Pushforward of a line bundle along a finite morphism of curves
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Let $f:Xrightarrow Y$ be a finite morphism (a branched covering) of degree $n$ of smooth complex algebraic curves.
It is a known result that for any line bundle $L$ on $X$, the pushforward $f_* L$ is a vector bundle of rank $n$ (i.e. the pushforward sheaf of the sheaf of sections of $L$ is locally free and of rank $n$).
Could anyone give a reference for a simple proof of this result? I know it can be derived as a particular case of more general results, but I am interested in this simple case.
algebraic-geometry algebraic-curves vector-bundles algebraic-vector-bundles
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
$endgroup$
add a comment |
$begingroup$
Let $f:Xrightarrow Y$ be a finite morphism (a branched covering) of degree $n$ of smooth complex algebraic curves.
It is a known result that for any line bundle $L$ on $X$, the pushforward $f_* L$ is a vector bundle of rank $n$ (i.e. the pushforward sheaf of the sheaf of sections of $L$ is locally free and of rank $n$).
Could anyone give a reference for a simple proof of this result? I know it can be derived as a particular case of more general results, but I am interested in this simple case.
algebraic-geometry algebraic-curves vector-bundles algebraic-vector-bundles
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
$endgroup$
$begingroup$
Being a vector bundle is a local property. So, restricting to open sets in $Y$, you may assume $L=mathcal{O}_X$.
$endgroup$
– Mohan
Jan 12 at 15:27
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You need to assume that $f$ is finite FLAT morphism, otherwise this is not true. For instance, if $X = mathbb{A}^2$, $Y = mathbb{A}^2/pm1$, and $f$ is the quotient morphism, then $f_*O_X$ is not locally free.
$endgroup$
– Sasha
Jan 12 at 17:14
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@Sasha Yes, but the OP assumed $X$ and $Y$ are smooth, so that $f:Xto Y$ is flat by Miracle Flatness.
$endgroup$
– Ariyan Javanpeykar
Jan 12 at 18:42
$begingroup$
@AriyanJavanpeykar: Right.
$endgroup$
– Sasha
Jan 12 at 19:26
add a comment |
$begingroup$
Let $f:Xrightarrow Y$ be a finite morphism (a branched covering) of degree $n$ of smooth complex algebraic curves.
It is a known result that for any line bundle $L$ on $X$, the pushforward $f_* L$ is a vector bundle of rank $n$ (i.e. the pushforward sheaf of the sheaf of sections of $L$ is locally free and of rank $n$).
Could anyone give a reference for a simple proof of this result? I know it can be derived as a particular case of more general results, but I am interested in this simple case.
algebraic-geometry algebraic-curves vector-bundles algebraic-vector-bundles
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
$endgroup$
Let $f:Xrightarrow Y$ be a finite morphism (a branched covering) of degree $n$ of smooth complex algebraic curves.
It is a known result that for any line bundle $L$ on $X$, the pushforward $f_* L$ is a vector bundle of rank $n$ (i.e. the pushforward sheaf of the sheaf of sections of $L$ is locally free and of rank $n$).
Could anyone give a reference for a simple proof of this result? I know it can be derived as a particular case of more general results, but I am interested in this simple case.
algebraic-geometry algebraic-curves vector-bundles algebraic-vector-bundles
algebraic-geometry algebraic-curves vector-bundles algebraic-vector-bundles
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
asked Jan 12 at 9:05
G. GallegoG. Gallego
727
727
$begingroup$
Being a vector bundle is a local property. So, restricting to open sets in $Y$, you may assume $L=mathcal{O}_X$.
$endgroup$
– Mohan
Jan 12 at 15:27
$begingroup$
You need to assume that $f$ is finite FLAT morphism, otherwise this is not true. For instance, if $X = mathbb{A}^2$, $Y = mathbb{A}^2/pm1$, and $f$ is the quotient morphism, then $f_*O_X$ is not locally free.
