Mixing
Definition of a Picard Group
$begingroup$
Let $X$ be a smooth variety. Then we know that the picard group of $X$ is the quotient $text{Div}(X)/text{Princ}(X)$ of the group of divisors of $X$ by the subgroup of principal divisors of $X$.
Now suppose in the case where $X$ is a complex manifold, one has the exponential sequence: $$ 0mapsto 2pi i mathbb{Z} mapsto mathcal{O}_X mapsto mathcal{O}_X^ast mapsto 0$$
Now we know from usual cohomology that one has the induced morphiam $$H^1(X, mathcal{O}_X^ast) mapsto H^2(X, mathbb{Z})$$ I've also read that in this case $H^1(X, ,mathcal{O}_X^ast)$ is the Picard Group of $X$ as well.
My question is how are the two definitions of the Picard Group I've stated above equivalent? Moreover, how does is image of morphism identified as the Neron-Severi group of $X$? The usual definition I have for the Neron-Severi group is the group of Cartier divisors modulo numerical equivalence. Would be thankful if anyone can provide some answers.
algebraic-geometry complex-geometry
asked Feb 2 at 5:26
thedilatedthedilated
7801617
$endgroup$
|
show 1 more comment
$begingroup$
Let $X$ be a smooth variety. Then we know that the picard group of $X$ is the quotient $text{Div}(X)/text{Princ}(X)$ of the group of divisors of $X$ by the subgroup of principal divisors of $X$.
Now suppose in the case where $X$ is a complex manifold, one has the exponential sequence: $$ 0mapsto 2pi i mathbb{Z} mapsto mathcal{O}_X mapsto mathcal{O}_X^ast mapsto 0$$
Now we know from usual cohomology that one has the induced morphiam $$H^1(X, mathcal{O}_X^ast) mapsto H^2(X, mathbb{Z})$$ I've also read that in this case $H^1(X, ,mathcal{O}_X^ast)$ is the Picard Group of $X$ as well.
My question is how are the two definitions of the Picard Group I've stated above equivalent? Moreover, how does is image of morphism identified as the Neron-Severi group of $X$? The usual definition I have for the Neron-Severi group is the group of Cartier divisors modulo numerical equivalence. Would be thankful if anyone can provide some answers.
algebraic-geometry complex-geometry
asked Feb 2 at 5:26
thedilatedthedilated
7801617
$endgroup$
1
$begingroup$
You may wish to consult Hartshorne exercise III.4.5, which leads you through a proof $H^1(X,mathcal{O}_X^ast)cong Pic(X)$ for any ringed space $X$. I think there might be more going on with the text you cite - the given map into $H^2$ you describe need not be injective, which would only give you a quotient of the Picard group rather than the whole thing.
$endgroup$
– KReiser
Feb 2 at 6:36
$begingroup$
@KReiser Oh my bad. It is the Neron-Severi group* not the Picard Group! I have edited the question. Sorry about it.
$endgroup$
– thedilated
Feb 2 at 7:13
$begingroup$
You can use the following exact sequence : $$0to mathcal{O}_X^*to mathcal{K}_X^*to mathcal{K}_X^*/mathcal{O}_X^*to 0$$ where $mathcal{K}_X^*$ is the sheaf of rational functions. This induces $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)to H^1(X,mathcal{O}_X^*)$ which is onto under some very general hypotheses. Almost by definition, $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)=operatorname{Div}(X)$ and the image of $H^0(X, mathcal{K}_X^*)$ is $operatorname{Princ}(X)$. Hence the isomorphism $operatorname{Div}(X)/operatorname{Princ}(X)=H^1(X,mathcal{O}_X^*)$.
$endgroup$
– Roland
Feb 2 at 8:42
$begingroup$
@Roland Thank you very much for the response. I would like to clarify the following: The definition of $text{Div}(X)$ I know is the (collection of) formal sums of prime divisors over $X$. However I interpret $H^0(X, mathcal{K}_X^ast/mathcal{O}_X^ast)$ as the global section of $X$, i.e. all regular functions $f$ vanishing at some point. How are these two equal by definition?
$endgroup$
– thedilated
Feb 2 at 9:44
2
$begingroup$
Ok I was using the notion of Cartier divisor and you are using the notion of Weil divisor. There is always a map $operatorname{CaDiv}(X)tooperatorname{Div}(X)$. On a smooth scheme this map is an isomorphism.
