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Problem of the Week #262 - Dec 19, 2017

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Euge

MHB Global Moderator
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Jun 20, 2014
1,894
Happy Holidays, everyone! Here is this week's POTW:

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Consider the open unit disk $\Bbb D\subset \Bbb C$ with Riemannian metric $ds^2 = \dfrac{\lvert dz\rvert^2}{(1 - \lvert z\rvert^2)^2}$. Find a formula for the (Riemannian) distance between two points in $\Bbb D$, and use it to find the distance between $-\frac{1}{2}e^{i\pi/4}$ and $\frac{1}{2}e^{i\pi/4}$.

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Remember to read the POTW submission guidelines to find out how to submit your answers!
 
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Euge

MHB Global Moderator
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Jun 20, 2014
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Due to the holiday break I’m giving an extra week to solve this problem. Happy New Year, everyone!
 
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Euge

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Jun 20, 2014
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I like Serena gets honorable mention for his correct calculation for the Riemannian distance from $-\frac{1}{2}e^{i\pi/4}$ to $\frac{1}{2}e^{i\pi/4}$. You can read my solution below.



Given $z,w\in \Bbb D$, the Riemannian distance between $z$ and $w$ is given by

$$d(z,w) = \frac{1}{2}\log\frac{1 + \left\lvert\dfrac{z - w}{1 - \bar{z}w}\right\rvert}{1 - \left\lvert\dfrac{z - w}{1 - \bar{z}w}\right\rvert}$$ If $\gamma$ is a smooth curve in $\Bbb D$, set $\displaystyle\rho(\gamma) := \int_\gamma \frac{\lvert dz\rvert}{1 -\lvert z\rvert^2}$. For every isometry $\phi\in \operatorname{Isom}(\Bbb D)$, $\rho(\phi\circ \gamma) = \rho(\gamma)$. For by the Schwarz lemma, $\dfrac{\lvert \phi'(z)\rvert}{1 - \lvert z\rvert^2} \le \dfrac{1}{1 - \lvert z\rvert^2}$ for all $z\in \Bbb D$; hence

$$\rho(\phi\circ \gamma) = \int_\gamma \frac{\lvert \phi'(w)\rvert}{1 - \lvert \phi(w)\rvert^2}\lvert dw\rvert \le \int_\gamma \frac{\lvert dw\rvert}{1 - \lvert w\rvert^2} = \rho(\gamma)$$

and similarly $\rho(\gamma) = \rho(\phi^{-1}\circ \phi \circ \gamma) \le \rho(\phi\circ \gamma)$. Consequently, $d$ is invariant under the action of Möbius transformations on $\Bbb D$. Using the Möbius transformation $\phi : \Bbb D\to \Bbb D$ given by $\phi(c) := e^{i\theta}\dfrac{z - c}{1 - \bar{z}c}$ (where $\theta$ is chosen so that $\phi(w) \ge 0$), we have $d(z,w) = d(0, \phi(w))$. A geodesic joining two points on the real axis is a segment on the axis between the points, so $$d(z,w) = \int_0^{\phi(w)} \frac{dx}{1 - x^2} = \frac{1}{2}\log\frac{1 + \phi(w)}{1 - \phi(w)} = \frac{1}{2}\log \frac{1 + \left\lvert \dfrac{z - w}{1 - \bar{z}w}\right\rvert}{1 - \left\lvert \dfrac{z - w}{1 - \bar{z}w}\right\rvert}$$

Using this formula, we compute

$$d\left(-\frac{1}{2}e^{i\pi/4}, \frac{1}{2}e^{i\pi/4}\right) = \frac{1}{2}\log\frac{1 + \frac{4}{5}}{1 - \frac{4}{5}} = \frac{1}{2}\log 9 = \log 3$$
 
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