1 Introduction
The aim of this paper is to study a family of unitary representations of the Higman–Thompson groups
F
n
⊂
T
n
⊂
V
n
that arise from *-representations of the Cuntz algebra
O
n
, which can be recovered by an action of
V
n
on the orbit of each point of the 1-dimensional dynamical system given by the interval map
Φ
(
x
)
=
n
x
(
mod
1
)
.
In the 1960s, R. J. Thompson first introduced the groups 𝐹, 𝑇 and 𝑉 as groups of piecewise linear bijections of the half open interval
[
0
,
1
)
onto itself which satisfy
F
⊂
T
⊂
V
(see [4]), now called the Thompson groups – notably
[
F
,
F
]
, 𝑇 and 𝑉 are the first examples of infinite, finitely presented simple groups.
Based on the work of G. Higman [8], K. S. Brown extended these groups to a family of triples
F
n
⊂
T
n
⊂
V
n
, the so-called Higman–Thompson groups [3], where
F
2
is Thompson’s group 𝐹,
T
2
is Thompson’s group 𝑇 and
V
2
is Thompson’s group 𝑉.
These are countable and discrete groups and fairly easy to define as certain piecewise linear maps from the interval
[
0
,
1
)
onto itself with non-differentiable points living in the 𝑛-adic set
Z
[
1
n
]
, as we shall review below.
Almost every question related to these groups is a challenge, typically harder for the smaller groups.
For example, it is still an open problem whether
F
n
is an amenable group or not (or sofic or weakly amenable), whereas the others contain copies of free groups thus being non-amenable; see e.g. [7].
Part of the problem sits in the lack of understanding of
Rep
(
F
n
)
, and for that, we need more explicit examples of unitary representations of
F
n
.
This paper is a contribution to this since we yield an uncountable family of explicit, irreducible and inequivalent unitary representations of the Higman–Thompson groups.
Indeed, Nekrashevych [10] introduced a realisation of the group
V
n
as a subgroup of the unitary group of the Cuntz algebra
O
n
, which served as the foundation for a family of (unitary) representations of the Higman–Thompson groups on Hilbert spaces introduced in [2, 1].
Every representation
ρ
x
of
V
n
acts on a Hilbert space built on the (generalised) orbit
orb
(
x
)
of a point 𝑥 of the interval maps
Φ
(
x
)
=
n
x
(
mod
1
)
.
By exploiting the results for the representations of the Cuntz algebras from [5], similar results were obtained for a family
{
ρ
x
:
x
∈
[
0
,
1
)
}
of representations of the Higman–Thompson group
V
n
for a given
n
∈
N
with
n
>
1
; see [2, 1], where it was proved that
f
⋅
y
=
f
(
y
)
with
f
∈
V
n
and
y
∈
orb
(
x
)
is a well-defined action of
V
n
on the set
orb
(
x
)
.
Naturally, the restrictions of these representations to
F
n
and
T
n
yield families
{
τ
x
:
x
∈
[
0
,
1
)
}
and
{
σ
x
:
x
∈
[
0
,
1
)
}
of representations of
F
n
and
T
n
, respectively.
Unlike for
V
n
, very little is known on the unitary equivalence or irreducibility of these unitary representations of
F
n
and
T
n
.
We tackle this problem in this paper for the Higman–Thompson groups
F
n
and
T
n
.
For
V
n
, this problem has been solved in [1, Theorem 4.26].
Nevertheless, the techniques developed in this paper provide another proof for [1, Theorem 4.26].
The plan for the rest of the paper is as follows.
In Section 2, we prepare some necessary background for the rest of the paper.
In Section 3, we lay the groundwork which will allow us to solve the unitary equivalence and irreducibility problem for the aforementioned representations.
The techniques employed rely on the construction of elements of the Higman–Thompson groups with certain useful properties or to universal observations regarding the elements of these groups.
In Section 4 and Section 5, we attain the main results of the paper.
Theorem 4.2 and Corollary 4.3 disclose the condition (see Definition 2.2) under which two representations are unitarily equivalent, denoted by
∼
, namely
τ
x
∼
τ
y
⇔
orb
(
x
)
=
orb
(
y
)
⇔
σ
x
∼
σ
y
,
and Theorem 5.2 proves their irreducibility.
Theorem 5.3 provides a characterisation of irreducible
F
n
-invariant subspaces of
τ
x
.
As a consequence, we see in Corollary 5.5 that, for
F
2
, all the representations
τ
x
are irreducible except
τ
0
.
The situation changes for the other groups
F
n
with
n
>
2
, where we can find many reducible representations, see Example 5.6, and some irreducible, see Proposition 5.7, among the representations
τ
x
.
2 Preliminaries
Let us start by defining the Higman–Thompson groups.
Out of several equivalent realisations of these groups (see, for instance, [9]), we will consider them as groups of functions defined on the unit interval and satisfying certain properties.
Definition 2.1
Let
n
∈
N
,
n
>
1
.
The Higman–Thompson group
V
n
is the group of piecewise linear bijections
f
:
[
0
,
1
)
→
[
0
,
1
)
such that
𝑓 is non-differentiable at most in a finite subset of
Z
[
1
n
]
;
the derivative of 𝑓, where it exists, is equal to
n
r
for some
r
∈
Z
.
F
n
and
T
n
are subgroups of
V
n
such that
f
∈
F
n
if it is continuous;
f
∈
T
n
if it has at most one discontinuity.
Throughout the remainder of the paper, let us assume the value of 𝑛 is fixed, and we will solely work with
F
n
⊂
T
n
⊂
V
n
for fixed 𝑛, in order to simplify the required notation.
In Definition 2.1,
Z
[
1
n
]
=
{
a
n
b
:
a
,
b
∈
Z
}
, usually called the set of 𝑛-adic rationals.
This set is dense in
[
0
,
1
)
, which will be a useful property later.
Note as well that
f
∈
F
n
must obey
f
(
0
)
=
0
and
lim
x
→
1
f
(
x
)
=
1
.
The representations we are interested in require us to partition the unit interval into the different orbits of the dynamical system described by the interval map
Φ
:
[
0
,
1
)
→
[
0
,
1
)
such that
Φ
(
x
)
=
n
x
mod
1
.
