# One-dimensional Actions of Higman’s Group

Cristóbal Rivas Partially supported by FONDECYT 1181548. Michele Triestino
Partially supported MathAmSud RGSD (Rigidity and Geometric Structures in
Dynamics) 19-MATH-04, the project ANR Gromeov (ANR-19-CE40-0007), and the
project “Jeunes Géométres” of F. Labourie (financed by the Louis D.
Foundation).

###### Abstract

We build a faithful action of Higman’s group on the line by homeomorphisms,
answering a question of Yves de Cornulier. As a by-product we obtain many
quasimorphisms from Higman’s group into the reals. We also show that every
action by $C^{1}$-diffeomorphisms of Higman’s group on the line or the circle
is trivial.

title = One-dimensional Actions of Higman’s Group, author = Cristóbal Rivas
and Michele Triestino, plaintextauthor = Cristóbal Rivas, Michele Triestino,
keywords = Higman’s group, orderable groups, one-dimensional dynamics,
Kaplansky’s zero divisors conjecture, year=2019, number=20, received=16 May
2019, published=23 December 2019, doi=10.19086/da.11151,

[classification=text]

##  1 Introduction

The first example of an infinite, finitely generated and simple group was
built by Higman [11] as a quotient of the group with the following
presentation

$\langle\,a_{i}\;\;(i\in\mathbb{Z}/4\mathbb{Z})\mid
a_{i}a_{i+1}a_{i}^{-1}=a_{i+1}^{2}\;\rangle=H.$ (1)

Indeed, Higman observed that $H$ is infinite, in fact torsion free, and
moreover that if four elements in a group satisfy the relations above and each
of them has finite order, then they all have to be the group identity. In
particular $H$ has no non-trivial finite quotient, and the quotient of $H$ by
a maximal normal subgroup is infinite, simple and finitely generated. With
time, the name _Higman’s group_ was attached to the group with the above
presentation.

The fact that $H$ has no non-trivial finite quotient implies that every finite
dimensional linear representation $\rho:H\to\mathsf{GL}(V)$ is trivial, since
finitely generated groups of matrices are residually finite by a theorem of
Mal´cev [14]. It also implies that every $C^{1}$-action
$\rho:H\to\mathsf{Diff}^{1}(M)$ having a periodic orbit is trivial if $M$ is a
connected manifold. Indeed, since $H$ has no finite quotient other than the
trivial, the finite orbit must be a global fixed point and, since $H$ has only
trivial linear representations, the derivative of every $h\in H$ must be
trivial at that point. It then follows from the Thurston Stability Theorem
[20] that if $\rho(H)$ is not the trivial group, then $\rho(H)$ (and hence
$H$) must admit a surjective homomorphism onto $\mathbb{Z}$, which is
certainly impossible since $H$ has trivial abelianization111The fact that $H$
is finitely generated and has trivial abelianization implies that $H$ is _not_
(locally) indicable. Curiously, this notion was introduced by Higman to study
zero divisors and units in group ring [12]. Compare with our Corollary B. The
fact that locally indicable groups are left-orderable is now well known and
was first proved by Burns and Hale in [3].. This immediately implies, for
instance, that every representation $\rho:H\to\mathsf{Diff}^{1}([0,1])$ is
trivial since $0$ is always in a finite orbit.

The main purpose of this work is to show that, despite the constrains
described above, Higman’s group can be faithfully represented as a group of
orientation-preserving homeomorphisms of the real line222Note that, since $H$
has no finite index subgroup, every action of $H$ by homeomorphisms of an
oriented manifold must preserve the orientation.. This implies, in particular,
that $H$ admits faithful representations inside the group of homeomorphisms of
any given manifold. Indeed, choose a ball $B^{d}$ in a manifold $M^{d}$, the
complement of a point $p\in B^{d}$ in $B^{d}$ is homeomorphic to
$\mathbb{R}\times\mathbb{S}^{d-1}$; given an action of $H$ on $\mathbb{R}$,
extend it to an action on $\mathbb{R}\times\mathbb{S}^{d-1}$, so that the
action on the $\mathbb{S}^{d-1}$-factor is trivial: this gives an action of
$H$ on $B^{d}$, which fixes $p$ and the boundary $\partial B^{d}$, so that it
can be extended as the identity outside $B^{d}$.

We start by showing, in Section 2, that $H$ admits a non-trivial action on the
real line. This amounts to find four homeomorphisms $a$, $b$, $c$, $d$ of the
line satisfying

$aba^{-1}=b^{2}\;,\;bcb^{-1}=c^{2}\;,\;cdc^{-1}=d^{2}\;,\;dad^{-1}=a^{2}.$

In our construction, $b$ and $d$ will be _lifts_ to the real line of strong
parabolic homeomorphisms of the circle, so that the subgroup $\langle
b,d\rangle$ display ping-pong dynamics and acts without global fixed points.
The choice of the generators $a$ and $c$ is more subtle and it relies on the
fact that the Baumslag-Solitar group
$\mathsf{BS}(1,2)=\langle\alpha,\beta\mid\alpha\beta\alpha^{-1}=\beta^{2}\rangle$
admits two different and easy-to-describe actions on the real line (see the
beginning of Section 2). We will build $a$ and $c$ as the limit of sequences
of homeomorphisms $(a_{n})_{n\geq 0}$ and $(c_{n})_{n\geq 0}$ respectively,
where the relations

$a_{n}ba_{n}^{-1}=b^{2}\quad\text{and}\quad c_{n}dc_{n}^{-1}=d^{2},$

hold for even $n$’s, and the relations

$bc_{n}b^{-1}=c_{n}^{2}\quad\text{and}\quad da_{n}d^{-1}=a_{n}^{2},$

hold for odd $n$’s. At each step, we will perform slight modifications of the
homeomorphisms so that, _in the limit_ , both sets of relations will be
satisfied. This provides a non-trivial representation

$\varphi:H\to\mathsf{Homeo}_{+}(\mathbb{R}).$

The fact that $\varphi:H\to\mathsf{Homeo}_{+}(\mathbb{R})$ is non-trivial
entails that $\varphi(H)$ is a left-orderable group, that is $\varphi(H)$
admits a total order $\preceq$ such that $\varphi(f)\preceq\varphi(g)$ implies
that $\varphi(hf)\preceq\varphi(hg)$ for every $f,g,h\in H$. In fact, for
countable groups being left-orderable is equivalent to admit an injective
representation into $\mathsf{Homeo}_{+}(\mathbb{R})$ (see, for instance, [5,
9]). In our case, we are unable to decide whether $\varphi$ is injective or
not, but, using Kurosh Theorem [19], in Section 3 we will show that the kernel
of $\varphi$ must be a free group, which is one of the basic examples of left-
orderable groups [5, 9]. In particular, we can lexicographically extend a
left-order on $\varphi(H)$ to a left-order on $H$ (see Section 3) so that we
can deduce:

######  Theorem A.

Higman’s group $H$ is left-orderable and hence it admits a faithful action on
the real line by homeomorphisms.

As a direct consequence of left-orderability (see [5, §1.8]), we get that the
Kaplasky Zero-Divisor Conjecture holds for Higman’s group:

######  Corollary B.

If $R$ is a ring without zero-divisors, then the group ring $R[H]$ of Higman’s
group has no zero-divisor.

