Does every hyperbolic, almost-transitive, triangulation of $\mathbb{R}^n$ have boundary homeomorphic with $\mathbb{S}^{n-1}$?

Question

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 $\mathbb{H}^4$?

Answer 1

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.