Mixing
Sheaves on a GIT quotient
$begingroup$
As stated in the title, my question regards sheaves on a GIT quotient. Let me fix the notation: $G$ is the group scheme acting on the scheme $X$ and both $X$ and $G$ are $k$-schemes. Searching online I found the theorem that states that in nice cases (e.g. if the action is free), then sheaves on $X/G$ are the same thing of equivariant sheaves on $X$. Let us consider this case. The notion of an equivariant sheaf is difficult to deal with, so I tried to reduce to the affine case and work out what happens, but I wasn't able to do it.
I know that if $M$ is a $R-Mod$ and $G$ is a group acting both on $M$ and $R$, for $M$ to be equivariant means that the multiplication map $R otimes M rightarrow M$ is $G$-equivariant. I expected to recover this notion for sheaves up to pass to the affine case. If $G = text{Spec} R'$ and $X = text{Spec} R$ then an action of $G$ on $X$ is the same as a map $phi : R rightarrow R otimes_{k} R'$. Therefore, for an $R$-module $M$ (thought as a sheaf on $X$) to be equivariant means that there exists an isomorphism of $R otimes_k R'$-modules
$$
M otimes_R left( R otimes_k R'right) rightarrow M otimes_R left( R otimes_k R' right)
$$
where the tensor product on the left is the one given via $phi$ and the other via left multiplication on $R$. How do this relate to the notion of equivariant module?
Moreover, once I know that a module is equivariant, is there a way to understand which is the module $overline{M}$ on $X/G$ such that $overline{M} otimes_{R^{G}} R simeq M$?
I didn't find references for these things, so any reference would be very welcome.
algebraic-geometry sheaf-theory quotient-spaces
asked Jan 20 at 17:00
FedericoFederico
893313
$endgroup$
add a comment |
$begingroup$
As stated in the title, my question regards sheaves on a GIT quotient. Let me fix the notation: $G$ is the group scheme acting on the scheme $X$ and both $X$ and $G$ are $k$-schemes. Searching online I found the theorem that states that in nice cases (e.g. if the action is free), then sheaves on $X/G$ are the same thing of equivariant sheaves on $X$. Let us consider this case. The notion of an equivariant sheaf is difficult to deal with, so I tried to reduce to the affine case and work out what happens, but I wasn't able to do it.
I know that if $M$ is a $R-Mod$ and $G$ is a group acting both on $M$ and $R$, for $M$ to be equivariant means that the multiplication map $R otimes M rightarrow M$ is $G$-equivariant. I expected to recover this notion for sheaves up to pass to the affine case. If $G = text{Spec} R'$ and $X = text{Spec} R$ then an action of $G$ on $X$ is the same as a map $phi : R rightarrow R otimes_{k} R'$. Therefore, for an $R$-module $M$ (thought as a sheaf on $X$) to be equivariant means that there exists an isomorphism of $R otimes_k R'$-modules
$$
M otimes_R left( R otimes_k R'right) rightarrow M otimes_R left( R otimes_k R' right)
$$
where the tensor product on the left is the one given via $phi$ and the other via left multiplication on $R$. How do this relate to the notion of equivariant module?
Moreover, once I know that a module is equivariant, is there a way to understand which is the module $overline{M}$ on $X/G$ such that $overline{M} otimes_{R^{G}} R simeq M$?
I didn't find references for these things, so any reference would be very welcome.
algebraic-geometry sheaf-theory quotient-spaces
asked Jan 20 at 17:00
FedericoFederico
893313
$endgroup$
$begingroup$
I think you have the right idea. One condition on an equivariant sheaf $mathcal{F}$ on $X$ is that you have an isomorphism $sigma^* mathcal{F} cong p_2^* mathcal{F}$, where $sigma,p_2 : G times X to X$ is the action map and projection respectively (there is another condition, but that's more complicated). If you reduce to the affine case and pass to the module category (assuming $mathcal{F}$ is an $mathcal{O}_X$-module) you should get something like the map you have above between the triple tensor product. I don't have pen and paper nearby, otherwise I would work it for you...
$endgroup$
– DKS
Jan 20 at 23:22
$begingroup$
A good elementary treatment of equivariant sheaves can be found in Chriss and Ginzburg's Representation Theory and Complex Geometry, I think it's at the beginning of the chapter on equivariant K-theory.
$endgroup$
– DKS
Jan 20 at 23:23
$begingroup$
@DKS Thank you for your answer. However, I know that’s the local description of the definition of an equivariant sheaf, my question was how to relate that local description to the intuitive idea that an equivariant sheaf should give, at least locally, an equivariant module.
$endgroup$
– Federico
Jan 21 at 10:07
$begingroup$
After thinking about this myself I'm also not sure. If you're willing to wait a bit I can ask someone who knows a few things about this stuff.
$endgroup$
– DKS
Jan 21 at 14:23
$begingroup$
@DKS That would be great, thank you.
$endgroup$
– Federico
Jan 21 at 14:26
add a comment |
$begingroup$
As stated in the title, my question regards sheaves on a GIT quotient. Let me fix the notation: $G$ is the group scheme acting on the scheme $X$ and both $X$ and $G$ are $k$-schemes. Searching online I found the theorem that states that in nice cases (e.g. if the action is free), then sheaves on $X/G$ are the same thing of equivariant sheaves on $X$. Let us consider this case. The notion of an equivariant sheaf is difficult to deal with, so I tried to reduce to the affine case and work out what happens, but I wasn't able to do it.
