1 Introduction
This paper concerns proving finitely presented groups hyperbolic [7].
In Section 2, we describe a curvature distribution method which very much mirrors the approach taken in [13, 17] and, in particular, [9, Section 5].
This involves for a given van Kampen diagram 𝑲, say, a detailed description of a so-called vertex discharge scheme which will be quite general, and a curvature distribution scheme which, unlike the vertex discharge scheme, is dependent on the group presentation being studied.
The key lemma (Lemma 2.1) states roughly that if there exists
0
<
ε
<
1
for which there is a vertex discharge scheme followed by a curvature distribution scheme for 𝑲 resulting in specific upper bounds on the curvature of the regions of 𝑲, then the number of regions of 𝑲 is bounded above linearly by the number of boundary edges of 𝑲 with constant terms dependent on 𝜀.
Some consequences of the vertex discharge scheme are also given.
In Section 3, we apply the method and results from Section 2 to presentations of finite extensions of the generalised Fibonacci groups.
Recall that the generalised Fibonacci group
F
(
r
,
n
)
is the group defined by the cyclic presentation
⟨
x
1
,
…
,
x
n
∣
x
i
x
i
+
1
…
x
i
+
(
r
−
1
)
=
x
i
+
r
(
1
≤
i
≤
n
)
⟩
,
where
r
>
1
,
n
>
1
and subscripts are taken
mod
n
.
The reader is referred to [18, 6] and references therein for further details on what is known about these groups.
The question of hyperbolicity for
F
(
2
,
n
)
was settled by the first-named author in [2], and in [4], it is shown that if
n
≥
8
, then
F
(
2
,
n
)
is non-elementary hyperbolic and therefore SQ-universal.
Recall that a hyperbolic group is elementary hyperbolic if it is virtually cyclic and non-elementary hyperbolic otherwise.
Our main results for these groups are the following.
Theorem 1.1
If
r
≥
3
and
n
≥
6
r
+
1
, then
F
(
r
,
n
)
is hyperbolic.
Theorem 1.2
The following statements hold.
If
n
∈
{
2
,
3
,
5
,
6
}
, then
F
(
3
,
n
)
is finite.
If
n
∈
{
4
,
8
}
, then
F
(
3
,
n
)
is not hyperbolic.
If
n
∉
{
2
,
3
,
4
,
5
,
6
,
8
}
, then
F
(
3
,
n
)
is non-elementary hyperbolic.
For other recent results on the hyperbolicity of cyclically presented groups, see [11, Section 7], [5] and [4].
In fact, the bulk of Section 3 is given to proving that, under the same conditions on 𝑟 and 𝑛, the group
G
(
r
,
n
)
defined by the presentation
P
(
r
,
n
)
:=
⟨
a
t
∣
t
n
,
t
r
a
r
t
−
1
a
−
1
⟩
is hyperbolic.
The proof of Theorem 1.1 then follows from observing that
F
(
r
,
n
)
has finite index in
G
(
r
,
n
)
.
For
F
(
3
,
n
)
, we show that
F
(
3
,
4
)
and
F
(
3
,
8
)
are not hyperbolic, and the remaining values of 𝑛 not covered by Theorem 1.1 are shown to be hyperbolic either by observing that they are known to be finite or by using KBMAG [8].
We finish by showing how non-elementary hyperbolicity follows from previously known results.
2 Diagrams and method
In this section, we outline a method for showing finitely presented groups hyperbolic.
To this end, let 𝒫 denote the group presentation
P
:=
⟨
X
∣
R
⟩
, where 𝑋 and 𝑅 are finite sets; let
F
(
X
)
denote the free group on 𝑋, so 𝑅 is a finite subset of
F
(
X
)
; and let
N
(
R
)
denote the normal closure of 𝑅 in
F
(
X
)
.
Let
G
(
P
)
denote the group defined by 𝒫.
Then
G
(
P
)
≅
F
(
X
)
/
N
(
R
)
.
Let 𝑲 be a van Kampen diagram over 𝒫 which is planar, connected and simply connected and with cyclically reduced boundary label
w
∈
F
(
X
)
∩
N
(
R
)
.
We seek to study the diagram 𝑲, and the reader is referred to [16, Chapter 5] for the basic theory of van Kampen diagrams.
The region Δ of 𝑲 is called a boundary region if Δ shares an edge with the boundary of 𝑲; otherwise, Δ is called interior.
It is important to note that we allow interior regions to have vertices on the boundary of 𝑲.
We introduce some notation for the diagram 𝑲.
L
=
number of boundary edges
,
F
=
number of regions
,
B
=
number of boundary regions
,
I
=
number of interior regions
,
V
=
number of vertices
,
E
=
number of edges
.
Note then that
V
−
E
+
F
=
1
,
F
=
B
+
I
and
B
≤
L
.
The first step we introduce is as follows: assign the amounts
2
π
to each vertex,
2
π
to each region and
−
2
π
to each edge of the diagram 𝑲.
Make the following definitions:
V
c
,
E
c
,
F
c
:=
total amount contained in the vertices, edges, regions, respectively
.
So then
V
c
+
E
c
+
F
c
=
2
π
V
−
2
π
E
+
2
π
F
=
2
π
.
Let
c
(
v
)
denote the total amount contained by the vertex 𝑣, so, at present,
c
(
v
)
=
2
π
, and let
c
(
Δ
)
denote the total amount contained by the region Δ, so again, at present,
c
(
Δ
)
=
2
π
.
We will refer to
c
(
v
)
,
c
(
Δ
)
as the curvature of the vertex 𝑣, region Δ (respectively).
The second step we introduce is as follows: transfer
−
π
from each edge of 𝑲 to each of its endpoints.
This means that we now have
E
c
=
0
and
V
c
+
F
c
=
2
π
.
Furthermore, the new curvature at a vertex 𝑣 is given by
c
(
v
)
=
(
2
−
d
(
v
)
)
π
≤
0
,
where
d
(
v
)
denotes the degree of the vertex 𝑣.
