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Riesz group
Let $G$ be a pogroup and $G^{+}$ the positive cone of $G$. The following are equivalent:
1. $G$, as a poset, sastisfies the Riesz interpolation property;
2. if $x,y_{1},y_{2}\in G^{+}$ and $x\leq y_{1}y_{2}$, then $x=z_{1}z_{2}$ with $z_{i}\leq y_{i}$ for some $z_{i}\in G^{+}$, $i=1,2$.
The second property above, put it plainly, says that any positive element that is bounded from above by a product of positive elements, can be “decomposed” as a product of positive elements. This property is known as the Riesz decomposition property.
Proof.
$(1\Rightarrow 2)$. Given $x\leq y_{1}y_{2}$ and $e\leq x,y_{1},y_{2}$. Set $r=y_{1}^{{1}}x$. Then we have four inequalities, which can be abbreviated as $\{r,e\}\leq\{x,y_{2}\}$, where each of the elements in the first set is less than or equal to each of the elements in the second set. By the Riesz interpolation property, we can insert an element between the sets: $\{r,e\}\leq z_{2}\leq\{x,y_{2}\}$. From this it is clear that $e\leq z_{2}\leq y_{1}$. Set $z_{1}=xz_{2}^{{1}}$. Since $z_{2}\leq x$, we have $e\leq xz_{2}^{{1}}=z_{1}$. Also, since $y_{1}^{{1}}x=r\leq z_{2}$, $z_{2}^{{1}}\leq x^{{1}}y_{1}$, so that $z_{1}\leq x(x^{{1}}y_{1})=y_{1}$.
$(2\Rightarrow 1)$. Suppose $\{a,b\}\leq\{c,d\}$. Set $x=a^{{1}}c$, $y_{1}=a^{{1}}d$ and $y_{2}=b^{{1}}c$. Then $x,y_{1},y_{2}\in G^{+}$. Since $e\leq db^{{1}}$, we have $x=a^{{1}}c=a^{{1}}ec\leq a^{{1}}(db^{{1}})c=(a^{{1}}d)(b^{{1}}c)=y_{1}y_% {2}$. By the Riesz decomposition property, $a^{{1}}c=x=z_{1}z_{2}$ for some $z_{1},z_{2}\in G$ with $e\leq z_{1}\leq y_{1}=a^{{1}}d$ and $e\leq z_{2}\leq y_{2}=b^{{1}}c$. The decomposition equality can be rewritten as $c=az_{1}z_{2}$, and the last two inequalities can be rewritten as $az_{1}\leq d$ and $bz_{2}\leq c$. Set $s=az_{1}$, so we have $a\leq az_{1}=s\leq az_{1}z_{2}=c$. Furthermore, since $bz_{2}\leq c=az_{1}z_{2}$, we get $b\leq az_{1}=s$. Finally from $z_{1}\leq a^{{1}}d$, we have $s=az_{1}\leq d$. Gather all the inequalities, we have finally $\{a,b\}\leq s\leq\{c,d\}$. ∎
Definitions. Let $G$ be a pogroup.

$G$ is called an interpolation group if $G$ satisfies one of the two equivalent conditions in the theorem above.

$G$ is a Riesz group if $G$ is a directed interpolation group. By directed we mean that $G$, as a poset, is a directed set.

$G$ is an antilattice if $G$ is a Riesz group with the property that if $a,b\in G$ have a greatest lower bound, then $a$ and $b$ are comparable.
Any latticeordered group is an antilattice. Here is an interpolation group that is not an lgroup. Let $G=\mathbb{Z}\times\mathbb{Z}$. Define $(a,b)\leq(c,d)$ iff $(c,d)(a,b)=(0,n)$ for some nonnegative integer $n$. This order is a partial order. But $G$ is not a lattice, since $(1,0)\vee(0,0)$ does not exist. However, if any two elements in $G$ have either an upper bound or a lower bound, then the elements are in fact comparable. Therefore, $\{a,b\}\leq\{c,d\}$ means that $a,b,c,d$ form a chain. So any element in the interval $[a\vee b,c\wedge d]$ “interpolates” $\{a,b\}$ and $\{c,d\}$. Note that $G$ is not a Riesz group, for otherwise it would be a chain.
Mathematics Subject Classification
06F20 no label found20F60 no label found Forums
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