# A problem on automatic groups and geodesic paths on the Cayley graph

Let $\Gamma = \langle S \mid R \rangle$ be a finitely generated group, with the neutral element $e \not \in S= S^{-1}$.
Let $\ell : \Gamma \to \mathbb{N}$ be the world length related to $S$.

For any $g \in \Gamma$ and for any $s \in S$, let $p_s(g)$ be the number of geodesic paths from $g$ to $e$ beginning by the edge $[g,gs]$ in the Cayley graph $\mathcal{G}(S,R)$. Next, let $p(g) = \sum_{s \in S}p_s(g)$.

The marked group $(\Gamma, S)$ belongs to the class $\mathcal{C}$ if:
$\forall g \in \Gamma$, $\exists \lambda_g > 0$ such that $\forall s \in S$, $\forall h \in \Gamma \setminus \{e,g^{-1} \}$
$$| \frac{p_s(g \cdot h)}{p(g \cdot h)}-\frac{p_s(h)}{p(h)}| \le \frac{\lambda_g}{\ell(h)}$$ Meaning: this property means that the left multiplication by $g$, almost preserves the proportion of geodesics from a given element $h$ to $e$, for a given direction $[h,hs]$.

Question: Are the automatic groups of class $\mathcal{C}$?

Note that the property of being automatic is independent on the choice of the set of generators.
Bonus question: Is the property of being $\mathcal{C}$, also independent on the choice of the set of generators?

Motivation: the class $\mathcal{C}$ appears naturally when we try to define a noncommutative geometry from a finitely generated group. We will show that (for their usual presentation) the automatic groups $\mathbb{Z}^2$ and $\mathbb{F}_2$ belong to $\mathcal{C}$, whereas the non-automatic Baumslag-Solitar group $B(2,1)$ not.

• First, $\mathbb{Z}^2 = \langle a^{\pm 1},b^{\pm 1} \mid aba^{-1}b^{-1} \rangle$. Let $h = a^nb^m$ (without lose of generality, take $m>0$)
Then $\ell(h) = m+n$, $p(h) = {n+m \choose m}$, $p_{b^{-1}}(g)={n+m-1 \choose m-1}$ and $\frac{p_{b^{-1}}(h)}{p(h)}\ell(h) = m$.
We finally observe that $\lambda_g = \ell(g)^2$ should be enough to have the property.
• Next, $\mathbb{F}_2 = \langle a^{\pm 1},b^{\pm 1} \mid \emptyset \rangle$. Let $h \in \mathbb{F}_2$, then $\exists! s \in S$ such that $p_s(h) \neq 0$ and $p(h) = 1$. We observe that $\lambda_g = \ell(g)$ is enough to have the property.
• Finally, $B(2,1) = \langle a^{\pm 1},b^{\pm 1} \mid a^2ba^{-1}b^{-1} \rangle$. Then $a^{2^n} = b^{n-1}a^2b^{1-n}$ and $\ell(a^{2^n}) = 2n$.
For $n>1$, we observe that $| \frac{p_b(a \cdot a^{2^n})}{p(a \cdot a^{2^n})}-\frac{p_b(a^{2^n})}{p(a^{2^n})}| = | \frac{2}{4}-\frac{2}{2}| = \frac{1}{2}$. So the property fails.

Remark: The Baumslag-Solitar $B(2,1)$ is amenable and solvable but not of polynomial growth. So the class $\mathcal{C}$ contains neither all the amenable marked groups, nor all the solvable marked groups.

Bonus question: Are the polynomial growth groups of class $\mathcal{C}$?
Gromov's theorem: a group is of polynomial growth iff it is virtually nilpotent.