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Let $G$ be a finitely generated group. Fix some symmetric finite generating set $S$ for $G$, and write $\Gamma$ for the Cayley graph of $G$ with respect to $S$.

Given finite subsets $X,S,Y$ of $G$, we say that $S$ separates $X$ from $Y$, denoted $(X:S:Y)$, if $X$ and $Y$ lie in distinct connected components of $\Gamma\setminus S$. If $B_n$ denotes the $n$-ball in $G$, we say that $g\in G$ is $n$-axial if

$(g^a B_n: g^b B_n : g^c B_n)$

whenever $a<b<c$ are integers. Hopefully, it is easy to imagine what such elements look like in $\mathbb{Z}$ or $\mathbb{Z}\ast\mathbb{Z}$, where some power of any nontrivial element will be $n$-axial. On the other hand, there are no $n$-axial elements in a one ended group like $\mathbb{Z}^2$.

My PhD thesis claims that, when $G$ has at least two ends, $n$-axial elements exist for $n$ sufficiently large. Sadly, Ville Salo and Ilkka Törmä recently pointed out to me that my proof is wrong.

Before returning my degree, it seems prudent to ask:

If $G$ has at least two ends, must $G$ contain an $n$-axial element for some $n$?

I also wonder about the Bass-Serre theory interpretation of this problem. That is, suppose that $G$ acts nontrivially (without global fixed point) with finite edge stabilizers on a tree $T$. If $g\in G$ acts hyperbolically on $T$, must some power of $g$ be $n$-axial?

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  • $\begingroup$ I guess it can be proved without Stallings' theorem. Freudenthal proved in the 40s that there exists some $g$ acting on the end boundary with a north-south dynamics. (without the language, he essentially proved the action on the boundary is convergence). I believe one could deduce this property (possibly for some power of $g$). I haven't checked details. $\endgroup$ – YCor Sep 19 at 7:35
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Let $G$ be a multi-ended finitely generated group.

  • We know from Stallings' theorem that $G$ splits over a finite subgroup. So $G$ can be thought of as the fundamental group of a (compact) graph of spaces $X$ whose edge-spaces have the form $Y \times [0,1]$ where $Y$ is a compact $2$-dimensional CW-complex such that $\pi_1(Y)$ is finite.

  • The universal cover $\widetilde{X}$ (on which $G$ acts geometrically) is now a tree of spaces whose edge-spaces have the form $\widetilde{Y} \times [0,1]$, where $\widetilde{Y}$ is the universal cover of $Y$; notice that $\widetilde{Y}$ must be a compact $2$-dimensional CW-complex.

  • In fact, by collapsing these edge-spaces to $[0,1]$ and the vertex-groups to points, you get a tree $T$ which coincides with the Bass-Serre tree associted to the initial splitting.

  • Consequently, if $g \in G$ acts on the Bass-Serre tree as a loxodromic isometry, then up to replacing $g$ with a sufficiently large power, we have $(g^p \widetilde{Y} : g^q \widetilde{Y} : g^r \widetilde{Y})$ for every $p<q<r$. (Here we are just saying that an edge in a tree always separates.) As $\widetilde{Y}$ is bounded in $\widetilde{X}$ the desired conclusion follows.

Details about graphs of spaces/groups can be found in Scott and Wall's paper Topological methods in group theory.

Working directly in the Cayley graph of $G$ should also be possible. An argument could be extracted from this answer.

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  • $\begingroup$ People are so fast, I had already basically written my answer after asking Cohen (I just had to add the axial corollary) and Cohen personally told me about this post, yet I was 15 minutes slower. If the cited answer works, it sounds much simpler than mine. $\endgroup$ – Ville Salo Sep 19 at 7:01
  • $\begingroup$ If I understand correctly, in your other answer, you find $a \in G$ such that a branch $B$ outside some separating set $A$, which we can take to be an $n$-ball, properly contains the set $aB$. Then $a$ indeed has infinite order, and we also obtain that $a$ is $n$-axial: any path from $1$ to $a^2$ in particular has to step into $aB$, since $a^2B$ is inside $aB$. Translate and apply the lemma that $(B_i : B_{i+1} : B_{i+2})$ implies $(B_i : B_j : B_k)$ for $i < j < k$. $\endgroup$ – Ville Salo Sep 19 at 8:37
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For $>2$-ended groups. By geodesics I mean paths in the Cayley graph whose length is equal to the word metric between their initial and terminal vertex.