I know that if $M$ is a $R-Mod$ and $G$ is a group acting both on $M$ and $R$, for $M$ to be equivariant means that the multiplication map $R otimes M rightarrow M$ is $G$-equivariant. I expected to recover this notion for sheaves up to pass to the affine case. If $G = text{Spec} R'$ and $X = text{Spec} R$ then an action of $G$ on $X$ is the same as a map $phi : R rightarrow R otimes_{k} R'$. Therefore, for an $R$-module $M$ (thought as a sheaf on $X$) to be equivariant means that there exists an isomorphism of $R otimes_k R'$-modules
$$
M otimes_R left( R otimes_k R'right) rightarrow M otimes_R left( R otimes_k R' right)
$$
where the tensor product on the left is the one given via $phi$ and the other via left multiplication on $R$. How do this relate to the notion of equivariant module?
Moreover, once I know that a module is equivariant, is there a way to understand which is the module $overline{M}$ on $X/G$ such that $overline{M} otimes_{R^{G}} R simeq M$?
I didn't find references for these things, so any reference would be very welcome.
algebraic-geometry sheaf-theory quotient-spaces
asked Jan 20 at 17:00
FedericoFederico
893313
$endgroup$
As stated in the title, my question regards sheaves on a GIT quotient. Let me fix the notation: $G$ is the group scheme acting on the scheme $X$ and both $X$ and $G$ are $k$-schemes. Searching online I found the theorem that states that in nice cases (e.g. if the action is free), then sheaves on $X/G$ are the same thing of equivariant sheaves on $X$. Let us consider this case. The notion of an equivariant sheaf is difficult to deal with, so I tried to reduce to the affine case and work out what happens, but I wasn't able to do it.
I know that if $M$ is a $R-Mod$ and $G$ is a group acting both on $M$ and $R$, for $M$ to be equivariant means that the multiplication map $R otimes M rightarrow M$ is $G$-equivariant. I expected to recover this notion for sheaves up to pass to the affine case. If $G = text{Spec} R'$ and $X = text{Spec} R$ then an action of $G$ on $X$ is the same as a map $phi : R rightarrow R otimes_{k} R'$. Therefore, for an $R$-module $M$ (thought as a sheaf on $X$) to be equivariant means that there exists an isomorphism of $R otimes_k R'$-modules
$$
M otimes_R left( R otimes_k R'right) rightarrow M otimes_R left( R otimes_k R' right)
$$
where the tensor product on the left is the one given via $phi$ and the other via left multiplication on $R$. How do this relate to the notion of equivariant module?
Moreover, once I know that a module is equivariant, is there a way to understand which is the module $overline{M}$ on $X/G$ such that $overline{M} otimes_{R^{G}} R simeq M$?
I didn't find references for these things, so any reference would be very welcome.
algebraic-geometry sheaf-theory quotient-spaces
algebraic-geometry sheaf-theory quotient-spaces
asked Jan 20 at 17:00
FedericoFederico
893313
asked Jan 20 at 17:00
FedericoFederico
893313
asked Jan 20 at 17:00
FedericoFederico
893313
asked Jan 20 at 17:00
FedericoFederico
893313
asked Jan 20 at 17:00
FedericoFederico
893313
893313
$begingroup$
I think you have the right idea. One condition on an equivariant sheaf $mathcal{F}$ on $X$ is that you have an isomorphism $sigma^* mathcal{F} cong p_2^* mathcal{F}$, where $sigma,p_2 : G times X to X$ is the action map and projection respectively (there is another condition, but that's more complicated). If you reduce to the affine case and pass to the module category (assuming $mathcal{F}$ is an $mathcal{O}_X$-module) you should get something like the map you have above between the triple tensor product. I don't have pen and paper nearby, otherwise I would work it for you...
$endgroup$
– DKS
Jan 20 at 23:22
$begingroup$
A good elementary treatment of equivariant sheaves can be found in Chriss and Ginzburg's Representation Theory and Complex Geometry, I think it's at the beginning of the chapter on equivariant K-theory.
$endgroup$
– DKS
Jan 20 at 23:23
$begingroup$
@DKS Thank you for your answer. However, I know that’s the local description of the definition of an equivariant sheaf, my question was how to relate that local description to the intuitive idea that an equivariant sheaf should give, at least locally, an equivariant module.
$endgroup$
– Federico
Jan 21 at 10:07
$begingroup$
After thinking about this myself I'm also not sure. If you're willing to wait a bit I can ask someone who knows a few things about this stuff.
$endgroup$
– DKS
Jan 21 at 14:23
$begingroup$
@DKS That would be great, thank you.
$endgroup$
– Federico
Jan 21 at 14:26
add a comment |
$begingroup$
I think you have the right idea. One condition on an equivariant sheaf $mathcal{F}$ on $X$ is that you have an isomorphism $sigma^* mathcal{F} cong p_2^* mathcal{F}$, where $sigma,p_2 : G times X to X$ is the action map and projection respectively (there is another condition, but that's more complicated). If you reduce to the affine case and pass to the module category (assuming $mathcal{F}$ is an $mathcal{O}_X$-module) you should get something like the map you have above between the triple tensor product. I don't have pen and paper nearby, otherwise I would work it for you...
$endgroup$
– DKS
Jan 20 at 23:22
$begingroup$
A good elementary treatment of equivariant sheaves can be found in Chriss and Ginzburg's Representation Theory and Complex Geometry, I think it's at the beginning of the chapter on equivariant K-theory.
$endgroup$
– DKS
Jan 20 at 23:23
$begingroup$
@DKS Thank you for your answer. However, I know that’s the local description of the definition of an equivariant sheaf, my question was how to relate that local description to the intuitive idea that an equivariant sheaf should give, at least locally, an equivariant module.
$endgroup$
– Federico
Jan 21 at 10:07
$begingroup$
After thinking about this myself I'm also not sure. If you're willing to wait a bit I can ask someone who knows a few things about this stuff.
$endgroup$
– DKS
Jan 21 at 14:23
$begingroup$
@DKS That would be great, thank you.
