5
$\begingroup$

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.

$\endgroup$
1

1 Answer 1

2
$\begingroup$

It is not true. Consider the following point at infinity:

$ \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.

$\endgroup$
2
  • 1
    $\begingroup$ 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). $\endgroup$ Jun 2, 2023 at 18:57
  • 1
    $\begingroup$ Thanks again for your counter-example. Since I do not have enough bounties in MO but could not figure out my second question, I will make another post in MS and include the link in the chat. $\endgroup$ Jun 6, 2023 at 21:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.