Definition 2.2
For
x
∈
[
0
,
1
)
, the orbit of 𝑥 is
orb
(
x
)
=
{
y
∈
[
0
,
1
)
:
Φ
a
(
x
)
=
Φ
b
(
y
)
for some
a
,
b
∈
N
}
.
Since Φ is not injective,
orb
(
x
)
does not include only forward images and inverse images of 𝑥 under Φ, but also inverse images of forward images of 𝑥.
For this reason, it is also called the generalised orbit of 𝑥.
For every
x
∈
[
0
,
1
)
, we intend to build a representation of the Higman–Thompson group
V
n
on
ℓ
2
(
orb
(
x
)
)
, the space of square-summable complex sequences over
orb
(
x
)
.
In the interest of simplicity, we will introduce the relevant representations without mention of the Cuntz algebras representations they can be obtained from.
To achieve this, we will instead define the representations from one of their properties.
Proposition 2.3 restates a result relating
V
n
to
orb
(
x
)
, which will be crucial when defining the representations of the Higman–Thompson groups.
Proposition 2.3
Proposition 2.3 ([1, Theorem 4.6])
Let
x
∈
[
0
,
1
)
,
f
∈
V
n
.
Then
f
(
x
)
∈
orb
(
x
)
.
For
y
∈
orb
(
x
)
, let
δ
y
∈
ℓ
2
(
orb
(
x
)
)
be the sequence valued 1 at 𝑦 and 0 everywhere else.
The set
{
δ
y
:
y
∈
orb
(
x
)
}
is an orthonormal basis for
ℓ
2
(
orb
(
x
)
)
.
Furthermore, let
B
(
ℓ
2
(
orb
(
x
)
)
)
be the space of bounded linear operators from
ℓ
2
(
orb
(
x
)
)
to itself.
Proposition 2.4 defines the intended representations based on another property asserted by [1, Theorem 4.6.] and further proves these representations to be well-defined.
Proposition 2.4
Let
x
∈
[
0
,
1
)
and
ρ
x
:
V
n
→
B
(
ℓ
2
(
orb
(
x
)
)
)
such that, for
f
∈
V
n
and
y
∈
orb
(
x
)
,
ρ
x
(
f
)
δ
y
=
δ
f
(
y
)
,
and extend
ρ
x
(
f
)
by linearity and continuity to
ℓ
2
(
orb
(
x
)
)
.
Then
ρ
x
is a well-defined unitary representation of
V
n
on
ℓ
2
(
orb
(
x
)
)
.
Proof
Observe that
ρ
x
(
f
)
δ
y
is well-defined since
δ
f
(
y
)
∈
ℓ
2
(
orb
(
x
)
)
due to Proposition 2.3.
Given
f
,
g
∈
V
n
and
y
,
z
∈
orb
(
x
)
,
ρ
x
(
f
)
ρ
x
(
g
)
δ
y
=
ρ
x
(
f
)
δ
g
(
y
)
=
δ
f
(
g
(
y
)
)
=
ρ
x
(
f
g
)
δ
y
and also
⟨
ρ
x
(
f
)
δ
y
,
ρ
x
(
f
)
δ
z
⟩
=
⟨
δ
f
(
y
)
,
δ
f
(
z
)
⟩
=
⟨
δ
y
,
δ
z
⟩
,
where the last equality holds because 𝑓 is bijective.
The conditions are satisfied on the orthonormal basis of
ℓ
2
(
orb
(
x
)
)
, so the result follows from the linearity and continuity of
ρ
x
(
f
)
and
ρ
x
(
g
)
, as well as the linearity in the first argument, conjugate symmetry and continuity of the inner product.
∎
By construction, it is clear that these representations are, in fact, those which are built in [1] from the representations of the Cuntz algebras since the property used in said construction uniquely defines them.
We are also able to define representations of
F
n
and
T
n
by restriction.
Proposition 2.5
Let
x
∈
[
0
,
1
)
.
The restrictions
τ
x
=
ρ
x
|
F
n
and
σ
x
=
ρ
x
|
T
n
are unitary representations of
F
n
and
T
n
on
ℓ
2
(
orb
(
x
)
)
, respectively.
Note that [1, Theorem 4.26] provides a full answer to the unitary equivalence and irreducibility problem for the representations of
V
n
, namely
ρ
x
∼
ρ
y
⇔
orb
(
x
)
=
orb
(
y
)
and every
ρ
x
is irreducible.
The result was originally obtained in [1] by exploiting the properties of the corresponding representations of the Cuntz algebras derived in [5, Theorem 6].
In Section 4 and Section 5, we will provide another proof for [1, Theorem 4.26].
3 Properties of elements of
F
n
In this section, we aim to establish results which will prove fundamental in attaining the results regarding the unitary equivalence and irreducibility of the representations of the Higman–Thompson groups.
This section is twofold.
The first goal is to construct elements of
F
n
which enjoy properties that will be useful to establish certain restrictive conditions later on.
The second goal is to refine the assertion of Proposition 2.3, which will be possible by establishing properties which are universal across elements of the Higman–Thompson groups.
Our first objective is to construct
f
∈
F
n
with certain restrictions on its fixed points, which we will achieve in Proposition 3.3.
Lemma 3.1
Let
a
,
n
∈
N
,
n
>
1
.
Then 𝑎 can be written as two sums of non-zero 𝑛-adic rationals,
a
=
∑
i
=
1
5
a
a
1
,
i
=
∑
i
=
1
5
a
a
2
,
i
,
such that
a
1
,
i
≠
a
2
,
i
and
a
2
,
i
a
1
,
i
is a power of 𝑛 for every
i
∈
{
1
,
…
,
5
a
}
.
Proof
Take, for instance,
a
1
,
i
=
{
1
n
if
i
≡
0
mod
5
,
n
-
2
n
if
i
≡
1
mod
5
,
n
-
1
n
2
if
i
≡
2
mod
5
,
1
n
3
if
i
≡
3
mod
5
,
n
-
1
n
3
if
i
≡
4
mod
5
,
a
2
,
i
=
{
1
n
3
if
i
≡
0
mod
5
,
n
-
2
n
2
if
i
≡
1
mod
5
,
n
-
1
n
3
if
i
≡
2
mod
5
,
1
n
2
if
i
≡
3
mod
5
,
n
-
1
n
if
i
≡
4
mod
5
.
