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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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    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 ! – 2017-01-18
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    How could I use this Cartan involution? – 2017-01-19
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    You compute its fixed point. – 2017-01-19
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    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.. – 2017-01-20
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    No you compute the unique Cartan involution wich intertwines $p_1, p_2$, then its fixed point. – 2017-01-20
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    More precisely given a point $p$ compute $\sigma _p$ in terms of the coordiantes of $p$, and write $\sigma _p p_1=p_2$ – 2017-01-20
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    I think I understand now what you're saying, but I'm not sure this procedure can be turned in a closed formula to find the coordinates of $m$, which is what I need – 2017-02-10
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    Ti this end, the best model is the Klein model. In the Poincaré model, or you can compute $\sigma _{p_0}$ for some $p_0$ (eg $(0,0,1)$) and use a conjugation by any isometry such that $p=gp_0$ (eg compose a homothety by $c$ if $p=(a,b,c)$ $(cx,cy,cz)$ by a translation by $(a,b,0,)$). – 2017-02-10

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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 $$\frac{p+q}{\sqrt{}}.$$ 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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    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? – 2017-02-10
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    @user00169: See the link in the edited answer. – 2017-02-10
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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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    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 $p_1p_2$? – 2017-01-19
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    @user00169 $k=(0,0,1)$. It is just an euclidean easy way to find the center of the circle. – 2017-01-20
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    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) – 2017-02-10