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Kahler Form Corresponding to Hyperplane Line Bundle?
Clash Royale CLAN TAG#URR8PPP
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Let $Ssubset mathbb P_mathbb C^3$ be a cubic hypersurface. By exponential sequence we know the natrual map
$$Pic(S)cong H^1(S,mathcal O_S^*)to H^2(S,mathbb Z)$$
is an isomorphism. We denote this isomorphism by $c_1$.
Now the primitive part $H^2(S,mathbb Z)_pr$ is defined to be the kernel of the map $-mapsto -wedge omega$ where $omega$ is the kahler form on $S$. I want to understand this primitive part. I read from somewhere (for example here) that it seems $c_1(mathcal O_S(1))=omega$. I don't know why this is true?
algebraic-geometry homology-cohomology complex-geometry intersection-theory
asked Jul 21 at 10:06
Akatsuki
8571623
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Let $Ssubset mathbb P_mathbb C^3$ be a cubic hypersurface. By exponential sequence we know the natrual map
$$Pic(S)cong H^1(S,mathcal O_S^*)to H^2(S,mathbb Z)$$
is an isomorphism. We denote this isomorphism by $c_1$.
Now the primitive part $H^2(S,mathbb Z)_pr$ is defined to be the kernel of the map $-mapsto -wedge omega$ where $omega$ is the kahler form on $S$. I want to understand this primitive part. I read from somewhere (for example here) that it seems $c_1(mathcal O_S(1))=omega$. I don't know why this is true?
algebraic-geometry homology-cohomology complex-geometry intersection-theory
asked Jul 21 at 10:06
Akatsuki
8571623
For degree at least 4, the above map (from Pic to $H^2$) is rarely surjective, so I think you are making a mistake.
– Mohan
Jul 21 at 12:08
@Mohan Yes I made a mistake. I have edited it. (only consider the case of cubic)
– Akatsuki
Jul 21 at 12:28
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $Ssubset mathbb P_mathbb C^3$ be a cubic hypersurface. By exponential sequence we know the natrual map
$$Pic(S)cong H^1(S,mathcal O_S^*)to H^2(S,mathbb Z)$$
is an isomorphism. We denote this isomorphism by $c_1$.
Now the primitive part $H^2(S,mathbb Z)_pr$ is defined to be the kernel of the map $-mapsto -wedge omega$ where $omega$ is the kahler form on $S$. I want to understand this primitive part. I read from somewhere (for example here) that it seems $c_1(mathcal O_S(1))=omega$. I don't know why this is true?
algebraic-geometry homology-cohomology complex-geometry intersection-theory
asked Jul 21 at 10:06
Akatsuki
8571623
Let $Ssubset mathbb P_mathbb C^3$ be a cubic hypersurface. By exponential sequence we know the natrual map
$$Pic(S)cong H^1(S,mathcal O_S^*)to H^2(S,mathbb Z)$$
is an isomorphism. We denote this isomorphism by $c_1$.
Now the primitive part $H^2(S,mathbb Z)_pr$ is defined to be the kernel of the map $-mapsto -wedge omega$ where $omega$ is the kahler form on $S$. I want to understand this primitive part. I read from somewhere (for example here) that it seems $c_1(mathcal O_S(1))=omega$. I don't know why this is true?
algebraic-geometry homology-cohomology complex-geometry intersection-theory
asked Jul 21 at 10:06
Akatsuki
8571623
edited Jul 21 at 12:28
asked Jul 21 at 10:06
Akatsuki
8571623
asked Jul 21 at 10:06
Akatsuki
8571623
asked Jul 21 at 10:06
Akatsuki
8571623
8571623
For degree at least 4, the above map (from Pic to $H^2$) is rarely surjective, so I think you are making a mistake.
– Mohan
Jul 21 at 12:08
@Mohan Yes I made a mistake. I have edited it. (only consider the case of cubic)
– Akatsuki
Jul 21 at 12:28
add a comment |Â
For degree at least 4, the above map (from Pic to $H^2$) is rarely surjective, so I think you are making a mistake.
– Mohan
Jul 21 at 12:08
@Mohan Yes I made a mistake. I have edited it. (only consider the case of cubic)
– Akatsuki
Jul 21 at 12:28
For degree at least 4, the above map (from Pic to $H^2$) is rarely surjective, so I think you are making a mistake.
– Mohan
Jul 21 at 12:08
For degree at least 4, the above map (from Pic to $H^2$) is rarely surjective, so I think you are making a mistake.
