A new method for verifying the hyperbolicity of finitely presented - - PowerPoint PPT Presentation

A new method for verifying the hyperbolicity of finitely presented groups

Derek Holt

University of Warwick

Stevens Institute, Conference on Groups and Computation, Paul Schupp Celebration, 30th June 2017

Derek Holt (University of Warwick) June, 2017 1 / 28

Contents

1 Group presentations 2 Programs to verify hyperbolicity 3 van Kampen diagrams and the Dehn function 4 Small cancellation 5 Curvature functions 6 The RSym curvature distribution scheme 7 Pregroups and coloured diagrams 8 Examples 9 Future plans

Derek Holt (University of Warwick) June, 2017 2 / 28

Group presentations

Throughout the talk G = X | R will be a group defined by a presentation with X and R finite. We put A := X ±1. So R is the set of defining relators of G, and R ⊆ A∗. We assume throughout that the words in R are cyclically reduced.

Derek Holt (University of Warwick) June, 2017 3 / 28

Group presentations

Throughout the talk G = X | R will be a group defined by a presentation with X and R finite. We put A := X ±1. So R is the set of defining relators of G, and R ⊆ A∗. We assume throughout that the words in R are cyclically reduced. We are interested in attempting to decide whether G is hyperbolic. It is known that there is no general algorithm for this purpose, but the hyperbolicity of G can be verified.

Derek Holt (University of Warwick) June, 2017 3 / 28

Group presentations

Throughout the talk G = X | R will be a group defined by a presentation with X and R finite. We put A := X ±1. So R is the set of defining relators of G, and R ⊆ A∗. We assume throughout that the words in R are cyclically reduced. We are interested in attempting to decide whether G is hyperbolic. It is known that there is no general algorithm for this purpose, but the hyperbolicity of G can be verified.

There are many equivalent conditions for hyperbolicity of G. These include

• Geodesic triangles in the Cayley graph Γ(G, A) are δ-slim for some fixed

δ > 0;

• Two infinite geodesic paths in Γ(G, A) from a common vertex either do not

diverge, or they diverge exponentially;

• The Dehn function ∆G of G is linear;
• G has a Dehn presentation (so the word-problem of G is solvable in linear

time).

Derek Holt (University of Warwick) June, 2017 3 / 28

Verifying hyperbolicity

The programs in the KBMAG package (available as a standalone program

• r via GAP or Magma) can verify hyperbolicity.

They do this by first finding a shortlex automatic structure for G and verifying that geodesic bigons in Γ(G, A) are uniformly thin. A result of Papasoglu then enables us to conclude that the group is hyperbolic.

Derek Holt (University of Warwick) June, 2017 4 / 28

Verifying hyperbolicity

The programs in the KBMAG package (available as a standalone program

• r via GAP or Magma) can verify hyperbolicity.

They do this by first finding a shortlex automatic structure for G and verifying that geodesic bigons in Γ(G, A) are uniformly thin. A result of Papasoglu then enables us to conclude that the group is hyperbolic. The alternative methods we are discussing today are based on generalizations of small cancellation theory. The project was initiated by Richard Parker in about 2008. Significant contributions have been made by Roney-Dougal, Neunh¨

• ffer, Linton and others.

Experimental programs have been written by Parker and Neunh¨

• ffer, there

is a new implementation in GAP by Markus Pfeiffer, and a recent Magma implementation by the speaker.

Derek Holt (University of Warwick) June, 2017 4 / 28

Advantages and disadvantages of KBMAG

In theory, given enough resources, KBMAG method can verify the hyperbolicity of any hyperbolic group G. It has been successful on difficult examples, such as the Fibonacci group F(2, 9). and the Heineken group x, y, z | [[x, [x, y]] = z, [y, [y, z]] = x, [z, [z, x]] = y.

Derek Holt (University of Warwick) June, 2017 5 / 28

Advantages and disadvantages of KBMAG

In theory, given enough resources, KBMAG method can verify the hyperbolicity of any hyperbolic group G. It has been successful on difficult examples, such as the Fibonacci group F(2, 9). and the Heineken group x, y, z | [[x, [x, y]] = z, [y, [y, z]] = x, [z, [z, x]] = y. It can calculate the growth series of G as a rational function.

Derek Holt (University of Warwick) June, 2017 5 / 28

Advantages and disadvantages of KBMAG

In theory, given enough resources, KBMAG method can verify the hyperbolicity of any hyperbolic group G. It has been successful on difficult examples, such as the Fibonacci group F(2, 9). and the Heineken group x, y, z | [[x, [x, y]] = z, [y, [y, z]] = x, [z, [z, x]] = y. It can calculate the growth series of G as a rational function. It provides no reasonable estimate of the slimness constant δ or of the Dehn function of G.