∎
Lemma 3.2
Let
p
∈
Z
[
1
n
]
∩
(
0
,
1
)
.
There is
f
∈
F
n
such that
f
(
x
)
=
x
if and only if
x
∈
{
0
}
∪
[
p
,
1
)
.
There is
g
∈
F
n
such that
g
(
x
)
=
x
if and only if
x
∈
[
0
,
1
-
p
]
.
Proof
(1) As
p
∈
Z
[
1
n
]
∩
(
0
,
1
)
, let
p
=
a
n
r
for some
a
,
r
∈
N
.
By Lemma 3.1, let
p
=
a
n
r
=
∑
i
=
1
5
a
a
1
,
i
n
r
=
∑
i
=
1
5
a
a
2
,
i
n
r
,
c
i
=
a
2
,
i
a
1
,
i
≠
1
,
where
a
1
,
i
,
a
2
,
i
∈
Z
[
1
n
]
and
c
i
is a power of 𝑛 for
i
∈
{
1
,
…
,
5
a
}
.
Let
t
1
,
i
=
∑
j
=
1
i
a
1
,
j
n
r
,
t
2
,
i
=
∑
j
=
1
i
a
2
,
j
n
r
,
where
t
1
,
i
,
t
2
,
i
∈
Z
[
1
n
]
for
i
∈
{
1
,
…
,
5
a
}
.
Without loss of generality, suppose
c
i
≤
c
i
+
1
for every
i
∈
{
1
,
…
,
5
a
-
1
}
.
If this is not the case, simply take an appropriate permutation of
{
1
,
…
,
5
a
}
for which the requirement is satisfied.
Take
f
(
x
)
=
{
c
1
x
if
x
∈
[
0
,
t
1
,
1
]
,
c
2
(
x
-
t
1
,
1
)
+
t
2
,
1
if
x
∈
[
t
1
,
1
,
t
1
,
2
]
,
⋮
⋮
c
i
(
x
-
t
1
,
i
-
1
)
+
t
2
,
i
-
1
if
x
∈
[
t
1
,
i
-
1
,
t
1
,
i
]
,
⋮
⋮
c
5
a
(
x
-
t
1
,
5
a
-
1
)
+
t
2
,
5
a
-
1
if
x
∈
[
t
1
,
5
a
-
1
,
t
1
,
5
a
]
,
x
if
x
∈
[
p
,
1
)
.
It is immediate that
f
∈
F
n
, and the only non-trivial property to check is that
f
(
x
)
≠
x
for
x
∈
(
0
,
p
)
.
Since
c
i
≠
1
for every
i
∈
{
1
,
…
,
5
a
}
, consider two cases for
x
∈
(
0
,
p
)
.
If
x
∈
[
t
1
,
i
-
1
,
t
1
,
i
]
with
c
i
<
1
, since the
c
i
are non-decreasing,
f
(
x
)
≤
c
i
x
<
x
for
x
>
0
.
If
x
∈
[
t
1
,
i
-
1
,
t
1
,
i
]
with
c
i
>
1
and
f
(
x
)
=
x
, since the
c
i
are non-decreasing,
f
(
p
)
≥
f
(
x
)
+
c
i
(
p
-
x
)
=
p
+
(
c
i
-
1
)
(
p
-
x
)
>
p
for
x
<
p
, which is absurd because
f
(
p
)
=
p
.
(2) Let
f
∈
F
n
be the element yielded by 1.
Take
g
(
x
)
=
{
0
if
x
=
0
,
1
-
f
(
1
-
x
)
if
x
∈
(
0
,
1
)
.
∎
Proposition 3.3
Let
x
,
y
∈
[
0
,
1
)
such that
y
∉
{
0
,
x
}
.
There is
f
∈
F
n
such that
f
(
x
)
=
x
and
f
(
y
)
≠
y
.
Proof
If
y
<
x
, as
Z
[
1
n
]
is dense in
[
0
,
1
)
, there is some
p
∈
Z
[
1
n
]
∩
[
0
,
1
)
such that
0
<
y
<
p
<
x
<
1
.
By Lemma 3.2 1, there is some
f
∈
F
n
which satisfies the required properties.
Similarly, if
y
>
x
, take 𝑝 such that
0
<
x
<
1
-
p
<
y
<
1
, and the result follows from Lemma 3.2 2.
∎
Our second objective concerns the values of elements of
F
n
at
Z
[
1
n
]
.
Our aim is to find a necessary and sufficient condition for the existence of
f
∈
F
n
such that
f
(
p
)
=
q
, where
p
,
q
∈
Z
[
1
n
]
.
We will achieve this in Lemma 3.8 and Lemma 3.9.
In order to accomplish our goal, we will prove somewhat similar results to Lemma 3.1 and Lemma 3.2, employing identical ideas.
Lemma 3.4
Let
a
,
b
∈
N
such that
n
-
1
divides
b
-
a
.
Then 𝑎 and 𝑏 can both be written as the sum of
max
{
a
,
b
}
powers of 𝑛.
Proof
Without loss of generality, assume
b
≥
a
.
We will show by induction that 𝑎 can be written as a sum of
a
+
k
(
n
-
1
)
powers of 𝑛 for
k
∈
N
0
.
Since
a
=
∑
i
=
1
a
1
,
then it can be written as a sum of 𝑎 powers of 𝑛, proving the base case,
k
=
0
.
For the induction step, notice
∑
m
=
1
j
+
k
(
n
-
1
)
n
a
m
=
n
⋅
n
a
j
+
k
(
n
-
1
)
-
1
+
∑
m
=
1
j
+
k
(
n
-
1
)
-
1
n
a
m
,
implying that a sum of
j
+
k
(
n
-
1
)
powers of 𝑛 can also be written as a sum of
n
+
j
+
k
(
n
-
1
)
-
1
=
j
+
(
k
+
1
)
(
n
-
1
)
powers of 𝑛, establishing the induction step.
Take
k
=
b
-
a
n
-
1
∈
N
, and the result follows.
∎
Lemma 3.4 provides information on how to write two integers as sums of powers of 𝑛.
However, in order to prove a result like Lemma 3.2, we are interested not only in powers of 𝑛 and integers, but rather in 𝑛-adic rationals.
