A geometric proof of Picard's little theorem
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I'm preparing a presentation where I'd like to present a proof of Picard's little theorem using hyperbolic geometry. Picard's little theorem states that the range of an entire function can omit at most $1$ point.
I can see that we can write $U:=mathbb{C}P^1setminus{infty,0,1}$ as a hyperbolic manifold, and as a quotient of $mathbb{D}$ through the Farey tesselation. I would like to say therefore that any conformal map $:mathbb{C}rightarrow U$ would factor conformally through $mathbb{D}$.
I can see that it should factor through the covering map, but can I even say that the covering map is conformal?
I know that Ahlfors' Lemma helps, but all the statements I've obtained of said lemma hold for Hermite metrics. Can I say that the hyperbolic metric on $mathbb{C}P^1setminus{infty,0,1}$ is Hermite?
complex-analysis reference-request complex-geometry hyperbolic-geometry
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
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add a comment |
$begingroup$
I'm preparing a presentation where I'd like to present a proof of Picard's little theorem using hyperbolic geometry. Picard's little theorem states that the range of an entire function can omit at most $1$ point.
I can see that we can write $U:=mathbb{C}P^1setminus{infty,0,1}$ as a hyperbolic manifold, and as a quotient of $mathbb{D}$ through the Farey tesselation. I would like to say therefore that any conformal map $:mathbb{C}rightarrow U$ would factor conformally through $mathbb{D}$.
I can see that it should factor through the covering map, but can I even say that the covering map is conformal?
I know that Ahlfors' Lemma helps, but all the statements I've obtained of said lemma hold for Hermite metrics. Can I say that the hyperbolic metric on $mathbb{C}P^1setminus{infty,0,1}$ is Hermite?
complex-analysis reference-request complex-geometry hyperbolic-geometry
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
$endgroup$
$begingroup$
Of course to both. Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds; any metric $f(z)|dz|^2$ on a $1$-dimensional complex manifold (for $f>0$) is a hermitian metric. And covering maps are local biholomorphisms, hence conformal, yes.
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– Ted Shifrin
Dec 7 '18 at 18:33
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Yes. Thanks a lot. +1 for pointing out that 'Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds'.
$endgroup$
– Teebro Prokash
Dec 12 '18 at 11:25
add a comment |
$begingroup$
I'm preparing a presentation where I'd like to present a proof of Picard's little theorem using hyperbolic geometry. Picard's little theorem states that the range of an entire function can omit at most $1$ point.
I can see that we can write $U:=mathbb{C}P^1setminus{infty,0,1}$ as a hyperbolic manifold, and as a quotient of $mathbb{D}$ through the Farey tesselation. I would like to say therefore that any conformal map $:mathbb{C}rightarrow U$ would factor conformally through $mathbb{D}$.
I can see that it should factor through the covering map, but can I even say that the covering map is conformal?
I know that Ahlfors' Lemma helps, but all the statements I've obtained of said lemma hold for Hermite metrics. Can I say that the hyperbolic metric on $mathbb{C}P^1setminus{infty,0,1}$ is Hermite?
complex-analysis reference-request complex-geometry hyperbolic-geometry
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
$endgroup$
I'm preparing a presentation where I'd like to present a proof of Picard's little theorem using hyperbolic geometry. Picard's little theorem states that the range of an entire function can omit at most $1$ point.
I can see that we can write $U:=mathbb{C}P^1setminus{infty,0,1}$ as a hyperbolic manifold, and as a quotient of $mathbb{D}$ through the Farey tesselation. I would like to say therefore that any conformal map $:mathbb{C}rightarrow U$ would factor conformally through $mathbb{D}$.
I can see that it should factor through the covering map, but can I even say that the covering map is conformal?
I know that Ahlfors' Lemma helps, but all the statements I've obtained of said lemma hold for Hermite metrics. Can I say that the hyperbolic metric on $mathbb{C}P^1setminus{infty,0,1}$ is Hermite?
complex-analysis reference-request complex-geometry hyperbolic-geometry
complex-analysis reference-request complex-geometry hyperbolic-geometry
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
asked Dec 4 '18 at 11:15
Teebro ProkashTeebro Prokash
738
738
$begingroup$
Of course to both. Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds; any metric $f(z)|dz|^2$ on a $1$-dimensional complex manifold (for $f>0$) is a hermitian metric. And covering maps are local biholomorphisms, hence conformal, yes.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 18:33
$begingroup$
Yes. Thanks a lot. +1 for pointing out that 'Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds'.
$endgroup$
– Teebro Prokash
Dec 12 '18 at 11:25
add a comment |
$begingroup$
Of course to both. Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds; any metric $f(z)|dz|^2$ on a $1$-dimensional complex manifold (for $f>0$) is a hermitian metric. And covering maps are local biholomorphisms, hence conformal, yes.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 18:33
$begingroup$
Yes. Thanks a lot. +1 for pointing out that 'Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds'.
$endgroup$
– Teebro Prokash
Dec 12 '18 at 11:25
$begingroup$
Of course to both. Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds; any metric $f(z)|dz|^2$ on a $1$-dimensional complex manifold (for $f>0$) is a hermitian metric. And covering maps are local biholomorphisms, hence conformal, yes.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 18:33
$begingroup$
Of course to both. Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds; any metric $f(z)|dz|^2$ on a $1$-dimensional complex manifold (for $f>0$) is a hermitian metric. And covering maps are local biholomorphisms, hence conformal, yes.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 18:33
$begingroup$
Yes. Thanks a lot. +1 for pointing out that 'Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds'.
$endgroup$
– Teebro Prokash
Dec 12 '18 at 11:25
$begingroup$
Yes. Thanks a lot. +1 for pointing out that 'Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds'.
$endgroup$
– Teebro Prokash
Dec 12 '18 at 11:25
add a comment |
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Of course to both. Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds; any metric $f(z)|dz|^2$ on a $1$-dimensional complex manifold (for $f>0$) is a hermitian metric. And covering maps are local biholomorphisms, hence conformal, yes.
$endgroup$
– Ted Shifrin
Dec 7 '18 at 18:33
$begingroup$
Yes. Thanks a lot. +1 for pointing out that 'Hermitian metrics are the analog on complex manifolds of Riemannian metrics on real manifolds'.
$endgroup$
– Teebro Prokash
Dec 12 '18 at 11:25