We point out that all the homeomorphisms in our construction of
$\varphi:H\to\mathsf{Homeo}(\mathbb{R})$ actually commute with the translation
$T:x\mapsto x+2$. Thus, we get a faithful action of $H$ on the circle
$\mathbb{R}/_{T(x)\sim x}$ without global fixed points. See Proposition 3.2.
With the same idea of construction, we are able to describe countably many
different (i.e. pairwise non-semi-conjugate333Recall that two actions
$\psi_{1}:G\to\mathsf{Homeo}_{+}(\mathbb{R})$ and
$\psi_{2}:G\to\mathsf{Homeo}_{+}(\mathbb{R})$ are semi-conjugate, if there is
a proper, non-decreasing map $\alpha:\mathbb{R}\to\mathbb{R}$ such that
$\alpha\circ\psi_{1}(g)=\psi_{2}(g)\circ\alpha$ for all $g\in G$. Two actions
on the circle are semi-conjugate if they have lifts to the real line which are
semi-conjugate. See [9, 15, 2] for further details.) actions of $H$ on the
circle, and uncountably many different actions on the real line, all without
global fixed points. See Section 4.

######  Remark 1.1.

We have already mentioned that Higman’s group $H$ is perfect (that is, its
abelianization is trivial), so every homomorphism $\Phi:H\to\mathbb{R}$ is
trivial. A weaker notion is that of _quasimorphism_ to $\mathbb{R}$, that is,
a map $\Phi:H\to\mathbb{R}$ for which there exists $D>0$ such that for every
$g,h\in H$, one has

$|\Phi(gh)-\Phi(g)-\Phi(h)|\leq D.$

The set of all quasimorphisms $H\to\mathbb{R}$ defines the real vector space
$\mathcal{QM}(H)$. By cohomological considerations, one deduces that the
dimension of $\mathcal{QM}(H)$ is the cardinal of continuum: indeed, for
perfect groups, the quotient $\mathcal{QM}(H)/\\{\text{bounded functions}\\}$
is isomorphic to the second bounded cohomology group
$H^{2}_{b}(H;\mathbb{R})$, whose dimension is the cardinal of continuum since
$H$ is a non-trivial amalgam [8]. Similarly, one can consider the second
bounded cohomology group $H^{2}_{b}(H;\mathbb{Z})$ with integer coefficients.
Every action on the circle without global fixed points determines a non-
trivial cocycle in $H^{2}_{b}(H;\mathbb{Z})$, by the so-called _bounded Euler
class_. Moreover, two actions of $H$ are semi-conjugate if and only if their
bounded Euler classes in $H^{2}_{b}(H;\mathbb{Z})$ are the same [9].
Therefore, the actions constructed in Section 4 provide countably many
linearly independent classes in $H^{2}_{b}(H;\mathbb{Z})$. See the references
[9, 7] for more about bounded cohomology related to group actions on the
circle.

In contrast, although our actions have no fixed points neither periodic
orbits, we can still show that they cannot be made differentiable. In fact we
have:

######  Theorem C.

Every representation $\rho:H\to\mathsf{Diff}^{1}(\mathbb{R})$ is trivial. The
same holds for every representation
$\rho:H\to\mathsf{Diff}^{1}(\mathbb{S}^{1})$.

Our proof relies on the knowledge of $C^{1}$-actions of the Baumslag-Solitar
group

$\mathsf{BS}(1,2)=\langle\alpha,\beta\mid\alpha\beta\alpha^{-1}=\beta^{2}\rangle$

on the line [1] together with an analysis of the possible combinatorics of
fixed points of elements of $H$ when acting on the line by homeomorphisms. The
proof is given in Section 5 and it is independent of the rest of the paper.

##  2 The construction of $\varphi$

###  2.1 Preliminaries

Higman’s group contains four canonical copies of the Baumslag-Solitar group

$\mathsf{BS}(1,2)=\langle\alpha,\beta\mid\alpha\beta\alpha^{-1}=\beta^{2}\rangle.$

This latter group acts on the real line via its _standard affine action_
$\alpha:x\mapsto 2x$, $\beta:x\mapsto x+1$ (see Figure 1, left), which is a
faithful action since every proper quotient of $\mathsf{BS}(1,2)$ is abelian.
_Non-standard affine_ actions of $\mathsf{BS}(1,2)$ on the line can be
obtained from the standard one by blowing up an orbit, following the method of
Denjoy (see Figure 1, right). In fact, it is not very hard to see that _if
there is_ a non-trivial action of $H$ on the line, then a non-standard affine
action must appear at least locally (cf. the proof of Theorem C in Section 5).
The following example is the one that we will use.

######  Example 2.1 (Denjoy trick).

Consider the affine maps of the real line $f(x)=2x$ and $g(x)=x+1$, which
generate a group isomorphic to $\mathsf{BS}(1,2)$. The orbit of $0$ coincides
with the dyadic rationals $\mathbb{Z}[1/2]=\\{n/2^{m}\mid n\in\mathbb{Z},m\geq
0\\}$. We blow up the orbit of $0$ to build a _new_ real line, that is, we
replace every point $o\in\mathbb{Z}[1/2]$ of the orbit of $0$ by a closed
interval $J_{o}$ in such a way that, for every compact subset $K$, the sum
$\sum_{o\in K}|J_{o}|$ is bounded (we are denoting by $|J|$ the length of the
interval $J$). For our purpose, if $n/2^{m}$ is a reduced fraction, we set
$|J_{n/2^{m}}|=\kappa/2^{2^{m}}$, for some constant $\kappa>0$ to be fixed
later. By convention the reduced fraction of $0$ is $0/1$, so
$|J_{0}|=\kappa/4$.

The maps $f$ and $g$ act naturally on the boundary points of the collection of
intervals $\\{J_{o}\\}$, and this action can be extended to the _new_ real
line by taking linear interpolation and passing to the closure of the union of
intervals. We put the new origin (still denoted by $0$) at the midpoint of
$J_{0}$. The resulting maps are called $\bar{f}$ and $\bar{g}$. Observe that
since $n/2^{m}$ is a reduced fraction if and only if
$1+n/2^{m}=(2^{m}+n)/2^{m}$ is a reduced fraction, we have that the
homeomorphism $\bar{g}$ preserves the length of the intervals it permutes and
hence $\bar{g}$ is still a translation. By adjusting $\kappa$, and possibly
replacing $\bar{g}$ by some power of it, we can assume that $\bar{g}$ is the
translation by 2 and that $|J_{0}|\leq 1/4$.

Finally, we let $M\subseteq\mathbb{R}$ be the minimal invariant set for
$\langle\bar{f},\bar{g}\rangle$ action on the line. That is, $M$ is closed,
$\langle\bar{f},\bar{g}\rangle$-invariant, and contains no proper, closed,
$\langle\bar{f},\bar{g}\rangle$-invariant subsets. In the present case, since
$J_{0}$ is a wandering interval, and no $\langle\bar{f},\bar{g}\rangle$-orbit
is discrete, the set $M$ is locally a Cantor set which intersects trivially
the $\langle\bar{f},\bar{g}\rangle$-orbit of $J_{0}$ (see [9]).