It can be assumed without any loss that, as well as 𝑤, each
r
∈
R
is also cyclically reduced.
Therefore,
d
(
v
)
≥
2
and, in particular,
V
c
≤
0
.
Assume from now on that the two initial steps described above have been applied to 𝑲.
We require the following notions.
A vertex discharge scheme for 𝑲 is a set of rules each listing transfer of a stated amount of curvature from vertices of 𝑲 to regions of 𝑲 such that if
τ
(
v
)
denotes the total amount of curvature transferred from any given vertex 𝑣, then the following condition is satisfied:
c
(
v
)
≤
τ
(
v
)
≤
0
.
A curvature distribution scheme for 𝑲 is a set of rules each listing transfer of a stated amount of curvature from regions of 𝑲 to regions of 𝑲.
Observe that, since a vertex discharge scheme moves curvature from vertices to regions and a curvature distribution scheme from regions to regions, it follows that, having applied both schemes to 𝑲, we still have
V
c
+
F
c
=
2
π
.
Assume then that a vertex discharge scheme has been applied to 𝑲, and let the resulting curvature of the region Δ be
c
(
Δ
)
.
We define
c
∗
(
Δ
)
to equal the curvature of Δ that results from then applying a curvature distribution scheme.
Thus
c
∗
(
Δ
)
=
c
(
Δ
)
minus all the curvature distributed from
c
(
Δ
)
plus all the curvature distributed to
c
(
Δ
)
according to the list of transfer rules.
We can now state the key lemma.
Lemma 2.1
If there exists
0
<
ε
<
1
such that there is a vertex discharge scheme for 𝑲 followed by a curvature distribution scheme yielding the conditions
c
∗
(
Δ
)
≤
−
ε
π
for each interior region Δ and
c
∗
(
Δ
)
≤
2
π
for each boundary region Δ,
then
F
≤
(
(
2
+
ε
)
/
ε
)
L
+
(
−
2
/
ε
)
.
Proof
As noted, after the two initial steps have been applied,
V
c
≤
0
, and condition (C0) guarantees that this still holds after any vertex discharge scheme, and clearly,
V
c
≤
0
is preserved by any curvature distribution scheme.
Therefore, we have
2
π
≤
2
π
−
V
c
=
F
c
.
Let Δ be a region of 𝑲.
If Δ is interior, then we have
c
∗
(
Δ
)
≤
−
ε
π
by (C1), or if Δ is a boundary region, then
c
∗
(
Δ
)
≤
2
π
by (C2), and so
2
π
≤
F
c
≤
2
π
B
−
ε
π
I
=
2
π
B
−
ε
π
(
F
−
B
)
=
π
B
(
2
+
ε
)
−
ε
π
F
≤
π
L
(
2
+
ε
)
−
ε
π
F
,
and the result follows.
∎
We now define a vertex discharge scheme.
It will be quite general in the sense that it does not depend on the presentation 𝒫 under consideration and so allows for changes, if required, for specific families of presentations.
As we will see, the curvature distribution scheme relies entirely on 𝒫.
We require the following definitions.
Let Δ be an interior region of 𝑲.
Then Δ is an
Int
(
1
)
-region if Δ does not share an edge with a boundary region of 𝑲, and if Δ does share an edge with a boundary region, then it is called an
Int
(
2
)
-region.
This partitions the interior regions of 𝑲, and we will often simply write
Δ
∈
Int
(
1
)
or
Δ
∈
Int
(
2
)
.
A corner of a region Δ is called a boundary, interior,
Int
(
1
)
,
Int
(
2
)
corner if Δ is a boundary, interior,
Int
(
1
)
,
Int
(
2
)
region (respectively), and a corner incident at a boundary vertex is called exterior if it is exterior to 𝑲, that is, it is not contained in any region of 𝑲.
For the vertex 𝑣 of 𝑲, let
β
(
v
)
denote the number of boundary corners incident at 𝑣, and let
ϵ
(
v
)
denote the number of exterior corners incident at 𝑣.
If 𝑣 is an interior vertex, then
β
(
v
)
≥
0
and
ϵ
(
v
)
=
0
, or if 𝑣 is a boundary vertex, then
β
(
v
)
≥
2
and
ϵ
(
v
)
≥
1
.
The proposed vertex discharge scheme is as follows.
At each interior vertex 𝑣 of 𝑲, transfer
c
(
v
)
d
(
v
)
=
(
2
−
d
(
v
)
d
(
v
)
)
π
=
(
−
1
+
2
d
(
v
)
)
π
≤
0
from 𝑣 to each of the
d
(
v
)
−
β
(
v
)
interior corners of 𝑲 incident at 𝑣.
Note that 𝑣 retains
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
not transferred to the boundary corners.
Now let 𝑣 be a boundary vertex.
If
d
(
v
)
≤
3
, then each corner incident at 𝑣 is either a boundary or exterior corner, and no transfer takes place.
If
d
(
v
)
=
4
, then there are no
Int
(
1
)
corners and at most one
Int
(
2
)
corner, and this is illustrated in Figure 1 (i).
If
d
(
v
)
=
5
, then again, there are no
Int
(
1
)
corners and at most two
Int
(
2
)
corners as in Figure 1 (ii).
If
d
(
v
)
=
6
, then there is at most one
Int
(
1
)
corner as shown in Figure 1 (iii), and there can be up to three
Int
(
2
)
corners as shown in Figure 1 (iv).
In general, the maximum number of
Int
(
2
)
corners is 0 if
d
(
v
)
≤
3
and is
d
(
v
)
−
3
for
d
(
v
)
>
3
, and the maximum number of
Int
(
1
)
corners is 0 if
d
(
v
)
≤
5
and is
d
(
v
)
−
5
for
d
(
v
)
>
5
.
Observe that if (D1) had been applied to a boundary vertex 𝑣, then 𝑣 would retain
β
(
v
)
≥
2
amounts of
c
(
v
)
/
d
(
v
)
not transferred to the
β
(
v
)
boundary corners incident at 𝑣 and retain
ϵ
(
v
)
≥
1
amounts of
c
(
v
)
/
d
(
v
)
not transferred to the
ϵ
(
v
)
exterior corners incident at 𝑣.