Definition 3.5 extends the usual notion of divisibility over the integers to
Z
[
1
n
]
, which will be useful later.
Definition 3.5
Let
a
,
b
∈
Z
[
1
n
]
.
We say that 𝑏 divides 𝑎, and write
b
∣
a
, if there is
c
∈
Z
[
1
n
]
such that
a
=
b
c
.
Throughout the remainder of the paper, unless explicitly stated otherwise, all mentions of divisibility refer to Definition 3.5 and divisibility over
Z
[
1
n
]
.
It is easy to check that the properties of divisibility in Proposition 3.6 hold over
Z
[
1
n
]
.
These properties resemble or generalise usual properties of divisibility over ℤ.
Proposition 3.6
For
a
,
b
,
c
∈
Z
[
1
n
]
, with
a
≠
0
,
a
∣
b
and
a
∣
c
implies
a
∣
b
+
c
;
a
∣
b
implies
a
∣
b
c
;
(
n
-
1
)
∣
n
r
-
1
for
r
∈
Z
and
n
∈
N
.
We are now capable of extending Lemma 3.4 to the 𝑛-adic rationals.
Lemma 3.7
Let
p
,
q
∈
Z
[
1
n
]
such that
(
n
-
1
)
∣
q
-
p
.
For some
m
∈
N
, 𝑝 and 𝑞 can both be written as a sum of 𝑚 powers of 𝑛.
Proof
For some
r
∈
Z
[
1
n
]
, we must have
q
-
p
=
(
n
-
1
)
r
.
Therefore, for a large enough
M
∈
N
, we have
q
n
M
-
p
n
M
=
(
n
-
1
)
r
n
M
,
where
p
n
M
,
q
n
M
and
r
n
M
are integers.
It follows that
(
n
-
1
)
∣
q
n
M
-
p
n
M
(regular divisibility, over ℤ).
Lemma 3.4 implies that
p
n
M
and
q
n
M
can both be written as a sum of
m
=
max
{
p
n
M
,
q
n
M
}
powers of 𝑛, and dividing by
n
M
, we conclude that 𝑝 and 𝑞 can also both be written as a sum of 𝑚 powers of 𝑛.
∎
It is now possible to prove a result, similar in nature to Lemma 3.2, regarding the existence of elements of
F
n
which map certain 𝑛-adic rationals to other 𝑛-adic rationals.
Lemma 3.8
Let
p
,
q
∈
Z
[
1
n
]
∩
(
0
,
1
)
such that
(
n
-
1
)
∣
q
-
p
.
There is
f
∈
F
n
such that
f
(
p
)
=
q
.
Proof
Since
(
n
-
1
)
∣
(
1
-
q
)
-
(
1
-
p
)
as well, by Lemma 3.7, let
p
=
∑
i
=
1
m
p
i
,
q
=
∑
i
=
1
m
q
i
,
1
-
p
=
∑
i
=
m
+
1
m
′
p
i
,
1
-
q
=
∑
i
=
m
+
1
m
′
q
i
,
where
p
i
and
q
i
are powers of 𝑛 for
i
∈
{
1
,
…
,
m
′
}
.
For
i
∈
{
1
,
…
,
m
′
}
, let
c
i
=
q
i
p
i
,
t
p
,
i
=
∑
j
=
1
i
p
j
,
t
q
,
i
=
∑
j
=
1
i
q
j
.
Take
f
(
x
)
=
{
c
1
x
if
x
∈
[
0
,
t
p
,
1
]
,
c
2
(
x
-
t
p
,
1
)
+
t
q
,
1
if
x
∈
[
t
p
,
1
,
t
p
,
2
]
,
⋮
⋮
c
i
(
x
-
t
p
,
i
-
1
)
+
t
q
,
i
-
1
if
x
∈
[
t
p
,
i
-
1
,
t
p
,
i
]
,
⋮
⋮
c
m
′
(
x
-
t
p
,
m
′
)
+
t
q
,
m
′
if
x
∈
[
t
p
,
m
′
-
1
,
t
p
,
m
′
)
.
∎
Lemma 3.8 shows that there is some
f
∈
F
n
that maps a given 𝑛-adic rational to another if
n
-
1
divides their difference.
Lemma 3.9 proves the converse:
n
-
1
always divides the difference between an 𝑛-adic rational and its image under an element of
F
n
.
Lemma 3.9
Let
f
∈
F
n
.
For
p
∈
Z
[
1
n
]
∩
[
0
,
1
)
,
(
n
-
1
)
∣
f
(
p
)
-
p
.
Proof
Suppose
a
0
<
a
1
<
⋯
<
a
m
are the points of non-differentiability of 𝑓.
We claim the property holds for
a
0
,
a
1
,
…
,
a
m
.
Observe that
a
0
=
f
(
a
0
)
=
0
and hence
(
n
-
1
)
∣
f
(
a
0
)
-
a
0
.
Suppose that the property holds for
a
j
.
We intend to show that it holds for
a
j
+
1
as well.
Notice that 𝑓 must be linear on
[
a
j
,
a
j
+
1
)
, and so
f
(
a
j
+
1
)
=
f
(
a
j
)
+
n
r
(
a
j
+
1
-
a
j
)
for some
r
∈
Z
.
It follows that
f
(
a
j
+
1
)
-
a
j
+
1
=
(
n
r
-
1
)
(
a
j
+
1
-
a
j
)
+
(
f
(
a
j
)
-
a
j
)
.
Since
(
n
-
1
)
∣
f
(
a
j
)
-
a
j
, by hypothesis, and
(
n
-
1
)
∣
n
r
-
1
, the claim follows.
For
p
∈
Z
[
1
n
]
, suppose
p
∈
[
a
j
,
a
j
+
1
)
.
Then, for some
r
∈
Z
,
f
(
p
)
=
n
r
(
p
-
a
j
)
+
f
(
a
j
)
,
and hence,
f
(
p
)
-
p
=
(
n
r
-
1
)
(
p
-
a
j
)
+
(
f
(
a
j
)
-
a
j
)
.
Since
(
n
-
1
)
∣
f
(
a
j
)
-
a
j
and
(
n
-
1
)
∣
n
r
-
1
, and because
p
-
a
j
∈
Z
[
1
n
]
, the result follows.
∎
We have now got all the tools needed to refine Proposition 2.3.
