Sharpness of the $1/6$-constant in the Cancellation Theorem

I originally posted this question over at Stackexchange, before realising it is much better suited for Overflow:

Let $\langle \; S \; | \; R \; \rangle$ be a presentation of a group $G$ with a set $R = R^{-1}$ of freely and cyclically reduced relators, and let $\Lambda$ be the girth of $\langle \; S \; | \; R \; \rangle$. Suppose that

1) the set $R$ of relators contains no proper power, and

2) for any triple $x,y,z \in F(S)$ of reduced words, such that $yx$ and $zx$ are two distinct (reduced) cyclic conjugates of elements of $R$, we have $$|x| \leq 1/6 \Lambda,$$ where $|x|$ denotes the length of $x$.

Then, the famous Cancellation Theorem says that the corresponding presentation complex $P(S,R)$ is aspherical.

Does anybody know if (and why) the constant $1/6$ in the statement is sharp ?

To be more precise, I am not looking for an aspherical presentation that fails to meet condition $2$ (such are easy to construct). I am looking for a group with a non-aspherical presentation (in the above sense) that satisfies both conditions, but with constant $c > 1/6$.

Ideally, I am looking for an answer to the following question: For every $\epsilon > 0$, does there exist a presentation $P_{\epsilon}$ of a group $G_{\epsilon}$, satisfying both conditions, but with constant $1/6 + \epsilon$, such that $P_{\epsilon}$ is not aspherical ?

• The dodecahedron. – HJRW Sep 5 '18 at 22:05
• Can you elaborate ? – Berni Waterman Sep 6 '18 at 5:46
• Sure. Take a dodecahedron. Identify all the vertices. The result is a presentation complex which is clearly not aspherical, and where each 2-cell (ie relator) overlaps its neighbours in precisely 1/5 of its length. – HJRW Sep 6 '18 at 7:35
• If $D\cong S^2$ is the dodecahedron (clearly non-aspherical, I hope), then the presentation complex $P=D/\sim$ is obtained by identifying the vertices of $D$, so we have a quotient map $\pi:D\to P$. As I think you say, this defines the "obvious" non-trivial element of $\pi_2(P)$. There are various ways to see that it's non-trivial. For instance, the universal cover $\widetilde{P}$ is an infinite tree of copies of $D$, and $\pi$ lifts to an inclusion $\tilde{\pi}:D\hookrightarrow\widetilde{P}$. There is also a retraction $\rho:\widetilde{P}\to D$. If $\pi$ were null-homotopic, by the... – HJRW Sep 6 '18 at 8:57
• ... Homotopy Lifting Lemma, $\tilde{\pi}$ would also be null-homotopic, and so, applying $\rho$ to the homotopy, we get that the identity map of $S^2$ is null-homotopic, which is a contradiction. – HJRW Sep 6 '18 at 9:00