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Is it possible to derive some general properties for Cayley graphs of hyperbolic triangle groups, presented as $$ \langle a,b,c | a^2=b^2=c^2=(ab)^p=(bc)^q=(ca)^r=1\rangle \text{, with } \frac1p+\frac1q+\frac1r<1 $$

Although I found this and this reference, I don't have an idea, how to think about these graphs...

Since triangle groups relate to tilings a lot, this looks helpful

Also, fun coincidence: if you create the dual tree to the tiling by putting a vertex inside each triangle and connecting two vertices by a line if one triangle is the image of another under one of the reflections, you get something that looks a lot like the Cayley graph of the reflection triangle group. The only difference is that each edge needs to be two edges (like a little loop) to reflect that each generator has order 2.

But despite the loopy edges, wouldn't that mean that my Cayley graph are infinite?

And looking at the list of tilings for example $\Delta(4,4,4)$, I count 10 duals. Does this mean that these 10 are all valid Cayley graphs (maybe depending on the generators)?

draks ...
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  • The dual is a graph, not a tree. I also do not understand how you got 10 duals, there is only one tiling for each hyperbolic Coxeter group $W$ and one dual to this tiling. It is obviously infinite since $W$ is infinite and it acts on the dual graph with finite vertex-stabilizers. – Moishe Kohan Aug 28 '20 at 19:06
  • @MoisheKohan 1. The "tree"-statement is a citation. What's makes difference in your opinion? Trees are graphs in the end 2. When you uncollapse the table at the linked wiki page I see 10 tilings. This confused me as well. Which one is the one connected to the group? 3. I construct these groups from finite graphs (e.g. here). I don't see why (in my case) the groups should be finite, when they act a finite set of vertices. Isn't that just permutations? – draks ... Aug 29 '20 at 12:21
  • By definition, a tree is a simply-connected graph. I guess, whoever wrote "baking with math" web blog, did not know the difference. – Moishe Kohan Aug 29 '20 at 14:13
  • My suggestion is to take a look at, say, "Office Hours with a Geometric Group Theorist." This probably will not answer your specific questions (I am not sure what they are), but will get you a better idea of the field than the sources you are currently reading. – Moishe Kohan Aug 29 '20 at 15:46
  • @MoisheKohan right, I should have realised the tree thing myself. I'll get a hand on the book, but I don't see why these groups/graphs (by my construction) need to be infinite. Are hyperbolic triangle groups necessarily infinite? – draks ... Aug 31 '20 at 09:04
  • Yes, all hyperbolic triangle groups are infinite since they act cocompactly on the hyperbolic plane. – Moishe Kohan Aug 31 '20 at 12:37
  • @MoisheKohan I see, so my graph-inspired representations represent finite subgroup of these infinite groups, right? – draks ... Aug 31 '20 at 12:54
  • I do not know what are the "graph-inspired representations", but all the pictures of tilings you will find, show only a finite portion of an infinite tiling and, accordingly, only a finite part of the dual graph. – Moishe Kohan Aug 31 '20 at 12:56
  • @MoisheKohan the one I gave here: https://math.stackexchange.com/questions/1622906/mapping-delta2-2-2-mapsto-delta4-4-2 – draks ... Aug 31 '20 at 14:42
  • I guess, you are describing a way to construct homomorphisms between Coxeter groups by changing labels. However, your old question was based on a wrong premise: The homomorphism should be in the opposite direction. It will be also sending a Euclidean triangle group to a finite group. One can use a similar construction to construct homomorphisms to finite groups from some hyperbolic triangle groups. – Moishe Kohan Aug 31 '20 at 14:54
  • @Moishe from Wiki "An action of a group G on a locally compact space X is cocompact if there exists a compact subset A of X such that GA = X. For a properly discontinuous action, cocompactness is equivalent to compactness of the quotient space X/G." What is the compact subset A in this case? Is it one single triangle, because G would (freely) generate the whole hyperbolic plane X from it? – draks ... Sep 11 '20 at 06:14
  • Yes, it is the single triangle. – Moishe Kohan Sep 11 '20 at 13:17
  • @MoisheKohan I hope you don't mind my beginners questions: 1. Since there are infinitely many points to turn the triangles, the group size is (countably?) infinite, right? 2. If I enumerate all the triangles, every group element permutes the triangles. Doesn't this give an infinite dimensional regular (no fixed points) representation of the group? – draks ... Sep 11 '20 at 13:32
  • Yes on all three questions. – Moishe Kohan Sep 11 '20 at 13:37

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