Inexistence of equivalent vectors in hyperbolic geometry.

Question

A translation defined by vector $\vec{v}=\vec{AB}$, $\tau=\tau_{\vec{v}}$, is, by definition, the composite of the reflections $\rho_2 \circ \rho_1 $, where $\rho_1$ is the reflection in the perpendicular bissector of $AB$ and $\rho_2$ is the reflection in the perpendicular line to $AB$ through $B$.

In euclidean geometry I can find two equivalent vectors that will define the same translation, but that does'nt happen in hyperbolic geometry. I'm needing help on proving this: $\tau_{\vec{AB}}=\tau_{\vec{CD}} \iff \vec{AB}, \vec{CD}$ are collinear vectors with same orientation and magnitude.

Thanks.

Answer 1

HINT

Every isometry (and translation is one of them) has the property that if $A'$ is the image of $A$ and $B'$ is the image of $B$ then $d(AB) = d(A'B')$

But the question is what happens with $d(AA')$ and $d(BB')$ ?

In Euclidean geometry rotations do not have the property that the distance between a point and its image in the translation is the same. so $ d(AA') \not= d(BB')$

In Euclidean geometry translations have the property that the distance between a point and its image in the translation is the same. so $ d(AA') = d(BB')$

After some reflection and testing I realised this is also not the case with hyperbolic translations.

The distance between a point and its image differs, they are the shortest along the line that contains the vector an the further away the larger the distance (the points form a Saccheri quadrilateral https://en.wikipedia.org/wiki/Saccheri_quadrilateral )

From this alone you can prove your statement

Answer 2

Try to use PROPOSITION 9.11 (GreenBerg 1993,p.331). it says:

If T is a translation along t and m is any line perpendicular to t, then there is a unique line l perpendicular to t such that T is the composeition of reflection of m and reflection of l.