GAGA for stacks











up vote
7
down vote

favorite
1












I am curious about stacky generalizations of the following GAGA theorem:




If $X, U$ are complex algebraic varieties of finite type, $X$ is proper and $f:Xto U$ is an analytic map then $f$ is algebraic.




There is an established theory of analytic stacks (as well as higher analytic stacks). I am curious about the following question: for what stacks $U$ does the above theorem still hold (with $X$ still assumed to be a proper scheme). One case that is known to hold since the original GAGA is $U = Bmathbb{G}_m$ and I believe it is also true more general affine reductive groups. I'm interested in whether this holds for more exotic stacks, for example $BA$ for $A$ an abelian variety.



In this special case (which I am most curious about) the question can be formulated more classically: suppose $X$ is a proper scheme (say, a curve), $A$ is a polarized abelian variety, and $mathcal{A}to X$ is a complex-analytic principal $A$-bundle over $X$. Is the total space $mathcal{A}$ also algebraic?






ag.algebraic-geometry complex-geometry stacks gaga







share|cite|improve this question






















  • If $U$ is an open substack of a proper algebraic stack with finite inertia over $mathbb{C}$ and $X$ is a proper scheme over $mathbb{C}$, then probably any morphism $X^{an}to U^{an}$ is algebraic.
    – Ariyan Javanpeykar
    1 hour ago















up vote
7
down vote

favorite
1












I am curious about stacky generalizations of the following GAGA theorem:




If $X, U$ are complex algebraic varieties of finite type, $X$ is proper and $f:Xto U$ is an analytic map then $f$ is algebraic.




There is an established theory of analytic stacks (as well as higher analytic stacks). I am curious about the following question: for what stacks $U$ does the above theorem still hold (with $X$ still assumed to be a proper scheme). One case that is known to hold since the original GAGA is $U = Bmathbb{G}_m$ and I believe it is also true more general affine reductive groups. I'm interested in whether this holds for more exotic stacks, for example $BA$ for $A$ an abelian variety.



In this special case (which I am most curious about) the question can be formulated more classically: suppose $X$ is a proper scheme (say, a curve), $A$ is a polarized abelian variety, and $mathcal{A}to X$ is a complex-analytic principal $A$-bundle over $X$. Is the total space $mathcal{A}$ also algebraic?






ag.algebraic-geometry complex-geometry stacks gaga







share|cite|improve this question






















  • If $U$ is an open substack of a proper algebraic stack with finite inertia over $mathbb{C}$ and $X$ is a proper scheme over $mathbb{C}$, then probably any morphism $X^{an}to U^{an}$ is algebraic.
    – Ariyan Javanpeykar
    1 hour ago













up vote
7
down vote

favorite
1









up vote
7
down vote

favorite
1






1





I am curious about stacky generalizations of the following GAGA theorem:




If $X, U$ are complex algebraic varieties of finite type, $X$ is proper and $f:Xto U$ is an analytic map then $f$ is algebraic.




There is an established theory of analytic stacks (as well as higher analytic stacks). I am curious about the following question: for what stacks $U$ does the above theorem still hold (with $X$ still assumed to be a proper scheme). One case that is known to hold since the original GAGA is $U = Bmathbb{G}_m$ and I believe it is also true more general affine reductive groups. I'm interested in whether this holds for more exotic stacks, for example $BA$ for $A$ an abelian variety.



In this special case (which I am most curious about) the question can be formulated more classically: suppose $X$ is a proper scheme (say, a curve), $A$ is a polarized abelian variety, and $mathcal{A}to X$ is a complex-analytic principal $A$-bundle over $X$. Is the total space $mathcal{A}$ also algebraic?






ag.algebraic-geometry complex-geometry stacks gaga







share|cite|improve this question













I am curious about stacky generalizations of the following GAGA theorem:




If $X, U$ are complex algebraic varieties of finite type, $X$ is proper and $f:Xto U$ is an analytic map then $f$ is algebraic.




There is an established theory of analytic stacks (as well as higher analytic stacks). I am curious about the following question: for what stacks $U$ does the above theorem still hold (with $X$ still assumed to be a proper scheme). One case that is known to hold since the original GAGA is $U = Bmathbb{G}_m$ and I believe it is also true more general affine reductive groups. I'm interested in whether this holds for more exotic stacks, for example $BA$ for $A$ an abelian variety.



