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To be specific, there is a logical way to prove that this metric satisfies the triangle inequality without using any magical identity ?

$ \displaystyle{d(z,z') = \begin{cases} \displaystyle\frac{2|z - z'|}{[(1 + |z|^2)^(1 + |z'|^2)]^{\frac{1}{2}}}, & \mbox{ if } z, z' \in \mathbb{C} \\ \displaystyle\frac{2}{(1 + |z|^2)}, & \mbox{ if } z \in \mathbb{C} \mbox{ and } z' = \infty \end{cases} } $

Thanks!

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    $\begingroup$ Let $\sigma : \mathbb{C} \cup \{\infty\} \to S^2$ be the stereographic projection (you might know this one as the inverse): $$ \sigma(z) = \left\{ \begin{array}{cc} \displaystyle {\frac{1}{\lvert z\rvert^2+1} \left(\overline{z}+z, \ i(\overline{z}-z), \ \lvert z\rvert^2-1 \right)} & \text{if } z\in \Bbb{C} \\ \\ (0,0,1) & \text{if } z = \infty \end{array} \right. $$ Now notice that $d(z,z')=\|\sigma(z) - \sigma(z')\|$, where $\|\cdot\|$ is the usual norm in $\mathbb{R}^3$. Thus, the triangle inequality follows at once from the one for $\|\cdot\|$ . $\endgroup$ – Alonso Delfín Mar 21 '17 at 22:13
  • $\begingroup$ Very nice! Thanks! $\endgroup$ – Richard Clare Mar 21 '17 at 23:00
  • $\begingroup$ You are welcome ! $\endgroup$ – Alonso Delfín Mar 21 '17 at 23:00
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Just for reference, I completed the answer to my question provided by Alfonso Delfin

$d(z,z') \leq d(z,z'') + d(z'',z')$

Let $\sigma: \mathbb{C}\cup\{\infty\} \to S$ be the stereographic projection

\begin{cases} (\displaystyle\frac{2\Re(z)}{|z|^2+1},\displaystyle\frac{2\Im(z)}{|z|^2+1},\displaystyle\frac{|z|^2 - 1}{|z|^2 + 1}), & \mbox{ if } z \in \mathbb{C}\\ (0,0,1), & \mbox{ if } z=\infty \end{cases}

Notice that

$d(z,z')=||\sigma(z)-\sigma(z')||$,

where $||\cdot||$ is the usual norm in $\mathbb{R}^3$. Thus, the triangle inequality follows from the one for $||\cdot||.$

Let $\textbf{u}, \textbf{v} \in \mathbb{R}^n$, then

$||\textbf{u} + \textbf{v}||^2 = (\textbf{u} + \textbf{v})\cdot(\textbf{u} + \textbf{v}) = ||\textbf{u}||^2 + 2(\textbf{u}\cdot\textbf{v}) + ||\textbf{v}||^2$

By Cauchy-Schwartz inequality we have

$ \textbf{u}\cdot\textbf{v} \leq ||\textbf{u}|| \mbox{ } ||\textbf{v}|| $

Hence,

$||\textbf{u} + \textbf{v}||^2 \leq ||\textbf{u}||^2 + 2||\textbf{u}|| \mbox{ } ||\textbf{v}|| + ||\textbf{v}||^2 = (||\textbf{u}|| + ||\textbf{v}||)^2 $

Taking square roots both sides yields

$||\textbf{u} + \textbf{v}|| \leq ||\textbf{u}|| + ||\textbf{v}||$

Now consider

$ d(z,z')= ||[\sigma(z)- \sigma(z'')] + [\sigma(z'') - \sigma(z')]||$

Let $\textbf{u}= \sigma(z) - \sigma(z''), \textbf{v} = \sigma(z'') - \sigma(z')$, then

$d(z,z')= ||\textbf{u} + \textbf{v}|| \leq ||\textbf{u}|| + ||\textbf{v}|| = d(z,z'') + d(z'',z') $

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