Cech Cohomology and module structure
I just came across a question I had never thought about and that could be simple, but I can't answer it. If we consider a scheme $X$ and we pick a sheaf of $mathcal{O}_{X}$-modules $mathcal{F}$, under some assumptions we can compute its cohomology either via an injective resolution or with Cech cohomology. Both the answers carry an $mathcal{O}_X$-module structure, but the theorem that states that the answers we get with the two approaches is the same is always stated saying that the abelian groups are the same. Therefore my question, is the isomorphism between Cech cohomology and cohomology as a derived functor an isomorphism of $mathcal{O}_{X}$-modules?
algebraic-geometry sheaf-cohomology
asked 2 days ago
FedericoFederico
820213
add a comment |
I just came across a question I had never thought about and that could be simple, but I can't answer it. If we consider a scheme $X$ and we pick a sheaf of $mathcal{O}_{X}$-modules $mathcal{F}$, under some assumptions we can compute its cohomology either via an injective resolution or with Cech cohomology. Both the answers carry an $mathcal{O}_X$-module structure, but the theorem that states that the answers we get with the two approaches is the same is always stated saying that the abelian groups are the same. Therefore my question, is the isomorphism between Cech cohomology and cohomology as a derived functor an isomorphism of $mathcal{O}_{X}$-modules?
algebraic-geometry sheaf-cohomology
asked 2 days ago
FedericoFederico
820213
add a comment |
I just came across a question I had never thought about and that could be simple, but I can't answer it. If we consider a scheme $X$ and we pick a sheaf of $mathcal{O}_{X}$-modules $mathcal{F}$, under some assumptions we can compute its cohomology either via an injective resolution or with Cech cohomology. Both the answers carry an $mathcal{O}_X$-module structure, but the theorem that states that the answers we get with the two approaches is the same is always stated saying that the abelian groups are the same. Therefore my question, is the isomorphism between Cech cohomology and cohomology as a derived functor an isomorphism of $mathcal{O}_{X}$-modules?
algebraic-geometry sheaf-cohomology
asked 2 days ago
FedericoFederico
820213
I just came across a question I had never thought about and that could be simple, but I can't answer it. If we consider a scheme $X$ and we pick a sheaf of $mathcal{O}_{X}$-modules $mathcal{F}$, under some assumptions we can compute its cohomology either via an injective resolution or with Cech cohomology. Both the answers carry an $mathcal{O}_X$-module structure, but the theorem that states that the answers we get with the two approaches is the same is always stated saying that the abelian groups are the same. Therefore my question, is the isomorphism between Cech cohomology and cohomology as a derived functor an isomorphism of $mathcal{O}_{X}$-modules?
algebraic-geometry sheaf-cohomology
algebraic-geometry sheaf-cohomology
asked 2 days ago
FedericoFederico
820213
asked 2 days ago
FedericoFederico
820213
asked 2 days ago
FedericoFederico
820213
asked 2 days ago
FedericoFederico
820213
asked 2 days ago
FedericoFederico
820213
820213
add a comment |
add a comment |
1 Answer
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How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $mathcal F = mathcal O_X = underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $mathcal F$ with an injective resolution. One injective resolution is $phi: underline{k} to underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $mathcal O$-linear by design because it coincides with $phi$ itself.
So, if you want an $mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $Gamma(mathcal O,X)$-linear.
answered 2 days ago
BenBen
2,593616
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
add a comment |
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How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $mathcal F = mathcal O_X = underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $mathcal F$ with an injective resolution. One injective resolution is $phi: underline{k} to underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $mathcal O$-linear by design because it coincides with $phi$ itself.
So, if you want an $mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $Gamma(mathcal O,X)$-linear.
answered 2 days ago
BenBen
2,593616
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
add a comment |
How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $mathcal F = mathcal O_X = underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $mathcal F$ with an injective resolution. One injective resolution is $phi: underline{k} to underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $mathcal O$-linear by design because it coincides with $phi$ itself.
So, if you want an $mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $Gamma(mathcal O,X)$-linear.
answered 2 days ago
BenBen
2,593616
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
add a comment |
How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $mathcal F = mathcal O_X = underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $mathcal F$ with an injective resolution. One injective resolution is $phi: underline{k} to underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $mathcal O$-linear by design because it coincides with $phi$ itself.
So, if you want an $mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $Gamma(mathcal O,X)$-linear.
answered 2 days ago
BenBen
2,593616
How do we prove that sheaf cohomology and cech cohomology are the same anyway? For reference see a nice note on this here: pub.math.leidenuniv.nl/~edixhovensj/teaching/2011-2012/AAG/lecture_14.pdf or see Hartshorne.
Basically you can build a resolution of $mathcal F$ by making a sheaf-version of the Cech complex (whose global sections is the usual Cech complex and which is a sheaf of $mathcal O_X$-modules). After you check this is a resolution, it is a fact that any resolution will map to an injective resolution. This involves making choices, but since the choices are unique up to homotopy you get the well-defined map from Cech cohomology to sheaf cohomology. But what if you switch the category from $mathcal O_X$-modules to sheaves of abelian groups - are the choices still unique up to homotopies taken from the other category?
The answer is no. Consider the case $X$ is a point. Let $mathcal F = mathcal O_X = underline{k}$ be the constant sheaf. If we compute in the category of sheaves of abelian groups then we have the freedom to replace $mathcal F$ with an injective resolution. One injective resolution is $phi: underline{k} to underline{k}$ taken to be any isomorphism chosen specifically not to be $k$-linear but only additive. Then in this case the map between sheaf cohomology and cech cohomology is not $mathcal O$-linear by design because it coincides with $phi$ itself.
So, if you want an $mathcal O$ linear map you can get one for free, and I think it is pretty standard to assume this map is $Gamma(mathcal O,X)$-linear.
answered 2 days ago
BenBen
2,593616
answered 2 days ago
BenBen
2,593616
answered 2 days ago
BenBen
2,593616
answered 2 days ago
BenBen
2,593616
2,593616
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
add a comment |
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
So you're saying that the isomorphism is for sure an isomorphism of $mathcal{O}$ modules only if I pick an injective resolution of $mathcal{O}$ modules and make the Cech complex map into this resolution. If I pass to the category of sheaves of abelian groups and then pick a resolution by flasque sheaves the resulting isomorphism might not be an isomorphism of $mathcal{O}$-modules, even if the target (cohomology as a derived functor) is an $mathcal{O}$-module. Did I understand correctly?
– Federico
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
Actually If the cohomology is an $mathcal O$ module I think this means you must have taken a resolution in the category of $mathcal O$ modules. In which case you may choose a map from the cech sheaf to it which is $mathcal O$ linear.
– Ben
2 days ago
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
In other words, if you acquire an $mathcal O$ module structure on the sheaf cohomology by some other means than inheriting it from the injective resolution, then there is no guarantee the map will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
I think in practice this won’t happen though, the module structure will always come from a resolution and the map on cohomology will be $mathcal O$ linear.
– Ben
yesterday
add a comment |
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