2
$\begingroup$

I am looking for a reference to a proof of the following well-know fact (cited for example by B.Farb in ``Relatively hyperbolic groups'', Geom. Funct. Anal. 8 (1998), no. 5, 810--840); MR1650094, DOI:10.1007/s000390050075.

Suppose $X$ is the universal covering of a negatively curved Riemannian manifold, let $O$ be an open horoball in $X$ and let $H=\partial O$ be the horospherical boundary of $O$. Also suppose that $\gamma\colon [0,1]\to X\setminus O$ is a rectifiable path such that $d(\gamma (t), H)\geq k>0$ for every $t\in [0,1]$, and let $\pi\colon X\setminus O\to H$ the (well-defined) nearest-point projection. Then, there exists $\alpha>0$ (only depending on the curvature of $X$) such that the length $L(\pi\circ\gamma)$ of $\pi\circ\gamma$ is bounded above by $e^{-\alpha k} L(\gamma)$.

Of course, this fact can be reduced to the computation of the Lipschitz constant of the projection of a horosphere onto another horosphere having the same basepoint. When $X$ is the real hyperbolic $n$-space, such a computation is very easy, and it is likely that the variable curvature case can be reduced to the hyperbolic case via some comparison theorem. However, I was wondering if there is some standard reference I could rely on.

$\endgroup$
  • $\begingroup$ I would write "Applying the comparison for triangle one which vertex running to infinity, we get ..." $\endgroup$ – Anton Petrunin May 31 '10 at 18:15
  • $\begingroup$ Yes, probably it is not too difficult to make such an argument work. Anyway, a little issue arises since one egde of the comparison triangles involved is not geodesic, but lies on a horosphere... $\endgroup$ – Roberto Frigerio Jun 1 '10 at 7:50
  • $\begingroup$ Dear Anton, on second thoughts I think your approach can easily lead to a solution. Even if the edge staying far from the infinity is not geodesic, one can approximate $\gamma$ and $\pi\circ\gamma$ with suitable ``polygonal'' paths approximating the length, then use your argument on the small segments, and finally put the estimates together. $\endgroup$ – Roberto Frigerio Jun 7 '10 at 12:34
6
$\begingroup$

Try Geometry of horospheres by Heintze and Im Hof.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ Thank you for the reference. In the meantime, a student of mine has written down a detailed proof using comparison theorems and looking at how one can pass to the limit when one vertex tends to infinity. This approach has the advantage that it works is general for CAT(-1) spaces. $\endgroup$ – Roberto Frigerio Jun 15 '10 at 18:07
  • $\begingroup$ Heintze-Im Hof's paper certainly treats similar issues but does it contain an answer to your specific question? Where exactly? You may find relevant the paper of Bowditch "Geometric finiteness with variable negative curvature" where in section 2 he discusses similar issues, but again not your specific question. $\endgroup$ – Igor Belegradek Jun 15 '10 at 18:51
  • $\begingroup$ It seems that the link intlpress.com/JDG/archive/1977/12-4-481.pdf is not working (at least not for me). Here are some other links: projecteuclid.org/euclid.jdg/1214434219 dx.doi.org/10.4310/jdg/1214434219 mathscinet.ams.org/mathscinet-getitem?mr=512919 $\endgroup$ – Martin Sleziak Dec 1 '17 at 13:56
1
$\begingroup$

I think you can find this in Chapter II.8 of Bridson and Haefliger.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ I gave a look therebut I didn't find the required estimate. It seems to me that in II.8 Bridson and Haefliger treat the general case of CAT(0) metric spaces, while I would need a result about CAT(-1) metric spaces. $\endgroup$ – Roberto Frigerio May 31 '10 at 16:57
  • $\begingroup$ I thought it followed from what they do there, possibly throwing in strict convexity of the distance function. Perhaps I'm mistaken. $\endgroup$ – Matthew Stover May 31 '10 at 17:54
1
$\begingroup$

I am now also reading Farb´s article your cited, and I would like to propose an answer of mine for your question and help that we can discuss it.

Let $\gamma$ be a geodesic such that $d(γ(t),H)≥k>0$ for every $t∈[0,1] $ where H is a horosphere. Let $p = \gamma(0), q = \gamma(1)$, then length of $\gamma$,that is $L(\gamma)$, equals to $d(p,q)$. Let $p^\prime, q^\prime$ be the projection points of p,q respectively on horosphere. Suppose that $d(p,p^\prime)=k, d(q,q^\prime) \ge k$. Then choose a point $m$ in geodesic connecting $q,q^\prime$ so that $d(q^\prime,m)=k$ and connect $m$ to p by geodesic by curve $\delta$ which constructed by push geodesic $\gamma$ along geodesic with distance k.

By Proposition 4.1 in Farb´s``Relatively hyperbolic groups'', Geom. Funct. Anal. 8 (1998), no. 5, 810--840, $d(p^\prime,q^\prime) \le L(\delta ) \times e^{-ak}$. On the other hand, $L(\delta) \le L(\gamma) $ which can prove use the definition of the length and the fact that $\gamma$ is a geodesic.(I don´t know whether $\delta$ in this case is a geodesic, but it not affect the result).

| cite | improve this answer | |
$\endgroup$

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.