1 point compactification and Cross-ratio

 

Def 1

$\overline{\mathbb{C}} = \mathbb{C} \cup \left\{\infty \right\}$

is 1 point compactification of $\mathbb{C}$ .

$\infty$ behave like the infinity we know. Therefore we can extend $f(z) = \dfrac{az + b}{cz + d}$ to

$f : \overline{\mathbb{C}} \to \overline{\mathbb{C}}$ by

$f(-\dfrac{d}{c}) = \infty$,     $f(\infty) = \dfrac{a}{c}$

 

 

 

Cor 1

Let $f : \overline{\mathbb{C}} \to \overline{\mathbb{C}}$ by

$f(z) = \dfrac{az+b}{cz+d}$     ($ad - bc \neq 0$)

Then

$f$ is 1-1 onto

 

 

 

Thm 1

Let $f : \overline{\mathbb{C}} \to \overline{\mathbb{C}}$ by

$f(z) = \dfrac{az+b}{cz+d}$     ($ad - bc \neq 0$)

Let $\mathscr{C}$ be collection of circles and lines in complex plane. Then $\forall A \in \mathscr{C}$,

$f(A) \in \mathscr{C}$

 

  Special Case : $f(z) = \dfrac{1}{z}$

  $\forall A \in \mathscr{C}$,

  Since $A$ is circle or line, there exists $\alpha, \beta, \gamma, \delta \in \mathbb{R}$  s.t.

  $A = \left\{(x,y) \; | \; \alpha (x^{2} + y^{2}) + \beta x + \gamma y + \delta = 0  \right\} $

  ( $\alpha = 0$  $\Rightarrow$  line,  $\alpha \neq 0$  $\Rightarrow$  circle )

 

  $\therefore$  $f(A) = \left\{(u, v) \; | \; \alpha (x^{2} + y^{2}) + \beta x + \gamma y + \delta = 0  \right\} $

  ( $x = \dfrac{u}{u^{2} + v^{2}}$,  $v = - \dfrac{v}{u^{2} + v^{2}}$ )

 

  $\therefore$  $f(A) = \left\{(u, v) \; | \; \alpha + \beta u - \gamma v + \delta (u^{2} + v^{2}) = 0  \right\}$

  ( $\delta = 0$  $\Rightarrow$  line,  $\delta \neq 0$  $\Rightarrow$  circle )

 

  $\therefore$  $f(A) \in \mathscr{C}$

 

  end

 

 

  General Case

  $f(z) = \dfrac{az+b}{cz+d}$ is composition of translatation, $\dfrac{1}{z}$, and linear function. 

  For all three, the image of a circle or a line remains a circle or a line.

  $\therefore$  $f(A) \in \mathscr{C}$

 

  end

 

 

 

Prop 1

$\overline{\mathbb{C}}$ is 1-1 onto with $S^{2}$

( $S^{2}$ is $2$-sphere )

 

  Let $S^{2} = \left\{(x, y, h) \; | \; x^{2} + y^{2} + h^{2} = 1 \right\}$ and $z = a + ib = (a, b, 0)$

  Define a line $l$ that connects points $(a, b, 0)$ and $(0, 0, 1)$.

  $\therefore$  $l = \left\{(x, y, h) \; | \; \dfrac{x}{a} = \dfrac{y}{b} = \dfrac{h-1}{-1} \right\} $

 

 

  Claim 1 : $S^{2} \cap l = \left\{(0, 0, 1), \; (\dfrac{2a}{a^{2} + b^{2} + 1}, \dfrac{2b}{a^{2} + b^{2} + 1}, \dfrac{a^{2} + b^{2} - 1}{a^{2} + b^{2} + 1}) \right\} $

  Let $\dfrac{x}{a} = \dfrac{y}{b} = \dfrac{h-1}{-1} = t$

  $\therefore$  $(at, bt, -t + 1) \in l$

 

  From $(at)^{2} + (bt)^{2} + (-t+1)^{2} = 1$,

  $(at, bt, -t+1) \in S^{2}$     $\Leftrightarrow$     $t = 0$  or  $t = \dfrac{2}{a^{2} + b^{2} + 1}$

 