$endgroup$
– Federico
Jan 21 at 14:26
$begingroup$
I think you have the right idea. One condition on an equivariant sheaf $mathcal{F}$ on $X$ is that you have an isomorphism $sigma^* mathcal{F} cong p_2^* mathcal{F}$, where $sigma,p_2 : G times X to X$ is the action map and projection respectively (there is another condition, but that's more complicated). If you reduce to the affine case and pass to the module category (assuming $mathcal{F}$ is an $mathcal{O}_X$-module) you should get something like the map you have above between the triple tensor product. I don't have pen and paper nearby, otherwise I would work it for you...
$endgroup$
– DKS
Jan 20 at 23:22
$begingroup$
I think you have the right idea. One condition on an equivariant sheaf $mathcal{F}$ on $X$ is that you have an isomorphism $sigma^* mathcal{F} cong p_2^* mathcal{F}$, where $sigma,p_2 : G times X to X$ is the action map and projection respectively (there is another condition, but that's more complicated). If you reduce to the affine case and pass to the module category (assuming $mathcal{F}$ is an $mathcal{O}_X$-module) you should get something like the map you have above between the triple tensor product. I don't have pen and paper nearby, otherwise I would work it for you...
$endgroup$
– DKS
Jan 20 at 23:22
$begingroup$
A good elementary treatment of equivariant sheaves can be found in Chriss and Ginzburg's Representation Theory and Complex Geometry, I think it's at the beginning of the chapter on equivariant K-theory.
$endgroup$
– DKS
Jan 20 at 23:23
$begingroup$
A good elementary treatment of equivariant sheaves can be found in Chriss and Ginzburg's Representation Theory and Complex Geometry, I think it's at the beginning of the chapter on equivariant K-theory.
$endgroup$
– DKS
Jan 20 at 23:23
$begingroup$
@DKS Thank you for your answer. However, I know that’s the local description of the definition of an equivariant sheaf, my question was how to relate that local description to the intuitive idea that an equivariant sheaf should give, at least locally, an equivariant module.
$endgroup$
– Federico
Jan 21 at 10:07
$begingroup$
@DKS Thank you for your answer. However, I know that’s the local description of the definition of an equivariant sheaf, my question was how to relate that local description to the intuitive idea that an equivariant sheaf should give, at least locally, an equivariant module.
$endgroup$
– Federico
Jan 21 at 10:07
$begingroup$
After thinking about this myself I'm also not sure. If you're willing to wait a bit I can ask someone who knows a few things about this stuff.
$endgroup$
– DKS
Jan 21 at 14:23
$begingroup$
After thinking about this myself I'm also not sure. If you're willing to wait a bit I can ask someone who knows a few things about this stuff.
$endgroup$
– DKS
Jan 21 at 14:23
$begingroup$
@DKS That would be great, thank you.
$endgroup$
– Federico
Jan 21 at 14:26
$begingroup$
@DKS That would be great, thank you.
$endgroup$
– Federico
Jan 21 at 14:26
add a comment |
2 Answers
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$begingroup$
So unfortunately my source had forgotten some of the details, so this is probably not perfect. I also found that a discussion of this is indeed in Representation Theory and Complex Geometry by Chriss and Ginzburg; with their help I was able to figure it out (I think, I'd be happy to discuss this further). Most of my answer below is paraphrased from that.
As you say in the description, from an affine group $G$ acting on an affine scheme $X = operatorname{Spec} A$ (assuming both are over a field $k$ for simplicity), an $mathcal{O}_X$-module $mathcal{M}$ comes from an $A$-module $M$, and the two maps $G times X rightrightarrows X$ (the action map and projection respectively) give two maps $tilde sigma, tilde p_2: A rightrightarrows mathcal{O}_G (G)otimes_k A$. Now if we assume that $mathcal{M}$ was equivariant, we have an isomorphism $sigma^* mathcal{M} cong p_2^* mathcal{M}$,
which when we take global sections, means that $phi: Gamma(Gtimes X, sigma^*mathcal{M}) to Gamma(Gtimes X, p_2^*mathcal{M})$
Now I claim that $M$ admits a structure of a $G$-module. First though, note that one can formulate the notion of an action differently. Let $operatorname{Map}(G,M)$ denote the vector space of arbitrary maps (this is verbatim from Chriss and Ginsburg, but probably 'arbitrary' is not right), then an action of $G$ on $M$ can be given as a map $M to operatorname{Map}(G,M)$. Indeed an action (in the ordinary sense) is a map $f:G times M to M$, so simply defining $m mapsto f(-,m)$ gives a map $M to operatorname{Map}(G,M)$. Conversely, suppose we have a map $varphi:M to operatorname{Map}(G,M)$. Then we can define an action by $g.m = varphi(m)(g^{-1})$.
Now define the action of $G$ on $M = Gamma(X,mathcal{M})$ to be the composition $$M overset{sigma^*}{to} Gamma(Gtimes X, sigma^*mathcal{M}) overset{phi}{to} Gamma(Gtimes X, p_2^*mathcal{M}) cong mathcal{O}_G(G) otimes_k Gamma(X,mathcal{M}).$$ Note that $mathcal{O}_G(G) otimes_k M$ lives in the space of maps as $sum f_i otimes m_i (g) = sum f_i(g) m_i$ is a map from $G to M$.
That this is compatible with the multiplication and such, I am told comes from the cocycle condition (I didn't verify this though).
answered Jan 22 at 22:08
DKSDKS
652412
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This is may be not a complete answer, this is just an example in an easy case, but I hope it will give you an idea of your problem. (I started to wrote this before @DKS
commented but decided to post anyway).
So let us have a concrete example : let $L/K$ a finite Galois extension with group $G$. Consider $X=operatorname{Spec} L$ which is endowed with an action of $G=operatorname{Spec}K[G]$. Here by $K[G]$ is not the group algebra, instead, this is the algebra $bigoplus_{gin G}K.g$ where the product is given component wise. In particular the identity element is $1=sum_{gin G}g$.
