hyperbolic-geometry

hyperbolic-geometry

- Mobius transform
- Half plane
- Cayley transform, Hyperbolic stereographic projection

hyperbolic-geometry-00-mobius-transform

$$$
% From AsciiMath
\newcommand{\bb}[0]{\mathbf}
\newcommand{\bbb}[0]{\mathbb}
\newcommand{\cc}[0]{\mathcal}
\newcommand{\RR}[0]{\bbb{R}}
\newcommand{\CC}[0]{\bbb{C}}
\newcommand{\xx}[0]{\times}
\newcommand{\del}[0]{\partial}
% Quick begin/end
\newcommand{\Ba}[0]{\begin{aligned}}
\newcommand{\Ea}[0]{\end{aligned}}
\newcommand{\Bc}[0]{\begin{cases}}
\newcommand{\Ec}[0]{\end{cases}}
\newcommand{\Bm}[0]{\begin{matrix}}
\newcommand{\Em}[0]{\end{matrix}}
\newcommand{\Bmp}[0]{\begin{pmatrix}} % round bracket
\newcommand{\Emp}[0]{\end{pmatrix}}
\newcommand{\Bms}[0]{\begin{bmatrix}} % square bracket
\newcommand{\Ems}[0]{\end{bmatrix}}
\newcommand{\Bmc}[0]{\begin{Bmatrix}} % curley bracket
\newcommand{\Emc}[0]{\end{Bmatrix}}
% Others
\newcommand{\t}[0]{\text}
\newcommand{\Abs}[1]{\left|#1\right|}
\newcommand{\Paren}[1]{\left(#1\right)}
\newcommand{\defiff}[0]{\overset{\text{def}}{\iff}}
\newcommand\ddfrac[2]{{\displaystyle\frac{\displaystyle #1}{\displaystyle #2}}}
$$$


## Mobius transform

### Def. 1

$m : \CC \to \CC$ is **Mobius transform** if $m(z) = \dfrac{a z + b}{c z + d}$
for some $a, b, c, d \in \CC$ s.t. $ad - bc \ne 0$.

**N.B.**

- We denote such $m$ by $m \equiv \Bms a & b \\ c & d \Ems \in \cc{M}$.
- Since $m(z) = \dfrac{a z + b}{c z + d} = \dfrac{\mu a z + \mu b}{\mu c z + \mu d}$,
  we have $\Bms a & b \\ c & d \Ems = \Bms \mu a & \mu b \\ \mu c & \mu d \Ems$.
  
  So, by taking $\mu = (ad - bc)^{-\frac{1}{2}}$, we have
  $(\mu a) (\mu d) - (\mu b) (\mu c) = \mu^2 (ad - bc) = 1$.
  
  Therefore, from here we often implicitly assume $ad - bc = 1$.

- Define $\cc{M}_{\RR}$ by: $m \in \cc{M}_{\RR} \iff a, b, c, d \in \RR$.

- We often implicitly use the isomorphism: 
  $\CC \leftrightarrow \RR^2 : x + i y \leftrightarrow (x, y)$.

### Prop. 2

$$
\Bms a & b \\ c & d \Ems
  \circ
    \Bms a' & b' \\ c' & d' \Ems
      = 
        \Bms a a' + b c' & a b' + b d' \\ c a' + d c' & c b' + d d' \Ems.
$$

***Proof***

We see:

$$
\frac{
  a \dfrac{a' z + b'}{c' z + d'} + b
}{
  c \dfrac{a' z + b'}{c' z + d'} + d
}
  = 
    \frac{
      a (a' z + b') + b (c' z + d')
    }{
      c (a' z + b') + d (c' z + d')
    }
  = 
    \frac{
      (a a' + b c') z + a b' + b d'
    }{
      (c a' + d c') z + c b' + d d'
    }.
$$


### Prop. 3

$$
\Ba
m(z)
  &= \dfrac{a z + b}{c z + d}
  = \dfrac{(a z + b) (c z + d)^\dag}{|c z + d|^2}
  = \dfrac{1}{|c z + d|^2}
      ( a c^\dag |z|^2 + b d^\dag + a d^\dag z + b c^\dag z^\dag ) \\
  &= \dfrac{1}{|c z + d|^2}
      \Paren{
        (a c^\dag |z|^2 + b d^\dag + (a d^\dag + b c^\dag) x)
        + i (a d^\dag - b c^\dag) y }.
\Ea
$$

