By Theorem 11.3, we have a decomposition of
$\operatorname{\mathsf{Der}}_{\mu}(F)$ into $\Phi$-invariant subspaces

$\operatorname{\mathsf{Der}}_{\mu}(F)=\mathcal{N}\sqcup\mathcal{I}\sqcup\widehat{\mathcal{I}}\sqcup\mathcal{P},$

defined as follows:

  * •

we let $\mathcal{N}\subset\operatorname{\mathsf{Der}}_{\mu}(F)$ be the subset
consisting of actions by translations of $F^{ab}$;

  * •

we fix a representative $\iota\in\operatorname{\mathsf{Der}}_{\mu}(F)$ of the
standard action on $(0,1)$, and let $\mathcal{I}=\\{\Phi^{t}(\iota)\colon
t\in\mathbb{R}\\}$ denote its $\Phi$-orbit, whilst $\widehat{\mathcal{I}}$
denotes the $\Phi$-orbit of the reversed action $\widehat{\iota}$;

  * •

we let $\mathcal{P}=\mathcal{P_{+}}\sqcup\mathcal{P}_{-}$ be the subset of
$\operatorname{\mathsf{Der}}_{\mu}(F)$ of $\mathbb{R}$-focal actions, where
$\mathcal{P}_{+}$ (respectively, $\mathcal{P}_{-}$) is the subset of
$\mathbb{R}$-focal actions which are increasingly (respectively, decreasingly)
horograded by the standard action on $(0,1)$. For $\varphi\in\mathcal{P}$ we
let $\xi_{\varphi}$ be the unique fixed point of $\varphi(f)$. We say that
$\varphi\in\mathcal{P}$ is $f$-centered if $\xi_{\varphi}=0$, and let
$\mathcal{P}^{0}\subset\mathcal{P}$ be the subset of $f$-centered
$\mathbb{R}$-focal actions. Finally we set
$\mathcal{P}^{0}_{\pm}=\mathcal{P}^{0}\cap\mathcal{P}_{\pm}$.

Let us now give a closer description of the topology of these subsets and of
the dynamics of the flow $\Phi$ on them.

First of all, from the elementary structure of the maximal quotient
$F/[F_{c},F_{c}]\cong\mathbb{Z}^{2}$ we obtain an explicit description of the
subset $\mathcal{N}$ in this case. Indeed $\mathcal{N}$ is homeomorphic to
$\operatorname{\mathsf{Der}}_{\bar{\mu}}(\mathbb{Z}^{2})$, where $\bar{\mu}$
is the projection of $\mu$ to $F^{ab}\cong\mathbb{Z}^{2}$. Every element of
$\operatorname{\mathsf{Der}}_{\bar{\mu}}(\mathbb{Z}^{2})$ corresponds to a
$\mathbb{Z}^{2}$-action by translations given by a non-trivial homomorphism
$\varphi\colon\mathbb{Z}^{2}\to(\mathbb{R},+)$ up to rescaling by a positive
real, so that $\operatorname{\mathsf{Der}}_{\bar{\mu}}(\mathbb{Z}^{2})$ is
homeomorphic to the circle $\mathbb{S}^{1}$. We deduce that $\mathcal{N}$ is a
closed subset of $\operatorname{\mathsf{Der}}_{\mu}(F)$ homeomorphic to a
circle, and it consists of points which are fixed by the translation flow
$\Phi$.

Let now $\tau_{0},\tau_{1}\colon F\to\mathbb{Z}$ be the two natural
homomorphisms obtained by identifying the groups of germs
$\operatorname{\mathsf{Germ}}(F,0)$ and $\operatorname{\mathsf{Germ}}(F,1)$
with $\mathbb{Z}$, with the convention that $\tau_{x}(g)>0$ if and only if the
corresponding endpoint $x\in\\{0,1\\}$ is an attracting fixed point of $g$.
Explicitly,

(11.2)
$\tau_{0}(g)=-\log_{2}D^{+}g(0)\quad\text{and}\quad\tau_{1}(g)=-\log_{2}D^{-}g(1).$

We identify $\tau_{0}$ and $\tau_{1}$ with the elements of $\mathcal{N}$ given
by their corresponding cyclic action. One readily shows that the actions
$\iota$ and $\widehat{\iota}$ satisfy the following (compare with Lemma
10.18):

$\
\lim_{t\to+\infty}\Phi^{t}(\iota)=\tau_{1},\lim_{t\to-\infty}\Phi^{t}(\iota)=\widehat{\tau_{0}},\quad\quad\lim_{t\to+\infty}\Phi^{t}(\widehat{\iota})=\tau_{0},\lim_{t\to-\infty}\Phi^{t}(\widehat{\iota})=\widehat{\tau_{1}}.$

Thus $\mathcal{I}$ is a copy of $\mathbb{R}$ inside
$\operatorname{\mathsf{Der}}_{\mu}(F)$ which connects the points
$\widehat{\tau_{0}}$ and $\tau_{1}$, while $\widehat{\mathcal{I}}$ connects
$\tau_{0}$ to $\tau_{1}$ as shown in Figure 10.

Recall that by Lemma 10.16 each subset $\mathcal{P}_{\pm}$ is open, and
$\mathcal{P}^{0}_{\pm}$ is a cross section for $\Phi$ inside
$\mathcal{P}_{\pm}$. Moreover, in this setting we have the following more
precise version of Proposition 10.17.

######  Proposition 11.10.

The subsets $\mathcal{P}_{-}^{0}$ and $\mathcal{P}_{+}^{0}$ are closed in
$\operatorname{\mathsf{Der}}_{\mu}(F)$. Moreover, for every
$\varphi\in\mathcal{P}^{0}_{\pm}$ the limits
$\lim_{t\to\pm\infty}\Phi^{t}(\varphi)$ exist and the following holds:

  1. (1)

if $\varphi\in\mathcal{P}_{+}^{0}$ then

$\lim_{t\to+\infty}\Phi^{t}(\varphi)=\tau_{1}\quad\text{and}\quad\lim_{t\to-\infty}\Phi^{t}(\varphi)=\widehat{\tau}_{1},$

where the convergence is uniform on $\varphi\in\mathcal{P}^{0}_{+}$;

  2. (2)

if $\varphi\in\mathcal{P}_{-}^{0}$ then

$\lim_{t\to+\infty}\Phi^{t}(\varphi)=\tau_{0}\quad\text{and}\quad\lim_{t\to-\infty}\Phi^{t}(\varphi)=\widehat{\tau}_{0},$

where the convergence is uniform on $\varphi\in\mathcal{P}^{0}_{-}$.

###### Proof.

We already know by Lemma 10.16 that
$\overline{\mathcal{P}^{0}}_{\pm}\subset\mathcal{P}^{0}_{\pm}\cup\mathcal{N}$.
Let us show that in this case the sets $\mathcal{P}_{\pm}^{0}$ are actually
closed. Assume by contradiction that
$(\varphi_{n})_{n\in\mathbb{N}}\subset\mathcal{P}^{0}_{+}$ is a sequence
converging to $\psi\in\mathcal{N}$. Since $\varphi_{n}(f)$ fixes $0$ for every
$n\in\mathbb{N}$, so does $\psi(f)$. Since $\psi$ is an action by
translations, this is possible only if $\psi(f)=\mathsf{id}$. Now fix
$x\in(0,1)$ and choose any $h\in F_{(0,x)}$ with non-trivial germ at $0$.
Recall the notation $\operatorname{I^{\varphi}}(x,\xi)$ from Definition 10.7.
After the claim in Lemma 10.19, there exists $D=D(x,\mu)>0$ such that
$|\operatorname{I^{\varphi_{\mathit{n}}}}(x,0)|\leq D$ for every
$n\in\mathbb{N}$. Thus, upon extracting a subsequence we can suppose that
$\eta_{n}:=\sup\operatorname{I^{\varphi_{\mathit{n}}}}(x,0)$ converges to a
limit $\eta\in\mathbb{R}$ as $n\to\infty$. The choice of $h$ implies that
$\varphi_{n}(h)$ fixes $\eta_{n}$ for every $n$, so that $\psi(h)$ fixes
$\eta$, and we deduce again that $\psi(h)=\mathsf{id}$. However note that in
the standard action the element $f$ has non-trivial derivatives at both
endpoints, while $h$ has non-trivial germ only at $0$. Thus $h$ and $f$
project to two linearly independent elements of $F^{ab}\cong\mathbb{Z}^{2}$.
Since $\psi$ is an action by translations and it vanishes on both, we have
that $\psi$ is trivial, which is a contradiction. This shows that
$\mathcal{P}^{0}_{+}$ is closed, and the proof for $\mathcal{P}^{0}_{-}$ is
closed is similar.

Let us now show that $\lim_{t\to+\infty}\Phi^{t}(\varphi)=\tau_{1}$ uniformly
on $\varphi\in\mathcal{P}^{0}_{+}$. Assume that
$(\varphi_{n})_{n\in\mathbb{N}}\subset\mathcal{P}^{0}_{+}$ and
$(t_{n})_{n\in\mathbb{N}}\subset\mathbb{R}$ are sequences such that $t_{n}$
increases to $+\infty$, and let $\psi$ be a cluster point of
$\psi_{n}:=\Phi^{t_{n}}(\varphi_{n})$. Repeating the proof of Proposition
10.17 in this case, one sees that
$\operatorname{\mathsf{Fix}}^{\psi}(F_{+})\neq\varnothing$, hence
$\psi(F_{+})=\\{\operatorname{\mathsf{id}}\\}$. This is only possible when
$\psi\in\\{\tau_{1},\widehat{\tau}_{1}\\}$. However note that $\psi_{n}(f)$ is
an expanding homothety with fixed point $-t_{n}$ for every $n\in\mathbb{N}$,
so that $\psi_{n}(f)(0)>0$ for sufficiently large $n$. Passing to the limit we
obtain that $\psi(f)(0)\geq 0$. Thus necessarily $\psi=\tau_{1}$. ∎

An immediate consequence of the uniform convergence in Proposition 11.10 is
the following. Given a topological space $K$, we define the _double cone_ over
$K$ to be the quotient $K\times[-1,1]/\sim$ where $(x,-1)\sim(y,-1)$ and
$(x,1)\sim(y,1)$ for every $x,y\in K$.

######  Corollary 11.11.

The closures of the subsets $\mathcal{P}_{+}$ and $\mathcal{P}_{-}$ in
$\operatorname{\mathsf{Der}}_{\mu}(F)$ are given by
$\overline{\mathcal{P}}_{+}=\mathcal{P}_{+}\cup\\{\tau_{1},\widehat{\tau}_{1}\\}$
and
$\overline{\mathcal{P}}_{-}=\mathcal{P}_{-}\cup\\{\tau_{0},\widehat{\tau}_{0}\\}$,
and both are homeomorphic to the double cone over the corresponding cross
section $\mathcal{P}^{0}_{\pm}$.

With this discussion in mind, illustration of
$\operatorname{\mathsf{Der}}_{\mu}(F)$ is provided by Figure 10. Note that in
particular it is visible from this description that
$\overline{\mathcal{P}}\cap\mathcal{I}=\varnothing$, so that $\iota$ is
transversely isolated. In particular the standard action of $F$ is locally
rigid (Theorem 11.5 above).

To conclude this discussion, we observe that it is a tantalizing problem to
obtain further results on the topology of the compact cross sections
$\mathcal{P}^{0}_{+}$ and $\mathcal{P}^{0}_{-}$, which is at the moment quite
mysterious. Note that these do not depend up to homeomorphism on the choice of
the generator $f\in F$ made above, and by symmetry they are homeomorphic one
to the other. The constructions of $\mathbb{R}$-focal actions which will
appear in the rest of the section show that the spaces $\mathcal{P}^{0}_{+}$
and $\mathcal{P}^{0}_{-}$ are uncountable, and contain homeomorphic copies of
a Cantor set. However we were not able to construct a non-trivial connected
subset of $\mathcal{P}^{0}_{+}$, and we do not know whether they are totally
disconnected. We also do not know the answer to the following question.

######  Question 11.12.

Do the cross sections $\mathcal{P}^{0}_{+}$ and $\mathcal{P}^{0}_{-}$ admit
isolated points?

By Corollary 3.4 this is equivalent to the question whether $F$ admits minimal
$\mathbb{R}$-focal actions which are locally rigid.

###  11.3. Simplicial $\mathbb{R}$-focal actions of $F$ and minimality of
$[F,F]$

Before discussing examples of minimal $\mathbb{R}$-focal actions of $F$, we
would like to single out an important question to address about any such
action: whether the commutator subgroup $[F,F]$ acts minimally or not. By
Lemma 7.12, the second possibility is equivalent to the fact that it does not
admit any non-empty closed minimal invariant subset of $\mathbb{R}$. The
$\mathbb{R}$-focal actions of $F$ with this property turn out to be very
special due to the following result, which elaborates on Proposition 8.33.

######  Proposition 11.13 (Simplicial $\mathbb{R}$-focal actions of $F$).

Let $\varphi:F\to\operatorname{\mathsf{Homeo}}_{+}(\mathbb{R})$ be a minimal
$\mathbb{R}$-focal action which can be increasingly horograded by the standard
action on $(0,1)$. Then, the following are equivalent:

  1. (i)

The image $\varphi(F_{+})$ admits no non-empty closed minimal invariant set.

  2. (ii)

The image $\varphi([F,F])$ admits no non-empty closed minimal invariant set.

  3. (iii)

$\varphi$ is represented by a focal action on a planar directed tree
$(\mathbb{T},\triangleleft,\prec)$, such that $\mathbb{T}$ is a simplicial
tree of countable degree, and the action of $F$ on $\mathbb{T}$ is by
simplicial automorphisms.

  4. (iv)

$\varphi$ is represented by a focal action on a planar directed tree
$(\mathbb{T},\triangleleft,\prec)$ such that $F$ acts by isometries with
respect to a compatible $\mathbb{R}$-tree metric on $\mathbb{T}$.

  5. (v)

Every pseudohomothety in the image of $\varphi$ is a homothety.

  6. (vi)

There exist bounded open intervals $I,J\subset\mathbb{R}$ such that for every
$g\in F_{+}$ the image $g.I$ does not contain $J$.

######  Remark 11.14.

The fact that some minimal $\mathbb{R}$-focal actions of $F$ can be encoded by
a simplicial action on a simplicial tree can seem in contradiction with the
fact that they can also be horograded by the standard action of $F$, which is
highly not isometric. However the tree provided by the proof Proposition 11.13
is not the same directed tree as the one provided by the proof of Theorem 10.3
(it arises from a different CF-cover). This is an illustration of the fact
that a planar directed tree encoding an $\mathbb{R}$-focal action is not
unique, and identifying a tree with good properties may be important for some
purposes. We note however that the focal germ representations associated to
both trees are semi-conjugate to the germ homomorphism
$F\to\operatorname{\mathsf{Germ}}(F,1)\simeq\mathbb{Z}$ associated to the
standard action (recall Lemma 8.25).

###### Proof of Proposition 11.13.

Let us first show that (i) and (ii) are equivalent. If $[F,F]$ admits a closed
minimal invariant subset, then by Lemma 7.12 it acts minimally on
$\mathbb{R}$, and thus so does $F_{+}$. Conversely, assume by contradiction
that $[F,F]$ does not act minimally on $\mathbb{R}$, but $F_{+}$ does. Then we
can apply Proposition 8.33 to $G=F_{+}$ and $N=[F,F]$ and we deduce that
$\varphi\restriction_{F_{+}}$ is $\mathbb{R}$-focal and is the dynamical
realization of an action of $F_{+}$ on an planar directed simplicial tree
$(\mathbb{T},\triangleleft,\prec)$. Moreover, the argument in its proof shows
that the focal germ representation of the action on
$(\mathbb{T},\triangleleft,\prec)$ coincides with the projection to
$F_{+}/[F,F]\cong\mathbb{Z}$. Since the quotient $F_{+}/[F,F]$ is simply the
group of germs $\operatorname{\mathsf{Germ}}(F_{+},0)$ we deduce that the
$\varphi$-image of every $g\in F_{+}$ with a non-trivial germ at $0$ must be a
pseudohomothety. However we were assuming that $\varphi$ is an
$\mathbb{R}$-focal action of $F$ increasingly horograded by the standard
action on $(0,1)$, so that the image of every element $g\in F_{+}$ must be
totally bounded. This is a contradiction.

The implication (i)$\Rightarrow$(iii) follows from Proposition 8.33, and
(iii)$\Rightarrow$(iv) is obvious. Let us prove that (iv) implies (v). Note
that the image of an element $g\in F$ is a pseudohomothety if and only if $g$
has a non-trivial germ at $1$. As in the proof of Proposition 8.33 we can
consider a horofunction $\pi\colon\mathbb{T}\to\mathbb{R}$ given by
$\pi(v)=d_{\mathbb{T}}(v\wedge v_{0},v_{0})-d_{\mathbb{T}}(v\wedge v_{0},v)$,
where $v_{0}$ is some basepoint, and we see that this provides a horograding
by an action by translations on $\mathbb{R}$ coming from the germ homomorphism
$F\to\operatorname{\mathsf{Germ}}(F,1)\cong\mathbb{Z}\to(\mathbb{R},+)$. Then
from Proposition 8.27 we obtain that every element with a non-trivial germ at
$1$ must act as a homothety under $\varphi$.

To show that (v) implies (vi), we show that (v) actually implies the following
more explicit condition, which clearly implies (vi).

  * _(vi’)_

_For every element $h\in F$ which in the standard action satisfies $h(x)>x$
for every $x\in(0,1)$ (so that $\varphi(h)$ is an expanding homothety) there
exists an open bounded interval $I\subset\mathbb{R}$ such that $I\subset h.I$
and such that for every $g\in F_{+}$, the interval $h.I$ is not contained in
$g.I$. _

Indeed assume that $h$ is such that $h(x)>x$ for every $x\in X$ and let
$\xi\in\mathbb{R}$ be the unique fixed point of $\varphi(h)$. Fix $x\in X$ and
let $I:=\operatorname{I^{\varphi}}(x,\xi)$ (note that this connected component
is defined for every $x\in X$). As $\varphi(h)$ is an expanding homothety, we
have $I\subset h.I$. Suppose by contradiction that there exists $g\in F_{+}$
such that $h.I\subset g.I$. Note that we have the equalities

$g.I=g.\operatorname{I^{\varphi}}(x,\xi)=\operatorname{I^{\varphi}}(g(x),g.\xi)=\operatorname{I^{\varphi}}(g(x),\xi)$

(the last equality follows from the assumption that $\xi\in g.I$). Moreover we
have that $h.I=\operatorname{I^{\varphi}}(h(x),\xi)$ and therefore we conclude
that
$\operatorname{I^{\varphi}}(h(x),\xi)\subset\operatorname{I^{\varphi}}(g(x),\xi)$
implying that $g(x)\geq h(x)$. Since $g\in F_{+}$ this implies that for some
$y\geq x$ it holds that $g(y)=h(y)$. Therefore,

$g.\operatorname{I^{\varphi}}(y,\xi)=\operatorname{I^{\varphi}}(g(y),\xi)=\operatorname{I^{\varphi}}(h(y),\xi)=h.\operatorname{I^{\varphi}}(y,\xi)$

(the first equality follows from the fact that
$\operatorname{I^{\varphi}}(y,\xi)\supset I$). Finally, this implies that
$\varphi(g^{-1}h)$ preserves $\operatorname{I^{\varphi}}(y,\xi)$. Since
$g^{-1}h$ has the same germ as $h$ at $1$, it acts as a pseudohomothety, so
that by (v) it is a homothety, and this is a contradiction with the fact that
it preserves $\operatorname{I^{\varphi}}(y,\xi)$.

