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Via Tesselation of the upper half plane via Ford Circles I was introduced to Ford circles ( https://en.wikipedia.org/wiki/Ford_circle note the wikipedia article has been updated since that question)

Ford circles made it possible for me to investigate an apeirogon, ( an infinite sided polygon see https://en.wikipedia.org/wiki/Apeirogon#Apeirogons_in_hyperbolic_plane )

Unfortunedly I could not yet investigate the regular apeirogon that circumscribes a horocycle, I have not(yet?) developed the nessessary formulas for that, but I did manage the formulas for the apeirogon that is circumscribed by a horocycle.

But when I did this and looked what I had made I realised that an apeirogon is not a (closed) polygon at all (the sides don't form a closed circuit)

Starting from a vertex and going in two direction of the sides at the vertex, following the sides, at the end the vertices of a polygon must coincide again to make a closed circuit.

But what I noticed that while in most models of hyperbolic geometry it looks that the vertices of an apeirogon come together again in reality they get further and further away from eachother. so they don't form a circuit so an apeirogon is no (closed) polygon

My construction in detail:

  • Poincare half-plane model

  • The circumscribing horocycle is the horocycle centered at $(0 , 0)$ going through $ (0 , 1)$ (so Euclidianly centered at $(0 , \frac{1}{2} )$ and is inscribed by the apeirgon that i am investigating

  • The vertices of the apeirgon are $P_0 = (0,1)$ and the points where Ford circles C(1,i) and C(-1,i) $ (i= 1,2, ... , \infty)$ are tangent to the circumscribing horocycle.

  • Ford circle C(1,n) is the circle Euclidianly centered at $(\frac{1}{n} , \frac{1}{2n^2})$ going trough $(\frac{1}{n} , 0)$ , so is hyperbolicly a horocycle. the point where this horocycle and the circumscribing horocycle meet is $(\frac{n}{n^2+1} ,\frac{1}{n^2+1} )$ and this point is vertex $P_n$ of the apeirogon.

  • Ford circle C(-1,n) is the circle Euclidianly centered at $(-\frac{1}{n} , \frac{1}{2n^2})$ going trough $(-\frac{1}{n} , 0)$ , this is the horocycles at the other side of the circumscribing horocycle. The point where they meet is vertex $P_{-n}$ of the apeirogon.

  • There is a deduction that shows that side has a length of $\operatorname{arcosh}(1.5)$ and every two vertices that are one vertex apart are $\operatorname{arcosh}(3)$ apart (n just drops out of the formula) this proofs that the apeirogon is regular (all sides have the same length and all angles are equal)

But then:

  • The hyperbolic distance between the vertices $P_n$ and $P_{-n}$ is $\operatorname{arcosh}(1+2n^2)$ (sometimes formulas are simple)

  • As n grows to infinity the distance between the vertices $P_n$ and $P_{-n}$ also grows to infinity and $P_n$ and $P_{-n}$ or $P_n$ and $P_{-n+1}$ are not going to coincide.

  • Thus the apeirogon sides don't form a circuit so the apeirogon is no (closed) polygon.

Did I do something wrong or did I overlook something?

Also what does this all mean for the circumscribing horocycle? Are the ends of the circumscribing horocycle also an infinite distance apart, but that contradicts that lines that don't cut horocycles orthogonal will cut the horocycle twice.

I am getting completely confused.

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1 Answer 1

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I thought, earlier, that things might make more sense if I emphasized the horocycles in the upper half plane model. The line $y=1$ in the upper half plane is a horocycle. Notice that it is very obviously infinite.

I think you have been troubled by curves approaching the $x$ axis in the upper half plane. Any such curve is of infinite length and does not "close up" in any reasonable sense. All horocycles are of infinite length. It is convenient to draw and calculate in the upper half plane or the unit disc, but anything approaching the "boundaries" is of infinite length.

If you have not done much with Bolyai's intrinsic descriptions, which I like, maybe it would help if you worked out things in the hyperboloid model. The nice part is that curves of infinite length also appear infinite in the hyperboloid. For example, a geodesic is just the intersection of the hyperboloid with a plane that passes through the origin. Let's see, a horocycle is the intersection of the hyperboloid with a plane at exactly $45^\circ$ from the $z$ axis (but cannot then pass through the origin).

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  • $\begingroup$ Any model where "anything approaching the "boundaries" is of infinite length." is not very enlightening and I am not very familiar with the hyperboloid model. But I am intrigued by " Bolyai's intrinsic descriptions" can you explain or give a reference? (reference is maybe beter) $\endgroup$
    – Willemien
    Commented Apr 22, 2016 at 9:50
  • $\begingroup$ @Willemien see Roberto Bonola, Non-Euclidean Geometry, which includes translations of Bolyai and Lobachevsky. Also George E. Martin, The Foundations of Geometry and the Non-Euclidean Plane. $\endgroup$
    – Will Jagy
    Commented Apr 22, 2016 at 17:19
  • $\begingroup$ @Willemien perhaps i should add that I was preparing an article on compass and straightedge constructions in the hyperbolic plane. Bolyai tended to emphasize that, and Martin's book has a very good chapter on it. zakuski.utsa.edu/~jagy/papers/Intelligencer_1995.pdf $\endgroup$
    – Will Jagy
    Commented Apr 22, 2016 at 17:33
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    $\begingroup$ Glad to help. I wanted to let the OP know that understanding this point can actually be very enlightening. $\endgroup$
    – Lee Mosher
    Commented Apr 23, 2016 at 17:33
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    $\begingroup$ A horocycle, or an apeirogon which closely tracks the horocycle, are both infinite segments in the hyperbolic plane itself, not in the circle at infinity. In the compactification, both endpoints of both of those segments approach the same point of the circle at infinity. This is not a contradiction, because the circle at infinity compactifies the hyperbolic plane in a topological sense but not in a metric sense: any path in the hyperbolic plane which approaches a point on the circle at infinity has infintie length. $\endgroup$
    – Lee Mosher
    Commented Apr 24, 2016 at 13:51

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