Then $K[G]$ is a coalgebra with coproduct given by $gmapsto sum_{hin G} gh^{-1}otimes h$. This coproduct induces $Gtimes Gto G$ when taking Specs. We have an coaction of $K[G]$ on $L$ given by $Lto Lotimes_K K[G]$ such that $lmapsto sum_{gin G} g(l)otimes g$. This coaction gives again the action of $Gtimesoperatorname{Spec}Ltooperatorname{Spec} L$ when taking Specs.
Now, an equivariant module is a vector space $V$ over $L$ with an isomorphism
$$phi:Votimes_L(Lotimes_K K[G])to Votimes_L(Lotimes_K K[G])$$
satisfying a cocycle condition (and remember that this is not the same tensor structure on the left and on the right).
I claim that this is the same thing as an action of $G$ on $V$ such that $g.(l v)=g(l)g(v)$. Indeed, given $phi$, $phi(votimes 1otimes 1)$ can be written $phi(votimes 1otimes 1)=sum_{gin G}v_gotimes 1otimes g$. We then put $g.v=v_g$.
Note that $phi(lvotimes 1otimes 1)=phi(sum_{gin G} votimes g(l)otimes h)$ since the tensor structure on the left is given by the coaction. Now we have by linearity
$$phi(votimes g(l)otimes g)=(sum_{hin G} v_hotimes 1otimes h)(g(l)otimes g)=v_gotimes g(l)otimes g=g(l)v_gotimes 1otimes g$$
(Remember that $gh=0$ in $K[G]$ if $gneq h$ from the definition of the algebra structure on $K[G]$). It follows that $phi(lvotimes 1otimes 1)=sum_{gin G}g(l)v_gotimes 1otimes g$, so $g.(lv)=g(l)g(v)$.
I let you check that this is an action, in other words that $(gh)v=g(h(v))$. You will need to use the cocycle condition. It is a bit tedious...
Conversely, the data of a semi-linear action of $G$ on $V$ is equivalent to the data of $phi$ by letting $phi(votimes 1otimes 1)=sum_{gin G} (g.v)otimes 1otimes g$ and extend by linearity. I let you check that $phi$ satisfy all the cocycle condition.
So now, what are the invariant ? Well this is "obviously" the one which are fixed by the action of $G$. In other words, this is the $vin V$ such that $g(v)=v$ for all $v$. And there is an easy way to express this in term of $phi$ : the invariant are exactly the $vin V$ such that $phi(votimes 1otimes 1)=votimes 1otimes 1$.
This is general : using your notation, we have a "fixed" module $overline{M}$ which consists of the $min M$ such that $phi(motimes 1otimes 1)=motimes 1otimes 1$.
Note that I never used the fact that the action of $G$ on $X$ is free or nice, and indeed, I did not prove that $overline{M}otimes_{R^G} Rsimeq M$.
answered Jan 21 at 20:53
RolandRoland
7,34411015
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Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
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– Federico
Jan 23 at 21:48
add a comment |
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2 Answers
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$begingroup$
So unfortunately my source had forgotten some of the details, so this is probably not perfect. I also found that a discussion of this is indeed in Representation Theory and Complex Geometry by Chriss and Ginzburg; with their help I was able to figure it out (I think, I'd be happy to discuss this further). Most of my answer below is paraphrased from that.
As you say in the description, from an affine group $G$ acting on an affine scheme $X = operatorname{Spec} A$ (assuming both are over a field $k$ for simplicity), an $mathcal{O}_X$-module $mathcal{M}$ comes from an $A$-module $M$, and the two maps $G times X rightrightarrows X$ (the action map and projection respectively) give two maps $tilde sigma, tilde p_2: A rightrightarrows mathcal{O}_G (G)otimes_k A$. Now if we assume that $mathcal{M}$ was equivariant, we have an isomorphism $sigma^* mathcal{M} cong p_2^* mathcal{M}$,
which when we take global sections, means that $phi: Gamma(Gtimes X, sigma^*mathcal{M}) to Gamma(Gtimes X, p_2^*mathcal{M})$
Now I claim that $M$ admits a structure of a $G$-module. First though, note that one can formulate the notion of an action differently. Let $operatorname{Map}(G,M)$ denote the vector space of arbitrary maps (this is verbatim from Chriss and Ginsburg, but probably 'arbitrary' is not right), then an action of $G$ on $M$ can be given as a map $M to operatorname{Map}(G,M)$. Indeed an action (in the ordinary sense) is a map $f:G times M to M$, so simply defining $m mapsto f(-,m)$ gives a map $M to operatorname{Map}(G,M)$. Conversely, suppose we have a map $varphi:M to operatorname{Map}(G,M)$. Then we can define an action by $g.m = varphi(m)(g^{-1})$.
Now define the action of $G$ on $M = Gamma(X,mathcal{M})$ to be the composition $$M overset{sigma^*}{to} Gamma(Gtimes X, sigma^*mathcal{M}) overset{phi}{to} Gamma(Gtimes X, p_2^*mathcal{M}) cong mathcal{O}_G(G) otimes_k Gamma(X,mathcal{M}).$$ Note that $mathcal{O}_G(G) otimes_k M$ lives in the space of maps as $sum f_i otimes m_i (g) = sum f_i(g) m_i$ is a map from $G to M$.
That this is compatible with the multiplication and such, I am told comes from the cocycle condition (I didn't verify this though).
answered Jan 22 at 22:08
DKSDKS
652412
$endgroup$
add a comment |
$begingroup$
So unfortunately my source had forgotten some of the details, so this is probably not perfect. I also found that a discussion of this is indeed in Representation Theory and Complex Geometry by Chriss and Ginzburg; with their help I was able to figure it out (I think, I'd be happy to discuss this further). Most of my answer below is paraphrased from that.
