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When a holomorphic function between hyperbolic surfaces is a covering map.
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I'm studying Milnor's book "Dynamics in one Complex Variable", and he states this problem to the reader in the middle of the proof of Pick's Theorem:
If $S$ and $S'$ are two hyperbolic Riemann surfaces (that is, they are both universally covered by the unitary disk $mathbbD$) and let $f: S longrightarrow S'$ be a holomorphic function between them. Let $phi_1: mathbbD longrightarrow S$ and $phi_2 : mathbbD longrightarrow S'$ be their universal covering maps.
Making some choice of points, we can lift $f$ to a function $F: mathbbD longrightarrow mathbbD$, such that the diagram below commutes:
$requireAMScd$
beginCD
mathbbD @>F>> mathbbD\
@Vphi_1VV @VVphi_2V\
S @>f>> S'
endCD
The statement from Milnor's book is:
f is a covering map if and only if F is a conformal automorphism.
Supposing that f is a covering map, we can use the universal property of $phi_2 $ to conclude that F must be a conformal automorphism.
The other side of this is bugging me. I've tried doing some arguments using that $phi_1$ is a local homeomorphism or that $f$ is open (since it is holomorphic), but i couldn't get it right.
Hence, the question is how to prove this fact: If $F$ is a conformal automorphism then $f$ is a covering map.
Accepting any suggestions and insights to prove this.
Note: I don't know if this result is generalizable for greater dimensions or for the smooth case ($S$, $S'$ smooth manifolds and $f$, $F$ being differentiable). Some counterexamples in those directions would be nice too.
general-topology differential-topology riemann-surfaces covering-spaces
asked Aug 3 at 3:54
Felipe Monteiro
384
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up vote
7
down vote
favorite
I'm studying Milnor's book "Dynamics in one Complex Variable", and he states this problem to the reader in the middle of the proof of Pick's Theorem:
If $S$ and $S'$ are two hyperbolic Riemann surfaces (that is, they are both universally covered by the unitary disk $mathbbD$) and let $f: S longrightarrow S'$ be a holomorphic function between them. Let $phi_1: mathbbD longrightarrow S$ and $phi_2 : mathbbD longrightarrow S'$ be their universal covering maps.
Making some choice of points, we can lift $f$ to a function $F: mathbbD longrightarrow mathbbD$, such that the diagram below commutes:
$requireAMScd$
beginCD
mathbbD @>F>> mathbbD\
@Vphi_1VV @VVphi_2V\
S @>f>> S'
endCD
The statement from Milnor's book is:
f is a covering map if and only if F is a conformal automorphism.
Supposing that f is a covering map, we can use the universal property of $phi_2 $ to conclude that F must be a conformal automorphism.
The other side of this is bugging me. I've tried doing some arguments using that $phi_1$ is a local homeomorphism or that $f$ is open (since it is holomorphic), but i couldn't get it right.
Hence, the question is how to prove this fact: If $F$ is a conformal automorphism then $f$ is a covering map.
Accepting any suggestions and insights to prove this.
Note: I don't know if this result is generalizable for greater dimensions or for the smooth case ($S$, $S'$ smooth manifolds and $f$, $F$ being differentiable). Some counterexamples in those directions would be nice too.
general-topology differential-topology riemann-surfaces covering-spaces
asked Aug 3 at 3:54
Felipe Monteiro
384
add a comment |Â
up vote
7
down vote
favorite
up vote
7
down vote
favorite
I'm studying Milnor's book "Dynamics in one Complex Variable", and he states this problem to the reader in the middle of the proof of Pick's Theorem:
If $S$ and $S'$ are two hyperbolic Riemann surfaces (that is, they are both universally covered by the unitary disk $mathbbD$) and let $f: S longrightarrow S'$ be a holomorphic function between them. Let $phi_1: mathbbD longrightarrow S$ and $phi_2 : mathbbD longrightarrow S'$ be their universal covering maps.
Making some choice of points, we can lift $f$ to a function $F: mathbbD longrightarrow mathbbD$, such that the diagram below commutes:
$requireAMScd$
beginCD
mathbbD @>F>> mathbbD\
@Vphi_1VV @VVphi_2V\
S @>f>> S'
endCD
The statement from Milnor's book is:
f is a covering map if and only if F is a conformal automorphism.
Supposing that f is a covering map, we can use the universal property of $phi_2 $ to conclude that F must be a conformal automorphism.
The other side of this is bugging me. I've tried doing some arguments using that $phi_1$ is a local homeomorphism or that $f$ is open (since it is holomorphic), but i couldn't get it right.
Hence, the question is how to prove this fact: If $F$ is a conformal automorphism then $f$ is a covering map.
Accepting any suggestions and insights to prove this.
Note: I don't know if this result is generalizable for greater dimensions or for the smooth case ($S$, $S'$ smooth manifolds and $f$, $F$ being differentiable). Some counterexamples in those directions would be nice too.
general-topology differential-topology riemann-surfaces covering-spaces
asked Aug 3 at 3:54
Felipe Monteiro
384
I'm studying Milnor's book "Dynamics in one Complex Variable", and he states this problem to the reader in the middle of the proof of Pick's Theorem:
If $S$ and $S'$ are two hyperbolic Riemann surfaces (that is, they are both universally covered by the unitary disk $mathbbD$) and let $f: S longrightarrow S'$ be a holomorphic function between them. Let $phi_1: mathbbD longrightarrow S$ and $phi_2 : mathbbD longrightarrow S'$ be their universal covering maps.
