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Kahler form lies in $H^2(X,mathbb Z)$?
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$X$ is a Kahler manifold. Then is it true that the class of Kahler form $[omega]$ lies in $H^2(X,mathbb Z)$?
In fact I am not sure I understand $H^2(X,mathbb Z)$ correctly. Why can we talk about $H^2_dR(X,mathbb Z)$? Because I don't think "forms with integer coefficients" is well-defined.
Edit
I didn't make question clear. As Tsemo Aristide's answer suggests, if $omega$ is a kahler form, then so is $c omega$. So I really want to ask is: can we always find a $c$ such that $comega$ lies in $H^2(X,mathbb Z)$?
algebraic-geometry complex-geometry kahler-manifolds
edited Jul 24 at 20:07
Jake
9241621
asked Jul 23 at 10:11
User X
848
 |Â
show 3 more comments
up vote
2
down vote
favorite
$X$ is a Kahler manifold. Then is it true that the class of Kahler form $[omega]$ lies in $H^2(X,mathbb Z)$?
In fact I am not sure I understand $H^2(X,mathbb Z)$ correctly. Why can we talk about $H^2_dR(X,mathbb Z)$? Because I don't think "forms with integer coefficients" is well-defined.
Edit
I didn't make question clear. As Tsemo Aristide's answer suggests, if $omega$ is a kahler form, then so is $c omega$. So I really want to ask is: can we always find a $c$ such that $comega$ lies in $H^2(X,mathbb Z)$?
algebraic-geometry complex-geometry kahler-manifolds
edited Jul 24 at 20:07
Jake
9241621
asked Jul 23 at 10:11
User X
848
TypIcally these types of questions really mean: 1. Show that $omega$ is closed and 2 Show that $[omega] neq 0$. The Kahler form is closed (it is symplectic), and if you look at the class of the top wedge power of $omega$ this must be nonzero. Thus $[omega] neq 0$.
– Rellek
Jul 23 at 10:16
@Rellek But does it lie in $H^n(X,mathbb Z)$?
– User X
Jul 23 at 10:39
2
No, it is a $2$-form. Elements of $H^n (X , mathbbZ)$ are classes of $n$-forms (assuming $n neq 2$).
– Rellek
Jul 23 at 10:41
2
Not in general. When it does it is sometimes called a Hodge Manifold. It is an important special case because of the Kodaira embedding theorem.
– Jake
Jul 24 at 20:03
2
@UserX Yes. One way to get an idea of how they are related is to consider the exponential sequence $0 to mathbbZ to mathcalO_Xmathop to limits^exp mathcalO_X^ times to 0$. Taking the LES of sheaf cohomology, and using the identification $H^1left( X,mathcalO_X^ times right) cong Picleft( X right)$, we get a map $Picleft( X right)mathop to limits^c_1 H^2left( X,mathbbZ right)$. The image of that positive line bundle under this map should be the same as the cohomology class of the Kahler form.
– Jake
Jul 24 at 20:22
 |Â
show 3 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$X$ is a Kahler manifold. Then is it true that the class of Kahler form $[omega]$ lies in $H^2(X,mathbb Z)$?
In fact I am not sure I understand $H^2(X,mathbb Z)$ correctly. Why can we talk about $H^2_dR(X,mathbb Z)$? Because I don't think "forms with integer coefficients" is well-defined.
Edit
I didn't make question clear. As Tsemo Aristide's answer suggests, if $omega$ is a kahler form, then so is $c omega$. So I really want to ask is: can we always find a $c$ such that $comega$ lies in $H^2(X,mathbb Z)$?
algebraic-geometry complex-geometry kahler-manifolds
edited Jul 24 at 20:07
Jake
9241621
asked Jul 23 at 10:11
User X
848
$X$ is a Kahler manifold. Then is it true that the class of Kahler form $[omega]$ lies in $H^2(X,mathbb Z)$?
In fact I am not sure I understand $H^2(X,mathbb Z)$ correctly. Why can we talk about $H^2_dR(X,mathbb Z)$? Because I don't think "forms with integer coefficients" is well-defined.
