GAGA for stacks
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
asked Dec 17 at 20:20
Dmitry Vaintrob
2,7101431
add a comment |
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
asked Dec 17 at 20:20
Dmitry Vaintrob
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
add a comment |
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
asked Dec 17 at 20:20
Dmitry Vaintrob
2,7101431
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
asked Dec 17 at 20:20
Dmitry Vaintrob
2,7101431
asked Dec 17 at 20:20
Dmitry Vaintrob
2,7101431
asked Dec 17 at 20:20
Dmitry Vaintrob
2,7101431
asked Dec 17 at 20:20
Dmitry Vaintrob
2,7101431
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
add a comment |
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
add a comment |
1 Answer
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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 $(g(B)=1)$.
answered Dec 17 at 21:09
abx
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
add a comment |
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1 Answer
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1 Answer
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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 $(g(B)=1)$.
answered Dec 17 at 21:09
abx
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
add a comment |
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 $(g(B)=1)$.
answered Dec 17 at 21:09
abx
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
add a comment |
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 $(g(B)=1)$.
answered Dec 17 at 21:09
abx
23.2k34783
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 $(g(B)=1)$.
answered Dec 17 at 21:09
abx
23.2k34783
answered Dec 17 at 21:09
abx
23.2k34783
answered Dec 17 at 21:09
abx
23.2k34783
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
add a comment |
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
add a comment |
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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