Timeline for Starting letters of equivalent infinite geodesic paths of hyperbolic Coxeter groups
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
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| Aug 14, 2019 at 18:21 | comment | added | Jim Belk | Note: In my previous comment, "path of length two as an isometrically embedded subgraph" should be "path of length two as an induced subgraph". | |
| Aug 14, 2019 at 13:02 | vote | accept | worldreporter | ||
| Aug 14, 2019 at 13:02 | comment | added | worldreporter | Thank you very much! | |
| Aug 14, 2019 at 10:49 | comment | added | Jim Belk | (continued) Indeed, it is not hard to show that any graph $\Gamma$ which does not have a path of length two as an isometrically embedded subgraph must be a disjoint union of complete graphs. It follows that a Gromov hyperbolic, right-angled Coxeter group $W_\Gamma$ has the property you want if and only if it is a free product of finite groups. | |
| Aug 14, 2019 at 10:41 | comment | added | Jim Belk | The group $G$ given at the end above corresponds to the right-angled Coxeter group $W_\Gamma$ for which $\Gamma$ is a path of length two. More generally, if $\Gamma$ is any graph which has a path of length two as an induced subgraph, then $W_\Gamma$ will have $G$ as an isometrically embedded subgroup, and the same phenomenon will occur. For example, $\mathbb{Z}_2* G$ is hyperbolic and irreducible (not a direct product -- I'm assuming that's what you mean by "irreducible") , and its Cayley graph has the Cayley graph of $G$ as an isometrically embedded subgraph. | |
| Aug 14, 2019 at 6:18 | comment | added | worldreporter | Thank you for your great answer! Is it possible that at least in the right-angled, irreducible case my question has a positive answer? | |
| Aug 13, 2019 at 21:41 | history | edited | Jim Belk | CC BY-SA 4.0 |
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| Aug 13, 2019 at 20:40 | history | edited | Jim Belk | CC BY-SA 4.0 |
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| Aug 13, 2019 at 20:27 | history | answered | Jim Belk | CC BY-SA 4.0 |