Definition of $operatornamePic^0(V)$ for $V$ a singular variety
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How does one define the $operatornamePic^0(V)$ for $V$ being a singular, not necessarily normal variety?
Until now the approach I found by searching Google is to prove that the Picard functor is representable but this is not understandable for me at all.
I thought one might define $operatornamePic^0(V)$ as the degree zero line bundles on $V$. But then 1) I am unsure whether I can just copy paste the usual definition of a line bundle in the case of a singular variety and 2) I am uncertain about the definition of the degree of such a bundle because to define the degree one takes the associated Weil divisor. Here, I don't know if the line-bundle-divisor-correspondence still holds and I was also reading that there are problems with defining both Weil and Cartier divisors on singular and in particular non-normal varieties.
Any comment will be appreciated. But please try to keep the arguments simple since I am still beginner on this subject.
algebraic-geometry vector-bundles singularity-theory picard-scheme
edited Aug 14 at 12:20
Bernard
111k635103
asked Aug 14 at 11:44
James
17211
add a comment |Â
up vote
1
down vote
favorite
How does one define the $operatornamePic^0(V)$ for $V$ being a singular, not necessarily normal variety?
Until now the approach I found by searching Google is to prove that the Picard functor is representable but this is not understandable for me at all.
I thought one might define $operatornamePic^0(V)$ as the degree zero line bundles on $V$. But then 1) I am unsure whether I can just copy paste the usual definition of a line bundle in the case of a singular variety and 2) I am uncertain about the definition of the degree of such a bundle because to define the degree one takes the associated Weil divisor. Here, I don't know if the line-bundle-divisor-correspondence still holds and I was also reading that there are problems with defining both Weil and Cartier divisors on singular and in particular non-normal varieties.
Any comment will be appreciated. But please try to keep the arguments simple since I am still beginner on this subject.
algebraic-geometry vector-bundles singularity-theory picard-scheme
edited Aug 14 at 12:20
Bernard
111k635103
asked Aug 14 at 11:44
James
17211
1
For singular varieties I like to think of Cartier divisors rather than line bundles for intuition. In general (for, say, $V$ projective over a field) $textPic(V)$ is itself a group scheme (in fact consisting of a union of connected components that are themselves increasing unions of quasi-projective varieties). $text Pic^0(V)$ is the connected component of the identity element of this group -- roughly, those line bundles that can be "continuously deformed" to the trivial bundle.
– John Brevik
Aug 14 at 13:46
Can we make "line bundles that can be continuously deformed" precise in some way? E.g. by using sheaves or anything else? Is there some reference that I can look at?
– James
Aug 14 at 15:41
1
Say that two line $L_0,L_1$ bundles on $V$ are prealgebraically equivalent if there is a connected scheme $T$ and a line bundle $mathcal L$ on $Vtimes T$, flat over $T$, and two closed points points $t_0, t_1in C$ such that $mathcal L_t_0 = L_0, mathcal L|t_1 = L_1$. Algebraic equivalence is the associated equivalence relation, and that's what I meant. Kleiman's intro to the Picard scheme discusses this; for surfaces, see Hartshorne V, Exercise. 1.7.
– John Brevik
Aug 14 at 16:26
After reading a lot in Hartshorne and Kleiman (the precise statement is Prop. 5.10), this provided me some clarification. Thanks a lot!
– James
Aug 15 at 8:45
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
How does one define the $operatornamePic^0(V)$ for $V$ being a singular, not necessarily normal variety?
Until now the approach I found by searching Google is to prove that the Picard functor is representable but this is not understandable for me at all.
I thought one might define $operatornamePic^0(V)$ as the degree zero line bundles on $V$. But then 1) I am unsure whether I can just copy paste the usual definition of a line bundle in the case of a singular variety and 2) I am uncertain about the definition of the degree of such a bundle because to define the degree one takes the associated Weil divisor. Here, I don't know if the line-bundle-divisor-correspondence still holds and I was also reading that there are problems with defining both Weil and Cartier divisors on singular and in particular non-normal varieties.
Any comment will be appreciated. But please try to keep the arguments simple since I am still beginner on this subject.
algebraic-geometry vector-bundles singularity-theory picard-scheme
edited Aug 14 at 12:20
Bernard
111k635103
asked Aug 14 at 11:44
James
17211
How does one define the $operatornamePic^0(V)$ for $V$ being a singular, not necessarily normal variety?
Until now the approach I found by searching Google is to prove that the Picard functor is representable but this is not understandable for me at all.
