It is well known fact that a (f.g.) group is hyperbolic if and only if it admits a (finite) Dehn presentation.

My question concerns something I'm struggling with since the first time I read the proof of this Theorem (in Bridson, Haefliger, Metric spaces of non-positive curvature). Are there a lot of known examples of (hyperbolic) groups with Dehn presentations (except trivial ones, like free groups)?

It seems (at least to me) that "no one" wants to "really compute" such presentations explicitly. (I know there are implemented algorithms for computing hyperbolic constants of hyperbolic groups)

The secondary reason for my question may be quite silly. I was talking to friend of mine (also mathematician) about this beautiful proof and he asked me: "Can you give me some examples?" .... I was stunned.

  • 3
    $\begingroup$ 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. $\endgroup$ – YCor Apr 25 '15 at 18:49
  • 1
    $\begingroup$ 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. $\endgroup$ – Derek Holt Apr 25 '15 at 18:59
  • $\begingroup$ @ Derek Holt: What if one asks if a given finite presentation satisfies $C^{'}(1/6)$? This should be possible right? $\endgroup$ – M.U. Apr 25 '15 at 20:51
  • 1
    $\begingroup$ 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? $\endgroup$ – Dylan Thurston Apr 26 '15 at 4:26
  • 2
    $\begingroup$ 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). $\endgroup$ – user1729 Apr 27 '15 at 8:04

I'll turn YCor's comment (above) into an answer. Surface groups are the simplest (and historically first) non-free example. Let $S_g$ be the surface of genus $g$. Then the fundamental group has presentation

$\pi_1(S_g) \cong \langle a_1, a_2, \ldots, a_{2g-1}, a_{2g} \mid a_1 a_2 \ldots a_{2g} A_1 A_2 \ldots A_{2g} \rangle$.

This presentation is a Dehn presentation. A variant of this fact is due to Dehn. See the following post for discussion and references.

| cite | improve this answer | |
  • $\begingroup$ I'm not sure what you mean by "variant", maybe "stronger variant"? anyway this result is 100% due to Dehn in the case of the standard presentation with relator $[a_1,a_2]\dots [a_{2g-1},a_{2g}]$. $\endgroup$ – YCor Apr 26 '15 at 9:01
  • $\begingroup$ The argument for this relation is the same, because it and the one you give yield identical tilings of the hyperbolic plane. $\endgroup$ – Sam Nead Apr 26 '15 at 19:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.