For these reasons, we distinguish between
Int
(
1
)
and
Int
(
2
)
for boundary vertices.
At each boundary vertex 𝑣 of 𝑲 with
d
(
v
)
≥
7
, transfer
c
(
v
)
d
(
v
)
=
(
−
1
+
2
d
(
v
)
)
π
≤
0
from 𝑣 to each
Int
(
2
)
corner incident at 𝑣, and transfer
c
(
v
)
d
(
v
)
+
(
ϵ
(
v
)
d
(
v
)
−
5
)
(
c
(
v
)
d
(
v
)
)
=
(
−
1
+
(
2
d
(
v
)
−
10
)
−
ϵ
(
v
)
(
d
(
v
)
−
2
)
d
(
v
)
(
d
(
v
)
−
5
)
)
π
≤
(
−
1
+
d
(
v
)
−
8
d
(
v
)
(
d
(
v
)
−
5
)
)
π
from 𝑣 to each
Int
(
1
)
corner incident at 𝑣.
Since
d
(
v
)
−
5
is the maximum number of
Int
(
1
)
corners incident at 𝑣, the number of times
c
(
v
)
/
d
(
v
)
is transferred from 𝑣 is at most
|
Int
(
1
)
|
+
|
Int
(
2
)
|
+
ϵ
(
v
)
=
d
(
v
)
−
β
(
v
)
, and so at least
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
are retained by 𝑣.
Let 𝑣 be a boundary vertex of degree 6.
Then there are at most three interior corners which will include at most one
Int
(
1
)
corner incident at 𝑣.
Indeed, if there is an
Int
(
1
)
corner, then
ϵ
(
v
)
=
1
as in Figure 1 (iii); if there are three
Int
(
2
)
corners, then
ϵ
(
v
)
=
1
as in Figure 1 (iv); if there are two
Int
(
2
)
corners, then
ϵ
(
v
)
≤
2
, and the case
ϵ
(
v
)
=
2
is shown in Figure 1 (v); or if there is exactly one
Int
(
2
)
corner, then
ϵ
(
v
)
≤
3
, and the case
ϵ
(
v
)
=
3
is shown in Figure 1 (vi).
Again, we use (not necessarily all of) the
ϵ
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
retained by 𝑣, this time for both
Int
(
1
)
and
Int
(
2
)
corners.
At each boundary vertex 𝑣 of 𝑲 with
d
(
v
)
=
6
, transfer
c
(
v
)
d
(
v
)
+
1
2
(
c
(
v
)
d
(
v
)
)
=
−
π
from 𝑣 to the
Int
(
1
)
corner incident at 𝑣 (if there is one), and transfer
c
(
v
)
d
(
v
)
+
ϵ
(
v
)
4
(
c
(
v
)
d
(
v
)
)
=
(
−
1
+
8
−
4
ϵ
(
v
)
24
)
π
≤
(
−
1
+
1
6
)
π
from 𝑣 to each
Int
(
2
)
corner incident at 𝑣.
Inspection shows that 𝑣 retains at least
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
, and if exactly
β
(
v
)
amounts, then 𝑣 is given by Figure 1 (iii) in which
β
(
v
)
.
c
(
v
)
/
d
(
v
)
=
−
4
π
/
3
.
Let 𝑣 be a boundary vertex of degree 5.
Then there are no
Int
(
1
)
corners incident at 𝑣 and at most two
Int
(
2
)
corners.
Or if
d
(
v
)
=
4
, then there are no
Int
(
1
)
corners and at
most one
Int
(
2
)
corner.
At each boundary vertex 𝑣 of 𝑲 with
d
(
v
)
=
5
, transfer
c
(
v
)
d
(
v
)
+
ϵ
(
v
)
2
(
c
(
v
)
d
(
v
)
)
=
(
−
1
+
4
−
3
ϵ
(
v
)
10
)
π
≤
(
−
1
+
1
10
)
π
from 𝑣 to each
Int
(
2
)
corner incident at 𝑣.
Inspection shows that 𝑣 retains at least
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
, and if exactly
β
(
v
)
amounts, then 𝑣 is given by Figure 1 (ii) in which
β
(
v
)
.
c
(
v
)
/
d
(
v
)
=
−
6
π
/
5
.
At each boundary vertex 𝑣 of 𝑲 with
d
(
v
)
=
4
, transfer
c
(
v
)
d
(
v
)
+
c
(
v
)
d
(
v
)
=
−
π
from 𝑣 to the
Int
(
2
)
corner incident at 𝑣 (if there is one), and again, it is straightforward to see that 𝑣 retains at least
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
.
There is one more discharge rule, but before that, we will need a further definition.
Consider the 2-segment of Figure 2 with endpoints 𝑢 and 𝑣, where, for ease of presentation, we have not included any edge labels.
Thus
d
(
u
)
>
2
,
d
(
v
)
>
2
and the degree of each intermediate vertex equals 2.
Then 𝑣 is called a 𝜌-vertex of the region Δ of Figure 2 if Δ is an
Int
(
2
)
-region and
Δ
^
is a boundary region.
Note that 𝑣 may be a boundary vertex.
Thus an interior region of 𝑲 is an
Int
(
2
)
-region precisely when it has at least one 𝜌-vertex.
Now note that, having applied (D1)–(D5), the vertex 𝑣 in Figure 2 each time retains at least
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
.
If Δ is an
Int
(
2
)
-region and 𝑣 is a 𝜌-vertex of Δ, then transfer
c
(
v
)
d
(
v
)
≤
−
π
3
(
v
interior
)
or
c
(
v
)
d
(
v
)
≤
−
π
2
(
v
boundary
)
from 𝑣 to Δ.
Lemma 2.2
The list of rules (D1)–(D6) is a vertex discharge scheme.
Proof
We need to show that condition (C0) is satisfied.
