Existence of Riemann surface, holomorphic mapsLinks between Riemann surfaces and algebraic geometryAnalogues...
Existence of Riemann surface, holomorphic maps
Links between Riemann surfaces and algebraic geometryAnalogues of the Weierstrass p function for higher genus compact Riemann surfacesPower series for meromorphic differentials on compact Riemann surfacesWhich Riemann surfaces arise from the Riemann existence theorem?Reference for openness of stable locus of holomorphic family of vector bundles on a compact riemann surface.Analytic curve on Riemann surfaceWhat is a branched Riemann surface with cuts?Is there a complex surface into which every Riemann surface embeds?Embed a bordered Riemann surface into punctured Riemann surfaces?The cotangent bundle of a non-compact Riemann surface
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Say I have compact Riemann surfaces $X$, $Y$. Is there necessarily a Riemann surface $Z$ which maps holomorphically onto both $X$, $Y$?
ag.algebraic-geometry
asked Feb 26 at 9:11
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$begingroup$
Say I have compact Riemann surfaces $X$, $Y$. Is there necessarily a Riemann surface $Z$ which maps holomorphically onto both $X$, $Y$?
ag.algebraic-geometry
asked Feb 26 at 9:11
user136313user136313
363
New contributor
user136313 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Say I have compact Riemann surfaces $X$, $Y$. Is there necessarily a Riemann surface $Z$ which maps holomorphically onto both $X$, $Y$?
ag.algebraic-geometry
asked Feb 26 at 9:11
user136313user136313
363
New contributor
user136313 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Say I have compact Riemann surfaces $X$, $Y$. Is there necessarily a Riemann surface $Z$ which maps holomorphically onto both $X$, $Y$?
ag.algebraic-geometry
ag.algebraic-geometry
asked Feb 26 at 9:11
user136313user136313
363
New contributor
user136313 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Feb 26 at 9:11
user136313user136313
363
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user136313 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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edited Feb 26 at 10:53
user136313
asked Feb 26 at 9:11
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363
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asked Feb 26 at 9:11
user136313user136313
363
asked Feb 26 at 9:11
user136313user136313
363
363
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user136313 is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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2 Answers
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Here is some construction. $X$ and $Y$ are smooth projective algebraic complex curves. So the complex surface $X times Y$ is projective (e.g. by Segre embedding) and so admits a very ample line bundle $L$. By Bertini theorem, the vanishing locus of a general section of $L$ is a smooth projective complex curve $Z$ (so a compact Riemann surface) in $X times Y$. The composition of the inclusion of $Z$ in $X times Y$ with the projection on $X$ (or $Y$) is not constant: if it were constant, $Z$ would be contained in some $Y$ fiber, having zero intersection with a generic $Y$ fiber, and contradicting ampleness of $L$.
answered Feb 26 at 10:50
user25309user25309
4,6322239
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Choose non-constant meromorphic functions $f:Xto mathbb P^1$ and $g:Yto mathbb P^1$ and denote by $Z$ the normalization of the fiber product $Xtimes_{mathbb P^1} Y={(x,y)in Xtimes Y; f(x)=g(y)}$. It comes equipped with two maps $tilde f:Zto X$ and $tilde g:Zto Y$ such that the following diagramm commutes
$require{AMScd}$
begin{CD}
Z @>{tilde f} >> X\
@V {tilde g} V V @VV f V\
Y @>>{ g}> mathbb P^1
end{CD}
Note that $Z$ may be disconnected, but the restriction of $tilde f$ and $tilde g$ to any of its connected components are surjective (because the fibers of those two maps are finite).
answered Feb 26 at 10:56
HenriHenri
2,08211213
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2 Answers
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2 Answers
2
active
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active
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$begingroup$
Here is some construction. $X$ and $Y$ are smooth projective algebraic complex curves. So the complex surface $X times Y$ is projective (e.g. by Segre embedding) and so admits a very ample line bundle $L$. By Bertini theorem, the vanishing locus of a general section of $L$ is a smooth projective complex curve $Z$ (so a compact Riemann surface) in $X times Y$. The composition of the inclusion of $Z$ in $X times Y$ with the projection on $X$ (or $Y$) is not constant: if it were constant, $Z$ would be contained in some $Y$ fiber, having zero intersection with a generic $Y$ fiber, and contradicting ampleness of $L$.
answered Feb 26 at 10:50
user25309user25309
4,6322239
$endgroup$
add a comment |
$begingroup$
Here is some construction. $X$ and $Y$ are smooth projective algebraic complex curves. So the complex surface $X times Y$ is projective (e.g. by Segre embedding) and so admits a very ample line bundle $L$. By Bertini theorem, the vanishing locus of a general section of $L$ is a smooth projective complex curve $Z$ (so a compact Riemann surface) in $X times Y$. The composition of the inclusion of $Z$ in $X times Y$ with the projection on $X$ (or $Y$) is not constant: if it were constant, $Z$ would be contained in some $Y$ fiber, having zero intersection with a generic $Y$ fiber, and contradicting ampleness of $L$.
