I want to know if a LFT, $F$, is conformal on the hyperbolic plane $\mathbb H^2$ , that is if we have the curves $\Gamma_1$ and $\Gamma_2$ that intersect at a point $P$ making the an angle $X$, then $F(\Gamma_1)$ and $F(\Gamma_2)$ should also make the angle $X$ at $F(P)$.

Any proofs on this? How would one do this?


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    $\begingroup$ What did you try? $\endgroup$ – Moishe Kohan Apr 4 at 0:50

Any real fractional linear transformation of $\mathbb R^2$ is conformal, if we consider on $\mathbb R^2$ the Euclidean metric. You can prove this by showing that any such map is a composition of translations, homothecies and the Euclidean inversion $x\mapsto x/||x||^2$, and then proving these functions are conformal with this metric.

I guess you're taking the half plane model of $\mathbb H^2$. This model is just the manifold $\{(x_1,x_2)\in\mathbb R^2:x_2>0\}$ with Riemannian metric given by $$ds^2=\frac{dx^2+dy^2}{y^2}.$$

Thus if you have $u,v\in T_{\overline x}\mathbb H^2$ for some $\overline{x}\in\mathbb H^2$, we get that their angle is given by $$\arccos\left(\frac{\langle u,v\rangle_{\mathbb H^2}}{||u||_{\mathbb H^2}||v||_{\mathbb H^2}}\right)=\arccos\left(\frac{\langle u,v\rangle/x_2^2}{(||u||/x_2)(||v||/x_2)}\right)=\arccos\left(\frac{\langle u,v\rangle}{||u||\cdot||v||}\right);$$

here $\overline x=(x_1,x_2)$ with $x_2>0$, so the hyperbolic angle between these two vectors is their Euclidean angle, and as we have proved that real fractional linear transformations are conformal with the Euclidean metric, they are conformal in the hyperbolic metric too.

You could also prove that your functions are actually isometries of $\mathbb H^2$, but this is a little bit more complicated, showing that the Euclidean inversion $x\mapsto x/||x||^2$ is an isometry is not that trivial.