Let $G$ be a partially ordered abelian group (written additively). I want to add $\mathbb{R}$-multiples to $G$ in a "free" way ,thus extending $G$ to an ordered vector space.
Construction:
To this end, let $\mathbb{R} G$ be the quotient of the free $\mathbb{R}$-vector space generated by the set $G$, subject to the relation $$ 1 \cdot g + 1 \cdot h = 1 \cdot (g + h) $$ for $g, h \in G$. Define the ordering on $\mathbb{R} G$ by $x \le y :\Leftrightarrow y - x \in P$ where $$ P := \left\{ \left[\alpha_1 g_1 + \ldots + \alpha_n g_n \right] \mid \alpha_i \in \mathbb{R}^+ ,\, g_i \in G,\, g_i \ge 0 \; \forall i \right\} . $$
Define the inclusion map $$ \iota : G \to \mathbb{R}G $$ by $\iota(g) := [1 \cdot g]$. It is easy to see that $\iota$ is a homo of ordered groups. One can further check that $\mathbb{R} G$ indeed has the usual "free" property: If $A$ is an ordered vsp and $f: G \to A$ is a morphism of ordered groups, then $f$ factors over $\iota$ to a morphism of ordered vsp $\bar{f}: \mathbb{R} G \to A$.
Questions:
- Does this construction have a well-known name? I couldn't find any material on it.
I am trying to prove that $\iota$ is strictly monotonic if $G$ is, say, lattice-ordered. Any hints?
- It feels to me like one only needs semi-closedness of $G$, i.e., that $x \ge 0$ whenever $nx = x + \ldots + x \ge 0$ for some $n$.
- The statement is wrong if $G$ is not semi-closed because $\mathbb{R} G$ is always semi-closed (one can literally multiply with $\frac 1 n$ on both sides).