Combinatorial and Geometric Group Theory ||

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Trends in Mathematics

Combinatorial and Geometric Group TheoryDortmund and Ottawa-Montreal Conferences

Oleg BogopolskiInna BumaginOlga KharlampovichEnric VenturaEditors

Birkhäuser

2000 Mathematics Subject Classification 20A, 20E, 20F, 20H, 20M, 20P, 03B, 03D, 05C, 08A, 51F, 57M, 57S, 60B, 68Q

Library of Congress Control Number: 2010926413

Bibliographic information published by Die Deutsche Bibliothek. Die Deutsche Bibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data is available in the Internet at http://dnb.ddb.de

ISBN 978-3-7643-9910-8

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© 2010 Springer Basel AG P.O. Box 133, CH-4010 Basel, SwitzerlandPart of Springer Science+Business MediaPrinted on acid-free paper produced from chlorine-free pulp. TCF ∞Cover Design: Alexander Faust, Basel, SwitzerlandPrinted in Germany

ISBN 978-3-7643-9910-8 e-ISBN 978-3-7643-9911-5

9 8 7 6 5 4 3 2 1 www.birkhauser.ch

Editors:

Oleg BogopolskiMathematisches Institut derHeinrich-Heine-Universität DüsseldorfUniversitätsstr. 140225 DüsseldorfGermanye-mail: [email protected]

Inna BumaginSchool of Mathematics and StatisticsCarleton University1125 Colonel By DriveOttawa, Ontario, K1S 5B6Canadae-mail: [email protected]

Olga KharlampovichDepartment of Mathematics and StatisticsMcGill University805 Sherbrooke St., WestMontreal, Quebec, H3A 2K6Canadae-mail: [email protected]

Enric VenturaDepartament de Matematica Aplicada IIIEPSEM – Universitat Politècnica de CatalunyaAv. Bases de Manresa 61–7308242 Manresa, BarcelonaSpaine-mail: [email protected]

Contents

Preface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . vii

O. Bogopolski and R. VikentievSubgroups of Small Index in Aut(Fn) andKazhdan’s Property (T) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1

P. BrinkmannDynamics of Free Group Automorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19

V. Diekert, A.J. Duncan and A.G. MyasnikovGeodesic Rewriting Systems and Pregroups . . . . . . . . . . . . . . . . . . . . . . . . . . 55

E. Frenkel, A.G. Myasnikov and V.N. RemeslennikovRegular Sets and Counting in Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . 93

D. Goncalves and P. WongTwisted Conjugacy for Virtually Cyclic Groups andCrystallographic Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 119

M. Hock and B. TsabanSolving Random Equations in Garside Groups UsingLength Functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 149

A. JuhaszAn Application of Word Combinatorics to Decision Problemsin Group Theory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 171

O. Kharlampovich and A.G. MyasnikovEquations and Fully Residually Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . 203

M. LustigThe FN -action on the Product of the Two Limit Treesfor an Iwip Automorphism . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 243

F. MatucciMather Invariants in Groups of Piecewise-linearHomeomorphisms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 251

vi Contents

P.V. Morar and A.N. ShevlyakovAlgebraic Geometry over the Additive Monoid of Natural Numbers:Systems of Coefficient Free Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261

D. SavchukSome Graphs Related to Thompson’s Group F . . . . . . . . . . . . . . . . . . . . . . 279

R. WeidmannGenerating Tuples of Virtually Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 297

R. ZarzyckiLimits of Thompson’s Group F . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 307

Combinatorial and Geometric Group Theory

Trends in Mathematics, vii–viiic© 2010 Springer Basel AG

Preface

We are pleased to present the book “Geometric Group Theory, Dortmund andCarleton Conferences”, a selection of the best research articles from two stronglyrelated 2007 international conferences:

• “Combinatorial and Geometric Group Theory with Applications” (GAGTA),the University of Dortmund (Germany) from August 27th to 31st;

• “Fields Workshop in Asymptotic Group Theory and Cryptography”, Car-leton University (Ottawa, Canada) from December 14th to 16th, followed by“Workshop on Actions on Trees, Non-Archimedian Words, and AsymptoticCones”, Saint Sauveur (Montreal) from December 17th to 21st.

The book contains a selection of refereed papers on Combinatorial and Geomet-ric Group Theory. The breadth of topics included will assure the interest of allspecialists and researchers in this area of mathematics; they will also prove tobe valuable to graduate students and mathematicians in other areas who wish toexplore deeper into this exciting and very active field of research.

The articles largely fall into five categories:

• equations and algebraic geometry over groups; Tarski problems,

• algorithmic problems in groups,

• groups of automorphisms of non-abelian free groups,

• groups of transformations of the unit interval and Thompson’s group F ,

• questions motivated by group-based cryptography.

Readers interested in the first topic may choose to look first at the excellent expos-itory paper by O. Kharlampovich and A.G. Myasnikov. Here, the authors explaintheir multifaceted techniques (part of them on algebraic geometry over groups) forsolving two of Tarski’s famous problems on elementary theories of free groups. Thepaper of P. Morar and A. Shevlyakov initiates investigations of algebraic geometryover some intriquing classes of monoids.

One can also learn a lot about dynamics of automorphisms of free groupsvia train tracks and actions on trees, by reading the thought-provoking papersof P. Brinkmann and M. Lustig. In a similar direction, R. Weidmann shows howMakanin-Razborov diagrams and Stallings foldings can be used to solve the rankproblem for virtually free groups.

viii Preface

The paper by O. Bogopolski and A. Vikentiev describes some particularlyuseful finite index subgroups of the automorphism group of a finitely generated freegroup. One of their uses may be to attack the problem on the Kazhdan property(T ) for these groups. The paper of A. Juhasz contains a solution of the difficultmembership problem in a subclass of one-relator groups.

Papers of F. Matucci, D. Savchuk and R. Zarzycki will attract the attentionof those who want to know more about groups of transformations of the unitinterval [0, 1], in particular about the famous Thompson’s group F and its limitproperties.

The paper by A.J. Duncan, V. Dieckert and A.G. Myasnikov contains a verythorough survey on rewriting systems with new issues on infinite rewriting systems.The paper by L. Frenkel, A.G. Myasnikov and V.N. Remeslennikov is devoted tothe problem of how to measure some subsets in free groups by using random walks.The results of this paper may be used for designing algorithms that run fast onalmost all inputs. This paper as well as the paper by M. Hock and B. Tsaban arehighly recommended to specialists in cryptography.

Finally, the paper by D. Goncalves and P. Wong is devoted to the twistedconjugacy in 2-dimensional crystallographic groups.

We are very grateful to the organizations that supported these two conferences:

• The conference in Dortmund was organized by O. Bogopolski, M.-T. Bochnig,G. Rosenberger, V. Shpilrain and E. Ventura. This conference was financiallysupported by DAAD (Deutscher Akademischer Austauschdienst), by DFG(Deutsche Forschungsgemeinschaft), and by the Universitat Dortmund. TheURL address for its homepage ishttp://www.mathematik.uni-dortmund.de/∼gcgta/.

• The workshops in Canada were co-organized by I. Bumagin, O. Kharlampov-ich and A.G. Myasnikov. The workshops could not have been held withoutthe generous support of the Fields Institute. The organizers also gratefullyacknowledge the financial support provided by the Faculty of Science of Car-leton University and by McGill University. More information about the work-shops can be found at the URLhttp://www.fields.utoronto.ca/programs/

scientific/07-08/asympotic/index.html

Finally, we wish to thank the contributors to this volume, and the anonymousreferees who ensured the high quality of its contents. Our thanks also go to ThomasHempfling at Birkhauser for his assistance in the typsetting and preparation of thisvolume. Without these joint efforts, this book would never have appeared.

The editors, O. Bogopolski,I. Bumagin,O. Kharlampovich,E. Ventura


Trends in Mathematics, 1–17c© 2010 Springer Basel AG

Subgroups of Small Index in Aut(Fn)and Kazhdan’s Property (T)

O. Bogopolski and R. Vikentiev

Abstract. We introduce a series of interesting subgroups of finite index inAut(Fn). One of them has index 42 in Aut(F3) and infinite abelianization.This implies that Aut(F3) does not have Kazhdan’s property (T); see [15]and [5] for other proofs. We prove also that every subgroup of finite indexin Aut(Fn), n � 3, which contains the subgroup of IA-automorphisms has afinite abelianization.

We introduce a subgroup K(n) of finite index in Aut(Fn) and show,that its abelianization is infinite for n = 3, and it is finite for n � 4. We ask,whether the abelianization of its commutator subgroup K(n)′ is infinite forn � 4. If so, then Aut(Fn) would not have Kazhdan’s property (T) for n � 4.

Mathematics Subject Classification (2000). 20F28, 20E05, 20E15.

Keywords. Automorphisms, free groups, Kazhdan’s property (T), congruencesubgroups.

1. Definitions, problems and motivations

In the mid 60’s, D. Kazhdan defined his property (T) for locally compact groupsand used it as a tool to demonstrate that many lattices in these groups are finitelygenerated [8]. Later this property found various surprising applications, in partic-ular, in the first explicit construction of expander graphs by G. Margulis [14], seealso the book of A. Lubotzky [9] and the paper of A. Lubotzky and I. Pak [10]. Werecommend to the reader the very informative book of B. Bekka, P. de la Harpeand A. Valette [2] and the lecture of Y. Shalom [21] on the property (T) and itsapplications.

There are several equivalent definitions of the property (T) for topologicalgroups. Below we give one of them in case, where the group is finitely generated.We will assume that the group is endowed with the discrete topology.

Let G be a finitely generated group and π : G→ U(H) a unitary representa-tion of G on a Hilbert space H. If S is a finite generating set of G and ε > 0, then

2 O. Bogopolski and R. Vikentiev

a unit vector v in H is called an (S, ε)-invariant vector if ||π(s)(v)− v|| < ε for alls ∈ S.

Definition 1.1. A finitely generated group G has the property (T) iff for everyfinite generating subset S ⊆ G there exists some ε > 0, such that the following issatisfied:

For any unitary representation π : G → U(H) on a Hilbert space H, theexistence of an (S, ε)-invariant vector implies the existence of a non-zeroπ(G)-invariant vector.

Any such ε is called a Kazhdan constant for G with respect to S. It is easyto show, that in this definition the words “for every finite generating set” can bereplaced by “for some finite generating set”.

Definition 1.2. A finitely generated group G has property (FH) if any action of Gby affine isometries on a Hilbert space has a fixed point.

By a theorem of Delorme–Guichardet (see Theorem 2.12.4 in [2]), the prop-erties (T) and (FH) are equivalent for finitely generated groups. There are strongconsequences on several types of actions: for a group with the property (FH), anyisometric action on a tree has a fixed point or a fixed edge (this is the property(FA) of Serre), any isometric action on a real or complex hyperbolic space has afixed point.

Definition 1.3. A group G has Serre’s property (FA) if acting simplicially andwithout inversions of edges on any simplicial tree, G has a global fixed point.

If G is finitely generated, this property can be reformulated in purely alge-braic terms.

Theorem 1.4. (J.-P. Serre; 1974). A finitely generated group G has the property(FA) if and only if the following two statements hold:

(1) G is not a nontrivial amalgamated product, that is G � A ∗C B with C �= Aand C �= B.

(2) G does not have a quotient isomorphic to Z.

Theorem 1.5. (Y. Watatani; 1982). Let G be a countable group. If G has theproperty (T) of Kazhdan, then it has the property (FA) of Serre.

Let Fn be the free group of rank n with basis x1, . . . , xn. In this paper we willconcentrate on the group Aut(Fn), the automorphism group of Fn. There is thecanonical epimorphism Φ : Aut(Fn) → GLn(Z), which sends an automorphismα ∈ Aut(Fn) to the matrix, whose ij-entry equals to the sum of exponents ofxj in α(xi), i, j = 1, . . . , n. The full preimage of SLn(Z) is called the specialautomorphism group of Fn and is denoted by SAut(Fn). The kernel of Φ is denotedby IA(Fn).

Subgroups of Small Index in Aut(Fn) 3

In the following table we summarize known facts on the (T) and (FA) prop-erties for SLn(Z) and SAut(Fn), n � 3.

(T) ⇒ (FA)

SLn(Z), n � 3 + +(Kazhdan, 1967) (Serre, 1974)

SAut (F3), − +(McCool, 1989) (Bogopolski, 1987)

SAut (Fn), n � 4, ? +(Bogopolski, 1987)

Remark 1.6. The property (T) is preserved under taking subgroups of finite index,the property (FA) is not preserved. Both properties are preserved under takingovergroups of finite index. Groups with the property (T) have no subgroups offinite index, which can be mapped onto Z.

Problems 1.7.

1) Does the group Aut(Fn), n � 4, have the property (T)?2) Does every finite index subgroup of Aut(Fn), n � 4, have the property (FA)?3) Characterize (in terms of actions on trees or algebraically) those finitely gen-

erated groups, whose subgroups of finite index have the (FA) property.

In [23], K. Vogtmann formulated the Out-versions of the problems 1) and 2)(see Problems 14 and 15 there). The first problem is also formulated as Problem(7.1) in the list of open questions in [2].

Due to A. Lubotzky and I. Pak [10], the presence of the Kazhdan propertyfor Aut(Fn) would imply a very elegant construction of an infinite series of ε-expanders. The later have applications to theoretical computer science, design ofrobust computer networks, and the theory of error-correcting codes [7].

Definition 1.8. Let ε be a positive real number. A finite d-regular graph Γ is calledan ε-expander, if for every subset of vertices B ⊂ Γ0 with |B| � |Γ0|/2, we have|∂B| � ε|B|, where

∂B = {v ∈ Γ0 | v /∈ B, but v is adjacent to a vertex in B}.Definition 1.9 (Graph Γn(G)). Fix a natural number n and a finite group G,which can be generated by n elements. The vertices of the graph Γn(G) are alltuples (g) = (g1, . . . , gn) such that 〈g1, . . . , gn〉 = G. Two tuples (g) and (g′) areconnected by an edge if (g′) can be obtained from (g) by applying one of thefollowing replacement operations:

R±i,j : (g1, . . . , gi, . . . , gn)→ (g1, . . . , gi · g±1j , . . . , gn),

L±i,j : (g1, . . . , gi, . . . , gn)→ (g1, . . . , g±1j · gi, . . . , gn),where i �= j.


Clearly, the graph Γn(G) is 4n(n− 1)-regular.The following theorem is a partial case of a more general Theorem 3.1 in [10].

It explains why establishing the Kazhdan property (T) for Aut(Fn) is important.

Theorem 1.10 (A. Lubotzky and I. Pak; 2001). If the group Aut(Fn) (or equiva-lently SAut(Fn)) has the property (T), then there exists an ε > 0, such that everyconnected component of the graph Γn(G) is an ε-expander for any n-generatedfinite group G.

The group Aut(F2) does not have the property (T), since it can be representedas a nontrivial amalgamated product. (The uniqueness of such representation upto conjugacy was shown by O. Bogopolski in [3]). The following two theoremsimply that Aut(F3) does not have the property (T). In this paper we suggest someideas for studying this problem in rank n � 4.

Theorem 1.11 (J. McCool; 1989). There is a subgroup of finite index in Out(F3),which can be approximated by torsion-free nilpotent groups. In particular, this sub-group can be mapped onto Z. Therefore Out(F3) and Aut(F3) do not have theKazhdan property (T).

Theorem 1.12 (F. Grunewald and A. Lubotzky; 2006). There exists a subgroup ofindex 168 in Aut(F3) which can be mapped onto F2. In particular, Aut(F3) has noKazhdan’s property (T).

The proof of this theorem has existed for a long time in a folklore form. Weknow variants of the proof from private talks with A. Casson (2000) and M. Bridson(2004). To our best knowledge, the first written proof (based on the same idea)appeared in the paper of F. Grunewald and A. Lubotzky [5], see Corollary 1.3 there.This corollary is deduced there from a more general Theorem 1.1 on representationsof Aut(Fn).

The structure of the paper is as follows. In Section 2 we give a short expo-sition of the proof of Theorem 1.12. Some elements of this proof motivated us tofurther constructions. In Section 3 we introduce some useful automorphisms. InSection 4 we prove that any finite index subgroup of Aut(Fn) containing IA(Fn)has finite abelianization (Theorem 4.1). Thus, to construct a finite index subgroupin Aut(Fn) with infinite abelianization, one should avoid IA-automorphisms.

In Section 5 we introduce and investigate congruence subgroups Cong(n,m)and SCong(n,m) in Aut(Fn) and SAut(Fn) respectively. Both congruence sub-groups contain IA(Fn).

In Section 6 we define a subgroup K(n) of index 2 in SCong(n, 2), so thatit does not contain IA(Fn). In Section 7 we show that K(3) has infinite abelian-ization. This gives an alternative proof of Theorem 1.12. Further we construct aseries of overgroups of K(3):

Aut(F3) �42C(3) �

2B(3) �

2A(3) �

2K(3).

The largest one, C(3), has index 42 in Aut(F3) and infinite abelianization. Weconjecture, that this is the minimal possible index for a subgroup in Aut(F3)


with infinite abelianization. We compute the abelianizations of these subgroupsexplicitly:

K(3)/K(3)′ ∼= Z142 × Z× Z,

A(3)/A(3)′ ∼= Z72 × Z× Z,

B(3)/B(3)′ ∼= Z42 × Z,

C(3)/C(3)′ ∼= Z32 × Z4 × Z.

Here we use the notation Zkn = Z/nZ× · · · × Z/nZ︸︷︷︸

k times

.

In Section 8 we prove that if n � 4, then the abelianization of K(n) is a finite2-group. In particular, the abelianization of K(4) is Z382 . We would like to know,what is the abelianization of the commutator subgroup K(n)′ for n � 4. If it isinfinite, the group Aut(Fn) does not have Kazhdan’s property (T) for n � 4.

In this paper we use the following notation for the commutator: [x, y] =xyx−1y−1.

2. A sketch of the proof of F. Grunewald and A. Lubotzkythat Aut(F3) has no Kazhdan’s property (T)

Let F3 = F (a, b, c) be the free group on free generators a, b, c. There exist exactly 7epimorphisms F (a, b, c)→ Z2. Therefore there exist exactly 7 subgroups of index2 in F3. Denote them by F (1)5 , . . . , F

(7)5 . Clearly, every such subgroup has rank 5.

We will work with one of them, F (1)5 = 〈a, b, c2, c−1ac, c−1bc〉. It is easy to checkthat Aut(F3) acts transitively on the set of these subgroups. Therefore the indexof St(F (1)5 ) in Aut(F3) is 7, where

St(F (1)5 ) = 〈α ∈ Aut(F3) |α(F (1)5 ) = F(1)5 〉.

Clearly, the restriction map Res : St(F (1)5 ) → Aut(F (1)5 ), α �→ α|F(1)5, is an

embedding. Now we introduce an important inner automorphism τ : x �→ c−1xc,x ∈ F3.

Lemma 2.1.

1) τ ∈ St(F (1)5 ),2) τ |

F(1)5

/∈ Inn(F (1)5 ),

3) (τ−1ϕ−1τϕ)|F(1)5∈ Inn(F (1)5 ) for every ϕ ∈ St(F (1)5 ).

Proof. The first two statements are straightforward. We prove the third one. Letx ∈ F

(1)5 and ϕ ∈ St(F (1)5 ). An easy computation shows that xτ−1ϕ−1τϕ =

ϕ(c−1)cxc−1ϕ(c). Thus we need to show that c−1ϕ(c) ∈ F(1)5 . The last is evident,

since F (1)5 and hence the second coset cF (1)5 are ϕ-invariant. �


Consider the composition of two homomorphisms

Ψ : St(F (1)5 ) Res−→ Aut(F (1)5 ) −→ GL5(Z),

where the second homomorphism sends an automorphism of F (1)5 to the automor-phism induced on the abelianization of F (1)5 (we identify the last automorphismwith the corresponding matrix, using the prescribed basis of F (1)5 ).

One can easily compute, that

Ψ(τ) =

⎛⎜⎜⎜⎜⎝0 0 0 1 00 0 0 0 10 0 1 0 01 0 0 0 00 1 0 0 0

⎞⎟⎟⎟⎟⎠ .

Since (Ψ(τ))2 = Id, we have Z5 ⊃ V+ ⊕ V−, where

V+ = Ker(Ψ(τ) + Id), V− = Ker(Ψ(τ) − Id).

The Z-submodule V+ has the basis {(1, 0, 0,−1, 0), (0, 1, 0, 0,−1)}.By Lemma 2.1.3), the matrices Ψ(τ) and Ψ(ϕ) commute for all ϕ ∈ St(F (1)5 ).Hence the submodule V+ is Ψ(ϕ)-invariant for every ϕ ∈ St(F (1)5 ). Thus, there isthe natural homomorphism

θ : St(F (1)5 )→ GL(V+) ∼= GL2(Z),

θ(ϕ) = Ψ(ϕ)∣∣V+.

This homomorphism is onto, since the automorphisms ϕ1 : a �→ a, b �→ ba, c �→ c

and ϕ2 : a �→ b, b �→ a, c �→ c belong to St(F (1)5 ) and are mapped onto the matrices(1 01 1

),

(0 11 0

),

which generate GL2(Z)Notice that GL2(Z) ∼= D4 ∗D2 D6, where Dm denotes the dihedral group

of order 2m (see [4] for example). Therefore there exists an epimorphism μ :GL2(Z)→ D12. The kernel of μ is a free group of rank 2, we denote it by F2. Thuswe have the following chain of embeddings and epimorphisms:

Aut(F3) � St(F (1)5 ) θ→ GL2(Z)μ→ D12.

Let H = Ker(θμ). Then H has index 24 in St(F (1)5 ). Hence H has index 168in Aut(F3). Moreover, θ(H) = Ker(μ) = F2. In particular, H can be homomorphi-cally mapped onto Z. Hence Aut(F3) does not have the Kazhdan property (T).

Remark 2.2. The group H is not normal in Aut(F3).

The above construction cannot be generalized for Aut(Fn), where n � 4,since in this case GLn−1(Z) (contrary to GL2(Z)) does not contain a subgroup offinite index with infinite abelianization.


3. Some notations and useful automorphisms

Let Fn be the free group on free generators x1, x2, . . . , xn. First we define someautomorphisms of Fn. We will write the image of xi only if it differs from xi.

1) For any i, j, k ∈ {1, 2, . . . , n}, where k �= i, j, we define the automorphism

αijk : xi → xi · x−1j x−1k xjxk.

In particular,αiik : xi → x−1k xixk.

Note that αijk = α−1ikj for distinct i, j, k. We say that the automorphism αijk

is of the first kind if i, j, k are distinct, and of the second kind if i = j.2) For any i, j ∈ {1, 2, . . . , n}, where i �= j, we define the automorphism

Eij : xi → xixj .

3) For any i ∈ {1, 2, . . . , n} we define the automorphismNi : xi → x−1i .

We denote Nij = NiNj for i �= j.The kernel of the canonical epimorphism Aut(Fn) → GLn(Z) is denoted by

IA(Fn). It is known, that IA(F2) = Inn(F2) and IA(Fn) is strictly larger thanInn(Fn) for n � 3. J. Nielsen for n � 3 [17] and W. Magnus for all n [12] provedthat IA(Fn) is generated by all automorphisms αijk (see also [11]).

4. Finite index subgroups of Aut(Fn) containing IA(Fn)

Theorem 4.1. Let n � 3. Any subgroup of finite index in Aut(Fn), containingIA(Fn), has a finite abelianization.

To prove this theorem we need to introduce more automorphisms of Aut(Fn)and to formulate a theorem of B. Sury and T.N. Venkataramana (see below) ongenerators of congruence subgroups of SLn(Z).4) For any i ∈ {1, 2, . . . , n} we define the automorphism

Ti : xi �→ x−1i , xi+1 �→ x−1i+1xi.

5) For any i, j ∈ {1, 2, . . . , n}, where i �= j, we define the automorphism

Tij : xi �→ xj , xj �→ x−1i .

Denote

T = {Tk | k = 1, . . . , n− 1} ∪ {Tij | i, j = 1, 2, . . . , n; i �= j} ∪ {I},where I is the identity automorphism of Fn.

Let ¯ : Aut(Fn) → GLn(Z) be the canonical epimorphism. For any α ∈Aut(Fn) we denote by α its canonical image in GLn(Z).


Let m be a natural number. The kernel of the canonical epimorphismSLn(Z) → SLn(Zm) is denoted by SLn(Z,mZ) and is called the congruence sub-group of SLn(Z) modulom. Of course, SLn(Z,mZ) is normal and has a finite indexin SLn(Z). The following theorem is called the congruence subgroup theorem forSLn(Z). It was proved by H. Bass, M. Lazard and J.-P. Serre [1] and independentlyby J. Mennicke [16].

Theorem 4.2. Let n � 3 be a natural number. Any subgroup of finite index inSLn(Z) contains a congruence subgroup SLn(Z,mZ) for some m.

Theorem 4.3. (B. Sury and T.N. Venkataramana; 1994). Let n � 3, m � 2. Thecongruence subgroup SLn(Z,mZ) is generated by the following set of matrices

{(α)(Eij)m(α)−1 |α ∈ T ; i, j ∈ {1, 2, . . . , n}; i �= j}.Corollary 4.4. The group SLn(Z, 2Z), n � 3, is generated by the set

X = {E2ij , N ij | i, j ∈ {1, 2, . . . , n}; i �= j}.

Proof. By Theorem 4.3, SLn(Z, 2Z) is the normal closure of all E2

ij in SLn(Z).Since SLn(Z) is generated by all transvections Epq, it is sufficient to verify that

for every x ∈ X and ε ∈ {−1, 1} holds E ε

pqxE−ε

pq ∈ 〈X〉. We leave this to thereader. �Proof of Theorem 4.1. Let G be a subgroup of finite index in Aut(Fn) containingIA(Fn). We show that G/G′ is finite. Let G be the image of G is GLn(Z). The indexof G ∩ SLn(Z) in SLn(Z) is finite. Hence, by the congruence subgroup theorem,there exists an m � 2, such that SLn(Z,mZ) � G. Let H be the full preimageof SLn(Z,mZ) in G. Since the index of H in G is finite, it is sufficient to showthat H/H ′ is finite. Since H contains IA(Fn), the results of Nielsen–Magnus andSury–Venkataramana imply that H is generated by the union of two sets:

{αijk | i, j, k ∈ {1, 2, . . . , n}; k �= i, j},{αEm

ij α−1 |α ∈ T ; i, j ∈ {1, 2, . . . , n}; i �= j}.

It is sufficient to prove that the mth power of each of these generators lies in[H,H ]. But this follows from the following formulas:1) [αiij , E

mjk] = αm

iik for distinct i, j, k ∈ {1, 2, . . . , n};2) [αiij , E

mik ] = (αikjα

−1jjk)

m−1αikjαm−1jjk ≡ αm

ikj mod[IA(Fn), IA(Fn)] for dis-tinct i, j, k ∈ {1, 2, . . . , n};

3) Em2

ik =(∏m−2

s=0 [αsiij , E

m2−(s+1)mik ][Em2−(s+1)m

ik , αs+1iij ]

)[Em

ij , Emjk]. �

5. Congruence subgroups SCong(n, k) in SAut(Fn)

Let G be a group and H be a normal subgroup in G. We denote

Aut(G;H) = {ϕ ∈ Aut(G) | ϕ(H) = H}.


andIA(G;H) = {ϕ ∈ Aut(G) | ∀g ∈ G : ϕ(gH) = gH}.

Equivalently

IA(G;H) = {ϕ ∈ Aut(G) | ∀g ∈ G∃xg ∈ H : ϕ(g) = g · xg}. (1)

Clearly, IA(G;H) is normal in Aut(G;H) and the corresponding factor groupis naturally embeddable into Aut(G/H). In particular, if |G : H | = 2, then we haveIA(G;H) = Aut(G;H).

Proposition 5.1. Let {Hi| i ∈ I} be a set of normal subgroups of a group G. Then

IA(G; ∩

i∈IHi

)= ∩

i∈IIA

(G;Hi

).

Proof. The proof is straightforward from description (1). �

Now we return to automorphisms of Fn = F (x1, . . . , xn). Let k � 2 be anatural number. Consider the standard epimorphisms

Fnπ−→ Zn ε−→ Zn

k .

They induce the epimorphisms

Aut(Fn)π−→ Aut(Zn) ε−→ Aut(Zn

k ).

Using usual identifications we may write

Aut(Fn)π−→ GLn(Z)

ε−→ GLn(Zk).

We denoteCong(n; k) = Ker(π ε),

SCong(n; k) = Cong(n; k) ∩ SAut(Fn)

and call these groups the congruence subgroups of Aut(Fn) and of SAut(Fn) mod-ulo k, respectively. In this paper we will work only with congruence subgroupsmodulo 2.

Let {σi | i = 1, . . . , 2n − 1} be the set of all epimorphisms Fn → Z2. Thekernel of σi is a free group of rank 2n − 1; we denote it by F

(i)2n−1. We fix σ1 by

the rule σ1(x1) = · · · = σ1(xn−1) = 0 and σ1(xn) = 1. Thus, F (1)2n−1 has the basis{x1, . . . , xn−1, x2n, x

−1n x1xn, . . . , x

−1n xn−1xn}. For brevity we denote

St(F (i)2n−1) = Aut(Fn, F(i)2n−1).

Proposition 5.2. For n � 3 holds:

1) Aut(Fn)/Cong(n, 2) ∼= GLn(Z2),2) |Cong(n, 2) : SCong(n, 2)| = 2,

3) Cong(n, 2) =2n−1⋂i=1

St(F (i)2n−1),


4) SCong(n, 2) is generated by the set

X = {αijk | i, j, k ∈ {1, 2, . . . , n}, k �= i, j}⋃

{E2ij , Nij | i, j ∈ {1, 2, . . . , n}, i �= j}.5) The abelianization of SCong(n, 2) is a finite 2-group.

Proof. 1) This statement follows from the definition of the congruence subgroup.2) We define the automorphism ϕ ∈ Aut(Fn) by the rule x1 �→ x1x

22, x2 �→

x1x22x1x2, xi �→ xi for i = 1, . . . , n. It is easy to see, that ϕ ∈ Cong(n, 2) \

SCong(n, 2).3) This statement follows from the chain of identities (with k = 2):

Ker(π ε) = IA(Fn; Ker(πε)

)=2n−1⋂i=1

IA(Fn; Ker(σi)

)=2n−1⋂i=1

Aut(Fn; Ker(σi)

).

The first identity is evident, the second one follows from Proposition 5.1 and the

fact that Ker(πε) =2n−1⋂i=1

Ker(σi). The third identity follows from the fact that

|Fn : Ker (σi)| = 2.4) This statement follows from Corollary 4.4 and Nielsen–Magnus result.5) The abelianization of SCong(n, 2) is finite by Theorem 4.1. The following

computations (where i, j, k are distinct) show, that it is a finite 2-group:

[αiij , E2jk] = α2iik,

[αiij , E2ik] = αikjα

−1jjkαikjαjjk ≡ α2ikj mod IA(Fn)′,

[E2ij , Njk] = E4ij ,

N2ij = 1. �Remark 5.3. Using first Johnson homomorphism and some homological methods,T. Satoh proved in [19], that for n � 2 and k � 2 holds

Cong(n, k)′ ∼= IA(Fn)′ ⊗Z Zk ⊕ Γ(n, k)′,

where Γ(n, k) is the congruence subgroup of GLn(Z) modulo k.

6. A subgroup K(n) of index 2 in SCong(n, 2)

We have the following chain of canonical embeddings and epimorphisms.

SCong(n, 2)θ1↪→ St(F (1)2n−1)

θ2↪→ Aut(F2n−1)

θ3→ GL2n−1(Z).

Here θ1 is the embedding due to Proposition 5.2.3, θ2 is the homomorphism,which sends every automorphism α of Fn with α(F

(1)2n−1) = F

(1)2n−1 to its restriction

on F (1)2n−1. In fact, θ2 is injective. The epimorphism θ3 is standard.Now we set θ = θ1θ2θ3 and define the subgroup

K(n) = θ−1(SL2n−1(Z)

).


Proposition 6.1. For n � 3 holds

1) |SCong(n, 2) : K(n)| = 2,2) |IA(Fn) : IA(Fn) ∩K(n)| = 2,3) |Inn(Fn) : Inn(Fn) ∩K(n)| is 1 if n is odd, and is 2 if n is even,4) K(n) is generated by the set

Y = X \ {αiin, Nin | i = 1, 2, . . . , n− 1} ∪ {N1nαiin | i = 1, 2, . . . , n− 1},where

X = {αijk | i, j, k ∈ {1, 2, . . . , n}, k �= i, j}⋃{E2ij , Nij | i, j ∈ {1, 2, . . . , n}, i �= j}

is the generating set for SCong(n, 2) defined in Proposition 5.2.

Proof. 1) Clearly, |SCong(n, 2) : K(n)| � 2. One can check that

N1n ∈ SCong(n, 2) \K(n).

Therefore this index is indeed 2.2) Since IA(Fn) � SCong(n, 2), we have by 1), that |IA(Fn) : IA(Fn) ∩

K(n)| � 2. One can verify, that α11n ∈ IA(Fn) \ K(n). Therefore this index isindeed 2.

3) By 2), this index does not exceed 2. For x ∈ Fn let x be the conjugationof Fn by x, i.e., x(y) = x−1yx for y ∈ Fn. To prove the statement, it is sufficientto check that x1, . . . , xn−1 ∈ K(n) and that xn ∈ K(n) if and only if n is odd.

4) We take {1, N1n} as the set of representatives of the cosets of K(n) inSCong(n, 2). For g ∈ SCong(n, 2), we denote by g the representative of the cosetK(n)g. One can easily check, that for g ∈ X holds

g =

{N1n, if g ∈ {Nin, αiin | i = 1, . . . , n− 1}1, if g ∈ X \ {Nin, αiin | i = 1, . . . , n− 1}.

By the Reidemeister–Schreier method, K(n) is generated by the elements

N1nxN1nx−1, where x runs through X . We show, that these elements can be

expressed as products of elements of Y ±1. Consider the following cases.

I. Let x = αiin, where i ∈ {1, . . . , n− 1}.Then N1nxN1nx

−1= N1nαiin ∈ Y .

II. Let x = αijk, where i, j, k ∈ {1, . . . , n} are distinct or x = αiik with k �= n (thecase x = αiik with k = n was considered above).

Then N1nxN1nx−1

= N1nαijkN−11n .

We consider several cases and rewrite this element as a product of elementsof Y ±1. We will use the fact that αijk = α−1ikj if i, j, k are distinct.

Case 1. {1, n} ⊆ {i, j, k}.Subcase 1.1. αijk is of the first kind, i.e., i, j, k are different.


a) i = 1. Then n ∈ {j, k} and we may assume that n = j. Then

N1nα1nkN−11n = N1kα

−111kE

−21n α1knα11kE

21nN

−11k .

b) i = n. Then 1 ∈ {j, k} and we may assume that j = 1. Then

N1nαn1kN−11n = α−1n1kαnnkα

−1nn1α

−1nnkαnn1αkk1αn1kα

−1kk1αn1k.

c) i �= 1, n. Then {1, n} = {j, k} and we may assume that j = 1 and k = n.Then

N1nαi1nN−11n = E−2in α−1ii1αin1E

2inαii1.

Subcase 1.2. αijk is of the second kind, i.e., it is αiik. By our assumption in II wehave k �= n. Therefore i = n and k = 1, and we have

N1nαnn1N−11n = α−1nn1.

Case 2. n ∈ {i, j, k}, 1 /∈ {i, j, k}. Then we choose t ∈ {i, j, k} \ {n} and write

N1nαijkN−11n = N1t(NtnαijkN

−1tn )N−11t .

The expression in the brackets can be computed as in Case 1 (by replacing 1 by t).

Case 3. n /∈ {i, j, k}, 1 ∈ {i, j, k}.Subcase 3.1. αijk is of the first kind, i.e., i, j, k are different.

a) i = 1. Then

N1nα1jkN−11n = α1kj · α11kα11jα−111kα−111j .

b) j = 1. Then

N1nαi1kN−11n = α−1ii1αik1αii1.

c) k = 1. This case reduces to Case b), since αijk = α−1ikj .

Subcase 3.2. αijk is of the second kind, i.e., it is αiik.a) i = 1. Then

N1nα11kN−11n = α11k.

b) k = 1. Then

N1nαii1N−11n = α−1ii1 .

Case 4. {1, n} ∩ {i, j, k} = ∅.

N1nαijkN−11n = αijk.


III. Let x = E2ij .

Then N1nxN1nx−1

= N1nE2ijN

−11n . The following formulas show, that this

element belongs to 〈Y 〉.N1nE

2njN

−11n = α2nnjE

−2nj (j �= 1, n),

N1nE21jN

−11n = α211jE

−21j (j �= 1, n),

N1nE2inN

−11n = E−2in (i �= 1, n),

N1nE2i1N

−11n = E−2i1 (i �= 1, n),

N1nE2n1N

−11n = α−2nn1E

2n1,

N1nE21nN

−11n = N12E

−21n N

−112 .

IV. Let x = Nij .

If i, j �= n, then N1nxN1nx−1

= Nij . If, say j = n, then i �= n and

N1nxN1nx−1

= N1i. �

7. K(3) and some its overgroups with infinite abelianization

The congruence subgroup SCong(3, 2) has index 2 · 168 in Aut(F3) and containsIA(F3). Therefore, by Theorem 4.1, it has a finite abelianization. Moreover, byProposition 5.2, this abelianization is a finite 2-group.

The subgroup K(3) has index 2 in SCong(3, 2) and does not contain IA(F3)by Proposition 6.1. This was an indication for us to check that the abelianizationof K(3) is infinite. In this section we also construct some overgroups of K(3),namely A(3), B(3) and C(3), with infinite abelianizations.

In computing the abelianizations of these groups, we used their generators,Nielsen’s presentation of Aut(F3) (see [18] or [13]) and the Reidemeister–Schreiermethod implemented in the GAP package. By Proposition 6.1 we have the follow-ing 16

Generators of K(3):(1) α112, α221, α331, α332, α123, α213, α312,(2) E212, E

213, E

221, E

223, E

231, E

232,

(3) N13α113, N13α223, N12.

Theorem 7.1.

1) K(3) has index 4 · 168 in Aut(F3).2) K(3)/K(3)′ ∼= Z142 × Z× Z.

Corollary 7.2. The above 16 automorphisms form a minimal generating set ofK(3). All of them, except of α123, α213, E212, E

221 have finite order in K(3)/K(3)′.

These four automorphisms have infinite order in K(3)/K(3)′.Moreover, E−412 ≡ α2123(modK(3)′) and E−421 ≡ α2213(modK(3)′). In particu-

lar, the image of the group 〈E212, E221〉 in K(3)/K(3)′ is isomorphic to Z× Z.


Proof. The first statement of this corollary follows straightforward from Theo-rem 7.1. The second statement follows from the formulas

[α221, N12] = α2221,

[α112, N12] = α2112,

[α331, E212] = α2332,

[α332, E221] = α2331,

[α331, E232] = α321α−1112α321α112 ≡ α2321 = α−2312 (modK(3)′),

[E231, N12] = E431,

[E232, N12] = E432,

[α221, E213][E223, N12] = E423,

[α112, E223][E213, N12] = E413,

(N13α113)2 = 1,

(N13α223)2 = 1,

N212 = 1.

This statement and Theorem 7.1.2) imply that the image of

〈α123, α213, E212, E221〉 in K(3)/K(3)′

can be mapped onto Z × Z. Therefore all remaining statements of the corollarywill follow, if we prove the congruences E−412 ≡ α2123(modK(3)′) and E−421 ≡α2213(modK(3)′). The first congruence follows from the identity

[α113N23, E212] = α332α−1123α

−1332α

−2112α

−1123α

2112E

−412 ≡ α−2123E

−412 (modK(3)′)

and the fact that the commutator [α113N23, E212] belongs to K(3)′; indeed, we haveα113N23 ∈ K(3) and E212 ∈ K(3). The second congruence follows similarly if weexchange indices 1 and 2. �

Still, the index of K(3) in Aut(F3) is large. We will enlarge K(3) (and sodecrease the index) by adding special generators. In this way we construct thefollowing chain of subgroups:

Aut(F3) � C(3) � B(3) � A(3) � K(3),

where

A(3) = 〈K(3), E31, E32〉,B(3) = 〈K(3), E31, E32, E21〉,C(3) = 〈K(3), E31, E32, E21, N3〉.


Theorem 7.3.

1) A(3) has index 168 in Aut(F3).2) A(3)/A(3)′ ∼= Z72 × Z× Z.3) A(3) has the following minimal set of generators:

α112, α221, α123, α213, E213, E

223, E31, E32, N12.

Theorem 7.4.

1) B(3) has index 84 in Aut(F3).2) B(3)/B(3)′ ∼= Z42 × Z.

Remark. We do not know a minimal generating set of B(3).

Theorem 7.5.

1) C(3) has index 42 in Aut(F3).2) C(3)/C(3)′ ∼= Z32 × Z4 × Z.3) C(3) has the following minimal set of generators:

α123, E213, E21, E32, N3.

Questions 7.6. Does there exist a subgroup of Aut(F3) of index smaller than 42,which can be mapped onto Z?

8. The group K(n) for n � 4

Theorem 8.1. If n � 4, then the abelianization of K(n) is a finite 2-group.

Proof. By Proposition 6.1, K(n) is generated by the set

Y = X \ {αiin, Nin | i = 1, 2, . . . , n− 1} ∪ {N1nαiin | i = 1, 2, . . . , n− 1},where

X = {αijk | i, j, k ∈ {1, 2, . . . , n}, k �= i, j}⋃{E2ij , Nij | i, j ∈ {1, 2, . . . , n}, i �= j}.

We show, that the square of each y ∈ Y is trivial modulo K(n)′.1. We show that E4ij ≡ 1 modK(n)′.

For j �= n, this follows from the identity

[E−2ij , Njk] = E−4ij ,

where we choose k /∈ {i, j, n}.For j = n, this follows from the identity

[αiik, E2kn][E

2in, Nik] = E4in,

where we choose k /∈ {i, n}.2. We show that α2iij ≡ 1modK(n)′.

This follows from the identity

[αiik, E2kj ] = α2iij ,


where we choose k /∈ {i, j, n}. Note an interesting fact, that αiin /∈ K(n), butα2iin ∈ K(n)′.3. We show that α2ijk ≡ 1modK(n)′ for different i, j, k.

If n /∈ {j, k}, then this follows from[αiik, E

2ij ] = αijkα

−1kkjαijkαkkj ≡ α2ijk modK(n)′

If n ∈ {j, k}, we may assume that k = n and then the required congruence followsfrom

[αiinNln, E2ij ] = αijnαiijαijnα

−1iij ≡ α2ijnmodK(n)′,

where we take l /∈ {i, j, n}. Note, that in this case αiinNln lies in K(n).4. One can easily check, that (N1nαiin)2 = 1 holds for all i = 1, . . . , n− 1. Finally,it is clear that N2ij = 1. �

Remark 8.2. Using the GAP package we have proved that the abelianization ofK(4) is isomorphic to Z382 .

Questions 8.3. Is it true that K(n)′/K(n)′′ is infinite for n � 4?

If this is true, then Aut(Fn) has no the Kazhdan property (T) for n � 4.

References

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[12] W. Magnus, Uber n-dimensionale Gittertransformationen, Acta Math., 64 (1935),353–367,

[13] W. Magnus, A. Karrass, D. Solitar, Combinatorial group theory, New York: Wiley,1966.

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O. BogopolskiInstitute of Mathematics ofSiberian Branch of Russian Academy of SciencesNovosibirsk, Russia

and

University of Dusseldorf, Germanye-mail: Oleg [email protected]

R. VikentievInstitute of Mathematics ofSiberian Branch of Russian Academy of SciencesNovosibirsk, Russia

and

Novosibirsk State Universitye-mail: [email protected]



Dynamics of Free Group Automorphisms

Peter Brinkmann

Abstract. We present a coarse convexity result for the dynamics of free groupautomorphisms: Given an automorphism φ of a finitely generated free groupF , we show that for all x ∈ F and 0 ≤ i ≤ N , the length of φi(x) is boundedabove by a constant multiple of the sum of the lengths of x and φN(x), withthe constant depending only on φ.

Mathematics Subject Classification (2000). 37E30.

Keywords. Free group automorphisms, train tracks.

Introduction

The following theorem is the main result of this paper. It follows from a technicalresult (Theorem 1.9) that uses the machinery of improved relative train track mapsof Bestvina, Feighn, and Handel [BFH00].

Theorem 0.1. Let φ : F → F be an automorphism of a finitely generated freegroup. Then there exists a constant K ≥ 1 such that for any pair of exponents N, isatisfying 0 ≤ i ≤ N , the following two statements hold:

1. If w is a cyclic word in G, then

||φi#(w)|| ≤ K

(||w||+ ||φN

#(w)||),

where ||w|| is the length of the cyclic reduction of w with respect to some wordmetric on F .

2. If w is a word in F , then

|φi#(w)| ≤ K

(|w|+ |φN (w)|

),

where |w| is the length of w.

Given an improved relative train track representative of some power of φ, theconstant K can be computed.

Remark 0.2 (A note on computability). Given an automorphism φ : F → F, we cancompute a relative train track representative of φ [BH92, DV96]. The construction

20 P. Brinkmann

of improved relative train track maps, however, involves a compactness argumentin a universal cover [BFH00, Proof of Proposition 5.4.3] that is not constructive.A number of algorithmic improvements of relative train tracks appear in [Bri], inthe context of an algorithm that detects automorphic orbits in free groups.

The statement of the theorem does not depend on the choice of generators ofF . The intuitive meaning of the theorem is that the map i �→ |φi(w)| is coarselyconvex for all words w ∈ F . Klaus Johannson informed me that a similar result isa folk theorem in the case of surface homeomorphisms. Also, while free-by-cyclicgroups are not, in general, CAT(0)-groups [Ger94], Theorem 0.1 suggests that theirdynamics mimics that of CAT(0)-groups. Theorem 0.1 complements the followingstrong convexity result in an important special case.

Theorem 0.3 ([Bri00]). If φ : F → F is an atoroidal automorphism, i.e., φ hasno nontrivial periodic conjugacy classes, then φ is hyperbolic, i.e., there exists aconstant λ > 1 such that

λ|x| ≤ max{|φ±1(x)|

}for all x ∈ F .

I originally set out to prove Theorem 0.1 because it immediately implies thatin a free-by-cyclic group

Γ = F �φ Z = 〈 x1, . . . , xn, t | t−1xit = φ(xi) 〉,words of the form t−kwtkφk(w−1) satisfy a quadratic isoperimetric inequality.(Note, however, that Theorem 0.1 is stronger than the mere existence of a qua-dratic isoperimetric inequality for such words.) Natasa Macura previously proveda quadratic isoperimetric inequality for mapping tori of automorphisms of polyno-mial growth [Mac00]. Martin Bridson and Daniel Groves have since proved that allfree-by-cyclic groups satisfy a quadratic isoperimetric inequality [BG]. They alsoobtain a new proof of Theorem 0.1 as an application of their techniques.

In Section 1, we review the pertinent definitions and results regarding traintrack maps from [BFH00]. We also state the main technical result, Theorem 1.9,and we show how Theorem 0.1 follows from Theorem 1.9. Section 2 provides somemore results on train tracks and automorphisms of free groups. Section 3 introducessome notation and terminology and lists a number of examples that illustrate someof the issues and subtleties that need to be addressed in the proof of Theorem 1.9.Section 4 establishes a technical proposition that may be of independent interest.Section 5 and Section 6 contain the proof of Theorem 1.9. Finally, the glossarylists some of the technical definitions for the convenience of the reader.

I would like to express my gratitude to Ilya Kapovich for many helpful discus-sions, to Mladen Bestvina for patiently answering my questions, to Steve Gerstenfor encouraging me to write up this result for its own sake, to the University ofOsnabruck for their hospitality, and to Swarup Gadde and the University of Mel-bourne as well as the Max-Planck-Institute of Mathematics for their hospitalityand financial support. Klaus Johannson and Richard Weidmann kindly served asa sounding board while I was working on the exposition of this paper.

Dynamics of Free Group Automorphisms 21

1. Improved relative train track maps

In this section, we review the theory of train tracks developed in [BH92, BFH00].We will restrict our attention to the collection of those results that we will use inthis paper.

Given an automorphism φ ∈ Aut(F ), we can find a based homotopy equiva-lence f : G→ G of a finite connected graph G such that π1(G) ∼= F and f inducesφ. This observation allows us to apply topological techniques to automorphismsof free groups. In many cases, it is convenient to work with outer automorphisms.Topologically, this means that we work with homotopy equivalences rather thatbased homotopy equivalences.

Oftentimes, a homotopy equivalence f : G → G will respect a filtration ofG, i.e., there exist subgraphs G0 = ∅ ⊂ G1 ⊂ · · · ⊂ Gk = G such that for eachfiltration element Gr, the restriction of f to Gr is a homotopy equivalence of Gr.The subgraph Hr = Gr \Gr−1 is called the rth stratum of the filtration. We saythat a path ρ has nontrivial intersection with a stratum Hr if ρ crosses at leastone edge in Hr.

If E1, . . . , Em is the collection of edges in some stratum Hr, the transitionmatrix of Hr is the nonnegative m×m-matrixMr whose ijth entry is the numberof times the f -image of Ej crosses Ei, regardless of orientation. Mr is said to beirreducible if for every tuple 1 ≤ i, j ≤ m, there exists some exponent n > 0 suchthat the ijth entry of Mn

r is nonzero. If Mr is irreducible, then it has a maximalreal eigenvalue λr ≥ 1 [Gan59]. We call λr the growth rate of Hr.

Given a homotopy equivalence f : G→ G, we can always find a filtration ofG such that each transition matrix is either a zero matrix or irreducible. A stratumHr in such a filtration is called zero stratum if Mr = 0. Hr is called exponentiallygrowing if Mr is irreducible with λr > 1, and it is called polynomially growing ifMr is irreducible with λr = 1.

An unordered pair of edges in G originating from the same vertex is calleda turn. A turn is called degenerate if the two edges are equal. We define a mapDf : {turns in G} → {turns in G} by sending each edge in a turn to the first edgein its image under f . A turn is called illegal if its image under some iterate of Dfis degenerate, legal otherwise.

An edge path ρ = E1E2 · · ·Es is said to contain the turns (E−1i , Ei+1) for1 ≤ i < s. ρ is said to be legal if all its turns are legal, and a path ρ ⊂ Gr is r-legalif no illegal turn in α involves an edge in Hr.

Let ρ be a path in G. In general, the composition fk ◦ ρ is not an immersion,but there is exactly one immersion that is homotopic to fk ◦ ρ relative endpoints.We denote this immersion by fk

#(ρ), and we say that we obtain fk#(ρ) from fk ◦ ρ

by tightening. If σ is a circuit in G, then fk#(σ) is the immersed circuit homotopic

to fk ◦ σ.

Remark 1.1. A path is tightened by cancelling adjacent pairs of inverse edges untilno inverse pairs are left. The result of such a sequence of cancellations is uniquely

22 P. Brinkmann

determined, but the sequence is not. For instance, EE−1E may be tightened asE(E−1E) or (EE−1)E.

Convention 1.2. Let ρi, i = 1, . . . , k be paths that can be concatenated to form apath ρ = ρ1ρ2 · · · ρk. When tightening f(ρ) to obtain f#(ρ), we adopt the conven-tion that we first tighten the images of ρi to f#(ρi). In a second step, we tightenthe concatenation f#(ρ1) · · · f#(ρk) to f#(ρ).

In many situations, the length of a subpath ρi will be greater than the numberof edges that cancel at either end, in which case it makes sense to talk about edgesin f#(ρ) originating from ρi.

A path ρ is a (periodic) Nielsen path if fk#(ρ) = ρ for some k > 0. In this

case, the smallest such k is the period of ρ. A Nielsen path ρ is called indivisible ifit cannot be expressed as the concatenation of shorter Nielsen paths. A path ρ isa pre-Nielsen path if fk

#(ρ) is Nielsen for some k ≥ 0.A decomposition of a path ρ = ρ1 · ρ2 · · · · · ρs into subpaths is called a k-

splitting if fk#(ρ) = fk

#(ρ1) · · · fk#(ρs). Such a decomposition is a splitting if it is

a k-splitting for all k > 0. We will also use the notion of k-splittings of circuitsσ = ρ1 · ρ2 · · · · · ρs, which requires, in addition, that there be no cancellationbetween fk

#(ρs) and fk#(ρ1).

The following theorem was proved in [BH92].

Theorem 1.3 ([BH92, Theorem 5.12]). Every outer automorphism O of F is repre-sented by a homotopy equivalence f : G→ G such that each exponentially growingstratum Hr has the following properties:1. If E is an edge in Hr, then the first and last edges in f(E) are contained

in Hr.2. If β is a nontrivial path in Gr−1 with endpoints in Gr−1 ∩Hr, then f#(β) is

nontrivial.3. If ρ is an r-legal path, then f#(ρ) is an r-legal path.

We call f a relative train track map.A path ρ in G is said to be of height r if ρ ⊂ Gr and ρ �⊂ Gr−1. If Hr = {Er}

is a polynomially growing stratum, then basic paths of height r are of the formErγ or ErγE

−1r , where γ is a path in Gr−1. If τ is a closed Nielsen path in Gr−1

and f(Er) = Erτl for some l ∈ Z, then paths of the form Erτ

k and ErτkE−1r are

exceptional paths of height r. Moreover, if s < r, τ ⊂ Gs−1, and f(Es) = Esτm,

then ErτkE−1s is also a exceptional path of height r.

For our purposes, the properties of relative train track maps are not strongenough, so we will use the notion of improved train track maps constructed in[BFH00]. We only list the properties used in this paper.

Theorem 1.4 ([BFH00, Theorem 5.1.5, Lemma 5.1.7, and Proposition 5.4.3]). Forevery outer automorphism O of F , there exists an exponent k > 0 such that Ok isrepresented by a relative train track map f : G→ G with the following additionalproperties:


1. If Hr is a zero stratum, then Hr+1 is an exponentially growing stratum, andthe restriction of f to Hr is an immersion. Hr is a zero stratum if and onlyif it is the union of the contractible components of Gr.

2. If v is a vertex, then f(v) is a fixed vertex. If Hr is a polynomially growingstratum and G′ is the collection of noncontractible components of Gr−1, thenall vertices in Hr ∩G′ are fixed.

3. If Hr is an exponentially growing stratum, then there is at most one indivisibleNielsen path τ of height r. If τ is not closed and if it starts and ends atvertices, then at least one endpoint of τ is not contained in Hr ∩Gr−1.

4. If Hr is a polynomially growing stratum, then Hr consists of a single edgeEr, and f(Er) = Er · ur for some closed path ur ⊂ Gr−1 whose base point isfixed by f .If σ ⊂ Gr is a basic path of height r that does not split as a concatenation oftwo basic paths of height r or as a concatenation of a basic path of height rwith a path contained in Gr−1, then either fk

#(σ) = Er · σ′ for some k ≥ 0,or ur is a Nielsen path and fk

#(σ) is an exceptional path of height r for somek ≥ 0.

We call f an improved relative train track map.Finally, we state a lemma from [BFH00] that simplifies the study of paths

intersecting strata of polynomial growth.

Lemma 1.5 ([BFH00, Lemma 4.1.4]). Let f : G → G be an improved train trackmap with a polynomially growing stratum Hr. If ρ is a path in Gr, then it splitsas a concatenation of basic paths of height r and paths in Gr−1.

Remark 1.6. In fact, part 4 of Theorem 1.4 implies that subdividing ρ at the initialendpoints of all occurrences of Er and at the terminal endpoints of all occurrencesof E−1r yields a splitting of ρ into basic paths of height r and paths in Gr−1.

Observe that ifHr = {Er} is a polynomially growing stratum, then fk#(Er) =

Er · ur · f#(ur) · · · · · fk−1# (ur). Each subpath of the form f i

#(ur) is called a blockof fk

#(Er). Since there is no cancellation between successive blocks, it makes senseto refer to the infinite path

Rr = ur · f#(ur) · f2#(ur) · · · · (1)

as the eigenray of Er.

Remark 1.7 (A note on terminology). The notion of a polynomially growing stra-tum Hr = {Er} first appeared in [BH92]. Polynomially growing strata are callednonexponentially growing strata in [BFH00]. Both terms are somewhat misleadingbecause the function k �→ |fk

#(Er)| may grow exponentially (see Lemma 2.4).

Given an improved train track map f : G → G, we construct a metric onG. If Hr is an exponentially growing stratum, then its transition matrix Mr hasa unique positive left eigenvector vr (corresponding to λr) whose smallest entry

24 P. Brinkmann

equals one [Gan59]. For an edge Ei in Hr, the eigenvector vr has an entry li > 0corresponding to Ei. We choose a metric on G such that Ei is isometric to aninterval of length li, and such that edges in zero strata or in polynomially growingstrata are isometric to an interval of length one. For a path ρ, we denote its lengthby L(ρ). Note that if the endpoints of ρ are vertices, then the number of edges inρ provides a lower bound for L(ρ). Moreover, if f is an absolute train track map,then f expands the length of legal paths by the factor λ.

Remark 1.8. We merely choose this metric for convenience. All statements hereare invariant under bi-Lipschitz maps, but our metric of choice simplifies the pre-sentation of our arguments.

We are now ready to state the main technical result of this paper.

Theorem 1.9. Let φ : F → F be an an automorphism. Then there exists an im-proved relative train track map representing some positive power of φ for whichthere exists a constant K ≥ 1 with the following property: For any pair of expo-nents N, i satisfying 0 ≤ i ≤ N , the following two statements hold:1. If σ is a circuit in G, then

L(f i#(σ)

)≤ K

(L(σ) + L

(fN# (σ)

)).

2. If ρ is a path in G that starts and ends at vertices, then

L(f i#(ρ)

)≤ K

(L(ρ) + L

(fN# (ρ)

)).

Given the improved relative train track map f : G → G, the constant K can becomputed.

We will present the proof of Theorem 1.9 in Section 5 and Section 6. Rightnow, we show how Theorem 0.1 follows from Theorem 1.9.

Proof of Theorem 0.1. Let φ : F → F be an automorphism of a finitely generatedfree group F = 〈x1, . . . , xn〉. The first part of Theorem 1.9 immediately impliesthat the first part of Theorem 0.1 holds for some positive power φk, i.e., thereexists some K ′ ≥ 1 such that for all 0 ≤ i ≤ N and w ∈ F , we have

||φik# (w)|| ≤ K ′ (||w||+ ||φNk

# (w)||),

where we compute lengths with respect to the generators x1, . . . , xn.Let L = max{|φ±1(xi)|}. Then, for 0 ≤ j < k, we have

L−k||φik+j(w)|| ≤ ||φik(w)|| ≤ Lk||φik+j(w)||for all w ∈ F . We conclude that for all 0 ≤ i ≤ N and w ∈ F , we have

L−k||φi#(w)|| ≤ K ′ (||w|| + Lk||φN

#(w)||),

so that the first part of Theorem 0.1 holds with K = L2kK ′.In order to prove the second assertion, we modify a trick from [BFH97]. Let

F ′ be the free group generated by x1, . . . , xn and an additional generator a. We


define an automorphism ψ : F ′ → F ′ by letting ψ(xi) = φ(xi) for all 1 ≤ i ≤ n,and ψ(a) = a.

By the previous step, the first part of Theorem 0.1 holds for ψ, with someconstantK ′ ≥ 1. Let w be some word in F . Then, for all i ≥ 0, ψi(aw) is a cyclicallyreduced word in F ′, so that we have |φi(w)| + 1 = ||ψi(aw)||. We conclude that

|φi(w)| + 1 ≤ K ′(|w| + |φN (w)|+ 2),

for all 0 ≤ i ≤ N . Now the second assertion of Theorem 0.1 holds with K =2K ′. �

2. More on train tracks

Thurston’s bounded cancellation lemma is one of the fundamental tools in thispaper. We state it in terms of homotopy equivalences of graphs.

Lemma 2.1 (Bounded cancellation lemma [Coo87]). Let f : G→ G be a homotopyequivalence. There exists a constant Cf , depending only on f , with the propertythat for any tight path ρ in G obtained by concatenating two paths α, β, we have

L(f#(ρ)) ≥ L(f#(α)) + L(f#(β)) − Cf .An upper bound for Cf can easily be read off from the map f [Coo87]. Let

f : G→ G be an improved relative train track map with an exponentially growingstratum Hr with growth rate λr. The r-length of a path ρ in G, Lr(ρ), is the totallength of ρ ∩Hr.

If β is an r-legal path in G whose r-length satisfies λrLr(β) − 2Cf > Lr(β)and α, γ are paths such that the concatenation αβγ is an immersion, then ther-length of the segment in fk

#(αβγ) corresponding to β (Convention (1.2)) willtend to infinity as k tends to infinity. The critical length Cr of Hr is the infimumof the lengths satisfying the above inequality, i.e.,

Cr =2Cf

λr − 1. (2)

We now list some additional technical results about improved train trackmaps. The following lemma is an immediate consequence of [Bri00, Prop. 6.2].If Hr is an exponentially growing stratum, and ρ is a path of height r, we let n(ρ)denote the number of r-legal segments in ρ.

Lemma 2.2. Let f : G → G be a relative train track map, and let Hr be an expo-nentially growing stratum. For each L > 0, there exists some computable exponentM > 0 such that if ρ is a path or circuit in Gr containing at least one full edge inHr, one of the following three statements holds:1. fM

# (ρ) has an r-legal segment of r-length greater than L.2. n(fM

# (ρ)) < n(ρ).3. ρ can be expressed as a concatenation τ1ρ′τ2, where τ1, τ2 each contain at most

one r-illegal turn, the r-length of the r-legal segments of τ1, τ2 is at most L,

26 P. Brinkmann

and ρ′ splits as a concatenation of pre-Nielsen paths (with one r-illegal turneach) and segments in Gr−1. Moreover, fM

# (ρ′) is a concatenation of Nielsenpaths of height r and segments in Gr−1.

Remark 2.3.

• The statement of Lemma 2.2 in [Bri00] does not explicitly mention the com-putability of M . The proof, however, only uses counting arguments, fromwhich the constant M can be computed.

• The presence of the subpaths τ1, τ2 in Part 2.2 is an artifact of the fact thatρ need not start or end at fixed points if it is a path. If ρ starts at a fixedpoint, then τ1 will be trivial, and if ρ ends at a fixed point, then τ2 will betrivial.

• The actual statement of [Bri00, Proposition 6.2] does not mention circuitssince they were not a concern in the context of [Bri00]. The proof, however,works for circuits as well as paths. If the first two statements of Lemma 2.2do not hold, than the third statement will hold with τ1 and τ2 trivial.

From now on, we assume that f : G → G that f is an improved train trackmap. Throughout the rest of this section, let M be the constant from Lemma 2.2for some fixed L > Cr (Equation 2).

Let Hr = {Er} be a polynomially growing stratum. We say that Hr is trulypolynomial if ur is trivial or, inductively, if ur is a concatenation of truly polyno-mial edges and Nielsen paths in exponentially growing strata. Clearly, Er is trulypolynomial if and only if the map k �→ |fk

#(Er)| grows polynomially. We say thata polynomially growing stratum is fast if it is not truly polynomial.

The following lemma give us an understanding of the growth of fast polyno-mial strata.

Lemma 2.4. There exists an exponent k0 with the following property: For all fastpolynomial strata Hr = {Er} there exists some s < r such that Hs is of expo-nential growth and fk0

# (Er) contains an s-legal subpath of height s whose s-lengthexceeds Cs.

In particular, this lemma implies that fast polynomial strata grow exponen-tially. Given an improved relative train track map, we can find k0 by successivelyevaluating f#, f2#, . . . until we see long legal segments in all images of fast polyno-mial edges.

Proof. We introduce classes of fast polynomial edges. Let Hr = {Er} be a fastpolynomial edge such that f(Er) = Erur. We say that Hr has class 1 if thereexists some s < r such that Hs is an exponentially growing stratum, ur ∩ Hs

does not only consist of Nielsen paths and paths of height less than s, and if ur

contains any polynomial edges Et for some t > s, then Et is truly polynomial. Werecursively define a fast polynomial edge Er to have class k if the highest class ofedges in ur is k − 1.


If Hr has class 1, then ur contains a subpath ρ of height s such that fk#(ρ)

contains a long s-legal segment for some sufficiently large k (Lemma 2.2). If ur

contains any subpaths whose height exceeds s, then by definition those subpathswill grow at most polynomially, so that eventually, the exponential growth of ρwill prevail.

In order to prove the lemma for an edge of class k, k > 1, we observe that noedges of class k − 1 are cancelled when fm(ur) is tightened to fm

# (ur). Now thelemma follows by Theorem 1.4, Part 4, and induction. �

Assume that Hr is an exponentially growing stratum, and let ρ be a path ofheight r. If Hr does not support a closed Nielsen path, then we let N(ρ) = n(ρ).If Hr supports a closed Nielsen path, then we let N(ρ) equal the number of legalsegments in ρ that do not overlap with a closed Nielsen subpath of ρ.

The following lemma is a generalization of [Bri00, Lemma 6.4].

Lemma 2.5. Assume that Hr is an exponentially growing stratum. There existcomputable constants λ > 1, N0 with the following property: If fM

# (ρ) does notcontain a legal segment of length at least L, and if N(ρ) > N0, then

N(fM# (ρ)) ≤ λ−1N(ρ).

Regardless of N(ρ), we have

N(fM# (ρ)) ≤ λ−1N(ρ) + 1.

Proof. If Hr does not support a closed Nielsen path, then the proof of [Bri00,Lemma 6.4] goes through unchanged. We repeat the argument here because theideas of the proof show up more clearly in this case.

If Hr does not support a closed Nielsen path, then the proof is based onthe following observation: If N(ρ) = 6 and f#(ρ) does not contain a long legalsegment, thenN(fM

# (ρ)) < 6. Suppose otherwise, i.e.,N(fM# (ρ)) = N(ρ). Then, by

Lemma 2.2, fM# (ρ) = τ1γτ2, where γ is a concatenation of three indivisible Nielsen

paths of height r and paths in Gr−1. This is impossible because by Theorem 1.4,Part 3, we can concatenate no more than two indivisible Nielsen paths of height rwith paths in Gr−1.

Hence, of every six consecutive legal segments in ρ, at least one cancels com-pletely when fM (ρ) is tightened to fM

# (ρ). This implies that if N(ρ) ≥ 6, thenN(fM

# (ρ)) ≤ 1011N(ρ). In order to see why this choice of λ works, we just observe

that if ρ consists of eleven legal segments and the sixth one cancels in fM# (ρ), then

there are no six consecutive legal segments that survive in fM# (ρ).

This completes the proof of the first inequality, with λ = 1110 and N0 = 5, if

Hr does not support a closed Nielsen path. Regarding the second inequality, weremark that if N(ρ) ≤ N0, then N(fM

# (ρ)) ≤ N(ρ) ≤ λ−1N(ρ) + 1.We now assume that Hr supports a closed indivisible Nielsen path σ. The

proof in this case is based on the following consequence of Lemma 2.2. If γ a path ofheight r, n(γ) = n(fM

# (γ)) = 4, and fM# (γ) does not contain a long legal segment,

28 P. Brinkmann

then fM# (γ) = τ1σ

±1τ2, where τ1 and τ2 are as in Lemma 2.2. Intuitively, thismeans that if few legal segments disappear, then many Nielsen paths will appear.Since N(ρ) only counts those legal segments that do not overlap with a Nielsenpath, this observation will yield the desired estimate.

First, consider a path γ of height r that does not contain any Nielsen sub-paths, i.e., we have N(γ) = n(γ). If N(γ) ≥ 4, then for every four consecutivelegal segments whose images do not cancel completely in fM

# (γ), fM# (γ) contains

at least one Nielsen subpath, so that we have N(fM# (γ)) ≤ 6

7N(γ), using the samereasoning as above.

We claim that if γ starts and ends at fixed points, then, by Remark 2.3, wehave N(fM

# (γ)) ≤ 67N(γ) regardless of N(γ). To this end, we first argue that if γ

starts and ends at fixed points, then n(fM# (γ)) < n(γ). If this were not true, then,

by Lemma 2.2 we would have fM# (γ) = σm for somem ∈ Z, which would imply that

γ = σm because γ starts and ends at fixed points. This is a contradiction since weassumed that γ does not contain any Nielsen subpaths. Now, if n(γ) = N(γ) < 4,then we conclude that N(fM

# (γ)) ≤ n(fM# (γ)) < n(γ). Now n(γ) < 4 implies that

67n(γ) ≥ n(γ)− 1, which implies that N(fM

# (γ)) ≤ 67n(γ) =

67N(γ).

After these preparations, we express ρ as a concatenation

ρ = ρ1σn1ρ2σ

n2ρ3 · · ·ρkσnkρk+1,

where n1, . . . , nk ∈ Z, and none of the subpaths ρi contains a Nielsen subpath.Note that the subpaths ρ2, . . . , ρk start and end at the base point v of the

Nielsen path σ, which is fixed by f . Hence, for 2 ≤ i ≤ k, we have N(fM# (ρi)) ≤

67N(ρi), and we have N(f

M# (ρ1)) ≤ 6

7N(ρ1) (resp. N(fM# (ρk+1)) ≤ 6

7N(ρk+1)) ifN(ρ1) ≥ 4 (resp. N(ρk+1) ≥ 4).

If N(ρ1) < 4 and N(ρk+1) < 4, we have

N(fM# (ρ)) ≤ N(ρ1) +N(ρk+1) +

67(N(ρ)−N(ρ1)−N(ρk+1))

≤ 6 +67(N(ρ)− 6) ≤ 6

7(1 +N(ρ)).

Similar estimates yield that N(fM# (ρ)) ≤ 6

7 (1 + N(ρ)) regardless of N(ρ1) andN(ρk+1).

If N(ρ) > 11, then 67 (1 + N(ρ)) ≤ 13

14N(ρ), which implies that N(fM# (ρ)) ≤

1314N(ρ) if N(ρ) > 11, so that the first inequality of the lemma holds with λ = 14

13

and N0 = 11. As for the second inequality, we remark that N(fM# (ρ)) ≤ N(ρ) and,

if N(ρ) ≤ 11, then N(ρ) ≤ λ−1N(ρ) + 1. �

The next lemma is a statement about the (absence of) cancellation betweeneigenrays of polynomially growing strata. It is a stronger version of [BFH00, Sub-lemma 1, Page 587].

Lemma 2.6. Let Hi = {Ei} and Hj = {Ej} be polynomially growing strata. LetSi (resp. Sj) be an initial segment of EiRi (resp. EjRj, see Equation 1) such that


the concatenation SiSj is a path. If Ei grows faster than linearly and if an entireblock of Rj is cancelled in fk

#(SiSj) for some k ≥ 0, then no entire block of Ri

will be cancelled in f l#(SiSj) for any l ≥ 0.

Proof. Suppose that at least one block of both Si and Sj cancels. Then there arepaths α, β, and γ such that fk

#(ui) = βγ, f l#(uj) = αβ for some k, l ≥ 0, and

f#(α) = γ (see Figure 1).

Ei

Ej

αββ

γγ

fk#(ui)

f l#(uj)

Figure 1. The idea of the proof of Lemma 2.6.

In particular, we have

Ri = uif#(ui) · · · fk−1# (ui)βf#(α)f#(β)f2#(α) . . . ,

andRj = ujf#(uj) · · · f l−1

# (uj)αβf#(α)f#(β)f2#(α) . . . .

In particular, the path ρ = EiRk−1i αRl−1

j Ej does not split. By Theorem 1.4, ρ isa exceptional path, and both Ei and Ej grow linearly. �

3. Terminology and examples

In this section, we discuss some examples that illustrate some of the main issuesthat we need to address in the proof of Theorem 1.9. Although we are not primarilyconcerned with free-by-cyclic groups in this article, the language of free-by-cyclicgroups will streamline the exposition.

Given a free group Fn = 〈x1, . . . , xn〉 and an automorphism φ of Fn, themapping torus of φ is the free-by-cyclic group

Mφ = 〈x1, . . . , xn, t | t−1xitφ(x−1i )〉.The letter t is called the stable letter of Mφ.

Definition 3.1. A reduced word w in the generators of Mφ is a hallway if w repre-sents the trivial element of Mφ and if w can be expressed as w = w1w2 such thatw1 only contains negative powers of t and w2 only contains positive powers of t[BF92]. Hallways of the form t−kxtkφk(x−1), for x ∈ Fn, are said to be smooth.