As you say in the description, from an affine group $G$ acting on an affine scheme $X = operatorname{Spec} A$ (assuming both are over a field $k$ for simplicity), an $mathcal{O}_X$-module $mathcal{M}$ comes from an $A$-module $M$, and the two maps $G times X rightrightarrows X$ (the action map and projection respectively) give two maps $tilde sigma, tilde p_2: A rightrightarrows mathcal{O}_G (G)otimes_k A$. Now if we assume that $mathcal{M}$ was equivariant, we have an isomorphism $sigma^* mathcal{M} cong p_2^* mathcal{M}$,
which when we take global sections, means that $phi: Gamma(Gtimes X, sigma^*mathcal{M}) to Gamma(Gtimes X, p_2^*mathcal{M})$
Now I claim that $M$ admits a structure of a $G$-module. First though, note that one can formulate the notion of an action differently. Let $operatorname{Map}(G,M)$ denote the vector space of arbitrary maps (this is verbatim from Chriss and Ginsburg, but probably 'arbitrary' is not right), then an action of $G$ on $M$ can be given as a map $M to operatorname{Map}(G,M)$. Indeed an action (in the ordinary sense) is a map $f:G times M to M$, so simply defining $m mapsto f(-,m)$ gives a map $M to operatorname{Map}(G,M)$. Conversely, suppose we have a map $varphi:M to operatorname{Map}(G,M)$. Then we can define an action by $g.m = varphi(m)(g^{-1})$.
Now define the action of $G$ on $M = Gamma(X,mathcal{M})$ to be the composition $$M overset{sigma^*}{to} Gamma(Gtimes X, sigma^*mathcal{M}) overset{phi}{to} Gamma(Gtimes X, p_2^*mathcal{M}) cong mathcal{O}_G(G) otimes_k Gamma(X,mathcal{M}).$$ Note that $mathcal{O}_G(G) otimes_k M$ lives in the space of maps as $sum f_i otimes m_i (g) = sum f_i(g) m_i$ is a map from $G to M$.
That this is compatible with the multiplication and such, I am told comes from the cocycle condition (I didn't verify this though).
answered Jan 22 at 22:08
DKSDKS
652412
$endgroup$
add a comment |
$begingroup$
So unfortunately my source had forgotten some of the details, so this is probably not perfect. I also found that a discussion of this is indeed in Representation Theory and Complex Geometry by Chriss and Ginzburg; with their help I was able to figure it out (I think, I'd be happy to discuss this further). Most of my answer below is paraphrased from that.
As you say in the description, from an affine group $G$ acting on an affine scheme $X = operatorname{Spec} A$ (assuming both are over a field $k$ for simplicity), an $mathcal{O}_X$-module $mathcal{M}$ comes from an $A$-module $M$, and the two maps $G times X rightrightarrows X$ (the action map and projection respectively) give two maps $tilde sigma, tilde p_2: A rightrightarrows mathcal{O}_G (G)otimes_k A$. Now if we assume that $mathcal{M}$ was equivariant, we have an isomorphism $sigma^* mathcal{M} cong p_2^* mathcal{M}$,
which when we take global sections, means that $phi: Gamma(Gtimes X, sigma^*mathcal{M}) to Gamma(Gtimes X, p_2^*mathcal{M})$
Now I claim that $M$ admits a structure of a $G$-module. First though, note that one can formulate the notion of an action differently. Let $operatorname{Map}(G,M)$ denote the vector space of arbitrary maps (this is verbatim from Chriss and Ginsburg, but probably 'arbitrary' is not right), then an action of $G$ on $M$ can be given as a map $M to operatorname{Map}(G,M)$. Indeed an action (in the ordinary sense) is a map $f:G times M to M$, so simply defining $m mapsto f(-,m)$ gives a map $M to operatorname{Map}(G,M)$. Conversely, suppose we have a map $varphi:M to operatorname{Map}(G,M)$. Then we can define an action by $g.m = varphi(m)(g^{-1})$.
Now define the action of $G$ on $M = Gamma(X,mathcal{M})$ to be the composition $$M overset{sigma^*}{to} Gamma(Gtimes X, sigma^*mathcal{M}) overset{phi}{to} Gamma(Gtimes X, p_2^*mathcal{M}) cong mathcal{O}_G(G) otimes_k Gamma(X,mathcal{M}).$$ Note that $mathcal{O}_G(G) otimes_k M$ lives in the space of maps as $sum f_i otimes m_i (g) = sum f_i(g) m_i$ is a map from $G to M$.
That this is compatible with the multiplication and such, I am told comes from the cocycle condition (I didn't verify this though).
answered Jan 22 at 22:08
DKSDKS
652412
$endgroup$
So unfortunately my source had forgotten some of the details, so this is probably not perfect. I also found that a discussion of this is indeed in Representation Theory and Complex Geometry by Chriss and Ginzburg; with their help I was able to figure it out (I think, I'd be happy to discuss this further). Most of my answer below is paraphrased from that.
As you say in the description, from an affine group $G$ acting on an affine scheme $X = operatorname{Spec} A$ (assuming both are over a field $k$ for simplicity), an $mathcal{O}_X$-module $mathcal{M}$ comes from an $A$-module $M$, and the two maps $G times X rightrightarrows X$ (the action map and projection respectively) give two maps $tilde sigma, tilde p_2: A rightrightarrows mathcal{O}_G (G)otimes_k A$. Now if we assume that $mathcal{M}$ was equivariant, we have an isomorphism $sigma^* mathcal{M} cong p_2^* mathcal{M}$,
which when we take global sections, means that $phi: Gamma(Gtimes X, sigma^*mathcal{M}) to Gamma(Gtimes X, p_2^*mathcal{M})$
Now I claim that $M$ admits a structure of a $G$-module. First though, note that one can formulate the notion of an action differently. Let $operatorname{Map}(G,M)$ denote the vector space of arbitrary maps (this is verbatim from Chriss and Ginsburg, but probably 'arbitrary' is not right), then an action of $G$ on $M$ can be given as a map $M to operatorname{Map}(G,M)$. Indeed an action (in the ordinary sense) is a map $f:G times M to M$, so simply defining $m mapsto f(-,m)$ gives a map $M to operatorname{Map}(G,M)$. Conversely, suppose we have a map $varphi:M to operatorname{Map}(G,M)$. Then we can define an action by $g.m = varphi(m)(g^{-1})$.
