I see the following theorem which characterizes the set of automorphisms of the upper half plane:
Every automorphism $f$ of the upper half plane is of the form
$$
f(z)=\frac{a z+b}{c z+d}
$$
where $a, b, c, d \in \mathbb{R} \text { and } a d-b c=1$.
My question is: $f(z)=2z$ is an automorphism but it can not be written as
$$
f(z)=\frac{a z+b}{c z+d}
$$
where $a, b, c, d \in \mathbb{R} \text { and } a d-b c=1$. Am I wrong?
Best Answer
It is more convenient to just require reals and $ad-bc>0.$ To get $1,$ your case uses $a=\sqrt 2,$ $d = 1/\sqrt 2,$ $b=c=0$