30 P. Brinkmann

Any hallway w can be expressed as

w = t−1uk−1t−1uk−2t−1 · · · t−1u1t−1w0tv1tv2t · · · tvk−1tw−1k ,

where w0, wk, u1, . . . , uk−1, v1, . . . , vk−1 are elements of Fn. The words ui and vimay be empty. In fact, a hallway is smooth if and only if all the ui and vi are trivial.For 1 ≤ i < k, we define wi to be the word obtained by tightening uiφ(wi−1)vi.Since w represents the identity, we have wk = φ(wk−1). We call wi the ith slice ofw. The number k is the duration D(w) of the hallway. Figure 2 illustrates thesenotions.

kw0

wiwk

u1

uk−1

v1

v2

vk−1

Figure 2. A hallway.

We say that the instances of letters of Fn that occur in the spelling of w arevisible. Theorem 0.1 states that if w is a smooth hallway, then the length of eachwi is bounded by a constant multiple of the number of visible edges in w.

The following examples illustrate the main issues that arise in the proof. Forthe remainder of this section, let F6 = 〈a, b, c, d, x, y〉, and define φ by letting

a �→ a

b �→ ba

c �→ caa

d �→ dc

x �→ y

y �→ xcy.

This automorphism admits the stratification H1 = {a}, H2 = {b}, H3 = {c},H4 = {d}, and H5 = {x, y}. The restriction of f to the filtration element G3 =H1 ∪ H2 ∪ H3 grows linearly, the restriction to G4 grows quadratically, and thestratum H5 is of exponential growth.


The first example illustrates the behavior of smooth hallways in linearly grow-ing filtration elements.

Example 3.2. Let w0 be a word from the list am, bamb−1, camc−1, for some integerm. Then φ(w0) = w0, so that the length of any slice of the hallway t−kw0t

kw−10 isthe same as the length of w0. Now, let w0 be a word from the list bam, cam, camb−1.Ifm ≥ 0, then |φk+1(w0)| = |φk(w0)|+1 for any k ≥ 0. Ifm < 0, then |φk+1(w0)| =|φk(w0)| − 1 for 0 ≤ k < −m (Figure 3). Hence, the length of each slice of thehallway t−kw0t

kφk(w−10 ) is bounded by the number of visible letters.

b

b

c c

a2a3

Figure 3. Illustration of Example 3.2.

If w0 is an arbitrary word in 〈a, b, c〉, then, by Remark 1.6, it splits as aconcatenation of words from the above lists and their inverses, which implies thatthe lengths of slices of smooth hallways is bounded by the number of visible letters,so that Theorem 0.1 holds with K = 1.

The next example shows that hallways that are not smooth may have sliceswhose length is not bounded in terms of a constant multiple of the number ofvisible edges.

Example 3.3. Let w = t−kct−kb−1t2kbc−1. For i < k, we have wi = a−ib−1, andfor k ≤ i ≤ 2k, we have wi = ca2k−ib−1 (Figure 4). In particular, there is a sliceof length k + 2 although there are only four visible edges in w. Informally, onemight say that hallways of this form bulge in the middle. A similar bulge occursfor hallways of the form w = t−kb−1t−kbtkb−1tkb.

The next example shows that we need to control the size of such bulges whenproving Theorem 1.9.

Example 3.4. First, note that for k ≥ 1, the last letter in the words f−k(xc)is always one of x, y, x−1, y−1, so that words of the form w0 = φ−k(xc)b−1 arereduced, and we have φk(w0) = xcakb−1 and φ2k(w0) = φk(x)b−1.

Hence, the smooth hallway w = t−2kw0t2kφ2k(w−10 ) contains a bulge like thefirst one in the previous example (Figure 5). The presence of this bulge does not

32 P. Brinkmann

b

b

b

b

b b

c

c

akak


b b

c

c

x

ak

φ−k(xc) φk(x)


contradict Theorem 0.1 because w contains a large number of visible instancesof the letters x and y. This example shows that we cannot consider the strataseparately when proving Theorem 1.9.

Example 3.5. If we let w0 = φ−k(dc)b−1, then the smooth hallway

w = t−2kw0t2kφ2k(w−10 )

contains a bulge like in Example 3.4. This does not contradict Theorem 0.1 as wcontains a large number of visible instances of the letter c.

Our final example illustrates a subtlety regarding linearly growing strata.

Example 3.6. Let F4 = 〈a, b, c, d〉 and define ψ by letting

a �→ a

b �→ ba

c �→ ca

d �→ dcb−1.


The map ψ is a linearly growing automorphism, so in particular the letter d is oflinear growth, although the image of d contains letters of linear growth other thand itself.

Letters of linear growth may thus behave in two different ways; they maycontribute to the growth of images under successive applications of ψ, or theymay remain inert as parts of a fixed word. In the proof of Theorem 1.9, we willneed to distinguish letters of linear growth according to their role.

Example 3.7. Let F3 = 〈a, x, y〉 and define φ by letting

a �→ axyx−1y−1

x �→ y−1

y �→ yx.

The stratum {x, y} grows exponentially, and we have ψ(xyx−1y−1) = xyx−1y−1.This means that a grows linearly although it maps across an exponentially growingstratum. This is another phenomenon that we need to consider when analyzingstrata of linear growth.

The notion of hallways naturally extends to mapping tori of homotopy equiv-alences of finite graphs. Specifically, a hallway ρ in the mapping torus of f : G→ Gis a sequence of paths of the form

ρ = (μk−1, μk−2, . . . , μ1, ρ0, ν1, ν2, . . . , νk−1, ρk),

where ρ0, ρk, μ1, . . . , μk−1, ν1, . . . , νk−1 are paths in G, satisfying f(τ(ρ0)) = ι(ν1),f(τ(νi)) = ι(νi+1), f(τ(νk−1)) = τ(ρk), f(ι(ρ0)) = τ(μ1), f(ι(μi)) = τ(μi+1), andf(ι(μk−1)) = ι(ρk), where ι(.) is the initial point of a path, and τ(.) is the terminalpoint.

The paths μi and νi are called notches. Some or all of the notches maybe trivial. For 1 ≤ i < k, we define ρi to be the path obtained by tighteningμif(ρi−1)νi. Since ρ is a closed path, we have ρk = f#(ρk−1). As before, we callρi the ith slice of ρ, and the number k is the duration D(ρ).Definition 3.8. The visible length of ρ is

V(ρ) = L(ρ0) + L(ρk) +k−1∑i=1

(L(μi) + L(νi)) .

Finally, we introduce quasi-smooth hallways: Given some C ≥ 0, we say thatw hallway ρ is C-quasi-smooth if the length of all the notches is bounded by C.

4. Strata of superlinear growth

Throughout this section, let f : G→ G be an improved relative train track map.In order to track images of edges through the slices of a hallway ρ, we assign

a marking to each edge. This assignment will, in general, involve arbitrary choices,but our arguments will not be affected by these choices.

34 P. Brinkmann

Definition 4.1. We begin by marking all edges in the initial slice ρ0 and in allnotches μi, νi with their height. Assume inductively that all edges in a slice ρi−1have been marked, and let E be an edge of height r in ρi−1, with marking s. Now,consider an edge E′ in f(E). If the height of E′ is r, or if Hs is a zero stratum,then we keep the marking s. If the height of E′ is less than r, then we mark E′ byr. This gives us a marking for all edges in μif(ρi−1)νi.

Note that, as we tighten μif(ρi−1)νi to obtain ρi, different choices in cancel-lation (Remark 1.1) may give rise to different possible markings, but this will notbe a problem.

We say that an edge E is marked by a linear/polynomial/exponential stratumif its marking is s and Hs is linear/polynomial/exponential.

The following proposition goes a long way toward proving Theorem 1.9.

Proposition 4.2. There exists some constant K ≥ 1 such that for every hallway ρand every slice ρi of ρ, the number of edges in ρi that are not marked by strata oflinear growth is bounded by KV(ρ).

Given the improved relative train track map f : G→ G, the constant K canbe computed.

If f has no edges of linear growth, then Proposition 4.2 immediately impliesTheorem 1.9: For a smooth hallway ρ of duration k, we have V(ρ) = L(ρ0) +L(ρk) = L(ρ0) + L(fk

#(ρ0)) and ρi = f i#(ρ0), so that Proposition 4.2 implies

L(ρi) ≤ K(L(ρ0) + L(fk#(ρ0)) in this case.

In order to streamline the exposition, we will not always make the choice ofK explicit. However, it will turn out that K can be chosen to be the product ofnumbers that can easily be read off from the train track map.

The intuition of the proof is that once significant growth occurs, it will bedue to the presence of long legal subpaths in exponentially growing strata or longsubsegments of eigenrays of polynomially growing strata that grow faster thanlinearly. Lemma 2.1 and Lemma 2.6 imply that there is hardly any cancellationbetween such subpaths and their surroundings, so that any significant growth thatoccurs in a slice will eventually be accounted for by visible edges.

The following definition will help us understand cancellation in hallways. Forevery stratum Hr, we define a number h(Hr) in the following way:

• If Hr is a constant stratum, then h(Hr) = 0.• If Hr is a nonconstant polynomially growing stratum, i.e., Hr = {Er} andf(Er) = Erur, then h(Hr) is the height of ur.

• If Hr is of exponential growth and Hr−1 is not a zero stratum, then h(Er)is the height of f(Hr) ∩ Gr−1, unless this intersection does not contain anyedges, in which case we let h(Hr) =∞.

• If Hr is of exponential growth and Hr−1 is a zero stratum, then h(Er) is theheight of f(Hr ∪Hr−1) ∩Gr−2. We also let h(Hr−1) = h(Hr).


Essentially, h(Hr) is the index of the highest stratum crossed by the imageof Hr, other than Hr itself. We may permute the strata of G (while preserving theimproved train track properties) such that h(Hr) > h(Hs) implies r > s.

Given a stratum Hs, we say that the set S(Hs) = {Hr|h(Hr) = s} is theleague of Hs, the motivation being that they, in a sense, “play at the same level.”If h(Hr) =∞, then Hr does not belong to any league.

Proof of Proposition 4.2. First of all, we note that if a slice ρi has a subpath ina zero stratum Hr, then this subpath is of uniformly bounded length, and it issurrounded by edges in higher strata (Theorem 1.4, Part 1), so that we have alinear estimate of the number of edges in Hr in ρi in terms of the number of edgesin higher strata.

Let q be the largest (finite) number for which the league S(q) is nonempty.Fix some stratum Hr for r > q. We want to find a linear bound on the number ofedges in ρi ∩Hr in terms of visible edges. By definition of S(q) and choice of r,edges in ρi ∩Hr never cancel with edges from other strata or their images.

If Hr = {Er} is of polynomial growth, then any occurrence of Er in ρi isthe image of a visible copy of Er, and ρi contains at most one copy of Er foreach visible copy of Er. Hence, the number of edges in ρi ∩Hr is bounded by thenumber of visible edges.

Now, assume that Hr is an exponentially growing stratum. A slice ρi de-composes into r-legal subpaths with r-illegal turns in between. By Lemma 2.1,a subpath whose r-length is greater than Cr (Equation 2) will eventually be ac-counted for by visible edges since it will not be shortened by cancellation withinslices.

Edges in Hr whose r-distance from an illegal turn is less than Cr

2 may canceleventually, and ρi contains at most Cr of them per r-illegal turn, so that we onlyneed to find a bound of the number of r-illegal turns in terms of the number ofvisible edges. Since the improved train track map f does not create any r-illegalturns, any r-illegal turn in ρi can be traced back to a visible illegal turn in ρ (oran illegal turn created by appending a notch to the image of a slice). This impliesthat the number of r-illegal turns in ρi is bounded by the number of visible edgesin ρ.

Summing up, we have bounded the number of edges in ρi∩(Hq+1∪Hq+2∪· · · )by a multiple of the number of visible edges. This establishes the base case of theproof.

We now assume inductively that the number of edges in S(p)∪S(p+1)∪· · ·has been bounded as a constant multiple of V(ρ). We need to find a bound on thenumber of edges in ρi ∩Hp.

We first assume that Hp = {Ep} is of polynomial growth. By definition ofS(p), an edge in ρi ∩Hp has one of four possible markings:• Its marking may be p, indicating that it is the image of a visible edge, or• it may be marked by an exponentially growing stratum in S(q), for someq ≥ p, or

36 P. Brinkmann

• it may be marked by a superlinear polynomially growing stratum in S(q), q ≥p, or

• it may be marked by a stratum of linear growth.

We are not concerned with edges of the fourth kind.As before, the number of edges of the first kind in ρi ∩Hp is bounded by the

number of visible edges. Let C be the largest number of copies of Ep that occurin the image of a single edge in an exponentially growing stratum Hs, for s > p.Then the number of edges of the second kind in ρi ∩ Hp is bounded by C timesthe number of exponentially growing edges in ρi−1 ∩ (S(p)∩S(p+1)∩ · · · ), whichin turn is bounded by a multiple of the number of visible edges.

We have no immediate bound on the number of edges of the third kind. Aswe trace the image of such an edge through subsequent slices, one of three possibleevents will occur:

• Either, it eventually maps to a visible edge, or• it cancels with an edge of the first or second kind, or• it cancels with an edge in the image of a polynomially growing (possiblylinearly growing) edge in S(p).

Note that these events may depend on choices in tightening (Remark 1.1), butonce again our estimates will not be affected by these choices.

The number of edges for which one of the first two events occurs is clearlybounded by a multiple of the number of visible edges. We only need to find abound on the number of edges in an eigenray that eventually cancel with edges inanother eigenray.

Lemma 2.6 implies that there is a uniform bound on the number of edges inHp that cancel when two rays meet, so that we only need to find a bound on thenumber of meetings between two rays. Clearly, any two rays meet at most once.

If an eigenray cancels with segments from more than one other ray (this isconceivable since a slice may be of the form ρi = ErS1S2, where Er is a polyno-mially growing edge in S(p) and S1, S2 are short segments from rays of edges inS(p) such that the ray of Er successively cancels with S1 and S2), then all exceptpossibly one of these segments cancel completely, so that they are no longer avail-able for subsequent cancellation. This implies that the number of meetings of raysis bounded by two times the number of pieces of rays available for cancellation,which in turn is bounded by the number of visible edges.

This completes our estimate of the number of edges in ρi ∩Hp when Hp is ofpolynomial growth. We now assume that Hp is of exponential growth.

The number of subpaths of height p of ρi is bounded by the number of edgesof height greater than p in ρi plus one. The contribution of p-legal subpaths ofp-length less than or equal to Cp is bounded by Cp times the number of subpathsof height p, so that we do not need to consider them here. Any p-legal subpathsof length greater than Cp will eventually show up in the visible part of ρ, so thatwe do not need to consider them, either. The remaining edges in ρi ∩Hp are at


p-distance less than Cp

2 from a p-illegal turn. Hence, we only need to find a boundon the number of p-illegal turns in ρi.

As before, we trace illegal turns in ρi back to their origin:• An illegal turn may be the image of a visible illegal turn (this case includesillegal turns created by appending notches μi, νi to the image f#(ρi−1) of aslice), or

• it may come from a illegal turn in the image of an exponentially growing edgein S(p), or

• it may be contained in the ray of a polynomially growing edge in S(p), or• it may be contained in a Nielsen path marked by a linear stratum (Exam-ple 3.7).

We are not concerned with illegal turns of the fourth type.The same arguments that we used for polynomially growing Hp yield that

the number of illegal turns of the first and second kind is bounded by a multipleof the number of visible edges.

Now, let C be the maximum of the number of illegal turns in the images ofpolynomially growing edges in S(p). Lemma 2.4 yields an exponent k0 such thatfor polynomially growing edge Er in S(p), fk0

# (ur) contains a long legal segment.This means, in particular, that if ρ contains a block fk

#(ur), k ≥ k0, then thisblock contains no more than C illegal turns per long legal segment. Since longlegal segments eventually show up as visible edges, the number of illegal turns insuch blocks is bounded by CV(ρ).

The remaining illegal turns are contained in initial subpaths of rays thatcontain no more than the first C + 1 blocks, i.e., there are at most C(C + 1)illegal turns of this kind per ray. Since we already know that the number of raysis bounded in terms of the number of visible edges, we are done in this case.

We have now obtained the desired estimate for edges in ρi of height p andhigher. In particular, this includes all strata in S(p − 1), which completes theinductive step. �

5. Polynomially growing automorphisms

In this section, we establish Theorem 1.9 in the case of polynomially growing auto-morphisms. Specifically, we find estimates for the contribution of linearly growingedges that we ignored in Proposition 4.2. As usual, let f : G→ G be an improvedrelative train track map. Since f is of polynomial growth, every stratum Hr con-tains only one edge Er, and we have f(Er) = Er ·ur, where ur is some closed pathin Gr−1. Note that all vertices of G are fixed.

We first record an obvious lemma.

Lemma 5.1. Let μ1, μ2 be Nielsen paths in G, and let ν be some path in G.• If μ1 and μ2 can be concatenated, then the path obtained from μ1μ2 by tight-ening relative endpoints is also a Nielsen path.

38 P. Brinkmann

• If μ1 and ν can be concatenated, let γ be the path obtained by tightening μ1ν,and let Δ = L(γ)− L(ν). Then, for all k ≥ 0, we have

L(fk#(γ)

)= L

(fk#(ν)

)+Δ and − L(μ1) ≤ Δ ≤ L(μ1). �

We now establish Theorem 1.9 for automorphisms of linear growth. Thislemma will provide the base case of our inductive proof of Theorem 1.9.

Lemma 5.2. Assume that f : G→ G is of linear growth. If ρ is a smooth hallway,and if ρ0 starts and ends at vertices, then the lengths of slices of ρ are bounded byV(ρ), i.e., Theorem 1.9 holds with K = 1.

Proof. The proof proceeds by induction up through the strata of G. The bottomstratum H1 is constant, so that the lemma trivially holds for the restriction of fto H1. We now assume that Hr is a linearly growing stratum, and that the lemmaholds for the restriction of f to Gp−1.

Consider the initial slice ρ0. Remark 1.6 yields a splitting of ρ0 into basicpaths of height p and paths in Gp−1. The splitting of ρ0 induces a decompositionof ρ into smooth hallways, so that it suffices to prove the claim for hallways whoseinitial slice is a basic path of height p or a path in Gp−1.

By induction, we only need to prove the claim if ρ0 is a basic path of heightp. If the basic path ρ0 is, in fact, an exceptional path, then the reasoning ofExample 3.2 proves our claim, so that we may assume that ρ0 is not an exceptionalpath.

Assume that ρ0 is a basic path of the form Epγ. Then, by Theorem 1.4,Part 4, there exists some smallest exponent m ≥ 0 for which fm+1

# (Epγ) splits asEp · γ′. Using Remark 1.6 once more, we conclude that Epγ can be expressed asEpu

−mp ν.If D(ρ) ≤ m, then ρ0 k-splits as Epu

−mp · ν. We can consider the subpaths

Epu−mp and ν separately, so that we are done in this case.Now assume that D(ρ) > m. For 0 ≤ i ≤ m, we have ρi = Epu

i−mp f i

#(ν)

and L(ρi) = 1 + (m − i)L(up) + L(f i#(ν)

). For m + 1 ≤ i ≤ D(ρ), we have

ρi = Epui−(m+1)p f i

#(upν) and L(ρi) = 1 + (i − m − 1)L(up) + L(f i#(ν)

)+ Δ,

where Δ is defined as in Lemma 5.1.We have V(ρ) = 2+(D(ρ) − 1)L(up)+L(ν)+L

(fD(ρ)# (ν)

)+Δ. By induction,

we have L(f i#(ν)

)≤ L(ν) +L

(fD(ρ)# (ν)

)for all 0 ≤ i ≤ D(ρ). This immediately

implies that L(ρi) ≤ V(ρ) for all 0 ≤ i ≤ D(ρ).If ρ0 = EpγE

−1p , we essentially repeat the same argument. Once more, we

can write ρ0 = Epu−mp ν, and in order to use the previous argument, we only need

to know that the lemma holds for ν. This, however, follows from the previous step,so that we are done. �

We now find estimates on the number of edges emitted by linearly growingedges, the quantity we ignored in Proposition 4.2. The idea is to take a hallway


and decompose it into smaller and smaller pieces until all remaining pieces onlyinvolve linearly growing edges and their rays. Simple counting arguments will giveus bounds on the number of the remaining pieces as well as the lengths of theirslices.

Let ρ be a hallway, and assume that there is a visible edge Er that doesnot cancel within ρ, i.e., we can trace its image through the slices of ρ until itreappears as another visible edge. Then ρ can be expressed as ρ = αErβE

−1r , and

we define two new hallways ρ′, ρ′′ by tightening t−kErβE−1r and αtk. We say that

ρ′ and ρ′′ are obtained from ρ by cutting along the trajectory of Er (Figure 6).The exponent k is the length of the cut. We say that a hallway ρ is indecomposableif it does not admit any cuts of length D(ρ).

α

β

ErEr ErEr

ur

tk

cut

sawtooth construction

Figure 6. Cutting and the sawtooth construction.

Now we obtain a new hallway σ from ρ′ by repeatedly replacing subwords ofthe form t−1Er by f(Er)t−1 and tightening (Figure 6). We refer to this operationas the sawtooth construction along the trajectory of Er.

IfM is a collection of hallways, we let

V(M) =∑σ∈M

V(σ).

The following lemma lists some basic properties of our two operations. Wesay that an edge is of degree d if fk

#(E) grows polynomially of degree d.

Lemma 5.3. Fix some C ≥ max{L(ur)}. Let ρ be a C-quasi-smooth hallway in G.Choose d > 1 such that the fastest growing edge crossed by ρ grows polynomiallyof degree d.

Obtain a collection M of hallways by cutting along all trajectories of edgesE in ρ of degree d. Let M1 be the collection of smooth elements of M, and letM2 consist of hallways obtained by performing the sawtooth construction along alltrajectories of E of degree d in those elements of M that are not smooth. Then1. The duration of all elements of M1 and M2 is at most D(ρ).2. None of the elements of M2 crosses edges of degree d, i.e., they only cross

edges of degree at most d− 1.3. All elements of M2 are 2C-quasi-smooth.

40 P. Brinkmann

4. The number of elements of M2 is bounded by 2CD(ρ).5. We have

V(M1) + V(M2) ≤ V(ρ) + (2CD(ρ))2 .Proof. The first four properties follow immediately from definitions. In order toprove the fifth property, we just remark that each element of M2 has at most2CD(ρ) visible edges that do not appear in ρ itself. Since M2 contains at most2CD(ρ) hallways, the estimate follows. �

Lemma 5.4. There exists a (computable) constant C with the following property:Let γ be a path of height r, starting and ending at vertices, and assume that

Er is of degree d > 1. Then, for all k ≥ 0,

L(γ) + L(fk#(γ)) ≥ Ckd.

Proof. It suffices to prove the lemma if either γ = Erγ′, or γ = Erγ

′E−1s , whereγ′ only involves edges of degree less than d, and Es is of degree d.

In the first case, the claim is obvious. In the second case, we remark thatLemma 2.6 guarantees that there is hardly any cancellation between the rays ofEr and Es, so that the lemma follows. �

The following proposition implies the second part of Theorem 1.9 in the caseof polynomially growing automorphisms. In particular, it provides bounds on thenumber of edges emitted by linearly growing edges. This is the quantity that weignored in Proposition 4.2.

Proposition 5.5. Assume that f represents an automorphism that grows polyno-mially of degree q. Fix some C ≥ max{L(ur)}. There exist computable constantsK1 ≤ K2 ≤ · · · ≤ Kq and K ′

1(C), . . . ,K′q(C) such that

1. If ρ is a smooth hallway whose fastest growing edge is of degree d, and if ρ0starts and ends at vertices, then

L(ρi) ≤ KdV(ρ)for all slices ρi of ρ.

2. If ρ is a C-quasi-smooth hallway whose fastest growing edge is of degree d,then in every slice ρi, the number of edges emitted by linearly growing edgesis bounded by

KdV(ρ) +K ′d(C)Dd+1(ρ),

so that we have

L(ρi) ≤ (K +Kd)V(ρ) +K ′d(C)Dd+1(ρ),

where K is the constant from Proposition 4.2.

Proof. We prove the proposition by induction on d. For d = 1, the first part holdswith K1 = 1 because of Lemma 5.2. Now, assume that ρ is a C-quasi-smoothhallway whose fastest growing edge grows of degree d = 1. Obtain a collectionMof hallways by cutting ρ along the trajectories of all linearly growing edges that


do not cancel within ρ. If σ is a smooth element ofM, then the first part impliesthat the number of edges in each σi emitted by linearly growing edges is boundedby V(σ).

If σ is not smooth, then in every slice σi, the number of edges emitted bylinearly growing edges is bounded by V(σ) + 2CD(ρ) (It is helpful to keep Exam-ple 3.3 in mind). Lemma 5.3 yields thatM contains no more than 2CD(ρ) piecesthat are not smooth. Summing up, we conclude that every slice of ρ contains atmost V(ρ)+(2CD(ρ))2 edges emitted by linearly growing edges, so that the secondstatement follows with K1 = 1 and K ′

1(C) = 4C2.Now, let K be the constant from Proposition 4.2, and assume inductively

that the proposition holds for some d ≥ 1. We want to find some Kd+1 such thatfor all hallways ρ whose fastest growing edge is of degree d+ 1, we have

L(ρi) ≤ Kd+1V(ρ).

for all slices ρi. It suffices to prove this with the assumption that ρ is indecom-posable. Then we can perform the sawtooth construction along all trajectories ofedges of degree d + 1. Since ρ is indecomposable, we obtain one C-quasi-smoothpiece σ that only crosses edges of degree d or lower, so that by induction, we con-clude that the number of edges in σi that were emitted by linearly growing edgesis bounded by

KdV(σ) +K ′d(C)Dd+1(ρ).

We conclude that

L(ρi) ≤ KV(ρ) +KdV(σ) +K ′d(C)Dd+1(ρ)

≤ (K +Kd)V(ρ) + (2C +K ′d(C))Dd+1(ρ).

Using Lemma 5.4, we can find some constant M such that

MV(ρ) ≥ (2C +K ′d(C))Dd+1(ρ)

for all indecomposable hallways ρ involving edges of degree d + 1. We concludethat the first statement of the proposition holds with Kd+1 = K +Kd +M .

We now prove the second assertion. Let ρ be a C-quasi-smooth hallway. Weobtain two collections M1,M2 of hallways by performing cutting and sawtoothoperations as in Lemma 5.3.

The elements of M1 are smooth hallways, so that for any σ ∈ M1, theprevious step yields

L(σi) ≤ Kd+1V(σ).

If σ is an element ofM2, then it is a 2C-quasi-smooth hallway, and inductionyields that in every slice of σ, the number of edges emitted by linearly growingedges is bounded by

KdV(σ) +K ′d(2C)Dd+1(σ).

42 P. Brinkmann

Summing over all elements of M1 and M2, we conclude that every slice ρiof ρ contains at most

Kd+1V(M1) +KdV(M2) + 2CD(ρ) ·K ′d(2C)Dd+1(ρ)

≤ Kd+1V(ρ) + 4C2KdD2(ρ) + 2CK ′d(2C)Dd+2(ρ)

edges emitted by linearly growing edges, so that the second statement of the propo-sition holds with

K ′d+1(C) = 4C2Kd + 2CK ′

d(2C). �

Remark 5.6. The estimates of Proposition 5.5 are rather crude; lots of edges arecounted several times rather than just once. I opted to present the most straight-forward estimates rather than tightest ones.

6. Proof of the main result

We now extend the techniques and results of Proposition 5 to arbitrary automor-phisms. The presence of exponentially growing strata will turn out to be a mixedblessing. On the one hand, they make for rather simple counting arguments aspolynomial contributions as in Proposition 5.5 are easily dwarfed by exponentialgrowth. On the other hand, we will need to consider more complicated decompo-sitions of hallways.

As usual, let f : G→ G be an improved relative train track map. Any state-ments regarding the computability of constants assume that we are given such amap. After permuting the strata as necessary, we may assume that if Hr and Hs

are truly polynomial strata and r > s, then the degree of Hr is at least as large asthat of Hs. Throughout this section, let K be the constant from Proposition 4.2.

If Hr is an exponentially growing stratum, then we fix some L > Cr, and wereplace f by fM , where M is the exponent from Lemma 2.2 for this choice of L.After replacing f by a power yet again if necessary, we may assume that the imageof each edge in Hr contains at least L edges in Hr. If Hr supports a closed Nielsenpath τ , then the initial and terminal edges of τ are partial edges in Hr, and wemay assume that the image of each of them also contains at least L edges in Hr.We say that a legal path of height r is long if it contains at least L edges in Hr.

We first record an exponential version of Lemma 5.4.

Lemma 6.1. Let Hr be an exponentially growing stratum or a fast polynomialstratum. Then there exists a computable constant λ > 1 such that if σ is a circuit inGr or a path starting and ending at fixed vertices, then either σ is a concatenationof Nielsen paths of height r and subpaths in Gr−1, or we have

L(σ) + L(fk#(σ)) ≥ λk

for all k ≥ 0.


Proof. IfHr is an exponentially growing stratum, we need to distinguish two cases:First, assume that for some i ≥ 0, f i

#(σ) is a concatenation of Nielsen paths andsubpaths in Gr−1. Since σ starts and ends a fixed vertices, we conclude that σitself is a concatenation of Nielsen paths and subpaths in Gr−1, so that there isnothing to show in this case.

Let λ−, N0 be the constants from Lemma 2.5, and assume that for all i ≥ 0,f i#(σ) is not a concatenation of Nielsen paths and subpaths in Gr−1. Let i0 bethe smallest index for which f i0

# (σ) contains a long legal segment. Then, usingLemma 2.5 and Lemma 2.2, we see that L(σ) ≥ λi0− . Moreover, we have L(fk

#(σ)) ≥λk−i0r .

If we let λ =√min{λ−, λr}, then we have λi0− + λk−i0

r ≥ λk. Hence, we haveL(σ) + L(fk

#(σ)) ≥ λi0− + λk−i0r ≥ λk.

If Hr = {Er} is a fast polynomial stratum, then we argue similarly, usingLemma 2.4 and Theorem 1.4, Part 4. �

If Hr is an exponentially growing stratum, we let Tr equal the length of thelongest path in f(Hr) ∩Gr−1. We fix another constant Sr > 0 with the followingproperty: Let γ be a path in Gr−1. If L(γ) ≥ Sr, then L(f#(γ)) > 3Tr andL(f2#(γ)) > 3Tr, and if L(γ) ≤ Tr, then L(f#(γ)) < Sr and L(f2#(γ)) < Sr. Wecan easily compute a suitable value Sr given the train track map f . We say thata path γ in Gr−1 is r-significant if L(γ) ≥ Sr.

If Hr is an exponentially growing stratum, and ρ is a C-quasi-smooth hallwayof height r, then we need to develop an understanding of the lengths of componentsof ρi∩Gr−1, i.e., we need to study subpaths inGr−1. Intuitively, we will accomplishthis by carving out subhallways in Gr−1.

Consider a maximal subpath γ ⊂ Gr−1 of some slice ρa, i.e., ρa can beexpressed as αγβ, and α (resp. β) is either trivial or ends (resp. starts) with a(possibly partial) edge in Hr. We begin the construction of a new hallway ρ′ byletting ρ′0 = γ.

Now, assume inductively that we have defined the slice ρ′i−1 such that ρ′i−1

is a maximal subpath of ρa+i−1 in Gr−1 (we write ρa+i−1 = αρ′i−1β), and recallthat the slice ρa+i is obtained by tightening μa+if(αρ′i−1β)νa+i. We define thenotch μ′i by taking the maximal terminal subpath in Gr−1 of the path obtainedfrom μa+if(α) by tightening. Similarly, we define the notch ν′1 by tightening themaximal initial subpath in Gr−1 of the path obtained from f(β)νa+i by tightening.Observe that tightening μ′iρ

′i−1ν

′i yields a maximal subpath in Gr−1 of ρa+i, and

that the length of μ′i and ν′i is bounded by C +Tr. We iterate this procedure until

we reach a point where tightening μ′i+1f(ρ′i)ν

′i+1 yields a trivial path.

By applying this construction wherever possible, we obtain a fan of C + Tr-quasi-smooth hallways in Gr−1. Let M be the set of maximal elements of thisfan. We letM1 be the collection of smooth hallways inM, and we letM′

2 be thecollection of hallways inM that are not smooth.

44 P. Brinkmann

Let σ be an element ofM′2, and assume that there exists some 0 < i < D(σ)

such that L(σi) < Sr. Then we obtain two new hallways σ′, σ′′ from σ by lettingσ′j = σj for 0 ≤ j ≤ i and σ′′j = σi+j for 0 ≤ j ≤ D(σ) − i; we may think ofthis operation as cutting σ along σi. We obtain a collection of hallways M2 byperforming all possible cuts of this kind on all elements ofM′

2.If σ ∈ M1 ∪M2, we say that σ intersects a slice ρi if one of the slices of σ

is a subpath of ρi. When looking for bounds on the lengths of a slice ρi, we needto find bounds on the lengths of slices of hallways σ that intersect ρi.

Definition 6.2. Fix some stratumHr. We say that the map f satisfies Condition Ar

if for any C ≥ 0, there exist computable constants Kr, K ′r(C), and an exponent

d ≥ 1, such that the following two conditions hold:

• If ρ is a smooth hallway in Gr such that the slice ρ0 starts and ends at fixedvertices, then

L(ρi) ≤ KrV(ρ)

for all slices ρi.• If ρ is a C-quasi-smooth hallway in Gr , then

L(ρi) ≤ KrV(ρ) +K ′r(C)Dd(ρ).

If Hr is an exponentially growing stratum, then a hallway of height r isadmissible if all its slices start and end at fixed vertices or at points in Hr.

Lemma 6.3. Let Hr be an exponentially growing stratum, and assume that Con-dition Ar−1 holds. Then, given some C ≥ 0, there exist computable constantsC1, C2 ≥ 1 with the following property: If ρ is an admissible C-quasi-smooth hall-way of height r, then

L(ρi) ≤ C1V(ρ) + C2∑

σ∈M2σ intersects ρi

in an r-significantsegment

Dd(σ)

for every slice ρi of ρ.

Proof. Since ρ is admissible, all slices of σ ∈ M1 start and end at fixed verticesunless σ0 is contain in a zero stratum, in which case all slices σi for i > 0 startand end at fixed vertices. Moreover, if σ0 is contained in a zero stratum, thenL(σ1) = L(σ0). By Condition Ar−1, we have

L(σi) ≤ Kr−1V(σ)

for all slices σi of σ ∈ M1.


Fix some slice ρi of ρ. Using Proposition 4.2 and Condition Ar−1, we see that

L(ρi) ≤ KV(ρ) +∑


Kr−1V(σ)

+∑


(Kr−1V(σ) +K ′

r−1(C + Tr)Dd(σ)).

Consider some σ ∈ M1 that intersects ρi. If the initial slice of σ is not visiblein ρ, then, as we noted before, its length is bounded by Tr. Similarly, if the terminalslice of σ is not visible in ρ, then its length is also bounded by Tr. The number ofelements ofM1 that intersect ρi is bounded by KV(ρ). Putting it all together, weconclude that ∑


V(σ) ≤ (2KTr + 1)V(ρ).

Similarly, using the fact that elements ofM2 are C + Tr-quasi-smooth, and thattheir initial and terminal slices are either visible in ρ or of length less than Sr, wesee that ∑


V(σ) ≤ (2KSr + 1)V(ρ) + 2(C + Tr)∑


D(σ).

Since ρi contains at most KV(ρ) subpaths in Gr−1, the total contribution ofsubpaths in Gr−1 that are not r-significant is bounded by KSrV(ρ). LettingC1 = K +2Kr−1(K(Sr +Tr)+ 1)+KSr and C2 = K ′

r−1(C +Tr) + 2(C +Tr), weconclude that

L(ρi) ≤ C1V(ρ) + C2∑


in an r-significantsegment

Dd(σ).

�Lemma 6.3 shows that from now on, we may focus on the polynomial con-

tribution of nonsmooth hallways in Gr−1 that intersect a given slice ρi in anr-significant subpath. In particular, if the initial slice ρ0 happens to be an r-legalpath, then

L(ρi) ≤ C1V(ρ)for all slices ρi sinceM2 is empty in this case.

Lemma 6.4. Let Hr be an exponentially growing stratum, and assume that Condi-tion Ar−1 holds. Given some C > 0, there exist computable constants C1, C2 withthe following property: If ρ is an admissible C-quasi-smooth hallway of height r,such that for every slice ρi except possibly the last one, f#(ρi) does not contain alegal segment of length at least L, then

L(ρi) ≤ C1V(ρ) + C2Dd+1(ρ)

for all slices ρi.

46 P. Brinkmann

Proof. By Lemma 6.3, we may restrict our attention to elements of M2 thatintersect a given slice ρi in an r-significant subpath. Let

D =∑

σ∈M2σ intersects ρi in anr-significant segment

Dd(σ).

We first claim that the number of r-significant subpaths in Gr−1 in a slice ρiis bounded by N(ρi). By choice of Sr, an r-significant subpath in Gr−1 will notcancel completely when f(ρi) is tightened to f#(ρi).

If there were two such subpaths in one legal segment of ρi, then there wouldbe a legal segment in Hr in between. Since we assumed that L(f(E) ∩Hr) ≥ Lfor each edge in Hr, the r-length of the image of this legal segment is at leastL, which means that the slice ρi+1 contains a legal segment of length at least L,contradicting our assumption. This proves the claim ifHr does not support a closedNielsen path, as in this case, the number of legal segments in ρi equals N(ρi).

If Hr supports a closed Nielsen path, then a legal segment of ρi that is ad-jacent to an illegal turn contained in a Nielsen subpath of ρi cannot contain anr-significant subpath in Gr−1. If such a segment contained an r-significant subpathin Gr−1, then f#(ρi) would contain a legal segment of r-length L because both theinitial and terminal partial edge of the Nielsen path of Hr map to legal segmentsof r-length at least L. This implies that the number of r-significant subpaths inGr−1 is bounded by N(ρi).

Now, fix some slice ρi. We make the worst-case assumption that every legalsegment of ρ that is not adjacent to an illegal turn contained in a Nielsen subpathcontains an r-significant subpath in Gr−1 that is a slice of a hallway σ ∈ M2 ofduration j ≥ i. The number of such hallways whose duration is a given numberj ≥ i is bounded by N(ρj) + 1. We conclude that

D ≤D(ρ)∑j=i

N(ρj)jd.

Choosing λ according to Lemma 2.5, we conclude thatN(ρi+1) ≤ λ−1N(ρi)+1 + 2C, as ρ is C-quasi-smooth. This implies, inductively, that

N(ρi) ≤ λ−iN(ρ0) + 2(1 + C)i−1∑j=0

λ−j ≤ λ−iN(ρ0) +λ

λ− 1(1 + 2C).

We choose some B ≥∑∞j=0 λ

−jjd, and we conclude that

D ≤D(ρ)∑j=0

N(ρj)jd ≤ BV(ρ) + λ

λ− 1(1 + 2C)Dd+1(ρ),

since N(ρ0) ≤ V(ρ).If C′1, C

′2 are the constants from Lemma 6.3, then the lemma holds with

C1 = C′1 + C′2B and C2 = λλ−1 (1 + 2C)C′2. �


Definition 6.5. LetHr be an exponentially growing stratum, and letN0 be the con-stant from Lemma 2.5. We say that an admissible smooth hallway ρ of height r hasProperty B if for all slices ρi, ρi contains no long r-legal segment, or N(ρi−1) < N0.

Lemma 6.6. Let Hr be an exponentially growing stratum, and assume that Condi-tion Ar−1 holds. Let N0 be the constant from Lemma 2.5. There exist computableconstants C1, C2 with the following property: If ρ is an admissible smooth hallwayof height r that satisfies Property B, then

L(ρi) ≤ C1V(ρ) + C2Dd+1(ρ)

for all slices ρi.

Proof. If no slice of ρ contains a long legal segment, then the claim follows fromLemma 6.4. Otherwise, let i0 be the smallest index for which ρi0 contains a longlegal segment. By choice of i0, ρi0−1 does not contain a long legal segment, andby hypothesis, we have N(ρi0−1) < N0. If i < i0, then, choosing D as in the proofof Lemma 6.4, we conclude that

D ≤(i0−1∑

j=0

N(ρj)jd)+N0Dd(ρ)

≤ BV(ρ) +N0Dd+1(ρ),

so that the lemma holds for all ρi with i < i0.For i ≥ i0, ρi splits as a concatenation of long r-legal paths and subpaths

that contain illegal turns and no long legal subpaths. Each slice may, conceiv-ably, contain slices of N(ρi0−1) < N0 hallways of duration D(ρ). The polynomialcontribution of these hallways is bounded by N0Dd(ρ).

In addition, the number of short legal segments around illegal turns is atmost 2N0. Each of them contains not more than one r-significant subpath in Gr−1,belonging to a hallway of duration at most D(ρ)− i0. The polynomial contributionof these paths is bounded by 2N0(D(ρ)− i0)d.

Now, since ρi0 contains a long legal segment, the length of

ρD(ρ) = fD(ρ)−i0# (ρi0)

is at least λD(ρ)−i0r . We can easily find some B′ > 0 such that B′λk

r ≥ 2N0kd forall k ≥ 0. We conclude that for the sum of all polynomial contributions in ρi, wehave

2N0(D(ρ) − i0)d +N0Dd(ρ) ≤ B′V(ρ) +N0Dd(ρ),

which completes the proof of the lemma. �

The remaining two lemmas deal with arbitrary smooth hallways of height ras well as quasi-smooth hallways by essentially decomposing them into pieces ofthe kind that we analyzed in the previous lemmas.

48 P. Brinkmann

Lemma 6.7. Let Hr be an exponentially growing stratum, and assume that Condi-tion Ar−1 holds. Then there exist computable constants C1, C2 with the followingproperty: If ρ is a smooth admissible hallway of height r, then

L(ρi) ≤ C1V(ρ) + C2Dd+1(ρ)

for all slices ρi.

Proof. Let λ−, N0 be the constants from Lemma 2.5. As in the proof of Lemma 6.1,we let λ =

√min{λ−, λr}, and we remark that for 0 ≤ j ≤ k, we have λj

−+λk−jr ≥

λk. This basic estimate will be crucial in the proof of this lemma. We choose someB > 0 such that Bλk > kd+1 for all k ≥ 0.

Let C′1, C′2 be the maximum of the corresponding constants from the previous

lemmas. We will see that the lemma holds with C1 = C′1 + 3BC′2 and C2 = C′2.We first observe that if ρ satisfies Property B, then the lemma follows from

Lemma 6.6. If ρ0 contains long legal segments, we can split ρ0 into long r-legalsubpaths and neighborhoods of illegal turns (i.e., illegal turns surrounded by le-gal paths whose length is at most Cr

2 ). Split ρ0 as ρ0 = α0;1β0;1α0;2 · · ·α0;mβ0;m,where all subpaths α0;i are long legal segments, and all subpaths β0;i are neigh-borhoods of illegal turns. Such a decomposition of ρ0 induces a decompositionof ρ into hallways, and we can choose the decomposition such that all resultingpieces are admissible, and that the legal segments are as long as possible, subjectto admissibility. We write αj;i = f j

#(α0;i) and βj;i = f j#(β0;i).

Let k = D(ρ). For each long legal subpath α0;i, Lemma 6.3 yields thatL(αj;i) ≤ C1(L(α0;i) + L(αk;i)), for all 0 ≤ j ≤ k. Since α0;i is a long legalsegment, we have L(αk;i) ≥ λk

r ≥ λk.If the hallway defined by β0;i satisfies Property B, then we have L(βj;i) ≤

C′1(L(β0;i) + L(βk;i)) + C′2kd+1, and we have kd+1 ≤ BL(αk;i), hence

L(βj;i) ≤ C′1(L(β0;i) + L(βk;i)) + BC′2L(αk;i),

i.e., we can find a legal segment adjacent to β0;i whose contribution to the visibleedges of ρ dominates the possible polynomial contribution of β0;i. This takes careof the long legal segments in ρ0 as well as the subpaths that satisfy Property B.Hence, we only need to deal with those paths that do not satisfy Property B.Assume that for some 0 ≤ i ≤ m, β0;i is one of them.

Then there exists some j0 such that βj0;i contains a long legal segment, butβj0−1;i does not, and N(βj0−1;i) ≥ N0.

As before, we split βj0;i into long legal segments and neighborhoods of ille-gal turns, obtaining a decomposition βj0,i = αj0;i,0βj0;i,0 · · ·αj0;i,mβj0;i,m, whereαj0;i,k are r-legal subpaths, and βj0;i,k are neighborhoods of illegal turns. Wecan find splittings βj;i = αj;i,0βj;i,0 · · ·αj;i,mβj;i,m for all 0 ≤ j ≤ k, such thatf#(αj;i,k) = αj+1;i,k and f#(βj;i,k) = βj+1;i,k. We may choose those splitting suchthat the resulting pieces are admissible, and such that the legal segments αj0;i,k

are as long as possible, subject to admissibility.


Now, fix on one subpath αj0;i,k. If N is the number of r-significant subpathsin Gr−1 in αj0;i,k, then αj0−1;i,k contains at least N legal segments containingr-significant subpaths in Gr−1. By Lemma 2.5, we have L(β0;i) ≥ N(βj0;i) ≥λj0−1− N(βj0−1;i), so that we can find λj0−1

− N illegal turns in β0;i, and we can findλk−j0r edges in βk,i. Using our earlier estimate, we see that (λ

j0−1− + λk−j0

r )N ≥λ−1− λkN .

The polynomial contribution of the r-significant subpaths in Gr−1 of αj0;i,k

is bounded by K ′r−1(Tr)Nkd+1 ≤ BK ′

r−1(Tr)Nλk, i.e., it is dominated by corre-sponding visible edges.

This leaves us to deal with the adjacent subpaths βj0;i,k and βj0;i,k−1. If β0;i,ksatisfies PropertyB, then its polynomial contribution is bounded byC′2kd+1, whichin turn is bounded by BC′2λ

k.This takes care of the legal segments αj0;i,k as well as those neighborhoods

of illegal turns that satisfy Property B. We apply the previous reasoning to theremaining paths βj0;i,k, completing the proof of the lemma. �

Lemma 6.8. Let Hr be an exponentially growing stratum, and assume that Condi-tion Ar−1 holds. Given some C > 0, there exist computable constants C1, C2 withthe following property: If ρ is an admissible C-quasi-smooth hallway of height r,then

L(ρi) ≤ C1V(ρ) + C2Dd+3(ρ)for all slices ρi.

Proof. The idea of this proof is to decompose the hallway ρ into pieces that areeither smooth or C-quasi-smooth satisfying the hypothesis of Lemma 6.4.

In order to find this decomposition, we introduce trajectories of points in Hr.This definition may be affected by the choices made when tightening (Remark 1.1).In order to avoid ambiguities, for each index 1 ≤ i < D(ρ), we fix a sequence ofelementary cancellations that turn μiρi−1νi into ρi.

If p is a point in ρi ∩Hr, we consider its image f(p) in f(ρi). We say thatp survives if f(p) is contained in Hr and if f(p) is not contained in an edge thatcancels when f(ρi) is tightened to f#(ρi). If p survives, then f(p) is contained inρi+1, or it is contained in the parts of f#(ρi) that cancel when μi+1f#(ρi)νi+1 istightened to ρi+1.

Thinking of the hallway ρ as spanning a (possibly singular) disk, we draw aline segment (in this disk) from the surviving points in each slice to their images. Ifp is a point in a visible edge such that p and all its images survive, then p defines aline starting and ending in visible edges, called the trajectory of p. The trajectoriesof two points need not be disjoint, but that does not concern us here.

We say that two trajectories are parallel if their initial points are both con-tained in ρ0 or both contained in the same notch, and if their terminal points areboth contained in ρD(ρ) or both contained in the same notch. The crucial obser-vation is that equivalence classes of parallel trajectories are closed subsets of thedisk spanned by ρ, so that in every equivalence class, we can find trajectories of

50 P. Brinkmann

two points p1, p2 that are extremal in the following sense: If p is a point whosetrajectory is parallel to those of p1 and p2, then p is located between p1 and p2.

We now cut ρ along the extremal trajectories of all equivalence classes ofparallel trajectories, obtaining pieces that are either smooth or C-quasi-smooth.Moreover, all the resulting pieces are admissible. Let M1 be the collection ofsmooth pieces and M2 the collection of pieces that are not smooth. Note thatV(M1) + V(M2) = V(ρ).

We now claim that all elements ofM2 satisfy the hypothesis of Lemma 6.8.Suppose otherwise, i.e., there exists some σ ∈ M2 such that for some slice σi,f#(σi) contains a legal segment of length at least L. Within the interior of this legalsegment, we can find some point p such that all images of p survive in subsequentslices. Since p is the image of surviving points, we obtain a trajectory along whichwe can cut σ, contradicting the fact that we obtained σ by cutting ρ along extremaltrajectories.

By Lemma 6.7, there are constants C′1, C′2 such that for every σ ∈ M1 andevery slice σi of σ, we have

L(σi) ≤ C′1V(σ) + C′2Dd+1(σ),

and by Lemma 6.8, there are constants C ′′1 , C′′2 such that

L(σi) ≤ C′′1 V(σ) + C′′2Dd+1(σ)

for every slice σi of every σ ∈M2.There are at most 2(D(ρ) − 1) notches, so that the number of equivalence

classes of parallel trajectories is bounded by (2(D(ρ)−1)+1)2 (another extremelycrude estimate, but it’ll do). Since we cut along no more than two trajectories perequivalence class, we obtain no more than

2 (2 (D(ρ)− 1) + 1)2 + 1 ≤ 8D2(ρ)pieces. Letting C1 = max{C′1, C′′1 } and C2 = 8max{C′1, C′′2 }, we conclude that

L(ρi) ≤ C1V(ρ) + C2Dd+3(ρ)

for all slices of ρ. �We now have all the ingredients that we need to prove Theorem 1.9.

Proof of Theorem 1.9. We first show that Condition Ar holds for all strata Hr.This implies, in particular, that the second statement of Theorem 1.9 holds forpaths starting and ending at fixed vertices. If ρ is a path starting and ending atarbitrary vertices, then Theorem 1.4, Part 2 yields that f#(ρ) starts and ends atfixed vertices, so that, in fact, the second statement of Theorem 1.9 follows fromCondition Ar in this case as well.

We note that Condition A0 holds trivially, and we assume inductively thatCondition Ar−1 holds for some r. We want to prove Condition Ar.

Assume that Hr is an exponentially growing stratum, and let ρ be a smoothhallway of height r such that ρ0 starts and ends at fixed vertices. If ρ0 is a con-catenation of Nielsen paths of height r and paths in Gr−1, then we can split ρ0 at


the endpoints of its subpaths in Gr−1, and Condition Ar−1 completes the proof.We now assume that ρ0 is not a concatenation of Nielsen paths and paths in Gr−1.

By Lemma 6.7, we have constants C1, C2 such that

L(ρi) ≤ C1V(ρ) + C2Dd+1(ρ)

for all slices ρi. Moreover, by Lemma 6.1, there exists some C > 0 and λ > 1,independently of ρ, such that

V(ρ) ≥ CλD(ρ).

We can easily find some constant B such that BCλk ≥ C2kd+1 for all k ≥ 0. Now

the first part of Condition Ar follows, with Kr = C1 + B. Lemma 6.8 yields thesecond part of Condition Ar, so that Condition Ar holds.

We now assume thatHr is a polynomially growing stratum. Because of Propo-sition 5.5, we only need to consider the following situation: Either Hr is fast, or Hr

is truly polynomial, but ρ contains fast polynomial edges or non-Nielsen subpathsin exponentially growing strata.

In order to see that the second part of ConditionAr holds for a C-quasismoothhallway ρ of height r, we apply cutting and sawtooth constructions to ρ, obtaininga collection of (C + |ur|)-quasismooth hallways of height r− 1 or less, so that thesecond part of Condition Ar immediately follows from the second part of Condi-tion Ar−1.

Now, given a smooth hallway ρ of height r, we apply cutting and sawtoothconstructions again, obtaining a collection of 2|ur|-quasismooth hallways. For eachslice ρi, the second part of Condition Ar−1 yields a polynomial bound on thenumber of edges marked by linear strata (Definition 4.1). Now, since either Hr isfast or ρ contains fast polynomial edges or non-Nielsen subpaths in exponentiallygrowing strata, Lemma 6.1 provides an exponential lower bound for the numberof visible edges. As before, the exponential lower bound for visible edges easilydominates the polynomial lower bound for edges marked by linear strata, whichcompletes the proof of Condition Ar.

Finally, in order to prove the first part of Theorem 1.9, we need to understandthe dynamics of circuits. Let σ be a circuit of height r. If Hr is a polynomiallygrowing stratum, then Remark 1.6 yields that σ splits, at fixed vertices, into basicpaths of height r and paths in Gr−1, so that Condition Ar proves the claim.

Assume that Hr is an exponentially growing stratum. If σ is a concatenationof Nielsen paths of height r and paths in Gr−1, then we can split σ at the endpointsof its subpaths in Gr−1, so that Condition Ar−1 completes the proof in this case.We now assume that σ is not a concatenation of Nielsen paths and subpaths inGr−1. Then σ splits at a point p in Hr, so that we may interpret σ as a pathstarting and ending at v. Let ρ be a smooth hallway with ρ0 = σ. Then, byLemma 6.7, we can find constants C1, C2 such that

L(ρi) ≤ C1V(ρ) + C2Dd(ρ)

52 P. Brinkmann

for all slices ρi. Moreover, by Lemma 6.1, we can find constants C, λ such that

V(ρ) ≥ CλD(ρ).

As before, we find some constant B such that BCλk ≥ C2kd for all k ≥ 0, so that

the first statement of Theorem 1.9 holds with Kr = C1 +B.Finally, if ρ0 is a Nielsen path of height r, then there is nothing to show. This

completes the proof. �

Glossary

For the convenience of the reader, we briefly list some of the more technical notionsused in this paper.

L(ρ) length of a path ρ in a conveniently chosen metric (Remark 1.8)λr growth rate of an exponentially growing stratum Hr

Cr critical length of an exponentially growing stratum Hr (Equa-tion 2)

Rr eigenray in a polynomially growing stratum Er (Equation 1)n(ρ) number of legal segments in a path ρ in an exponentially growing

stratumN(ρ) number of legal segments that do not overlap with closed Nielsen

subpath of ρM exponent from Lemma 2.2V(w) visible length (Definition 3.8) of a hallway (Definition 3.1)D(w) duration of a hallwayCondition Ar technical condition (Definition 6.2)Property B technical condition (Definition 6.5)

References

[BF92] M. Bestvina and M. Feighn. A combination theorem for negatively curvedgroups. J. Differential Geom., 35(1):85–101, 1992.

[BFH97] M. Bestvina, M. Feighn, and M. Handel. Laminations, trees, and irreducibleautomorphisms of free groups. Geom. Funct. Anal., 7(2):215–244, 1997.

[BFH00] Mladen Bestvina, Mark Feighn, and Michael Handel. The Tits alternative forOut(Fn). I. Dynamics of exponentially-growing automorphisms. Ann. of Math.(2), 151(2):517–623, 2000.

[BG] Martin R. Bridson and Daniel P. Groves. The quadratic isoperimetric inequalityfor mapping tori of free group automorphisms. arXiv:0802.1323.

[BH92] Mladen Bestvina and Michael Handel. Train tracks and automorphisms of freegroups. Ann. of Math. (2), 135(1):1–51, 1992.

[Bri] Peter Brinkmann. Detecting automorphic orbits in free groups.arXiv:0806.2889v1.


[Bri00] Peter Brinkmann. Hyperbolic automorphisms of free groups. Geom. Funct.Anal., 10(5):1071–1089, 2000. arXiv:math.GR/9906008.

[Coo87] Daryl Cooper. Automorphisms of free groups have finitely generated fixed pointsets. J. Algebra, 111(2):453–456, 1987.

[DV96] Warren Dicks and Enric Ventura. The group fixed by a family of injective en-domorphisms of a free group. American Mathematical Society, Providence, RI,1996.

[Gan59] F.R. Gantmacher. The theory of matrices. Vols. 1, 2. Chelsea Publishing Co.,New York, 1959. Translated by K.A. Hirsch.

[Ger94] S.M. Gersten. The automorphism group of a free group is not a CAT(0) group.Proc. Amer. Math. Soc., 121(4):999–1002, 1994.

[Mac00] N. Macura. Quadratic isoperimetric inequality for mapping tori of polynomi-ally growing automorphisms of free groups. Geom. Funct. Anal., 10(4):874–901,2000.

Peter BrinkmannDepartment of MathematicsThe City College of CUNYNew York, NY 10031, USAe-mail: [email protected]



Geodesic Rewriting Systems and Pregroups

Volker Diekert, Andrew J. Duncan and Alexei G. Myasnikov

Abstract. In this paper we study rewriting systems for groups and monoids,focusing on situations where finite convergent systems may be difficult to findor do not exist. We consider systems which have no length increasing rulesand are confluent and then systems in which the length reducing rules lead togeodesics. Combining these properties we arrive at our main object of studywhich we call geodesically perfect rewriting systems. We show that these arewell behaved and convenient to use, and give several examples of classes ofgroups for which they can be constructed from natural presentations. Wedescribe a Knuth-Bendix completion process to construct such systems, showhow they may be found with the help of Stallings’ pregroups and converselymay be used to construct such pregroups.

Mathematics Subject Classification (2000). 68Q42, 20F05, 20M32, 20E06.

Keywords. String rewriting systems, Geodesically Perfect, Knuth-Bendix,Stallings pregroups.

1. Introduction

A presentation of a group or monoid may be thought of as a rewriting systemwhich, in certain cases, may give rise to algorithms for solving classical algorithmicproblems. For example if the rewriting system is finite and convergent (that is,confluent and terminating) then it can be used to solve the word problem and tofind normal forms for elements of the group. This is one reason for the importanceof convergent rewriting systems in group theory. However there are many wellbehaved, algorithmically tractable, groups for which natural presentations do notgive rise to convergent rewriting systems. In this paper we investigate propertiesof rewriting systems, which are not in general finite or terminating, but which allthe same give algorithms for such tasks as solving the word problem, computationof normal forms or computation of geodesic representatives of group elements. We

Part of this work was begun in 2007 when the first and third author were at the CRM (CentroRecherche Matematica, Barcelona) at the invitation of Enric Ventura.

56 V. Diekert, A.J. Duncan and A.G. Myasnikov

contend that the resulting algorithms are often more convenient and practical thanthose arising from more conventional finite convergent systems.

Rewriting methods in algebra have a very long and rich history. In groups andsemigroups they are usually related to the word problem and take their roots in theground breaking works of Dehn and Thue (not to mention the classical Euclideanand Gaussian elimination algorithms!). Several famous algorithms in group theoryare in fact particular types of string rewriting processes: the Nielsen method infree groups, Hall collection in nilpotent and polycyclic groups, the Dehn algorithmin small cancellation and hyperbolic groups, Tits rewriting in Coxeter groups,convergent rewriting systems for finite groups, and so on. In rings and algebrasrewriting methods appear as a major tool in computing normal forms of elements[40, 45, 10], for solving the word and ideal-membership problems. These techniquesemerged independently in various branches of algebra at different times and underdifferent names (the diamond lemma, Grobner or Shirshov bases, Buchberger’salgorithm and S-polynomials, for instance). They have gained prominence withthe progress of practical computing, as real applications have become available.Notably, crucial developments in methods of computational algebra originated incommutative algebra and algebraic geometry, with Buchberger’s celebrated algo-rithm and related computational techniques, which revolutionised the whole areaof applications. We refer to [11], and the references therein, for more details.

From the theoretical view point the main shift in the paradigm came withthe seminal paper of Knuth and Bendix [33]. In this paper they introduced aprocess, now known as the Knuth-Bendix (KB) procedure, which unified the fieldof rewriting techniques in (universal) algebra. The KB procedure gives a solidtheoretical basis for practical implementations, even though the procedure itselfmay lead to non-optimal algorithms for solving word problems.

Roughly speaking a KB procedure takes as input a finite system of identities(between terms) and a computable (term) ordering such that the identities can beread as a finite set of directed rewrite rules. Using the crucial concept of criticalpairs the procedure adds in each round more and more rules, and it stops only ifthe system is completed. Thus the KB procedure attempts to construct an equiva-lent convergent (term) rewriting system: which in particular allows unique normalforms to be found by a simple strategy. When the procedure terminates we obtaina solvable word problem.

In the case of commutative algebra this concept can be viewed as Buch-berger’s algorithm and termination is guaranteed. In case of algebraic structureslike groups or monoids we have a special case of a term rewriting system sincethe rewriting process is based on strings. (Formally, monoid generators are readas unary function symbols, and the neutral element is read as a constant.)

As has been mentioned above, the history of rewriting systems in monoidsand groups is about one hundred years old, with the main focus on convergentrewriting systems and algorithms for computing normal forms. Any presentationM = 〈Γ | �i= ri(i∈I)〉 of a monoid M gives a rewriting system S={�i→ri,i∈I}

Geodesic Rewriting Systems and Pregroups 57

which defines M via the congruence relation it generates on the free monoid Γ∗.Every rule � → r ∈ S allows one to rewrite a word u�v into the word urv andthis gives a (non-deterministic) word rewriting procedure associated with S. Ifthe system S is convergent (see Section 2) then this rewriting system describesa deterministic algorithm which computes the normal forms of elements, thussolving the word problem in the monoidM . This yields the major interest in finiteconvergent systems. Many groups are known to allow finite convergent systems(for example Coxeter groups, polycyclic groups, some small cancellation groups:see books [30, 46, 34] for more examples and details). The primary task here is tofind a finite convergent system for a given finitely presented monoid, assuming thatsuch a system exists. In principle the KB procedure performs this task. However,several obstacles may present themselves. By design, to start the KB procedureone has to fix in advance an ordering on Γ∗, with particular properties, as describedin Section 2.3. This may seem like a minor hurdle but the difficulty is that, evenfor well-understood groups with two orderings which look very much alike, it mayhappen that using the first the KB process halts and outputs a convergent systemwhile with respect to the second there exists no finite convergent system: seeExample 2.5 below. Furthermore, the existence of a finite convergent system alsodepends on a choice of the set of generators of the group. This means that forKB to succeed one has to make the right choice of a set of generators Γ and ofan ordering on Γ∗. In fact [42] in general the problem of whether or not a givenfinitely presented group can be defined by a finite convergent rewriting system isundecidable. In addition, even when restricted to instances where the generatorsand the order have been chosen so that the KB process will halt giving a finiteconvergent rewriting system, there may be no effectively computable upper boundon the running time of the KB procedure. To make things even more interesting,having a finite convergent rewriting system S does not guarantee a fast solution ofthe word problem in the monoid M (see Section 2.3). All these results show thatthe KB process for finite convergent systems, while being an important theoreticaltool, is not a panacea for problems in computational algebra.

As a first step towards resolving some of these difficulties we consider, inSection 4, the class of preperfect rewriting systems: that is those which are con-fluent and have no length increasing rules. These restrictions are enough to allowsolution of the word problem and to find geodesic representatives and, as exam-ples show, such systems are common in geometric group theory. In fact in Sec-tion 7 we describe preperfect rewriting systems for Coxeter groups, graph groups,HNN-extensions and free products with amalgamation. One disadvantage of thesesystems is that it is undecidable whether a finite rewriting system is preperfect ornot [38] (see Theorem 4.6).

Another desirable property of rewriting systems is that they should be geo-desic; meaning that shortest representatives of elements can be found by applyingonly the length reducing rules of the system. A group defined by a finite geodesicrewriting system has solvable word problem and in [24] these groups are char-acterised as the finitely generated virtually free groups. However, as we show in


Section 5.1, the question of a whether or not a finite rewriting system is geodesicis undecidable.

Combining properties of preperfect and geodesic rewriting systems we arriveat geodesically perfect rewriting systems (defined in Section 5.2). These were firstinvestigated by Nivat and Benois [41] where they were called quasi-parfaites. Else-where these rewriting systems are also known as almost confluent, see for example[6] but here we prefer the notation geodesically perfect since these systems are de-signed to deal with geodesics in groups and monoids. In [41] it was shown that theproperty of being geodesically perfect is decidable for finite systems. This leads toa new Knuth-Bendix completion procedure for constructing geodesically perfectsystems as we explain below. One advantage of this KB process is that it requiresno choice of ordering, using only the partial order given by word length in Γ∗.

Among the examples of Section 7 are rewriting systems for amalgamatedproducts and HNN-extensions. As several important frameworks have been de-veloped to unify the studies of such groups (Bass-Serre Theory, pregroups andrelatively hyperbolic groups, for example) it is natural to look for a unified theoryof rewriting systems covering HNN-extensions and amalgamated products. In thispaper, following Stallings [49, 50], we approach this unification question from acombinatorial view-point via pregroups and their universal groups: which seem tolend themselves naturally to algorithmic and model theoretic problems. Intuitively,a pregroup can be viewed as a “partial group”, that is, a set P with a partial (noteverywhere defined) multiplication m : P × P → P , or a piece of the multiplica-tion table of some group, that satisfies some particular axioms. In this case theuniversal group U(P ) can be described as the group defined by the presentationwith a generating set P and a set of relations xy = z for all x, y ∈ P such thatm(x, y) is defined and equal to z. On the other hand, Stallings proved that U(P )can be realized constructively as the set of all P -reduced forms (reduced sequencesof elements of P ) modulo a suitable equivalence relation and a naturally definedmultiplication. We discuss these definitions in detail in Section 8.

In Section 8.1 we show how the existence of a pregroup allows us to construct apreperfect rewriting system for the universal group. Moreover, we show in Theorem8.4 that this system is geodesically perfect. In this way pregroups may play arole in clarifying completion procedures of KB type. In particular, completing agiven presentation (in terms of generators and relators) of a group G to a largerpresentation, which is a pregroup, amounts to a construction of a geodesicallyperfect rewriting system for G.

As an application of these results we obtain a slight strengthening of theresult of [24]. It is known that a group G is virtually free if and only if G = U(P )for a finite pregroup P [44] and combining this result with Theorem 8.4 we seethat a finitely generated group is virtually free if and only if it is defined by ageodesically perfect rewriting system (Corollary 8.7).


2. Rewriting techniques

2.1. Basics

In this section we recall the basic concepts from string rewriting. We use rewrit-ing techniques as a tool to prove that certain constructions have the expectedproperties.

A rewriting relation over a set X is a binary relation =⇒⊆ X×X . We denoteby ∗=⇒ the reflexive and transitive closure of =⇒, by ⇐⇒ its symmetric closure

and by ∗⇐⇒ its symmetric, reflexive, and transitive closure. We also write y ⇐= x

whenever x =⇒ y, and we write x≤k

=⇒ y whenever we can reach y in at most ksteps from x.

Definition 2.1. The relation =⇒⊆ X ×X is called:

i) strongly confluent, if y⇐=x=⇒z implies y ≤1=⇒ w≤1⇐= z for some w,

ii) confluent, if y ∗⇐= x∗=⇒ z implies y ∗=⇒ w

∗⇐= z for some w,

iii) Church-Rosser, if y ∗⇐⇒ z implies y ∗=⇒ w∗⇐= z for some w,

iv) locally confluent, if y⇐=x=⇒z implies y ∗=⇒ w∗⇐= z for some w.

The following facts are well known and can be found in several text books(see for example [6, 31]).

1) Strong confluence implies confluence.2) Confluence is equivalent to Church-Rosser.3) Confluence implies local confluence but the converse is false, in general.

2.2. Rewriting in monoids

Rewriting systems over monoids (and in particular over groups) play an importantpart in algebra. LetM be a monoid. A rewriting system overM is a binary relationS ⊆M ×M . It defines the rewriting relation =⇒

S⊆M ×M such that

x=⇒Sy if and only if x = p�q, y = prq for some (�, r) ∈ S.

The relation ∗⇐⇒S⊆M ×M is a congruence onM , hence the quotient setM/

∗⇐⇒S

forms a monoid with respect to the multiplication induced from M . We denote itby M/ { � = r | (�, r) ∈ S } or, simply by M/S or MS . Two rewriting systems Sand T over a monoid M are termed equivalent if ∗⇐⇒

S= ∗⇐⇒

T, i.e., MS =MT .

We say that a rewriting system S is strongly confluent (or confluent, etc.)if the relation =⇒

Shas the corresponding property. Instead of (�, r) ∈ S we also

write �−→r ∈ S and �←→r ∈ S in order to indicate that both (�, r) and (r, �) arein S.


We say that a word w is S-irreducible (sometimes we omit S here), if noleft-hand side � of S occurs in w as a factor. Thus, if w is irreducible then w ∗=⇒

Sw′

implies w = w′. The set of all irreducible words is denoted by IRR(S).In order to compute with monoids (in particular, groups) we usually specify

a choice of monoid generators Γ, sometimes called an alphabet. For groups we oftenassume that Γ is closed under inversion, so Γ = Σ∪Σ−1 where Σ is a set of groupgenerators. For an alphabet Γ we denote by Γ∗ the free monoid with basis Γ.Throughout, 1 denotes the neutral element in monoids or groups. In particular, 1is also used to denote the empty word in a free monoid Γ∗. If we can write w = xuy,then we say that u is a factor of w. For free monoids a factor is sometimes also calleda subword, but this might lead to confusion because other authors understand bya subword simply a subsequence or scattered subword.

Rewriting systems S over a free monoid Γ∗ are sometimes called string rewrit-ing systems or semi-Thue systems. In this case the quotient Γ∗/S has the standardmonoid presentation 〈Γ | { � = r | (�, r) ∈ S }〉. We say that a string rewritingsystem S defines a monoid M if Γ∗/S is isomorphic to M . In addition, if P is aproperty of rewriting systems (Church-Rosser, strongly confluent, confluent, etc.)we say that a monoid M has a P -presentation if it can be defined by a systemwith property P .

For groups two types of presentations via generators and relators arise:monoid presentations, described above, and group presentations, typical in com-binatorial group theory and topology. More precisely, we say that G = Γ∗/S is amonoid presentation of a group G if the alphabet Γ is of the form Γ = Σ ∪ Σ−1,where Σ is a set of group generators, and Σ−1 = {σ−1 | σ ∈ Σ} is the set of formalinverses of Σ (in which case Γ∗ is a the free monoid with an involution σ → σ−1).Given a group presentation 〈X | R〉 of a group G one can easily obtain a monoidpresentation of G by adding the formal inverses X−1 to the set of generators X ofG and the “trivial” relations xx−1 = 1, x−1x = 1, x ∈ X} to the relators of G. Weconsider here monoid presentations of groups, except where explicitly indicatedotherwise.

2.3. Convergent rewriting systems

In this section we briefly discuss convergent (or complete) rewriting systems, whichplay an important role in algebra due to their relation to normal forms.

A relation =⇒⊆ X×X is called terminating (or Noetherian) if every infinitechain

x0∗=⇒x1

∗=⇒· · ·xi−1∗=⇒xi

∗=⇒· · ·becomes stationary.

There are two typical sources of terminating string rewriting systems S ⊆Γ∗ × Γ∗. Systems of the first type are length-reducing, i.e., for any rule �→ r ∈ Sone has |�| > |r|, where |x| is the length of a word x ∈ Γ∗. Systems of the secondtype are compatible with a given reduction ordering � on Γ∗, which means that if� → r ∈ S then � � r. Recall that a reduction ordering on Γ∗ is a well-ordering


preserving left and right multiplication (i.e., if u � v then aub � avb for anya, b ∈ Γ∗). Clearly, such systems are terminating. In fact, the condition that Sis compatible with some partial order, �, preserving left and right multiplicationis just a reformulation of the terminating property. Indeed, if S is terminatingthen there is a binary relation �S on Γ∗ defined by u �S v if and only if u ∗=⇒

Sv.

In this case �S is a partial well-founded ordering (no infinite descending chains),such that � �S r for any rule � → r ∈ S. Moreover, the converse is also true.(The condition that � is total is not needed here but is required in running theKnuth-Bendix completion procedure, see below).

A relation =⇒⊆ X ×X is called convergent (or complete) if it is locally con-fluent and terminating. The following properties are crucial. Let S be a convergentrewriting system.1) S is confluent (see for example [6, 31]).2) Every ∗⇐⇒

Sequivalence class in Γ∗ contains a unique S-reduced word (a word

to which no rule from S is applicable).3) If S is finite then for a given word w ∈ Γ∗ one can effectively find its unique

S-reduced form (just by subsequently rewriting the word w until the resultis S-reduced).The results above show that if a monoid M has a finite convergent presen-

tation then the word problem in M , as well as the problem of finding the normalforms, is decidable. This explains the popularity of convergent systems in algebra.There are many examples of groups that have finite convergent presentations: finitegroups, polycyclic groups, free groups and some geometric groups (see [46, 20, 34]for details).

