Let $G$ be a finitely generated group and $H$ a hyperbolic group in the sense of Gromov, with a fixed word metric $d$.
Let $\varphi : G\to H$ be a quasi-isometry. Does there exist uniform constant $C> 0$ such that for any $g,h \in G$, $$d(\varphi(g)\varphi(h), \varphi(gh)) \leq C? \ \ \ \ \ \ \ \ \ \ \ \ \ (*)$$
Discussion:
It is not too hard to see that there are quasi-isometries of non-hyperbolic spaces which do not coarsely preserve the group operation. For example, one can take the map $f : \mathbb{R}^2 \to \mathbb{R}^2$, $(r, \theta) \mapsto (r, \log(1+r))$, where $(r,\theta)$ are polar coordinates on $\mathbb{R}^2$ and $(\mathbb{R}^2, +)$ is considered as the additive group.
On the other hand, the Morse Lemma states that any quasi-geodesic in a hyperbolic space is a uniform distance away from a geodesic connecting its endpoints ("uniform" for a given choice of quasi-isometry constants and the hyperbolicity constant). Again, our example above shows that this fails badly in non-hyperbolic spaces.
Questions:
Is Morse Lemma enough to coarsely preserve the group operation under quasi-isometries? Or is there an insightful example of a quasi-isometry of hyperbolic groups where $(*)$ fails miserably?
Is there a natural condition that one can put on groups $G,H$ which would force $(*)$ to hold? (Short of something trivial like requiring one – and thus both – of $G$ and $H$ to be finite.)
Best Answer
Here is a class of examples of infinite hyperbolic groups $G$ such that every self-quasi-isometry $G\to G$ is at finite distance from a left-translation by an element of $G$.
Start with a compact connected $n$-dimensional manifold $M$ with nonempty boundary which admits a hyperbolic metric such that the boundary is totally-geodesic. Then the universal cover $\tilde{M}$ of $M$ is isometric to a certain convex subset $C\subset {\mathbb H}^n$ in the hyperbolic $n$-space, $n\ge 3$. The fundamental group $\pi(M)$ of $M$ acts on $C$ isometrically as the group of covering transformations $\Pi$ for $\tilde{M}\to M$. Let $G$ denote the maximal subgroup of isometries of ${\mathbb H}^n$ preserving $C$. Since $C$ has infinitely many boundary components, the subgroup $G$ is discrete; it acts on $C$ cocompactly since $\Pi< G$. Thus, $G$ is a hyperbolic group. As it turns out, every quasi-isometry $C\to C$ is within finite distance from an element of $G$, see
FrÃgerio, Roberto, Commensurability of hyperbolic manifolds with geodesic boundary, Geom. Dedicata 118, 105-131 (2006). ZBL1096.32014.
Thus, $G$ satisfies the required property.
As for the original question, the answer is negative even when $G=H$, unless $G$ is finite. I'll prove it only for nonelementary hyperbolic groups $G$. Let $x\in G$ be an element of infinite order and $L_x: G\to G$ be the left translation of $G$ by $x$ ($L_x(y)=xy$). Then $L_x$ is an isometry of $G$. However, $L_x$ does not satisfy (*) for any constant $C$. Indeed, apply $(*)$ with $h=g^{-1}$, then you get $$ d(gxg^{-1}, 1) = d(xgxg^{-1}, x)=d(xgxg^{-1}, xgg^{-1})\le C $$ for all $g\in G$. But if $x, g$ do not commute and $g$ has infinite order, then $$ \lim_{n\to\infty} d(g^nxg^{-n}, 1)=\infty. $$
As for your "secret question", it sounds very difficult. I am sure it has negative answer (in general) but I do not know how to construct an example.