0
$\begingroup$

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?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Narasimham Nov 14 '18 at 20:55
0
$\begingroup$

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 $a = \sinh(x/\cosh(b))$).

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.