One of the major results on convergent systems concerns the Knuth-Bendixprocedure (KB) (see [6] for general rewriting systems and [46, 20] for groups), whichcan be stated as follows. Let � be a reduction well-ordering on Γ∗ and S ⊆ Γ∗×Γ∗a finite rewriting system compatible with �. If there exists a finite convergentrewriting system T ⊆ Γ∗ × Γ∗ compatible with � which is equivalent to S, then,in finitely many steps, the Knuth-Bendix procedure KB finds a finite convergentrewriting system S′ ⊆ Γ∗ × Γ∗ compatible with � which is also equivalent to S.

There are three principle remarks due here.

Remark 2.2. The time complexity of the word problem in a monoid MS definedby a finite convergent system S may be of an arbitrarily high complexity [43].

Remark 2.3. It may happen that the word problem in a monoid MS defined by afinite convergent system S is decidable in polynomial time, whereas the complexityof the standard rewriting algorithm that finds the S-reduced forms of words canbe of an arbitrarily high complexity [43].

These remarks show that convergent rewriting systems may not be the besttool to deal with complexity issues related to word problems and normal forms inmonoids.


Remark 2.4. The Knuth-Bendix procedure really depends on the chosen ordering�. The following example shows that in a free Abelian group of rank two the KBprocedure relative to one length-lexicographic ordering results in a finite conver-gent presentation, while another length-lexicographic ordering does not allow anyfinite convergent presentations for the same group.

Example 2.5 ([21], page 127). Let G be the free Abelian group given by the fol-lowing monoid presentation.

〈x, y, x−1, y−1 | xy = yx, xx−1 = x−1x = yy−1 = y−1y = 1〉.Then the KB procedure with respect to the length-lexicographic ordering inducedby the ordering x < x−1 < y < y−1 of the generators outputs a finite convergentsystem defining G:

xx−1 =⇒ 1, x−1x =⇒ 1, yy−1 =⇒ 1, y−1y =⇒ 1,

yx =⇒ xy, y−1x =⇒ xy−1, yx−1 =⇒ x−1y, y−1x−1 =⇒ x−1y−1.However, there are no finite convergent systems defining G and compatible withthe length-lexicographic ordering x < y < x−1 < y−1.

Therefore, even if a finite convergent presentation for a monoid M exists itmight be hard to find it using the Knuth-Bendix procedure. In addition O’Dun-laing [42] has shown that the problem of whether or not a given finitely presentedgroup can be defined by a finite convergent rewriting system is undecidable.

It is not hard to see that all finitely generated commutative monoids have afinite convergent presentation, [13]. However, this demands enough generators, ingeneral. For example, a free Abelian group of rank k can be generated as a monoidby an alphabet of size k + 1, but in order to find a finite convergent system for itwe need at least 2k generators, see [14]. Another nice example of this kind is thenon-commutative semi-direct product of Z by Z. Even as a monoid we need justtwo generators a and b and one relation abba = 1. There is no finite convergentsystem S ⊆ { a, b }∗ × { a, b }∗ such that { a, b }∗ / { abba = 1 } = { a, b }∗ /S, butclearly such systems exist if we allow more generators. See [31] for more detailsabout this example.

We finish the section with a few open problems.

Problem 2.6. Is it true that every hyperbolic group has a finite convergent pre-sentation?

It is known that some hyperbolic groups have finite convergent presentations,for example, surface groups [34].

Problem 2.7. Is it true that every finitely generated fully residually free group hasa finite convergent presentation?

The next two problems are from [43].

Problem 2.8. Do all automatic groups have finite convergent presentations?


Problem 2.9. Do all one-relator groups have finite convergent presentations?

Notice that all the groups above satisfy the homological condition FP∞;which is the main known condition necessary for a group to have a finite convergentpresentation, see [48, 47].

2.4. Computing with infinite systems

In this section we discuss computing with infinite systems. An infinite string rewrit-ing system S ⊆ Γ∗ × Γ∗ can be used in computation if it satisfies some naturalconditions. Firstly, one has to be able to recognize if a given pair (u, v) ∈ Γ∗ × Γ∗

gives a rule u → v ∈ S or not, i.e., the system S must be a recursive subset ofΓ∗ × Γ∗. We call such systems recursive. Secondly, to rewrite with S one has tobe able to check if for a given u ∈ Γ∗ there is a rule � → r ∈ S with � = u,so we assume that the set L(S) of left-hand sides of the rules in S is a recursivesubset of Γ∗. Systems satisfying these two conditions are termed effective rewritingsystems. Clearly, every finite system is effective. Notice also, that every recursivenon-length-increasing system S (i.e., |�| ≥ |r| for every rule �→ r ∈ S) is effective.Indeed, given u ∈ Γ∗ one can check if a rule u → v is in S or not for all words vwith |v| ≤ |u|, thus effectively verifying whether u ∈ L(S) or not.

The argument above shows that for a recursive non-length-increasing systemS one can effectively enumerate all the rules in S in such a way

�0 → r0, �1 → r1, . . . , �i → ri, . . . (1)

that if i < j then �i � �j in the length-lexicographical ordering � and also if�i = �j then ri � rj . We call this enumeration of S standard.

Proposition 2.10. Let S be an infinite effective convergent system. Then the wordproblem in the monoid MS defined by S is decidable.

Proof. Given a word u ∈ Γ∗ one can start the rewriting process by applying rulesfrom S. Indeed, for a given factor w of u one can check if w ∈ L(S) or not, thus,enumerating all factors of u, one can either find a factor w of u with w ∈ L(S) orprove that u is S-irreducible. If such w exists one can enumerate all pairs (w, v)with v ∈ Γ∗ and check one by one if (w, v) ∈ S or not. This procedure eventuallyterminates with a rule w → v ∈ S. Now one can apply this rule to u and rewriteu into u1. Applying again this procedure to u1 one eventually arrives at a uniqueS-irreducible word u. To check if two words are equal in the monoid MS one canfind their S-irreducibles and check whether they are equal or not. �

There are various modifications of the algorithm described above that workfor other types of, not necessarily convergent, infinite systems. We consider someof these below.


3. Length-reducing and Dehn systems

3.1. Finite length-reducing systems

In this section we study a very particular type of rewriting system, called length-reducing, where, for every rule � → r one has |�| > |r|. The main interest inlength-reducing systems comes from the fact that, contrary to the case of finiteconvergent systems, the algorithm for computing the reduced forms is fast.

Lemma 3.1. [7] If S is a finite length-reducing string rewriting system then irre-ducible descendants of a given word can be computed in linear time (in the lengthof the word).

This result is well known, we use it in many parts of the paper, and it canbe seen easily as follows.

Proof. First, we choose some ε > 0 such that (1−ε)|�| ≥ |r| for all rules (�, r) ∈ S.Now, consider an input w ∈ Γ∗ of length n = |w|. For the moment let a

configuration be a pair (u, v) such that (i) w ∗⇐⇒S

uv and (ii) u is irreducible. The

goal is to transform the initial configuration (1, w) in O(n) steps into some finalconfiguration (w, 1).

Say we are in the configuration (u, v). The goal is achieved if v = 1. So assumethat v = av′ where a is a letter. If ua is irreducible then we replace (u, av′) by(ua, v′), and (ua, v′) is the next configuration. If however ua is reducible then wecan write ua = u′� for some (�, r) ∈ S; and u′ is irreducible. So, we replace (u, av′)by (u′, rv′), and (u′, rv′) is the next configuration. The algorithm is obviouslycorrect. Defining the weight γ of configurations by γ(u, v) = (1− ε)|u|+ |v| we seethat γ reduces from one configuration to the next by at least ε. Hence we havetermination in linear time. �

In fact, length reducing rewriting systems arise naturally in the class of smallcancellation groups and more generally hyperbolic groups, which we might regardas a paradigm for groups with easily solvable word problem. To be precise: a groupG is hyperbolic if and only if there is a finite generating set Γ for G and a finitelength-reducing system S ⊆ Γ × Γ (so G = Γ/S) such that a word w representsthe trivial element of G if and only if w can be S-reduced to the empty word,see [2]. In other words a group is hyberbolic if and only if there exists a finitelength-reducing system which is confluent on the empty word.

Definition 3.2. A length-reducing string rewriting system which is confluent onthe empty word is called a Dehn system.

If a group is defined by a finite length-reducing Dehn rewriting system thenthe rewriting algorithm is known in group theory as the Dehn algorithm. Moregeneral definitions of Dehn algorithms, for rewriting systems over a larger alphabetthan the generators of the group, have been studied by Goodman and Shapiro [25]and Kambites and Otto [32]. In particular in [25] it is shown that such generalised


Dehn algorithms solve the word problem in finitely generated nilpotent groups andmany relatively hyperbolic groups.

It is known that, given a finite presentation of a hyperbolic group G, one canproduce a finite Dehn presentation of G by adding, to a given presentation, all newrelators of G up to some length (which depends on the hyperbolicity constant ofG). However, this algorithm is very inefficient and the following questions remain.

Problem 3.3. Is there a Knuth-Bendix type completion process that, given a finitepresentation of a hyperbolic group G, finds a finite Dehn presentation of G?

Problem 3.4. Is there an algorithm that, given a finite presentation of a hyperbolicgroup, determines whether or not this presentation is Dehn?

Notice that some partial answers to this question are known. Namely, in [3]Arzhantseva has shown that there is an algorithm that, given a finite presentationof a hyperbolic group and α ∈ [3/4, 1), detects whether or not this presentationis an α-Dehn presentation. Here a presentation 〈X | R〉 of a group G is calledan α-Dehn presentation if any non-empty freely reduced word w ∈ (X ∪ X−1)∗

representing the identity in G contains as a factor a word u which is also a factorof a cyclic shift of some r ∈ R±1 with |u| > α|r|.

3.2. Infinite length-reducing systems

Let us discuss some algorithmic aspects of rewriting with infinite length-reducingsystems.

Proposition 3.5. Let S ⊆ Γ∗ × Γ∗ be an infinite recursive string rewriting system.Then the following hold.

1) If S is length-reducing then an irreducible descendant of a given word can becomputed.

2) If S is Dehn and MS is a group, then the word problem in MS is decidable.

Proof. The system S is effective since it is recursive and length-reducing (seeremark before Proposition 2.10). Now the argument in the proof of Proposition2.10 shows that for a givenw one can effectively find an S-irreducible representativeof w, so 1) and 2) follow. �

In the case of length-reducing systems one can try to estimate the time com-plexity of the algorithms involved. To this end we need the following definition.Let S be an effective non-length increasing rewriting system and

�0 → r0, �1 → r1, . . . , �i → ri, . . .

its standard enumeration (see Section 2.4). If there is an algorithm A and a poly-nomial p(n) such that for every n ∈ N the algorithm A writes out the initial partof the standard enumeration of S with |�i| ≤ n in time p(n) then the system S iscalled enumerable in time p(n) or Ptime enumerable. In particular, we say that Sis linear (quadratic) time enumerable if the polynomial p(n) is linear (quadratic).


Proposition 3.6. Let S ⊆ Γ∗×Γ∗ be an infinite non-length increasing string rewrit-ing system, which is enumerable in time p(n). Then the following hold.

1) If S is length-reducing then an irreducible descendant of a given word w canbe computed in polynomial time.

2) If S is Dehn and MS is a group then the word problem in MS is decidable inpolynomial time.

Proof. Given a wordw one can list in time p(|w|) all the rules �→ r of the standardenumeration of S with |�| ≤ |w|. Now in time O(p(|w|)|w|2) one can check whetherone of the listed rules can be applied to w or not. This proves 1) and 2). �

3.3. Weight-reducing systems

Many results above can be generalised to weight-reducing systems. A weight γassigns to each generator a a positive integer γ(a) with the obvious extensions towords by γ(a1 · · · an) =

∑ni=1 γ(ai). A system is called weight-reducing if, for every

rule � → r, one has γ(�) > γ(r). The following statements in this paragraph aretaken from [16]. It is decidable whether a finite system is weight-reducing by linearinteger programming. The reason to consider weight-reducing systems is that thereare monoids like { a, b, c }∗ /ab = c2 having an obvious finite convergent weight-reducing presentation, but where no finite convergent length-reducing presentationexists.

For groups the situation is unclear. Actually, the following conjecture hasbeen stated.

Conjecture 3.7. Let G be a finitely generated group. Then the following assertionsare equivalent.

1) G is a plain group, i.e., G is a free product of free and finite groups.2) G has a finite convergent length-reducing presentation.3) G has a finite convergent weight-reducing presentation.

The implications 1) =⇒ 2) =⇒ 3) are trivial, and 2) =⇒ 1) is known asthe Gilman conjecture and was stated first in [23].

It is clear that the conjugacy problem can be decided in plain groups and thisholds for groups G having a finite convergent weight-reducing presentation too. Infact, for s, t ∈ G the set Rs,t =

{g ∈ G

∣∣ gsg−1 = t}is an effectively computable

rational subset of G.

4. Preperfect systems

4.1. General results

In this section we discuss preperfect rewriting systems, which play an importantpart in solving their word problem and finding geodesics (shortest representativesin the equivalence classes) in groups.


Definition 4.1. A Thue system is a rewriting system S ⊆ Γ∗ × Γ∗ such that thefollowing conditions hold.

i) If � −→ r ∈ S then |�| ≥ |r|.ii) If � −→ r ∈ S, with |�| = |r|, then r −→ � ∈ S too.

To every rewriting system is associated an equivalent Thue system. In orderto specify a Thue system which is equivalent to a rewriting system S one can do thefollowing: symmetrize S by adding all the rules r −→ � whenever � −→ r ∈ S, thenthrow out all the length increasing rules. The new system, denoted T (S), is calledthe Thue resolution of S. It follows that every monoid has a Thue presentation.

Definition 4.2. A confluent Thue system is called preperfect.

The main interest in preperfect systems in algebra comes from the followingknown (and easy) complexity result: for which we require the following definition.

Definition 4.3. A word w ∈ Γ∗ is termed S-geodesic, with respect to a stringrewriting system S, if it has minimal length in its ∗⇐⇒

S-equivalence class (and

simply geodesic where no ambiguity arises).

Clearly, S-geodesic words are precisely the geodesic words in the monoidΓ/S relative to the generating set Γ, i.e., they have minimal length among all thewords in Γ∗ that represent the same element in Γ/S. Sometimes we say that aword w ∈ Γ∗ is a geodesic of a word u ∈ Γ∗ if w is S-geodesic and ∗⇐⇒

S-equivalent

to u.

Proposition 4.4. If a rewriting system S is finite and preperfect, then one candecide the word problem in the monoid defined by S in polynomial space and hencein exponential time. Moreover, along the way one can find an S-geodesic of a givenword w as well as all S-geodesics of w.

A locally confluent (strictly) length-reducing system is convergent, hence,from the above, preperfect. However the Thue resolution of an arbitrary finiteconvergent rewriting system may fail to be terminating or confluent as simpleexamples show. (Let Γ = {a, b, c, d, u, v} and S be the system with rules ab−→u,

bc−→ v, uc−→ d3 and av−→ d3. Then T (S) is not confluent. The system with one

rule a−→ b has non-terminating Thue resolution.) It is also easy to see that T (S)may be preperfect when S is not confluent. On the other hand, if a confluent systemS has no length-increasing rules then the Thue resolution can be constructedby symmetrizing S relative to all length preserving rules in S (by adding therule r −→ � for each length preserving rule � −→ r ∈ S) and a straightforwardargument shows that in this case T (S) is confluent, so preperfect.

Lemma 4.5. If S is a confluent rewriting system with no length-increasing rulesthen the Thue resolution T (S) is preperfect.


For a system S (for example a Thue system) where all rules � → r ∈ Sare either length-reducing |�| > |r|, or length-preserving |�| = |r|, it is convenientto split S into a length reducing part SR and a length preserving part SP , soS = SR ∪SP . If S is a Thue system then all S-geodesic words that lie in the sameequivalence class have the same length and any two of them are SP -equivalent(can be transformed one into other by a sequence of rules from SP ). Therefore, theword length in Γ∗ induces a well-defined length on the factor-monoidM = Γ∗/SP

(application of relations from SP does not change the length). Hence, one can viewSR as a length reducing rewriting system over the monoid M = Γ∗/SP , in whichcase we assume that SR ⊆ M ×M . Note that if SP is finite and if there is aneffective way to perform reduction steps with SR then the word problem in M isdecidable.

Decidability of the word problem in M = Γ∗/SP allows one to test whethera given rule from SR is applicable to an element of M . Since SR ⊆ M ×M isterminating it suffices to show local confluence to ensure convergence. This maytempt one to introduce an analogue of the Knuth-Bendix completion. However,in general an infinite number of critical pairs may appear in the Knuth-Bendixprocess, and one needs to be able to recognize when the current system becomespreperfect. Unfortunately, this is algorithmically undecidable. More precisely, thefollowing result holds.

Theorem 4.6 ([38]). The problem of verifying whether a finite Thue system ispreperfect or not is undecidable.

In fact in [39] this problem is shown to be undecidable even in the case of aThue system whose length-preserving part SP consists only of a single rewritingrule of the form ab←→ ba. On the other hand, under some additional assumptions

such a procedure can yield useful results ([17, 18]) – good examples in our contextare graph groups, c.f. Section 7.1.

In the final part of this section we discuss some complexity issues in com-puting with preperfect systems. By Proposition 4.4 finite preperfect systems allowone to solve the word problem and find geodesics in at most exponential time.

Proposition 4.7. Let S be an infinite preperfect rewriting system. Then:1) if S is recursive then the word problem in the monoid MS defined by S is

decidable;2) if S is Ptime enumerable then one can solve the word problem in MS and

find a geodesic of a given word in exponential time.

Proof. Since preperfect rewriting systems are non-length-increasing it follows thatrecursive preperfect systems are effective (see the remark before Proposition 2.10).Therefore, given a word w one can effectively list all the rules in S with the left-hand sides of length at most |w|. Denote this subsystem of S by Sw. Now rewritingw using S is exactly the same as using Sw, so 1) and 2) follow from the argumentin the proof of Proposition 4.4 for finite preperfect systems. �


5. Geodesically perfect rewriting systems

In this section we consider a subclass of Thue systems which are designed to dealwith geodesics in groups or monoids. In particular, we study confluent geodesicsystems, which form a subclass of preperfect string rewriting systems, and whichbehave better in many ways than general preperfect systems. We call these systemsgeodesically perfect, as this indicates their essential properties and fits with theterminology of preperfect systems. However as discussed in Section 1 they arealso known in the literature as almost confluent or quasi-perfect. The motivationfor the study of geodesically perfect systems in group theory comes mainly fromattempts to solve the, algorithmically difficult, geodesics problem: that is, given afinite presentation of a group G and a word w in the generators, find a word ofminimal length representing w as an element of G.

5.1. Geodesic systems

We consider first a somewhat larger, less well-behaved, class of rewriting systems.

Definition 5.1. A string rewriting system S ⊆ Γ∗ × Γ∗ is called geodesic if S-geodesic words are exactly those words to which no length reducing rule from Scan be applied.

Note that if S is a geodesic rewriting system then its Thue resolution isalso geodesic. This allows us to assume, without loss of generality, that geodesicsystems are Thue systems.

Remark 5.2. Dehn rewriting systems are not in general geodesic: they need onlyrewrite words that represent the identity to (empty) geodesics in Γ∗/S.

A finite geodesic system gives a linear time algorithm to find a geodesic ofa given word u ∈ Γ∗. The following algebraic characterisation of finite geodesicsystems in groups is given in [24]. (The definition of geodesic in [24] is slightlymore restrictive than ours, however this makes no difference to the result.)

Theorem 5.3 ([24]). A group G is defined by a finite geodesic system S if and onlyif G is a finitely generated virtually free group.

From the result of Rimlinger quoted above finitely generated virtually freegroups are precisely the universal groups of finite pregroups. It follows that everyfinite length reducing geodesic system can be transformed to the length reducingpart of the rewriting system (see Section 8.1) associated with a finite pregroup.

The following result follows from Proposition 3.6.

Proposition 5.4. Let S be a geodesic Ptime enumerable string rewriting systemsuch that the monoid MS is a group. Then the word problem in the group MS isdecidable in polynomial time.

Very little is known about geodesic systems which do not present groups. Inparticular, it is not clear whether the word problem remains decidable: that is,given u, v ∈ Γ∗ decide whether or not u ∗⇐⇒

Sv.


Problem 5.5. Does there exist a finite geodesic system S for which the word prob-lem is undecidable?

The following result demonstrates one of the principal difficulties of workingwith geodesic systems.

Theorem 5.6. It is undecidable whether a finite rewriting system is geodesic.

Proof. The proof is a modification of the proof by Narendran and Otto [39] whichshowed undecidability of preperfectness in presence of a single commutation rule.

We need some notation and we adhere as far as possible to that of [39].We shall define the computation of a Turing machine by a set of rewriting rules.A configuration of the machine is then a particular form of word over the tapealphabet, the states and the end markers. In detail let Σ be a finite set, the tapealphabet, let Σ be a disjoint copy of Σ, let Q be a finite set of states, and α and βbe special symbols representing end markers. There are two marked states q0 andqf , the initial and final states. The computation of the machine can be describedby a finite set of rules which fall into the following categories, where we use thenotation p, q ∈ Q, p �= qf , a, a′, b ∈ Σ:1) pa−→ a′q.

(Read a in state p, write a′, move one step to the right, switch to state q.)2) bpa−→ qba′. (As above, but move one step to the left.)

3) pβ−→ paβ. (Create new space before the right end marker.)

These rewriting rules constitute the rewriting system associated to M . We assumethat the machine is deterministic, so there are no overlapping rules. A configurationof a (deterministic) Turing machine is then a word αuqvβ with u ∈ Σ

∗, v ∈ Σ+,

and q ∈ Q. The initial configuration on input x ∈ Σ∗ is the word αq0xβ. Weassume that the machine stops if and only if it reaches the state qf .

Now let M be a Turing machine for which it is undecidable whether or notcomputation halts on input x ∈ Σ∗. Using this machine we are going to construct,for each x ∈ Σ∗, a new length reducing rewriting system Sx, which is geodesic ifand only if the machine M does not stop on input x.

The alphabet Γ of each such system is to consist of the symbols of Σ ∪ Σ ∪Q ∪ {α, β} and new additional symbols d, e, γ, δ, I, C. The system Sx will consistof rules which simulate the computation of M on input x, with some additionalcontrol on the number of steps of the computation carried out. Let x ∈ Σ∗. Tobegin with, we introduce rules leading to two different initial configurations. Letm = |x|+ 5. We define the two rules

αq0xβγ←−(1)

ICm−→(2)

αq0xβδ.

Next we introduce rules, involving d, e, γ and δ, to control the number of stepsof the simulation. Symbols γ and δ convert d’s to e’s. The latter act as tokens tocontrol the number of steps performed by the simulation ofM . Both γ and δ moveright consuming three d’s and producing two e’s, the difference being that γ may


move to the right arbitrarily far from β whereas δ is forced to remain very closeto β. The effect of each rule on the length of a word is the same.

Explicitly, we add new rules of the form:

γddd−→ eeγ, βδddd−→βeeδ.

Note that, using rule (1), all words in ICm(ddd)∗ now reduce as follows

ICmd3n=⇒(1)

αq0xβγd3n ∗=⇒αq0xβe

2nγ.

However, using the rule ICm−→(2)

αq0xβδ in the first step we can only do

ICmd3n=⇒(2)

αq0xβδd3n ∗=⇒αq0xβeeδd

3n−3

and then, for n ≥ 2, we are stuck.Now we bring e into the game. The letter e is used to enable a computation

step of M . It can move to the left until it is at distance one to the right of a statesymbol. The generic rules for e allow e to move left and are as follows:

abee−→ aeb, aebee−→ aeeb for a ∈ Σ, b ∈ Σ ∪ {β}.

Let us describe the effect of these rules on words of the form

αupa1 · · · akβδd3n

where n is huge (and k is viewed as constant k ≥ 0), ai ∈ Σ, u ∈ Σ∗. The maximal

possible reduction leads to a word of the form

αupa1ee · · ·akeeβeeδd3n′ .

In this case, if n is large enough, then n′ > 0, no further reduction is possible andactually n− n′ ∈ O(1).

At this point we introduce rules to simulate the computation of the machineM . There is one simulation rule corresponding to each rule in the rewriting systemassociated to M . More precisely we introduce a rule uee−→ v for each rewriting

rule u−→ v of M : so we have simulation rules of three types (where again we use

the notation p, q ∈ Q, a, a′, b ∈ Σ)paee−→ a′q, bpaee−→ qba′, pβee−→ paβ.

The system Sx consists of the rules defined so far, which we list in Figure 1, so islength reducing.

Now assume that the machine M halts on input x. This implies that onlyfinitely many computation steps t can be performed. Again choose n huge and viewt and |x| as constants. Consider a word of the form ICmd3n. Starting a reductionwith the second rule we get stuck at an irreducible word when the simulationreaches state qf :

ICmd3n−→(2)

αq0xβδd3n ∗=⇒αuqfa1ee · · ·akeeβeeδd

3n′


I) Initial rules:αq0xβγ←−

(1)ICm−→

(2)αq0xβδ.

II) Step control rules, for a ∈ Σ, b ∈ Σ ∪ {β}:γddd−→ eeγ,

βδddd−→βeeδ,

abee−→ aeb,

aebee−→ aeeb.

III) Simulation rules, for a, b ∈ Σ, p ∈ Q\{qf}:paee−→a′q,

bpaee−→ qba′,

pβee−→ paβ.

Figure 1. The system Sx.

at which point n−n′ ∈ O(1). The system Sx cannot be geodesic because with theother initial rule we can first move Γ to the right of all the d’s thereby losing nletters immediately:

ICmd3n−→(1)

αq0xβγd3n ∗=⇒αq0xβd

2nγ

and then when the simulation reaches the state qf the resulting irreducible wordwill end e2n

′γ instead of δd3n

′(and otherwise will be the same).

It remains to cover the case when the machine does not halt on input x.We shall show that in this case the system Sx is geodesic. Note that, as M neverreaches state qf , for all n > 0

αq0xβe2n ∗=⇒

Sx

αupyβ,

where u ∈ Σ∗, p ∈ Q and y ∈ (Σ∪{e})∗. For technical reasons we define a sequenceof words wi for i ≥ 0 as follows. We let w0 = q0x and let αwi+1β be defined to bethe irreducible descendant of αwiβee. The sequence of words wi is infinite becausethe machine does not stop on input x. Thus

∀i ≥ 0 : αwiβee∗=⇒Sx

αwi+1β ∈ IRR(Sx).

Now we add infinitely many rules to Sx to form a new system Tx as follows:

∀i ≥ 0 : αwiβδ−→αwiβγ.

As the rules of Tx are generated by steps of the Knuth-Bendix completionprocedure applied to Sx the congruences generated by Sx and Tx are the same.


IV.) Completion rules:∀i ≥ 0 : αwiβδ−→αwiβγ

where αwiβee∗=⇒Sx

αwi+1β,∀i ≥ 0, and w0 = q0x.

Figure 2. The additional rules of system Tx.

To summarise, the system Tx consists of the rules of Figure 1 and those listed inFigure 2. Thus Tx is terminating and local confluence can be checked directly.

Each word w ∈ Γ∗ has a unique factorisation where we choose k and all nj

to be maximal:u0(ICmdn1)u1 · · · (ICmdnk)uk.

The benefit of the system Tx is that it provides us with canonical geodesics. Ageodesic of w is given by:

u0(αwi1βγdm1)u1 · · · (αwikβγd

mk)uk,

where mj = nj mod 3. The crucial observation is that allowing only rules from Sx

we achieve exactly the same form with the exception that some γ’s are still δ’s.Thus, the system is geodesic. �

5.2. Geodesically perfect systems

Definition 5.7. A string rewriting system S ⊆ Γ∗×Γ∗ is called geodesically perfect if

i) S is geodesic andii) if u, v ∈ Γ∗ are S-geodesics then u

∗⇐⇒S

v if and only if u ∗⇐⇒SP

v, where SP is

the length-preserving part of S.

Again, it follows directly that if S is a geodesically perfect system then so isits Thue resolution, so we can assume that geodesically perfect systems are Thue.If S is a geodesically perfect Thue system then we write it as S = SR ∪ SP whereSR is its length reducing part and SP its length preserving part. It also followsfrom the definition that a geodesically perfect system is confluent.

There is a simple procedure to describe geodesics of elements in the monoidΓ∗/S defined by a geodesically perfect Thue system S. Namely, the geodesics of agiven word w ∈ Γ∗ are the SR-reduced forms of w and any two such geodesics canbe obtained from one another by applying finitely many rules from SP . Moreoverit is shown in [6] that the word problem for monoids defined by finite geodesicallyperfect rewriting systems is PSPACE complete.

The following result relates geodesically perfect to preperfect Thue systems.

Proposition 5.8. Let S ⊆ Γ∗ × Γ∗ be a Thue system. Then

1) if S is geodesically perfect then it is preperfect and2) if S is preperfect and geodesic then it is geodesically perfect.


Proof. 1) follows from the observation that geodesically perfect implies confluent.To see 2) observe that S is confluent, hence Church-Rosser. Therefore, if u, v aretwo geodesics with u

∗⇐⇒S

v then u∗=⇒S

w and w∗⇐=S

v for some w ∈ Γ∗. Since u, v

are S-geodesics the only rules that could be applied in u∗=⇒S

w and w∗⇐=S

v are

length preserving, hence u ∗⇐⇒SP

v, as required. �

In Section 8.1 we will describe a general tool to construct geodesically perfectsystems defining groups: based on the fact that rewriting systems associated withpregroups are always geodesically perfect.

In Corollary 8.7 we prove that groups defined by finite geodesic systems areexactly the groups defined by finite geodesically perfect systems.

Obviously, every geodesic rewriting system S contains the length-reducingpart TR of some (infinite) geodesically perfect Thue system T defining the samemonoid. Indeed, one can obtain T by first constructing the Thue resolution T ′ ofS and then adding length-preserving rules to T ′ to make it confluent. But it is nottrue that every finite geodesic rewriting system S is the length-reducing part of afinite geodesically perfect system defining the same monoid. To see this considerthe following example.

Example 5.9. The following system is geodesic, and it is not the length-reducingpart of any finite geodesically perfect system defining the same quotient monoid.

add−→ ab, add−→ac, bdd−→ eb, cdd−→ ec.

Indeed, let S be the system above, and let T = S ∪{ b←→ c }. The new system Tis geodesically perfect by Proposition 6.1. But T -geodesics are computed by usingrules from S. As ∗⇐⇒

S⊆ ∗⇐⇒

Twe see that S is a geodesic system.

Let us show that S is not the length-reducing part of any equivalent, finite,geodesically perfect system. For a contradiction, assume that a finite set T of non-trivial symmetric rules can be added to S such that S ∪ T becomes geodesicallyperfect and is equivalent to S. Assume T involves a new letter, say f . Then f isequal to some word uf over {a, b, c, d, e} which is irreducible with respect to S. If uf

is empty then we do not need f , hence uf is nonempty and we have uf∗=⇒T

f . The

rules of T are symmetric (hence length preserving), so f is accompanied by a rule,say f ←→ a, and f is redundant. So, actually we may assume T ⊆ {a, b, c, d, e}∗×{a, b, c, d, e}∗. Clearly, aenb ∗⇐=

Sad2n+2

∗=⇒S

aenc, hence aenb ∗⇐⇒S

aenc and so aenb

and aenc are in the same class and are S-reduced. Because T is finite, some left-hand side of T must contain a word u ∈ ae∗ ∪ e∗ ∪ e∗b∪ e∗c. But all these words uare S-reduced, hence geodesic. Moreover, for any such u there is no other word vin the same class as u and of the same length. So, for large enough n the rules of Tcannot be applied to either aenb or aenc. As T is the length preserving part of thesupposedly geodesically perfect system S ∪ T , this is the required contradiction.


Remark 5.10. LetM = {a, b, c, d, e}∗/S be the quotient monoid as in Example 5.9.The proof above can be modified in order to show that actually there is no finitesystem T ⊆ {a, b, c, d, e}∗ × {a, b, c, d, e}∗ which is geodesically perfect and whichdefines M . However, if we use an additional letter f then the following systemdefines M , too.

dd−→ f, af←→ ab, af←→ ac, bf←→ eb, cf←→ ec.

The system is geodesically perfect, by Proposition 6.1 again.

We note that Example 5.9 illustrates a general fact: namely that if S is arewriting system and there exists a set T of symmetric rules such that S ∪ T isgeodesically perfect (but not necessarily equivalent to S) then S itself is geodesic.Since geodesic systems are undecidable whereas geodesically perfect systems aredecidable this could prove to be a useful test for a geodesic system.

6. Knuth-Bendix completion for geodesically perfect systems

A classical result of Nivat and Benois (stated in Proposition 6.1) shows that it isdecidable whether a finite Thue system is geodesically perfect. In order to explainthe criterion we need the notion of critical pair. All rewriting systems S in thissubsection are viewed as Thue systems and split into a length reducing part SR

and a length preserving part of symmetric rules SP . By definition, a critical pairis a pair (x, y) arising from the situation

x ⇐=(1,r1)

z =⇒(2,r2)

y

subject to the following conditions.1. (�1, r1) ∈ SR is length reducing but (�2, r2) ∈ S can be any rule.2. z = �iui = uj�j , with |ui| < |�j| and i, j ∈ {1, 2} such that i = j, implies

ui = uj = 1.

Proposition 6.1 ([41]). A finite Thue system S is geodesically perfect if and only if,for all critical pairs (x, y), there are words x′ and y′ such that with length reducingreductions we have:

x′ ∗⇐=SR

x, y∗=⇒

SR

y′,

and with length preserving reductions we have:

x′ ∗⇐⇒SP

y′.

Proof. The proof is not very difficult and can be found, for example, in the book[6, Thm. 3.6.4]. �

Remark 6.2. Note that the words x′ and y′ in Proposition 6.1 need not be irre-ducible w.r.t. the length reducing subsystem SR. This fact is actually used in theproof of Proposition 6.3.


This criterion leads to the following version of the Knuth-Bendix procedure.Consider a finite Thue system S0. We shall construct a series of Thue systemsS0 ⊆ S1 ⊆ S2 ⊆ · · · such that the union over all Si is geodesically perfect and wehave Si = Si+1 (i.e., the completion procedure stops) if and only if there existsa finite Thue system T which is geodesically perfect and equivalent to S0, that is∗⇐⇒

S0= ∗⇐⇒

T. We divide the procedure into phases. We assume that in phase i

a Thue system Si = SR ∪ SP has been defined such that SR contains the lengthreducing rules, SP contains the length preserving rules, and ∗⇐⇒

S0= ∗⇐⇒

Si

.

We begin phase i + 1 by computing a list of all critical pairs of the systemSi (which were not already considered in phases 1 to i). For each such pair (x, y)choose words x, y, irreducible with respect to the subsystem SR, such that

x∗⇐=

SR

x, y∗=⇒

SR

y.

Define new rules as follows.• If |x| > |y| then add the rule x−→ y to SR.

• If |y| > |x| then add the rule y−→ x to SR.

• If |y| = |x| then test whether or notx

∗⇐⇒SP

y.

If the answer is negative then add the symmetric rule x←→ y to SP .

The system Si+1 is defined to be Si together with all new rules which have beenadded to resolve all critical pairs of Si. On a formal level we define Si for all i ≥ 0but, of course, the procedure stops as soon as Si = Si+1, i.e., no new rules areneeded to resolve critical pairs of Si. Thus, if it stops with Si = Si+1 then Si isa finite geodesically perfect Thue system, which is equivalent to S0 (and we haveSi = Sj for all i ≤ j). However, what we really wish is stated in the followingproposition.

Proposition 6.3. Let S0 = SR ∪ SP be a finite Thue system with length reducingrules SR and length preserving rules SP . Let

S0 ⊆ S1 ⊆ · · ·Si ⊆ · · ·be the sequence of Thue systems which are computed by the Knuth-Bendix comple-tion as described above. Let S =

⋃i≥0 Si. Then the system S is geodesically perfect

and we have ∗⇐⇒S0

= ∗⇐⇒Si

= ∗⇐⇒s

for all i ≥ 0. Moreover the following statements

are equivalent.1) We have Si = Si+1 for some i ≥ 0.2) The Thue system S is finite and geodesically perfect.3) There exists some finite geodesically perfect Thue system T such that

∗⇐⇒S0

= ∗⇐⇒T

.


Proof. If Si = Si+1, for some i ≥ 0, then clearly S is finite. It is geodesicallyperfect by the criterion of Nivat and Benois, cf. Proposition 6.1. So, assume thereexists some finite geodesically perfect Thue system T with ∗⇐⇒

S0= ∗⇐⇒

T. We

have to show that the procedure stops. We let m be large enough that m ≥max { |�| | (�, r) ∈ T }. Next we consider i large enough that Si contains all rulesfrom S where the left-hand side has length of at most m. Clearly, an index i ∈N with this property exists. We will show that Si is geodesically perfect, andProposition 6.1 immediately implies Si = Si+1. For technical reasons, in a firststep we remove from T all length preserving rules (�, r) ∈ T where we can applyto � a length reducing rule of T . It is clear that the new and smaller system T ′

is still geodesically perfect and ∗⇐⇒S0

= ∗⇐⇒T ′

; so we replace T with T ′. Since

now (�, r) ∈ T with |�| = |r| implies that � and r are geodesics and since S isgeodesically perfect and m is large enough, we see that ∗⇐⇒

TP

⊆ ∗⇐⇒(Si)P

.

Next consider some word x which is irreducible with respect to the lengthreducing rules in Si. The claim is that x is a geodesic. Indeed assume the contrary.Then a length reducing rule (�, r) ∈ T can be applied to x. Since � is not geodesic,there is a length reducing rule in S which can be applied to � but due to thedefinition of m this rule is in Si, too. Thus, we have a contradiction and so Si isgeodesic. Now suppose that x and y are geodesic and that x ∗⇐⇒

Si

y. Then x∗⇐⇒T

y

so x ∗⇐⇒TP

y. As ∗⇐⇒TP

⊆ ∗⇐⇒(Si)P

this implies x ∗⇐⇒(Si)P

y so Si is geodesically perfect. �

A finite geodesic (or geodesically perfect) rewriting system S ⊆ Γ∗×Γ∗ allowsone to find S-geodesics in linear time. In particular, if the monoid M = Γ∗/Sdefined by S is a group one can solve the word problem in M in linear time.However, in general there seems to be no linear time reduction from the wordproblem in a monoid M to the geodesic problem.

7. Examples of preperfect systems in groups

7.1. Graph groups

Let Δ = (Σ, E) be an undirected graph. The graph group (or right angled Artingroup, or partially commutative group) defined by Δ is the group G(Δ) given bythe presentation

G(Δ) = F (Σ)/ { ab = ba | (a, b) ∈ E } ,

where F (Σ) is the free group with basis Σ. The group G(Δ) has a monoid pre-sentation given by a preperfect rewriting system SΔ. Indeed, let Γ = Σ ∪Σ where


Σ is a disjoint copy of Σ. The rules of SΔ are:

aa −→ 1ab ←→ ba if

{(a, b), (a, b), (a, b), (a, b)

}∩ E �= ∅

where a, b ∈ Γ and a = a for all a ∈ Γ.If the graph Δ is finite the system SΔ provides us with a decision algorithm

for solving the word problem in G(Δ), though not the fastest one (WP in graphgroups can be solved in linear time, see [53, 18]). However, the system SΔ is veryintuitive and simple and it gives the geodesics in G(Δ), which are precisely thewords whose length cannot be reduced by SΔ.

Although it is preperfect the system SΔ is not geodesically perfect. Howeverevery graph group may be constructed by a sequence of HNN-extensions and freeproducts with amalgamation, starting with infinite cyclic groups, and so, from theresults of Section 8 below, it follows that these groups may be defined by (infinite)geodesically perfect systems. Moreover finite convergent rewriting systems for thesegroups have been found by Hermiller and Meier [27] (see also [5, 22, 52]).

7.2. Coxeter groups

Let D3 = {a, b}∗/{a2 = 1, b2 = 1, (ab)3 = 1} be a dihedral group. Define apreperfect system S by the following rules

aa −→ 1,bb −→ 1,aba ←→ bab.

More generally, a Coxeter group on n generators a1, . . . , an is given by asymmetric n × n matrix (mij) with entries in N and 1’s on the diagonal. Thedefining relations are given by:

(aiaj)mij = 1 for all 1 ≤ i, j ≤ n.

Note that this implies a2i = 1 since mii = 1; and if mij = 0 then the equation(aiaj)0 = 1 is trivial. (Therefore it is also common to write (aiaj)∞ = 1, becauseaiaj turns out to be an element of infinite order in this case.)

The word problem of Coxeter groups can be solved by the preperfect Titssystem [51] (see also ([9, 1, 12]) of rewriting rules:

a2i −→ 1, for 1 ≤ i ≤ n,

(aiajaiaj · · · )←→ (ajaiajai · · · ) for 1 ≤ i, j ≤ n and

|(aiajaiaj · · · )| = |(ajaiajai · · · )| = mij .

The classical proof that this system is preperfect relies on the fact that Cox-eter groups are linear [4]. Of course this system is not geodesically perfect. Forvirtually free Coxeter groups Corollary 8.7 guarantees the existence of a finitegeodesically perfect rewriting system. It is shown in [26] that every Coxeter groupis either virtually free or contains a surface group; but the question of whether the


latter can be defined by a geodesically perfect system (necessarily infinite) remainsopen.

Convergent rewriting systems for Coxeter groups have been constructed, us-ing the Knuth-Bendix procedure, by le Chenadec [34], but in general these are notfinite. Finite convergent rewriting systems for certain classes of Coxeter groupshave been found by Hermiller [28] (see also [19, 8]).

7.3. HNN-extensions

Let G be any group with isomorphic subgroups A and B. Let Φ : A → B anisomorphism and let t be a fresh letter. By 〈G, t〉 we mean the free product ofG with the free group F (t) over t. The HNN-extension of G by (A,B,Φ) is thequotient group

HNN(G;A,B,Φ) = 〈G, t〉 /{t−1at = Φ(a)

∣∣ a ∈ A}.

There is a normal form theorem for elements in HNN(G;A,B,Φ), which impliesthat G embeds into HNN(G;A,B,Φ) and shows under which restrictions decid-ability of the word problem for G transfers to HNN-extensions. Usually the normalform theorem is shown by appeal to a combination of arguments of Higman, Neu-mann and Neumann and Britton, see [35, Chapter IV, Theorem 2.1].

Another option is to define a convergent string rewriting system. To see this,let Γ =

{t, t−1

}∪G \ {1} and view Γ as a possibly infinite alphabet. We identify

1 ∈ G with the empty word 1 ∈ Γ∗. We choose transversals for cosets of A and B.This means we choose X,Y ⊆ G such that there are unique decompositions

G = AX = BY.

We may assume that 1 ∈ X ∩ Y .The system S ⊆ Γ∗ × Γ∗ is now defined by the following rules with the

convention that [gh] denotes gh ∈ G (as a single letter or the empty word).

t−1t −→ 1; tt−1 −→ 1; gh −→ [gh], for all g, h ∈ G;tg −→ aty, if a ∈ A, a �= 1, y ∈ Y, Φ(a)y = g in G;t−1g −→ bt−1x, if b ∈ B, b �= 1, x ∈ X, Φ−1(b)x = g in G.

Proposition 7.1. The system S above is convergent and defines the HNN-extensionof G by (A,B,Φ). Every irreducible normal form admits a unique decomposition as

g = g0tε1g1 · · · tεngn

with n minimal such that n ≥ 0, g0 ∈ G \ {1}, and either εi = −1 with gi ∈ X orεi = 1 with gi ∈ Y , for all 1 ≤ i ≤ n.

Proof. Obviously, Γ∗/S defines the HNN-extension of G by by (A,B,Φ). Althoughthe system has length-increasing rules it is not too difficult to prove termination.Local confluence is straightforward, so S is indeed convergent. Since all elementsof G are irreducible we see that G embeds into the HNN-extension. Moreover, itis also clear that we obtain the normal form as stated in the proposition. �


This convergent system also leads to the following well-known classical fact.

Corollary 7.2. Assume that have the following properties: H is finitely generatedand has a decidable word problem, membership problems for A and B are solvable,and the isomorphism Φ : A→ B is effectively calculable. Then the HNN-extensionof G by by (A,B,Φ) has a decidable word problem.

Proof. We may represent all group elements in H by length-lexicographic first ele-ments (i.e., choose among all geodesics the lexicographical first one). The transver-sal X (resp. Y ) may be chosen to consist of the length-lexicographic first elementof each coset Ag (resp. Bg), where g runs over G. Given g we can compute therepresentative of Ag in X (resp. Bg in Y ), because membership is decidable for Aand B. Now, given b ∈ B, the ability to compute Φ allows us to find a ∈ A withΦ(a) = b. Thus, all steps in computing normal forms are effective. �

It should be clear however that the purpose of the system S above is notto decide the word problem effectively; but rather to facilitate straightforwardproofs of other results, such as Britton’s lemma. Consider the following system Bof Britton reduction rules.

t−1t −→ 1; tt−1−→ 1; gh−→ f, if gh = f in G;

t−1at −→ Φ(a) if a ∈ A;

tbt−1 −→ Φ−1(b) if b ∈ B.

The system B is length reducing, but not confluent. However, =⇒B⊆ ∗=⇒

H,

hence we can think of B as as subsystem of H . Britton’s lemma says that B isconfluent on all words which represent 1 in the HNN-extension. Here is a proofusing our system S. Consider any Britton reduced word g. It has the form g =g0t

ε1g1 · · · tεngn. Applying rules from H does not destroy the property of beingBritton reduced and neither t nor t−1 can vanish. Thus, if g reduces to the emptyword using H , then g is already the empty word.

Observe that B is not a geodesic system, because atΦ(a)−1 is Britton re-duced, but atΦ(a)−1 = t. In Example 8.2 below we construct a geodesically perfectrewriting system for an HNN-extension.

7.4. Free products with amalgamation

There is a natural convergent (resp. geodesically perfect) rewriting system whichdefines amalgamated products. Let A and B be groups intersecting in a commonsubgroup H . This time we choose transversals for cosets of H in A and in B; thatis X ⊆ A and Y ⊆ B with 1 ∈ X ∩ Y such that there are unique decompositionsA = HX and B = HY . We let Γ = (A∪B)\{1} and we identify 1 with the emptyword in Γ∗.

We use the convention of writing [ab] for the product ab whenever it is defined.This means [ab] is viewed as a letter in Γ or [ab] = 1 and it is defined if eithera, b ∈ A or a, b ∈ B.


The system S ⊆ Γ2 × ({1} ∪ Γ ∪ Γ2) is now defined by the following rules:

ab −→ [ab] if [ab] is defined,ab −→ [ah]y if 1 �= a ∈ A, h ∈ H, b �= y ∈ Y, and b = [hy],ba −→ [bh]x if 1 �= b ∈ B, h ∈ H, a �= x ∈ X, and a = [hx].

The system defines the amalgamated product G = A ∗H B. It is terminating bya length lexicographical ordering. Local confluence follows by a direct inspection,whence convergence. Again we obtain the normal form theorem (cf. [36, Corollary4.4.1]): every element g of G has a unique decomposition as

g = [hg0]g1 · · · gn,where h ∈ H , gi is a non-trivial element of X ∪ Y and gi and gi+1 do not lie inthe same factor. However, in practice we may not wish to compute transversalsexplicitly. So let us apply only length reducing rules ab −→ [ab] only until we endup with a word g = g0 · · · gn, to which no length reducing rule may be applied.Since we cannot apply length reducing rules to g we obtain that

∀0 ≤ i < n : gi ∈ A ⇐⇒ gi+1 ∈ B \H ∧ gi ∈ B ⇐⇒ gi+1 ∈ A \H.Further applications of the rules of S preserve this property. Thus, S is geodesicallyperfect, even if we use the length preserving rules only in the direction indicatedabove. Moreover, if we cannot apply length reducing rules to g = g0 · · · gn then wehave g = 1 if and only if both n = 0 and g0 = 1.

8. Stallings’ pregroups and their universal groups

We now turn to the notion of pregroups in the sense of Stallings, [49], [50]. Apregroup P is a set P with a distinguished element ε, equipped with a partialmultiplication m : D → P , (a, b) �→ ab, where D ⊆ P × P , and an involution (orinversion) i : P → P , a �→ a−1, satisfying the following axioms for all a, b, c, d ∈ P .(By “ab is defined” we mean to say that (a, b) ∈ D and m(a, b) = ab.)(P1) aε and εa are defined and aε = εa = a;(P2) a−1a and aa−1 are defined and a−1a = aa−1 = ε;(P3) if ab is defined then so is b−1a−1 and (ab)−1 = b−1a−1;(P4) if ab and bc are defined then (ab)c is defined if and only if a(bc) is defined, in

which case(ab)c = a(bc);

(P5) if ab, bc, and cd are all defined then either abc or bcd is defined.It is shown in [29] that (P3) follows from (P1), (P2), and (P4), hence can beomitted.

The universal group U(P ) of the pregroup P can be defined as the quotientmonoid

U(P ) = Γ∗/ { ab = c | m(a, b) = c } ,


where Γ = P \ {ε} and ε ∈ P is identified again with the empty word 1 ∈ Γ∗.The elements of U(P ) may therefore represented by finite sequences (a1, . . . , an) ofelements from Γ such that aiai+1 is not defined in P for 1 ≤ i < n: such sequencesare called P -reduced sequences or reduced sequences. Since every element in U(P )has an inverse, it is clear that U(P ) forms a group.

If Σ is any set then the disjoint union P = {ε} ∪ Σ ∪ Σ, where Σ is a copyof Σ, yields a pregroup with involution given by ε = ε, a = a, for all a ∈ Σ, suchthat pp = ε, for all p ∈ P . In this case the universal group U(P ) is nothing butthe free group F (Σ).

The universal property of U(P ) holds trivially, namely the canonical mor-phism of pregroups P → U(P ) defines the left-adjoint functor to the forgetfulfunctor from groups to pregroups.

Stallings [49] showed that composition of the inclusion map P → P ∗ withthe standard quotient map P ∗ → U(P ) is injective, where P ∗ is the free monoidon P . The first step of his proof establishes reduced forms of elements of U(P ),up to an equivalence relation ≈ which, for completeness, we describe here. Definefirst a binary relation ∼ on the set of finite sequences of elements of P by

(a1, . . . , ai, ai+1, . . . , an) ∼ (a1, . . . , aic, c−1ai+1, . . . , an),

provided (ai, c), (c−1, ai+1) ∈ D. Then Stallings’ equivalence relation ≈ is thetransitive closure of ∼.

Guiding examples are again amalgamated products and HNN-extensions.

Example 8.1. As in Section 7.4, let A and B be groups intersecting in a commonsubgroup H . Consider the subset P = A ∪ B ⊆ G = A ∗H B. Define a partialmultiplication p · q in the obvious way; that is p · q is defined if and only if eitherp, q are both in A or p, q both in B. Then P is a pregroup where D = A×A∪B×B.We obtain the following geodesically perfect rewriting system (where the length iscomputed w.r.t. P , thus elements of P are viewed as letters).

1 −→ εp · q −→ r if (p, q) ∈ D, pq = r ∈ Ga · b ←→ ah · h−1b if a ∈ A \H, b ∈ B \H, h ∈ H.

Example 8.2. Let H be the HNN-extension HNN(G;A,B,Φ) as defined in Sec-tion 7.3 and, as before, let X and Y be transversals for A and B in G withX ∩ Y = {1}. Consider the subset

P = G ∪GtY ∪Gt−1X ⊂ H.

We define a partial multiplication by the obvious rules (left to the reader)according to the following table.

G×G −→ G

G×GtY −→ GtY

G×Gt−1X −→ Gt−1X


GtY ×G −→ GtY

Gt−1X ×G −→ Gt−1X

Gt−1X ×GtY −→ G if the inner part XG is in A

GtY ×Gt−1X −→ G if the inner part Y G is in B.

This defines a pregroup P for H , where

D = G×G ∪G×GtY ∪G×Gt−1X ∪GtY ×G ∪Gt−1X ×G ∪ S,where S is the subset of Gt−1X×GtY ∪GtY ×Gt−1X where inner partsXG or Y Gbelong to A or B, as appropriate. The partial multiplication table can be directlyread from the convergent system we used in Section 7.3. As we shall see below,it defines an (infinite) geodesically perfect rewriting system, where again we viewelements of P as letters. Note also that we could replace X and Y by X = Y = Gthroughout the definition of our pregroup P in which the multiplication table couldbe slightly more simply described, but would be unnecessarily large.

In [49] an alternative pregroup forH is defined with underlying set consistingof equivalence classes of elements of G ∪ t−1G∪Gt∪ t−1Gt under the equivalencerelation generated by t−1at ∼ Φ(a), for a ∈ A. However we feel that the resultingrewriting rules are obscured by the equivalence relation on the underlying set.

The following is the principal result on the universal groups of pregroups.

Theorem 8.3 (Stallings [49]). Let P be a pregroup. Then the following hold.1) Every element of U(P ) can be represented by a P -reduced sequence;2) any two P -reduced sequences representing the same element are ≈ equivalent,

in particular they have the same length, and3) P embeds into U(P ).

8.1. Rewriting systems for universal groups

The result of [49] cited above may be regarded as showing that composition of theinclusion map P → P ∗ with the standard quotient map P ∗ → U(P ) is injective,where P ∗ is the free monoid on P . We show here how to achieve this with the helpof a geodesically perfect Thue system. Since this approach may be new we workout the details.

It is convenient to work over P ∗ and view each element of P as a letter. Wehave to distinguish whether a product is taken in the free monoid P ∗ or in P ,and we introduce the following convention. Whenever we write [ab] we mean that(a, b) ∈ D ⊆ P ×P with m(a, b) = [ab] ∈ P : that is the product ab is defined in Pand yields a letter.

The system S = S(P ) ⊆ P ∗ × P ∗ is now defined by the following rules.

ε −→ 1 (= the empty word)ab −→ [ab] if (a, b) ∈ Dab ←→ [ac][c−1b] if (a, c), (c−1, b) ∈ D


Theorem 8.4. Let P be a pregroup. Then the following hold.1) P ∗/S(P ) U(P ).2) S is a geodesically perfect Thue system.

Proof. Obviously, P ∗/S defines U(P ) which proves 1). To prove 2) we show firstthat the system S is strongly confluent. For this we have to consider two rules suchthat the left-hand sides overlap. Strong confluence involving only symmetric rulesis trivial. Thus, we may assume that one rule is length-reducing. If one of the rulesis ε −→ 1, then (by symmetry) the other rule is either εb −→ b or εb −→ c[c−1b].Since (c−1, b) ∈ D implies (c, c−1b) ∈ D and [c(c−1b)] = b [49], both situationslead to b in at most one step. The next situation is:

[ab]⇐=Sab=⇒

S[ac][c−1b]

Since (a, b) and (c−1, b) both belong to D we have (a, c(c−1b)) ∈ D, as above,and (P4) implies that (ac, c−1b) ∈ D, so we can apply the rule [ac][c−1b] −→ [ab].Finally, we have to consider:

yd⇐=Sabd=⇒

Saz

with a, b, d ∈ P and y, z ∈ P ∗. We may assume that one rule is length-reducingof type ab −→ y = [ab]. The other rule is either of type bd −→ [bd] or of typebd←→ [bc][c−1d]. Assume first that (b, d) ∈ D, then in both cases we can use:

[ab]d=⇒S[abb−1][bd] = a[bd]⇐=

Sa[bc][c−1d].

The remaining case is that (b, d) /∈ D and the situation is:

[ab]d⇐=Sabd=⇒

Sa[bc][c−1d].

Since (a, b), (b, c) and (c, c−1d) are inD, (P5) implies that either abc or bcc−1d = bdis defined in P . But bd is not defined, therefore abc is defined. We obtain:

[ab]d=⇒S[abc][c−1d]⇐=

Sa[bc][c−1d].

Now we show that S is geodesic, from which it follows that it is geodesi-cally perfect. Start with a sequence w ∈ P ∗ and apply only length-reducingrules until this is no longer possible. Clearly, the resulting sequence is P -reduced:w

∗=⇒S

a1 · · · an ∈ Γ∗ such that aiai+1 is not defined in P for 1 ≤ i < n. Possibly,

one can still apply the symmetric rules, but we claim that any application of thesymmetric rules gives again a P -reduced system. Indeed, assume u ∈ Γ∗ is P -reduced, but it is not P -reduced after one application of a length-preserving rulefrom S(P ). Then there are four consecutive elements abde in u and an elementc ∈ P , such that neither ab nor bd nor de is defined, but bc, c−1d are definedand either a(bc) or (bc)(c−1d) or (c−1d)e is defined. Assume the product a(bc) isdefined. Then the sequence a, bc, c−1, d satisfies the premise of the axiom (P5), soeither a(bc)c−1 = ab or (bc)c−1d = bd must be defined, contradicting the assump-tion that u is P -reduced. Similarly, (c−1d)e cannot be defined. Suppose now that


(bc)(c−1d) is defined. Then the sequence b, c, c−1d satisfies the premise of (P4),since (bc) and c(c−1d) are defined. Since (bc)(c−1d) is defined (P4) implies thatb(c−1(cd)) = b(1d) = bd is defined, in contradiction with P -reducibility of u. �Remark 8.5. Stallings’ normal form Theorem 8.3 is now a consequence of The-orem 8.4 because elements from P are irreducible and the rewriting system isgeodesically perfect. Thus, P -reduced sequences that define the same elements inU(P ) are ≈ equivalent.

Remark 8.6. As above let Γ = P \{ε}. Since S(P ) ⊆ P ∗×P ∗ is strongly confluentand geodesic, we obtain a geodesically perfect presentation of the universal groupU(P ). In some sense it is however nicer to have such a presentation over Γ. So, letus put S′(P ) ⊆ Γ∗ × Γ∗ defined by the following rules:

aa−1 −→ 1 if a ∈ Γab −→ c if (a, b) ∈ D, a �= b−1, [ab] = cab ←→ [ac][c−1b] if (a, c), (c−1, b) ∈ D

The difference is that a rule aa−1 −→ ε ∈ S (ε ∈ P is a letter) is replaced byaa−1 −→ 1 ∈ S′(P ). This rule of S′(P ) needs two steps of S(P ), but in S(P ) wewin strong confluence, whereas S′(P ) is not strongly confluent. However confluenceof S(P ) transfers to S′(P ). Hence, both systems S(P ) and S′(P ) are geodesicallyperfect.

Using the geodesically perfect system S(P ) for U(P ) where P is finite we seethat the result of Rimlinger [44] leads to the following statement which is slightlystronger than the result of [24].

Corollary 8.7. Let G be a finitely generated group. The following conditions areequivalent.1) G is virtually free.2) G can be presented by some finite geodesically perfect system.3) G can be presented by some finite geodesic system.

Proof. By Rimlinger [44], a finitely generated virtually free group is the universalgroup U(P ) of some finite pregroup P . By Theorem 8.4 it has a presentation by the(finite) geodesically perfect system S(P ). In our setting every geodesically perfectsystem is geodesic, so we get the implication from 2) to 3) for free. In order topass from 3) to 1) one has to show that the set of words which are equivalentto 1 ∈ G forms a context-free language. This is can be demonstrated using anargument from [15], which has also been used in [24]. Consider a word w and writeit as as w = uv such that u is geodesic. The prefix u is kept on a push downstack. Suppose that v = av′, for some letter a. Push a onto the top of the stack:so the stack becomes ua. There is no reason to suppose that ua is geodesic and weperform length reducing reduction steps on it to produce an equivalent geodesicword u. Suppose this requires k steps:

uak=⇒

SR

u


Let us show that we can bound k by some constant depending only on S. Indeedfor all letters a we may fix a word wa such that awa

∗=⇒SR

1. But this means

uwa∗=⇒

SR

u,

where u is geodesic and u represents the same group element as u did. But u wasgeodesic too. Hence |u| = |u|. Therefore |u| ≥ |u| − |wa| and this tells us k ≤ |wa|.Since k is bounded by some constant we see that the whole reduction process in-volves a bounded suffix of the word ua, only. This means we can factorise ua = pqand u = pr, where the length of q is bounded by some constant depending onS only. Moreover, q k=⇒

SR

r. Since the length of q is bounded this reduction can be

performed using the finite control of the pushdown automaton. The automatonstops once the input has been read and then the stack gives us a geodesic corre-sponding to the input word w. In particular, the set of words which represent 1 inthe group is context-free. Thus, the group presented is context-free; and using aresult of Muller and Schupp [37] we see that G is virtually free. �

8.2. Characterisation of pregroups in terms of geodesic systems

In this section we consider Thue systems S ⊆ Γ∗ × Γ∗ corresponding to grouppresentations, i.e., Γ = X ∪X−1 and S contains all the rules xx−1 → 1, x−1x →1, x ∈ X . We shall refer to these as group rewriting systems. We say that a rewritingsystem S ⊆ Γ∗ × Γ∗ is triangular if each rule � → r ∈ S satisfies the “triangular”condition: |�| = 2, |r| ≤ 1, so every rule in S is of the form ab→ c where a, b ∈ Γand c ∈ Γ ∪ {1}. Observe that a triangular system is length-reducing.

We also say that S is almost triangular if S = S′ ∪S◦, where S′ is triangularand all rules in S◦ are trivial, i.e., of the form a → 1, for some a ∈ Γ. Non-trivial examples of triangular systems come from triangulated presentations ofgroups. Namely, if 〈X | R〉 is a presentation of a group then one can triangulatethis presentation by adding new generators and replacing old relations by finitelymany triangular ones.

Another type of example arises from pregroups. Let P be a pregroup. InSection 8.1 we defined two rewriting systems S(P ) and S′(P ) associated with Pthat define the universal group U(P ). Notice that the length-reducing part S′(P )Rof S′(P ) is triangular (here Γ = P \ {ε}):

S′(P )R = {aa−1 → 1, ab→ c | a, b, c ∈ Γ, (a, b) ∈ D, [ab] = c, a �= b−1},meanwhile, the length reducing part S(P )R of S(P ) is almost triangular, since itcontains the trivial rule ε→ 1.

Theorem 8.4 implies the following result.

Corollary 8.8. Let P be a pregroup. Then S′(P )R is a triangular geodesic sys-tem, S(P ) is an almost triangular geodesic system and U(P ) = Γ∗/S′(P )R =P ∗/S(P )R.


Proof. It suffices to observe that S′(P )R and S′(P ) define the same equivalencerelation on Γ∗. Indeed, every rule of the type ab→ [ac][c−1b], where [ac] and [c−1b]are defined, can be realized as the following rewriting sequence in S′(P )R:

ab← acc−1b→ [ac]c−1b→ [ac][c−1b],

which shows that S′(P )R and S′(P ) are equivalent. The rest follows from Theorem8.4 and Remark 8.6. �

To prove the converse of this corollary we need some notation. Let S ⊆ Γ∗×Γ∗be a triangular group rewriting system, where Γ = X ∪X−1. The congruence ∗⇐⇒

S

on Γ∗ induces an equivalence relation on the subset Γ ∪ {1}, which we denote by≈. Define PS to be the quotient (Γ ∪ {1})/ ≈ and write [z] for the equivalenceclass of the element z ∈ Γ ∪ {1} and in addition ε for the equivalence class of 1.Define an involution p → p−1 on PS by setting [x]−1 = [x−1] and [x−1]−1 = [x],for x ∈ X , and setting ε−1 = ε. (Note that, since S is a group rewriting system,x ≈ 1 if and only if x−1 ≈ 1, so this involution is well defined.) Now we define a“partial multiplication” on PS as follows.• For p, q ∈ PS \ {ε} the product pq is defined and equal to s if there existx, y ∈ Γ such that p = [x], q = [y] and there is a rule xy → z ∈ S, withz ∈ Γ ∪ {1} and s = [z].

• For all p ∈ PS we put pε = εp = p and pp−1 = p−1p = ε.It is not hard to see that the partial multiplication on PS is well defined.

Lemma 8.9. Let S be a geodesic triangular group rewriting system. Then the fol-lowing hold.1) PS is a pregroup.2) U(PS) is isomorphic to the group Γ∗/S.

Proof. Clearly, the axioms P1) and P2) hold in PS by construction. It suffices toshow that P4) and P5) hold in PS , in which case P3) follows.

P4). If any one of p, q, r = ε then P4) holds trivially, so we may assume thatp, q, r ∈ PS \ {ε}. Suppose then p = [a], q = [b], r = [c] ∈ PS and the products pqand qr are defined, i.e., S contains rules ab→ x and bc→ y for some x, y ∈ Γ∪{1}.Suppose also that (pq)r is defined in PS , so either [x] = [z] and zc → t ∈ S forsome z, t ∈ Γ ∪ {1}, or [x] = ε, in which case let us define t = c. This meansthat abc ∗⇐⇒

St, for some t ∈ Γ∪ {1} and also abc ∗⇐⇒

Syc. As S is geodesic either S

contains a rule yc→ u, for some u ∈ Γ∪ {1}, or y = 1, in which case let us defineu = c. Then (pq)r = [t] = [u] = p(qr) in PS . It follows, by symmetry, that P4)holds.

P5). Again we may assume we have p, q, r, s ∈ PS \ {ε} such that p = [a], q =[b], r = [c], s = [d] and the products pq, qr, rs are defined; so there are rules ab →x, bc→ y, cd→ z ∈ S. We need to show that either pqr or qrs is defined. Assumepqr is not defined. This means in particular that y �= 1 and that S contains norule with left-hand side ay.


We may rewrite abcd in two different ways: abcd → xcd → xz and abcd →ayd. As S is geodesic either S must contain a rule which can be applied to ayd orone of a, y, d must be 1. Given our assumptions this means that S contains a rulewith left-hand side yd. Thus we have (qr)s defined, so P5) holds. This proves thefirst statement.

The second statement follows from Theorem 8.4, Remark 8.6 and Corollary8.8. Indeed, it suffices to note that, by construction, the system S is the lengthreducing part of the system S′(PS) associated with the pregroup PS . �

Combining Corollary 8.8 and Lemma 8.9 one gets the following character-isation of pregroups and their universal groups in terms of triangular geodesicsystems.

Theorem 8.10. Let P be a pregroup. Then the reduced part of the rewriting sys-tem S′(P ) is a geodesic triangular group system which defines the universal groupU(P ). Conversely, if S is a triangular geodesic group system then PS is a pregroup,whose universal group is that defined by S.

This result gives a method of constructing a potentially useful pregroup fora group given by a presentation in generators and relators. It would be helpful tohave a KB like procedure for finding such pregroups.

Problem 8.11. Design an (KB-like) algorithm that for a given finite triangularrewriting system finds an equivalent triangular geodesic system.

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Volker DiekertUniversitat StuttgartUniversitatsstr. 38D-70569 Stuttgart, Germany

Andrew J. DuncanNewcastle UniversityNewcastle upon TyneNE1 7RU, United Kingdom

Alexei G. MyasnikovMcGill UniversityMontreal, Canada, H3A 2K6



Regular Sets and Counting in Free Groups

Elizaveta Frenkel, Alexei G. Myasnikov andVladimir N. Remeslennikov

Abstract. In this paper we study asymptotic behavior of regular subsets ina free group F of finite rank, compare their sizes at infinity, and developtechniques to compute the probabilities of sets relative to distributions on Fthat come naturally from random walks on the Cayley graph of F . We applythese techniques to study cosets, double cosets, and Schreier representativesof finitely generated subgroups of F with an eye on complexity of algorithmicproblems in free products with amalgamation and HNN extensions of groups.

Mathematics Subject Classification (2000). 20E05.

Keywords. Geometric group theory, regular set, measures on free groups,Schreier transversals, generic and negligible sets.

1. Introduction

During the last decade it has been realized that a natural set of algebraic objectsusually can be divided into two parts. The large one (the regular part) consistsof typical, “generic” objects; and the smaller one (the “singular” part) is madeof “exceptions”. Essentially, this idea appeared first in the form of zero-one lawsin probability theory, number theory, and combinatorics. It became popular afterseminal works of Erdos, that shaped up the so-called Probabilistic Method (see,for example, [1]). In finite group theory the idea of genericity can be traced downto a series of papers by Erdos and Turan in 1960-70’s (for recent results see, forexample, [55]). In combinatorial group theory the concept of generic behavior isdue to Gromov. His inspirational works [27, 28] turned the subject into an areaof very active research, see for example, [2, 3, 4, 14, 7, 8, 13, 18, 15, 16, 17, 8, 9,10, 32, 34, 35, 36, 37, 38, 39, 42, 50, 51, 53, 59, 61]. It turned out that the genericobjects usually have much simpler structure, while the exceptions provide mostof the difficulties. For instance, generic finitely generated groups are hyperbolic[27, 51], generic subgroups of hyperbolic groups are free [26], generic cyclicallyreduced elements in free groups are of minimal length in their automorphic orbits[39], generic automorphisms of a free group are strongly irreducible [52], etc.

94 E. Frenkel, A.G. Myasnikov and V.N. Remeslennikov

In practice, the generic-case analysis of algorithms is usually more importantthan the worst-case one. For example, knowing generic properties of objects onecan often design simple algorithms that work very fast on generic elements. Incryptography, many successful attacks exploit the generic properties of randomelements from a particular class, ignoring the existing hard instances [47, 48, 49,54]. In the precise form the generic complexity of algorithmic problems appearedfirst in the papers [34, 35, 14, 13]. We refer the reader to a comprehensive survey[25] on generic complexity of algorithms.

In this paper we lay down some techniques that allow one to measure setswhich appear naturally when computing with infinite finitely presented groups.Our main idea is to approximate a given set by some regular subsets and estimatethe asymptotic sizes of the regular sets using powerful tools of random walks ongraphs and generating functions. The particular applications we have in mindconcern with the generic complexity of the Word and Conjugacy problems in freeproducts with amalgamation and HNN extensions. In general, such problems canbe extremely hard. In [45] Miller described a free product of free groups withfinitely generated amalgamation where the Conjugacy problem is undecidable;while in [45] he gave similar examples in the class of generalized HNN-extension offree groups. However, it has been proven in [15, 16] that on a precisely describedset RP of “regular elements” in amalgamated free products and HNN extensionsG the Conjugacy problem is decidable (under some natural conditions on thefactors), furthermore, it is decidable in polynomial time. Namely, it was shown in[15, 16] that the group G (satisfying some natural assumptions) can be stratifiedinto two parts with respect to the “hardness” of the conjugacy problem:

• the Regular Part RP consists of so-called regular elements for which the con-jugacy problem is decidable in polynomial time by the standard algorithms(described in [44, 43]). Moreover, one can decide whether or not a givenelement is regular in G;

• the Black Hole BH (the complement of RP in G) consists of elements in Gfor which either the standard algorithms do not work at all, or they are slow,or the situation is not quite clear yet.

The missing piece is to show that the set RP is, indeed, generic in G. Thisis not easy, the complete proof, which will appear in [24], relies on the techniquesdeveloped in the present paper. Now, a few words on the structure of the paper.In Section 2, following [14], we describe some techniques for measuring subsets ina free group F , the asymptotic classification of large and small sets, and approxi-mations via context-free and regular sets.

In Section 3 we study, using graph techniques, Schreier system of represen-tatives (transversal) of a finitely generated subgroup C in a free group F of finiterank. If S is a fixed Schreier transversal of C then s ∈ S is called stable (on theright) if sc ∈ S for any c ∈ C. Intuitively, the stable representatives are “regular”,they are easy to deal with.

Regular Sets and Counting in Free Groups 95

In Section 4 we estimate the sizes of various subsets of F . In particular, weshow that S is regular and thick (see definitions in Section 2), meanwhile the setSnst of non-stable representatives from S is exponentially negligible. Furthermore,the set Snst is exponentially negligible even relative to the set S. Our approachhere is to “approximate” the sets in question by regular sets and to measure sizesof the regular sets using tools of random walks on graphs and Perron-Frobeniustechniques.

In Section 5 we develop a technique to compare sizes of different regular setsat “infinity” and give an asymptotic classification of regular subsets of F relativeto a fixed prefix-closed regular subset L ⊆ F . The main result describes whenregular subsets of L are “large” or “small” at infinity in comparison to L. Notice,in the case when L = F , this result has been proven in [14] (Theorem 3.2).

2. Preliminaries

In this section, following [14], we describe some techniques for measuring sub-sets in a free group F , the asymptotic classification of large and small sets, andapproximations via context-free and regular sets.

2.1. Asymptotic densities

Let F = F (X) be a free group with basis X = {x1, . . . , xm}. We use this notationthroughout the paper.

Let R be a subset of the free group F and Sk = {w ∈ F | |w| = k } thesphere of radius k in F . The fraction

fk(R) =|R ∩ Sk||Sk|

is the frequency of elements from R among the words of length k in F. The asymp-totic density ρ(R) of R is defined by

ρ(R) = lim supk→∞

fk(R).