######  Remark 2.2.

Observe that $\bar{f}$ from Example 2.1 is the identity on $J_{0}$. For our
purpose it will be important to replace $\bar{f}$ by another map which is not
the identity on $J_{0}$, while preserving the action on the minimal set $M$
from the original action (Figure 1, right). This will be done in Lemma 2.4
below.

Figure 1: The standard affine action of $\mathsf{BS}(1,2)$ (left) and an
action _à la Denjoy_ (right).

Crucially, $\mathsf{BS}(1,2)$ admits a different family of actions without
fixed points on the real line. These are the so-called _Conradian actions_ ,
which are actions where $\alpha$ acts without fixed points and $\beta$ fixes a
fundamental domain of $\alpha$ 444For the purpose of this note, this can be
taken as a definition of Conradian action of $\mathsf{BS}(1,2)$. In general, a
Conradian action of a group is a faithful action where every induced order is
Conradian. Equivalently, it is an action without crossed elements, see [17]
for details. In [18] it is shown that any faithful action of
$\mathsf{BS}(1,2)$ without global fixed points is either of affine type or
Conradian. (see Figure 2). The following lemma formalizes the idea that for
building a Conradian action one just need pick
$\alpha\in\mathsf{Homeo}_{+}(\mathbb{R})$ and _any other_ homeomorphism
defined on a fundamental domain of $\alpha$ which is then extended to a
homeomorphism of the line using the relation of $\mathsf{BS}(1,2)$. For the
statement, we denote by $\mathsf{Homeo}_{0}(I)$ the group of homeomorphisms of
an open subset $I\subset\mathbb{R}$ which are isotopic to the identity. Also,
given a homeomorphism $h:I\to I$, we define its _support_ as the open subset
$\mathsf{supp}(h)=\\{x\in I\mid h(x)\neq x\\}$.

Figure 2: An example of Conradian action of $\mathsf{BS}(1,2)$ on the real
line.

######  Lemma 2.3.

Given $f\in\mathsf{Homeo}_{+}(\mathbb{R})$, an open set $I$ such that
$f(I)\cap I=\emptyset$, and $h\in\mathsf{Homeo}_{0}(I)$, there exists a map
$R_{1}:(f,I,h)\mapsto g_{h}\in\mathsf{Homeo}_{+}(\mathbb{R})$ such that

  1. (i)

$g_{h}(x)=h(x)$ for every $x\in I$,

  2. (ii)

$g_{h}$ is the identity outside $\bigcup_{n\in\mathbb{Z}}f^{n}(I)$,

  3. (iii)

$fg_{h}f^{-1}(x)=g_{h}^{2}(x)$ for every $x\in\mathbb{R}$.

###### Proof.

We need to define $g_{h}$. Observe that the condition $f(I)\cap I=\emptyset$
implies that the images $f^{n}(I)$ are pairwise disjoint. On $f^{-n}(I)$
($n>0$) we define $g_{h}$ as $f^{-n}h^{2^{n}}f^{n}$ and on $f^{n}(I)$ we want
to define $g_{h}$ as $f^{n}\,\mathsf{2root}^{(n)}(h)f^{-n}$, where
$\mathsf{2root}(h)$ is the square root of $h$ and
$\mathsf{2root}^{(n)}=\mathsf{2root}\circ\cdots\circ\mathsf{2root}$ $n$-times.

Certainly, every $h\in\mathsf{Homeo}_{+}(\mathbb{R})$ has a square root in
$\mathsf{Homeo}_{+}(\mathbb{R})$. For instance, if $h$ has no fixed points,
then it is conjugate to a translation and $\mathsf{2root}(h)$ can be then
defined as half of the translation conjugated back, while for general $h$ one
restricts the attention to connected components of the support of $h$ and
repeats the argument locally. The problem is that the square root is highly
non-unique (for instance, the translation to which a homeomorphism $h$ without
fixed points is conjugate to is not unique), so $\mathsf{2root}$ is not well
defined (so far).

To give a precise definition of
$\mathsf{2root}:\mathsf{Homeo}_{0}(I)\to\mathsf{Homeo}_{0}(I)$ (and hence a
precise definition of $g_{h}$) we proceed as follows. For every connected
component $C$ of the support of $h\in\mathsf{Homeo}_{0}(I)$, let $x_{0}\in C$
be its midpoint (this is well-defined, as from $f(I)\cap I=\emptyset$, every
component $C$ is a bounded interval). Assume $h(x_{0})>x_{0}$ (the case
$h(x_{0})<x_{0}$ is analogous). Then define a measure $\mu_{0}$ on
$[x_{0},h(x_{0}))$, as the push forward of the Lebesgue measure $\mathsf{Leb}$
on $[0,1)$ by the unique affine map $A$ sending $[0,1)$ to $[x_{0},h(x_{0}))$
bijectively. So $\mu_{0}(X)=\mathsf{Leb}(A^{-1}(X))$ for every Borel subset
$X\subset[x_{0},h(x_{0}))$. Then there is a unique way of extending $\mu_{0}$
to an $h$-invariant measure over $C$. The resulting measure $\mu_{C}$ is fully
supported on $C$, gives finite mass to compact subsets of $C$, and has not
atoms. We can now use $\mu_{C}$ to conjugate the restriction of $h$ to $C$ to
the translation by 1 on the real line. Indeed if we let
$F(x)=\mathsf{sign}(x-x_{0})\mu([x_{0},x))$, then $FhF^{-1}$ is the
translation by 1. Then we can take the square root inside the group of
translations and conjugate back. This is our preferred $\mathsf{2root}$. ∎

###  2.2 Choosing $b$ and $d$

Let $J_{0}$ and $\bar{g}:x\mapsto x+2$ be as in Example 2.1, and let
$\psi:\mathbb{R}\to(-1,1)$ be the homeomorphism given by the linear
interpolation of

$\psi(n)=\left\\{\begin{array}[]{cr}0&\text{if }n=0,\\\
\sum_{i=1}^{n}2^{-i}&\text{if }n>0,\\\ -\sum_{i=1}^{-n}2^{-i}&\text{if
}n<0,\end{array}\right.$

for $n\in\mathbb{Z}$. Let
$\beta_{0}=\psi\circ\bar{g}\circ\psi^{-1}\in\mathsf{Homeo}_{+}(-1,1)$, and
call $I_{0}$ the image of $J_{0}$ under $\psi$. Call $I_{1}=I_{0}+1$ and
define $\beta_{1}\in\mathsf{Homeo}_{+}(0,2)$ by
$\beta_{1}(x)=\beta_{0}(x-1)+1$. We note that $I_{0}$ and $I_{1}$ are disjoint
intervals, one centered at $0$, and the other centered at $1$; we note also
that the maps $\beta_{0}$ and $\beta_{1}$ are piecewise linear. Finally, we
extend $\beta_{0}$ and $\beta_{1}$ to homeomorphisms of the real line by
imposing that they commute with the translation $T:x\mapsto x+2$. The
resulting homeomorphisms are still denoted $\beta_{0}$ and $\beta_{1}$. We
also let $\mathcal{I}_{0}=\bigcup_{n\in\mathbb{Z}}I_{0}+2n$ and
$\mathcal{I}_{1}=\bigcup_{n\in\mathbb{Z}}I_{1}+2n$, which are disjoint open
sets, and choose $N>0$ such that