It is clear that each transfer from a given vertex 𝑣 to any corner incident at 𝑣 involves a negative amount of curvature, and so
τ
(
v
)
≤
0
.
Now observe that, for (D1)–(D5), the vertex 𝑣 retains at least
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
, and it follows that
(
d
(
v
)
−
β
(
v
)
)
.
c
(
v
)
/
d
(
v
)
≤
τ
(
v
)
.
But then a total of at most
β
(
v
)
amounts of
c
(
v
)
/
d
(
v
)
is transferred out of 𝑣 by (D6), and so
c
(
v
)
=
(
d
(
v
)
−
β
(
v
)
)
.
c
(
v
)
/
d
(
v
)
+
β
(
v
)
)
.
c
(
v
)
/
d
(
v
)
≤
τ
(
v
)
as required.
∎
We end this section with some further observations that will be useful later.
Remark 2.1
Observe that no curvature has been transferred to any boundary region Δ in (D1)–(D6), and so
c
(
Δ
)
=
2
π
still holds.
Let Δ be an interior region of 𝑲 and let 𝑣 be a vertex of Δ.
We define the contribution of 𝑣 to
c
(
Δ
)
at a given corner 𝑐 of Δ incident at 𝑣 to be 𝜅, where
−
π
+
κ
equals the amount transferred from 𝑣 to the corner 𝑐 according to (D1)–(D5), and it is denoted by
cont
(
v
,
Δ
)
.
Lemma 2.3
Let 𝑣 be a vertex of the region Δ.
If 𝑣 is an interior vertex, then
cont
(
v
,
Δ
)
=
2
π
/
d
(
v
)
.
If 𝑣 is a boundary vertex and Δ is an
Int
(
1
)
-region, then either
d
(
v
)
=
6
and
cont
(
v
,
Δ
)
=
0
or
d
(
v
)
≥
7
, in which case
cont
(
v
,
Δ
)
=
(
(
2
d
(
v
)
−
10
)
−
ϵ
(
v
)
(
d
(
v
)
−
2
)
d
(
v
)
(
d
(
v
)
−
5
)
)
π
≤
d
(
v
)
−
8
d
(
v
)
(
d
(
v
)
−
5
)
π
.
If 𝑣 is a boundary vertex and Δ an
Int
(
2
)
-region, then either
d
(
v
)
=
4
and
cont
(
v
,
Δ
)
=
0
,
d
(
v
)
=
5
and
cont
(
v
,
Δ
)
=
4
−
3
ϵ
(
v
)
10
π
≤
π
10
,
d
(
v
)
=
6
and
cont
(
v
,
Δ
)
=
8
−
4
ϵ
(
v
)
24
π
≤
π
6
,
or
d
(
v
)
≥
7
and
cont
(
v
,
Δ
)
=
2
π
d
(
v
)
.
If 𝑣 is a boundary vertex and Δ an
Int
(
1
)
-region, then
cont
(
v
,
Δ
)
≤
5
π
/
104
, or if Δ is an
Int
(
2
)
-region, then
cont
(
v
,
Δ
)
≤
2
π
/
7
.
Proof
Statements (i)–(iii) are immediate from the definitions of rules (D1)–(D5).
For (iv), the maximum value of
(
d
(
v
)
−
8
)
/
(
d
(
v
)
(
d
(
v
)
−
5
)
for
d
(
v
)
≥
7
is the value
5
π
/
104
obtained when
d
(
v
)
=
13
, and the maximum
2
π
/
7
is clear from inspecting statement (iii).
∎
Let Δ be an interior region of 𝑲 and let
d
(
Δ
)
=
k
.
Let
v
j
(
1
≤
j
≤
k
) be the 𝑘 vertices of Δ such that
d
(
v
j
)
>
2
.
Assume further that each
v
j
is an interior vertex of 𝑲. If
Δ
∈
Int
(
1
)
, then the new total curvature
c
(
Δ
)
of Δ following (D1)–(D5) satisfies
c
(
Δ
)
=
2
π
+
∑
j
=
1
k
c
(
v
j
)
d
(
v
j
)
=
2
π
+
∑
j
=
1
k
2
−
d
(
v
j
)
d
(
v
j
)
π
=
(
2
−
k
)
π
+
∑
j
=
1
k
2
π
d
(
v
j
)
=
(
2
−
k
)
π
+
∑
j
=
1
k
cont
(
v
j
,
Δ
)
.
If
Δ
∈
Int
(
2
)
, then the new total curvature
c
(
Δ
)
of Δ following (D1)–(D5) and then taking into account (D6) satisfies
c
(
Δ
)
≤
(
2
−
k
)
π
+
∑
j
=
1
k
cont
(
v
j
,
Δ
)
−
π
3
.
3 Generalised Fibonacci groups
In this section, we first consider the group presentation
P
(
r
,
n
)
:=
⟨
a
,
t
∣
t
n
,
t
r
a
r
t
−
1
a
−
1
⟩
,
where
r
≥
3
and
n
≥
6
r
+
1
, and we consider the groups
G
(
r
,
n
)
they define.
Let 𝑲 be a reduced and connected van Kampen diagram over
P
(
r
,
n
)
and assume that the vertex discharge scheme of Section 2 has been applied to 𝑲.
Then the regions Δ of 𝑲 are given (up to cyclic permutation and inversion) by Figure 3 (i) and (ii).
The region Δ of Figure 3 (ii) will sometimes be referred to as a 𝑡-region.
Much use will be made of the fact that, in Figure 3 (i), there are at least three 𝑡-edges between vertices
v
1
and
v
4
, and at least three 𝑎-edges between vertices
v
4
and
v
3
.
Note that 𝑲 being reduced implies that no pair of 𝑡-regions share an edge.
It follows from this and
n
≥
6
r
+
1
that if Δ of Figure 3 (ii) is an interior region, then
d
(
Δ
)
≥
7
.
So if Δ is an interior 𝑡-region, then
c
(
Δ
)
≤
(
2
−
7
)
π
+
14
π
3
=
−
π
3
.
