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Consider the hyperbolic 3-space $(\mathbb{H}^3,ds^2)$ with

$$\mathbb{H}^3:=\{(x,y,z)\in\mathbb{R}^3|z>0\}, \quad ds^2=\frac{dx^2+dy^2+dz^2}{z^2}$$

Geodesics for this space are circular arcs normal to $\{z=0\}$ and vertical rays normal to $\{z=0\}$.

Given any two points, for example $p1=(2,-1,3),p_2=(1,2,4)$ (I've chosen them so they don't lie on a vertical ray) is it possible to obtain a closed formula for the coordinates of the midpoint $m$ of $p_1$ and $p_2$? (i.e. the point $m$ is such that $d(m,p_1)=d(m,p_2)=d(p_1,p_2)/2$ where $d$ is the metric induced by $ds^2$)

I just can think of this method: given $p_1$ and $p_2$ find the plane $H$ orthogonal to $\{z=0\}$ and which contains both $p_1$ and $p_2$. Then in $H$ find the circular arc $g:[0,d(p_1,p_2)]\rightarrow H$ orthogonal to $\{z=0\}$ with $g(0)=p_1$, $g(1)=p_2$ and $|\dot g(t)|_{ds^2}=1$. Finally $m=g(d(p_1,p_2)/2)$.

But my method is rather long and complicated. Do you know of a better one which could be written directly as a formula? I'm interested in the case of $(\mathbb{H}^3,ds^2)$ to find a method which could be generalized for other riemannian and metric spaces.

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  • $\begingroup$ You can use the fact that there exists a unique isometry of $H^3$ which is an involution with one fixed point : the middle of $p_1,p_2$. Computing the (Cartan) involution whch fixes a point $p=(x,y,z)$ is quite a nice exercice ! $\endgroup$ – Thomas Jan 18 '17 at 9:16
  • $\begingroup$ How could I use this Cartan involution? $\endgroup$ – user00169 Jan 19 '17 at 16:48
  • $\begingroup$ You compute its fixed point. $\endgroup$ – Thomas Jan 19 '17 at 21:36
  • $\begingroup$ I don't understand... I should compute the unique isometry of $H^3$ which fixes the middlepoint $m$ of $p_1$ and $p_2$ and then compute its fixed point? It will be $m$ of course. The point is that I don't know $m$ in the first place.. $\endgroup$ – user00169 Jan 20 '17 at 11:59
  • $\begingroup$ No you compute the unique Cartan involution wich intertwines $p_1, p_2$, then its fixed point. $\endgroup$ – Thomas Jan 20 '17 at 16:16
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A clean way to do this is to work with the hyperboloid model of the hyperbolic space where $H^n$. Then, given two points $p, q$ in the (upper) hyperboloid $H^n\subset R^{1,n}$, their midpoint is $$\frac{p+q}{\sqrt{<p+q, p+q>}}.$$ If you know how to go back and forth between the upper half space and the hyperboloid model, you can translate this formula into the upper half space model. (I think it will not be pretty.) See also this question.

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  • $\begingroup$ thank you, this seems what I need. May I ask you how to go from the upper half space model to the hyperboloid model? Or maybe a reference to it? $\endgroup$ – user00169 Feb 10 '17 at 22:27
  • $\begingroup$ @user00169: See the link in the edited answer. $\endgroup$ – Moishe Kohan Feb 10 '17 at 22:29
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Hint

  • Assume that the geodesic passing through the given points $p_1$ and $p_2$ is the circle with center at $p=(a,b,0)$ and radius $r$. Then
    $$(x-a)^2+(y-b)^2+z^2=r^2$$ The vectors $u=pp_1$ and $v=pp_2$ satisfy $$(u\times v).k=0$$ Using these two equations, you can evaluate $a,b$ and $r$. I think(I am not sure) the rest is easy.
  • I remembered that the Euclidean bisector of the segment $p_1p_2$ intersects the plane $z=0$ at the center of the required geodesic.
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  • $\begingroup$ what is $k$ in the first point of your answer? I suspect it's $k=p_1p_2$. Also, what do you mean by your second point? What is the Euclidean bisector of the segment $p_1p_2$? $\endgroup$ – user00169 Jan 19 '17 at 17:09
  • $\begingroup$ @user00169 $k=(0,0,1)$. It is just an euclidean easy way to find the center of the circle. $\endgroup$ – Semsem Jan 20 '17 at 13:48
  • $\begingroup$ this is a good way to find the equation of the arc of the circle between the two points, but it doesn't really tell me how to find the mid point (the difficult part is finding the arc parametrization) $\endgroup$ – user00169 Feb 10 '17 at 18:32

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