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Let $f:\mathbb{D} \rightarrow \mathbb{D}$ be a local isometry. I want to show that is an automorphism (i.e. a Mobius map).

Here's one brief argument I've come across:

"By the Pick Theorem (in Milnor - Dynamics in One Complex Variable page 23), we have that $f:\mathbb{D} \rightarrow \mathbb{D}$ is a covering map, therefore it must be an automorphism of $\mathbb{D}$."

I was wondering if anyone could clarify this statement a little bit, or provide an alternative argument? Why must local isometric covering maps of the disk to itself to automorphisms?

Any help would be greatly appreciated.

OllyT777
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    By a local isometry, do you mean a local isometry with respect to the hyperbolic metric on $\mathbb{D}$? – Deane Aug 10 '23 at 13:50
  • Yes, that's right I do :) – OllyT777 Aug 10 '23 at 14:07
  • Because $\mathbb D$ is simply-connected, any connected covering must be an isomorphism. – Kenta S Aug 10 '23 at 14:44
  • Sorry, but what do you mean by a connected covering? – OllyT777 Aug 10 '23 at 15:26
  • First, verify that Mobius transformations are isometry. Suppose $f$ is an isometry and $f(0) = a$. Find a Mobius transformation $M$ such that $M(a) = 0$. $M\circ f$ is an isometry that maps $0$ to $0$. Show that this has to be a rotation and therefore a Mobius transformation. Conclude that $f$ is a Mobius transformation. – Deane Aug 10 '23 at 17:35
  • Do we use the Schwarz lemma to show this is a rotation? If so, I'm not sure how to show that $\vert (M \circ f)'(0) \vert$ = 1 @Deane – OllyT777 Aug 11 '23 at 09:12
  • I’m not using any complex analysis at all. I never use the fact that $f$ is holomorphic. But I am assuming $f$ preserves orientation. – Deane Aug 11 '23 at 13:26
  • Ah ok, I think I see. For any $z \in \mathbb{D}$, we have the distance from $z$ to 0 is the same as the distance from $f(z)$ to 0, and this is only possible if each point is rotated, is that right? – OllyT777 Aug 11 '23 at 16:24
  • @Deane Sorry for the bother, but I think I've confused myself. How are Mobius maps isometries? For example, the map $z \rightarrow z/2 $ is a Mobius map, but here distance is not preserved. – OllyT777 Aug 12 '23 at 10:29
  • Sorry. Not all mobius transformations are isometries of the disk. My comment is correct if you restrict to the right subset. See https://math.stackexchange.com/questions/209308/can-we-characterize-the-m%c3%b6bius-transformations-that-maps-the-unit-disk-into-itse – Deane Aug 12 '23 at 14:18
  • @Deane Just to check, are you saying that Mobius maps of the form $\frac{e^{i\theta}(z-a)}{1-\bar{a} z}$ are isometries? If so, I'm a little unsure to see how. Thanks – OllyT777 Aug 12 '23 at 14:32
  • Could you say what you know about the hyperbolic plane and when a map is an isometry? – Deane Aug 12 '23 at 14:48
  • @Deane I know that the hyperbolic plane is usually given the poincare metric which measures the distance between two points. A map is an isometry if it is bijective and if it preserves the distance between any two points i.e. d(f(z_1)-f(z_2))=d(z_1 - z_2). Admittedly, the field is still a little new to me, so I thank you for bearing with me. – OllyT777 Aug 12 '23 at 15:07
  • OK. You could try looking at this: http://library.msri.org/books/Book31/files/cannon.pdf – Deane Aug 12 '23 at 17:03

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