Edit
I didn't make question clear. As Tsemo Aristide's answer suggests, if $omega$ is a kahler form, then so is $c omega$. So I really want to ask is: can we always find a $c$ such that $comega$ lies in $H^2(X,mathbb Z)$?
algebraic-geometry complex-geometry kahler-manifolds
edited Jul 24 at 20:07
Jake
9241621
asked Jul 23 at 10:11
User X
848
edited Jul 24 at 20:07
Jake
9241621
edited Jul 24 at 20:07
Jake
9241621
edited Jul 24 at 20:07
Jake
9241621
9241621
asked Jul 23 at 10:11
User X
848
asked Jul 23 at 10:11
User X
848
asked Jul 23 at 10:11
User X
848
848
TypIcally these types of questions really mean: 1. Show that $omega$ is closed and 2 Show that $[omega] neq 0$. The Kahler form is closed (it is symplectic), and if you look at the class of the top wedge power of $omega$ this must be nonzero. Thus $[omega] neq 0$.
– Rellek
Jul 23 at 10:16
@Rellek But does it lie in $H^n(X,mathbb Z)$?
– User X
Jul 23 at 10:39
2
No, it is a $2$-form. Elements of $H^n (X , mathbbZ)$ are classes of $n$-forms (assuming $n neq 2$).
– Rellek
Jul 23 at 10:41
2
Not in general. When it does it is sometimes called a Hodge Manifold. It is an important special case because of the Kodaira embedding theorem.
– Jake
Jul 24 at 20:03
2
@UserX Yes. One way to get an idea of how they are related is to consider the exponential sequence $0 to mathbbZ to mathcalO_Xmathop to limits^exp mathcalO_X^ times to 0$. Taking the LES of sheaf cohomology, and using the identification $H^1left( X,mathcalO_X^ times right) cong Picleft( X right)$, we get a map $Picleft( X right)mathop to limits^c_1 H^2left( X,mathbbZ right)$. The image of that positive line bundle under this map should be the same as the cohomology class of the Kahler form.
– Jake
Jul 24 at 20:22
 |Â
show 3 more comments
TypIcally these types of questions really mean: 1. Show that $omega$ is closed and 2 Show that $[omega] neq 0$. The Kahler form is closed (it is symplectic), and if you look at the class of the top wedge power of $omega$ this must be nonzero. Thus $[omega] neq 0$.
– Rellek
Jul 23 at 10:16
@Rellek But does it lie in $H^n(X,mathbb Z)$?
– User X
Jul 23 at 10:39
2
No, it is a $2$-form. Elements of $H^n (X , mathbbZ)$ are classes of $n$-forms (assuming $n neq 2$).
– Rellek
Jul 23 at 10:41
2
Not in general. When it does it is sometimes called a Hodge Manifold. It is an important special case because of the Kodaira embedding theorem.
– Jake
Jul 24 at 20:03
2
@UserX Yes. One way to get an idea of how they are related is to consider the exponential sequence $0 to mathbbZ to mathcalO_Xmathop to limits^exp mathcalO_X^ times to 0$. Taking the LES of sheaf cohomology, and using the identification $H^1left( X,mathcalO_X^ times right) cong Picleft( X right)$, we get a map $Picleft( X right)mathop to limits^c_1 H^2left( X,mathbbZ right)$. The image of that positive line bundle under this map should be the same as the cohomology class of the Kahler form.
– Jake
Jul 24 at 20:22
TypIcally these types of questions really mean: 1. Show that $omega$ is closed and 2 Show that $[omega] neq 0$. The Kahler form is closed (it is symplectic), and if you look at the class of the top wedge power of $omega$ this must be nonzero. Thus $[omega] neq 0$.
– Rellek
Jul 23 at 10:16
TypIcally these types of questions really mean: 1. Show that $omega$ is closed and 2 Show that $[omega] neq 0$. The Kahler form is closed (it is symplectic), and if you look at the class of the top wedge power of $omega$ this must be nonzero. Thus $[omega] neq 0$.
– Rellek
Jul 23 at 10:16
@Rellek But does it lie in $H^n(X,mathbb Z)$?
– User X
Jul 23 at 10:39
@Rellek But does it lie in $H^n(X,mathbb Z)$?
– User X
Jul 23 at 10:39
2
2
No, it is a $2$-form. Elements of $H^n (X , mathbbZ)$ are classes of $n$-forms (assuming $n neq 2$).
– Rellek
Jul 23 at 10:41
No, it is a $2$-form. Elements of $H^n (X , mathbbZ)$ are classes of $n$-forms (assuming $n neq 2$).