R is called generic if ρ(R) = 1, and negligible if ρ(R) = 0. If, in addition, thereexists a positive constant δ < 1 such that

1− δk < fk(R) < 1

for all sufficiently large k then R is called exponentially generic. Meanwhile, iffk(R) < δk for large enough k then R is exponentially negligible. In both the caseswe refer to δ as a rate upper bound. Sometimes such sets are also called stronglygeneric or strongly negligible, but we refrain from this.

The Cesaro limit

ρc(R) = limn→∞

1n(f1 + · · ·+ fn) . (1)

gives another asymptotic characteristic, called Cesaro density, or asymptotic av-erage density. Sometimes, it is more sensitive then standard asymptotic density ρ


(see, for example, [14], [60]). However, if limk→∞ fk(R) exists (hence is equal toρ(R)) then ρc(R) also exists and ρc(R) = ρ(R). We will have to say more aboutρc(R) below.

Asymptotic density gives the first coarse classification of large (small) subsets:

Coarse classification

1) Generic sets;2) visible or thick sets: the set R is visible if ρ(R) > 0;3) negligible sets.

Unfortunately, this classification is very coarse, it does not distinguish many setswhich, intuitively, have different sizes.

All our results in this paper concern with the strong version of the asymp-totic density ρ, when the actual limit limk→∞ fk(R) exists. This allows one todifferentiate sets with the same asymptotic density with respect to their growthrates. Thus generic sets R divide into subclasses of exponential, subexponential,superpolynomial, polynomial generic sets, with respect to the convergence rates oftheir frequency sequences {fk(R)}k∈N. The same holds for negligible sets as well.

2.2. Generating random elements and multiplicative measures

One can use a no-return random walkWs (s ∈ (0, 1]) on the Cayley graph C(F,X)of F with respect to the generating set X , as a random generator of elements of F(see [14]). We start at the identity element 1 and either do nothing with probabilitys (and return value 1 as the output of our random word generator), or move toone of the 2m adjacent vertices with equal probabilities (1− s)/2m. If we are at avertex v �= 1, we either stop at v with probability s (and return the value v as theoutput), or move, with probability 1−s

2m−1 , to one of the 2m − 1 adjacent verticeslying away from 1, thus producing a new freely reduced word vx±1i . Since theCayley graph C(F,X) is a tree and we never return to the word we have alreadyvisited, it is easy to see that the probability μs(w) for our process to terminate ata word w is given by the formula

μs(w) =s(1 − s)|w|

2m · (2m− 1)|w|−1for w �= 1 (2)

andμs(1) = s. (3)

For R ⊆ F its measure μs(R) is defined by μs(R) =∑

w∈R μs(w). Recalcu-lating μs(R) in terms of s, one gets

μs(R) = s

∞∑k=0

fk(1− s)k,

and the series on the right-hand side is convergent for all s ∈ (0, 1). The ensembleof distributions {μs} can be encoded in a single function

μ(R) : s ∈ (0, 1)→ μs(R) ∈ R.


The argument above shows that for every subset R ⊆ F , μ(R) is an analyticfunction of s.

It has been shown in [14] that μ(R) contains a lot of information about theasymptotic behaviour of the set R. To see where this information comes fromrenormalise the measures μs and consider the parametric family μ∗ = {μ∗s} ofadjusted measures

μ∗s(w) =(

2m2m− 1

· 1s

)· μs(w). (4)

This new measure μ∗s is multiplicative in the sense that

μ∗s(u ◦ v) = μ∗s(u)μ∗s(v), (5)

where u◦v denotes the product of non-empty words u and v such that |uv| = |u|+|v|(no cancellation in the product uv). Moreover, if we denote

t = μ∗s(x±1i ) =

1− s

2m− 1(6)

thenμ∗s(w) = t|w| (7)

for every non-empty word w. Assume now, for the sake of minor technical conve-nience, that R does not contain the identity element 1. It is easy to see that

μ∗s(R) =∞∑

k=0

nk(R)tk

is the generating function of the spherical growth sequence

nk(R) = |R ∩ Sk|of the set R in variable t which is convergent for each t ∈ [0, 1).

The distribution μs has the uncomfortably big standard deviation σ =√1−ss ,

which reflects the fact that μs is strongly skewed towards “short” elements. Themean length of words in F distributed according to μs is equal to Ls = 1

s − 1, soLs → ∞ when s → 0. This shows that the asymptotic behaviour of the set R at“infinity” (when Ls → ∞) depends on the behaviour of the function μ(R) whens→ 0+.

Following [14], for a subset R of F we define a numerical characteristic

μ0(R) = lims→0+

μ(R) = lims→0+

s ·∞∑

k=0

fk(1− s)k.

If μ(R) can be expanded as a convergent power series in s at s = 0 (and hence insome neighborhood of s = 0):

μ(R) = m0 +m1s+m2s2 + · · · ,

thenμ0(R) = lim

s→0+μs(R) = m0,


and an easy corollary from a theorem by Hardy and Littlewood [30, Theorem 94]asserts that μ0(R) is precisely the Cesaro limit ρc(R).

A subset R ⊆ F is called smooth [14] if μ(R) can be expanded as a convergentpower series in s at s = 0.

2.3. The frequency measure

In this section we discuss the frequency measure, introduced in [14].LetW0 be the no-return non-stop random walk on the Cayley graph C(F,X)

of F (like Ws with s = 0), where the walker moves from a given vertex to anyadjacent vertex away from the initial vertex 1 with equal probabilities 1/2m. Inthis event, the probability λ(w) that the walker hits an element w ∈ F in |w| steps(which is the same as the probability that the walker ever hits w) is equal to

λ(w) =1

2m(2m− 1)|w|−1, if w �= 1, and λ(1) = 1.

This gives rise to a measure called the frequency measure on F , or Boltzmanndistribution, defined for subsets R ⊆ F by

λ(R) =∑w∈R

λ(w),

if the sum above is finite. One can view λ(R) as the cumulative frequency of Rsince

λ(R) =∞∑

k=0

fk(R).

This measure is not probabilistic, since, for instance, λ(F ) = ∞, moreover, λ isadditive, but not σ-additive.

A subset R ⊆ F is called λ-measurable, or simply measurable (since we donot consider any other measures in this paper) if λ(R) <∞. Every measurable setis negligible.

Linear approximation. If the set R is smooth then the linear term in the expansionof μ(R) gives a linear approximation of μ(R):

μ(R) = m0 +m1s+O(s2).

In this case, m0 = μ0(R) is the Cesaro density of R. An easy corollary of [30,Theorem 94] shows that if μ0(R) = 0 then

m1 =∞∑

k=1

fk(R) = λ(R).

On the other hand, even without assumption that R is smooth, if R is mea-surable, then

μ0(R) = 0 and μ1 = lims→0+

μ(s)s

= λ(R).


2.4. Asymptotic classification of subsets

In this section we describe a classification of subsets R in F , according to theasymptotic behavior of the functions μ(R).

Recall, that the function μ(R) is analytic on (0, 1) for every subset R of F .R is smooth if μ(R) can be analytically extended to a neighborhood of 0. Thesubset R is called rational, algebraic, etc, with respect to μ if the function μ(R)is rational, algebraic, etc.

Asymptotic classification of sets. The following subtler classification of sets in F(based on the linear approximation of μ(R)) was introduced in [14]:

• Thick subsets: μ0(R) exists, μ0(R) > 0 and

μ(R) = μ0(R) + α0(s), where lims→0+

α0(s) = 0.

• Negligible subsets of intermediate density: μ0(R) = 0 but μ1(R) does notexist.

• Sparse negligible subsets: μ0(R) = 0, μ1(R) exists and

μ(R) = μ1(R)s+ α1(s) where lims→0+

α1(s)s

= 0.

• Exponentially negligible subsets.• Singular subsets. μ0(R) does not exist.

For sparse sets, the values of μ1 provide a further and more subtle discrimi-nation by size.

Lemma 2.1. [14] A subset is sparse in F if and only if it is measurable.

2.5. Context-free and regular languages as a measuring tool

The simple observation in Section 2.2 that μ(R) is the generating function of thegrows sequence {nk(R)}k∈N allows one to apply a well-established machinery ofgenerating functions of regular and context-free languages to estimate asymptoticsizes of subsets R in F . We refer to [31] on regular and context-free languages, andto [21] on regular languages in groups.

Algebraic sets and context free languages. If the set R is an (unambiguous) contextfree language then, by a classical theorem of Chomsky and Schutzenberger [19],the generating function μ∗(R) =

∑nk(R)tk, and hence the function μ(R), are

algebraic functions of s. Moreover, if R is regular then μ(R) is a rational functionwith rational coefficients [22, 57].

It is well known that singular points of an algebraic function are either polesor branching points. Since μ(R) is bounded for s ∈ (0, 1), this means that, for acontext-free set R, the function μ(R) has no singularity at 0 or has a branchingpoint at 0. A standard result on analytic functions allows us to expand μs(R) asa fractional power series:

μs(R) = m0 +m1s1/n +m2s

2/n + · · · ,


n being the branching index. This technique was used in [16, 17] for numericalestimates of generic complexity of algorithms.

If R is regular, than we actually have the usual power series expansion:

μs(R) = m0 +m1s+m2s2 + · · · ;

in particular, μ(R) can be analytically extended in the neighborhood of 0, so R issmooth.

The following gives an asymptotic classification of regular subsets of F .

Theorem 2.2. [14, 6]1) Every negligible regular subset of F is strongly negligible.2) A regular subset of F is thick if and only if its prefix closure contains a cone.3) Every regular subset of F is either thick or strongly negligible.

3. Schreier systems of representatives

3.1. Subgroup and coset graphs

In this section for a given finitely generated subgroup of a free group we discussits subgroup and coset graphs.

Let F = F (X) be a free group with basis X = {x1, . . . , xn}. We identifyelements of F with reduced words in the alphabet X ∪ X−1. Fix a subgroupC = 〈h1, . . . , hm〉 of F generated by finitely many elements h1, . . . , hm ∈ F .

Following [33], we associate with C two graphs: the subgroup graph Γ = ΓC

and the coset graph Γ∗ = Γ∗C . We freely use definitions and results from [33] inthe rest of the paper.

Recall, that Γ is a finite connected digraph with edges labeled by elementsfrom X and a distinguished vertex (based-point) 1, satisfying the following twoconditions. Firstly, Γ is folded, i.e., there are no two edges in Γ with the same labeland having the same initial or terminal vertices. Secondly, Γ accepts precisely thereduced words in X ∪X−1 that belong to C. To explain the latter observe, thatwalking along a path p in Γ one can read a word �(p) in the alphabet X ∪X−1,the label of p (reading x in passing an edge e with label x along the orientation ofe, and reading x−1 in the opposite direction). We say that Γ accepts a word w ifw = �(p) for some closed path p that starts at 1 and has no backtracking. One candescribe Γ as a deterministic finite state automata with 1 as the unique startingand accepting state.

For example, the graph Γ for the subgroup generated by x1x2x−11 is shown

in Figure 1 below.

1� �x1 �

��

x2

Figure 1


Given the generators h1, . . . , hm of the subgroup C, as words from F (X), onecan effectively construct the graph Γ in time O(n log∗ n) [58].

The coset graph (also known as the Schreier graph) Γ∗ = Γ∗C of C is aconnected labeled digraph with the set {Cu | u ∈ F} of the right cosets of C in Fas the vertex set, and such that there is an edge from Cu to Cv with a label x ∈ Xif and only if Cux = Cv. One can describe the coset graph Γ∗ as obtained from Γby the following procedure. Let v ∈ Γ and x ∈ X such that there is no outgoing orincoming edge at v labeled by x. We call such v a boundary vertex of Γ and denotethe set of such vertices by ∂Γ. For every such v ∈ ∂Γ and x ∈ X we attach to va new edge e (correspondingly, either outgoing or incoming) labeled x with a newterminal vertex u (not in Γ). Such vertices u are called frontier vertices, we denotethe set of frontier vertices of Γ by ∂+Γ. Then we attach to u the Cayley graphC(F,X) of F relative to X (identifying u with the vertex 1 of C(F,X)), and thenwe fold the edge e with the corresponding edge in C(F,X) (that is labeled x andis incoming to u). Observe, that for every vertex v ∈ Γ∗ and every reduced wordw in X ∪X−1 there is a unique path Γ∗ that starts at v and has the label w. Bypw we denote such a path that starts at 1, and by vw the end vertex of pw. Hereis the fragment of the graph Γ∗ for C =

⟨x1x2x

−11

⟩:

1� �

��

�

x1 � ��

x2

x1 � ��

��

�

�

� ��

x2

x2

�x1

��

Figure 2

Lemma 3.1. Γ∗C is the coset graph of C in F .

Proof. See, for example, [33]. �Notice that Γ = Γ∗ if and only if the subgroup C has finite index in F .

Indeed, Γ = Γ∗ if and only if for every vertex v of Γ and every label x ∈ X , thereis an edge in Γ labeled by x which exits from v, and an edge with label x whichenters v, but this is precisely the characterization of subgroups of finite index inF [33, Proposition 8.3].


A spanning subtree T of Γ with the root at the vertex 1 is called geodesic iffor every vertex v ∈ V (Γ) the unique path in T from 1 to v is a geodesic path inΓ. For a given graph Γ one can effectively construct a geodesic spanning subtreeT (see, for example, [33]).

From now on we fix an arbitrary spanning subtree T of Γ. It is easy to seethat the tree T uniquely extends to a spanning subtree T ∗ of Γ∗.

Let V (Γ∗) be the set of vertices of Γ∗. For a subset Y ⊆ V (Γ∗) and a subgraphΔ of Γ∗, we define the language accepted by Δ and Y as the set L(Δ, Y, 1) of thelabels �(p) of paths p in Δ that start at 1 and end at one of the vertices in Y , andhave no backtracking. Notice that the words �(p) are reduced since the graph Γ∗

is folded. Notice, that F = L(Γ∗, V (Γ∗), 1) and C = L(Γ, {1}, 1) = L(Γ∗, {1}, 1).Sometimes we will refer to a set of right (left) representatives of C as the right

(left) transversal of C. Furthermore, to simplify terminology, a transversal willusually mean a right transversal, if not said otherwise. Recall, that a transversal Sof C is termed Schreier if every initial segment of a representative from S belongsto S.

Proposition 3.2. Let C be a finitely generated subgroup of F . Then:

1) for every spanning subtree T of Γ the set ST∗ = L(T ∗, V (Γ∗), 1) is a Schreiertransversal of C.

2) For every Schreier transversal S of C there exists a spanning subtree T of Γsuch that S = ST∗ .

Proof. The statement 1) follows directly from Lemma 3.1. To prove 2) noticethat every reduced path p ∈ Γ∗ can be decomposed as p = pint ◦ pout, wherepint is a maximal reduced path in Γ, and pout is the tail of p outside of Γ. Thisdecomposition is unique. Moreover:

• if v ∈ V (Γ) and p is a reduced path from 1 to v in Γ∗ then p passes onlythrough vertices of Γ;

• if v ∈ V (Γ∗) \ V (Γ) and p′ and p′′ are two paths from 1 to v, where p′ =p′int ◦ p′out, and p′′ = p′′int ◦ p′′out, then p′out = p′′out.

Let S be a Schreier transversal of C in F and s ∈ S. Suppose that the reducedpath ps ends at some vertex vs in Γ. Then the whole path ps lies in Γ. Let T bea subgraph of Γ generated by the union of all paths ps, where s ∈ S and vs ∈ Γ.Since S is a Schreier transversal, T is a maximal subtree of Γ. It is clear thatS = ST∗ . Hence, the result. �

Proposition 3.2 allows one to identify elements from a given Schreier transver-sal S of C with the vertices of the graph Γ∗, provided a maximal subtree of Γ isfixed. We use this frequently in the sequel.

Corollary 3.3. The number of distinct Schreier transversals of C in F is finite andequal to the number of spanning subtrees of Γ.


3.2. Schreier transversals

In this section we introduce various types of representatives of C in F relative toa fix basis X of F .

Definition 3.4. Let S be a transversal of C.• A representative s ∈ S is called internal if the path ps ends in Γ, i.e., vs ∈V (Γ). By Sint we denote the set of all internal representatives in S. Elementsfrom Sext = S � Sint are called the external representatives in S.

• A representative s ∈ S is called geodesic if it has minimal possible length inits coset Cs. The transversal S is geodesic if every s ∈ S is geodesic.

• A representative s ∈ S is called singular if it belongs to the generalizednormalizer of C:

N∗F (C) = {f ∈ F |f−1Cf ∩ C �= 1}.All other representatives from S are called regular. By Ssin and, respectively,Sreg we denote the sets of singular and regular representatives from S.

• A representative s ∈ S is called stable (on the right) if sc ∈ S for any c ∈ C.By Sst we denote the set of all stable representatives in S, and Snst = S�Sstis the set of all non-stable representatives from S.

In the following proposition we collect some basic properties of various typesof representatives. Recall that the cone defined by (or based at) an element u ∈ Fis the set C(u) of all reduced words in F that have u as an initial segment. In thiscase C(u) = {w ∈ F | w = u ◦ v, v ∈ F}.Proposition 3.5. Let S be a Schreier transversal for C, so S = ST∗ for somespanning subtree T ∗ of Γ∗. Then the following hold:1) Sint is a basis of C, in particular, |Sint| = |V (Γ)|.2) Sext is the union of finitely many coni C(u), where vu ∈ ∂+Γ.3) Ssin is contained in a finite union of double cosets Cs1s

−12 C of C, where

s1, s2 ∈ Sint.4) Snst is a finite union of left cosets of C of the type s1s−12 C, where s1, s2 ∈ Sint.

Proof. 1) is well known, see [12], for example. 2) follows immediately from theconstruction. To see 3) notice first that Ssin ⊆ N∗F (C) and N∗F (C) is the unionof finitely many double cosets CsC, where s ∈ Ssin, and furthermore, every suchcoset has the form CsC = Cs1s

−12 C, where s1, s2 ∈ Sint (see Lemma 5 in [12], or

Propositions 9.8 and 9.11 in [33], or Theorem 2 in [32]).To see 4) assume that s ∈ S is not stable, so there exists an element c ∈ C

such that sc �∈ S. Then s = s1 ◦ t, c = t−1 ◦ d, sc = s1 ◦ d. We claim, that theterminal vertex of s1 lies in Γ (viewing s as a path in Γ∗). Indeed, if not, then s1,as well as s1 ◦ d, is in Sext – contradiction. Hence, s1 ∈ Sint. Since c = t−1 ◦ dthere is a closed path in Γ with the label t−1 ◦ d, starting at 1C . Let s2 ∈ Sint bethe representative of t−1. Then s2d = c1 ∈ C, hence sc = s1 ◦ d = s1s

−12 c1, so

s ∈ s1s−12 C, as claimed. �


Proposition 3.6. Let S be a Schreier transversal for C, so S = ST∗ for somespanning subtree T ∗ of Γ∗. Then the following hold:1) If T ∗ is a geodesic subtree of Γ∗ (and hence T is a geodesic subtree of Γ) then

S is a geodesic transversal.2) If C is a malnormal subgroup of F then Ssin = ∅.3) Ssin ⊆ Snst.

Proof. 1) is straightforward (see also [33]).2) If C is malnormal then N∗F (C) = 1, so Ssin = ∅.3) If s ∈ Ssin then c = s−1c1s for some non-trivial c, c1 ∈ C, so c1s = sc.

Since sc �= s we conclude that sc �∈ S, hence s ∈ Snst. �

4. Measuring subsets of F

Recall that a finite automaton A is a finite labeled oriented graph (possibly withmultiple edges and loops). We refer to its vertexes as states. Some of the statesare called initial states, some accept or final states. We assume further that everyedge of the graph is labeled by one of the symbols x±1, x ∈ X, where F = F (X) isa free group of finite rank m. The language accepted by an automaton A is the setL = L(A) of labels on paths from initial to accept states. An automaton is saidto be deterministic if, for any state there is at most one arrow with a given labelexiting the state. A regular set is a language accepted by a finite deterministicautomaton.

The following facts about regular sets are well known. LetA andB are regularsubsets in F . Then:• the sets A ∪B, A ∩B and A � B are regular.• The prefix closure A of a regular set A is regular. Here, the prefix closure Ais the set of all initial segments of all words in A.

• If ab = a ◦ b for any a ∈ A, b ∈ B then AB is regular.• If ab = a ◦ b for any a, b ∈ A then A∗ is regular.The following notation is useful. For u, v ∈ F define

c(u, v) = 12 (|u|+ |v| − |uv|),

the amount of cancellation in the product uv.

Proposition 4.1. Let R1 and R2 be subsets of F . Then the following statementshold:1) If R1 ⊆ R2 and R2 is negligible (exponentially negligible) then so is R1.2) If R1, R2 are negligible (exponentially negligible) then so is R1 ∪R2.3) If R1 and R2 are negligible (exponentially negligible) then so is the set

R1 ◦R2 = {r1r2 | ri ∈ Ri, c(r1, r2) = 0}.4) If R1 and R2 are negligible (exponentially negligible) then so is the set

R1 ◦tR2 = {r1r2 | ri ∈ Ri, c(r1, r2) ≤ t}.


Proof. The proof is straightforward. �

Definition 4.2. Let R1 and R2 be subsets of F and f : R1 → R2 a map. Then:

• f is called d-isometry, where d is a non-negative real number, if for anyw ∈ R1

|w| − d ≤ |f(w)| ≤ |w|+ d.

• f has uniformly bounded fibers if there exists a constant c such that everyelement w ∈ R2 has at most c pre-images in R1.

Proposition 4.3. Let R1 and R2 be subsets of F . Then the following statementshold:

1) If f : R1 → R2 is a surjective d-isometry and R1 is negligible (exponentiallynegligible) then so is R2.

2) If f : R1 → R2 is a d-isometry with uniformly bounded fibers and R2 isnegligible (exponentially negligible) then so is R1.

Proof. Notice that for k > d

fk(R2) ≤k+d∑

j=k−d

fj(R1),

and 1) follows. Similarly,

fk(R1) ≤ c

k+d∑j=k−d

fj(R2)

for k > d and 2) follows. �

Proposition 4.4. Let C be a finitely generated subgroup of infinite index of a freegroup F . Then every Schreier transversal of C in F is regular and thick.

Proof. By definition S = Sint ∪ Sext. By Proposition 3.5 the set Sint is finite andthe set Sext is a non-empty finite union of cones. By Theorem 2.2 each cone inSext is thick. Therefore, the set Sext, as well as the set S, is thick. Clearly, everycone is regular, so is the set S. �

Proposition 4.5. Let C be a finitely generated subgroup of infinite index in F . Thenthe following hold:

1) C is exponentially negligible in F and one can find some upper bound δ < 1for the growth rate of C.

2) Every coset of C in F is exponentially negligible in F .

Proof. 1) follows from Proposition 1 and Corollary 1 in [6].2) follows from 1) above and 4) from Proposition 4.1. �


Proposition 4.6. Let C be a finitely generated subgroup of infinite index in F . Thenthe following hold:1) C∗ =

⋃f∈F

Cf is exponentially negligible in F .

2) For every c ∈ C the set conjugacy class cF = {f−1cf |f ∈ F} is exponentiallynegligible in F .

Proof. The statement 1) has been shown in [14] and also in Proposition 1.10 in [5].The statement 2) is shown in Proposition 1.11 in [5]. �Proposition 4.7. Let C be a finitely generated subgroup of infinite index in F andS is a Schreier transversal of C in F . If S0 ⊆ S is a exponentially negligible subsetof F , then the set

⋃s∈S0

Cs is exponentially negligible in F.

Proof. By Proposition 3.5 S = Sint ∪ Sext, where Sint is a finite set and Sext is aunion of finitely many cones C(u), where u ends at ∂+Γ, i.e., vu ∈ ∂+Γ. It sufficesto prove the result for S0 ∩C(u) for a fixed vu ∈ ∂+Γ. To this end we may assumefrom the beginning that S0 ⊆ C(u). If s is the representative of u in S, then everyword from C(u) contains s as an initial segment. Since s is not readable in Γ theamount of cancellation c(w, t) in the product wt, where w ∈ C and t ∈ C(u) doesnot exceed the length of s. Hence

CS0 = C ◦|s|S0

and the result follows from the statement 4) of Proposition 4.1. �Proposition 4.8. Let A and B be finitely generated subgroups of infinite index inF . Then for any w ∈ F the double coset AwB is exponentially negligible in F.

Proof. Observe, that AwB = ABw−1w, so by the statement 2) of Proposition4.5 it suffices to show that ABw−1 is exponentially negligible. Since Bw−1 is justanother finitely generated subgroup of infinite index in F one can assume from thebeginning that w = 1. Let S be a geodesic Schreier transversal for A in F. Then

AB =⋃

s∈S0

As

for some subset S0 ⊆ S. By Proposition 4.7 it suffices to show that the subset S0 isexponentially negligible. Since the set Sint is finite we may assume that S0 ⊆ Sext.Now we construct an r-isometry α : S0 → B. Let TA be the spanning subtree ofΓA such that S = STA

∗ and TB be a spanning geodesic subtree of ΓB. Denote byd the maximum of the diameters of the trees TA and TB. To describe the mapα choose an arbitrary element s ∈ S0. Without loss of generality assume that|s| ≥ d, because there are only finitely many such s that have smaller length andby Proposition 4.5 they will not extremely change asymptotic size of AB since Aof infinite index in F. Then as = b for some a ∈ A and b ∈ B. We claim that thereexists an element bs ∈ B such that |sb−1s | ≤ 2d. Indeed, the cancellation in theproduct as is at most d (see the argument in Proposition 4.7). Hence s and b have a


common terminal segment t of length at least |s|−d (recall that |s| ≥ d). It followsthat in the graph ΓB there exists a path from some vertex v to 1B with the label tv.Then bs = tvt ∈ B and |sb−1s | = |st−1t−1v | ≤ 2d. Hence s and bs has a long commonterminal segment and differ only on the initial segment of length at most 2d. Itfollows that the map α : s→ bs gives a 2d-isometry α : S0 → B. Notice that α hasuniformly bounded fibers. Indeed, if α(s1) = α(s2) = b then s1 and s2 differ fromb, hence from each other, only on the initial segment of length at most 2d. So thereare at most (2d)2|X| such distinct elements. Since B is exponentially negligible byProposition 4.3 the set S0 is also exponentially negligible, as claimed. This provesthe result. Notice, that the property being geodesic for Schreier transversal S forA in F is not crucial for our prove. Namely, for arbitrary Schreier transversal Sall conclusions can be repeated with slightly different constant. �

Now we can state the main result of the section.

Theorem 4.9. Let C be a finitely generated subgroup of infinite index in F and Sa Schreier transversal for C. Then the following hold:1) The generalized normalizer N∗F (C) of C in F is exponentially negligible in F .2) The set of singular representatives Ssin is exponentially negligible in F .3) The set Snst of unstable representatives is exponentially negligible in F .

Proof. To see 1) recall that the generalized normalizer N∗F (C) of C in F is a finiteunion of double cosets of C in F. Therefore N∗F (C) is exponentially negligible inF by Proposition 4.8.

2) follows immediately from 1).To prove 3) observe that Snst is a finite union of left cosets of C (see 3) in

Proposition 3.5). Now the result follows from Proposition 4.5. �Theorem 4.9 can be strengthen as follows.

Corollary 4.10. Let C be a finitely generated subgroup of infinite index in F. Thenthe sets

Sin (C) =⋃S

Ssin, Uns (C) =⋃S

Snst,

where S runs over all Schreier transversals of C, are exponentially negligible.

Proof. By Corollary 3.3 there are only finitely many Schreier transversals of C.Now the result follows from Theorem 4.9 and Proposition 4.1. �


5. Comparing sets at infinity

5.1. Comparing Schreier representatives

In this section we give another version of Theorem 4.9. To explain we need a fewdefinitions.

For subsets R,L of F we define their size ratio at length k by

fk(R,L) =fk(R)fk(L)

=|R ∩ Sk||L ∩ Sk|

.

The size ratio ρ(R,L) at infinity of R relative to L (or the relative asymptoticdensity) is defined by

ρ(R,L) = lim supk→∞

fk(R,L).

By rL(R) we denote the cumulative size ratio of R relative to L:

rL(R) =∞∑

k=1

fk(R,L).

We say that R is L-measurable, if rL(R) is finite. R is called negligible relative toL if ρ(R,L) = 0. Obviously, an L-measurable set is L-negligible. A set R is termedexponentially negligible relative to L (or exponentially L-negligible) if fk(R,L) ≤ qk

for all sufficiently large k.The following result is simple, but useful.

Proposition 5.1. Let R be an exponentially negligible set in F .1) For any w ∈ F the set R is a exponentially negligible relative to the cone

C(w).2) The set R is exponentially negligible relative to any exponentially generic

subset T of F .

Proof. Observe, that fk(C(w)) = 1/2m(2m− 1)|w|−1 is a constant. Since

fk(R,C(w)) =fk(R)

fk(C(w))

it follows that R is exponentially negligible relative to C(w). This proves 1).To prove 2) denote by p and q the corresponding rate bounds for R and T ,

so fk(R) ≤ pk, fk(T ) ≥ 1− qk for sufficiently large k. Then, for such k,

fk(R, T ) =fk(R)fk(T )

≤ pk

1− qk=(

p

(1 − qk)1k

)k

.

Sincelimk→∞

p

(1 − qk)1k

= p

it follows that for any ε > 0

fk(R, T ) ≤ (p+ ε)k

for sufficiently large k, as claimed. �


Corollary 5.2. Let C be a finitely generated subgroup of infinite index in F and Sa Schreier transversal for C. Then the following hold:1) The set of singular representatives Ssin is exponentially negligible in S.2) The set Snst of unstable representatives is exponentially negligible in S.

Proof. The statements of this corollary follow immediately from Theorem 4.9 andPropositions 3.5, 3.6 and 5.1. �

5.2. Comparing regular sets

In this section we give an asymptotic classification of regular subsets of F relativeto a fixed prefix-closed regular subset L ⊆ F .

For this purpose we are going to describe how one can use a random walkon the finite automaton B recognizing regular subset R ⊆ L similar to the one inSection 2.3. It will be convenient to further put B to special form consistent to L.

Recall Myhill-Nerode’s theorem on regular languages (see, for example, [21],Theorem 1.2.9). For a language R over an alphabet A consider an equivalencerelation ∼ on A∗ defined as follows: two strings w1 and w2 are equivalent if andonly if for each string u over A the words w1u and w2u are either simultaneouslyin R or not in R. Then R is regular if and only if there are only finitely many∼-equivalence classes.

Now, let R ⊆ L. Define an equivalence relation ∼ on L such that w1 ∼ w2if and only if for each u ∈ F the following condition holds: w1u = w1 ◦ u andw1u ∈ R if and only if w2u = w2 ◦ u and w2u ∈ R.

The following is an analog of Myhill-Nerode’s theorem for free groups.

Lemma 5.3. Let R ⊆ L ⊆ F and L prefix-closed and regular. Then R is regular ifand only if there are only finitely many ∼-equivalence classes in L.

Proof. The proof is similar to the original one. We give a short sketch of the mostinteresting part of it. If the set of the equivalence classes is finite one can definean automaton B on the set of equivalence classes as states. If x ∈ X ∪X−1 and[w] is the equivalence class of some w such that w ◦ x ∈ R then one connects thestate [w] with an edge labeled by x to the state [wx]. The class [ε], where ε is theempty word, is the initial state, while a state [w] is an accepting state if and onlyif w ∈ R. In this case L(B) = R. �

Since R is regular, we suppose that B as in Lemma 5.3 and modify it in thenext way. Without loss of generality we can assume that A is in the normal form,i.e., it has only one initial state I and doesn’t contain inaccessible states.

Let S = [w] be a state of B. Denote by Spr the uniquely defined state in Awhich is the terminal state of the path with the label w in A, starting at [ε]. Thestate Spr is well defined, it does not depend on the choice of w. We call Spr theprototype of S.

Since B accepts only reduced words in X ∪ X−1 one can transform B to aform where the following hold:


a) B has only one initial state I and one accepting state Z.b) For any state S of B, all arrows which enter S have the same label x ∈ X∪X−1

and arrows exiting from S cannot have label x−1 (this can be achieved bysplitting the states of B, see Figure 1). We shall say in this situation that Shas type x.

c) For every state S of B there is a direct path from S to the accept state Z.d) There are no arrows entering the initial state I.

=⇒

�

�

�

��

�

�

�

�

�

�

�

�

��

�

��

��

��

��

�

��

A

A′′

A′

b

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a

d

c

b

a

a

d

d

c

c

Figure 3. Splitting the states of the automaton B.

The final version of obtained automaton B we will call an automaton consis-tent with A.

Now we are ready to define a no-return random walk on B as it was claimedabove. Namely, let B be consistent with A and let S be a state in B. Denote byν = ν(Spr) the number of edges exiting from the prototype state Spr in A. Thewalker moves from S along some outgoing edge with the uniform probability 1

ν .In this event, the probability that the walker hits an element w ∈ R in |w| steps(when starting at [ε]) is the product of frequencies of arrows in a direct path fromthe initial state I to the accept state Z with the label w. It is not hard to see thatthis probability is equal to λL(w). This gives rise to the measure λL on R:

λL(R) =∑w∈R

λL(w) =∞∑

k=0

f ′k(R,L),

wheref ′k(R,L) =

∑w∈R∩Sk

λL(w).

Note that, generally speaking, f ′k(R,L) differs from fk(R,L) defined in Sec-tion 5.1. Indeed, walking in B we have different number of possibilities to con-tinue our walk on the next step depending on way we chose. On the other hand,f ′k(R,F (X)) = fk(R,F (X)).

Now we can use the tools of random walks to compute λL(R). Notice, thatλL is multiplicative, i.e.,

λL(uv) = λL(u)λL(v)


for any u, v ∈ R such that uv = u◦v and uv ∈ R. We say that R is λL-measurable,if λL(R) is finite. A set R is termed exponentially λL-measurable, if f ′k(R,L) ≤ qk

for all sufficiently large k.The following result is simple, but useful.Let w ∈ F. The set CL(w) = L∩C(w) is called an L-cone. Obviously, CL(w)

is a regular set. We say that CL(w) is L-small, if it is exponentially λL-measurable.The following is the main result of this section.

Theorem 5.4. Let R be a regular subset of a prefix-closed regular set L in a freegroup F. Then either the prefix closure R of R in L contains a non-small L-coneor R is exponentially λL-measurable.

Before proving the theorem we establish a few preliminary facts. We fix aprefix-closed regular subset L of F .

Proposition 5.5. Let R1 and R2 be subsets of F . Let also P be one of the properties{ “to be L-measurable”, “to be exponentially L-negligible”, “to be λL-measurable”,“to be exponentially λL-measurable”}. Then the following hold:1) If R1 ⊆ R2 and R2 has property P then so is R1.2) If R1, R2 have property P then so is R1 ∪R2.3) If R1 and R2 have property P then so is the set

R1 ◦R2 = {r1r2 | ri ∈ Ri, c(r1, r2) = 0}.Proof. The proofs are easy. �

To strengthen the last statement in Proposition 5.5 we need the followingnotation. For a subset T ⊆ F put T ◦1 = T and define recursively T ◦k+1 = T ◦k ◦ T .Denote

T ◦∞ =∞⋃

k=1

T ◦k .

Lemma 5.6. Let T be a regular set and a number q, 0 < q < 1, such that f ′k(T, L) ≤qk for every positive integer k. Then the set T ◦∞ is exponentially λL-measurable.

Proof. Every word w ∈ T ◦∞ of length k comes in the form w = u1 ◦ u2 ◦ · · · ◦ ut,where ui’s are non-trivial elements from T and k = |u1| + · · · + |ut|. On theother hand, if k = k1 + · · · + kt is an arbitrary partition of k into a sum ofpositive integers and u1, . . . , ut are words in T such that |ui| = ki, then w =u1 · · ·ut ∈ T ◦∞. Since λL is multiplicative every partition of k adds to f ′k(T

◦∞, L) a

number f ′k1(T◦∞, L) . . . f ′kt

(T ◦∞, L), which is bounded from above by qk1+···+kt = qk.If p(k) is the number of all partitions of k into a sum of positive integers thenf ′k(T

◦∞, L) ≤ p(k)qk. It is known (Hardy and Ramanujan) that

p(k) ∼ eπ√

2k3

4k√3.

Hence f ′k(T◦∞, L) < qk1 , for some 0 < q < q1 < 1 and all sufficiently large k, so T ◦∞

is exponentially λL-measurable, as claimed. �


Proof of Theorem 5.4. In the most part we follow the proof of Theorem 2.2 from[14]. Suppose that all L-cones in R are non-small. Since R ⊆ R by Proposition5.5 we can assume that R itself is prefix-closed in L. We have to prove that Ris exponentially λL-measurable. Let R = L(B) and B consistent to A (where Arecognize L).

It is convenient to further split B into two parts. Denote by B1 the automatonobtained from B by removing all arrows exiting from Z.

Figure 4. The automaton B. Figure 5. The automaton B1.

Let B2 be the automaton formed by all states in B that are accessible fromthe state Z, with the same arrows between them as in B; Z is the only initial andaccepting state of B2.

Figure 6. The automaton B2.

We assign to arrows in B1 and B2 the same frequencies as to the correspondingarrows in B. If R1 and R2 are the languages accepted by B1 and B2 then, obviously,R = R1 ◦R2. By Proposition 5.5 to prove the theorem it suffices to show that R1and R2 are exponentially λL-measurable.


Claim. The set R2 is exponentially λL-measurable.

Proof of the claim. Notice, that for every w ∈ R1 w ◦R2 ⊆ L(A) = R and w ◦R2is an L-cone. It is easy to see, that R2 is exponentially λL-measurable if and onlyif so w ◦R2 is.

Let R3 ⊆ R2 be the subset consisting of those non-trivial words w ∈ R2,whose paths pw visit the state Z of B2 only once. The set R3 is regular – it isaccepted by an automaton B3, which is obtained from B2 by splitting the state Zinto two separate states: the initial state Z1 and an accepting state Z2, in sucha way that the edges exiting from Z in B2 are now exiting from Z1 and there noingoing edges at Z1, while there are no edges exiting from Z2 and all those arrowsincoming in Z in B2 are now incoming into Z2.

Figure 7. The automaton B3.It follows immediately from the construction, that

R2 = {ε}⋃

R3⋃(R3 ◦R3)

⋃(R3 ◦R3 ◦R3)

⋃· · · = (R3)◦∞

soλL(R2) = λL(R3) + (λL(R3))2 + (λL(R3))3 + · · · . (8)

By Lemma 5.6 it suffices to show that there is a number q, 0 < q < 1, suchthat f ′k(R3, L) ≤ qk for every k (not only for sufficiently large k). It is not hard tosee that this condition holds if R3 is exponentially λL-measurable and λL(R3) < 1,so it suffices to prove the latter two statements.

By our assumption all L-cones in R = R are L-small. If for every state [w] = Sin B2 and every x ∈ X ∪X−1 there is an outgoing edge labeled by x at [w] if andonly if the same holds for the state Spr in A (i.e., B2 is X-complete relative to A)then for every given w ∈ R1 one has C(w) ∩ R = w ◦ R2 = C(w) ∩ L, so w ◦ R2is an L-cone. Hence it is L-small, i.e., exponentially λL-measurable, but then theset R2, hence R2, is exponentially λL-measurable, as claimed.

This implies that for some state S there are less then ν = ν(Spr) arrowsexiting from S. Consider a finite Markov chain M with the same states as in B3


together with an additional dead state D. We set transition probabilities from Z2to Z2 and from D to D being equal 1. Every arrow from a state S in B3 givesthe corresponding transition from the state S inM which we assign the transition

probability1ν. If at some state S of type x in B3 there is no exiting arrow labeled

y ∈ (X∪X−1)�{x−1}, inM we make a transition from S to D with the transition

probability1ν. This describesM.

Figure 8. The automatonM.

The states Z2 and D of Markov chainM are absorbing, and all other statesare transient. The probability distribution onM concentrated at the initial stateZ1, converges to the steady state P which is zero everywhere with the exception ofthe two absorbing states Z2 and D. Obviously, P (Z2) = λL(R3). Since P (D) �= 0we have λL(R3) = P (Z2) < 1, so one of the required conditions on R3 holds (formore details on this proof we refer to [14, 41]). The other one follows directlyfrom Corollary 3.1.2 in [41], which claims that in this case R3 is exponentiallyλL-measurable. This proves the claim.

A similar argument shows that R1 is exponentially λL-measurable. Thisproves the theorem. �

Acknowledgement

The authors thank Alexandre Borovik for very fruitful discussions.


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Elizaveta FrenkelMoscow, Russiae-mail: [email protected]

Alexei G. MyasnikovDepartment of Mathematics and StatisticsMcGill UniversityMontreal, Quebec, Canadae-mail: [email protected]

Vladimir N. RemeslennikovOmsk Branch of Mathematical Institute SB RAS13 Pevtsova StreetOmsk 644099, Russiae-mail: [email protected]



Twisted Conjugacy for Virtually Cyclic Groupsand Crystallographic Groups

Daciberg Goncalves and Peter Wong

Abstract. A group is said to have the property R∞ if every automorphismhas an infinite number of twisted conjugacy classes. In this paper, we classifyall virtually cyclic groups with the R∞ property. Furthermore, we determinewhich of the 17 crystallographic groups of rank 2 have this property.

Mathematics Subject Classification (2000). Primary: 20E45;Secondary: 55M20.

Keywords. Reidemeister number, elementary groups, Gromov hyperbolicgroups, fixed point theory.

1. Preliminaries

Let ϕ : G → G be a group endomorphism and let R(ϕ) denote the Reidemeisternumber of ϕ, or equivalently the cardinality of the set of ϕ-twisted conjugacyclasses, i.e., the number of orbits of the (left) action σ · α �→ σαϕ(σ)−1 whereσ, α ∈ G. Our primary interest is in the (in)finiteness of R(ϕ) for automorphismsϕ. We say that G has the property R∞ for automorphisms, in short G has theproperty R∞, if for every automorphism ϕ : G → G we have R(ϕ) = ∞. Byelementary groups, we mean groups that are finite extensions of cyclic groups, i.e.,these are the virtually cyclic groups.

This work is motivated by Levitt and Lustig [7] who showed implicitly thatautomorphisms of finitely generated non-elementary Gromov hyperbolic groupshave infinite Reidemeister numbers (see also [3]). These results leave out the classof elementary Gromov hyperbolic groups. It is the purpose of this paper to studythe R∞ property for the class of (infinite) elementary groups or equivalently the

This work was initiated during the first author’s visit to Bates College April 8–30, 2006. Researchwas continued during the first author’s visit to Bates College in the period March 12–16, 2008and the second author’s subsequent visits to IME-USP in 2006, 2007, and 2008. The visits of the

second author were supported by a grant from the National Science Foundation OISE-0334814.The second author was supported in part by the National Science Foundation DMS-0805968.

120 D. Goncalves and P. Wong

class of virtually Z groups and their generalizations, namely, the family of virtuallyfree abelian groups. The study of the infiniteness of R(ϕ) has application in fixedpoint theory. For instance, examples of finitely generated torsion-free nilpotentgroups with property R∞ have been constructed in [5] so that for every positiveinteger n ≥ 5, there exists a compact n-dimensional nilmanifold Mn such thatevery homeomorphism of Mn is isotopic to a fixed point free map.

Our basic technique in computing the Reidemeister number R(ϕ) in thispaper is by means of short exact sequences as stated in the following lemma whoseproof can be found in [4], for instance.

Consider the following commutative diagram of short exact sequences ofgroups and endomorphisms.

1 −−−−→ Ai−−−−→ B

p−−−−→ C −−−−→ 1

ϕ′⏐⏐$ ϕ

⏐⏐$ ϕ

⏐⏐$1 −−−−→ A

i−−−−→ Bp−−−−→ C −−−−→ 1

(1.0.1)

There is a short exact sequence of sets and corresponding functions i and p:

R(ϕ′) i→R(ϕ) p→R(ϕ) (1.0.2)

where R(ψ) denotes the set of ψ-twisted conjugacy classes. Denote by R(ψ) thecardinality of R(ψ).

Lemma 1.1. Given the commutative diagram labeled (1.0.1) above,

(1) if R(ϕ) =∞ then R(ϕ) =∞,(2) R(ϕ) =

∑[α]∈R(ϕ)#i(R(ταϕ′)) where τα(β) = αβα−1 and p(α) = α,

(3) if either Fix(ταϕ′) = 1 for every [α] ∈ R(ϕ), in which case i is injective, orC is finite, and if R(α · ϕ′) = R(ταϕ′) = ∞ for some [α] ∈ R(ϕ) then wehave R(ϕ) =∞,

(4) if C = Z, ϕ(t) = t−1 and either R(ϕ′) =∞ or R(t ·ϕ′) =∞ then R(ϕ) =∞.

This paper is organized as follows. In Section 2, we classify the infinite vir-tually cyclic groups with the R∞ property. Every infinite virtually cyclic group Gadmits a short exact sequence 0 → Z → G → Q → 1 for some finite group Q.We show (cf. Prop. 2.8) that if this extension is not central then G has the R∞property. On the other hand, if the extension is central, then we give necessaryand sufficient conditions (cf. Prop. 2.5) for G to have R∞ property. In Section 3,we first recall the 17 two-dimensional crystallographic groups in the order as listedin [1]. In Sections 4–7, we study the R∞ property for these groups. As it turnsout, of the 17 cases, only cases 1, 2 (Section 3) and 13 (Section 6) do not possessthe R∞ property. Concluding remarks are made in Section 8.

Twisted Conjugacy Classes 121

2. Elementary groups

In this section we study the R∞ property for the elementary groups. We willconsider an alternative description of such groups and then we show that there areexamples of such groups with the R∞ property. This is somewhat surprising sincesuch groups are virtually Z and Z does not have the property R∞ for R(−id) = 2where −id : Z → Z is the non-trivial automorphism. Such groups G have torsion,except the case when it is isomorphic to Z. So except in this case we do not havea manifold which has the homotopy type of a finite K(G, 1).

Recall that the elementary groups are the ones which are virtually cyclic. IfG is an infinite then we have a short exact sequence

0→ Z → G→ Q→ 1 (2.0.3)

where Q is a finite group. Such groups have been completely classified (see, e.g., [2,Theorem 6.12, p. 129]. Next, we show that this classification falls into two classesaccording to whether the extension (2.0.3) is central or not.

Theorem 2.1. Let G be an infinite virtually cyclic group which admits the extension(2.0.3). Then(i) the extension (2.0.3) is central iff G ∼= F �Z for some finite normal subgroup

F ;(ii) the extension (2.0.3) is not central iff there exists a finite normal subgroup F

such that G/F ∼= D∞ ∼= Z2 ∗ Z2.

Proof. It follows from [2, Theorem 6.12, p. 129] that for every infinite virtuallycyclic group G, there exists a finite normal subgroup F such that G/F is either Zor the infinite dihedral group D∞. It suffices to show that G/F ∼= Z iff (2.0.3) iscentral.

Suppose G/F ∼= Z. Then G ∼= F �θ Z. Since F is finite, the automorphismθ(t) has finite order, say r, where Z ∼= 〈t〉. Now the infinite cyclic subgroup H =〈(1F , tr)〉 is central in G since

(1F , tr)(α, tj) = ((θ(t))r(α), tr+j) = (α, tr+j) = (α, tj)(1F , tr).

Moreover,G/H is finite and hence G admits a central extension of the form (2.0.3).Conversely, suppose G admits a central extension of the form (2.0.3). We

write Q = {1Z, g1Z, . . . , gkZ} with 1 = g0. For any g ∈ G, there exists a uniquei, 0 ≤ i ≤ k such that g = gin where n ∈ Z. This gives rise to a map π : G → Zwhich is a well-defined group homomorphism since Z is a central subgroup in G.Evidently, π is surjective and F = Kerπ is finite. Since Z is free, π has a sectionand thus G ∼= F � Z. �

We will divide the family of infinite elementary groups into two families. LetF1 be the family of the groups such that the extension (2.0.3) is central and F2otherwise.

Corollary 2.2. Let G be an infinite elementary group. Then either G = Z or G hastorsion.


Proof. Follows immediately from Theorem 2.1. �

Now we move to the case where the extension is not central, i.e., G ∈ F2.The following simple lemma gives an alternative description of such groups.

Lemma 2.3. Let G ∈ F2. Then G is the middle term of a short exact sequence ofthe form

1→ G1 → Gp→ Z2 → 0

where G1 ∈ F1. Furthermore, if w ∈ G such that p(w) is the nontrivial element ofZ2 then w2 belongs to the torsion subgroup of G1.

Proof. Given an extension

0→ Z → G→ Q→ 1,

consider Q1 ⊂ Q the subgroup consisting of elements which act trivially on Z.Then the pullback of this inclusion is a subgroup G1 of G of index 2 so that thesequence 1 → G1 → G → Z2 → 0 is exact. Moreover, the subgroup G1 yields ashort exact sequence of the form 0 → Z → G1 → Q1 → 1 and thus G1 ∈ F1.For the second part, consider the short exact sequence given from the first part.By Theorem 2.1(i), G1 = H �φ Z. Now w2 is of the form (h, t). We will showthat � = 0. Suppose that wk ∈ Z for some k �= 0. Since w commutes with wk

it follows that w acts trivially on the subgroup (k�)Z, which is a contradictionif � �= 0 because w acts as multiplication non-trivially on some infinite subgroupof Z. So it remains to show that wk ∈ Z for some k �= 0. Let r be the order ofthe automorphism φ(t) ∈ Aut(H) where Z = 〈t〉. A simple calculation shows thatw2r = (h · φ(t)(h) · · · · · φ(t)(r−1)(h), tr). Note that the element h · φ(t)(h) ·· · · · φ(t)(r−1)(h) ∈ H has finite order, say s. Since φ(t)r = id it follows thatw2rs = (h · φ(t)(h) · · · · · φ(t)(r−1)(h))s, trs) = (1, trs) so that wk ∈ Z for somek �= 0. Now the proof is complete. �

The above lemma provides the following result:

Proposition 2.4. If G ∈ F2 then there exist a finite subgroup H ⊂ G, an embeddingZ ⊂ G, such that H∩Z = 1, and H and Z generate G. Further, there is a subgroupH1 ⊂ H of index 2 and a nontrivial subgroup H ′ ⊂ Z which is normal and theaction G→ Aut(Z) is nontrivial.

Proof. Straightforward. �

Now we give a necessary and sufficient condition for a group in F1 to havethe R∞ property.

Proposition 2.5. Let Z = 〈t〉, H a finite group, θ(t) ∈ Aut(H), and G = H �θ Z.Then G has the R∞ property if, and only if, θ(t) is not conjugated to its inversein Aut(H).


Proof. Assume that G has the R∞ property. Suppose that there is an automor-phism φ ∈ Aut(H) such that φ ◦ θ(t) ◦φ−1 = θ(t−1). Let Φ(h, tr) = (φ(h), t−r) forh ∈ H and tr ∈ Z. Note that

Φ((1, t)(h, 1)) = Φ(θ(t)(h), t) = (φ(θ(t)(h)), t−1)

= (θ(t−1)(φ(h)), t−1)

= (1, t−1)(θ(h), 1) = Φ(1, t)Φ(h, 1).

(2.0.4)

It follows that Φ ∈ Aut(G) such that φ = Φ|H and the induced automorphism Φis given by t �→ t−1. Since H is finite and R(Φ) = 2, it follows from Lemma 1.1(2)that R(Φ) <∞, a contradiction. Hence, such an automorphism φ cannot exist.

Conversely, we assume that there is no automorphism φ ∈ Aut(H) withφ ◦ θ(t) ◦ φ−1 = θ(t−1). Let Φ : G → G be an automorphism. Since H is theunique maximal torsion subgroup, it is characteristic in G. Let ϕ = Φ|H . Theinduced homomorphism Φ on G/H = Z is either the identity or given by t �→ t−1.If Φ(t) = t−1 then the calculation in (2.0.4) shows that for all h ∈ H , ϕ(θ(t)(h)) =θ(t−1)(ϕ(h)) since Φ is a group homomorphism. This violates the assumption onAut(H) and thus Φ must be the identity on G/H = Z and hence R(Φ) = ∞. Itfollows from Lemma 1.1(1) that G has the R∞ property. �

As before let G ∈ F1.

Corollary 2.6. If Aut(H) is abelian and θ is neither trivial nor of order 2, then Ghas the R∞ property.

Proof. If Aut(H) is abelian the equality φ ◦ θ(t) ◦ φ−1 = θ(t−1) implies θ is eithertrivial or of order 2. Hence the result follows. �

Example 2.7. Consider the group G = Z5�Z where the action Z → Aut(Z5) = Z4is the natural projection. This group is also an extension of Z by a finite group (itis an elementary hyperbolic group), namely

0→ 4Z → Z5 � Z → Z5 � Z4 → 0

where the first map is the inclusion in the second component. By the corollaryabove the group G has the R∞ property.

Now we will consider groups in F2. We first show an example of a group inF2 which has the property R∞. This group is Z � Z2 which is also isomorphic toZ2 ∗ Z2 and this group has been shown to have the R∞ property [5, Theorem 2].Because this group is the fundamental group of the closed 3-manifold RP 3#RP 3,it follows that for every homotopy equivalence f of RP 3#RP 3 the Reidemeisternumber of f is always infinite.

Now we derive the main result about the groups in F2.

Proposition 2.8. Every group G ∈ F2 has the R∞ property.


Proof. Let p : G → Q be the projection given by a short exact sequence whichdefines G and Q1 ⊂ Q the subgroup of index 2 of the elements which acts triviallyon Z. By Lemma 2.3, we have the following short exact sequence

0→ G1 = p−1(Q1)→ G→ Z2 → 0,

where G1 ∈ F1 is of the form H � Z for some finite H . Let us assume that G1is characteristic. If so, given any automorphism φ : G → G we have that therestriction of φ to G1 is also an automorphism and so φ(H) = H . Thereforeφ : G → G induces an automorphism on the quotient φ : G/H → G/H whereG/H = Z � Z2. Since Z � Z2 has the property R∞, it follows that G also has theproperty and the result follows.

It remains to show that G1 is characteristic. For this, let w ∈ G be an elementwhich projects to the non-trivial element in Z2. It suffices to show that φ(w) alsoprojects to the non-trivial element. Suppose not, i.e., φ(w) = v ∈ G1, and letk ∈ Z(G1) ∩ Z(φ(G1)) be a nontrivial element of infinite order (which is possiblebecause both groups have finite index in G). Then applying the automorphism φto the equation t ◦ k ◦ t−1 = −k leads to a contradiction. �

3. Crystallographic groups

The crystallographic groups of rank 2, or the so-called wallpaper groups, are themiddle term of extensions of Z2 by a finite group. So they are special cases ofvirtually free abelian groups of rank 2. The main references that we are going touse for these groups are Coexter and Moser [1] pages 40 to 51 and [8]. For eachof the crystallographic groups we will provide the presentation given by [1] andin most cases also the one given by [8]. The label of the item corresponds to theenumeration given in [1]. For our computation, it is sometimes more convenientto use alternative presentations some of which are given in [8]. The 17 wallpapergroups are listed as follows.

1. 〈X,Y | XY = Y X〉. This group corresponds to the group G1 and the presen-tation given in [8] is the same.

2. 〈X,Y, T | XY = Y X, T 2 = 1, TXT = X−1, TY T = Y −1〉. This groupcorresponds to the group G2 and the presentation given in [8] is the same.

3. 〈X,Y,R| XY = Y X, R2 = 1, RXR = X−1, RY R = Y 〉. This groupcorresponds to the group G11 and the presentation given in [8] is the same.

4. 〈X,Y, P | XY = Y X, P 2 = Y, P−1XP = X−1〉. This group corresponds tothe group G31 and the presentation given in [8] is the same.

5. 〈P,Q,R| P 2 = Q2, R2 = 1, RPR = Q〉. This group corresponds to the groupG21. We need several steps to show that the two presentations are equivalents.

6. 〈X1, X2, X3, Y |X21 = X22 = X23 = 1, X1Y = Y X1, X2Y = Y X2, X1X3 =X3X1, X2X3 = X3X2, X3Y X

−13 = Y −1〉. According to [1], this group is

isomorphic to D∞ ×D∞ where D∞ is the infinite dihedral group.


7. 〈P,Q,R| P 2 = Q2, R2 = 1, RPR = P−1, RQR = Q−1〉. By letting P = αβand Q = β, this group can be given the following presentation:

〈α, β,R|βαβ−1 = α−1, αR = α, βR = β−1, R2 = 1〉.

8. 〈P,Q, T | P 2 = Q2, T 2 = 1, TPT = Q−1〉. By letting P = αβ and Q = β,this group can be given the following presentation:

〈α, β, T |βαβ−1 = α−1, αT = α−1, βT = αβ−1, T 2 = 1〉.9. 〈R1, R2, T | R21 = R22 = T 2 = (R1R2)2 = (R1TR2T )2 = 1〉. By letting α =

R2TR1 and β = R1R2T , this group can be given the following presentationand is a finite extension of Z2:⟨

α, β,R1, R2|αβ = βα,αR1 = β, βR1 = α, αR2 = β−1,

βR2 = α−1, R21 = R22 = (R1R2)2 = 1⟩.

10. 〈S, T | S4 = T 2 = (ST )4 = 1〉. This group corresponds to the group G4.To see that the two presentations are equivalent, consider the map σ → S,α → TS2 and β → STS. It is a straightforward calculation to see that thisbijection between the sets of generators extends to an automorphism of thegroup.

11. 〈R,R1, R2| R2 = R21 = R22 = (RR1)4 = (R1R2)2 = (R2R)4 = 1〉. This groupcorresponds to the group G14 and the isomorphism is given by σ → RR1,α→ R2RR1R, β → RR2RR1, and ρ→ R1.

12. 〈R,S| R2 = S4 = (RS−1RS)2 = 1〉. This group corresponds to the group G24and the isomorphism is given by σ → S, ρ → RS−1, α → RS−1RS−1, andβ → SRS−1RS−2.

13. 〈S1, S2, S3| S31 = S32 = S33 = S1S2S3 = 1〉. This group corresponds to thegroup G3. To see that the two presentations are equivalent, consider the mapS1 → σ, S2 → ασα and S3 → βσβ. It is a straightforward calculation to seethat this bijection between the sets of generators extends to an automorphismof the group.

14. 〈R,S| R2 = S3 = (RS−1RS)3 = 1〉. This group corresponds to the group G13and the isomorphism is given by R→ ρ and S → σαβ−1.

15. 〈R1, R2, R3| R21 = R22 = R23 = (R1R2)3 = (R2R3)3 = (R3R1)3 = 1〉. Thisgroup corresponds to the group G23. The isomorphism between them is givenby R1 → ρσβ, R2 → σρ, R3 → ρ.

16. 〈S, T | S3 = T 2 = (ST )6 = 1〉. This group corresponds to the group G6. Tosee that the two presentations are equivalent, consider the map σ → ST ,α → TSTSTST and β → S2TST . It is a straightforward calculation to seethat this bijection between the sets of generators extends to an automorphismof the group. One can also construct the inverse map S → σ4α−1, T → ασ3.

17. 〈R,R1, R2| R2 = R21 = R22 = (R1R2)3 = (R2R)2 = (RR1)6 = 1〉. This groupcorresponds to the group G16. The correspondence is given by σ → R1R,α→R2R(R1R)3, β → R1R2(R1R)2, and ρ→ R.


4. Cases 1–5

For Cases 1–5, these are the wallpaper groups that can be expressed as extensionsof Z2 by Z2 or by the trivial group.

4.1. Case 1.

It is well known (see, e.g., [6]) that for endomorphisms ψ : Zn → Zn, R(ψ) < ∞iff det(1− ψ) �= 0 and in the case R(ψ) <∞, R(ψ) = | det(1 − ψ)|. The group Z2

does not have property R∞ since one can easily find automorphism ϕ such thatdet(1 − ϕ) �= 0. This implies that R(ϕ) <∞.

4.2. Case 4.

This groupK is the fundamental group of the Klein bottle which is doubly-coveredby the torus. It has been shown in [5] that K has the property R∞.

For cases 2, 3, and 5, each can be expressed as a semi-direct Z2 �θ Z2. Sinceevery element in a semi-direct product can be uniquely written as a product ofelements from the kernel and from the quotient, every automorphism ϕ can berepresented by an array of the form

ϕ =

⎡⎣a c rb d sε δ γ

⎤⎦ (4.2.1)

where ϕ(α) = αaβbtε;ϕ(β) = αcβdtδ;ϕ(t) = αrβstγ . Here, Z2 ∼= 〈α, β|αβ = βα〉and Z2 = 〈t|t2 = 1〉.

4.3. Case 2.

This group corresponds to G2 in [8] and has the following presentation

G = 〈α, β, t|αβ = βα, αt = α−1, βt = β−1, t2 = 1〉.Note that G ∼= Z2 �θ Z2 where θ : Z2 → Aut(Z2) is given by

θ(t) =(−1 00 −1

).

Let ϕ ∈ Aut(G) be given by

ϕ =

⎡⎣ 1 2 0−1 −1 00 0 1

⎤⎦ .It is easy to see that ϕ is an automorphism and it induces the identity ϕ = id on Z2with two Reidemeister classes [1] and [t]. Over the class [1], R(ϕ′) = | det(1−ϕ′)| =2 where ϕ′ =

[1 2−1 −1

]. Over the class [t], R(t · ϕ′) = | det(1 − θ(t)ϕ′)| = 2.

It follows from Lemma 1.1(3) that R(ϕ) = 4 < ∞. Hence, G does not haveproperty R∞.


4.4. Case 3.


G = 〈α, β, t|αβ = βα, αt = α, βt = β−1, t2 = 1〉.Thus, G ∼= Z2 �θ Z2 where θ : Z2 → Aut(Z2) is given by

θ(t) =(1 00 −1

).

Let ϕ ∈ Aut(G) be given by (4.2.1). Since t is of order 2, it follows that γ = 1.Moreover, ϕ(t2) = 1 implies that (αrβst)2 = αrβsαrβ−s = α2r = 1. Thus, r = 0.Since ϕ(βt) = ϕ(β−1), we have

(βst)(αcβdtδ)(βst) = t−δβ−dα−c. (4.4.1)

Since α is central in G, by equating the exponents of α on both sides of (4.4.1)yields that c = 0. Now c = 0 and r = 0, the subgroup

H = 〈β, t|βt = β−1, t2 = 1〉 ∼= Z � Z2 ∼= D∞

is characteristic in G. By [5], H = D∞ has the property R∞. It follows fromLemma 1.1(3) that G also has property R∞.

4.5. Case 5.


G = 〈α, β, t|αβ = βα, αt = β, βt = α, t2 = 1〉.Thus, G ∼= Z2 �θ Z2 where θ : Z2 → Aut(Z2) is given by

θ(t) =(0 11 0

).

Let ϕ ∈ Aut(G). Again, since t is of order 2, we have γ = 1. Since ϕ(t2) = 1, itfollows that (αrβst)2 = αrβsβrαs = (αβ)r+s = 1. Thus, r + s = 0. The equalityϕ(αt) = ϕ(β) yields

αrβstαaβbtεαrβst = αcβdtδ. (4.5.1)By equating the exponents of t, we have ε = δ.

If ε = δ = 1, then ϕ(αβ) = ϕ(βα) yields

αaβbtαcβdt = αcβdtαaβbt

⇒ αaβbβcαd = αcβdβaαb

⇒ a+ d = b+ c.

In fact, (4.5.1) yields αrβsβaαbαrβs = αcβd so that b+2r = c and a+2s = d. Sincer + s = 0, we have a+ b = c+ d. It follows that a = c and b = d, a contradictionto the assumption that ϕ is an automorphism. The case ε = δ = 1 cannot occur.Suppose ε = δ = 0. In this case, (4.5.1) becomes αrβstαaβbαrβst = αcβd orαrβsβaαbβrαs = αcβd. It follows that r + b + s = c and s + a + r = d so thatb = c and a = d. Thus, the subgroup generated by α and β, which is isomorphic


to Z2, is characteristic and ϕ′ = ϕ|Z2 is given by the matrix[a bb a

]. Since ϕ′

is an automorphism, we have a2 − b2 = ±1. This happens when a = ±1, b = 0or a = 0, b = ±1. In the latter case, | det(1 − ϕ′)| = 0 so that R(ϕ′) = ∞ andhence R(ϕ) = ∞ by Lemma 1.1(3). In the former case, i.e., a = ±1, b = 0,| det(1− t · ϕ′)| = 0 so that R(t · ϕ′) =∞ and hence R(ϕ) =∞ by Lemma 1.1(3).Hence, G has property R∞.

5. Cases 6–9

Next, we deal with cases 6–9. First, we study the cases 7 and 8 since in each case,the group is a semi-direct product ofK, the fundamental group of the Klein bottle,and Z2.

5.1. Case 7.

This group can be given the following presentation

G = 〈α, β, t|βαβ−1 = α−1, αt = α, βt = β−1, t2 = 1〉

and G = K � Z2 where K = 〈α, β|βαβ−1 = α−1〉 and Z2 = 〈t|t2 = 1〉. Letϕ ∈ Aut(G). Since K is also a semi-direct product Z � Z, it follows that ϕ canalso be represented by

ϕ =


⎤⎦ . (5.1.1)

Note that γ = 1 since ϕ(t) has order 2. Moreover, (ϕ(t))2 = 1 implies thatαrβstαrβst = 1 or αrβsαrβ−s = αrα(−1)

sr = 1. If s is even then r = 0.Suppose (ε, δ) = (0, 1). The equality ϕ(αt) = ϕ(α) yields

αrβsαaβ−bβ−sα−r = αaβb

⇒ αr(βsαaβ−s)β−bα−r = αrα(−1)saβ−bα−r = αaβb

⇒ αr+(−1)saβ−b = αaβbαr = α(−1)br+aβb.

It follows that b = 0 and either a = 0 or s is even. If a = 0, then we have ϕ(α) = 1,a contradiction.

This means that s is even and a �= 0. In this case, the equality ϕ(βαβ−1) =ϕ(α−1) yields

αcβdtαaβbt−1β−dα−c = β−bα−a

⇒ αcβdαaβ−bβ−dα−c = β−bα−a

⇒ αcβdαaβ−dβ−bα−c = β−bα−a

⇒ αcα(−1)daα−c = α−a (here b = 0).


It follows that d is odd since a �= 0. Now ϕ(βt) = ϕ(β−1) implies that

αrβstαcβdtt−1β−sα−r = t−1β−dα−c

⇒ αrβsαcβ−dβsα−r = βdα−c

⇒ αrαcβ2s−dα−r = βdα−c (here s is even)

⇒ αr+cβ2s−d = α(−1)d(r−c)βd = αc−rβd (here d is odd).

It follows that s = d, a contradiction to the fact that s is even and d is odd.Therefore, the case (ε, δ) = (0, 1) cannot occur.

Suppose that (ε, δ) = (1, 0). The equality ϕ(βαβ−1) = ϕ(α−1) yields

αcβdαaβbtβ−dα−c = t−1β−bα−a

⇒ αcβdαaβbβdα−c = βbα−a

⇒ αcβdαaβdβb = βbαc−a

⇒ αcβdαaβd = βbαc−aβ−b

⇒ αcα(−1)daβ2d = α(−1)

b(c−a).

This implies that d = 0 and c+(−1)da = (−1)b(c− a). If b is even then a = 0 andif b is odd then c = 0, which in this case implies that ϕ(β) = 1, a contradiction.Thus, d = 0, b is even, and a = 0. Now ϕ(βt) = ϕ(β−1) implies that

αrβstαct−1β−sα−r = α−c

⇒ αrβsαcβ−sα−r = α−c

⇒ αrα(−1)scα−r = α−c.

Since c �= 0, it follows that s must be odd. Now, ϕ(αt) = ϕ(α) yields

αrβstβbβ−sα−r = βbt

⇒ αrβsβ−bβsα−r = βb

⇒ αrβ2s−bα−r = βb.

It follows that s = b. But b is even and s is odd, a contradiction. Thus, the case(ε, δ) = (1, 0) cannot occur.

Suppose (ε, δ) = (1, 1). The equality ϕ(αt) = ϕ(α) yields

αrβstαaβbβ−sα−r = αaβbt

⇒ αrβsαaβ−bβsα−r = αaβb

⇒ αr+(−1)saβ2s−b = αaβbαr = αa+(−1)brβb.

It follows that s = b and r+(−1)sa = (−1)br+a. If s is odd then r = a. This casecannot occur otherwise ϕ(α) = ϕ(t), a contradiction. Thus, s = b and s is even.


Now, ϕ(βt) = ϕ(β−1) implies that

αrβstαcβdtt−1β−sα−r = t−1β−dα−c

⇒ αrβsαcβ−dβsα−r = βdα−c

⇒ αrβsαcβ−sβ2s−d = βdαr−c

⇒ αrα(−1)scβ2s−d = βdαr−c

⇒ αr+cβ2s−2d = α(−1)d(r−c).

It follows that s = d and c = 0. Thus, in this case, we have b = d = s is even andc = 0.

The equality ϕ(βαβ−1) = ϕ(α−1) yields

βdtαaβbtt−1β−d = t−1β−bα−a

⇒ βdαaβ−bβdα−c = βbαaβ−bβbα−c = βbα−a

⇒ α(−1)baβb = α(−1)

b(−a)βb.

It follows that a = 0. This means that ϕ(α) = ϕ(β), a contradiction.Hence, (ε, δ) = (0, 0) and K is characteristic. Since K has property R∞, it

follows that G also has property R∞.

5.2. Case 8.

This group can be given the following presentation

G = 〈α, β, t|βαβ−1 = α−1, αt = α−1, βt = αβ−1, t2 = 1〉and G = K � Z2 where K = 〈α, β|βαβ−1 = α−1〉 and Z2 = 〈t|t2 = 1〉. Letϕ ∈ Aut(G). As in Case 7, the automorphism ϕ can be represented by (5.1.1). Sincet2 = 1, we may assume that γ = 1. Moreover, ϕ(t2) = 1 implies that (αrβst)2 = 1so that αrβsα−rβ−sβs(αβ−1)s = 1. When s is odd, we have (αβ−1)s = β1−sαβ−1.It follows that αrβsα−rβ−s = α and so 2r = 1, a contradiction. Thus, s must beeven, in which case, (αβ−1)s = β−s.

The equality ϕ(βt) = ϕ(αβ−1) yields

αrβstαcβdtδt−1β−sα−r = αaβbtεt−δβ−dα−c. (5.2.1)

By equating the exponents of t, we have δ = ε− δ which implies that ε = 0.If δ = 0, then K is characteristic and G will have property R∞ following the

same argument as in Case 7. Thus, we assume δ = 1.The equality ϕ(αt) = ϕ(α−1) yields

αrβstαaβbt−1β−sα−r = β−bα−a

which implies that

αrβsα−a(αβ−1)bβ−sα−r = β−bα−a.

If b is odd then (αβ−1)b = αβ−b so that αrβsα−aαβ−bβ−sα−r = β−bα−a. Since bis odd and s is even, it follows that αr−a+1β−bαa−r = β−b such that 2a = 2r+ 1,a contradiction. Thus, b must be even.


Now, the equality ϕ(βαβ−1) = ϕ(α−1) yields

αcβdtαaβbt−1β−dα−c = β−bα−a

⇒ αcβdα−a(αβ−1)bβ−dα−c = β−bα−a

⇒ αcβdα−aβ−b−dα−c = β−bα−a (since b is even)

⇒ αcβdα−aβ−d = αc−a

⇒ βdα−aβ−d = α−a.

If a �= 0 then d must be even.Suppose d is even. Then ϕ(β2) = (αcβdt)2 = αcβdα−cβ−d = αcα−c = 1, a

contradiction to the assumption that ϕ is an automorphism and β is of infiniteorder. Thus, d must be odd and hence a = 0.

Now that a = 0, (5.2.1) becomes

αrβstαcβd−sα−r = βbt−1β−dα−c

⇒ αrβsα−c(αβ−1)d−sαrt = βb(αβ−1)−dαct

⇒ αrβsα−c(αβ−1)d−s−1(αβ−1)αr = βb(αβ−1)−d−1(αβ−1)αc

⇒ αrβsα−cβ1+s−dαβ−1αr = βb+dβαβ−1αc

(both d− s− 1 and −d− 1 are even)

⇒ αrα−cβ2s−dβαβ−1αr = βb+dαc−1

⇒ αr−cβ2s−b−2d = αr−c (since b is even and d is odd).