$\beta_{0}^{\pm
N}(\mathbb{R}\setminus\mathcal{I}_{1})\subseteq\mathcal{I}_{1}\quad\text{and}\quad\beta_{1}^{\pm
N}(\mathbb{R}\setminus\mathcal{I}_{0})\subseteq\mathcal{I}_{0}.$ (2)

By possibly enlarging $N$, we can (and will) also assume that for all $n\geq
N$ we have

$(\beta_{i}^{\pm n})^{\prime}(x)\leq 1/2\quad\text{for all
$x\notin\mathcal{I}_{i+1}$,}$ (3)

where the indices are taken in $\mathbb{Z}/2\mathbb{Z}$. Note that by (2) and
the ping-pong lemma, the group $\langle\beta_{0}^{N},\beta_{1}^{N}\rangle$ is
a free group inside $\mathsf{Homeo}_{+}(\mathbb{R})$. We let $b=\beta_{0}^{N}$
and $d=\beta_{1}^{N}$.

###  2.3 Choosing $a$ and $c$

For $f$ and $g$ homeomorphisms of the line, we let

$|f-g|_{\infty}=\sup_{x\in\mathbb{R}}|f(x)-g(x)|+\sup_{x\in\mathbb{R}}|f^{-1}(x)-g^{-1}(x)|.$

This “norm” may be infinite but in many cases it is bounded, so it defines a
true distance. This happens for instance in restriction to the subgroup

${\mathsf{Homeo}_{+,T}(\mathbb{R})}=\\{f\in\mathsf{Homeo}_{+}(\mathbb{R})\mid
T\circ f=f\circ T\\},$

where $T$ is a translation (which is our case). Endowed with
$|\,\cdot\,|_{\infty}$, ${\mathsf{Homeo}_{+,T}(\mathbb{R})}$ becomes a
complete metric space, which makes it a Polish group (see [10, Proposition
I.2.4]). We will build $a$ and $c$ as the limit (in this topology) of
converging sequences of homeomorphisms $(a_{n})_{n\geq 0}$ and $(c_{n})_{n\geq
0}$ respectively; for every $n\geq 0$, the homeomorphism $a_{n}$ (resp.
$c_{n}$) will be defined as a perturbation of $a_{n-1}$ (resp. $c_{n-1}$). We
keep the notation from §2.2.

Let $\bar{f}$ and $\bar{g}$ be the homeomorphisms from Example 2.1 and $M$ its
minimal set. Let
$a_{-1}=\psi\circ\bar{f}\circ\psi^{-1}\in\mathsf{Homeo}_{+}(-1,1)$, and
$c_{-1}\in\mathsf{Homeo}_{+}(0,2)$ be defined by $c_{-1}(x)=a_{-1}(x-1)+1$. We
extend $a_{-1}$ and $c_{-1}$ inside ${\mathsf{Homeo}_{+,T}(\mathbb{R})}$ by
imposing that they commute with the translation $T:x\mapsto x+2$. We denote
the resulting elements by $a_{0}$ and $c_{0}$ respectively. Observe that
$a_{0}ba_{0}^{-1}(x)=b^{2}(x)$ for all $x\in\mathbb{R}$ and in the same way
$c_{0}dc_{0}^{-1}(x)=d^{2}(x)$ for all $x\in\mathbb{R}$. Note also that
$a_{0}$ is supported on $\mathbb{R}\setminus\mathcal{I}_{0}$ and $c_{0}$ is
supported on $\mathbb{R}\setminus\mathcal{I}_{1}$ (because $\bar{f}$ is the
identity on $J_{0}$).

We now explain how to obtain $a_{n}$ and $c_{n}$ for $n\geq 1$. The
construction of the sequences $(a_{n})$ and $(c_{n})$, as well as the proof of
its convergence are independent and completely analogous upon exchanging the
roles of $b$ and $d$ (and replacing $0$ indices by $1$ indices). We only
explain the case of $a_{n}$. Let
$\mathcal{M}_{0}=\bigcup_{n\in\mathbb{Z}}\psi(M)+2n$. We will perform
successive perturbations on $a_{0}$ on the complement of $\mathcal{M}_{0}$.

######  Lemma 2.4.

Let $T:\mathbb{R}\to\mathbb{R}$ denote the translation by $2$. There exists a
map $R_{2}:h\mapsto a_{h}\in{\mathsf{Homeo}_{+,T}(\mathbb{R})}$ defined on
elements of $\mathsf{Homeo}_{+,T}(\mathbb{R})$ supported on $\mathcal{I}_{0}$
with the following properties.

  1. (i)

$a_{h}(x)=h(x)$ for every $x\in\mathcal{I}_{0}$.

  2. (ii)

$a_{h}(x)=a_{0}(x)$ for $x\in\mathcal{M}_{0}$ and
$a_{h}ba_{h}^{-1}(x)=b^{2}(x)$ for every $x\in\mathbb{R}$.

  3. (iii)

$R_{2}$ is an isometry with respect to the $|\,\cdot\,|_{\infty}$-norm: for
any two $h,h^{\prime}\in\mathsf{Homeo}_{+,T}(\mathbb{R})$ supported on
$\mathcal{I}_{0}$, one has
$|a_{h}-a_{h^{\prime}}|_{\infty}=|h-h^{\prime}|_{\infty}$.

###### Proof.

Suppose that $h\in{\mathsf{Homeo}_{+,T}(\mathbb{R})}$ is supported on
$\mathcal{I}_{0}$. Since $\mathcal{I}_{0}$ is contained in a fundamental
domain of $b$ (i.e. $b^{n}(\mathcal{I}_{0})\cap\mathcal{I}_{0}=\emptyset$ for
$n\neq 0$), the infinite composition
$h_{1}:=\prod_{n\in\mathbb{Z}}b^{n}hb^{-n}$ is a composition of homeomorphisms
with pairwise disjoint support, all commuting with $T$. Hence $h_{1}$ is a
well-defined element in $\mathsf{Homeo}_{+,T}(\mathbb{R})$, supported on
$\bigcup_{n\in\mathbb{Z}}b^{n}(\mathcal{I}_{0})$, which furthermore commutes
with $b$. We can then define $R_{2}(h)=a_{h}=a_{0}h_{1}$ which satisfies item
(i).

For (ii) we observe that $h_{1}(x)=x$ for all $x\in\mathcal{M}_{0}$ since
$b^{n}(\mathcal{I}_{0})$ and $\mathcal{M}_{0}$ are disjoint (see the end of
Example 2.1) and $h_{1}$ is supported on $\bigcup b^{n}(\mathcal{I}_{0})$. In
particular $a_{h}(x)=a_{0}(x)$ for all $x\in\mathcal{M}_{0}$. We can also
check that
$a_{h}b(x)=a_{0}h_{1}b(x)=a_{0}bh_{1}(x)=b^{2}a_{0}h_{1}(x)=b^{2}a_{h}(x)$ for
all $x\in\mathbb{R}$.