If $G$ is a group with finite generating set $S$, $A$ finite and $G \setminus A$ has all components infinite and has at least three components (in the Cayley graph w.r.t. $S$). Suppose $h \notin A$, then there exists $k$ such that $A \cap kA = \emptyset$ and $kh$ is connected to $h$ in $G \setminus (A \cup kA)$.

Proof. Consider the set $P \subset G$ consisting of all geodesics in $G \setminus A$ that begin from $\partial A$ (the boundary of $A$, i.e. elements adjacent to $A$ but not inside it), in the component of $h$. Write $P_n$ for the set of nodes first reached by some geodesic in $P$ after exactly $n$ steps. Since every translate of $A$ splits $G$ into at least three components, it is easy to show that for some $\alpha > 1$, we have $|P_n| \geq \alpha^n$ for all large enough $n$. For all $g \in P_n$, fix a geodesic representation $w_g \in S^n$, representing a the path from some element $e_g \in \partial A$ to $g$ arbitrarily.

Now pick $h'$ in $G \setminus A$ in $P_m$ for large $m$. Suppose that for all $g \in P_n$, the set $g(h')^{-1}A$ intersects the range of the geodesic $w_g$ (in the obvious sense of partial products starting from $e_n$). Then, intuitively, we can use this fact to compress the information in the paths in $P_n$. More precisely, if $g \in P_n$ then $g = g'a^{-1}h'$ for some $a \in A$ where $g'$ lies on the range of $w_g$. Because $h' \in P_m$ has word norm at least $m - |e_g|$ and $w_g$ is a geodesic, $g'$ must be in $P_{n-m+k}$ where $k \in [-3r,3r]$ where $r$ is the maximal word norm among elements of $A$. In particular, the distance of $g$ from $\partial A$ is reduced by at least $m-3r$ when we multiply it by a suitable $h^{-1}a$. We can now describe any element of $P_n$ by listing at most $\lceil n/(m-3r) \rceil$ elements $a$, encountered when iteratively applying this observation, starting from $g$, and listing also the element of $P_{n'}$ for $n' < m+3r$ obtained in the end. Now, $C |S|^{m+3r+1} |A|^{\lceil n/(m-r) \rceil}$ (where $C |S|^{m+3r}$ is an upper bound for the number of elements obtained in the end, $C$ is the boundary size of $A$) has slower asymptotic growth in $n$ than $\alpha^n$, as soon as $|A|^{1/(m-3r)} < \alpha$, a contradiction if $m$ was picked large enough.

This means we can necessarily find some $g \in P_n$ (for any sufficiently large $n$) such that $g(h')^{-1} A$ does not intersect the geodesic $w_g$. Assuming $n$ and $m$ are large enough, we have both $A \cap g(h')^{-1}A = \emptyset$, and that the connected component of $g$ in $G \setminus (A \cup g(h')^{-1}A)$ contains both $h$ and $g(h')^{-1}h$. The first fact is obvious, and for the second, observe that as $h$ and $g(h')^{-1}h$ are, respectively, in the same connected components as $h'$ and $g(h')^{-1}h' = g$ in $G \setminus A$ and $G \setminus g(h')^{-1}A$ respectively, it is suffices that $n$ and $m$ be larger than the length of a minimal path between $h$ and $h'$ in $G \setminus A$. This means we can pick $k = g(h')^{-1}$. End square.

Let $G$ have at least $3$ ends. Then $G$ has an $n$-axial element for some $n$.

