*Note: originally posted on [math.SE](https://math.stackexchange.com/questions/3132992/valid-metric-on-a-hyperbolic-space).*


I'm looking at the distance that's defined in [this paper on Poincaré Embeddings](https://papers.nips.cc/paper/7213-poincare-embeddings-for-learning-hierarchical-representations.pdf):


$d(\mathbf{u}, \mathbf{v}) = \operatorname{arccosh} \left(1 + 2\frac{\left\| \mathbf{u} - \mathbf{v} \right\|^2}{(1 - \left\| \mathbf{u} \right\|^2)(1 - \left\| \mathbf{v} \right\|^2)} \right)$


for $\mathbf{u}, \mathbf{v} \in \mathcal{B}^d$ using the Poincaré ball model of hyperbolic space.


Does this define a valid metric? It's easy to see that 


* $d(\mathbf{u}, \mathbf{v}) \geq 0$ since $\left\| \mathbf{u} \right\| \leq 1$ and $\left\| \mathbf{v} \right\| \leq 1$ and so the rhs inside the parentheses is positive ($\operatorname{arccosh}(x)$ is an increasing function for $x \geq 0$),
* $\mathbf{u} = \mathbf{v}$, then $d(\mathbf{u}, \mathbf{v}) = 0$, and
* $d(\mathbf{u}, \mathbf{v}) = d(\mathbf{v}, \mathbf{u})$


However I'm having a hard time trying to prove the triangle inequality in this case. I've tried using the logarithmic form but that didn't get me anywhere. I also found the identity ${\displaystyle \operatorname {arcosh} u\pm \operatorname {arcosh} v=\operatorname {arcosh} \left(uv\pm {\sqrt {(u^{2}-1)(v^{2}-1)}}\right)}$ under "Addition Formulae (sic.)" [on Wikipedia](https://en.wikipedia.org/wiki/Inverse_hyperbolic_functions#Inverse_hyperbolic_cosine) but that also seemed to be a dead end. Am I missing something obvious?



Valid metric on a hyperbolic space
I'm not sure about directly proving the triangle inequality, but it *does* follow from the fact that the distance metric is induced by a Riemannian metric.


**In general a Riemannian metric induces a distance metric (necessarily satisfying the triangle inequality)**


See for example: <https://en.wikipedia.org/wiki/Riemannian_manifold#Riemannian_manifolds_as_metric_spaces_2>


In the case of Hyperbolic space (for which the Poincaré disc is a model), given any two points, there is a unique curve joining the two points with length realising the distance (a length minimising geodesic). One way to get the formula you gave is to work out equations for geodesics and then calculate their length. But as in the comments, this is probably easiest in the single-sheeted hyperboloid model. To get the disc model formula, use the explicit isometry between the hyperboloid model and the disc model. I think it should be do-able directly in the disc model as well but I haven't tried.


For details, see the following references for the Poincaré disc model, hyperbolic geometry and the relation between various models:


* <https://en.wikipedia.org/wiki/Hyperbolic_geometry>
* <https://en.wikipedia.org/wiki/Poincar%C3%A9_metric>
* <https://en.wikipedia.org/wiki/Poincar%C3%A9_disk_model>


