Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

A bijective holomorphic map from unit disk to itself will be rotation? That mean $f(z)=e^{i\alpha}z$? How do I approach to solve this problem?In addition I want to know how one can remember the conformal maps which sends unit disk to upper half plane or conversely,and all possible known place to other places like that?

share|cite|improve this question
up vote 5 down vote accepted

For $w\in\mathbb{D}$, consider the following function defined on $\mathbb{D}$:

$$B_w(z) = \frac{z-w}{1-\overline{w}z}.$$

This function is called a "Blaschke factor." It defines a bijective holomorphism from $\mathbb{D}$ to $\mathbb{D}$, but it is not a rotation if $w\neq 0$. This demonstrates there are many such mappings that are not rotations.

To see this, first note that that by an easy computation, $B_{w}$ is its own inverse, so it is biholomorphic. To see it maps $\mathbb{D}\rightarrow \mathbb{D}$, just take the modulus and do the standard manipulations, noting that $|z|<1$ and $|w|<1$.

However, if you make the assumption that $f(0)=0$, your guess is correct. All biholomorphic maps of the unit disk that fix the origin are rotations. We use the Schwartz lemma to show this. If you are not familiar with this, leave a comment below and I will edit this answer to include a proof and discussion.

By the lemma, we have $|f'(0)|\le 1$. Because $f^{-1}$ is another origin preserving biholomorphism, we also have $|f^{-1}$'$(0)|\le 1$. Because we know by the differentiation rule for inverse functions that $f'(0)=\frac{1}{f^{-1}\text{'}(0)}$, we must have $|f'(0)|=|f^{-1}\text{'}(0)|=1$, and considering the equality case in the Schwartz lemma, we see $f$ must be a rotation.

We know that $f^{-1}$ is differentiable. Then the chain rule gives

$$f(f^{-1}(x))=x\rightarrow f'(f^{-1}(x))=\frac{1}{f^{-1}\text{'}(x)}$$

and noting here that $f^{-1}(0)=0$ gives the result.

To answer the second part of your question, I suggest you use Google and the literature available to you. The magic search terms are "automorphism group of the disk" and "automorphism group of the upper half plane." This will tell you about all possible biholomorphisms of each space. Now, because the half-plane is conformally equivalent to the unit disk, knowing the automorphism group of the unit disk is enough to know of all the the maps from the unit disk to the upper half-plane.

share|cite|improve this answer
I know about S Lemma, and thank you very much for your answer, could you just write me one line why $|f'(0)|=|f^{-1}(0)|$?and also do you know any back ground of that B-factor?any detail? – La Belle Noiseuse Jun 13 '12 at 9:12
I hope that helps. If you need more detail, just let me know. – Potato Jun 13 '12 at 9:21
thank you again. – La Belle Noiseuse Jun 13 '12 at 9:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.