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.


(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.

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

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