By the theorem, we know that if is bijective, it is biholomorphic. So that we can study the group of conformal transformations of some domain.
Group of Conformal Transformations
The group of conformal transformations of some domain is defined as
Proposition
Suppose and , and let Then .
Proof Since Möbius transformations map generalized disks to generalized disks, it is enough to show that and . We have , and since , we have . For , we have $$\begin{aligned}|f(z)|&=\left|\frac{z-w}{-\overline{w}z+1}\right|\&=\frac1{|\overline{z}|}\left|\frac{z-w}{-\overline{w}z+1}\right|\&=\left|\frac{z-w}{-\overline{w}+\overline{z}}\right|\&=\frac{|z-w|}{|\overline{z}-\overline{w}|}=1\end{aligned}$$$\square$
The Schwarz Lemma
The Schwarz Lemma
Let with for all and . Then for all , and .
Moreover, if or for some non-zero , then is a rotation for some with .
Proof Since , has a removable singularity at , and . Hence the function defined by satisfies . Fix , and let . Apply the weak maximum principle to on , we obtain:This holds for any , so we may take limit as to obtain for all . Thus for all . Since , we have .
Moreover, if or for some non-zero , then for some . Thus by the strong maximum principle, is constant, with constant value having modulus . Thus for all as required.
Lemma
Let and suppose . Then is a rotation, i.e. there is some with such that for all .
Proof Since for all and is biholomorphic, the chain rule gives In particular, . Since , we have for all . By the Schwarz lemma, we have . We can apply the same reasoning to to obtain . But , so we must have . By the Schwarz lemma again, for some .
Theorem
Proof By the proposition, it suffices to prove the inclusion. Let , . Recall that and . Then , and is a conformal transformation of that fixes . By the previous lemma, for some . Thus as required.
Remark
We can now describe the geometric meaning of a conformal transformation of the disk. is the point that being sent to , and is the rotation factor.