Now define the action of $G$ on $M = Gamma(X,mathcal{M})$ to be the composition $$M overset{sigma^*}{to} Gamma(Gtimes X, sigma^*mathcal{M}) overset{phi}{to} Gamma(Gtimes X, p_2^*mathcal{M}) cong mathcal{O}_G(G) otimes_k Gamma(X,mathcal{M}).$$ Note that $mathcal{O}_G(G) otimes_k M$ lives in the space of maps as $sum f_i otimes m_i (g) = sum f_i(g) m_i$ is a map from $G to M$.
That this is compatible with the multiplication and such, I am told comes from the cocycle condition (I didn't verify this though).
answered Jan 22 at 22:08
DKSDKS
652412
edited Jan 23 at 0:46
answered Jan 22 at 22:08
DKSDKS
652412
answered Jan 22 at 22:08
DKSDKS
652412
answered Jan 22 at 22:08
DKSDKS
652412
652412
add a comment |
add a comment |
$begingroup$
This is may be not a complete answer, this is just an example in an easy case, but I hope it will give you an idea of your problem. (I started to wrote this before @DKS
commented but decided to post anyway).
So let us have a concrete example : let $L/K$ a finite Galois extension with group $G$. Consider $X=operatorname{Spec} L$ which is endowed with an action of $G=operatorname{Spec}K[G]$. Here by $K[G]$ is not the group algebra, instead, this is the algebra $bigoplus_{gin G}K.g$ where the product is given component wise. In particular the identity element is $1=sum_{gin G}g$.
Then $K[G]$ is a coalgebra with coproduct given by $gmapsto sum_{hin G} gh^{-1}otimes h$. This coproduct induces $Gtimes Gto G$ when taking Specs. We have an coaction of $K[G]$ on $L$ given by $Lto Lotimes_K K[G]$ such that $lmapsto sum_{gin G} g(l)otimes g$. This coaction gives again the action of $Gtimesoperatorname{Spec}Ltooperatorname{Spec} L$ when taking Specs.
Now, an equivariant module is a vector space $V$ over $L$ with an isomorphism
$$phi:Votimes_L(Lotimes_K K[G])to Votimes_L(Lotimes_K K[G])$$
satisfying a cocycle condition (and remember that this is not the same tensor structure on the left and on the right).
I claim that this is the same thing as an action of $G$ on $V$ such that $g.(l v)=g(l)g(v)$. Indeed, given $phi$, $phi(votimes 1otimes 1)$ can be written $phi(votimes 1otimes 1)=sum_{gin G}v_gotimes 1otimes g$. We then put $g.v=v_g$.
Note that $phi(lvotimes 1otimes 1)=phi(sum_{gin G} votimes g(l)otimes h)$ since the tensor structure on the left is given by the coaction. Now we have by linearity
$$phi(votimes g(l)otimes g)=(sum_{hin G} v_hotimes 1otimes h)(g(l)otimes g)=v_gotimes g(l)otimes g=g(l)v_gotimes 1otimes g$$
(Remember that $gh=0$ in $K[G]$ if $gneq h$ from the definition of the algebra structure on $K[G]$). It follows that $phi(lvotimes 1otimes 1)=sum_{gin G}g(l)v_gotimes 1otimes g$, so $g.(lv)=g(l)g(v)$.
I let you check that this is an action, in other words that $(gh)v=g(h(v))$. You will need to use the cocycle condition. It is a bit tedious...
Conversely, the data of a semi-linear action of $G$ on $V$ is equivalent to the data of $phi$ by letting $phi(votimes 1otimes 1)=sum_{gin G} (g.v)otimes 1otimes g$ and extend by linearity. I let you check that $phi$ satisfy all the cocycle condition.
So now, what are the invariant ? Well this is "obviously" the one which are fixed by the action of $G$. In other words, this is the $vin V$ such that $g(v)=v$ for all $v$. And there is an easy way to express this in term of $phi$ : the invariant are exactly the $vin V$ such that $phi(votimes 1otimes 1)=votimes 1otimes 1$.
This is general : using your notation, we have a "fixed" module $overline{M}$ which consists of the $min M$ such that $phi(motimes 1otimes 1)=motimes 1otimes 1$.
Note that I never used the fact that the action of $G$ on $X$ is free or nice, and indeed, I did not prove that $overline{M}otimes_{R^G} Rsimeq M$.
answered Jan 21 at 20:53
RolandRoland
7,34411015
$endgroup$
$begingroup$
Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
$endgroup$
– Federico
Jan 23 at 21:48
add a comment |
$begingroup$
This is may be not a complete answer, this is just an example in an easy case, but I hope it will give you an idea of your problem. (I started to wrote this before @DKS
commented but decided to post anyway).
So let us have a concrete example : let $L/K$ a finite Galois extension with group $G$. Consider $X=operatorname{Spec} L$ which is endowed with an action of $G=operatorname{Spec}K[G]$. Here by $K[G]$ is not the group algebra, instead, this is the algebra $bigoplus_{gin G}K.g$ where the product is given component wise. In particular the identity element is $1=sum_{gin G}g$.
Then $K[G]$ is a coalgebra with coproduct given by $gmapsto sum_{hin G} gh^{-1}otimes h$. This coproduct induces $Gtimes Gto G$ when taking Specs. We have an coaction of $K[G]$ on $L$ given by $Lto Lotimes_K K[G]$ such that $lmapsto sum_{gin G} g(l)otimes g$. This coaction gives again the action of $Gtimesoperatorname{Spec}Ltooperatorname{Spec} L$ when taking Specs.