Proposition 3.10 will allow us to identify what elements of
orb
(
y
)
actually can be the image of 𝑦 under elements of
F
n
, while Proposition 3.11 will assure us that, for elements of
T
n
and
V
n
, every element in
orb
(
x
)
can be such an image.
Proposition 3.10
Let
y
∈
(
0
,
1
)
.
Then
{
f
(
y
)
:
f
∈
F
n
}
=
{
n
r
y
+
(
n
-
1
)
s
n
t
:
r
,
s
,
t
∈
Z
}
∩
(
0
,
1
)
.
Proof
(1) For
f
∈
F
n
, as
Z
[
1
n
]
is dense in
[
0
,
1
)
, there must be some
a
,
b
∈
Z
[
1
n
]
such that
0
<
a
≤
y
<
b
<
1
and 𝑓 is linear on
[
a
,
b
)
, satisfying
f
(
b
)
-
f
(
a
)
b
-
a
=
n
r
for some
r
∈
Z
.
In this case,
f
(
y
)
=
n
r
(
y
-
a
)
+
f
(
a
)
=
n
r
y
-
(
n
r
-
1
)
a
+
(
f
(
a
)
-
a
)
.
By Lemma 3.9,
(
n
-
1
)
∣
f
(
a
)
-
a
, and because
(
n
-
1
)
∣
n
r
-
1
, this implies
f
(
y
)
=
n
r
y
+
(
n
-
1
)
s
n
t
for some
s
,
t
∈
Z
.
Since
f
(
y
)
∈
[
0
,
1
)
, but
f
(
y
)
≠
0
(because
y
≠
0
and 𝑓 is bijective),
f
(
y
)
∈
(
0
,
1
)
as well.
Hence,
{
f
(
y
)
:
f
∈
F
n
}
⊆
{
n
r
y
+
(
n
-
1
)
s
n
t
:
r
,
s
,
t
∈
Z
}
∩
(
0
,
1
)
.
(2) Suppose
y
,
z
∈
(
0
,
1
)
such that, for some
r
,
s
,
t
∈
Z
,
z
=
n
r
y
+
(
n
-
1
)
s
n
t
.
We aim to find
f
∈
F
n
such that
f
(
y
)
=
z
.
Take some
a
,
b
∈
Z
[
1
n
]
such that
0
<
a
<
y
,
0
<
n
r
a
+
(
n
-
1
)
s
n
t
<
z
and
y
<
b
<
1
,
z
<
n
r
b
+
(
n
-
1
)
s
n
t
<
1
,
which is possible since
Z
[
1
n
]
is dense in
[
0
,
1
)
.
Since
(
n
-
1
)
∣
(
n
r
a
+
(
n
-
1
)
s
n
t
)
-
a
and similarly for 𝑏, by Lemma 3.8, there are
f
1
,
f
2
∈
F
n
such that
f
1
(
a
)
=
n
r
a
+
(
n
-
1
)
s
n
t
,
f
2
(
b
)
=
n
r
b
+
(
n
-
1
)
s
n
t
.
Take
f
∈
F
n
such that
f
(
x
)
=
{
f
1
(
x
)
if
x
∈
[
0
,
a
]
,
n
r
x
+
(
n
-
1
)
s
n
t
if
x
∈
[
a
,
b
]
,
f
2
(
x
)
if
x
∈
[
b
,
1
)
,
which fulfils the necessary requirements.
Hence,
{
n
r
y
+
(
n
-
1
)
s
n
t
:
r
,
s
,
t
∈
Z
}
∩
(
0
,
1
)
⊆
{
f
(
y
)
:
f
∈
F
n
}
.
∎
Proposition 3.11
Let
y
∈
[
0
,
1
)
.
Then
{
f
(
y
)
:
f
∈
T
n
}
=
{
f
(
y
)
:
f
∈
V
n
}
=
orb
(
y
)
.
Proof
The inclusion
{
f
(
y
)
:
f
∈
T
n
}
⊆
{
f
(
y
)
:
f
∈
V
n
}
⊆
orb
(
y
)
is immediate since
T
n
⊂
V
n
and by Proposition 2.3.
Given
g
∈
F
n
and
p
∈
Z
[
1
n
]
,
h
:
[
0
,
1
)
→
[
0
,
1
)
such that
h
(
x
)
=
g
(
(
x
-
p
)
mod
1
)
satisfies
h
∈
T
n
since it is merely a horizontal translation of 𝑔 with at most one discontinuity, preserving the remaining relevant properties.
For some
k
∈
{
0
,
…
,
n
-
1
}
, we have
y
∈
[
k
n
,
k
+
1
n
)
, so
Φ
(
y
)
=
n
y
-
k
.
Take
p
=
k
n
∈
Z
[
1
n
]
, in which case,
(
y
-
p
)
mod
1
=
y
-
k
n
.
Proposition 3.10 ensures there is
g
∈
F
n
such that
g
(
y
-
k
n
)
=
n
(
y
-
k
n
)
=
n
y
-
k
=
Φ
(
y
)
,
yielding
h
∈
T
n
satisfying
h
(
y
)
=
Φ
(
y
)
.
Since, for
z
∈
orb
(
y
)
, there are
a
,
b
∈
N
such that
Φ
a
(
y
)
=
Φ
b
(
z
)
, for some
h
1
,
h
2
∈
T
n
,
h
1
a
(
y
)
=
Φ
a
(
y
)
=
Φ
b
(
z
)
=
h
2
b
(
z
)
,
and so
z
=
h
2
-
b
h
1
a
(
y
)
with
h
2
-
b
h
1
a
∈
T
n
and
orb
(
y
)
⊆
{
f
(
y
)
:
f
∈
T
n
}
.
∎
Proposition 3.12 provides a characterisation of
orb
(
y
)
similar to that given by Proposition 3.10 for
{
f
(
y
)
:
f
∈
F
n
}
.
Proposition 3.12
Let
y
∈
[
0
,
1
)
.
Then
orb
(
y
)
=
{
n
r
y
+
s
n
t
:
r
,
s
,
t
∈
Z
}
∩
[
0
,
1
)
.
Proof
Suppose
z
∈
orb
(
y
)
.
