Distance of a point in the upper half plane to the unit circle in the upper half plane

Question

Let $$f:\mathbb{H}\to \mathbb{R}, (x,y)\mapsto \left(d_\mathrm{hyp}\left((x,y),\mathbb{S}^1\cap \mathbb{H}\right)\right)^2,$$ where $\mathbb{H}=\{(x,y)\in \mathbb{R}^2: y>0\}$ and $\mathbb{S}^1$ is the unit circle centered at origin.

In order to compute the distance, we need to compute the following: \begin{align*} d_\mathrm{hyp}\left((x,y),\mathbb{S}^1\cap \mathbb{H}\right) &=\inf\left\{ d_\mathrm{hyp}((x,y),(u,v)):(u,v)\in \mathbb{S}^1\cap\mathbb{H}\right\}\\ & = \inf_{(u,v)\in \mathbb{S}^1\cap \mathbb{H}}\inf\left\{l(\gamma):\gamma \text{ is a piecewise smooth curve joining }(x,y) \text{ to } (u,v)\right\} \end{align*} In order to compute all these things, we need tho compute the infimum of the length of piecewise smooth curves joining tose two points. I claim that the geodesic will be the circle passes trough the point $(x,y)$ whose center is on the $x$-axis and hits the unit cirlce perpendiculalry. I tried to prove the claim, but the expression is too much complicated to handle with.

Edit The main concern of the problem is why does the geodesics (circle) hit the line or the unit circle perpendicularly.

Answer 1

You should confirm these: the geodesics in the upper half plane are of just two types ( parametrized by arc length), with real constants $A,B$ and $B > 0$

$$ A + i e^t $$ $$ A + B \tanh t + i B \operatorname{sech} t $$ The latter are semicircles with center on the real axis

Next, given your point at $z_0$ with positive imaginary part, we can map your diagram isometrically using $$ f(z) = \frac{z+1}{-z+1} $$ where the original unit circle is mapped to the (positive) imaginary axis. Your point $z_0$ is mapped to some $f(z_0).$ The geodesic that the imaginary axis at a right angle (and passes through your point) is centered at the origin with $A = 0,$ so it is just $ B \tanh t + i B \operatorname{sech} t$ with $B$ the norm of $f(z_0).$ The geodesic distance is just the absolute value of $t$ that takes you to $f(z_0)$

Let's see, given real numbers $a,b,c,d$ with $ad-bc > 0,$ you should confirm that the Mobius transformation $$ m(z) = \frac{az+b}{cz+d} $$ maps the upper half plane to itself, and is an isometry. This can be split into three special cases, $\frac{-1}{z} \; , \;$ $az \; , \;$ $z+b \; . \;$

In case you want it, the inverse mapping uses the adjoint matrix, call it $$ g(z) = \frac{z-1}{z+1} $$

Answer 2

Doing this directly by the manner you've outlined in your post is indeed somewhat difficult.

But you can do it indirectly if you are willing to use the fact that the the group of orientation preserving isometries of $\mathbb H$ can be expressed using a complex coordinate $z=x+iy$ as the group of fractional linear transformations $f(z)=\frac{az+b}{cz+d}$ with $a,b,c,d \in \mathbb R$ and $ad-bc=1$. Here's a brief outline.

First do the special case where your point lies on the $x$-axis and so has the form $(0,y)$. It's not hard to use the symmetry of the metric across the $y$-axis to give a simple analytic proof that the shortest path between $(0,y)$ and the hyperbolic line $\mathbb S \cap \mathbb H$ is the segment on the $y$-axis connecting $(0,y)$ to $(0,1)$; and the length of this path is $|\ln(y)|$. This is very similar to using the Euclidean metric to prove analytically that the shortest path between two points $p=(0,p_2)$ and $q = (0,q_2)$ is the straight line whose length is $|p_2-q_2|$.

To handle the general case, you will need to work out the general form of fractional linear transformations $f(z)=\frac{az+b}{cz+d}$ which preserve the hyperbolic line $\mathbb H$ and preserve its orientation. You can work do this solving the system of equations $ad-bc=1$, $f(1)=1$, $f(-1)=-1$, or by other methods. There will be one free parameter $r$ to your solution, which illustrates the fact that the set of such isometries acts transitively on the hyperbolic line $\mathbb S^1 \cap \mathbb H$. Let me use $f_r$ to denote this family of isometries (I could give you the formula for $f_r$ but that would deprive you of most of the fun).

Once you have the general form of such isometries, you can write out the equation $$f_r(x+iy)=0+iy' $$ then take the real part of that equation and solve for $r$ in terms of $x$ and $y$, and then plug that value of $r$ into the imaginary part of the equation to obtain $y'$ in terms of $x$ and $y$. Your desired distance will therefore be $|\ln(y')|$.