4
$\begingroup$

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 $\mathbb{H}^4$?

$\endgroup$

1 Answer 1

3
$\begingroup$

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.

$\endgroup$
4
  • $\begingroup$ I added the asphericity condition. Suggestions for additional conditions are welcome, as long as the answer may be helpful towards Question 1. $\endgroup$
    – Agelos
    May 20, 2022 at 15:39
  • 3
    $\begingroup$ As you wrote before the edit, uniform complex hyperbolic lattices aren't qi to real hyperbolic ones so the answer to Q3 is no. This follows from qi rigidity of rank one symmetric spaces. There are other examples, e.g. fundamental groups of some Gromov-Thurston closed negatively curved 4-manifolds are not qi to the real hyperbolic space. $\endgroup$ May 20, 2022 at 15:52
  • $\begingroup$ I could not find a reference for the existence of uniform examples! I’ll add it back. $\endgroup$
    – Sam Nead
    May 20, 2022 at 16:44
  • 3
    $\begingroup$ Examples of uniform complex hyperbolic lattices? It is in Borel's Compact Clifford-Klein forms of symmetric spaces, Topology 1963, sciencedirect.com/science/article/pii/0040938363900260. And of course one needs qi rigidity for the complex hyperbolic space. $\endgroup$ May 20, 2022 at 17:34

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.