Mixing
Néron-Severi group definition
$begingroup$
Let $X$ be a smooth projective variety defined over $mathbb{C}$. Hartshorne defines the Néron-Severi group as the group of divisors modulo algebraic equivalence. In Lazarsfeld's book "Positivity in algebraic geometry", he defines it is as group of divisors modulo numeric equivalence.
So do the two equivalence coincide? Is this true in general?
algebraic-geometry
edited Feb 2 at 19:53
Pedro
2,9441721
asked Sep 20 '16 at 18:01
user349424user349424
34317
$endgroup$
add a comment |
$begingroup$
Let $X$ be a smooth projective variety defined over $mathbb{C}$. Hartshorne defines the Néron-Severi group as the group of divisors modulo algebraic equivalence. In Lazarsfeld's book "Positivity in algebraic geometry", he defines it is as group of divisors modulo numeric equivalence.
So do the two equivalence coincide? Is this true in general?
algebraic-geometry
edited Feb 2 at 19:53
Pedro
2,9441721
asked Sep 20 '16 at 18:01
user349424user349424
34317
$endgroup$
$begingroup$
A bit late, but the answer in this MO thread partially answers the question: mathoverflow.net/questions/15001/…
$endgroup$
– user347489
Jan 18 '18 at 0:32
add a comment |
$begingroup$
Let $X$ be a smooth projective variety defined over $mathbb{C}$. Hartshorne defines the Néron-Severi group as the group of divisors modulo algebraic equivalence. In Lazarsfeld's book "Positivity in algebraic geometry", he defines it is as group of divisors modulo numeric equivalence.
So do the two equivalence coincide? Is this true in general?
algebraic-geometry
edited Feb 2 at 19:53
Pedro
2,9441721
asked Sep 20 '16 at 18:01
user349424user349424
34317
$endgroup$
Let $X$ be a smooth projective variety defined over $mathbb{C}$. Hartshorne defines the Néron-Severi group as the group of divisors modulo algebraic equivalence. In Lazarsfeld's book "Positivity in algebraic geometry", he defines it is as group of divisors modulo numeric equivalence.
So do the two equivalence coincide? Is this true in general?
algebraic-geometry
algebraic-geometry
edited Feb 2 at 19:53
Pedro
2,9441721
asked Sep 20 '16 at 18:01
user349424user349424
34317
edited Feb 2 at 19:53
Pedro
2,9441721
asked Sep 20 '16 at 18:01
user349424user349424
34317
edited Feb 2 at 19:53
Pedro
2,9441721
edited Feb 2 at 19:53
Pedro
2,9441721
edited Feb 2 at 19:53
Pedro
2,9441721
2,9441721
asked Sep 20 '16 at 18:01
user349424user349424
34317
asked Sep 20 '16 at 18:01
user349424user349424
34317
asked Sep 20 '16 at 18:01
user349424user349424
34317
34317
$begingroup$
A bit late, but the answer in this MO thread partially answers the question: mathoverflow.net/questions/15001/…
$endgroup$
– user347489
Jan 18 '18 at 0:32
add a comment |
$begingroup$
A bit late, but the answer in this MO thread partially answers the question: mathoverflow.net/questions/15001/…
$endgroup$
– user347489
Jan 18 '18 at 0:32
$begingroup$
A bit late, but the answer in this MO thread partially answers the question: mathoverflow.net/questions/15001/…
$endgroup$
– user347489
Jan 18 '18 at 0:32
$begingroup$
A bit late, but the answer in this MO thread partially answers the question: mathoverflow.net/questions/15001/…
$endgroup$
– user347489
Jan 18 '18 at 0:32
add a comment |
1 Answer
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Bad news: No, the two definitions do not agree when you work with integer coefficients. Here is an example:
Let $S$ be an Enriques surface over $mathbb{C}$. Then $H^{1}(S,mathcal{O}_{S})=0$, so by the exponential sequence the first Chern class is an injective homomorphism
$$ c_{1}colon operatorname{Pic}(S)hookrightarrow H^{2}(S,mathbb{Z}) $$
We have $omega_{S}notcong mathcal{O}_{S}$, so $c_{1}(omega_{S})neq 0$. This means that $omega_{S}$ is not algebraically trivial (you can find this in Griffiths and Harris, Principles of Algebraic Geometry, page 462). But $omega_{S}otimes omega_{S}cong mathcal{O}_{S}$, hence $2c_{1}(omega_{S})=0$. In particular $omega_{S}$ is numerically trivial (this follows immediately from Lazarsfeld's definition of intersection product).
Good news: The two definitions do agree up to some torsion, as pointed out by Lazarsfeld in Remark 1.1.21 of Positivity in Algebraic Geometry I. In particular, if you work with rational or real coefficients the torsion disappears and the two definitions become the same. For example, when you talk about cones of curves and ample and nef cones (as Lazarsfeld does later on in the book), you work with real or rational coefficients, so the two definitions agree.
answered Feb 2 at 19:47
PedroPedro
2,9441721
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
Bad news: No, the two definitions do not agree when you work with integer coefficients. Here is an example:
Let $S$ be an Enriques surface over $mathbb{C}$. Then $H^{1}(S,mathcal{O}_{S})=0$, so by the exponential sequence the first Chern class is an injective homomorphism
$$ c_{1}colon operatorname{Pic}(S)hookrightarrow H^{2}(S,mathbb{Z}) $$
We have $omega_{S}notcong mathcal{O}_{S}$, so $c_{1}(omega_{S})neq 0$. This means that $omega_{S}$ is not algebraically trivial (you can find this in Griffiths and Harris, Principles of Algebraic Geometry, page 462). But $omega_{S}otimes omega_{S}cong mathcal{O}_{S}$, hence $2c_{1}(omega_{S})=0$. In particular $omega_{S}$ is numerically trivial (this follows immediately from Lazarsfeld's definition of intersection product).