For some
a
,
b
∈
N
and
j
,
k
∈
Z
, we have
n
a
y
-
j
=
Φ
a
(
y
)
=
Φ
b
(
z
)
=
n
b
z
-
k
,
and so
z
=
n
a
-
b
y
+
k
-
j
n
b
,
which proves
orb
(
y
)
⊆
{
n
r
y
+
s
n
t
:
r
,
s
,
t
∈
Z
}
∩
[
0
,
1
)
.
On the other hand, for
r
,
s
,
t
∈
Z
, for some
j
,
k
∈
Z
,
Φ
t
(
n
r
y
+
s
n
t
)
=
n
r
+
t
y
-
j
and
Φ
r
+
t
(
y
)
=
n
r
+
t
y
-
k
.
Since
Φ
t
(
n
r
y
+
s
n
t
)
,
Φ
t
+
r
(
y
)
∈
[
0
,
1
)
and
j
,
k
are integers, it follows that
j
=
k
and
Φ
t
(
n
r
y
+
s
n
t
)
=
Φ
t
+
r
(
y
)
showing that
{
n
r
y
+
s
n
t
:
r
,
s
,
t
∈
Z
}
∩
[
0
,
1
)
⊆
orb
(
y
)
.
∎
4 Unitary equivalence
In this section, we will make use of the tools developed in Section 3 to establish necessary and sufficient conditions for the unitary equivalence of representations of the Higman–Thompson groups.
Note that, in Proposition 4.1, if
0
∉
orb
(
y
)
or
z
∉
orb
(
y
)
, the corresponding coefficient
a
z
or
b
z
must be equal to 0, evidently.
Proposition 4.1 is simply asserting that, if any coefficients are different from 0, they must be the coefficients of
δ
0
or
δ
z
.
Proposition 4.1
Let
x
,
y
∈
[
0
,
1
)
, and let
U
:
ℓ
2
(
orb
(
x
)
)
→
ℓ
2
(
orb
(
y
)
)
be a bounded linear operator such that, for every
f
∈
F
n
,
U
τ
x
(
f
)
=
τ
y
(
f
)
U
.
For every
z
∈
orb
(
x
)
, there are
a
z
,
b
z
∈
C
such that
U
δ
z
=
a
z
δ
0
+
b
z
δ
z
.
Proof
Let
z
∈
orb
(
x
)
.
Since
U
δ
z
∈
ℓ
2
(
orb
(
y
)
)
, it can be written as
U
δ
z
=
∑
w
∈
orb
(
y
)
c
w
δ
w
.
Let
k
∈
orb
(
y
)
\
{
0
,
z
}
.
By Proposition 3.3, there is
f
∈
F
n
such that
f
(
z
)
=
z
and
f
(
k
)
≠
k
.
Therefore,
U
τ
x
(
f
)
δ
z
=
U
δ
f
(
z
)
=
U
δ
z
=
∑
w
∈
orb
(
y
)
c
w
δ
w
and also
τ
y
(
f
)
U
δ
z
=
τ
y
(
f
)
∑
w
∈
orb
(
y
)
c
w
δ
w
=
∑
w
∈
orb
(
y
)
c
w
δ
f
(
w
)
.
But since
U
τ
x
(
f
)
δ
z
=
τ
y
(
f
)
U
δ
z
, then
c
f
(
w
)
=
c
w
for every
w
∈
orb
(
y
)
.
Since
f
∈
F
n
, it is a strictly increasing function, and thus
f
(
k
)
≠
k
implies that the set
K
=
{
f
m
(
k
)
:
m
∈
N
0
}
is infinite.
But since
c
w
=
c
f
(
w
)
for every
w
∈
orb
(
y
)
, this inductively implies that, for
j
∈
K
, we have
c
j
=
c
k
.
On the other hand, by Proposition 2.3,
K
⊆
orb
(
y
)
, and since 𝐾 is infinite,
∑
w
∈
orb
(
y
)
|
c
w
|
2
≥
∑
j
∈
K
|
c
j
|
2
=
∑
j
∈
K
|
c
k
|
2
=
{
+
∞
if
c
k
≠
0
,
0
if
c
k
=
0
.
Since
U
δ
z
∈
ℓ
2
(
orb
(
y
)
)
, the sum must be finite, so we have
c
k
=
0
for every
k
∈
orb
(
y
)
\
{
0
,
z
}
.
Therefore,
U
δ
z
=
a
z
δ
0
+
b
z
δ
z
for some
a
z
,
b
z
∈
C
.
∎
Theorem 4.2 now allows us to establish unitary equivalence between the representations of
F
n
.
Theorem 4.2
Let
x
,
y
∈
[
0
,
1
)
.
Then
τ
x
and
τ
y
are unitarily equivalent if and only if
orb
(
x
)
=
orb
(
y
)
.
Proof
If
orb
(
x
)
=
orb
(
y
)
,
τ
x
and
τ
y
are the same representation and hence unitarily equivalent.
To prove the converse, suppose that
τ
x
and
τ
y
are unitarily equivalent.
There is some unitary operator
U
:
ℓ
2
(
orb
(
x
)
)
→
ℓ
2
(
orb
(
y
)
)
such that
U
τ
x
(
f
)
=
τ
y
(
f
)
U
for every
f
∈
F
n
.
This also means
U
-
1
τ
y
(
f
)
=
τ
x
(
f
)
U
-
1
for every
f
∈
F
n
.
By Proposition 4.1, for every
z
∈
orb
(
x
)
,
U
δ
z
=
a
z
δ
0
+
b
z
δ
z
for some
a
z
,
b
z
∈
C
.
Consider two cases.
(1) If
0
∉
orb
(
y
)
, then
a
z
=
0
and
U
δ
z
=
b
z
δ
z
∈
ℓ
2
(
orb
(
y
)
)
.
Since 𝑈 is unitary,
b
z
≠
0
, so
δ
z
∈
ℓ
2
(
orb
(
y
)
)
, which implies
z
∈
orb
(
y
)
, so
orb
(
x
)
=
orb
(
y
)
.
