Find an example of $f \in PSL(2, \mathbb{C})$ that gives an isometry of $\mathbb{H}$ with Riemannian metric

Question

I started studying on differential and projective geometry and I came across the following question on which I have diffuculties to solve/find examples:

Let $\mathbb{H}$ = { $z \in \mathbb{C}$ | $Im(z)$} > $0$

Consider the 1D complex projective space $P_1(\mathbb{C})$. Recall that the group $PSL(2, \mathbb{C})$ of projective transformations of $P_1( \mathbb{C})$ consists of the maps

$f = \displaystyle{\frac{az + b}{cz + d}}$, where $a, b, c, d \in \mathbb{C}$ and $ad - bc = 1$

• Find an example of $f \in PSL(2, \mathbb{C})$ that gives an isometry of $\mathbb{H}$ with Riemannian metric $\displaystyle{\frac{dx^2 + dy^2}{y^2}}$ but is not an isometry of $\mathbb{C}$ with the Euclidean metric $dx^2+ dy^2$

• Find a converse example.

So my question is how can I find such examples and which steps do I need to prove that its a valid example.

I tried $f(z) = -z$ as an example but I am stuck at proving an isometry in $\mathbb{H}$

Answer 1

How about scaling ? $f(z) = az/(1/a) = a^2 z$ i.e., $d = 1/a$, $b=c=0$ and $a>0$ and $a \neq 1$ is a positive real number, is an isometry in the Riemannian metric but not an isometry in Euclidean metric.

Answer 2

Take any element of $PSL(2, \mathbb{R})$, it will work. You need real coefficients, and positive determinant of the rational transformation to invariate $\mathbb{H}$ at the least.