This is a followup question related to this question. Recall that a left-invariant partial order on a finitely generated group $G$ is called a partial word order if for every $a\le b\le c$ we have $|b|\le C(|a|+|c|)$ for some constant $C$ (where $|x|$ is the word length of $x$). For example, the following partial order on ${\mathbb Z}^2$: $(m,n)\le (k,l)$ iff $m\le k, n\le l$ is a word partial order. Certainly the empty (trivial) partial order ($a\le b$ iff $a=b$) is always a word order.

Question 1. Is there a canonical way to construct non-trivial partial word orders on groups?

Update 2 A sub-question: is there a general algebraic property of a group that guaranties existence of such non-trivial partial orders? (Being free Abelian of finite rank is such a property, but I am looking for non-trivial answers.)

Update 1 One possible generalization. I call a partial order on a group quasi-invariant if for every $g,a,b$ if $a\lt b$, then there exist two elements $c,d$ such that $gac\lt gbd$ and $|c|,|d|\lt C$ for some constant $C$.

Question 2. Did anybody study quasi-invariant partial orders on groups?

Motivation The reason I want to study such things is to introduce an extra structure on the asymptotic cones of groups. If one tries to define a partial order on an asymptotic cone, one would need a quasi-(left)-invariant word partial order on the group.


If G is a torsion group, then no elements may be comparable, so the only left-invariant order is the trivial one. That seems to rule out a canonical construction.

  • $\begingroup$ I did not say that the canonical construction should always be non-trivial. If the group is trivial, all orders on it are also trivial. But your remark does rule out lots of groups. $\endgroup$ – Mark Sapir Nov 1 '10 at 8:20
  • $\begingroup$ In general, question 1 is formulated in such a way that a negative answer seems impossible. How do you prove that a canonical way does not exist? The problem is that "canonical" is not precisely defined. When you see it, you can say that it is, but proving that it does not exist is not possible. I have reformulated the question to make it more concrete. Still "no" is not possible, but there are more variations for "yes". $\endgroup$ – Mark Sapir Nov 1 '10 at 8:52

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