Markdown TeX Demo

markdown-tex-demo.txt

$$$
% Quick begin/end
\newcommand{\Ba}[0]{\begin{aligned}}
\newcommand{\Ea}[0]{\end{aligned}}
\newcommand{\Bc}[0]{\begin{cases}}
\newcommand{\Ec}[0]{\end{cases}}
\newcommand{\Bm}[0]{\begin{matrix}}
\newcommand{\Em}[0]{\end{matrix}}
\newcommand{\Bmp}[0]{\begin{pmatrix}}
\newcommand{\Emp}[0]{\end{pmatrix}}
% Others
\newcommand{\bb}[0]{\mathbf}
\newcommand{\del}[0]{\partial}
\newcommand{\xx}[0]{\times}
\renewcommand{\ll}[0]{\left}
\newcommand{\rr}[0]{\right}
\newcommand{\RR}[0]{\mathbb{R}}
\newcommand\ddfrac[2]{{\displaystyle\frac{\displaystyle #1}{\displaystyle #2}}}
$$$

## Stereographic projection

### Def. 1

We define $F$ as below:

$$
\begin{aligned}
F : \RR^{n + 1} \supset
        S^n \setminus \{ (1, \bb{0}) \}  &\to     \RR^n  \\
    (x_0, \bb{x}) \quad\quad             &\mapsto \bb{y} = \frac{\bb{x}}{1 - x_0}.
\end{aligned}
$$


### Prop. 2

The inverse of \(F\) is:

\[
\begin{aligned}
F^{-1}(\bb{y}) =
    \left(
        \frac{|\bb{y}|^2 - 1}{|\bb{y}|^2 + 1},
        \frac{2 \bb{y}}{|\bb{y}|^2 + 1}
    \right)
    \equiv G(\bb{y}).
\end{aligned}
\]

***Proof***

Given $x_0^2 + |\bb{x}|^2 = 1$, we see:

$$
\begin{aligned}
\bb{y} = F(x_0, \bb{x}) = \frac{\bb{x}}{1 - x_0}
    &\implies
        |\bb{y}|^2
            = \frac{|\bb{x}|^2}{(1 - x_0)^2}
            = \frac{1 - x_0^2}{(1 - x_0)^2}
            = \frac{1 + x_0}{1 - x_0} \\
    &\implies
        (1 - x_0) |\bb{y}|^2 = 1 + x_0 \\
    &\implies
        x_0 = \frac{|\bb{y}|^2 - 1}{|\bb{y}|^2 + 1}
            = 1 - \frac{2}{|\bb{y}|^2 + 1}.
\end{aligned}
$$

And, we also have:

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

So, we have proved $F^{-1} \circ F = \text{id}$.


For the converse $F \circ F^{-1} = \text{id}$, we have:

$$
\begin{aligned}
\frac{\bb{x}}{1 - x_0}
    =
        \ddfrac
            { \frac{2 \bb{y}}{|\bb{y}|^2 + 1} }
            { 1 - \frac{|\bb{y}|^2 - 1}{|\bb{y}|^2 + 1} }
    = 
        \dfrac
            {  2 \bb{y} }
            { |\bb{y}|^2 + 1 - (|\bb{y}|^2 - 1)  }
    = \bb{y}.
\end{aligned}
$$

### Prop. 3

For the Jacobian $J \equiv \bb{D}G|_{y} \in \RR^{(n + 1) \xx n}$,
we have $J^T J = \ll( \dfrac{2}{|\bb{y}|^2 + 1} \rr)^2 I$.

***Proof***

First we note:

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

Thus, separating components as:

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

we can see:

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

Therefore:

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