Finally to show that (vi) implies (i), assume by contradiction (using Lemma
7.12) that the action of $F_{+}$ is minimal. As we are assuming (vi), the
action cannot be proximal, so by Theorem 2.17 the centralizer of
$\varphi(F_{+})$ in $\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ must be
infinite cyclic generated by an element $T$ without fixed point. Since $F_{+}$
is normal, we deduce that the whole group $\varphi(F)$ must normalize $\langle
T\rangle$ and thus centralize it, contradicting that a minimal
$\mathbb{R}$-focal action is always proximal (Proposition 7.7). ∎

######  Definition 11.15.

We will say that a minimal $\mathbb{R}$-focal action $\varphi\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ is simplicial if it
satisifies one of the equivalent conditions in Proposition 11.13.

For many constructions of $\mathbb{R}$-focal actions of $F$ discussed below
one can easily check conditions (i) or (ii), and thus they turn out to be
simplicial (although a simplicial tree does not always appear naturally in the
construction, and it might also be not obvious to check directly condition
(v)). It is more delicate to construct $\mathbb{R}$-focal actions of $F$ which
are _not_ simplicial: this will be done in §11.6.

###  11.4. A plethora of $\mathbb{R}$-focal actions I: restriction preorders

Starting from now we will present various constructions of $\mathbb{R}$-focal
actions of the group $F$ and study some of their properties. We will focus on
some examples, and indicate how they can be modified to obtain more.

######  Remark 11.16.

The attentive reader will notice that a common feature of all our
constructions of $\mathbb{R}$-focal actions of $F$ is the choice of a closed
subset $K\subset(0,1)$, which is invariant under the generator $f$ given by
(11.1). These sets appear quite naturally with the point of view of focal
actions. To understand this, take a minimal $\mathbb{R}$-focal action
$\varphi\colon F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$. By Theorem
11.3 we know that it can be horograded by the standard action on $(0,1)$,
meaning that one can find a focal action
$\Phi:G\to\operatorname{\mathsf{Aut}}(\mathbb{T},\triangleleft,\prec)$ on a
directed planar tree which is horograded by the standard action, and whose
dynamical realization is (conjugate to) $\varphi$. By Proposition 11.4, the
element $f$ fixes a unique end $\xi_{0}\in\partial^{*}\mathbb{T}$, so that it
preserves the axis $]\xi_{0},\omega[\subset\mathbb{T}$, which is naturally
identified with the interval $(0,1)$ via the horograding map
$\pi:\mathbb{T}\to(0,1)$. In particular, the $\pi$-image of the closure
$\overline{]\xi_{0},\omega[\cap\operatorname{\mathsf{Br}}(\mathbb{T})}$ of the
subset of branching points on this axis defines an $\varphi(f)$-invariant
closed subset $K\subset(0,1)$. Although for some choices of the action $\Phi$
(as the one in the proof of Theorem 10.3) the subset $K$ is the whole interval
$(0,1)$, in most examples it is not the case with correct choice of $\Phi$.

####  11.4.1. A reinterpretation in terms of preorders

We start with a simple observation which is useful to understand
$\mathbb{R}$-focal actions of $F$. Recall that we write $\tau_{1}\colon
F\to\mathbb{Z}\cong\operatorname{\mathsf{Germ}}(F,1)$ for the germ
homomorphism given by (11.2), and $f\in F$ for the element given by (11.1). We
will also write $1_{F}$ for the trivial element of $F$ (and we will simply
denote it by $1$ when there is no risk of confusion). Since $\tau_{1}(f)$ is a
generator of $\operatorname{\mathsf{Germ}}(F,1)$, we have a splitting

$F=F_{+}\rtimes\langle f\rangle,$

where as usual we write $F_{+}=\operatorname{\mathsf{ker}}\tau_{1}$. Then we
can make $F$ act on $F_{+}$ “affinely” by letting $F_{+}$ act on itself by
left translations and $f$ act on $F_{+}$ by conjugation. In formula, for
$g=hf^{n}\in F$, with $h\in F_{+}$ and $n\in\mathbb{Z}$ and for $r\in F_{+}$,
this action is given by $g\cdot r=hf^{n}rf^{-n}$.

Assume that $\preceq$ is a left preorder on $F_{+}$ which is invariant under
conjugation by $f$. In particular its residual subgroup $H=[1]_{\preceq}$ is
normalized by $f$, so that the action of $F$ on $F_{+}$ descends to an order-
preserving action on $(F_{+}/H,\prec)$, where $\prec$ is the total order
induced by $\preceq$. Then, we can consider the dynamical realization
$\varphi\colon F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ of this
action. We have the following equivalence.

######  Proposition 11.17.

Let $\varphi\colon F\to\operatorname{\mathsf{Homeo}}_{+}(\mathbb{R})$ be an
action. The following are equivalent.

  1. (i)

$\varphi$ is a minimal $\mathbb{R}$-focal action increasingly horograded by
the standard action on $(0,1)$.

  2. (ii)

There exists a preorder $\preceq$ on $F_{+}$ invariant under conjugation by
$f$, such that, writing $H=[1]_{\preceq}$, the map $f$ acts as a homothety on
$(F_{+}/H,\prec)$, and $\varphi$ is conjugate to the dynamical realization of
the action of $F$ on $(F_{+}/H,\prec)$.

Moreover two distinct preorders as in (ii) give rise to non-conjugate minimal
$\mathbb{R}$-focal actions.

###### Proof.

Let us prove that (ii) implies (i). Assume that $\preceq$ verifies the
conditions, and let $\varphi$ be the dynamical realization of the action of
$F$ on $(F_{+}/H,\prec)$. Since $f$ is a homothety on $(F_{+}/H,\prec)$,
Proposition 2.43 implies that $\varphi$ is minimal, and $\varphi(f)$ is a
homothety. Since moreover $\varphi$ must fall into one of the cases of Theorem
11.3, the only possibility is that $\varphi$ is $\mathbb{R}$-focal,
increasingly horograded by the standard action on $(0,1)$.

For the converse, let $\varphi$ be is as in (i). Then $\varphi(f)$ is an
expanding homothety (see Proposition 11.4); let $\xi\in\mathbb{R}$ be its
unique fixed point, and consider the preorder $\preceq$ on $F_{+}$ associated
with this point: $g\precneq h$ if and only if $g.\xi<h.\xi$. Using that $\xi$
is fixed by $f$, we see that $\preceq$ is invariant under conjugation by $f$,
and that the natural action of $F$ on $(F_{+}/[1]_{\preceq},\prec)$ can be
identified with the action of $F$ on the orbit of $\xi$, showing the claim.

Finally note that these two constructions are inverse to each other, and since
$\xi$ is the unique fixed point of $f$, the preorder $\prec$ is uniquely
determined by the conjugacy class of the action. ∎

####  11.4.2. Restriction preorders on $F_{+}$

We now explain a concrete construction of preorders on $F_{+}$ satisfying (ii)
in Proposition 11.17. This yields a family of $\mathbb{R}$-focal actions of
$F$ which contains as special cases the constructions in §§9.1–9.2.

Let $K\subseteq(0,1)$ be a closed $f$-invariant subset. To the subset $K$ we
associate a preorder $\preceq^{K}$ on $F_{+}$ which is obtained by looking at
the restriction of elements of $F_{+}$ to $K$, as follows. We first consider
the subgroup $H=\\{g\in F_{+}\colon g(x)=x\text{ for every }x\in K\\}$ of
elements which fix $K$ pointwise, and for $g\in F_{+}$ define

$x_{g}=\left\\{\begin{array}[]{lr}0&\text{if }g\in H,\\\ \sup\\{x\in K\colon
g(x)\neq x\\}&\text{if }g\in F_{+}\smallsetminus H.\end{array}\right.$

We immediately observe that $x_{g}=x_{g^{-1}}$ for every $g\in F_{+}$.
Moreover, we have the following behavior when considering compositions.

######  Lemma 11.18.

Let $K\subseteq(0,1)$ be a non-empty closed subset, and take $g,h\in F_{+}$.
Then we have the inequality $x_{gh}\leq\max\\{x_{g},x_{h}\\}$, and when
$x_{h}\neq x_{g}$ the equality $x_{gh}=\max\\{x_{g},x_{h}\\}$ holds.

###### Proof.

Note that if $x\in K$ is such that $x>\max\\{x_{g},x_{h}\\}$, then
$gh(x)=g(x)=x$. This gives the inequality $x_{gh}\leq\max\\{x_{g},x_{h}\\}$.

Assume now $x_{h}<x_{g}$ and assume first we are in the case $g(x_{g})\neq
x_{g}$. Then $gh(x_{g})=g(x_{g})\neq x_{g}$, proving that $x_{g}\leq x_{gh}$,
hence $x_{gh}=x_{g}$ (using the previous inequality). When $g(x_{g})=x_{g}$,
then $x_{g}$ is accumulated from the left by points of $K$ which are moved by
$g$; in particular for every such point $x$ with $x_{h}<x<x_{g}$, we have
$gh(x)=g(x)\neq x$, giving $x\leq x_{gh}$. Taking the supremum we obtain the
desired equality $x_{g}=x_{gh}$. Note also that the same assumption
$x_{h}<x_{g}$ (which is equivalent to $x_{h^{-1}}<x_{g^{-1}}$) gives
$x_{g^{-1}h^{-1}}=x_{g^{-1}}=x_{g}$. As $x_{hg}=x_{g^{-1}h^{-1}}$, we deduce
from the previous case that $x_{hg}=x_{g}$. This concludes the proof. ∎

We next introduce the subset

(11.3) $P_{K}=\left\\{g\in F_{+}\smallsetminus H\colon\text{either
}g(x_{g})>x_{g},\text{ or }g(x_{g})=x_{g}\text{ and }D^{-}g(x_{g})>1\right\\}$

and observe the following.

######  Lemma 11.19.

For any non-empty closed subset $K\subseteq(0,1)$, the subset $P_{K}$ defines
a positive cone in $F_{+}$.

###### Proof.

We have to verify the conditions in Remark 2.34. Let us first prove that
$F_{+}=P_{K}\sqcup H\sqcup P_{K}^{-1}$. For this notice that, since
$x_{g}=x_{g^{-1}}$, we have

$P_{K}^{-1}=\\{g\in F_{+}\smallsetminus H\colon\text{either
}g(x_{g})<x_{g},\text{ or }g(x_{g})=x_{g}\text{ and }D^{-}g(x_{g})<1\\}.$

Thus, we automatically get that $H\cap\big{(}P_{K}\cup
P_{K}^{-1}\big{)}=\varnothing$ and $P_{K}^{-1}\cap P_{K}=\varnothing$. It only
remains to shows that $F_{+}\subseteq P_{K}\sqcup H\sqcup P_{K}^{-1}$. For
this, take $g\in F_{+}\smallsetminus H$, so that $x_{g}>0$. If $x_{g}\neq
g(x_{g})$ we are done. In the complementary case, $x_{g}$ must be accumulated
from the left by points that are moved by $g$. Since $g$ is piecewise linear
we must have $D^{-}g(x_{g})\neq 1$ showing that $g\in P_{K}\sqcup P_{K}^{-1}$.
Next, let us check that $P_{K}$ is a semigroup and $HP_{K}H\subseteq P_{K}$.

Take $g,h\in P_{K}$, and assume first $x_{h}<x_{g}$. Then Lemma 11.18 gives
$x_{gh}=x_{g}$ and $gh(x_{gh})=g(x_{g})$. If $g(x_{g})>x_{g}$, we deduce
immediately $gh\in P_{K}$; otherwise $x_{g}$ is accumulated from the left by
points of $K$, which must be fixed by $h$, so that $D^{-}h(x_{g})=1$. Then
$D^{-}(gh)(x_{gh})=D^{-}g(x_{g})\,D^{-}h(x_{g})>1$, and we conclude that
$gh\in P_{K}$.

Assume now that $x_{g}<x_{h}$, so that $x_{gh}=x_{h}$ by Lemma 11.18. Consider
first the case $h(x_{h})=x_{h}$. Then $gh(x_{gh})=g(x_{h})=x_{h}=x_{gh}$, and
as in the previous case we see that $D^{-}g(x_{h})=1$, so that
$D^{-}(gh)(x_{gh})=D^{-}h(x_{h})>1$. When $h(x_{h})>x_{h}$, then
$gh(x_{gh})>g(x_{h})=x_{h}$. In both cases we have $gh\in P_{K}$.

Note that the previous argument works also when one of the two elements is in
the residue $H$, proving that $HP_{K}H\subseteq P_{K}$.

Finally, consider the case $x_{g}=x_{h}$. As $h(x_{h})\geq x_{h}$ and
$g(x_{g})\geq x_{g}$, then if any of the two inequalities is strict we deduce
$gh(x_{g})>x_{g}$, and thus $x_{gh}=x_{g}$ (by the inequality of Lemma 11.18)
and $gh\in P_{K}$. Otherwise, assume that both $g$ and $h$ fix $x_{g}=x_{h}$.
Then we have the relation $D^{-}(gh)(x_{gh})=D^{-}g(x_{g})\,D^{-}h(x_{h})>1$,
showing that $x_{gh}=x_{g}$ (again by Lemma 11.18) and $gh\in P_{K}$ also in
this case. ∎

The previous lemma leads to the following definition.

######  Definition 11.20.

Given a closed subset $K\subseteq(0,1)$, the preorder $\preceq^{K}$ on $F_{+}$
defined by the positive cone $P_{K}$ in (11.3) will be called the _restriction
preorder_ associated with $K$. We will always write $H=[1]_{\preceq^{K}}$ for
its residue.

Let us describe some elementary properties related to the preorder
$\preceq^{K}$ that will be useful in the sequel.

######  Lemma 11.21.

Let $K\subseteq(0,1)$ be a non-empty closed subset, and let $\preceq^{K}$ be
the corresponding restriction preorder on $F_{+}$. Then the following hold.

  1. (i)

For $g,h\in F_{+}$ with $1\preceq^{K}g\preceq^{K}h$, we have $x_{g}\leq
x_{h}$.

  2. (ii)

For $x\in(0,1)$, the subset $L_{x}:=\\{g\in F_{+}\colon x_{g}\leq x\\}$ is a
$\preceq^{K}$-convex subgroup.

###### Proof.

We first prove (i). We can assume $g\in P_{K}$ otherwise $x_{g}=0$ and the
result follows. Assume for contradiction that $x_{g}>x_{h}$. Then from Lemma
11.18 we have $x_{g^{-1}h}=x_{g}$. Consider first the case $g(x_{g})>x_{g}$,
then $g^{-1}h(x_{g})=g^{-1}(x_{g})<x_{g}$, so that $g^{-1}h{}^{K}1_{F}$,
contradicting the assumption $g\preceq^{K}h$. Consider next the case
$g(x_{g})=x_{g}$, so that $D^{-}g(x_{g})>1$ and $D^{-}h(x_{g})=1$ (as in this
case $x_{g}$ is accumulated from the left by points of $K$). Then
$g^{-1}h(x_{g})=x_{g}$ and $D^{-}(g^{-1}h)(x_{g})=D^{-}g(x_{g})^{-1}<1$,
giving again the contradiction $g^{-1}h{}^{K}1_{F}$.

The inequality $x_{gh}\leq\max\\{x_{g},x_{h}\\}$ from Lemma 11.18 shows that
the subset $L_{x}$ in (ii) is a subgroup, whilst (i) proves that $L_{x}$ is
$\preceq^{K}$-convex. ∎

Note that the coset space $F_{+}/H$ can be identified with the set of
restrictions $\\{g\restriction_{K}\colon g\in F_{+}\\}$, so that two elements
$g,h\in F_{+}$ are equivalent for $\preceq^{K}$ if and only if their
restrictions to $K$ coincide.

######  Lemma 11.22.

For every $f$-invariant closed set $K\subseteq(0,1)$, the restriction preorder
$\preceq^{K}$ on $F_{+}$ is invariant under conjugation by $f$, and the
conjugacy induces a homothety on $(F_{+}/H,\prec^{K})$ fixing $H$.

###### Proof.

The verification that $\preceq^{K}$ is invariant under conjugation follows
easily from $f$-invariance of $K$. Let us check that $f$ acts a homothety on
$(F_{+}/H,\prec^{K})$. It is clear that it fixes the point corresponding to
$H$. We next verify that conjugation by $f$ preserves the positive cone
$P_{K}$. Take $h\in P_{K}$, write $x_{*}=x_{h}$ and note that
$f(x_{h})=x_{fhf^{-1}}$; when $h(x_{h})>x_{h}$, we have
$fhf^{-1}(x_{fhf^{-1}})=fh(x_{h})>x_{h}$, otherwise we have $h(x_{h})=x_{h}$
and

$D^{-}(fhf^{-1})(x_{fhf^{-1}})=D^{-}h(x_{h})>1.$

Hence $fhf^{-1}\in P_{K}$, as wanted.

More generally, for $n\in\mathbb{N}$, consider $h_{n}=f^{n}hf^{-n}$ and
observe that the point $x_{h_{n}}=f^{n}(x_{h})$ tends to $1$ as $n\to\infty$.
Take $r\in P_{K}$ and let $y\in(0,1)$ be such that $r$ acts trivially on
$(y,1)$. If $n$ is large enough so that $h_{n}(x_{h_{n}})=f^{n}h(x_{h})$ and
$x_{h_{n}}$ are greater than $y$, we have that $x_{r^{-1}h_{n}}=x_{h_{n}}$ and
$r^{-1}h_{n}$ coincides with $h_{n}$ on a neighborhood of $x_{h_{n}}$. Since
$h_{n}\in P_{K}$, and this depends only on the behavior of $h_{n}$ on some
neighborhood of $x_{h_{n}}$, we must have $r^{-1}h_{n}\in P_{K}$ for $n$ large
enough, and thus $h_{n}{}^{K}r$. Since $h$ and $r$ were arbitrary
$\preceq^{K}$-positive elements and we can repeat the same reasoning for
arbitrary $h,r\in P_{K}^{-1}$, this shows that the conjugation by $f$ is a
homothety. ∎

For every $f$-invariant subset $K\subseteq(0,1)$ let us denote by
$\psi_{K}\colon F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ the
dynamical realization of the action of $F$ on $(F_{+}/H,\prec^{K})$ defined
above. By Proposition 11.17 this action is $\mathbb{R}$-focal and increasingly
horograded by the standard action on $(0,1)$. Note also that since the residue
$H$ is the fixator of $K$, and two distinct closed subsets of $(0,1)$ have
different fixators, then $\preceq^{K}$ determines $K$ completely. In
particular, by the last part of Proposition 11.17, if $K_{1}\neq K_{2}$ their
associated $\mathbb{R}$-focal actions $\psi_{K_{1}}$ and $\psi_{K_{2}}$ are
not conjugate.

####  11.4.3. Some properties of the actions arising from restriction
preorders

Given a non-empty $f$-invariant closed subset $K\subseteq(0,1)$, we keep
denoting by $\psi_{K}:F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ the
action constructed above. We want to point out some dynamical properties of
this family of actions. Recall that a minimal action of a group $G$ on a
locally compact space $Y$ is _topologically free_ if the set of fixed points
$\operatorname{\mathsf{Fix}}(g)$ has empty interior for every $g\in G$. By
Baire’s theorem this is equivalent to the requirement that a dense
$G_{\delta}$-set of points in $Y$ have a trivial stabilizer in $G$.