**N.B.**

For $m = \Bms a & b \\ c & d \Ems \in \cc{M}_\RR$ with $ad - bc = 1$, we have

$$
\Ba
m(z)
  &= \dfrac{1}{|c z + d|^2}
      \Paren{
        (a c |z|^2 + b d  + (a d + b c) x) + i y}.
\Ea
$$

### Def. 4

We define stereographic projection
$
\t{SP} : S^2 \to \RR^2 \simeq \CC
  : (x_0, x, y) \mapsto \dfrac{(x, y)}{1 - x_0} \mapsto \dfrac{x + i y}{1 - x_0}
$.

### Prop. 5 (Mobius transform and $S^2$ rotation)

$$
\Bm
  & S^2        & \xrightarrow{\t{SP}}   & \CC & \\
\Bmp \t{Rot}_{2 \phi} & 0 \\ 0 & 1 \Emp
  & \downarrow &                        & \downarrow & 
    [ \t{Rot}_\phi ] \\
  & S^2        & \xrightarrow[\t{SP}]{} & \CC &
\Em
$$

i.e.

 $$
\t{SP} \circ \Bmp \t{Rot}_{2 \phi} & 0 \\ 0 & 1 \Emp
  = [ \t{Rot}_\phi ] \circ \t{SP}
~~:~ S^2 \to \CC
$$

where
$\t{Rot}_{\phi}
  \equiv \Bmp \cos(\phi) & - \sin(\phi) \\ \sin(\phi) & \cos(\phi) \Emp$.

**N.B.**

- On the LHS, $\Bmp \t{Rot}_{\phi} & 0 \\ 0 & 1 \Emp \in \RR^3$ represents usual linear transform
  by matrix.
- On the RHS, $[ \t{Rot}_\phi ]$ represents mobius transform with the entry
  given by $\t{Rot}$ matrix.
  
***Proof***

We will show
$
\t{SP} \circ \Bmp \t{Rot}_{2 \phi} & 0 \\ 0 & 1 \Emp \circ \t{SP}^{-1}
  = [ \t{Rot}_\phi ]
~:~ \CC \to \CC
$.

We note that:

$$
\Ba
[ \t{Rot}_\phi ] (z)
  &=
    \dfrac{
      \cos(t) \sin(t) (|z|^2 - 1 )
        + (\cos^2(t) - \sin^2(t)) x + i y
    }{|\sin(t) z + \cos(t)|^2} \\
  &=
    \dfrac{
      \frac{\sin(2t)}{2} (|z|^2 - 1 )
        + \cos(2t) x + i y
    }{
      \sin^2(t) |z|^2 + \cos^2(t) + 2 \sin(t) \cos(t) x
    } \\
  &=
    \dfrac{
      \frac{1}{2} \sin(2t) (|z|^2 - 1 )
        + \cos(2t) x + i y
    }{
      \frac{1}{2}
        \Paren{
          |z|^2 + 1 - \cos(2t) (|z|^2 - 1)
        }
      + \sin(2t) x
    } \\
  &=
    \dfrac{
      \sin(2t) (|z|^2 - 1 )
        + \cos(2t) 2 x + i 2 y
    }{
      |z|^2 + 1 - \cos(2t) (|z|^2 - 1)
      + \sin(2t) 2 x
    }.
\Ea
$$

