Unique complex structure on the modular curve $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$
1
$begingroup$
Is the complex structure on the modular curve coming from the quotient $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ unique? (Here $mathbb{H}$ is the upper half plane in $mathbb{C}$)
According to the first comment here, it should be, but the brief description of how this is proved is not so clear to me. I am looking to better understand this argument (or another), or locate a reference discussing this. Does it use the fact that $S^2$ has a unique complex structure?
Edit: Changed terminology from "modular surface" to "modular curve" -- thanks @reuns
algebraic-geometry complex-geometry riemann-surfaces quotient-spaces complex-manifolds
asked Jan 14 at 15:46
Paul W.Paul W.
62
$endgroup$
|
show 2 more comments
1
$begingroup$
Is the complex structure on the modular curve coming from the quotient $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ unique? (Here $mathbb{H}$ is the upper half plane in $mathbb{C}$)
According to the first comment here, it should be, but the brief description of how this is proved is not so clear to me. I am looking to better understand this argument (or another), or locate a reference discussing this. Does it use the fact that $S^2$ has a unique complex structure?
Edit: Changed terminology from "modular surface" to "modular curve" -- thanks @reuns
algebraic-geometry complex-geometry riemann-surfaces quotient-spaces complex-manifolds
asked Jan 14 at 15:46
Paul W.Paul W.
62
$endgroup$
$begingroup$
You meant the modular curve (modular surfaces are quotients of $mathbb{H} times mathbb{H} $). If you change the complex structure on $mathbb{H}$ by the one of $rho(mathbb{H}), rho(x+iy)=-x+iy$ you get another complex structure on $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ (making $overline{f}$ meromorphic for $rho$ iff $f$ is meromorphic for $rho^2$). Your linked post says given a complex structure on $mathbb{H}$ you have some holomorphic charts of the modular curves so a unique complex structure on it.
$endgroup$
– reuns
Jan 15 at 5:41
$begingroup$
Thanks. By unique complex structure on $Omega$ I mean that if $(Omega, J_1)$ and $(Omega, J_2)$ are complex structures on $Omega$, then there exists a biholomorphism $g:(Omega, J_1)rightarrow(Omega, J_2)$. The map $rho$ above doesn't seem to give a different complex structure in that sense, since $(x,y)mapsto(-x,y)$ gives a biholomorphism between the complex structures induced by $rho$ and $rho^2$.
$endgroup$
– Paul W.
Jan 15 at 6:57
$begingroup$
You are right that the discussion in the linked post also involves a holomorphic quotient map; here I am just interested in uniqueness of the complex structure on the specific quotient space $mathbb{H}/PSL(2,mathbb{Z})$ without any restrictions on the quotient map.
$endgroup$
– Paul W.
Jan 15 at 6:58
$begingroup$
$rho$ is a complex conjugaison, it is not a biholomorphism, but yes $mathbb{H}/PSL(2,mathbb{Z}),rho^2$ and $mathbb{H}/PSL(2,mathbb{Z}),rho$ are isomorphic as Riemann surfaces. The complex torus $mathbb{C}/(mathbb{Z}+imathbb{Z})$ when changing its complex structure becomes any complex torus $mathbb{C}/(mathbb{Z}+taumathbb{Z})$ (this time the two are not isomorphic).
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
The same happen should happen with the compactified level $n$ modular curve $mathbb{H}/Gamma_0(n)$, changing its complex structure will give any curve of the same genus. The full modular curve has genus 0, so it is isomorphic to every curves of genus 0. The modular curve without the point at $infty$ is isomorphic to $mathbb{C}$, topologically it is just a simply connected domain, changing its complex structure may give something isomorphic to the unit disk, by the Riemann mapping theorem all the others will be isomorphic to one of them.
$endgroup$
– reuns
Jan 15 at 8:37
|
show 2 more comments
1
1
1
1
$begingroup$
Is the complex structure on the modular curve coming from the quotient $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ unique? (Here $mathbb{H}$ is the upper half plane in $mathbb{C}$)
According to the first comment here, it should be, but the brief description of how this is proved is not so clear to me. I am looking to better understand this argument (or another), or locate a reference discussing this. Does it use the fact that $S^2$ has a unique complex structure?
Edit: Changed terminology from "modular surface" to "modular curve" -- thanks @reuns
algebraic-geometry complex-geometry riemann-surfaces quotient-spaces complex-manifolds
asked Jan 14 at 15:46
Paul W.Paul W.
