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Does every possible Kähler metric on a projective variety arise from the Fubini-Study metric for some embedding?
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Every projective variety inherits a Kähler structure from a projective embedding, by restriction of the Fubini-Study metric. They will generally admit many Kähler structures though. I was wondering if every Kähler structure, maybe only up to cohomology, can be obtained in this way from some projective embedding?
Thanks!
algebraic-geometry complex-geometry kahler-manifolds
asked Sep 1 at 22:36
doetoe
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up vote
4
down vote
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Every projective variety inherits a Kähler structure from a projective embedding, by restriction of the Fubini-Study metric. They will generally admit many Kähler structures though. I was wondering if every Kähler structure, maybe only up to cohomology, can be obtained in this way from some projective embedding?
Thanks!
algebraic-geometry complex-geometry kahler-manifolds
asked Sep 1 at 22:36
doetoe
1,132711
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Every projective variety inherits a Kähler structure from a projective embedding, by restriction of the Fubini-Study metric. They will generally admit many Kähler structures though. I was wondering if every Kähler structure, maybe only up to cohomology, can be obtained in this way from some projective embedding?
Thanks!
algebraic-geometry complex-geometry kahler-manifolds
asked Sep 1 at 22:36
doetoe
1,132711
Every projective variety inherits a Kähler structure from a projective embedding, by restriction of the Fubini-Study metric. They will generally admit many Kähler structures though. I was wondering if every Kähler structure, maybe only up to cohomology, can be obtained in this way from some projective embedding?
Thanks!
algebraic-geometry complex-geometry kahler-manifolds
asked Sep 1 at 22:36
doetoe
1,132711
asked Sep 1 at 22:36
doetoe
1,132711
asked Sep 1 at 22:36
doetoe
1,132711
asked Sep 1 at 22:36
doetoe
1,132711
1,132711
add a comment |Â
add a comment |Â
1 Answer
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6
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The answer is no: If $omega$ is a Kahler metric, then so is $comega$ for all $c>0$. But lots of them are not integral: $comega in H^2(X, mathbb Z)$, so lots of them cannot be presented by pullback of Fubini-Study metric (which has to be of integral class).
Even if $[omega]$ is an integral class, the assertion might not be true. Note that since $[omega]$ is integral, it is the first Chern class of some holomorphic line bundle $L$. Indeed your assumption is that $L$ is positive, which is equivalent to that $L$ is ample, via the Kodaira embedding.
$L$ is ample if $L^k$ induces a projective embedding for some large $k$. If $k=1$ for your $L$, then $[omega]$ is represented by pullback of Fubini-study metric. In general, there are lots of ample line bundle which are not very ample, so all these will be counterexamples (which can be found here).
answered Sep 1 at 22:56
John Ma
37.9k93669
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
The answer is no: If $omega$ is a Kahler metric, then so is $comega$ for all $c>0$. But lots of them are not integral: $comega in H^2(X, mathbb Z)$, so lots of them cannot be presented by pullback of Fubini-Study metric (which has to be of integral class).
Even if $[omega]$ is an integral class, the assertion might not be true. Note that since $[omega]$ is integral, it is the first Chern class of some holomorphic line bundle $L$. Indeed your assumption is that $L$ is positive, which is equivalent to that $L$ is ample, via the Kodaira embedding.
$L$ is ample if $L^k$ induces a projective embedding for some large $k$. If $k=1$ for your $L$, then $[omega]$ is represented by pullback of Fubini-study metric. In general, there are lots of ample line bundle which are not very ample, so all these will be counterexamples (which can be found here).
answered Sep 1 at 22:56
John Ma
37.9k93669
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
add a comment |Â
up vote
6
down vote
accepted
The answer is no: If $omega$ is a Kahler metric, then so is $comega$ for all $c>0$. But lots of them are not integral: $comega in H^2(X, mathbb Z)$, so lots of them cannot be presented by pullback of Fubini-Study metric (which has to be of integral class).
Even if $[omega]$ is an integral class, the assertion might not be true. Note that since $[omega]$ is integral, it is the first Chern class of some holomorphic line bundle $L$. Indeed your assumption is that $L$ is positive, which is equivalent to that $L$ is ample, via the Kodaira embedding.
$L$ is ample if $L^k$ induces a projective embedding for some large $k$. If $k=1$ for your $L$, then $[omega]$ is represented by pullback of Fubini-study metric. In general, there are lots of ample line bundle which are not very ample, so all these will be counterexamples (which can be found here).
answered Sep 1 at 22:56
John Ma
37.9k93669
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
add a comment |Â
up vote
6
down vote
accepted
up vote
6
down vote
accepted
The answer is no: If $omega$ is a Kahler metric, then so is $comega$ for all $c>0$. But lots of them are not integral: $comega in H^2(X, mathbb Z)$, so lots of them cannot be presented by pullback of Fubini-Study metric (which has to be of integral class).
Even if $[omega]$ is an integral class, the assertion might not be true. Note that since $[omega]$ is integral, it is the first Chern class of some holomorphic line bundle $L$. Indeed your assumption is that $L$ is positive, which is equivalent to that $L$ is ample, via the Kodaira embedding.
$L$ is ample if $L^k$ induces a projective embedding for some large $k$. If $k=1$ for your $L$, then $[omega]$ is represented by pullback of Fubini-study metric. In general, there are lots of ample line bundle which are not very ample, so all these will be counterexamples (which can be found here).
answered Sep 1 at 22:56
John Ma
37.9k93669
The answer is no: If $omega$ is a Kahler metric, then so is $comega$ for all $c>0$. But lots of them are not integral: $comega in H^2(X, mathbb Z)$, so lots of them cannot be presented by pullback of Fubini-Study metric (which has to be of integral class).
Even if $[omega]$ is an integral class, the assertion might not be true. Note that since $[omega]$ is integral, it is the first Chern class of some holomorphic line bundle $L$. Indeed your assumption is that $L$ is positive, which is equivalent to that $L$ is ample, via the Kodaira embedding.
$L$ is ample if $L^k$ induces a projective embedding for some large $k$. If $k=1$ for your $L$, then $[omega]$ is represented by pullback of Fubini-study metric. In general, there are lots of ample line bundle which are not very ample, so all these will be counterexamples (which can be found here).
answered Sep 1 at 22:56
John Ma
37.9k93669
edited Sep 1 at 23:18
answered Sep 1 at 22:56
John Ma
37.9k93669
answered Sep 1 at 22:56
John Ma
37.9k93669
answered Sep 1 at 22:56
John Ma
37.9k93669
37.9k93669
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
add a comment |Â
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
I knew it! Couldn't prove it, though . . .
– Robert Lewis
Sep 1 at 23:13
add a comment |Â
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