Moreover, if Δ of Figure 3 (i) is interior, then, since none of
a
t
±
1
,
a
−
1
t
±
1
,
t
±
1
a
,
t
±
1
a
−
1
are pieces, it follows that
d
(
v
i
)
>
2
(
1
≤
i
≤
4
)
, and since
a
±
r
is not a piece, the 𝑎-segment between vertices
v
4
and
v
3
must split.
All of this implies that
d
(
Δ
)
≥
5
.
Lemma 3.1
If Δ is an interior region of 𝑲, then either
c
(
Δ
)
≤
−
π
/
6
or
c
(
Δ
)
=
0
in which case Δ shares an edge with at least one interior 𝑡-region.
Proof
If Δ is a 𝑡-region, then
c
(
Δ
)
≤
−
π
/
3
, so assume otherwise.
Suppose that Δ has at least one boundary vertex.
Then the two cases are
Δ
∈
Int
(
1
)
:
c
(
Δ
)
≤
(
2
−
5
)
π
+
8
π
3
+
5
π
104
=
−
89
π
312
,
Δ
∈
Int
(
2
)
:
c
(
Δ
)
≤
(
2
−
5
)
π
+
8
π
3
+
2
π
7
−
π
3
=
−
8
π
21
by Lemma 2.3 (iv).
Assume that Δ does not contain any boundary vertices.
If
d
(
Δ
)
≥
7
, then we have
c
(
Δ
)
≤
−
π
/
3
, so this leaves
5
≤
d
(
Δ
)
≤
6
to be considered.
Suppose first that the 𝑡-segment between vertices
v
1
and
v
4
splits forcing
d
(
Δ
)
=
6
.
We claim that Δ contains at least one vertex of degree at least 4.
Suppose that
d
(
u
3
)
=
2
and
d
(
v
4
)
=
3
.
The fact that
t
2
is followed (in the relator up to cyclic permutation and inversion) by 𝑡 or 𝑎 forces the vertex
v
4
to be as shown in Figure 3 (iii).
Since
t
−
1
a
2
is not a subword, this forces
d
(
u
2
)
>
2
, and since
t
−
1
a
is followed by 𝑡, if
d
(
u
2
)
=
3
, then
u
2
is as shown in Figure 3 (iii).
But
r
≥
3
, and
d
(
Δ
)
=
6
implies that there are at least two edges in the 2-segment between vertices
u
2
and
v
3
, and so we obtain the subword
t
−
1
a
2
, a contradiction.
Now suppose that
d
(
u
3
)
=
d
(
v
4
)
=
3
.
Since
t
2
is followed by 𝑡 or 𝑎, it follows that the vertex
u
3
is given by Figure 3 (iv).
Since
a
−
1
t
is followed by 𝑡, the vertex
v
4
is given by Figure 3 (iv), and as shown above, this forces
d
(
u
2
)
>
2
.
But, again as before,
d
(
u
2
)
=
3
yields a contradiction.
This proves the claim, and so
c
(
Δ
)
≤
(
2
−
6
)
π
+
10
π
3
+
2
π
4
=
−
π
6
.
Now suppose that the 𝑡-segment between vertices
v
1
and
v
4
does not split and Δ is given by Figure 3 (v) in which
Δ
t
is a 𝑡-region.
Suppose that
d
(
Δ
)
=
6
.
If Δ is an
Int
(
2
)
region, then
c
(
Δ
)
≤
(
2
−
6
)
π
+
12
π
/
3
−
π
/
3
=
−
π
/
3
, or if Δ is an
Int
(
1
)
region, then
c
(
Δ
)
≤
0
, but
Δ
t
is then interior.
Suppose that
d
(
Δ
)
=
5
.
If
d
(
v
4
)
=
3
, then the labelling of the edges incident at
v
4
is given by Figure 3 (iv), and as before, this forces
d
(
u
2
)
>
3
.
Suppose that
d
(
v
j
)
=
3
(
1
≤
j
≤
3
).
Then the labelling of
v
j
(
1
≤
j
≤
3
) in Figure 3 (vi) is forced.
Since
a
2
t
is not a subword, we must have
d
(
u
1
)
>
3
again with labelling as shown in Figure 3 (vi).
If Δ is an
Int
(
2
)
region, then
c
(
Δ
)
≤
(
2
−
5
)
π
+
6
π
/
3
+
4
π
/
4
−
π
/
3
=
−
π
/
3
, or if Δ is an
Int
(
1
)
region, then
c
(
Δ
)
≤
(
2
−
5
)
π
+
6
π
/
3
+
4
π
/
4
=
0
, but the region
Δ
t
of Figure 3 (vi) is an interior 𝑡-region, and this completes the proof.
∎
If Δ is an interior region and
c
(
Δ
)
=
0
, then, by Lemma 3.1, Δ shares an edge with at least one 𝑡-region
Δ
t
, and we will say that Δ is associated with each such
Δ
t
.
Given this, the curvature distribution scheme is as follows.
Transfer
c
(
Δ
t
)
k
+
1
from each interior 𝑡-region
Δ
t
having at least one associated region to each of its
k
≥
1
associated regions.
Recall that
c
∗
(
Δ
)
equals
c
(
Δ
)
minus all the curvature transferred from Δ plus all the curvature transferred to Δ according to the distribution scheme given immediately above.
Thus if Δ is a boundary region, then
c
∗
(
Δ
)
=
c
(
Δ
)
≤
2
π
and condition (C2) of Lemma 2.1 holds, so assume that Δ is an interior region.
If
c
(
Δ
)
<
0
and either Δ is not a 𝑡-region or Δ is a 𝑡-region but having no associated regions, then
c
∗
(
Δ
)
=
c
(
Δ
)
≤
−
π
/
6
.
This leaves the case when either
c
(
Δ
)
=
0
or
Δ
=
Δ
t
is a 𝑡 region having
k
≥
1
associated regions.