– Mohan
Jul 21 at 12:08
@Mohan Yes I made a mistake. I have edited it. (only consider the case of cubic)
– Akatsuki
Jul 21 at 12:28
@Mohan Yes I made a mistake. I have edited it. (only consider the case of cubic)
– Akatsuki
Jul 21 at 12:28
add a comment |Â
1 Answer
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If $omega_mathbb P^3$ is the standard Fubini-Study Kahler form on $mathbb P^3$, then it is a standard fact that $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3$.
[Indeed, $H^1,1(mathbb P^3)$ is one-dimensional, so $c_1(mathcal O_mathbb P^3(1))$ must be proportional to $omega_mathbb P^3$. In fact, it must be a positive multiple, since $mathcal O_mathbb P^3(1)$ is positive line bundle. The fact that the constant of proportionality is one is a matter of convention. We conventionally normalise $omega_mathbb P^3$ so that $int_mathbb P^3 omega_mathbb P^3^3 = 1.$ Meanwhile, we know that $int_mathbb P^3 c_1(mathcal O_mathbb P^3(1))^3 = 1$ because three linearly-independent hyperplanes in $mathbb P^3$ intersect at a single point. Hence $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3.$]
Now suppose $i : S to mathbb P^3$ is the embedding of the cubic $S$ into $mathbb P^3$. Note that:
The standard Kahler form on the cubic $S$ is defined as the pullback $omega_S := i^star (omega_mathbb P^3)$, which is the same thing as $i^star (c_1 (mathcal O_mathbb P^3(1)))$.
The line bundle $mathcal O_S(1)$ is defined as the pullback $O_S(1) := i^star mathcal O_mathbb P^3(1)$.
So in view of the fact of the naturality property of Chern classes, which gives $$i^star c_1(mathcal O_mathbb P^3(1)) = c_1 (i^star mathcal O_mathbb P^3(1)),$$
we have
$$ omega_S=c_1(mathcal O_S(1)).$$
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
If $omega_mathbb P^3$ is the standard Fubini-Study Kahler form on $mathbb P^3$, then it is a standard fact that $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3$.
[Indeed, $H^1,1(mathbb P^3)$ is one-dimensional, so $c_1(mathcal O_mathbb P^3(1))$ must be proportional to $omega_mathbb P^3$. In fact, it must be a positive multiple, since $mathcal O_mathbb P^3(1)$ is positive line bundle. The fact that the constant of proportionality is one is a matter of convention. We conventionally normalise $omega_mathbb P^3$ so that $int_mathbb P^3 omega_mathbb P^3^3 = 1.$ Meanwhile, we know that $int_mathbb P^3 c_1(mathcal O_mathbb P^3(1))^3 = 1$ because three linearly-independent hyperplanes in $mathbb P^3$ intersect at a single point. Hence $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3.$]
Now suppose $i : S to mathbb P^3$ is the embedding of the cubic $S$ into $mathbb P^3$. Note that:
The standard Kahler form on the cubic $S$ is defined as the pullback $omega_S := i^star (omega_mathbb P^3)$, which is the same thing as $i^star (c_1 (mathcal O_mathbb P^3(1)))$.
The line bundle $mathcal O_S(1)$ is defined as the pullback $O_S(1) := i^star mathcal O_mathbb P^3(1)$.
So in view of the fact of the naturality property of Chern classes, which gives $$i^star c_1(mathcal O_mathbb P^3(1)) = c_1 (i^star mathcal O_mathbb P^3(1)),$$
we have
$$ omega_S=c_1(mathcal O_S(1)).$$
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
add a comment |Â
up vote
1
down vote
accepted
If $omega_mathbb P^3$ is the standard Fubini-Study Kahler form on $mathbb P^3$, then it is a standard fact that $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3$.
[Indeed, $H^1,1(mathbb P^3)$ is one-dimensional, so $c_1(mathcal O_mathbb P^3(1))$ must be proportional to $omega_mathbb P^3$. In fact, it must be a positive multiple, since $mathcal O_mathbb P^3(1)$ is positive line bundle. The fact that the constant of proportionality is one is a matter of convention. We conventionally normalise $omega_mathbb P^3$ so that $int_mathbb P^3 omega_mathbb P^3^3 = 1.$ Meanwhile, we know that $int_mathbb P^3 c_1(mathcal O_mathbb P^3(1))^3 = 1$ because three linearly-independent hyperplanes in $mathbb P^3$ intersect at a single point. Hence $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3.$]
Now suppose $i : S to mathbb P^3$ is the embedding of the cubic $S$ into $mathbb P^3$. Note that:
The standard Kahler form on the cubic $S$ is defined as the pullback $omega_S := i^star (omega_mathbb P^3)$, which is the same thing as $i^star (c_1 (mathcal O_mathbb P^3(1)))$.