Also, we see:

$$
\Ba
\t{SP}^{-1}(z)
  &= \dfrac{1}{|z|^2 + 1} \Bmp |z|^2 - 1 \\ 2 z \Emp,
  = \dfrac{1}{|z|^2 + 1} \Bmp |z|^2 - 1 \\ 2  x \\ 2 y \Emp,  
\\
\Bmp \t{Rot}_{2 \phi} & 0 \\ 0 & 1 \Emp \circ \t{SP}^{-1} (z)
  &=
     \dfrac{1}{|z|^2 + 1}
       \Bmp
         \cos(2t) (|z|^2 - 1) - \sin(2t) 2 x \\
         \sin(2t) (|z|^2 - 1) + \cos(2t) 2 x \\
         2 y
       \Emp,
\Ea
$$

$$
\Ba
\t{SP} &\circ \Bmp \t{Rot}_{2 \phi} & 0 \\ 0 & 1 \Emp \circ \t{SP}^{-1} (z) \\
  &=
     \dfrac{1}{
       1 - \frac{1}{|z|^2 + 1} (\cos(2t) (|z|^2 - 1) - \sin(2t) 2 x)
     }  
     \dfrac{1}{|z|^2 + 1}
       \Bmp
         \sin(2t) (|z|^2 - 1) + \cos(2t) 2 x \\
         2 y
       \Emp \\
  &=
     \dfrac{1}{
       |z|^2 + 1 - (\cos(2t) (|z|^2 - 1) - \sin(2t) 2 x)
     }  
     \Bmp
       \sin(2t) (|z|^2 - 1) + \cos(2t) 2 x \\
       2 y
     \Emp.
\Ea
$$

hyperbolic-geometry-01-half-plane

$$$
% From AsciiMath
\newcommand{\bb}[0]{\mathbf}
\newcommand{\bbb}[0]{\mathbb}
\newcommand{\cc}[0]{\mathcal}
\newcommand{\RR}[0]{\bbb{R}}
\newcommand{\CC}[0]{\bbb{C}}
\newcommand{\xx}[0]{\times}
\newcommand{\del}[0]{\partial}
% Quick begin/end
\newcommand{\Ba}[0]{\begin{aligned}}
\newcommand{\Ea}[0]{\end{aligned}}
\newcommand{\Bc}[0]{\begin{cases}}
\newcommand{\Ec}[0]{\end{cases}}
\newcommand{\Bm}[0]{\begin{matrix}}
\newcommand{\Em}[0]{\end{matrix}}
\newcommand{\Bmp}[0]{\begin{pmatrix}} % round bracket
\newcommand{\Emp}[0]{\end{pmatrix}}
\newcommand{\Bms}[0]{\begin{bmatrix}} % square bracket
\newcommand{\Ems}[0]{\end{bmatrix}}
\newcommand{\Bmc}[0]{\begin{Bmatrix}} % curley bracket
\newcommand{\Emc}[0]{\end{Bmatrix}}
% Others
\newcommand{\t}[0]{\text}
\newcommand{\Abs}[1]{\left|#1\right|}
\newcommand{\Paren}[1]{\left(#1\right)}
\newcommand{\defiff}[0]{\overset{\text{def}}{\iff}}
\newcommand\ddfrac[2]{{\displaystyle\frac{\displaystyle #1}{\displaystyle #2}}}
$$$


## Poincare's half-plane model

**N.B.**

- In this section, we assume $m \in \cc{M}_\RR$.

### Def. 1

We define $\bbb{H} \equiv \RR \xx i \RR_{\gt 0} \subset \CC$
and metric tensor $g_{x, y} = \dfrac{1}{y^2} I$.


### Prop. 2

$\t{Range}_m(\bbb{H}) = \bbb{H}$.

***Proof***

Given $y \gt 0$, from previously obtained formula, we see: 

$$
\t{Im}(m(z)) = \dfrac{y}{|c z + d|^2} \gt 0.
$$


### Prop. 3

Writing Jacobian $J \equiv \del_{x,y} m$,
then $J^T g_{m(x, y)} J = g_{x, y}$ (i.e. $m$ is isometry).

***Proof***

First, note that the complex derivative of $m$ is:

$$
\del_z m
  = \frac{1}{(c z + d)^2} ( -c (a z + b) + (c z + d) a)
  = \frac{ad - bc}{(c z + d)^2} = \frac{1}{(c z + d)^2}.
$$

Second, we see the complex derivative relates to the Jacobian as:

$$
J = \del_{x, y} m
  =
    \Bmp
      \t{Re}(\del_z m) & - \t{Im}(\del_z m) \\
      \t{Im}(\del_z m) & \t{Re}(\del_z m)
    \Emp 
$$

thus, 

$$
J^T g_{m(z)} J
  = \frac{1}{(\t{Im}(m(z)))^2} J^T J
  = \frac{1}{(\t{Im}(m(z)))^2} |\del_z m|^2 I
  = \frac{1}{\dfrac{y^2}{|cz + d|^2}} \dfrac{1}{|cz + d|^2} I = g_z.
$$


