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The chordal metric between two points $z,w \in \hat{\mathbb{C}} (=\mathbb{C}\cup{\{\infty\}})$ is defined as , \begin{align*} d(z,w)=\displaystyle\frac{2|z-w|}{\sqrt{(1+|z|^2)(1+|w|^2)}} \end{align*} \begin{align*} d(z,\infty)=\displaystyle\frac{2}{\sqrt{(1+|z|^2)}} \end{align*} The spherical metric is defined as goes , $\gamma \colon [0,1] \to \mathbb{C}\cup{\{\infty\}}$; The spherical length of the curve $\gamma$ in $ \hat{\mathbb{C}} $ is defined as, \begin{align*} \Lambda(\gamma)= \int_\gamma ds = \int_\gamma \displaystyle\frac{2|dz|}{1+|z|^2} \end{align*} The spherical distance between two points is the infimum of the spherical lengths of paths joining the two points: \begin{align*} \rho (z,w) := inf_\gamma \Lambda(\gamma) \end{align*} $\gamma$ varies over all paths joining $z$ and $w$. In "Normal Families" by schiff, it's written that , \begin{align*} d(z,w) \leq \rho(z,w) \leq \dfrac{\pi}{2} d(z,w) \end{align*} I cannot figure out why is this true .

confused
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  • Hint: See the alternative description of the chordal metric here. Can you think of a similar description of the spherical distance? – Moishe Kohan Sep 28 '21 at 11:35
  • If you mean of finding a function to represent spherical distance two points as distance between image of of those points under the function… i could not find anything . Do you mean in that way ? – confused Sep 28 '21 at 12:05
  • I could not understand the sentence you wrote. Did you understand the linked answer? – Moishe Kohan Sep 28 '21 at 15:11
  • Sorry for being imprecise. The idea in linked answer was to transfer the chordal metric to euclidean metric using the function F ,right ? In that case are you asking me to do something similar ? – confused Sep 28 '21 at 15:43
  • Right, for the spherical metric, using exactly the same map. – Moishe Kohan Sep 28 '21 at 15:50

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Let me first introduce my notations $d_c(z,w)=$ Chordal distance between $z$ and $w$. $d_s(z,w)=$ Spherical distance between $z$ and $w$ and $S=$ Unit sphere in $\mathbb{R^3}$. Now suppose that $z$ and $w$ maps to $Z$ and $W$ in the sphere under the stereographic map.

Now I can always intersect the $S$ with the plane passing through the points $Z$ and $W$ such that the intersection is great circle, say $C$. Suppose that chord joining point $Z$ and $W$ makes angle $\theta$ to center (origin $O$) of $C$. During this process we will get triangle $OZW$ lies in that plane, where $OZ = 1$ and $OW = 1$ and we know that Euclidean distance between $Z$ and $W$ is $d_c(z,w)$.

Now using the Arc Length Formula of circle we have,

\begin{align} \theta = d_s(z,w) \end{align}

Also we know that if we draw angle bisector of $\theta$, say $OM$,we get (N0te that $OZW$ is isosceles triangle),

\begin{align} 2Sin(\frac{\theta}{2}) = d_c(z,w) \end{align}

We have \begin{align} 2Sin(\frac{d_s(z,w)}{2}) = d_c(z,w) \end{align}

Now using the simple trigonometric fact $\frac{2\theta}{\pi} \leq Sin\theta \leq \theta$, We have

\begin{align} \frac{2}{\pi} d_s(z,w) \leq d_c(z,w) \leq d_s(z,w) \end{align}

Hence from this we proved that spherical metric $d_s$ and chordal metric $d_c$ are equivalent. If you want to get the same inequality as mentioned in the question you just need to play around the above trigonometric fact and above metric equivalent inequality.

N_Set
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