Convex lattice subgroup

A convex subgroup of an ordered group that is also closed under join and meet. A normal convex lattice subgroup is the kernel of the quotient lattice-ordered group homomorphism.

Assume $uv\le v^2{u}^2$ for all $u,v\ge 1$ in a lattice-ordered group. For a convex lattice subgroup $H$ and $a\ge 1$, the displayed bounds allow $h\in H$ with $h\ge 1$ and integers $n\ge 0$. To prove the first set is a subgroup, use $(h{a}^n)(k{a}^m)\le h{k}^2{a}^{2n+m}$. Products and inverse products are both bounded because $xy\le |x||y|$ and $(xy{)}^{-1}\le |y||x|$. Taking the larger exponent and a join of the two bounds in $H$ bounds their absolute value in a lattice-ordered group. Inversion preserves that absolute value; common bounds using $h\vee k$ prove lattice closure and convexity. The set contains $H$ and $a$, and any convex lattice subgroup containing them contains the entire set. The second equality follows by applying the proof to the opposite group, where the required inequality is the original one with the variables interchanged.

The displayed relation makes the right cosets into a lattice, with $Pg\vee Ph=P(g\vee h)$ and $Pg\wedge Ph=P(g\wedge h)$. Antisymmetry uses convexity: $g\le ph$ and $h\le qg$ give $1\le ph{g}^{-1}\le pq\in P$, hence $h{g}^{-1}\in P$. A prime convex lattice subgroup makes this order total even if the subgroup is not normal.

The convex lattice subgroups of a lattice-ordered group form a distributive lattice, with intersection as meet and the convex lattice subgroup generated by a union as join. A positive element of a join is bounded by a finite product of positive elements from the generating subgroups. Riesz decomposition in a lattice-ordered group splits such an element into positive factors from those subgroups. If the original element belongs to another convex subgroup $E$, every factor does too. Consequently $E\cap (C\vee D)=(E\cap C)\vee (E\cap D)$.

For $u\ge 1$ in a lattice-ordered group, the displayed set is the smallest convex lattice subgroup containing $u$. Products of two such bounds are bounded by a larger power of $u$; inversion and the lattice operations preserve the bounds. This gives an explicit test for membership.

A proper convex lattice subgroup $P$ is prime when $a,b\ge 1$ and $a\wedge b=1$ imply $a\in P$ or $b\in P$. Equivalently the ordered right cosets of $P$ form a totally ordered set. Prime refers to this order property, and does not require a normal subgroup or prime index.

A value of $g\ne 1$ is a convex lattice subgroup maximal among those omitting $g$. Zorn's lemma supplies a value: the union of a chain of convex lattice subgroups omitting $g$ still omits $g$. Such subgroups are prime. Their unique cover is the smallest convex lattice subgroup properly containing them.

Let $p,q,x\ge 1$ in a lattice-ordered group. Under the displayed hypothesis, suppose $d=(x{q}^{-2}{p}^{-2})\vee 1>1$ and choose a value $P$ of $d$. If $x\in P$, then $1\le d\le x$ contradicts its omission. Otherwise extend $P$ to a value $V$ of $x$. The right-coset test is $Vu\le Vv⟺(u{v}^{-1})\vee 1\in V$. If $p\notin V$, put $c=(x{q}^{-2}{p}^{-1})\vee 1\in V$. Then $(p{c}^{-1})\vee 1\notin V$, while its inverse positive part is $(c{p}^{-1})\vee 1=d$. These parts have meet one, so primeness of $P$ forces $d\in P$. If $p\in V$, put $c=(x{q}^{-1})\vee 1\in V$ and $t=p^{2}q\notin V$; the disjoint pair $(t{c}^{-1})\vee 1$ and $(c{t}^{-1})\vee 1=d$ gives the same contradiction. No normality assumption is used in this separation argument.

If $V$ is a value in a lattice-ordered group of $t>1$, every convex lattice subgroup strictly containing $V$ contains $t$. Thus the convex lattice subgroup generated by $V$ and $t$ is the unique least subgroup properly containing $V$. It is called the cover of $V$. There is no intermediate convex lattice subgroup. A normal-valued lattice-ordered group requires $V$ to be a normal subgroup of this cover, not necessarily of the whole group. Values are prime convex lattice subgroups: maximality omitting $t$ makes them meet-irreducible in the lattice of convex lattice subgroups, so their ordered right cosets form a chain.

Under the inequality in one-sided bounds for adjoining a positive element, if $a\ge 1$ belongs to the cover of a value in a lattice-ordered group $V^{\ast }$ but not to $V$, then $⟨V,a{⟩}_c=V^{\ast }$. Thus every $b\ge 1$ in the cover is bounded above by both $v{a}^n$ and $a^{m}w$ for suitable nonnegative $v,w\in V$ and integers $m,n\ge 0$. The corresponding right and left coset bounds give the displayed cofinality, without first assuming that $V$ is normal. Left-coset order is obtained by inversion of right-coset order.