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Why is the map $f(z)={z-c\over z-\overline{c}}$ for any $c\in$ {the upper half plane} $\subset C$ necessarily an isometry from the the upper half plane to the disc models of the hyperbolic plane? I know that the Cayley-transform is an isometry, but for an arbitrary $c$? I am not sure that it even maps the entire $H$ to the entire $D$...

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Not sure if it's a duplicate, but a previous question that I asked is very similar to this... – The Chaz 2.0 Feb 26 '12 at 13:59
Write down $df$ and then check that it preserves the hyperbolic metric ... – Neal Feb 26 '12 at 14:06
The point $c$ should lie in the upper half plane. – Christian Blatter Feb 26 '12 at 14:32
@ChristianBlatter: You are very right. If we add in that condition, it the statement then true? – Miller Feb 26 '12 at 14:47
@Neal: I may be a bit confused, but aren't the metrics in $H$ and $D$ different? – Miller Feb 26 '12 at 14:53
up vote 1 down vote accepted

The given map $f$ is a Moebius transformation; therefore it maps circles (:=circles or lines) onto circles. When $z\in{\mathbb R}$ then $|z-c|=| z-\bar c|$, so $|f(z)|=1$. Together with $c\in H$ this shows that $f$ maps the upper half plane $H$ conformally onto the unit disk $D$ in the $w$-plane.

The hyperbolic metric in $H:=\{z=x+iy\ |\ y > 0\}$ is defined by $$ds ={1\over y}\,|dz|\ ,$$ whereas the hyperbolic metric in $D=\{w\in{\mathbb C}\ |\ |w|<1\}$ is defined by $$d\sigma ={\rho\over 1-|w|^2}\,|dw|$$ with a constant $\rho>0$ depending on the author. We now have to check whether $f^*(d\sigma)=ds$ for a suitable choice of $\rho$. Let $c=a+ib$ with $b>0$. Then $$f'(z)={c-\bar c \over (z-\bar c)^2}\ ;$$ therefore $${|dw|\over 1-|w|^2}={|f'(z)|\ |dz|\over1-|f(z)|^2}={|c-\bar c|\over |z- \bar c|^2-|z-c|^2} ={2 b\over 4 b y}|dz|={1\over 2y}\ |dz|\ .$$ This shows that $f$ is a hyperbolic isometry iff we put $\rho=2$ in the definition of the hyperbolic metric in $D$.

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Thank you! Your argument is very clear. – Miller Feb 26 '12 at 16:57

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