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I want to prove the following basic property of hyperbolic lines in $IR^{2,1}$.

If x $\in$ $H^2$ and l is a line in $H^2$ then there is a unique line l' through x orthogonal to l.

I want to prove this in the hyperboloid model. Let $\langle\,,\rangle$ denote the Lorentz scalar product. Consider a line l. l is the intersection of $H^2$ with a 2-dimensional linear subspace

$\begin{align*} U= & \; \{y \in IR^{2,1} \mid | \langle\,y,n \rangle=0\} \end{align*}$

where $\langle\,n,n \rangle=1$. I have to construct a normal vector n' such that the associated plane U' contains x (that is $\langle\,x,n'\rangle=0$) and $\langle\,n',n\rangle=0$. Then the line l'=$H^2 \cap U'$ should intersect l orthogonally. The problem is that I don't see how to construct n' or prove the uniqueness. I cannot apply the Gram-Schmidt-Algorithm since this would also change x. Is there an elementary way to show this?

Polymorph
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  • Is the scenario relevant? If radical axis of two orthogonal circles in the Euclidean plane passes through origin, then it can be converted to a Poincare polar model. A circle cutting them orthogonality can be taken as the hyperbolic boundary.Two orthogonal (interior segments) circles are possible.A sketch needs to be added. – Narasimham Nov 14 '18 at 20:55

1 Answers1

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I would do it as follows.

Assume that $l$ is the X axis, i.e., the set of points with the $y$ coordinate equal to 0.

Let $C=(0,0,1)$, $X_a = \left(\begin{array}{ccc} \cosh a&0&\sinh a\\0&1&0\\\sinh a&0&\cosh a\end{array}\right)$ be the isometry which shifts $C$ $a$ units to the right, and $Y_a$ be the similar isometry which shifts $C$ $a$ units up. Hence $l = \{X_aC: a \in \mathbb{R}\}$.

The orthogonal line at point $X_aC$ is $\{X_a Y_b C: b \in \mathbb{R}\}$. If you multiply the matrices it is clear that $a$ and $b$ exist and are unique (if I remember correctly, for the point $(x,y,z)$ we have $b = \sinh(y)$ and $a = \sinh(x/\cosh(b))$).

Zeno Rogue
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