######  Proposition 11.23 (Freeness and non-freeness).

Let $K\subseteq(0,1)$ be a non-empty $f$-invariant closed subset. Then the
$\mathbb{R}$-focal action $\psi_{K}\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ defined above is
topologically free if and only if $K=(0,1)$. In particular, $F$ admits both
topologically free and non-topologically free minimal $\mathbb{R}$-focal
actions.

###### Proof.

Assume $K=(0,1)$. We claim that the action $\psi:=\psi_{K}$ is topologically
free. Indeed in this case the preorder $\preceq^{K}$ is actually a total order
on $F_{+}$. Thus there is a dense subset of points in $\mathbb{R}$ with a
trivial stabilizer for $\psi(F_{+})$, which implies that the action of $F_{+}$
is topologically free. Assume by contradiction that $g\in F$ is such that
$\operatorname{\mathsf{Fix}}^{\psi}(g)$ has non-empty interior, and let $I$ be
a connected component of its interior. Note that $g\notin F_{+}$ so that by
Propositions 11.17 and 11.4, the image $\psi(g)$ must be a pseudohomothety; in
particular $I$ is bounded. As the action $\psi$ is proximal (see for instance
Lemma 5.13), there exists $h\in F$ such that $\psi(h)(I)\Subset I$. Then it is
not difficult to see that the commutator $[g,h]=ghg^{-1}h^{-1}$ is non-
trivial, belongs to $F_{+}$ and fixes $\psi(h)(I)$ pointwise. This is a
contradiction since we have already shown that the action of $F_{+}$ is
topologically free.

Now consider the case $K\neq(0,1)$. We can take a connected component
$U=(y,z)$ of the complement $(0,1)\smallsetminus K$, and consider a non-
trivial element $h\in F_{+}$ whose support is contained in $U$. Fix $x<y$ and
consider the $\preceq^{K}$-convex subgroup $L_{x}$ from Lemma 11.21. Take an
element $g\in L_{x}$, and let us prove that the conjugate $g^{-1}hg$ belongs
to $H$. For this, note that the condition $x_{g}<x$ implies $g^{-1}(U)=U$, so
that the restriction of $g^{-1}hg$ to the complement $(0,1)\smallsetminus U$
is trivial. This immediately implies that $g^{-1}hg$ fixes every point of $K$,
so that $g^{-1}hg$ belongs to the residue $H$. This proves that $hgH=gH$ for
any element $g\in L_{x}$, so that the element $h$ fixes the $\prec^{K}$-convex
subset $L_{x}/H$ pointwise. We deduce that $\psi_{K}(h)$, which is non-trivial
as the action $\psi_{K}$ is faithful, fixes a non-trivial interval. ∎

######  Remark 11.24.

Proposition 11.23 should be compared with the fact that many groups arising
via a micro-supported action by homeomorphisms satisfy rigidity results for
their non-topologically free actions on _compact_ spaces, as shown in [LBMB-
sub, bon2018rigidity, boudec2020commutator] using results on uniformly
recurrent subgroups (URS) and confined subgroups. As an example tightly
related to this setting consider Thompson’s group $F$ and its sibling $T$
acting on the circle. Then every minimal action of $T$ on any compact space is
either topologically free or factors onto its standard action on the circle,
while every faithful minimal action of $F$ on a compact space is topologically
free [LBMB-sub]. Proposition 11.23 shows that actions on the line behave very
differently from this perspective, and the notion of topological freeness is
much less relevant.

Another feature of this family of actions is that they are all simplicial in
the sense of §11.3.

######  Proposition 11.25 (Simpliciality).

Let $K\subset(0,1)$ be a non-empty $f$-invariant closed subset and consider
the corresponding action $\psi_{K}\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$, as constructed above. Then
the image of $F_{+}$ does not act minimally on $\mathbb{R}$. In particular
every action $\psi_{K}$ is simplicial.

###### Proof.

Fix $x\in(0,1)$ and consider the $\preceq^{K}$-convex subgroup $L_{x}=\\{g\in
F_{+}\colon x_{g}\leq x\\}$ (Lemma 11.21). In the dynamical realization
$\psi_{K}$ of the action
$F\to\operatorname{\mathsf{Aut}}\left(F_{+}/H,\prec^{K}\right)$, the cosets of
$L_{x}$ span an $F_{+}$-invariant family of disjoint open intervals, showing
that the $\psi_{K}$-action of $F_{+}$ is not minimal. In particular neither is
the action of $[F,F]\subseteq F_{+}$. ∎

One way to analyze finer properties of $\mathbb{R}$-focal actions of the group
$F$ is to apply Theorem 11.3 iteratively, by exploiting the self-similarity of
$F$. Namely assume that $\varphi\colon
F\to\operatorname{\mathsf{Homeo}}_{+}(\mathbb{R})$ is a minimal
$\mathbb{R}$-focal action increasingly horograded by the standard action on
$(0,1)$. Recall that for every dyadic $x\in(0,1)$ the group $F_{(0,x)}$ is
isomorphic to $F$, and its image under $\varphi$ is totally bounded, that is
$\operatorname{\mathsf{Fix}}^{\varphi}(F_{(0,x)})$ accumulates on both
$\pm\infty$. Thus we can apply Theorem 11.3 to the action of $F_{(0,x)}$ on
every connected component $J$ of
$\operatorname{\mathsf{Supp}^{\varphi}}(F_{(0,x)})$. It follows that this
action still falls into one of the three cases up to semi-conjugacy: action by
translations, the standard action, and $\mathbb{R}$-focal actions. In the
third case, this analysis can of course be iterated. We will speak of
“sublevels” of the action $\varphi$ to refer to the actions of the subgroups
$F_{(0,x)}$ obtained in this way. From this point of view the actions
$\psi_{K}$ arising from restriction preorder are very special: indeed they are
not exotic on any sublevel (in contrast with other $\mathbb{R}$-focal actions
of $F$; see Proposition 11.33 below).

######  Proposition 11.26 (Absence of exotic sublevels).

Let $K\subset(0,1)$ be a non-empty $f$-invariant closed subset and consider
the corresponding action $\psi_{K}\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$, as constructed above. Let
$x\in X$ be a dyadic point, and let $J$ be a connected component of
$\operatorname{\mathsf{Supp}}^{\psi}(F_{(0,x)})$. Then the $\psi$-action of
$F_{(0,x)}$ on $J$ is semi-conjugate either to its standard action on $(0,x)$,
or to a cyclic action by translations induced from group of germs
$\operatorname{\mathsf{Germ}}(F_{(0,x)},x)\cong\mathbb{Z}$.

###### Proof.

Let $\xi_{0}$ be the unique fixed point of $\psi(f)$. Let us first show the
claim for the action of $F_{(0,x)}$ on $J=\operatorname{I^{\psi}}(x,\xi_{0})$
(the connected component of $\operatorname{\mathsf{Supp}}^{\psi}(F_{(0,x)})$
containing $\xi_{0}$). The semi-conjugacy type of this action is determined by
the preorder $\preceq_{\xi_{0}}\in\operatorname{\mathsf{LPO}}(F_{(0,x)})$
induced by the point $\xi_{0}$ on $F_{(0,x)}$, which coincides with the
restriction of $\preceq^{K}$ to $F_{(0,x)}$. Now we distinguish two cases.

First assume that $K\cap(0,x)$ does not accumulate on $x$. Write
$y=\sup\\{K\cap(0,x)\\}<x$ and let
$\preceq_{y}\in\operatorname{\mathsf{LPO}}(F_{(0,x)})$ be its induced preorder
on $F_{(0,x)}$. By definition of $\preceq^{K}$, it follows that
$\preceq_{\xi_{0}}$ is dominated (in the sense of Definition 3.6) by
$\preceq_{y}$. Since the dynamical realization of $\preceq_{y}$ is the
standard action of $F_{(0,x)}$, the conclusion in this case follows from Lemma
3.16.

Assume now that $\sup\\{K\cap(0,x)\\}=x$. In this case, by definition of
$\preceq^{K}$ we get that $\preceq_{\xi_{0}}$ is dominated by a preorder
obtained as the pull-back of one of the two non-trivial preorders on
$\operatorname{\mathsf{Germ}}(F_{(0,x)},x)\cong\mathbb{Z}$. This shows the
conclusion for $\xi=\xi_{0}$.

If now $\xi\in\operatorname{\mathsf{Supp}}^{\psi}(F_{(0,x)})$ is arbitrary,
then by minimality we can choose $h\in F$ such that
$\psi(h)(\xi_{0})\in\operatorname{I^{\psi}}(x,\xi)$. Then the conclusion
follows from the previous case applied to the action of
$F_{(0,h^{-1}(x))}=h^{-1}F_{(0,x)}h$ on
$\operatorname{I^{\psi}}(h^{-1}(x),\xi_{0})$. ∎

####  11.4.4. Some variations on the restriction preorder construction

The restriction preorder construction can be modified in multiple ways to
produce new families of minimal $\mathbb{R}$-focal actions, which are not
conjugate to the actions $\psi_{K}$ defined above. We indicate some of them,
without detailed exploration nor attempting to include them all in a unified
family.

  1. (1)

_Twisting with sign choices._ In addition to the $f$-invariant set
$K\subseteq(0,1)$ consider an $f$-invariant choice of signs $u\colon
K\to\\{+1,-1\\}$. We proceed to define a preorder $\preceq^{(K,u)}$ in
$F_{+}$. For this, given $g\in F_{+}$ we say that $g{}^{(K,u)}1_{F}$ if either
$u(x_{g})=1$ and $g{}^{K}1_{F}$ or $u(x_{g})=-1$ and $g{}^{K}1_{F}$. It is
direct to check (following the proof of Lemma 11.19) that $\preceq^{(K,u)}$ is
an invariant preorder on $F_{+}$ and that the $f$-invariance of $u$ makes
$\preceq^{(K,u)}$ invariant under conjugation by $f$. Of course, when $u\equiv
1$ the preorders $\preceq^{(K,u)}$ and $\preceq^{K}$ coincide. There are some
straightforward variations to this twist. For instance one may consider two
different $f$-invariant functions $u,v\colon K\to\\{\pm 1\\}$ to determine the
sign in the two different cases $g(x_{g})\neq x_{g}$ and $g(x_{g})=x_{g}$.

  2. (2)

_Twisting with derivative morphisms._ In this case, in addition to the
$f$-invariant set $K\subseteq(0,1)$ consider a total left order $<_{0}$ on the
abelian group

$A=\\{(2^{n},2^{m})\colon n,m\in\mathbb{Z}\\}\cong\mathbb{Z}^{2}$

(note that $A$ can be though as the set of derivatives that an element of $F$
can take at a dyadic point). As before, we will define a preorder on $F_{+}$
which is invariant under conjugacy by $f$. For this, consider a different
definition of $x_{g}$, namely define

$x^{\prime}_{g}:=\sup\left\\{x\in K:g(x)\neq x,\text{ or }g(x)=x\text{ and
}(D^{-}g(x),D^{+}g(x))\neq(1,1)\right\\}.$

Then, set $\preceq^{K}_{0}\in\operatorname{\mathsf{LPO}}(F_{+})$ so that
$g{}^{K}_{0}1_{F}$ if either $g(x^{\prime}_{g})>x^{\prime}_{g}$, or
$g(x^{\prime}_{g})=x^{\prime}_{g}$ and
$(D^{-}g(x^{\prime}_{g}),D^{+}g(x^{\prime}_{g}))>_{0}(1,1)$. Again, it is
straightforward to check (following the proof of Lemma 11.19) that the
preorder $\preceq^{K}_{0}$ is left invariant and also invariant under
conjugation by $f$.

To compare these preorders with the preorders of the form $\preceq^{(K,u)}$,
consider $p\in(0,1)\cap\mathbb{Z}[1/2]$ and the closed subset
$K_{p}=\\{f^{n}(p):n\in\mathbb{Z}\\}$. In this case all the twists
$\preceq^{(K_{p},u)}$ given by sign choices coincide with $\preceq^{K_{p}}$,
while the preorder $\preceq^{K_{p}}_{0}$ just defined does not. The interested
reader can also check that in this case, the dynamical realization of
$\preceq^{K_{p}}_{0}$ has sublevels (in the sense of Proposition 11.26) semi-
conjugate to $\varphi\circ\pi^{ab}$ where $\pi^{ab}:F_{(0,x)}\to
F_{(0,x)}^{ab}\cong\mathbb{Z}^{2}$ is the abelianization of $F_{(0,x)}$ (i.e.
$x$ is dyadic) and $\varphi$ is the dynamical realization of $<_{0}$.

  3. (3)

_Twisting with new orderings of $(0,1)$._ In the construction of the preorder
$\preceq^{K}$ one can modify the definition of the point $x_{g}$ by taking the
supremum with respect to an order $\prec_{0}$ on $K$ which is different from
the order induced from the embedding $K\subseteq(0,1)$. The whole construction
will still be well-defined provided $\prec_{0}$ is $f$-invariant and satisfies
suitable assumptions, which are not difficult to figure out but are rather
technical to state. Instead of discussing this in general, let us give an
example.

Take $0<x_{0}<p_{1}<p_{2}<f(x_{0})<1$ and define $K$ as the union of the
orbits of $p_{1}$ and $p_{2}$. Then, we define the total order $\prec_{0}$ on
$K$ so that $f^{m}(p_{i})\prec_{0}f^{n}(p_{j})$ if either $m+i<n+j$, or
$m+i=n+j$, $i=1$ and $j=2$. More explicitly we have

$\cdots\prec_{0}f^{-2}(p_{2})\prec_{0}p_{1}\prec_{0}f^{-1}(p_{2})\prec_{0}f(p_{1})\prec_{0}\cdots.$

It is clear that $\prec_{0}$ is $f$-invariant. We can then define a preorder
$\preceq^{K,\prec_{0}}$ in the same way as the restriction preorder
$\preceq^{K}$, except that we replace the point $x_{g}$ by the point
$x^{\prime\prime}_{g}$ consisting on the $\prec_{0}$-greatest element of the
subset $\\{x\in K:g(x)\neq x\\}$. It is straightforward to check that
$\preceq^{K,\prec_{0}}$ is left invariant and also invariant under conjugation
by $f$, inducing an order-preserving action
$F\to\operatorname{\mathsf{Aut}}\left(F_{+}/[1]_{\preceq^{K,\prec_{0}}},\prec^{K,\prec_{0}}\right)$
as above. Denote by $\Psi_{0}=\Psi_{K,\prec_{0}}$ the dynamical realization of
this action and assume that its associated good embedding satisfies
$\iota([1]_{\preceq^{K,\prec_{0}}})=0$. It can be shown that different choices
of $p_{1}$ and $p_{2}$ produce non-conjugate actions. On the other hand, the
interested reader can check that the semi-conjugacy classes of the sublevels
$F_{(0,x)}\curvearrowright\mathrm{I}^{\Psi_{0}}(x,0)$ only depend on the
choice of $p_{2}$ but not of $p_{1}$. This shows that exotic actions cannot be
reconstructed with the information of the semi-conjugacy classes of its sub-
levels as defined in Proposition 11.26.

Again there are some obvious variations of this, such as considering preorders
on $A$ instead of orders, and modifying the definition of the point
$x^{\prime}_{g}$ accordingly.

Of course one can consider appropriate combinations of the variants defined
above. However whether such combinations make sense or not depends on the
choice of the parameters, and a unified treatment would be obscure and
pointless. All constructions obtained using these methods yield simplicial
actions.

###  11.5. A plethora of $\mathbb{R}$-focal actions II: ordering the orbit of
a closed subset of $(0,1)$

We now describe another method to construct $\mathbb{R}$-focal actions of $F$.
The starting ingredient of this method is again a non-empty closed subset
$K\subset(0,1)$ which is invariant under the generator $f$ given by (11.1). We
assume now $K\neq(0,1)$, and consider the $F$-orbit of $K$ among closed
subsets of $(0,1)$, and denote it by

$\mathcal{O}_{K}:=\\{g(K)\colon g\in F\\}.$

As $K\subset(0,1)$ is a proper subset, we clearly have that the orbit
$\mathcal{O}_{K}$ is infinite. It is then natural to try to define an
$F$-invariant order $\prec$ on $\mathcal{O}_{K}$, and then consider its
dynamical realization. While this may seem similar to the construction just
discussed in §11.4.2, it turns out to be quite different and to produce
actions with more exotic dynamical properties. Note that we are not aware of
any general receipt to build $F$-invariant orders on $\mathcal{O}_{K}$ which
works _for all_ $K$: the way such orders arise depend subtly on the properties
of the subset $K$. However, there is a general strategy which is conveniently
described in the language of directed trees. We will first describe this
strategy in general, and then illustrate it in practice with a concrete choice
of subset $K$. More examples of actions obtained using this method will appear
later in §11.6.

####  11.5.1. A strategy to order $\mathcal{O}_{K}$

Assume that $K\subset(0,1)$ is an $f$-invariant closed subset. Since the germ
of $f$ at $1$ generates the group of germs
$\operatorname{\mathsf{Germ}}(F,1)\cong\mathbb{Z}$ and $K$ is $f$-invariant,
it follows that every $K_{1}=g(K)\in\mathcal{O}_{K}$ must coincide with $K$ on
an interval of the form $(1-\varepsilon,1)$, with $\varepsilon>0$. Thus, it
follows that for every pair $K_{1},K_{2}\in\mathcal{O}_{K}$, the subsets
$K_{1}$ and $K_{2}$ coincide on some interval of the form $(1-\varepsilon,1)$,
so that we can define

(11.4) $\alpha(K_{1},K_{2})=\inf\\{x\in(0,1)\colon
K_{1}\cap[x,1)=K_{2}\cap[x,1)\\}.$

As $K$ is closed, we have $\alpha(K_{1},K_{2})\in K_{1}\cap K_{2}$, unless
$K_{1}=K_{2}$ (in which case $\alpha(K_{1},K_{2})=0$). Moreover, in light of
the previous discussion, we get that $\alpha(K_{1},K_{2})<1$ for every
$K_{1},K_{2}\in\mathcal{O}_{K}$. It is clear from the definition that for
every $K_{1},K_{2},K_{3}\in\mathcal{O}_{K}$ with
$\alpha(K_{1},K_{3})\leq\alpha(K_{2},K_{3})$, we have
$\alpha(K_{1},K_{2})\leq\alpha(K_{2},K_{3})$ (indeed, when $K_{2}\neq K_{3}$,
the three intersections $K_{i}\cap[\alpha(K_{2},K_{3}),1)$ for
$i\in\\{1,2,3\\}$ coincide). This gives the ultrametric inequality

$\alpha(K_{1},K_{2})\leq\max\\{\alpha(K_{2},K_{3}),\alpha(K_{1},K_{3})\\}.$

Note also that for given $g\in F$ and $K_{1},K_{2}\in\mathcal{O}_{K}$, we have

$\displaystyle\alpha(g(K_{1}),g(K_{2}))$ $\displaystyle=\inf\\{x\in(0,1)\colon
K_{1}\cap[g^{-1}(x),1)=K_{2}\cap[g^{-1}(x),1)\\}$
$\displaystyle=\inf\\{g(y)\in(0,1)\colon
K_{1}\cap[y,1)=K_{2}\cap[y,1)\\}=g\left(\alpha(K_{1},K_{2})\right).$

In the terminology of §8.6, we have just verified that the map
$\alpha:\mathcal{O}_{K}\times\mathcal{O}_{K}\to[0,1)$ is an $F$-equivariant
ultrametric kernel with respect to the standard $F$-actions on
$\mathcal{O}_{K}$ and $(0,1)$ respectively. Moreover, we have the following.

