Existence of Riemann surface, holomorphic maps
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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
user136313user136313
363
$endgroup$
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
$endgroup$
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
$endgroup$
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
asked Feb 26 at 9:11
user136313user136313
363
edited Feb 26 at 10:53
user136313
asked Feb 26 at 9:11
user136313user136313
363
asked Feb 26 at 9:11
user136313user136313
363
asked Feb 26 at 9:11
user136313user136313
363
363
add a comment |
add a comment |
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
4,7322239
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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
$requireAMScd$
beginCD
Z @>tilde f >> X\
@V tilde g V V @VV f V\
Y @>> g> mathbb P^1
endCD
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
2,09211214
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2 Answers
2
active
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2 Answers
2
active
oldest
votes
active
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active
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votes
$begingroup$
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
4,7322239
$endgroup$
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
4,7322239
$endgroup$
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
4,7322239
$endgroup$
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
4,7322239
answered Feb 26 at 10:50
user25309user25309
4,7322239
answered Feb 26 at 10:50
user25309user25309
4,7322239
answered Feb 26 at 10:50
user25309user25309
4,7322239
4,7322239
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
$requireAMScd$
beginCD
Z @>tilde f >> X\
@V tilde g V V @VV f V\
Y @>> g> mathbb P^1
endCD
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
2,09211214
$endgroup$
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
$requireAMScd$
beginCD
Z @>tilde f >> X\
@V tilde g V V @VV f V\
Y @>> g> mathbb P^1
endCD
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
2,09211214
$endgroup$
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
$requireAMScd$
beginCD
Z @>tilde f >> X\
@V tilde g V V @VV f V\
Y @>> g> mathbb P^1
endCD
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
2,09211214
$endgroup$
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
$requireAMScd$
beginCD
Z @>tilde f >> X\
@V tilde g V V @VV f V\
Y @>> g> mathbb P^1
endCD
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
2,09211214
answered Feb 26 at 10:56
HenriHenri
2,09211214
answered Feb 26 at 10:56
HenriHenri
2,09211214
answered Feb 26 at 10:56
HenriHenri
2,09211214
2,09211214
add a comment |
add a comment |
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