Mixing
Pushforward of a line bundle along a finite morphism of curves
$begingroup$
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
$endgroup$
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
$begingroup$
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 |
$begingroup$
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 |
0
active
oldest
votes
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070721%2fpushforward-of-a-line-bundle-along-a-finite-morphism-of-curves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3070721%2fpushforward-of-a-line-bundle-along-a-finite-morphism-of-curves%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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