p. 3 and the abstract — the volume exponent of the first Grigorchuk group exists and equals α_0: log log v_{G,S}(n)/log n → α_0

The statement has no hypotheses and no free variables. It is a single closed assertion about one specific group of permutations, one specific four-element generating set of it, and one specific real number $\alpha_0$. All of these are spelled out below.

The ambient group of tree automorphisms

Let $W = \{0,1\}^{*}$ be the set of all finite words over the two-letter alphabet $\{0,1\}$, including the empty word $\varnothing$. Write $|v|$ for the length of a word $v$, and $xv$ for the word with first letter $x$ followed by $v$.

Let $\mathrm{Aut}$ be the set of all bijections $\sigma : W \to W$ such that, for all words $v, w$,

$$|\sigma(v)| = |v| \qquad\text{and}\qquad \bigl(v \text{ is a prefix of } w\bigr) \iff \bigl(\sigma(v) \text{ is a prefix of } \sigma(w)\bigr).$$

Here "prefix" allows $v = w$ and $v = \varnothing$. $\mathrm{Aut}$ is a group under composition: $(\sigma\tau)(v) = \sigma(\tau(v))$. Its identity $e$ is the identity map, and inverses are inverse bijections.

The four generators

The first generator $a$ flips the first letter of a word and does nothing else:

$$a(\varnothing) = \varnothing, \qquad a(xw) = \bar{x}\,w \quad (\bar 0 = 1,\ \bar 1 = 0).$$

The other three are built from a sequence. Let $\kappa : \{0,1,2\} \to \{b,c,d\}$ be the labelling

$$\kappa(0) = d, \qquad \kappa(1) = c, \qquad \kappa(2) = b .$$

Take $\gamma \in \{b, c, d\}$ and a sequence $\theta = (\theta_0, \theta_1, \theta_2, \dots)$ with entries in $\{0,1,2\}$. Write $\theta' = (\theta_1, \theta_2, \dots)$ for the shifted sequence. Then the map $g_{\theta,\gamma} : W \to W$ is defined by recursion on the word:

$$g_{\theta,\gamma}(\varnothing) = \varnothing, \qquad g_{\theta,\gamma}(0w) = \begin{cases} 0\,a(w) & \text{if } \gamma \neq \kappa(\theta_0),\\ 0\,w & \text{if } \gamma = \kappa(\theta_0), \end{cases} \qquad g_{\theta,\gamma}(1w) = 1\,g_{\theta',\gamma}(w).$$

Now fix the specific sequence

$$\omega = (\omega_0, \omega_1, \omega_2, \dots), \qquad \omega_n = n \bmod 3, \qquad\text{that is}\qquad \omega = 0,1,2,0,1,2,\dots,$$

and set $b = g_{\omega,b}$, $c = g_{\omega,c}$ and $d = g_{\omega,d}$.

Explicit description. Every word is either $1^k$ for some $k \ge 0$, or has the form $1^k 0 w$: $k \ge 0$ leading $1$'s, then a $0$, then an arbitrary word $w$. Each $\gamma \in \{b,c,d\}$ fixes every word $1^k$. On words of the second form it acts by

$$\gamma(1^k 0 w) = \begin{cases} 1^k 0\, a(w) & \text{if } \gamma \neq \kappa(k \bmod 3),\\ 1^k 0\, w & \text{if } \gamma = \kappa(k \bmod 3). \end{cases}$$

So $\gamma$ flips the letter immediately after the first $0$, if there is such a letter, with these exceptions:

• $b$ does not flip when $k \equiv 2 \pmod 3$;
• $c$ does not flip when $k \equiv 1 \pmod 3$;
• $d$ does not flip when $k \equiv 0 \pmod 3$.