Proof. Let $A$ finite and $G \setminus A$ has all components infinite and has at least three components, you can always find such by flood filling the finitely many finite ones. Let $h_1 \notin A$ and $h_2 \notin A$ be in distinct components of $G \setminus A$. Now, use the previous lemma to find $k_1$ such that $h_1$ and $k_1h_1$ are in the same component of $G \setminus (A \cup k_1A)$ and $A$ and $k_1A$ are disjoint. Then find $k_2$ so that $A$ and $k_2A$ are disjoint and $h_2$ is in the same component of $G \setminus (A \cup k_2A)$ as $k_2h_2$. Then $k_2^{-1}k_1$ is $n$-axial, if $n$ is bigger than the size of $A$.

Now consider the sequence $B_0 = A, B_1 = k_2^{-1}A, B_2 = k_2^{-1} k_1 A, B_3 = k_2^{-1} k_1 k_2^{-1} A, B_4 = k_2^{-1}k_1k_2^{-1}k_1 A...$ i.e. you multiply the left translation element from the right alternately by $k_2^{-1}$ and $k_1$.

We have $(B_0 : B_1 : B_2)$: The separation $(k_2A : A : k_1A)$ is clear, and gives $(A : k_2^{-1}A : k_2^{-1}k_1A)$ by translating.

We have $(B_1 : B_2 : B_3)$: We prove that $(A : k_1A : k_1k_2^{-1}A)$ holds: suppose there is a path from $A$ to $k_1k_2^{-1}A$ that does not go through $k_1A$. Then, there is a path from $k_1h_1$ to $k_1h_2$ that does not go through $k_1A$ (which would be a contradiction). To see this, take a path from $k_1h_1$ to $h_1$ (using the assumption on $k_1$) that does not visit any of the three sets, then move to $k_1k_2^{-1}A$ and then $k_1k_2^{-1}h_2$ without visiting $k_1A$ (using the counterassumption that $(A : k_1A : k_1k_2^{-1}A)$ does not hold). Now, observe that since $h_2$ and $k_2h_2$ are in the same connected component of $G \setminus (A \cup k_2A)$, by translating by $k_1k_2^{-1}$ we get that $k_1k_2^{-1}h_2$ and $k_1h_2$ are in the same connected component of $G \setminus (k_1k_2^{-1} A \cup k_1A)$, so we have found a path from $k_1h_1$ to $k_1h_2$ that avoids $k_1A$. From $(A : k_1A : k_1k_2^{-1}A)$, we get $(B_1 : B_2 : B_3) = (k_2^{-1}A : k_2^{-1}k_1A : k_2^{-1}k_1k_2^{-1}A)$ by translating, as desired.

We have shown $(B_0 : B_1 : B_2)$ and $(B_1 : B_2 : B_3)$, and by translating by $k_2^{-1}k_1$ from the left we get $(B_i : B_{i+1} : B_{i+2}) \implies (B_{i+2} : B_{i+3} : B_{i+4})$, so by induction $(B_i : B_{i+1} : B_{i+2})$ holds for all $i$. By a basic connectivity lemma (see e.g. Cohen's thesis) we have $(B_i : B_j : B_k)$ for all $i < j < k$. The sequence $B_{2i}$ proves that $k_2^{-1}k_1$ is axial. End square.

Edit: The axial part had a chirality issue. That's what you get for trying to be fast, I should maybe have drawn a picture. Fixed some typos as well.

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  • $\begingroup$ I prove this in two parts above because the first lemma was what I needed myself and the second is just to show you can get an axial from it. But I think the first lemma could be strengthened so axial elements drop more directly, by instead for any $A$ finding $k$ so $h_1$ is connected to $kh_2$ in $G \setminus (A \cup kA)$. The proof should be exactly the same, and then $k$ seems clearly axial if $h_1$ and $h_2$ are in distinct components of $G \setminus A$. $\endgroup$ – Ville Salo Sep 19 at 10:33

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