# Is Thompson's group definably orderable?

Is Thompson's group $$F$$ definably left-orderable? definably bi-orderable?

Orderability definitions: Recall that a group $$G$$ is left-orderable (resp. bi-orderable) if it admits a left-invariant (resp. bi-invariant) total order. If $$S$$ is a submonoid of $$G$$ such that $$S\cup S^{-1}=G$$ and $$S\cap S^{-1}=\{1_G\}$$ (call this a cone in $$G$$), then $$\le_S$$ defined by: $$g\le_S h\Leftrightarrow g^{-1}h\in S$$ is a left-invariant total order, and $$S$$ is conjugation-invariant iff $$\le_S$$ is bi-invariant. Conversely if $$\le$$ is a left-invariant total order then $$S_\le=\{g:g\ge 1\}$$ is a cone (which is conjugacy-invariant iff $$\le$$ is bi-invariant. We have bi-orderable $$\Rightarrow$$ left-orderable $$\Rightarrow$$ torsion-free.

Definability: a subset of a group $$G$$ is definable if it can be described using logical Boolean operators, quantifiers on group elements, group operations, and parameters in the group: for instance for $$a,b\in G$$, one can consider the definable subset of $$G$$: $$\{x\in G:\forall y\in G,\exists z\in G:[a,z][b,z^3]=[x,y^5] \vee (x^2=y^2=1)\}$$ (this example is totally random). Similarly one can define a definable subset of $$G^n$$ for every $$n$$.

A group $$G$$ is definably (left/bi)-orderable if it admits a (left/bi)-invariant total order that is a definable subset of $$G^2$$, or equivalently whose positive cone is a definable subset of $$G$$.

One motivation is that such a group satisfies a single sentence $$\Phi$$ such that every group satisfying $$\Phi$$ is also (left/bi)-orderable (and hence torsion-free). Another is whether one can interpret an infinite total order in Thompson's group.

Examples:

(a) The trivial group is definably bi-orderable (this is the only obvious example).

(b) The cyclic group $$\mathbf{Z}$$ is not definably left-orderable, and more generally any group with a nontrivial abelian direct factor is not definably left-orderable. (Indeed, for such torsion-free $$G$$, the theory of $$G\times\mathbf{Z}/p\mathbf{Z}$$ tends to the theory of $$G$$ when the prime $$p$$ tends to infinity, so no $$\Phi$$ as above can exist).

(c) The Heisenberg group over $$\mathbf{Z}$$ (or $$\mathbf{R}$$) is definably bi-orderable. This is not hard but a bit tricky.

(d) The free group $$F_n$$, $$n\ge 2$$, which is bi-orderable, is not definably left-orderable (according to an expert I asked, every definable submonoid of a free group is a subgroup).

Concerning the question: actually, the bi-invariant total orders on Thompson's group are classified (by Navas and Rivas, arXiv link; GGD 2010) and this is very explicit: I don't know if any of those is definable.

• Did you check that your formula doesn't define a biorder on it? (j/k) – Ville Salo Mar 12 at 14:21
• Do you allow constants from G – Benjamin Steinberg Mar 12 at 15:00
• @BenjaminSteinberg yes (this is explicit in the definition I gave of definable: "and parameters in the group"). – YCor Mar 12 at 15:25
• Sorry i missed that – Benjamin Steinberg Mar 12 at 17:20