# Cancellation of elements in the Gromov boundary of a free group

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 $$[\gamma]_{\leq i} = \gamma_1 \cdots \gamma_i$$. Obviously, for each $$\gamma\in\partial F$$, we have $$\gamma = \lim_i [\gamma]_{\leq i}$$. Fix $$\gamma\in\partial F$$ such that $$\gamma \neq \lim_n gh^n$$ for any $$g, h\in F$$. Then define:

$$C = \overline{ \big\{ [\gamma]_n^{-1} \big\}_{n\in\mathbb{N}} }$$

$$C\cap\partial F$$ 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 is: is it true that for each $$x\in C$$, $$\lim_n [\gamma]_n x = \gamma$$? If not, is the set of exceptions countable or uncountable?

Update: Thanks for Sam's answer, the answer to the question above is negative. Now I wonder, if the set of exceptions is always countable regardless of $$\gamma$$ that does not have the form $$\lim_n gh^n$$. One of my attempts shows the following statement is true:

$$\forall k\in\mathbb{N}\, \forall N\in\mathbb{N}\, \exists\,m>N \hspace{0.3cm} \text{ suh that } \vert\, [\gamma]_N \wedge [\gamma]_N[\gamma]_m^{-1}\,\vert \geq k$$ This tells for a fixed $$k\in\mathbb{N}$$ and for any subsequence of $$\{[\gamma]_n\}$$, there exists another subsequence of $$\{ [\gamma]_m^{-1} \}$$ such that the latter subsequence will not cancel all and leave at least the first $$k$$ letters of $$\gamma$$. I believeed applying the diagonalization method could give me something but then failed.

$$\gamma = a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbbbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdot bbb \cdot a \cdot b \cdot a \cdot bb \cdot a \cdot b \cdot a \cdots$$
That is, $$\gamma$$ is the limit of the sequence $$(\gamma_n)$$ defined by $$\gamma_0 = a$$ and $$\gamma_{n+1} = \gamma_n \cdot b^{n+1} \cdot \gamma_n$$. It is an exercise to show that $$C$$ contains $$x = \lim \gamma_n^{-1}$$. It is also an exercise to show that the sequence $$([\gamma]_k x)$$ does not converge.
I will further guess that (for carefully choosen $$\gamma$$) there will be uncountably many $$x \in C$$ where the sequence $$([\gamma]_k x)$$ does not converge.
• A very nice counterexample! I believe your example also indicates the set of $\gamma$ that has exception points is uncountable. If we have a hitting measure that arises from a random walk on $F$, l also wonder if this set has measure zero (which may depend on the support of the step distribution). Jun 2, 2023 at 18:57