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I read a solution to a conformal mapping problem that made the claim, "Schwarz's lemma implies that any analytic conformal map taking the upper half-plane to the upper half-plane must be a fractional linear transformation." I had not heard this, and I wonder why this is true.

If you don't care to write out the whole proof, could you perhaps provide me with a reference?

Apologies if this is a repeat, I wasn't able to find this question on the site.

Eric Auld
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    For clarity's sake: "taking the upper half-plane onto the upper half-plane". Consider conjugation $\phi\circ f\circ \phi^{-1}$ where $\phi$ is a FLT from the halfplane to unit disk. – 40 votes Jul 29 '13 at 16:49
  • So we want to say that if $f:D\to D$ (open disk) is surjective and analytic, then it must be $f(z) = az$ for $|a|=1$? How do we know that $f:D\to D$ surjective implies that $|f(z)|=|z|$ for some $z\in D$? – Eric Auld Jul 29 '13 at 17:07
  • Brouwer's Fixed Point Theorem. – Trevor J Richards Jul 29 '13 at 17:11
  • @Trevor How do we get around the fact that $D$ is not compact, and we don't know for sure that any $K\subset D$ compact is mapped to itself? – Eric Auld Jul 29 '13 at 17:14
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    Compose $f$ with an FLT of the disk to make sure $f(0)=0$. Observe that both $f$ and $f^{-1}$ satisfy the assumptions of the Schwarz lemma. Conclude that $|f'(0)|=1$. Use the equality case of the Schwarz lemma. (@Trevor no, that's not going to help.) – 40 votes Jul 29 '13 at 17:14
  • @40votes So we must assume injectivity of $f$ as well? – Eric Auld Jul 29 '13 at 17:15
  • The invertibility of $f$ follows from the Monodromy theorem, since $f^{-1}$ exists locally and can be continued analytically along any path in the disk. I'll try to come up with something more elementary. – 40 votes Jul 29 '13 at 17:23
  • @EricAuld You are right, I was wrong. – Trevor J Richards Jul 29 '13 at 17:28
  • @40votes I just discovered in Ahlfors, the problem is stated as follows: "Prove by use of Schwarz's lemma that every one-to-one conformal mapping of a disk onto another (or a half-plane) is given by a linear transformation." So it appears that the 1:1 assumption is likely necessary. – Eric Auld Jul 29 '13 at 17:31
  • I rather think it's necessary in order to prove the result with the tools available so far in the book. – 40 votes Jul 29 '13 at 17:45
  • Also, see this question: http://math.stackexchange.com/questions/632091/on-the-structure-of-operatornameaut-im-z0 – EthanAlvaree Apr 29 '15 at 09:27

2 Answers2

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Let $\mathbb H$ be a halfplane, and $\mathbb D$ the unit disk. We know there is a Möbius map $\phi:\mathbb H\to\mathbb D$. Given a holomorphic map $f:\mathbb H\to\mathbb H$, we can form $F=\phi\circ f\circ \phi^{-1}$, which is a holomorphic map $F:\mathbb D\to\mathbb D$. This is a very common thing to do, called the conjugation of $f$ by $\phi$.

There is a Möbius map of $\mathbb D$ that sends $F(0)$ to $0$. Compose it with $F$ to get $G:\mathbb D\to\mathbb D$ such that $G(0)=0$. So far we did not need any special assumptions on $f$.

If $f$ is a bijection, then so is $G$. Applying the Schwarz lemma to $G$ and $G^{-1}$, we see that $|G(z)|\equiv |z|$; consequently, $G$ is a rotation. Unwinding the transformations that led from $f$ to $G$, we see that $f$ is a Möbius map.

Remark. It suffices to assume that $f$ is a surjection with nonvanishing derivative. In fact, if $f:\Omega\to \Omega'$ is a surjective holomorphic map with $f'\ne 0$, and $\Omega'$ is simply-connected, then $f$ is a bijection. This is actually a topological fact: a covering map onto a simply-connected space is a homeomorphism. In terms of holomorphic functions, this is given by the Monodromy theorem: since the (potentially multivalued) function $f^{-1}$ can be analytically continued in simply-connected domain $\Omega'$, it is single-valued there.

40 votes
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As a reference, consider Bak and Newman, Theorem 13.17, which I summarize here.

Essentially, you need to prove that all automorphisms of the upper half-plane are of the form $$f(z) = \frac{az+b}{cz+d}, ad-bc > 0, a,b,c,d \in \mathbb{R}.$$

It is easy to see that if $z$ is real-valued, then so too must be $f$.

Next, using $\textrm{Im} z = \frac{z-\overline{z}}{2i}$, we have $$\textrm{Im}(f(z)) = \frac{ad-bc}{c^2+d^2} > 0,$$ so $f$ must map $i$ into the upper half-plane. This is sufficient to show that $f$ is an automorphism of the upper half-plane.

To show uniqueness, consider an automorphism from the upper half-plane onto the unit disc:

$$g(z) = e^{i\theta}\left(\frac{z-\alpha}{1-\overline{\alpha}z}\right), |\alpha| < 1$$

and consider the most basic automorphism, just discovered,

$$h(z) = \frac{z-i}{z+i}.$$

Then, $h^{-1} \circ g \circ h$ is an automorphism of the upper-half plane (this is a very trivial lemma).

Simply carrying out the arithmetic, and you will find that any automorphism of the upper half-plane must be of the desired form.

Emily
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  • The contributions of Schwarz lemma are used implicitly in the step where we assume all automorphisms of the unit disc are of the form described by $g$ above. – Emily Jul 29 '13 at 17:43
  • But the original question didn't state automorphism! :) – Ted Shifrin Jul 29 '13 at 19:56
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    But it did state it's a conformal map of a region onto itself! Or did I infer "onto"... not that it matters! – Emily Jul 29 '13 at 20:01
  • Hi Arkamis, fantastic answer! However I don't see how you simplified $\textrm{Im}(f(z)) = \frac{ad-bc}{c^2+d^2}$. I created a question about it here: http://math.stackexchange.com/questions/1257236/simplify-im-left-fracazbczd-right – EthanAlvaree Apr 29 '15 at 08:46