### Prop. 4

$$
m(i) = x + i y
  \iff
    m =
      \Bms
        y^{1/2} \Paren{\cos(t) + \frac{x}{y} \sin(t)} &
          y^{1/2} \Paren{- \sin(t) + \frac{x}{y} \cos(t)} \\
        y^{-1/2} \sin(t) & y^{-1/2} \cos(t)
      \Ems.
$$

**N.B.**

- This means $\forall z \in \bbb{H}. \exists m \in \cc{M}_\RR. m(i) = z$
  (i.e. $\cc{M}_\RR$ acts transitively on $\bbb{H}$).
- Especially for $m(i) = i$ (aka stabilizer of $i$), we have $m = \Bms \t{Rot}_t \Ems$,
  which can be seen from **Mobius transform: Prop. 5**.


***Proof***

From previous formula, we know:

$$
m(i)
  = \dfrac{1}{|c z + d|^2}
      \Paren{
        (a c |z|^2 + b d  + (a d + b c) x) + i y}
  = \dfrac{1}{c^2 + d^2} (ac + bd + i).
$$

Therefore, noting $ad - bc = 1$, we have:

$$
\Ba
m(i) = x + i y
  &\iff
    \Bc
      y &= \dfrac{1}{c^2 + d^2} \\
      x &= \dfrac{ac + bd}{c^2 + d^2} \\
    \Ec
  \iff
    \Bc
      \dfrac{1}{y} &= c^2 + d^2 \\
      \dfrac{x}{y} &= ac + bd. \\
    \Ec \\\\
  &\iff
    \Bc
      \dfrac{1}{y} &= c^2 + d^2 \\ 
      \Bmp 1 \\ \frac{x}{y} \Emp
        &=
          \Bmp d & -c \\ c & d \Emp
          \Bmp a \\ b \Emp.
    \Ec    
\Ea
$$

So, by writing $(c, d) = (y^{-1/2} \sin(t), y^{-1/2} \cos(t))$, we see:

$$
\Ba
\Bmp 1 \\ \frac{x}{y} \Emp
  &=
    \Bmp d & -c \\ c & d \Emp
    \Bmp a \\ b \Emp \\
  &\iff
    \Bmp 1 \\ \frac{x}{y} \Emp
      =
        y^{-1/2} \; \t{Rot}_t
          \Bmp a \\ b \Emp \\
  &\iff
    \Bmp a \\ b \Emp
      =
        y^{1/2}
          \t{Rot}_{-t}
            \Bmp 1 \\ \frac{x}{y} \Emp
      =
        y^{1/2}
          \Bmp
            \cos(t) + \frac{x}{y} \sin(t) \\
            - \sin(t) + \frac{x}{y} \cos(t)
          \Emp.
\Ea
$$


### Prop. 6 

Hyperbolic line (geodesic curve) from geodesic equation

???


### Prop. 7

Constant velocity geodesic trajectory from $i$ xx

???

hyperbolic-geometry-02-misc

$$$
% From AsciiMath
\newcommand{\bb}[0]{\mathbf}
\newcommand{\bbb}[0]{\mathbb}
\newcommand{\cc}[0]{\mathcal}
\newcommand{\RR}[0]{\bbb{R}}
\newcommand{\CC}[0]{\bbb{C}}
\newcommand{\xx}[0]{\times}
\newcommand{\del}[0]{\partial}
% Quick begin/end
\newcommand{\Ba}[0]{\begin{aligned}}
\newcommand{\Ea}[0]{\end{aligned}}
\newcommand{\Bc}[0]{\begin{cases}}
\newcommand{\Ec}[0]{\end{cases}}
\newcommand{\Bm}[0]{\begin{matrix}}
\newcommand{\Em}[0]{\end{matrix}}
\newcommand{\Bmp}[0]{\begin{pmatrix}} % round bracket
\newcommand{\Emp}[0]{\end{pmatrix}}
\newcommand{\Bms}[0]{\begin{bmatrix}} % square bracket
\newcommand{\Ems}[0]{\end{bmatrix}}
\newcommand{\Bmc}[0]{\begin{Bmatrix}} % curley bracket
\newcommand{\Emc}[0]{\end{Bmatrix}}
% Others
\newcommand{\t}[0]{\text}
\newcommand{\Abs}[1]{\left|#1\right|}
\newcommand{\Paren}[1]{\left(#1\right)}
\newcommand{\defiff}[0]{\overset{\text{def}}{\iff}}
\newcommand\ddfrac[2]{{\displaystyle\frac{\displaystyle #1}{\displaystyle #2}}}
$$$


