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I am studying (on my own) some random group theory, and using this primer. The book focuses on finitely presented groups, and the main definition of a hyperbolic group there is "word-hyperbolic", meaning: given a word $w$ representing the trivial element in the group, the minimal number of relators needed to write $w$ is bounded by some constant times its length $|w|$ (see page 18).

In page 20 the author explains the connection between tilings and groups, but he does so very informally. Specifically, he says that the hyperbolic group $\langle a,b,c,d \mid aba^{-1}b^{-1} cdc^{-1}d^{-1} \rangle$ corresponds to octagon tiling of the hyperbolic plane (tiles correspond to the relation), and that because there's a tiling of the Euclidean plane with hexagons, there's a group which satisfies the $C'(1/6+\varepsilon)$ condition without being hyperbolic.

I understand the intuition - the area of hexagon tiling grows linearly while the perimeter quadratically, and in hyperbolic octagon they both grow at the same rate. Still, the correspondence lacks "explicitation" for me.

My questions are:

  1. Can we generally construct hyperbolic (resp. non-hyperbolic) groups from tiling of hyperbolic (resp. non-hyperbolic) space? My first guess would be to take the symmetry group of a tiling, but will it Cayley graph be really similar to the tiling? Will it be hyperbolic if we started with hyperbolic space?
  2. Suppose I take the group $\langle S \mid R \rangle$ where $R$ is a set of one relation of length 6. When will it necessarily be non-hyperbolic (in the sense of word-hyperbolic)? For example consider $S=\{ abcabc \}$, $S= \{ aba^{-1} cdc^{-1}\}$ or $S= \{abc a^{-1} b^{-1} c^{-1} \}$. (EDIT: I'm pretty confident that the last example is indeed non-hyperbolic, since I chose the generator to give a valid tiling)
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Hyperbolicity, as I understand it, is a property of a finitely presented group that only depends on its Cayley graph as a metric space up to quasi-isometry. Consequently, you can study it by studying any metric space quasi-isometric to the Cayley graph. An important theorem is that if $M$ is a closed Riemannian manifold, then $\pi_1(M)$ is quasi-isometric to the universal cover $\widetilde{M}$. In particular, if $M$ is a closed hyperbolic surface, then $\pi_1(M)$ is quasi-isometric to the hyperbolic plane. $\widetilde{M}$ is tiled by copies of $M$ so that's where the tiling comes into play.

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    $\begingroup$ Thanks. I'll go on reading about this theorem (does it have a name, btw?). I've already seen the notion of quasi-isometry and its relation to Cayley graphs (actually, the primer also includes the $\delta$ definition of hyperbolicity). The more I read, the more I understand that the word-hyperbolic def. is useful for proving some theorems (say, "Density 1/2" of Gromov, which I recommend reading), but is not the most intuitive and especially not the most geometric. $\endgroup$ – Ofir Jun 3 '15 at 18:08
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    $\begingroup$ (cont.) To get full understanding one should see what Gromov's motivation was for the word-hyperbolic definition, and read his books. I guess the theorem you mentioned will make an appearance there. $\endgroup$ – Ofir Jun 3 '15 at 18:09
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    $\begingroup$ @Ofir: Here's a sketch. If you put a Riemannian metric on $M$, the covering transformations define a cocompact action on $\tilde M$ by isometries. Now apply the Svarc-Milnor lemma. Note that the cocompactness condition is essential. If $M$ is a closed hyperbolic manifold, it is in particular complete with curvature $-1$; its universal cover is, too, and thus must be the hyperbolic space. $\endgroup$ – user98602 Jun 3 '15 at 18:10

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