Good news: The two definitions do agree up to some torsion, as pointed out by Lazarsfeld in Remark 1.1.21 of Positivity in Algebraic Geometry I. In particular, if you work with rational or real coefficients the torsion disappears and the two definitions become the same. For example, when you talk about cones of curves and ample and nef cones (as Lazarsfeld does later on in the book), you work with real or rational coefficients, so the two definitions agree.
answered Feb 2 at 19:47
PedroPedro
2,9441721
$endgroup$
add a comment |
$begingroup$
Bad news: No, the two definitions do not agree when you work with integer coefficients. Here is an example:
Let $S$ be an Enriques surface over $mathbb{C}$. Then $H^{1}(S,mathcal{O}_{S})=0$, so by the exponential sequence the first Chern class is an injective homomorphism
$$ c_{1}colon operatorname{Pic}(S)hookrightarrow H^{2}(S,mathbb{Z}) $$
We have $omega_{S}notcong mathcal{O}_{S}$, so $c_{1}(omega_{S})neq 0$. This means that $omega_{S}$ is not algebraically trivial (you can find this in Griffiths and Harris, Principles of Algebraic Geometry, page 462). But $omega_{S}otimes omega_{S}cong mathcal{O}_{S}$, hence $2c_{1}(omega_{S})=0$. In particular $omega_{S}$ is numerically trivial (this follows immediately from Lazarsfeld's definition of intersection product).
Good news: The two definitions do agree up to some torsion, as pointed out by Lazarsfeld in Remark 1.1.21 of Positivity in Algebraic Geometry I. In particular, if you work with rational or real coefficients the torsion disappears and the two definitions become the same. For example, when you talk about cones of curves and ample and nef cones (as Lazarsfeld does later on in the book), you work with real or rational coefficients, so the two definitions agree.
answered Feb 2 at 19:47
PedroPedro
2,9441721
$endgroup$
add a comment |
$begingroup$
Bad news: No, the two definitions do not agree when you work with integer coefficients. Here is an example:
Let $S$ be an Enriques surface over $mathbb{C}$. Then $H^{1}(S,mathcal{O}_{S})=0$, so by the exponential sequence the first Chern class is an injective homomorphism
$$ c_{1}colon operatorname{Pic}(S)hookrightarrow H^{2}(S,mathbb{Z}) $$
We have $omega_{S}notcong mathcal{O}_{S}$, so $c_{1}(omega_{S})neq 0$. This means that $omega_{S}$ is not algebraically trivial (you can find this in Griffiths and Harris, Principles of Algebraic Geometry, page 462). But $omega_{S}otimes omega_{S}cong mathcal{O}_{S}$, hence $2c_{1}(omega_{S})=0$. In particular $omega_{S}$ is numerically trivial (this follows immediately from Lazarsfeld's definition of intersection product).
Good news: The two definitions do agree up to some torsion, as pointed out by Lazarsfeld in Remark 1.1.21 of Positivity in Algebraic Geometry I. In particular, if you work with rational or real coefficients the torsion disappears and the two definitions become the same. For example, when you talk about cones of curves and ample and nef cones (as Lazarsfeld does later on in the book), you work with real or rational coefficients, so the two definitions agree.
answered Feb 2 at 19:47
PedroPedro
2,9441721
$endgroup$
Bad news: No, the two definitions do not agree when you work with integer coefficients. Here is an example:
Let $S$ be an Enriques surface over $mathbb{C}$. Then $H^{1}(S,mathcal{O}_{S})=0$, so by the exponential sequence the first Chern class is an injective homomorphism
$$ c_{1}colon operatorname{Pic}(S)hookrightarrow H^{2}(S,mathbb{Z}) $$
We have $omega_{S}notcong mathcal{O}_{S}$, so $c_{1}(omega_{S})neq 0$. This means that $omega_{S}$ is not algebraically trivial (you can find this in Griffiths and Harris, Principles of Algebraic Geometry, page 462). But $omega_{S}otimes omega_{S}cong mathcal{O}_{S}$, hence $2c_{1}(omega_{S})=0$. In particular $omega_{S}$ is numerically trivial (this follows immediately from Lazarsfeld's definition of intersection product).
Good news: The two definitions do agree up to some torsion, as pointed out by Lazarsfeld in Remark 1.1.21 of Positivity in Algebraic Geometry I. In particular, if you work with rational or real coefficients the torsion disappears and the two definitions become the same. For example, when you talk about cones of curves and ample and nef cones (as Lazarsfeld does later on in the book), you work with real or rational coefficients, so the two definitions agree.
answered Feb 2 at 19:47
PedroPedro
2,9441721
answered Feb 2 at 19:47
PedroPedro
2,9441721
answered Feb 2 at 19:47
PedroPedro
2,9441721
answered Feb 2 at 19:47
PedroPedro
2,9441721
2,9441721
add a comment |
add a comment |
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$begingroup$
A bit late, but the answer in this MO thread partially answers the question: mathoverflow.net/questions/15001/…
$endgroup$
– user347489
Jan 18 '18 at 0:32