$endgroup$
– Roland
Feb 2 at 11:07
|
show 1 more comment
$begingroup$
Let $X$ be a smooth variety. Then we know that the picard group of $X$ is the quotient $text{Div}(X)/text{Princ}(X)$ of the group of divisors of $X$ by the subgroup of principal divisors of $X$.
Now suppose in the case where $X$ is a complex manifold, one has the exponential sequence: $$ 0mapsto 2pi i mathbb{Z} mapsto mathcal{O}_X mapsto mathcal{O}_X^ast mapsto 0$$
Now we know from usual cohomology that one has the induced morphiam $$H^1(X, mathcal{O}_X^ast) mapsto H^2(X, mathbb{Z})$$ I've also read that in this case $H^1(X, ,mathcal{O}_X^ast)$ is the Picard Group of $X$ as well.
My question is how are the two definitions of the Picard Group I've stated above equivalent? Moreover, how does is image of morphism identified as the Neron-Severi group of $X$? The usual definition I have for the Neron-Severi group is the group of Cartier divisors modulo numerical equivalence. Would be thankful if anyone can provide some answers.
algebraic-geometry complex-geometry
asked Feb 2 at 5:26
thedilatedthedilated
7801617
$endgroup$
Let $X$ be a smooth variety. Then we know that the picard group of $X$ is the quotient $text{Div}(X)/text{Princ}(X)$ of the group of divisors of $X$ by the subgroup of principal divisors of $X$.
Now suppose in the case where $X$ is a complex manifold, one has the exponential sequence: $$ 0mapsto 2pi i mathbb{Z} mapsto mathcal{O}_X mapsto mathcal{O}_X^ast mapsto 0$$
Now we know from usual cohomology that one has the induced morphiam $$H^1(X, mathcal{O}_X^ast) mapsto H^2(X, mathbb{Z})$$ I've also read that in this case $H^1(X, ,mathcal{O}_X^ast)$ is the Picard Group of $X$ as well.
My question is how are the two definitions of the Picard Group I've stated above equivalent? Moreover, how does is image of morphism identified as the Neron-Severi group of $X$? The usual definition I have for the Neron-Severi group is the group of Cartier divisors modulo numerical equivalence. Would be thankful if anyone can provide some answers.
algebraic-geometry complex-geometry
algebraic-geometry complex-geometry
asked Feb 2 at 5:26
thedilatedthedilated
7801617
asked Feb 2 at 5:26
thedilatedthedilated
7801617
edited Feb 2 at 7:15
thedilated
asked Feb 2 at 5:26
thedilatedthedilated
7801617
asked Feb 2 at 5:26
thedilatedthedilated
7801617
asked Feb 2 at 5:26
thedilatedthedilated
7801617
7801617
1
$begingroup$
You may wish to consult Hartshorne exercise III.4.5, which leads you through a proof $H^1(X,mathcal{O}_X^ast)cong Pic(X)$ for any ringed space $X$. I think there might be more going on with the text you cite - the given map into $H^2$ you describe need not be injective, which would only give you a quotient of the Picard group rather than the whole thing.
$endgroup$
– KReiser
Feb 2 at 6:36
$begingroup$
@KReiser Oh my bad. It is the Neron-Severi group* not the Picard Group! I have edited the question. Sorry about it.
$endgroup$
– thedilated
Feb 2 at 7:13
$begingroup$
You can use the following exact sequence : $$0to mathcal{O}_X^*to mathcal{K}_X^*to mathcal{K}_X^*/mathcal{O}_X^*to 0$$ where $mathcal{K}_X^*$ is the sheaf of rational functions. This induces $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)to H^1(X,mathcal{O}_X^*)$ which is onto under some very general hypotheses. Almost by definition, $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)=operatorname{Div}(X)$ and the image of $H^0(X, mathcal{K}_X^*)$ is $operatorname{Princ}(X)$. Hence the isomorphism $operatorname{Div}(X)/operatorname{Princ}(X)=H^1(X,mathcal{O}_X^*)$.
$endgroup$
– Roland
Feb 2 at 8:42
$begingroup$
@Roland Thank you very much for the response. I would like to clarify the following: The definition of $text{Div}(X)$ I know is the (collection of) formal sums of prime divisors over $X$. However I interpret $H^0(X, mathcal{K}_X^ast/mathcal{O}_X^ast)$ as the global section of $X$, i.e. all regular functions $f$ vanishing at some point. How are these two equal by definition?
