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I have seen that in the book that (where $\operatorname{Aut}(U)$ is the set of all univalent bi-holomorphic mappings from $U$ onto $U$) $$\operatorname{Aut}(\Im z>0)=\left\{\frac{az+b}{cz+d}, a,b,c,d\in\mathbb{R}, ad-bc>0\right\}.\tag1$$ And it was derived from the fact that the set of all bi-holomorphic mappings from $\Im z>0$ to $D(0,1)$ (the unit disk) is $$w=e^{i\theta}\frac{z-a}{z-\bar a},\ \theta\in \mathbb{R},\ \Im a>0.$$

I want to show (1) as follows. Let $f\in \operatorname{Aut}(\Im z>0)$, then $$\frac{f(z)+i}{f(z)-i}$$ should also be bi-holomorphic from $\Im z>0$ to $D(0,1)$, and thus $$\frac{f(z)+i}{f(z)-i}=e^{i\theta}\frac{z-a}{z-\bar a}$$ for some $\theta\in\mathbb{R}, \Im a>0$. From which, I deduce $$f(z)=\frac{i(1+e^{i\theta})z-iae^{i\theta}-i\bar a}{(1-e^{i\theta})z+ae^{i\theta}-\bar a}.$$ By multiplying both denumeraotr and numerator with $1-e^{-i\theta}$, we get the desired form $$\frac{az+b}{cz+d}$$ with the coefficients $a,b,c,d$ all to be real. However, I compute $$ad-bc=2i(a-\bar a)(e^{i\theta}-1)$$ is not real, using pencil-paper or Mathematica. What's wrong with it?

Michael Hardy
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  • You must have gone wrong somewhere in your computation, because if $a$, $b$, $c$ and $d$ are all real, then so is $ad -bc$. Maybe you should write down some details. – Seub Jan 09 '14 at 09:19
  • At least in your write-up here, you have swapped numerator and denominator in some places. Since $f(z)$ is in the upper half plane, it is closer to $i$ than to $-i$, and the biholomorphic map to the unit disk is $$\frac{f(z)-i}{f(z)+i}.$$ That seems to be only the write-up, however, since $$f(z) = \frac{i(1+e^{i\theta})z -iae^{i\theta}-i\overline{a}}{(1-e^{i\theta})z+ae^{i\theta}-\overline{a}}$$ is correct. I suspect some of your confusion is due to using $a$ for two different things. And possibly multiplying numerator and denominator with $ie^{-i\theta/2}$ works better. – Daniel Fischer Jan 09 '14 at 14:40
  • Also, see this question: http://math.stackexchange.com/questions/454963/schwarzs-lemma-rightarrow-an-analytic-conformal-map-uhp-touhp-must-be-an-f – EthanAlvaree Apr 29 '15 at 09:27

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