Let $\mathbb{H}^n$ be the hyperboloid model for hyperbolic space and $\text{Isom}(\mathbb H^n) = PO(n,1)$.

Let $\rho: \Gamma \rightarrow PO(n,1)$ be a representation of finitely generated group $\Gamma$ and $x_0 \in \mathbb H^n$. We define the orbit map given by $\tau_\rho (\gamma): = \rho (\gamma) \cdot x_0$.

Definition: We say that $\rho$ is convex cocompact iff $\Gamma$ is finitely generated and $\tau_\rho$ is a quasi-isometric embedding.

I want to prove that this definition does not depend on the choice of base point $x_0$ for the orbit map.

Since $\tau_\rho : \Gamma \to \mathbb H^n$ is a qi-embedding, we have the following: there exist $C\geq 0$ and $K \geq 1$ such that $$-C+\frac{1}{K}d_S(\gamma, \gamma') \leq d(\rho(\gamma)(x_0), \rho(\gamma')(x_0))\leq Kd_S(\gamma, \gamma')+C$$ where $d_S$ is the word metric on $\Gamma$.

I think we need to prove that $\tau_\rho$ is a qi-embedding for any $x_0 \in \mathbb H^n$. Let $y_0$ be another base point.

But I have no idea how to involve $y_0$ in the above inequalities. Could it be the right approach? Let $y_0 \in \mathbb{H}^n$, then there exists $A\in PO(n,1)$ such that $y_0 = Ax_0$.

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    $\begingroup$ As written your post is a bare problem statement with little context. But I suspect it would be easy for you to hit the edit button and add some context. For example, do you know how to write out the inequalities that define what it means for $\tau_\rho : \Gamma \to \mathbb H^n$ to be a quasi-isometric embedding? And if so, do you know how to rewrite those inequalities when $x_0$ is replaced by a different base point? And can you describe more precisely where in these steps you get stuck in your efforts? $\endgroup$
    – Lee Mosher
    Commented Jul 9 at 18:01
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    $\begingroup$ You might also wish to look at our general guidelines for providing context and for avoiding "I cannot" questions. $\endgroup$
    – Lee Mosher
    Commented Jul 9 at 18:03
  • $\begingroup$ This has nothing to do with hyperbolic space and will work for any isometric group action on any metric space. $\endgroup$ Commented Jul 9 at 19:05
  • $\begingroup$ @LeeMosher I have edited. can you please advise me about how can I rewrite the inequalities involving another base point? $\endgroup$
    – yyffds
    Commented Jul 10 at 11:18
  • $\begingroup$ Hint: Prove first that if $f: X\to Y$ is a qi embedding of metric spaces and $f': X\to Y$ is a map at finite distance from $f$, then $f'$ is also a qi embedding (you can even keep the same multiplicative qi constant). $\endgroup$ Commented Jul 10 at 14:37

1 Answer 1


The proof is a bunch of applications of the triangle inequality, and some substitutions which make use of the fact that the group acts on the metric space by isometries.

You start by writing out the inequalities that you assume to be true for all $\gamma,\gamma' \in \Gamma$. These are the inequalities you have already written out in your post, with given constants $K$ and $C$ and the given base point $x_0$.

Next, you write out the inequalities that you must prove for all $\gamma,\gamma' \in \Gamma$. These new inequalities that you must prove are obtained by making substitutions into your old inequalities. The main thing is to simply substitute the given new base point $y_0$ in place of the given old base point $x_0$. But you must also be careful with constants: you will have to discover some new constants in place of the old, given constants $K$ and $C$. Using $K'$ and $C'$ to denote the new, unknown constants we get: $$(*) \qquad -C'+\frac{1}{K'}d_S(\gamma, \gamma') \leq d(\rho(\gamma)(y_0), \rho(\gamma')(y_0))\leq K' d_S(\gamma, \gamma')+C' $$ So our job is to discover and define appropriate values of the new constants $C'$ and $K'$, and to prove the new inequalities (*).

Let's do this first with the right hand inequality of $(*)$. Simply apply the triangle inequality, and define $K'$ and $C'$ as shown: $$\begin{align*} d(\rho(\gamma)(y_0),\rho(\gamma')(x_0)) & \le \underbrace{d(\rho(\gamma)(y_0),\rho(\gamma)(x_0))}_{=d(x_0,y_0)} + \underbrace{d(\rho(\gamma)(x_0),\rho(\gamma')(x_0))}_{\le K \, d_S(\gamma,\gamma')+C} \\ &\qquad\qquad\qquad + \underbrace{d(\rho(\gamma')(x_0),d(\rho(\gamma')(y_0))}_{=d(x_0,y_0)} \\ &\le \underbrace{K}_{=K'} \, d_S(\gamma,\gamma') + \underbrace{2 \, d(x_0,y_0) + C}_{=C'} \end{align*} $$

For proving the left hand inequality of $(*)$, I had to rewrite this part a couple of times before realizing that it turns out best if you first rewrite the given left hand inequality from your post, before applying the triangle inequality: $$\begin{align*} \frac{1}{K} d_S(\gamma,\gamma') &\le d(\rho(\gamma)(x_0),\rho(\gamma')(x_0)) + C \\ &\le \underbrace{d(\rho(\gamma)(x_0),\rho(\gamma)(y_0))}_{d(x_0,y_0)} + d(\rho(\gamma)(y_0),\rho(\gamma')(y_0)) \\ & \qquad\qquad\qquad\qquad + \underbrace{d(\rho(\gamma')(y_0),\rho(\gamma')(x_0))}_{d(x_0,y_0)} + C\\ &= d(\rho(\gamma)(x_0),\rho(\gamma')(x_0)) + \underbrace{2 \, d(x_0,y_0)+C}_{= \, C'} \end{align*} $$ and we are done.

One final comment to emphasize the point made by @MoisheKahan: notice that nowhere in this whole proof did we use anything special about $\mathbb H^n$ and $\rho$, other than that $\mathbb H^n$ is a metric space and the group $\Gamma$ acts on that metric space by isometries.

  • $\begingroup$ Much much thanks. $\endgroup$
    – yyffds
    Commented Jul 10 at 16:40

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