######  Lemma 11.27.

For every closed $f$-invariant subset $K\subsetneq(0,1)$, the action of $F$ on
$\mathcal{O}_{K}$ expands $\alpha$-balls.

###### Proof.

As the action of $F$ on $\mathcal{O}_{K}$ is transitive, it is enough to check
that there exists a sequence of elements $(g_{n})\subset F$ such that the
sequence of balls $g_{n}.B_{\alpha}(K,x)$ defines an increasing exhaustion of
$\mathcal{O}_{K}$. For this, note that by $f$-invariance we have
$f^{n}.B_{\alpha}(K,x)=B_{\alpha}(K,f^{n}(x))$ and thus
$\mathcal{O}_{K}=\bigcup_{n\geq 0}f^{n}.B_{\alpha}(K,x)$, as desired. ∎

As a consequence of the discussion above and Corollary 8.41, for every
$\alpha$-convex order $<$ on $\mathcal{O}_{K}$ we get a minimal
$\mathbb{R}$-focal action increasingly horograded by the standard $F$-action.
However, it is not clear a priori that for a given subset $K$ such an
$\alpha$-convex order exists, and this is why what we have just described is
simply a _strategy_. We will see in §11.5.2 and §11.6, that for some choices
of $K$, $\alpha$-convex orders actually exist, although this is false in
general (see Example 11.29 below).

######  Remark 11.28.

In practice, $\alpha$-convex orders on $\mathcal{O}_{K}$ are such that the
order relation between $K_{1}$ and $K_{2}$ only depends on how $K_{1},K_{2}$
behave “right before” the point $\alpha(K_{1},K_{2})$, in an $F$-invariant
manner. For a formal presentation of this correspondence see Proposition 8.40.

######  Example 11.29 (Non planarly orderable actions).

Recall from Definition 8.36, that there is a natural construction of directed
tree associated with the ultrametric kernel $\alpha$ on $\mathcal{O}_{K}$.
More precisely, we obtain an action

$\Phi:F\to\operatorname{\mathsf{Aut}}(\mathbb{T}_{K},\triangleleft)$

together with a $F$-equivariant injective map
$i:\mathcal{O}_{K}\to\partial^{\ast}\mathbb{T}_{K}$. Roughly speaking, the
directed tree $(\mathbb{T}_{K},\triangleleft)$ is obtained by taking a copy of
$(0,1)$ for each $K_{1}\in\mathcal{O}_{K}$, and by gluing the two copies
corresponding to $K_{1}$ and $K_{2}$ along the interval
$[\alpha(K_{1},K_{2}),1)$. We denote by
$p:\mathcal{O}_{K}\times(0,1)\to\mathbb{T}_{K}$ the quotient projection and
$[K_{1},x]:=p(K_{1},x)$. Then, two points $v,w\in\mathbb{T}_{K}$ satisfy
$v\triangleleft w$ (that is, $v$ lies below $w$) if and only there exists
$K_{1}\in\mathcal{O}_{K}$ and $x,y\in(0,1)$ so that $v=[K_{1},x]$,
$w=[K_{1},y]$, and $x<y$. The diagonal action of $F$ on
$\mathcal{O}_{K}\times(0,1)$ descends to an action on
$(\mathbb{T}_{K},\triangleleft)$ and the projection to the second coordinate
descends to an increasing $F$-equivariant horograding
$\pi\colon\mathbb{T}_{K}\to(0,1)$. Finally, the embedding
$i:\mathcal{O}_{K}\to\partial^{\ast}\mathbb{T}_{K}$ is defined so that each
$K_{1}\in\mathcal{O}_{K}$ is sent to the infimum of the $\triangleleft$-chain
$\\{[K_{1},x]:x\in(0,1)\\}$, which naturally belongs to
$\partial^{\ast}\mathbb{T}_{K}$.

Although non-strictly necessary for the sequel, this point of view is well-
suited for understanding in a more conceptual way whether for given $K$ the
action $\Phi:F\to\operatorname{\mathsf{Aut}}(\mathbb{T}_{K},\triangleleft)$
admits a $\Phi$-invariant planar order, and this is the same (after
Proposition 8.40) to the condition that $\mathcal{O}_{K}$ admit an
$\alpha$-convex order. This turns out to depend on the local geometry of $K$
relatively to dilations by 2. We explain this with an example, but first we
introduce some terminology to discuss the local geometry. We say that two
closed sets $K_{1},K_{2}\subseteq(0,1)$ have equivalent left-germs at $x$ if
for some $\varepsilon>0$ it holds
$K_{1}\cap(x-\varepsilon,x]=K_{2}\cap(x-\varepsilon,x]$. We denote by
$K^{-}_{x}$ the left-germ class of the subset $K$ at $x$. Notice that the
group $\operatorname{\mathsf{Germ}}_{-}(x)$ of left-germs of homeomorphisms
fixing $x$ naturally acts on the set of left-germs of closed sets at $x$. We
denote by $h_{x}\in\operatorname{\mathsf{Germ}}_{-}(x)$ the germ of the
homothety that fixes $x$ and has derivative $2$.

Recall from Remark 8.16 that the existence of such invariant planar ordering
boils down to the existence, for each branching point $v\in\mathbb{T}_{K}$, of
an ordering of the set of connected components $E^{-}_{v}$ below $v$, which is
invariant under the action of the stabilizer
$\operatorname{\mathsf{Stab}}^{\Phi}(v)$. An obstruction for this is clearly
given by finite orbits. With this in mind, consider an $f$-invariant closed
subset $K\subseteq(0,1)$ containing a dyadic point $x\in
K\cap\mathbb{Z}[\frac{1}{2}]$ such that $h_{x}(K^{-}_{x})\neq K^{-}_{x}$ but
$h_{x}^{n}(K^{-}_{x})=K^{-}_{x}$ for some $n>1$. Write $v=[K,x]$ and let
$e_{v}(K)$ be the component of $E_{v}^{-}$ corresponding to the ray
$\\{[K,x]:x\in(0,1)\\}$. Since $\operatorname{\mathsf{Germ}}_{-}(F,x)=\langle
h_{x}\rangle$, it holds that the component $e_{v}(K)$ has a finite orbit which
is not a fixed point, so that there exists no
$\mathsf{Stab}^{\Phi}(v)$-invariant total order on $E_{v}^{-}$.

####  11.5.2. A concrete example

We now illustrate the flexibility of the method described in §11.5.1, with an
explicit example of subset $K$. More precisely we will construct a subset
$K\subset(0,1)$ with the following property: _there is an explicit
(continuous) injective map from the set $\mathsf{O}(\mathbb{N})$ of orders on
the natural numbers $\mathbb{N}$ to the set of $F$-invariant orders on the
orbit $\mathcal{O}_{K}$_. This will provide a family of $\mathbb{R}$-focal
actions of $F$ which are naturally indexed by orders on $\mathbb{N}$.

We start by choosing an irrational point $x_{0}\in(0,1)$, and consider the
interval $I=(f^{-1}x_{0},x_{0}]$, which is a fundamental domain for $f$. Next
we choose a sequence of open intervals $(J_{n})_{n\geq 1}$ with dyadic
endpoints such that $J_{n}\Subset J_{n+1}\subset I$ for every $n\geq 1$, and
such that $\bigcup_{n\geq 1}J_{n}=(f^{-1}x_{0},x_{0})$. For every $n\geq 1$,
write $y_{n}=\sup J_{n}$ and choose an element $h_{n}\in F_{J_{n}}$ with the
following properties:

  * •

$h_{n}(x)>x$ for every $x\in J_{n}$,

  * •

$h_{n}(J_{n-1})\cap J_{n-1}=\varnothing$ for $n\geq 2$.

  * •

$D^{-}h_{n}(y_{n})=1/2$ (in other words, the germ of $h_{n}$ at $y_{n}$
generates the group of germs
$\operatorname{\mathsf{Germ}}(F_{(0,y_{n})},y_{n})$).

Choose now a dyadic point $z_{0}\in J_{0}$, and let
$\Sigma_{0}=\\{h_{1}^{n}(z_{0})\colon n\in\mathbb{N}\\}$ be its forward orbit
under $h_{1}$. By construction we have the inclusion $\Sigma_{0}\subset J_{0}$
and equality $\overline{\Sigma_{0}}=\Sigma_{0}\cup\\{y_{1}\\}$. Set
$\Sigma_{1}=\bigcup_{n\geq 0}h_{2}^{n}(\overline{\Sigma_{0}})$, so that
$\overline{\Sigma_{1}}=\Sigma_{1}\cup\\{y_{2}\\}$. Continue in this way by
defining for every $i\geq 1$ a subset $\Sigma_{i}=\bigcup_{n\geq
0}h_{i+1}^{n}(\overline{\Sigma_{i-1}})$. Set
$\Sigma_{\omega}=\bigcup_{i\in\mathbb{N}}\Sigma_{i}$, and note that
$\overline{\Sigma_{\omega}}=\Sigma_{\omega}\cup\\{x_{0}\\}$. Note also that
$\overline{\Sigma_{\omega}}$ is contained in the fundamental domain $I$ on
$f$. Thus we obtain an $f$-invariant closed subset $K$ as

(11.5) $K=\bigcup_{n\in\mathbb{Z}}f^{n}(\overline{\Sigma_{\omega}}).$

By construction the subset $\overline{\Sigma_{\omega}}$ is invariant under the
semigroup $S:=\langle h_{n}\colon n\geq 1\rangle_{+}$, in the sense that
$s(\overline{\Sigma_{\omega}})\subset\overline{\Sigma_{\omega}}$ for every
$s\in S$. See Figure 11

Figure 11. Construction of the compact set $K$.

The subset $\overline{\Sigma_{\omega}}$ is countable and compact, and its
points can be classified according to their _Cantor–Bendixson rank_ (see
[Kechris, §6]), as follows. Points of rank 0 are the isolated points: these
are exactly points in the orbit of $z_{0}$ under the semigroup $S$. Points of
rank 1 are those that are not isolated, but become isolated after removing the
isolated points: these are exactly points in the $S$-orbit of $y_{1}$.
Continuing in this way, points of rank $n$ are precisely points in the
$S$-orbit of $y_{n}$. Finally there is a unique point whose rank is the first
countable ordinal $\omega$, namely the point $x_{0}$. This discussion can be
directly extended to the subset $K$. We write
$\operatorname{\mathsf{rk}}_{K}(x)$ for the Cantor–Bendixson rank of a point
$x\in K$. Note that for every $g\in F$ and $x\in K$, we have the relation
$\operatorname{\mathsf{rk}}_{g(K)}(g(x))=\operatorname{\mathsf{rk}}_{K}(x)$.

We next consider the $F$-equivariant ultrametric kernel
$\alpha:\mathcal{O}_{K}\times\mathcal{O}_{K}\to[0,1)$ defined as in (11.4) and
the key observation is that the particular choice of the subset $K$ allows to
directly relate $\alpha$ with the Cantor–Bendixson rank.

######  Lemma 11.30.

Let $K\subset(0,1)$ be the subset defined at (11.5). For every
$K_{1},K_{2}\in\mathcal{O}_{K}$, the point $x=\alpha(K_{1},K_{2})$ is such
that $\operatorname{\mathsf{rk}}_{K_{1}}(x)$ and
$\operatorname{\mathsf{rk}}_{K_{2}}(x)$ are both finite, and moreover
$\operatorname{\mathsf{rk}}_{K_{1}}(x)\neq\operatorname{\mathsf{rk}}_{K_{2}}(x)$
unless
$\operatorname{\mathsf{rk}}_{K_{1}}(x)=\operatorname{\mathsf{rk}}_{K_{2}}(x)=0$.

Conversely, for every distinct $n,m\in\mathbb{N}$ there exist
$K_{1},K_{2}\in\mathcal{O}_{K}$ such that the point $x=\alpha(K_{1},K_{2})$
satisfies $\operatorname{\mathsf{rk}}_{K_{1}}(x)=n$ and
$\operatorname{\mathsf{rk}}_{K_{2}}(x)=m$.

###### Proof.

We first need some observations.

######  Claim 1.

For every $x\in K$ and every $g\in F$ such that $g(x)=x$, there exists
$\varepsilon>0$ such that $g(K)\cap(x-\varepsilon,x]=K\cap(x-\varepsilon,x]$.

###### Proof of claim.

Up to replace $g$ by its inverse, we can assume $D^{-}g(x)\leq 1$. Also, upon
conjugating by powers of $f$, we can assume $x\in\overline{\Sigma_{\omega}}$.
If $x=x_{0}$ then this follows from the fact that we chose $x_{0}$ to be
irrational, so that every element of $F$ that fixes $x_{0}$ must actually fix
a neighborhood of it. If $x$ is isolated in $K$ the conclusion is obvious.
Finally assume that
$n:=\operatorname{\mathsf{rk}}_{K}(x)\notin\\{0,\omega\\}$. Then $x$ is in the
$S$-orbit of the point $y_{n}$, so that it is fixed by a conjugate $h$ of
$h_{n}$, which has therefore the property that $D^{-}h(x)=1/2$. Hence the
restriction of $g$ to a left-neighborhood of $x$ must coincide with the
restriction of some non-negative power of $h$, so that we can conclude from
the fact that $K$ is forward invariant under $h$. ∎

######  Claim 2.

For every pair of points $x,y\in K$ with
$\operatorname{\mathsf{rk}}_{k}(x)=\operatorname{\mathsf{rk}}_{k}(y)$, there
exist an element $h\in F$ and $\varepsilon>0$ such that $h(x)=y$ and
$h(K)\cap(y-\varepsilon,y]=K\cap(y-\varepsilon,y]$.

###### Proof of claim.

Upon replacing $x,y$ with $f^{m}(x),f^{n}(y)$ for suitable $n,m$ we can assume
that $x,y\in\overline{\Sigma_{\omega}}$. Then $x$ and $y$ are in the same
$S$-orbit, and so it is enough to observe that elements of $S$ and their
inverses have this property. ∎

With this in mind, let us prove the lemma. We can assume without loss of
generality that $K_{1}=K$. Take $g\in F$ such that $K_{2}=g(K)$ and set
$x=\alpha(K,K_{2})$ and $y=g^{-1}(x)\in K$, so that
$\operatorname{\mathsf{rk}}_{K_{2}}(x)=\operatorname{\mathsf{rk}}_{K}(y)$.
Assume by contradiction that
$\operatorname{\mathsf{rk}}_{K}(x)=\operatorname{\mathsf{rk}}_{K}(y)\geq 1$.
After Claim 2, we can choose $h\in F$ such that $h(x)=y$ and $\varepsilon>0$
such that $h(K)\cap(y-\varepsilon,y]=K\cap(y-\varepsilon,y]$. Then the element
$g^{\prime}=hg$ is such that $g(y)=y$, so that upon taking a smaller
$\varepsilon$, by Claim 1 we also have
$g^{\prime}(K)\cap(y-\varepsilon,y]=K\cap(y-\varepsilon,y]$. Applying $h^{-1}$
we deduce that there is $\varepsilon^{\prime}>0$ such that
$g(K)\cap(x-\varepsilon^{\prime},x]=K\cap(x-\varepsilon^{\prime},x]$, and the
latter intersection is not reduced to $\\{x\\}$ since we assume that
$\operatorname{\mathsf{rk}}_{K}(x)\geq 1$. This contradicts the definition of
$x=\alpha(g(K),K)$. Thus
$\operatorname{\mathsf{rk}}_{K}(x)\neq\operatorname{\mathsf{rk}}_{g(K)}(x)$
unless both ranks are 0. Finally this also implies that we cannot have
$\operatorname{\mathsf{rk}}_{K}(x)=\omega$, indeed since points of rank
$\omega$ are the only non-dyadic points in $K$ this would imply that
$\operatorname{\mathsf{rk}}_{g(K)}(x)=\omega$ as well, contradicting the
previous reasoning. ∎

Now let $\mathsf{O}(\mathbb{N})$ be the set of total orders on the natural
numbers. To every order $\prec$ in $\mathsf{O}(\mathbb{N})$ we associate an
$F$-invariant order $\prec^{*}$ on $\mathcal{O}_{K}$, as follows. Given
distinct $K_{1},K_{2}\in\mathcal{O}_{K}$, set
$n_{1}=\operatorname{\mathsf{rk}}_{K_{1}}(\alpha(K_{1},K_{2}))$ and
$n_{2}=\operatorname{\mathsf{rk}}_{K_{2}}(\alpha(K_{1}.K_{2}))$. If $n_{1}\neq
n_{2}$, then we declare $K_{1}\prec^{*}K_{2}$ if and only if $n_{1}\prec
n_{2}$. Else, by Lemma 11.30 we have $n_{1}=n_{2}=0$, namely the point
$\alpha(K_{1},K_{2})$ is isolated in both $K_{1}$ and $K_{2}$. In this case
set

(11.6) $x_{i}=\max\\{x\in K_{i}\colon x<\alpha(K_{1},K_{2})\\}\quad\text{for
}i\in\\{1,2\\}.$

Then for $i\in\\{1,2\\}$ we must have $x_{i}<\alpha(K_{1},K_{2})$ and
$x_{1}\neq x_{2}$ by definition of $\alpha(K_{1},K_{2})$. In this case we
declare $K_{1}\prec^{*}K_{2}$ if and only if $x_{1}<x_{2}$. It is routine to
verify that this defines indeed a total order relation, and it is clear from
the construction, and $F$-equivariance of the ultrametric kernel $\alpha$ and
the Cantor–Bendixson rank, that this order is $F$-invariant.

Denote by $\varphi_{\prec}:F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$
the dynamical realization of the action of $F$ on
$(\mathcal{O}_{K},\prec^{\ast})$. We want to prove that $\varphi_{\prec}$ is a
minimal $\mathbb{R}$-focal action, increasingly horograded by the standard
action of $F$. After the discussion in §11.5.1, this is equivalent to the
property that the $\alpha$-balls are $\prec^{*}$-convex. This is what we
verify next.

######  Lemma 11.31.

With notation as above, the $\alpha$-ball

$B_{\alpha}(L,x)=\left\\{L^{\prime}\in\mathcal{O}_{K}:\alpha(L,L^{\prime})\leq
x\right\\}$

is $\prec^{\ast}$-convex for every $L\in\mathcal{O}_{K}$ and $x\in(0,1)$.

###### Proof.

First notice that the $\prec^{\ast}$-order relation between
$K_{1},K_{2}\in\mathcal{O}_{K}$ is determined by the intersections
$K_{1}\cap[x,1)$ and $K_{2}\cap[x,1)$ for any $x\in(0,1)$ such that these
intersections do not coincide.

Now take elements $K_{1},K_{2}\in B_{\alpha}(L,x)$ for some
$L\in\mathcal{O}_{K}$ and $x\in(0,1)$. This is equivalent to the condition
that

(11.7) $K_{1}\cap[x,1)=K_{2}\cap[x,1)=L\cap[x,1).$

Consider next an element $K_{3}$ between $K_{1}$ and $K_{2}$ (with respect to
$\prec^{*}$) and assume by contradiction that $K_{3}\notin B_{\alpha}(L,x)$.
This means that $K_{3}\cap[x,1)\neq L\cap[x,1)$ and therefore, considering the
equalities (11.7), the $\prec^{\ast}$-order relation between $K_{i}$ and
$K_{3}$ is determined by the intersections $K_{3}\cap[x,1)$ and $L\cap[x,1)$
for every $i\in\\{1,2\\}$. Hence we conclude that the $\prec^{\ast}$-order
relation between $K_{1}$ and $K_{3}$ coincides with that of $K_{2}$ and
$K_{3}$. As we are assuming that $K_{3}$ lies between $K_{1}$ and $K_{2}$, we
necessarily have $K_{1}=K_{2}=K_{3}$, but this contradicts the assumption
$K_{3}\notin B_{\alpha}(L,x)$. ∎

As a conclusion of our discussion, we have the following.