Finally, to show (iii) we note that

$\sup_{x\in\mathbb{R}}|a_{0}h_{1}(x)-a_{0}h_{1}^{\prime}(x)|=\sup_{n\in\mathbb{Z}}\sup_{x\in\mathcal{I}_{0}}|a_{0}b^{n}h(x)-a_{0}b^{n}h^{\prime}(x)|=\sup_{n\in\mathbb{Z}}\sup_{x\in\mathcal{I}_{0}}|b^{2n}a_{0}h(x)-b^{2n}a_{0}h^{\prime}(x)|.$
(4)

Further, for $x\in\mathcal{I}_{0}$, the images $h(x)$ and $h^{\prime}(x)$
belong to $\mathcal{I}_{0}$, on which $a_{0}$ is the identity and, as
$\mathcal{I}_{0}$ and $\mathcal{I}_{1}$ are disjoint, condition (3) implies
that the supremum in the last term is achieved for $n=0$. After these
considerations, equality (4) reduces to

$\sup_{x\in\mathbb{R}}|a_{0}h_{1}(x)-a_{0}h_{1}^{\prime}(x)|=\sup_{n\in\mathbb{Z}}\sup_{x\in\mathcal{I}_{0}}|b^{2n}h(x)-b^{2n}h^{\prime}(x)|=\sup_{x\in\mathcal{I}_{0}}|h(x)-h^{\prime}(x)|.$

Similarly we control the difference of inverses: we have

$\sup_{x\in\mathbb{R}}|(a_{0}h_{1})^{-1}(x)-(a_{0}h_{1}^{\prime})^{-1}(x)|=\sup_{x\in\mathbb{R}}|(h_{1})^{-1}(x)-(h_{1}^{\prime})^{-1}(x)|=\sup_{n\in\mathbb{Z}}\sup_{x\in\mathcal{I}_{0}}|b^{n}(h)^{-1}(x)-b^{n}(h^{\prime})^{-1}(x)|$

which, by (3), equals
$\sup_{x\in\mathcal{I}_{0}}|(h)^{-1}(x)-(h^{\prime})^{-1}(x)|$. Putting the
two pieces together, we obtain the equality
$|a_{h}-a_{h^{\prime}}|_{\infty}=|h-h^{\prime}|_{\infty}$. ∎

We will use sequentially Lemma 2.3 and Lemma 2.4 to perturb $a_{0}$ outside
$\mathcal{M}_{0}$. To this extent we let

$\operatorname{\mathsf{Dom}}_{0}=\\{f\in{\mathsf{Homeo}_{+,T}(\mathbb{R})})\mid
f(x)=a_{0}(x)\text{ for every $x\in\mathcal{M}_{0}$}\\}.$

Observe that $\operatorname{\mathsf{Dom}}_{0}$ is a closed subset of
${\mathsf{Homeo}_{+,T}(\mathbb{R})}$ and for
$f\in\operatorname{\mathsf{Dom}}_{0}$, we have that
$f(\mathcal{I}_{0})=\mathcal{I}_{0}$, and $fbf^{-1}(x)=b^{2}(x)$ for every
$x\in\mathcal{M}_{0}$. In particular, for
$f\in\operatorname{\mathsf{Dom}}_{0}$, we can define
$\widetilde{R_{2}}(f)=R_{2}(f\restriction_{\mathcal{I}_{0}})$, where
$f\restriction_{\mathcal{I}_{0}}$ denotes the restriction of $f$ to
$\mathcal{I}_{0}$, extended as to be the identity outside $\mathcal{I}_{0}$.
Certainly $\widetilde{R_{2}}$ is an idempotent of
$\operatorname{\mathsf{Dom}}_{0}$ (i.e.
$\widetilde{R_{2}}^{2}=\widetilde{R_{2}}$) and, from Lemma 2.4.iii, it also
satisfies

$|\widetilde{R_{2}}(f)-\widetilde{R_{2}}(g)|_{\infty}=|f\restriction_{\mathcal{I}_{0}}-g\restriction_{\mathcal{I}_{0}}|_{\infty}.$
(5)

Note also that by (2),
$d^{n}(\mathbb{R}\setminus\mathcal{I}_{0})\subset\mathcal{I}_{0}$ for all
$n\neq 0$ (and hence $d^{n}(\mathcal{M}_{0})\subset\mathcal{I}_{0}$ for all
$n\neq 0$). In particular, for $f\in\operatorname{\mathsf{Dom}}_{0}$ we can
apply $R_{1}$ from Lemma 2.3 to define
$\widetilde{R_{1}}(f)=R_{1}(d,\mathbb{R}\setminus\mathcal{I}_{0},f\restriction_{\mathbb{R}\setminus\mathcal{I}_{0}})$.
Certainly $\widetilde{R_{1}}$ is also an idempotent of
$\operatorname{\mathsf{Dom}}_{0}$ which, by (3), satisfies that

$|\widetilde{R_{1}}(f)\restriction_{\mathcal{I}_{0}}-\widetilde{R_{1}}(g)\restriction_{\mathcal{I}_{0}}|_{\infty}\leq\tfrac{1}{2}\;|f\restriction_{\mathbb{R}\setminus\mathcal{I}_{0}}-g\restriction_{\mathbb{R}\setminus\mathcal{I}_{0}}|_{\infty}.$
(6)

Suppose that $a_{n}$ is defined for $n\leq 2k$ ($k\geq 0$). We define
$a_{2k+1}=\widetilde{R_{1}}(a_{2k})$ and
$a_{2k+2}=\widetilde{R_{2}}(a_{2k+1})$.

######  Main Claim.

The sequence $a_{n}$ converges in $\mathsf{Homeo}_{+,T}(\mathbb{R})$.

Before giving the proof note that if $a_{n}$ converges to $a$, then $a$ is
both $\widetilde{R_{2}}$ and $\widetilde{R_{1}}$ invariant so, by Lemma
2.3.iii and Lemma 2.4.ii, it satisfies $aba^{-1}=b^{2}$ and $dad^{-1}=a^{2}$
over $\mathbb{R}$ as desired.

###### Proof of Claim.

First observe that, from (5) and (6), it follows that
$\widetilde{R_{2}}\widetilde{R_{1}}$ and $\widetilde{R_{1}}\widetilde{R_{2}}$
are contractions (by $1/2$) of the complete metric space
$(\mathsf{Homeo}_{+,T}(\mathbb{R}),|\,\cdot\,|_{\infty})$, so by the classical
Banach Fixed Point Theorem, both of them have unique (attracting) fixed points
in $\operatorname{\mathsf{Dom}}_{0}\subset\mathsf{Homeo}_{+,T}(\mathbb{R})$;
call them $p=\widetilde{R_{2}}\widetilde{R_{1}}(p)$ and
$q=\widetilde{R_{1}}\widetilde{R_{2}}(q)$. Observe that since
$\widetilde{R}_{1}$ and $\widetilde{R}_{2}$ are idempotents, if follows by
uniqueness that $\widetilde{R}_{1}(q)=q$ and $\widetilde{R}_{2}(p)=p$. We need
to show that $p=q$.