Thus, 2s = b+ 2d. Since d is odd, we have ϕ(β2) = (αcβdt)2 = αcβdα−c(αβ−1)d

which in turn is equal to α2c−1.Now, we have

ϕ =

⎡⎣ 0 c r2(s− d) d s

0 1 1

⎤⎦ .Since ϕ(β2) = α2c−1, the subgroup H generated by α and β2 is ϕ-invariant and

H ∼= Z2. If ϕ′ = ϕ|H then ϕ′ =[0 2c− 1κ 0

]where κ = s − d. Note that H is

a subgroup of index 2 in K so H is a subgroup of finite index in G. Since His generated by α and β2, it is easy to see that αη and (β2)η belong to H forη = α, β, t. This means that H is a normal subgroup of finite index in G so thatthere is a short exact sequence

0→ H → G→ F → 1

for some finite group F .Since ϕ′ is an automorphism detϕ′ = ±1 which means that (2c− 1)κ = ±1,

we have (c, κ) ∈ {(0, 1), (0,−1), (1, 1), (1,−1)}. When (c, κ) = (0,−1) or (c, κ) =(1, 1), it is easy to see that det(1−ϕ′) = 0 so that R(ϕ′) =∞. Note that β ∈ G−Hso the projection β of β in F is non-trivial. Over the Reidemeister class [β], we


consider R(τβϕ′) where τβϕ′ is simply ϕ′ conjugated by β. It follows that τβϕ′ =[0 1− 2cκ 0

]. Now, for (c, κ) = (0, 1) or (c, κ) = (1 − 1), det(1− τβϕ

′) = 0 so that

R(τβϕ′) = ∞ for either case and hence R(ϕ) = ∞ by Lemma 1.1(3). Thus whenδ = 1, R(ϕ) =∞. We can now conclude that G has property R∞.

5.3. Case 6.

According to [1], this group is isomorphic toD∞×D∞. Write G = D1∞×D2∞ whereDi∞ ∼= D∞, D1∞ = 〈α, s|αs = α−1, s2 = 1〉, and D2∞ = 〈β, t|βt = β−1, t2 = 1〉.

Consider the subgroup H = 〈α, β|αβ = βα〉 ⊂ D1∞ ×D2∞. Let ϕ ∈ Aut(G).Every element in D1∞ can be uniquely written as αisε, ε ∈ {0, 1} and likewise everyelement in D2∞ is of the form βjtδ, δ ∈ {0, 1}. Suppose ϕ(α, 1) = (αasε, βbtδ)and ϕ(1, β) = (αcsε, βdtδ). We may assume that ε + δ �= 0, ε + δ �= 0, εδ = 0,εδ = 0. If (ε, δ) = (1, 0) then (ε, δ) = (1, 0) otherwise ϕ(α, β) = (αAs, βBt) forsome A,B. However, αAs and βBt are both of finite order which contradicts thatH is torsion-free. In this case, ϕ(α2, 1) = (1, β2b) and ϕ(1, β2) = (1, β2d) so thesubgroup generated by ϕ(α2, 1) and ϕ(1, β2) is of rank 1, a contradiction to thefact that ϕ is an automorphism and (α2, 1) and (1, β2) together generate Z2.Similarly, the case where (ε, δ) = (0, 1) = (ε, δ) cannot occur. Note that the case(ε, δ) = (1, 1) = (ε, δ) cannot occur since αAs and βBt are both of finite order forany A and B. This implies that (ε, δ) = (0, 0) = (ε, δ) and thus H is characteristic.

Now, G ∼= H �θ (Z2 × Z2) where

Z2 × Z2 = 〈s|s2 = 1〉 × 〈t|t2 = 1〉and

θ(s, 1) =(−1 00 1

)θ(1, t) =

(1 00 −1

).

Let ϕ be the induced automorphism on the quotient Z2 × Z2. Supposeϕ(s, 1) = (1, t) and ϕ(1, t) = (s, 1). The equality ϕ(αs) = ϕ(α−1) yields

(1, t)(αa, βb)(1, t) = (α−a, β−b)⇒ a = 0.

Similarly, the equality ϕ(βt) = ϕ(β−1) yields d = 0. This implies that ϕ′ =[0 cb 0

]. It follows that (b, c) ∈ {(1, 1), (−1,−1), (1,−1), (−1, 1)}. If (b, c) = (1, 1)

or (−1,−1), det(1−ϕ′) = 0⇒ R(ϕ′) =∞. If (b, c) = (1,−1) or (−1, 1), θ(s, 1)ϕ′ =[0 −cb 0

]and so det(1 − θ(s, 1)ϕ′) = 0 ⇒ R(θ(s, 1)ϕ′) = ∞. Either case implies

that R(ϕ) =∞ by Lemma 1.1(3).Suppose ϕ(s, 1) = (s, 1) and ϕ(1, t) = (1, t). The equality ϕ(βt) = ϕ(β−1)

yields(1, t)(αc, βd)(1, t) = (α−c, β−d)⇒ c = 0.

Similarly, using the equality ϕ(αs) = ϕ(α−1), we can show that b = 0 so that

ϕ′ =[a 00 d

].


In this case, ad = detϕ′ = ±1 so that (a, d) ∈ {(1, 1), (−1,−1), (1,−1), (−1, 1)}.Similar calculations as above show that R(ϕ) =∞.

Suppose ϕ(s, 1) = (s, t). The equality ϕ(βs) = ϕ(β) implies that

(s, t)(αc, βd)(s, t) = (αc, βd)⇒ c = 0 = d.

This leads to a contradiction.Likewise, if ϕ(1, t) = (s, t) then the equality ϕ(αt) = ϕ(α) implies that

(s, t)(αa, βb)(s, t) = (αa, βb)⇒ a = 0 = b, a contradiction.

Hence, we conclude that R(ϕ) =∞ and G has the property R∞.

5.4. Case 9.

This group has the following presentation:

G = 〈R1, R2, T | R21 = R22 = T 2 = (R1R2)2 = (R1TR2T )2 = 1〉.Consider the subgroup H generated by R1, R2, R1TR1T,R2TR2T . If we let σ =R1, ρ = R2, α = R1TR1T and β = R2TR2T , then

H = 〈α, β, σ, ρ|αβ = βα, ασ = α−1, βσ = β, αρ = α,

βρ = β−1, (σρ)2 = 1 = σ2 = ρ2〉.Note that H ∼= D∞ × D∞ and G = H �θ 〈T |T 2 = 1〉 where θ(T ) is given byα �→ α−1;β �→ β−1;σ �→ α−1σ; and ρ �→ β−1ρ. Since G = H �θ Z2, we have

G = 〈α, β, σ, ρ, T |αβ = βα, ασ = α−1, βσ = β, αρ = α, βρ = β−1,

σ2 = ρ2 = (σρ)2 = T 2 = 1, αT = α−1,

βT = β−1, σT = α−1σ, ρT = β−1ρ〉Let ϕ ∈ Aut(G). We can represent ϕ by

ϕ =

⎡⎢⎢⎢⎢⎣a c r x Ab d s y Bε δ γ z Cm n p q Df g u v E

⎤⎥⎥⎥⎥⎦ ,where the columns correspond to the images under ϕ of the generators α, β, σ, ρand T , respectively.

The equality ϕ(σT ) = ϕ(α−1σ) yields

αAβBσCρDTEαrβsσγρpT uαAβBσCρDTE = T−fρ−mσ−εβ−bα−aαrβsσγρpT u.

Equating the exponents of T , we have

E + u+ E = −f + u ⇒ −f = 2E = 0 ⇒ f = 0.

Similarly, ϕ(ρT ) = ϕ(β−1ρ) yields

αAβBσCρDTEαxβyσzρqT vαAβBσCρDTE = T−gρ−nσ−δβ−dα−cαxβyσzρqT v.


Equating the exponents of T , we have

E + v + E = −g + v ⇒ −g = 2E = 0 ⇒ g = 0.

Suppose u = 1. The equality ϕ(σ2) = 1 yields

αrβsσγρpTαrβsσγρpT = 1

⇒ αrβsσγρpα−rβ−s(α−1σ)γ(β−1ρ)p = 1.(5.4.1)

Case (γ, p) = (0, 1): equation (5.4.1) becomes αrβsρα−rβ−sβ−1ρ = 1 which im-plies that αrβsα−rβs+1 = 1 or 2s+ 1 = 0, a contradiction.Case (γ, p) = (1, 0): equation (5.4.1) becomes αrβsσα−rβ−sα−1σ = 1 which im-plies that αrβsαrβ−sα = 1 or 2r + 1 = 0, a contradiction.Case (γ, p) = (1, 1): equation (5.4.1) becomes αrβsσρα−rβ−sα−1σβ−1ρ = 1 whichimplies that αrβsσα−rβsα−1σβ = 1 or αrβsαrβsαβ = 1 so that 2r + 1 = 0, acontradiction.

Thus, when u = 1, we must have (γ, p) = (0, 0). With (γ, p, u) = (0, 0, 1), theequality ϕ(ασ) = ϕ(α−1) yields

αrβsTαaβbσερmαrβsT = ρ−mσ−εβ−bα−a

⇒ αrβsα−aβ−b(α−1σ)ε(β−1ρ)mα−rβ−s = ρ−mσ−εβ−bα−a.(5.4.2)

Case (ε,m) = (0, 1): equation (5.4.2) becomes

αrβsα−aβ−bβ−1ρα−rβ−s = ρβ−bα−a

which implies that αrβsα−aβ−bβ−1α−rβs = βbα−a. Thus, s − b − 1 + s = b. Itfollows that 2s = 2b+ 1, a contradiction.Case (ε,m) = (1, 0): equation (5.4.2) becomes

αrβsα−aβ−bα−1σα−rβ−s = σβ−bα−a

which implies that αrβsα−aβ−bα−1αrβ−s = β−bαa. Thus, r − a − 1 + r = a or2r = 2a+ 1, a contradiction.Case (ε,m) = (1, 1): equation (5.4.2) becomes

αrβsα−aβ−bα−1σβ−1ρα−rβ−s = ρσβ−bα−a

which implies that

αrβsα−aβ−bα−1σβ−1α−rβs = σβbα−a

orαrβsα−aβ−bα−1β−1αrβs = βbαa.

Thus, r − a− 1 + r = a or 2r = 2a+ 1, a contradiction.Thus, when u = 1, γ = p = ε = m = 0.Again, when u = 1, the equality ϕ(βσ) = ϕ(β) yields

αrβsTαcβdσδρnTβ−sα−r = αcβdσδρn

⇒ αrβsα−cβ−d(α−1σ)δ(β−1ρ)nβ−sα−r = αcβdσδρn.(5.4.3)


Case (δ, n) = (0, 1): equation (5.4.3) becomes

αrβsα−cβ−dβ−1ρβ−sα−r = αcβdρ

which implies that αrβsα−cβ−dβ−1βsα−r = αcβd. Thus, s − d − 1 + s = d or2s = 2d+ 1, a contradiction.


αrβsα−cβ−dα−1σβ−sα−r = αcβdσ

which implies that αr−c−1βs−dσβ−sα−r = αcβdσ or αr−c−1βs−dβ−sαr = αcβd.Thus, r − c− 1 + r = c or 2r = 2c+ 1, a contradiction.


αrβsα−cβ−dα−1σβ−1ρβ−sα−r = αcβdσρ

which implies that αr−c−1βs−dσβ−1βsα−r = αcβdσ or αr−c−1βs−dβs−1αr =αcβd. Thus, r − c− 1 + r = c or 2r = 2c+ 1, a contradiction.

Thus, if u = 1 then γ = p = ε = m = δ = n = f = g = 0. Assuming u = 1,the equality ϕ(αρ) = ϕ(α) yields

αxβyσzρqT vαaβbαxβyσzρqT v = αaβb. (5.4.4)

Case (q, v) = (0, 0): Note that z �= 0 otherwise ϕ(ρ) would have infinite order.Now z = 1, equation (5.4.4) becomes αxβyσαa+xβb+yσ = αaβb which impliesthat αxβyα−a−xβb+y = αaβb. It follows that a = 0 and y = 0. In this case,ϕ(βρ) = ϕ(β−1) yields

αxσαcβdαxσ = β−dα−c

⇒ αxα−cβdαx = β−dα−c ⇒ d = 0.

Now ϕ(βσ) = ϕ(β) yields

αrβsTαcTβ−sα−r = αc

⇒ αrβsα−cβ−sα−r = αc ⇒ c = 0.

It follows that ϕ(β) = 1, a contradiction. Thus, the case (q, v) = (0, 0) cannotoccur when u = 1.

Suppose (q, v) = (1, 0). Then (5.4.4) becomes αxβyσzραaβbαxβyσzρ = αaβb

which implies that αxβyσzαaβ−bαxβ−yσz = αaβb.Case z = 0: We have αxβyαaβ−bαxβ−y = αaβb which implies that x = 0

and b = 0. Now ϕ(βρ) = ϕ(β−1) yields

βyραcβdβyρ = β−dα−c

⇒ βyαcβ−dβ−y = β−dα−c ⇒ c = 0.


Now ϕ(βσ) = ϕ(β) yields

αrβsTβdαrβsT = βd

⇒ αrβsβ−dα−rβ−s = βd ⇒ d = 0.

It follows that ϕ(β) = 1, a contradiction.Case z = 1: We have

αxβyσαaβ−bαxβ−y = αaβb

which implies that αxβyα−aβ−bα−xβ−y = αaβb. It follows that a = 0 and b = 0and hence ϕ(α) = 1, a contradiction.

Thus, the case (q, v) = (1, 0) cannot occur when u = 1. In other words, whenu = 1, v = 1.

Suppose v = 1. Case (z, q) = (0, 0): The equality ϕ(σρ)2 = 1 yields

(αrβsTαxβyT )2 = 1

⇒ (αrβsα−xβ−y)2 = 1

⇒ r = x, s = y ⇒ ϕ(σ) = ϕ(ρ), a contradiction.

(5.4.5)

Consider the equality ϕ(ρ)2 = 1. This implies that

αxβyσzρqT vαxβyσzρqT v = 1. (5.4.6)

Case (z, q) = (0, 1): equation (5.4.6) becomes

αxβyρTαxβyρT = 1

⇒ αxβyρα−xβ−yβ−1ρ = 1

⇒ αxβyα−xβy+1 = 1 ⇒ 2y + 1 = 0, a contradiction.

(5.4.7)


αxβyσTαxβyσT = 1

⇒ αxβyσα−xβ−yα−1σ = 1

⇒ αxβyαxβ−yα = 1 ⇒ 2x+ 1 = 0, a contradiction.

(5.4.8)


αxβyσρTαxβyσρT = 1

⇒ αxβyσρα−xβ−yα−1σβ−1ρ = 1

⇒ αxβyσα−xβyα−1σβ = 1

⇒ αxβyαxβyαβ = 1 ⇒ 2x+ 1 = 0, 2y + 1 = 0 a contradiction.

(5.4.9)

Thus, when u = 1, v �= 1. Hence, we conclude that u = 0. Moreover, usingthe equalities ϕ(σρ)2 = 1 and ϕ(ρ)2 = 1, the equations (5.4.7), (5.4.8), (5.4.9), and(5.4.5) imply that v = 0. Now, f = g = u = v = 0 implies that H is characteristic.Since H = D∞×D∞, by Case 6, H has the property R∞. It follows from Lemma1.1(3) that G has the property R∞.


6. Cases 10, 13, 16

In this section, we show that the wallpaper groups in cases 10 and 16 have the R∞property but case 13 does not. In each of these cases, the group is a semi-directproduct of the form Z2 � Zn for n = 3, 4, 6.

6.1. Case 13.


G = 〈α, β, t|αβ = βα, αt = α−1β, βt = α−1, t3 = 1〉.Note that G ∼= Z2 �θ Z3 where

θ(t) =(−1 −11 0

)and θ(t2) =

(0 1−1 −1

).

Consider the subgroup Z2 = H = 〈α, β|αβ = βα〉 and the automorphism

α �→ α−1;β �→ β−1 given by ϕ′ =[−1 00 −1

]. Since αβ = βα, ϕ′ extends to an

automorphism ϕ of G by sending t �→ t. Moreover,

t · ϕ′ = θ(t)ϕ′ =[1 1−1 0

]and t2 · ϕ′ = θ(t2)ϕ′ =

[0 −11 1

].

Therefore, R(ϕ′) < ∞, R(t · ϕ′) < ∞, and R(t2 · ϕ′) < ∞ since det(1 − ϕ′) �=0, det(1− t · ϕ′) �= 0, and det(1− t2 · ϕ′) �= 0. Since Z3 = G/H is finite, it followsfrom Lemma 1.1(2) that R(ϕ) <∞ and thus G does not have the property R∞.

6.2. Case 10.


G = 〈α, β, t|αβ = βα, αt = β, βt = α−1, t4 = 1〉.Note that G ∼= Z2 �θ Z4 where

θ(t) =(0 −11 0

)θ(t2) =

(−1 00 −1

)and θ(t3) =

(0 1−1 0

).

Let ϕ ∈ Aut(G) and be represented by

ϕ =


⎤⎦ . (6.2.1)

The equality ϕ(αt) = ϕ(β) yields

αrβstγαaβbtεt−γβ−sα−r = αcβdtδ.

By equating the exponents of t, we have ε = δ. The equality ϕ(αβ) = ϕ(βα) yields

αaβbtεαcβdtδ = αcβdtδαaβbtε. (6.2.2)


Case δ = 1 = ε: Equation (6.2.2) becomes

αaβbtαcβdt = αcβdtαaβbt

⇒ αaβbβcα−d = αcβdβaα−b

⇒ a− d = c− b and b+ c = a+ d or a = c, b = d.

It follows that ϕ(α) = ϕ(β), a contradiction.


αaβbt2αcβdt2 = αcβdt2αaβbt2

⇒ αaβbα−cβ−d = αcβdα−aβ−b

⇒ a− c = c− a and b− d = d− b or a = c, b = d.

It follows that ϕ(α) = ϕ(β), a contradiction.


αaβbt3αcβdt3 = αcβdt3αaβbt3

⇒ αaβbβ−cαd = αcβdβ−aαb

⇒ a+ d = c+ b and b− c = d− a or a = c, b = d.

It follows that ϕ(α) = ϕ(β), a contradiction.Thus, δ = 0 = ε and H = 〈α, β|αβ = βα〉 is characteristic in G.The equality ϕ(αt) = ϕ(β) yields

αrβstγαaβbt−γβ−sα−r = αcβd. (6.2.3)

Case γ = 1: Equation (6.2.3) becomes

αrβstαaβbt−1β−sα−r = αcβd

⇒ αrβsβaα−bβ−sα−r = αcβd.

It follows that r − b − r = c and s + a − s = d so that a = d and b = −c. Therestriction ϕ′ = ϕ|H is given by

ϕ′ =[a −bb a

].

Since ϕ′ is an automorphism, detϕ′ = a2 + b2 = 1. Since a, b ∈ Z, it follows that

ϕ′ = (i)[1 00 1

], (ii)

[−1 00 −1

], (iii)

[0 11 0

], (iv)

[0 −1−1 0

].

For (i), (iii) and (iv), det(1 − ϕ′) = 0 so R(ϕ′) = ∞. For (ii), t2 · ϕ′ = θ(t2)ϕ′ =[1 00 1

]so that R(t2 · ϕ′) = ∞. Thus, we can conclude from Lemma 1.1(3) that

R(ϕ) =∞.


Case γ = 2: Equation (6.2.3) becomes

αrβst2αaβbt−2β−sα−r = αcβd

⇒ αrβsα−aβ−bβ−sα−r = αcβd.

It follows that r − a − r = c and s − b− s = d so that a = −c and b = −d. Therestriction ϕ′ = ϕ|H is given by

ϕ′ =[a −ab −b

].

This contradicts the fact that ϕ′ is an automorphism so this case cannot occur.Case γ = 3: Equation (6.2.3) becomes


⇒ αrβsβ−aαbβ−sα−r = αcβd.

It follows that r + b − r = c and s − a − s = d so that a = −d and b = c. Therestriction ϕ′ = ϕ|H is given by

ϕ′ =[a bb −a

].

Since ϕ′ is an automorphism, detϕ′ = a2 + b2 = 1. It follows that

ϕ′ = (i)[1 00 −1

], (ii)

[−1 00 1

], (iii)

[0 11 0

], (iv)

[0 −1−1 0

].

For all of these four cases, det(1− ϕ′) = 0 so R(ϕ′) =∞.Hence by Lemma 1.1(3) the group G has the property R∞.

6.3. Case 16.


G = 〈α, β, t|αβ = βα, αt = β, βt = α−1β, t6 = 1〉.Note that G ∼= Z2 �θ Z6 where

θ(t) =(0 −11 1

)θ(t2) =

(−1 −11 0

)θ(t3) =

(−1 00 −1

)θ(t4) =

(0 1−1 −1

)and θ(t5) =

(1 1−1 0

).

Let ϕ ∈ Aut(G) and be represented by (6.2.1). The equality ϕ(αt) = ϕ(β) yields

αrβstγαaβbtεt−γβ−sα−r = αcβdtδ.

By equating the exponents of t, we have γ + ε − γ = δ or ε = δ. Since ϕ(t) is oforder 6, it follows that γ = 1 or 5.

The equality ϕ(βt) = ϕ(α−1β) yields

αrβstγαcβdtδt−γβ−sα−r = t−εβ−bα−aαcβdtδ.


By equating the exponents of t, we have

γ + δ − γ = −ε+ δ ⇒ δ = 0⇒ ε = 0.

Now,

ϕ =

⎡⎣a c rb d s0 0 γ

⎤⎦so that H is characteristic in G.

Case γ = 1: the equality ϕ(αt) = ϕ(β) yields

αrβstαaβbt−1β−sα−r = αcβd

⇒ αrβsβaα−bβbβ−sα−r = αcβd.

It follows that r − b− r = c and s + a + b− s = d so that b = −c and a+ b = d.

Now ϕ′ = ϕ|Z2 =[a −bb a+ b

]. Since ϕ′ is an automorphism, a2 + ab+ b2 = ±1.

Case γ = 5: the equality ϕ(αt) = ϕ(β) yields


⇒ αrβsαaβ−aαbβ−sα−r = αcβd.

It follows that r + a+ b− r = c and s− a− s = d so that a = −d and a+ b = c.

Now ϕ′ = ϕ|Z2 =[a a+ bb −a

]. Since ϕ′ is an automorphism, a2 + ab+ b2 = ±1.

If |a| �= |b| (say |b| < |a|) and ab < 0 then (a+ b)2 < a2+ b2+ ab < a2+ b2, acontradiction. Similarly, a2+ b2+ab > 1 if ab > 0. Thus, either ab = 0 or |a| = |b|.It follows that

(a, b) ∈ {(1, 0), (0, 1), (1,−1), (−1, 0), (0,−1), (−1, 1)}.

When γ = 1, we have

ϕ′ ∈{[

1 00 1

],

[−1 00 −1

],

[0 −11 1

],

[0 1−1 −1

],

[1 1−1 0

],

[−1 −11 0

]}.

Note that ϕ′ = θ(tj) for j = 0, . . . , 5. It follows that for ϕ′, there exists some j suchthat det(1 − θ(tj)ϕ′) = 0. This implies that R(tj · ϕ′) =∞ and hence R(ϕ) = ∞by Lemma 1.1(3).

When γ = 5, we have

ϕ′ ∈{[

1 10 −1

],

[−1 −10 1

],

[0 11 0

],

[0 −1−1 0

],

[1 0−1 −1

],

[−1 01 1

]}.

If ϕ′ =[−1 −10 1

]then θ(t)ϕ′ =

[0 −1−1 0

]and thus det(1 − θ(t)ϕ′) = 0 or

R(t · ϕ′) =∞.


Likewise, if ϕ′ =[1 0−1 −1

]then θ(t)ϕ′ =

[1 10 −1

]and thus R(t · ϕ′) =∞.

For the remaining four cases for ϕ′, we proceed as in the previous situation whenγ = 1 and conclude that R(tj · ϕ′) =∞ for some j.

Hence, we conclude from Lemma 1.1(3) that R(ϕ) = ∞ and G has theproperty R∞.

7. Cases 11, 12, 14, 15, and 17

In this section, we show that groups in the remaining five cases all have the prop-erty R∞. In each of these cases, we use an appropriate presentation so that everyautomorphism can be represented by

ϕ =

⎡⎢⎢⎣a c r xb d s yε δ γ zm n p q

⎤⎥⎥⎦ . (7.0.1)

First, we will analyze cases 14 and 15. Although the group G3 for case 13 doesNOT have the property R∞ and the groups in these two cases are finite extensionsof G3, it turns out that there are very few automorphisms in these situations.

7.1. Case 14.

This group is G13 in [8] and has the following presentation

G = 〈α, β, σ, ρ|αβ = βα, ασ = α−1β, βσ = α−1, αρ = α,

βρ = αβ−1, ρ2 = σ3 = (σρ)2 = 1〉.Note that G ∼= Z2 �θ (Z3 � Z2) where

Z3 � Z2 ∼= 〈σ, ρ|ρ2 = σ3 = (σρ)2 = 1〉.Note also that σρ = σ−1 = σ2. Let ϕ ∈ Aut(G).

Since ϕ(σ3) = 1,

αrβsσγρpαrβsσγρpαrβsσγρp = 1.

By equating the exponents of ρ, we have 3p = 0 so p = 0. This means that γ �= 0otherwise ϕ(σ) would have infinite order.

The equality ϕ(ρ2) = 1 implies that (z, q) �= (0, 0). Similarly, (z, q) �= (1, 0)and (z, q) �= (2, 0). Thus, q = 1.

Moreover, ϕ(ρ2) = 1 implies that αxβyσzραxβyσzρ = 1 or

αxβyσzαx(αβ−1)yσ−z = 1 (7.1.1)

When z = 0, (7.1.1) yields y = −2x. When z = 1, (7.1.1) yields x = −2y. Whenz = 2, (7.1.1) gives x = y.

The equality ϕ(ασ) = ϕ(α−1β) implies that

αrβsσγαaβbσερmσ−γβ−sα−r = ρ−mσ−εβ−bα−aαcβdσδρn. (7.1.2)


By equating the exponents of ρ, we have m = −m + n or n = 2m = 0 so n = 0.The equality ϕ(βσ) = ϕ(α−1) yields

αrβsσγαcβdσδσ−γβ−sα−r = ρ−mσ−εβ−bα−a. (7.1.3)

By equating the exponents of ρ, we have m = 0.The equality ϕ(αρ) = ϕ(α) implies that

αxβyσzραaβbσερ−1σ−zβ−yα−x = αaβbσε

⇒ αxβyσzαa(αβ−1)bσ−εσ−zβ−yα−x = αaβbσε.

By equating the exponents of σ, we have ε = 0. Now (7.1.3) becomes

αrβsσγαcβdσδσ−γβ−sα−r = β−bα−a.

By equating the exponents of σ, we have γ + δ − γ = 0 so δ = 0. Thus,

ϕ =

⎡⎢⎢⎣a c r xb d s y0 0 γ z0 0 0 1

⎤⎥⎥⎦ .Now, equation (7.1.2) becomes αrβsσγαaβbσ−γβ−sα−r = αc−aβd−b. When

γ = 1, we have αrβs(α−1β)aα−bβ−sα−r = αc−aβd−b so that c = −b and d = a+b.When γ = 2, we have αrβsβ−a(αβ−1)bβ−sα−r = αc−aβd−b so that d = −a andc = a+ b. In either case, detϕ′ = ±(a2 + ab+ b2) = ±1.

For γ = 2, we have

ϕ′ ∈{[

1 10 −1

],

[0 11 0

],

[1 0−1 −1

],

[−1 −10 1

],

[0 −1−1 0

],

[−1 01 1

]}.

For ϕ′ =[1 10 −1

],

[1 0−1 −1

],

[−1 −10 1

], or

[−1 01 1

], det(1 − ϕ′) = 0 so that

R(ϕ′) = ∞. For ϕ′ =[0 11 0

], or

[0 −1−1 0

], θ(σ)ϕ′ =

[−1 −10 1

], or

[1 10 −1

].

In either case, det(1 − θ(σ)ϕ′) = 0 so that R(θ(σ)ϕ′) =∞.For γ = 1, we have

ϕ′ ∈{[

1 00 1

],

[0 −11 1

],

[1 1−1 0

],

[−1 00 −1

],

[0 1−1 −1

],

[−1 −11 0

]}.

For ϕ′ =[1 00 1

], det(1− ϕ′) = 0. For ϕ′ =

[0 −11 1

],

[−1 00 −1

], or

[0 1−1 −1

],

it is easy to see that det(1 − θ(ρ)ϕ′) = 0 where θ(ρ) =[1 10 −1

]. For ϕ′ =[

1 1−1 0

], or

[−1 −11 0

], det(1− θ(σρ)ϕ′) = 0.

It follows from Lemma 1.1(3) that G has the property R∞.


7.2. Case 15.

This group is G23 in [8] and has the following presentation

G = 〈α, β, σ, ρ|αβ = βα, ασ = α−1β, βσ = α−1, αρ = β,

βρ = α, ρ2 = σ3 = (σρ)2 = 1〉.

Note that G ∼= Z2 �θ (Z3 � Z2) where

Z3 � Z2 ∼= 〈σ, ρ|ρ2 = σ3 = (σρ)2 = 1〉.

Note also that σρ = σ−1 = σ2. Let ϕ ∈ Aut(G).Since ϕ(σ3) = 1,

αrβsσγρpαrβsσγρpαrβsσγρp = 1.

By equating the exponents of ρ, we have 3p = 0 so p = 0. This means that γ �= 0otherwise ϕ(σ) would have infinite order. The equality ϕ(ρ2) = 1 and the fact thatσ has order 3 imply that q = 1.

The equality ϕ(ασ) = ϕ(α−1β) yields

αrβsσγαaβbσερmσ−γβ−sα−r = ρ−mσ−εβ−bα−aαcβdσδρn.

By equating the exponents of ρ, we have m = −m + n or n = 2m = 0 so n = 0.The equality ϕ(βσ) = ϕ(α−1) yields

αrβsσγαcβdσδσ−γβ−sα−r = ρ−mσ−εβ−bα−a.

By equating the exponents of ρ, we have m = 0. Since m = 0, we can equate theexponents of σ to obtain γ + δ − γ = −ε or δ = −ε. Now, we have

ϕ =

⎡⎢⎢⎣a c r xb d s yε −ε γ z0 0 0 1

⎤⎥⎥⎦ .Consider ϕ(α3) = (αaβbσε)3. When ε = 1, 2, it is straightforward to see thatϕ(α3) = 1. However, ϕ(α) is of infinite order thus we conclude that ε = 0. Thismeans that Z2 is characteristic.

Following Case 14, we use the equality ϕ(ασ) = ϕ(α−1β) to conclude thatthe restriction ϕ′ has detϕ′ = ±(a2 + ab+ b2) = ±1 so that

ϕ′ ∈{[

1 10 −1

],

[0 11 0

],

[1 0−1 −1

],

[−1 −10 1

],

[0 −1−1 0

],

[−1 01 1

]}.

It follows that either det(1 − ϕ′) = 0 or det(1 − θ(σ)ϕ′) = 0. We conclude fromLemma 1.1(3) that G has the property R∞.


7.3. Case 11.


G = 〈α, β, σ, ρ|αβ = βα, ασ = β, βσ = α−1, αρ = α,

βρ = β−1, ρ2 = σ4 = (σρ)2 = 1〉.Note that G ∼= G4 �θ Z2 where G4 is the wallpaper group in case 10. Let ϕ ∈Aut(G).

The equality ϕ(ασ) = ϕ(β) yields

αrβsσγρpαaβbσερmρ−pσ−γβ−sα−r = αcβdσδρn.

Equating the exponents of ρ yields p+m−p = n or m = n. The equality ϕ(βσ) =ϕ(α−1) yields

αrβsσγρpαcβdσδρnρ−pσ−γβ−sα−r = ρ−mσ−εβ−bα−a.

Equating the exponents of ρ yields p + n − p = −m or m = −n. It follows thatm = 0 = n.

Since ϕ(σ)4 = 1, we have (αrβsσγρp)4 = 1. Suppose p = 1. If γ = 0,(αrβsσγρp)4 = 1 becomes α4r = 1, a contradiction. If γ = 1, then ϕ(σ)4 =(αrβsσραrβsσρ)2 = (αr+sβr+s)2 = 1. It follows that r + s = 0 but this meansthat ϕ(σ) has order 2, a contradiction. If γ = 2, then (αrβsσγρp)4 = 1 becomesβ4s = 1, again a contradiction. Finally if γ = 3, ϕ(σ)4 = (αrβsσ3αrβ−sσ−3)2 =(αr−sβs−r)2 = 1. It follows that r = s and ϕ(σ) has order 2, a contradiction.Moreover, the case r = 0 = s cannot occur for in this case ϕ(σ) = σγρ is of order2, again a contradiction.

We conclude that the case when p = 1 cannot occur. Thus, we have m = n =p = 0 which means that G4 is characteristic in G. Since G4 has the property R∞,it follows that G also has the property.

7.4. Case 12.


G = 〈α, β, σ, ρ|αβ = βα, ασ = β, βσ = α−1, αρ = α,

βρ = β−1, ρ2 = α, σ4 = (σρ)2 = 1〉.Note that G ∼= G4 �θ Z2 where G4 is the wallpaper group in case 10 and Z2 =〈t|t2 = 1〉. Here, the projection G → Z2 is given by sending α, β, σ to 1, ρ �→ twith kernel G4 and the section Z2 → G is given by t �→ σρ. Here, the action θ(t)is given by α �→ β, β �→ α, σ �→ σ−1β−1. Let ϕ ∈ Aut(G) be given by

ϕ =

⎡⎢⎢⎣a c r xb d s yε δ γ zm n p q

⎤⎥⎥⎦where the columns represent the images of the generators α, β, σ, and t (not ρ).


Suppose p = 1. The equality ϕ(σ4) = 1 yields

(αrβsσγtαrβsσγt)2 = 1. (7.4.1)

If γ = 0 then (7.4.1) becomes (αrβstαrβst)2 = (αrβsβrαs)2 = 1 which impliesthat r+ s = 0. This means that ϕ(σ2) = 1 or ϕ(σ) has order 2, a contradiction. Ifγ = 1 then (7.4.1) becomes

(αrβsσtαrβsσt)2 = (αrβsσβrαsσ−1β−1)2 = (αrβsα−rβsβ−1)2 = (β2s−1)2 = 1,

a contradiction.Similarly, a straightforward calculation shows that if γ = 2 then (7.4.1) be-

comes (αr−s+1βs−r−1)2 = 1 or r − s+ 1 = 0 so that ϕ(σ) has order 2 and that ifγ = 3 then (7.4.1) becomes (α2r+1)2 = 1, a contradiction. Thus, we conclude thatp = 0.

Now p = 0, ϕ(σ4) = 1 yields (αrβsσγαrβsσγ)2 = 1. Since p = 0, γ �= 0 orelse ϕ(σ) would be of infinite order. If γ = 2, then we have (αrβsσ2αrβsσ2)2 =(αrβsα−rβ−s)2 = 1 so ϕ(σ) has order 2, a contradiction. Thus, γ is either 1 or 3.

The equality ϕ(ασ) = ϕ(β), with p = 0, yields

αrβsσγαaβbσεtmσ−γβ−sα−r = αcβdσδtn.

Equating the exponents of t yields m = n.Suppose (m,n) = (1, 1). Note that σt = σ−1β−1 thus the equality ϕ(σt) =

ϕ(σ−1β−1) yields

αxβyσztqαrβsσt−qσ−zβ−yα−x = σ−1β−sα−st−1σ−δβ−dα−c.

Equating the exponents of t leads to a contradiction so the case (m,n) = (1, 1)cannot occur.

Thus, m = n = p = 0 and G4 is characteristic in G. Hence G has propertyR∞ since G4 does.

7.5. Case 17.


G = 〈α, β, σ, ρ|αβ = βα, ασ = β, βσ = α−1β, αρ = α,

βρ = αβ−1, ρ2 = σ6 = (σρ)2 = 1〉.Note that G ∼= G6 � Z2 where G6 is the wallpaper group in case 16 and Z2 =〈ρ|ρ2 = 1〉. Let ϕ ∈ Aut(G) be given by (7.0.1).

The equality ϕ(βσ) = ϕ(α−1β) yields

αrβsσγρpαcβdσδρnρ−pσ−γβ−sα−r = ρ−mσ−εβ−bα−aαcβdσδρn. (7.5.1)

Equating exponents of ρ yields p+ n− p = −m+ n which implies that m = 0.The equality ϕ(ασ) = ϕ(β) yields

αrβsσγρpαaβbσερ−pσ−γβ−sα−r = αcβdσδρn. (7.5.2)

Equating exponents of ρ yields p− p = n which implies that n = 0.


Suppose p = 1. The equality ϕ(σ6) = 1 becomes

(αrβsσγρ)6 = (αrβsσγραrβsσγρ)3 = 1

or(αrβsσγαr(αβ−1)sσ−γ)3 = 1. (7.5.3)

Note that if r = 0 = s then (7.5.3) shows that ϕ(σ) = σγρ has order 2, a contra-diction. We assume that (r, s) �= (0, 0).Caseγ = 0: (7.5.3) reduces to (α2r+s)3 = 1. That means 2r + s = 0 and so ϕ(σ)has order 2, a contradiction.Caseγ = 1: (7.5.3) reduces to (αr+sβr+s)3 = 1. That means r+ s = 0 and so ϕ(σ)has order 2, a contradiction.Caseγ = 2: (7.5.3) reduces to (βr+2s)3 = 1. That means r + 2s = 0 and so ϕ(σ)has order 2, a contradiction.Caseγ = 3: (7.5.3) reduces to (α−sβ2s)3 = 1. That means s = 0 and so ϕ(σ) hasorder 2, a contradiction.Caseγ = 4: (7.5.3) reduces to (αr−sβr+s)3 = 1. That means r = s = 0, a contra-diction.Caseγ = 0: (7.5.3) reduces to (α2rβ−r)3 = 1. That means r = 0 and so ϕ(σ) hasorder 2, a contradiction.

Hence, we conclude that the case when p = 1 cannot occur and so p = 0.Since m = 0 = n, the subgroup G6 is characteristic. Since G6 has the propertyR∞, so does G.

8. Concluding remarks

Given a short exact sequence of groups 1 → Zn → G → F → 1 where F is afinite group, F acts on the kernel by conjugation in G and the action is given byθ : F → GLn(Z). If this action is injective and n = 2, then the isomorphism classesof such groups G are precisely the 17 wallpaper groups. It is natural to investigatethe R∞ property for such extensions with injective actions in higher dimensions.There are some partial results in this direction based upon by our knowledge ofthe cases n = 2. However the complete solution of the problems is much morecomplex and we intend to investigate this problem in the sequel.

References

[1] H.S.M. Coxeter and W.O.J. Moser, Generators and relations for discrete groups.Springer-Verlag, Berlin-Gottingen-Heidelberg, 1957. viii + 155 pp.

[2] W. Dicks and M. Dunwoody, Groups acting on graphs. Cambridge Studies in Ad-vanced Mathematics, 17. Cambridge University Press, Cambridge, 1989. xvi+283 pp.

[3] A.L. Fel’shtyn, The Reidemeister number of any automorphism of a Gromov hyper-bolic group is infinite, Zapiski Nauchnych Seminarov POMI 279 (2001), 229–241.


[4] D. Goncalves and P. Wong, Twisted conjugacy classes in exponential growth groups,Bull. London Math. Soc. 35 (2003), 261–268.

[5] D. Goncalves and P. Wong, Twisted conjugacy classes for nilpotent groups, J. ReineAngew. Math. 633 (2009), 11–27.

[6] B. Jiang, “Lectures on Nielsen Fixed Point Theory,” Contemporary Mathematicsvol. 14, Amer. Math. Soc., Providence, Rhode Island, 1983.

[7] G. Levitt and M. Lustig, Most automorphisms of a hyperbolic group have very simple

dynamics, Ann. Scient. Ec. Norm. Sup. 33 (2000), 507–517.

[8] R. Lyndon, Groups and Geometry. LMN Lecture Note Series, 101, Cambridge Uni-versity press, 1987.

Daciberg GoncalvesDept. de Matematica – IME – USPCaixa Postal 66.281 – CEP 05314-970Sao Paulo – SP, BrasilFAX: 55-11-30916183e-mail: [email protected]

Peter WongDepartment of MathematicsBates CollegeLewiston, ME 04240, USAFAX: 1-207-7868331e-mail: [email protected]



Solving Random Equations in Garside GroupsUsing Length Functions

Martin Hock and Boaz Tsaban

Abstract. We give a systematic exposition of memory-length algorithms forsolving equations in noncommutative groups. This exposition clarifies somepoints untouched in earlier expositions. We then focus on the main ingredientin these attacks: Length functions.

After a self-contained introduction to Garside groups, we describe lengthfunctions induced by the greedy normal form and by the rational normal formin these groups, and compare their worst-case performances.

Our main concern is Artin’s braid groups, with their two known Gar-side presentations, due to Artin and due to Birman-Ko-Lee (BKL). We showthat in B3 equipped with the BKL presentation, the (efficiently computable)rational normal form of each element is a geodesic, i.e., is a representativeof minimal length for that element. (For Artin’s presentation of B3, Bergersupplied in 1994 a method to obtain geodesic representatives in B3.)

For arbitrary BN , finding the geodesic length of an element is NP-hard,by a 1991 result of by Paterson and Razborov. We show that a good esti-mation of the geodesic length of an element of BN in Artin’s presentationis measuring the length of its rational form in the BKL presentation. This isproved theoretically for the worst case, and experimental evidence is providedfor the generic case.

Mathematics Subject Classification (2000). 05E15, 94A60.

Keywords. Random equations, Garside groups, length functions, braid group,Artin presentation, Birman-Ko-Lee presentation, minimal length, geodesics.

1. Solving random equations

All groups considered in this paper are multiplicative noncommutative groups,with an efficiently solvable word problem, that is, there is an efficient algorithmfor deciding whether two given (finite products of) elements in the group are equalas elements of the group. Throughout this paper, G denotes such a group.

The second author was partially supported by the Koshland Center for Basic Research.

150 M. Hock and B. Tsaban

Problems involving solutions of equations in groups have a long history, andare nowadays also explored towards applications in public-key cryptography [14].We mention some of the more elegant problems of this type.

Problem 1 (Conjugacy Search). Given conjugate a, b ∈ G, find x ∈ G such thatb = xax−1.

Problem 2 (Root Search). Given a ∈ G, find x ∈ G such that a = x2, providedthat such x exists.

Problem 3 (Decomposition Search). Let H be a proper subgroup of G. Givena, b ∈ G, find x, y ∈ H such that b = xay, provided that there exist such x, y.

We will discuss the meaning of the terms “given” and “find”, appearing inProblems 1–3, later.

Problems 1–3, as well as many additional ones, can be stated generally asfollows. By free-group word w(t1, . . . , tk) we mean a product of variables tε1i1 · t

ε2i2·

· · · · tεn

infor any choice of a positive integer n and elements i1, . . . , in ∈ {1, . . . , k}

and ε1, . . . , εn ∈ {1,−1}, such that no cancellation is possible, that is, for eachj = 1, . . . , n, if ij = ij+1, then εj �= −εj+1.Problem 4 (Solution Search). Fix H1, . . . , Hk ≤ G and a free-group word w(t1,. . . , tk+n). Given parameters p1, . . . , pn ∈ G and an element c ∈ G, find x1 ∈H1, . . . , xk ∈ Hk such that c = w(x1, . . . , xk, p1, . . . , pn), provided that there existsuch x1, . . . , xk.

Problem 4 deals with the solution of a single solvable equation (with pa-rameters). It can also be stated for systems of several equations. The algorithmsproposed here easily generalize to cover this case, cf. [10].

1.1. Making the problems meaningful

It suffices to discuss Problem 4.First, all given information must be coded in some compact form. For exam-

ple, the subgroups H1, . . . , Hk of G may be described by lists of generators andrelations, all (the list, the generators, and the relations) of manageable length.

Second, the problem may require that it be possible to find a solution foreach possible instance of the problem, or for a certain portion of the instances.Already in the case of free groups, the problem of solving equations in this sense isextremely difficult. For example, the problem of solving quadratic equations overfree groups is known to be NP-hard.

Alternatively, the instances of the problem may be chosen according to acertain distribution D, and we may require that a solution can be found with ahigh-enough probability (a probabilistic model).

Finally, by “find” we mean “find efficiently”, i.e., use an algorithm with a fea-sible running time. Otherwise, in most cases of interest the problems are solvable.E.g., if G is a finitely generated group with solvable word problem, then we cansolve Problem 4 by enumerating Gk recursively, and trying all possible solutions

Random Equations in Garside Groups 151

until one is found. This algorithm always succeeds in a finite running time, butusually this running time is infeasible.

In this discussion, all quantitative terms (compact, efficient, significant, etc.)have two natural interpretations: Concrete (e.g., of size less than 1GB) or asymp-totic (e.g., polynomial in the size of the input).

1.2. The probabilistic model

With an eye towards applications, we will always use the probabilistic versionof the problems, where we wish to find (efficiently) a solution with a significantprobability, provided that the instances of the problem are chosen according to acertain known distribution D.

More precisely, in Problem 4 we fix a distribution D on Gk+n such that foreach (x1, . . . , xk, p1, . . . , pn) in the support of D, we have that x1 ∈ H1, . . . , xk ∈Hk. An instance of the problem is generated as follows: A secret tuple (x1, . . . , xk,p1, . . . , pn) ∈ Gk+n is chosen according to the distribution D, and we are givenp1, . . . , pn and an element c ∈ G equal to w(x1, . . . , xk, p1, . . . , pn) in G. We mustthen search for elements x1 ∈ H1, . . . , xk ∈ Hk such that with a significant proba-bility, c = w(x1, . . . , xk, p1, . . . , pn) in G.

By peeling off known parameters on the left of the given word w(x1, . . . , xk,p1, . . . , pn), we may assume that it begins with a variable xi (possibly inverted).If we are able to find xi (with a significant probability), we can treat it as aparameter henceforth, and proceed to the next leading variable after peeling offall parameters on the left. Continuing in this manner, we find suggestions for allvariables, and can check whether we obtained a solution.

Thus, it is natural to consider the following problem.

Problem 5 (Leading-Variable Search). Fix H1, . . . , Hk ≤ G and a free-group wordt1 · w(t1, . . . , tk+n). Given parameters p1, . . . , pn ∈ G and an element c = x1 ·w(x1, . . . , xk, p1, . . . , pn) ∈ G such that x1 ∈ H1, . . . , xk ∈ Hk, find x1 ∈ H1, suchthat there are x2 ∈ H2, . . . , xk ∈ Hk with c = x1 · w(x1, . . . , xk, p1, . . . , pn).

Clearly, any algorithm solving Problem 4 also solves Problem 5, with at leastthe same probability of success. On the other hand, an algorithm for Problem5 can be iterated, as explained above, to obtain a solution for Problem 4 (witha smaller probability of success, which also depends on its performance on theinduced distributions along the iteration).

1.3. Decision problems

All mentioned problems also have a decision version. For example, the CongugacyProblem is: Given a, b ∈ G, are they conjugate? If we only consider algorithmswith bounded running time, then a solution to the search version also implies asolution to the decision version, in the following sense.

Assume that A is an algorithm searching for solutions of equations of a certaintype (e.g., b = xax−1), and that its running time is bounded, say by a certainfunction of the length of its input. We define a decision algorithm A′ with running


time bounded by the same function: Given an instance of the equation to bechecked, run A on this instance until the running time reaches its bound, and thenterminate it if it did not terminate already. If a solution was found, the decisionof A′ is Yes. Otherwise, it is No.

Assume that the instances of the equation are distributed according to somedistribution E. This induces a distribution D on the solvable equations, by condi-tioning that the chosen equation be solvable. Let p be the probability that A findsa solution to (necessarily, solvable) equations distributed according to D.

For each specific instance of the equation, A′ is correct in probability at leastp: If this instance has a solution, it will be found by A in probability p, in whichcase A′ decides “Yes”. And if this instance has no solution, then in probability 1,A will not find a solution (because there is none), and A′ decides “No”.

This can also be viewed as follows: Let q = 1 − p. The probability that A′

comes up with a wrong answer is:

P (Wrong decision) == P (Decision = Yes | �Solution) · P (�Solution) ++ P (Decision = No | ∃Solution) · P (∃Solution) == 0 · P (�Solution) + q · P (∃Solution) == q · P (∃Solution).

In particular, this probability is at most q, and the worst case is whenP (∃Solution) is 1, in which the distribution may be assumed to be supportedby solvable instances, and we are actually in the search version of the problem.

This justifies, to some extent, restricting attention to search problems whenworking in the probabilistic model, with algorithms of bounded running time.

2. The memory-length approach

The potential usefulness of length functions for solving the conjugacy search prob-lem was identified in [11]. In [9, 10], it was pointed out that this approach can beused to solve arbitrary (systems of) equations.

Let H ≤ G be generated by elements a1, . . . , am of G. Assume that an in-stance x ·w(x, x2, . . . , xk, p1, . . . , pn) of Problem 5 is chosen according to a certaindistribution D, with H1 = H , and we are given c which is equal to it in G. Letw = w(x, x2, . . . , xk, p1, . . . , pn).

Let A = {a1, . . . , am}±1. Assume that the shortest expression of x as a prod-uct of elements of A has length n. Let COR(x) be the set of all a ∈ A which appearfirst in an expression of x as a product of n generators, i.e., {a ∈ A : x ∈G a·An−1}.For each a ∈ COR(x), a−1x has an expression of length n − 1, whereas fora /∈ COR(x), a−1xmay in general not have an expression shorter than n+1. In par-ticular, we expect a−1x to be “shorter” when a ∈ COR(x) than when a /∈ COR(x).Heuristically, this expectation is extended to xw.


Often, we cannot compute the length of a shortest expression of a groupelement, and we only assume that we have an efficiently computable function� : G→ R≥0, which approximates the above situation, i.e., such that �(abw) tendsto be greater than �(w) for w ∈ G, a, b ∈ {a1, . . . , am}±1.

By standard arguments, we may for convenience assume that n is known[10, 16].1 One may then try all a ∈ A, and pick one with �(a−1xw) minimal.Hopefully, a ∈ COR(x), and we can continue with the peeled-off element a−1xw.After n steps, we hopefully have (a shortest expression for) x.

In cases of interest this approach does not work as stated [9], and the followingimprovement was proposed in [10].

2.1. The memory-length algorithm

Using the above-mentioned notation, the algorithm generates an ordered list ofMsequences of length n, with the aim that with a significant probability, a sequence

((j1, ε1), (j2, ε2), . . . , (jn, εn)),

such that x = aε1j1aε2j2. . . aεn

jnin G, appears in the list, and tends to be among its

first few members. It consists of the following steps:

Step 1. For each j = 1, . . . ,m and each ε ∈ {1,−1}, compute a−εj c = a−ε

j xy, andgive (j, ε) the score �(a−ε

j c). Keep in memory the M elements (j, ε) with the best(=lowest) scores.

Steps s = 2, 3, . . . , n. For each sequence ((j1, ε1), . . . , (js−1, εs−1)) out of the Msequences stored in the memory, each js = 1, . . . ,m, and each εs ∈ {1,−1}, com-pute

�(a−εs

js(a−εs−1

js−1 · · · a−ε1j1

c)) = �(a−εs

jsa−εs−1js−1 · · · a−ε1

j1xy),

and assign this score to the sequence ((j1, ε1), . . . , (js, εs)). Keep in memory onlythe M sequences with the best scores.

The algorithm terminates after n steps, withM proposals for ((j1, ε1), (j2, ε2),. . . , (jn, εn)).

It is not difficult to see that the complexity of this algorithm is n(n+ 4m+1)M/2 group operations and evaluations of �.

It is interesting to note that this algorithm may also be useful for solving thefollowing.

Problem 6 ((Shortest) Subgroup Membership Search). Given a1, . . . , am ∈ G andx ∈ 〈a1, . . . , am〉, find a (shortest possible) expression of x as a product of elementsfrom the set {a1, . . . , am}±1.

1This has a computational cost, so we cannot assume that we know the lengths of shortestexpressions of many elements.


2.2. Sufficiency for the general problem

Assume that the algorithm succeeds, with a significant probability, to have theleading element x in the final list. Then we have the following.

If there is only one unknown variable in the equation (e.g., Problems 1–3),then we can check (in running time M) all elements in the list and find one whichis a solution to the problem.

In the general case (Problem 4) there are several unknown variables, andwe can iterate the algorithm by checking each suggestion in the list. The overallcomplexity is in principle Mk. However, the suggestions for each variable are or-dered more or less according to their likelihood, and it suffices to check, for someN "M , the N most likely solutions. This reduces the complexity to Nk, or moreprecisely to N1 · N2 · · ·Nk, where Nk is the number of elements required at thekth step, and it is likely that Ni+1 " Ni for each i.

2.3. Improvements

Certain simple modifications in the memory-length algorithm increase its successrates. We refer the reader to [16] for details.

2.4. The length function

For this algorithm to be meaningful and useful, one must have a good and ef-ficiently computable length function on the group G. Our introduction of thememory-length algorithm suggests a natural model for comparing length func-tions for appropriateness to this method. We explore this below, after introducinga new proposal for a length function on the braid group. The braid group is,thus far, the most popular in applications related to cryptography [14]. Most ofthese cryptographic applications give rise to an equation, whose solution wouldimply the insecurity of the application. Thus, it is natural to look for good lengthfunctions on this group. See [14] for more details.

3. Excursion: Garside groups

We are going to consider two Garside structures on the braid group (to be defined).This section is an essentially self-contained introduction to Garside groups, andmay be skipped by readers who are familiar with this concept, and by readers whodo not insist on understanding all details of this paper.

Garside groups were introduced by Dehornoy and Paris [6], and later in amore general form by Dehornoy [5]. We treat the latter, more general case. Allunproved assertions, as well as most of the proved ones, are from [6].

3.1. Garside monoids and groups

Let M be a monoid with cancellation. x ∈M is an atom if x �= 1, and x = ab fora, b ∈M implies a = 1 or b = 1. M is atomic if M is generated by its atoms, andfor each a ∈M , the maximum number of atoms in an expression of a as a productof atoms, denoted ‖a‖, exists. It follows that ‖ab‖ ≥ ‖a‖ + ‖b‖ for all a, b ∈ M .


In particular, as 1 = 1 · 1, we have that ‖1‖ ≥ ‖1‖ + ‖1‖, and thus ‖1‖ = 0. Fora �= 1, ‖a‖ > 0.

Let M be an atomic monoid. For a, b ∈ M , a is a left divisor of b if there isc ∈ M such that ac = b. Similarly, a is a right divisor of b if there is c ∈ M suchthat ca = b. a ∈M is a Garside element of M if its left divisors and right divisorscoincide, and include all atoms of M .

M is a Garside monoid if it is atomic, has a Garside element, and for alla, b ∈M , a greatest common divisor a ∧ b and a least common multiple a ∨ b of aand b exist in M , both with respect to left divisibility.

For a, b ∈M , the complement a \ b is the unique c ∈M such that ac = a∨ b.The closure of the set of atoms under the operations of complement and leastcommon multiple is the set S of simple elements ofM . The least common multipleof all elements of S, if it exists (e.g., if M is finitely generated), is called thefundamental element of M and denoted δ. δ, if it exists, is the least Garsideelement of M .

G is a Garside group if it is the group of fractions of a Garside monoidM . Inthis case, the elements ofM are called the positive elements of G. In the remainderof this section, M is a Garside group with a fundamental element δ, and G is theGarside group of fractions of M .

3.2. Greedy normal form

For x ∈ M with x �= 1, the simple element δ ∧ x �= 1. Define ∂(x) = (δ ∧ x)−1x.Then ∂(x) ∈M , and as x = (δ ∧ x)∂(x), ‖x‖ ≥ ‖δ ∧ x‖+ ‖∂(x)‖ > ‖∂(x)‖. Definesimple elements s1, s2, . . . , as follows. Set x1 = x, and for each i = 1, . . . , r, letsi = δ ∧ xi, and xi+1 = ∂(xi). ‖x‖ = ‖x1‖ > ‖x2‖ > · · · ≥ 0, and thus there is aminimal n such that xn+1 = 1. x = s1 · · · sn. Let k ≥ 0 be maximal with si = δ,and define pi = sk+i, i = 1, . . . , r, r = n− k. The expression

x = δkp1 · · · pris called the greedy normal form of x.

Consider now x ∈ G\M . If x = δks and s ∈M , then k < 0. Take the maximalinteger k such that x = δks for some s ∈ M . Fix such s, and let δ0p1 · · · pr =p1 · · · pr be the greedy normal form of s. The greedy normal form of x is thenagain defined to be δkp1 · · · pr.

By the construction, we have that pi+1 ∧ p−1i δ = (pi+1 · · · pr ∧ δ) ∧ p−1i δ =pi+1 · · · pr ∧ (δ ∧ p−1i δ) = xi+1 ∧ p−1i δ = 1 for all i = 1, . . . , r− 1, and that pr �= 1.We say in such cases that the sequence p1, . . . , pr is left-weighted.

3.3. Rational normal form

Following Thurston [7, Chapter 9], Dehornoy and Paris define the rational normalform2 of an element x ∈ G. To this end, we need the following.

Theorem 7 (Dehornoy-Paris [6]). For each x ∈ G, there is a unique pair (u, v) inM ×M such that x = u−1v and u ∧ v = 1.

2Also called mixed or symmetric normal form.


Let x ∈ G, and let u, v ∈ M be as in Theorem 7. Let s1 · · · sk, p1 · · · pl bethe greedy normal form of u, v, respectively. The rational normal form of x is theexpression

x = (s1 · · · sk)−1(p1 · · · pl).All si, pj are simple, s1 ∧ p1 = 1, and the sequences s1, . . . , sk and p1, . . . , pl areboth left-weighted. (The special cases where k = 0 or l = 0 are also allowed.)

For each a ∈ G, define τ(a) = aδ = δ−1aδ. τ is an inner automorphism of G,and its nth iterate at a is τn(a) = aδn

. τ maps simple elements to simple elements:For each simple s, let p be such that sp = δ. Then p is simple, and thus there is asimple q with pq = δ. Then

sδ = spq = δq,

and thus sδ = q is simple. In particular,M is invariant under τ . Any automorphismof G mapping positive elements to positive elements, maps atoms to atoms. Itfollows that τ is a permutation of the atoms of M .

One can obtain the rational normal form from the greedy normal form. Tosee this, we use the following.

Lemma 8. If s, p are simple and sp is left-weighted, then so are sδpδ and sδ−1pδ−1.

Proof. If ac = b are all positive, then aδ±1cδ±1

= (ac)δ±1

= bδ±1, and cδ

±1 ∈ M .Thus, τ±1 both map left divisors to left divisors, and therefore

(a ∧ b)δ±1 = aδ±1 ∧ bδ±1

for all a, b ∈M . Now, assume that sp is left-weighted. Then

(sδ±1)−1δ ∧ pδ±1 = (s−1δ)δ

±1 ∧ pδ±1 = (s−1δ ∧ p)δ±1 = 1δ±1

= 1,

showing that sδ±1pδ±1

is left-weighted. �

Proposition 9. If s, p are simple and sp is left-weighted, then so are

((pδk

)−1δ)((sδk+1

)−1δ),

for all integer k.

Proof. Assume that sp is left-weighted. Then so is (p−1δ)((sδ)−1δ):

(p−1δ)−1δ ∧ ((sδ)−1δ) = pδ ∧ (s−1δ)δ = (p ∧ (s−1δ))δ = 1δ = 1.

By Lemma 8, ((pδk

)−1δ)((sδk+1

)−1δ) = ((p−1δ)((sδ)−1δ))δk

is also left-weighted.�

Let δkp1 · · · pr be the greedy normal form of x. Consider three possible cases.

Case 1: k ≥ 0. Then δkp1 · · · pr is already a rational normal form (with a trivialnegative part).


Case 2: k = −m < 0 and m ≥ r. By definition, δ−na = aδn

δ−n for all a and all n.Using this, we have that

δ−mp1 · · · pr = δ−1pδm−11 · δ−1pδm−2

2 · · · · · δ−1pδm−r

r · δ−(m−r)

=(δm−r · (pδm−r

r )−1δ · · · · · (pδm−22 )−1δ · (pδm−1

1 )−1δ)−1

.

By Proposition 9, the last inverted expression is left-weighted, and thus we havea rational form, with a trivial positive part.

Case 3: k = −m < 0 and m < r. In the same manner, we have that

δ−mp1 · · · pr = δ−1pδm−11 · δ−1pδm−2

2 · · · · · δ−1pm · pm+1 · · · · · pr=

(p−1m δ · · · · · (pδm−2

2 )−1δ · (pδm−11 )−1δ

)−1(pm+1 · · · · · pr),

By Proposition 9, each of the bracketed expressions is left-weighted. Thus, thisexpression is in rational normal form.

4. Several length functions on Garside groups

Let M be a Garside monoid with fundamental element δ, and G be its group ofquotients.

Assumption 10. We assume that for each simple s ∈ M , the minimal length �(s)of an expression of s as a product of atoms can be efficiently computed.

There is always an algorithm for computing �(s): Enumerate all words oflength 1, 2, 3, . . . , until one equal to s is found. The running time is bounded byk(a) ≤ k‖a‖, where k is the number of atoms. But this is in general infeasible.When Assumption 10 fails, one may use in applications an estimation of � insteadof the true function.

Fortunately, in the specific monoids in which we are interested, all relationsare length-preserving, and thus �(s) is just the length of any expression of s as aproduct of atoms. Thus, Assumption 10 is true in our applications.

Example 11 (Artin’s presentation of BN ). Consider the monoid B+N generated byσ1, . . . , σN−1, subject to the relations

σiσi+1σi = σi+1σiσi+1;σiσj = σjσi when |i− j| > 1.

The quotient group of this monoid is the braid group BN on N strings. B+N is aGarside monoid with atoms σ1, . . . , σN−1, and fundamental element

δ = (σ1 · · ·σN−1)(σ1 · · ·σN−2) · · · (σ1σ2)σ1.The positive elements of BN are the words in σ1, . . . , σN−1 not involving inversesof generators. As the relations are length preserving, all expressions of a positive


element as a product of atoms have the same length. Thus, for a ∈M , ‖a‖ is thelength of a (any) presentation of a.

Elements of BN can be identified with braids having N strings, where eachgenerator σi performs a half-twist on the ith and i + 1st strings. This way, δ isa half-twist of the full set of strings. The simple elements correspond to positivebraids in which any two strings cross at most once. A simple element is describeduniquely by the permutation it induces on the strings, and every permutation ofthe N strings corresponds to a simple element.

Example 12 (BKL presentation of BN ). Generalizing the geometric interpretationin Example 11 to allow half-twists of the ith and the jth string for arbitrary i, j,Birman, Ko, and Lee [3] introduced the following presentation of the braid groupBN . The monoid BKL+N is generated by at,s, 1 ≤ s < t ≤ N , subject to therelations

at,sar,q = ar,qat,s if (t− r)(t − q)(s− r)(s− q) > 0;at,sas,r = at,rat,s = as,rat,r if t > s > r.

Also here, the relations are length preserving, and thus the norm is equal to thenumber of atoms in any expression of the element.

This monoid also has the braid group BN as its quotient group. In terms ofArtin’s presentation (Example 11), the Birman-Ko-Lee (BKL) generators can beexpressed by

at,s = (σt−1 · · ·σs+1)σs(σ−1s+1 · · ·σ−1t−1).

BKL+n is a Garside monoid with fundamental element

δ = an,n−1an−1,n−2 · · · a2,1.Here too, a simple element is described uniquely by the permutation it induceson the strings. However, not every permutation of the n strings corresponds to asimple element.

Definition 13. Let M be a Garside monoid with Garside group G, and let x ∈ G.

1. �(x), the minimal length of x, is the minimal length of an expression of x asa product of elements of A±1, where A is the set of atoms of M .

2. �G(x), the greedy length of an x, is the sum of the minimal lengths of allsimple elements (including the inverted ones) in the greedy normal form ofx. Similarly:

3. �R(x), the rational length of x, is the sum of the minimal lengths of all simpleelements (including the inverted ones) in the rational normal form of x.

Specifically, if the greedy normal form of x is δks1 · · · sr, then �G(x) =k · �(δ) + �(s1) + · · · + �(sr), and if the rational normal form of length of x is(s1 . . . sk)−1p1 . . . pl, then �R(x) = �(s1) + · · ·+ �(sk) + �(p1) + · · ·+ �(pl).

Proposition 14. For each a ∈M , �(aδ) = �(a).


Proof. Let n = �(a), and a = a1 · · · an with a1, . . . , an atoms. Then aδ = aδ1 · · · aδ

n.As conjugation by δ moves atoms to atoms, �(aδ) ≤ n = �(a). Similarly, if m =

�(aδ) and aδ = b1 · · · bm with b1, . . . , bm atoms, then a = aδδ−1

= bδ−11 · · · bδ−1m , and

as conjugation by δ moves atoms to atoms, �(a) ≤ m = �(aδ). �

The presentation in the previous section of the rational normal form in termsof the greedy normal form gives the following.

Corollary 15. The rational length of an element with greedy normal form δ−ms1· · · sr, where 0 < m ≤ r, is

�(s−11 δ) + · · ·+ �(s−1m δ) + �(sm+1) + · · ·+ �(sr),

and similarly for the cases where m ≤ 0 or 0 < r < m.

Corollary 16. If the relations of M are length-preserving, then the rational lengthof an element with greedy normal form δks1 · · · sr can be obtained by removing2∑min(r,k)

i=1 �(si) from its greedy normal length.

Proof. If the relations ofM are length-preserving, we have that �(ab) = �(a)+�(b)for all a, b ∈ M , and thus for simple s, �(δ) = �(s) + �(s−1δ), that is, �(s−1δ) =�(δ)− �(s). �

This shows, in particular, that the length function considered in [9, 10] in thecase of the Artin presentation of BN is in fact the rational length for the Artinpresentation of BN . This was first pointed out to us by Dehornoy.

4.1. Quasi-geodesics in Garside groups

Even when the relations are length-preserving, it is generally not the case thatan efficient algorithm for computing the minimal length �(x) is available. Even ifthe monoid relations are length-preserving, finding �(x) for x not in the monoid(nor in its inverse) may be a difficult task. Indeed, assuming P �= NP , there is nopolynomial-time algorithm computing �(x) with respect to the Artin presentationof BN , for arbitrary N and x ∈ BN [15]. Fortunately, in Garside groups �(x) canbe approximated. For simplicity, we treat the case of length-preserving relations,so that � is easy to compute on positive elements.

Theorem 17. Let M be a Garside monoid with length preserving relations andfundamental element δ, and let G be its fractions group. For each x ∈ G:1. If x ∈M , then �G(x) = �R(x) = �(x).2. If x ∈M−1, then �R(x) = �(x).3. �(x) ≤ �R(x) ≤ �G(x) ≤ (2�(δ)− 1)�(x).4. �R(x) ≤ (�(δ)− 1)�(x).

Moreover, these bounds in (3) cannot be improved.

Proof. (1) For x ∈ M , each normal form gives some positive presentation of x,and thus the corresponding length is the same as the minimal length.


(2) Fix x ∈ M−1. Then �R(x) = �R(x−1), and by (1), �R(x−1) = �(x−1) =�(x).

(3) The first inequality is clear. The second follows from Corollary 16. Weprove the third. Let

x = aε11 · · ·aεm

m (1)with m = �(x), a1, . . . , am atoms, and ε1, . . . , εm ∈ {1,−1}. For each atom a,let a be the simple element such that aa = δ. Then a−1 = δ−1a. Rewrite eachnegative atom in the equation 1 in this form, and move all occurrences of δ−1 tothe left, using the relation aδ−1 = δ−1aδ−1 . Let n = |{i : εi = −1}|. We obtain apresentation

x = δ−nb1 · · · bm,with each bi being (up to an application of τ an integer number of times, whichpreserves length by Proposition 14) ai if εi = 1, and ai otherwise. In particular,�(bi) = 1 if εi = 1, and �(ai) = �(δ)− 1 otherwise.

Let δks1 · · · sj be the left-weighted form of b1 · · · bm. Then the greedy nor-mal form of x is δ−n+ks1 · · · sj , which cannot be longer than δ−nδks1 · · · sj . Asexpressions of positive elements all have the same length, the length of δks1 · · · sjis exactly that of b1 · · · bm. Thus,

�G(x) ≤ n�(δ) + �(b1 · · · bm) = n�(δ) + �(b1 · · · bm)= n�(δ) + n(�(δ)− 1) + (m− n)= n(2�(δ)− 2) +m ≤ (2�(δ)− 1)m,

as n ≤ m.3

(4) This can be proved as in the proof of (3). Alternatively, one can useCharney’s Theorem [4], extended to general Garside groups by Dehornoy and Paris[6], that the number of simple elements in the rational normal form is minimalamongst presentations of x as a product of simple elements (possibly inverted):If x ∈ M±1, we can use (1) or (2) and there is nothing to prove. Otherwise, letx = aε1

1 · · · aεmm be a minimal presentation of x. In particular each aε1

i is a (possiblyinversed) simple element. Thus, the number n of simple elements in the rationalform of x is at most m. As x /∈M±1, no simple element in the rational form of xis δ. It follows that �R(x) ≤ (�(δ)− 1)m.

(1) shows that the lower bounds cannot be improved. To see that the upperbounds in (3) cannot be improved, consider �G(a−m) form positive and an atom a.

�

The following corollary of Theorem 17 is of special interest. In 1994, Bergersupplied an efficient method to compute a minimal length representative of anelement of B3, in terms of Artin generators [2]. We show that the same is true

3The step before last is added to emphasize that for random words, the upper bound is far from

being optimal. Indeed, in this case we have n ≈ m/2, which gives roughly half of the mentionedbound. There is an elbow room for improvements in the random case.


for the BKL presentation. Indeed, a minimal length representative for the BKLpresentation is supplied by the rational normal form.

Corollary 18. Consider the BKL presentation of B3. For each x ∈ B3, �R(x) =�(x).

Proof. Here, �(δ) = 2. By Theorem 17, �(x) ≤ �R(x) ≤ (�(δ)− 1)�(x) = �(x). �

Remark 19. Let M be a Garside monoid, and G be its fractions group. Dehornoyand Paris [6] proved that for each x ∈ G, there is a unique pair (u, v) ∈ M2,such that x = u−1v. It follows that for each braid x, the rational normal form ofx belongs to BN with the smallest possible N . In fact, if we define the supportof a braid as the set of strands that cross in every braid representative, then therational normal form of x detects its support. This is another reason why rationalnormal forms approximate the minimal length.

Remark 20. We do not know whether the upper bound in (4) of Theorem 17 canbe improved. At first it seems that for positive m and distinct non-commutingatoms a, b, �R(amb−m) = (�(δ) − 1)�(amb−m), but this is not the case: Considerσ22σ

−21 in the Artin presentation of BN . Its rational normal form in B3 (and thus

by Remark 19 in BN for all N) is (σ−11 σ−12 ) · (σ−12 σ−11 ) · (σ2σ1) · (σ1σ2), and thus�R(x) = 8 = 2�(x). But �(Δ)− 1 = 2 only when N = 3.

Theorem 17 shows that �R gives a better approximation than �G, and givesa theoretical motivation for the results described in [9]. Having both experimen-tal [9] and theoretical evidence for the superiority of �R over �G, we concentratehenceforth on the former.

4.2. Quasi-geodesics in embedded Garside groups

We need not stop here, and may consider, as in the case of BN , two distinctGarside structures of the same group, such that one of them embeds in the other.LetM1,M2 be Garside monoids with fundamental elements Δ, δ, respectively, suchthat each atom of M1 is also an atom of M2, and the group of fractions of M1

coincides with that of M2. Then we may take a length in one Garside structure asan estimation for the length in the other. We will denote the used structure by asuperscripted index. By Theorem 17,

�2R(x) ≤ (�2(δ)− 1)�2(x) ≤ (�2(δ)− 1)�1(x);�1R(x) ≤ (�1(Δ)− 1)�1(x).

Thus, if �2(δ) < �1(Δ), �2R(x) has a smaller approximation factor at its upperbound.

For the lower bound, let A2 be the set of atoms of M2, and set

α = max{�1(a) : a ∈ A2}.Then �1(x) ≤ α�2(x), and thus

�1(x) ≤ α�2(x) ≤ α�2R(x).


This gives the following.

Theorem 21. In the above notation,

1α�1(x) ≤ �2R(x) ≤ (�2(δ)− 1)�1(x). �

The advantage of Theorem 21 is that the distortion factors are symmetrizedaround the used length function �2R(x). Our main application is the following.

4.3. The case of the braid group

Consider the braid group as generated by the Artin monoid B+N as well as bythe BKL monoid BKL+N (Examples 11–12), and let Δ and δ be their respectivefundamental elements. Consider the minimal lengths �1 for the Artin structure,and �2 for the BKL structure of BN , respectively.

�1(Δ) = N(N − 1)/2, whereas �2(δ) = N − 1. For each atom at,s of BKL+N ,�1(at,s) ≤ 2(t− s− 1)+1 = 2(t− s)− 1. In particular, the maximum α of all theselengths satisfies

α ≤ 2N − 3.