Derek Holt (University of Warwick) June, 2017 5 / 28

Advantages and disadvantages of KBMAG

In theory, given enough resources, KBMAG method can verify the hyperbolicity of any hyperbolic group G. It has been successful on difficult examples, such as the Fibonacci group F(2, 9). and the Heineken group x, y, z | [[x, [x, y]] = z, [y, [y, z]] = x, [z, [z, x]] = y. It can calculate the growth series of G as a rational function. It provides no reasonable estimate of the slimness constant δ or of the Dehn function of G. It enables only a quadratic-time solution of the word problem.

Derek Holt (University of Warwick) June, 2017 5 / 28

Advantages and disadvantages of KBMAG

In theory, given enough resources, KBMAG method can verify the hyperbolicity of any hyperbolic group G. It has been successful on difficult examples, such as the Fibonacci group F(2, 9). and the Heineken group x, y, z | [[x, [x, y]] = z, [y, [y, z]] = x, [z, [z, x]] = y. It can calculate the growth series of G as a rational function. It provides no reasonable estimate of the slimness constant δ or of the Dehn function of G. It enables only a quadratic-time solution of the word problem. It can only be applied to individual group presentations - not to infinite families.

Derek Holt (University of Warwick) June, 2017 5 / 28

Disadvantages and advantages of the new methods

They are not guaranteed to succeed on all presentations of hyperbolic

• groups. KBMAG is more likely to succeed on short hard examples.

Derek Holt (University of Warwick) June, 2017 6 / 28

Disadvantages and advantages of the new methods

They are not guaranteed to succeed on all presentations of hyperbolic

• groups. KBMAG is more likely to succeed on short hard examples.

They run in polynomial time (returning true or fail). In practice, when they work they do so much more quickly than KBMAG.

Derek Holt (University of Warwick) June, 2017 6 / 28

Disadvantages and advantages of the new methods

They are not guaranteed to succeed on all presentations of hyperbolic

• groups. KBMAG is more likely to succeed on short hard examples.

They run in polynomial time (returning true or fail). In practice, when they work they do so much more quickly than KBMAG. Unlike KBMAG, they can be used on presentations with large numbers of generators or relators.

Derek Holt (University of Warwick) June, 2017 6 / 28

Disadvantages and advantages of the new methods

They are not guaranteed to succeed on all presentations of hyperbolic

• groups. KBMAG is more likely to succeed on short hard examples.

They run in polynomial time (returning true or fail). In practice, when they work they do so much more quickly than KBMAG. Unlike KBMAG, they can be used on presentations with large numbers of generators or relators. They can sometimes be applied by hand, and to infinite families of group presentations.

Derek Holt (University of Warwick) June, 2017 6 / 28

Disadvantages and advantages of the new methods

They are not guaranteed to succeed on all presentations of hyperbolic

• groups. KBMAG is more likely to succeed on short hard examples.

They run in polynomial time (returning true or fail). In practice, when they work they do so much more quickly than KBMAG. Unlike KBMAG, they can be used on presentations with large numbers of generators or relators. They can sometimes be applied by hand, and to infinite families of group presentations. They provide a reasonable estimate of the Dehn function of G (which can be used to estimate the slimness constant).

Derek Holt (University of Warwick) June, 2017 6 / 28

Disadvantages and advantages of the new methods

They are not guaranteed to succeed on all presentations of hyperbolic

• groups. KBMAG is more likely to succeed on short hard examples.

They run in polynomial time (returning true or fail). In practice, when they work they do so much more quickly than KBMAG. Unlike KBMAG, they can be used on presentations with large numbers of generators or relators. They can sometimes be applied by hand, and to infinite families of group presentations. They provide a reasonable estimate of the Dehn function of G (which can be used to estimate the slimness constant). If the presentation is itself a Dehn presentation, then the programs may be able to verify this property, and hence enable a fast linear-time solution of the word problem.

Derek Holt (University of Warwick) June, 2017 6 / 28

van Kampen diagrams

For w ∈ A∗ with w =G 1, there is a van Kampen diagram for w. This is a simply connected planar diagram ∆ with directed edges labelled by group generators, in which the internal faces are labelled by the group relators or their inverses, and the boundary of ∆ (i.e. the external face) is labelled by w.

Derek Holt (University of Warwick) June, 2017 7 / 28

van Kampen diagrams

For w ∈ A∗ with w =G 1, there is a van Kampen diagram for w. This is a simply connected planar diagram ∆ with directed edges labelled by group generators, in which the internal faces are labelled by the group relators or their inverses, and the boundary of ∆ (i.e. the external face) is labelled by w. The area Area(∆) is the number of its internal faces. For w ∈ A∗ with w =G 1, we define Area(w) = min{Area(∆) : ∆ van Kampen diagram for w}, and then define the define the Dehn function DG : N → N by DG(n) = max{Area(w) : w =G 1, |w| ≤ n}.

