I am presently a learner of Hyperbolic Geometry and am using J. W. Anderson's book $Hyperbolic$ $Geometry$. Now the author presents a sketch proof of why every circle preserving homeomorphism in $\overline{\mathbb{C}}$ is an element of the general Möbius group, which is what I am struggling to understand.

First, a brief outline.

Let $f$ be an element of the set of all circle preserving homeomorphisms which we denote Homeo$^{C}(\overline{\mathbb{C}})$ and let $p$ be a Möbius transormations that maps the triples $(f(0),f(1),f(\infty))$ to $(0,1,\infty)$

Then we see that $p\circ f(0) = 0, p\circ f(1) = 1$ and $p \circ f(\infty)=\infty$, and since $p \circ f(\mathbb{R}) = \mathbb{R}$, either such a composition maps the upper half of the complex plane, $\mathbb{H}$ to itself or the lower half of the complex plane.

If $p \circ f(\mathbb{H}) = \mathbb{H}$, we take $ m =p$, while if $m \circ (\mathbb{H})$ goes to the lower half, we just take $m = W \circ p$, where $W(z) = \overline{z}$

Now here is the thing I don't understand:

  1. Let $A$ be an euclidean circle in $\mathbb{C}$ with euclidean centre $\frac{1}{2}$ and radius $\frac{1}{2}$. Let $V(0), V(1)$ be the vertical lines through the points $x=0$ and $x=1$.

Can anybody explain why as $V(0)$ and $V(1)$ are vertical tangents to the circle, then these lines under the map $m \circ f(z)$, namely $m \circ f\Big(V(0)\Big)$ and $m \circ f\Big(V(1)\Big)$ are again vertical tangents to the circle $m \circ f(A)$ at $m \circ f(0) = 0$ and $m \circ f(1) = 1$?

I am trying to conclude from here that $m \circ f = Id_z$ , the identity transformation.



  • $\begingroup$ Hi David, Möbius transformations are conformal mappings. If you take a look at wikipedia, you will see that those transformations preserve angles, exactly what you need ! To see that they are indeed conformal I guess you could use Conway, Complex analysis. $\endgroup$ – Leandro Mar 29 '11 at 3:57
  • $\begingroup$ One more detail, m will be a möbius transformation even if it map the upper half plane to the lower half plane. $\endgroup$ – Leandro Mar 29 '11 at 3:59
  • $\begingroup$ @Leandro, I am just trying to show that every circle preserving homeomorphism is a mobius transform, please help me in trying to understand the above! $\endgroup$ – user38268 Mar 29 '11 at 6:03

I would assume that "circle-preserving" here means "preserving circles and straight lines"; that is, if $L$ is a straight line in the plane, then $L \cup \lbrace\infty \rbrace$ is a circle in $\overline{\mathbb{C}}$ (this is the usual definition in this context). I think that assuming this, all you really then need is that these maps are bijections: $m \circ f (V(0))$ and $m \circ f (V(1))$ are circles/straight lines, and since $m \circ f$ fixes $0$, $1$ and $\infty$, we see that $m \circ f (V(0))$ is a straight line through $0$, and $m \circ f (V(1))$ is a straight line through $1$. Then if $m \circ f (V(0))$ were not vertical, it would intersect $A$ at some point $w \neq 0$, which is a contradiction, because $(m \circ f)^{-1}(w) \neq 0$ would be another point of $V(0) \cap A$. Then $m \circ f (V(1))$ is vertical for similar reasons, or because it must be parallel to $m \circ f (V(0))$.

  • $\begingroup$ Ok thanks for your answer that makes things clear. $\endgroup$ – user38268 Apr 20 '11 at 0:31

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