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.

  • $\begingroup$ 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 $X$. $\endgroup$
    – HJRW
    Jan 23, 2018 at 13:44
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    $\begingroup$ @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. $\endgroup$
    – M. Dus
    Jan 23, 2018 at 13:47
  • $\begingroup$ You're quite right -- my bad! I had forgotten where hyperbolicity ends up playing a role. $\endgroup$
    – HJRW
    Jan 23, 2018 at 13:52
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    $\begingroup$ 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 $G*G\to G$. $\endgroup$
    – HJRW
    Jan 23, 2018 at 13:55
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    $\begingroup$ 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. $\endgroup$
    – YCor
    Jan 23, 2018 at 20:38

1 Answer 1


As already mentioned in the comments, there exist finitely generated groups having non-discrete translation spectra. (There seems to be only a few examples though, it would be interesting to have more.) About relatively hyperbolic groups, we have the following statement:

Proposition. Let $G$ be a relatively hyperbolic group. There exists some $\epsilon>0$ such that the translation number of every hyperbolic element is a rational $>\epsilon$.

The fact that the translation number is rational is actually quite general. See in my other answer here for the sketch of a proof that Morse elements in finitely generated groups have rational translation numbers. In order to bound below the translation number, we can consider the action of $G$ on the hyperbolic space $X$ obtained by coning-off $G$ over its parabolic subgroups, i.e. $X$ is the graph obtained from (the Cayley graph of) $G$ by adding an edge between any two vertices that belong to a common parabolic subgroup. The action of $G$ on $X$ is acylindrical and an element of $G$ is loxodromic in $X$ if and only if it is hyperbolic in $G$. Therefore, the translation length in $X$ of a hyperbolic $g \in G$ cannot be arbitrarily small. (See for instance Lemma 2.1 in Fujiwara's article Subgroups generated by two pseudo-Anosov elements in a mapping class group I.) But the obvious map $G \to X$ is $1$-Lipschitz, so the translation length of $g$ in $X$ is at most the translation number of $g$ in $G$. The desired conclusion follows.

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    $\begingroup$ Thank you for this answer. I eventually managed to find a proof of your stated Proposition. This is Proposition 7.8 in this paper arxiv.org/pdf/1811.10849.pdf. I should have answered my own question. Anyway, I'm going to accept yours since it answers a more general question. $\endgroup$
    – M. Dus
    Sep 7, 2022 at 8:30

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