$endgroup$
– thedilated
Feb 2 at 9:44
2
$begingroup$
Ok I was using the notion of Cartier divisor and you are using the notion of Weil divisor. There is always a map $operatorname{CaDiv}(X)tooperatorname{Div}(X)$. On a smooth scheme this map is an isomorphism.
$endgroup$
– Roland
Feb 2 at 11:07
|
show 1 more comment
1
$begingroup$
You may wish to consult Hartshorne exercise III.4.5, which leads you through a proof $H^1(X,mathcal{O}_X^ast)cong Pic(X)$ for any ringed space $X$. I think there might be more going on with the text you cite - the given map into $H^2$ you describe need not be injective, which would only give you a quotient of the Picard group rather than the whole thing.
$endgroup$
– KReiser
Feb 2 at 6:36
$begingroup$
@KReiser Oh my bad. It is the Neron-Severi group* not the Picard Group! I have edited the question. Sorry about it.
$endgroup$
– thedilated
Feb 2 at 7:13
$begingroup$
You can use the following exact sequence : $$0to mathcal{O}_X^*to mathcal{K}_X^*to mathcal{K}_X^*/mathcal{O}_X^*to 0$$ where $mathcal{K}_X^*$ is the sheaf of rational functions. This induces $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)to H^1(X,mathcal{O}_X^*)$ which is onto under some very general hypotheses. Almost by definition, $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)=operatorname{Div}(X)$ and the image of $H^0(X, mathcal{K}_X^*)$ is $operatorname{Princ}(X)$. Hence the isomorphism $operatorname{Div}(X)/operatorname{Princ}(X)=H^1(X,mathcal{O}_X^*)$.
$endgroup$
– Roland
Feb 2 at 8:42
$begingroup$
@Roland Thank you very much for the response. I would like to clarify the following: The definition of $text{Div}(X)$ I know is the (collection of) formal sums of prime divisors over $X$. However I interpret $H^0(X, mathcal{K}_X^ast/mathcal{O}_X^ast)$ as the global section of $X$, i.e. all regular functions $f$ vanishing at some point. How are these two equal by definition?
$endgroup$
– thedilated
Feb 2 at 9:44
2
$begingroup$
Ok I was using the notion of Cartier divisor and you are using the notion of Weil divisor. There is always a map $operatorname{CaDiv}(X)tooperatorname{Div}(X)$. On a smooth scheme this map is an isomorphism.
$endgroup$
– Roland
Feb 2 at 11:07
1
1
$begingroup$
You may wish to consult Hartshorne exercise III.4.5, which leads you through a proof $H^1(X,mathcal{O}_X^ast)cong Pic(X)$ for any ringed space $X$. I think there might be more going on with the text you cite - the given map into $H^2$ you describe need not be injective, which would only give you a quotient of the Picard group rather than the whole thing.
$endgroup$
– KReiser
Feb 2 at 6:36
$begingroup$
You may wish to consult Hartshorne exercise III.4.5, which leads you through a proof $H^1(X,mathcal{O}_X^ast)cong Pic(X)$ for any ringed space $X$. I think there might be more going on with the text you cite - the given map into $H^2$ you describe need not be injective, which would only give you a quotient of the Picard group rather than the whole thing.
$endgroup$
– KReiser
Feb 2 at 6:36
$begingroup$
@KReiser Oh my bad. It is the Neron-Severi group* not the Picard Group! I have edited the question. Sorry about it.
$endgroup$
– thedilated
Feb 2 at 7:13
$begingroup$
@KReiser Oh my bad. It is the Neron-Severi group* not the Picard Group! I have edited the question. Sorry about it.
$endgroup$
– thedilated
Feb 2 at 7:13
$begingroup$
You can use the following exact sequence : $$0to mathcal{O}_X^*to mathcal{K}_X^*to mathcal{K}_X^*/mathcal{O}_X^*to 0$$ where $mathcal{K}_X^*$ is the sheaf of rational functions. This induces $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)to H^1(X,mathcal{O}_X^*)$ which is onto under some very general hypotheses. Almost by definition, $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)=operatorname{Div}(X)$ and the image of $H^0(X, mathcal{K}_X^*)$ is $operatorname{Princ}(X)$. Hence the isomorphism $operatorname{Div}(X)/operatorname{Princ}(X)=H^1(X,mathcal{O}_X^*)$.