Derek Holt (University of Warwick) June, 2017 7 / 28

van Kampen diagrams

For w ∈ A∗ with w =G 1, there is a van Kampen diagram for w. This is a simply connected planar diagram ∆ with directed edges labelled by group generators, in which the internal faces are labelled by the group relators or their inverses, and the boundary of ∆ (i.e. the external face) is labelled by w. The area Area(∆) is the number of its internal faces. For w ∈ A∗ with w =G 1, we define Area(w) = min{Area(∆) : ∆ van Kampen diagram for w}, and then define the define the Dehn function DG : N → N by DG(n) = max{Area(w) : w =G 1, |w| ≤ n}. An internal face of ∆ is called a boundary face if it has an edge on the boundary of ∆. A vertex or edge is boundary if it is on the boundary.

Derek Holt (University of Warwick) June, 2017 7 / 28

A van Kampen Diagram

x x x y x y y x x y y

Figure: van Kampen diagram for x3 in x, y | y −1xyx−2, y 2

Derek Holt (University of Warwick) June, 2017 8 / 28

Consolidated edges and faces

We call a path of edges in a van Kampen diagram separated by vertices of valency 2 and connecting two vertices of valency at least 3 a consolidated edge, and we call the enclosed faces consolidated faces. a b ab

Figure: Consolidated edges and consolidated face

Derek Holt (University of Warwick) June, 2017 9 / 28

Consolidated edges and faces

We call a path of edges in a van Kampen diagram separated by vertices of valency 2 and connecting two vertices of valency at least 3 a consolidated edge, and we call the enclosed faces consolidated faces. a b ab

Figure: Consolidated edges and consolidated face

The diagram is reduced if no two adjacent internal faces are labelled by r = uv and r−1 = v−1u−1 with common consolidated edge labelled u.

In an unreduced diagram ∆, we can collapse two such faces and decrease Area(∆).

Derek Holt (University of Warwick) June, 2017 9 / 28

Small cancellation conditions

Let ˆ R be the set of cyclic conjugates of the elements of R ∪ R−1. If uv, uw ∈ ˆ R with v = w, then u is called a piece.

Derek Holt (University of Warwick) June, 2017 10 / 28

Small cancellation conditions

Let ˆ R be the set of cyclic conjugates of the elements of R ∪ R−1. If uv, uw ∈ ˆ R with v = w, then u is called a piece. For p > 0, the presentation satisfies C ′(1/p) if |u| < |uv|/p for all uv ∈ ˆ R with u a piece.

Derek Holt (University of Warwick) June, 2017 10 / 28

Small cancellation conditions

Let ˆ R be the set of cyclic conjugates of the elements of R ∪ R−1. If uv, uw ∈ ˆ R with v = w, then u is called a piece. For p > 0, the presentation satisfies C ′(1/p) if |u| < |uv|/p for all uv ∈ ˆ R with u a piece. For k > Z>0, it satisfies C(k) if no r ∈ ˆ R is a product of fewer than k pieces.

Derek Holt (University of Warwick) June, 2017 10 / 28

Small cancellation conditions

Let ˆ R be the set of cyclic conjugates of the elements of R ∪ R−1. If uv, uw ∈ ˆ R with v = w, then u is called a piece. For p > 0, the presentation satisfies C ′(1/p) if |u| < |uv|/p for all uv ∈ ˆ R with u a piece. For k > Z>0, it satisfies C(k) if no r ∈ ˆ R is a product of fewer than k pieces. So C ′(1/p) ⇒ C(p + 1).

Derek Holt (University of Warwick) June, 2017 10 / 28

Small cancellation conditions

Let ˆ R be the set of cyclic conjugates of the elements of R ∪ R−1. If uv, uw ∈ ˆ R with v = w, then u is called a piece. For p > 0, the presentation satisfies C ′(1/p) if |u| < |uv|/p for all uv ∈ ˆ R with u a piece. For k > Z>0, it satisfies C(k) if no r ∈ ˆ R is a product of fewer than k pieces. So C ′(1/p) ⇒ C(p + 1). Note that C(k) is equivalent to the property that all internal non-boundary consolidated faces in reduced van Kampen diagrams for G have degree at least k.

Derek Holt (University of Warwick) June, 2017 10 / 28

Small cancellation conditions

Let ˆ R be the set of cyclic conjugates of the elements of R ∪ R−1. If uv, uw ∈ ˆ R with v = w, then u is called a piece. For p > 0, the presentation satisfies C ′(1/p) if |u| < |uv|/p for all uv ∈ ˆ R with u a piece. For k > Z>0, it satisfies C(k) if no r ∈ ˆ R is a product of fewer than k pieces. So C ′(1/p) ⇒ C(p + 1). Note that C(k) is equivalent to the property that all internal non-boundary consolidated faces in reduced van Kampen diagrams for G have degree at least k. The presentation satisfies T(k) if all internal vertices in reduced consolidated van Kampen diagrams for G have degree at least k.