By the definition of $\operatorname{\mathsf{Dom}}_{0}$, we have
$p(x)=q(x)=a_{0}(x)$ for all $x\in\mathcal{M}_{0}$ and hence

$|p\restriction_{\mathbb{R}\setminus\mathcal{I}_{0}}-q\restriction_{\mathbb{R}\setminus\mathcal{I}_{0}}|_{\infty}\leq|J|,$

where $J$ is the largest connected component of the complement of
$\mathcal{M}_{0}\cup\mathcal{I}_{0}$. In particular, applying
$\widetilde{R_{1}}$, we get from (6) that

$|\widetilde{R_{1}}(p)\restriction_{\mathcal{I}_{0}}-q\restriction_{\mathcal{I}_{0}}|_{\infty}\leq\tfrac{1}{2}|J|,$

and then, applying $\widetilde{R_{2}}$, (5) gives

$|p-\widetilde{R_{2}}(q)|_{\infty}\leq\tfrac{1}{2}|J|.$

If we now repeat $k$ times this process we get

$|\widetilde{R_{1}}(p)\restriction_{\mathcal{I}_{0}}-q\restriction_{\mathcal{I}_{0}}|_{\infty}\leq|J|/2^{k}\quad\text{and}\quad|p-\widetilde{R_{2}}(q)|_{\infty},\leq|J|/2^{k}.$

It follows that $p=\widetilde{R_{2}}(q)$ and hence $\widetilde{R_{1}}(p)=q$.
But, respectively, this implies that $p=q$ over $\mathcal{I}_{0}$ (see Lemma
2.4.i) and over $\mathbb{R}\setminus\mathcal{I}_{0}$ (see Lemma 2.3.i). So
$p=q$, and we declare $a=p=q$ as needed. ∎

######  Remark 2.5.

Observe that the four homeomorphisms $a,b,c,d$ all commute with the
translation $T$, so they define homeomorphisms of the circle
$\mathbb{S}^{1}\cong\mathbb{R}/_{T(x)\sim x}$, and hence a representation of
$H$ into $\mathsf{Homeo}(\mathbb{S}^{1})$. Further, since $c$ and $d$ do not
share fixed points, the resulting action on the circle is without global fixed
points. See Figure 3.

Figure 3: The action of Higman’s group on the circle
$\mathbb{S}^{1}\cong\mathbb{R}/\langle T\rangle$.

##  3 A faithful action

In the action $\varphi:H\to{\mathsf{Homeo}_{+,T}(\mathbb{R})}$ built in
Section 2, the subgroup $F=\langle b,d\rangle$ acts as a lift of a ping pong
action on the circle (with parabolic elements) so $\varphi$ is an embedding in
restriction to $\langle b,d\rangle$. This remark, together with Kurosh Theorem
[19], give the following:

######  Lemma 3.1.

The kernel of the action $\varphi$ is a free group and hence is left-
orderable.

###### Proof.

The group $H$ is the amalgamated product

$H=\langle d,a,b\mid dad^{-1}=a^{2}\;,\,aba^{-1}=b^{2}\rangle*_{\langle
b,d\rangle}\langle b,c,d\mid bcb^{-1}=c^{2}\;,\,cdc^{-1}=d^{2}\rangle.$ (7)

By the preceding remark, $\mathsf{ker}(\varphi)$ has trivial intersection with
the edge group $F=\langle b,d\rangle$. Therefore we can apply Kurosh Theorem
to $\mathsf{ker}(\varphi)$ in the amalgamated product (7) (in the case of a
normal subgroup, it is enough to look at the intersection with the edge
group): it gives that $\mathsf{ker}(\varphi)$ is the free product of a free
group with intersections of $\mathsf{ker}(\varphi)$ with conjugates of the
vertex groups

$H_{1}=\langle d,a,b\mid
dad^{-1}=a^{2}\;,\,aba^{-1}=b^{2}\rangle\quad\text{and}\quad H_{2}=\langle
b,c,d\mid bcb^{-1}=c^{2}\;,\,cdc^{-1}=d^{2}\rangle.$

It is then enough to prove that the intersection of $\mathsf{ker}(\varphi)$
with every conjugate of $H_{1}$ and $H_{2}$ is also a free group. Observe that
by normality of $\mathsf{ker}(\varphi)$, it is enough to prove that the
intersections $\mathsf{ker}(\varphi)\cap H_{1}$ and $\mathsf{ker}(\varphi)\cap
H_{2}$ are free. For this, we repeat nearly the same argument.

Indeed, the group $H_{1}=\langle d,a,b\mid
dad^{-1}=a^{2}\;,\,aba^{-1}=b^{2}\rangle$ is also an amalgamated product

$H_{1}=\langle d,a\mid dad^{-1}=a^{2}\rangle*_{\langle a\rangle}\langle
a,b\mid aba^{-1}=b^{2}\rangle,$ (8)

and the two vertex groups $\langle d,a\mid dad^{-1}=a^{2}\rangle$ and $\langle
a,b\mid aba^{-1}=b^{2}\rangle$ are two copies of $\mathsf{BS}(1,2)$ which act
faithfully by construction of the action (every proper quotient of
$\mathsf{BS}(1,2)$ is abelian, and certainly this is not the case for this
action). Therefore the intersection of $\mathsf{ker}(\varphi)$ with the vertex
group in $H_{1}$ is trivial and this implies that $\mathsf{ker}(\varphi)\cap
H_{1}$ is a free group. The same argument works for the intersection with
$H_{2}$. This shows that $\mathsf{ker}(\varphi)$ is a free group (potentially
trivial).

Finally, the fact that free groups are left-orderable is well known and can be
found in [5] or [9]. ∎

###### Proof of Theorem A.

As a consequence, the group $H$ fits into the short exact sequence

$1\to\mathsf{ker}(\varphi)\to H\to\varphi(H)\to 1,$

where $\mathsf{ker}(\varphi)$ and $\varphi(H)$ are both left-orderable after
Lemma 3.1. This implies that $H$ is left-orderable as well: denote by
$\preceq_{\varphi}$ and $\preceq_{\mathsf{ker}}$ the left-invariant orders on
$\varphi(H)$ and $\mathsf{ker}(\varphi)$ respectively; then define the order
$\preceq$ on $H$ by declaring

$id\preceq h\quad\text{if and only
if}\quad\begin{cases}h\in\mathsf{ker}(\varphi)\text{ and
}id\preceq_{\mathsf{ker}}h,\text{ or}\\\ h\notin\mathsf{ker}(\varphi)\text{
and }\varphi(id)\preceq_{\varphi}\varphi(h),\end{cases}$

and extending it so that it is left-invariant (declaring $h\preceq g$ if and
only if $id\preceq h^{-1}g$; the fact that $\preceq_{\varphi}$ and
$\preceq_{\mathsf{ker}}$ are left-invariant implies that the definition is
consistent, see [5] for more details). Finally, every countable left-orderable
group is isomorphic to a subgroup of $\mathsf{Homeo}_{+}(\mathbb{R})$ (see [5]
or [9]), therefore $H$ admits faithful actions on the real line. ∎

We can also show the analogous result for the circle.

######  Proposition 3.2.

$H$ admits a faithful action on the circle without global fixed points.

###### Scketch of proof.

The proof follows the same lines as for Theorem A.

  1. 1.

For countable groups, acting faithfully on the circle by orientation
preserving homeomorphisms is equivalent to admitting a circular order
invariant under left-multiplications [4, 16].

  2. 2.

