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Do you know any proof of the fact that $\mathbb H^n$ is Rips-hyperbolic (i.e., geodesic triangles are $\delta$-slim for some $\delta$, also called "Gromov-hyperbolic" in some contexts), which makes no use of the Klein model, the ball model or the upper halfspace model for $\mathbb H^n$, but just of the hyperboloid model (i.e.: points in $\mathbb R^{n+1}$ whose Minkowski quadratic form is $-1$, with distance $d(u,v)=arcosh(-\langle u,v\rangle)$ )?

Thank you in advance.

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    $\begingroup$ If you like, you could take the proof in the upper half plane model and transfer it over the the hyperboloid model pretty much unchanged. In fact I'm pretty sure that the proof could be done abstractly, starting only from the assumption that one has a complete, simply connected Riemannian manifold of sectional curvatures $\le -1$. $\endgroup$ – Lee Mosher May 10 '16 at 22:40
  • $\begingroup$ Thank you very much. In fact, I've found that Ratcliffe (Fundaments on Hyperbolic Manifolds) uses the Klein disc model to show that the area of geodesic triangles in $\mathbb H^2$ is always $\leq\pi$, and then the proof may follow Bridson-Haefliger's hint (a ball's area tends to $\infty$ with the radius). Ratcliffe's proof remains also very close to the hyperboloid model. $\endgroup$ – W. Rether May 12 '16 at 8:19

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