Cancellation of elements in the Gromov boundary of a finitely generated free group

Question

This question is inspired by this post from MO

Let $A$ be a finite set of free generators and their inverses and $F$ the free group generated by elements in $A$ (some call $A$ the alphabet of $F$). For each $g\in F$, use $\vert\,g\,\vert$ to denote the length of $g$. The Gromov boundary of $F$, denoted by $\partial F$, can be viewed as the set of words that consists of letters from $A$ and has infinite length starting from the left. For instance, $a_1 a_2 \cdots $ is the general form of elements in $\partial F$ where $a_i\in A$ and $a_i a_{i+1}\neq e$ for each $i\in\mathbb{N}$.

Set $\overline{F} = F\cup\partial F$. Given $x, y\in\overline{F}$, if $x\neq y$, let $x\wedge y$ denote the common part between $x$ and $y$ starting from the left. For instance: $ (a_1 \cdots a_n )\wedge (a_1 \cdots a_n b_{n+1} b_{n+2} \cdots) = a_1 \cdots a_n$. Define a function $d: \overline{F}\times\overline{F}\rightarrow(0, 1)$ as follows: $d(x, y)=0$ if $x=y$ and $d(x, y)= \operatorname{exp}(-\vert\,x\wedge\,y\,\vert)$. One can check $(\overline{F}, d)$ is a compact metric space and $d$ is an ultrametric. From now on, we assume $\overline{F}$ is equipped with the $d$-metric topology. An action of $F$ on $\partial F$ can be defined as follows: given $g\in F$ and $x\in\partial F$, $g\cdot x = gx$ (after cancellation).

Given $\gamma\in\partial F$, for each $i\in\mathbb{N}$, let $\gamma_i$ denote the $i$-th letter of $\gamma$ and define $[\gamma]_i = \gamma_1 \cdots \gamma_i$. Obviously, for each $\gamma\in\partial F$, we have $\gamma = \lim_i [\gamma]_i$. From now on we only consider $\gamma\in\partial F$ such that $\gamma \neq \lim_n gh^n$ for any $g, h\in F$. Then define:

$$ C_{\gamma} = \overline{ \big\{ [\gamma]_n^{-1} \big\}_{n\in\mathbb{N}} } \cap\partial F $$

which is the set of clustered points of the sequence $\big\{ [\gamma]_n^{-1} \big\}$ and, by compactness of $\overline{F}$, is non-empty. My question are:

1. Is $C_{\gamma}$ countable regardless of the choice of $\gamma$? If not, is the set $\big\{ \gamma\in\partial F \,\vert\, C_{\gamma} \text{ uncountable } \big\}$ countable?

Answer 1

There are a lot of questions in your post, each one of them being quite substantial. Let me suggest, again, one question per post please. I'll answer your first question.

In fact there is an example where $C_\gamma = \partial F$, and so no, $C_\gamma$ need not be countable.

To describe the example, start with a list of the reduced words representing all nontrivial elements of $F$: $$\delta_1, \delta_2, \delta_3, ... $$ Let's choose them so that $\delta_{i} \delta_{i+1}$ has no cancellations, for all $i \ge 1$. I believe this is possible, with a little care.

Define $\gamma$ to be the infinite concatenation of these elements: $$\gamma = \delta_1 \delta_2 \delta_3 \cdots $$ The sequence $[\gamma]_n^{-1}$ contains the following subsequence: $$\underbrace{\delta_1^{-1}}_{\theta_1} \quad \underbrace{\delta_2^{-1}\delta_1^{-1}}_{\theta_2} \quad \underbrace{\delta_3^{-1}\delta_2^{-1}\delta_1^{-1}}_{\theta_2} \quad \cdots $$ Notice that the sequence $\delta_1^{-1}$, $\delta_2^{-2}$, $\delta_3^{-3}$ also lists the reduced words representing all nontrivial elements of $F$.

From this it follows that the set $C_\gamma$ contains every element of $\partial F$. To prove this, consider any element of $\partial F$ write it as an infinite word $a_1a_2a_3...$ with no cancellation. Each initial segment $a_1a_2a_3...a_i$ is equal to some $\delta_{n_i}^{-1}$, and so the sequence $\theta_{n_1},\theta_{n_2},\theta_{n_3},...$ converges to $a_1a_2a_3...$.

I believe that one can tailor this method to show that for every closed subset $C \subset \partial F$ there exists a $\gamma$ such that $C=C_\gamma$. It would be quite a bit trickier to carefully choose the appropriate list $\delta_1,\delta_2,\delta_3$,..., but this should be possible.

This gives a very strongly negative answer to the sub-question of your first question: there are uncountably many uncountable closed subsets of $\partial F$, so the set $\{\gamma \mid C_\gamma \, \text{is uncountable}\}$ is uncountable (that was a hard sentence to write! ... and, a day later, I realized that a missed one "uncountable").