MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $(X,d)$ be a unique geodesic metric space and $Y\subseteq X$ a subset, equipped with the induced path metric $d'$. We say that the inclusion $i\colon Y\hookrightarrow X$ is uniformly proper if $\forall A>0\ \exists B>0$ such that $d(x,y)< A$ implies $d'(x,y)< B$ for every $x,y\in Y$. Now consider the case of a compact hyperbolic surface $S$ uniformly properly embedded in a hyperbolic $3$-manifold $N$ with isomorphic fundamental group $G=\pi_1(S)\cong\pi_1(N)$. I'd like to prove that in this situation the inclusion $j\colon \tilde{S}\to\tilde{N}$ between the correspond universal covers is still uniformly proper. This is my idea: since $S$ is compact, then $G$ is finitely generated and its Cayley graph may be embedded in $\mathbb{H}^2$ via a map $\varphi\colon \mathcal{G}_G\to\mathbb{H}^2$ such that $\mathcal{G}_G(G)$ is quasi-isometric to $\mathbb{H}^2$. Now, if $S$ were not uniformly properly embedded in $N$, there would exist a positive constant $A$ and a sequence $(x_i,y_i)_i\subseteq S\times S$ such that $d_N (x_i,y_i)< A$ but $d_S (x_i,y_i)\to\infty$. Thus, in particular, there exists a sequence $( g_i, h_i ) \subseteq G \times G $ such that $d_{\mathbb{H}^2}\(g_i,h_i\) \to \infty$, i.e. for every $\gamma\in G$ there exists a sequence $k^\gamma_i\subseteq G$, where $k^\gamma_i=\gamma g_i^{-1}h_i$ such that $k^\gamma_i$ eventually exits any bounded ball centred at $\gamma$. What I'd like to do is to use this argument to prove my claim, but I don't understand how this should work in the opposite direction. Suppose that $\tilde{S}\hookrightarrow\tilde{N}$ is not uniformly proper. Then there exists a sequence of points such that the distance in $\tilde{S}$ goes to infinity whereas it is bounded in $\tilde{N}$. And then? How can I say that this property actually passes to the quotient? I think there should be a way of using an argument similar to the one above... Thanks in advance!

share|cite|improve this question
Eek! Fix the TeX errors! – Lee Mosher May 21 '12 at 14:58
That's what I'm trying to do but for some strange reason in the preview everything is shown correctly... I'll try again! Sorry! – Damiano Lupi May 21 '12 at 15:01
OK... done... sorry! – Damiano Lupi May 21 '12 at 15:04
First, the hypothesis of "the case" that you consider is trivially true: every map from a compact metric space to another metric space is uniformly proper. Second, the set of pairs $(x,y) \in \tilde S$ with $d_S(x,y) \le A$ is cocompact under the diagonal action of $\pi_1(S)$. So $max d_{H^2}(x,y)$ on this set is bounded. – Lee Mosher May 21 '12 at 15:27
I understand why the hypothesis is trivially true in this case, but thanks for pointing it out. I'm still confused about the second part: why do we have to consider the set of pairs with $d_S(x,y)< A$? I mean: the inequality holds when we look at the distance in the bigger space, so in $\tilde{N}$, and we are assuming that it doesn't hold in $\tilde{S}$, by contradiction. Perhaps, using your hint, we may say the following: if $\tilde{S}\hookrightarrow \tilde{N}$ were not uniformly proper, than we would have this sequence of pairs of points $(x_i,y_i)$ in $\tilde{S}$ such that their... – Damiano Lupi May 21 '12 at 16:10

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.