– Rellek
Jul 23 at 10:41
2
2
Not in general. When it does it is sometimes called a Hodge Manifold. It is an important special case because of the Kodaira embedding theorem.
– Jake
Jul 24 at 20:03
Not in general. When it does it is sometimes called a Hodge Manifold. It is an important special case because of the Kodaira embedding theorem.
– Jake
Jul 24 at 20:03
2
2
@UserX Yes. One way to get an idea of how they are related is to consider the exponential sequence $0 to mathbbZ to mathcalO_Xmathop to limits^exp mathcalO_X^ times to 0$. Taking the LES of sheaf cohomology, and using the identification $H^1left( X,mathcalO_X^ times right) cong Picleft( X right)$, we get a map $Picleft( X right)mathop to limits^c_1 H^2left( X,mathbbZ right)$. The image of that positive line bundle under this map should be the same as the cohomology class of the Kahler form.
– Jake
Jul 24 at 20:22
@UserX Yes. One way to get an idea of how they are related is to consider the exponential sequence $0 to mathbbZ to mathcalO_Xmathop to limits^exp mathcalO_X^ times to 0$. Taking the LES of sheaf cohomology, and using the identification $H^1left( X,mathcalO_X^ times right) cong Picleft( X right)$, we get a map $Picleft( X right)mathop to limits^c_1 H^2left( X,mathbbZ right)$. The image of that positive line bundle under this map should be the same as the cohomology class of the Kahler form.
– Jake
Jul 24 at 20:22
 |Â
show 3 more comments
1 Answer
1
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Consider the $2$-torus endowed with its standard Kahler form $omega $. Let $i$ be any irrational number $iomega$ is also a Kahler for the same complex structure but is not in $H^2(mathbbT^2,mathbbZ)$.
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
Consider the $2$-torus endowed with its standard Kahler form $omega $. Let $i$ be any irrational number $iomega$ is also a Kahler for the same complex structure but is not in $H^2(mathbbT^2,mathbbZ)$.
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
add a comment |Â
up vote
2
down vote
Consider the $2$-torus endowed with its standard Kahler form $omega $. Let $i$ be any irrational number $iomega$ is also a Kahler for the same complex structure but is not in $H^2(mathbbT^2,mathbbZ)$.
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
add a comment |Â
up vote
2
down vote
up vote
2
down vote
Consider the $2$-torus endowed with its standard Kahler form $omega $. Let $i$ be any irrational number $iomega$ is also a Kahler for the same complex structure but is not in $H^2(mathbbT^2,mathbbZ)$.
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
Consider the $2$-torus endowed with its standard Kahler form $omega $. Let $i$ be any irrational number $iomega$ is also a Kahler for the same complex structure but is not in $H^2(mathbbT^2,mathbbZ)$.
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
answered Jul 23 at 11:29
Tsemo Aristide
51k11243
51k11243
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
add a comment |Â
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
Thanks for your answer! But can we find a $c$ such that $c omega$ lies in $H^2(X,mathbb Z)$? (I edited the question)
– User X
Jul 24 at 19:57
add a comment |Â
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TypIcally these types of questions really mean: 1. Show that $omega$ is closed and 2 Show that $[omega] neq 0$. The Kahler form is closed (it is symplectic), and if you look at the class of the top wedge power of $omega$ this must be nonzero. Thus $[omega] neq 0$.
– Rellek
Jul 23 at 10:16
@Rellek But does it lie in $H^n(X,mathbb Z)$?
– User X
Jul 23 at 10:39
2
No, it is a $2$-form. Elements of $H^n (X , mathbbZ)$ are classes of $n$-forms (assuming $n neq 2$).
– Rellek
Jul 23 at 10:41
2
Not in general. When it does it is sometimes called a Hodge Manifold. It is an important special case because of the Kodaira embedding theorem.
– Jake
Jul 24 at 20:03
2
@UserX Yes. One way to get an idea of how they are related is to consider the exponential sequence $0 to mathbbZ to mathcalO_Xmathop to limits^exp mathcalO_X^ times to 0$. Taking the LES of sheaf cohomology, and using the identification $H^1left( X,mathcalO_X^ times right) cong Picleft( X right)$, we get a map $Picleft( X right)mathop to limits^c_1 H^2left( X,mathbbZ right)$. The image of that positive line bundle under this map should be the same as the cohomology class of the Kahler form.
– Jake
Jul 24 at 20:22