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I need to proof (Under Hilbert axiomatization) that hyperbolic rigid motions, with respect to the metric $ d:\mathbb{H}^2 \times\mathbb{H}^2\rightarrow\mathbb{R}: d(A,B) =\left| \log \left( \frac{|AA_{\infty}| \ |BB_{\infty}|}{|AB_{\infty}| \ |BA_{\infty}|} \right) \right| $, does preserve paralelism.

I tried to show that two hyperbolic lines that share an ideal point will share another ideal point, but I suppose this isn't the best way to think parallelism since I'm stuck at the formalization.

Can you please give me a hint? Thanks.

Willemien
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  • If you know a proof for Euclidean geometry, it should be the same proof, because this fact should be independent of the parallel postulate. – Lee Mosher Feb 17 '16 at 02:31
  • I think " two hyperbolic lines that share an ideal point will share another ideal point" is just false, lines cannot share two points ideal or not – Willemien Feb 17 '16 at 09:29
  • But the congruence definition here is diferent. AB is congruence to CD iff there is a Lobatchevski transformation f such as f(A)=C and f(B)=D. So I believe the euclidean proof doesn't work here.

    "Share an ideal point" here isn't the same as "the ideal point belongs to the two lines" (An ideal point isn't even a point of the plane), but I mean that the euclidean semicircles that represent the two lines have a common euclidean point in the line $r_{\infty}$..

     –  Feb 18 '16 at 00:10

1 Answers1

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In order to apply the metric to lines, you likely consider these lines as sets of points. Two lines intersect if they have a point in common, ore phrased in terms of the metric, if there exists a point on the first line and a point on the second line which have distance zero to one another.

If you have a pair of parallel lines, they don't intersect, so for any point $A$ on the first line and $B$ on the second line, the distance will be non-zero. If you apply a rigid motion to them, the corresponding image points will still have that property, so you don't get any at distance zero. Therefore the resulting lines must still be parallel.

You can generalize this to say that rigid motions preserve incidence, since two sets of points are incident if they have a point in common.

If you care about the distinction between asymptotically parallel lines ans ultraparallel lines, as your comment indicates, then you'd observe that the asymptotically parallel lines are those of the set of all parallel lines which are closest to a given line. If distances are preserved, then so is the property of being closest. Together with the above this guarantees the preservation of asymptotically parallel lines.

MvG
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  • MvG, thanks for your reply. But there are also ultraparallel hyperbolic lines, which doesn't intersect neither share a common ideal point. –  Feb 17 '16 at 22:13
  • @Alnitak: In my book, both asymptotic parallel and ultraparallel would be called parallel. Proving that the distinction between these two classes is preserved requires another step. – MvG Feb 18 '16 at 07:16