By Theorem 21, we have that �2R, the length in BKL generators of the rationalnormal form in the BKL structure of BN , is quite symmetrically close to theminimal Artin length:

Corollary 22. For each x ∈ BN :

12N − 3

�1(x) ≤ �2R(x) ≤ (N − 2)�1(x). �

For comparison, measuring the minimal Artin length by working solely withthe Artin structure of BN , we only have (by Theorem 17):

�1(x) ≤ �1R(x) ≤ (�1(Δ)− 1)�1(x) =N2 −N − 2

2�1(x).

The gain may be viewed as follows: In the latter case, we have a constant (in N)error factor from below, and quadratic error from above. In Corollary 22, botherrors are linear, that is, the errors are symmetrized by dividing by O(N) terms.

Another matter, which we cannot prove at present, is that the lower boundin Corollary 22 seems to be a big underestimate in the generic case. It seems to usthat in the generic case, the lower bound factor should not be much smaller than1 (indeed, it may be greater than 1).

In summary, we have theoretical evidence suggesting that estimating theminimal length in Artin generators by using rational BKL normal form should bebetter than the same estimation using rational Artin normal form. We now turnto experimental results concerning the random case.


5. Experimental results

5.1. Initial experiments

For the Artin presentation, it is shown in [9] that the rational Artin length is muchbetter than greedy Artin length, at least with regards to solving random equationswith difficult parameters. Our initial experiments showed that this is also the casefor the BKL presentation: The rational BKL length is better than greedy BKLlength.

In the initial phase of this project, we have compared various length functionsinduced by various alternative ways of measuring lengths of elements, and foundout that only the rational BKL length outperforms the rational Artin length whenthe problem’s parameters are getting difficult. The remainder of this report istherefore dedicated to the comparison of the these two leading candidates.

5.2. A detailed comparison

We adopt the basic framework of [1, 10, 9]: The equations are in a finitely generatedgroup G = 〈a1, . . . , ang〉 ≤ Bns, where ns denotes the number of strings andng denotes the number of generators of G. Each generator ai is a word in Bns

obtained by multiplyingwl (word length) independent uniformly random elementsof {σ1, . . . , σns−1}±1. In G, we build a sentence X of length sl (sentence length):

X = a1a2 · · · asl

(For the while, we restrict sl ≤ ng.) Some of the ai-s may be equal, but we didnot force that intentionally.

We begin with a description of a test suitable for groups G which are closeto being free. For each i ∈ {1, . . . ,ng} and each ε ∈ {1,−1}, we give the generatoraεi the score

�(a−εi X),

sort the generators according to their scores (position 1 is for the shortest length),and reorder each block of identical scores by applying a random permutation. Wethen keep in a histogram the position of a1. We do one such computation for eachsample of G and X .

While a1a2 · · · asl is not the way a random sl sentence in G was defined, thisdoes not make the problem easier: We use each group G to produce only one suchsentence.

To partially compensate for the fact that G need not be free, we do the follow-ing. There could be several i ∈ {1, . . . ,ng} such that X = aia1 · · · ai−1ai+1 · · ·asl.Let COR denote the set of these ai, the correct first generators. After sorting allgenerators as above, instead of looking for the position of a1, we look at the lowestposition an element of COR attained.

Remark 23. A more precise, but infeasible, way to construct COR would be to findall shortest presentations of X as a product of elements from {a1, . . . , am}±1, andlet COR be the set of the first generators in these presentations. For the parameters


we have checked, we believe that this should not make a big difference. The resultsin Section 5.6 support this hypothesis.

We have also checked one set of cases where sl > ng. In these cases wedefined

X = ai1ai2 · · · aisl ,

where ij = (j−1 mod ng)+1 for j = 1, . . . , sl, and made the obvious adjustments.In summary, for each set of parameters (ns,wl,ng, sl) mentioned below, and

for � being either the rational Artin or the rational BKL length, we have repeatedthe following at least 1, 000 times: Choose a1, . . . , ang, compute X , compute COR,sort all generators aε

i according to the lengths �(a−εi X), find the lowest position

attained by an element of COR, and store this position number in the histogram.After dividing the numbers in the histogram by the numbers of samples made,

we obtain the distribution of the best position of a correct generator. In light ofthe intended application described in the first two sections, a natural measure tothe effectiveness of � is the graph of the accumulated probability, showing for eachx = 1, . . . , 2ng the probability that some correct generator attained a position≤ x.

The results of our experiments are divided into 4 sets such that in each set ofexperiments, only one parameter varies. This shows the effect of that parameteron the difficulty of the problem. The varying parameter takes 3 possible values,so we have 3 pairs (since there are two length functions) of graphs. Each pair ofgraphs has its own line style, so to allow plotting all 6 graphs on the same figure.

For all pairs, one of the graphs is always above or almost the same as theother. Fortunately, in all cases, it is the rational BKL length which is above therational Artin length, so there is no need to supply this information in the figure.

Finally, since the accumulated distributions all reach 1 for x = 2ng, thegraphs are more interesting for the smaller values of x. We therefore plot only thefirst 35 values of x.

5.3. When the sentence length varies

Fix ns = 64,wl = 8,ng = 128. Figure 1 shows the accumulated probabilities forsl ∈ {32, 64, 128}.5.4. When the word length varies

For ns = sl = 64,ng = 128, and wl ∈ {8, 16, 32}, we obtain the graphs inFigure 2. The problem gets easier when wl increases, since this way G gets closerto a free group (where the length approach is optimal). The remarkable observationis that the harder the problem becomes (by making wl smaller), the greater theimprovement of the rational BKL length over the rational Artin length becomes.

5.5. When the number of generators varies

Now set ns = sl = 64,wl = 8, and let ng ∈ {32, 64, 128}. The graphs appearin Figure 3. Here too, the more difficult the problem becomes (by increasing thenumber of generators), the greater the advantage of BKL over Artin is. Moreover,


SL=32

SL=64

SL=128

Y

X0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 5 10 15 20 25 30

Figure 1. When sl varies

WL=8

WL=16

WL=32

Y

X0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 5 10 15 20 25 30

Figure 2. When wl varies


NG=32

NG=64

NG=128

Y

X0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 5 10 15 20 25 30

Figure 3. When ng varies

NS=16

NS=32

NS=64

Y

X0.10

0.20

0.30

0.40

0.50

0.60

0.70

0.80

0.90

1.00

0 5 10 15 20 25 30

Figure 4. When ns varies


the graphs show that doubling ng has little influence on the performance of therational BKL length, whereas it seriously degrades the performance of the rationalArtin length.

5.6. When the number of strings varies

Finally, setwl = 8, sl = 64,ng = 128, and let ns ∈ {16, 32, 64}. Here, the problembecomes easier when we increase ns (Figure 4). This is not in accordance withearlier results in [9, 10], and is perhaps due to the fact that we allow any correctgenerator, whereas in the earlier works we only counted a1 a success. Indeed, themore strings there are, the greater the chances are that words of length 8 commute.On the other hand, the graphs show that while the BKL approach benefits a greatdeal when the number of strings is doubled, this is not quite so for the Artinapproach. This means that the improvement in success rates due to commutinggenerators is not substantial.

6. Concluding remarks and proposed future research

Memory-length algorithms give a powerful heuristic method to solve arbitraryequations in noncommutative groups, and consequently a variety of otherwise in-tractable problems. These algorithms rely on a good length function on the groupin question. In the past, greedy Artin length was used as a length function on thebraid group, and it was realized that rational Artin length gives better results.

In this paper, we suggested to use rational BKL length to measure the min-imal Artin length, and gave theoretical as well as experimental evidence for theadvantage of the new function over rational Artin length, at least when random-ization is modelled as in [1].

The main drawback in our estimations is that they give much larger lengthsthan the minimal length. Some interesting directions for possible improvementsare:1. As we have seen, the rational form can be computed from the greedy nor-

mal from by “removing” δ-s from the leading simple elements. We may bemore greedy, and remove the available δ-s from the (leftmost) longest simpleelements in the greedy normal form.4 This gives a new normal form in BN ,which has shorter length in terms of atoms. The resulting length functionmay be yet better than the one proposed here.

2. For each x and each proposal for a length function of x, we can take theminimum of the lengths of several elements whose minimal length is notsmaller than that of x, including: x, x−1, xδk

for each k = 1, . . . ,m − 1,where m is the minimal with δm central.

3. Since we use left-oriented normal forms in our estimations, we can also trythe corresponding right-oriented normal forms, and take the minimum.

4This was suggested to us by Uzi Vishne.


4. We can iterate conjugation by δ and inverses (and other operations which arenot increasing the minimal length) with shortening heuristics like Dehornoyhandle-reduction. In [13] this was done only to a very limited extent.

5. In [13], Dehornoy handle-reduction was applied to the greedy normal formto obtain an estimation of the minimal length. We conjecture that applyingDehornoy handle-reduction to the rational normal form would give betterestimations.

Acknowledgement

We thank Joan Birman and Dima Ruinskiy for their comments on earlier versionsof the paper. We also thank Patrick Dehornoy and Sang Jin Lee for informativediscussions concerning our notation, and the referees for their useful comments. Aspecial thanks is owed to Arkadius Kalka for useful discussions and suggestions.

References

[1] I. Anshel, M. Anshel and D. Goldfeld, An algebraic method for public-key cryptog-raphy, Math. Res. Lett. 6 (1999), 287–291.

[2] M. Berger, Minimum crossing numbers for 3-braids, Journal of Physics A: Mathe-matical and General 27 (1994), 6205–6213.

[3] J. Birman, K.H. Ko, J.S. Lee, A new approach to the word and conjugacy problemsin the braid groups, Advances in Mathematics 139 (1998), 322–353.

[4] R. Charney, Geodesic automation and growth functions for Artin groups of finitetype, Mathematische Annalen 301 (1995), 307–324.

[5] P. Dehornoy, Groupes de Garside, Annales Scientifiques de l’Ecole Normale Supe-rieure 35 (2002), 267–306.

[6] P. Dehornoy and L. Paris, Gaussian groups and Garside groups, two generalisationsof Artin groups, Proceedings of the London Mathematical Society 79 (1999), 569–604.

[7] D. Epstein, J. Cannon, D. Holt, S. Levy, M. Paterson, and W. Thurston, WordProcessing in Groups, Jones and Bartlett Publishers, Boston: 1992.

[8] D. Garber, Braid group cryptography,www.ims.nus.edu.sg/Programs/braids/files/david.pdf

[9] D. Garber, S. Kaplan, M. Teicher, B. Tsaban, and U. Vishne, Length-based conjugacysearch in the Braid group, Contemporary Mathematics 418 (2006), 75–87.

[10] D. Garber, S. Kaplan, M. Teicher, B. Tsaban, and U. Vishne, Probabilistic solutionsof equations in the braid group, Advances in Applied Mathematics 35 (2005), 323–334.

[11] J. Hughes and A. Tannenbaum, Length-based attacks for certain group based en-cryption rewriting systems, Workshop SECI02 Securite de la Communication surInternet, September 2002.

[12] K.H. Ko, S.J. Lee, J.H. Cheon, J.W. Han, S.J. Kang and C.S. Park, New Public-key Cryptosystem using Braid Groups, CRYPTO 2000, Lecture Notes in ComputerScience 1880 (2000), 166–183.


[13] A. Myasnikov, V. Shpilrain, and A. Ushakov, A practical attack on some braid groupbased cryptographic protocols, in: CRYPTO 2005, Lecture Notes in Computer Science3621 (2005), 86–96.

[14] A. Myasnikov, V. Shpilrain, and A. Ushakov, Group-based cryptography, AdvancedCourses in Mathematics – CRM Barcelona, Birkhauser, 2008.

[15] M. Paterson and A. Razborov, The set of minimal braids is co-NP-complete, Journalof Algorithms 12 (1991), 393–408.

[16] D. Ruinskiy, A. Shamir, and B. Tsaban, Length-based cryptanalysis: The case ofThompson’s Group, Journal of Mathematical Cryptology 1 (2007), 359–372.

Martin HockDepartment of Computer ScienceUniversity of WisconsinMadison, WI 53706, USAe-mail: [email protected]

Boaz TsabanDepartment of MathematicsBar-Ilan UniversityRamat-Gan 52900, Israel

and

Department of MathematicsWeizmann Institute of ScienceRehovot 76100, Israele-mail: [email protected]: http://www.cs.biu.ac.il/~tsaban



An Application of Word Combinatorics toDecision Problems in Group Theory

Arye Juhasz

Abstract. In this work we develop a graph theoretical test on graphs corre-sponding to subgroups of one-relator groups with small cancellation conditionwhich, if successful, implies that the subgroup under consideration has solv-able membership problem with a simple solution. The proof of the solvabilityof the membership problem relies on word combinatorics in an essential way.

Mathematics Subject Classification (2000). Primary 20E06; Secondary 20F05,20F06.

Keywords. One-relator groups, small cancellation conditions, membershipproblem, Whithead graphs.

Introduction

In his seminal paper [M], Wilhelm Magnus solved the word problem for one relatorgroups. Briefly, he showed that one-relator groups which are generated by at leasttwo elements have the structure of HNN-extensions with Magnus subgroups asspecial subgroups (Magnus subgroups are groups generated by proper subsets ofthe given generators of the group). Then he solved the membership problem forMagnus subgroups. From this one easily gets a solution of the word problem. Recallthat the membership problem (M.P.) asks for an algorithm to decide whether agiven element of the group belongs to a given subgroup. In spite of the long periodpassed from the occurrence of [M], very few new results are in the literature aboutthe membership problem in one-relator groups. This problem is not known to besolvable even in hyperbolic one-relator groups.

In the present work we exhibit a class of subgroups in one-relator groups withsmall cancellation, which are not Magnus subgroups and for which the membershipproblem is solvable. Whether a given subgroup belongs to this class is determinedby a graph attached to the subgroup. Thus, let G be a one-relator group givenby finite presentation P = 〈X |R〉, where R is the symmetric closure of a singlecyclically reduced word R. Let F (X) be the free group freely generated by X . Let

172 A. Juhasz

H be a subgroup of F (X) and let H be its image in G. Recall that the Whiteheadgraph of a word W in F (X) is, by definition, the graph on 2|X | vertices whichare labeled by elements of X ∪ X−1 in which two vertices, say v1, with label a1and v2 with label a2 (a1, a2 ∈ X ∪X−1), are connected by an edge if a−11 a2 is asubword of W or of W−1 [L-S]. We shall denote by Wh(W ) the graph obtainedfrom the Whitehead graph by identifying edges with the same endpoints. If T isa set of words, we denote by Wh(T ) the graph with vertices X ∪ X−1 and twovertices u and v are connected by an edge if and only if u−1v or v−1u is a subwordof a word of T . Our Main Theorem loosely says that if Wh(H) does not contain alarge portion of Wh(R), then H has solvable membership problem.

Denote by E(W ) the set of edges of Wh(W ) and similarly let E(T ) the set ofedges of Wh(T ). Denote by R the cyclic word corresponding to R.

Main Theorem. Let P = 〈X |R〉 be a one-relator presentation of a one-relatorgroup G with |R| ≥ 5 which satisfies the small cancellation conditions C ′(1/5)and T (4). Let F = F (X) and let θ : F → G be the canonical homomorphism. LetY ⊆ F be a finite subset of F , let H be the subgroup of F generated by Y and letH = θ(H). Let Z = E(R) and let K = E(H). If

|Z ∩K| < |Z| − 3 (∗)then H has solvable Membership Problem in G.

Two remarks are in order here. First, with the cost of more work on the sideof small cancellation theory, we can replace the metric condition C ′(1/5) by thecombinatorial condition C(6). Next, when checking condition (∗) we do not haveto check each element of H when forming K, because there is a standard wayby folding edges with a common endpoint and the same label in the graph whichcorresponds to H (see [L-S], p. 118) to construct a graph from which K can beeasily read off.

Example 0.1. Let F = 〈a, b, c, d| − 〉, R = P 2U1U2, where

P = ab−2a2c3d−1acb−1d−2a, U1 = d−1a14d−1b−1, U2 = c2b−1d−1cb−1.

Let H = 〈U1, U2〉. ThenH := θ(H) has solvable membership problem, by the MainTheorem. To see this, consider first the Whitehead graphWh(R) of the cyclic wordR. It is depicted in Fig. 1, where a denotes a−1 and similarly for b, c and d.

Thus,Wh(R) is a 4-regular graph with8 · 42

= 16 edges. Next, we consider the

Whitehead graphs of the elements of H . Since U1 and U2 have length greater thantwo and since the only cancellation in the products Uεi

i Uεj

j , where εi, εj ∈ {1,−1},i, j ∈ {1, 2} and i = j ⇒ εi = εj are in U1U

−12 , in which case only one letter is

cancelled out (b), hence the Whitehead graph of every element in H is contained inthe union of the Whitehead graphs of Ui and Uεi

i Uεj

j , εi, εj ∈ {1,−1}, i, j ∈ {1, 2}.These are marked in Fig. 1. by a tilde. Thus, we see that K ≤ 8. But |Wh(R)| = 16and 8 < 16− 3 = 13. Consequently, if R satisfies C(6)&T (4) then H has solvable

An Application of Word Combinatorics to Decision Problems 173

Figure 1

membership problem in G, by the Main Theorem (see remark above). But sinceWh(R) contains no closed curves of length less than or equal three, it follows thatR satisfies T (4) and it is easy to check that it satisfies the condition C(6) as well.Thus, H has solvable membership problem in G, by the Main Theorem and thefirst remark following the theorem.

We finish this introduction with a few remarks on the method of proof. Thiswork extends the methods and results of [J2], which we recall in Section 1 andthe beginning of Section 2. We use small cancellation theory. A central ingredi-ent of the theory is Greendlinger’s Lemma, which guaranties the existence of atleast two Greendlinger regions in every van Kampen diagram M which has atleast two regions. (For definitions of van Kampen diagrams and regions see 1.1.)These are regions with the property that their boundary has a large common por-tion with the boundary of M . Since the label of the boundary ∂M of M is aconsequence of R and for every consequence there is such a diagram, this meansin algebraic terms that every consequence of R contains a large portion of oneof the defining relations. (See [L-S, Ch. V].) If the small cancellation conditionis strong enough, this result alone is enough to solve the word problem. How-ever, the solution of our problem requires a more precise information than just alarge portion of the defining relator. In the present work we extend Greendlinger’sLemma in order to get the more precise information by describing the relationshipbetween Whitehead graphs of defining relators and their consequences, using mas-sive word combinatorics. The appearance of word combinatorics in the context ofsmall cancellation theory is quite natural: for example, there are standard results,like Lyndon-Schutzenberger’s Lemma for periodic words, which guarantees thatlong subwords which occur more than once in the word, occur in a very specialconfigurations, which can be avoided by an appropriate small cancellation theory.

174 A. Juhasz

Yet, the application of word combinatorics in the present work is of a differentnature: we apply word combinatorics in order to improve Greendlinger’s Lemmafor the one-relator case. We exemplify this by the proof of Magnus’ Freiheitssatz forone-relator free products with several components; G = 〈G1 ∗ · · · ∗Gm |R〉, m ≥ 3as given in [J2]. (This problem is similar to ours, but much simpler.) We have toshow that every consequence of R contains a letter from each Gi, i = 1, . . . ,m.Let C be a consequence of R and let M be a van Kampen diagram with C as aboundary label. (See 1.1.) It follows by standard small cancellation theory thatunder the condition C ′(1/5)&T (4) a diagram M has a Greendlinger region Dwhich either has only one neighbouring region E in M or it has two neighboursEr and E which have common boundary paths with D, having labels P1 andP2, respectively. (See Fig. 8(a) and Fig. 9, respectively.) Suppose for simplicitythe first and let P be the label of the common boundary path ∂E ∩ ∂D of Eand D and let Q be the label of the common boundary path of D and M . ThusR1 := QP is a boundary label of D, hence a cyclic conjugate of Rε, ε ∈ {1,−1}.We claim that every letter in P necessarily occurs in Q. Now P is a common labelof ∂D and ∂E. Therefore, P occurs as a subword of R1 and also as a subword ofa boundary label R2 of E. But since R is the only defining relator of G, R1 andR2 are cyclic conjugates of R±1. Therefore, in addition to the above-mentionedoccurrence of P in R1, P ε, ε ∈ {1,−1}, has another occurrence in R1, which wedenote by P ′; it comes from the occurrence of P as a subword of R2 which is acyclic conjugate of R1. (These occurrences are different because we assume thatour diagrams are reduced.) Now, since R1 = QP and P ′ is a subword of the cyclicword R1, either P ′ is a subword of Q in which case we are done, or else P ′ overlapsnon-trivially with P . See Fig. 8 (b). In this case ε = 1 and we have the followingword equations: P ′ = XY , P = AX andQ = Y Q1. In particular,XY = AX(= P )hence by the well-known (and easy) result from word combinatorics we get thatP = (KL)αK, α ≥ 1 and Y = KL, for certain subwords K and L of P . But thenQ = Y Q1 = KLQ1 and since all the different letters occurring in P already occurin KL, we get that all the letters in P occur in Q, as required. Observe that weused here word combinatorics in order to shift the letters of P which occur insidethe diagram into Q, which occurs on the boundary of the diagram.

In the case when D has two neighbours Er and E it is not always true thatQ contains all the letters of P1 and P2. In fact we prove that

QQQr contains every letter of R, where Qr and Q are tails andhead of the labels of ∂Er ∩ ∂M and ∂E ∩ ∂M , respectively. (∗)

This requires the development of a rather complicated combinatorial machin-ery on words which we develop in Section 2 of the present paper. See Appendixfor full details. We describe it very briefly. First, observe that in the above wordequations we were interested not so much in their solution but rather the values ofthe function Supp : F → L, on the solutions where L is the set of all the subsetsof {1, . . . ,m} and for a word W in F, Supp(W ) is the set of the indices i for


which Gi contributes a letter in W . We rely very heavily on this observation whenproving (∗). In the present work we deal with Whitehead graphs hence we have tointroduce a new function σ : F → K, where K is the set of all the subsets of the setof words in F with length two. (This corresponds to the edges of the correspondingWhitehead graph.)

Finally, we point out that the above-mentioned improved version of Greend-linger’s Lemma holds true under the small cancellation condition C(6), with a fewknown exceptions, however, the corresponding word combinatorics is incomparablymuch more complicated.

The paper is organised as follows.In Section 1 we introduce the necessary preliminaries from small cancellationtheory.In Section 2 we develop the word combinatorics needed.In Section 3 we prove the Main Theorem.

1. Preliminaries on small cancellation theory

1.1. Diagrams

One of our main tools is van Kampen diagrams. These are planar complexes in-troduced by van Kampen in 1933, which give precise description of the way can-cellation carried out when forming the right-hand side of (1.2) below.

For basic results on diagrams see [L-S, Ch.V]. We recall here some of the basicdefinitions and results from [L-S, pp. 237–239 and pp. 274–276], for convenience.Let E2 denote the Euclidean plane. If S ⊆ E2 then ∂S will denote the boundary ofS, the topological closure of S will be denoted by S. A vertex is a point of E2. Anedge is a bounded subset of E2 homeomorphic to the open unit interval. A regionis a bounded set homeomorphic to the open unit disc. A map is a finite collectionof vertices, edges and regions which are pairwise disjoint and satisfy:(i) If e is an edge of M , there are vertices u and v (not necessarily distinct) in

M such that e = e ∪ {u} ∪ {v}.(ii) The boundary, ∂D, of each region D of M is connected and there is a set of

edges e1, . . . , en such that ∂D = e1 ∪ · · · ∪ en.A diagram over a group F is an oriented map M and a function Φ assigning toeach oriented edge e of M as a label an element Φ(e) of F such that if e is anoriented edge of M and e−1 the oppositely oriented edge, then Φ(e−1) = Φ(e)−1,and if μ = e1v1e2v2 · · · ek is a path in M then Φ(μ) = Φ(e1)Φ(e2) · · ·Φ(ek). Wedenote by ΦM the labelling function of M over F . If M is fixed we shall write Φfor ΦM . If M is a diagram then its underlying map is the map obtained from Mby ignoring the labels. For a closed path μ in M denote by h(μ) its initial vertexand by t(μ) its terminal vertex.

Now, the connection between diagrams and consequences of relators is givenby the following two theorems.

176 A. Juhasz

Theorem 1.1. [L-S, p. 237] If F is a free group and C1, . . . , Cn, n ≥ 0 is a se-quence of non-trivial elements of F (the conjugates of relators), there is a diagramM(C1, . . . , Cn) which satisfies each of the following:(i) if e is an edge of M then Φ(e) �= 1;(ii) M is connected and simply connected, with a distinguished vertex 0 on ∂M ;(iii) there is boundary cycle e1 · · · · · et of M beginning at 0 such that the product

Φ(e1) · · · · · Φ(et) is reduced as written and in the free group we have

Φ(e1) · · · · ·Φ(et) = C1 · · · · · Cn (1.1)

Diagrams satisfying conditions (i), (ii) and (iii), are called van Kampen dia-grams.

The next theorem provides a converse of Theorem 1.1.

Theorem 1.2. (Normal Subgroup Theorem) [L-S, p. 239] Let M be a connected,simply connected diagram with regions D1, . . . , Dm. Let α be a boundary cycle ofM beginning at a vertex v0 ∈ ∂M and let W = Φ(α). Then there exist labels Ri ofDi and elements fi of F for 1 ≤ i ≤ m such that

W = f−11 R1f1 · · · · · f−1m Rmfm (1.2)

Example 1.3. (See [Jo, p. 214].) Consider the (by now familiar) group

G =⟨x, y |x2yxy3, y2xyx3

⟩.

The non-obvious fact that x7 = e in G is embodied in the following diagram, whereW = x7.

Figure 2

Definitions 1.4. Let M be a diagram over F .(a) Two regions D1 and D2 in M are neighbours if ∂D1 ∩ ∂D2 �= ∅. They are

proper neighbours if ∂D1 ∩ ∂D2 contains a non-empty edge. For example inFig. 2. D1 and D2 are proper neighbours, while D2 and D3 are neighbours,but not proper neighbours.


(b) A region D is a boundary region if ∂D ∩ ∂M �= ∅. A region D is a properboundary region if ∂D∩∂M contains a non-empty edge. A region ofM whichis not a boundary region is an inner region. For example in Fig. 2. D2 is aninner region, D3 is a proper boundary region, while D1 is a boundary region,but not proper.

(c) A neighbour E of a boundary region D is an inner-neighbour if E is an innerregion.

Definition 1.5. Let M be a connected, simply connected map. M is a simple one-layer map, if the dual map M∗, obtained from M by putting in each region Da vertex D∗ and connecting two vertices D∗1 and D∗2 by an edge if D1 and D2are proper neighbours, is a straight line. (See Fig. 3(b).) In particular, M hasconnected interior, every region is a boundary region, each region has at mosttwo proper neighbours and if M contains more than one region then M containsexactly two regions (D1 and Dr on Fig. 3(b)) which have exactly one neighboureach.M is a one-layer map if it is composed from simple one-layer maps and pathsin the way shown on Fig. 3(a).

(a)

(b)

Figure 3

If M is a simple one layer map consisting of regions D1, . . . , Dt which occurin this order from left to right or from right to left then we shall denote this by〈D1, . . . , Dt〉.

1.2. Diagrams with small cancellation conditions

We recall the following result from [L-S, p. 274].

Lemma 1.6. Let P = 〈X |R〉 be a presentation and letM be a van Kampen diagram,the boundary label of which is a consequence of R.

178 A. Juhasz

(a) If P satisfies the condition C(p) for p ≥ 3, then every inner region of M hasat least p proper neighbours.

(b) If P satisfies the condition T (q) for q ≥ 3, then every inner vertex of M hasvalency at least q.

By analogy, we shall say that a map satisfies the condition C(p) if every innerregion has at least p proper neighbours and a map satisfies the condition T (q) ifevery inner vertex has valency at least q. Thus if M is the underlying map of adiagram, the boundary label of which is a consequence of R and P satisfies thecondition C(p) and T (q) then M also satisfies these conditions.

We recall the main structure theorem for C ′(1/5) and T (4) maps from [J1],where it is proved in a more general setting. (Observe that conditionC ′(1/5)&T (4)implies condition W (6) in [J1].)

(a) (b)

Figure 4

Theorem 1.7 (Layer Decomposition). [J1] (See Fig. 4.) LetM be a simply connectedmap (diagram) with connected interior and let D0 be a region of M . Assume thatM satisfies the condition C(6)&T (4).Define St0(D0) = D0 and let Sti(D0) = Sti−1(D) ∪ Li(D0) for i ≥ 1, whereL0(D0) = {D0},

Li(D0) =⟨D in M\Sti−1(D0)

∣∣∂D ∩ ∂Sti−1(D0) �= ∅ ⟩.Let p be the smallest number such that Stp(D0) =M and assume that p > 0 (i.e.,M contains more than one region). Then each of the following holds:(a) Every regular submap of Sti+1(D0) containing Sti(D0) for 0 ≤ i ≤ p is

simply connected. (A submap is regular if every edge is on the boundary of aregion.)

(b) Every connected and simply connected submap of Li(D0) is a one-layer map.When D0 is fixed, we shall abbreviate Li(D0) by Li and call Λ(D0) = (L0, . . . ,Lp)a layer decomposition of M . We call D0 the center of the layer decomposition.


(c) For a region D ∈ Li, i ≥ 1 denote by A(D) the set of regions E in Li−1,which have a non-trivial common edge with D, denote by B(D) the set ofregions S in Li with ∂S ∩ ∂D �= ∅. Also, let a(D) = |A(D)|, b(D) = |B(D)|.Then a(D) ≤ 1 and b(D) ≤ 2. In other words, D has at most two neighboursin Li and at most one neighbour in Li−1.

(d) If v ∈ ∂Sti(D0) then v has valency at most three in Sti(D0).(e) For regions D,E in M with ∂D∩∂E �= ∅ we have that ∂D∩∂E is connected.

Remark 1.8. Let M be a connected simply connected map (diagram) with con-nected interior and let D be a region in M . Let Λ(D) be a layer decomposition ofM with center D. Suppose that D is a boundary region of M with a non-emptyedge on ∂M . (See Fig. 5.) Then it follows from the above theorem that L1(D) isnot annular, hence simply connected, though not necessarily with connected inte-rior. (See Fig. 5(a), where the interior of L1 is simply connected and connectedand see Fig. 5(b), where the interior of L1 is not connected.) But then due to thesimply connectedness of M , Li is simply connected for every i.

(a) (b)

Figure 5

Remark 1.9. We could start the construction of the layers not with a region D0,but with a vertex v and defining St0(v) = {v}, St1(v) is the submap consistingof {D

∣∣ v ∈ ∂D} and for k ≥ 2 define Stk(v) like in Theorem 1.7. Moreover, wecan define a layer decomposition relative to a path inM , under certain conditions.Thus, a center may be either an arbitrary vertex, or an arbitrary region, or a pathof a special type which we call a transversal. (See Definition 1.19.)

Remark 1.10. The notion of one-layer map is independent of any given layer de-composition. For example the map on Fig. 3(b) is a one-layer map, by definition.However, it also has a layer decomposition Λ with center D1 for which L1 consistsof D2 and D3. Another typical example is Mk on Fig. 3(a), which is a one-layermap. In its layer decomposition with center D0 every layer consists of a singleregion, hence certainly each Li is a one-layer map, however these layers Li are“perpendicular” to the natural one-layer structure of Mk.

180 A. Juhasz

In the next definition we introduce special subdiagrams and regions, theboundaries of which share a large portion with the boundary of M .

Definition 1.11. Let Λ(D0) be a layer decomposition of M with center D0 on theboundary of M such that if D0 is a region then it has a non-empty edge on ∂M .If Λ(D0) = (L0,L1, . . . ,Lp) then the closure of every connected component P ofthe interior of Lp is a peak. We say that P is a peak relative to D0. (See Fig. 4(a),where p = 2 and L2(D0) is a peak and See Fig. 4(b), where Lp(D0) is a peak.)By Theorem 1.7 every layer structure has a peak.Related to peaks is the following notion.

Definition 1.12. A boundary region D of M is a k-corner region for k = 1, 2 ifeach of the following holds:1) ∂D ∩ ∂M is connected and2) D has k inner proper neighbours.

Definition 1.13. Let μ be a boundary path of M , let P be a peak of M relativeto D0 and let D be a k-corner region with k ≥ 2. Say that μ contains P if∂P ∩ ∂M ⊆ μ. Similarly, μ contains D if ∂D ∩ ∂M ⊆ μ.

Lemma 1.14. Let D be a k-corner region in a diagram M with k ≤ 2, whichsatisfies the condition C ′(1/5). Let θ = ∂D ∩ ∂M and let η be the complement ofθ on ∂D. (Thus, vθuη is a boundary cycle of D with v and u endpoints of θ andη.) Then |Φ(θ)| > |Φ(η)|.Proof. Since k ≤ 2, the region D has at most two proper neighbours, hence Φ(η)

is the product of at most two pieces. Therefore |Φ(η)| < 2 · 15|∂D| < 1

2|∂D|. Since

|Φ(θ)| = |R| − |Φ(η)|, |Φ(θ)| > |R| − 12|R| = 1

2|R| > |Φ(η)|, i.e., |Φ(θ)| > |Φ(η)|,

as required.The lemma is proved. �

Lemma 1.15. Let M be a map with a layer decomposition Λ(D0) and let P be apeak relative to D0. Then P contains a k-corner region for some k ≤ 2.

Proof. Let P = 〈D1, . . . , Dk〉, k ≥ 1. If k ≥ 2 then its extremal regions D1 and Dk

are 2-corner regions because a(D1) ≤ 1 and b(D1) ≤ 1, due to being extremal andclearly ∂D1 ∩ ∂M is connected. Also, if a(Dk−1) = 0 then Dk−1 in Fig. 6(a) is a

(a) (b)

Figure 6


2-corner region. If P is a peak consisting of a single region E, then E is a 1-cornerregion due to Theorem 1.7.

The lemma is proved. �

The following lemma is an immediate consequence of Lemmas 1.14 and 1.15,hence we omit its proof.

Lemma 1.16. Let N be the normal closure of R in F . Let M be a diagram for aconsequence of R, which satisfies the condition C ′(1/5) and let μ be a boundarypath of M . Suppose that μ contains a peak relative to D0. Then Φ(μ) is not ashortest representative of the elements of Φ(μ)N .

Definition 1.17. LetM be a connected, simply connected diagram with no verticesof valency one. Let μ be a boundary path of M and let v ∈ μ be a vertex. Saythat v is a double point of μ if v is met more than once when traversing along μ.(Actually it may be a multiple point, but from our point of view all we need isthat it is not regular.)

Remark 1.18. It is well known (see [L-S, Ch.V]) that if μ has a double point thenμ encloses one or more connected components of the interior of M . Hence, if μcontains a double point v then μ contains a peak relative to v, due to Theorem 1.7and Remark 1.9.

An immediate consequence of the last remark and Lemma 1.16 is the following:

If Φ(μ) is a shortest representative of Φ(μ)N in Gthen μ contains no double points.

(∗)

We introduce now one of our main tools concerning diagrams, the transver-sals. Let Λ(D) be a layer decomposition of M and let ωi = ∂Li ∩ ∂Li+1 fori = 1, . . . , p− 1, see Fig. 7(a) showing ωi and ωi+1. Then due to Theorem 1.7 ωi

has the property that ∂E ∩ ωi is either a vertex or an edge, for every region E ofLi+1. (Observe that this is not true for regions in Li. If K is a region of Li then∂K ∩ωi may contain more than one edge.) It turns out that this property of ωi isresponsible for the existence of peaks in M . We develop this idea below in orderto produce peaks on specific places along ∂M .

1.3. Transversals in diagrams with Small Cancellation Conditions

Let Λ(D) be a layer decomposition of M , where D may be a single vertex. Anyvertex in M with valency at least three not in the last layer is on the commonboundary ωi of Li(D) and Li+1(D) for some i. Suppose that Li+1(D) contains avertex w1 ∈ ωi which has valency at least four in Li+1(D). (See Fig. 7(a).) Forexample, if Li+1(D) contains at least three regions, then, due to the C ′(1/5)&T (4)condition, it follows from Theorem 1.7(c) that Li+1(D) contains such a vertex w1.Let u be the initial vertex of ωi and let uθ1w1 be the subpath of ωi which startsat u and terminates at w1. Then, as observed above, due to Theorem 1.7(c) θ1satisfies condition (∗∗) below, with μ = θ1

182 A. Juhasz

(a) (b)

Figure 7

when traversing along μ, for every region E to the left of μ with∂E ∩ μ nontrivial, (i.e., contains an edge) we have that ∂E ∩ μ isconnected and ∂E ∩ μ is an edge (i.e., does not contain a vertexwith valency at least three in M).

(∗∗)

Since w1 has valency at least four in M , hence Li+1(D) contains a region Ewith ∂E∩ωi = {w1}. Let θ2 = ∂E∩∂F , where F is the right-hand side neighbourof E in Li+1(D). Then uθ1w1θ2w2 satisfies condition (∗∗) above, where w2 is theendpoint of θ2, different from w1. Assume that w2 is not a boundary vertex ofM .Then w2 ∈ ωi+1 and there are left most adjacent regions E1 and F1 in Li+2(D)which contain w2 on their boundary. Define θ3 = ∂E1 ∩ ∂F1 and keep definingθi and wi for i ≥ 3 until reaching ∂M . Let θ = uθ1w1θ2w2 . . . θev where v is aboundary vertex of M and wi are inner vertices of M for 1 ≤ i ≤ e − 1. Thenv ∈ ∂M and each initial segment of θ satisfies condition (∗∗). This leads us to thefollowing definition.

Definition 1.19. Let notation be as above. Let θ be a simple path in M withinitial and terminal vertices u and v, respectively. Suppose that u, v ∈ ∂M andθ∩∂M = ∅ (i.e., the open path θ does not intersect ∂M). Call θ a left transversal ifit satisfies condition (∗∗) above. If μ is a left transversal, then μ defines a submapMμ ofM by its boundary uμvζ−1 where ζ is the boundary path ofM which startsat u, terminates at v and is to the left of μ. (See Fig. 7(a).)

The proof of Theorem 1.7 (see [J1]) can be easily adopted to prove the nextproposition. We omit its proof here.

Proposition 1.20. Let μ be a left transversal in M . Then Mμ has a layer structurerelative to μ. In particular, Mμ has a peak relative to μ. More precisely, define


L0(μ) = μ. Let L1(μ) be the submap of Mμ consisting of those regions of Mμ, theboundary of which intersects μ. Then L1(μ) is connected and simply connected, andμ is a boundary path of L1(μ). Let μ′1 be its complement on ∂L1(μ). Then μ′1 =α1μ1α

−12 , where α1 and α2 are boundary paths of M and μ1 is an inner path of M .

Define L2(μ) = L1(μ1) and repeat this process until the last layer Lk(μ) is reached.(See Fig. 7(b).) Then L0(μ), . . . ,Lk(μ) is a layer structure of Mμ relative to μ.

Clearly, by the same method we may construct right transversals. Hence, ifLi+1(D) contains at least two vertices with valency at least four in Li+1(D) orone vertex with valency at least five in Li(D) and u, u′ are endpoints of ωi, sayu to the left of u′, then we may construct a left transversal μ with initial vertexu and a right transversal ν with initial vertex u′. (See Fig. 7(a).) It follows byan easy induction on the number of layers that Mμ ∩Mν = ∅. Now, due to theC ′(1/5)&T (4) condition and Theorem 1.7(c), if Li+1(D) contains at least fourregions which have a non-empty common edge with ωi, then Li+1(D) contains atleast two vertices with valency at least four. We summarize this in the followingproposition.

Proposition 1.21. Let M be a diagram, let D be either a boundary region of M suchthat ∂D∩∂M contains a non-empty edge or a boundary vertex, and let Λ(D) be alayer decomposition of M with center D. Suppose that Li+1(D), (i ≥ 1) containsat least four regions which have a non-empty edge on ωi := Li(D) ∩ Li+1(D) or avertex with valency at least five in Li+1(D). Then

(a) M contains a right transversal ν and a left transversal μ with Mμ ∩Mν = ∅such that ν starts at t(ωi) and μ starts at h(ωi). (See Fig. 7(a).)

(b) Let ζμ be the boundary path of M with initial vertex h(μ) and terminal vertext(μ) which is to the left of Mμ and similarly, let ζν be the boundary path ofM which starts at h(ν) and ends at t(ν) and is to the right of Mν . Theneach of ζμ and ζν contains a peak.

1.4. The Main Theorem for almost σ-complete presentations

We are now in a position to prove the Main Theorem under some additionalassumptions, defined below. The main result of Section 3, which is mostly wordcombinatorics is that this assumptions are satisfied under the conditions of thetheorem.

Definitions 1.22. Let R be a cyclically reduced word, let R be the symmetricclosure of R and let P = 〈X |R〉 be a one-relator presentation. Let M be anR-diagram with layer decomposition Λ and let P be a peak relative to Λ, withα = ∂P ∩ ∂M .

(a) For every word W denote by σ(W ) the set of all the subwords of W of lengthtwo.

(b) P is almost σ-complete if σ(Φ(α)

)⊇ Γ for some subset Γ of σ(R), with

|Γ| ≥ |σ(R)| − 3.

184 A. Juhasz

(c) M is almost σ-complete if every peak relative to every layer decompositionof M is almost σ-complete.

(d) P is almost σ-complete if every R-diagram M is almost σ-complete.

The next lemma shows the significance of Definition 1.22.

Lemma 1.23. Let notation be as in Definition 1.22. Let H be a subgroup of F (X)which satisfies condition (∗) of the Main Theorem. Let M be a van Kampen dia-gram for a consequence of R and let ν be a boundary path of M with V := Φ(ν) ∈H. If M is almost σ-complete then ν contains no peaks. In particular, ν containsno double points.

Proof. If ν contains no peaks then it cannot contains double points, due to Theo-rem 1.7. (See Remark 1.18.) Hence it is enough to show that ν contains no peaks.It follows from the assumption of the theorem that

if σ(V ) contains a non-empty subset Γ of σ(R) then |Γ| < |σ(R)|−3for every reduced word V in H.

(∗)

If P is a peak relative to a layer decomposition Λ and α = ∂P ∩ ∂M , then by theadditional assumption that P is almost σ-complete it follows that for some non-empty subset Γ of σ

(Φ(α)

)which is contained in σ(R), we have |Γ| ≥ |σ(R)| − 3.

Since Φ(α) contains a subword of R of length greater than |R|− 25|R| = 3

5|R| ≥ 3,

due to the C ′(1/5) and T (4) conditions and Lemma 1.14, hence σ(Φ(α)

)contains

a non-empty subset Γ of σ(R). But since α is a subpath of ν by assumption, thiscontradicts the fact that M is almost σ-complete, proving the lemma. �

In this subsection we prove the Main Theorem for almost σ-complete one-relator presentations with small cancellation condition C ′(1/5)&T (4).

Theorem 1.24. Let notation and assumptions be as in the Main Theorem. If P isalmost σ-complete then the result of the Main Theorem holds true.

Proof. We keep the notation of the Introduction and the Main Theorem. Also weassume P is almost σ-complete. We start with the remark that G has solvableword problem by [L-S, p.262]

Our proof is based on the following proposition.

Proposition 1.25. Let M be a connected, simply connected R-diagram. Let μ bea boundary path of M and let ν be its complement on ∂M . (Thus, uμvν−1 is aboundary cycle of M , where u and v are vertices.) Let Φ(μ) = U and let Φ(ν) = V .Let U = u1 · · ·up and let R = r1 · · · rq, reduced as written, ui, ri ∈ X ∪X−1. LetS =

{u±1i , r±1j , 1 ≤ i ≤ p, 1 ≤ j ≤ q

}∪{1} and for every natural number n let Sn

be the set of all the products of n elements from S. Suppose that U is a shortestrepresentative of θ(U) and V ∈ H. Then each of the following holds:(a) If η is a simple boundary path of M then Φ(η) ∈ S|η|.(b) |V | ≤ 9|U ||R| and V is a word on S, effectively computable from U and R.


Proof. Since U is a shortest representative of θ(U), μ contains no double points,by (∗) in Remark 1.18. Also, by Lemma 1.23, ν contains no peaks. Consequently,M has the form like the diagram in Fig. 3(a).

(a) If M1, . . . ,Mk are the connected components of Int(M) then the subpaths ζi

for i = 0, . . . , k, k+1, which are the connected components of M \k⋃

i=1

{M i

},

are subpaths of μ and ν. Hence, if e is an edge ofM then one of the followingholds for e:(i) e ⊆ ζi for some i, i = 0, . . . , k + 1;(ii) e is an edge of Mi, for some i, i = 1, . . . , k;We propose to show that in each of these cases Φ(e) ∈ S. This will clearlyprove part (a). Now, in case (i) Φ(e) = u±1j for some j, 1 ≤ j ≤ p, henceΦ(e) ∈ S. In case (ii) Φ(e) = r±1j for some j, 1 ≤ j ≤ q, hence in particularΦ(e) ∈ S, as required.

(b) Let Λ be the layer decomposition ofM with center u. Assume Λ has s layers.For i = 1, . . . , s− 1 let ωi = ∂Li ∩ ∂Li+1 and for i = 1, . . . , s let μi = ∂Li ∩μand νi = ∂Li ∩ ν. Then μ = μ1 · · ·μs, ν = ν1 · · · νs and by the definition oflayers we have

1 ≤ |μi| and 1 ≤ |νi|. (1.1)

Moreover, due to (1.1)

s ≤ |μ| and s ≤ |ν|. (1.2)

Suppose M has a layer Li+1 which contains at least four regions with a non-empty edge on ωi or a vertex with valency at least five in Li+1. Then byProposition 1.20,M has transversals α and β starting at the endpoints of ωi,respectively, such thatMα∩Mβ = ∅. But then it follows from Proposition 1.21thatMα andMβ contain disjoint peaks Pα and Pβ , respectively. Since uμvν−1

is a boundary cycle of M , this implies that either ∂Pα ∩ ∂M ⊆ μ or ∂Pβ ∩∂M ⊆ ν, whereMα is to the left ofMβ. This however violates the assumption,that μ contains no peaks. Therefore, every vertex of M has valency at mostfive and every layer Li of M contains at most four regions, which have acommon edge with ωi. We compute the number of regions in Li+1. Let abe the number of regions of Li+1 which have a common edge with ωi andlet b be the number of regions of Li+1 which have only a common vertexwith ωi. Then |Li+1| = a + b. Now, if ωi = w0θ1w1θ2 · · ·wrθrwr+1, wherewj vertices and θi edges, then r = a and each vertex contributes at mostd(w) − 3 = 1 to b, if w �= w0 and w �= wr and w0 and wr contribute at mostd(w) − 2 = 2. Consequently, b ≤ 1 · (r − 1) + 2 · 2 ≤ r + 3. Since r ≤ 3,|Li+1| = a+ b ≤ 3 + 3 + 3 = 9.

Hence, for every layer Li of Λ, we have |μi| ≤ 9|R| and |νi| ≤ 9|R|.Together with (1.1) we get

1 ≤ |μi| ≤ 9|R| and 1 ≤ |νi| ≤ 9|R|. (1.3)

186 A. Juhasz

But then by (1.2) and (1.3) we have s ≤ |μ| ≤ 9s|R| and s ≤ |ν| ≤ 9s|R|.Consequently, |ν| ≤ 9s|R| ≤ 9|μ||R| = 9|U ||R|.

Now, denote S =|ν|⋃i=0

Si. Then S is a set of words computable from S

since |ν| < c|μ| and c|μ| is given. Let S0 the subset of all the elements of Swhich are equal to U in G. Since G has solvable word problem due to thecondition C ′(1/5)&T (4), S0 is computable from S. Finally, let S1 be thesubset of S0 of the elements of H in S1. By assumption, S1 is non-empty,hence S1 can be constructed from S.

The proposition is proved. �

Completion of the proof. Let U be a reduced word in F , U �= 1 with θ(U) ∈ θ(H).Then there exists a reduced word V ∈ H with θ(UV −1) = 1. Therefore, by Theo-rem 1.1 there exists a van Kampen diagramM with boundary cycle uμvν−1 withΦ(μ) = U and Φ(ν) = V , where u and v are vertices. Since the word problem for Gis solvable, we may assume that U is a shortest element among the representativesof θ(U). Hence Proposition 1.25 applies. Applying Proposition 1.25 it follows that|V | ≤ 9|U ||R| and V can be written down effectively, knowing U . This solves theMembership Problem for H .

The theorem is proved. �

2. Word combinatorics

The aim of this section is to introduce the basic results in word combinatoricswhich enable us to show in Section 3 that P is almost σ-complete.

2.1. Words

Let F be a free group, freely generated by a set X and let W be a reduced wordin F . Denote by H(W ) the set of initial subwords of W and by T (W ) the set ofterminal subwords of W . Also, for a reduced non-empty word W we denote byh(W ) the first letter of W and by t(W ) the last letter of W . Denote by Supp(W )the set of all the elements of X which occur in W or W−1.

We have the following well-known result.

Lemma 2.1. Let A,B and C be reduced words such that AB and BC are reducedas written. If AB = BC then A = KL, C = LK and B = (KL)βK, β ≥ 0.

We introduce below the key notion of the work.

Definitions and notations.

(a) Let W1 and W2 be reduce words in F . W2 majorises W1 if Supp(W2) ⊇Supp(W1). In this case write W2 � W1. If W1 � W2 and also W1 � W3 weshall write W1 � W2,W3. Also, if Supp(W1) ∪ Supp(W2) ⊇ Supp(W3) weshall write W1,W2 �W3.


(b) For W1 and W2 in part (b) define W1 ∼W2 if W1 ≺W2 and W2 ≺W1. ThusW1 ∼W2 if and only if Supp(W1) = Supp(W2).

Clearly “∼” is an equivalence relation, which contains the equality ofelements in F .

The following lemma is immediate from the definition, hence its proof isomitted.

Lemma 2.2.

(a) If A is a subword of B then A ≺ B.(b) If A ≺ B then A±1 ≺ B±1.(c) If A ∼ B and A ≺ C then B ≺ C.(d) If A = P1 . . . Pm, reduced as written and Pi ∼ Q for i = 1, . . . ,m then A ∼ Q.(e) If A � P1, . . . , Pm then A �W (P1, . . . , Pm), for every wordW on P1, . . . , Pm.

Parts (a) and (b) of the following lemma are immediate corollaries of Lem-ma 2.1 and Lemma 2.2, hence we omit their proofs.

Lemma 2.3.

(a) Let A,B and C be as in Lemma 2.1 (a). Then B ≺ A ∼ C ∼ AB ∼ BC. Ifβ ≥ 1 then B ∼ A.

(b) If AB = KAC with AB and KAC, reduced as written then B � A, B � Cand K � A.

(c) Let K,Q,U, V and S be non-empty words such that KQ, UV, V U and KSare reduced as written, of length at least two. If KQ = UV and KS = V Uthen Q ∼ S � K,U, V .

(d) Let B,Q,L, U and V be non-empty words such that BQ, UV, LB and V Uare reduced as written. If BQ = UV and LB = V U then one of the followingholds:(i) B = U, Q = L = V or(ii) Q � B,U, V, L and L � B,U, V,Q (hence L ∼ Q ∼ UV ).

(e) Let L, K, Q1, M and N be non-empty reduced words, such that KQ1, MN ,Q1M and LK are reduced as written with length at least two. If KQ1 =MNand Q1M = LK, then one of the following holds:(i) Q1 = N = L and K =M or(ii) Q1 � K,L,M,N .

Proof. We prove here only part (c), because the proofs of parts (d) and (e) followthe same line of proof. We prove part (c) by induction on |UV |. If |UV | = 2 thenK = U = S and Q = V = K, hence the claim of part (c) of the lemma holds true.Assume |UV | > 2 and consider the first equation, KQ = UV . Then one of thefollowing holds:

Case 1. K = U . Then Q = V . Substitution in the second equation gives US = V U ,hence the result follows by part (a).

188 A. Juhasz

Case 2. K = UK1, K1 �= 1. Then V = K1Q, hence UK1S = K1QU . ByLemma 2.1 K1S =MN , K1Q = NM , U = (MN)uM , u ≥ 0. Since |MN | < |UV |and K1 �= 1, we may apply the induction hypothesis on these equations to giveS � K1,M,N, Q � K1,M,N and Q ∼ S. Consequently tracing back K and Vwe get S � K,U, V .

Case 3. U = KU1, U1 �= 1. Then Q = U1V . Substitution in the second equationyields KS = V KU1. The result follows by part (b) and Lemma 2.2.

The lemma is proved. �Recall from Definition 1.22 that σ(W ) denotes the set of all the subwords of

W with length two. We need some further definitions.

Definitions 2.4. Let W be a reduced word.(a) A decomposition of W is a partition of W into subwords. More precisely, a

decomposition of W into k parts (subwords) is a function δk : F → (F )k(:=

F × · · · × F︸︷︷︸k times

)such that if εk : (F )k → F is the function εk(x1, . . . , xk) =

x1 · · ·xk for xi ∈ F , then εk ◦ δk(W ) = W . If δk(W ) = (W1, . . . ,Wk) thenwe shall write this by W =W1 ∗W2 ∗ · · ·∗Wk. By definition,W1 · · ·Wk =W ,

reduced as written. Denote F∞ =∞⋃

k=1

(F )k and δ : F → F∞. We shall use

this notation when k is not important.(b) Let δ : W =W1 ∗ · · · ∗Wk be a decomposition of W .

Define τδ(W ) ={t(Wi)h(Wi+1)

∣∣i = 1, . . . , k− 1}. Thus τδ(W ) is a subset of

σ(W ) which contains at most k − 1 elements.When δ is clear from the context we shall write τ(W ) for τδ(W ).

(c) We shall need to consider cyclic words. Assume W is cyclically reducedand denote by W the cyclic word which corresponds to W . Then σ(W ) =σ(W 2) = σ(W ) ∪ {t(W )h(W )} and if W = W1 ∗ · · · ∗ Wk then τ(W ) =τ(W ) ∪ {t(Wk) · h(W1)}. We shall also denote σ(W ) by σ(W ) and τ(W ) byτ (W ), when convenient.

The following two lemmas are immediate from the definition.

Lemma 2.5. Let W be a reduced word in F and let δ :W =W1 ∗ · · · ∗Wk, k ≥ 2,a decomposition of W . Then

(a) σ(W ) =

(k⋃

i=1

σ(Wi)

)∪ τδ(W ).

(b) If W is cyclically reduced then σ(W ) = σ(W ) ∪ {t(W )h(W )}.Lemma 2.6. Let A and B be reduced non-empty words.(a) If A is a subword of B then σ(A) ⊆ σ(B).(b) If A1 ∈ T (A) and B0 ∈ H(B) non-empty words, then τ(A ∗B) = τ(A1 ∗B0).


(c) Let in addition C be a word with decomposition C1 ∗ · · · ∗ Ck.If σ(A) = σ(B) ∪ τ(C1 ∗ · · · ∗ Ck) then |σ(A)| ≥ |σ(B)| − k.

Lemma 2.7. Let AB = BC reduced as written, A �= 1, B �= 1. Then(a) σ(ABC) = σ(AB) = σ(BC).(b) σ(AB) = σ(A) ∪ τ(A ∗B), σ(BC) = σ(C) ∪ τ(B ∗ C).

Proof. (a) Clearly, σ(AB) ⊆ σ(ABC) and σ(BC) ⊆ σ(ABC), by Lemma 2.6(a).We show that σ(ABC) ⊆ σ(AB).

σ(ABC) =by Lemma 2.5(a)

σ(AB) ∪ τ(AB ∗ C)

=by Lemma 2.6(b)

σ(AB) ∪ τ(B ∗ C)

⊆by Lemma 2.5(a)

σ(AB) ∪ σ(BC)

=since AB=BC

σ(AB) = σ(BC).

(b) We have A = KL, B = (KL)bK, b ≥ 0 and C = LK by Lemma 2.1. Henceσ(AB) = σ(A)∪σ(B)∪τ(A∗B) by Lemma 2.5(a). But σ(B) = σ

((KL)bK

)⊆

σ(KL) ∪ τ(L ∗K) = σ(A) ∪ τ(L ∗K). Therefore,

σ(AB) ⊆ σ(A) ∪ τ(L ∗K). (2.4)

Since LK and A are subwords of AB, hence by Lemma 2.6(a)

σ(A) ∪ τ(L ∗K) ⊆ σ(AB). (2.5)

From (2.4) and (2.5) we get

σ(AB) = σ(A) ∪ τ(L ∗K). (2.6)

By Lemma 2.6(b) we have τ(L ∗K) = τ(A ∗ B), since L ∈ T (A) andK ∈ H(B). Substituting this in (2.6) gives σ(AB) = σ(A) ∪ τ(A ∗ B). Theremaining equation of part (b) follows by the same argument.The lemma is proved. �

Corollary 2.8. Suppose W = A∗B ∗C. If AB = BC then σ(W ) = σ(A)∪τ(A∗B)and also σ(W ) = σ(C) ∪ τ(B ∗ C).

Finally, we need the following basic notions.

Definition 2.9.

(a) Let R be a cyclically reduced word in F and let P be a subword of a cyclicconjugate of R. P is a piece in R (or a piece relative to the symmetricclosure R of R) if R has distinct cyclic conjugates R1 and R2 such thatR1 = PR1, R

ε2 = PR2, reduced as written, for some ε ∈ {1,−1}. We call the

two occurrences of P in R1 and Rε2, respectively, a piece-pair and denote it

by (P, P ′), where P ′ = P ε is the occurrence of P ε in R2. We shall deliber-ately use both notations ε(P ′i ) and εi for ε, if there are several pieces Pi, asconvenient.

190 A. Juhasz

(b) A piece pair (P, P ′) as in part (a) of the definition is right normalized if R−11 R2is reduced as written.

Let R1, R2 ∈ R. Then by definition R2 is a cyclic conjugate of Rε1, for some

ε ∈ {1,−1}. We can express this fact by the following two word equations: R1 =KL andRε

2 = LK for some words L andK, reduced as written. Hence the fact thatP is a piece, as in Definition 2.9, we can express via the following set of equations:PR′1 = KL and PR′2 = (LK)ε. More generally, if D is a region inM with adjacentneighbours E1 and E2 then we can write R = P1P2Q, where P1 = Φ(∂D ∩ ∂E1),P2 = Φ(∂D ∩ ∂E2). Clearly, P1 and P2 are pieces, hence as above we can writefor them a system of word equations. In Section 3 we are going to consider thissystem of word equations, together with the following two equations UXP1 = R1and P2V Y = R2, where X and Y are pieces and R1 and R2 cyclic conjugatesof R±. We are not interested in solving this equations, but rather use them ascarriers of the set theoretical information we need, namely σ(X), σ(Y ), Supp(X)and Supp(Y ). Thus, σ : Fact(R)→ S, where Fact(R) is the set of all the subwordsof R and S is the set of all the subsets of subwords of length two of R and wewould like to show that σ(Y QX) contains every subword of R of length two, withthe exception of at most three. The point is that this is a set theoretical statementwhich is much easier to prove that solve first the set of equation and then for eachsolution to show that σ(Y QX) is as written above.

3. Piece configurations of 1-corner regions and 2-corner regions

In this section we assume the conditions of the Main Theorem are satisfied.

3.1. 1-corner regions

Let D be a 1-corner region in M with inner neighbour E. Denote α = ∂D ∩ ∂E,let P = Φ(α) and let Q = Φ(∂D ∩ ∂M). Thus, σ(Q) ⊆ σ(R), see Fig. 8(a). P is apiece and vPuQ is a boundary label of D with vertices u, v.

(a) (b)

Figure 8


Lemma 3.1. Let notation be as above. Then Q � P and∣∣σ(Q)∣∣ ≥ ∣∣σ(R)∣∣− 2.

Proof. Let (P, P ′) be the corresponding piece pair. Then one of the following holds:(1) P ′ is a subword of Q;(2) P ′ contains u as an inner vertex;(3) P ′ contains v as an inner vertex.In case (1) Q � P , by Lemma 2.2 (a). In cases (2) and (3) we assume that |P | ≥ 2.In case (2) we have P = AX, P ′ = XY, Q = Y Q1, reduced as written,Q1 ∈ T (Q).See Fig. 8(b). Applying Lemma 2.3(a) to the first two of these equations andremembering that P−1 cannot overlap P , we get A ∼ Y � X and hence P ∼ Y ,by Lemma 2.2(d). Applying Lemma 2.2(a) to the last equation implies Q � P .Finally, Case 3 is dual to Case 2. Hence in all the cases Q � P .

Also, it follows from Lemma 2.7(b) that

σ(P ) = σ(XY ) = σ(Y ) ∪ τ(X ∗ Y ). (3.7)

Since X ∈ T (P ) and Y ∈ H(Q), by Lemma 2.6(b),τ(X ∗ Y ) = τ(P ∗Q). (3.8)

Since Y is a subword of Q, by Lemma 2.6(a),

σ(Y ) ⊆ σ(Q). (3.9)

Combining (3.8) and (3.9) with (3.7) we get

σ(P ) ⊆ σ(Q) ∪ τ(P ∗Q). (3.10)

Now,

σ(R) =by Lemma 2.5

σ(P ) ∪ σ(Q) ∪ τ(P ∗Q) ∪ τ(Q ∗ P )

=by (3.10)

σ(Q) ∪ τ(P ∗Q) ∪ τ(Q ∗ P ).

Therefore, by Lemma 2.6(c)∣∣σ(R)∣∣ ≥ |σ(Q)| − 2, as required.

The lemma is proved. �3.2. 2-corner regions

Let D be a 2-corner region in M with inner neighbours Er and E. See Fig. 9.Denote α1 = ∂D ∩ ∂Er and denote α2 = ∂D ∩ ∂E. Let v0 = α1 ∩ ∂M , letv2 = α2 ∩ ∂M and let v1 = α1 ∩ α2. Denote P1 = Φ(α1), P2 = Φ(α2) andQ = Φ(∂D ∩ ∂M). It is convenient and harmless to identify Pi with αi and,similarly, P ′i with α′i, i = 1, 2. Then v2Qv0P1v1P2v2 is a boundary label of D,which we may assume to coincide with R, without loss of generality, hence P1 andP2 are pieces.

Let (P1, P ′1) and (P2, P′2) be the corresponding piece pairs. Then P ′1 and P ′2

are subwords of the cyclic word R or R−1, hence one of the following holds foreach of P ′1 and P

′2:

Case 1 v0 is an inner vertex of P ′1 Case 1′ v2 is an inner vertex of P ′2

192 A. Juhasz



Case 4 P ′1 is a subword of P2 Case 4′ P ′2 is a subword of P1

Case 5 P ′1 is a subword of Q Case 5′ P ′2 is a subword of Q

Figure 9

We propose to show that in most of the cases Q � P1P2. (See precise state-ment below.) We see that for i = 1, 2, 3, 4 and 5, Case i′ is the dual of Case iobtained by exchanging P1 with P2 and v0 with v2. Hence out of the 25 cases(i, j′), 1 ≤ j′, i ≤ 5, only those with j′ ≥ i have to be checked, because the rest isobtained by duality.

The following is the main result of this section.

Proposition 3.2. Let the notation be as above and assume that R satisfies the as-sumptions of the Main Theorem. Assume that the piece pairs (P1, P ′1) and (P2, P

′2)

are right normalised. Let Qr = ∂Er ∩ ∂M and let Q = ∂E ∩ ∂M . Then σ(Qr),σ(Q) and σ(Q) are subsets of σ(R) and one of the following holds:

(a) Q � P1P2 and |σ(Q)| ≥∣∣σ(R)∣∣− 3.

(b) If both v0 and v2 have valency three and both Qr and Q are not pieces (i.e.,the products of at least two pieces) then either QλQ � R or QQρ � R. Alsoeither |σ(QλQ)| ≥

∣∣σ(R)∣∣− 3 or |σ(QQρ)| ≥∣∣σ(R)∣∣− 3.

(c) Moreover,Q � P1 in cases (1, j′), j′ = 1, 2, 3, 4, 5 and Qρ � P1 in cases (2, j′), j′ =2, 3, 4, 5;Q � P1, P2 in cases (3, j), j = 3, 4, 5 and cases (4, 5) and (5, 5);Qρ � P1, P2 and Qλ � P1, P2 in case (4, 4).Dually,Qλ � P2 in cases (i, 1), i = 1, 2, 3, 4, 5 and cases (i, 2), i = 2, 3, 4, 5Q � P1, P2 in cases (i, 3), i = 3, 4, 5 and case (5, 4).


Proof. As explained above it is enough to check the 15 cases (i, j), 1 ≤ i ≤ j′ ≤ 5.We shall check here only the case (i, j′) = (1, 1). The rest can be completed byborrowing the results from [J2, Section 2.2, pp. 183–189], together with Lemmas2.5–2.8. For the sake of completeness we present the details in the Appendix.

Case (1, 1 ′).Since P1 and P ′1 overlap and F contains no elements of order two hence ε(P ′1) = 1.Therefore we have P ′1 = Q1X, Q1 ∈ τ(Q) and P1 = XY . Hence P1 = Q1X = XYand by Lemma 2.3 Q1 ∼ X ∼ Y . Therefore by Lemma 2.2 Q � Q1 � P1and P1 ≺ Q. Also, Q1P1 = Q1XY , hence σ(Q1P1) = σ(Q1) ∪ τ(Q1 ∗ P1), byLemma 2.6(b) and Corollary 2.8. Hence we have:

P1 ≺ Q and σ(Q1P1) = σ(Q1) ∪ τ(Q1 ∗ P1). (3.11)

We show thatP2 ≺ Q and σ(QP1P2) ≥ |σ(Q)| − 3. (3.12)

This follows from the above argument with P1 and P2 interchanged henceP2 ≺ Q follows from (3.11). It also follows from (3.11) that

σ(P2Q0) = σ(Q0) ∪ τ(P2 ∗Q0) (3.13)

for an initial subword Q0 of Q such that Q = Q0Q2Q1. Hence, by Lemma 2.5

σ(R) = σ(QP1P2)= σ(Q) ∪ σ(P1) ∪ σ(P2) ∪ τ(Q ∗ P1 ∗ P2)=

by (3.11) and (3.13)σ(Q) ∪ σ(Q0) ∪ σ(Q1) ∪ τ (Q ∗ P1 ∗ P2)

∪ τ(P2 ∗Q0) ∪ τ(Q1 ∗ P1)=

by Lemma 2.6(a)σ(Q) ∪ τ (Q ∗ P1 ∗ P2) ∪ τ(P2 ∗Q0) ∪

∪ τ(Q1 ∗ P1).But by Lemma 2.6(b) τ(P2 ∗ Q0) = τ(P2 ∗ Q) and τ(Q1 ∗ P1) = τ(Q ∗ P1).Hence, τ (Q ∗ P1 ∗ P2) ∪ τ(P2 ∗Q0) ∪ τ(Q1 ∗ P2) = τ (Q ∗ P1 ∗ P2). Consequently,σ(R) = σ(Q)∪ τ (Q∗P1∗P2) and by Lemma 2.6(c) |σ(R)| ≥ |σ(Q)|−3, as required.

We close this section with the following consequence of the proposition.

Proposition 3.3. Let M be an R-diagram. Let P be a peak relative to a layerdecomposition Λ with Lp �= L1. Let α = ∂P ∩ ∂M . Then SuppΦ(α) ⊇ Supp(R)and |σ

(Φ(μ)

)| ≥ |R| − 3.

Proof. Let P = 〈D1, . . . , Dk〉. If k = 1 then the result follows from Lemma 3.1.If k ≥ 3 then it follows from Theorem 1.7(d) and the T (4) condition that eitherP contains a 1-corner region or a 2-corner region D with two neighbours Er andE such that ∂D ∩ ∂Er ∩ ∂M and ∂D ∩ ∂E ∩ ∂M are vertices with valency threeand ∂Er ∩ ∂M and ∂E ∩ ∂M are not pieces (due to the C ′(1/5)-condition). SeeFig. 6(a), where D1 is E, D2 is D and D3 is Er. In both cases the result follows bypart (b) of Proposition 3.2. Hence, if Int(Lp) has a component P such that |P | �= 2

194 A. Juhasz

then we are done. Therefore, assume that every peak in Lp contains exactly tworegions. We show that then (∂P ∩ ∂M)∪(∂Lp−1 ∩ ∂M) contains a subpath μ suchthat SuppΦ(μ) ⊇ Supp(R) and

∣∣σ(Φ(μ))∣∣ ≥ |σ(R)| − 3.Let P = 〈D1, D2〉 be an extremal peak, say left extremal. (See Fig. 11(b).)

Then both D1 and D2 are 2-corner regions, and D1 has one neighbour E in Lp−1.Since D1 is a left extremal region in Lp, D1 is the only neighbour of E1 in Lp anddM (E) ≤ 3+1 = 4. Due to the C(6)-condition ∂E∩∂M is the product of at leasttwo pieces and since D1 is left extremal the vertex v := ∂D1∩∂E∩∂M has valencythree. It follows from Theorem 1.7(d) that w := ∂D1∩∂D2 ∩∂M also has valencythree in M . Therefore Proposition 3.2(b) applies and it implies that ∂Lp ∩ ∂Mcontains a subpath μ with SuppΦ(μ) ⊇ Supp(R) and

∣∣σ(Φ(μ))∣∣ ≥ |σ(R)| − 3, asrequired.

The proposition is proved. �

Proposition 3.4. Let M be a connected simply connected R diagram with connectedinterior. Let Λ be a layer decomposition of M with center v, where v is a boundaryvertex. Then ∂M has a subpath μ with v /∈ μ such that SuppΦ(μ) ⊇ Supp(R) and∣∣σ(Φ(μ))∣∣ ≥ |σ(R)|−3. In other words, if P = 〈X |R〉 then P is almost σ-complete.

Proof. Suppose first that M consists of a single region and let ω a boundary cyclewith o(ω) = t(ω) = v. Then clearly SuppΦ(ω) ⊇ Supp(R) and

∣∣σ(Φ(ω))∣∣ ≥|σ(R)| − 1. So assume M consists of more than one region. If M = L1(v) thenv is a common boundary vertex of all the regions of M by definition of L1(v).Therefore eitherM contains a (boundary) region E with dM (E) = 1 or it containsat least three regions each of which has at most two neighbours and one of whichsatisfies the conditions of Proposition 3.2(b). In this case the result follows by

Proposition 3.2(b). Finally assume M =p⋃

i=0

Li, p ≥ 2. Then the statement of the

proposition follows by Proposition 3.3.The proposition is proved. �

3.3. Proof of the Main Theorem

By Theorem 1.24 it is enough to show that P is almost σ-complete. But P isalmost σ-complete by Proposition 3.4.

The theorem is proved. �

Appendix

Case 2 ′.Consider two subcases according as P ′2 is a subword of P1P2 or is not a subwordof P1P2. In both cases P2 and P ′2 overlap, hence ε(P

′2) = 1.

Subcase 1. P ′2 is a subword of P1P2.We have P1 = UX, P2 = XY = Y V . Consequently, P1 � X, X ∼ V ∼ P2 byLemma 2.3 and hence P1 ≺ P2 by Lemma 2.2. Hence (3.11) implies Q � P2. Also


σ(P2) = σ(X) ∪ τ(X ∗ Y ) by Lemma 2.7(b). Since X ∈ T (P1) and Y ∈ H(P2),hence τ(X∗Y ) = τ(P1∗P2) by Lemma 2.6(b) and hence σ(P2) = σ(X)∪τ(P1∗P2).Since X is a subword of P1, hence σ(X) ⊆ σ(P1) by Lemma 2.6(a), and thereforeσ(P2) ⊆ σ(P1)∪τ(P1 ∗P2). Combining this with (3.11) and using Lemma 2.5 givesσ(QP1P2) = σ(Q)∪τ (Q∗P1∗P2), hence |σ(QP1P2)| ≥ |σ(Q)|−3 by Lemma 2.6(c),as required.

Subcase 2. P ′2 is not a subword of P1P2.We have P ′2 = Q2P1X, Q2 ∈ τ(Q), P2 = XY . Hence (Q2P1)X = XY andQ2P1 ∼ Y ∼ P2. But Q2 ≺ Q by Lemma 2.2, and P1 ≺ Q by (3.11), henceP2 ∼ Q2P1 implies that P2 ≺ Q. Also, σ(P2) = σ(XY ) = σ(Q2P1) ∪ τ(Q2P1 ∗X)by Lemma 2.7(b). SinceX ∈ H(P2), hence τ(Q2P1∗X) = τ(P1∗P2), by Lemma 2.7and hence

σ(P2) = σ(Q2P1) ∪ τ(P1 ∗ P2). (3.14)

Since Q1 and Q2 are subwords of Q it follows from (3.11) that σ(QP1) = σ(Q) ∪τ(Q ∗ P1) and from (3.14) that σ(P2) ⊆ σ(QP1) ∪ τ(P1 ∗ P2). Hence σ(QP1P2) =σ(Q) ∪ τ(Q ∗ P1) ∪ τ(P1 ∗ P2) ∪ τ(P2 ∗Q), by Lemma 2.5. But then σ(QP1P2) ≥|σ(Q)| − 3, by Lemma 2.6(c).