The line bundle $mathcal O_S(1)$ is defined as the pullback $O_S(1) := i^star mathcal O_mathbb P^3(1)$.
So in view of the fact of the naturality property of Chern classes, which gives $$i^star c_1(mathcal O_mathbb P^3(1)) = c_1 (i^star mathcal O_mathbb P^3(1)),$$
we have
$$ omega_S=c_1(mathcal O_S(1)).$$
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
If $omega_mathbb P^3$ is the standard Fubini-Study Kahler form on $mathbb P^3$, then it is a standard fact that $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3$.
[Indeed, $H^1,1(mathbb P^3)$ is one-dimensional, so $c_1(mathcal O_mathbb P^3(1))$ must be proportional to $omega_mathbb P^3$. In fact, it must be a positive multiple, since $mathcal O_mathbb P^3(1)$ is positive line bundle. The fact that the constant of proportionality is one is a matter of convention. We conventionally normalise $omega_mathbb P^3$ so that $int_mathbb P^3 omega_mathbb P^3^3 = 1.$ Meanwhile, we know that $int_mathbb P^3 c_1(mathcal O_mathbb P^3(1))^3 = 1$ because three linearly-independent hyperplanes in $mathbb P^3$ intersect at a single point. Hence $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3.$]
Now suppose $i : S to mathbb P^3$ is the embedding of the cubic $S$ into $mathbb P^3$. Note that:
The standard Kahler form on the cubic $S$ is defined as the pullback $omega_S := i^star (omega_mathbb P^3)$, which is the same thing as $i^star (c_1 (mathcal O_mathbb P^3(1)))$.
The line bundle $mathcal O_S(1)$ is defined as the pullback $O_S(1) := i^star mathcal O_mathbb P^3(1)$.
So in view of the fact of the naturality property of Chern classes, which gives $$i^star c_1(mathcal O_mathbb P^3(1)) = c_1 (i^star mathcal O_mathbb P^3(1)),$$
we have
$$ omega_S=c_1(mathcal O_S(1)).$$
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
If $omega_mathbb P^3$ is the standard Fubini-Study Kahler form on $mathbb P^3$, then it is a standard fact that $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3$.
[Indeed, $H^1,1(mathbb P^3)$ is one-dimensional, so $c_1(mathcal O_mathbb P^3(1))$ must be proportional to $omega_mathbb P^3$. In fact, it must be a positive multiple, since $mathcal O_mathbb P^3(1)$ is positive line bundle. The fact that the constant of proportionality is one is a matter of convention. We conventionally normalise $omega_mathbb P^3$ so that $int_mathbb P^3 omega_mathbb P^3^3 = 1.$ Meanwhile, we know that $int_mathbb P^3 c_1(mathcal O_mathbb P^3(1))^3 = 1$ because three linearly-independent hyperplanes in $mathbb P^3$ intersect at a single point. Hence $c_1(mathcal O_mathbb P^3(1)) = omega_mathbb P^3.$]
Now suppose $i : S to mathbb P^3$ is the embedding of the cubic $S$ into $mathbb P^3$. Note that:
The standard Kahler form on the cubic $S$ is defined as the pullback $omega_S := i^star (omega_mathbb P^3)$, which is the same thing as $i^star (c_1 (mathcal O_mathbb P^3(1)))$.
The line bundle $mathcal O_S(1)$ is defined as the pullback $O_S(1) := i^star mathcal O_mathbb P^3(1)$.
So in view of the fact of the naturality property of Chern classes, which gives $$i^star c_1(mathcal O_mathbb P^3(1)) = c_1 (i^star mathcal O_mathbb P^3(1)),$$
we have
$$ omega_S=c_1(mathcal O_S(1)).$$
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
answered Jul 23 at 8:35
Kenny Wong
15.6k21135
15.6k21135
add a comment |Â
add a comment |Â
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For degree at least 4, the above map (from Pic to $H^2$) is rarely surjective, so I think you are making a mistake.
– Mohan
Jul 21 at 12:08
@Mohan Yes I made a mistake. I have edited it. (only consider the case of cubic)
– Akatsuki
Jul 21 at 12:28