Observe that
c
(
Δ
t
)
k
+
1
≤
c
(
Δ
t
)
d
(
Δ
t
)
+
1
≤
(
2
−
d
(
Δ
t
)
)
π
3
d
(
Δ
t
)
+
3
≤
−
π
24
since
d
(
Δ
t
)
≥
7
.
It follows that condition (C1) of Lemma 2.1 holds with
ε
=
π
/
24
.
Therefore,
F
(
K
)
≤
49
L
(
K
)
−
48
.
Since 𝑲 was arbitrary, we have proved (see, for example, [7]) the following theorem.
Theorem 3.2
The group
G
(
r
,
n
)
is hyperbolic for
r
≥
3
and
n
≥
6
r
+
1
.
Proof of Theorem 1.1
The group
F
(
r
,
n
)
admits an automorphism which permutes the generators cyclically
x
i
→
x
i
+
1
(subscripts
mod
n
).
Forming the resulting semi-direct product
E
(
r
,
n
)
of
F
(
r
,
n
)
with the cyclic group of order 𝑛 acting in this way, we obtain (see, for example, [14, Section 10.2])
E
(
r
,
n
)
=
⟨
x
1
,
…
,
x
n
,
t
∣
t
n
=
1
,
x
i
x
i
+
1
…
x
i
+
(
r
−
1
)
=
x
i
+
r
,
t
−
1
x
i
t
=
x
i
+
1
(
1
≤
i
≤
n
)
⟩
,
where the subscripts are taken
mod
n
.
Eliminating the generators
x
2
,
…
,
x
n
and replacing
x
1
with 𝑥 yields
E
(
r
,
n
)
=
⟨
x
,
t
∣
t
n
,
(
x
t
−
1
)
r
x
−
1
t
r
⟩
=
⟨
x
,
t
,
a
∣
t
n
,
(
x
t
−
1
)
r
x
−
1
t
r
,
a
−
1
x
t
−
1
⟩
=
⟨
a
,
t
∣
t
n
,
t
r
a
r
t
−
1
a
−
1
⟩
.
Therefore,
E
(
r
,
n
)
≅
G
(
r
,
n
)
, and since
F
(
r
,
n
)
has index 𝑛 in
E
(
r
,
n
)
, we have
F
(
r
,
n
)
is hyperbolic if and only if
G
(
r
,
n
)
is hyperbolic, and the result follows.
∎
We turn now to the proof of Theorem 1.2 for which the following lemmas are required.
Lemma 3.3
The group
F
(
3
,
8
)
=
⟨
a
,
b
,
c
,
d
,
e
,
f
,
g
,
h
∣
a
b
c
=
d
,
b
c
d
=
e
,
c
d
e
=
f
,
d
e
f
=
g
,
e
f
g
=
h
,
f
g
h
=
a
,
g
h
a
=
b
,
h
a
b
=
c
⟩
is not hyperbolic.
Proof
Using KBMAG, it can be shown that
F
(
3
,
8
)
is shortlex automatic and that the generator 𝑎 has infinite order.
(The reader is referred to [10, Section 5], for example, for definitions of the terms used in this proof.)
The first author has written an extension [3] to KBMAG which allows the calculation of word differences [10, page 127] of various geodesic equations to be traced.
We specifically look for word differences in this trace which are powers of the generator 𝑎.
This yielded the geodesic equation
U
1
=
V
1
, where
U
1
=
h
−
1
e
−
1
(
f
−
1
a
−
1
b
−
1
e
−
1
)
2
f
−
1
a
−
1
(
f
e
b
a
)
f
e
b
,
V
1
=
g
−
1
f
−
1
b
(
a
b
−
1
d
−
1
c
−
1
)
2
a
(
b
−
1
e
−
1
f
−
1
a
−
1
)
b
−
1
e
−
1
f
−
1
,
and moreover that the 12th word difference
U
1
(
12
)
−
1
V
1
(
12
)
, that is,
(
h
−
1
e
−
1
(
f
−
1
a
−
1
b
−
1
e
−
1
)
2
f
−
1
)
−
1
g
−
1
f
−
1
b
(
a
b
−
1
d
−
1
c
−
1
)
2
a
reduces to the power
a
14
.
This has led us to consider the words
U
m
=
h
−
1
e
−
1
(
f
−
1
a
−
1
b
−
1
e
−
1
)
2
m
f
−
1
a
−
1
(
f
e
b
a
)
m
f
e
b
,
V
m
=
g
−
1
f
−
1
b
(
a
b
−
1
d
−
1
c
−
1
)
2
m
a
(
b
−
1
e
−
1
f
−
1
a
−
1
)
m
b
−
1
e
−
1
f
−
1
.
The second word
V
m
is accepted as a shortlex reduced word by the word acceptor,
F
(
3
,
8
)
.
w
a
, calculated by KBMAG, and so, in particular,
V
m
is geodesic.