Note that all presentations satisfy T(3).

Derek Holt (University of Warwick) June, 2017 10 / 28

Small cancellation conditions (ctd)

Dehn, Greendlinger, and Lyndon & Schupp used curvature arguments on (planar) van Kampen diagrams to deduce properties of the word and conjugacy problems of G from various small cancellation hypotheses.

Derek Holt (University of Warwick) June, 2017 11 / 28

Small cancellation conditions (ctd)

Dehn, Greendlinger, and Lyndon & Schupp used curvature arguments on (planar) van Kampen diagrams to deduce properties of the word and conjugacy problems of G from various small cancellation hypotheses. In particular, Greendlinger proved that if G satisfies C ′(1/6), then it is a Dehn presentation (and hence G is hyperbolic).

Derek Holt (University of Warwick) June, 2017 11 / 28

Small cancellation conditions (ctd)

Dehn, Greendlinger, and Lyndon & Schupp used curvature arguments on (planar) van Kampen diagrams to deduce properties of the word and conjugacy problems of G from various small cancellation hypotheses. In particular, Greendlinger proved that if G satisfies C ′(1/6), then it is a Dehn presentation (and hence G is hyperbolic). More recently, Gersten and Short proved that any one of the hypotheses C(7), C(5) + T(4), C(4) + T(5), or T(7) + C(3) implies that G is hyperbolic.

Derek Holt (University of Warwick) June, 2017 11 / 28

Small cancellation conditions (ctd)

Dehn, Greendlinger, and Lyndon & Schupp used curvature arguments on (planar) van Kampen diagrams to deduce properties of the word and conjugacy problems of G from various small cancellation hypotheses. In particular, Greendlinger proved that if G satisfies C ′(1/6), then it is a Dehn presentation (and hence G is hyperbolic). More recently, Gersten and Short proved that any one of the hypotheses C(7), C(5) + T(4), C(4) + T(5), or T(7) + C(3) implies that G is hyperbolic. Intuitively, you cannot tile the plane with polygons all of which have at least 7 sides. So in a diagram ∆ for a presentation satisfying C(7), a nonzero proportion of the faces must be boundary faces, and hence the Dehn function is linear.

Derek Holt (University of Warwick) June, 2017 11 / 28

Curvature functions

For a van Kampen diagram ∆, let F = F∆, V = V∆ and E = E∆ be respectively the internal faces, the vertices, and the edges of ∆.

Derek Holt (University of Warwick) June, 2017 12 / 28

Curvature functions

For a van Kampen diagram ∆, let F = F∆, V = V∆ and E = E∆ be respectively the internal faces, the vertices, and the edges of ∆. A curvature function κ = κ∆ on ∆ is a function κ : F ∪ V ∪ E → R with

• f ∈F

κ(f ) +

• v∈V

κ(v) +

• e∈E

κ(e) = 1.

Derek Holt (University of Warwick) June, 2017 12 / 28

Curvature functions

For a van Kampen diagram ∆, let F = F∆, V = V∆ and E = E∆ be respectively the internal faces, the vertices, and the edges of ∆. A curvature function κ = κ∆ on ∆ is a function κ : F ∪ V ∪ E → R with

• f ∈F

κ(f ) +

• v∈V

κ(v) +

• e∈E

κ(e) = 1. For example, we could define κ(f ) = κ(v) = 1 and κ(e) = −1 for all f ∈ F, v ∈ V , e ∈ E, and the condition holds by Euler’s formula.

Derek Holt (University of Warwick) June, 2017 12 / 28

Curvature functions

For a van Kampen diagram ∆, let F = F∆, V = V∆ and E = E∆ be respectively the internal faces, the vertices, and the edges of ∆. A curvature function κ = κ∆ on ∆ is a function κ : F ∪ V ∪ E → R with

• f ∈F

κ(f ) +

• v∈V

κ(v) +

• e∈E

κ(e) = 1. For example, we could define κ(f ) = κ(v) = 1 and κ(e) = −1 for all f ∈ F, v ∈ V , e ∈ E, and the condition holds by Euler’s formula. A curvature distribution scheme on G is an assignment of a curvature function κ∆ to each reduced diagram ∆ for G.

