# 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) Mar 12 '20 at 14:21
• Do you allow constants from G Mar 12 '20 at 15:00
• @BenjaminSteinberg yes (this is explicit in the definition I gave of definable: "and parameters in the group").
– YCor
Mar 12 '20 at 15:25
• Sorry i missed that Mar 12 '20 at 17:20
• (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.)
– YCor
Aug 6 '20 at 13:27

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.

• 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 $t' < t$. Aug 6 '20 at 7:46
• 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.
– YCor
Aug 6 '20 at 8:09
• @VilleSalo just use that if they commute then the support of each one is invariant by the other one.
– YCor
Aug 6 '20 at 8:11
• @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. Aug 6 '20 at 12:30
• @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 $a,b>1$).
– YCor
Aug 6 '20 at 13:20