Plane Hyperbolic Geometry

Assume

$\mathbb{D}=\{ z: |z|<1\}$ is the unit disc on the complex plane $\mathbb{C},$

$\mathbb{H}=\{z: \Im{z}>0\}$ is the upper half plane on the complex plane,

$\mathbb{B}=\{ z: |\Im{z}|<\pi/2\}$ is the band between $y=-\pi/2$ and $y=\pi/2$ .

Definition 1. Hyperbolic metric on the unit disc.

The hyperbolic metric on the unit disc $\mathbb{D}$ is defined as

$\rho_{\mathbb{D}}(z)=\frac{2}{1-|z|^{2}} |dz|$ for all $z \in \mathbb{D} .$

If $\phi : U \rightarrow \mathbb{D}$ is a conformal mapping, where $U \subseteq \mathbb{C}$ , then we can also define the hyperbolic metric on the domain U,

$\rho_{U}(z)=\frac{2 |\phi^{'}(z)|}{1-|\phi(z)|^{2}} |dz|$ for all $z\in U.$

From above and $\phi(z)=(z-i)/(z+i)$ is a conformal mapping which maps the upper half plane $\mathbb{H}$ onto the unit disc $\mathbb{D}.$ From above formula, we can calculate the hyperbolic metric on $\mathbb{H}$ is

$\rho_{\mathbb{H}}(z)=\frac{1}{\Im{z}} |dz|$ for all $z\in \mathbb{H}.$

The hyperbolic metric on the band $\mathbb{B}$ is

$\rho_{\mathbb{B}}(z)=\frac{1}{\cos \Im{z} } |dz|$ for all $z\in \mathbb{B}.$

Similarly, we can define the one dimensional hyperbolic metric. On the real line $\mathbb{R}$ , if the interval $I=(-1,1)$ , then the restriction of the hyperbolic metric on the unit disc $\mathbb{D}$ is

$\rho_{I}(x)= \frac{2}{1-x^{2}} dx$ for all $x \in (-1,1).$

This is called the hyperbolic metric of the interval I.

Using the same idea, we can extend the definition of hyperbolic metric on any real interval $I=(a,b)$ . Since there exists a linear map $\phi$ which maps a to -1 and b to 1, i.e. $\phi(x)=(2x-b-a)/(b-a)$ . Its derivative is $\phi^{'}(x)= 2/(b-a)$ . Therefore, the hyperbolic metric on the interval I is

$\rho_{(a,b)}(x)=\frac{2|\phi^{'}(x)|}{1-|\phi(x)|^{2}} dx= \frac{b-a}{(x-a)(b-x)} dx= (\frac{1}{x-a}+ \frac{1}{b-x}) dx$ for all $x\in (a,b).$

Moreover, assume $(c,d) \subseteq (a,b)$ , then the hyperbolic distance between c and d is

$\int_{c}^{d} \rho_{(a,b)}(x) dx = \int_{c}^{d} (\frac{1}{x-a} + \frac{1}{b-x}) dx = (\ln\frac{x-a}{b-x}) |_{x=c}^{x=d} = \ln \frac{(d-a)(b-c)}{(b-d)(c-a)}.$

If we use the notation of cross ratio, then assume $l=(a,c), j=(c,d), r=(d,b),$ $t=(a,b)$ . Therefore, the hyperbolic distance between c and d in the interval (a,b) equals to

$\ln \frac{(|l|+|j|)\cdot (|j|+|r|)}{|l| \cdot |r|} = \ln (1+ \frac{|t|\cdot |j|}{|l| \cdot |r|}) = \ln (1+ Cr(t,j)),$

where $Cr(t,j)= (|t|\cdot |j|) / (|l| \cdot |r|).$

Definition 2. (Curvature of conformal metric)

Let $\rho$ be a $C^{2}$ positive function on an open subset $U \subseteq \mathbb{C}$ . Then the curvature of the metric $\rho(z)|dz|$ is given by

$K(z)=-\frac{(\Delta \ln \rho)(z)}{\rho^{2}(z)},$

where $\Delta$ is the Laplacian operator $\Delta= \frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}}.$

Remark. Use the identities

$\frac{\partial}{\partial \overline{z}} =\frac{1}{2} (\frac{\partial}{\partial x} + i \frac{\partial}{\partial y}),$

$\frac{\partial}{\partial z} =\frac{1}{2} (\frac{\partial}{\partial x} - i \frac{\partial}{\partial y}),$

we get

$\Delta=\frac{\partial^{2}}{\partial x^{2}} + \frac{\partial^{2}}{\partial y^{2}} = 4 \frac{\partial^{2}}{\partial z \partial \overline{z}}.$

Theorem 1.

The curvature of hyperbolic metric of the unit disc $\mathbb{D}$ , the upper half plane $\mathbb{H}$ and the band $\mathbb{B}$ is -1.

Theorem 2.

If $\phi: U\rightarrow \mathbb{D}$ is a conformal mapping, where $U \subseteq \mathbb{C}$ is an open subset of the complex plane $\mathbb{C}$ . From above, the hyperbolic metric on U is