Derek Holt (University of Warwick) June, 2017 12 / 28

Curvature functions

For a van Kampen diagram ∆, let F = F∆, V = V∆ and E = E∆ be respectively the internal faces, the vertices, and the edges of ∆. A curvature function κ = κ∆ on ∆ is a function κ : F ∪ V ∪ E → R with

• f ∈F

κ(f ) +

• v∈V

κ(v) +

• e∈E

κ(e) = 1. For example, we could define κ(f ) = κ(v) = 1 and κ(e) = −1 for all f ∈ F, v ∈ V , e ∈ E, and the condition holds by Euler’s formula. A curvature distribution scheme on G is an assignment of a curvature function κ∆ to each reduced diagram ∆ for G. Our methods are based on the following results, which can be proved by a straightforward counting argument using Euler’s formula.

Derek Holt (University of Warwick) June, 2017 12 / 28

Hyperbolicity proved by curvature

Theorem

Suppose that there exists a constant ǫ > 0 and a curvature distribution scheme on G such that, for all reduced van Kampen diagrams ∆ for G, κ = κ∆ satisfies the following conditions.

1 κ(v) ≤ 0 for all v ∈ V ; 2 κ(e) ≤ 0 for all e ∈ E; 3 κ(f ) ≤ 1 for all f ∈ FB; 4 κ(f ) ≤ −ǫ for all f ∈ F \ FB.

where FB denote the sets of internal boundary faces of ∆.

Derek Holt (University of Warwick) June, 2017 13 / 28

Hyperbolicity proved by curvature

Theorem

Suppose that there exists a constant ǫ > 0 and a curvature distribution scheme on G such that, for all reduced van Kampen diagrams ∆ for G, κ = κ∆ satisfies the following conditions.

1 κ(v) ≤ 0 for all v ∈ V ; 2 κ(e) ≤ 0 for all e ∈ E; 3 κ(f ) ≤ 1 for all f ∈ FB; 4 κ(f ) ≤ −ǫ for all f ∈ F \ FB.

where FB denote the sets of internal boundary faces of ∆. Then G is hyperbolic and, for n ∈ N, DG(n) ≤ (1 + 1 ǫ )n.

Derek Holt (University of Warwick) June, 2017 13 / 28

The RSym scheme

We now describe a specific curvature distribution scheme. The curvature function κ = κ∆ is defined on reduced diagrams ∆ by the following procedure.

Derek Holt (University of Warwick) June, 2017 14 / 28

The RSym scheme

We now describe a specific curvature distribution scheme. The curvature function κ = κ∆ is defined on reduced diagrams ∆ by the following procedure.

RSym

Step 1 Initially κ(f ) = κ(v) = 1 and κ(e) = −1 for all f ∈ F, v ∈ V , e ∈ E. Step 2 Each edge donates −1/2 to each of its two incident vertices. So now κ(e) = 0 and κ(v) ≤ 0 for all e ∈ E and v ∈ V . Step 3 Each vertex divides its (negative) curvature equally among its incident internal faces. Now κ certainly satisfies Conditions 1, 2 and 3 in the hypothesis of the theorem, so only Condition 4 remains to be verified.

Derek Holt (University of Warwick) June, 2017 14 / 28

The RSym scheme (ctd)

We say that RSym succeeds on G if Condition 4 is satisfied, which then implies that G is hyperbolic. Our main procedure attempts to verify that RSym must succeed by examining the possible consolidated edges (labelled by pieces) and vertices in reduced diagrams for G. It can also attempt to verify that the presentation is Dehn.

Derek Holt (University of Warwick) June, 2017 15 / 28

The RSym scheme (ctd)

We say that RSym succeeds on G if Condition 4 is satisfied, which then implies that G is hyperbolic. Our main procedure attempts to verify that RSym must succeed by examining the possible consolidated edges (labelled by pieces) and vertices in reduced diagrams for G. It can also attempt to verify that the presentation is Dehn. It is straightforward to show that RSym succeeds when G satisfies any one

• f the small cancellation conditions C(7), C(5) + T(4) or C(4) + T(5)

with ǫ = 1/6, 1/3 and 1/5, respectively.

Derek Holt (University of Warwick) June, 2017 15 / 28

The RSym scheme (ctd)

We say that RSym succeeds on G if Condition 4 is satisfied, which then implies that G is hyperbolic. Our main procedure attempts to verify that RSym must succeed by examining the possible consolidated edges (labelled by pieces) and vertices in reduced diagrams for G. It can also attempt to verify that the presentation is Dehn. It is straightforward to show that RSym succeeds when G satisfies any one

• f the small cancellation conditions C(7), C(5) + T(4) or C(4) + T(5)

with ǫ = 1/6, 1/3 and 1/5, respectively. If we assume C(3) + T(7), then a weaker version of RSym, in which Condition 3 is satisfied for all faces at dual distance at least 2 from the boundary, is satisfied with ǫ = −1/14. This is still enough to prove hyperbolicity but with a weaker Dehn function estimate.