I thought one might define $operatornamePic^0(V)$ as the degree zero line bundles on $V$. But then 1) I am unsure whether I can just copy paste the usual definition of a line bundle in the case of a singular variety and 2) I am uncertain about the definition of the degree of such a bundle because to define the degree one takes the associated Weil divisor. Here, I don't know if the line-bundle-divisor-correspondence still holds and I was also reading that there are problems with defining both Weil and Cartier divisors on singular and in particular non-normal varieties.
Any comment will be appreciated. But please try to keep the arguments simple since I am still beginner on this subject.
algebraic-geometry vector-bundles singularity-theory picard-scheme
edited Aug 14 at 12:20
Bernard
111k635103
asked Aug 14 at 11:44
James
17211
edited Aug 14 at 12:20
Bernard
111k635103
edited Aug 14 at 12:20
Bernard
111k635103
edited Aug 14 at 12:20
Bernard
111k635103
111k635103
asked Aug 14 at 11:44
James
17211
asked Aug 14 at 11:44
James
17211
asked Aug 14 at 11:44
James
17211
17211
1
For singular varieties I like to think of Cartier divisors rather than line bundles for intuition. In general (for, say, $V$ projective over a field) $textPic(V)$ is itself a group scheme (in fact consisting of a union of connected components that are themselves increasing unions of quasi-projective varieties). $text Pic^0(V)$ is the connected component of the identity element of this group -- roughly, those line bundles that can be "continuously deformed" to the trivial bundle.
– John Brevik
Aug 14 at 13:46
Can we make "line bundles that can be continuously deformed" precise in some way? E.g. by using sheaves or anything else? Is there some reference that I can look at?
– James
Aug 14 at 15:41
1
Say that two line $L_0,L_1$ bundles on $V$ are prealgebraically equivalent if there is a connected scheme $T$ and a line bundle $mathcal L$ on $Vtimes T$, flat over $T$, and two closed points points $t_0, t_1in C$ such that $mathcal L_t_0 = L_0, mathcal L|t_1 = L_1$. Algebraic equivalence is the associated equivalence relation, and that's what I meant. Kleiman's intro to the Picard scheme discusses this; for surfaces, see Hartshorne V, Exercise. 1.7.
– John Brevik
Aug 14 at 16:26
After reading a lot in Hartshorne and Kleiman (the precise statement is Prop. 5.10), this provided me some clarification. Thanks a lot!
– James
Aug 15 at 8:45
add a comment |Â
1
For singular varieties I like to think of Cartier divisors rather than line bundles for intuition. In general (for, say, $V$ projective over a field) $textPic(V)$ is itself a group scheme (in fact consisting of a union of connected components that are themselves increasing unions of quasi-projective varieties). $text Pic^0(V)$ is the connected component of the identity element of this group -- roughly, those line bundles that can be "continuously deformed" to the trivial bundle.
– John Brevik
Aug 14 at 13:46
Can we make "line bundles that can be continuously deformed" precise in some way? E.g. by using sheaves or anything else? Is there some reference that I can look at?
– James
Aug 14 at 15:41
1
Say that two line $L_0,L_1$ bundles on $V$ are prealgebraically equivalent if there is a connected scheme $T$ and a line bundle $mathcal L$ on $Vtimes T$, flat over $T$, and two closed points points $t_0, t_1in C$ such that $mathcal L_t_0 = L_0, mathcal L|t_1 = L_1$. Algebraic equivalence is the associated equivalence relation, and that's what I meant. Kleiman's intro to the Picard scheme discusses this; for surfaces, see Hartshorne V, Exercise. 1.7.
– John Brevik
Aug 14 at 16:26
After reading a lot in Hartshorne and Kleiman (the precise statement is Prop. 5.10), this provided me some clarification. Thanks a lot!
– James
Aug 15 at 8:45
1
1
For singular varieties I like to think of Cartier divisors rather than line bundles for intuition. In general (for, say, $V$ projective over a field) $textPic(V)$ is itself a group scheme (in fact consisting of a union of connected components that are themselves increasing unions of quasi-projective varieties). $text Pic^0(V)$ is the connected component of the identity element of this group -- roughly, those line bundles that can be "continuously deformed" to the trivial bundle.
– John Brevik
Aug 14 at 13:46
For singular varieties I like to think of Cartier divisors rather than line bundles for intuition. In general (for, say, $V$ projective over a field) $textPic(V)$ is itself a group scheme (in fact consisting of a union of connected components that are themselves increasing unions of quasi-projective varieties). $text Pic^0(V)$ is the connected component of the identity element of this group -- roughly, those line bundles that can be "continuously deformed" to the trivial bundle.
