On this page in Schlag's book on complex analysis, he is discussing the upper half-plane model of $\mathbb{H}^2$. He says
for all $z_0\in \mathbb{H}$ $$\{T'(z_0) \mid T \in PSL(2, \mathbb{R}) \cap \text{Stab}(z_0)\} = SO(2,\mathbb{R}),$$ which means that the stabilizer subgroup in $PSL(2, \mathbb{R})$ at any point $z_0$ in the upper half-plane acts on the tangent space at $z_0$ by arbitrary rotations. Therefore, the geodesics of $\mathbb{H}$ are all circles that intersect the real line at a right angle.
I understand both of these facts in isolation, but could someone explain to me why the first fact implies the second? Thanks
I believe there is a fact in Riemannian geometry that (oriented) geodesics through a given point are determined by the tangent vector it makes at that point. Thus, in particular, given any point in the hyperbolic plane, the vertical line through it is the unique geodesic with vertical tangent vector.
Now, given any other geodesic through a point, one may apply a rotation from the point-stabilizer to make a new geodesic whose tangent line is vertical - this must be the vertical line. As such, any geodesic is the image of a vertical line under a Mobius transformations, which are vertical lines and semicircles. Since the vertical line intersects the real axis at a right angle and hyperbolic isometries are conformal, all the semicircles must intersect the real axis at right angles.