Universal covering of $n$ punctured 2-sphere Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For an elliptic curve E, does there exist a cofinite Fuchsian group without elliptic elements with quotient E minus a finite subsetFuchsian groups and surfacesFundamental group and Riemann surfacesExplicitly realizing Riemann surfaces as a quotient of the upper-half planeAction of modular group on complex upper half plane. Quotient space is $P^1$. [Reference request]Galois groupoid of covering map: are endomorphisms of universal covering spaces automorphisms?Fundamental group and the universal covering space for $X$ which is obtained by attaching a Mobius band to a torus.Covering Maps Abelian Lie groupsA connected manifold $N$ can be identified with its universal covering quotient a discrete groupClassification of covering spaces for spaces that are not locally path connected: counterexamples?
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Universal covering of $n$ punctured 2-sphere
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 17/18, 2019 at 00:00UTC (8:00pm US/Eastern)For an elliptic curve E, does there exist a cofinite Fuchsian group without elliptic elements with quotient E minus a finite subsetFuchsian groups and surfacesFundamental group and Riemann surfacesExplicitly realizing Riemann surfaces as a quotient of the upper-half planeAction of modular group on complex upper half plane. Quotient space is $P^1$. [Reference request]Galois groupoid of covering map: are endomorphisms of universal covering spaces automorphisms?Fundamental group and the universal covering space for $X$ which is obtained by attaching a Mobius band to a torus.Covering Maps Abelian Lie groupsA connected manifold $N$ can be identified with its universal covering quotient a discrete groupClassification of covering spaces for spaces that are not locally path connected: counterexamples?
$begingroup$
Actually, I just want to understand first sentence of Here. It says that
Let $X$ be the $mathbbCP_1$ with $n$ points deleted. Let $n geq 3$. If I understand correctly, the universal covering of $X$ is isomorphic to the upper half plane as a complex analytic space.
And the answer is related to Fuchsian group, but I still do not find a introductory paper about that. Could you explain this statement, or just give some reference about why Fuchsian group is related to universal covering of $X$?
algebraic-topology complex-geometry riemann-surfaces
asked Mar 27 at 3:55
user124697user124697
640515
$endgroup$
add a comment |
$begingroup$
Actually, I just want to understand first sentence of Here. It says that
Let $X$ be the $mathbbCP_1$ with $n$ points deleted. Let $n geq 3$. If I understand correctly, the universal covering of $X$ is isomorphic to the upper half plane as a complex analytic space.
And the answer is related to Fuchsian group, but I still do not find a introductory paper about that. Could you explain this statement, or just give some reference about why Fuchsian group is related to universal covering of $X$?
algebraic-topology complex-geometry riemann-surfaces
asked Mar 27 at 3:55
user124697user124697
640515
$endgroup$
2
$begingroup$
For starters, see en.wikipedia.org/wiki/Uniformization_theorem.
$endgroup$
– Qiaochu Yuan
Mar 27 at 4:08
$begingroup$
@QiaochuYuan Thank you very much!
$endgroup$
– user124697
Mar 27 at 14:54
1
$begingroup$
Uniformization theorem reduces the problem to proving that the universal cover of $X$ is not biholomorphic to $mathbb C$. This follows, for instance, that the fundamental group of $X$ is free nonabelian since the group $Aff(mathbb C)$ contains no discrete free nonabelian subgroups.
$endgroup$
– Moishe Kohan
Mar 30 at 1:33
add a comment |
$begingroup$
Actually, I just want to understand first sentence of Here. It says that
Let $X$ be the $mathbbCP_1$ with $n$ points deleted. Let $n geq 3$. If I understand correctly, the universal covering of $X$ is isomorphic to the upper half plane as a complex analytic space.
And the answer is related to Fuchsian group, but I still do not find a introductory paper about that. Could you explain this statement, or just give some reference about why Fuchsian group is related to universal covering of $X$?
algebraic-topology complex-geometry riemann-surfaces
asked Mar 27 at 3:55
user124697user124697
640515
$endgroup$
Actually, I just want to understand first sentence of Here. It says that
Let $X$ be the $mathbbCP_1$ with $n$ points deleted. Let $n geq 3$. If I understand correctly, the universal covering of $X$ is isomorphic to the upper half plane as a complex analytic space.