The KBMAG binary wordreduce establishes the following identities in
F
(
3
,
8
)
:
e
h
g
−
1
f
−
1
b
=
b
a
2
;
e
b
a
f
b
=
b
f
e
b
a
;
f
e
b
a
2
=
a
2
f
e
b
;
f
e
d
−
1
c
−
1
=
a
2
;
Using these identities shows that
U
m
(
8
m
+
3
)
−
1
V
m
(
8
m
+
3
)
=
f
(
e
b
a
f
)
2
m
e
h
g
−
1
f
−
1
b
(
a
b
−
1
d
−
1
c
−
1
)
2
m
=
f
(
e
b
a
f
)
2
m
−
1
e
b
a
f
b
a
2
a
b
−
1
d
−
1
c
−
1
(
a
b
−
1
d
−
1
c
−
1
)
2
m
−
1
(
1
)
=
f
(
e
b
a
f
)
2
m
−
1
b
f
e
b
a
a
2
a
b
−
1
d
−
1
c
−
1
(
a
b
−
1
d
−
1
c
−
1
)
2
m
−
1
(
2
)
=
f
(
e
b
a
f
)
2
m
−
1
b
a
4
f
e
b
b
−
1
d
−
1
c
−
1
(
a
b
−
1
d
−
1
c
−
1
)
2
m
−
1
(
3
)
=
f
(
e
b
a
f
)
2
m
−
1
b
a
6
(
a
b
−
1
d
−
1
c
−
1
)
2
m
−
1
(
4
)
=
f
b
a
8
m
+
2
(2), (3), (4)
2
m
−
1
times
=
a
4
(
2
m
+
1
)
(
5
)
and, using
U
m
(
−
k
)
to denote the last 𝑘 letters of
U
m
,
U
m
(
−
(
4
m
+
3
)
)
V
m
(
−
(
4
m
+
3
)
)
−
1
=
(
f
e
b
a
)
m
f
e
b
f
e
b
(
a
f
e
b
)
m
=
(
f
e
b
a
)
m
−
1
f
e
b
a
a
6
a
f
e
b
(
a
f
e
b
)
m
−
1
(
6
)
=
(
f
e
b
a
)
m
−
1
a
8
f
e
b
f
e
b
(
a
f
e
b
)
m
−
1
(
3
)
=
a
4
(
2
m
+
1
)
+
2
(6), (3)
m
−
1
times
These two expressions for
a
4
(
2
m
+
1
)
establish that
U
m
(
8
m
+
3
)
a
−
1
U
m
(
−
(
8
m
+
3
)
)
=
V
m
(
8
m
+
3
)
a
V
m
(
−
(
8
m
+
3
)
)
,
and so
h
−
1
e
−
1
(
f
−
1
a
−
1
b
−
1
e
−
1
)
2
m
f
−
1
a
−
1
(
f
e
b
a
)
m
f
e
b
=
g
−
1
f
−
1
b
(
a
b
−
1
d
−
1
c
−
1
)
2
m
a
(
b
−
1
e
−
1
f
−
1
a
−
1
)
m
b
−
1
e
−
1
f
−
1
is an equation between geodesic words whose
4
(
2
m
+
1
)
th word difference is
a
U
m
(
8
m
+
3
)
−
1
V
m
(
8
m
+
3
)
a
=
a
4
(
2
m
+
1
)
+
2
=
a
8
m
+
6
.
Since 𝑚 is arbitrary, the result follows from [10, Proposition 5.8.3 and Theorem 6.3.2].
Alternatively, observe that we have constructed a family of arbitrarily non-slim geodesic bigons (which can be viewed as geodesic triangles with two equal vertices), and this proves non-hyperbolicity.
∎
Lemma 3.4
The group
E
(
3
,
4
)
=
⟨
a
,
b
∣
a
4
=
1
,
a
b
a
=
b
3
⟩
is not hyperbolic.
Proof
The result follows from the following five claims:
(
b
a
−
1
)
n
(
a
−
1
b
)
n
=
(
b
a
2
b
)
n
(
b
4
)
x
n
, where
x
1
=
0
,
x
2
=
1
and
x
n
=
(
n
−
1
)
+
x
n
−
1
(
n
≥
3
)
.
(
b
a
2
b
)
n
(
a
b
−
1
)
n
=
(
a
−
1
b
)
n
(
b
4
)
y
n
, where
y
1
=
1
,
y
2
=
3
and
y
n
=
1
+
2
y
n
−
1
−
y
n
−
2
(
n
≥
3
)
.
(
b
−
1
a
)
n
(
b
a
2
b
)
n
=
(
b
a
−
1
)
n
(
b
4
)
y
n
, where
y
1
=
1
,
y
2
=
3
and
y
n
=
1
+
2
y
n
−
1
−
y
n
−
2
(
n
≥
3
)
.
w
=
(
a
b
−
1
)
n
(
b
−
1
a
)
2
n
(
a
b
−
1
)
n
is a geodesic (
n
≥
1
).
The generator 𝑏 has infinite order.
To see this, first observe that
x
n
=
n
(
n
−
1
)
/
2
and that
y
n
=
n
(
n
+
1
)
/
2
.
Now combining (C1) and (C2) yields
(
b
a
−
1
)
n
(
a
−
1
b
)
n
(
a
b
−
1
)
n
(
b
−
1
a
)
n
=
(
b
4
)
x
n
(
b
4
)
y
n
=
(
b
4
)
n
2
,
and combining (C1) and (C3) yields
(
a
b
−
1
)
n
(
b
−
1
a
)
n
(
b
a
−
1
)
n
(
a
−
1
b
)
n
=
(
b
4
)
n
2
.
These two expressions for
(
b
4
)
n
2
yield the equation
(
a
b
−
1
)
n
(
b
−
1
a
)
2
n
(
a
b
−
1
)
n
=
(
b
−
1
a
)
n
(
a
b
−
1
)
2
n
(
b
−
1
a
)
n
whose
2
n
th word difference is
(
b
4
)
n
2
, and again, the result follows from [10, Proposition 5.8.3 and Theorem 6.3.2] together with (C4) and (C5) or once more observing that a family of arbitrarily non-slim geodesic bigons has been constructed.
It remains to prove claims (C1)–(C5).
To prove (C1), consider the rewrite rules
b
a
b
→
a
−
1
b
4
;
w
1
b
4
w
2
→
w
1
w
2
b
4
.
Using these, the following identities are easy to verify:
(
n
=
1
)
(
b
a
−
1
)
(
a
−
1
b
)
=
b
a
2
b
(
x
1
=
0
)
,
(
n
=
2
)
(
b
a
−
1
)
2
(
a
−
1
b
)
2
=
(
b
a
2
b
)
b
4
(
x
2
=
1
)
.
Assume
n
>
2
and by way of induction that (C1) is true for all values less than 𝑛.
