Does every hyperbolic, almost-transitive, triangulation of $\mathbb{R}^n$ have boundary homeomorphic with $\mathbb{S}^{n-1}$? Question 1: Let $T$ be a triangulation of $\mathbb{R}^n$. Suppose that the 1-skeleton of $T$ endowed with the graph-metric (i.e. each 1-cell is given length 1) is Gromov-hyperbolic. Suppose moreover that $T$ is almost-transitive, i.e. its group of automorphisms $Aut(T)$ (i.e. cellular self-homeomorphisms) acts on $T$ with finitely many orbits of vertices. Must the hyperbolic boundary of $T$ be homeomorphic with $\mathbb{S}^{n-1}$?

Here is a proof sketch for $n=3$ (and $n=2$):
Notice that any element of $Aut(T)$ that fixes a 3-cell pointwise must fix its neighbouring 3-cells, hence all of $T$. Therefore, each point of $T$ has a finite stabiliser. Thus the induced action of $Aut(T)$ on $\mathbb{R}^3$ is properly discontinuous. It is also co-compact by almost-transitivity. Moreover, it is smoothable. Apply the Geometrization Theorem to the quotient 3-orbifold. It seems to me (experts please correct me) that it follows that $Aut(T)$ acts (still properly discontinuously and co-compactly) by isometries on $\mathbb{H}^3$ (see From topological actions on $\mathbb{R}^3$ to isometric actions). By the Svarc-Milnor Lemma, $Aut(T)$ is quasi-isometric with $\mathbb{H}^3$, and also with $T$. Thus the hyperbolic boundary of $T$ coincides with that of $\mathbb{H}^3$, i.e. with $\mathbb{S}^{2}$.
(I've been a bit sloppy about the definition of $Aut(T)$; we want to consider a subgroup each element of which is determined by its action on the vertices of $T$.)

How about $n=4$ or higher? One way to attack this question is via the next ones:
Question 2: Is there a finite list $\mathcal{L}$ of metric spaces, such that any fundamental group $G$ of an aspherical, compact 4-manifold (without boundary) is quasi-isometric with an element of $\mathcal{L}$? Restrict to 1-ended $G$ if it helps.  (Edited to add the aspherical hypothesis.)
Question 3: If $G$ is as in Question 2, and in addition 1-ended and hyperbolic, must it be quasi-isometric with $\mathbb{H}^4$?
 A: Question 1 is interesting but, I'll guess, very difficult?
The answer to question 2 is "no".  This is because any finitely presented group appears as the fundamental group of some compact four-manifold (without boundary).  I suppose that I also need to display an infinite collection of quasi-isometry types of finitely presented groups so... $\mathbb{Z}^n$.
The answer to question 3 is "no".  Consider $S^2 \times S_2$, the two-sphere crossed with the surface of genus two.

Perhaps you want to add the hypothesis that the four-manifold is aspherical: that is, the four-manifold is a $K(G, 1)$.  I suspect that the answers to 2 and 3 will remain "no"?

As Igor points out (in the comments below) there are (torsion free) uniform lattices in the complex hyperbolic plane.  The resulting quotients of the complex hyperbolic plane give compact four-manifolds without boundary which have word-hyperbolic fundamental groups.  However, as Igor also points out, the complex hyperbolic plane is not quasi-isometric to real hyperbolic four-space.  So the answer to 3 is "no" even with the aspherical assumption.
