# Are all groups with relators whose exponents sum to $0$ totally orderable?

I apologize for the lengthy preamble, I have no idea how much of what I am relying on is common knowledge (or even correct to be honest). Let $$G=\langle S\mid R\rangle$$ be a finitely presented group of rank $$n$$, and define $$\sigma:\mathbb{F}_n\to \mathbb{Z}$$ to be the sum of the exponents of an element of $$\mathbb{F}_n,$$ with the usual basis for $$\mathbb{F}_n$$. We see $$\sigma$$ is well defined, since $$\mathbb{F}_n$$ is a group of equivalence classes on $$n$$ letters (and $$n$$ "inverse" letters) where $$x\sim y$$ if and only if $$x=ww'$$ and $$y=waa^{-1}w'$$ where $$w,w'$$ are strings on $$n$$ letters and $$a$$ is a letter. Now, suppose $$\sigma(r)=0$$ for all $$r\in R.$$ Then $$\sigma$$ may be defined similarly on $$G$$, since $$G$$ is a group of equivalence classes of elements of $$G$$ where $$x\sim y$$ if and only if $$x=ww'$$ and $$y=wpw',$$ where either $$p=ss^{-1}$$ for $$s\in S$$ or $$p\in R.$$

Examples of groups of this type include Right Angled Artin Groups, Free Groups (since $$R=\{\}$$ in this case), and Braid Groups.

My question is this: are all of these groups totally orderable?

• Your $\sigma$ is only well-defined if you fix a basis. For example, the free group of rank $2$ can be generated by either $x$ and $y$; or by $xy$ and $y$. The sum of exponents of $xy^2$ relative to the first basis is $3$, but the sum of exponents relative to the second basis is $2$, since $xy^2 = (xy)y$ is a product of just two basis elements, raised to the first power. So you need to specify your free basis for $\sigma$ to work. Jan 14, 2020 at 7:24
• The title is misleading: I think you mean "with relators whose exponents sum to $0$".
– YCor
Jan 14, 2020 at 19:54
• You should define "totally orderable". It sometimes means: admits a left-invariant total order; sometimes, admits a bi-invariant total order, which is strictly stronger. Anyway this doesn't matter since the answer shows that you can even arrange the group to be non-torsion-free, say with $\langle x,y|[x,y]^2\rangle$.
– YCor
Jan 14, 2020 at 19:55

## 1 Answer

Any group with a presentation of the form $$\langle \mathbf{x}\mid R^n\rangle$$, with $$R\in [F(\mathbf{x}), F(\mathbf{x})]$$ and $$n>1$$, satisfies the exponent-sum condition and contains torsion elements$$^{[1]}$$. However, these groups are not totally-ordered as any totally-ordered group is torsion-free.

The issue with your idea is that it is simply saying that the group $$G$$ surjects onto $$\mathbb{Z}^{|S|}$$, which is a "global" property of the group. For orderability you need "local" properties, so properties of (finitely generated) subgroups. In the above example, although the group itself maps onto $$\mathbb{Z}$$, the subgroup $$\langle R\rangle$$ does not.

A condition which does imply orderability is local indicability. A group is locally indicable if every finitely generated subgroup subject into $$\mathbb{Z}$$. These groups were first studied by Higman in his PhD thesis (regarding zero divisors of group rings), and they admit a total order (although not necessarily a two-sided order). It is a theorem of Brodskii, and independently Howie$$^{[2]}$$, that torsion-free one-relator groups are locally indicable (so groups $$\langle \mathbf{x}\mid R\rangle$$ where $$R\neq S^n$$ for any $$n>1$$, and no restriction on $$R$$ only being taken from the derived subgroup!).

For the specific groups you mention: free groups, and more generally RAAGS, are bi-orderable$$^{[3]}$$, while the Braid groups $$B_n$$ are not for $$n\geq3^{[4]}$$. However, the braid groups $$B_n$$ are all right-orderable, $$n\geq1^{[5]}$$.

[1] Theorem 4.12 of the book Wilhelm Magnus, Abraham Karrass, and Donald Solitar. Combinatorial group theory. Dover Publications, 1976.

[2] Howie, James. "On locally indicable groups." Mathematische Zeitschrift 180.4 (1982): 445-461.

[3] Duchamp, Gérard, and Jean-Yves Thibon. "Simple Orderings for Free Partially Commutative Groups." IJAC 2.3 (1992): 351-356.

[4] Neuwirth, Lo Po. "The status of some problems related to knot groups." Topology conference. Springer, Berlin, Heidelberg, 1974.

[5] Dehornoy, Patrick. "Braid groups and left distributive operations." Transactions of the American Mathematical Society 345.1 (1994): 115-150.

• Surely both $B_1$ and $B_2$ are right-orderable, for silly reasons Jan 15, 2020 at 15:35
• @RyleeLyman Yes, they are also bi-orderable! The $n\geq3$ was attached to the wrong statement. I've correct it now. Thanks. Jan 15, 2020 at 16:14