Equivalent recursive description. Because $\omega$ has period $3$, the same three maps are characterised by the following identities, valid for every word $w$:

$$\begin{aligned} b(0w) &= 0\,a(w), & b(1w) &= 1\,c(w),\\ c(0w) &= 0\,a(w), & c(1w) &= 1\,d(w),\\ d(0w) &= 0\,w, & d(1w) &= 1\,b(w), \end{aligned}$$

together with $b(\varnothing) = c(\varnothing) = d(\varnothing) = \varnothing$.

Each of $a, b, c, d$ is a bijection of $W$, lies in $\mathrm{Aut}$, and is its own inverse.

The group $G$ and the generating set $S$

$G$ is the subgroup of $\mathrm{Aut}$ generated by $a, b, c, d$, that is, the smallest subgroup containing all four. It carries the group law of $\mathrm{Aut}$, which is composition.

$S$ is a subset of $G$: it is the set of those elements of $G$ whose underlying permutation is one of $a, b, c, d$. All four lie in $G$, so $S$ consists of exactly these four elements of $G$.

Balls and their sizes

For each integer $n \ge 0$, let $B(n)$ be the set of $g \in G$ that can be written as

$$g = s_1 s_2 \cdots s_m \qquad\text{with}\qquad 0 \le m \le n,$$

where each factor $s_i$ satisfies $s_i \in S$ or $s_i^{-1} \in S$. The product is taken in $G$, so it is the composition $s_1 \circ s_2 \circ \cdots \circ s_m$. The empty product ($m = 0$) is the identity $e$, so $B(0) = \{e\}$. Each generator is its own inverse, so the condition on the factors is simply $s_i \in \{a,b,c,d\}$.

$B(n)$ is finite, with at most $1 + 4 + 4^2 + \cdots + 4^n$ elements. Write $|B(n)|$ for its number of elements; this is a genuine count, not a default value.

Formally, the ball is evaluated at the real radius $n$ and that radius is rounded down to an integer. This rounding returns $n$ itself, so it changes nothing.

The number $\alpha_0$

Let

$$\lambda_0 = \sup\{\, x \in \mathbb{R} : x > 0 \text{ and } x^3 - x^2 - 2x - 4 = 0 \,\}.$$

The cubic $x^3 - x^2 - 2x - 4$ has exactly one positive real root. Its value is $\approx 2.46750$, between $2$ and $3$. So the set is a singleton and $\lambda_0$ is that root.

The supremum convention in use assigns $0$ to an empty or unbounded set of reals. That case does not arise here.

Then

$$\alpha_0 = \frac{\log 2}{\log \lambda_0} \approx 0.76743,$$

with natural logarithms. The ratio is the same in any base.

The assertion

For $n = 0, 1, 2, \dots$ define the real number

$$q_n = \frac{\log\bigl(\log |B(n)|\bigr)}{\log n}.$$

The statement asserts that this sequence converges to $\alpha_0$:

$$\lim_{n \to \infty} \frac{\log\bigl(\log |B(n)|\bigr)}{\log n} = \alpha_0 .$$

This is an ordinary limit along the natural numbers: for every $\varepsilon > 0$ there is an $N$ such that $|q_n - \alpha_0| < \varepsilon$ for all $n \ge N$. It is a full limit, not merely a bound on the $\limsup$ or the $\liminf$.

Conventions for small $n$. The logarithm here is defined on all of $\mathbb{R}$, with $\log 0 = 0$ and $\log x = \log|x|$ for $x < 0$. Division by zero gives $0$. These conventions affect only the first two terms:

• For $n = 0$: $|B(0)| = 1$, so the numerator is $\log(\log 1) = \log 0 = 0$. The denominator is $\log 0 = 0$, so $q_0 = 0$.
• For $n = 1$: the denominator is $\log 1 = 0$, so $q_1 = 0$.
• For $n \ge 2$: $\log n > 0$. Also $|B(n)| \ge 2$, because $a \ne e$ ($a$ sends the word $0$ to $1$). So $\log|B(n)| > 0$, and the numerator is the ordinary logarithm of a positive number.

Since a limit ignores finitely many terms, these conventions do not affect the assertion.