Then
(
b
a
−
1
)
n
(
a
−
1
b
)
n
=
(
b
a
−
1
)
(
b
a
−
1
)
n
−
1
(
a
−
1
b
)
n
−
1
(
a
−
1
b
)
=
(
b
a
−
1
)
(
b
a
2
b
)
n
−
1
(
a
−
1
b
)
(
b
4
)
x
n
−
1
induction and (R5)
=
b
a
3
b
a
2
b
(
b
a
2
b
)
n
−
2
(
a
−
1
b
)
(
b
4
)
x
n
−
1
(R1)
=
b
a
2
b
3
a
b
(
b
a
2
b
)
n
−
2
(
a
−
1
b
)
(
b
4
)
x
n
−
1
(R2)
=
(
b
a
2
b
)
(
b
a
−
1
)
(
b
a
2
b
)
n
−
2
(
a
−
1
b
)
(
b
4
)
x
n
−
1
+
1
(R3), (R5)
=
(
b
a
2
b
)
n
−
2
(
b
a
−
1
)
(
b
a
2
b
)
(
a
−
1
b
)
(
b
4
)
x
n
−
1
+
(
n
−
2
)
(R1), (R2), (R3), (R5)
repeated
n
−
3
times
=
(
b
a
2
b
)
n
−
2
(
b
a
−
1
)
2
(
a
−
1
b
)
2
(
b
4
)
x
n
−
1
+
(
n
−
2
)
inverse of (R4)
=
(
b
a
2
b
)
n
−
2
(
b
a
2
b
)
2
(
b
4
)
x
n
−
1
+
(
n
−
1
)
induction and (R5)
.
The proof for both (C2) and (C3) is similar.
It is straightforward to establish the base cases
n
=
1
and
n
=
2
, and for
n
>
2
again, we apply a sequence of the rewrite rules and induction to the left-hand side of the equation to obtain the right-hand side.
In fact, the sequence is the same in each case: induction, (R5), (R1), (R2), (R3), induction, (R5), induction.
We omit the details.
For (C4), consider the image of 𝑤 under the natural map from
E
(
3
,
4
)
onto
D
(
4
,
4
,
2
)
=
⟨
a
,
b
∣
a
4
,
b
4
,
(
a
b
)
2
⟩
.
Now the Cayley graph of
D
(
4
,
4
,
2
)
is an infinite grid on the Euclidean plane consisting of directed square loops whose boundaries are labelled by the relators.
Part of this grid is shown (without vertex labels) in Figure 4 (i).
It is straightforward to check that 𝑤 yields a geodesic word (indeed, the start and end points of the edge path traced out by 𝑤 are the opposite corners of a
4
n
by
4
n
square) and therefore is a geodesic in
E
(
3
,
4
)
.
For (C5), we build a graph Γ as follows.
First glue together two loops each having label
a
b
a
b
−
3
as shown in Figure 4 (ii) and then glue together four copies of the resulting directed graph as indicated in Figure 4 (iii), in which the 𝑏-segments
(
u
1
,
u
2
)
and
(
v
1
,
v
2
)
are identified, to form a four-sided square vertical column of length 4 having the label
a
4
, read in an anti-clockwise direction, at the top and bottom.
Gluing such columns along the label
a
4
yields a four-side vertical column of infinite length.
We call the top and bottom edges in Figure 4 (ii) horizontal𝑎-edges, and the middle edge labelled by 𝑎 is a diagonal𝑎-edge.
We can glue together a pair of infinite columns so that they then share a common side, by shifting vertically the 𝑏-edges on one of the sides in such a way that the slopes and lengths of the horizontal, diagonal (respectively) 𝑎-edges of the first column’s side then match the slopes and lengths of the diagonal, horizontal (respectively) 𝑎-edges of the second column’s side.
The vertical shifting that cause this matching between the diagonal and horizontal 𝑎-edges of the two sides still leaves the 𝑏-edges of the two sides coincident with each other, and so we can identify the 𝑏-edges as well as the matching 𝑎-edges of the two column sides to complete the gluing together of the two columns.
We can continue gluing columns together in this way so that, eventually, every column is joined by its four sides to four other columns and every vertical 𝑏-edge axis is shared by four column sides.
In this way, an infinite directed grid Γ is formed that fills out Euclidean 3-space.
In Figure 5, we illustrate the four column sides that are arranged
N
,
E
,
S
,
W
clockwise around an infinite 𝑏-edge axis in Γ, where each set of four vertices
(
i
,
j
)
are identified for
i
∈
Z
and
0
≤
j
≤
3
.
This completes the construction, and we can make the following observations: each vertex in Γ has degree 4 with an incoming 𝑎-edge and 𝑏-edge, and an outgoing 𝑎-edge and 𝑏-edge; each 𝑎-edge is shared by three loops whose labels, starting from the 𝑎-edge, are
a
b
a
b
−
3
,
a
b
−
3
a
b
and
a
4
; each 𝑏-edge is shared by four loops whose labels, starting from the 𝑏-edge, are
b
a
b
−
3
a
,
b
−
3
a
b
a
,
b
−
2
a
b
a
b
−
1
and
b
−
1
a
b
a
b
−
2
; and Γ is reduced in the sense that there is no pair of loops sharing at least one edge and having inverse labels.
It follows that the boundary label of any reduced van Kampen diagram over the given presentation traces out a closed path in Γ.
Since all the edges in Γ labelled by 𝑏 are vertical, this shows that 𝑏 has infinite order.
∎
Proof of Theorem 1.2
If
n
≥
19
, then
F
(
3
,
n
)
is hyperbolic by Theorem 1.1.
If
n
∈
{
2
,
3
,
5
,
6
}
, then
F
(
3
,
n
)
is finite.
If
n
=
7
or
9
≤
n
≤
18
, then
F
(
3
,
n
)
can be confirmed to be hyperbolic using KBMAG [8].
If
n
∈
{
4
,
8
}
, then
F
(
3
,
n
)
is not hyperbolic by Lemmas 3.3 and 3.4.
If
n
∉
{
2
,
3
,
4
,
5
,
6
,
8
}
, then
F
(
3
,
n
)
is infinite [6], aspherical [1] and therefore torsion-free [12].
Furthermore,
F
(
3
,
n
)
has finite Abelianisation [15], and being non-elementary hyperbolic now follows.
∎
