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 $\alpha.$

  • 3
    $\begingroup$ What have you tried? What definition of a geodesic are you using? (Edit your post with this information so that those answering can do so more effectively.) $\endgroup$ – Dan Rust Nov 13 '13 at 2:12
  • $\begingroup$ From your given definition this is just a computation: parametrise the curves by length and compute the covariant acceleration $\nabla_{\alpha '} \alpha'$. $\endgroup$ – Anthony Carapetis Nov 13 '13 at 8:28

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.

  • $\begingroup$ Thanks! If I were to use $\nabla_{\alpha '} \alpha'$ how will I go about doing that? $\endgroup$ – Lays Nov 13 '13 at 19:04
  • $\begingroup$ First you need an explicit formula for the nabla in terms of the $\Gamma$ symbols. Then parametrize the vertical line by the exponential function (this gives a unit speed parametrisation with respect to the hyperbolic metric). Applying your formula to that will give $0$. $\endgroup$ – Mikhail Katz Nov 14 '13 at 13:27

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