How do I show that each Möbius transformation that preservers the open unit circle (maps it to itself) must be of the form: $c \frac{z-z_0}{\bar{z_0}z-1}, |c|=1, |z_0|<1$ ? I've seen previous answers saying that the fact that a Möbius transformation is determined by its values on the points $0, 1, \infty$ can be used to easily show this, although I didn't manage to use this in order to solve the problem.

Help would be appreciated!

  • $\begingroup$ Start with looking at Möbius transformations that additionally have $T(0) = 0$. These are easily determined by applying the Schwarz lemma. Then compose an arbitrary automorphism of the unit disk with an appropriate Möbius transformation to get into the aforementioned case. $\endgroup$ – Daniel Fischer Dec 9 '14 at 19:12
  • $\begingroup$ The automorphisms of the unit disk are exactly of the form I wrote. I can't say that an automorphism is of this form because that's what I need to prove. $\endgroup$ – Bob Dec 9 '14 at 19:35
  • $\begingroup$ Yes. You can't assume what you want to prove. Show what you need to show in steps. First, find all automorphisms $f$ of $\mathbb{D}$ with $f(0) = 0$. Then, take an arbitrary automorphism $g$ of $\mathbb{D}$, and compose it with an appropriate Möbius transformation $T$ so that $g\circ T$ or $T\circ g$ is an automorphism - call it $f$ - with fixed point $0$. Then you have $g = f\circ T^{-1}$ or $g = T^{-1}\circ f$ and thus you obtain the structure of $g$. $\endgroup$ – Daniel Fischer Dec 9 '14 at 19:43
  • $\begingroup$ Thanks a lot for your explanation I managed to prove it (using Schwarz lemma as you said), pretty elegant and really short proof. Although I was given this exercise as HW before we studied Schwarz lemma so its interesting how to prove it directly. NONETHELESS your proof is sufficient for me thanks a lot! $\endgroup$ – Bob Dec 9 '14 at 19:52
  • $\begingroup$ Without the Schwarz lemma, I think the easiest way is to determine the automorphisms of the upper half-plane - use the reflection principle to deduce that they are actually all restrictions of automorphisms of the Riemann sphere, hence Möbius transformations. Then it follows that all automorphisms of the unit disk are Möbius transformations, and once you know that, finding the structure is not too hard. Well, it's still easier to prove the Schwarz lemma ;) $\endgroup$ – Daniel Fischer Dec 9 '14 at 20:02

Let $Tz=\frac{az+b}{cz+d}$ map the unit disc onto itself. Let $z_0$ be the point which is mapped by $T$ to $0$. By the symmetry principle, $\frac{1}{\overline{z_0}}$ is mapped to $\infty$. Thus $T$ is of the form $Tz = c \frac{z-z_0}{\overline{z_0}z-1}$. Since $1$ is mapped onto the unit circle, we have $1 = |T1| = |c||\frac{1-z_0}{\overline{z_0}-1}|=|c|$ (where $|\frac{1-z_0}{\overline{z_0}-1}|=1$ by an isosceles triangle argument.

  • $\begingroup$ how did you deduce the form by the two points? $\endgroup$ – Bob Dec 15 '14 at 13:55
  • $\begingroup$ $z_0$ must be a root of the numerator and $1/\overline{z_0}$ must be a root of the denominator. $\endgroup$ – Frost Boss Dec 15 '14 at 20:32
  • $\begingroup$ How is that true? definitely not true for $z/(z-1)$ $\endgroup$ – Marcos G Neil Feb 10 '20 at 5:07

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