# Rational stable translation length

Let $G$ be a finitely generated group and $S$ a finite generating set and consider the word metric associated to $S$.

If $g\in G$, define its stable translation length as $l(g)=\lim_n \frac{d(e,g^n)}{n}$.

This number can actually be defined in a more general context: if $G$ acts by isometries on a set $X$, define $l(g)=\lim_n \frac{d(x,g^n\cdot x)}{n}$ and this do not depend on the point $x$, but we restrict our attention to a word metric in the following.

If $G$ is hyperbolic, then there exists $C\in \mathbb{R}$, such that for every $g\in G$, $l(g)\in C\mathbb{Z}$.

My question is the following: are there examples of groups not satisfying this property for the word metric ? More precisely, fixig a word metric on a group $G$, can we find two elements $g,h\in G$ such that $l(g)$ and $l(h)$ are arbitrarily close ? (settled, see the comment of YCor below).

I am specially interested with hyperbolic elements in relatively hyperbolic groups, so another related question is the following: If $G$ is relatively hyperbolic, can one find two hyperbolic elements $g,h$ such that $l(g)$ and $l(h)$ are arbitrarily close ?

As noticed by YCor, it would be enough to find either a loxodromic element with irrational translation length, or to find a relatively hyperbolic group with loxodromic elements of rational translation length but arbitrarily large denominator.

• The proof that $l(g)$ is independent of the basepoint uses hyperbolicity of the space. So I don't think it's true that the number makes sense for any action of a hyperbolic group $G$ on a metric space $X$. – HJRW Jan 23 '18 at 13:44
• @HJRW I don't think it does. Fix $x$ and $y$ in $X$. Then, $d(x,g^n\cdot x)\leq d(x,y)+d(y,g^n\cdot y)+d(g^n\cdot y, g^n\cdot x)=2d(x,y)+d(y,g^n\cdot )$. Thus, $l(g)$ computed with $x$ is lower or equal to $l(g)$ computed with $y$, and by symmetry, the two $l(g)$ coincide. – M. Dus Jan 23 '18 at 13:47
• You're quite right -- my bad! I had forgotten where hyperbolicity ends up playing a role. – HJRW Jan 23 '18 at 13:52
• Anyway, your question about relatively hyperbolic groups is more or less the same as for arbitrary groups. If I have an example of a group $G$ with your property, then $G*G$ will be a relatively hyperbolic group with the same property, using the obvious retraction $G*G\to G$. – HJRW Jan 23 '18 at 13:55
• And googling yields G. Conner, Arch. Math. 69.4 265-274 1997 A class of finitely generated groups with irrational translation numbers (link.springer.com/article/10.1007/s000130050120): polycyclic with an element with irrational stable length. We there's a polycyclic group and generating subset and an irrational $r$ such that the set of stable translation lengths contains $\mathbf{N}\cup\mathbf{N}r$, which contains arbitrary close points. – YCor Jan 23 '18 at 20:38