answered Feb 26 at 10:50
user25309user25309
4,6322239
$endgroup$
add a comment |
$begingroup$
Here is some construction. $X$ and $Y$ are smooth projective algebraic complex curves. So the complex surface $X times Y$ is projective (e.g. by Segre embedding) and so admits a very ample line bundle $L$. By Bertini theorem, the vanishing locus of a general section of $L$ is a smooth projective complex curve $Z$ (so a compact Riemann surface) in $X times Y$. The composition of the inclusion of $Z$ in $X times Y$ with the projection on $X$ (or $Y$) is not constant: if it were constant, $Z$ would be contained in some $Y$ fiber, having zero intersection with a generic $Y$ fiber, and contradicting ampleness of $L$.
answered Feb 26 at 10:50
user25309user25309
4,6322239
$endgroup$
Here is some construction. $X$ and $Y$ are smooth projective algebraic complex curves. So the complex surface $X times Y$ is projective (e.g. by Segre embedding) and so admits a very ample line bundle $L$. By Bertini theorem, the vanishing locus of a general section of $L$ is a smooth projective complex curve $Z$ (so a compact Riemann surface) in $X times Y$. The composition of the inclusion of $Z$ in $X times Y$ with the projection on $X$ (or $Y$) is not constant: if it were constant, $Z$ would be contained in some $Y$ fiber, having zero intersection with a generic $Y$ fiber, and contradicting ampleness of $L$.
answered Feb 26 at 10:50
user25309user25309
4,6322239
answered Feb 26 at 10:50
user25309user25309
4,6322239
answered Feb 26 at 10:50
user25309user25309
4,6322239
answered Feb 26 at 10:50
user25309user25309
4,6322239
4,6322239
add a comment |
add a comment |
$begingroup$
Choose non-constant meromorphic functions $f:Xto mathbb P^1$ and $g:Yto mathbb P^1$ and denote by $Z$ the normalization of the fiber product $Xtimes_{mathbb P^1} Y={(x,y)in Xtimes Y; f(x)=g(y)}$. It comes equipped with two maps $tilde f:Zto X$ and $tilde g:Zto Y$ such that the following diagramm commutes
$require{AMScd}$
begin{CD}
Z @>{tilde f} >> X\
@V {tilde g} V V @VV f V\
Y @>>{ g}> mathbb P^1
end{CD}
Note that $Z$ may be disconnected, but the restriction of $tilde f$ and $tilde g$ to any of its connected components are surjective (because the fibers of those two maps are finite).
answered Feb 26 at 10:56
HenriHenri
2,08211213
$endgroup$
add a comment |
$begingroup$
Choose non-constant meromorphic functions $f:Xto mathbb P^1$ and $g:Yto mathbb P^1$ and denote by $Z$ the normalization of the fiber product $Xtimes_{mathbb P^1} Y={(x,y)in Xtimes Y; f(x)=g(y)}$. It comes equipped with two maps $tilde f:Zto X$ and $tilde g:Zto Y$ such that the following diagramm commutes
$require{AMScd}$
begin{CD}
Z @>{tilde f} >> X\
@V {tilde g} V V @VV f V\
Y @>>{ g}> mathbb P^1
end{CD}
Note that $Z$ may be disconnected, but the restriction of $tilde f$ and $tilde g$ to any of its connected components are surjective (because the fibers of those two maps are finite).
answered Feb 26 at 10:56
HenriHenri
2,08211213
$endgroup$
add a comment |
$begingroup$
Choose non-constant meromorphic functions $f:Xto mathbb P^1$ and $g:Yto mathbb P^1$ and denote by $Z$ the normalization of the fiber product $Xtimes_{mathbb P^1} Y={(x,y)in Xtimes Y; f(x)=g(y)}$. It comes equipped with two maps $tilde f:Zto X$ and $tilde g:Zto Y$ such that the following diagramm commutes
$require{AMScd}$
begin{CD}
Z @>{tilde f} >> X\
@V {tilde g} V V @VV f V\
Y @>>{ g}> mathbb P^1
end{CD}
Note that $Z$ may be disconnected, but the restriction of $tilde f$ and $tilde g$ to any of its connected components are surjective (because the fibers of those two maps are finite).
answered Feb 26 at 10:56
HenriHenri
2,08211213
$endgroup$
Choose non-constant meromorphic functions $f:Xto mathbb P^1$ and $g:Yto mathbb P^1$ and denote by $Z$ the normalization of the fiber product $Xtimes_{mathbb P^1} Y={(x,y)in Xtimes Y; f(x)=g(y)}$. It comes equipped with two maps $tilde f:Zto X$ and $tilde g:Zto Y$ such that the following diagramm commutes
$require{AMScd}$
begin{CD}
Z @>{tilde f} >> X\
@V {tilde g} V V @VV f V\
Y @>>{ g}> mathbb P^1
end{CD}
Note that $Z$ may be disconnected, but the restriction of $tilde f$ and $tilde g$ to any of its connected components are surjective (because the fibers of those two maps are finite).
answered Feb 26 at 10:56
HenriHenri
2,08211213
answered Feb 26 at 10:56
HenriHenri
2,08211213
answered Feb 26 at 10:56
HenriHenri
2,08211213
answered Feb 26 at 10:56
HenriHenri
2,08211213
2,08211213
add a comment |
add a comment |
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