Question on Kahler geometry: Kahler form and $mathcalO_X(1)$

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Given a projective surface (2 complex dimensions) $X$ we can equip it with the line bundle $mathcalO_X(1)$. Let us fix this line bundle. Recall that all projective surfaces are Kahler surfaces. Then, I heard in a seminar that for the class of the associated Kahler form $J$ it is true that
$$
[J] = c_1(mathcalO_X(1))
$$
Why is this the case (if I formulated correctly)? Could you explain such a relation?
algebraic-geometry complex-geometry vector-bundles
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up vote
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down vote
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Given a projective surface (2 complex dimensions) $X$ we can equip it with the line bundle $mathcalO_X(1)$. Let us fix this line bundle. Recall that all projective surfaces are Kahler surfaces. Then, I heard in a seminar that for the class of the associated Kahler form $J$ it is true that
$$
[J] = c_1(mathcalO_X(1))
$$
Why is this the case (if I formulated correctly)? Could you explain such a relation?
algebraic-geometry complex-geometry vector-bundles
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given a projective surface (2 complex dimensions) $X$ we can equip it with the line bundle $mathcalO_X(1)$. Let us fix this line bundle. Recall that all projective surfaces are Kahler surfaces. Then, I heard in a seminar that for the class of the associated Kahler form $J$ it is true that
$$
[J] = c_1(mathcalO_X(1))
$$
Why is this the case (if I formulated correctly)? Could you explain such a relation?
algebraic-geometry complex-geometry vector-bundles
Given a projective surface (2 complex dimensions) $X$ we can equip it with the line bundle $mathcalO_X(1)$. Let us fix this line bundle. Recall that all projective surfaces are Kahler surfaces. Then, I heard in a seminar that for the class of the associated Kahler form $J$ it is true that
$$
[J] = c_1(mathcalO_X(1))
$$
Why is this the case (if I formulated correctly)? Could you explain such a relation?
algebraic-geometry complex-geometry vector-bundles
algebraic-geometry complex-geometry vector-bundles
asked Sep 10 at 13:31
Gorbz
24419
24419
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Because you're restricting the hyperplane class (which corresponds to the sheaf $mathscr O_Bbb P^n(1)$) in $Bbb P^n$ to $X$ to get the the sheaf $mathscr O_X(1)$, and $c_1$ of the hyperplane class bundle is precisely the Kähler form on $Bbb P^n$. So it's just naturality, pulling back by the inclusion map in both cases.
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Because you're restricting the hyperplane class (which corresponds to the sheaf $mathscr O_Bbb P^n(1)$) in $Bbb P^n$ to $X$ to get the the sheaf $mathscr O_X(1)$, and $c_1$ of the hyperplane class bundle is precisely the Kähler form on $Bbb P^n$. So it's just naturality, pulling back by the inclusion map in both cases.
add a comment |Â
up vote
0
down vote
Because you're restricting the hyperplane class (which corresponds to the sheaf $mathscr O_Bbb P^n(1)$) in $Bbb P^n$ to $X$ to get the the sheaf $mathscr O_X(1)$, and $c_1$ of the hyperplane class bundle is precisely the Kähler form on $Bbb P^n$. So it's just naturality, pulling back by the inclusion map in both cases.
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Because you're restricting the hyperplane class (which corresponds to the sheaf $mathscr O_Bbb P^n(1)$) in $Bbb P^n$ to $X$ to get the the sheaf $mathscr O_X(1)$, and $c_1$ of the hyperplane class bundle is precisely the Kähler form on $Bbb P^n$. So it's just naturality, pulling back by the inclusion map in both cases.
Because you're restricting the hyperplane class (which corresponds to the sheaf $mathscr O_Bbb P^n(1)$) in $Bbb P^n$ to $X$ to get the the sheaf $mathscr O_X(1)$, and $c_1$ of the hyperplane class bundle is precisely the Kähler form on $Bbb P^n$. So it's just naturality, pulling back by the inclusion map in both cases.
answered Sep 20 at 18:57


Ted Shifrin
61k44388
61k44388
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