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, 2020 at 14:21
  • 1
    $\begingroup$ Do you allow constants from G $\endgroup$ Mar 12, 2020 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, 2020 at 15:25
  • $\begingroup$ Sorry i missed that $\endgroup$ Mar 12, 2020 at 17:20
  • $\begingroup$ (About $F_{n\ge 2}$: this group is model-theoretically stable by Sela, so can't interpret an infinite total order, so there's no definable total order, even without invariance assumption.) $\endgroup$
    – YCor
    Aug 6, 2020 at 13:27

1 Answer 1


Yes, Thompson's group $F$ is definably bi-orderable.

Let $a$ be some element of $F$ with the support of $a$ equal to $(0,1/2)$. Let $b$ be some element of $F$ with the support of $b$ equal to $(1/2,1)$.

We will rely upon the following facts

  1. If $g$ and $h$ are in $F$ then $[a^g,b^h] = 1_F$ if and only if $(1/2)g \leq (1/2)h$.
  2. $F$ acts transitively on the dyadic rationals.

The set $S_1:= \left\{ f \in F \mid [a,b^f] = 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f \geq 1/2$.

The set $S_2:= \left\{ f \in F \mid [a^f,b] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which $(1/2)f > 1/2$.

The set $S_3:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ (here $d = (1/2)g$).

The set $S_4:= \left\{ f \in F \mid \exists g \in F \text{ with } [a^{gf},b^g] \neq 1_F \text{ and } \forall h \in F \text{ we have } [a^h,b^{hf}] = 1_F \vee [a^g,b^h] = 1_F\right\}$ is the set of elements $f$ of $F$ for which there is some dyadic rational $d \in (0,1)$ with $(d)f>d$ and for all dyadic rationals $e \in (0,1)$ either $(e)f \geq e$ or $e \geq d$ (here $d = (1/2)g$ and $e = (1/2)h$.

Equivalently $S_4$ is the set of non-identity elements $f$ of $F$ for which the right gradient at the infimum of the support of $f$ is strictly greater than $1$.

The union $\{1_F\} \cup S_4$ forms the cone of a bi-order on $F$. Specifically $\{1_F\} \cup S_4$ is the cone of $\preceq^+_{x^-}$ from the article of Navas and Rivas linked to in the question.

EDIT: Since people seem to be interested in the case of $[F,F]$ and Chehata's group I have added below a slightly stronger argument that applies to them.

Let $G$ satisfy the following

  1. Every element of $G$ has only finitely many components of support.
  2. For any $0 < u < v < 1$ and $0 < w < x < y < z < 1$ there is $g \in G$ with $w < (u)g < x < y < (v)g < z$.
  3. There is some element $a$ (that we now fix) of $G$ with a single component of support $(p,q)$ bounded away from $0$ and $1$.

$G$ could be either of $[F,F]$ and Chehata's group.

Fix a non-identity elements $b$ of $G$ with the support of $b$ a proper subset of the support of $a$.

Let $S_5$ be the set of $g$ in $G$ such that $[g,a] = 1_G = [g,b]$.

For $h \in G$ write $\bar{h}$ for the boundary of the support of $h$. If $h$ is in $S_5$ then $\mathrm{supt}(h) \cap \bar{b} = \varnothing = \bar{h} \cap \mathrm{supt}(a)$. It follows that $\mathrm{supt}(h) \cap \mathrm{supt}(a) = \varnothing$. An element of $G$ whose support is disjoint from the support of $a$ is easily in $S_5$ so $S_5$ is the set of elements of $G$ whose supports do not intersect the support of $a$.

Fix a non-identity element $c$ of $G$ with the support of $c$ a subset of $(q,1)$.

Let $S_6$ be the set of $g$ in $G$ such that there exists $h \in G$ with $[a^h,a] = 1_G = [a^h,b]$ and $[a^h,c] \neq 1_G \neq [a^h,a^g]$.

We will now argue that $S_6$ is the set of those $g$ in $G$ with $q < (q)g$.

Let $g$ be in $S_6$. There exists a conjugate $a^h$ of $a$ whose support does not intersect $\mathrm{supt}(a)$ but does intersect $\mathrm{supt}(c)$ and does intersect $\mathrm{supt}(g)$.

Any conjugate of $a$ must have a single component of support so either $\mathrm{supt}(a^h) \subseteq (0,p)$ or $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^h$ intersects the support of $c$ we must have $\mathrm{supt}(a^h) \subseteq (q,1)$. Since the support of $a^g$ intersects the support of $a^h$ it follows that the support of $a^g$ intersects $(q,1)$. Now it follows that $q < (q)g$.

If $g$ is in $G$ and $q < (q)g$ then $g$ is in $S_6$ by condition 2. above.

$S_6$ now corresponds to $S_1$ from the original proof and the rest of the construction works similarly.

  • $\begingroup$ For the non-trivial direction of 1. (I'm sure it's simpler than this though): If $(1/2)g = t$ and $(1/2)h < t$ then $(t)b^h \neq t$. If $(t)b^h < t$ then for small $\epsilon > 0$ we have $(t+\epsilon) a^g b^h = (t+\epsilon) b^h = x < t$ so $(t+\epsilon) b^h a^g = x a^g \neq x$ because $x$ is in the support $[0,t]$ of $a^g$. If $(t)b^h > t$, then let $t'$ be such that $(t')b^h = t$ and we have $(t'+\epsilon) b^h a^g = (t'+\epsilon) b^h$ because $(t'+\epsilon) b^h > t$, while $(t'+\epsilon) a^g b^h = (t'+\epsilon) b^h$ would imply $(t'+\epsilon) a^g = t'+\epsilon$, contradicting $t' < t$. $\endgroup$
    – Ville Salo
    Aug 6, 2020 at 7:46
  • $\begingroup$ Great! It seems that this definition works under the bare assumption of a subgroup $G$ of $\mathrm{Homeo}^+([0,1])$ with a dense orbit $Gx$ on the open interval $]0,1[$, such that there exists $a,b$ with support $]0,x[$ and $]x,1[$. This defines $S_4$ as the (conjugation-invariant) set of $g\in G$ such that for some $y$ with $yg>y$ and $zg\ge z$ for all $z\le y$. That this defines a (strict) total order requires something on $G$. It seems enough that $G$ acts piecewise analytically. $\endgroup$
    – YCor
    Aug 6, 2020 at 8:09
  • $\begingroup$ @VilleSalo just use that if they commute then the support of each one is invariant by the other one. $\endgroup$
    – YCor
    Aug 6, 2020 at 8:11
  • $\begingroup$ @YCor So if I've followed correctly, this argument implies that Chehata's group $G(I)$ (which contains his example of a simple bi-orderable group) is definably bi-orderable. $\endgroup$ Aug 6, 2020 at 12:30
  • $\begingroup$ @shane.orourke yes, this works for a wealth of known "rich" subgroups of $\mathrm{Homeo}^+([0,1])$. Actually, with only assumption existence of $a,b$ and density, one already gets a definable partial bi-ordering (which is not trivial, since $a,b>1$). $\endgroup$
    – YCor
    Aug 6, 2020 at 13:20

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