– John Brevik
Aug 14 at 13:46
Can we make "line bundles that can be continuously deformed" precise in some way? E.g. by using sheaves or anything else? Is there some reference that I can look at?
– James
Aug 14 at 15:41
Can we make "line bundles that can be continuously deformed" precise in some way? E.g. by using sheaves or anything else? Is there some reference that I can look at?
– James
Aug 14 at 15:41
1
1
Say that two line $L_0,L_1$ bundles on $V$ are prealgebraically equivalent if there is a connected scheme $T$ and a line bundle $mathcal L$ on $Vtimes T$, flat over $T$, and two closed points points $t_0, t_1in C$ such that $mathcal L_t_0 = L_0, mathcal L|t_1 = L_1$. Algebraic equivalence is the associated equivalence relation, and that's what I meant. Kleiman's intro to the Picard scheme discusses this; for surfaces, see Hartshorne V, Exercise. 1.7.
– John Brevik
Aug 14 at 16:26
Say that two line $L_0,L_1$ bundles on $V$ are prealgebraically equivalent if there is a connected scheme $T$ and a line bundle $mathcal L$ on $Vtimes T$, flat over $T$, and two closed points points $t_0, t_1in C$ such that $mathcal L_t_0 = L_0, mathcal L|t_1 = L_1$. Algebraic equivalence is the associated equivalence relation, and that's what I meant. Kleiman's intro to the Picard scheme discusses this; for surfaces, see Hartshorne V, Exercise. 1.7.
– John Brevik
Aug 14 at 16:26
After reading a lot in Hartshorne and Kleiman (the precise statement is Prop. 5.10), this provided me some clarification. Thanks a lot!
– James
Aug 15 at 8:45
After reading a lot in Hartshorne and Kleiman (the precise statement is Prop. 5.10), this provided me some clarification. Thanks a lot!
– James
Aug 15 at 8:45
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
$mathrmPic^0$ is meaningful only for complete varieties. If the variety is not normal, this is defined as an abelian group, but not as a scheme (see Variete du Picard, by CS Seshadri). To define as a group, one can do the following. Line bundles can be defined for any scheme the usual way, no mention of divisors necessary. For any non-singular projective curve $C$ and a morphism $f:Cto V$, one has the pull back map $mathrmPic, Vto mathrmPic, C$ and thus the inverse image of $mathrmPic^0C$ in $mathrmPic, V$ (Notice that $mathrmPic^0C$ makes sense since $C$ is a non-singular projective curve). The intersection of all these subgroups as $C$ (and $f$) varies is called $mathrmPic^0 V$.
answered Aug 14 at 13:54
Mohan
11k1816
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
1
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
 |Â
show 1 more comment
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$mathrmPic^0$ is meaningful only for complete varieties. If the variety is not normal, this is defined as an abelian group, but not as a scheme (see Variete du Picard, by CS Seshadri). To define as a group, one can do the following. Line bundles can be defined for any scheme the usual way, no mention of divisors necessary. For any non-singular projective curve $C$ and a morphism $f:Cto V$, one has the pull back map $mathrmPic, Vto mathrmPic, C$ and thus the inverse image of $mathrmPic^0C$ in $mathrmPic, V$ (Notice that $mathrmPic^0C$ makes sense since $C$ is a non-singular projective curve). The intersection of all these subgroups as $C$ (and $f$) varies is called $mathrmPic^0 V$.
answered Aug 14 at 13:54
Mohan
11k1816
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
1
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
 |Â
show 1 more comment
up vote
1
down vote
accepted
$mathrmPic^0$ is meaningful only for complete varieties. If the variety is not normal, this is defined as an abelian group, but not as a scheme (see Variete du Picard, by CS Seshadri). To define as a group, one can do the following. Line bundles can be defined for any scheme the usual way, no mention of divisors necessary. For any non-singular projective curve $C$ and a morphism $f:Cto V$, one has the pull back map $mathrmPic, Vto mathrmPic, C$ and thus the inverse image of $mathrmPic^0C$ in $mathrmPic, V$ (Notice that $mathrmPic^0C$ makes sense since $C$ is a non-singular projective curve). The intersection of all these subgroups as $C$ (and $f$) varies is called $mathrmPic^0 V$.
answered Aug 14 at 13:54
Mohan
11k1816
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
1
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
 |Â
show 1 more comment
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$mathrmPic^0$ is meaningful only for complete varieties. If the variety is not normal, this is defined as an abelian group, but not as a scheme (see Variete du Picard, by CS Seshadri). To define as a group, one can do the following. Line bundles can be defined for any scheme the usual way, no mention of divisors necessary. For any non-singular projective curve $C$ and a morphism $f:Cto V$, one has the pull back map $mathrmPic, Vto mathrmPic, C$ and thus the inverse image of $mathrmPic^0C$ in $mathrmPic, V$ (Notice that $mathrmPic^0C$ makes sense since $C$ is a non-singular projective curve). The intersection of all these subgroups as $C$ (and $f$) varies is called $mathrmPic^0 V$.
