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Does a projective variety have only finitely many associated Hilbert polynomials?
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Let $X$ be a projective variety over $mathbbC$. If $L$ is an ample line bundle, then $h_L$ denotes the Hilbert polynomial.
Is it true that, if $L$ and $L'$ are ample line bundles which are equal in the Neron-Severi group, then $h_L = h_L'$?
Does this imply, together with finite generation of Neron-Severi groups, that the set of polynomials $ L $ ample line bundle on $X $ is finite?
Can anyone recommend a text (book or article) in which these things are explained (to some extent)?
ag.algebraic-geometry complex-geometry ample-bundles
asked 1 hour ago
Paolo Avi
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up vote
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down vote
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Let $X$ be a projective variety over $mathbbC$. If $L$ is an ample line bundle, then $h_L$ denotes the Hilbert polynomial.
Is it true that, if $L$ and $L'$ are ample line bundles which are equal in the Neron-Severi group, then $h_L = h_L'$?
Does this imply, together with finite generation of Neron-Severi groups, that the set of polynomials $ L $ ample line bundle on $X $ is finite?
Can anyone recommend a text (book or article) in which these things are explained (to some extent)?
ag.algebraic-geometry complex-geometry ample-bundles
asked 1 hour ago
Paolo Avi
82
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group.
– R. van Dobben de Bruyn
1 hour ago
@R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)?
– Paolo Avi
45 mins ago
1
Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way.
– R. van Dobben de Bruyn
43 mins ago
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $X$ be a projective variety over $mathbbC$. If $L$ is an ample line bundle, then $h_L$ denotes the Hilbert polynomial.
Is it true that, if $L$ and $L'$ are ample line bundles which are equal in the Neron-Severi group, then $h_L = h_L'$?
Does this imply, together with finite generation of Neron-Severi groups, that the set of polynomials $ L $ ample line bundle on $X $ is finite?
Can anyone recommend a text (book or article) in which these things are explained (to some extent)?
ag.algebraic-geometry complex-geometry ample-bundles
asked 1 hour ago
Paolo Avi
82
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Let $X$ be a projective variety over $mathbbC$. If $L$ is an ample line bundle, then $h_L$ denotes the Hilbert polynomial.
Is it true that, if $L$ and $L'$ are ample line bundles which are equal in the Neron-Severi group, then $h_L = h_L'$?
Does this imply, together with finite generation of Neron-Severi groups, that the set of polynomials $ L $ ample line bundle on $X $ is finite?
Can anyone recommend a text (book or article) in which these things are explained (to some extent)?
ag.algebraic-geometry complex-geometry ample-bundles
ag.algebraic-geometry complex-geometry ample-bundles
asked 1 hour ago
Paolo Avi
82
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Paolo Avi
82
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Paolo Avi
82
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 1 hour ago
Paolo Avi
82
asked 1 hour ago
Paolo Avi
82
82
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Paolo Avi is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
1
As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group.
– R. van Dobben de Bruyn
1 hour ago
@R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)?
– Paolo Avi
45 mins ago
1
Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way.
– R. van Dobben de Bruyn
43 mins ago
add a comment |Â
1
As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group.
– R. van Dobben de Bruyn
1 hour ago
@R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)?
– Paolo Avi
45 mins ago
1
Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way.
– R. van Dobben de Bruyn
43 mins ago
1
1
As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group.
– R. van Dobben de Bruyn
1 hour ago
As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group.
– R. van Dobben de Bruyn
1 hour ago
@R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)?
– Paolo Avi
45 mins ago
@R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)?
– Paolo Avi
45 mins ago
1
1
Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way.
– R. van Dobben de Bruyn
43 mins ago
Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way.
– R. van Dobben de Bruyn
43 mins ago
add a comment |Â
1 Answer
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up vote
4
down vote
accepted
Yes for the first question, by Riemann--Roch.
No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
answered 1 hour ago
Sasha
19.3k22550
2
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
2
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
@abx: Of course.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
accepted
Yes for the first question, by Riemann--Roch.
No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
answered 1 hour ago
Sasha
19.3k22550
2
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
2
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
@abx: Of course.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
add a comment |Â
up vote
4
down vote
accepted
Yes for the first question, by Riemann--Roch.
No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
answered 1 hour ago
Sasha
19.3k22550
2
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
2
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
@abx: Of course.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
add a comment |Â
up vote
4
down vote
accepted
up vote
4
down vote
accepted
Yes for the first question, by Riemann--Roch.
No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
answered 1 hour ago
Sasha
19.3k22550
Yes for the first question, by Riemann--Roch.
No for the second --- even in the simplest case of a projective line, the polynomial $td + 1$ is the Hilbert polynomial (with respect to $L = O(d)$).
answered 1 hour ago
Sasha
19.3k22550
edited 41 mins ago
answered 1 hour ago
Sasha
19.3k22550
answered 1 hour ago
Sasha
19.3k22550
answered 1 hour ago
Sasha
19.3k22550
19.3k22550
2
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
2
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
@abx: Of course.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
add a comment |Â
2
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
2
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
@abx: Of course.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
2
2
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
You don't really need Riemann-Roch, only that $chi$ is constant in an algebraic family.
– abx
1 hour ago
2
2
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
Should that be $td +1$? I thought the degree of the Hilbert polynomial always equals the dimension of the variety.
– Paolo Avi
46 mins ago
@abx: Of course.
– Sasha
40 mins ago
@abx: Of course.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
@Paolo Avi: Right, I corrected this.
– Sasha
40 mins ago
add a comment |Â
Paolo Avi is a new contributor. Be nice, and check out our Code of Conduct.
Paolo Avi is a new contributor. Be nice, and check out our Code of Conduct.
Paolo Avi is a new contributor. Be nice, and check out our Code of Conduct.
Paolo Avi is a new contributor. Be nice, and check out our Code of Conduct.
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1
As Sasha rightly points out, the answer to the second question is negative because you can take linear combinations of line bundles. My answer to this question contains a slightly modified statement you can deduce from finite generation of the Néron–Severi group.
– R. van Dobben de Bruyn
1 hour ago
@R.vanDobbendeBruyn Do I understand correctly that your answer says that, given a choice of generators, one obtains a "universal" polynomial $p$ (in the sense that any other Hilbert polynomial is a very specific type of specialization of this one)?
– Paolo Avi
45 mins ago
1
Yes, that is the conclusion. (It's universal in a weak sense, because it depends on some choices.) I also worked out one non-trivial example, to show that it's not so easy to get a complete parametrisation of all Hilbert polynomials obtained this way.
– R. van Dobben de Bruyn
43 mins ago