######  Proposition 11.32.

With notation as above, for any $\prec\in\mathsf{O}(\mathbb{N})$, the
dynamical realization
$\varphi_{\prec}:F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ of the
action of $F$ on $(\mathcal{O}_{K},\prec^{*})$ is a minimal $\mathbb{R}$-focal
action increasingly horograded by the standard action of $F$.

Moreover, if $\prec_{1}$ and $\prec_{2}$ are distinct orders on $\mathbb{N}$,
then the actions $\varphi_{\prec_{1}}$ and $\varphi_{\prec_{2}}$ are not
conjugate.

###### Proof.

After Lemma 11.31, the $\alpha$-balls are $\prec^{*}$-convex, so the first
statement is a consequence of Corollary 8.41.

For a given order $\prec\in\mathsf{O}(\mathbb{N})$, observe that by definition
of dynamical realization, the $F$-action on $(\mathcal{O}_{K},\prec^{*})$ can
be identified with the $\varphi_{\prec}$-action on the orbit of the unique
fixed point $\xi$ of $\varphi_{\prec}(f)$ with the order induced by
$\mathbb{R}$, so that the order $\prec^{*}$ can be reconstructed from
$\varphi_{\prec}$. Finally the order $\prec$ on $\mathbb{N}$ can be
reconstructed from $\prec^{*}$ by the last statement in Lemma 11.30. ∎

We now point out a qualitative difference which distinguishes the family of
$\mathbb{R}$-focal actions constructed here from the one obtained via the
restriction preorder construction as in §11.4.2. Indeed, in this case the
actions of the subgroups $F_{(0,x)}\cong F$ on the components of their support
can remain exotic (compare this with Proposition 11.26).

######  Proposition 11.33 (Presence of exotic sublevels).

Fix an order $\prec$ on $\mathbb{N}$ and let $\varphi:=\varphi_{\prec}\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ be the $\mathbb{R}$-focal
action constructed above. Then there exist a dyadic point $x\in(0,1)$ and a
connected component $J$ of $\operatorname{\mathsf{Supp}^{\varphi}}(F_{(0,x)})$
such that the action of $F_{(0,x)}$ on $J$ is semi-conjugate to a faithful
$\mathbb{R}$-focal action.

###### Proof.

Note first that for every $x\in(0,1)$ and $g\in F$ such that
$g(K\cap[x,1))=K\cap[g(x),1)$ then $g.B_{\alpha}(K,x)=B_{\alpha}(K,g(x))$. In
particular, the $\alpha$-ball $B_{\alpha}(K,x)$ is preserved by the subgroup
$F_{(0,x)}$. Let $\iota\colon(\mathcal{O}_{K},\prec^{*})\to\mathbb{R}$ be an
equivariant good embedding associated with $\varphi$ (in the terminology of
Definition 2.37) and let $I_{x}$ be the open interval spanned by
$\iota(B_{\alpha}(K,x))$, namely $I_{x}$ is the interior of the closure of
$\iota(B_{\alpha}(K,x))$ (using minimality of the action and that the
$\alpha$-balls are $\prec^{*}$-convex after Lemma 11.31). Consider the element
$h_{1}$ from the construction of $K$, and consider the points $z_{0}\in J_{1}$
and $y_{1}=\sup J_{1}$ as in the construction; for $n\geq 1$ set
$z_{n}=h_{1}^{n}(z_{0})$, which by construction is an increasing sequence
converging to $y_{1}$. For every $n\geq 0$ we have
$h_{1}^{n}(K\cap[z_{0},1))=K\cap[z_{n},1)$, so that
$h_{1}^{n}.B_{\alpha}(K,z_{0})=B_{\alpha}(K,z_{n})$. The corresponding
intervals $I_{z_{n}}$ satisfy $I_{z_{n}}\Subset I_{z_{n+1}}$ and
$h_{1}.I_{z_{n}}=I_{z_{n+1}}$. Set $B:=\bigcup_{n\geq
0}B_{\alpha}(K,h_{1}^{n}(z_{0}))$, and let $J=\bigcup_{n\geq 0}I_{z_{n}}$ be
the interval spanned by $\iota(B)$. Then $B$ is preserved by $F_{(0,y_{1})}$
and the action of $F_{(0,y_{1})}$ on $B$ is cofinal (with respect to the order
$\prec^{*}$ restricted to $B$). As a consequence, $F_{(0,y_{1})}$ preserves
$J$ and acts on it without fixed points, so that $J$ is a connected component
of $\operatorname{\mathsf{Supp}^{\varphi}}(F_{(0,y_{1})})$. Since moreover
$h_{1}.I_{z_{n}}=I_{z_{n+1}}$, we deduce that $\varphi(h_{1})$ acts on $J$ as
a pseudohomothety. This cannot happen if the action of $F_{(0,y_{1})}$ on
$\operatorname{I^{\varphi}}(y_{1},\xi)$ is semi-conjugate to an action by
translations, nor if it is semi-conjugate to the standard action on
$(0,y_{1})$. Thus by Theorem 11.3 the action of $F_{(0,y_{1})}$ must be semi-
conjugate to an $\mathbb{R}$-focal action. ∎

Nonetheless, this family of examples still turns out to produce simplicial
$\mathbb{R}$-focal actions in the sense of §11.3.

######  Proposition 11.34 (Simpliciality).

For an order $\prec$ on $\mathbb{N}$, let $\varphi_{\prec}\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ be the $\mathbb{R}$-focal
action constructed above. Then $\varphi_{\prec}(F_{+})$ does not act minimally
on $\mathbb{R}$. In particular each action $\varphi_{\prec}$ is simplicial

###### Proof.

We keep the same notation as in the proof of Proposition 11.33. Let $x_{0}\in
K$ be the point as in the construction of $K$. We claim that the $\alpha$-ball
$B_{\alpha}(K,x_{0})\subset\mathcal{O}_{K}$ has the property that for every
$g\in F_{+}$ we have either $g(B_{\alpha}(K,x_{0}))=B_{\alpha}(K,x_{0})$ or
$g(B_{\alpha}(K,x_{0}))\cap B_{\alpha}(K,x_{0})=\varnothing$. It then follows
that the interval $\operatorname{I^{\varphi}}(x_{0},\xi)$ has the same
property for $\varphi_{\prec}(F_{+})$, so that the union of its translates
defines a proper invariant open subset, contradicting minimality. Indeed,
suppose that $g\in F_{+}$ and $K_{1}\in B_{\alpha}(K,x_{0})$ are such that
$g(K_{1})\in B_{\alpha}(K,x_{0})$, namely we assume

$K_{1}\cap[x_{0},1)=g(K_{1})\cap[x_{0},1)=K\cap[x_{0},1).$

The key observation is that this implies that $g$ must actually fix
$K\cap[x_{0},1)$. First of all observe that $g$ must send points of rank
$\omega$ in $K_{1}$ to points of rank $\omega$ in $g(K_{1})$, and the set of
such points in both $K_{1}\cap[x_{0},1)$ and $g(K_{1})\cap[x_{0},1)$ consists
precisely of the sequence $x_{n}:=f^{n}(x_{0})$ for $n\geq 0$. This is a
discrete increasing sequence and $g(x_{n})=x_{n}$ for $n$ large enough, hence
we deduce from the condition $g\in F_{+}$ that $g(x_{n})=x_{n}$ for every
$n\geq 0$. As a consequence the cyclic subgroup $\langle g\rangle$ must
preserve every interval $[x_{n},x_{n+1}]$, with $n\geq 0$, and thus every
intersection $K\cap[x_{n},x_{n+1}]$ for $n\geq 0$. Assume by contradiction
that there exist $n\geq 0$ and a point $t\in K\cap[x_{n},x_{n+1}]$ which is
not fixed by $g$, and consider the orbit $\Omega=\\{g^{m}(t)\colon
m\in\mathbb{Z}\\}$, which is a subset of $K$. As $K\cap[x_{n},x_{n+1}]$ is
compact, the point $\inf\Omega$ is in $K\cap[x_{n},x_{n+1}]$, and it is
accumulated by points of $\Omega$ (and hence of $K$) from the right. This is
in contradiction with the choice of $K$, as by construction every point of $K$
is isolated from the right hand side. Hence $g$ fixes $K\cap[x_{0},1)$, which
implies that $g.B_{\alpha}(K,x_{0})=B_{\alpha}(K,x_{0})$. ∎

###  11.6. A plethora of $\mathbb{R}$-focal actions III: existence of non-
simplicial $\mathbb{R}$-focal actions

All the example of $\mathbb{R}$-focal actions of $F$ discussed so far are
simplicial in the sense of Definition 11.15, so that it is tempting to try to
prove that all $\mathbb{R}$-focal actions of $F$ must be simplicial. However,
we build an exotic action of Thompson’s group $F$ which fails to have this
property. With Proposition 11.13 in mind, we will prove the following.

######  Theorem 11.35.

There exist faithful $\mathbb{R}$-focal actions $\varphi\colon
F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ such that $\varphi([F,F])$
acts minimally (and thus are not simplicial). More precisely, there exist
uncountably many such actions which are pairwise non-conjugate, and whose
restriction to $[F,F]$ yield pairwise non-conjugate actions of $[F,F]$.

Note the following consequence of independent interest.

######  Corollary 11.36.

The group $[F,F]$ admits uncountably many, pairwise non-conjugate minimal
actions $\varphi\colon[F,F]\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$.

This should be compared with the general constructions of exotic actions of
groups of compactly supported homeomorphisms described in §5.3, which provide
actions without any closed minimal invariant subset. The construction given
here relies on the classical symbolic coding of the the standard action of $F$
by binary sequences, which is specific to Thompson’s groups.

For the proof it will be convenient to see $F$ as a group of homeomorphisms of
$X=\mathbb{R}$ rather than of the interval $(0,1)$. Namely we realize
$F\subseteq\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ as the group of
piecewise linear maps of the line, with dyadic breakpoints, slopes in the
group $\langle 2^{n}:n\in\mathbb{Z}\rangle$ and which coincide with integer
translations near $\pm\infty$. It is well-known that this action is conjugate
to the natural action of $F$ on $(0,1)$ (see e.g. [BieriStrebel, Lemma
E18.4]).

_From now and until the end of this subsection, the term _standard action_
will refer to the action of $F$ on $\mathbb{R}$ described above. We will
denote by $f\in F$ the translation $f(x)=x+1$. _ (Note that the element $f$
corresponds to the element given by (11.1) in the action on $(0,1)$).

The proof of Theorem 11.35 employs the strategy described in §11.5.1, namely
we will start with a closed $f$-invariant subset $K\subset\mathbb{R}$, and
define an invariant order on its orbit $\mathcal{O}_{K}:=\\{g(K)\colon g\in
F\\}$. The main difficulty is that we need to construct a subset $K$
satisfying a somewhat delicate combination of properties. We begin with a
definition.

######  Definition 11.37.

We say that a subset $K\subset\mathbb{R}$ has property $(O)$ if it is proper,
non-empty, closed, $f$-invariant, and moreover $g(K)\cap K$ is open in $K$ for
every $g\in F$.

######  Remark 11.38.

Note that the last condition for property $(O)$ is actually equivalent to that
$K_{1}\cap K_{2}$ be open in $K_{1}$ and $K_{2}$ for every
$K_{1},K_{2}\in\mathcal{O}_{K}$.

######  Example 11.39.

Property $(O)$ is clearly satisfied when $K$ is a non-empty $f$-invariant
discrete subset, as for example the $f$-orbit of a point. However, this is not
a good example for the construction described in this subsection, as the
stabilizer of such $K$ in $[F,F]$ is trivial (cf. Proposition 11.41).

Assume that $K$ is a subset with property $(O)$ (many examples are exhibited
by Lemma 11.42). We consider the $F$-equivariant ultrametric kernel
$\alpha:\mathcal{O}_{K}\times\mathcal{O}_{K}\to\mathbb{R}\cup\\{-\infty\\}$
defined as in §11.5.1, namely

$\alpha(K_{1},K_{2}):=\inf\left\\{x\in\mathbb{R}:[x,+\infty)\cap
K_{1}=[x,+\infty)\cap K_{2}\right\\}.$

Reasoning as in the example of §11.5.2, we proceed to construct an
$F$-invariant order on $\mathcal{O}_{K}$ and then prove that the dynamical
realization of the action of $F$ on $(\mathcal{O}_{K},\prec)$ is minimal and
$\mathbb{R}$-focal. This is the content of the next result.

######  Proposition 11.40.

If a subset $K\subset\mathbb{R}$ has property $(O)$, the relation $\prec$ on
$\mathcal{O}_{K}$ defined by $K_{1}\prec K_{2}$ if and only if

$\max\left\\{x\in K_{1}\colon x<\alpha(K_{1},K_{2})\right\\}<\max\left\\{x\in
K_{2}\colon x<\alpha(K_{1},K_{2})\right\\},$

is an $F$-invariant total order on $\mathcal{O}_{K}$. Moreover, the dynamical
realization $\varphi_{K}:F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ of
the action of $F$ on $(\mathcal{O},\prec)$ is a minimal $\mathbb{R}$-focal
action.

###### Proof.

Recall that $\alpha(K_{1},K_{2})\in K_{1}\cap K_{2}$ whenever $K_{1}$ and
$K_{2}$ are different elements of $\mathcal{O}_{K}$. As $K$ satisfies property
$(O)$, we have that $K_{1}\cap K_{2}$ is open inside both $K_{1}$ and $K_{2}$
and hence $K_{1}\cap K_{2}$ is an open neighborhood of $\alpha(K_{1},K_{2})$
inside $K_{1}$ and $K_{2}$. Therefore $\alpha(K_{1},K_{2})$ is isolated from
the left hand side in both $K_{1}$ and $K_{2}$. As for (11.6), we deduce that
the points

$x_{i}:=\max\\{x\in K_{i}\colon x<\alpha(K_{1},K_{2})\\}\quad\text{for
}i\in\\{1,2\\}$

are distinct, so we can declare $K_{1}\prec K_{2}$ if and only if
$x_{1}<x_{2}$. As for the order $\prec^{*}$ from §11.5.2, it is routine to
check that this defines indeed an $F$-invariant total order on
$\mathcal{O}_{K}$.

Similarly one proceeds as in §11.5.2 to check that $\varphi_{K}$ is minimal
and $\mathbb{R}$-focal. Namely, one verifies that the order $\prec$ is
$\alpha$-convex, and the proof of Lemma 11.31 can be adapted verbatim to this
case (just replacing $\prec^{\ast}$ with $\prec$ and $(0,1)$ with
$\mathbb{R}$). Then Corollary 8.41 gives the desired conclusion. ∎

The main difference from the construction in §11.5.2 is the way that the
commutator subgroup $[F,F]$ acts in the actions $\varphi_{K}$.

######  Proposition 11.41.

Given a subset $K\subset(0,1)$ with property $(O)$, let
$\varphi_{K}:F\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ be the
corresponding minimal $\mathbb{R}$-focal action from Proposition 11.40. Then
the following hold.

  1. (i)

If $K$ has property $(O)$, then $\varphi_{K}([F,F])$ acts minimally provided
that the stabilizer of $K$ in $[F,F]$ (with respect to the standard action)
acts on $K$ without fixed points. Moreover in this case the induced action of
$[F,F]$ is minimal and $\mathbb{R}$-focal.

  2. (ii)

If two distinct subsets $K,K^{\prime}\subset\mathbb{R}$ have property $(O)$,
then the restrictions of $\varphi_{K}$ and $\varphi_{K^{\prime}}$ to $[F,F]$
are not conjugate actions of $[F,F]$. In particular $\varphi_{K}$ and
$\varphi_{K^{\prime}}$ are not conjugate.

###### Proof.

To prove (i), assume that the stabilizer of $K$ in $[F,F]$ acts without fixed
points on $K$. Fix $x\in(0,1)$ and choose a sequence of elements of $[F,F]$
which preserve $K$ and such that $g_{n}(x)$ tends to $+\infty$. Then
$g_{n}.B_{\alpha}(K,x)=B_{\alpha}(K,g_{n}(x))$, so that
$\mathcal{O}_{K}=\bigcup_{n\geq 0}g_{n}.B_{\alpha}(K,x)$. Then by Proposition
7.6, the subgroup $[F,F]$ admits a unique minimal invariant set
$\Lambda\subset\mathbb{R}$, which is preserved by $F$ because $[F,F]$ is a
normal subgroup. We deduce that the action of $[F,F]$ is also minimal.

To prove (ii) take $K\neq K^{\prime}$ with property $(O)$ and assume without
loss of generality that $K^{\prime}\not\subset K$. Write
$\alpha:\mathcal{O}_{K}\times\mathcal{O}_{K}\to\mathbb{R}\cup\\{-\infty\\}$
and
$\beta:\mathcal{O}_{K^{\prime}}\times\mathcal{O}_{K^{\prime}}\to\mathbb{R}\cup\\{-\infty\\}$
for the corresponding ultrametric kernels. Fix $x\in K$ and let $D$ be the
subgroup of $[F,F]_{(x,+\infty)}$ which fixes $K$ pointwise. Then for every
$g\in D$ we have $g.B_{\alpha}(K,x)=B_{\alpha}(K,x)$ and actually $g$ fixes
$B_{\alpha}(K,x)$ pointwise, so that the dynamical realization $\varphi_{K}$
fixes a non-empty open interval pointwise. On the contrary, for any $y\geq x$
with $y\in K^{\prime}\smallsetminus K$, we can consider an element $h\in D$
such that $h(y)\notin K^{\prime}$ and $h(y)>y$. Let us show that for such
choices we have $h.B_{\beta}(K^{\prime},y)\cap
B_{\beta}(K^{\prime},y)=\varnothing$. Indeed, assume there exists $L\in
h.B_{\beta}(K^{\prime},y)\cap B_{\beta}(K^{\prime},y)$; then, as
$h.B_{\beta}(K^{\prime},y)=B_{\beta}(h(K^{\prime}),h(y))$, we have

$L\cap[h(y),+\infty)=h(K^{\prime})\cap[h(y),+\infty)$

and in particular $h(y)\in L$. However, if $L\in B(K^{\prime},y)$, then
$L\cap[y,+\infty)=K^{\prime}\cap[y,+\infty)$ and thus $h(y)\notin L$, which is
an absurd.

By $f$-invariance of $K^{\prime}\smallsetminus K$, we can find arbitrarily
large points $y$, and thus elements $h\in D$, satisfying such properties. As
$\mathcal{O}_{K^{\prime}}=\bigcup_{y}B_{\beta}(K^{\prime},y)$, this implies
that $D$ acts without fixed points, so that the actions $\varphi_{K}$ and
$\varphi_{K^{\prime}}$ cannot be conjugate. ∎

After the previous proposition, in order to prove Theorem 11.35 we need to
show the existence of subsets $K\subset\mathbb{R}$ with property $(O)$ and
with the additional property that the stabilizer of $K$ in $[F,F]$ does not
have fixed points. For this, we are going to use the symbolic description of
real numbers by binary expansions.