And the answer is related to Fuchsian group, but I still do not find a introductory paper about that. Could you explain this statement, or just give some reference about why Fuchsian group is related to universal covering of $X$?
algebraic-topology complex-geometry riemann-surfaces
algebraic-topology complex-geometry riemann-surfaces
asked Mar 27 at 3:55
user124697user124697
640515
asked Mar 27 at 3:55
user124697user124697
640515
asked Mar 27 at 3:55
user124697user124697
640515
asked Mar 27 at 3:55
user124697user124697
640515
asked Mar 27 at 3:55
user124697user124697
640515
640515
2
$begingroup$
For starters, see en.wikipedia.org/wiki/Uniformization_theorem.
$endgroup$
– Qiaochu Yuan
Mar 27 at 4:08
$begingroup$
@QiaochuYuan Thank you very much!
$endgroup$
– user124697
Mar 27 at 14:54
1
$begingroup$
Uniformization theorem reduces the problem to proving that the universal cover of $X$ is not biholomorphic to $mathbb C$. This follows, for instance, that the fundamental group of $X$ is free nonabelian since the group $Aff(mathbb C)$ contains no discrete free nonabelian subgroups.
$endgroup$
– Moishe Kohan
Mar 30 at 1:33
add a comment |
2
$begingroup$
For starters, see en.wikipedia.org/wiki/Uniformization_theorem.
$endgroup$
– Qiaochu Yuan
Mar 27 at 4:08
$begingroup$
@QiaochuYuan Thank you very much!
$endgroup$
– user124697
Mar 27 at 14:54
1
$begingroup$
Uniformization theorem reduces the problem to proving that the universal cover of $X$ is not biholomorphic to $mathbb C$. This follows, for instance, that the fundamental group of $X$ is free nonabelian since the group $Aff(mathbb C)$ contains no discrete free nonabelian subgroups.
$endgroup$
– Moishe Kohan
Mar 30 at 1:33
2
2
$begingroup$
For starters, see en.wikipedia.org/wiki/Uniformization_theorem.
$endgroup$
– Qiaochu Yuan
Mar 27 at 4:08
$begingroup$
For starters, see en.wikipedia.org/wiki/Uniformization_theorem.
$endgroup$
– Qiaochu Yuan
Mar 27 at 4:08
$begingroup$
@QiaochuYuan Thank you very much!
$endgroup$
– user124697
Mar 27 at 14:54
$begingroup$
@QiaochuYuan Thank you very much!
$endgroup$
– user124697
Mar 27 at 14:54
1
1
$begingroup$
Uniformization theorem reduces the problem to proving that the universal cover of $X$ is not biholomorphic to $mathbb C$. This follows, for instance, that the fundamental group of $X$ is free nonabelian since the group $Aff(mathbb C)$ contains no discrete free nonabelian subgroups.
$endgroup$
– Moishe Kohan
Mar 30 at 1:33
$begingroup$
Uniformization theorem reduces the problem to proving that the universal cover of $X$ is not biholomorphic to $mathbb C$. This follows, for instance, that the fundamental group of $X$ is free nonabelian since the group $Aff(mathbb C)$ contains no discrete free nonabelian subgroups.
$endgroup$
– Moishe Kohan
Mar 30 at 1:33
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First, you need to know the Uniformization Theorem:
Theorem. Every simply-connected Riemann surface $X$ is biholomorphic to a model surface which is either $S^2$ or $mathbb C$ of the hyperbolic plane $H^2$ (the upper half-plane if you like).
Depending on which of the three models occur, one calls $X$ spherical, euclidean or hyperbolic. (Sadly, this terminology is inconsistent with the AG terminology, where "elliptic" means "euclidean" while "rational" means "spherical" and "general type" is "hyperbolic".) The same applies to (connected) surfaces $Y$ which are covered by $X$: Such a surface is biholomorphic to the quotient of a model surface by a discrete group $G$, of biholomorphic automorphisms, $Gcong pi_1(Y)$, acting freely (without fixed points) on $X$. Now, the question becomes which groups $G$ can act in such a way on the model surfaces. The group $Aut(mathbb C)$ of conformal automorphisms of $mathbb C$ consists of complex affine transformations
$$
zmapsto az+b.
$$
(One usually proves this in an undergraduate CA class.) Since $G<Aut(mathbb C)$ is supposed to act freely, it follows that $a=1$ for all elements of $G$. Thus, $G$ is free abelian of rank $le 2$. This implies that all surfaces of euclidean type have abelian fundamental group. Since $pi_1$ of $n$ times punctured sphere is free of rank $n-1$, if $nge 3$ then this group is nonabelian, hence, the surface cannot have parabolic type. It obviously cannot have elliptic type. Thus, we conclude that such a surface is the quotient of the hyperbolic plane by a group of automorphisms (hyperbolic isometric) acting properly discontinuously and freely. The same argument shows that every connected RS of euclidean type is either $T^2$ (aka an "ellipic curve"), or is conformal to $mathbb C^times$ or to $mathbb C$.
