lexicographic ordering. To fix notations, we use the *little endian* convention: the leftmost factor will have the greatest weight. More precisely, if $\Lambda_1$ and $\Lambda_2$ are totally ordered abelian groups, the lexicographic ordering on $\Lambda_1\oplus\Lambda_2$ is defined by $(x_1,x_2)\leq (y_1,y_2)$ if $x_1<y_1$ or ($x_1=y_1$ and $x_2\leq y_2$).

A morphism ${\varphi}\co \Lambda{\rightarrow}\Lambda'$ between two totally-ordered abelian groups is a non-decreasing group morphism. Given $a,b\in\Lambda$, the subset $[a,b]=\{x\in\Lambda \,|\, a\leq x\leq b\}$ is called the *segment* between $a$ and $b$. A subset $E\subset\Lambda$ is *convex* if for all $a,b\in E$, $[a,b]\subset E$. The kernel of a morphism is a convex subgroup, and if $\Lambda_0\subset \Lambda$ is a convex subgroup, then $\Lambda/\Lambda_0$ has a natural structure of totally ordered abelian group. By *proper* convex subgroup of $\Lambda$, we mean a convex subgroup strictly contained in $\Lambda$.

The set of convex subgroups of $\Lambda$ is totally ordered by inclusion. The height of $\Lambda$ is the (maybe infinite) number of proper convex subgroups of $\Lambda$. Thus, the height of ${{\mathbb{R}}}^n$ is $n$. $\Lambda$ is *archimedean* if its height is at most 1. It is well known that a totally ordered abelian group is archimedean if and only if it is isomorphic to a subgroup of 