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 $x\le {y}_{1}{y}_{2}$, then $x={z}_{1}{z}_{2}$ with ${z}_{i}\le {y}_{i}$ 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 $x\le {y}_{1}{y}_{2}$ and $e\le x,{y}_{1},{y}_{2}$. Set $r={y}_{1}^{1}x$. Then we have four inequalities, which can be abbreviated as $\{r,e\}\le \{x,{y}_{2}\}$, 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: $\{r,e\}\le {z}_{2}\le \{x,{y}_{2}\}$. From this it is clear that $e\le {z}_{2}\le {y}_{1}$. Set ${z}_{1}=x{z}_{2}^{1}$. Since ${z}_{2}\le x$, we have $e\le x{z}_{2}^{1}={z}_{1}$. Also, since ${y}_{1}^{1}x=r\le {z}_{2}$, ${z}_{2}^{1}\le {x}^{1}{y}_{1}$, so that ${z}_{1}\le x({x}^{1}{y}_{1})={y}_{1}$.
$(2\Rightarrow 1)$. Suppose $\{a,b\}\le \{c,d\}$. 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 $e\le d{b}^{1}$, we have $x={a}^{1}c={a}^{1}ec\le {a}^{1}(d{b}^{1})c=({a}^{1}d)({b}^{1}c)={y}_{1}{y}_{2}$. By the Riesz decomposition property, ${a}^{1}c=x={z}_{1}{z}_{2}$ for some ${z}_{1},{z}_{2}\in G$ with $e\le {z}_{1}\le {y}_{1}={a}^{1}d$ and $e\le {z}_{2}\le {y}_{2}={b}^{1}c$. The decomposition equality can be rewritten as $c=a{z}_{1}{z}_{2}$, and the last two inequalities can be rewritten as $a{z}_{1}\le d$ and $b{z}_{2}\le c$. Set $s=a{z}_{1}$, so we have $a\le a{z}_{1}=s\le a{z}_{1}{z}_{2}=c$. Furthermore, since $b{z}_{2}\le c=a{z}_{1}{z}_{2}$, we get $b\le a{z}_{1}=s$. Finally from ${z}_{1}\le {a}^{1}d$, we have $s=a{z}_{1}\le d$. Gather all the inequalities, we have finally $\{a,b\}\le s\le \{c,d\}$. ∎
Definitions. Let $G$ be a pogroup.
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$G$ is a Riesz group if $G$ is a directed interpolation group. By directed we mean that $G$, as a poset, is a directed set^{}.

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$G$ 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 $(a,b)\le (c,d)$ 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, $\{a,b\}\le \{c,d\}$ 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.
Title  Riesz group 

Canonical name  RieszGroup 
Date of creation  20130322 17:09:18 
Last modified on  20130322 17:09:18 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  7 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 06F20 
Classification  msc 20F60 
Defines  Riesz decomposition property 
Defines  interpolation group 
Defines  antilattice 