- Describe all the automorphisms of the right half-plane \(H=\{z\in\mathbb C:\operatorname{Re}z>0\}\), meaning holomorphic bijections from \(H\) onto itself.
- Describe all the automorphisms of C (holomorphic bijective maps from C onto C).
Solution
Write \(D=\{w:|w|<1\}\) and \[ T(z)=\frac{z-1}{z+1},\qquad T^{-1}(w)=\frac{1+w}{1-w}. \] The answers are \[ \boxed{\operatorname{Aut}(H)= \left\{z\longmapsto T^{-1}\!\left( \lambda\frac{T(z)-a}{1-\overline a T(z)}\right): a\in D,\ |\lambda|=1\right\},} \] and \[ \boxed{\operatorname{Aut}(\mathbb C)= \{z\longmapsto az+b:a,b\in\mathbb C,\ a\ne0\}.} \]
1. Part (a) reduces to the disk automorphisms.
The displayed maps \(T,T^{-1}\) are inverse by substitution. For \(z\in H\) and \(w\in D\), \[ 1-|T(z)|^2=\frac{4\operatorname{Re}z}{|z+1|^2}>0, \qquad \operatorname{Re}T^{-1}(w)=\frac{1-|w|^2}{|1-w|^2}>0. \] Thus they are biholomorphic maps between \(H\) and \(D\). Conjugation by \(T\) gives a bijection between their automorphism groups.
For a disk automorphism \(g\), take \(a=g^{-1}(0)\) and let \(\phi_a(w)=(w-a)/(1-\overline a w)\). The inverse is \((v+a)/(1+\overline a v)\); the identity \[ 1-|\phi_a(w)|^2= \frac{(1-|a|^2)(1-|w|^2)}{|1-\overline a w|^2} \] and the corresponding inverse identity show that this is a disk automorphism. Both \(h=g\circ\phi_a^{-1}\) and \(h^{-1}\) fix zero. Schwarz’s lemma gives \(|h(w)|\leq|w|=|h^{-1}(h(w))|\leq|h(w)|\). Its equality case implies \(h(w)=\lambda w\) with \(|\lambda|=1\) [@SS03]. Hence \(g=\lambda\phi_a\). Conversely each such map is a rotation composed with a disk automorphism. Conjugating proves exactly the formula for part (a), including its converse.
2. A plane automorphism must be a polynomial.
Let \(F:\mathbb C\to\mathbb C\) be a holomorphic bijection. Its inverse is holomorphic, hence continuous, by the nonvanishing derivative of an injective holomorphic function and the inverse function theorem [@SS03]. For every \(M>0\), the set \(F^{-1}(\{w:|w|\leq M\})\) is compact: it is the continuous image of this compact disk under the inverse. It is therefore bounded. Consequently \(|F(z)|\to\infty\) as \(|z|\to\infty\).
For small nonzero \(\zeta\), put \(h(\zeta)=1/F(1/\zeta)\). It is holomorphic there and tends to zero. The removable singularity theorem extends it with \(h(0)=0\) [@SS03]. Since it is nonzero off zero, its Taylor series has a first nonzero term, so \(h(\zeta)=\zeta^m u(\zeta)\) for some \(m\geq1\) and holomorphic \(u\) with \(u(0)\ne0\). Thus, for all sufficiently large \(|z|\), \(|F(z)|\leq C|z|^m\) for a constant \(C\). Cauchy’s coefficient estimate for the entire Taylor series of \(F\) gives \(|c_k|\leq CR^{m-k}\) for large \(R\). Letting \(R\to\infty\) shows \(c_k=0\) for every \(k>m\) [@SS03]. Hence \(F\) is a polynomial.
3. The polynomial has degree one, and all such maps are automorphisms.
The polynomial cannot be constant. If its degree were at least two, its derivative would be a nonconstant polynomial, hence have a complex zero by the fundamental theorem of algebra [@SS03]. This contradicts the nonvanishing derivative of an injective holomorphic function. Therefore \(F(z)=az+b\) with \(a\ne0\). Conversely this map is holomorphic and has the entire inverse \(w\mapsto(w-b)/a\). This proves part (b).