Let $\mathbb{H}^2$ be the hyperbolic space in the model $$\mathbb{H}^2=(\mathbb{R}\times\mathbb{R}_+,g=\frac{1}{y^2}(dx^2+dy^2)).$$
It is known that the Mobius transformations, with $ad-bc=1$, are all the isometries of $\mathbb{H}^2$.
Let $\mathcal{H}$ be $\mathbb{R}\times\mathbb{R}_+$ endowed with a complete metric $h$ with intrinsic curvature $-1$. Since $\mathbb{R}\times\mathbb{R}_+$ is simply connected, it is known that the $\mathbb{H}^2$ is isometric to $\mathcal{H}$.
My questions, possibly dumbs, are:
The isometries groups of $\mathbb{H}^2$ and $\mathcal{H}$ are the same? That is, all isometries of $\mathcal{H}$ are the Mobius transformations, with $ad-bc=1$?
When the metric $h$ is only orthogonal, $h=Edx^2+Gdy^2$, how can I to determine the isometries of $\mathcal{H}$? Cause, I think with a metric $h$ like that, it is complicated to use conformal parameters.
Any help I appreciate.
No, the isometries of $\mathbb H^2$ and $\cal H$ are not the same.
What one knows is that there exists a diffeomorphism $f : \cal H \to \mathbb H^2$ which is an isometry from the $\cal H$ metric to the $\mathbb H^2$ metric.
This does tell you how the isometries correspond: there is a bijection $A_f : \text{Isom}(\cal H) \to \text{Isom}(\mathbb H^2)$ given by the formula $$A_f(\phi) = f \circ \phi \circ f^{-1} $$