Rationality of translation lengths in hyperbolic groups Recall that the translation length $\tau(g)$ of an element $g \in G$ is the limit $d(1, g^n)/n$, where $d$ is the word metric on $G$ with resepct to some generating set.
It is a theorem of Gromov that the translation length of any hyperbolic element in a hyperbolic group is a rational number with denominator uniformly bounded.
This result is a bit surprising to me that the translation length is such a delicate thing. This also recalls me that the growth series of a hyperbolic group is a rational function (a theorem of Cannon). 
Question 1: I'm wondering whether anyone could give some explaination to these such type of results? For example, what is the motivation of Gromov's theorem? any analogous results on Riemaninan manifolds?
Quesion 2: I would like to see whether there is certain relation over the spectums of translation lengths of all elements. An analogous result comes to my mind at this moment is the Mcshane identity over the lengths of geodesics on a punctured surfaces. See http://en.wikipedia.org/wiki/McShane%27s_identity
For the translation lengths, if we take the reciprocal of $\tau(g)^2$ and sum all of them, we see that it is a finite number. It is not clear how this finite number depends on the choices of generating sets. Can such type of an idea be further improved? 
I saw some negative result in "On the absence of McShane-type identities for the outer space" by Ilya Kapovicha, Igor Rivinb. But I did not have a chance to loot at it very carefully, maybe their paper also gives a negative answer to my question here.
 A: Regarding question 1, this theorem of Gromov is a two-fold phenomenon related to the very special and very "discrete" nature of the word metric. First, that metric takes on only integer values. Second, the special effect of hyperbolicity of the Cayley graph is that the expression $d(1,g^n)/n$ achieves its limit at some finite value $n=N$ independent of $g$, and so the translation length equals an integer divided by $N$. 
It is interesting to see what happens when you make some other assignment to the lengths of the generators. You could choose them to be rational numbers, for instance. But since there are only finitely many generators, then in computing any translation length $d(1,g^n)$ one is summing rational numbers with bounded denominator, and all such sums have the same denominator bound. The discrete nature of the Cayley graph still implies that ratio $d(1,g^n)/n$ achieves its limit at some finite $n=N$ independent of $g$, and so the translation lengths of all elements are still rational with bounded denominators. Or, you could choose the translation lengths of the generators to be in the field $\mathbb{Q}(\sqrt{2})$. Again the translation numbers of all elements of the group would be elements of the field $\mathbb{Q}(\sqrt{2})$ with bounded denominators.
You ask whether there are analogous results in Riemannian manifolds. If $M$ is a closed Riemannian manifold with word hyperbolic fundamental group, it seems possible to me that either of the two "discrete" phenomena mentioned earlier are candidates for failure. First, I am not sure whether it is still true that $d(1,g^n)/n$ achieves its limiting value at some finite $n=N$ independent of $g$. But even if it does, $d(1,g^n)$ is no longer computed by summing integers. 
A: I would like to add some clarification about Question 1. The fact that all the translation numbers of a (Cayley graph of a) group belong to $\frac{1}{N}\mathbb{Z}$ for some $N \geq 1$ is a rather common phenomenon, and it is often a consequence of the following dichotomy:

There exists some $N_0 \geq 1$ such that, for every $g \in G$ of infinite order, $g^k$ admits an axis, i.e. a bi-infinite geodesic line on which it acts as a translation, for some $k \leq N_0$.

(Here, $\tau(g^k)$ must be an integer, so $\tau(g)=\tau(g^k)/k$ must belong to $\frac{1}{N_0!}\mathbb{Z}$.)
The dichotomy holds for hyperbolic groups, Garside groups, and groups admitting Cayley graphs in specific families of graphs such as median graph (a.k.a. one-skeleta of CAT(0) cube complexes), bridged graphs (a.k.a. one-skeleta of systolic complexes), or Helly graphs. The last source of examples includes right-angled Artin groups and braid groups, among others.
A useful criterion which can be used to prove the dichotomy can be extracted from Delzant's argument for hyperbolic groups:

Assume that $g \in G$ acts cocompactly on an isometrically embedded quasi-line $A(g) \subset G$. Then $g^k$ admits an axis for some $k \leq \mathrm{width}(A(g))$.

(The width of a quasi-line is the minimal size of a set of vertices separating the two ends of the quasi-line.)
The idea is to fix an arbitrary total order on the edges of $A(g)/\langle g \rangle$ and to extend it lexicographically to an order on the bi-infinite geodesics in $A(g)$. Then the minimal geodesics have to be pairwise disjoint, which implies that that there must be at most $\mathrm{width}(A(g))$ of them. This proves that $g^{\mathrm{width}(A(g))!}$ preserves these bi-infinite geodesics, and so admits an axis. (In order to get some $k \leq \mathrm{width}(A(g))$ as the power, there is some extra-work to do.)
Therefore, in order to show that an element has a rational translation number with a controlled denominator, it suffices to find a stabilised quasi-line with a controlled width. In hyperbolic groups, this is rather easy thanks to the Morse property. In fact, the same argument works for Morse elements in arbitrary groups. (An element $g \in G$ is $M$-Morse if any $(A,B)$-quasigeodesics between two points in $\langle g \rangle$ stays in the $M(a,b)$-neighbourhood of $\langle g \rangle$. The map $M$ is the Morse gauge.)

Proposition. Let $G$ be a finitely generated group. For every $M$, there exists some $N \geq 1$ such that the translation number of every $M$-Morse element is a rational number with a denominator $\leq N$.

It suffices to set $A(g):= \text{convex hull of } \langle g \rangle$. The width of $A(g)$ is controlled by the localy finiteness of (the Cayley graph of) $G$ and the Hausdorff distance between $\langle g \rangle$ and $A(g)$ (itself controlled by the Morse gauge $M$). Then the criterion applies.
Because every infinite-order element in a hyperbolic group is $M$-Morse for some $M$ depending only on the hyperbolicity constant, the translation numbers are rationals with uniformly bounded denominators.