Now, an equivariant module is a vector space $V$ over $L$ with an isomorphism
$$phi:Votimes_L(Lotimes_K K[G])to Votimes_L(Lotimes_K K[G])$$
satisfying a cocycle condition (and remember that this is not the same tensor structure on the left and on the right).
I claim that this is the same thing as an action of $G$ on $V$ such that $g.(l v)=g(l)g(v)$. Indeed, given $phi$, $phi(votimes 1otimes 1)$ can be written $phi(votimes 1otimes 1)=sum_{gin G}v_gotimes 1otimes g$. We then put $g.v=v_g$.
Note that $phi(lvotimes 1otimes 1)=phi(sum_{gin G} votimes g(l)otimes h)$ since the tensor structure on the left is given by the coaction. Now we have by linearity
$$phi(votimes g(l)otimes g)=(sum_{hin G} v_hotimes 1otimes h)(g(l)otimes g)=v_gotimes g(l)otimes g=g(l)v_gotimes 1otimes g$$
(Remember that $gh=0$ in $K[G]$ if $gneq h$ from the definition of the algebra structure on $K[G]$). It follows that $phi(lvotimes 1otimes 1)=sum_{gin G}g(l)v_gotimes 1otimes g$, so $g.(lv)=g(l)g(v)$.
I let you check that this is an action, in other words that $(gh)v=g(h(v))$. You will need to use the cocycle condition. It is a bit tedious...
Conversely, the data of a semi-linear action of $G$ on $V$ is equivalent to the data of $phi$ by letting $phi(votimes 1otimes 1)=sum_{gin G} (g.v)otimes 1otimes g$ and extend by linearity. I let you check that $phi$ satisfy all the cocycle condition.
So now, what are the invariant ? Well this is "obviously" the one which are fixed by the action of $G$. In other words, this is the $vin V$ such that $g(v)=v$ for all $v$. And there is an easy way to express this in term of $phi$ : the invariant are exactly the $vin V$ such that $phi(votimes 1otimes 1)=votimes 1otimes 1$.
This is general : using your notation, we have a "fixed" module $overline{M}$ which consists of the $min M$ such that $phi(motimes 1otimes 1)=motimes 1otimes 1$.
Note that I never used the fact that the action of $G$ on $X$ is free or nice, and indeed, I did not prove that $overline{M}otimes_{R^G} Rsimeq M$.
answered Jan 21 at 20:53
RolandRoland
7,34411015
$endgroup$
$begingroup$
Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
$endgroup$
– Federico
Jan 23 at 21:48
add a comment |
$begingroup$
This is may be not a complete answer, this is just an example in an easy case, but I hope it will give you an idea of your problem. (I started to wrote this before @DKS
commented but decided to post anyway).
So let us have a concrete example : let $L/K$ a finite Galois extension with group $G$. Consider $X=operatorname{Spec} L$ which is endowed with an action of $G=operatorname{Spec}K[G]$. Here by $K[G]$ is not the group algebra, instead, this is the algebra $bigoplus_{gin G}K.g$ where the product is given component wise. In particular the identity element is $1=sum_{gin G}g$.
Then $K[G]$ is a coalgebra with coproduct given by $gmapsto sum_{hin G} gh^{-1}otimes h$. This coproduct induces $Gtimes Gto G$ when taking Specs. We have an coaction of $K[G]$ on $L$ given by $Lto Lotimes_K K[G]$ such that $lmapsto sum_{gin G} g(l)otimes g$. This coaction gives again the action of $Gtimesoperatorname{Spec}Ltooperatorname{Spec} L$ when taking Specs.
Now, an equivariant module is a vector space $V$ over $L$ with an isomorphism
$$phi:Votimes_L(Lotimes_K K[G])to Votimes_L(Lotimes_K K[G])$$
satisfying a cocycle condition (and remember that this is not the same tensor structure on the left and on the right).
I claim that this is the same thing as an action of $G$ on $V$ such that $g.(l v)=g(l)g(v)$. Indeed, given $phi$, $phi(votimes 1otimes 1)$ can be written $phi(votimes 1otimes 1)=sum_{gin G}v_gotimes 1otimes g$. We then put $g.v=v_g$.
Note that $phi(lvotimes 1otimes 1)=phi(sum_{gin G} votimes g(l)otimes h)$ since the tensor structure on the left is given by the coaction. Now we have by linearity
$$phi(votimes g(l)otimes g)=(sum_{hin G} v_hotimes 1otimes h)(g(l)otimes g)=v_gotimes g(l)otimes g=g(l)v_gotimes 1otimes g$$
(Remember that $gh=0$ in $K[G]$ if $gneq h$ from the definition of the algebra structure on $K[G]$). It follows that $phi(lvotimes 1otimes 1)=sum_{gin G}g(l)v_gotimes 1otimes g$, so $g.(lv)=g(l)g(v)$.
I let you check that this is an action, in other words that $(gh)v=g(h(v))$. You will need to use the cocycle condition. It is a bit tedious...
Conversely, the data of a semi-linear action of $G$ on $V$ is equivalent to the data of $phi$ by letting $phi(votimes 1otimes 1)=sum_{gin G} (g.v)otimes 1otimes g$ and extend by linearity. I let you check that $phi$ satisfy all the cocycle condition.
So now, what are the invariant ? Well this is "obviously" the one which are fixed by the action of $G$. In other words, this is the $vin V$ such that $g(v)=v$ for all $v$. And there is an easy way to express this in term of $phi$ : the invariant are exactly the $vin V$ such that $phi(votimes 1otimes 1)=votimes 1otimes 1$.
This is general : using your notation, we have a "fixed" module $overline{M}$ which consists of the $min M$ such that $phi(motimes 1otimes 1)=motimes 1otimes 1$.
