Question 2.1. Determine all holomorphic automorphisms of the upper half plane \(u =\) \(\lbrace z : I m z > 0 \rbrace\)
Solution
Let \(\HH=\{z:\Im z>0\}\) and let \(\phi(z)=\frac{z-i}{z+i}\), the Cayley transform, a biholomorphism \(\HH\to\DD\).
1. Every map \(T(z)=\frac{az+b}{cz+d}\) with \(a,b,c,d\in\RR\) and \(ad-bc=1\) is an automorphism of \(\HH\).
For real coefficients, \(\Im T(z)=\frac{(ad-bc)\Im z}{\abs{cz+d}^2}=\frac{\Im z}{\abs{cz+d}^2}\), so \(T\) maps \(\HH\) into \(\HH\). Its inverse \(\frac{dz-b}{-cz+a}\) has the same form.
2. Every automorphism \(T\) of \(\HH\) has this form.
\(\phi\circ T\circ\phi^{-1}\) is an automorphism of \(\DD\), hence a Möbius map \(w\mapsto e^{i\theta}\frac{w-a}{1-\bar aw}\) by the Schwarz lemma, so \(T\) is a Möbius map. It extends to a homeomorphism of \(\overline\HH\cup\{\infty\}\), so it maps \(\RR\cup\{\infty\}\) onto itself. A Möbius map is determined by the images of \(0\), \(1\) and \(\infty\), and when these are real, the linear conditions on \((a,b,c,d)\) have real coefficients and a one-dimensional solution space; so \(T\) has a real representative, which can be scaled to \(ad-bc=\pm1\). Since \(\Im T(i)=\frac{ad-bc}{c^2+d^2}>0\), the sign is \(+1\).
3. Q.E.D.
By steps 1 and 2, \(\operatorname{Aut}(\HH)=\boxed{\Bigl\{z\mapsto\frac{az+b}{cz+d}:a,b,c,d\in\RR,\ ad-bc=1\Bigr\}}\cong\operatorname{PSL}_2(\RR)\).