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I've stumbled across a proof of geodesic stability in hyperbolic space, located in the following blog post:

https://lamington.wordpress.com/2010/05/19/hyperbolic-geometry-notes-5-mostow-rigidity/

Maybe it would be wiser to comment there directly, but I figured I would ask here first. Is the proof of lemma 5 valid? I am really not convinced of it. I know where to find a proof in the setting of $\delta$-hyperbolic spaces, but it would be nice to use this one since it's quite simple in comparison.

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  • $\begingroup$ Can you precise the point you find uncorrect? $\endgroup$ – Seirios Mar 16 '16 at 8:55
  • $\begingroup$ @Seirios I'm not sure about how they're picking the C neighborhood after realizing that the length of sigma is bounded. Or why this really proves the statement. I thought they would try to uniformly bound the image of sigma under the QI (but I guess that could be done by multiply by K and epsilon, as they mention). Also the preceding lemma isn't convincing either, but that might be a separate issue. $\endgroup$ – Hanna Mar 16 '16 at 15:31
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Let $f : \mathbb{H}^n \to \mathbb{H}^n$ be a $(k,\epsilon)$-quasi-isometry and let $[p,q]$ denote the unique geodesic between two points $p,q \in \mathbb{H}^n$.

  • Step 1: Fix some $C$ satisfying $k^2e^{-C} <1$.

  • Step 2: If $[r,s] \subset [p,q]$ is a maximal subsegment such that $f([r,s])$ is outside the $C$-neighborhood of $[p,q]$, then $$d(r,s) \leq \frac{2kC + k \epsilon + k \epsilon e^{-C}}{1-k^2 e^{-C}}.$$

  • Step 3: Let $t \in [p,q]$. If $d(f(t),[f(p),f(q)]) \leq C$ there is nothing to prove, so suppose $d(f(t),[f(p),f(q)]) > C$. Then $t$ belongs a maximal subsegment $[r,s]$ similar to the previous one. Now, $$d(f(t),[f(p),f(q)]) \leq d(f(t),f(r)) + d(f(r),[f(p),f(q)]).$$ But $d(f(r),[f(p),f(q)])=C$ and $$d(f(t),f(r)) \leq k d(t,r)+ \epsilon \leq k d(r,s) + \epsilon \leq k \cdot \frac{2kC + k \epsilon + k \epsilon e^{-C}}{1-k^2 e^{-C}} + \epsilon.$$ Therefore, $d(f(t),[f(p),f(q)])$ is bounded by a constant which depends only on $k$ and $\epsilon$.

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  • $\begingroup$ Great, thanks for clarifying this. It makes more sense now! $\endgroup$ – Hanna Mar 18 '16 at 20:25

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