Geodesic hyperbolic metric

1.2k Views Asked by At

For a hyperbolic metric on the upper half plane $H = \{(u,v)\in \mathbb{R}^2 \ | \ v>0\},$ how can I prove that the vertical lines are geodesics and that the intersection of any circle centered on $x-\text{axis}$ with $H$ is a geodesic?

The definition I have for geodesic is:

Let $\alpha : [a, b] \to \Sigma$ be a regular parameterized curve then we call it geodesic if its tangent vector is parallel along $\alpha.$

1

There are 1 best solutions below

2
On BEST ANSWER

The problem can be solved synthetically as follows. The vertical ray, e.g. the $y$-axis, is invariant under the reflection $(x,y)\mapsto (-x,y)$. The reflection is an isometry and therefore the vertical line must be a geodesic.

To show that the semicircles are geodesics, use an isometry sending one of its endpoints to infinity. The result is a vertical ray hence a geodesic as already shown.

If the semicircle passes through the origin, just use $z\mapsto 1/z$. Otherwise use a horisontal translation first.