62
$endgroup$
Is the complex structure on the modular curve coming from the quotient $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ unique? (Here $mathbb{H}$ is the upper half plane in $mathbb{C}$)
According to the first comment here, it should be, but the brief description of how this is proved is not so clear to me. I am looking to better understand this argument (or another), or locate a reference discussing this. Does it use the fact that $S^2$ has a unique complex structure?
Edit: Changed terminology from "modular surface" to "modular curve" -- thanks @reuns
algebraic-geometry complex-geometry riemann-surfaces quotient-spaces complex-manifolds
algebraic-geometry complex-geometry riemann-surfaces quotient-spaces complex-manifolds
asked Jan 14 at 15:46
Paul W.Paul W.
62
asked Jan 14 at 15:46
Paul W.Paul W.
62
edited Jan 15 at 7:10
Paul W.
asked Jan 14 at 15:46
Paul W.Paul W.
62
asked Jan 14 at 15:46
Paul W.Paul W.
62
asked Jan 14 at 15:46
Paul W.Paul W.
62
62
$begingroup$
You meant the modular curve (modular surfaces are quotients of $mathbb{H} times mathbb{H} $). If you change the complex structure on $mathbb{H}$ by the one of $rho(mathbb{H}), rho(x+iy)=-x+iy$ you get another complex structure on $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ (making $overline{f}$ meromorphic for $rho$ iff $f$ is meromorphic for $rho^2$). Your linked post says given a complex structure on $mathbb{H}$ you have some holomorphic charts of the modular curves so a unique complex structure on it.
$endgroup$
– reuns
Jan 15 at 5:41
$begingroup$
Thanks. By unique complex structure on $Omega$ I mean that if $(Omega, J_1)$ and $(Omega, J_2)$ are complex structures on $Omega$, then there exists a biholomorphism $g:(Omega, J_1)rightarrow(Omega, J_2)$. The map $rho$ above doesn't seem to give a different complex structure in that sense, since $(x,y)mapsto(-x,y)$ gives a biholomorphism between the complex structures induced by $rho$ and $rho^2$.
$endgroup$
– Paul W.
Jan 15 at 6:57
$begingroup$
You are right that the discussion in the linked post also involves a holomorphic quotient map; here I am just interested in uniqueness of the complex structure on the specific quotient space $mathbb{H}/PSL(2,mathbb{Z})$ without any restrictions on the quotient map.
$endgroup$
– Paul W.
Jan 15 at 6:58
$begingroup$
$rho$ is a complex conjugaison, it is not a biholomorphism, but yes $mathbb{H}/PSL(2,mathbb{Z}),rho^2$ and $mathbb{H}/PSL(2,mathbb{Z}),rho$ are isomorphic as Riemann surfaces. The complex torus $mathbb{C}/(mathbb{Z}+imathbb{Z})$ when changing its complex structure becomes any complex torus $mathbb{C}/(mathbb{Z}+taumathbb{Z})$ (this time the two are not isomorphic).
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
The same happen should happen with the compactified level $n$ modular curve $mathbb{H}/Gamma_0(n)$, changing its complex structure will give any curve of the same genus. The full modular curve has genus 0, so it is isomorphic to every curves of genus 0. The modular curve without the point at $infty$ is isomorphic to $mathbb{C}$, topologically it is just a simply connected domain, changing its complex structure may give something isomorphic to the unit disk, by the Riemann mapping theorem all the others will be isomorphic to one of them.
$endgroup$
– reuns
Jan 15 at 8:37
|
show 2 more comments
$begingroup$
You meant the modular curve (modular surfaces are quotients of $mathbb{H} times mathbb{H} $). If you change the complex structure on $mathbb{H}$ by the one of $rho(mathbb{H}), rho(x+iy)=-x+iy$ you get another complex structure on $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ (making $overline{f}$ meromorphic for $rho$ iff $f$ is meromorphic for $rho^2$). Your linked post says given a complex structure on $mathbb{H}$ you have some holomorphic charts of the modular curves so a unique complex structure on it.
$endgroup$
– reuns
Jan 15 at 5:41
$begingroup$
Thanks. By unique complex structure on $Omega$ I mean that if $(Omega, J_1)$ and $(Omega, J_2)$ are complex structures on $Omega$, then there exists a biholomorphism $g:(Omega, J_1)rightarrow(Omega, J_2)$. The map $rho$ above doesn't seem to give a different complex structure in that sense, since $(x,y)mapsto(-x,y)$ gives a biholomorphism between the complex structures induced by $rho$ and $rho^2$.