To each infinite sequence $(a_{n})_{n\geq
1}\in\\{\mathtt{0},\mathtt{1}\\}^{\mathbb{N}}$ we associate the real number
$\operatorname{\mathsf{ev}}((a_{n})):=\sum_{n\geq 1}a_{n}2^{-n}\in[0,1]$. Note
that this association is continuous if we endow
$\\{\mathtt{0},\mathtt{1}\\}^{\mathbb{N}}$ with the product topology. If
$z,w_{1},w_{2}$ are finite binary sequences (_binary words_ for short), we
consider the _cylinder_ over $z$, defined by

$C_{z}=\\{w\in\\{\mathtt{0},\mathtt{1}\\}^{\mathbb{N}}:z\text{ is a prefix of
}w\\},$

and denote by
$\tilde{K}_{0}(w_{1},w_{2})\subseteq\\{\mathtt{0},\mathtt{1}\\}^{\mathbb{N}}$
the subset of all infinite concatenations of $w_{1}$’s and $w_{2}$’s. Clearly,
both images $\operatorname{\mathsf{ev}}(C_{z})$ and
$K_{0}(w_{1},w_{2}):=\operatorname{\mathsf{ev}}(\tilde{K}_{0}(w_{1},w_{2}))$
are closed subsets of $[0,1]$, and the former is a closed interval with dyadic
endpoints (a _dyadic interval_ for short). Note that, conversely, any closed
dyadic interval is the union of (the real numbers represented by) finitely
many cylinders.

With this in mind, if $z_{1}$ and $z_{2}$ are binary words, the _substitution
map_ $S(z_{1},z_{2}):C_{z_{1}}\to C_{z_{2}}$ defined by
$S(z_{1},z_{2})(z_{1}w)=z_{2}w$ represents an affine map
$\overline{S(z_{1},z_{2})}$ between dyadic intervals of $[0,1]$. Therefore, in
the action of $F$ on $\mathbb{R}$, every element of $F$ locally coincides
(except at breakpoints, which are finitely many dyadic rationals) with
transformations of the form $f^{n}\circ\overline{S(z_{1},z_{2})}\circ f^{m}$,
for some powers $n,m\in\mathbb{Z}$ and some finite sequences $z_{1},z_{2}$.

We say that a pair of binary words $w_{1},w_{2}$ has the _cancellation
property_ if whenever $zw=w^{\prime}$ for
$w,w^{\prime}\in\tilde{K}_{0}(w_{1},w_{2})$, it holds that $z$ is a finite
concatenation of $w_{1}$’s and $w_{2}$’s. As a concrete example of a pair of
words with the cancellation property, we may take $w_{1}=\mathtt{0}$ and
$w_{2}=\mathtt{1}$ but these are _constant_ binary words (i.e. made of a
single repeated bit). As a concrete example of non-constant binary words with
the cancellation property we can take $w_{1}=\mathtt{10001}$ and
$w_{2}=\mathtt{01110}$.

######  Lemma 11.42.

Let $w_{1}$ and $w_{2}$ be non-constant binary words satisfying the
cancellation property and write $K_{0}:=K_{0}(w_{1},w_{2})$. Then, the subset
$K:=\bigcup_{n\in\mathbb{Z}}f^{n}(K_{0})$ has property $(O)$.

###### Proof.

Since $K_{0}$ is a closed subset of $[0,1]$, the subset $K$ is a closed and
$f$-invariant subset of $\mathbb{R}$. Also, since $w_{1}$ and $w_{2}$ are non-
constant, the set $\tilde{K}_{0}(w_{1},w_{2})$ has no eventually constant
sequences, and so the subset $K_{0}$ contains no dyadic points. It follows
that $\operatorname{\mathsf{ev}}\colon\tilde{K}_{0}(w_{1},w_{2})\to K_{0}$ is
a homeomorphism onto its image, and that the set of intersections of the form
$(p/2^{n},(p+1)/2^{n})\cap K$ with $p\in\mathbb{Z}$ and $n\in\mathbb{N}$,
forms a basis of its topology. The restriction of every element of $F$ to $K$
is locally given by maps of the form $f^{n}\circ\overline{S(z_{1},z_{2})}\circ
f^{m}$, and since $K$ is $f$-invariant, in order to check property $(O)$ it is
enough to check that
$\overline{S(z_{1},z_{2})}(K_{0}\cap\operatorname{\mathsf{ev}}(C_{z_{1}}))$ is
open in $K_{0}$ for every pair of finite binary words $z_{1},z_{2}$.

For this, consider two binary finite words $z_{1},z_{2}$ and also
$w\in\tilde{K}_{0}(w_{1},w_{2})\cap C_{z_{1}}$ so that
$S(z_{1},z_{2})(w)\in\tilde{K}_{0}(w_{1},w_{2})$. We need to check that
$S(z_{1},z_{2})(\tilde{K}_{0}(w_{1},w_{2})\cap C_{z_{1}})$ contains a
neighborhood of $S(z_{1},z_{2})(w)$. Since the pair $w_{1},w_{2}$ has the
cancellation property, we can write $w=z_{1}^{\prime}w^{\prime}$ with
$z_{1}^{\prime}=z_{1}z_{1}^{\prime\prime}$ and
$w^{\prime}\in\tilde{K}_{0}(w_{1},w_{2})$. Since $S(z_{1},z_{2})(w)$ equals
$z_{2}z_{1}^{\prime\prime}w^{\prime}$ and belongs to
$\tilde{K}_{0}(w_{1},w_{2})$, again by the cancellation property we conclude
that $z_{2}z_{1}^{\prime\prime}$ is a finite concatenation of $w_{1}$’s and
$w_{2}$’s. Therefore

$S(z_{1},z_{2})\left(C_{z_{1}^{\prime}}\cap\tilde{K}_{0}(w_{1},w_{2})\right)=C_{z_{2}z_{1}^{\prime\prime}}\cap\tilde{K}_{0}(w_{1},w_{2}),$

showing that $S(z_{1},z_{2})(\tilde{K}_{0}(w_{1},w_{2}))$ contains a
neighborhood of $S(z_{1},z_{2})(w)$ inside $\tilde{K}_{0}(w_{1},w_{2})$. This
concludes the proof. ∎

In order to ensure that the stabilizer of $K$ in $[F,F]$ has no fixed points,
we need to to impose one last extra condition on $K$.

Say that a map $h:I\to J$ is a _dyadic affine map_ between intervals if $I$ is
a dyadic interval and $h$ is of the form $x\mapsto ax+b$, where
$a\in\\{2^{n}\colon n\in{\mathbb{Z}}\\}$, and $b\in\mathbb{Z}[1/2]$. Consider
now a compact subset $K_{0}\subseteq(0,1)$. We say that $K_{0}$ admits a _self
similar decomposition_ if there exists a pair of dyadic affine maps
$h_{1},h_{2}:I\to(0,1)$ such that

  * •

$h_{1}(I)\cap h_{2}(I)=\varnothing$,

  * •

$h_{1}(K_{0})\cup h_{2}(K_{0})=K_{0}$.

For example, the subset $K_{0}(w_{1},w_{2})$ admits a self similar
decomposition provided the words $w_{1}$ and $w_{2}$ are such that
$\operatorname{\mathsf{ev}}(C_{w_{1}})$ and
$\operatorname{\mathsf{ev}}(C_{w_{2}})$ are disjoint. Indeed in this case we
have that for $i\in\\{1,2\\}$, the symbolic maps $w\mapsto w_{i}w$ correspond
to dyadic affine maps $h_{i}:[0,1]\to[0,1]$ with disjoint images and such that
$K_{0}=h_{1}(K_{0})\sqcup h_{2}(K_{0})$.

######  Lemma 11.43.

Let $K_{0}\subset(0,1)$ be a closed subset admitting a self similar
decomposition, and let $K=\bigcup_{n\in\mathbb{Z}}f^{n}(K_{0})$. Then, the
action of $H=\\{g\in[F,F]\colon g(K)=K\\}$ on $K$ has no fixed points.

To show Lemma 11.43 we use its self-similarity to build elements in $H$ moving
points of the real line arbitrarily far away. But before giving the formal
proof, let us see how this ends the proof of Theorem 11.35.

###### Proof of Theorem 11.35 given Lemma 11.43.

Let $K\subsetneq\mathbb{R}$ be a subset satisfying property $(O)$ and such
that $K\cap(0,1)$ admits a self similar decomposition. As concrete example we
may take $K=\bigcup_{n\in\mathbb{Z}}f^{n}(K_{0})$ with
$K_{0}=K_{0}(w_{1},w_{2})$ for $w_{1}=\mathtt{10001}$ and
$w_{2}=\mathtt{01110}$. By Lemma 11.42 we may consider the $\mathbb{R}$-focal
action $\varphi_{K}$, and by Lemma 11.43 and Proposition 11.41 we have that
$\varphi_{K}([F,F])$ acts minimally. To finish the proof, we show that from
the existence of one subset $K$ with these properties, we can deduce the
existence of uncountably many. Let $K$ be one such subset. Clearly $K$ is
locally a Cantor set, so $\mathbb{R}\smallsetminus K$ is a countable union of
open intervals, that we call the _gaps_ of $K$. Pick $\beta\in(0,1)$. For each
gap $I$ of $K$, consider the point $p_{I}(\beta)$ where $p_{I}:(0,1)\to I$ is
the unique order-preserving affine map. We let $K^{\beta}$ be the subset
resulting from adding to $K$ all the points of the form $p_{I}(\beta)$ where
$I$ runs over gaps of $K$. Clearly $K^{\beta}$ is still closed and
$f$-invariant. Moreover, $K^{\beta}$ still admits a self similar decomposition
since the maps involved in the definition of self similar decomposition are
affine maps sending gaps of $K$ to gaps of $K$, so in particular they preserve
the proportion of the subdivision we have introduced in the gaps. We claim,
that for uncountably many $\beta\in(0,1)$, the subset $K^{\beta}$ also
satisfies (the last condition of) property $(O)$, that is $g.K^{\beta}\cap
K^{\beta}$ is open in $K^{\beta}$ for every $g\in F$.

Fix $g\in F$. The only problem that may arise is that a point of the form
$p_{I}(\beta)$ (which is an isolated point) might land inside $K$ under the
action of $g$. But if we fix a gap $I$ of $K$, the set of parameters $\beta$
such that $g(p_{I}(\beta))$ does not belong to $K$ is open and dense in
$(0,1)$ since $K$ has empty interior. In particular, since there are only
countably many gaps and $F$ is also countable, with a Baire argument we obtain
that $K^{\beta}$ satisfies property $(O)$ for a generic choice of
$\beta\in(0,1)$. ∎

We conclude this subsection with the proof of Lemma 11.43. For this we need
the following elementary interpolation lemma. Its proof follows from the
transitivity of the action of $[F,F]$ on unordered $n$-tuples of dyadic
numbers (see for instance [BieriStrebel]), and details are left to reader. To
simplify the statement, given (possibly unbounded) intervals $I,J$, we write
$I<J$ whenever $\sup I<\inf J$.

######  Lemma 11.44.

Consider intervals $I_{1}<I_{2}<\cdots<I_{k}$ and $J_{1}<J_{2}<\cdots<J_{k}$
with dyadic endpoints such that $I_{1}=J_{1}=(-\infty,p]$,
$I_{k}=J_{k}=[q,+\infty)$, and such that $h_{n}:I_{n}\to J_{n}$ are dyadic
affine maps for $n\in\\{1,\ldots,k\\}$. Assume moreover that $h_{1}$ and
$h_{k}$ are restrictions of the identity. Then, there exists $g\in[F,F]$ such
that $g\restriction_{I_{n}}=h_{n}$ for $n\in\\{1,\ldots,k\\}$.

###### Proof of Lemma 11.43.

Consider the dyadic affine maps $h_{1},h_{2}:I_{0}\to(0,1)$ given by the self
similar decomposition of $K_{0}$. Since $K_{0}$ is a closed subset of $(0,1)$,
we can assume that $I_{0}$ is a closed dyadic interval inside $(0,1)$. For
$i\in\\{1,2\\}$, write $K_{0}^{i}=h_{i}(K_{0})$ and $I_{0}^{i}=h_{i}(I_{0})$.
Note that $I_{0}^{1}\cap I_{0}^{2}=\varnothing$ and $K_{0}=K_{0}^{1}\sqcup
K_{0}^{2}$.

Now, for $n\in\mathbb{Z}$ and $i\in\\{1,2\\}$, write $I_{n}=f^{n}(I_{0})$,
$K_{n}^{i}=f^{n}(K_{0}^{i})$, and $I_{n}^{i}=f^{n}(I_{0}^{i})$. Then consider
the following locally dyadic affine maps.

  * •

$a:I_{0}^{1}\sqcup I_{0}^{2}\to I_{0}\sqcup I_{1}$ defined by

$a(x)=\left\\{\begin{array}[]{lc}h_{1}^{-1}(x)&\text{if }x\in I_{0}^{1},\\\
f\circ h_{2}^{-1}(x)&\text{if }x\in I_{0}^{2},\end{array}\right.$

  * •

$b:I_{3}\sqcup I_{4}\to I_{4}^{1}\sqcup I_{4}^{2}$ defined by

$b(x)=\left\\{\begin{array}[]{lc}f^{4}\circ h_{1}\circ f^{-3}(x)&\text{if
}x\in I_{3},\\\ f^{4}\circ h_{2}\circ f^{-4}&\text{if }x\in
I_{4},\end{array}\right.$

  * •

$c:[1,2]\to[2,3]$ defined by $c(x)=x+1$.

Then, we can apply Lemma 11.44 to construct $h\in[F,F]$ which simultaneously
extends $a,b,c$ and $\mathsf{id}\restriction_{(-\infty,0]\cup[4,+\infty)}$. By
construction $h$ preserves $K$ and has no fixed points in $[1,2]$. Thus the
subgroup $H=\\{h\in[F,F]\colon h(K)=K\\}$ has no fixed points inside $[1,2]$.
Finally, note that $f$ normalizes $H$, so that it preserves its set of fixed
points. Since $\bigcup_{n\in\mathbb{Z}}f^{n}([1,2])=\mathbb{R}$, we deduce
that $H$ has no fixed points on $\mathbb{R}$, whence on $K$. ∎

##  12\. Finitely generated locally moving groups with few exotic actions

###  12.1. Bieri–Strebel groups over the real line

In this section we study actions of Bieri–Strebel groups $G(X;A,\Lambda)$ in
the case $X=\mathbb{R}$ (see Definition 2.44). These are close relatives of
Thompson’s group $F$, yet their minimal $\mathbb{R}$-exotic actions turn out
to be much more rigid (and, in some cases, there are only finitely many such
actions). Here we denote by $\langle S\rangle_{*}$ the multiplicative group
generated by a subset $S\subset\mathbb{R}_{+}^{*}$.

The results of [BieriStrebel, §§A–B] allow to characterize under which
conditions the group $G(\mathbb{R};A,\Lambda)$ is in the class $\mathcal{F}$.
(Note that since they all contain translation, this is the same as asking
whether they belong to $\mathcal{F}_{0}$.) As for many properties of the
groups $G(X;A,\Lambda)$, this depends on the $\mathbb{Z}[\Lambda]$-submodule

$I\Lambda\cdot A:=\langle(\lambda-1)a\colon\lambda\in\Lambda,a\in A\rangle.$

Indeed, we have the following.

######  Lemma 12.1.

The Bieri–Strebel group $G(\mathbb{R};A,\Lambda)$ is in the class
$\mathcal{F}$ if and only if the following conditions are satisfied:

  1. (BS1)

$\Lambda$ is finitely generated as a group,

  2. (BS2)

$A$ is finitely generated as $\mathbb{Z}[\Lambda]$-module,

  3. (BS3)

the quotient $A/I\Lambda\cdot A$ is finite.

###### Proof.

Conditions (BS1)–(BS2) characterize when the group $G(\mathbb{R};A,\Lambda)$
is finitely generated [BieriStrebel, Theorem B7.1]. On the other hand,
Bieri–Strebel groups of the form $G((-\infty,x];A,\Lambda)$ and
$G([x,+\infty);A,\Lambda)$ are finitely generated if and only if $x\in A$ and
all three conditions (BS1)–(BS3) are satisfied (this is the statement of
[BieriStrebel, Theorem B8.2]); this immediately implies that
$G(\mathbb{R};A,\Lambda)$ is in the class $\mathcal{F}$. For the converse,
recall that two points $p,q\in A\cap(-\infty,x)$ are in the same orbit under
the group $G((-\infty,x];A,\Lambda)$ if and only if $p-q\in I\Lambda\cdot A$
[BieriStrebel, Corollary A.5.1]. If $G$ is in the class $\mathcal{F}$ there
exists a finitely generated subgroup $H\subseteq G((-\infty,x];A,\Lambda)$
which contains a subgroup of the form $G((-\infty,y];A,\Lambda)$ with $y<x$.
Since all points of $A\cap(-\infty,y)$ occur as breakpoints for elements in
$G((-\infty,y];A,\Lambda)$, they must all belong to the $H$-orbit of one of
the finitely many breakpoints of a finite generating set of $H$ (see
[BieriStrebel, §B6]). Thus condition (BS3) is also necessary. ∎

As an important special case, the reader can have in mind the following
example.

######  Example 12.2.

For $\lambda>1$, we denote by $G(\lambda)$ the Bieri–Strebel group
$G(\lambda):=G(\mathbb{R};A,\Lambda)$ corresponding to the cyclic group
$\Lambda=\langle\lambda\rangle_{*}$ and to
$A:=\mathbb{Z}[\lambda,\lambda^{-1}]$. One has $I\Lambda\cdot
A=(\lambda-1)\,\mathbb{Z}[\lambda,\lambda^{-1}]$. It is not difficult to see
that the quotient $A/I\Lambda\cdot A$ is finite if and only if $\lambda$ is
algebraic (see [BieriStrebel, Illustration A4.3]). For instance, for
$\lambda=p/q$ rational (with $p$ and $q$ coprime), one has $|A/I\Lambda\cdot
A|=p-q$. Therefore, by Lemma 12.1, the group $G(\lambda)$ is in the class
$\mathcal{F}$ exactly for algebraic $\lambda$.

Thus if $G(\mathbb{R};A,\Lambda)$ satisfies conditions (BS1)–(BS3), its
actions on the real line satisfy Theorem 10.3, so that all its exotic actions
are semi-conjugate to a minimal $\mathbb{R}$-focal action horograded by the
standard action on $\mathbb{R}$. Our goal is to give an explicit description
of all such minimal $\mathbb{R}$-focal actions, thus yielding a complete
classification of the faithful actions of such groups up to semi-conjugacy.
For this, recall that in §9.3 we gave a construction of actions of
Bieri–Strebel groups $G(X;A,\Lambda)$ parameterized by preorders on the group
$\Lambda$. In what follows, we will denote by $\varphi_{+,\leq_{\Lambda}}$
(respectively, $\varphi_{-,\leq_{\Lambda}}$) the dynamical realization of the
right (respectively, left) jump preorder associated with
$\leq_{\Lambda}\in\operatorname{\mathsf{LPO}}(\Lambda)$ (see Definition 9.13).
The main result is that these are the only exotic actions of $G$.

######  Theorem 12.3.

Let $G=G(\mathbb{R};A,\Lambda)$ be a Bieri–Strebel group satisfying conditions
(BS1)–(BS3). Then every action $\varphi\colon
G\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ without fixed points, is
semi-conjugate to an action in one of the following families.

  1. (1)

_(Non-faithful)_ An action induced from the quotient $G/[G_{c},G_{c}]$.