  $\therefore$  $S^{2} \cap l = \left\{(0, 0, 1), \; (\dfrac{2a}{a^{2} + b^{2} + 1}, \dfrac{2b}{a^{2} + b^{2} + 1}, \dfrac{a^{2} + b^{2} - 1}{a^{2} + b^{2} + 1}) \right\} $

 

  end

 

  Define $F : \overline{\mathbb{C}} \to S^{2}$ by

  $F(z) = \begin{cases}
(\dfrac{ 2 \mathrm{Re}(z) }{\left|z \right|^{2} + 1}, \dfrac{ 2 \mathrm{Im}(z) }{\left|z \right|^{2} + 1}, \dfrac{ \left|z \right|^{2} - 1 }{\left|z \right|^{2} + 1}) & z \in \mathbb{C} \\
\\
(0, 0, 1) & z = \infty 
\end{cases} $

 

  By Claim 1, $F$ is well defined.

 

 

  Show 1 : $F$ is 1-1

  Suppose  $F(z_{1}) = F(z_{2})$

  By definition, the point where the line connecting $(0, 0, 1)$ and $F(z_{i})$ intersects $xy$-plane is $z_{i}$

  $\therefore$  $z_{1} = z_{2}$

  $\therefore$  $F$ is 1-1

 

  end

 

 

  Show 2: $F$ is onto

  $\forall A \in \mathscr{C}$,

  Let $(x, y, h) \in S^{2}$

  Let $z$ be the point where the line connecting $(0, 0, 1)$ and $(x, y, h)$ intersects $xy$-plane.

  By definition, $F(z) = (x, y, h)$

  $\therefore$  $F$ is onto

 

  end

 

  $\therefore$  $\overline{\mathbb{C}}$ is 1-1 onto with $S^{2}$

 

 

Def 2

For $z_{1}, z_{2}, z_{3}, z_{4} \in \overline{\mathbb{C}}$,

$(z_{1}, z_{2} ; z_{3}, z_{4}) = \dfrac{ (z_{3} - z_{1})(z_{4} - z_{2}) }{ (z_{3} - z_{2})(z_{4} - z_{1}) }$

is Cross-ratio of $z_{1}, z_{2}, z_{3}, z_{4}$

 

 

 

Thm 2

Let $f(z) = \dfrac{az + b}{cz + d}$

For $z_{1}, z_{2}, z_{3}, z_{4} \in \overline{\mathbb{C}}$,

$(f(z_{1}), f(z_{2}) ; f(z_{3}), f(z_{4})) = (z_{1}, z_{2} ; z_{3}, z_{4})$

 

  Since $f(z) - f(z_{i}) = \dfrac{(ad - bc)(z - z_{i})}{(cz + d)(cz_{i} + d)}$ ,

 

  $\dfrac{ f(z_{3}) - f(z_{1}) }{ f(z_{3}) - f(z_{2}) } = \dfrac{ (cz_{2} + d)(z_{3} - z_{1}) }{(cz_{1} + d)(z_{3} - z_{2})}$

 

  $\dfrac{ f(z_{4}) - f(z_{2}) }{ f(z_{4}) - f(z_{1}) } = \dfrac{(cz_{1} + d)(z_{4} - z_{2})}{(cz_{2} + d)(z_{4} - z_{1})}$

 

  $\therefore$  $\dfrac{ f(z_{3}) - f(z_{1}) }{ f(z_{3}) - f(z_{2}) } \cdot \dfrac{ f(z_{4}) - f(z_{2}) }{ f(z_{4}) - f(z_{1}) } = \dfrac{ z_{3} - z_{1}}{ z_{3} - z_{2} } \cdot \dfrac{ z_{4} - z_{2} }{ z_{4} - z_{1} }$

 

  $\therefore$  $(f(z_{1}), f(z_{2}) ; f(z_{3}), f(z_{4})) = (z_{1}, z_{2} ; z_{3}, z_{4})$

 

 

 

Cor 1

Let $f(z) = \dfrac{az + b}{cz + d}$  and

$f(z_{i}) = w_{i}$     for $i = 1, 2, 3$

Then

$\dfrac{ f(z) - w_{1} }{ f(z) - w_{3} } \cdot \dfrac{ w_{2} - w_{3} }{ w_{2} - w_{1} } = \dfrac{ z - z_{1}}{ z - z_{3} } \cdot \dfrac{ z_{2} - z_{3} }{ z_{2} - z_{1} }$