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$\def\abs#1{\left|#1\right|}$I would like to understand this example:

  • Why is the following set a hyperbolic manifold?

$X=\{[1:z:w]\in \mathbb{CP}_2\mid0<\abs z< 1, \abs w < \abs{\exp(1/z)}\}$

It's an examples given in the book Hyperbolic Manifolds and Holomorphic Mappings: An Introduction by Kobayashi, in order to give a counterexample of an optimistic generalization of the Big Picard Theorem. They claim that it is biholomorphic to $\mathbb{D}\times\mathbb{D}^*$. I dont understand why.

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1 Answer 1

up vote 2 down vote accepted

$\mathbb{CP}^2$ is a natural complex manifold where the chart are given by the maps : $\begin{array}{lclc} \varphi_i: & U_i:=\{[z_0:z_1:z_2]\in \mathbb{CP}^2 \ | \ z_i\neq 0\} & \longrightarrow & \mathbb{C}^2 \\ & {[z_0:z_1:z_2]} & \longmapsto & (\dfrac{z_j}{z_0},\dfrac{z_k}{z_0}) \end{array}$ where $j,k\neq i$.

However, you can write $U_1$ as $\{[1:z:w]\in \mathbb{CP}^2\}$ and by definition of a biholomorphic map between two complex manifolds, it tells you that $X$ is biholomorphic to $\varphi_1(X)$.

The map $(z,w)\in \varphi_1(X)\mapsto (z,we^{-\frac{1}{z}})\in \mathbb D^\star\times \mathbb D$ is clearly a biholomorphism. So $X$ is biholomorphic to $\mathbb D^\star\times \mathbb D$.

Since $\mathbb D$ and $\mathbb D^\star$ are hyperbolic manifolds then so is $\varphi_1(X)$ and consequently $X$ is a hyperbolic manifold.

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My only issue is: why is it clearly a biholomorphism? I cannot see it. –  Andrea Sep 23 '12 at 15:19
    
Let $\phi$ be the map from $\varphi_1(X)\subset \mathbb C^2$ with values in $\mathbb C^2$ given by $(z,w)\mapsto (\phi_1(z,w),\phi_2(z,w))=(z,we^{-\frac{1}z})$. Since each of the $\phi_k$ is a holomorphic function then so is $\phi$. Moreover $\phi$ is injective and its image is, by construction, $\mathbb D^\star\times \mathbb D$. On the other side, the map $\psi(z,w)=(z,we^{\frac{1}{z}})$ from $\mathbb D^\star\times \mathbb D$ onto $\varphi_1(X)$ is also holomorphic and you can check that $\phi\circ \psi=id$ and $\psi \circ \phi=id$. –  Bebop Sep 23 '12 at 17:06
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