Ive been thinking about the following question.

Let $p \in H^2,$ and $L$ be a complete geodesic in $H^2.$ Prove that there is a unique comlete geodesic $L'$ through $p$ and orthogonal to $L$ at some point $q \in L.$ Moreover, the line segment from $p$ to $q$ minimizes the distance from $p$ to any point on $L.$

The question is asking to build two complete geodesics that are orthogonal to each other at some point.

In my textbook, a complete geodesic is said to be an "open semi-circle" centered on the x-axis and delimited by two points on the x-axis.

Now, this seems curious. How can to semi-circles, both centered and delimited by the x-axis, intersect orthogonally?

Do angles work differently in hyperbolic space? Does a right-angle in hyperbolic space look different than a right angle in Euclidean space?


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    $\begingroup$ See, for instance, MathWorld's "Orthogonal Circles" entry. $\endgroup$ – Blue Sep 9 '19 at 17:34
  • $\begingroup$ “How can two semi-circles…” It’s really easy. I’m sure you can sketch some examples. If not, draw two circles of radius $1$, centers at $\pm\sqrt2/2$. $\endgroup$ – Lubin Sep 9 '19 at 20:09

The usual hyperbolic metric on $\mathbb{H}^2$ is conformal. It can be written as

$$\frac{dx^2+dy^2}{y^2} = e^{g(x,y)} (dx^2+dy^2),$$

with $g(x,y) = -2\ln(y)$. The metric tensor as each point is diagonal (with respect ot the usual coordinates), which means that is preserves the angles. In particular, right angle for the hyperbolic metric are right for the Euclidean metric.

  • $\begingroup$ Thanks Thomine! $\endgroup$ – Rafael Vergnaud Sep 9 '19 at 17:35

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