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
$endgroup$
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
add a comment |
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$begingroup$
First, you need to know the Uniformization Theorem:
Theorem. Every simply-connected Riemann surface $X$ is biholomorphic to a model surface which is either $S^2$ or $mathbb C$ of the hyperbolic plane $H^2$ (the upper half-plane if you like).
Depending on which of the three models occur, one calls $X$ spherical, euclidean or hyperbolic. (Sadly, this terminology is inconsistent with the AG terminology, where "elliptic" means "euclidean" while "rational" means "spherical" and "general type" is "hyperbolic".) The same applies to (connected) surfaces $Y$ which are covered by $X$: Such a surface is biholomorphic to the quotient of a model surface by a discrete group $G$, of biholomorphic automorphisms, $Gcong pi_1(Y)$, acting freely (without fixed points) on $X$. Now, the question becomes which groups $G$ can act in such a way on the model surfaces. The group $Aut(mathbb C)$ of conformal automorphisms of $mathbb C$ consists of complex affine transformations
$$
zmapsto az+b.
$$
(One usually proves this in an undergraduate CA class.) Since $G<Aut(mathbb C)$ is supposed to act freely, it follows that $a=1$ for all elements of $G$. Thus, $G$ is free abelian of rank $le 2$. This implies that all surfaces of euclidean type have abelian fundamental group. Since $pi_1$ of $n$ times punctured sphere is free of rank $n-1$, if $nge 3$ then this group is nonabelian, hence, the surface cannot have parabolic type. It obviously cannot have elliptic type. Thus, we conclude that such a surface is the quotient of the hyperbolic plane by a group of automorphisms (hyperbolic isometric) acting properly discontinuously and freely. The same argument shows that every connected RS of euclidean type is either $T^2$ (aka an "ellipic curve"), or is conformal to $mathbb C^times$ or to $mathbb C$.
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
$endgroup$
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
add a comment |
$begingroup$
First, you need to know the Uniformization Theorem:
Theorem. Every simply-connected Riemann surface $X$ is biholomorphic to a model surface which is either $S^2$ or $mathbb C$ of the hyperbolic plane $H^2$ (the upper half-plane if you like).
Depending on which of the three models occur, one calls $X$ spherical, euclidean or hyperbolic. (Sadly, this terminology is inconsistent with the AG terminology, where "elliptic" means "euclidean" while "rational" means "spherical" and "general type" is "hyperbolic".) The same applies to (connected) surfaces $Y$ which are covered by $X$: Such a surface is biholomorphic to the quotient of a model surface by a discrete group $G$, of biholomorphic automorphisms, $Gcong pi_1(Y)$, acting freely (without fixed points) on $X$. Now, the question becomes which groups $G$ can act in such a way on the model surfaces. The group $Aut(mathbb C)$ of conformal automorphisms of $mathbb C$ consists of complex affine transformations
$$
zmapsto az+b.
$$
(One usually proves this in an undergraduate CA class.) Since $G<Aut(mathbb C)$ is supposed to act freely, it follows that $a=1$ for all elements of $G$. Thus, $G$ is free abelian of rank $le 2$. This implies that all surfaces of euclidean type have abelian fundamental group. Since $pi_1$ of $n$ times punctured sphere is free of rank $n-1$, if $nge 3$ then this group is nonabelian, hence, the surface cannot have parabolic type. It obviously cannot have elliptic type. Thus, we conclude that such a surface is the quotient of the hyperbolic plane by a group of automorphisms (hyperbolic isometric) acting properly discontinuously and freely. The same argument shows that every connected RS of euclidean type is either $T^2$ (aka an "ellipic curve"), or is conformal to $mathbb C^times$ or to $mathbb C$.
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
$endgroup$
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
add a comment |
$begingroup$
First, you need to know the Uniformization Theorem:
Theorem. Every simply-connected Riemann surface $X$ is biholomorphic to a model surface which is either $S^2$ or $mathbb C$ of the hyperbolic plane $H^2$ (the upper half-plane if you like).
