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Line bundle on a curve is positive iff it has positive degree
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Let $C$ a complex curve and $L$ a holomorphic line bundle on it.
I want to show that $L$ is positive iff it has positive degree.
Here the degree is defined as $int_C c_1(L)$ and positive means that $c_1(L)$ can be represented by a positive closed real (1,1)-form, i.e. a Kähler form.
(I am not interested in a proof using Kodaira embedding thm)
One direction is easy: Let $L$ positive. Then $c_1(L)$ is a Kähler class and $deg(L)=Vol(C)>0$.
But what about the other direction? Why does $int_C c_1(L)>0$ imply that $c_1(L)$ is positive??
geometry differential-geometry complex-geometry vector-bundles riemann-surfaces
asked Jul 18 at 12:10
mathsta
1638
add a comment |Â
up vote
4
down vote
favorite
Let $C$ a complex curve and $L$ a holomorphic line bundle on it.
I want to show that $L$ is positive iff it has positive degree.
Here the degree is defined as $int_C c_1(L)$ and positive means that $c_1(L)$ can be represented by a positive closed real (1,1)-form, i.e. a Kähler form.
(I am not interested in a proof using Kodaira embedding thm)
One direction is easy: Let $L$ positive. Then $c_1(L)$ is a Kähler class and $deg(L)=Vol(C)>0$.
But what about the other direction? Why does $int_C c_1(L)>0$ imply that $c_1(L)$ is positive??
geometry differential-geometry complex-geometry vector-bundles riemann-surfaces
asked Jul 18 at 12:10
mathsta
1638
1
I suspect that if you don't use the fact that $C$ is projective, you'll have to do a Hodge-theoretic argument such as on pp. 148-150 on Griffiths-Harris. Be warned that the lemma on p. 149 is wrong: You need to assume $d$-exact (which holds in this case, of course) and the proof uses both $partial$- and $barpartial$-exactness.
– Ted Shifrin
Jul 19 at 0:04
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Let $C$ a complex curve and $L$ a holomorphic line bundle on it.
I want to show that $L$ is positive iff it has positive degree.
Here the degree is defined as $int_C c_1(L)$ and positive means that $c_1(L)$ can be represented by a positive closed real (1,1)-form, i.e. a Kähler form.
(I am not interested in a proof using Kodaira embedding thm)
One direction is easy: Let $L$ positive. Then $c_1(L)$ is a Kähler class and $deg(L)=Vol(C)>0$.
But what about the other direction? Why does $int_C c_1(L)>0$ imply that $c_1(L)$ is positive??
geometry differential-geometry complex-geometry vector-bundles riemann-surfaces
asked Jul 18 at 12:10
mathsta
1638
Let $C$ a complex curve and $L$ a holomorphic line bundle on it.
I want to show that $L$ is positive iff it has positive degree.
Here the degree is defined as $int_C c_1(L)$ and positive means that $c_1(L)$ can be represented by a positive closed real (1,1)-form, i.e. a Kähler form.
(I am not interested in a proof using Kodaira embedding thm)
One direction is easy: Let $L$ positive. Then $c_1(L)$ is a Kähler class and $deg(L)=Vol(C)>0$.
But what about the other direction? Why does $int_C c_1(L)>0$ imply that $c_1(L)$ is positive??
geometry differential-geometry complex-geometry vector-bundles riemann-surfaces
asked Jul 18 at 12:10
mathsta
1638
edited Jul 18 at 21:42
asked Jul 18 at 12:10
mathsta
1638
asked Jul 18 at 12:10
mathsta
1638
asked Jul 18 at 12:10
mathsta
1638
1638
1
I suspect that if you don't use the fact that $C$ is projective, you'll have to do a Hodge-theoretic argument such as on pp. 148-150 on Griffiths-Harris. Be warned that the lemma on p. 149 is wrong: You need to assume $d$-exact (which holds in this case, of course) and the proof uses both $partial$- and $barpartial$-exactness.
– Ted Shifrin
Jul 19 at 0:04
add a comment |Â
1
I suspect that if you don't use the fact that $C$ is projective, you'll have to do a Hodge-theoretic argument such as on pp. 148-150 on Griffiths-Harris. Be warned that the lemma on p. 149 is wrong: You need to assume $d$-exact (which holds in this case, of course) and the proof uses both $partial$- and $barpartial$-exactness.
– Ted Shifrin
Jul 19 at 0:04
1
1
I suspect that if you don't use the fact that $C$ is projective, you'll have to do a Hodge-theoretic argument such as on pp. 148-150 on Griffiths-Harris. Be warned that the lemma on p. 149 is wrong: You need to assume $d$-exact (which holds in this case, of course) and the proof uses both $partial$- and $barpartial$-exactness.
– Ted Shifrin
Jul 19 at 0:04
I suspect that if you don't use the fact that $C$ is projective, you'll have to do a Hodge-theoretic argument such as on pp. 148-150 on Griffiths-Harris. Be warned that the lemma on p. 149 is wrong: You need to assume $d$-exact (which holds in this case, of course) and the proof uses both $partial$- and $barpartial$-exactness.
– Ted Shifrin
Jul 19 at 0:04
add a comment |Â
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1
I suspect that if you don't use the fact that $C$ is projective, you'll have to do a Hodge-theoretic argument such as on pp. 148-150 on Griffiths-Harris. Be warned that the lemma on p. 149 is wrong: You need to assume $d$-exact (which holds in this case, of course) and the proof uses both $partial$- and $barpartial$-exactness.
– Ted Shifrin
Jul 19 at 0:04