Derek Holt (University of Warwick) June, 2017 15 / 28

Random presentations

Suppose that the group G has m generators. Then there are approximately (2m − 1)ℓ cylically reduced words of a given length ℓ. Let G be the group defined by a presentation with (2m − 1)dℓ random relators of length ℓ for some d ∈ [0, 1]. Gromov proved that, with probability approaching 1 as l → ∞, G is infinite hyperbolic when d < 1/2, and |G| ≤ 2 when d > 1/2.

Derek Holt (University of Warwick) June, 2017 16 / 28

Random presentations

Suppose that the group G has m generators. Then there are approximately (2m − 1)ℓ cylically reduced words of a given length ℓ. Let G be the group defined by a presentation with (2m − 1)dℓ random relators of length ℓ for some d ∈ [0, 1]. Gromov proved that, with probability approaching 1 as l → ∞, G is infinite hyperbolic when d < 1/2, and |G| ≤ 2 when d > 1/2. We can use the program to investigate this phenomenon, but we expect RSym to succeed only for smaller values of d and, for small m, this behaviour becomes apparent only when l is large. For large m and ℓ = 3, the probability that RSym succeeds appears to be a function of |R|/m. This can be explained using probability theory: we expect the success probability to tend to a Poisson distribution as m → ∞.

Derek Holt (University of Warwick) June, 2017 16 / 28

Pregroups

Problem

Many presentations involve relators xk for small values of k and, since x will typically be a piece, this means that C(k + 1) cannot be satisfied. RSym is unlikely to succeed on such presentations.

Derek Holt (University of Warwick) June, 2017 17 / 28

Pregroups

Problem

Many presentations involve relators xk for small values of k and, since x will typically be a piece, this means that C(k + 1) cannot be satisfied. RSym is unlikely to succeed on such presentations. In their celebrated book on Combinatorial Group Theory. Lyndon & Schupp handle this situation by developing theories of small cancellation in free products (with amalgamation) and in HNN-extensions. Our approach is based on Stallings’ theory of pregroups, which he developed in connection with his investigation of the fundamental groups

• f 3-dimensional manifolds.

Derek Holt (University of Warwick) June, 2017 17 / 28

Coloured relators

We split our relator set R into two disjoint sets Rred and Rgreen. The relators in in Rred (the pregroup relators) all have length 3, and the group X | Rred is always virtually free. Relators of the form x2 for x ∈ X are omitted. But the generator x is recorded as being self-inverse, and edges in diagrams labelled by x are undirected. We may decide to introduce new generators in this process.

Derek Holt (University of Warwick) June, 2017 18 / 28

Coloured relators

We split our relator set R into two disjoint sets Rred and Rgreen. The relators in in Rred (the pregroup relators) all have length 3, and the group X | Rred is always virtually free. Relators of the form x2 for x ∈ X are omitted. But the generator x is recorded as being self-inverse, and edges in diagrams labelled by x are undirected. We may decide to introduce new generators in this process.

Example

For G = x, y|x2, y4, (xy)m we would introduce a new generator z = y2, and put Rred = {y2z}, Rgreen = {(xy)m}, and mark x and z as involutory generators.

This group is hyperbolic when m > 4.

Derek Holt (University of Warwick) June, 2017 18 / 28

Coloured diagrams

In our van Kampen diagrams, we colour the faces labelled by relators in Rred and Rgreen red and green. So, in the example above with m = 4, we could have two adjacent faces as below:

y y x y y y y x x y x z

Derek Holt (University of Warwick) June, 2017 19 / 28

Coloured diagrams (ctd)

We also replace each edge by two directed half edges with mutually inverse labels, where each half edge is assigned the colour of the incident face on its right:

y −1 y

The external face and half-edges on the boundary of ∆ are coloured green.

Derek Holt (University of Warwick) June, 2017 20 / 28

Coloured diagrams (ctd)

We also replace each edge by two directed half edges with mutually inverse labels, where each half edge is assigned the colour of the incident face on its right:

y −1 y

The external face and half-edges on the boundary of ∆ are coloured green. A connected (by adjacent edges) set of red faces is called a red blob. Let EH be the set of half-edges of ∆. The theorem proving hyperbolicity and the RSym scheme are modified as follows.

Derek Holt (University of Warwick) June, 2017 20 / 28

Hyperbolicity proved by curvature 2

Theorem

Suppose that there exists a constant ǫ > 0 and a curvature distribution scheme on G such that, for all reduced van Kampen diagrams ∆ for G, κ = κ∆ satisfies the following conditions.

1 κ(v) ≤ 0 for all v ∈ V ; 2 κ(e) ≤ 0 for all e ∈ EH; 3 κ(f ) ≤ 1 for all f ∈ FB; 4 κ(f ) ≤ 0 for all f ∈ (F \ FB) ∩ Fred; 5 κ(f ) ≤ −ǫ for all f ∈ (F \ FB) ∩ Fgreen.

where FB denotes the sets of internal boundary faces of ∆.