## Cayley transform

### Def. 1

Define $\t{CT} = \Bms 1 & -i \\ 1 & i \Ems \in \cc{M}$, i.e.
$\t{CT}(z) \equiv \dfrac{z - i}{z + i}$.

***N.B.***

- $|\t{CT}(z)|^2 = \dfrac{|z - i|^2}{|z + i|^2} = \dfrac{x^2 + (y - 1)^2}{x^2 + (y + 1)^2}$,
  thus $y \gt 0 \iff |CT(z)| \lt 1$ and so $\bbb{H} \stackrel{\t{CT}}{\cong} \bbb{D}$.
  
- Via matrix inverse (up to scale), we have $\t{CT}^{-1} = \Bms 1 & 1 \\ i & -i \Ems$.
  

### Prop. 2 (Cayley transform is isometry)

Writing the Jacobian $J = \del_{(x, y)} \t{CT}$,
then we have $J^T g_{\bbb{D}} J = g_{\bbb{H}}$ i.e.

$$ \dfrac{4}{(|\t{CT}(z)|^2 - 1)^2} J^T J = \dfrac{1}{y^2} I. $$

***Proof***

(Cf. **Poincare's half-plane model : Prop. 3**)

Noting $\del_z m = \dfrac{ad - bc}{(cz + d)^2}$, we have $\del_z \t{CT} = \dfrac{2i}{(z + i)^2}$. Thus,

$$
\Ba
J^T J &= |\del_z \t{CT}|^2 I = \dfrac{4}{|z + i|^2} I, \\
\dfrac{4}{(|\t{CT}(z)|^2 - 1)^2} J^T J
  &= \dfrac{4}{(|\t{CT}(z)|^2 - 1)^2} \dfrac{4}{|z + i|^2} I
  = \dfrac{16}{(|z - i|^2 - |z + i|^2)^2} I \\
  &= \dfrac{16}{(4 y)^2} I = \dfrac{1}{y^2} I.
\Ea
$$


## Hyperbolic stereographic projection

(cf. [Stereographic projection](https://hi-ogawa.github.io/markdown-tex/?id=250f2363b93b83aec636bba353f183d4&filename=markdown-tex-demo.txt))

***Notations***

- $\bbb{B} \in \RR^3 \equiv \{ (x_0, \bb{x}) \mid x_0 \le -1 \;\wedge\; |x_0|^2 - |\bb{x}|^2 = 1 \}$
  : hyperboloid bottom sheet,
- $\bbb{D} \in \RR^2 \equiv \{ \bb{y} \mid |y|^2 \lt 1 \} $ : open unit disk,


### Def. 1

Define $F : \bbb{B} \to \bbb{D}$ by $F(x_0, \bb{x}) \equiv \dfrac{\bb{x}}{1 - x_0}$.

**N.B.** This is well-defined since:

$$
\Abs{ \dfrac{\bb{x}}{1 - x_0} }^2
  = \dfrac{\Abs{\bb{x}}^2}{(1 - x_0)^2}
  = \dfrac{x_0^2 - 1}{(x_0 - 1)^2}
  = \dfrac{x_0 + 1}{x_0 - 1}
  = 1 - \dfrac{2}{1 - x_0} \in (0, 1).
$$