Case 3 ′.We may assume that v1 /∈ P ′2, by Subcase 2 above. Then P ′2 = Q2X, Q2 ∈ τ(Q)and P1 = XY . Therefore Q2 ≺ Q and X ≺ Q due to (3.11). Consequently,P ′2 = Q2X ≺ Q, by Lemma 2.2 and (3.12) follows. Also, since P ′2 = Q2X , σ(P2) =σ(Q2) ∪ σ(X) ∪ τ(Q2 ∗ X). Since Q2 ∈ T (Q) and X ∈ H(P1), we have σ(P2) ⊆σ(Q) ∪ σ(X) ∪ τ(Q ∗ P1). But since X is a subword of P1, σ(X) ⊆ σ(P1), henceσ(P1) ∪ σ(P2) ∪ σ(Q) ⊆ σ(P1) ∪ σ(Q) ∪ τ(Q ∗ P1). Hence due to (3.11) we getσ(P1P2Q) = σ(Q) ∪ τ (P1 ∗ P2 ∗Q) and by Lemma 2.6(c) the result follows.

Case 4 ′.If ε(P ′2) = 1 then P2 ≺ P1, hence P2 ≺ Q, due to (3.11) and Lemma 2.2. Ifε(P ′2) = −1 then P−12 ≺ P1, hence we have P1 ≺ Q and P−12 ≺ Q. Also due to(3.11) we get σ(P1P2Q) = σ(Q) ∪ τ(P1 ∗ P2 ∗Q), and hence the result follows byLemma 2.7(c).

Case 5 ′.Similar to Case 4 ′.

Case (2, 2 ′).Consider 4 subcases according to whether v2 belongs or does not belong to P ′1 anddually, v0 belongs or does not belong to P ′2.Subcase 1. v2 /∈ P ′1 and v0 /∈ P ′2. (See Fig. 10(a).)Then P1 = XY, P ′1 = Y Z, P2 = ZT , hence τ(Y ∗ Z) ∈ σ(P1) and σ(P1) =σ(X) ∪ τ(X ∗ Y ). Now, τ(Y ∗Z) = τ(P1 ∗ P2), by Lemma 2.6(b) since Y ∈ T (P1)and Z = H(P2), hence

τ(P1 ∗ P2) ⊆ σ(P1). (3.15)

196 A. Juhasz

(a)

(b)

(c)

Figure 10

Suppose dM (v0) = 3. Fig. 10(a) shows the relative positions of P ′1 and P1 on ∂D,due to the assumption that v1 ∈ P ′1. (P ′1 has an occurrence on ∂D because Er andD have the same boundary labels.) We see that in order to get P1 from P ′1 we haveto turn P ′1 along ∂D anticlockwise by |X |. Since D and Er have the same boundarylabels (i.e., labels of boundary cycles and their inverses) we can reproduce P1 on∂Er by turning P ′1 clockwise along ∂Er by |X |. But P ′1 starts at v0 on ∂Er justas P1 starts at v0 on ∂D. Hence Qr has X−1 as a head, provided that |Qr| ≥ |X |.But X is a piece while Qr is not, hence X is a proper subword of Qr. We callthis occurrence of P1 on ∂Er the image of P1 on ∂Er and denote it by P ′′

1 .Hence X−1 ∈ H(Qr). Define Qρ = X . Then we get P1 = QρY = Y Z, P2 = ZT .Therefore σ(P1) = σ(Qρ) ∪ τ(X ∗ Y ). Since by (3.15) τ(P1 ∗ P2) ⊆ σ(P1), we getσ(P1P2) = σ(P1) ∪ σ(P2) ∪ τ(P1 ∗ P2) = σ(P1) ∪ σ(P2), hence

σ(P1P2) = σ(Qρ) ∪ σ(P2) ∪ τ(X ∗ Y ). (3.16)

Since ε(P ′1) = −1 hence QρY = Z−1Y , hence by Lemma 2.1(b) |Y | = 1 andQρ ∼ Z−1Y −1 = P−11 . Hence Qρ � P1. So assume ε(P1) = 1. Using Lemma 2.3(a),the first pair of equations gives

Qρ ∼ P1. (3.17)

But P2 and P ′2 overlap, since v1 is an inner vertex of P′2, hence P2 = LV, P ′2 =

KL and P1 = UK. Applying Lemma 2.3(a) to the first pair of equations, if |P2| ≥ 2and ε(P ′2) = 1, then

P2 ∼ K, (3.18)


while the last equation gives

K ≺ P1. (3.19)

Now, from (3.17), (3.18) and (3.19) we get P1 ∼ Qρ, P2 ≺ Qρ. Also σ(P2) =σ(K)∪τ(K∗L), by Lemma 2.7. SinceK ∈ T (P1) and L ∈ H(P2), hence τ(K∗L) =τ(P1 ∗ P2), by Lemma 2.6(b) and hence σ(P2) ⊆ σ(P1) ∪ τ(P1 ∗ P2). Thus, (3.16)implies σ(P1P2) = σ(Qρ) ∪ τ(X ∗ Y ). Consequently,

σ(QP1P2) = σ(Q) ∪ σ(P1P2) ∪ τ(P2 ∗Q) ∪ τ(Q ∗ P1)⊆ σ(Q) ∪ σ(Qρ) ∪ τ(X ∗ Y ) ∪ τ(P2 ∗Q) ∪ τ(Q ∗ P1).

Therefore, |σ(Q) ∪ σ(Qρ)| ≥ |σ(QP1P2)| − 3 and since σ(QQρ) = σ(Q) ∪ σ(Qρ) ∪τ(Q ∗Qρ), hence σ(QQρ) ≥ |σ(QP1P2)| − 3.

Subcase 2. v2 /∈ P ′1 and v0 ∈ P ′2. (See Fig. 10(b).)Then (3.17) above still holds, and for P2 and P ′2 we get P

′2 = HP1L, P2 = LK,

where H ∈ T (Q). Hence, by Lemma 2.3(a), HP1 ∼ P2, i.e., QQρ � P2, P1. Also,P2 = LK = HP1L implies by Lemma 2.7 that σ(P2) = σ(HP1)∪τ(P1∗L) and sinceL ∈ H(P2), hence τ(P1 ∗ L) = τ(P1 ∗ P2). Hence, σ(P2) = σ(HP1) ∪ τ(P1 ∗ P2) =σ(H) ∪ σ(P1)∪ τ(H ∗ P1)∪ τ(P1 ∗ P2), by Lemma 2.5(a). Now, H ∈ T (Q), hence,τ(H ∗ P1) = τ(Q ∗ P1) and hence σ(P2) ⊆ σ(QP1) ∪ τ(P1 ∗ P2). Consequently,

σ(QP1P2) = σ(Q) ∪ σ(P1) ∪ σ(P2) ∪ τ (Q ∗ P1 ∗ P2)= σ(Q) ∪ σ(P1) ∪ τ(Q ∗ P1) ∪ τ(P1 ∗ P2) ∪

∪ τ(P2 ∗Q)=

due to (3.17)σ(Q) ∪ σ(Qρ) ∪

∪ τ (Q ∗ P1 ∗ P2).As above, this implies |σ(QQ−1ρ )| ≥ |σ(QP1P2)| − 3.

Subcases 3 and 4. v2 ∈ P ′1. (See Fig. 10(c).)Then P1 = XY, P ′1 = Y P2T where T ∈ H(Q). Therefore,

σ(P1P2T ) = σ(XY P2T ) = σ(X) ∪ τ(X ∗ Y ),by Lemma 2.7. Consequently,

σ(P1P2Q) = σ(X) ∪ σ(Q) ∪ τ(X ∗ Y ) ∪ τ(Q ∗ P1) ∪ τ(P2 ∗Q)= σ(Qρ) ∪ σ(Q) ∪ τ(X ∗ Y ) ∪ τ(Q ∗ P1) ∪ τ(P2 ∗Q)

and consequently, |σ(QQ−1ρ )| ≥ |σ(QP1P2)|−3. Since ε(P ′1) = 1, by Lemma 2.3(a),X ∼ P1 ∼ P2T Now, X ∈ H(Qr) as in Subcase 1. Thus P1, P2 ≺ Qr.

Case (2, 3).v1 ∈ P ′1 and v0 ∈ P ′2. (See Fig. 11(a) and Fig. 11(b).) We consider three maincases according as v1 /∈ P ′2 and v2 /∈ P ′1 or v1 ∈ P ′2 or v1 /∈ P2 and v2 ∈ P ′1. In eachmain case we consider two cases according as ε2 = 1 or ε2 = −1.

198 A. Juhasz

(a) (b)

Figure 11

Main Case 1. v1 /∈ P ′2 and v2 /∈ P ′1. We have

P ′2 = Q1U,P1 = UV, P1 = XY,P ′1 = Y Z and P2 = ZW, h(W ) �= h(Z) (3.20)

where all subwords Q1, U, V,X, Y, Z and W are non-empty, Q1 ∈ τ(Q) and P1,P2, P ′1 and P ′2 are reduced as written. Also, h(W ) �= h(Z) because the piece-pair (P1, P ′1) is right normalized. Since F contains no elements of order two, mayassume ε1 = 1. Then

P2 = Q1U = ZW and P1 = XY = Y Z = UV. (3.20′)

Observe thatQ1 ∈ T (Q) and X−1 ∈ H(Qr). (3.20′′)

It follows from (3.20′) and Lemma 2.1 that

X = KL, Y = (KL)kK, k ≥ 0, Z = LK. (3.20′′′)

It follows from (3.20′) and Lemma 2.3(i) that X � Y, Z and by the second andfourth equations in (3.20′) also X � U, V . Hence if we define Qρ = X−1 then Qρ �Y, Z, U, V . Since P ′2 = Q1U hence QQρ � Q1, Qρ and hence QQρ � P2, P1. Also, itfollows from (3.20′) that σ(P1) = σ(X)∪τ(X ∗Y ) and σ(P2) = σ(Q1)∪τ(Q1 ∗U).Therefore, σ(QP1P2) = σ(Q) ∪ σ(X)∪

(τ(X ∗ Y )∪ τ(Q1 ∗U)∪ τ(Q ∗P1)∪ τ(P1 ∗

P2) ∪ τ(P2 ∗Q)). But τ(Q1 ∗ U) = τ(Q ∗ P1) and τ(P1 ∗ P2) = τ(Y ∗ Z) ∈ σ(P1),

by (3.20′) hence σ(QP1P2) = σ(Q) ∪ σ(X) ∪(τ(X ∗ Y ) ∪ τ(Q ∗ P1) ∪ τ(P2 ∗Q)

).

Since X−1 ∈ H(Qr), by (3.20′′), we get |σ(Q) ∪ σ(Qr)| ≥ |σ(QP1P2)| − 3, byLemma 2.6(c).

Main Case 2. v1 ∈ P ′2. Then P ′2 = Q1P1U, P2 = UV, Q1 ∈ τ(Q), all expressionsreduced as written. Assume that ε2 = −1. Then P2 = U−1P−11 Q−11 = UV , henceby Lemma 2.1(b)(i) |U | = 1 and P2 ∼ P−11 Q−11 . Therefore QQρ � Q1Qρ � P1, P2.May assume ε2 = 1. Then P ′2 = Q1P1U, P2 = UV . Consequently, v2 /∈ P ′1 since|P2| > |P ′1| and hence P1 = XY, P ′1 = Y Z and P2 = ZW . The equations P2 =UV = Q1P1U imply that V ∼ P2 ∼ Q1P1, by Lemma 2.3. Also, we observe thatV −1 ∈ τ(Q). Therefore taking Qλ = V −1, we get Qλ � P−11 , P−12 . Also, we have


σ(P1) ⊆ σ(P2), since P2 = Q1P1U . Compute σ(P2). We have P2 = UV = Q1P1U ,V −1 ∈ T (Q), hence σ(P2) = σ(V )∪τ(U ∗V ) ⊆ σ(Qλ)∪τ(U ∗V ) by Lemma 2.6(a)and Lemma 2.7. Consequently, σ(QP1P2) = σ(Q) ∪ σ(P2) ∪

(τ(Q ∗ P1) ∪ τ(P1 ∗

P2)∪τ(P2∗Q))= σ(Q)∪σ(Qλ)∪

(τ(U ∗V )∪τ(Q∗P1)∪τ(P1∗P2)∪τ(P2∗Q)

). But

τ(U ∗V ) ∈ σ(P2) and τ(Q∗P1) = τ(Q1 ∗P1) ∈ σ(P2). Also, τ(P1 ∗P2) = τ(P1 ∗U)by Lemma 2.6(b), as U ∈ H(P2), hence τ(P1 ∗ P2) ∈ σ(P2). Therefore, we getσ(QP1P2) = σ(Q) ∪ σ(Qλ) ∪ τ(P2 ∗ Q). Thus, |σ(Q) ∪ σ(Qλ)| ≥ |σ(R)| − 1 ≥|σ(R)| − 3.

Main Case 3. v1 /∈ P2 and v2 ∈ P ′1. Then

P1 = XY, P ′1 = Y P2Q0, Q0 ∈ H(Q),P ′2 = Q1U, P1 = UV, X−1 ∈ H(Qr). (3.21)

Since P1 and P ′1 overlap, ε(P′1) = 1 then the first two equations of (3.21) yield

X ∼ P2Q0 ∼ P1, hence taking Qρ = X−1, gives Qρ � P1, P2. Now, it followsfrom (3.21) that σ(P2) ⊆ σ(P1) and σ(P1) = σ(X) ∪ τ(X ∗ Y ). Also, τ(Q ∗ P1) =τ(Q1∗U) ∈ σ(P ′2), since Q1 ∈ τ(Q) and U ∈ H(P1), τ(P1∗P2) = τ(Y ∗P2) ∈ σ(P ′1)since Y ∈ T (P1) by (3.21) and finally τ(P2∗Q) = τ(P2∗Q0) ∈ σ(P ′1). Consequently,σ(QP1P2) = σ(Q)∪σ(X)∪τ(X ∗Y ). Hence, σ(QP1P2) ⊆ σ(Q)∪σ(Qr)∪τ(X ∗Y ),since X−1 ∈ H(Qr), by (3.21). But then |σ(Q) ∪ σ(Qr)| ≥ |σ(QP1P2)| − 1, byLemma 2.6(c), and the result follows.

Case (2, 4).P ′2 is a subword of P1.Assume first that v2 is not an inner vertex of P ′1. Then

P1 = XY, P ′1 = Y Z, P2 = ZW and P1 = UP ′2V. (3.22)

Since ε(P ′1) = 1 then the first two equations of (3.22) imply P1 ∼ X ∼ Z � Y andthe last equation of (3.22) implies P1 � P ′2. Consequently, X � P1, P

′2.

Define Qρ = X−1. Then Qρ � P1, P2. Assume now that v2 is an inner vertexof P ′1. Since P1 and P ′1 overlap, ε1 = 1. Hence,

P1 = XY, P ′1 = Y P2Q0, Q0 �= 1, Q0 ∈ H(Q) and P1 = UP ′2V. (3.23)

The first two equations of (3.23) imply X = KL, P2Q0 = LK and Y = (KL)αK,α ≥ 0. Consequently, Q0 � P2, P1, L,K,X, Y , hence Q � Q0 � P1, P2.

Now it follows from (3.22) that σ(P1) = σ(X)∪τ(X ∗Y ) and σ(P2) ⊆ σ(P1).Also, it follows from (3.23) and Lemma 2.6(b) that τ(P1 ∗P2) = τ(Y ∗Z) ∈ σ(P1)and τ(P2 ∗Q) = τ(P2 ∗Q0) ∈ σ(P1). Therefore, σ(QP1P2) = σ(Q)∪σ(P1)∪

(τ(X ∗

Y ) ∪ τ(Q ∗ P1))⊆ σ(Q) ∪ σ(Qr) ∪ τ(X ∗ Y ) ∪ τ(Q ∗ P1). Hence |σ(Q) ∪ σ(Qr)| ≥

|σ(R)| − 2, by Lemma 2.6(c) and the result follows.

Case (2, 5).Assume v2 /∈ P ′1.Since P1 and P ′1 overlap, ε(P

′1) = 1. Hence P1 = XY, P ′1 = Y Z and P2 = ZW .

200 A. Juhasz

By the first two equations, via Lemma 2.3(a) P1 ∼ Z and by the third equationP2 � Z. Hence P1 ∼ Z ≺ P2 ≺ Q. Consequently, Q � P2 � P1, hence P1, P2 ≺ Q.

Since P ′2 is a subword of Q, by Lemma 2.6(a), we have σ(P2) ⊆ σ(Q). Inparticular, σ(Z) ⊆ σ(Q). Now, σ(P1) = σ(Z) ∪ τ(Y ∗ Z) by Lemma 2.7 andτ(Y ∗Z) = τ(P1 ∗ P2), by Lemma 2.6(b), since Y ∈ T (P1) and Z ∈ H(P2). Henceσ(P1) ⊆ σ(Q) ∪ τ(P1 ∗ P2). Consequently, σ(QP1P2) = σ(Q) ∪ τ(P1 ∗ P2) ∪ τ(Q ∗P1) ∪ τ(P2 ∗Q), hence by Lemma 2.6(c) |σ(Q)| ≥ |σ(R)| − 3, as required.

Assume now that v2 ∈ P ′1.Then P1 = XY, P ′1 = Y P2Q0, Q0 �= 1, Q0 ∈ H(Q). Hence P2Q0 � P1, byLemma 2.2(a). If ε(P ′2) = 1 then Q � P2, hence Q � P2Q0 � P1, i.e., Q � P1, P2.Suppose ε(P ′2) = −1. By Lemma 2.1 X = KL, Y = (KL)αK, α ≥ 0 andP2Q0 = LK. Consequently, P2Q0 � P1P2 hence Q � P1, P2. Now, by Lemma 2.7

σ(P1) = σ(P2Q0) ∪ τ(Y ∗ P2)= σ(P2) ∪ σ(Q0) ∪ τ(P2 ∗Q)⊆ σ(Q) ∪ τ(P2 ∗Q),

since σ(P2) ⊆ σ(Q), by Lemma 2.6(a). Therefore, σ(P1P2Q) = σ(Q)∪τ (P1∗P2∗Q),hence |σ(Q)| ≥ |σ(R)| − 3, by Lemma 2.6(c).

Case (3, 3′).The cases when v1 ∈ P ′1 or v1 ∈ P ′2 can be dealt with in a way similar to Subcases 2and 3 of Case (2, 2′). So we concentrate on the case v1 /∈ P ′1 and v1 /∈ P ′2. We haveP1 = XU, P ′1 = Y Q0, P2 = V Y, P ′2 = Q1X where Q0 ∈ H(Q) and Q1 ∈ T (Q).Due to the second and fourth equations, it is enough to prove X,Y ≺ Q. Now,σ(P1) = σ(Y ) ∪ σ(Q0) ∪ τ(P2 ∗Q) and σ(P2) = σ(X) ∪ σ(Q1) ∪ τ(Q ∗ P1). Hence

σ(QP1P2) = σ(X) ∪ σ(Y ) ∪ σ(Q) ∪ τ(P1 ∗ P2 ∗Q). (3.24)

Subcase 1. ε1 = ε2 = 1.Then Q1X = V Y and XU = Y Q0. If X = Y then Q1 = V and U = Q0, henceh(U) = h(Q0) = h(Q), violating the right normalization of the piece pair (P2, P ′2)(see Definition 2.9). If X = Y Z, Z �= 1, then Q0 = ZU, Q1Y Z = V Y , hence V =Q1T, Y Z = TY and, by Lemma 2.3, Y ≺ Z ∼ T . Therefore X = Y Z implies Y ≺Z ∼ X ∼ T . But Q0 = ZU implies Q0 � Z. Consequently, Q � Q0 � X,Y . Also,σ(Z) ⊆ σ(Q0) andX = Y Z = TY imply σ(X) = σ(Z)∪τ(Y ∗Z) ⊆ σ(Q0)∪τ(Y ∗Z)⊆ σ(Q)∪ τ(P2 ∗Q), by Lemma 2.7. Since σ(Y ) ⊆ σ(X) it follows from (3.24) thatσ(QP1P2) = σ(Q)∪ τ(P1 ∗P2 ∗Q), hence |σ(Q)| ≥ |σ(R)| − 3. If Y = XZ, Z �= 1,then U = ZQ0, Q1X = V XZ. Hence Q1 = V T, TX = XZ and, as in the previouscase, T ∼ X ∼ Z ≺ Q1, hence Y = XZ ≺ Q1, by Lemma 2.2. Thus, X,Y ≺ Q.Since Y = TX = XZ, σ(Y ) = σ(XZ) = σ(T )∪ τ(T ∗X) = σ(T )∪ τ(Q ∗P1), i.e.,σ(Y ) = σ(T )∪τ(Q∗P1). But Q1 = V T implies σ(T ) ⊆ σ(Q1) ⊆ σ(Q), hence sinceσ(X) ⊆ σ(Y ) by (3.24) σ(R) = σ(X) ∪ σ(Y ) ∪ τ (QP1P2) = σ(Q) ∪ τ (Q ∗ P1 ∗ P2)and again the result follows, by Lemma 2.6(c).

Subcase 2. ε1 = 1 and ε2 = −1.Then XU = Y Q0 and V Y = X−1Q−11 . Suppose first X = Y . Then U = Q0 and


V X = X−1Q1. If V = X−1V1 then Q1 = V1X , hence Q1 � X,Y . If X−1 = V Z,then X = ZQ1, hence X = ZQ1 = Z−1V −1. Since R contains no elementsof order two, hence, this case can not occur. Hence V = X−1V1, V1X = Q−1.Consequently, σ(X) ⊆ σ(Q) and the result follows by (3.24), as above. If X = Y Z,Z ∈ H(Q0) then V Y = Z−1Y −1Q−11 , hence Y ∈ T (Q−11 ) and hence σ(Y ) ⊆ σ(Q)and σ(X) = σ(Y Z) ⊆ σ(Y ) ∪ σ(Q) ∪ τ(Y ∗ Z) = σ(Q) ∪ τ(P2 ∗ Q). Hencethe result follows from (3.24), as above. Finally, if Y = XU0, U = U0Q0 thenV Y = V XU0 = X−1Q−11 . Therefore V = X−1V1, Q−11 = V1XU0 = V1Y . Inparticular σ(Y ), σ(X) ⊆ σ(Q1) ⊆ σ(Q) and the result follows by (3.24).

Subcase 3. ε1 = −1 and ε2 = 1.This subcase dual to Subcase 2.

Subcase 4. ε1 = −1 and ε2 = −1.This subcase follows directly from the four equations at the beginning of Case(3, 3′).

Case (3, 4).We have P2 = UV, P ′1 = V Q0, P1 = HP ′2T , hence σ(P2) ⊆ σ(P1).There are four cases to check, according as ε1 = ±1 and ε2 = ±1. We shall checkonly the cases ε1 = 1 and ε2 = ±1. The other two cases are similar.Subcase 1. ε1 = 1, ε2 = 1.Then H(UV )T = V Q0. Hence Q0 = Q′0T and V Q′0 = HUV . Consequently,Q′0 � V,H,U , hence Q � P1, P2. Also, V Q′0 = HUV implies by Lemma 2.7 (withA = HU , B = V , C = Q′0) σ(V Q

′0) = σ(Q′0) ∪ τ(V ∗ Q′0). Since τ(V ∗ Q′0) =

τ(P2 ∗Q), we get σ(V ) ⊆ σ(Q) ∪ τ(P2 ∗Q). Now σ(R) = σ(P1) ∪ σ(P2) ∪ σ(Q) ∪τ (Q ∗ P1 ∗ P2) = σ(P1) ∪ σ(Q) ∪ τ (Q ∗ P1 ∗ P2). But σ(P1) = σ(P ′1) = σ(V Q0) =σ(V )∪σ(Q0)∪τ(V ∗Q0) ⊆ σ(V )∪σ(Q)∪τ(P2∗Q) ⊆ σ(Q)∪τ(P2∗Q). Consequently,σ(R) = σ(Q)∪ τ (Q∗P1∗P2) and the result follows by Lemma 2.6(c) as in previouscases.

Subcase 2. ε1 = 1, ε2 = −1.Then H(V −1U−1)T = V Q0, hence Q0 = Q′0U

−1T and V Q′0 = HV −1. SinceF has no elements of order two, H = VH1 and Q′0 = H1V

−1. Consequently,Q � V,H1, hence Q � U, V, P1, P2. Also, σ(P1) = σ(V ) ∪ σ(Q0) ∪ τ(V ∗ Q0) ⊆σ(Q) ∪ τ(P2 ∗ Q). Therefore, σ(QP1P2) = σ(Q) ∪ τ (Q ∗ P1 ∗ P2) and the resultfollows, by Lemma 2.6(c).

Case (3, 5′).Then P ′1 = V Q0, Q0 ∈ H(Q), P2 = UV , hence because P ′2 is a subword of Q,Q � P2. Consequently, Q � U, V , hence Q � P1, P2. Also, σ(P2) ⊆ σ(Q) andσ(P1) = σ(V )∪σ(Q0)∪ τ(V ∗Q0) ⊆ σ(Q)∪ τ(P2 ∗Q). Consequently, σ(QP1P2) =σ(Q) ∪ τ (Q ∗ P1 ∗ P2) and the result follows, by Lemma 2.6(c).Case (4, 4′).Then P ′1 is a subword of P2 and P ′2 is a subword of P1. Consequently, |P1| = |P2|and since R is reduced ε1 = ε2 = 1. Therefore, P1 = P2 and if v0 has valency

202 A. Juhasz

three in H then Qr has P−11 as a head. Therefore, σ(P1) = σ(P2) ⊆ σ(Qr) and asresult σ(R) ⊆ σ(Q) ∪ σ(Qr) ∪ τ (P1 ∗ P2 ∗Q). Similarly, if v2 has valency three inH then σ(R) ⊆ σ(Q)∪ σ(Q)∪ τ(P1 ∗P2 ∗Q). In both cases the result follows, byLemma 2.6(c).

Cases (4, 5′) and (5, 5′).It follows immediately that Q � P1, P2 and σ(R) = σ(Q) ∪ τ (Q ∗ P1 ∗ P2).

The proposition is proved. �Acknowledgement

I am grateful to the referee for his useful comments regarding the presentation ofthe work.

References

[J1] Juhasz, A. Small cancellation theory with unified small cancellation condition. J.London. Math. Soc. 2 (40):57–80 (1989).

[J2] Juhasz, A. Solution of the membership problem of Magnus subgroups of one-relator free products with small cancellation condition. In: Algebra and Geome-try in Geneva and Barcelona, “Asymptotic and Probabilistic Methods in GeometricGroup Theory”, Geneva, June 2005 and “Barcelona Group Theory Conference”,Barcelona, July 2005, 169–195, Birkhauser (2007).

[J3] Juhasz, A. A Freiheitssatz for Whitehead graphs of one-relator groups with smallcancellation. Communications in Algebra 37:8, 2714–2741 (2009).

[Jo] Johnson, D.L. Topics in the Theory of Group Presentations. LMSLNS, CambridgeUniversity Press, Cambridge. 42 (1980).

[L-S] Lyndon, R.C., Schupp, P.E. Combinatorial Group Theory, Springer-Verlag, Berlin-Heidelberg-New York, (1977).

[M] Magnus, W., Das Identitatsproblem fur Gruppen mit einer definierenden Relation,Math. Ann. 106 pp. 295–307 (1932).

Arye JuhaszDepartment of MathematicsTechnion – Israel Institute of Technology32000 Haifa, Israele-mail: [email protected]



Equations and Fully Residually Free Groups

Olga Kharlampovich and Alexei G. Myasnikov

Abstract. This paper represents notes of the mini-courses given by the authorsat the GCGTA conference in Dortmund (2007), Ottawa-Saint Sauveur con-ference (2007), Escola d’Algebra in Rio de Janeiro (2008) and Alagna (Italy,2008) conference on equations in groups. We explain here the Eliminationprocess for solving equations in a free group which has Makanin-Razborovprocess as a prototype. We also explain how we use this process to obtain thestructure theorem for finitely generated fully residually free groups and manyother results.

Mathematics Subject Classification (2000). 20-02.

Keywords. Equations, free groups.

1. Introduction

1.1. Motivation

Solving equations is one of the main themes in mathematics. A large part of thecombinatorial group theory revolves around the word and conjugacy problems –particular types of equations in groups. Whether a given equation has a solutionin a given group is, as a rule, a non-trivial problem. A more general and moredifficult problem is to decide which formulas of the first-order logic hold in a givengroup.

Around 1945 A. Tarski put forward two problems on elementary theories offree groups that served as a motivation for much of the research in group theory andlogic for the last sixty years. A joint effort of mathematicians of several generationsculminated in the following theorems, solving these Tarski’s conjectures.

Theorem 1 (Kharlampovich and Myasnikov [44], Sela [60]). The elementary theo-ries of non-Abelian free groups coincide.

Theorem 2 (Kharlampovich and Myasnikov [44]). The elementary theory of a freegroup F (with constants for elements from F in the language) is decidable.

Mini-course for the GCGTA conference in Dortmund (2007)and Escola d’Algebra in Rio deJaneiro (2008).

204 O. Kharlampovich and A.G. Myasnikov

We recall that the elementary theory Th(G) of a groupG is the set of all first-order sentences in the language of group theory which are true in G. A discussionof these conjectures can be found in several textbooks on logic, model and grouptheory (see, for example, [14], [22], [29]).

The work on the Tarski conjectures was rather fruitful – several areas ofgroup theory were developed along the way. It was clear from the beginning thatto deal with the Tarski’s conjectures one needs to have at least two principalthings done: a precise description of solution sets of systems of equations over freegroups and a robust theory of finitely generated groups which satisfy the sameuniversal (existential) formulas as a free non-Abelian group. In the classical case,algebraic geometry provides a unifying view-point on polynomial equations, whilecommutative algebra and the elimination theory give the required decision tools.Around 1998 three papers appeared almost simultaneously that address analogousissues in the group case. Basics of algebraic (or Diophantine) geometry over groupshas been outlined by Baumslag, Myasnikov and Remeslennikov in [5], while thefundamentals of the elimination theory and the theory of fully residually freegroups appeared in the works by Kharlampovich and Myasnikov [37], [38]. Thesetwo papers contain results that became fundamental for the proof of the abovetwo theorems, as well as in the theory of fully residually free groups. The goal ofthese lectures is to explain why these results are important and to give some ideasof the proof.

1.2. Milestones of the theory of equations in free groups

The first general results on equations in groups appeared in the 1960’s [31]. Aboutthis time Lyndon (a former student of Tarski) came up with several extremelyimportant ideas. One of these is to consider completions of a given group G byextending exponents into various rings (analogs of extension of ring of scalarsin commutative algebra) and use these completions to parameterize solutions ofequations in G. Another idea is to consider groups with free length functions withvalues in some ordered Abelian group. This allows one to axiomatize the classicalNielsen technique based on the standard length function in free groups and applyit to “non-standard” extensions of free groups, for instance, to ultrapowers offree groups. A link with the Tarski’s problems comes here by the Keisler-Shelahtheorem, that states that two groups are elementarily equivalent if and only if theirultrapowers (with respect to a non-principal ultrafilter) are isomorphic. The idea tostudy freely discriminated (fully residually free) groups in connection to equationsin a free group also belongs to Lyndon. He proved [30] that the completion FZ[t] of afree group F by the polynomial ring Z[t] (now it is called the Lyndon’s completionof F ) is discriminated by F . At the time the Tarski’s problems withstood theattack, but these ideas gave birth to several influential theories in modern algebra,which were instrumental in the recent solution of the problems. One of the mainingredients that was lacking at the time was a robust mechanism to solve equationsin free groups and a suitable description of the solution sets of equations. The main

Equations and Fully Residually Free Groups 205

technical goal of these lectures is to describe a host of methods that altogethergive this mechanism, that we refer to as Elimination Processes.

Also in 1960’s Malcev [34] described solutions of the equation

zxyx−1y−1z−1 = aba−1b−1

in a free group, which is the simplest non-trivial quadratic equation in groups. Thedescription uses the group of automorphisms of the coordinate group of the equa-tion and the minimal solutions relative to these automorphisms – a very powerfulidea, that nowadays is inseparable from the modern approach to equations. Thefirst break-through on Tarski’s problem came from Merzljakov (who was a part ofMalcev’s school in Novosibirsk). He proved [50] a remarkable theorem that any twonon-Abelian free groups of finite rank have the same positive theory and showedthat positive formulas in free groups have definable Skolem functions, thus givingquantifier elimination of positive formulas in free groups to existential formulas.Recall that the positive theory of a group consists of all positive (without nega-tions in their normal forms) sentences that are true in this group. These resultswere precursors of the current approach to the elementary theory of a free group.

In the eighties new crucial concepts were introduced. Makanin proved [48]the algorithmic decidability of the Diophantine problem over free groups, andshowed that both, the universal theory and the positive theory of a free groupare algorithmically decidable. He created an extremely powerful technique (theMakanin elimination process) to deal with equations over free groups.

Shortly afterwards, Razborov (at the time a PhD student of Steklov’s Insti-tute, where Makanin held a position) described the solution set of an arbitrarysystem of equations over a free group in terms of what is known now as Makanin-Razborov diagrams [54], [55].

A few years later Edmunds and Commerford [18] and Grigorchuck and Kur-chanov [27] described solution sets of arbitrary quadratic equations over freegroups. These equations came to group theory from topology and their role ingroup theory was not altogether clear then. Now they form one of the corner-stonesof the theory of equations in groups due to their relations to JSJ-decompositionsof groups.

1.3. New age

These are milestones of the theory of equations in free groups up to 1998. The lastmissing principal component in the theory of equations in groups was a generalgeometric point of view similar to the classical affine algebraic geometry. Back to1970’s Lyndon (again!) was musing on this subject [32] but for no avail. Finally,in the late 1990’s Baumslag, Kharlampovich, Myasnikov, and Remeslennikov de-veloped the basics of the algebraic geometry over groups [5, 37, 38, 39, 36], in-troducing analogs of the standard algebraic geometry notions such as algebraicsets, the Zariski topology, Noetherian domains, irreducible varieties, radicals andcoordinate groups, rational equivalence, etc.


With all this machinery in place it became possible to make the next crucialstep and tie the algebraic geometry over groups, Makanin-Razborov process forsolving equations, and Lyndon’s free Z[t]-exponential group F Z[t] into one closelyrelated theory. The corner stone of this theory is Decomposition Theorem from[38] (see Section 4.2 below) which describes the solution sets of systems of equa-tions in free groups in terms of non-degenerate triangular quasi-quadratic (NTQ)systems. The coordinate groups of the NTQ systems (later became also knownas residually free towers) play a fundamental role in the theory of fully residuallyfree groups, as well as in the elementary theory of free groups. The Decomposi-tion Theorem allows one to look at the processes of the Makanin-Razborov’s typeas non-commutative analogs of the classical elimination processes from algebraicgeometry. With this in mind we refer to such processes in all their variations asElimination Processes (EP).

In the rest of the notes we discuss more developments of the theory, focusingmostly on the elimination processes, fully residually free (limit) groups, and newtechniques that appear here.

2. Basic notions of algebraic geometry over groups

Following [5] and [39] we introduce here some basic notions of algebraic geometryover groups.

Let G be a group generated by a finite set A, F (X) be a free group withbasis X = {x1, x2, . . . , xn}, we defined G[X ] = G∗F (X) to be a free product of Gand F (X). If S ⊂ G[X ] then the expression S = 1 is called a system of equationsover G. As an element of the free product, the left side of every equation in S = 1can be written as a product of some elements from X ∪ X−1 (which are calledvariables) and some elements from A (constants). To emphasize this we sometimeswrite S(X,A) = 1.

A solution of the system S(X) = 1 over a group G is a tuple of elementsg1, . . . , gn ∈ G such that after replacement of each xi by gi the left-hand side ofevery equation in S = 1 turns into the trivial element of G. To study equationsover a given fixed group G it is convenient to consider the category of G-groups,i.e., groups which contain the group G as a distinguished subgroup. If H and Kare G-groups then a homomorphism φ : H → K is a G-homomorphism if gφ = gfor every g ∈ G, in this event we write φ : H →G K. In this category morphismsare G-homomorphisms; subgroups are G-subgroups, etc. A solution of the systemS = 1 over G can be described as a G-homomorphism φ : G[X ] −→ G such thatφ(S) = 1. Denote by ncl(S) the normal closure of S in G[X ], and by GS thequotient group G[X ]/ncl(S). Then every solution of S(X) = 1 in G gives rise toa G-homomorphism GS → G, and vice versa. By VG(S) we denote the set of allsolutions in G of the system S = 1, it is called the algebraic set defined by S. Thisalgebraic set VG(S) uniquely corresponds to the normal subgroup

R(S) = {T (x) ∈ G[X ] | ∀A ∈ Gn(S(A) = 1→ T (A) = 1}


of the group G[X ]. Notice that if VG(S) = ∅, then R(S) = G[X ]. The subgroupR(S) contains S, and it is called the radical of S. The quotient group

GR(S) = G[X ]/R(S)

is the coordinate group of the algebraic set V (S). Again, every solution of S(X) = 1in G can be described as a G-homomorphism GR(S) → G.

By HomG(H,K) we denote the set of all G-homomorphisms from H into K.It is not hard to see that the free product G∗F (X) is a free object in the categoryof G-groups. This group is called a free G-group with basis X , and we denote itby G[X ]. A G-group H is termed finitely generated G-group if there exists a finitesubset A ⊂ H such that the set G ∪ A generates H . We refer to [5] for a generaldiscussion on G-groups.

A group G is called a CSA group if every maximal Abelian subgroupM of Gis malnormal, i.e., Mg ∩M = 1 for any g ∈ G−M. The abbreviation CSA meansconjugacy separability for maximal Abelian subgroups. The class of CSA-groups isquite substantial. It includes all Abelian groups, all torsion-free hyperbolic groups,all groups acting freely on Λ-trees and many one-relator groups (see, for exam-ple, [26].

We define a Zariski topology on Gn by taking algebraic sets in Gn as a sub-basis for the closed sets of this topology. Namely, the set of all closed sets in theZariski topology on Gn can be obtained from the set of algebraic sets in two steps:

1) take all finite unions of algebraic sets;2) take all possible intersections of the sets obtained in step 1).

If G is a non-Abelian CSA group and we in the category of G-groups, thenthe union of two algebraic sets is again algebraic. Indeed, if {wi = 1, i ∈ I} and{uj = 1, j ∈ J} are systems of equations, then in a CSA group their disjunctionis equivalent to a system

[wi, uj] = [wi, uaj ] = [wi, u

bj] = 1, i ∈ I, j ∈ J

for any two non-commuting elements a, b from G. Therefore the closed sets in theZariski topology on Gn are precisely the algebraic sets.

A group G is called equationally Noetherian if every system S(X) = 1 withcoefficients from G is equivalent over G to a finite subsystem S0 = 1, whereS0 ⊂ S, i.e., VG(S) = VG(S0). It is known that linear groups (in particular,freely discriminated groups) are equationally Noetherian (see [25], [11], [5]). If Gis equationally Noetherian then the Zariski topology on Gn is Noetherian for everyn, i.e., every proper descending chain of closed sets in Gn is finite. This impliesthat every algebraic set V in Gn is a finite union of irreducible subsets (they arecalled irreducible components of V ), and such decomposition of V is unique. Recallthat a closed subset V is irreducible if it is not a union of two proper closed (inthe induced topology) subsets.


3. Fully residually free groups

3.1. Definitions and elementary properties

Finitely generated fully residually free groups (limit groups) play a crucial role inthe theory of equations and first-order formulas over a free group. It is remarkablethat these groups, which have been widely studied before, turn out to be the basicobjects in newly developing areas of algebraic geometry and model theory of freegroups. Recall that a group G is called fully residually free (or freely discriminated,or ω-residually free) if for any finitely many non-trivial elements g1, . . . , gn ∈ Gthere exists a homomorphism φ of G into a free group F , such that φ(gi) �= 1for i = 1, . . . , n. The next proposition summarizes some simple properties of fullyresidually free groups.

Proposition 1. Let G be a fully residually free group. Then G possesses the followingproperties.

1) G is torsion-free;2) Each subgroup of G is a fully residually free group;3) G has the CSA property;4) Each Abelian subgroup of G is contained in a unique maximal finitely gen-

erated Abelian subgroup, in particular, each Abelian subgroup of G is finitelygenerated;

5) G is finitely presented, and has only finitely many conjugacy classes of itsmaximal Abelian subgroups.

6) G has solvable word problem;7) G is linear;8) Every 2-generated subgroup of G is either free or Abelian;9) If rank(G) = 3 then either G is free of rank 3, free Abelian of rank 3, or

a free rank one extension of centralizer of a free group of rank 2 (that isG = 〈x, y, t|[u(x, y), t] = 1〉, where the word u is not a proper power).

Properties 1 and 2 follow immediately from the definition of an F -group. Aproof of property 3 can be found in [5]; property 4 is proven in [38]. Properties 4and 5 are proved in [38]. Solvability of the word problem follows from [49] or fromresidual finiteness of a free group. Property 9 is proved in [23]. Property 7 followsfrom linearity of F and property 6 in the next proposition. The ultraproduct ofSL2(Z) is SL2(∗Z), where ∗Z is the ultpaproduct of Z. (Indeed, the direct product∏SL2(Z) is isomorphic to SL2(

∏Z). Therefore, one can define a homomorphism

from the ultraproduct of SL2(Z) onto SL2(∗Z). Since the intersection of a finitenumber of sets from an ultrafilter again belongs to the ultrafilter, this epimorphismis a monomorphism.) Being finitely generated G embeds in SL2(R), where R is afinitely generated subring in ∗Z.

Proposition 2. (no coefficients) Let G be a finitely generated group. Then the fol-lowing conditions are equivalent:


1) G is freely discriminated (that is for finitely many non-trivial elementsg1, . . . , gn ∈ G there exists a homomorphism φ from G to a free groupsuch that φ(gi) �= 1 for i = 1, . . . , n);

2) [Remeslennikov]G is universally equivalent to F (in the language without constants);

3) [Baumslag, Kharlampovich, Myasnikov, Remeslennikov]G is the coordinate group of an irreducible variety over a free group.

4) [Sela] G is a limit group (to be defined in the proof of Proposition 3).5) [Champetier and Guirardel]

G is a limit of free groups in Gromov-Hausdorff metric (to be defined in theproof of Proposition 3).

6) G embeds into an ultrapower of free groups.

Proposition 3. (with coefficients) Let G be a finitely generated group containing afree non-Abelian group F as a subgroup. Then the following conditions are equiv-alent:

1) G is F -discriminated by F ;2) [Remeslennikov]

G is universally equivalent to F (in the language with constants);3) [Baumslag, Kharlampovich, Myasnikov, Remeslennikov]

G is the coordinate group of an irreducible variety over F .4) [Sela] G is a restricted limit group.5) [Champetier and Guirardel]

G is a limit of free groups in Gromov-Hausdorff metric.6) G F -embeds into an ultrapower of F .

We will prove Proposition 3, the proof of Proposition 2 is very similar. We willfirst prove the equivalence 1)⇔ 2). Let LA be the language of group theory withgenerators A of F as constants. Let G be a f.g. group which is F -discriminated byF . Consider a formula

∃X(U(X,A) = 1 ∧W (X,A) �= 1).

If this formula is true in F , then it is also true in G, because F ≤ G. If it istrue in G, then for some X ∈ Gm holds U(X, A) = 1 and W (X, A) �= 1. SinceG is F -discriminated by F , there is an F -homomorphism φ : G → F such thatφ(W (X, A)) �= 1, i.e., W (Xφ, A) �= 1. Of course U(Xφ, A) = 1. Therefore theabove formula is true in F . Since in F -group a conjunction of equations [inequali-ties] is equivalent to one equation [resp., inequality], the same existential sentencesin the language LA are true in G and in F .

Suppose now that G is F -universally equivalent to F . Let G = 〈X,A |S(X,A) = 1〉, be a presentation of G and w1(X,A), . . . , wk(X,A) nontrivial el-ements in G. Let Y be the set of the same cardinality as X . Consider a systemof equations S(Y,A) = 1 in variables Y in F . Since the group F is equationallyNoetherian, this system is equivalent over F to a finite subsystem S1(Y,A) = 1.


The formula

Ψ = ∀Y (S1(Y,A) = 1→(w1(Y,A) = 1 ∨ · · · ∨ wk(Y,A) = 1)

).

is false in G, therefore it is false in F . This means that there exists a set ofelements B in F such that S1(B,A) = 1 and, therefore, S(B,A) = 1 such thatw1(B,A) �= 1∧ · · · ∧wk(B,A) �= 1. The map X → B that is identical on F can beextended to the F -homomorphism from G to F .

1)⇔ 3) Let H be an equationally Noetherian CSA-group. We will provethat V (S) is irreducible if and only if HR(S) is discriminated in H by H-homo-morphisms.

Suppose V (S) is not irreducible and V (S) = V (S1) ∪ V (S2) is its decompo-sition into proper subvarieties. Then there exist si ∈ R(Si) \R(Sj), j �= i. The set{s1, s2} cannot be discriminated in H by H-homomorphisms.

Suppose now s1, . . . , sn are elements such that for any retract f : HR(S) → H

there exists i such that f(si) = 1; then V (S) =⋃m

i=1 V (S ∪ si). �Sela [57] defined limit groups as follows. Let G be a f.g. group and let {φj}

be a sequence of homomorphisms from G to a free group F belonging to dis-tinct conjugacy classes (distinct F -homomorphisms belong to distinct conjugacyclasses).

F acts by isometries on its Cayley graph X which is a simplicial tree. Hence,there is a sequence of actions of G on X corresponding to {φj}.

By rescaling metric on X for each φj one obtains a sequence of simplicialtrees {Xj} and a corresponding sequence of actions of G. {Xj} converges to areal tree Y (Gromov-Hausdorff limit) endowed with an isometric action of G. Thekernel of the action of G on Y is defined as

K = {g ∈ G | gy = y, ∀y ∈ Y }.Finally, G/K is said to be the limit group (corresponding to {φj} and rescal-

ing constants). We will prove now the equivalence 1)⇔ 4). A slight modificationof the proof below should be made to show that limit groups are exactly f.g. fullyresidually free groups.

Suppose that G = 〈g1, . . . , gk〉 is f.g. and discriminated by F . There existsa sequence of homomorphisms φn : G → F, so that φn maps the elements in aball of radius n in the Cayley graph of G to distinct elements in F . By rescalingthe metric on F , we obtain a subsequence of homomorphisms φm which convergesto an action of a limit group L on a real tree Y . In general, L is a quotient ofG, but since the homomorphisms were chosen so that φn maps a ball of radius nmonomorphically into F , G is isomorphic to L and, therefore, G is a limit group.

To prove the converse, we need the fact (first proved in [57]) that a f.g. limitgroup is finitely presented. We may assume further that a limit group G is non-Abelian because the statement is, obviously, true for Abelian groups. By definition,there exists a f.g. group H , an integer k and a sequence of homomorphisms hk :H → F , so that the limit of the actions of H on the Cayley graph of F via thehomomorphisms hk is a faithful action of G on some real tree Y . Since G is finitely


presented for all but finitely many n, the homomorphism hn splits through the limitgroup G, i.e., hn = φψn, where φ : H → G is the canonical projection map, andthe ψn’s are homomorphisms ψn : G → F . If g �= 1 in G, then for all but finitelymany ψn’s gψn �= 1. Hence, for every finite set of elements g1, . . . , gm �= 1 in G forall but finitely many indices n, gψn

1 , . . . , gψnm �= 1, so G is F discriminated. �

The equivalence 2)⇔ 6) is a particular case of general results in model theory(see for example [4] Lemma 3.8 Chap.9). �

5)⇔ 6). Champetier and Guirardel [16] used another approach to limit groups.A marked group (G,S) is a group G with a prescribed family of generators

S = (s1, . . . , sn).Two marked groups (G, (s1, . . . , sn)) and (G′, (s′1, . . . , s

′n)) are isomorphic as

marked groups if the bijection si ←→ s′i extends to an isomorphism. For example,(〈a〉, (1, a)) and (〈a〉, (a, 1)) are not isomorphic as marked groups. Denote by Gn

the set of groups marked by n elements up to isomorphism of marked groups.One can define a metric on Gn by setting the distance between two marked

groups (G,S) and (G′, S′) to be e−N if they have exactly the same relations oflength at most N (under the bijection S ←→ S′).

Finally, a limit group in their terminology is a limit (with respect to themetric above) of marked free groups in Gn.

It is shown in [16] that a group is a limit group if and only if it is a finitelygenerated subgroup of an ultraproduct of free groups (for a non-principal ultrafil-ter), and any such ultraproduct of free groups contains all the limit groups. Thisimplies the equivalence 5)⇔ 6). �

Notice that ultrapowers of a free group have the same elementary theory asa free group by Los’ theorem.

First non-free finitely generated examples of fully residually free groups, thatinclude all non-exceptional surface groups, appeared in [2], [3]. They obtained fullyresidually free groups as subgroups of free extensions of centralizers in free groups.

3.2. Lyndon’s completion F Z[t]

Studying equations in free groups Lyndon [33] introduced the notion of a groupwith parametric exponents in an associative unitary ring R. It can be defined asa union of the chain of groups

F = F0 < F1 < · · · < Fn < · · · ,where F = F (X) is a free group on an alphabet X , and Fk is generated by Fk−1and formal expressions of the type

{wα | w ∈ Fk−1, α ∈ R}.That is, every element of Fk can be viewed as a parametric word of the type

wα11 wα2

2 · · ·wαmm ,


where m ∈ N, wi ∈ Fk−1, and αi ∈ R. In particular, he described free exponentialgroups F Z[t] over the ring of integer polynomials Z[t]. Notice that ultrapowers offree groups are operator groups over ultraproducts of Z.

In the same paper Lyndon proved an amazing result that F Z[t] is fully resid-ually free. Hence all subgroups of F Z[t] are fully residually free. Lyndon showedthat solution sets of one variable equations can be described in terms of parametricwords. Later it was shown in [1] that coordinate groups of irreducible one-variableequations are just extensions of centralizers in F of rank one (see the definition inthe second paragraph below). In fact, this result is not entirely accidental, exten-sions of centralizers play an important part here. Recall that Baumslag [2] alreadyused them in proving that surface groups are freely discriminated.

Now, breaking the natural march of history, we go ahead of time and formu-late one crucial result which justifies our discussion on Lyndon’s completion FZ[t]

and highlights the role of the group FZ[t] in the whole subject.Theorem (The Embedding Theorem [38],[39]) Given an irreducible system S = 1over F one can effectively embed the coordinate group FR(S) into FZ[t].

A modern treatment of exponential groups was done by Myasnikov andRemeslennikov [35] who proved that the group FZ[t] can be obtained from F byan infinite chain of HNN-extensions of a very specific type, so-called extensions ofcentralizers:

F = G0 < G1 < · · · < · · · ∪Gi = FZ[t]

whereGi+1 = 〈Gi, ti | [CGi(ui), ti] = 1〉.

(extension of the centralizer CGi(ui), where ui ∈ Gi).This implies that finitely generated subgroups of FZ[t] are, in fact, subgroups

of Gi. Since Gi in an HNN-extension, one can apply Bass-Serre theory to describethe structure of these subgroups. In fact, f.g. subgroups of Gi are fundamentalgroups of graphs of groups induced by the HNN structure of Gi. For instance, itis routine now to show that all such subgroups H of Gi are finitely presented.Indeed, we only have to show that the intersections Gi−1 ∩ Hg are finitely gen-erated. Notice, that if in the amalgamated product amalgamated subgroups arefinitely generated and one of the factors is not, then the amalgamated productis not finitely generated (this follows from normal forms of elements in the amal-gamated products). Similarly, the base group of a f.g. HNN extension with f.g.associated subgroups must be f.g. Earlier Pfander [52] proved that f.g. subgroupsof the free Z[t]-group on two generators are finitely presented. Description of f.g.subgroups of F Z[t] as fundamental groups of graphs of groups implies immediatelythat such groups have non-trivial Abelian splittings (as amalgamated product orHNN extension with Abelian edge group which is maximal Abelian in one of thebase subgroups). Furthermore, these groups can be obtained from free groups byfinitely many free constructions (see next section).

The original Lyndon’s result on fully residual freeness of FZ[t] gives decid-ability of the Word Problem in FZ[t], as well as in all its subgroups. Since any


fully residually free group given by a finite presentation with relations S can bepresented as the coordinate group FR(S) of a coefficient-free system S = 1. TheEmbedding Theorem then implies decidability of WP in arbitrary f.g. fully resid-ually residually free group.

The Conjugacy Problem is also decidable in F Z[t] – but this was provedmuch later, by Ribes and Zalesski in [59]. A similar, but stronger, result is due toLyutikova who showed in [47] that the Conjugacy Problem in FZ[t] is residuallyfree, i.e., if two elements g, h are not conjugate in F Z[t] (or in Gi) then there isan F -epimorphism φ : F Z[t] → F such that φ(g) and φ(h) are not conjugate in F .Unfortunately, this does not imply immediately that the CP in subgroups of F Z[t]

is also residually free, since two elements may be not conjugated in a subgroupH ≤ FZ[t], but conjugated in the whole group F Z[t]. We discuss CP in arbitraryf.g. fully res. free groups in Section 5.

4. Main results in [38]

4.1. Structure and embeddings

In 1996 we proved the converse of the Lyndon’s result mentioned above, everyfinitely generated fully residually free group is embeddable into FZ[t].

Theorem 3 ([38], [39]). Given an irreducible system S = 1 over F one can effec-tively embed the coordinate group FR(S) into FZ[t], i.e., one can find n and anembedding FR(S) → Gn into an iterated centralizer extension Gn.

Corollary 1. For every freely indecomposable non-Abelian finitely generated fullyresidually free group one can effectively find a non-trivial splitting (as an amalga-mated product or HNN extension) over a cyclic subgroup.

Corollary 2. Every finitely generated fully residually free group is finitely presented.There is an algorithm that, given a presentation of a f.g. fully residually free groupG and generators of the subgroup H, finds a finite presentation for H.

Corollary 3. Every finitely generated residually free group G is a subgroup of adirect product of finitely many fully residually free groups; hence, G is embeddableinto FZ[t] × · · · × FZ[t]. If G is given as a coordinate group of a finite system ofequations, then this embedding can be found effectively.

Indeed, there exists a finite system of coefficient free equations S = 1 suchthat G is a coordinate group of this system, and ncl(S) = R(S). If V (S) =∪n

i=1V (Si) is a representation of V (S) as a union of irreducible components, thenR(S) = ∩n

i=1R(Si) and G embeds into a direct product of coordinate groups ofsystems Si = 1, i = 1, . . . , n.

This allows one to study the coordinate groups of irreducible systems ofequations (fully residually free groups) via their splittings into graphs of groups.This also provides a complete description of finitely generated fully residually freegroups and gives a lot of information about their algebraic structure. In particular,


they act freely on Zn-trees, and all these groups, except for Abelian and surfacegroups, have a non-trivial cyclic JSJ-decomposition.

Let K be an HNN-extension of a group G with associated subgroups A andB. K is called a separated HNN-extension if for any g ∈ G, Ag ∩B = 1.

Corollary 4. Let a group G be obtained from a free group F by finitely manycentralizer extensions. Then every f.g. subgroup H of G can be obtained from freeAbelian groups of finite rank by finitely many operations of the following type:free products, free products with Abelian amalgamated subgroups at least one ofwhich is a maximal Abelian subgroup in its factor, free extensions of centralizers,separated HNN-extensions with Abelian associated subgroups at least one of whichis maximal.

Corollary 5 (Groves, Wilton [28]). One can enumerate all finite presentations offully residually free groups.

Theorem 3 is proved as a corollary of Theorem 6 below.

Corollary 6. Every f.g. fully residually free group acts freely on some Zn-tree withlexicographic order for a suitable n.

Hence, a simple application of the change of the group functor shows thatH also acts freely on an Rn-tree. Recently, Guirardel proved the latter resultindependently using different techniques [24]. It is worthwhile to mention herethat free group actions on Zn-trees give a tremendous amount of information onthe group and its subgroups, especially with regard to various algorithmic problems(see Section 5).

Notice, that there are f.g. groups acting freely on Zn-trees which are notfully residually free (see conjecture (2) from Sela’s list of open problems). Thesimplest example is the group of closed non-orientable surface of genus 3. In fact,the results in [45, 46] show that there are very many groups like that – the class ofgroups acting freely on Zn-trees is much wider than the class of fully residually freegroups. This class deserves a separate discussion, for which we refer to [45, 46].Combining Corollary 4 with the results from [40] or [8] we proved in [38] thatf.g. fully residually free groups without subgroups Z × Z (or equivalently, withcyclic maximal Abelian subgroups) are hyperbolic. We will see in Section 4.2 thatthis has some implication on the structure of the models of the ∀∃-theory of agiven non-Abelian free group. Later, Dahmani [21] proved a generalization of this,namely, that an arbitrary f.g. fully residually free group is hyperbolic relative toits maximal Abelian non-cyclic subgroups.Recently N. Touikan described coordinate groups of two-variable equations [61].

4.2. Triangular quasi-quadratic systems

We use an Elimination Process to transform systems of equations. EliminationProcess EP is a symbolic rewriting process of a certain type that transforms for-mal systems of equations in groups or semigroups. Makanin (1982) introduced theinitial version of the EP. This gives a decision algorithm to verify consistency of a


given system – decidability of the Diophantine problem over free groups. He esti-mates the length of the minimal solution (if it exists). Makanin introduced the fun-damental notions: generalized equations, elementary and entire transformations,notion of complexity. Razborov (1987) developed EP much further. Razborov’sEP produces all solutions of a given system in F . He used special groups of auto-morphisms, and fundamental sequences to encode solutions.

We obtained in 1996 [38] an effective description of solutions of equationsin free (and fully residually free ) groups in terms of very particular triangularsystems of equations. First, we give a definition.Triangular quasi-quadratic (TQ) system is a finite system that has the followingform

S1(X1, X2, . . . , Xn, A) = 1,

S2(X2, . . . , Xn, A) = 1,· · ·

Sn(Xn, A) = 1where either Si = 1 is quadratic in variables Xi, or Si = 1 is a system of commu-tativity equations for all variables from Xi and, in addition, equations [x, u] = 1for all x ∈ Xi and some u ∈ FR(Si+1,...,Sn) or Si is empty.

A TQ system above is non-degenerate (NTQ) if for every i, Si(Xi, . . .,Xn, A) = 1 has a solution in the coordinate group FR(Si+1,...,Sn), i.e., Si = 1 (inalgebraic geometry one would say that a solution exists in a generic point of thesystem Si+1 = 1, . . . , Sn = 1).

We proved in [37] (see also [39]) that NTQ systems are irreducible and, there-fore, their coordinate groups (NTQ groups) are fully residually free. (Later Selacalled NTQ groups ω-residually free towers [57].)

We represented a solution set of a system of equations canonically as a unionof solutions of a finite family of NTQ groups.

Theorem 4 ([38], [39]). One can effectively construct EP that starts on an arbitrarysystem

S(X,A) = 1and results in finitely many NTQ systems

U1(Y1) = 1, . . . , Um(Ym) = 1

such thatVF (S) = P1(V (U1)) ∪ · · · ∪ Pm((Um))

for some word mappings P1, . . . , Pm. (Pi maps a tuple Yi ∈ V (Ui) to a tuple X ∈V (S). One can think about Pi as an A-homomorphism from FR(S)into FR(Ui), thenany solution ψ : FR(Ui) → F pre-composed with Pi gives a solution φ : FR(S) → F ).

Our elimination process can be viewed as a non-commutative analog of theclassical elimination process in algebraic geometry.

Hence, going “from the bottom to the top” every solution of the subsystemSn = 1, . . . , Si = 1 can be extended to a solution of the next equation Si−1 = 1.


Theorem 5 ([38], [39]). All solutions of the system of equations S = 1 in F (A)can be effectively represented as homomorphisms from FR(S) into F (A) encodedinto the following finite canonical Hom-diagram. Here all groups, except, maybe,the one in the root, are fully residually free, (given by a finite presentation) ar-rows pointing down correspond to epimorphisms (defined effectively in terms ofgenerators) with non-trivial kernels, and loops correspond to automorphisms ofthe coordinate groups.

FR(S)

��

��

FR(Ωv1 )

σ1

��

��

��FR(Ωv2 ) · · · FR(Ωvn )

FR(Ωv21 ) · · · FR(Ωv2m )

��

��

σ2

��

· · ·

· · · FR(Ωvk)

F (A) ∗ F (T )

F (A)

A family of homomorphisms encoded in a path from the root to a leaf of thistree is called a fundamental sequence or fundamental set of solutions (because eachhomomorphism in the family is a composition of a sequence of automorphisms andepimorphisms). Later Sela called such family a resolution. Therefore the solutionset of the system S = 1 consists of a finite number of fundamental sets. And eachfundamental set “factors through” one of the NTQ systems from Theorem 4. IfS = 1 is irreducible, or, equivalently, G = FR(S) is fully residually free, then,obviously, one of the fundamental sets discriminates G. This gives the followingresult.

Theorem 6 ([38], [39]). Finitely generated fully residually free groups are subgroupsof coordinate groups of NTQ systems. There is an algorithm to construct an em-bedding.

This corresponds to the extension theorems in the classical theory of elimina-tion for polynomials. In [37] we have shown that an NTQ group can be embeddedinto a group obtained from a free group by a series of extensions of centralizers.Therefore Theorem 3 follows from Theorem 6.

Since NTQ groups are fully residually free, fundamental sets correspondingto different NTQ groups in Theorem 4 discriminate fully residually free groups


which are coordinate groups of irreducible components of system S(X,A) = 1.This implies

Theorem 7 ([38], [39]). There is an algorithm to find irreducible components for asystem of equations over a free group.

Now we will formulate a technical result which is the keystone in the proof ofTheorems 4 and 5. An elementary Abelian splitting of a group is the splitting as anamalgamated product or HNN-extension with Abelian edge group. Let G = A∗CBbe an elementary Abelian splitting of G. For c ∈ C we define an automorphismφc : G→ G such that φc(a) = a for a ∈ A and φc(b) = bc = c−1bc for b ∈ B.

If G = A∗C = 〈A, t|ct = c′, c ∈ C〉 then for c ∈ C define φc : G → G suchthat φc(a) = a for a ∈ A and φc(t) = ct.

We call φc a Dehn twist obtained from the corresponding elementary Abeliansplitting of G. If G is an F -group, where F is a subgroup of one of the factors A orB, then Dehn twists that fix elements of the free group F ≤ A are called canonicalDehn twists.

If G = A ∗C B and B is a maximal Abelian subgroup of G, then everyautomorphism of B acting trivially on C can be extended to the automorphism ofG acting trivially on A. The subgroup of Aut(G) generated by such automorphismsand canonical Dehn twists is called the group of canonical automorphisms of G.

Let G and K be H-groups and A ≤ AutH(G) a group of H-automorphismsof G. Two H-homomorphisms φ and ψ from G into K are called A-equivalent(symbolically, φ ∼A ψ) if there exists σ ∈ A such that φ = σψ (i.e., gφ = gσψ forg ∈ G). Obviously, ∼A is an equivalence relation on HomH(G,K).

Let G be a fully residually F group (F = F (A) ≤ G) generated by a finite setX (over F ) andA the group of canonical F automorphisms of G. Let F = F (A∪Y )a free group with basis A∪Y (here Y is an arbitrary set) and φ1, φ2 ∈ HomF (G, F ).We write φ1 < φ2 if there exists an automorphism σ ∈ A and an F -endomorphismπ ∈ HomF (F , F ) such that φ2 = σ−1φ1π and∑

x∈X

|xφ1 | <∑x∈X

|xφ2 |.

F F

FR(S)

FR(S)

Figure 1. φ1 < φ2


An F -homomorphism φ : G→ F is called minimal if there is no φ1 such thatφ1 < φ. In particular, if S(X,A) = 1 is a system of equations over F = F (A) andG = FR(S) then X ∪ A is a generating set for G over F . In this event, one canconsider minimal solutions of S = 1 in F .

Definition 1. Denote by RA the intersection of the kernels of all minimal (withrespect to A) F -homomorphisms from HomF (G, F ). Then G/RA is called themaximal standard quotient of G and the canonical epimorphism η : G→ G/RA isthe canonical projection.

Theorem 8 ([38]). The maximal standard quotient of a finitely generated fullyresidually free group is a proper quotient and can be effectively constructed.

This result (without the algorithm) is called the “shortening argument” inSela’s approach.

5. Elimination process

Given a system S(X) = 1 of equations in a free group F (A) one can effectivelyconstruct a finite set of generalized equations

Ω1, . . . ,Ωk

(systems of equations of a particular type) such that:• given a solution of S(X) = 1 in F (A) one can effectively construct a reducedsolution of one of Ωi in the free semigroup with basis A ∪A−1.

• given a solution of some Ωi in the free semigroup with basis A∪A−1 one caneffectively construct a solution of S(X) = 1 in F (A).This is done as follows. First, we replace the system S(X) = 1 by a system of

equations, such that each of them has length 3. This can be easily done by addingnew variables. For one equation of length 3 we can construct a generalized equa-tion as in the example below. For a system of equations we construct it similarly(see [38]).Example. Suppose we have the simple equation xyz = 1 in a free group. Supposethat we have a solution to this equation denoted by xφ, yφ, zφ where is φ is a givenhomomorphism into a free group F (A). Since xφ, yφ, zφ are reduced words in thegeneratorsA there must be complete cancellation. If we take a concatenation of thegeodesic subpaths corresponding to xφ, yφ and zφ we obtain a path in the Cayleygraph corresponding to this complete cancellation. This is called a cancellationtree. Then xφ = λ1 ◦ λ2, yφ = λ−12 ◦ λ3 and zφ = λ−13 ◦ λ−11 , where u ◦ v denotesthe product of reduced words u and v such that there is no cancellation between uand v. In the case when all the words λ1, λ2, λ3 are non-empty, the combinatorialgeneralized equation is shown at the bottom of Fig. 2.

Given a generalized equation Ω one can apply elementary transformations(there are only finitely many of them) and get a new generalized equation Ω′. If σ


x y

z

1

2

xyz = 1

3

1 13 32 2

x y z

Figure 2. From the cancellation tree for the equation xyz = 1 to thegeneralized equation (xφ = λ1 ◦ λ2, yφ = λ−12 ◦ λ3, zφ = λ−13 ◦ λ−11 ).

is a solution of Ω, then elementary transformation transforms σ into σ′.

(Ω, σ)→ (Ω′, σ′).

Elimination process is a branching process such that on each step one of the finitenumber of elementary transformations is applied according to some precise rulesto a generalized equation on this step.

Ω0 → Ω1 → · · · → Ωk.

From the group theoretic view-point the elimination process tells somethingabout the coordinate groups of the systems involved.

This allows one to transform the pure combinatorial and algorithmic resultsobtained in the elimination process into statements about the coordinate groups.

5.1. Generalized equations

Definition 2. A combinatorial generalized equation Ω (which is convenient to vi-sualize as on the picture)

1 2 3 ρ ρ+ 1

�

��

�

λλ

μμ

consists of the following components:


1. A finite set of bases BS = BS(Ω). The set of bases M consists of 2n ele-ments M = {μ1, . . . , μ2n}. The set M comes equipped with two functions:a function ε :M→ {1,−1} and an involution Δ :M→M (that is, Δ is abijection such that Δ2 is an identity on M). Bases μ and μ are called dualbases. We denote bases by letters μ, λ, etc.

2. A set of boundaries BD = BD(Ω). BD is a finite initial segment of the setof positive integers BD = {1, 2, . . . , ρ+1+m}, where m is the cardinality ofthe basis A = {a1, . . . , am} of the free group F = F (A). We use letters i, j,etc. for boundaries. (The example above has nine boundaries, ρ = 8,m = 0.)

3. Two functions α : BS → BD and β : BS → BD. We call α(μ) and β(μ)the initial and terminal boundaries of the base μ (or endpoints of μ). Thesefunctions satisfy the following conditions for every base μ ∈ BS: α(μ) < β(μ)if ε(μ) = 1 and α(μ) > β(μ) if ε(μ) = −1. (In the example α(λ) = 1,β(λ) = 4.)

4. The set of boundary connections (p, λ, q), where p is a boundary on λ (thatis a number between α(λ) and β(λ)) and q is a boundary on λ. If (p, λ, q)is a boundary connection then (q, λ, p) is also a boundary connection. (Themeaning of the boundary connections will be explained in ET5. To our exam-ple we can add some boundary connection, say (2, λ, 6). For the generalizedequation to be consistent it is necessary that in the case ε(λ) = ε(λ), p1 > p2implies q1 > q2 and in the case ε(λ) = −ε(λ), p1 > p2 implies q1 < q2.) Aboundary p is λ-tied if there is a boundary connection (p, λ, q) for some q.

For a combinatorial generalized equation Ω, one can canonically associate asystem of equations in variables h1, . . . , hρ over F (A) (variables hi are sometimescalled items). This system is called a generalized equation, and (slightly abusingthe language) we denote it by the same symbol Ω. The generalized equation Ωconsists of the following three types of equations.1. Each pair of dual bases (λ, λ) provides an equation

[hα(λ)hα(λ)+1 . . . hβ(λ)−1]ε(λ)

= [hα(λ)hα(λ)+1 . . . hβ(λ)−1]ε(λ).

These equations are called basic equations.2. Every boundary connection (p, λ, q) gives rise to a boundary equation

[hα(λ)hα(λ)+1 · · ·hp−1] = [hα(λ)hα(λ)+1 · · ·hq−1],

if ε(λ) = ε(λ) and

[hα(λ)hα(λ)+1 · · ·hp−1] = [hqhq+1 · · ·hβ(λ)−1]−1,

if ε(λ) = −ε(λ).3. Constant equations: hρ+1 = a1, . . . , hρ+1+m = am.

Remark 1. We assume that every generalized equation comes associated with acombinatorial one.


Denote by FR(Ω) the coordinate group of the generalized equation.

Definition 3. Let Ω(h) = {L1(h) = R1(h), . . . , Ls(h) = Rs(h)} be a generalizedequation in variables h = (h1, . . . , hρ). A sequence of reduced nonempty words inan extended alphabet (A∪Z)±1, U = (U1(Z,A), . . . , Uρ(Z,A)) is a solution of Ω if:1. all words Li(U), Ri(U) are reduced as written,2. Li(U) = Ri(U), i ∈ [1, s].

If we specify a particular solution δ of a generalized equation Ω then we usea pair (Ω, δ).

5.2. Elementary transformations

In this section we describe elementary transformations of generalized equations.Let Ω be a generalized equation. An elementary transformation (ET ) associates toa generalized equation Ω a family of generalized equations ET (Ω) = {Ω1, . . . ,Ωk}and surjective homomorphisms πi : FR(Ω) → FR(Ωi) such that for any solution δof Ω and corresponding epimorphism πδ : FR(Ω) → F there exists i ∈ {1, . . . , k}and a solution δi of Ωi such that the following diagram commutes.

FR(Ω) FR(Ωi)

F ∗ F (Z)

�πi

�

πδ

��

πδi

(ET1) (Cutting a base (see Fig. 3)). Let λ be a base in Ω and p an internalboundary of λ with a boundary connection (p, λ, q). Then we cut thebase λ in p into two new bases λ1 and λ2 and cut λ in q into the basesλ1, λ2.

(ET2) (Transferring a base (see Fig. 4)). If a base λ of Ω contains a base μ(that is, α(λ) ≤ α(μ) < β(μ) ≤ β(λ)) and all boundaries on μ are λ-tied by boundary connections then we transfer μ from its location onthe base λ to the corresponding location on the base λ.

(ET3) (Removal of a pair of matched bases (see Fig. 5)). If the bases λ and λare matched (that is, α(λ) = α(λ), β(λ) = β(λ)) then we remove λ, λfrom Ω.

Remark 2. Observe, that for i = 1, 2, 3, we have k = 1, ETi(Ω) and Ω havethe same set of variables H , and the identity map F [H ] → F [H ] induces anisomorphism π : FR(Ω) → FR(Ω′). Moreover, δ is a solution of Ω if and only if δ isa solution of Ω′.