answered Aug 14 at 13:54
Mohan
11k1816
$mathrmPic^0$ is meaningful only for complete varieties. If the variety is not normal, this is defined as an abelian group, but not as a scheme (see Variete du Picard, by CS Seshadri). To define as a group, one can do the following. Line bundles can be defined for any scheme the usual way, no mention of divisors necessary. For any non-singular projective curve $C$ and a morphism $f:Cto V$, one has the pull back map $mathrmPic, Vto mathrmPic, C$ and thus the inverse image of $mathrmPic^0C$ in $mathrmPic, V$ (Notice that $mathrmPic^0C$ makes sense since $C$ is a non-singular projective curve). The intersection of all these subgroups as $C$ (and $f$) varies is called $mathrmPic^0 V$.
answered Aug 14 at 13:54
Mohan
11k1816
answered Aug 14 at 13:54
Mohan
11k1816
answered Aug 14 at 13:54
Mohan
11k1816
answered Aug 14 at 13:54
Mohan
11k1816
11k1816
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
1
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
 |Â
show 1 more comment
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
1
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for providing a more complete answer. I did not mean to suggest that line bundles didn't exist but just that to me the notion is less intuitive than divisors. I am not familiar with this formulation of $textPic^0$ and wonder whether it differs from what I am familiar with (EGA via Kleiman). Can you comment on the two constructions? I am genuinely curious here.
– John Brevik
Aug 14 at 14:11
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
Thank you for your answer. There are still some things unclear to me. First of all, why do we choose curves? Your definition is very abstract and I have the impression that it will be impossible for me to use it in practice. How is this related to line bundles on $V$, i.e. which line bundles on $V$ will we find in $operatornamePic^0(V)$?
– James
Aug 14 at 15:45
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@JohnBrevik I have not looked at EGA, Kleiman in thirty years, but I doubt the definitions really differ. Line bundles are somewhat pleasanter, since it allows you to deal with relative Picards (often the case in EGA) without invoking flat families.
– Mohan
Aug 14 at 16:26
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
@Mohan Thank you for the reply. My understanding was that once $textPic$ was representable, $textPic^0$ is a scheme, in particular for an integral projective variety, normal or not. I will have a look at your reference!
– John Brevik
Aug 14 at 16:28
1
1
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
@James Definitions tend to be abstract. The curve case is necessary (I hope that is clear to you) and so it makes sense to define the concept using curves. In practice, there are many ways of trying to see whether a line bundle has `degree' zero, but it will depend on what you have. For example, if your variety is projective (not just complete), then you can check that $LinmathrmPic^0V$ if and only if $Lcdot H^n-1=0$ where $H$ is an ample bundle and $n=dim V$. But, as you know, intersection products are quite abstract too, though intuitively, they may look easier.
– Mohan
Aug 14 at 16:31
 |Â
show 1 more comment
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1
For singular varieties I like to think of Cartier divisors rather than line bundles for intuition. In general (for, say, $V$ projective over a field) $textPic(V)$ is itself a group scheme (in fact consisting of a union of connected components that are themselves increasing unions of quasi-projective varieties). $text Pic^0(V)$ is the connected component of the identity element of this group -- roughly, those line bundles that can be "continuously deformed" to the trivial bundle.
– John Brevik
Aug 14 at 13:46
Can we make "line bundles that can be continuously deformed" precise in some way? E.g. by using sheaves or anything else? Is there some reference that I can look at?
– James
Aug 14 at 15:41
1
Say that two line $L_0,L_1$ bundles on $V$ are prealgebraically equivalent if there is a connected scheme $T$ and a line bundle $mathcal L$ on $Vtimes T$, flat over $T$, and two closed points points $t_0, t_1in C$ such that $mathcal L_t_0 = L_0, mathcal L|t_1 = L_1$. Algebraic equivalence is the associated equivalence relation, and that's what I meant. Kleiman's intro to the Picard scheme discusses this; for surfaces, see Hartshorne V, Exercise. 1.7.
– John Brevik
Aug 14 at 16:26
After reading a lot in Hartshorne and Kleiman (the precise statement is Prop. 5.10), this provided me some clarification. Thanks a lot!
– James
Aug 15 at 8:45