$endgroup$
– Roland
Feb 2 at 8:42
$begingroup$
You can use the following exact sequence : $$0to mathcal{O}_X^*to mathcal{K}_X^*to mathcal{K}_X^*/mathcal{O}_X^*to 0$$ where $mathcal{K}_X^*$ is the sheaf of rational functions. This induces $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)to H^1(X,mathcal{O}_X^*)$ which is onto under some very general hypotheses. Almost by definition, $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)=operatorname{Div}(X)$ and the image of $H^0(X, mathcal{K}_X^*)$ is $operatorname{Princ}(X)$. Hence the isomorphism $operatorname{Div}(X)/operatorname{Princ}(X)=H^1(X,mathcal{O}_X^*)$.
$endgroup$
– Roland
Feb 2 at 8:42
$begingroup$
@Roland Thank you very much for the response. I would like to clarify the following: The definition of $text{Div}(X)$ I know is the (collection of) formal sums of prime divisors over $X$. However I interpret $H^0(X, mathcal{K}_X^ast/mathcal{O}_X^ast)$ as the global section of $X$, i.e. all regular functions $f$ vanishing at some point. How are these two equal by definition?
$endgroup$
– thedilated
Feb 2 at 9:44
$begingroup$
@Roland Thank you very much for the response. I would like to clarify the following: The definition of $text{Div}(X)$ I know is the (collection of) formal sums of prime divisors over $X$. However I interpret $H^0(X, mathcal{K}_X^ast/mathcal{O}_X^ast)$ as the global section of $X$, i.e. all regular functions $f$ vanishing at some point. How are these two equal by definition?
$endgroup$
– thedilated
Feb 2 at 9:44
2
2
$begingroup$
Ok I was using the notion of Cartier divisor and you are using the notion of Weil divisor. There is always a map $operatorname{CaDiv}(X)tooperatorname{Div}(X)$. On a smooth scheme this map is an isomorphism.
$endgroup$
– Roland
Feb 2 at 11:07
$begingroup$
Ok I was using the notion of Cartier divisor and you are using the notion of Weil divisor. There is always a map $operatorname{CaDiv}(X)tooperatorname{Div}(X)$. On a smooth scheme this map is an isomorphism.
$endgroup$
– Roland
Feb 2 at 11:07
|
show 1 more comment
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$begingroup$
You may wish to consult Hartshorne exercise III.4.5, which leads you through a proof $H^1(X,mathcal{O}_X^ast)cong Pic(X)$ for any ringed space $X$. I think there might be more going on with the text you cite - the given map into $H^2$ you describe need not be injective, which would only give you a quotient of the Picard group rather than the whole thing.
$endgroup$
– KReiser
Feb 2 at 6:36
$begingroup$
@KReiser Oh my bad. It is the Neron-Severi group* not the Picard Group! I have edited the question. Sorry about it.
$endgroup$
– thedilated
Feb 2 at 7:13
$begingroup$
You can use the following exact sequence : $$0to mathcal{O}_X^*to mathcal{K}_X^*to mathcal{K}_X^*/mathcal{O}_X^*to 0$$ where $mathcal{K}_X^*$ is the sheaf of rational functions. This induces $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)to H^1(X,mathcal{O}_X^*)$ which is onto under some very general hypotheses. Almost by definition, $H^0(X, mathcal{K}_X^*/mathcal{O}_X^*)=operatorname{Div}(X)$ and the image of $H^0(X, mathcal{K}_X^*)$ is $operatorname{Princ}(X)$. Hence the isomorphism $operatorname{Div}(X)/operatorname{Princ}(X)=H^1(X,mathcal{O}_X^*)$.
$endgroup$
– Roland
Feb 2 at 8:42
$begingroup$
@Roland Thank you very much for the response. I would like to clarify the following: The definition of $text{Div}(X)$ I know is the (collection of) formal sums of prime divisors over $X$. However I interpret $H^0(X, mathcal{K}_X^ast/mathcal{O}_X^ast)$ as the global section of $X$, i.e. all regular functions $f$ vanishing at some point. How are these two equal by definition?
$endgroup$
– thedilated
Feb 2 at 9:44
2
$begingroup$
Ok I was using the notion of Cartier divisor and you are using the notion of Weil divisor. There is always a map $operatorname{CaDiv}(X)tooperatorname{Div}(X)$. On a smooth scheme this map is an isomorphism.
$endgroup$
– Roland
Feb 2 at 11:07