not module morphisms.
In general, the action of $G$ on $X$ will not induce an action of $G$ on a particular $calO_X$-module. For example, if we take a sky scraper sheaf over some point with nontrivial orbit, then clearly $G$ will act by sending the sky scraper sheaf to a sky scraper sheaf with a different base point.
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
$endgroup$
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
1
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
|
show 3 more comments
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1 Answer
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$begingroup$
Ok, this question makes more sense to me than your other (earlier) question.
$G$ doesn't act on local sections. $G$ isn't guaranteed to preserve open sets, (i.e. $gUne U$ necessarily) so how could it possibly act on $newcommandcalO{mathcal{O}}calO_X(U)$ for all open $U$?
Instead, if $U$ is $G$-invariant, $G$ descends to an action on $calO_X(U)$ in the obvious way. Note, however that $G$ acts by ring homomorphisms not module morphisms.
In general, the action of $G$ on $X$ will not induce an action of $G$ on a particular $calO_X$-module. For example, if we take a sky scraper sheaf over some point with nontrivial orbit, then clearly $G$ will act by sending the sky scraper sheaf to a sky scraper sheaf with a different base point.
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
$endgroup$
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
1
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
|
show 3 more comments
$begingroup$
Ok, this question makes more sense to me than your other (earlier) question.
$G$ doesn't act on local sections. $G$ isn't guaranteed to preserve open sets, (i.e. $gUne U$ necessarily) so how could it possibly act on $newcommandcalO{mathcal{O}}calO_X(U)$ for all open $U$?
Instead, if $U$ is $G$-invariant, $G$ descends to an action on $calO_X(U)$ in the obvious way. Note, however that $G$ acts by ring homomorphisms not module morphisms.
In general, the action of $G$ on $X$ will not induce an action of $G$ on a particular $calO_X$-module. For example, if we take a sky scraper sheaf over some point with nontrivial orbit, then clearly $G$ will act by sending the sky scraper sheaf to a sky scraper sheaf with a different base point.
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
$endgroup$
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
1
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
|
show 3 more comments
$begingroup$
Ok, this question makes more sense to me than your other (earlier) question.
$G$ doesn't act on local sections. $G$ isn't guaranteed to preserve open sets, (i.e. $gUne U$ necessarily) so how could it possibly act on $newcommandcalO{mathcal{O}}calO_X(U)$ for all open $U$?
Instead, if $U$ is $G$-invariant, $G$ descends to an action on $calO_X(U)$ in the obvious way. Note, however that $G$ acts by ring homomorphisms not module morphisms.
In general, the action of $G$ on $X$ will not induce an action of $G$ on a particular $calO_X$-module. For example, if we take a sky scraper sheaf over some point with nontrivial orbit, then clearly $G$ will act by sending the sky scraper sheaf to a sky scraper sheaf with a different base point.
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
$endgroup$
Ok, this question makes more sense to me than your other (earlier) question.
$G$ doesn't act on local sections. $G$ isn't guaranteed to preserve open sets, (i.e. $gUne U$ necessarily) so how could it possibly act on $newcommandcalO{mathcal{O}}calO_X(U)$ for all open $U$?
Instead, if $U$ is $G$-invariant, $G$ descends to an action on $calO_X(U)$ in the obvious way. Note, however that $G$ acts by ring homomorphisms not module morphisms.
In general, the action of $G$ on $X$ will not induce an action of $G$ on a particular $calO_X$-module. For example, if we take a sky scraper sheaf over some point with nontrivial orbit, then clearly $G$ will act by sending the sky scraper sheaf to a sky scraper sheaf with a different base point.
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
edited Dec 27 '18 at 3:02
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
answered Dec 27 '18 at 2:25
jgonjgon
15.3k32042
15.3k32042
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
1
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
|
show 3 more comments
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
1
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
So I guess this is the motivation to study stacks instead of schemes because one tends to run into trouble by trying to define a quotient structure for schemes. According to your explanations the object $X/G$ genereally isn't a scheme.
$endgroup$
– KarlPeter
Dec 27 '18 at 2:38
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
One more comment: But if we work with $G$-invariant schemes (so the action can indeed be extended to structure sheaf) does this action extends in reasoable way to all sheaves $mathcal{F}$ on X. The point is that I often see that one talks without calling the construction about "sheaf of invariants" $mathcal{F}^G$. So somehow it should be arrise. Or is this construction only possible with another group action eg. Galois as introduced in another thread of mine?
$endgroup$
– KarlPeter
Dec 27 '18 at 2:48
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter, while that is true, I'm not sure my answer quite gets at the issue there. In this answer I'm just trying to clarify precisely how the action works, and why it doesn't descend to a local action. It is certainly true though that forming correct quotients is one motivation for stacks.
$endgroup$
– jgon
Dec 27 '18 at 2:51
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
$begingroup$
@KarlPeter In response to your second comment. I'm not sure what a $G$-invariant scheme would be. Oh, where every open set is $G$-invariant? I suspect that would be a very strong requirement, perhaps even forcing the action of $G$ to be trivial. I don't think an action on the structure sheaf will naturally produce an action on modules over the structure sheaf. I could be wrong there, but an action would need to be produced. In the case of the Galois group action, since the sheaves came from $X$ rather than $X_L$ when you pulled them back to $X_L$ they got this nice $G$ action.
$endgroup$
– jgon
Dec 27 '18 at 2:56
1
1
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
$begingroup$
@KarlPeter, it depends on how precisely you define the subsheaf of invariants. One approach is to find invariant open subsets of a noninvariant open subset by taking the orbit of that open subset and intersecting those sets. You need finiteness of $G$ there (or rather finiteness of the orbit) to guarantee that the intersection is open.
$endgroup$
– jgon
Dec 27 '18 at 3:19
|
show 3 more comments
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