If $1\to L\to G\to C\to 1$, where $C$ is a circularly left-orderable group and
$L$ is a left-orderable group, then $G$ is circularly left-orderable [4].

In our case we have that $1\to\mathsf{ker}(\varphi)\to
H\to\overline{\varphi}(H)\to 1$, where
$\overline{\varphi}:H\to\mathsf{Homeo}(\mathbb{S}^{1})$ is the projection of
$\varphi$ to $\mathbb{S}^{1}\cong\mathbb{R}_{T(x)\sim x}$, see Remark 2.5.
Since $\overline{\varphi}(H)$ acts without global fixed points, the same is
true for actions coming from circular orders obtained from the preceding short
exact sequence [16]. ∎

##  4 Many inequivalent actions

For the construction of $\varphi$ in Section 2, the initial point was the
choice of $b$ and $d$ starting from $T$-periodic extensions of
$\beta_{0}\in\mathsf{Homeo}_{+}([-1,1])$ and
$\beta_{1}\in\mathsf{Homeo}_{+}([0,2])$. It is possible to get an analogous
construction of $\varphi$ by imposing milder periodicity conditions to the
extensions of $\beta_{0}$ and $\beta_{1}$ to the real line. Concretely, for
arbitrary sequences $(\epsilon_{n})$ and
$(\delta_{n})\in\\{-1,+1\\}^{\mathbb{Z}}$, we can consider extensions of
$\beta_{0}$ and $\beta_{1}$ verifying

$\begin{array}[]{lr}\beta_{0}(T^{n}(x))=T^{n}(\beta_{0}^{\epsilon_{n}}(x))&\text{for
every }n\in\mathbb{Z}\setminus\\{0\\}\text{ and every }x\in[-1,1],\\\
\beta_{1}(T^{n}(x))=T^{n}(\beta_{1}^{\delta_{n}}(x))&\text{for every
}n\in\mathbb{Z}\setminus\\{0\\}\text{ and every }x\in[0,2].\\\ \end{array}$

The effect in the corresponding choice of the new $b=\beta_{0}^{N}$ and
$d=\beta_{1}^{N}$ is that, locally, they could be switched to the inverse of
the old $b$ and $d$ (much in the spirit of [13]). This new $b$ and $d$ (and
hence the new $a$ and $c$) commute with a translation (and hence project to
homeomorphisms of the circle) if and only if $(\epsilon_{n})$ and
$(\delta_{n})$ have a common period. Moreover, any two distinct choices of
sequences (which are not shift-equivalent) give actions which are not semi-
conjugate one to another.

##  5 Smooth actions

In this section we give the proof of Theorem C. We will first establish the
result for actions on the real line.

######  Proposition 5.1.

Every representation $\rho:H\to\mathsf{Diff}^{1}(\mathbb{R})$ is trivial.

Our proof relies on the following:

######  Lemma 5.2.

Let $\psi:H\to\mathsf{Homeo}(\mathbb{R})$ be a non-trivial representation
without global fixed points. Then there is a copy $B$ of the Baumslag-Solitar
group $\mathsf{BS}(1,2)$ inside $H$ such that the restriction of $\psi$ to $B$
is faithful, $\psi(B)$ preserves a compact interval $I$, and the restriction
of the action of $\psi(B)$ to $I$ is not semi-conjugate to the standard affine
action.

Before proving the lemma, let us explain how to deduce the proposition.

###### Proof of Proposition 5.1.

As pointed out before, $H$ contains several copies of $\mathsf{BS}(1,2)$.
After [1] every action of $\mathsf{BS}(1,2)$ inside
$\mathsf{Diff}^{1}_{+}(I)$, where $I$ is a compact interval, must be
(topologically) conjugate to the standard affine action. By Lemma 5.2 there is
a copy of $\mathsf{BS}(1,2)$, call it $B\leq H$, that preserves a compact
interval $I$ and, moreover, the action of $B$ on $I$ is not conjugate to an
affine action (it is a Conradian action in the terminology of Section 2). In
particular, no action of $H$ on the line is conjugate to an action by
$C^{1}$-diffeomorphisms. ∎

Let us next prove the lemma.

###### Proof of Lemma 5.2.

For the proof it will be convenient to define $H=\langle a_{i}\mid
a_{i}a_{i+1}a_{i}^{-1}=a_{i+1}^{2}\rangle$ where the indices are taken over
$\mathbb{Z}/4\mathbb{Z}$. Let $\psi:H\to\mathsf{Homeo}_{+}(\mathbb{R})$ be a
non-trivial representation.

######  Claim 1.

The set of fixed points of each $\psi(a_{i})$ accumulates on $\pm\infty$.

###### Proof of Claim.

We argue by contradiction. We will assume that the fixed points of one
generator do not accumulate on $+\infty$; the case of fixed points not
accumulating on $-\infty$ is treated similarly. After re-indexation, we may
also assume that this generator is $a_{3}$. Since for orientation preserving
actions on the line finite orbits are necessarily fixed points, the relation
$a_{2}a_{3}a_{2}^{-1}=a_{3}^{2}$ implies that the subset

$\mathsf{Fix}(\psi(a_{3}))=\\{x\in\mathbb{R}\mid\psi(a_{3})(x)=x\\}$

is $\psi(a_{2})$-invariant. Therefore, $\psi(a_{2})$ maps any connected
component of the complement
$\mathsf{supp}(\psi(a_{3}))=\mathbb{R}\setminus\mathsf{Fix}(\psi(a_{3}))$ to a
(not necessarily different) connected component of
$\mathsf{supp}(\psi(a_{3}))$.

Let $I$ be an unbounded component of the support of $\psi(a_{3})$ accumulating
on $+\infty$. Then, by the previous remark, $\psi(a_{2})$ must either preserve
or move $I$ off itself. Since $I$ is unbounded, we conclude that $\psi(a_{2})$
preserves $I$. It follows from [18] that in this case the action of
$\langle\psi(a_{2}),\psi(a_{3})\rangle$ on $I$ is _semi-conjugate_ to the
standard affine action of $\mathsf{BS}(1,2)$ on $\mathbb{R}$ (i.e. where
$a_{3}$ acts as a translation and $a_{2}$ as dilation by 2), namely there
exists a continuous, non-decreasing and surjective map $\alpha:I\to\mathbb{R}$
such that

$\alpha\circ\psi(a_{2})(x)=2\alpha(x)\quad\text{and}\quad\alpha\circ\psi(a_{3})(x)=\alpha(x)\pm
1.$

The sign of translation depends on the action. If
$\alpha\circ\psi(a_{3})(x)=\alpha(x)+1$, then
$\psi(a_{2})(x)>\psi(a_{3})(x)>x$ for every $x$ large enough. If
$\alpha\circ\psi(a_{3})(x)=\alpha(x)-1$, then
$\psi(a_{2})(x)>x>\psi(a_{3})(x)$ for every $x$ large enough. In both cases,
the set of fixed points of $\psi(a_{2})$ does not accumulate on $+\infty$. We
can therefore repeat the argument with $a_{2}$ and $a_{1}$, then with $a_{1}$
and $a_{0}$ and then with $a_{0}$ and $a_{3}$ (this time $\psi(a_{3})$ acting
as dilation), to get

$\psi(a_{3})(x)<\psi(a_{2})(x)<\psi(a_{1})(x)<\psi(a_{0})(x)<\psi(a_{3})(x)$

for all $x$ large enough. Hence we obtain a contradiction. ∎

Observe that no $\psi(a_{i})$ can be the identity, otherwise the image
$\psi(H)$ would be trivial. Therefore every open support
$\mathsf{supp}(\psi(a_{i}))=\\{x\in\mathbb{R}\mid\psi(a_{i})(x)\neq x\\}$ is
non-empty, and by Claim 1 all connected components are bounded intervals.

