Talk:Hyperbolic group

Latest comment: 8 years ago by Jean Raimbault in topic Adjacent pages

Delta thin

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So is the definition that every geodesic triangle in the Cayley graph must be δ-thin for some δ ≥ 0? Do we have to fix a δ for which all geodesic triangles are δ-thin at once? This could be made more clear if this is the case. - Gauge 00:41, 19 August 2006 (UTC)Reply

Presumably this is the case so I changed the wording accordingly, and put the definition of δ-hyperbolic in bold so it stands out. - Gauge 00:45, 19 August 2006 (UTC)Reply
Yes, the definition is just that the Cayley graph is δ-hyperbolic. (Equivalently, the group itself with any word metric - the definition of δ-hyperbolicity does not require the space to be a length space. δ-hyperbolicity is preserved under quasi-isometries, since δ-hyperbolicity is equivalent to being quasi-geodesically stable (by the Morse Lemma)). But the definition about the Linear Isoperimetric Inequality is very important.

Future directions

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Have to say that it doesn't matter which word metric is used, since a change of generators is a quasi-isometry.

Most importantly, a group is hyperbolic iff it satisfied a linear isoperimetric inequality. A combinatorial and a geometric interpretation is possible and a picture would be a good idea. A hyperbolic group is in fact, an automatic group, that is, multiplication in it can be checked by a finite automaton. 146.186.132.188 06:21, 14 October 2007 (UTC)Reply

Computational Properties

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I added a couple of sentences on the word problem for hyperbolic groups, and the fact that they are automatic. I'm pretty sure that there is actually a linear time solution to the word problem (the automation gives you a quadratic time one), but with large constants which make it impractical. I think this was proved by David Epstein, but it's some time since I looked at this stuff, and my books are on the other side of the world. If anyone can help, that would be much appreciated. TheAstonishingBadger 23:48, 9 November 2007 (UTC)Reply

Thin bigons

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One surprising fact about hyperbolic groups is that any group where bigons (two-sided polygons) are δ-thin are hyperbolic. That is, thin-bigons implies thin triangles (although the triangles can a lot thicker than the bigons).

I think this was proved by Papasoglou. Again, I don't have any reference materials with me, but this would be a nice thing to include. TheAstonishingBadger 23:52, 9 November 2007 (UTC)Reply

Illustrations

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This article's appeal will increase at least by a factor of ten if we can put in a good picture illustrating the delta-thin triangle property. Help! Arcfrk (talk) 14:50, 9 February 2008 (UTC)Reply

Hyperbolic knot groups

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A new editor recently changed the article to say that the knot group of a hyperbolic knot is not a hyperbolic group, while previously the article claimed that hyperbolic knot always had a hyperbolic group as its fundamental group.

I have found references that indicate that knot groups that are not infinite cyclic contain free abelian groups of rank two (Cossey,Dey,Meskin 1971 attribute this to Neuwirth 1965), that a free abelian group of rank two is a Baumslag-Solitar group (Baumslag–Solitar group's presentation makes it clear B(1,1) is such), and that a group containing a Baumslag-Solitar group as a subgroup is not hyperbolic (Springer's EoM article, BS group, in the geometric group theory section). I also found a presentation of the knot group of the figure eight knot (MathWorld, Figure 8), and checked using GAP that it has a subquotient isomorphic to Z^2, so is not infinite cyclic. All of this taken together is a claim that the figure eight knot complement's fundamental group is not hyperbolic.

Can someone back up the previous claim in the text? Can someone give a shorter citation for the new counterclaim in the text? JackSchmidt (talk) 20:12, 12 July 2008 (UTC)Reply

Sorry, my mistake, and I've removed the wrong claim. What's true is that all but finitely many Dehn fillings of a hyperbolic knot complement are hyperbolic manifolds (by a famous theorem of Thurston); thus, if G = π1(S3K) is the fundamental group of a hyperbolic knot complement, H = Z2 is the cusp subgroup (the image of π1 of the boundary torus), and the cyclic subgroup <h> ≤ H represents the slope, then the group G/<<h>> is hyperbolic for all but finitely many slopes. I am not sure how much sense does it make to add this example at this stage, it's more natural in the context of relatively hyperbolic groups. The Z2 theorem is proved in Bridson and Haefliger. Arcfrk (talk) 05:22, 13 July 2008 (UTC)Reply
New text looks good. I was pretty sure the knot group was at least some sort of hyperbolic based on the sources I could find that more directly addressed the question. In Epstein, et al. knot groups of hyperbolic knots are automatic as a corollary of the fact that geometrically finite hyperbolic groups are automatic. The proof given is very brief, but mentioned being able to extend from a torus. JackSchmidt (talk) 12:44, 13 July 2008 (UTC)Reply

Interesting discussion, it's been a while since this article was on my watchlist and a lot longer since I added anything to it. I'm glad to see it's developed. Regarding the relevance of the correct/fixed example, it's basically already covered under the listed example of compact Riemannian manifolds of negative curvature. The cusped hyperbolic manifold case may be of some relevance here as a prototypical example of a group (its fundamental group) that isn't delta hyperbolic, but you want it to be, so you have to kill off that lattice in the cusps somehow, perhaps via electrification. This is what I believe Arcfrk is talking about when he mentioned relatively hyperbolic groups. Anyway, a more interesting example to include is probably electrified Teichmuller space, which is delta hyperbolic (even quasi isometric to the curve complex), but the original Teichmuller space is not. --C S (talk) 06:05, 15 July 2008 (UTC)Reply

Current text is incorrect. All knot complements except those of torus knots and satellite knots have hyperbolic fundamental groups by the work of Thurston(although they DO contain a nontrivial Z+Z subgroup, the peripheral structure). However, just having a Z+Z subgroup does not imply non-hyperbolicity. The idea here is that taking the standard Cayley graph of Z+Z, we can embed this into a graph in such a way that the new edges create new geodesics, which allow the delta-hyperbolicity to be satisfied in this larger space. I have included the old text here, but I am taking out the incorrect statement. Mad2Physicist (talk) 05:16, 1 November 2008 (UTC)Reply
You are mistaken, hyperbolic groups cannot contain Z^2 subgroups. I've reverted and added precise references. The centralizer of any element of a hyperbolic group is a quasi-convex subgroup. This rules out the "idea" that you've mentioned. For the relation with Thurston's work, see the discussion above. Arcfrk (talk) 03:40, 3 November 2008 (UTC)Reply

Definitions

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I was somewhat intrigued to see the phrase "Hyperbolic groups can be defined in several different ways. All definitions use the Cayley graph of the group..." in the Definitions section. This isn't true. For example, a group is Hyperbolic if and only if it's word problem is soluble via Dehn's algorithm, or (if I remember correctly), has linear-time word problem. So I've changed the "All definitions" to "Many definitions".

Farpov (talk) 22:43, 19 January 2012 (UTC)Reply


Adjacent pages

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I just made several substantial modifications to the page (fleshed out the examples section and added more properties). I think the page is now fairly complete but the individual sections could probably benefit from a few rewrites (ideally from somebody familiar with academic mathematics but not geometric group theory, but I'll probably do a bit of it myself if nobody else is interested). One thing that I think is still less than ideal: many pages linked in this one are little better than stubs. In particular having a good page on Gromov-hyperbolic spaces would be a good thing since their proper study is not the object of this page and the definition given here is probably a bit on the short side (in my opinion as it should be). Jean Raimbault (talk) 20:35, 21 May 2016 (UTC)Reply