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Note: cross-posted to mathoverflow.net

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?

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    $\begingroup$ I don't understand. If $u=v$ then the hyperbolic distance is zero and $f$ equal $1$ $\endgroup$ – user84976 Mar 2 at 23:48
  • $\begingroup$ sorry should have been the original function $d$ - have updated the question $\endgroup$ – tdc Mar 3 at 9:43

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