2
$\begingroup$

I'm trying to study Hyperbolic geometry, but I can not understand the following statement.

Let $X$ be a $δ$-hyperbolic space. Then, there exists $M > 0$ such that for any geodesic $γ$, and any ball $B_r(x)$ of arbitrary radius that is disjoint from $γ$, the closest-point projection of the ball to the geodesic has diameter $≤ M$.

I want to know the proof of the question. (Maybe $M = 5δ$)

If anyone could help, It would be really appreciated.Thanks.

$\endgroup$
1
$\begingroup$

Hint: Notice that the statement is easy to prove when $X$ is a real tree. Therefore, suppose the general statement fails in some hyperbolic space $X$ and take a sequence of balls $B_n$ and geodesics $\gamma_n$ such that the projection of $B_n$ on $\gamma_n$ has diameter at least $n$; then, deduce that your statement fails in some ultralimit $\left( \frac{1}{n} X , y_n \right) \to (Y,y)$. Because $Y$ is a real tree, you get a contradiction.

EDIT: In Géométrie et théorie des groupes, les groupes hyperboliques de Gromov (proposition 10.2.1 page 108), a more general statement is proved using a quantitative approximation of finite subspaces by simplicial trees. In particular, if $B_r$ is ball at distance at least $r$ from a geodesic $\gamma$, then the projection of $B_r$ on $\gamma$ has diameter at most $12 \delta$.

$\endgroup$
  • $\begingroup$ Thank you for answering. I can understand that the statement is true if X is a real tree. But, what's n and y_n ? Is n a radius of B ? $\endgroup$ – Tai Nov 23 '14 at 11:04
  • $\begingroup$ And why the lim X(δ-hyperbolic)/n = Y(a real tree)? $\endgroup$ – Tai Nov 23 '14 at 11:10
  • $\begingroup$ $n$ is just the index of the sequence; it corresponds also to a lower bound on the diameter of the projection of $B_n$ on $\gamma_n$, by definition. Then, $Y$ is a real tree because it is a $0$-hyperbolic geodesic space. Of course, you have to choose $y_n$ carefully, but I just gave a hint. $\endgroup$ – Seirios Nov 23 '14 at 13:09
  • 1
    $\begingroup$ I found a more quantitative solution in the book Géométrie et théorie des groupes, les groupes hyperboliques de Gromov. $\endgroup$ – Seirios Nov 24 '14 at 8:29
  • 1
    $\begingroup$ Yes, it is the main idea. But the conclusion does not immediately follow, further work is needed. $\endgroup$ – Seirios Feb 4 '15 at 19:57

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.