Let $p$ be a point in the $2$-dimensional hyperbolic space $H_2$. Consider a normal coordinate system $(x,y)$ at centered at $p$. Let $o_x\in H_2$ be the point of coordinate $(x,0)$ and let $C_x$ be the hyperbolic circle centered in $o$ passing through $p$.

I have difficulties with the $2$ following questions:

  • Given a point $q$ which has coordinates $(x_q>0,y_q)$, is there always a $x$ such that $q$ is in the interior of $C_x$? It is true in Euclidean geometry, I am having doubts for hyperbolic geometry...
  • In case it is not true, what it the curve $f(y)$ such that points $(x\leq f(y),y)$ are not in the interior of any $C_{x'}$ while $(x>f(y),y)$ are for some $x'$?

Thank you for your help.

  • 2
    $\begingroup$ I am having a little trouble parsing your question, but I think it belongs on MSE not MO. To answer the first part, consider the Poincare disk model D={(x,y)| x^2+y^2 < 1} ds_H^2 = ds_E^2/(1-r^2)^2, and look at circles through the origin with centers on the positive part of the axis. Notice here, Euclidean circles model hyperbolic circles (but the center of the hyperbolic circle is closer to the boundary than the center of the Euclidean circle). However, Euclidean circles can't have radius more than 1/2 and go through the origin. So consider points in set S= {(x,y)| x>0, (x-1/2)^2+y^2 >1/4}. $\endgroup$ Oct 20, 2021 at 14:15
  • $\begingroup$ @NeilHoffman, thank you very much, this is what I was looking for. I didn't think of seeing the problem like this. $\endgroup$
    – Chevallier
    Oct 20, 2021 at 14:54

1 Answer 1


Your question is actually best reformulated in terms of horospheres and horoballs. Indeed, the set of points described in normal coordinates as $\{(x,0), x\ge 0\}$ is precisely a geodesic ray $\gamma$ in the hyperbolic plane issued from the point $p$. Then the union of the interiors of all your $C_x$ is precisely the horoball in the hyperbolic plane centered at the limit point $\gamma_\infty\in\partial \mathbf H^2$ of the ray $\gamma$ and passing through the point $p$.

Now it is more convenient (at least for me) to use the upper hlaf space model, and take $p=i$ and $\gamma$ to be the vertical geodesic ray directed upwards, so that $\gamma_\infty=\infty$. Then the corresponding horosphere (or rather the horocycle, as its dimension is 1) is just the horizontal line $y=1$, and it is clear that even if the directing vector of a geodesic issued from $p$ has a positive vertical component, then the resulting endpoint still may very well end up to be below the horosphere $y=1$. Therefore, the answer to your first question is "no".

As for your second question, by applying some elementary hyperbolic geometry one can explicitly describe all the tangent vectors at the point $p=i$ (I continue talking about the upper half space model) such that the endpoints of the corresponding geodesic are within the horoball centered at $\infty$ and passing through $p$. Let $(x,y)$ be the coordinates of the tangent vector (since I am using the upper half space model, $x$ and $y$ are swapped here compared to your original formulation), and let $\gamma$ be the geodesic issued from $p$ in the direction $(x,y)$. Then the condition is that $\gamma(d)$ is inside the horoball (i.e., its vertical coordinate is $>1$), where $d=\sqrt{x^2+y^2}$ is the length of the tangent vector. In other words, $d$ should be less than the hyperbolic distance between $p$ and the point $p'$ where $\gamma$ intersects the horosphere $\{y=1\}$. The geodesic $\gamma$ is an arc of the circle centered at $(\frac{y}x, 0)$, whence $p'=(2\frac{y}x,1)$, so that $$ d(p,p') = arcosh \bigl( 1+ 2 (\tfrac{y}x)^2 \bigr) $$ (e.g., see). Thus, the condition is $$ \sqrt{x^2+y^2} > arcosh \bigl( 1+ 2 (\tfrac{y}x)^2 \bigr) $$ or $$ \cosh \sqrt{x^2+y^2} > \bigl( 1+ 2 (\tfrac{y}x)^2 \;. $$

You are asking whether, for a given $x$ there is a threshold value $y_0=y_0(x)$ separating "good" and "bad" point (once again, I remind, that $x$ and $y$ are swapped compared to your original question). I am not sure this is true though. For instance, for $x=1$ the corresponding equation seems to have two roots.


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