# For a group with one end does the property of connected spheres follow?

One of my friends is studying group actions on the circle, and he ended up with a question in geometrical group theory. Let us consider a finitely generated group $G$ with generators $g_1, \ldots g_n$. The notion of a length of an element $g$ can be given as a length of a minimal representation of $g$ in terms of generators.

Let me recall a classical definition of a group with one end. If $G$ is a connected, locally path connected, locally compact topological space. Then $G$ has one end if given a compact subset $K \subset G$, there is a compact set $L$: $K \subset L$ such that for any $x,y \in G \setminus L$ there is a path in $G \setminus K$ joining $x$ and $y$.

For a group with one end let us define a property, that rises from the studies of group actions on the circle, namely a property of connected spheres: For any ball $B_R$ let us take a nonbounded component of its compliment $(B_R)^c_{\infty}$: this component is unique since our group has one end.

Then we say, that a group has a property of connected spheres if there exists $C>0$ such that for any ball $B_R$ of radius $R$ the points in the fiber $(B_R)^c_{\infty} \cap B_{R+C}$ could be connected by the path in the group, i.e. for any $x,y$ in the fiber $(B_R)^c_{\infty} \cap B_{R+C}$ there exists a finite number of group elements, such that $x=gy$ and $g$ is a word in the alphabet $g_1, \ldots g_n$ and all the steps still lie in the fiber considered between the spheres.

The question is if for a group with one end the property of connected spheres holds automatically or not, and what are the examples in the case?

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The question does not make sense for me. For every two elements $x,y$ there exists a path in the left Cayley graph from $x$ to $y$. The label of that path is your $g$. Perhaps you wanted all elements $z_k=g_{k+1}...g_ny$ to satisfy $R<|z_k|<R+C$? In that case the answer is "no" because of "dead ends of arbitrary depth". – Mark Sapir Jul 4 '12 at 6:36
Thank you very much for your comment: for sure, $z_k$ have to lie in the fiber considered. Secondly, what is important, a case of dead ends is not an issue here - I corrected a definition of connected spheres after your remark, see a new one above. – Olga Jul 4 '12 at 7:53
My guess is that this might be true if $G$ is finitely presented, but should be false for finitely generated groups in general. Indeed, there are f.g. groups with one end that are limits (in the space of marked groups) of f.p. groups, each of which has infinitely many ends. One such example is the lamplighter group $(\mathbb{Z}/2\mathbb{Z}) wr \mathbb{Z}$. This is the group I would try to look at first. – Ashot Minasyan Jul 4 '12 at 11:34
I am sorry, who's this friend? Antoine Gournay put yesterday a paper in arxiv where he uses basically the same property. He has a sketch of proof that the property holds for all finitely presented group. We have had a chat right now and we are pretty convinced that it fails for every one-ended group with at least one non simply connected asymptotic cone. – Valerio Capraro Jul 4 '12 at 12:47
@Ashot: you are correct that this is true when $G$ is finitely presented. In the Cayley graph $\Gamma$ for $G$, if the "property of connected spheres" fails then for each $C$ there exists $R$ for which the set $(B_R)^c_\infty \cap B_{R+C}$ is disconnected; letting $P = \frac{R+C}{2}$ one can show that the 2-complex obtained from $\Gamma$ by attaching discs to all closed edge paths of length $\le P$ is not simply connected. Since $C$ can be arbitrarily large, $P$ is also arbitrarily large, so $G$ is not finitely presented. If this is useful I'll put more details in an answer, but I'm out of roo – Lee Mosher Jul 4 '12 at 12:51