Derek Holt (University of Warwick) June, 2017 21 / 28

Hyperbolicity proved by curvature 2

Theorem

Suppose that there exists a constant ǫ > 0 and a curvature distribution scheme on G such that, for all reduced van Kampen diagrams ∆ for G, κ = κ∆ satisfies the following conditions.

1 κ(v) ≤ 0 for all v ∈ V ; 2 κ(e) ≤ 0 for all e ∈ EH; 3 κ(f ) ≤ 1 for all f ∈ FB; 4 κ(f ) ≤ 0 for all f ∈ (F \ FB) ∩ Fred; 5 κ(f ) ≤ −ǫ for all f ∈ (F \ FB) ∩ Fgreen.

where FB denotes the sets of internal boundary faces of ∆. Then G is hyperbolic and, for m := max{|r| : r ∈ Rgreen}, we have DG(n) ≤ (6 + m + 3 + m 2ε )n − (3 + m) 2ε .

Derek Holt (University of Warwick) June, 2017 21 / 28

The RSym scheme - revised version

RSymV2

Step 1 Initially κ(f ) = κ(v) = 1 and κ(e) = −1/2 for all f ∈ F, v ∈ V , e ∈ EH. Step 2a Each green half-edge donates −1/2 to its target vertex. Step 2b Each red half-edge donates −1/2 to its incident red blob. Now κ(e) = 0 for all e ∈ EH and κ(f ) = −1/2 for all f ∈ Fred.

Derek Holt (University of Warwick) June, 2017 22 / 28

The RSym scheme - revised version

RSymV2

Step 1 Initially κ(f ) = κ(v) = 1 and κ(e) = −1/2 for all f ∈ F, v ∈ V , e ∈ EH. Step 2a Each green half-edge donates −1/2 to its target vertex. Step 2b Each red half-edge donates −1/2 to its incident red blob. Now κ(e) = 0 for all e ∈ EH and κ(f ) = −1/2 for all f ∈ Fred. Step 3 Each vertex divides its (negative) curvature equally among its incident green faces. Step 4 Each red blob divides its (negative) curvature equally among its incident green faces.

Derek Holt (University of Warwick) June, 2017 22 / 28

The RSym scheme - revised version (ctd)

It is no longer true in general that κ(v) ≤ 0 for all v ∈ V . But by making further technical assumptions and the properties of pregroups, we can assume that all vertices are incident with at least two green faces, which ensures that this condition holds.

Derek Holt (University of Warwick) June, 2017 23 / 28

The RSym scheme - revised version (ctd)

It is no longer true in general that κ(v) ≤ 0 for all v ∈ V . But by making further technical assumptions and the properties of pregroups, we can assume that all vertices are incident with at least two green faces, which ensures that this condition holds. We say that RSym succeeds on G if Condition 5 is satisfied, which implies that G is hyperbolic.

Derek Holt (University of Warwick) June, 2017 23 / 28

Triangle groups

Example

G = x, y | xℓ, ym, (xy)n. RSym proves easily that G is hyperbolic whenever 1/ℓ + 1/m + 1/n < 1.

Derek Holt (University of Warwick) June, 2017 24 / 28

Triangle groups

Example

G = x, y | xℓ, ym, (xy)n. RSym proves easily that G is hyperbolic whenever 1/ℓ + 1/m + 1/n < 1. We may assume that 2 ≤ ℓ ≤ m ≤ n, and we extend the (monoid) generating set to {xi : 1 ≤ i < ℓ} ∪ {yi : 1 ≤ i < m}. Then the first two relators are replaced by (several) red relators, and Rgreen = {(xy)n}.

Derek Holt (University of Warwick) June, 2017 24 / 28

Triangle groups

Example

G = x, y | xℓ, ym, (xy)n. RSym proves easily that G is hyperbolic whenever 1/ℓ + 1/m + 1/n < 1. We may assume that 2 ≤ ℓ ≤ m ≤ n, and we extend the (monoid) generating set to {xi : 1 ≤ i < ℓ} ∪ {yi : 1 ≤ i < m}. Then the first two relators are replaced by (several) red relators, and Rgreen = {(xy)n}. Suppose (for example) that ℓ ≥ 3 and hence n ≥ 4. Then there can be no internal edges with both adjacent faces green. In Step 4 of RSym, any f ∈ (F \ FB) ∩ Fgreen receives at most −1/6 curvature from each of its 2n neighbouring red faces, so it finishes with curvature at most −1/3, and hence RSym succeeds with ǫ = 1/3.

Derek Holt (University of Warwick) June, 2017 24 / 28

The groups (2, 3, m; n)

Example

The groups (ℓ, m, n; p) = x, y | xℓ, ym, (xy)n, [x, y]p were studied by Coxeter. It follows from a general result of Gromov that they are infinite and hyperbolic for large enough ℓ, m, n, p, but we can use RSym to get more precise results.

