# Example involving partially ordered Abelian groups

Definition 1: Let $(G,\leq)$ be a nonzero partially ordered Abelian group with order unit $u$. (Recall that $u\in G$ is a order unit if, for every $g\in G$, there exists $N\in\mathbb N$ such that $-Nu\leq g\leq Nu$.) For any $g\in G$, define the quantities (as it is done in proposition 4.7 of the book Partially Ordered Abelian Groups (by K.R. Goodearl)) \begin{align*} p(g)=\sup\{k/m:k\in\mathbb Z;m\in\mathbb N;ku\leq mg\} \end{align*} and \begin{align*} r(g)=\inf\{l/n:l\in\mathbb Z,n\in\mathbb N,ng\leq lu\}. \end{align*} Question: It is clear that $-\infty<p(g)\leq r(g)<\infty$ for any $g\in G$. What would constitute an example of a group with an order unit $(G_0,u_0)$ and an element $g_0\in G_0$ such that $p(g_0)<r(g_0)$.

Definition 2: We call a function $s:G\to\mathbb R$ a state on $G$ if $s$ is a group order-homomorphism such that $s(u)=1$.

Remark 1: It can be shown that for any $p(g)\leq q\leq r(g)$, there exists a state $s$ on $G$ such that $s(g)=q$. Therefore, there cannot exist a unique state on $(G_0,s_0)$ (as is the case for totally ordered Abelian groups, so any example involving $\mathbb R$ with its usual order is out of the question).

Remark 2: It can be shown that, for any state $s$ on $G$ and element $g\in G$, we have that $p(g)\leq s(g)\leq r(g)$. Therefore, it is enough to find a group with an order unit $(G_0,u_0)$, an element $g_0\in G_0$ and two states $s_0$ and $s_0'$ on $G_0$ such that $s_0(g)\neq s_0'(g)$.

Let $G=\mathbb{R}^2$ with order defined by
$(a_1,a_2)\leq (b_1,b_2) \iff a_1\leq b_1$ and $a_2\leq b_2$.
Let $u=(1,1)$ and let $g_0=(1,2)$.
Then $p(g_0)=1$ and $r(g_0)=2$.
• Also, the "states" on this group are all of the form $(a,b)\mapsto \lambda_1a+\lambda_2b$ where $\lambda_1,\lambda_2\geq 0$. Nov 24 '13 at 4:58