$endgroup$
– Sasha
Jan 12 at 17:14
$begingroup$
@Sasha Yes, but the OP assumed $X$ and $Y$ are smooth, so that $f:Xto Y$ is flat by Miracle Flatness.
$endgroup$
– Ariyan Javanpeykar
Jan 12 at 18:42
$begingroup$
@AriyanJavanpeykar: Right.
$endgroup$
– Sasha
Jan 12 at 19:26
add a comment |
$begingroup$
Being a vector bundle is a local property. So, restricting to open sets in $Y$, you may assume $L=mathcal{O}_X$.
$endgroup$
– Mohan
Jan 12 at 15:27
$begingroup$
You need to assume that $f$ is finite FLAT morphism, otherwise this is not true. For instance, if $X = mathbb{A}^2$, $Y = mathbb{A}^2/pm1$, and $f$ is the quotient morphism, then $f_*O_X$ is not locally free.
$endgroup$
– Sasha
Jan 12 at 17:14
$begingroup$
@Sasha Yes, but the OP assumed $X$ and $Y$ are smooth, so that $f:Xto Y$ is flat by Miracle Flatness.
$endgroup$
– Ariyan Javanpeykar
Jan 12 at 18:42
$begingroup$
@AriyanJavanpeykar: Right.
$endgroup$
– Sasha
Jan 12 at 19:26
$begingroup$
Being a vector bundle is a local property. So, restricting to open sets in $Y$, you may assume $L=mathcal{O}_X$.
$endgroup$
– Mohan
Jan 12 at 15:27
$begingroup$
Being a vector bundle is a local property. So, restricting to open sets in $Y$, you may assume $L=mathcal{O}_X$.
$endgroup$
– Mohan
Jan 12 at 15:27
$begingroup$
You need to assume that $f$ is finite FLAT morphism, otherwise this is not true. For instance, if $X = mathbb{A}^2$, $Y = mathbb{A}^2/pm1$, and $f$ is the quotient morphism, then $f_*O_X$ is not locally free.
$endgroup$
– Sasha
Jan 12 at 17:14
$begingroup$
You need to assume that $f$ is finite FLAT morphism, otherwise this is not true. For instance, if $X = mathbb{A}^2$, $Y = mathbb{A}^2/pm1$, and $f$ is the quotient morphism, then $f_*O_X$ is not locally free.
$endgroup$
– Sasha
Jan 12 at 17:14
$begingroup$
@Sasha Yes, but the OP assumed $X$ and $Y$ are smooth, so that $f:Xto Y$ is flat by Miracle Flatness.
$endgroup$
– Ariyan Javanpeykar
Jan 12 at 18:42
$begingroup$
@Sasha Yes, but the OP assumed $X$ and $Y$ are smooth, so that $f:Xto Y$ is flat by Miracle Flatness.
$endgroup$
– Ariyan Javanpeykar
Jan 12 at 18:42
$begingroup$
@AriyanJavanpeykar: Right.
$endgroup$
– Sasha
Jan 12 at 19:26
$begingroup$
@AriyanJavanpeykar: Right.
$endgroup$
– Sasha
Jan 12 at 19:26
add a comment |
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$begingroup$
Being a vector bundle is a local property. So, restricting to open sets in $Y$, you may assume $L=mathcal{O}_X$.
$endgroup$
– Mohan
Jan 12 at 15:27
$begingroup$
You need to assume that $f$ is finite FLAT morphism, otherwise this is not true. For instance, if $X = mathbb{A}^2$, $Y = mathbb{A}^2/pm1$, and $f$ is the quotient morphism, then $f_*O_X$ is not locally free.
$endgroup$
– Sasha
Jan 12 at 17:14
$begingroup$
@Sasha Yes, but the OP assumed $X$ and $Y$ are smooth, so that $f:Xto Y$ is flat by Miracle Flatness.
$endgroup$
– Ariyan Javanpeykar
Jan 12 at 18:42
$begingroup$
@AriyanJavanpeykar: Right.
$endgroup$
– Sasha
Jan 12 at 19:26