######  Claim 2.

There is a pair of consecutive generators $a_{i},a_{i+1}$ such that
$\psi(a_{i})$ moves off itself a connected component of the open support of
$\psi(a_{i+1})$.

###### Proof of Claim.

Let $I_{3}$ be a connected component of $\mathsf{supp}(\psi(a_{3}))$. Arguing
as in the proof of Claim 1, we see that if $\psi(a_{2})$ does not preserve
$I_{3}$, then it must move it off itself, so we are done. So suppose that
$\psi(a_{2})$ fixes $I_{3}$. Then the action of
$\langle\psi(a_{2}),\psi(a_{3})\rangle$ on $I_{3}$ is semi-conjugate to the
standard affine action of $\mathsf{BS}(1,2)$ (where $\psi(a_{3})$ is acting as
the translation), and hence there is a connected component of
$\mathsf{supp}(\psi(a_{2}))$ which is strictly contained in $I_{3}$. If
$\psi(a_{1})$ does not moves $I_{2}$ off itself, then we find a connected
component $I_{1}\subsetneq I_{2}$ of $\mathsf{supp}(\psi(a_{1}))$. Again, if
$\psi(a_{0})$ does not moves $I_{1}$ off itself, we find a connected component
$I_{0}\subsetneq I_{1}\subsetneq I_{2}\subsetneq I_{3}$ of
$\mathsf{supp}(\psi(a_{0}))$. But then
$I_{0}\subsetneq\mathsf{supp}(\psi(a_{3}))$, so $\psi(a_{3})$ cannot preserve
$I_{0}$, and therefore it must move $I_{0}$ off itself. ∎

Let $i$ be the index given by Claim 2, and let $J$ be a connected component of
$\mathsf{supp}(\psi(a_{i+1}))$ which is moved off itself by $\psi(a_{i})$. Let
$I\subset\mathbb{R}$ be the smallest interval containing all the images
$\psi(a_{i})^{n}(J)$, $n\in\mathbb{Z}$. Claim 1 applied to $a_{i}$ implies
that $I$ is bounded, and moreover its choice gives that it is preserved by
both $\psi(a_{i})$ and $\psi(a_{i+1})$. Clearly the restriction of the action
of $\langle\psi(a_{i}),\psi(a_{i+1})\rangle\cong\mathsf{BS}(1,2)$ to $I$ is
faithful, and is not semi-conjugate to the standard affine action. ∎

We can finally conclude.

###### Proof of Theorem C.

A direct consequence of Proposition 5.1 is that the analogous result holds for
the circle, namely that every representation
$\psi:H\to\mathsf{Diff}^{1}(\mathbb{S}^{1})$ is trivial. This follows from the
well-known fact that Higman’s group is acyclic [6], in particular
$H^{2}(H,\mathbb{Z})=0$ and every (continuous) action on the circle lifts to
an action on the real line (the obstruction to such a lift is represented by
the Euler class of the action, which is a $2$-cocycle [9]).

We can also use a more dynamical argument. For this, suppose such $\psi$ is
given.

######  Claim 3.

For every $i\in\mathbb{Z}/4\mathbb{Z}$, the group
$\langle\psi(a_{i}),\psi(a_{i+1})\rangle$ has a global fixed point in
$\mathbb{S}^{1}$.

###### Proof of Claim.

Observe that the group $\langle a_{i},a_{i+1}\rangle\cong\mathsf{BS}(1,2)$ is
amenable. This implies that there exists a Borel probability measure $\mu_{i}$
on $\mathbb{S}^{1}$ which is invariant by the group
$\langle\psi(a_{i}),\psi(a_{i+1})\rangle$, and moreover, the rotation number
restricts to a group homomorphism
$\operatorname{\mathsf{rot}}:\langle\psi(a_{i}),\psi(a_{i+1})\rangle\to\mathbb{R}/\mathbb{Z}$
(see [9]). We deduce that

$\operatorname{\mathsf{rot}}\psi(a_{i+1})=\operatorname{\mathsf{rot}}\psi(a_{i}a_{i+1}a_{i}^{-1})=\operatorname{\mathsf{rot}}\psi(a_{i+1}^{2})=2\operatorname{\mathsf{rot}}\psi(a_{i+1}),$

so that $\operatorname{\mathsf{rot}}\psi(a_{i+1})=0$, implying that
$\psi(a_{i+1})$ has a fixed point. Cycling over the indices, we get that also
$\psi(a_{i})$ has a fixed point, and thus the probability measure $\mu_{i}$
must be supported on points that are fixed by both $\psi(a_{i})$ and
$\psi(a_{i+1})$, and thus by $\langle\psi(a_{i}),\psi(a_{i+1})\rangle$. ∎

From [1] we know that for every maximal interval $I\subset\mathbb{S}^{1}$ on
which $\langle\psi(a_{i}),\psi(a_{i+1})\rangle$ has no global fixed point, we
either have

  1. 1.

$\psi(a_{i+1})$ is the identity in restriction to $I$ or

  2. 2.

the restriction to $I$ of the action of
$\langle\psi(a_{i}),\psi(a_{i+1})\rangle$ is conjugate to the standard affine
action, and moreover the derivative of $\psi(a_{i})$ at the internal fixed
point equals $2$.

Continuity of derivatives implies that there are only finitely many intervals
of the second type. Therefore $\mathsf{supp}(\psi(a_{i+1}))$ has finitely many
connected components. Repeating the argument with the subgroup
$\langle\psi(a_{i+1}),\psi(a_{i+2})\rangle$, we obtain that not only
$\mathsf{supp}(\psi(a_{i+2}))$ has finitely many components, but moreover,
they are strictly fewer than those of $\mathsf{supp}(\psi(a_{i+1}))$. Cycling
over the indices, one gets a contradiction. ∎

## Acknowledgments

C.R. would like to thank the organizers of Groups of Dynamical Origin II where
he learned about the question of orderability of Higman’s group. In
particular, he is grateful to Yves de Cornulier for sharing this question and
to Andrés Navas for his constant encouragement. M.T. acknowledges the
hospitality of USACH wile part of this project was carried on.

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[rivas] Cristóbal Rivas
Dpto. de Matemáticas y C.C., Universidad de Santiago de Chile
Alameda 3363, Estación Central, Santiago, Chile
cristobal.rivasusachcl
http://www.mat.usach.cl/index.php/2012-12-19-12-50-19/academicos/183-cristobal-
rivas [tries] Michele Triestino
Institut de Mathématiques de Bourgogne (IMB, UMR 5584)
9 av. Alain Savary, 21000 Dijon, France
michele.triestinou-bourgognefr
http://mtriestino.perso.cnrs.fr