Note that I never used the fact that the action of $G$ on $X$ is free or nice, and indeed, I did not prove that $overline{M}otimes_{R^G} Rsimeq M$.
answered Jan 21 at 20:53
RolandRoland
7,34411015
$endgroup$
This is may be not a complete answer, this is just an example in an easy case, but I hope it will give you an idea of your problem. (I started to wrote this before @DKS
commented but decided to post anyway).
So let us have a concrete example : let $L/K$ a finite Galois extension with group $G$. Consider $X=operatorname{Spec} L$ which is endowed with an action of $G=operatorname{Spec}K[G]$. Here by $K[G]$ is not the group algebra, instead, this is the algebra $bigoplus_{gin G}K.g$ where the product is given component wise. In particular the identity element is $1=sum_{gin G}g$.
Then $K[G]$ is a coalgebra with coproduct given by $gmapsto sum_{hin G} gh^{-1}otimes h$. This coproduct induces $Gtimes Gto G$ when taking Specs. We have an coaction of $K[G]$ on $L$ given by $Lto Lotimes_K K[G]$ such that $lmapsto sum_{gin G} g(l)otimes g$. This coaction gives again the action of $Gtimesoperatorname{Spec}Ltooperatorname{Spec} L$ when taking Specs.
Now, an equivariant module is a vector space $V$ over $L$ with an isomorphism
$$phi:Votimes_L(Lotimes_K K[G])to Votimes_L(Lotimes_K K[G])$$
satisfying a cocycle condition (and remember that this is not the same tensor structure on the left and on the right).
I claim that this is the same thing as an action of $G$ on $V$ such that $g.(l v)=g(l)g(v)$. Indeed, given $phi$, $phi(votimes 1otimes 1)$ can be written $phi(votimes 1otimes 1)=sum_{gin G}v_gotimes 1otimes g$. We then put $g.v=v_g$.
Note that $phi(lvotimes 1otimes 1)=phi(sum_{gin G} votimes g(l)otimes h)$ since the tensor structure on the left is given by the coaction. Now we have by linearity
$$phi(votimes g(l)otimes g)=(sum_{hin G} v_hotimes 1otimes h)(g(l)otimes g)=v_gotimes g(l)otimes g=g(l)v_gotimes 1otimes g$$
(Remember that $gh=0$ in $K[G]$ if $gneq h$ from the definition of the algebra structure on $K[G]$). It follows that $phi(lvotimes 1otimes 1)=sum_{gin G}g(l)v_gotimes 1otimes g$, so $g.(lv)=g(l)g(v)$.
I let you check that this is an action, in other words that $(gh)v=g(h(v))$. You will need to use the cocycle condition. It is a bit tedious...
Conversely, the data of a semi-linear action of $G$ on $V$ is equivalent to the data of $phi$ by letting $phi(votimes 1otimes 1)=sum_{gin G} (g.v)otimes 1otimes g$ and extend by linearity. I let you check that $phi$ satisfy all the cocycle condition.
So now, what are the invariant ? Well this is "obviously" the one which are fixed by the action of $G$. In other words, this is the $vin V$ such that $g(v)=v$ for all $v$. And there is an easy way to express this in term of $phi$ : the invariant are exactly the $vin V$ such that $phi(votimes 1otimes 1)=votimes 1otimes 1$.
This is general : using your notation, we have a "fixed" module $overline{M}$ which consists of the $min M$ such that $phi(motimes 1otimes 1)=motimes 1otimes 1$.
Note that I never used the fact that the action of $G$ on $X$ is free or nice, and indeed, I did not prove that $overline{M}otimes_{R^G} Rsimeq M$.
answered Jan 21 at 20:53
RolandRoland
7,34411015
answered Jan 21 at 20:53
RolandRoland
7,34411015
answered Jan 21 at 20:53
RolandRoland
7,34411015
answered Jan 21 at 20:53
RolandRoland
7,34411015
7,34411015
$begingroup$
Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
$endgroup$
– Federico
Jan 23 at 21:48
add a comment |
$begingroup$
Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
$endgroup$
– Federico
Jan 23 at 21:48
$begingroup$
Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
$endgroup$
– Federico
Jan 23 at 21:48
$begingroup$
Thank you very much for this detailed example! I will select the other answer as accepted because it's more in the spirit of the question. Thank you again.
$endgroup$
– Federico
Jan 23 at 21:48
add a comment |
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$begingroup$
I think you have the right idea. One condition on an equivariant sheaf $mathcal{F}$ on $X$ is that you have an isomorphism $sigma^* mathcal{F} cong p_2^* mathcal{F}$, where $sigma,p_2 : G times X to X$ is the action map and projection respectively (there is another condition, but that's more complicated). If you reduce to the affine case and pass to the module category (assuming $mathcal{F}$ is an $mathcal{O}_X$-module) you should get something like the map you have above between the triple tensor product. I don't have pen and paper nearby, otherwise I would work it for you...
$endgroup$
– DKS
Jan 20 at 23:22
$begingroup$
A good elementary treatment of equivariant sheaves can be found in Chriss and Ginzburg's Representation Theory and Complex Geometry, I think it's at the beginning of the chapter on equivariant K-theory.
$endgroup$
– DKS
Jan 20 at 23:23
$begingroup$
@DKS Thank you for your answer. However, I know that’s the local description of the definition of an equivariant sheaf, my question was how to relate that local description to the intuitive idea that an equivariant sheaf should give, at least locally, an equivariant module.
$endgroup$
– Federico
Jan 21 at 10:07
$begingroup$
After thinking about this myself I'm also not sure. If you're willing to wait a bit I can ask someone who knows a few things about this stuff.
$endgroup$
– DKS
Jan 21 at 14:23
$begingroup$
@DKS That would be great, thank you.
$endgroup$
– Federico
Jan 21 at 14:26