In this special case (which I am most curious about) the question can be formulated more classically: suppose $X$ is a proper scheme (say, a curve), $A$ is a polarized abelian variety, and $mathcal{A}to X$ is a complex-analytic principal $A$-bundle over $X$. Is the total space $mathcal{A}$ also algebraic?






ag.algebraic-geometry complex-geometry stacks gaga




ag.algebraic-geometry complex-geometry stacks gaga






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked 4 hours ago









Dmitry Vaintrob

2,6751430




2,6751430












  • If $U$ is an open substack of a proper algebraic stack with finite inertia over $mathbb{C}$ and $X$ is a proper scheme over $mathbb{C}$, then probably any morphism $X^{an}to U^{an}$ is algebraic.
    – Ariyan Javanpeykar
    1 hour ago


















  • If $U$ is an open substack of a proper algebraic stack with finite inertia over $mathbb{C}$ and $X$ is a proper scheme over $mathbb{C}$, then probably any morphism $X^{an}to U^{an}$ is algebraic.
    – Ariyan Javanpeykar
    1 hour ago
















If $U$ is an open substack of a proper algebraic stack with finite inertia over $mathbb{C}$ and $X$ is a proper scheme over $mathbb{C}$, then probably any morphism $X^{an}to U^{an}$ is algebraic.
– Ariyan Javanpeykar
1 hour ago




If $U$ is an open substack of a proper algebraic stack with finite inertia over $mathbb{C}$ and $X$ is a proper scheme over $mathbb{C}$, then probably any morphism $X^{an}to U^{an}$ is algebraic.
– Ariyan Javanpeykar
1 hour ago










1 Answer
1






active

oldest

votes

















up vote
12
down vote



accepted










For your "special" question, the answer is negative, already when $A$ is an elliptic curve. In fact, a principal $A$-bundle over a smooth projective curve $B$ which is not topologically trivial is never algebraic — see the book by Barth, Hulek, Peters, Van de Ven, ch. V, Proposition 5.3. There are many examples of this situation, for instance Hopf surfaces $(B=mathbb{P}^1)$ or Kodaira primary surfaces $(g(B)=1)$.






share|cite|improve this answer





















  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    3 hours ago






  • 4




    Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
    – François Brunault
    3 hours ago













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: "504"
};
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',
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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f318911%2fgaga-for-stacks%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes








up vote
12
down vote



accepted










For your "special" question, the answer is negative, already when $A$ is an elliptic curve. In fact, a principal $A$-bundle over a smooth projective curve $B$ which is not topologically trivial is never algebraic — see the book by Barth, Hulek, Peters, Van de Ven, ch. V, Proposition 5.3. There are many examples of this situation, for instance Hopf surfaces $(B=mathbb{P}^1)$ or Kodaira primary surfaces $(g(B)=1)$.






share|cite|improve this answer





















  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    3 hours ago






  • 4




    Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
    – François Brunault
    3 hours ago

















up vote
12
down vote



accepted










For your "special" question, the answer is negative, already when $A$ is an elliptic curve. In fact, a principal $A$-bundle over a smooth projective curve $B$ which is not topologically trivial is never algebraic — see the book by Barth, Hulek, Peters, Van de Ven, ch. V, Proposition 5.3. There are many examples of this situation, for instance Hopf surfaces $(B=mathbb{P}^1)$ or Kodaira primary surfaces $(g(B)=1)$.






share|cite|improve this answer





















  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    3 hours ago






  • 4




    Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
    – François Brunault
    3 hours ago















up vote
12
down vote



accepted







up vote
12
down vote



accepted






For your "special" question, the answer is negative, already when $A$ is an elliptic curve. In fact, a principal $A$-bundle over a smooth projective curve $B$ which is not topologically trivial is never algebraic — see the book by Barth, Hulek, Peters, Van de Ven, ch. V, Proposition 5.3. There are many examples of this situation, for instance Hopf surfaces $(B=mathbb{P}^1)$ or Kodaira primary surfaces $(g(B)=1)$.






share|cite|improve this answer












For your "special" question, the answer is negative, already when $A$ is an elliptic curve. In fact, a principal $A$-bundle over a smooth projective curve $B$ which is not topologically trivial is never algebraic — see the book by Barth, Hulek, Peters, Van de Ven, ch. V, Proposition 5.3. There are many examples of this situation, for instance Hopf surfaces $(B=mathbb{P}^1)$ or Kodaira primary surfaces $(g(B)=1)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered 3 hours ago









abx

23k34782




23k34782












  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    3 hours ago






  • 4




    Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
    – François Brunault
    3 hours ago




