Depending on which of the three models occur, one calls $X$ spherical, euclidean or hyperbolic. (Sadly, this terminology is inconsistent with the AG terminology, where "elliptic" means "euclidean" while "rational" means "spherical" and "general type" is "hyperbolic".) The same applies to (connected) surfaces $Y$ which are covered by $X$: Such a surface is biholomorphic to the quotient of a model surface by a discrete group $G$, of biholomorphic automorphisms, $Gcong pi_1(Y)$, acting freely (without fixed points) on $X$. Now, the question becomes which groups $G$ can act in such a way on the model surfaces. The group $Aut(mathbb C)$ of conformal automorphisms of $mathbb C$ consists of complex affine transformations
$$
zmapsto az+b.
$$
(One usually proves this in an undergraduate CA class.) Since $G<Aut(mathbb C)$ is supposed to act freely, it follows that $a=1$ for all elements of $G$. Thus, $G$ is free abelian of rank $le 2$. This implies that all surfaces of euclidean type have abelian fundamental group. Since $pi_1$ of $n$ times punctured sphere is free of rank $n-1$, if $nge 3$ then this group is nonabelian, hence, the surface cannot have parabolic type. It obviously cannot have elliptic type. Thus, we conclude that such a surface is the quotient of the hyperbolic plane by a group of automorphisms (hyperbolic isometric) acting properly discontinuously and freely. The same argument shows that every connected RS of euclidean type is either $T^2$ (aka an "ellipic curve"), or is conformal to $mathbb C^times$ or to $mathbb C$.
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
$endgroup$
First, you need to know the Uniformization Theorem:
Theorem. Every simply-connected Riemann surface $X$ is biholomorphic to a model surface which is either $S^2$ or $mathbb C$ of the hyperbolic plane $H^2$ (the upper half-plane if you like).
Depending on which of the three models occur, one calls $X$ spherical, euclidean or hyperbolic. (Sadly, this terminology is inconsistent with the AG terminology, where "elliptic" means "euclidean" while "rational" means "spherical" and "general type" is "hyperbolic".) The same applies to (connected) surfaces $Y$ which are covered by $X$: Such a surface is biholomorphic to the quotient of a model surface by a discrete group $G$, of biholomorphic automorphisms, $Gcong pi_1(Y)$, acting freely (without fixed points) on $X$. Now, the question becomes which groups $G$ can act in such a way on the model surfaces. The group $Aut(mathbb C)$ of conformal automorphisms of $mathbb C$ consists of complex affine transformations
$$
zmapsto az+b.
$$
(One usually proves this in an undergraduate CA class.) Since $G<Aut(mathbb C)$ is supposed to act freely, it follows that $a=1$ for all elements of $G$. Thus, $G$ is free abelian of rank $le 2$. This implies that all surfaces of euclidean type have abelian fundamental group. Since $pi_1$ of $n$ times punctured sphere is free of rank $n-1$, if $nge 3$ then this group is nonabelian, hence, the surface cannot have parabolic type. It obviously cannot have elliptic type. Thus, we conclude that such a surface is the quotient of the hyperbolic plane by a group of automorphisms (hyperbolic isometric) acting properly discontinuously and freely. The same argument shows that every connected RS of euclidean type is either $T^2$ (aka an "ellipic curve"), or is conformal to $mathbb C^times$ or to $mathbb C$.
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
answered Mar 31 at 16:53
Moishe KohanMoishe Kohan
49k344111
49k344111
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
add a comment |
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
$begingroup$
Wow, thank you very much for such a detailed explanation :) I will definitely study uniformization theorem well to understand your writing precisely.
$endgroup$
– user124697
Mar 31 at 20:48
add a comment |
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$begingroup$
For starters, see en.wikipedia.org/wiki/Uniformization_theorem.
$endgroup$
– Qiaochu Yuan
Mar 27 at 4:08
$begingroup$
@QiaochuYuan Thank you very much!
$endgroup$
– user124697
Mar 27 at 14:54
1
$begingroup$
Uniformization theorem reduces the problem to proving that the universal cover of $X$ is not biholomorphic to $mathbb C$. This follows, for instance, that the fundamental group of $X$ is free nonabelian since the group $Aff(mathbb C)$ contains no discrete free nonabelian subgroups.
$endgroup$
– Moishe Kohan
Mar 30 at 1:33