(2) If
0
∈
orb
(
y
)
, then, since
U
-
1
=
U
*
is a bounded linear operator satisfying
U
-
1
τ
y
(
f
)
=
τ
x
(
f
)
U
-
1
for every
f
∈
F
n
, it follows from Proposition 4.1 that
U
-
1
δ
0
=
(
a
0
+
b
0
)
δ
0
∈
ℓ
2
(
orb
(
x
)
)
for some
a
0
,
b
0
∈
C
, and because
U
-
1
is unitary, then
a
0
+
b
0
≠
0
implying
δ
0
∈
ℓ
2
(
orb
(
x
)
)
, and
0
∈
orb
(
x
)
as well as
orb
(
x
)
=
orb
(
y
)
.
∎
Corollary 4.3 is immediate since
σ
x
and
ρ
x
are extensions of
τ
x
.
Corollary 4.3
Let
x
,
y
∈
[
0
,
1
)
.
Then the representations
σ
x
and
σ
y
are unitarily equivalent if and only if
orb
(
x
)
=
orb
(
y
)
;
ρ
x
and
ρ
y
are unitarily equivalent if and only if
orb
(
x
)
=
orb
(
y
)
.
5 Decomposability and irreducibility
In this section, we will settle the questions regarding the decomposability of the representations in question.
Proposition 5.1 will be invaluable for this purpose, providing a necessary and sufficient condition for a unitary representation to be irreducible.
Proposition 5.1
Proposition 5.1 ([6, Lemma 3.5])
A unitary representation 𝜋 of a group 𝐺 on a Hilbert space 𝐻 is irreducible if and only if the commutant of
π
(
G
)
in
B
(
H
)
is
C
1
, where 𝟏 is the identity operator.
Theorem 5.2
Given
x
∈
[
0
,
1
)
, the representations
σ
x
and
ρ
x
are irreducible.
Proof
Suppose
U
∈
B
(
ℓ
2
(
orb
(
x
)
)
)
such that
U
σ
x
(
f
)
=
σ
x
(
f
)
U
for every
f
∈
T
n
.
By Proposition 4.1,
U
δ
y
=
a
y
δ
0
+
b
y
δ
y
for
y
∈
orb
(
x
)
.
If
0
∉
orb
(
x
)
, for any given
y
∈
orb
(
x
)
, by Proposition 3.11, take some
f
∈
T
n
such that
f
(
y
)
=
x
.
Then
b
x
δ
x
=
U
δ
f
(
y
)
=
U
σ
x
(
f
)
δ
y
=
σ
x
(
f
)
U
δ
y
=
σ
x
(
f
)
(
b
y
δ
y
)
=
b
y
δ
f
(
y
)
=
b
y
δ
x
.
Therefore,
b
x
=
b
y
for every
y
∈
orb
(
x
)
, so
U
δ
y
=
b
x
δ
y
for every
y
∈
orb
(
x
)
, meaning 𝑈 is a scalar multiple of the identity.
Similarly, if
0
∈
orb
(
x
)
, for any given
y
∈
orb
(
x
)
, by Proposition 3.11, take
f
∈
T
n
such that
f
(
y
)
=
0
.
Then
b
0
δ
0
=
U
δ
f
(
y
)
=
U
σ
x
(
f
)
δ
y
=
σ
x
(
f
)
U
δ
y
=
σ
x
(
f
)
(
a
y
δ
0
+
b
y
δ
y
)
=
a
y
δ
f
(
0
)
+
b
y
δ
0
,
so
a
y
=
0
,
b
y
=
b
0
for every
y
∈
orb
(
x
)
, and
U
δ
y
=
b
0
δ
y
for every
y
∈
orb
(
x
)
, implying 𝑈 is a scalar multiple of the identity.
Hence, any bounded linear operator which commutes with
σ
x
(
f
)
for every
f
∈
T
n
must be a scalar multiple of the identity, and by Proposition 5.1,
σ
x
is irreducible.
The irreducibility of
ρ
x
follows from it being an extension of
σ
x
.
∎
Theorem 5.2 asserts that the representations
σ
x
and
ρ
x
are irreducible.
However, this will not be the case for
τ
x
.
Note, for instance, that
C
δ
0
is a closed
F
n
-invariant subspace of
ℓ
2
(
orb
(
0
)
)
because
f
(
0
)
=
0
for every
f
∈
F
n
, and so
τ
0
is not irreducible.
Theorem 5.3
Let
y
∈
orb
(
x
)
.
The space of square-summable sequences in the closed linear span of
{
δ
f
(
y
)
:
f
∈
F
n
}
,
Span
¯
{
δ
f
(
y
)
:
f
∈
F
n
}
,
is a closed
F
n
-invariant subspace of
ℓ
2
(
orb
(
x
)
)
with no proper closed non-trivial
F
n
-invariant subspaces.
Proof
That
Span
¯
{
δ
f
(
y
)
:
f
∈
F
n
}
is a closed
F
n
-invariant subspace of
ℓ
2
(
orb
(
x
)
)
is immediate.
That it has no proper closed
F
n
-invariant subspaces follows by an argument similar to Theorem 5.2 and by Proposition 5.1.
∎
Theorem 5.3 provides a way to decompose
τ
x
into irreducible components.
This is aided by Proposition 3.10, which provides a characterisation for the set
{
f
(
y
)
:
f
∈
F
n
}
upon which the relevant subspaces are built.
In general, however, the structure of the sets
{
f
(
y
)
:
f
∈
F
n
}
inside
orb
(
x
)
seems to be quite difficult to characterise in terms of Φ.
Example 5.4
For
F
2
and
x
∈
orb
(
0
)
, as previously noted,
C
δ
0
is a closed
F
2
-invariant subspace of
τ
x
.
As a consequence of Proposition 3.10, for
y
≠
0
, the set
{
f
(
y
)
:
f
∈
F
n
}
contains the value of
Φ
(
y
)
(except if it is equal to 0) and both values of
Φ
-
1
(
y
)
.
Since
Φ
(
0
)
=
0
, the only value in
orb
(
x
)
and not in
{
f
(
y
)
:
f
∈
F
n
}
is precisely 0.
Since
C
δ
0
is
F
2
-invariant, immediately
(
C
δ
0
)
⟂
is also
F
2
-invariant, but there is no guarantee that it is irreducible.
In this case, however, Theorem 5.3 asserts that
Span
¯
{
δ
f
(
y
)
:
f
∈
F
n
}
=
(
C
δ
0
)
⟂
is a closed
F
2
-invariant subspace of
τ
x
which has no proper closed non-trivial invariant subspaces, yielding the decomposition
ℓ
2
(
orb
(
x
)
)
=
C
δ
0
⊕
(
C
δ
0
)
⟂
into closed
F
2
-invariant subspaces of
τ
x
.
