Timeline for Explicit examples of Dehn presentations of hyperbolic groups
Current License: CC BY-SA 3.0
16 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 16, 2023 at 13:40 | comment | added | seldom seen | Does anybody know how to do it for one-relator groups with negative immersions, après Linton arxiv.org/abs/2202.11324 ? | |
| Aug 18, 2015 at 17:44 | vote | accept | M.U. | ||
| May 5, 2015 at 8:15 | comment | added | M.U. | @ Derek Holt: I found a paper where an explicit algorithm is described which decides if a given finite presentation is a Dehn presentation. G.N. Arzhantseva, An algorithm detecting Dehn presentations, preprint, University of Geneva, 2000. | |
| Apr 27, 2015 at 8:04 | comment | added | user1729 | A rather neat class of examples are presentations of the form $\langle X; R^n\rangle$ where $n>1$ (so one relator groups with torsion). This follows from the B.B.Newman spelling theorem (which also implies the original surface groups result). | |
| Apr 26, 2015 at 9:41 | history | edited | Sam Nead |
added tag
|
|
| Apr 26, 2015 at 8:57 | answer | added | Sam Nead | timeline score: 3 | |
| Apr 26, 2015 at 6:07 | comment | added | YCor | To be even more precise, I understand "Dehn presentation" as a presentation for which every null-homotopic reduced word contains more than half of a relator (up to cyclic rearrangement and inversion of relators). Hence the obvious replacement algorithm works. | |
| Apr 26, 2015 at 4:26 | comment | added | Dylan Thurston | Just to be precise, is a "Dehn presentation" one for which the Dehn algorithm works, i.e., repeatedly replace a long subword of a relation with the complementary shorter subword? | |
| Apr 25, 2015 at 22:51 | comment | added | YCor | A nontrivial hyperbolic group with Kazhdan's Property T cannot be $C'(1/6)$ (D. Wise), and there are many such groups. | |
| Apr 25, 2015 at 22:22 | comment | added | M.U. | Is there any result towards a statement about "how many hyperbolic groups are not $C^{1/6}$"? I know this is kind of a "meta-question". | |
| Apr 25, 2015 at 21:20 | comment | added | Derek Holt | Yes, that's a straightforward check. And if it does, then the Dehn algorithm just consists of the reductions arising directly from the relations. | |
| Apr 25, 2015 at 20:51 | comment | added | M.U. | @ Derek Holt: What if one asks if a given finite presentation satisfies $C^{'}(1/6)$? This should be possible right? | |
| Apr 25, 2015 at 19:00 | history | edited | Derek Holt |
add gr-group-theory tag
|
|
| Apr 25, 2015 at 18:59 | comment | added | Derek Holt | I agree that this an interesting question computationally. The standard proofs involve taking something like all length reducing rules with LHS of length at most $4\delta$ (where $\delta$ is the thinness constant), and even if you know $\delta$, that could be a lot of rules. A major difficulty is that there appears to be no algorithm for checking whether a given set of rules is a Dehn algorithm. | |
| Apr 25, 2015 at 18:49 | comment | added | YCor | It's not that bad. The first example is given by surface groups with their standard presentation (with $\chi\le -2$), this is precisely Dehn's theorem. Next, a big source of examples is given by $C'(1/6)$ small cancelation presentations. I'd also guess that things can be made explicit for most Gromov-hyperbolic Coxeter groups but it's just a guess. | |
| Apr 25, 2015 at 18:23 | history | asked | M.U. | CC BY-SA 3.0 |