  2. (2)

_(Standard)_ The standard piecewise linear action of $G$ on $\mathbb{R}$.

  3. (3)

_( $\mathbb{R}$-focal)_ An action of the form $\varphi_{\pm,\leq_{\Lambda}}$,
obtained as the dynamical realization of a jump preorder (see Definition
9.13).

In particular, this provides a classification of all minimal faithful actions
of $G$ up to conjugacy.

######  Corollary 12.4.

Under the assumptions of Theorem 12.3, every minimal faithful action
$\varphi\colon G\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ is
topologically conjugate either to the standard action on $\mathbb{R}$ or to
the dynamical realization of a jump preorder.

The following special case gives Theorem 1.19 from the introduction.

######  Example 12.5.

Let us consider the special case $G=G(\lambda)$ as in Example 12.2, with
$\lambda>1$ algebraic. Since in this case the group $\Lambda$ is infinite
cyclic, it admits only two non-trivial preorders, namely the usual order
$<_{\Lambda}$ and its opposite. Thus, the jump preorder construction gives
exactly two actions $\varphi_{+}:=\varphi_{+,<_{\Lambda}}$ and
$\varphi_{-}:=\varphi_{-,<_{\Lambda}}$, which are the dynamical realizations
of the right and left jump preorders associated with
$<_{\Lambda}\in\operatorname{\mathsf{LPO}}(\Lambda)$. Note indeed, that after
Remark 9.21, the dynamical realization of the jump preorder corresponding to
$<_{\Lambda}^{op}$ is conjugate to that for $<_{\Lambda}$. Thus in this case
the group $G$ admits finitely many (more precisely, two) $\mathbb{R}$-focal
actions. Note that such actions $\varphi_{\pm}$ are both locally rigid, which
shows that groups in the class $\mathcal{F}_{0}$ may admit locally rigid
actions other than the standard action.

######  Remark 12.6.

In the setting of Theorem 12.3, let us comment on the actions coming from the
largest quotient $G/[G_{c},G_{c}]$. The structure of this quotient highly
depends on $A$ and $\Lambda$. Note that $G/G_{c}$ embeds in the product
$\operatorname{\mathsf{Aff}}(A,\Lambda)\times\operatorname{\mathsf{Aff}}(A,\Lambda)$
via the product $\mathcal{G}_{-\infty}\times\mathcal{G}_{+\infty}$ of the germ
homomorphisms, and thus $G/[G_{c},G_{c}]$ is solvable of derived length at
most 3. More precisely (see [BieriStrebel, Corollary A5.5]), writing
$\tau_{\pm}(f)\in A$ for the translation part of the germ of $f\in G$ at
$\pm\infty$, respectively, the image
$(\mathcal{G}_{-\infty}\times\mathcal{G}_{+\infty})(G)$ in
$\operatorname{\mathsf{Aff}}(A,\Lambda)\times\operatorname{\mathsf{Aff}}(A,\Lambda)$
is the subgroup

$\left\\{(f,g)\in\operatorname{\mathsf{Aff}}(A,\Lambda)\colon\tau_{+}(g)-\tau_{-}(f)\in
I\Lambda\cdot A\right\\}.$

The investigation of the abelianization $G_{c}/[G_{c},G_{c}]$ is much less
understood; partial results are discussed in [BieriStrebel, §C12]. If
$G_{c}/[G_{c},G_{c}]$ is trivial, then one must have $I\Lambda\cdot A=A$
[BieriStrebel, Theorem C12.14], but the converse is not true in general.

In the case $G=G(\lambda)$ for $\lambda>1$ algebraic (Example 12.2) one can
check that the image of $G/G_{c}$ under the map induced by
$\mathcal{G}_{-\infty}\times\mathcal{G}_{+\infty}$ has finite index in
$\operatorname{\mathsf{Aff}}(A,\Lambda)\times\operatorname{\mathsf{Aff}}(A,\Lambda)$
(see Example 12.2). Moreover if $\lambda=p/q$ is rational (with $p$ and $q$
coprime), the abelianization $G_{c}/[G_{c},G_{c}]$ is free of rank $p-q-1$
(see [BieriStrebel, Corollary C12.12]). For other algebraic values of
$\lambda$, up to our knowledge, only the case $\lambda=\frac{\sqrt{5}+1}{2}$
has been studied in detail: after work of Burillo, Nucinkis, and Reeves
[burillo2020irrationalslope], we know that
$G_{c}/[G_{c},G_{c}]\cong\mathbb{Z}_{2}$ (note that in this case
$I\Lambda\cdot A=A$). Note also that for any algebraic $\lambda>1$, the
abelianization $G/[G,G]$ is free of rank $|A/I\Lambda\cdot A|+1$ (see
[BieriStrebel, Corollary C12.2]).

The situation is particularly simple in the case $\lambda=2$, where we have

$G/[G_{c},G_{c}]\cong\operatorname{\mathsf{Aff}}(\mathbb{Z}[1/2],\langle
2\rangle_{*})\times\operatorname{\mathsf{Aff}}(\mathbb{Z}[1/2],\langle
2\rangle_{*})\cong\operatorname{\mathsf{BS}}(1,2)\times\operatorname{\mathsf{BS}}(1,2),$

where $\operatorname{\mathsf{BS}}(1,2)=\langle a,b\mid aba^{-1}=b^{2}\rangle$
is the solvable Baumslag–Solitar group (see Proposition 12.17 below).

Before getting to the proof of Theorem 12.3, we need further preliminary
discussions. Recall that the group $G$ contains the affine subgroup
$\operatorname{\mathsf{Aff}}(A,\Lambda)\cong A\rtimes\Lambda$ of
transformations of the form $x\mapsto\lambda x+a$, with $\lambda\in\Lambda$
and $a\in A$. Given $a\in A$ and $\lambda\in\Lambda$, we will denote by
$g(a,\lambda)$ the affine transformation $x\mapsto\lambda x+(1-\lambda)a$,
which is the unique element of $\operatorname{\mathsf{Aff}}(A,\Lambda)$ which
fixes $a$ and has slope $\lambda$. We will also consider the elements

(12.1)
$g_{+}(a,\lambda):x\mapsto\left\\{\begin{array}[]{lr}x&x\in(-\infty,a],\\\\[5.0pt]
g(a,\lambda)(x)&x\in[a,+\infty),\end{array}\right.$

and $g_{-}(a,\lambda):=g(a,\lambda)\,g_{+}(a,\lambda)^{-1}$. Note that
$g_{+}(a,\lambda)\in G_{(a,+\infty)}$ and $g_{-}(a,\lambda)\in
G_{(-\infty,a)}$. For $a\in A$ we also denote by $t_{a}$ the translation
$x\mapsto x+a$.

For every $a\in A$ and $\lambda\in\Lambda$, if $h\in G(\mathbb{R};A,\Lambda)$
is an element with no breakpoint on $(a,+\infty)$ (respectively on
$(-\infty,a)$), we have

$hg_{+}(a,\lambda)h^{-1}=g_{+}(h(a),\lambda)\quad\text{(respectively,
}hg_{-}(a,\lambda)h^{-1}=g_{-}(h(a),\lambda)).$

In particular we have the following relations for such elements (see
[BieriStrebel, §B7]):

(12.2) $h\,g_{\pm}(a,\lambda)\,h^{-1}=g_{\pm}(h(a),\lambda)\quad\text{for
every }h\in\operatorname{\mathsf{Aff}}(A,\Lambda),$

as well as

$g_{+}(a,\lambda)\,g_{+}(b,\mu)\,g_{+}(a,\lambda)^{-1}=g_{+}(g(a,\lambda)(b),\mu)\quad\text{for
every }a>b,$ (12.3)
$g_{-}(a,\lambda)\,g_{-}(b,\mu)\,g_{-}(a,\lambda)^{-1}=g_{-}(g(a,\lambda)(b),\mu)\quad\text{for
every }a<b.$

We also remark that the subset

$\left\\{t_{a}\right\\}_{a\in
A}\cup\left\\{g(0,\lambda),g_{+}(0,\lambda)\right\\}_{\lambda\in\Lambda}$

is generating for $G(\mathbb{R};A,\Lambda)$ (see [BieriStrebel, Theorem
B7.1]).

For what follows, the reader can keep in mind the following example.

######  Example 12.7.

For $\lambda>1$, the Bieri–Strebel group $G(\lambda)$ is generated by the
finite subset $\\{g(0,\lambda),g_{+}(0,\lambda),t_{1}\\}$.

######  Remark 12.8.

The group $\left\langle g(0,\lambda),g_{+}(0,\lambda),t_{1}\right\rangle$
appears in [BLT] (denoted $G_{\lambda}$), where it was shown that, for certain
algebraic numbers $\lambda>1$ (called Galois hyperbolic ibid.) it admits no
faithful $C^{1}$ action on the closed interval. The fact that this group
coincides with the Bieri–Strebel group $G(\lambda)$ was unnoticed in [BLT].

We also need some preliminary results, stated in the following more flexible
setting.

######  Assumption 12.9.

Fix a non-trivial multiplicative subgroup $\Delta\subseteq\mathbb{R}_{*}$, and
a $\Delta$-submodule $A\subseteq\mathbb{R}$, and let
$H=G(\mathbb{R};A,\Delta)$ be the corresponding Bieri–Strebel group. Moreover
we let $G\subseteq\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ be a subgroup
in the class $\mathcal{F}$ which contains $H$ as a subgroup. Finally we assume
that $\mathcal{L}=\\{L_{x}\\}_{x\in\mathbb{R}}$ is a family of subgroups of
$G$ with the following properties.

  1. (C1)

For each $x\in\mathbb{R}$ we have $\bigcup_{y<x}G_{(-\infty,y)}\subseteq
L_{x}\subseteq G_{(-\infty,x)}$.

  2. (C2)

For every $x\in\mathbb{R}$ and every $g\in G$ we have $gL_{x}g^{-1}=L_{g(x)}$.

  3. (C3)

For every $x\in A$ and every $\delta\in\Delta$ we have $g_{-}(x,\delta)\in
L_{x}$.

We write $L_{+}=\bigcup_{x\in\mathbb{R}}L_{x}$ which, by (C2), defines a
normal subgroup of $G$. Finally we assume that $\varphi\colon
G\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ is a faithful minimal
action of $G$ which is $\mathbb{R}$-focal and increasingly horograded by its
standard action on $\mathbb{R}$.

######  Remark 12.10.

For the proof of Theorem 12.3, the reader can have in mind the case where
$G=G(\mathbb{R};A,\Lambda)$ is itself a Bieri–Strebel group, $H$ is a subgroup
corresponding to some $\Delta\subseteq\Lambda$, and the $L_{x}$ are subgroups
of $G_{(-\infty,x)}$ consisting of elements whose left-derivative at $x$
belongs to some intermediate group
$\Delta\subseteq\Lambda_{1}\subseteq\Lambda$. However a different choice of
$G$ will be used later in §12.2.

We will write $X=\mathbb{R}$ for the real line on which the standard action of
$G$ is defined. We resume notation of §10.2.3. As introduced in Definition
10.7, for every $x\in X$ and $\xi\in\mathbb{R}$ with
$\xi\notin\operatorname{\mathsf{Fix}}^{\varphi}\left(G_{(-\infty,x)}\right)$,
we write $\operatorname{I^{\varphi}}(x,\xi)$ for the connected component of
$\operatorname{\mathsf{Supp}^{\varphi}}\left(G_{(-\infty,x)}\right)$
containing $\xi$. Recall that intervals of this form define a CF-cover.
Moreover we will consider the planar directed tree
$(\mathbb{T},\triangleleft,\prec)$ constructed in §10.2.3, whose dynamical
realization is $\varphi$. Recall that vertices of $\mathbb{T}$ are intervals
of the form
$\operatorname{{I}^{\varphi}_{\mathsf{out}}}(x,\xi)=\operatorname{\mathsf{Int}}\left(\bigcap_{y>x}\operatorname{I^{\varphi}}(y,\xi)\right)$
and that the map $\pi\colon\mathbb{T}\to\mathbb{R}$ given by
$\pi\left(\operatorname{{I}^{\varphi}_{\mathsf{out}}}(x,\xi)\right)=x$ is an
equivariant horograding. Recall also that we write
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(y,\xi)=\bigcup_{x<y}\operatorname{I^{\varphi}}(x,\xi)$.

######  Lemma 12.11.

Under Assumption 12.9, the group
$\varphi\left(\operatorname{\mathsf{Aff}}(A,\Delta)\right)$ has a unique fixed
point $\eta\in\mathbb{R}$. This point satisfies
$\eta\in\operatorname{\mathsf{Supp}^{\varphi}}\left(G_{(-\infty,x)}\right)$
for every $x\in X$.

Moreover, the map

$\begin{array}[]{cccc}q_{+}:&\mathbb{R}&\to&(\eta,+\infty)\\\
&x&\mapsto&\sup\operatorname{I^{\varphi}}(x,\eta)\end{array}$

is monotone increasing and
$\operatorname{\mathsf{Aff}}(A,\Delta)$-equivariant. In particular the
standard affine action of $\operatorname{\mathsf{Aff}}(A,\Delta)$ on
$X=\mathbb{R}$ is positively semi-conjugate to its action $\varphi$ on
$(\eta,+\infty)$. Similarly the map

$\begin{array}[]{cccc}q_{-}:&\mathbb{R}&\to&(-\infty,\eta)\\\
&x&\mapsto&\inf\operatorname{I^{\varphi}}(x,\eta)\end{array}$

defines a negative semi-conjugacy.

###### Proof.

Every translation $t_{a}$, with $a\in A$, acts without fixed points, therefore
by Proposition 8.32, its $\varphi$-image is a homothety. As the subgroup of
translations is abelian, its $\varphi$-image has a unique fixed point
$\eta\in\mathbb{R}$. Moreover, as such subgroup is normal in
$\operatorname{\mathsf{Aff}}(A,\Delta)$, this is a fixed point for
$\varphi\left(\operatorname{\mathsf{Aff}}(A,\Delta)\right)$.

Assume first that there exists $x\in X$ such that
$\eta\in\operatorname{\mathsf{Fix}}^{\varphi}\left(G_{(-\infty,x)}\right)$.
Then we would have
$\eta=g.\eta\in\operatorname{\mathsf{Fix}}^{\varphi}\left(G_{(-\infty,x)}\right)$
for every $g\in\operatorname{\mathsf{Aff}}(A,\Delta)$, so that by minimality
of the standard action of $\operatorname{\mathsf{Aff}}(A,\Delta)$ on $X$ we
get $\eta\in\operatorname{\mathsf{Fix}}^{\varphi}\left(G_{+}\right)$. Since
$G_{+}$ is a normal subgroup of $G$, and $\varphi$ is minimal, this implies
that $\varphi\left(G_{+}\right)$ acts trivially, contradicting faithfulness of
$\varphi$ in Assumption 12.9. Thus
$\eta\in\operatorname{\mathsf{Supp}^{\varphi}}\left(G_{(-\infty,x)}\right)$,
and the interval $\operatorname{I^{\varphi}}(x,\eta)$ is well-defined for
every $x\in X$.

The second statement is now a direct consequence of the properties of these
two families of intervals: the family of
$\\{\operatorname{I^{\varphi}}(x,\eta)\\}_{x\in X}$ is increasing with respect
to $x\in X$ and moreover one has the equivariance relation
$g.\operatorname{I^{\varphi}}(x,\eta)=\operatorname{I^{\varphi}}(g(x),g.\eta)$
for every $x\in X$ and $g\in G$ (see §10.2.2). ∎

In what follows we will always denote by $\eta$ the unique fixed point of
$\varphi\left(\operatorname{\mathsf{Aff}}(A,\Lambda)\right)$ provided by Lemma
12.11. For $x\in X$ and $\xi\in\operatorname{\mathsf{Supp}^{\varphi}}(L_{x})$
we will denote by $\operatorname{I^{\varphi}}(\mathcal{L},x,\xi)$ the
connected component of $\operatorname{\mathsf{Supp}^{\varphi}}(L_{x})$
containing $\xi$. Note that condition (C1) implies that
$\operatorname{I^{\varphi}}(\mathcal{L},x,\xi)$ is increasing with respect to
$x\in\mathbb{R}$, and moreover

$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\xi)\subset\operatorname{I^{\varphi}}(\mathcal{L},x,\xi)\subset\operatorname{I^{\varphi}}(x,\xi).$

The key point is to establish the following strict inclusion when $x\in A$.

######  Lemma 12.12.

Under Assumption 12.9, assume that there exists $g\in H$ such that
$g.\eta\neq\eta$. Then for every $x\in A$ we have a strict inclusion
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)\subsetneq\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$.

###### Proof.

Assume by contradiction that
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)=\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$
for some $x\in A$. Note that then this is automatically true for every $x\in
A$, since the group $\operatorname{\mathsf{Aff}}(A,\Delta)$ acts transitively
on $A$ (it contains all translations by elements in $A$) and fixes $\eta$, so
that for $h\in\operatorname{\mathsf{Aff}}(A,\Delta)$ we have
$h.\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)=\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(h(x),\eta)$
and

(12.4)
$h.\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)=\operatorname{I^{\varphi}}(\mathcal{L},h(x),\eta)$

(after condition (C2)).

Fix $a\in A$ and choose $\delta\in\Delta$ with $\delta>1$ such that the
element $g_{-}(a,\delta).\eta\neq\eta$. Such a $\delta$ exists because the
elements $g_{-}(a,\delta)$ together with
$\operatorname{\mathsf{Aff}}(A,\Delta)$ generate $H$, and we assume that
$\varphi(H)$ does not fix $\eta$. Note also that once such an element
$g_{-}(a,\delta)$ is found, it follows that $g_{-}(x,\delta).\eta\neq\eta$ for
every $x\in A$, since these elements are all conjugate to each other by
elements of $\operatorname{\mathsf{Aff}}(A,\Lambda)$.

Note that by condition (C3) the image $\varphi(g_{-}(x,\delta))$ must preserve
$\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$ for every $x\in A$. We claim
that it acts on it as an expanding homothety. To this end, we look at the
action of $H$ on the planar directed tree $(\mathbb{T},\triangleleft,\prec)$
constructed in §10.2.3. Consider the subset $\mathbb{T}_{0}$ of $\mathbb{T}$
given by

$\mathbb{T}_{0}=\left\\{\operatorname{{I}^{\varphi}_{\mathsf{out}}}(y,\xi)\colon
y<x,\xi\in\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)\right\\}.$

Then the equality
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)=\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$
implies that $\mathbb{T}_{0}$ is a directed planar subtree of $\mathbb{T}$
invariant under the subgroup $L_{x}$ (note that $\mathbb{T}_{0}$ is equal to
the direction below the vertex
$v=\operatorname{{I}^{\varphi}_{\mathsf{out}}}(x,\eta)$ corresponding to
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)$). Moreover the
restriction of the horograding $\pi\colon\mathbb{T}\to\mathbb{R}$ to
$\mathbb{T}_{0}$ takes values in $(-\infty,x)$ and is $L_{x}$-equivariant.
Finally since the intervals
$\operatorname{{I}^{\varphi}_{\mathsf{out}}}(y,\eta)$ for $y<x$ are relatively
compact inside $\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)$ and every
$I\in\mathbb{T}_{0}$ is contained in one such interval, the restriction of the
planar order on $\mathbb{T}_{0}$ is proper. Thus we can apply Proposition 8.26
to the action of $L_{x}$ on $\mathbb{T}_{0}$, and since the element
$g_{-}(x,\delta)$ satisfies $g_{-}(x,\delta)(y)>y$ for every
$y\in(-\infty,x)$, it follows that it has no fixed points in $\mathbb{T}_{0}$
and thus acts on $(\partial\mathbb{T}_{0},\prec)$ as an expanding homothety,
and we conclude as in the proof of Proposition 8.32 that its $\varphi$-image
is an expanding homothety on $\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$,
as desired.

