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I'm working with Gromov's density model of random groups, and a nice fact is that for a fixed density parameter $0 \leq d \leq 1$, a generic group in the density model satisfies the $C'(2d)$ small cancellation condition.

This is of particular interest when $d \leq \frac{1}{12}$ and we get the $C'(\frac{1}{6})$ small cancellation condition, because groups satisfying this condition have a number of nice properties:

  • They are hyperbolic.
  • They have a solvable word problem.
  • Greendlinger's lemma holds, saying roughly that a word in the free group which is trivial in our small cancellation group must contain some large piece of a relator.

Somewhat surprisingly I have not found examples of a $C'(\lambda)$ small cancellation condition, where $\lambda < \frac{1}{6}$, being important. I am interested to know if there are qualitative improvements to Greendlinger's lemma to be made when we have $C'(\lambda)$ small cancellation for $\lambda < \frac{1}{6}$, or if there are other theorems that make critical use of such a condition.

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  • $\begingroup$ @AGenevois Sorry for the bad notation, I am interested in the latter. I have updated the question. $\endgroup$
    – ckefa
    Commented Jul 21 at 4:37
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    $\begingroup$ Geometrically, Greendlinger's lemma amounts to saying that every reduced van Kampen diagram contains a spur or a shell. The property can be strengthened by counting the number of spurs and/or shells. See e.g. MacCammond and Wise's article Fans and ladders in small cancellation theory. This property is used in virtually every statement proved about C'(1/6)-groups, e.g. classification of finite subgroups, conjugacy problem, cubulation. It is worth mentioning that C'(1/6)-groups are hyperbolic of dimension 2 and linear over $\mathbb{Z}$. $\endgroup$
    – AGenevois
    Commented Jul 21 at 5:29
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    $\begingroup$ See for instance arxiv.org/abs/1212.5280 for other results using $C’(\lambda)$ for $\lambda=\frac18,\frac1{10}$ and $\frac1{12}$. Their “criss-cross decomposition” might be an instance of what you’re looking for. $\endgroup$
    – Corentin B
    Commented Jul 21 at 12:12
  • $\begingroup$ @AGenevois Thank you, the McCammond-Wise paper is very interesting. $\endgroup$
    – ckefa
    Commented Jul 24 at 1:46
  • $\begingroup$ @CorentinB Thank you, this is a great example. $\endgroup$
    – ckefa
    Commented Jul 24 at 1:47

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Greendlinger's Lemma is about the geometry of disk diagrams. If we instead consider annular diagrams, the assumption that $\lambda<1/8$ gives an improved structural result.

Annular diagrams encode the conjugacy problem: two words $u$ and $v$ represent conjugate group elements if and only if there exists an annular diagram with inner boundary labelled $u$ and outer boundary labelled $v$.

The geometry of annular diagrams for $C'(\lambda)$ groups is covered in Chapter V.5 of Lyndon and Schupp. The relevant results are Theorems 5.3 and 5.5. We will assume the $C'(1/6)$ condition throughout the following, and that the boundary labels are Dehn reduced (contain no more than half of a relator).

Briefly, Theorem 5.3 says that, under our ambient assumptions, if $M$ is a reduced annular diagram which has no regions ($2$-cells) containing both an edge of the inner and outer boundaries of $M$, then $M$ consists of two layers of regions: an inner layer, all intersecting the inner boundary, and an outer layer, all intersecting the outer boundary. This is illustrated nicely in Figure 5.2.1 of Lyndon and Schupp, which I've pasted below:

Annular diagram containing regions which do not intersect both boundaries

Theorem 5.5 of Lyndon and Schupp covers the complementary situation to 5.3, and says that, under our ambient assumptions, if $M$ is a reduced annular diagram which contains a region ($2$-cells) which has both an edge of the inner and outer boundaries of $M$, then every region has both an edge of the inner and outer boundaries of $M$, and contains at most two internal edges. This is illustrated nicely in Figure 5.3 of Lyndon and Schupp, which I've again pasted below:

Annular diagram containing regions all of which intersect both boundaries

We then have the following, which says that the $2$-layer case is not possible if $\lambda<1/8$. It is phrased to reflect Lyndon and Schupp's Theorem 5.5, so all definitions are contained in Lyndon and Schupp. Roughly though, $R$ is our (recursive, symmetrised) set of relators for a group presentation $\langle X\mid R\rangle$, and if $\gamma$ is an edge on the boundary of a region then $\phi(\gamma)$ is the label of this edge, and "$\phi(\gamma)$ is not $>\frac12R$" means that this label is less than half of the boundary label.

Proposition. Let $R$ satisfy the $C'(\lambda)$ condition for $\lambda<1/8$. Assume that:

  1. $M$ is a reduced annular diagram.
  2. (Dehn reduced boundaries.) Let $\sigma$ and $\tau$ be respectively the outer and inner boundaries of $M$. If $D$ is a region of $M$ with $\sigma_1=\partial D\cap\sigma$, then $\phi(\sigma_1)$ is not $>\frac12 R$. Assume the same hypothesis with $\sigma$ replaced with $\tau$.

Then every region $D$ of $M$ has edges on both $\sigma$ and $\tau$, and has at most $2$ internal edges.

Proof. Assume the situation, and hence the conclusion, of Theorem 5.3. Here, each region has $4$ inner pieces, and so this picture is only possible if $1-4\lambda\leq1/2$, i.e. if $\lambda\geq1/8$. This means that if $\lambda<1/8$, situation of Theorem 5.3 is not possible. The result now follows from Theorem 5.5.

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  • $\begingroup$ Can you formulate a statement of the form "Let $G$ admit a $C'(t)$ presentation, $t<1/8$, then X", where X does not refer to the presentation? (and X should be unclear or, better, false, for general $C'(1/6)$ groups) $\endgroup$
    – YCor
    Commented Jul 23 at 9:36
  • $\begingroup$ @YCor I'm not sure what you mean. I've updated the statement to make it more precise, in the language of Lyndon and Schupp. The statement is false for $C'(1.6)$ presentations, as the 2-layer case is possible here (I can add an example if it would be helpful). $\endgroup$
    – ADL
    Commented Jul 23 at 10:11
  • $\begingroup$ Here the conclusion is a conclusion about the presentation (regions, edges...). I'm asking whether you can get a conclusion not referring to the presentation (e.g., "has a solvable conjugacy problem" is such a conclusion — however, it is true for all Gromov-hyperbolic groups). $\endgroup$
    – YCor
    Commented Jul 23 at 10:14
  • $\begingroup$ @YCor Ah, right. This has no algebraic consequences, as far as I know. I've just always found it an interesting quirk :-) The original motivation was indeed the conjugacy problem, but this is of course superseded by hyperbolicity. (Moreover, the conjugacy problem follows from Lyndon and Schupp's Theorem 5.3, with Theorem 5.5 really just tying up a loose end.) $\endgroup$
    – ADL
    Commented Jul 23 at 10:37
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    $\begingroup$ Thank you, this is very helpful! $\endgroup$
    – ckefa
    Commented Jul 24 at 1:50

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