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I was trying to show in $\mathbb{H}$ -Poincaré half-plane model- that hypercycles defined for a given line $l$ and a given $a$ in $\mathbb{R}$ as $$H(l)=\{z \in \mathbb{H} : d_{\mathbb{H}}(z,l)=a \} $$ are not lines.

To prove something I tried to argue that given that $Mob(\mathbb{H})$ acts transitvely on the lines of $\mathbb{H}$ I can find the hypercycles of the imaginary axis and via the $\gamma \in Mob(\mathbb{H})$ that maps my line in the imaginary axis i can state $$H(l)=\gamma^{-1}(H(\gamma(l))$$

Where I'm using the fact that a Möbius transformations preserves distances. At this point I defined [already here i have doubts] $d_{\mathbb{H}}(z,l)$ for a given $z \in \mathbb{H}$ and a line $l$ as the inf in k of the distances $d_{\mathbb{H}}(z,w)$ with $w \in l$.

Given this and the general assumption $z=a+bi$ and $w=ki$ i'm left with trying to calculate the inf of $$arccosh(1+\frac{|a+(b-k)i|^{2}}{2bk})$$

EDITED FROM HERE TO SHOW MORE WORK:

given that arccosh is a monotone function i need to find the inf of $$\frac{a^{2}+{b}^{2}}{2bk}+\frac{k}{2b}-1$$

taking the derivative I'm looking to the zeroes of:

$$\frac{-2a^{2}-2b^{2}+k^{2}b}{2bk^{2}}$$

solving and remembering that i need $k>0$ i get $k=(\frac{2(a^{2}+b^{2})}{b})^{\frac{1}{2}}$

looking now for the $a+bi$ such that for a given $p$ we have: $$ d_{\mathbb{H}}(a+bi,(\frac{2(a^{2}+b^{2})}{b})^{\frac{1}{2}}i)=p $$

expanding this i got an expression of $z,z^{*}$ that is not a standard form for a line in the plane and it should end the exercise.

Is any of this correct? There is any "distance from a given point to a given line" standard formula in hyperbolic geometry? Thanks for any help and hint.

Augusto Matteini
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  • It is even simpler than this: Hypercycles are never connected, but lines are. – Moishe Kohan Jun 23 '21 at 13:59
  • Ok, the fact that arccosh should help in my tractation. About connection you mean as they are symmetric in respect to my line? – Augusto Matteini Jun 23 '21 at 14:01
  • Do you know what this means for a topological space to be connected? – Moishe Kohan Jun 23 '21 at 14:03
  • Yes but I think in this case is not a "real" answer. In the real cases i have two parallel lines, they are not connected but are "lines". I think the question should be answered with this pov. – Augusto Matteini Jun 23 '21 at 14:05
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    Then you should update your question asking for a proof that the two connected components of the hypercycles are non-geodesic (provided, of course, that $a>0$). And in this case the real answer lies in the description of hyperbolic geodesics in the upper half-plane model, from which it follows that any two points in the boundary circle lie in the closure of a unique geodesic, from what it follows that components of a hypercycle are never geodesic. – Moishe Kohan Jun 23 '21 at 14:48
  • I showed quite a bit of my work... – Augusto Matteini Jun 23 '21 at 15:21
  • If $a$ is a positive real constant, $f(z) = a z$ is also a Mobius transformation preserving the upper half plane and preserving distances. – Will Jagy Jun 23 '21 at 16:33
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    On that note, what is the distance of $3+4i$ from the imaginary axis? – Will Jagy Jun 23 '21 at 16:34
  • According to my computation it should be $arccos(1+\frac{3+16+25/2-sqrt(2)10}{sqrt(2)20})$. Besides it not looking good I don't understand your observation about f(z)=az, I'm sorry... – Augusto Matteini Jun 23 '21 at 17:04
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    Your formulation of the question suggests that you want an analytic answer (using calculus/analysis). But would you be open to a synthetic answer (using geometry/symmetry)? – Lee Mosher Jun 23 '21 at 17:50
  • I would like to know if my work is wrong or could be recovered but for sure any way to attach the problem is welcomed – Augusto Matteini Jun 23 '21 at 21:06
  • I did not check your computations but, from what is written: (1) your computations are insufficient for the proof, (2) there is a much more direct argument that I sketched above, which is based on the description of a hypercycle as the union of two disjoint circular arcs. – Moishe Kohan Jun 24 '21 at 15:48

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I will now answer to the questions, leaving my work in the original post to show my mistakes.

My $k$ was wrong, and the right calculations shows that the inf is taken in $$k=\sqrt{a^{2}+b^{2}}$$ Now I'm left trying do describe the locus of points in the form $a+bi$ such that for a given $p$ $$d_{\mathbb{H}}(a+bi,(\sqrt{a^{2}+b^{2}})i)=p$$ Now using the Poincare half-plane metric i get to show that $\frac{|z|}{Im z}=p$. Showing that my hypercycle is defined given an angle and it is an euclidean line passign through the origin and the given point, clearly not an hyperbolic line.

Augusto Matteini
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