  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    3 hours ago






  • 4




    Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
    – François Brunault
    3 hours ago


















Thanks! Is there a good criterion for when such a result does hold?
– Dmitry Vaintrob
3 hours ago




Thanks! Is there a good criterion for when such a result does hold?
– Dmitry Vaintrob
3 hours ago




4




4




Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
– François Brunault
3 hours ago






Other examples are given by complex tori: Shafarevich constructed an extension of elliptic curves $0 to E_1 to X to E_2 to 0$ where $X$ is not algebraic. More generally, given two abelian varieties $A_1,A_2$ of dim $>0$, almost all extensions $0 to A_1 to X to A_2 to 0$ are not abelian varieties.
– François Brunault
3 hours ago




















draft saved

draft discarded




















































Thanks for contributing an answer to MathOverflow!


  • 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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmathoverflow.net%2fquestions%2f318911%2fgaga-for-stacks%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Futebolista

Image
Ronaldo primeiro jogador a ser eleito por três vezes o melhor jogador do mundo pela FIFA. Ronaldinho Gaúcho atuando pelo Barcelona onde foi duas vezes eleito o melhor jogador do mundo pela FIFA. Kaká atuando pelo Milan onde foi eleito em 2007 o melhor jogador do mundo pela FIFA. Lothar Matthäus primeiro jogador a ser eleito o melhor jogador do mundo pela FIFA em 1991 pela Inter de Milão. Neymar durante partida contra a Escócia, em 27 de março de 2011. Crianças praticando futebol. Robinho, futebolista brasileiro, em 15 de novembro de 2006. Atletas do Chelsea, da Inglaterra. A venda de camisas com nome e número dos jogadores é uma grande fonte de arrecadação. Jürgen Klinsmann atuou como jogador e foi treinador da Alemanha. Estádio de futebol, principal local onde laboram os jogadores. O jogador de futebol , ou Futebolista , é um atleta profissional de futebol, e cuja prática do desporto coletivo é fundamental. Estes trabalhadores sã...
Read more

Lallio

Image
Coordenadas: 45° 40' N 9° 38' E Lallio      Comuna    Lallio Localização de Lallio na Itália Coordenadas 45° 40' N 9° 38' E Região Lombardia Província Bérgamo Área - Total 2 km² Altitude 216 m População  - Total 3 826     • Densidade 1 913 hab./km² Outros dados Comunas limítrofes Bergamo, Dalmine, Stezzano, Treviolo Código ISTAT 016123 Código cadastral E422 Código postal 24040 Prefixo telefônico 035 Website www.comune.lallio.bg.it Lallio é uma comuna italiana da região da Lombardia, província de Bérgamo, com cerca de 3.826 habitantes. Estende-se por uma área de 2 km², tendo uma densidade populacional de 1913 hab/km². Faz fronteira com Bergamo, Dalmine, Stezzano, Treviolo. [ 1 ] [ 2 ] [ 3 ] Demografia | Variação demográfica do município entre 1861 e 2011 [ 3 ] Fonte : Istituto Nazionale di Statistica (ISTAT) - Elabor...
Read more

Jornalista

Image
Uma repórter de televisão segurando um microfone na frente de um cinegrafista. Fotojornalistas no Campeonato Mundial de Atletismo 2013 Uma repórter entrevistando Boris Johnson, Prefeito de Londres. Jornalista é o profissional formado em Jornalismo. É a pessoa responsável pela apuração, investigação e apresentação de notícias, reportagens, entrevistas ou distribuição de notícias ou outra informação de interesse coletivo. O trabalho do jornalista é chamado jornalismo. Um jornalista pode trabalhar com questões gerais ou especializar-se em determinadas áreas. No entanto, a maioria dos jornalistas tendem a se especializar, e cooperando com outros jornalistas, produzir publicações que abrangem muitos tópicos. [ 1 ] Por exemplo, um jornalista esportivo cobre notícias dentro do mundo dos esportes, mas este jornalista pode ser uma parte de um jornal que cobre diversos temas. O exercício do Jornalismo é privativo de jornalista. Entre as áreas em que o jornalista trabalha estão o...
Read more