Now for $x\in A$ let us denote by
$\xi_{x}\in\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$ the unique fixed
point of
$\varphi(g_{-}(x,\delta))\restriction_{\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)}$.
Note that $\xi_{x}\neq\eta$, by the choice of $\delta$. Without loss of
generality, we assume that $\xi_{b}>\eta$ for some $b\in A$. Then we have the
following.

######  Claim.

For every $x\in A$ we have $\xi_{x}>\eta$ and the map $x\mapsto\xi_{x}$ is
monotone increasing.

###### Proof of claim.

The relations (12.2) and (12.4) give that the map $x\mapsto\xi_{x}$ is
$\operatorname{\mathsf{Aff}}(A,\Delta)$-equivariant. The conclusion follows
using that $\operatorname{\mathsf{Aff}}(A,\Delta)$ acts transitively on $A$
and that, after Lemma 12.11, we know that the action of
$\operatorname{\mathsf{Aff}}(A,\Delta)$ on $(\eta,+\infty)$ is positively
semi-conjugate to the standard affine action. ∎

Now fix $x\in X$. By the assumption that
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)=\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$
and by (C1) we can find $y\in A$ with $y<x$ such that
$\operatorname{I^{\varphi}}(\mathcal{L},y,\eta)$ contains $\xi_{x}$. After the
claim, we have $\eta<\xi_{y}<\xi_{x}$. Since $\xi_{x}$ is a repelling fixed
point for $\varphi(g_{-}(x,\delta))$, we have the inclusion
$g_{-}(x,\delta).\operatorname{I^{\varphi}}(\mathcal{L},y,\eta)\supset\operatorname{I^{\varphi}}(\mathcal{L},y,\eta)$,
and since the latter contains $\eta$ we have
$g_{-}(x,\delta).\operatorname{I^{\varphi}}(\mathcal{L},y,\eta)=\operatorname{I^{\varphi}}\left(\mathcal{L},g_{-}(x,\delta)(y),\eta\right)$.
Then (12.3) implies that $g_{-}(x,\delta).\xi_{y}=\xi_{g_{-}(x,\delta)(y)}$.
However, $g_{-}(x,\delta).\xi_{y}<\xi_{y}$ since $\xi_{y}$ lies to the left of
$\xi_{x}$. On the other hand $\xi_{g_{-}(x,\lambda)(y)}>\xi_{y}$ since
$g_{-}(x,\lambda)(y)>y$, and we know that the map $y\mapsto\xi_{y}$ is
increasing after the claim. This is a contradiction, giving end to the proof
of the lemma. ∎

######  Proposition 12.13.

Under the same assumptions as in Lemma 12.12, for every $x\in A$ the action of
$\varphi(L_{x})$ on $\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$ is semi-
conjugate to a non-faithful action induced from an action of the group of left
germs $\operatorname{\mathsf{Germ}}\left(L_{x},x\right)$.

###### Proof.

Lemma 12.12 implies that the normal subgroup
$\left(G_{(-\infty,x)}\right)_{+}=\bigcup_{y<x}G_{(-\infty,y)}$ of $L_{x}$ has
fixed points in $\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$ (namely the
endpoints of $\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)$). Thus the
action of $\varphi(L_{x})$ on $\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$
is semi-conjugate to an action induced from the quotient
$L_{x}/\left(G_{(-\infty,x)}\right)_{+}\cong\operatorname{\mathsf{Germ}}\left(L_{x},x\right)$.
∎

The next result is the only place where a particular choice of the family
$\mathcal{L}=\\{L_{x}\\}$ is needed.

######  Corollary 12.14.

Let $G=G(\mathbb{R};A,\Lambda)$ be a Bieri–Strebel group satisfying conditions
(BS1)–(BS3). Let $\varphi:G\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$
be a minimal faithful $\mathbb{R}$-focal action, increasingly horograded by
the standard action on $\mathbb{R}$. Assume there exists an element
$g=g_{-}(x,\lambda)\in G$ such that $g.\eta\neq\eta$. Then
$g.\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)\cap\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)=\varnothing$.

###### Proof.

Take $g=g_{-}(x,\lambda)$ not fixing $\eta$ and consider the Bieri–Strebel
group $H=G(\mathbb{R};A,\langle\lambda\rangle_{\ast})$, which is a subgroup of
$G$. We consider the family $\mathcal{L}=\\{L_{x}:x\in X\\}$ defined by

$L_{x}=\left\\{h\in G_{(-\infty,x)}\colon
D^{-}h(x)\in\langle\lambda\rangle_{\ast}\right\\}.$

It is straightforward to verify that Assumption 12.9 is fulfilled by such
choices. Therefore, by Proposition 12.13, the action of $L_{x}$ on
$\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$ is semi-conjugate to an
action that factors through the germ homomorphism
$\mathcal{G}_{x}:L_{x}\to\operatorname{\mathsf{Germ}}\left(L_{x},x\right)$.
Since in this case $\operatorname{\mathsf{Germ}}\left(L_{x},x\right)$ is
generated by $\mathcal{G}_{x}(g)$, we conclude that
$\operatorname{\mathsf{Fix}}^{\varphi}(g)\cap\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)=\varnothing$.

On the other hand, by Lemma 12.12, we get that
$\operatorname{I^{\varphi}}(\mathcal{L},x,\eta)$ strictly contains
$\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)$. Then, since
$\\{\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\xi):x,\xi\in\mathbb{R}\\}$
is a CF-cover preserved by the action, we must have
$g.\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)\cap\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(x,\eta)=\varnothing$
as desired. ∎

The next two lemmas analyse properties of the jump preorders. The first one
gives decompositions for elements in $G_{+}=\bigcup_{x<\infty}G_{(-\infty,x)}$
which are well suited for our purposes. The second one allows to identify
dynamical realizations of jump preorders.

######  Lemma 12.15.

Let $G=G(\mathbb{R};A,\Lambda)$ be a Bieri–Strebel group and let
$\leq_{\Lambda}$ be a preorder on $\Lambda$. Let $\preceq$ be the
corresponding right jump preorder and take $g\in G_{+}$ with
$\mathsf{id}\precneq g$. Then, there exist elements $h,k,g_{-}(y,\lambda)\in
G_{+}$ such that

  1. (1)

$g=kg_{-}(y,\lambda)h$,

  2. (2)

$h\in[1]_{\preceq}$,

  3. (3)

$1\lneq_{\Lambda}\lambda$, and

  4. (4)

$k\in G_{(-\infty,z)}$ for some $z<y$.

The analogous result holds for the left jump preorder.

###### Proof.

The condition $g\in G_{+}$ gives the equality $j^{+}(g,y)=D^{-}g(y)$ for every
$y\in X$. Write $\Lambda_{0}=[1]_{\leq_{\Lambda}}$. Since we are assuming
$\mathsf{id}\precneq g$, we can consider $x=x_{g,\Lambda_{0}}$ as in (9.9).
Thus the restriction of $g$ to $(x,+\infty)$ coincides with some element $h\in
G(\mathbb{R};A,\Lambda_{0})_{+}$. Consider the product $f=gh^{-1}$. The
rightmost point of $\operatorname{\mathsf{Supp}}(f)$ is $y=g(x)=h(x)$, and by
the chain rule we have

$D^{-}f(y)=D^{-}g(x)\,D^{-}h^{-1}(y)=D^{-}g(x)\,D^{-}h(x)^{-1}.$

Since $D^{-}h(x)\in\Lambda_{0}$ and
$D^{-}g(x)\in\Lambda\smallsetminus\Lambda_{0}$, we get
$1\lneq_{\Lambda}D^{-}f(y)$. Write $\lambda=D^{-}f(y)$. As before, we have
that the rightmost point of the support of $fg_{-}(y,\lambda)^{-1}$ coincides
with $g_{-}(y,\lambda)(z)$ where $z$ is the second largest breakpoint of $f$
(the one before $y$). Then write $k=fg_{-}(y,\lambda)^{-1}$. It is direct to
check that the decomposition $g=kg_{-}(y,\lambda)h$ satisfies conditions (1–4)
in the statement. ∎

######  Lemma 12.16.

Consider a Bieri–Strebel group $G=G(\mathbb{R};A,\Lambda)$ and a preorder
$\preceq\in\operatorname{\mathsf{LPO}}(G)$ containing
$\operatorname{\mathsf{Aff}}(A,\Lambda)$ in its residue subgroup. Assume
further that $\preceq^{\prime}\in\operatorname{\mathsf{LPO}}(G)$ is a right
(respectively, left) jump preorder coinciding with $\preceq$ over $G_{+}$
(respectively, $G_{-}$). Then $\preceq$ and $\preceq^{\prime}$ are the same
preorder.

###### Proof.

Assume that $\preceq^{\prime}$ is the right jump preorder associated with the
preorder $\leq_{\Lambda}\in\operatorname{\mathsf{LPO}}(\Lambda)$, the case
where $\preceq^{\prime}$ is a left jump preorder is analogous. Denote by
$\Lambda_{0}$ the residue of $\leq_{\Lambda}$ and recall from §9.3, that in
this case the residue of $\preceq^{\prime}$ is the subgroup

$H:=\\{g\in G:j^{+}(g,x)\in\Lambda_{0}\ \forall x\in\mathbb{R}\\}$

(see (9.8)). Since elements of $\operatorname{\mathsf{Aff}}(A,\Lambda)$ have
constant derivative it holds that $j^{+}(g,x)=1$ for every
$g\in\operatorname{\mathsf{Aff}}(A,\Lambda)$ and $x\in\mathbb{R}$. In
particular we have $\operatorname{\mathsf{Aff}}(A,\Lambda)\subseteq H$.

Note that $G$ decomposes as
$G=G_{+}\rtimes\operatorname{\mathsf{Aff}}(A,\Lambda)$. Then, for every $g\in
G$ we can write $g=g_{+}a_{g}$ with $g_{+}\in G_{+}$ and
$a_{g}\in\operatorname{\mathsf{Aff}}(A,\Lambda)$. Denote by $P$ and
$P^{\prime}$ the positive cones of $\preceq$ and $\preceq^{\prime}$
respectively. Since $\operatorname{\mathsf{Aff}}(A,\Lambda)$ is contained in
the residue of both $\preceq$ and $\preceq^{\prime}$, it holds that $g\in P$
if and only if $g_{+}\in G_{+}\cap P$ and also that $g\in P^{\prime}$ if and
only if $g_{+}\in G_{+}\cap P^{\prime}$. Finally, since by hypothesis it holds
$G_{+}\cap P=G_{+}\cap P^{\prime}$, the lemma follows. ∎

###### Proof of Theorem 12.3.

The assumptions on $G=G(\mathbb{R};A,\Lambda)$ ensure that $G$ is in the class
$\mathcal{F}$ (see Lemma 12.1). After Theorem 10.3, we only need to show that
a minimal faithful $\mathbb{R}$-focal action $\varphi\colon
G\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$, increasingly
(respectively, decreasingly) horograded by its standard action on
$\mathbb{R}$, is conjugate to an action of the form
$\varphi_{+,\leq_{\Lambda}}$ (respectively, $\varphi_{-,\leq_{\Lambda}}$) for
some preorder $\leq_{\Lambda}$. We will only discuss the case of increasing
horograding by the standard action, the decreasing case being totally
analogous.

We write $\eta\in\mathbb{R}$ for the unique fixed point of
$\varphi\left(\operatorname{\mathsf{Aff}}(A,\Lambda)\right)$ given by Lemma
12.11 (applied to the case $\Delta=\Lambda$). In what follows, let $\preceq$
be the preorder on $G$ induced by $\eta$, namely by declaring $g\precneq h$ if
and only if $g.\eta<h.\eta$. We will show that this preorder coincides with a
right jump preorder associated with some
$\leq_{\Lambda}\in\operatorname{\mathsf{LPO}}(\Lambda)$. Let us first identify
such preorder. For $x\in A$, the set of elements
$T_{x}:=\left\\{g_{-}(x,\lambda)\colon\lambda\in\Lambda\right\\}$ is a
subgroup of $G$ isomorphic to $\Lambda$, which is a section inside
$G_{(-\infty,x)}$ of the group of germs
$\operatorname{\mathsf{Germ}}\left(G_{(-\infty,x)},x\right)$. We put on
$\Lambda$ the preorder $\leq_{\Lambda}$ given by restricting $\preceq$ to this
subgroup, namely by setting $\lambda\lneq_{\Lambda}\mu$ if
$g_{-}(x,\lambda).\eta<g_{-}(x,\mu).\eta$. Note that this preorder does not
depend on the choice of $x\in A$, as for $x,y\in A$ the groups $T_{x}$ and
$T_{y}$ are conjugate by an element of
$\operatorname{\mathsf{Aff}}(A,\Lambda)$ (see (12.2)), which fixes $\eta$.
Denote $\preceq^{\prime}$ the right jump preorder in $G$ associated with
$\leq_{\Lambda}$. We proceed to show that $\preceq^{\prime}$ and $\preceq$
coincide on $G_{+}$ which, by Lemma 12.16 will conclude the proof.

Denote by $\Lambda_{0}$ the residue of the preorder $\leq_{\Lambda}$ and
recall that in this case the residue of $\preceq^{\prime}$ equals $H=\\{g\in
G:j^{+}(g,x)\in\Lambda_{0}\ \forall x\in\mathbb{R}\\}$ (see (9.8)). As
observed in the proof of Lemma 12.15, for $g\in G_{+}$ we have
$j^{+}(g,x)=D^{-}g(x)$ for every $x\in X$, so that we have the equality $H\cap
G_{+}=G(\mathbb{R};A,\Lambda_{0})_{+}$. Note that $H\cap G_{+}$ fixes $\eta$,
since it is generated by $\left\\{g_{-}(x,\lambda)\colon x\in
A,\lambda\in\Lambda_{0}\right\\}$ (this can be easily checked from
[BieriStrebel, §8.1]). Thus, we have

(12.5) $G_{+}\cap[1]_{\preceq^{\prime}}\subseteq G_{+}\cap[1]_{\preceq}.$

Assume now that $g\in G_{+}$ satisfies $\mathsf{id}{}^{\prime}g$. We proceed
to show that in this case $\eta<g.\eta$ which implies that
$\mathsf{id}\precneq g$. For this, consider the decomposition
$g=kg_{-}(y,\lambda)h$ given by Lemma 12.15. Then, by Lemma 12.15.(2) and
(12.5), we get that $h.\eta=\eta$ and therefore
$g.\eta=kg_{-}(y,\lambda).\eta$. On the other hand, Lemma 12.15.(3) together
with the definition of $\leq_{\Lambda}$ imply that
$g_{-}(y,\lambda).\eta>\eta$. Then, by Corollary 12.14, we get that
$g_{-}(y,\lambda).\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(y,\eta)\cap\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(y,\eta)=\varnothing$.
Finally, by Lemma 12.15.(4) we have that
$k.\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(y,\eta)=\operatorname{{I}^{\varphi}_{\mathsf{inn}}}(y,\eta)$
which, in light of what we have already done, shows that
$g.\eta=kg_{-}(y,\lambda).\eta>\eta$ as desired. Analogously one shows that if
$g{}^{\prime}\mathsf{id}$ then $g\precneq\mathsf{id}$. This shows that the
preorders $\preceq^{\prime}$ and $\preceq$ coincide over $G_{+}$ and concludes
the proof. ∎

###  12.2. A finitely generated group in the class $\mathcal{F}_{0}$ with no
$\mathbb{R}$-focal actions

Here we construct an example of a finitely generated locally moving group in
the class $\mathcal{F}_{0}$, which admits no faithful $\mathbb{R}$-focal
actions. In particular every minimal faithful action on $\mathbb{R}$ is
conjugate to the standard action.

The starting point of the construction is the Bieri–Strebel group
$G(2)=G(\mathbb{R};\mathbb{Z}[1/2],\langle 2\rangle_{*})$ of all finitary
dyadic PL homeomorphisms of $\mathbb{R}$, which we already proved to admit
only two $\mathbb{R}$-focal actions (Theorem 12.3 and Example 12.5). Moreover,
as discussed in Remark 12.6, we have that its largest proper quotient is
isomorphic to the direct product of two solvable Baumslag–Solitar groups
$\operatorname{\mathsf{BS}}(1,2)$. Therefore also the non-faithful case in
Theorem 12.3 is very restrictive. Indeed, we have the following result
(compare with [BLT, Theorem 6.12]).

######  Proposition 12.17.

Let $\Gamma_{1}$ and $\Gamma_{2}$ be two groups isomorphic to the
Baumslag–Solitar group $\operatorname{\mathsf{BS}}(1,2)$, and consider their
direct product $\Gamma:=\Gamma_{1}\times\Gamma_{2}$. Then every action
$\varphi\colon\Gamma\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ without
fixed points is semi-conjugate either to an action by translation of the
abelianization $\Gamma^{ab}=\mathbb{Z}^{2}$, or to an action obtained by
composing the projection to one of the factors $\Gamma_{i}$, $i\in\\{1,2\\}$,
with the action of $\Gamma_{i}$ on $\mathbb{R}$ by affine dyadic maps.

###### Proof.

The following proof is based on the classification of actions of the
Bausmlag–Solitar group $\operatorname{\mathsf{BS}}(1,2)$ on the real line,
without fixed points: up to semi-conjugacy, it is either an action of the
abelianization $\mathbb{Z}$, or the standard affine action (see e.g. [RivasBS,
BMNR]). Consider an action
$\varphi\colon\Gamma\to\operatorname{\mathsf{Homeo}}_{0}(\mathbb{R})$ without
fixed points. After Proposition 5.4, without loss of generality, we can assume
that the $\varphi$-image $[\Gamma_{1},\Gamma_{1}]$ has fixed points. Thus the
action $\varphi$ is semi-conjugate to an action of the quotient
$\Gamma_{1}^{ab}\times\Gamma_{2}$. Assume that the factor
$\Gamma_{1}^{ab}\cong\mathbb{Z}$ has no fixed point and that the action
$\varphi$ is not semi-conjugate to any action of the abelianization
$\Gamma^{ab}=\Gamma_{1}^{ab}\times\Gamma_{2}^{ab}$. This means that the action
of $\Gamma_{2}$ is semi-conjugate to the standard affine action. This is
however not possible, because elements in $\Gamma_{2}$ corresponding to
homotheties under the semi-conjugacy have a compact set of fixed points, which
must be fixed by $\Gamma_{1}$, an absurd. ∎

As a consequence of Corollary 9.10, if we want to avoid $\mathbb{R}$-focal
actions, we must leave the setting of finitary PL transformations. We will
consider groups whose elements are PL with a countably many breakpoints that
accumulate at some finite subset of “higher order” singularities (with some
control on these).

Given an open interval $X\subset\mathbb{R}$ we say that a homeomorphism
$f\in\operatorname{\mathsf{Homeo}}_{0}(X)$ is _locally
$\operatorname{\mathsf{PL}}$_ if there is a finite subset $\Sigma\subset X$
such that $f$ is (finitary) $\operatorname{\mathsf{PL}}$ in
$X\smallsetminus\Sigma$. For such an $f$, we denote by
$\operatorname{\mathsf{BP}}^{2}(f)\subset X$ the minimal subset such that $f$
is $\operatorname{\mathsf{PL}}$ in