GAGA for stacks












14














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
    Dec 17 at 23:02






  • 1




    ...or if $U=[Y/G]$ is the quotient stack for an action of a finite group $G$.
    – Piotr Achinger
    Dec 18 at 9:20










  • @PiotrAchinger Probably one needs some mild "separatedness" condition on the action of $G$ on $Y$...
    – Ariyan Javanpeykar
    Dec 18 at 13:41
















14














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
    Dec 17 at 23:02






  • 1




    ...or if $U=[Y/G]$ is the quotient stack for an action of a finite group $G$.
    – Piotr Achinger
    Dec 18 at 9:20










  • @PiotrAchinger Probably one needs some mild "separatedness" condition on the action of $G$ on $Y$...
    – Ariyan Javanpeykar
    Dec 18 at 13:41














14












14








14


3





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






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 17 at 20:20









Dmitry Vaintrob

2,7101431




2,7101431












  • 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
    Dec 17 at 23:02






  • 1




    ...or if $U=[Y/G]$ is the quotient stack for an action of a finite group $G$.
    – Piotr Achinger
    Dec 18 at 9:20










  • @PiotrAchinger Probably one needs some mild "separatedness" condition on the action of $G$ on $Y$...
    – Ariyan Javanpeykar
    Dec 18 at 13:41


















  • 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
    Dec 17 at 23:02






  • 1




    ...or if $U=[Y/G]$ is the quotient stack for an action of a finite group $G$.
    – Piotr Achinger
    Dec 18 at 9:20










  • @PiotrAchinger Probably one needs some mild "separatedness" condition on the action of $G$ on $Y$...
    – Ariyan Javanpeykar
    Dec 18 at 13:41
















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
Dec 17 at 23:02




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
Dec 17 at 23:02




1




1




...or if $U=[Y/G]$ is the quotient stack for an action of a finite group $G$.
– Piotr Achinger
Dec 18 at 9:20




...or if $U=[Y/G]$ is the quotient stack for an action of a finite group $G$.
– Piotr Achinger
Dec 18 at 9:20












@PiotrAchinger Probably one needs some mild "separatedness" condition on the action of $G$ on $Y$...
– Ariyan Javanpeykar
Dec 18 at 13:41




@PiotrAchinger Probably one needs some mild "separatedness" condition on the action of $G$ on $Y$...
– Ariyan Javanpeykar
Dec 18 at 13:41










1 Answer
1






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oldest

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20














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
    Dec 17 at 21:25






  • 7




    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
    Dec 17 at 21:27













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1 Answer
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1 Answer
1






active

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20














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
    Dec 17 at 21:25






  • 7




    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
    Dec 17 at 21:27


















20














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
    Dec 17 at 21:25






  • 7




    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
    Dec 17 at 21:27
















20












20








20






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 Dec 17 at 21:09









abx

23.2k34783




23.2k34783












  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    Dec 17 at 21:25






  • 7




    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
    Dec 17 at 21:27




















  • Thanks! Is there a good criterion for when such a result does hold?
    – Dmitry Vaintrob
    Dec 17 at 21:25






  • 7




    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
    Dec 17 at 21:27


















Thanks! Is there a good criterion for when such a result does hold?
– Dmitry Vaintrob
Dec 17 at 21:25




Thanks! Is there a good criterion for when such a result does hold?
– Dmitry Vaintrob
Dec 17 at 21:25




7




7




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
Dec 17 at 21:27






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
Dec 17 at 21:27




















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