### Prop. 2

Given $\bb{y} = F(x_0, \bb{x}) = \dfrac{\bb{x}}{1 - x_0} \in \bbb{D}$, we have:
  
$$
\Ba
F^{-1}(\bb{y})
  &= \Paren{ \dfrac{|\bb{y}|^2 + 1}{|\bb{y}|^2 - 1}, \dfrac{- 2 \bb{y}}{|\bb{y}|^2 - 1} }.
\Ea
$$

***Proof***

Given $\bb{y} = \dfrac{\bb{x}}{1 - x_0}$, we see:

$$
\Ba
\Abs{\bb{y}}^2
  = \Abs{ \frac{\bb{x}}{1 - x_0} }^2
  = \dfrac{x_0 + 1}{x_0 - 1}
  = 1 - \dfrac{2}{1 - x_0}
  \implies
    x_0
      = \frac{|\bb{y}|^2 + 1}{|\bb{y}|^2 - 1}
      = 1 + \dfrac{2}{|\bb{y}|^2 - 1}
\Ea
$$

Thus

$$
\bb{x} = (1 - x_0) \bb{y} = \frac{- 2}{|\bb{y}|^2 - 1} \bb{y}.
$$ 


### Prop. 3

Writing $G(\bb{y}) = F^{-1}(\bb{y})$ and its Jacobian $J = \del_y G$, we have:

$$
J^T \eta J = \dfrac{4}{(|\bb{y}|^2 - 1)^2} I,
$$

where $~\eta = \Bmp -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \Emp$.

**N.B.**  We obtained a metric on Poincare's disk by pull-backing Minkowski metric on Hyperboloid.

***Proof***

First we note:

$$
\del_y \dfrac{1}{|\bb{y}|^2 - 1}
    =  - 2 \dfrac{\bb{y}^T}{(|\bb{y}|^2 - 1)^2} \in \RR^{1 \xx n}.
$$

By separating components as:

$$
\del_y G
  = \Bmp
      \del_y x_0    &\in \RR^{1 \times n} \\
      \del_y \bb{x} &\in \RR^{n \times n}
    \Emp,
$$

we have:

$$
\Ba
\del_y x_0
    &= \del_y \Paren{ 1 + \frac{2}{|\bb{y}|^2 - 1} }
    = \frac{-1}{(|y|^2 - 1)^2} 4 \bb{y}^T \\
\del_y \bb{x}
    &= \del_y \frac{- 2 \bb{y}}{|\bb{y}|^2 - 1}
    = - 2
        \Paren{
          \frac{1}{|\bb{y}|^2 - 1} I + 
          \bb{y} \; \del_y \frac{1}{|\bb{y}|^2 - 1}
        } \\
    &= - 2
        \Paren{
          \frac{1}{|\bb{y}|^2 - 1} I
          - 2 \frac{\bb{y} \bb{y}^T}{(|\bb{y}|^2 - 1)^2}
        }, \\
    &= \frac{- 1}{(|y|^2 - 1)^2} 2
         \Paren{
           (|y|^2 - 1) I
           - 2 \bb{y} \bb{y}^T
         }, \\
J &=
  \frac{- 1}{(|y|^2 - 1)^2}
    \Bmp
      4 \bb{y}^T \\
      2 
        \Paren{
          (|y|^2 - 1) I
          - 2 \bb{y} \bb{y}^T
        }
    \Emp.
\Ea
$$   

Therefore:

$$
\Ba
J^T \mu J &=
  \Bmp
    (\del_y x_0)^T \;
    (\del_y \bb{x})^T
  \Emp
  \Bmp
    - \del_y x_0 \\
    \del_y \bb{x}
  \Emp
  = 
    - (\del_y x_0)^T \del_y x_0
    + (\del_y \bb{x})^T \del_y \bb{x}  \\
  &=
    \frac{1}{(|y|^2 - 1)^4}
      \Paren{
        - 16 \bb{y} \bb{y}^T
        + 4
          \Paren{
            (|y|^2 - 1)^2 I
            + 4 |y|^2 \bb{y} \bb{y}^T
            - 4 (|y|^2 - 1) \bb{y} \bb{y}^T
          }
      } \\
  &=
    \frac{1}{(|y|^2 - 1)^4} 4 (|y|^2 - 1)^2 I \\
  &=
    \frac{4}{(|y|^2 - 1)^2} I.
\Ea
$$