$endgroup$
– Paul W.
Jan 15 at 6:57
$begingroup$
You are right that the discussion in the linked post also involves a holomorphic quotient map; here I am just interested in uniqueness of the complex structure on the specific quotient space $mathbb{H}/PSL(2,mathbb{Z})$ without any restrictions on the quotient map.
$endgroup$
– Paul W.
Jan 15 at 6:58
$begingroup$
$rho$ is a complex conjugaison, it is not a biholomorphism, but yes $mathbb{H}/PSL(2,mathbb{Z}),rho^2$ and $mathbb{H}/PSL(2,mathbb{Z}),rho$ are isomorphic as Riemann surfaces. The complex torus $mathbb{C}/(mathbb{Z}+imathbb{Z})$ when changing its complex structure becomes any complex torus $mathbb{C}/(mathbb{Z}+taumathbb{Z})$ (this time the two are not isomorphic).
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
The same happen should happen with the compactified level $n$ modular curve $mathbb{H}/Gamma_0(n)$, changing its complex structure will give any curve of the same genus. The full modular curve has genus 0, so it is isomorphic to every curves of genus 0. The modular curve without the point at $infty$ is isomorphic to $mathbb{C}$, topologically it is just a simply connected domain, changing its complex structure may give something isomorphic to the unit disk, by the Riemann mapping theorem all the others will be isomorphic to one of them.
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
You meant the modular curve (modular surfaces are quotients of $mathbb{H} times mathbb{H} $). If you change the complex structure on $mathbb{H}$ by the one of $rho(mathbb{H}), rho(x+iy)=-x+iy$ you get another complex structure on $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ (making $overline{f}$ meromorphic for $rho$ iff $f$ is meromorphic for $rho^2$). Your linked post says given a complex structure on $mathbb{H}$ you have some holomorphic charts of the modular curves so a unique complex structure on it.
$endgroup$
– reuns
Jan 15 at 5:41
$begingroup$
You meant the modular curve (modular surfaces are quotients of $mathbb{H} times mathbb{H} $). If you change the complex structure on $mathbb{H}$ by the one of $rho(mathbb{H}), rho(x+iy)=-x+iy$ you get another complex structure on $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ (making $overline{f}$ meromorphic for $rho$ iff $f$ is meromorphic for $rho^2$). Your linked post says given a complex structure on $mathbb{H}$ you have some holomorphic charts of the modular curves so a unique complex structure on it.
$endgroup$
– reuns
Jan 15 at 5:41
$begingroup$
Thanks. By unique complex structure on $Omega$ I mean that if $(Omega, J_1)$ and $(Omega, J_2)$ are complex structures on $Omega$, then there exists a biholomorphism $g:(Omega, J_1)rightarrow(Omega, J_2)$. The map $rho$ above doesn't seem to give a different complex structure in that sense, since $(x,y)mapsto(-x,y)$ gives a biholomorphism between the complex structures induced by $rho$ and $rho^2$.
$endgroup$
– Paul W.
Jan 15 at 6:57
$begingroup$
Thanks. By unique complex structure on $Omega$ I mean that if $(Omega, J_1)$ and $(Omega, J_2)$ are complex structures on $Omega$, then there exists a biholomorphism $g:(Omega, J_1)rightarrow(Omega, J_2)$. The map $rho$ above doesn't seem to give a different complex structure in that sense, since $(x,y)mapsto(-x,y)$ gives a biholomorphism between the complex structures induced by $rho$ and $rho^2$.
$endgroup$
– Paul W.
Jan 15 at 6:57
$begingroup$
You are right that the discussion in the linked post also involves a holomorphic quotient map; here I am just interested in uniqueness of the complex structure on the specific quotient space $mathbb{H}/PSL(2,mathbb{Z})$ without any restrictions on the quotient map.
$endgroup$
– Paul W.
Jan 15 at 6:58
$begingroup$
You are right that the discussion in the linked post also involves a holomorphic quotient map; here I am just interested in uniqueness of the complex structure on the specific quotient space $mathbb{H}/PSL(2,mathbb{Z})$ without any restrictions on the quotient map.
$endgroup$
– Paul W.
