1
$\begingroup$

Note: originally posted on math.SE.

I'm looking at the distance that's defined in this paper on Poincaré Embeddings:

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

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 but that also seemed to be a dead end. Am I missing something obvious?

$\endgroup$
  • 3
    $\begingroup$ While this paper uses the Poincaré ball model and is even titled after it, they really should be using the Minkowski hyperboloid model instead (and Poincaré only for presentation). I suppose they chose the Poincaré ball model because they did not know much about computational hyperbolic geometry. See the followup paper by the same authors, it changes to the Minkowski hyperboloid model, and it works much better. Also it is much easier to understand and to work with. $\endgroup$ – Zeno Rogue Mar 4 at 11:53
  • $\begingroup$ Interesting ... thanks. Which paper are you referring to? $\endgroup$ – tdc Mar 4 at 11:58
  • 2
    $\begingroup$ This one: arxiv.org/pdf/1806.03417.pdf (they call it the Lorentz model, but it is more commonly called the hyperboloid model). $\endgroup$ – Zeno Rogue Mar 4 at 12:02
  • 1
    $\begingroup$ Agreed! Much simpler ... $\endgroup$ – tdc Mar 4 at 23:28
  • $\begingroup$ Also, it's good to have some context around hyperbolic space. The formula for distance follows from more geometrically intuitive descriptions of geodesics and Riemannian metric. $\endgroup$ – Neal Mar 6 at 3:12
2
$\begingroup$

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:

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.