On the other hand, if
x
∉
orb
(
0
)
, the set
{
f
(
y
)
:
f
∈
F
n
}
necessarily includes all values of
Φ
(
y
)
and
Φ
-
1
(
y
)
for every
y
∈
orb
(
x
)
, and Theorem 5.3 implies that
ℓ
2
(
orb
(
x
)
)
has no proper closed non-trivial
F
2
-invariant subspaces and
τ
x
is irreducible.
In particular, we have proved the following result.
Corollary 5.5
Let
x
∈
[
0
,
1
)
.
Then the representation
τ
x
of
F
2
is irreducible if and only if
orb
(
x
)
≠
orb
(
0
)
.
So we have only one reducible representation of
F
2
, namely
τ
0
.
The situation changes for
F
n
with
n
>
2
, where we can find many reducible representations among the
τ
x
’s as we see in the following example.
Example 5.6
For
n
>
2
and irrational 𝑥, the equation
n
r
x
+
(
n
-
1
)
s
n
t
=
Φ
(
x
)
has no integer solution
(
r
,
s
,
t
)
unless
Φ
(
x
)
=
n
x
or
Φ
(
x
)
=
n
x
-
(
n
-
1
)
.
Hence, for
x
∈
[
1
n
,
n
-
1
n
)
,
Φ
(
x
)
∉
{
f
(
x
)
:
f
∈
F
n
}
, so
Span
¯
{
δ
f
(
x
)
:
f
∈
F
n
}
and
Span
¯
{
δ
f
(
Φ
(
x
)
)
:
f
∈
F
n
}
are two distinct, closed
F
n
-invariant subspaces of
τ
x
.
We wrap up with Proposition 5.7, which provides a finite family of non-unitarily equivalent irreducible representations of
F
n
+
1
.
Proposition 5.7
Let
n
∈
N
,
n
>
1
.
The following set is a family of pairwise non-unitarily equivalent irreducible representations of
F
n
+
1
:
{
τ
m
n
:
m
n
∈
[
0
,
1
)
and
gcd
(
m
,
n
)
=
1
}
.
Proof
Since
Φ
(
m
n
)
=
(
(
n
+
1
)
m
n
)
mod
1
=
(
m
+
m
n
)
mod
1
=
m
n
,
the orbits in question are pairwise disjoint, and by Theorem 4.2,
the representations are pairwise non-unitarily equivalent.
To show the irreducibility, by Theorem 5.3, it suffices to show that
{
f
(
m
n
)
:
f
∈
F
n
+
1
}
=
orb
(
m
n
)
.
By Proposition 3.12, the elements of
orb
(
m
n
)
are, for some
a
,
b
,
c
∈
Z
, of the form
(
n
+
1
)
a
m
n
+
b
(
n
+
1
)
c
=
(
n
+
1
)
a
+
c
m
+
b
n
n
(
n
+
1
)
c
.
We may suppose that
(
n
+
1
)
a
+
c
m
+
b
n
is an integer, for if this is not the case, simply multiply both the numerator and denominator by an appropriate power of
(
n
+
1
)
and the numerator will still be of the same form.
In this case, notice that
gcd
(
m
,
n
)
=
1
if and only if
gcd
(
(
n
+
1
)
a
+
c
m
+
b
n
,
n
)
=
1
, and so every element in
orb
(
m
n
)
is of the form
p
n
(
n
+
1
)
c
for some
p
,
c
∈
N
0
and
gcd
(
p
,
n
)
=
1
.
For this reason, it suffices to show that
Φ
(
p
n
(
n
+
1
)
c
)
∈
{
f
(
p
n
(
n
+
1
)
c
)
:
f
∈
F
n
+
1
}
,
and the result will follow by inductive reasoning over the structure of
orb
(
m
n
)
.
By Proposition 3.10, this amounts to showing there is an integer solution
(
r
,
s
,
t
)
to the equation
(
n
+
1
)
r
p
n
(
n
+
1
)
c
+
n
s
(
n
+
1
)
t
=
(
n
+
1
)
p
n
(
n
+
1
)
c
-
j
since, for some
j
∈
{
0
,
…
,
n
-
1
}
,
Φ
(
p
n
(
n
+
1
)
c
)
=
(
n
+
1
)
p
n
(
n
+
1
)
c
-
j
.
Let
1
≤
p
-
1
≤
n
-
1
be a multiplicative inverse of 𝑝 modulo 𝑛.
Such an inverse exists since, by hypothesis,
gcd
(
p
,
n
)
=
1
.
Then
{
r
=
1
+
(
n
-
j
)
p
-
1
,
s
=
-
p
(
(
n
+
1
)
1
+
(
n
-
j
)
p
-
1
-
(
n
+
1
)
)
-
j
n
(
n
+
1
)
c
n
2
,
t
=
c
,
is a solution to the equation.
Immediately,
r
,
t
are integers.
We show that 𝑠 is an integer by working modulo
n
2
,
-
p
(
(
n
+
1
)
1
+
(
n
-
j
)
p
-
1
-
(
n
+
1
)
)
-
j
n
(
n
+
1
)
c
≡
-
p
(
∑
q
=
0
1
+
(
n
-
j
)
p
-
1
(
1
+
(
n
-
j
)
p
-
1
q
)
n
q
-
(
n
+
1
)
)
-
j
n
(
∑
q
=
0
c
(
c
q
)
n
q
)
≡
-
p
(
(
1
+
(
n
-
j
)
p
-
1
)
n
+
1
-
n
-
1
)
-
j
n
(
c
n
+
1
)
≡
-
p
p
-
1
n
(
n
-
j
)
-
j
n
≡
p
p
-
1
j
n
-
j
n
≡
(
p
p
-
1
-
1
)
j
n
(
mod
n
2
)
.
Observe that
p
p
-
1
-
1
≡
0
(
mod
n
)
, so
(
p
p
-
1
-
1
)
j
n
≡
0
(
mod
n
2
)
.
Therefore, 𝑠 is an integer.
It is also straightforward to check that
(
r
,
s
,
t
)
is, in fact, a solution to the equation.
∎