Jan 15 at 6:58
$begingroup$
$rho$ is a complex conjugaison, it is not a biholomorphism, but yes $mathbb{H}/PSL(2,mathbb{Z}),rho^2$ and $mathbb{H}/PSL(2,mathbb{Z}),rho$ are isomorphic as Riemann surfaces. The complex torus $mathbb{C}/(mathbb{Z}+imathbb{Z})$ when changing its complex structure becomes any complex torus $mathbb{C}/(mathbb{Z}+taumathbb{Z})$ (this time the two are not isomorphic).
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
$rho$ is a complex conjugaison, it is not a biholomorphism, but yes $mathbb{H}/PSL(2,mathbb{Z}),rho^2$ and $mathbb{H}/PSL(2,mathbb{Z}),rho$ are isomorphic as Riemann surfaces. The complex torus $mathbb{C}/(mathbb{Z}+imathbb{Z})$ when changing its complex structure becomes any complex torus $mathbb{C}/(mathbb{Z}+taumathbb{Z})$ (this time the two are not isomorphic).
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
The same happen should happen with the compactified level $n$ modular curve $mathbb{H}/Gamma_0(n)$, changing its complex structure will give any curve of the same genus. The full modular curve has genus 0, so it is isomorphic to every curves of genus 0. The modular curve without the point at $infty$ is isomorphic to $mathbb{C}$, topologically it is just a simply connected domain, changing its complex structure may give something isomorphic to the unit disk, by the Riemann mapping theorem all the others will be isomorphic to one of them.
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
The same happen should happen with the compactified level $n$ modular curve $mathbb{H}/Gamma_0(n)$, changing its complex structure will give any curve of the same genus. The full modular curve has genus 0, so it is isomorphic to every curves of genus 0. The modular curve without the point at $infty$ is isomorphic to $mathbb{C}$, topologically it is just a simply connected domain, changing its complex structure may give something isomorphic to the unit disk, by the Riemann mapping theorem all the others will be isomorphic to one of them.
$endgroup$
– reuns
Jan 15 at 8:37
|
show 2 more comments
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$begingroup$
You meant the modular curve (modular surfaces are quotients of $mathbb{H} times mathbb{H} $). If you change the complex structure on $mathbb{H}$ by the one of $rho(mathbb{H}), rho(x+iy)=-x+iy$ you get another complex structure on $mathbb{H}/operatorname{PSL}(2,mathbb{Z})$ (making $overline{f}$ meromorphic for $rho$ iff $f$ is meromorphic for $rho^2$). Your linked post says given a complex structure on $mathbb{H}$ you have some holomorphic charts of the modular curves so a unique complex structure on it.
$endgroup$
– reuns
Jan 15 at 5:41
$begingroup$
Thanks. By unique complex structure on $Omega$ I mean that if $(Omega, J_1)$ and $(Omega, J_2)$ are complex structures on $Omega$, then there exists a biholomorphism $g:(Omega, J_1)rightarrow(Omega, J_2)$. The map $rho$ above doesn't seem to give a different complex structure in that sense, since $(x,y)mapsto(-x,y)$ gives a biholomorphism between the complex structures induced by $rho$ and $rho^2$.
$endgroup$
– Paul W.
Jan 15 at 6:57
$begingroup$
You are right that the discussion in the linked post also involves a holomorphic quotient map; here I am just interested in uniqueness of the complex structure on the specific quotient space $mathbb{H}/PSL(2,mathbb{Z})$ without any restrictions on the quotient map.
$endgroup$
– Paul W.
Jan 15 at 6:58
$begingroup$
$rho$ is a complex conjugaison, it is not a biholomorphism, but yes $mathbb{H}/PSL(2,mathbb{Z}),rho^2$ and $mathbb{H}/PSL(2,mathbb{Z}),rho$ are isomorphic as Riemann surfaces. The complex torus $mathbb{C}/(mathbb{Z}+imathbb{Z})$ when changing its complex structure becomes any complex torus $mathbb{C}/(mathbb{Z}+taumathbb{Z})$ (this time the two are not isomorphic).
$endgroup$
– reuns
Jan 15 at 8:37
$begingroup$
The same happen should happen with the compactified level $n$ modular curve $mathbb{H}/Gamma_0(n)$, changing its complex structure will give any curve of the same genus. The full modular curve has genus 0, so it is isomorphic to every curves of genus 0. The modular curve without the point at $infty$ is isomorphic to $mathbb{C}$, topologically it is just a simply connected domain, changing its complex structure may give something isomorphic to the unit disk, by the Riemann mapping theorem all the others will be isomorphic to one of them.
$endgroup$
– reuns
Jan 15 at 8:37