Derek Holt (University of Warwick) June, 2017 25 / 28

The groups (2, 3, m; n)

Example

The groups (ℓ, m, n; p) = x, y | xℓ, ym, (xy)n, [x, y]p were studied by Coxeter. It follows from a general result of Gromov that they are infinite and hyperbolic for large enough ℓ, m, n, p, but we can use RSym to get more precise results. We consider the case ℓ = 2, m = 3 in more detail, and we take Rred = {y3} and Rgreen = {(xy)n, [x, y]p}.

Derek Holt (University of Warwick) June, 2017 25 / 28

The groups (2, 3, m; n)

Example

The groups (ℓ, m, n; p) = x, y | xℓ, ym, (xy)n, [x, y]p were studied by Coxeter. It follows from a general result of Gromov that they are infinite and hyperbolic for large enough ℓ, m, n, p, but we can use RSym to get more precise results. We consider the case ℓ = 2, m = 3 in more detail, and we take Rred = {y3} and Rgreen = {(xy)n, [x, y]p}. The basic version of RSym gives:

Theorem

(2, 3, n; p) is hyperbolic when n > 12 and p > 6.

Derek Holt (University of Warwick) June, 2017 25 / 28

The groups (2, 3, n; p) (ctd)

After running RSym on a diagram ∆, some internal green faces may have curvature less than −ǫ, and they could distribute this excess negative curvature to (neighbouring) green faces with curvature greater than −ǫ. This extra step may enable RSym to succeed. It is difficult to mechanise this process, but it can be used in hand calculations, and we can prove

Derek Holt (University of Warwick) June, 2017 26 / 28

The groups (2, 3, n; p) (ctd)

After running RSym on a diagram ∆, some internal green faces may have curvature less than −ǫ, and they could distribute this excess negative curvature to (neighbouring) green faces with curvature greater than −ǫ. This extra step may enable RSym to succeed. It is difficult to mechanise this process, but it can be used in hand calculations, and we can prove

Theorem

(2, 3, n; p) is hyperbolic when either (i) n ≥ 7 and p ≥ 18; or (ii) p ≥ 4 and n > 24. KBMAG together with coset enumeration and other techniques can be used to investigate individual remaining cases.

Derek Holt (University of Warwick) June, 2017 26 / 28

The groups (2, 3, n; p) (ctd)

Theorem

(2, 3, n; p) is infinite hyperbolic in each of the following cases: (i) n = 7, p ≥ 13; (ii) n = 8, p ≥ 8; (iii) n = 9, p ≥ 7; (iv) n = 10, p ≥ 6; (v) n = 11, p ≥ 5; (vi) n = 15, p ≥ 4.

Derek Holt (University of Warwick) June, 2017 27 / 28

The groups (2, 3, n; p) (ctd)

Theorem

(2, 3, n; p) is infinite hyperbolic in each of the following cases: (i) n = 7, p ≥ 13; (ii) n = 8, p ≥ 8; (iii) n = 9, p ≥ 7; (iv) n = 10, p ≥ 6; (v) n = 11, p ≥ 5; (vi) n = 15, p ≥ 4. It is automatic but not hyperbolic in the following cases: (i) n = 7, p = 12; (ii) n = 8, p = 7; (iii) n = 9, p = 6; (iv) n = 10, p = 5; (v) n = 14, p = 4.

Derek Holt (University of Warwick) June, 2017 27 / 28

The groups (2, 3, n; p) (ctd)

Theorem

(2, 3, n; p) is infinite hyperbolic in each of the following cases: (i) n = 7, p ≥ 13; (ii) n = 8, p ≥ 8; (iii) n = 9, p ≥ 7; (iv) n = 10, p ≥ 6; (v) n = 11, p ≥ 5; (vi) n = 15, p ≥ 4. It is automatic but not hyperbolic in the following cases: (i) n = 7, p = 12; (ii) n = 8, p = 7; (iii) n = 9, p = 6; (iv) n = 10, p = 5; (v) n = 14, p = 4. It is finite in the following cases only: (i) n ≤ 6; (ii) p ≤ 3; (iii) n = 7, 4 ≤ p ≤ 8; (iv) n = 8 or 9, p = 4 or 5; (v) n = 10, 11 or 13, p = 4;

Derek Holt (University of Warwick) June, 2017 27 / 28

Future plans

• Devise other more effective curvature distribution schemes
• Implement extensions of RSym that transfer negative curvature from

faces that have more than enough already, to other faces that have none.

• Implement Dehn algorithm testing and linear-time word problem

solvers.

• Extend methods to “flat” groups (when RSym succeeds with ǫ = 0).

Derek Holt (University of Warwick) June, 2017 28 / 28