Making some choice of points, we can lift $f$ to a function $F: mathbbD longrightarrow mathbbD$, such that the diagram below commutes:
$requireAMScd$
beginCD
mathbbD @>F>> mathbbD\
@Vphi_1VV @VVphi_2V\
S @>f>> S'
endCD
The statement from Milnor's book is:
f is a covering map if and only if F is a conformal automorphism.
Supposing that f is a covering map, we can use the universal property of $phi_2 $ to conclude that F must be a conformal automorphism.
The other side of this is bugging me. I've tried doing some arguments using that $phi_1$ is a local homeomorphism or that $f$ is open (since it is holomorphic), but i couldn't get it right.
Hence, the question is how to prove this fact: If $F$ is a conformal automorphism then $f$ is a covering map.
Accepting any suggestions and insights to prove this.
Note: I don't know if this result is generalizable for greater dimensions or for the smooth case ($S$, $S'$ smooth manifolds and $f$, $F$ being differentiable). Some counterexamples in those directions would be nice too.
general-topology differential-topology riemann-surfaces covering-spaces
asked Aug 3 at 3:54
Felipe Monteiro
384
asked Aug 3 at 3:54
Felipe Monteiro
384
asked Aug 3 at 3:54
Felipe Monteiro
384
asked Aug 3 at 3:54
Felipe Monteiro
384
384
add a comment |Â
add a comment |Â
1 Answer
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up vote
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I'll follow the proof as it is in Milnor's book. So we have Poincaré metric on the surfaces and the covering maps are Riemannian covering.
If $F$ is a conformal isomorphism, then $f$ is a local isometry.
Now, just use that a surjective local isometry with complete domain is always a Riemannian covering map.
Observation: This fact about Riemannian geometry used in the end can be found on Manfredo's book, for example, on the section about Hadamard theorem. The precise statement is the following.
Let $M$ be a complete Riemannian manifold and let $f:M to N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: For all $pin M$ and all $vin T_p M$, we have $|df_p(f)|geq |v|$. Then $f$ is a covering map.
answered Aug 3 at 13:53
Hugocito
1,6251019
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
I'll follow the proof as it is in Milnor's book. So we have Poincaré metric on the surfaces and the covering maps are Riemannian covering.
If $F$ is a conformal isomorphism, then $f$ is a local isometry.
Now, just use that a surjective local isometry with complete domain is always a Riemannian covering map.
Observation: This fact about Riemannian geometry used in the end can be found on Manfredo's book, for example, on the section about Hadamard theorem. The precise statement is the following.
Let $M$ be a complete Riemannian manifold and let $f:M to N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: For all $pin M$ and all $vin T_p M$, we have $|df_p(f)|geq |v|$. Then $f$ is a covering map.
answered Aug 3 at 13:53
Hugocito
1,6251019
add a comment |Â
up vote
2
down vote
accepted
I'll follow the proof as it is in Milnor's book. So we have Poincaré metric on the surfaces and the covering maps are Riemannian covering.
If $F$ is a conformal isomorphism, then $f$ is a local isometry.
Now, just use that a surjective local isometry with complete domain is always a Riemannian covering map.
Observation: This fact about Riemannian geometry used in the end can be found on Manfredo's book, for example, on the section about Hadamard theorem. The precise statement is the following.
Let $M$ be a complete Riemannian manifold and let $f:M to N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: For all $pin M$ and all $vin T_p M$, we have $|df_p(f)|geq |v|$. Then $f$ is a covering map.
answered Aug 3 at 13:53
Hugocito
1,6251019
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
I'll follow the proof as it is in Milnor's book. So we have Poincaré metric on the surfaces and the covering maps are Riemannian covering.
If $F$ is a conformal isomorphism, then $f$ is a local isometry.
Now, just use that a surjective local isometry with complete domain is always a Riemannian covering map.
Observation: This fact about Riemannian geometry used in the end can be found on Manfredo's book, for example, on the section about Hadamard theorem. The precise statement is the following.
Let $M$ be a complete Riemannian manifold and let $f:M to N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: For all $pin M$ and all $vin T_p M$, we have $|df_p(f)|geq |v|$. Then $f$ is a covering map.
answered Aug 3 at 13:53
Hugocito
1,6251019
I'll follow the proof as it is in Milnor's book. So we have Poincaré metric on the surfaces and the covering maps are Riemannian covering.
If $F$ is a conformal isomorphism, then $f$ is a local isometry.
Now, just use that a surjective local isometry with complete domain is always a Riemannian covering map.
Observation: This fact about Riemannian geometry used in the end can be found on Manfredo's book, for example, on the section about Hadamard theorem. The precise statement is the following.
Let $M$ be a complete Riemannian manifold and let $f:M to N$ be a local diffeomorphism onto a Riemannian manifold $N$ which has the following property: For all $pin M$ and all $vin T_p M$, we have $|df_p(f)|geq |v|$. Then $f$ is a covering map.
answered Aug 3 at 13:53
Hugocito
1,6251019
edited Aug 3 at 15:00
answered Aug 3 at 13:53
Hugocito
1,6251019
answered Aug 3 at 13:53
Hugocito
1,6251019
answered Aug 3 at 13:53
Hugocito
1,6251019
1,6251019
add a comment |Â
add a comment |Â
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