(ET4) (Removal of a lone base (see Fig. 6)). Suppose, a base λ in Ω does not in-tersect any other base, that is, the items hα(λ), . . . , hβ(λ)−1 are containedonly on the base λ. Suppose also that all boundaries in λ are λ-tied, i.e.,


21

12

1 2 3 4 5 6 7 8 9

1 2 3 4 5 6 7 8 9

"CUT BASE"

Figure 3. Elementary transformation (ET1).

for every i (α(λ) < i ≤ β(λ)− 1) there exists a boundary b(i) such that(i, λ, b(i)) is a boundary connection in Ω. Then we remove the pair ofbases λ and λ together with all the boundaries α(λ) + 1, . . . , β(λ) − 1(and rename the rest of the boundaries correspondingly).We define the homomorphism π : FR(Ω) → FR(Ω′) as follows:

π(hj) = hj if j < α(λ) or j ≥ β(λ)

π(hi) =

{hb(i) . . . hb(i+1)−1, if ε(λ) = ε(λ),

h−1b(i) . . . h−1b(i+1)−1, if ε(λ) = −ε(λ)

for α(λ) ≤ i ≤ β(λ)− 1. It is not hard to see that π is an isomorphism.(ET5) (Introduction of a boundary (see Fig. 7)). Suppose a boundary p in a

base λ is not λ-tied. The transformation (ET5) λ-ties it. To this end,suppose δ is a solution of Ω. Denote λδ by Uλ, and let U ′λ be the be-ginning of this word ending at p. Then we perform one of the followingtransformations according to where the end of U ′λ on λ might be situ-ated:

(a) If the end of U ′λ on λ is situated on the boundary q, we introducethe boundary connection 〈p, λ, q〉. In this case the correspondinghomomorphism πq : FR(Ω) → FR(Ωq) is induced by the identityisomorphism on F [H ]. Observe that θq is not necessary an isomor-phism.


01 2 3 4 5 6 7 8 9

"TRANSFER BASE"

1

01 2 3 4 5 6 7 8 9 1


(b) If the end of U ′λ on λ is situated between q and q + 1, we in-troduce a new boundary q′ between q and q + 1 (and rename allthe boundaries); introduce a new boundary connection (p, λ, q′).Denote the resulting equation by Ω′q. In this case the correspond-ing homomorphism πq′ : FR(Ω) → FR(Ωq′ ) is induced by the mapπq′(h) = h, if h �= hq, and πq′(hq) = hq′hq′+1. Observe that πq′ isan isomorphism.Obviously, the is only a finite number of possibilities such that forany solution δ one of them takes place.

5.3. Derived transformations and auxiliary transformations

In this section we define complexity of a generalized equation and describe severaluseful “derived” transformations of generalized equations. Some of them can berealized as finite sequences of elementary transformations, others result in equiva-lent generalized equations but cannot be realized by finite sequences of elementarymoves.

A boundary is open if it is an internal boundary of some base, otherwise it isclosed. A section is an interval σ = [i, . . . , i+k]. It is said to be closed if boundariesi and i+ k are closed and all the boundaries between them are open.

Sometimes it will be convenient to subdivide all sections of Ω into active andnon-active sections. Constant section will always be non-active. A variable hq is


ET3

01 2 3 4 5 6 7 8 9 1

01 2 3 4 5 6 7 8 9 1


called free if it meets no base. Free variables are transported to the very end ofthe interval behind all items in Ω and become non-active.(D1) (Deleting a complete base). A base μ of Ω is called complete if there exists a

closed section σ in Ω such that σ = [α(μ), β(μ)].Suppose μ is a complete base of Ω and σ is a closed section such that

σ = [α(μ), β(μ)]. In this case using (ET5), we transfer all bases from μ to μ;using (ET4), we remove the lone base μ together with the section σ(μ).

Complexity. Denote by n(σ) the number of bases in a closed section σ.The complexity of an equation Ω is the number

τ = τ(Ω) =∑

σ∈AΣΩ

max{0, n(σ)− 2},

where AΣΩ is the set of all active closed sections.(D2) (Linear elimination). Let γ(hi) denote the number of bases met by hi. A base

μ ∈ BS(Ω) is called eliminable if at least one of the following holds:(a) μ contains an item hi with γ(hi) = 1,(b) at least one of the boundaries α(μ), β(μ) is different from 1, ρ+1, is not

an endpoint of any other base, and is not connected by any boundaryconnection.

We denote this boundary by ε. A linear elimination for Ω works as follows.Suppose the base μ is removable because it satisfies condition (b). We first

cut μ at the nearest to ε μ-tied boundary and denote it by τ . If there is no such


"DELETE LONE BASE"

01 2 3 4 5 6 7 8 9 1

1 2 3 4 5 6


a boundary we denote by τ the other boundary of μ. Then we remove the baseobtained from μ between ε and τ together with its dual (maybe this part is thewhole base μ), and remove the boundary ε. Denote the new equation by Ω′.

Suppose the base is removable because it satisfies condition (a).Suppose first that γ(hi) = 1 for the leftmost item hi on μ. Denote by ε the left

boundary of hi. Let τ be the nearest to ε μ-tied boundary (or the other terminalboundary of μ if there are no μ-tied boundaries). We remove the base obtainedfrom μ between ε and τ together with its dual (maybe this part is the whole baseμ), and remove hi.

We make a mirror reflection of this transformation if γ(hi) = 1 for the right-most item hi on μ.

Suppose now that hi is not the leftmost or the rightmost item on μ. Let εand τ be the nearest to hi μ-tied boundaries on the left and on the right of hi

(each of them can be a terminal boundary of μ). We cut μ at the boundaries ε andτ , remove the base between ε and τ together with its dual and remove hi.

Lemma 1. Linear elimination does not increase the complexity of Ω, and the num-ber of items decreases. Therefore the linear elimination process stops after finitenumber of steps.

Proof. The input of the closed sections not containing μ into the complexity doesnot change. The section than contained μ could be divided into two. The totalnumber of bases can increase only if μ is eliminable according to case (a) and hi


ET5

ET

5

ET5

01 2 3 4 5 6 7 8 9 1 101 2 3 4 5 6 7 8 9 1 1

101 2 3 4 5 6 7 8 9 1 101 2 3 4 5 6 7 8 9 1


is not the leftmost or the rightmost item on μ. In this case the number of basesis increased by two, but the section is divided into two closed sections, and eachsection contains at least two bases. Therefore the complexity is the same. In allother cases the total number of bases does not increase, therefore the complexitycannot increase too. The number of items every time is decreased by one.

We repeat linear elimination until no eliminable bases are left in the equation.The resulting generalized equation is called a kernel of Ω and we denote it byKer(Ω). It is easy to see that Ker(Ω) does not depend on a particular linearelimination process. Indeed, if Ω has two different eliminable bases μ1, μ2, anddeletion of a part of μi results in an equation Ωi then by induction (on the numberof eliminations) Ker(Ωi) is uniquely defined for i = 1, 2. Obviously, μ1 is stilleliminable in Ω2, as well as μ2 is eliminable in Ω1. Now eliminating μ1 and μ2from Ω2 and Ω1 we get one and the same equation Ω0. By induction Ker(Ω1) =Ker(Ω0) = Ker(Ω2) hence the result.

The following statement becomes obvious.

Lemma 2. The generalized equation Ω (as a system of equations over F ) has asolution if and only if Ker(Ω) has a solution.

So linear elimination replaces Ω by Ker(Ω).Let us consider what happens on the group level in the process of linear elim-

ination. This is necessary only for the description of all solutions of the equation.


We say that a variable hi belongs to the kernel (hi ∈ Ker(Ω)), if either hi

belongs to at least one base in the kernel, or it is constant.Also, for an equation Ω by Ω we denote the equation which is obtained from

Ω by deleting all free variables. Obviously,

FR(Ω) = FR(Ω) ∗ F (Y )

where Y is the set of free variables in Ω.We start with the case when a part of just one base is eliminated. Let μ be an

eliminable base in Ω = Ω(h1, . . . , hρ). Denote by Ω1 the equation resulting from Ωby eliminating μ.

(a) Suppose hi ∈ μ and γ(hi) = 1. Let μ = μ1 . . . μk, where μ1, . . . , μk are theparts between μ-tied boundaries. Let hi ∈ μj . Replace the basic equationcorresponding to μ by the equations corresponding to μ1, . . . , μk. Then thevariable hi occurs only once in Ω – precisely in the equation sμj = 1 corre-sponding to μj . Therefore, in the coordinate group FR(Ω) the relation sμj = 1can be written as hi = w, where w does not contain hi. Using Tietze trans-formations we can rewrite the presentation of FR(Ω) as FR(Ω′), where Ω′ isobtained from Ω by deleting sμj and the item hi. It follows immediately that

FR(Ω1) FR(Ω′) ∗ 〈hi〉and

FR(Ω) FR(Ω′) FR(Ω1)∗ F (B) (1)

for some free or trivial group F (B).(b) Suppose now that μ satisfies case (b) in (D2) with respect to a boundary i.

Let μ = μ1 . . . μk. Replace the equation sμ = 1 and the boundary equationscorresponding to the boundary connections through μ by the equations sμi ,i = 1, . . . , k. Then in the equation sμk

= 1 the variable hi−1 either occursonly once or it occurs precisely twice and in this event the second occurrenceof hi−1 (in μ) is a part of the subword (hi−1hi)±1. In both cases it is easy tosee that the tuple

(h1, . . . , hi−2, sμk, hi−1hi, hi+1, . . . , hρ)

forms a basis of the ambient free group generated by (h1, . . . , hρ) and con-stants from A. Therefore, eliminating the relation sμk

= 1, we can rewrite thepresentation of FR(Ω) in generators Y = (h1, . . . , hi−2, hi−1, hi, hi+1, . . . , hρ).Observe also that any other basic or boundary equation sλ = 1 (λ �= μ) of Ωeither does not contain variables hi−1, hi or it contains them as parts of thesubword (hi−1hi)±1, that is, any such a word sλ can be expressed as a wordwλ(Y ) in terms of generators Y . This shows that

FR(Ω) G(Y )R(wλ(Y )|λ =μ) FR(Ω′),

where Ω′ is a generalized equation obtained from Ω1 by deleting the boundaryi. Denote by Ω′ an equation obtained from Ω′ by adding a free variable z to


the right end of Ω′. It follows now that

FR(Ω1) FR(Ω′′) FR(Ω) ∗ 〈z〉and

FR(Ω) FR(Ω′) ∗ F (Z) (2)

or some free group F (Z). Notice that all the groups and equations whichoccur above can be found effectively.By induction on the number of steps in an elimination process we obtain the

following lemma.

Lemma 3. FR(Ω) FR(Ker(Ω)) ∗ F (Z), where F (Z) is a free group on Z.

Proof. LetΩ = Ω0 → Ω1 → · · · → Ωl = Ker(Ω)

be a linear elimination process for Ω. It is easy to see (by induction on l) that forevery j ∈ [0, l− 1]

Ker(Ωj) = Ker(Ωj).Moreover, if Ωj+1 is obtained from Ωj as in the case (b) above, then (in thenotation above)

Ker(Ωj)1 = Ker(Ω′j).Now the statement of the lemma follows from the remarks above and equalities(1) and (2). �

5.4. Rewriting process for ΩIn this section we describe a rewriting process for a generalized equation Ω.

5.4.1. Tietze cleaning and entire transformation. In the rewriting process of gen-eralized equations there will be two main sub-processes:1. Titze cleaning. This process consists of repetition of the following four transfor-mations performed consecutively:(a) Linear elimination,(b) deleting all pairs of matched bases,(c) deleting all complete bases,(d) moving all free variables to the right and declare them non-active.2. Entire transformation. This process is applied if γ(hi) ≥ 2 for each hi in theactive sections. We need a few further definitions. A base μ of the equation Ω iscalled a leading base if α(μ) = 1. A leading base is said to be maximal (or a carrier)if β(λ) ≤ β(μ), for any other leading base λ. Let μ be a carrier base of Ω. Anyactive base λ �= μ with β(λ) ≤ β(μ) is called a transfer base (with respect to μ).

Suppose now that Ω is a generalized equation with γ(hi) ≥ 2 for each hi

in the active part of Ω. An entire transformation is a sequence of elementarytransformations which are performed as follows. We fix a carrier base μ of Ω. Wetransfer all transfer bases from μ onto μ. Now, there exists some i < β(μ) suchthat h1, . . . , hi belong to only one base μ, while hi+1 belongs to at least two bases.


Applying (ET1) we cut μ along the boundary i + 1. Finally, applying (ET4) wedelete the section [1, i+ 1].

Notice that neither process increases complexity.

5.4.2. Solution tree. Let Ω be a generalized equation. We construct a solutiontree T (Ω) (with associated structures), as a rooted tree oriented from the root v0,starting at v0 and proceeding by induction on the distance n from the root.

Letv → v1 → · · · → vs → u

be a path in T (Ω). If vi → vi+1 is an edge then there exists a finite sequenceof elementary or derived transformations from Ωvi to Ωvi+1 and the epimorphismπ(vi, vi+1) is the composition of the epimorphisms corresponding to these trans-formations. By π(v, u) we denote the composition of epimorphisms

π(v, u) = π(v, v1) · · ·π(vs, u).We also assume that active [non-active] sections in Ωvi+1 are naturally inher-

ited from Ωvi , if not said otherwise.Suppose a path in T (Ω) is constructed by induction up to a level n, and

suppose v is a vertex at distance n from the root v0. We describe now how toextend the tree from v.

We apply the Tietze cleaning at the vertex vn if it is possible. If it is im-possible (γ(hi) ≥ 2 for any hi in the active part of Ωv), we apply the entiretransformation. Both possibilities involve either creation of new boundaries andboundary connections or creation of new boundary connections without creation ofnew boundaries, and, therefore, addition of new relations to FR(Ωv). The boundaryconnections can be made in few different ways, but there is a finite number of pos-sibilities. According to this, different resulting generalized equations are obtained,and we draw edges from v to all the vertices corresponding to these generalizedequations.

Termination condition: 1. Ωv does not contain active sections. In this case thevertex v is called a leaf or an end vertex. There are no outgoing edges from v.

2. Ωv is inconsistent. There is a base λ such that λ is oriented the oppositeway and overlaps with λ, or the equation implies an inconsistent constant equation.

5.4.3. Quadratic case. Suppose Ωv satisfies the condition γi = 2 for each hi in theactive part. Then FR(Ωv) is isomorphic to the free product of a free group and acoordinate group of a standard quadratic equation (to be defined below) over thecoordinate group FR(Ω′) of the equation Ω′ corresponding to the non-active part.In this case entire transformation can go infinitely along some path in T (Ω), and,since the number of bases if fixed, there will be vertices v and w such that Ωv andΩw are the same. Then the corresponding epimorphism π : FR(Ωv) → FR(Ωw) isan automorphism of FR(Ωv) that decreases the total length of the interval. For aminimal solution of FR(Ωv) the process will stop.


Definition 4. A standard quadratic equation over the group G is an equation ofthe one of the following forms (below d, ci are nontrivial elements from G):

n∏i=1

[xi, yi] = 1, n > 0; (3)

n∏i=1

[xi, yi]m∏i=1

z−1i cizid = 1, n,m ≥ 0,m+ n ≥ 1; (4)

n∏i=1

x2i = 1, n > 0; (5)

n∏i=1

x2i

m∏i=1

z−1i cizid = 1, n,m ≥ 0, n+m ≥ 1. (6)

Equations (3), (4) are called orientable, equations (5), (6) are called non-orientable. Number n is called a genus of the equation (notation gen(S).)

The proof of the following fact can be found in [29].

Lemma 4. Let W be a strictly quadratic word over G (each variable occurs exactlytwice). Then there is a G-automorphism f ∈ AutG(G[X ]) such that W f is astandard quadratic word over G.

5.4.4. Entire transformation goes infinitely. Let now γ(hi) ≥ 2 for all hi in theactive part, and for some hi the inequality is strict. Let (Ω, δ) be a generalizedequation with a solution δ in the alphabet (A ∪ Z)±1. A base participates in theentire transformation if it is a leading base or a transfer base.

It is possible that the cleaning after the entire transformation decreases com-plexity. This occurs if some base is transferred onto its dual and removed by (ET3).Otherwise, we use the same name for a base of Ωi and the reincornation of thisbase in Ωi+1. If we cannot apply Tietze cleaning after the entire transformation,then we successively apply entire transformation. It is possible that the entiretransformation sequence for Ω goes infinitely. Suppose the entire transformationgoes infinitely along some path in T (Ω), then after a finite number of steps, everybase that participates, actually, participates infinitely often.

Define the excess ψ of (Ω, δ):

ψ = Σλ(λδ)− 2|Iδ|,where λ runs through the set of bases participating in the entire transformationsequence and I = [h1, . . . , hk] is the segment between the initial boundary of theinterval and the leftmost boundary k of the base that never participates (as carrieror transfer base).

It is possible that the entire transformation goes infinitely and the complex-ity does not decrease. If we apply the entire transformation to (Ω, δ) and thecomplexity does not decrease, then ψ does not change.


We say that bases μ and its dual of the equation Ω form an overlapping pairif μ intersects with its dual μ.

If φ1 and φ2 are two solutions of a generalized equation Ω in F (A,Z), thenwe define φ1 <A φ2 if φ2 = σφ1π, where σ is an automorphism in a subgroup A ofthe group of automorphisms of FR(Ω) and π is an endomorphism of F (A,Z), and∑ρ

i=1(hi)φ1 <∑ρ

i=1(hi)φ2 . Then we can define minimal solutions of a generalizedequation with respect to A. Solutions minimal with respect to the whole group ofautomorphisms of FR(Ω) are called minimal solutions.

Theorem 9 ([38]). Let (Ω, δ) be a generalized equation with a minimal solution.Suppose (Ω, δ) = (Ω0, δ0), (Ω1, δ1), . . . be the generalized equations (with solutions)formed by the entire transformation sequence. Then one can construct a numberN = N(Ω) such that the sequence ends after at most N steps.

We can temporary change generalized equation Ω in such a way that it con-sists of one or several quadratic closed sections (such that γ(hi) = 2 for any hi) andnon-quadratic sections (such that γ(hi) > 2 for any hi). Indeed, if σ is a quadraticsection of Ω, we can cut all bases in Ω through the end-points of σ. Moreover, wewill put the non-quadratic sections on the right part of the interval. Denote by Ω1this new generalized equation.

We will prove this theorem after proving key Lemmas 5-8.

Lemma 5. If δ is a solution minimal with respect to the subgroup A generated bythe canonical Dehn twists corresponding to the quadratic part of Ω, then one canconstruct a recursive function f = f(Ω) such that |Iδ| ≤ fψ.

This lemma shows that for a minimal solution the length of the participatingpart of the interval is bounded in terms of the excess. And the excess does notchange in the sequence of entire transformations when the complexity does notdecrease.

Proof. We apply the entire transformation to the pair (Ω1, δ1), where δ1 is obtainedfrom δ, and, therefore, minimal. We can find a number k(Ω) such that after ktransformations

(Ω1, δ1)→ · · · → (Ωk, δk)all bases situated on the quadratic part will either form matched pairs or will betransferred to the non-quadratic part. Indeed, while we transforming the quadraticpart we notice that:1) two equations Ωi and Ωj for i < j cannot be the same, because then δj would

be shorter than δi, contradicting the minimality.2) there is only a finite number of possibilities for the quadratic part since the

number of items and complexity does not increase.The sequence of consecutive quadratic carrier bases is bounded. Therefore

after a bounded number of steps, a quadratic coefficient base is carrier, and wetransfer a transfer base to the non-quadratic part. For a minimal solution, thelength of a free variable corresponding to a matching pair is 1. And for each base λ


transferred to the non-quadratic part, λδ is shorter than the interval correspondingto the non-quadratic part, and, therefore, shorter than ψ. This gives a hint how tocompute a function f(Ω). We can now return to the generalized equation Ω andreplace its solution by a minimal solution δ. �

The exponent of periodicity of a family of reduced words {w1, . . . , wk} in afree group F is the maximal number t such that some wi contains a subword ut forsome simple cyclically reduced word u. The exponent of periodicity of a solutionδ is the exponent of periodicity of the family {hδ

1, . . . , hδρ}.

We call a solution of a system of equations in the group F (A,Z) stronglyminimal if it is minimal and cannot be obtained from a shorter solution by asubstitution of reduced words from F (A,Z) instead of letters.

Lemma 6 (Bulitko’s lemma). Let S be a system of equations over a free group. Theexponent of periodicity of a strongly minimal solution can be effectively bounded.

Proof. Let P be a simple cyclically reduced word. A P -occurrence in a word w isan occurrence in w of a word P εt, ε = ±1, t ≥ 1. We call a P -occurrence v1 ·P εt ·v2stable if v1 ends with P ε and v2 starts with P ε. Clearly, every stable P -occurrencelies in a maximal stable P -occurrence. Two distinct maximal stable P -occurrencesdo not intersect.

A P -decomposition DP (w) of a word w is the unique representation of w asa product

v0 · P ε1r1 · v1 · · · · · P εmrm · vmwhere the occurrences of P εiri are all maximal stable u-occurrences in w. If w hasno stable P -occurrences then, by definition, its P -decomposition is trivial, that is,it has one factor which is w itself.

By adding new variables we can transform the system S to the triangularform, namely, such that each equation has length 3. If we have equation xyz = 1with solution xφ, yφ, zφ, then the cancellation table for this solution looks as thetriangle in Fig. 2.

Let

xφ = v10 · P ε11r11 · v11 · · · · · P ε1,mr1,m · v1,m,yφ = v20 · P ε21r21 · u21 · · · · · P ε2,nr2,n · v2,n,zφ = v30 · P ε31r31 · v31 · · · · · P ε3,kr3,k · v3,k

be corresponding P -decompositions. From the cancellation table we will have a sys-tem of equations on the natural numbers rij , i = 1, 2, 3, j = 1, . . . ,max{k,m, n}.All equations except, maybe, one will have form rij = rst for some pairs i, j ands, t and one equation may correspond to the middle of the triangle. If the middle ofthe triangle is inside a stable P -occurrence in zφ, then the equation would eitherhave form r1j + r2s+2 = r3t or r1j + r2s+3 = r3t. Notice that since xφ, yφ, zφ arereduced words, the middle of the triangle cannot be inside a stable P -occurrencefor more than one variable.


If we replace a solution rij , i=1,2,3, j = 1, . . . ,max{k,m, n} of this system ofequations by another positive solution, say qij , i = 1, 2, 3, j = 1, . . . ,max{k,m, n}and replace in the solution xφ, yφ, zφ stable P -occurrences P rij by P qij we willhave another solution of the equation xyz = 1.

Now, instead of one equation xyz = 1 we take a system of equations S.We obtain a corresponding linear system for natural numbers rij ’s. Let R be thefamily of variables rij ’s that occur in the linear equations of length 3. The numberof such equations is not larger than the number of triangles, that is the numberof equations in the system S. Therefore R is a finite family. Consider a system ofall linear equations on R. It depends on the particular solution of S, but there isa finite number of possible such systems. We now can replace values of variablesfrom R by a minimal positive solution, say {qij}, of the same linear system (ifrij does not appear in any linear equation we replace it by qij = 1) and replacein the solution of the system S stable P -occurrences P rij by P qij . We obtainanother solution of the system S. The length of a minimal positive solution {qij}of the linear system is bounded as in the formulation of the lemma. The lemma isproved. �

Let G be a group, we say that H is a maximal fully residually free quotient ofG if any other fully residually free quotient of G is a quotient of H . In particular, afully residually free group G is the maximal fully residually free quotient of itself.

Lemma 7. Suppose FR(Ω) is not a free product with an Abelian factor, and there aresolutions of Ω with unboundedly large exponent of periodicity. One can effectivelyfind a number M and a splitting with Abelian edge groups of a maximal fullyresidually free quotient of FR(Ω) as an amalgamated product with Abelian vertexgroup or as an HNN-extension (or both), such that the exponent of periodicityof a minimal solution of Ω with respect to the group generated by Dehn twistscorresponding to this splitting and the quadratic part (if exists) is bounded by M .

The proof of this lemma uses the notion of a periodic structure and can befound in ([41], Lemma 22) or in [38].

Consider an infinite path in T (Ω) corresponding to an infinite sequence inentire transformation

r = v1 → v2 → · · · → vm. (7)

Let δ be a solution of Ω. The following lemma gives the way to construct afunction f1 depending on Ω such that for any numberM , if the sequence of entiretransformations for (Ω, δ) has f1(M) steps, then |Iδ| > Mψ.

Denote by μi the carrier base of the equation Ωvi . The path (7) will be calledμ-reducing if μ1 = μ and one of the following holds:

1. μ2 does not overlap with its double and μ occurs in the sequence μ1, . . . , μm−1at least twice.

2. μ2 overlaps with its double and μ occurs in the sequence μ1, . . . , μm−1 atleast M + 2 times, where M is the exponent of periodicity of δ.


The following lemma is just an easy exercise.

Lemma 8. In a μ-reducing path the length of Iδ decreases at least by |μδ|/10.Proof. Case 1. μ = μ1 �= μ2, and not more than half of μ2 overlaps with its double.Then after two steps the leftmost boundary of the reincornation of μ will be tothe right of the middle of μ2. Therefore by the time when the reincornation of μbecomes a carrier, the part from the beginning of the interval to the middle of μ2will be cut and removed. This part is already longer than half of μ.

Case 2. μ = μ1 = μ2, and μ2 (second reincornation of μ) does not overlapwith its double. Then on the first step we cut the part of the interval that is longerthan half of μ.

Case 3. μ2 overlaps with its double. Denote by μδ(i) the value of the reincor-nation of μδ on step i and by [1, σ]δ(i) the word corresponding to the beginning ofthe interval until boundary σ on step i. Then [1, α(μ2))]δ(2) = P d for some cycli-cally reduced word P which is not a proper power and μδ(2), μδ(2)

2 are beginningsof [1, β(μ2)]δ(2) which is a beginning of P∞.

We haveμδ(2) = P rP1, r ≤M (8)

Let μi1 = μi2 = μ for i1 < i2 and μi �= μ for i1 < i < i2. If

|μδ(i1+1)i1+1

| ≥ 2|P | (9)

and [1, ρi1+1 + 1]δ(i1+1) begins with a cyclic permutation of P 3, then

|[1, α(μi1+1)]δ(i1+1| ≥ |P |.

The base μ occurs in the sequence μ1, . . . , μm−1 at least r + 1 times, so either (9)fails for some i1 ≤ m− 1 or the part of the interval that was removed after m− 1steps is longer than max{|r − 3||P |, |P |}.

If (9) fails, then |[1, α(μi1 )]δ(i1)| ≥ (r− 2)|P |. So everything is reduced to thecase when the part of the interval that was removed afterm−1 steps is longer thanmax{|r− 3||P |, |P |}. Together with (8) this implies that in m− 1 steps the lengthof the interval was reduced at least by 15 |μδ(2)| which is not less than 1

10 |μδ|. �

Proof of Theorem 9. Let L be the family of bases such that every base μ ∈ Loccurs infinitely often as a leading base. Suppose a number m is so big that forevery base μ in L, a μ-reducing path occurs more than 20nf times during thesem steps. Since Σ|μδm | ≥ ψ, where we sum over all the participating bases, at leastfor one base λ ∈ L, |λδm | ≥ ψ/2n. Moreover, |λδi | ≥ |λδm | ≥ ψ/2n for all i ≤ m.Since a λ-reducing path occurs more than 20nf times, the length of the intervalwould be decreased in m steps by more than it initially was. This gives a bound onthe number of steps in the entire transformation sequence for (Ω, δ) for a minimalsolution δ. Theorem 9 has been proved. �Proof of Theorem 4.We replace in the tree T (Ω) every infinite path correspondingto an infinite sequence of entire transformation of generalized equations beginning


at Ωvi by a loop corresponding to automorphisms of FR(Ωvi) and finite sequence

of transformations for a minimal solution of Ωvi . At the end of this sequenceof transformations we obtain a generalized equation Ωvj such that either it hassmaller complexity than Ωvi or FR(Ωvj

) is a proper quotient of FR(Ωvi). Any proper

chain of residually free quotients is finite. Therefore we obtain a finite graph (theonly cycles are loops corresponding to automorphisms) and its maximal subtree.The equation S(X) = 1 has a non-trivial solution if and only if we are able toconstruct Tsol(Ω) at least for one of the generalized equations corresponding tothe system S(X) = 1.

Let v0 → v1 → . . . vk be a path in Tsol(Ω) from the root to a leaf. Let vi−1(i ≥ 1) be the first vertex such that there is a loop corresponding to automorphismsof FR(Ωvi−1 ) attached to vi−1. And let vj be the next such vertex or (if there isno such a second vertex) vj = vk. All the homomorphisms from FR(Ω) to F inthe fundamental set corresponding to the path from v0 to vk factor through afree product of FR(Ωvi−1 ) and, maybe, some free group (that occurred as a resultof Titze cleaning when going from Ωv0 to Ωvi−1). All the homomorphisms fromFR(Ωvi−1 ) to F in the fundamental set corresponding to the path from vi−1 to vk areobtained by the composition of a canonical automorphism σ of FR(Ωvi−1 ), canonicalepimorphism π = πi . . . πj from FR(Ωvi−1 ) onto FR(Ωvj

) and a homomorphism fromthe fundamental set of homomorphisms from FR(Ωvj

) to F . The composition σπ

is a solution of some system of equations, denoted by S1(H1, H2, H3, Hπ, A) = 1,over FR(Ωvj

). (Notice that by H we denote a generating set of FR(Ωvi−1 ) moduloF (A)). Therefore Hπ and A are the sets of coefficients of this system.) The systemS1(H1, H2, H3, Hπ, A) = 1 consists of three types of subsystems:

1. Quadratic system in variables h ∈ H1, where H1 is the collection of itemsin the quadratic part of Ωvi−1 . This system is obtained from Ω by replacing ineach basic, and boundary equation each variable h in the non-quadratic part bythe coefficient hπ.

2. For each splitting of FR(Ωvi−1 ) as an amalgamated product with a freeAbelian vertex group of rank k from the second part of Lemma 6, we reservek variables x1, . . . , xk ∈ H2 and write commutativity equations [xi, xj ] = 1 fori, j = 1, . . . , k and, in addition, equations [xi, u

π] = 1 for each generator u of theedge group.

3. For each variable x ∈ H3 there is an equation xuπx−1 = vπ , where u is agenerator of an edge group corresponding to a splitting of FR(Ωvi−1 ) as an HNN-extension from the third part of Lemma 6 and x corresponds to the stable letterof this HNN-extension.

Notice that σπ is a solution of the system S1(H1, H2, H3, Hπ, A) = 1 overthe group FR(Ωvj

).4. For each x ∈ H3 and corresponding edge group we introduce a new variable

y and equations [y, uπ] = 1, where u is a generator of the edge group. Let H4 bethe family of these new variables.


Denote by S2(H1, H2, H4, Hπ, A) = 1 the system of equations 1,2 and 4. Thissystem is NTQ over FR(Ωvj

). Make a substitution x = xπy. Using this substitutionevery solution σπ of the system S1(H1, H2, H3, Hπ, A) = 1 in the group FR(Ωvj

)

can be obtained from a solution of the NTQ system S2(H1, H2, H4, Hπ, A) = 1.We made the induction step. Since Tsol(Ω) is finite, the proof of Theorem 4

can be completed by induction. �

It is clear that Theorem 9 is the main technical result required for the proof ofTheorem 8 or, in other terminology, the base for our “shortening argument”. Theproof of Theorem 9 is technically complicated because everything is done effectively(the algorithms are given). For comparison we will give a non-constructive proofof Theorem 9 using the following lemma.

Lemma 9. Let Ω0, Ω1, . . . be the generalized equations formed by the entire trans-formation sequence. Then one of the following holds.

1. the sequence ends,2. for some i we obtain the quadratic case on the interval I,3. we obtain an overlapping pair λ, λ such that λ is a leading base, λδ begins

with some nth power of the word [α(λ), α(λ)]δ and there are solutions δ of Ωi

with number n arbitrary large (with arbitrary large exponent of periodicity).

Proof. We assume cases 1 and 2 do not hold. Then, our sequence is infinite andwe may assume that every base that is carried is carried infinitely often, and thatevery base that carries does so infinitely often. So every base that participatesdoes so infinitely often. We also assume the complexity does not change and thatno base is moved off the interval.

Let Ω be a generalized equation (with solution) and let

B = Ω1, . . . ,Ωn, . . .

be an infinite branch. Let δ1, δ2, . . . be a set of solutions of Ω such that δi “factors”through Ωi.

If we rescale the metric so that Iδi has length 1, then each δi puts a lengthfunction on the items of Ω, in particular we assume that each base has lengthand midpoint between 0 and 1. This means that for each δi there is a point xi in[0, 1]m, where this point represents the lengths of the bases and the items as wellas their midpoints in the normalized metric.

We pass to a subsequence of δi (omit the double subscript) such that the xi

converge to a point x in [0, 1]m. We call the limit a metric on (Ω, and denote it δ∗).Normed excess denoted m(ψ) is a constant and we can apply the Bestv-

ina, Feighn argument [7] (toral case) on the generalized equation Ω with lengthsgiven by δ∗.

The argument goes as follows. Entire transformation is moving bases to theright and shortening them, and m(ψ) is a constant. During the process the initial


point of every base is only moved towards the final point of I, and the length of abase is never increased, therefore, every base has a limiting position. Since m(ψ) isa constant, there is a base λ of length not going to zero that participates infinitelyoften. If λ is eventually the only carrier, then we must have case 3 for the processto go unboundedly long. Suppose λ is carried infinitely often. Whenever λ is thecarrier, the midpoint of some base moves the distance between the midpoints ofλ and its dual. Since every base has a limiting position, it follows that λ and itsdual have the same limiting position.

The argument shows that after some finite number of steps we get an over-lapping initial section, i.e., carrier and dual have high length, but midpoints areclose.

It follows that for n sufficiently large doing the process with (Ω, δn) will givea similar picture.

This implies case 3. �

Case 3 can only happen is there are solutions of an arbitrary large exponent ofperiodicity. If we consider only minimal solutions, then the exponent of periodicitycan be effectively bounded, and entire transformation always stops after a boundednumber of steps.

On the group level, case 2 corresponds to the existence a QH vertex groupin the JSJ decomposition of FR(Ω) and case 3 corresponds to the existence of anAbelian vertex group in the Abelian JSJ decomposition of FR(Ω).

6. Elementary free groups

If an NTQ group does not contain non-cyclic Abelian subgroups we call it regularNTQ group. We have shown in [38] that regular NTQ groups are hyperbolic. (LaterSela called these groups hyperbolic ω-residually free towers [57].)

Theorem 10. [44], [60] Regular NTQ groups are exactly the f.g. models of theelementary theory of a non-Abelian free group.

7. Stallings foldings and algorithmic problems

A new technique to deal with F Z[t] became available when Myasnikov, Remeslen-nikov, and Serbin showed that elements of this group can be viewed as reducedinfinite words in the generators of F . It turned out that many algorithmic prob-lems for finitely generated fully residually free groups can be solved by the samemethods as in the standard free groups. Indeed, they introduces an analog of theStallings’ folding for an arbitrary finitely generated subgroup of FZ[t], which allowsone to solve effectively the membership problem in F Z[t], as well as in an arbitraryfinitely generated subgroup of it.


Theorem 11 (Myasnikov-Remeslennikov-Serbin, [51]). Let G be a f.g. fully residu-ally free group and G ↪→ G∗ the effective Nielsen completion. For any f.g. subgroupH ≤ G one can effectively construct a finite graph ΓH that in the group G∗ acceptsprecisely the normal forms of elements from H.

Theorem 12 ([43]). The following algorithmic problems are decidable in a f.g. fullyresidually free group G:• the membership problem,• the intersection problem (the intersection of two f.g. subgroups in G is f.g.and one can find a finite generated set effectively),

• conjugacy of f.g. subgroups,• malnormality of subgroups,• finding the centralizers of finite subsets.

It was proved by Chadas and Zalesski [15] that finitely generated fully resid-ually free groups are conjugacy separable.

Notice that the decidability of conjugacy problem also follows from the resultsof Dahmani and Bumagin. Indeed, Dahmani showed that G is relatively hyperbolicwith Abelian parabolics and Bumagin proved that the conjugacy problem is decid-able in a relatively hyperbolic group provided it is solvable in parabolic subgroups.We prove that for finitely generated subgroups H,K of G there are only finitelymany conjugacy classes of intersections Hg ∩ K in G. Moreover, one can find afinite set of representatives of these classes effectively. This implies that one caneffectively decide whether two finitely generated subgroups of G are conjugate ornot, and check if a given finitely generated subgroup is malnormal in G. Observe,that the malnormality problem is decidable in free groups, but is undecidable intorsion-free hyperbolic groups – Bridson and Wise constructed corresponding ex-amples. We provide an algorithm to find the centralizers of finite sets of elementsin finitely generated fully residually free groups and compute their ranks. In par-ticular, we prove that for a given finitely generated fully residually free group Gthe centralizer spectrum Spec(G) = {rank(C) | C = CG(g), g ∈ G}, where rank(C)is the rank of a free Abelian group C, is finite and one can find it effectively.

Theorem 13 ([13]). The isomorphism problem is decidable in f.g. fully residuallyfree groups.

We also have an algorithm to solve equations in fully residually free groupsand to construct the Abelian JSJ decomposition for them.

Recently Dahmani and Groves [19] proved

Theorem 14. The isomorphism problem is decidable in torsion-free relatively hy-perbolic groups with Abelian parabolics.

Dahmani [21] proved the decidability of the existential theory of a torsionfree relatively hyperbolic group with virtually Abelian parabolic subgroups. Thisimplies our result in [42] about the decidability of the existential theory of f.g.fully residually free groups.


8. Residually free groups

Any f.g. residually free group can be effectively embedded into a direct product ofa finite number of fully residually free groups [38].

Important steps towards the understanding of the structure of finitely pre-sented residually free groups were recently made in [9, 10].

There exists finitely generated subgroups of F × F (this group is residuallyfree but not fully residually free) with unsolvable conjugacy and word problem(Miller).

In finitely presented residually free groups these problems are solvable [9].

Theorem 15 ([10]). Let G < Γ0× · · · ×Γn be the subdirect product of limit groups.Then G is finitely presented iff it satisfies the virtual surjection to pairs (VSP)property:

∀ 0 ≤ i < j ≤ n |Γi × Γj : Pij(G)| <∞.

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[43] O. Kharlampovich, A. Myasnikov, V. Remeslennikov, D. Serbin. Subgroups of fullyresidually free groups: algorithmic problems,Group theory, Statistics and Cryptog-raphy, Contemp. Math., Amer. Math. Soc., 360, 2004, 61–103.

[44] Kharlampovich O., Myasnikov A., Elementary theory of free non-abelian groups, J.Algebra, 302, Issue 2, 451–552, 2006.

[45] O. Kharlampovich, A. Myasnikov, D. Serbin, Groups with free regular length functionon Zn.

[46] O. Kharlampovich, A. Myasnikov, D. Serbin Zn-free groups.

[47] E. Lioutikova, Lyndon’s group is conjugately residually free. Internat. J. AlgebraComput. 13 (2003), no. 3, 255–275.

[48] G.S. Makanin. Equations in a free group (Russian). Izv. Akad. Nauk SSSR, Ser.Mat., 46:1199–1273, 1982. transl. in Math. USSR Izv., V. 21, 1983; MR 84m:20040.

[49] G.S. Makanin. Decidability of the universal and positive theories of a free group(Russian). Izv. Akad. Nauk SSSR, Ser. Mat., 48(1):735–749, 1985. transl. in Math.USSR Izv., V. 25, 1985; MR 86c:03009.

[50] Ju.I. Merzljakov. Positive formulae on free groups. Algebra i Logika, 5(4):25–42,1966.

[51] Myasnikov A., Remeslennikov V., Serbin D., Fully residually free groups and graphslabeled by infinite words. to appear in IJAC.

[52] P. Pfander, Finitely generated subgroups of the free Z[t]-group on two generators,Model theory of groups and automorphism groups (Blaubeuren, 1995), 166–187,London Math. Soc. Lecture Note Ser., 244, Cambridge Univ. Press, Cambridge,1997.

[53] E. Rips and Z. Sela. Cyclic splittings of finitely presented groups and the canonicalJSJ decomposition. Annals of Math., 146, 53–109, 1997.

[54] A. Razborov. On systems of equations in a free group. Math. USSR, Izvestiya,25(1):115–162, 1985.

[55] A. Razborov. On systems of equations in a free group. PhD thesis, Steklov Math.Institute, Moscow, 1987.

[56] V. Remeslennikov, ∃-free groups, Siberian Math J., 30 (6):998–1001, 1989.


[57] Z. Sela. Diophantine geometry over groups I: Makanin-Razborov diagrams. Publica-tions Mathematiques de l’IHES 93(2001), 31–105.

[58] J.R. Stallings. Finiteness of matrix representation. Ann. Math., 124:337–346, 1986.

[59] L. Ribes, P. Zalesskii, Conjugacy separability of amalgamated free products of groups.J. Algebra, 1996, v. 179, 3, pp. 751–774

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Olga Kharlampovich and Alexei G. MyasnikovMcGill UniversityDepartment of Mathematics and StatisticsBurnside Hall, room 915805 Sherbrooke West, MontrealQuebec, Canada, H3A 2K6e-mail: [email protected]

[email protected]



The FN -action on the Product of theTwo Limit Trees for an Iwip Automorphism

Martin Lustig

Abstract. An elementary proof is given for the fact that, for every non-surfaceiwip automorphisms ϕ of a free group FN , the FN -action, on the cartesianproduct T+(ϕ) × T+(ϕ

−1) of the (non-simplicial) forward limit R-trees for ϕand ϕ−1, is properly discontinuous. Alternative proofs, derived from deeperresults, have been given by Bestvina-Feighn-Handel [3] and later by Levitt-Lustig [10]; compare also Guirardel [9].

Mathematics Subject Classification (2000). Primary 20F36,Secondary 20E36, 57M05.

Keywords. R-trees, discrete action on product, iwip automorphismsof free groups.

1. Introduction

Let ϕ ∈ Aut(FN ) be an irreducible outer automorphism, with irreducible positivepowers, of a finitely generated non-abelian free group FN (see §2). We also assumeϕ to be non-geometric, i.e., ϕ is not induced by a homeomorphism of a surfacewith boundary. Then there exists an R-tree T+(ϕ) with isometric FN -action thathas the following properties (compare §2):

– The FN -action on T is free, but not simplicial: the FN -orbit of every pointis dense in T+(ϕ).

– The tree T+(ϕ) is projectively invariant under ϕ, with stretching factor λ > 1,and T+(ϕ) is obtained from iterating some train track representative f : τ →τ of ϕ.

– The tree T+(ϕ) is uniquely determined by ϕ up to FN -equivariant homothety.

Analogous statements hold for ϕ−1 as well.

Theorem 1.1. The product action of FN on T+(ϕ)× T+(ϕ−1) is free and properlydiscontinuous.

244 M. Lustig

The first proofs of this result appeared in [3] and [12]. In [10] and [9] moregeneral results have been proved. The proof given here, a slightly elaborate versionof [12], is short and elementary (in the sense that it only uses standard train tracktechniques). More references and background are given below in the last section.

To prove this theorem we show that large powers of ϕ contract any conjugacyclass [w] ⊂ FN , provided that w has sufficiently small T+(ϕ)-translation length||w|| = min{d(x,wx) | x ∈ T+(ϕ)}. More precisely:

Proposition 1.2. For any basis A of FN there exists a number r(ϕ) ≥ 1 with theproperty that for any integer r ≥ r(ϕ) there is a bound K(r) > 0 such that, forany w ∈ FN with T+(ϕ)-translation length ||w|| ≤ K(r), the lengths (with respectto A) of the cyclically reduced words w and ϕr(w) satisfy the strict inequality

|ϕr(w)|A < |w|A .

This proposition implies directly the above Theorem 1.1, as, for r ≥ r(ϕ), r(ϕ−1),if w (and hence ϕr(w)) had small translation length on both, T+(ϕ) and T+(ϕ−1),we would get the contradiction

|w|A = | ϕ−r(ϕr(w)) |A < |ϕr(w)|A < |w|A .

2. The set-up

In this section we recall some known facts about train track representatives andthe limit tree T+(ϕ). This also serves to fix our notation. The reader who prefersa more expanded version of the presentation given in this section is referred to§§2–5 of [13].

An automorphism ϕ ∈ Aut(FN ) is irreducible with irreducible powers (iwip) ifno ϕt, for any t ≥ 1, fixes a proper free factor of FN up to conjugation. Every such ϕis represented by a train track map f : τ → τ with associated geometric transitionmatrixM(f) = (me,e′ )e,e′∈Edges(τ), where the coefficient me,e′ denotes the numberof times that the edge path f(e′) crosses over the edge e or its inverse (bothcounted positively). The integer matrix M(f) is non-negative and irreducible (inthe standard meaning for non-negative matrices), and thus has Perron-Frobeniuseigenvalue (= the spectral radius of M(f))

λ > 1 ,

see [1].

A path γ in τ is called reduced if there is no backtracking along γ (i.e., if γ,interpreted as map from part of R to τ , is locally injective). For any non-reduced γwe denote by [γ] the reduced path resulting from γ by cancelling all backtrackingsubpaths. Notice that one always has

[f(γ)] = [f([γ])] .

The FN -action on T+(ϕ)× T+(ϕ−1) 245

A reduced path γ is called legal if f t(γ) is reduced, for all t ≥ 1. In particular,every edge of τ is legal; in fact, this is precisely the defining property of a traintrack map.

If γ (possibly non-reduced) is not legal, then a point of γ (always a vertexof τ) which separates two maximal legal subpaths of γ is called an illegal turn inγ. We denote by ILT(γ) the number of illegal turns in γ. If γ is a closed path inτ , then by “reduced” or “legal” we always mean cyclically reduced or cyclicallylegal. Similarly, [γ] denotes the cyclically reduced path, and ILT(γ) denotes thenumber of illegal turns on the cyclic path γ.

Fact 2.1. The R-tree T+(ϕ), obtained from iterating some train track representa-tive f : τ → τ of ϕ, is projectively invariant:

||ϕ(w)|| = λ ||w|| for all w ∈ FN ,

with λ > 1 as given above (see [2], [8], [11]).

For the universal covering τ of τ there is an FN -equivariant map i : τ →T+(ϕ). One can define a PF-length for every edge of τ , given through an eigenvector�v of the Perron-Frobenius eigenvalue λ > 1 of M(f).

By an edge path we mean a path which starts and ends in a vertex. Two edgepaths are equal if the sequences of edges traversed, with orientation, are the same.Most paths in this papers are edge paths, even if this is not explicitly stated.

Fact 2.2. For every legal path γ in τ , one has:

PF-length(f(γ)) = λ PF-length(γ).

Fact 2.3. For every legal path γ in τ , and any lift γ of γ to τ , one has:

length(i(γ)) = PF-length(γ) .

If γ is a (not necessarily legal) edge path in τ we denote by the simpliciallength of γ the number of edges which are crossed by γ. As M(f) is irreducible,it follows that all entries of the eigenvector �v are strictly positive. Hence the twolengths are related as follows:

Fact 2.4. There exists a constant C > 0 such that

C−1 · PF-length(γ) ≤ simplicial-length(γ) ≤ C · PF-length(γ) .Fact 2.5. For every w ∈ FN , and every reduced loop γ in τ which represents theconjugacy class of w, one has:

||w|| = lims→∞

PF-length([fs(γ)]λs

.

From the finiteness of τ and the fact that f represents an automorphism onededuces:

Fact 2.6. There is an upper bound B ≥ 0 for the PF-length of a backtracking pathin the image f(γ) of any reduced path γ in τ (compare [8]).

246 M. Lustig

Fact 2.7. An irreducible ϕ ∈ Aut(FN ) with irreducible powers is geometric (=induced by a surface homeomorphism) if and only if there is a non-trivial ϕ-periodic conjugacy class in FN (see [1], §4).

3. The proof of Proposition 1.2

In order to prove Proposition 1.2 we now derive two lemmas:

Lemma 3.1. There are constants K > 0 and ε > 0 such that every closed reducedpath γ in τ with legal subpath γ0 of simplicial length bigger than K represents theconjugacy class of an element w ∈ FN with ‖ w ‖> ε.

Proof. Let γ0 be a (non-closed) legal path in τ with non-trivial subpath γ1, suchthat both components γ(i) (for i = 1, 2) of γ0 � γ1 satisfy

(λ− 1) PF-length(γ(i)) ≥ B

for B as in Fact 2.6. Calling such a pair (γ0, γ1) B-long, we observe, by Fact 2.2,that the pair of legal paths (f(γ0)∗, f(γ1)) is again B-long, where f(γ0)∗ denotesthe path obtained from f(γ0) after cancelling boundary subpaths of PF-length B.Hence, by Facts 2.2, 2.5 and 2.6, if the conjugacy class of any element w ∈ FN isrepresented by a reduced loop γ in τ with a B-long pair of subpaths (γ0, γ1), thenone has ||w|| ≥ PF-length(γ1) > 0. Thus the statement of Lemma 3.1 follows fromFact 2.4. �

For K as in Lemma 3.1 we denote by L(t) the set of closed or non-closedreduced edge paths γ in τ such that [f t′(γ)] does not contain any legal subpath ofsimplicial length bigger than K, for all 0 ≤ t′ ≤ t, with the possible exception, fort′ ≥ 1, of the initial and terminal maximal legal subpath, in case γ is non-closed.Notice that by definition one has

· · · ⊂ L(t+ 1) ⊂ L(t) ⊂ · · · ⊂ L(0) .

If γ′ is a subpath of a reduced path γ, we consider the subpath f(γ′) off(γ), and the corresponding reduced paths [f(γ′)] and [f(γ)]. It is possible that[f(γ′)] is not a subpath of [f(γ)] but bifurcates from [f(γ)] at initial and terminalsubpaths. We denote by [f(γ′)]γ the subpath of [f(γ′)] obtained by erasing theseinitial and terminal subpaths (admitting the degenerate cases where the leftoverpath [f(γ′)]γ consists of a single point only or is empty), and we note that, if γ isa concatenation γ = γ1γ2 . . . γr of reduced subpaths, it follows that [f(γ)] is theconcatenation [f(γ)] = [f(γ1)]γ [f(γ2)]γ . . . [f(γr)]γ .

Lemma 3.2. There exists a constant C ≥ 1 and an exponent q ≥ 1 with thefollowing properties:

(a) Every subpath γ′ of an edge path γ ∈ L(q), with ILT(γ′) = C, satisfiesILT([f q(γ′)]γ) < ILT(γ′).


(b) For any t ≥ 1, if γ ∈ L(tq) and ILT(γ) ≥ 2tC, then

ILT([f tq(γ)]) ≤(2C − 12C

)t

ILT(γ) .

Proof. (a) We first notice that for any constant K > 0 there exist only finitelymany distinct legal edge paths γk in τ with

simplicial-length(γk) ≤ K

(recall that two edge paths are equal iff the sequences of edges traversed, withorientation, are the same). It follows that there is a bound C ≥ 1 such that, ifγ ∈ L(0) and γ′ is a subpath of γ with ILT(γ′) = C, written as concatenationof maximal legal subpaths γ′ = γ1γ2 . . . γC+1, then there exist indices i < j in{1, . . . , C} with γi = γj and γi+1 = γj+1. Furthermore, it follows that there areonly finitely many closed edge paths γ ∈ L(0) with ILT(γ) ≤ C.

Now, the fact that the f -image of every legal path is again legal implies thatthe number of illegal turns in any path in τ will not be increased when applyingf . Let us assume that, for γ and γ′ as in the previous paragraph, and for someq ≥ 1, one has γ ∈ L(q) and ILT([f q(γ′)]γ) = ILT(γ′). It follows that for none ofthe maximal legal subpath γi of γ′ the subpath f q(γi) of f q(γ′) degenerates to asingle point or is cancelled entirely when passing from f q(γ′) to [f q(γ′)]γ . In otherwords, every maximal legal subpath γi of γ′ contains a point with f q-image thatdoes not lie on one of the backtracking subpaths of f q(γ) at the two endpointsof f q(γi). Hence, for any two adjacent maximal legal subpaths γi, γi+1 of γ′, thebacktracking subpath in f q(γ) at the illegal turn which connects f q(γi) to f q(γi+1)depends only on γi and γi+1, and not on the rest of the path γ.

In particular, as γ′ = γ1γ2 . . . γC+1 with γi = γj and γi+1 = γj+1 for 1 ≤ i <j ≤ C, we obtain a closed edge path γ, defined by cyclic concatenation of the pathγi+1γi+2 . . . γj , which satisfies γ ∈ L(q) and ILT(f t(γ)) = ILT(γ) = j − i ≤ C for1 ≤ t ≤ q.

As the number of closed edge paths in L(0) with C or less illegal turnsis finite, it follows that, if q is sufficiently big, the conjugacy classes given by thef t(γ) cannot be pairwise distinct for all 1 ≤ t ≤ q. Hence γ determines a non-trivialconjugacy class in FN which is ϕ-periodic. But this contradicts our assumption thatf represents a non-geometric irreducible automorphism with irreducible powers,see Fact 2.7. Hence there must be an upper bound q0 ≥ 1 to all exponents q ≥1 with ILT([f q(γ′)]γ) = ILT(γ′) for any subpath γ′ of a path γ ∈ L(q) withILT(γ′) = C. Thus any q > q0 satisfies the claim (a).

(b) Let q and C be as in part (a) and let γ be a path in L(q) with ILT(γ) ≥ C. Wecan subdivide γ at illegal turns into edge paths γi with 2C ≥ ILT(γi) ≥ C, andapply (a) to deduce that ILT([f q(γi)]γ) ≤ ILT(γi)−1 for each γi. As γ decomposesinto at least 1

2C ILT(γ) of such γi, this gives

ILT([f q(γ)]) ≤ ILT(γ)− ILT(γ)2C

=2C − 12C

ILT(γ) .

248 M. Lustig

Thus, if γ is a path in L(tq) for some t ≥ 1 and ILT(γ) ≥ 2tC, then eitherone has ILT([f t′q(γ)]) < C for some 1 ≤ t′ ≤ t, in which case it follows

ILT([f tq(γ)]) ≤ ILT([f t′q(γ)]) < C ≤(12

)t

ILT(γ) ≤(2C − 12C

)t

ILT(γ) .

Otherwise, we can apply the above consideration t times to γt′ = [f t′q(γ)], fort′ = t, t− 1, . . . , 1, thus computing

ILT([f tq(γ)]) ≤ 2C − 12C

ILT([f (t−1)q(γ)])

≤(2C − 12C

)2ILT([f (t−2)q(γ)]) ≤ · · · ≤

(2C − 12C

)t

ILT(γ) .

This proves our assertion. �Proof of the proposition. For any basis A of FN there is a constant d > 0 suchthat the A-length of any cyclically reduced word w ∈ FN and the simplicial lengthof the corresponding closed reduced path γ in τ are related by inequalities

d−1|w|A ≤ simplicial-length(γ) ≤ d |w|A .Furthermore, for K > 0 as in Lemma 3.1 and γ ∈ L(0) one has the inequalities

K−1 simplicial-length(γ) ≤ ILT(γ) ≤ simplicial-length(γ) .

Hence there is a constant D > 0 such that

D−1 |w|A ≤ ILT(γ) ≤ D |w|A .Now, choose t large enough such that ( 2C

2C−1 )t > D2, for C as in Lemma 3.2.

For any r ≥ tq we find, by Lemma 3.1 and Fact 2.1, a constant ε(r) > 0 such thatthe condition ||w|| ≤ ε(r) implies γ ∈ L(r), and furthermore ILT(γ) ≥ 2tC, as thenumber of non-trivial closed edge paths γ in L(0) with ILT(γ) < 2tC is finite. Butthen we obtain, using Lemma 3.2 (b), the inequalities

|ϕr(w)|A ≤ D · ILT([f r(γ)]) ≤ D · ILT([f tq(γ)])

≤ D ·(2C − 12C

)t

ILT(γ) < D−1 · ILT(γ) ≤ |w|A . �

4. A little history and some references

The statement of Theorem 1.1 was originally conjectured by Peter Shalen (inthe early ’90s) and communicated to me by Gilbert Levitt, who, in joint workwith Culler and Shalen [6], investigated consequences of the analogous result forpseudo-Anosov homeomorphisms of closed surfaces. Around the time that the firstwritten version [12] of the proof presented here was circulated, Bestvina, Feighnand Handel had obtained an independent proof as consequence of their deep workon the Tits alternative in Out(FN ), see Theorem 5.3 of [3]. The advantage of theproof presented in the preceding sections is that it is short and direct in that


it uses only standard facts about train track maps, essentially known since theirintroduction in [1]. Our result is also slightly more general than Theorem 5.3 of[3], as it does not just apply to irreducible automorphisms with irreducible powers(compare [4]). The precise conditions for this generalization are given as follows:

Remark 4.1. The statement of Proposition 1.2 and hence that of Theorem 1.1 arealso true for a more general class of automorphisms of FN . The proof given in§3 uses only the weaker assumption that there exists a train track representativef : τ → τ of ϕ with transition matrix M(f) which admits a strictly positive roweigenvector �v with eigenvalue strictly bigger than 1, and that there is no ϕ-periodicconjugacy class in FN �{1}. Examples of such automorphisms, which however arenot iwip, are given for example by attaching an extra edge e to a fixed point Pof a train track map f that represents a non-geometric iwip automorphism, if oneextends f by defining f(e) = eγ, for any non-trivial legal loop γ with endpointsat the fixed point P .

Shalen stated his original conjecture in the context of the question whetherthe groups FN �Z are coherent (meaning: “finitely generated subgroups are finitelypresented”), in analogy to 3-manifold groups. This question has been answeredin the positive by Feighn-Handel [7], but without using the above Theorem 1.1.However, apart from its natural appeal within the theory of group actions on R-trees, the proof presented here seems also to be of some interest on is own: forexample, Peter Brinkmann told me that it inspired him for a crucial part of histhesis (compare [5]), and the main idea employed in §3 is also used again in [14].

For arbitrary automorphisms of FN a strict generalization of Theorem 1.1 isclearly wrong. However, using methods related to the above proof, but in a muchmore technically loaded context, the following result has been proved in [10]:

Theorem 4.2 (Levitt-Lustig). For every ϕ ∈ Aut(FN ) there exist projectively in-variant R-trees T+ and T−, with stretching factors λ+ ≥ 1 and λ− ≤ 1 respectively,as well as a constant ε > 0 such that, if for any w ∈ FN the translation lengthson T+ and on T− are both smaller than ε (or equal to 0 on one of the two), thenw fixes a point in T+ × T−.

Finally, Vincent Guirardel has investigated in a much more general contextgroup actions on the cartesian product of two R-trees in [9]. In particular his §9contains results relevant to this paper.

References

[1] M. Bestvina and M. Handel, Train tracks for automorphisms of the free group, Annalsof Math. 135, pp. 1–51 (1992)

[2] M. Bestvina and M. Feighn, Outer limits, preprint 1992

[3] M. Bestvina, M. Feighn and M. Handel, Laminations, trees, and irreducible auto-morphisms of free groups, Geom. Funct. Anal. 7, pp. 215–244 (1998)

250 M. Lustig

[4] M. Bestvina, M. Feighn and M. Handel, Erratum to Laminations, trees, and irre-ducible automorphisms of free groups, Geom. Funct. Anal. 7, pp. 1143 (1998)

[5] P. Brinkmann, Hyperbolic automorphisms of free groups, Geom. Funct. Anal. 10, pp.1071–1089 (2000)

[6] M. Culler, G. Levitt, P. Shalen, unpublished manuscript

[7] M. Feighn and M. Handel, Mapping tori of free group automorphisms are coherent,Ann. of Math. 149, pp. 1061–1077 (1999)

[8] D. Gaboriau, A. Jager, G. Levitt and M. Lustig, An index for counting fixed pointsof automorphisms of free groups, Duke Math. J. 93, pp. 425–452 (1998)

[9] V. Guirardel, Coeur et nombre d’intersection pour les actions de groupes sur lesarbres, Annales de l’E.N.S. 38, pp. 847–888 (2005)[English version: Core and intersection number for group actions on trees, arXivmath. Gr 0407206]

[10] G. Levitt and M. Lustig, Automorphisms of free groups have asymtotically periodicdynamics, to appear in J. reine u. angew. Math. (arXiv math. GR 0407437)

[11] M. Lustig, Automorphismen von freien Gruppen, Habilitationsschrift 1992, Ruhr-Universitat Bochum

[12] M. Lustig, Discrete actions on the product of two non-simplicial R-trees, preprint1994

[13] M. Lustig, Conjugacy and centralizers for iwip automorphisms of free groups, Geo-metric Group Theory, Trends in Mathematics, pp. 197–224, Birkhauser, Basel 2007

[14] M. Lustig et al., Seven steps to happiness, preliminary preprint 2008

Martin LustigMathematiques (LATP)Universite Paul Cezanne – Aix Marseille IIIav. Escadrille Normandie-NiemenF-13397 Marseille 20, Francee-mail: [email protected]



Mather Invariants in Groups ofPiecewise-linear Homeomorphisms

Francesco Matucci

Abstract. We describe the relation between two characterizations of conju-gacy in groups of piecewise-linear homeomorphisms, discovered by Brin andSquier in [3] and Kassabov and Matucci in [6]. Thanks to the interplay be-tween the techniques, we produce a simplified point of view of conjugacy thatallows us to easily recover centralizers and lends itself to generalization.

Mathematics Subject Classification (2000). 20E45; 37E05; 37E10.

Keywords. Piecewise-linear homeomorphism groups, conjugacy invariant.

1. Introduction

We denote by PL+(I) the group of orientation-preserving piecewise-linear home-omorphisms of the unit interval I = [0, 1] with finitely many breakpoints. We willtreat only the case of PL+(I) even if all the results can be adapted to certainsubgroups of PL+(I) of homeomorphisms with certain requirements on the break-points and the slopes (for example, Thompson’s group F and the Thompson-Steingroups PLS,G(I) introduced in the works of Stein [8] and Bieri-Strebel [2]). Inparticular, it is sufficient to restrict our study to functions that do not intersectthe diagonal, except for the points 0 and 1 (see Section 2 for the motivation).

In their work [3] Brin and Squier define an invariant under conjugacy formaps of PL+(I) that do not intersect the diagonal. Their description is basedon similar earlier work by Mather [7] for diffeomorphisms of the unit interval andallows the classification of centralizers and the detection of roots of elements. Thesetechniques were originally introduced as an attempt to solve the conjugacy problemin Thompson’s group F (which was then proved to be solvable by Guba and Sapirin [5]). Later on this approach was refined by Gill and Short in [4] and Belkand Matucci [1] to give another proof of the solution to the conjugacy problem

The author gratefully acknowledges the Centre de Recerca Matematica (CRM) and its staff forthe support received during the development of this work.

252 F. Matucci

in Thompson’s group F . On the other hand, Kassabov and Matucci showed asolution to the simultaneous conjugacy problem in [6] by producing an algorithmto build all conjugators, if they exist. Similarly, these techniques can be used toobtain centralizers and roots as a byproduct.

The aim of this note is to show the connection between the techniques in [3]and [6] to characterize conjugacy in groups of piecewise-linear homeomorphisms.By defining a modified version of Brin and Squier’s invariant and using a mixtureof those points of view it is possible to produce a short proof of the description ofconjugacy and centralizers in PL+(I). In particular, the interplay between thesetwo points of view lends itself to generalizations giving a tool to study larger classof groups of piecewise-linear homeomorphisms.

This paper is organized as follows. In Section 2 we give a short account ofa key algorithm in [6] (the stair algorithm) to build a particular conjugator g fortwo elements y, z ∈ PL+(I). In Section 3 we define a conjugacy invariant (calledMather invariant) that essentially encodes the characterization of conjugacy in [3]for PL+(I). In Section 4 we show to use the stair algorithm to simplify the proofof the characterization of conjugacy of [3] using Mather invariants. In turn, inSection 5 we will show how Mather invariants allow us to shorten the argumentsin [6] to classify centralizers of elements. We finish by briefly describing possibleextensions of these tools.

2. The stair algorithm for functions in PL<+(I)

In this section we will discuss how to find a special conjugator g ∈ PL+(I) fortwo functions y, z ∈ PL+(I), if it exists. The idea will be to assume that such aconjugator g exists and obtain conditions that g must satisfy.

Definition 2.1. We denote by PL<+(I) the subset of PL+(I) of all functions that

lie below the diagonal, that is the maps z ∈ PL+(I) such that f(t) < t for allt ∈ (0, 1). Similarly, we define the subset PL>

+(I) of functions that lie above thediagonal. A function z ∈ PL+(I) is defined to be a one-bump function if eitherz ∈ PL<

+(I) or z ∈ PL>+(I).

We will restrict to study conjugacy for one-bump functions. The reason forthis assumption is easily explained: if two functions y, z ∈ PL+(I) are conjugatethrough g, then g−1(∂Fix(y)) = ∂Fix(g−1yg) = ∂Fix(z); since the boundary ofthe set of fixed points of either y or z is finite, the first step to verify conjugacyis to check if ∂Fix(y) and ∂Fix(z) have the same size. If this is the case, we canalways build a map h ∈ PL+(I) such that h−1 (∂Fix(y)) = ∂Fix(z), hence wereduce to check if h−1yh and z, which share the same boundary of the fixed set,are conjugate; this is true if, for any two consecutive points ti, ti+1 ∈ ∂Fix(z),we can find a conjugator gi ∈ PL+([ti, ti+1]) for the restrictions of h−1yh and zto [ti, ti+1], which are either identity maps or one-bump functions. By restrictingthe study of conjugacy to the intervals [ti, ti+1], we derive our assumption on themaps.

Mather Invariants 253

If z ∈ PL+(I), we define initial slope and final slope, respectively, to be thenumbers z′(0) and z′(1). It is clear that if two one-bump functions y and z areconjugate, their initial and final slope are the same. A more interesting fact is thata conjugator has to be linear in certain boxes around 0 and 1. This fact, togetherwith the ability to identify the two functions step by step, allows us to build aconjugator.

Lemma 2.2 (Kassabov and Matucci, [6]). Suppose y, z ∈ PL<+(I).

1. (initial box) Let g ∈ PL+(I) be such that g−1yg = z. Assume y(t) = z(t) =ct for t ∈ [0, α] and c < 1. Then the graph of g is linear inside the box[0, α]× [0, α]. A similar statement is true for a “final box”.

2. (identification trick) Let α ∈ (0, 1) be such that y(t) = z(t) for t ∈ [0, α]. Thenthere exists a g ∈ PL+(I) such that z(t) = g−1yg(t) for t ∈ [0, z−1(α)] andg(t) = t in [0, α]. The element g is uniquely defined up to the point z−1(α).

3. (uniqueness of conjugators) For any positive real number q there exists atmost one g ∈ PL+(I) such that g−1yg = z and g′(0) = q.

4. (conjugator for powers) Let g ∈ PL+(I) and n ∈ N. Then g−1yg = z if andonly if g−1yng = zn.

Proof. The proof of (1) is straightforward. To prove (2) we observe that, if such ag exists then, for t ∈ [0, z−1(α)]

y(g(t)) = g(z(t)) = z(t)

since z(t) ≤ α. Thus g(t) = y−1z(t) for t ∈ [0, z−1(α)]. To prove that such a gexists, define

g(t) :=

{t t ∈ [0, α]y−1z(t) t ∈ [α, z−1(α)]

and extend it to I as a line from the point (z−1(α), y−1(α)) to (1, 1). To prove (3),assume that there exist two conjugators g1, g2 with initial slope q. Since g−11 yg1 =g−12 yg2 we have that g := g1g

−12 centralizes y and it has initial slope 1. Assume,

by contradiction, that g is the identity on [0, α] for some α, but g′(α+) �= 1. Sincewe have

y(g(t)) = g(y(t)) = y(t)

for t ∈ [α, y−1(α)], this implies that g(t) = y−1y(t) = t on [α, y−1(α)], which is acontradiction. To prove the last statement we observe that if f := g−1yng = zn,then f is centralized by both g−1yg and z. Since g−1yg and z have the same initialslope, then by (3) we have g−1yg = z. �

Part (1) of the previous lemma tells us that any given conjugator g must belinear in two suitable boxes [0, α]2 and [β, 1]2, hence if we are given a point (p, g(p))in any of those boxes (say the final one), we can draw the longest segment containedin [β, 1]2 passing through (p, g(p)) and (1, 1) and obtain the map g in that box.We are now going to build a candidate conjugator with a given initial slope.

254 F. Matucci

Theorem 2.3 (Stair Algorithm, [6]). Let y, z ∈ PL<+(I), let [0, α]

2 be the initiallinearity box and let 0 < q < 1 be a real number. There is an N ∈ N such that theunique candidate conjugator with initial slope q is given by

g(t) = y−Ng0zN(t) ∀t ∈ [0, z−N(α)]

and linear otherwise, where g0 is any map in PL+(I) which is linear in the initialbox and such that g′0(0) = q.

By “unique candidate conjugator” we mean a function g such that, if thereexists a conjugator between y and z with initial slope q, then it must be equal to g.Hence we can test our candidate conjugator to verify if it is indeed a conjugator.

Proof. Let [β, 1]2 be the final box and N an integer big enough so that

min{z−N(α), y−N (qα)} > β.

We will build a candidate conjugator g between yN and zN (if it exists) as aproduct of two functions g0 and g1. We note that the linearity boxes for yN andzN are still given by [0, α]2 and [β, 1]2. By Lemma 2.2(1) g has to be linear on[0, α] and so we define an “approximate conjugator” g0 by:

g0(t) := qt t ∈ [0, α]and extend it to the whole I as a line through (1, 1). We then define y1 := g−10 yg0and look for a conjugator g1 of yN

1 and zN , noticing that yN1 and zN coincide on

[0, α]. By the proof of Lemma 2.2(3), we define

g1(t) :=

{t t ∈ [0, α]y−N1 zN(t) t ∈ [α, z−N (α)]

and extend it to I as a line through (1, 1) so that g−11 yN1 g1 = zN on [0, z−N(α)].

Finally, build a function g such that g(t) := g0g1(t) for t ∈ [0, z−N(α)] and extendit to I as a line through (1, 1) on [z−N(α), 1]. The map g is inside the final box att = z−N(α) > β, in fact

g(z−N(α)) = g0g−10 y−Ng0(α) = y−N (qα) > β.

We observe that, by construction, g is a conjugator for yN and zN on [0, z−N(α)],that is g = g0g1 = y−Ng0g1z

N on [0, z−N(α)]. Therefore

g(t) = y−Ng0g1zN(t) = y−Ng0z

N(t) ∀t ∈ [0, z−N(α)]

since g1zN(t) = zN(t) for t ∈ [0, z−N(α)].By parts (1) and (3) of Lemma 2.2, if there is a conjugator for yN and zN

with initial slope q, it must be equal to g. So we just check if g conjugates yN

to zN . Moreover, Lemma 2.2(4) tells us that g is a conjugator for yN and zN ifand only it is for y and z and so we are done. We remark that this proof does notdepend on the choice of g0. The only requirements on g0 are that it must be linearin the initial box and g′0(0) = q. �


3. Mather invariants for functions in PL>+(I)

In this section we will give an alternate description of Brin and Squier’s conjugacyinvariant in [3]. This reformulation was also used by Belk and Matucci in [1] tocharacterize conjugacy in Thompson’s group F : however, their proof relies onspecial kinds of diagrams peculiar to F and cannot be generalized to other groupsof homeomorphisms.

Roughly speaking, the Mather invariant of a map z ∈ PL>+(I) is defined by

taking a power of z large enough so that points very close to 0 get mapped topoints very close to 1.

We will now define it precisely. Consider a one-bump function z ∈ PL>+(I),

with initial slope m0 and final slope m1. In a neighborhood of zero, z acts asmultiplication by m0: for any sufficiently small t > 0 and sufficiently small powersof z, we have z(t) = m0t, z

2(t) = m20t, z3(t) = m30t, . . ., that is the interval [t,m0t]

is a “fundamental domain” for the action of z:

If we make the identification t ∼ m0t in the interval (0, ε), for a sufficientlysmall ε > 0, we obtain a circle C0, with natural projection map p0 : (0, ε) → C0.Similarly, if we identify (1−t) ∼ (1−m1t) on the interval (1−δ, 1), for a sufficientlysmall δ > 0, we obtain a circle C1, with natural projection map p1 : (1−δ, 1)→ C1.

Let ε′ > 0 be small enough so that (ε′, ε) surjects onto C0: if N is sufficientlylarge, then zN will take (ε′, ε) and map it to the interval (1− δ, 1). This induces amap z∞ : C0 → C1, making the following diagram commute:

(ε′, ε) (1− δ, 1)

C0 C1

.................................................................................................... ............zN

................................................................................................................

p0

................................................................................................................

p1

............. ............. ............. ......................... ............z∞

The map z∞ defined above is called the Mather invariant for z. We note that z∞

does not depend on the specific value of N chosen. Any map zm, for m ≥ N ,induces the same map z∞. This is because z “acts as the identity on C1”: we canwrite zm(t) as zm−N(zN (t)), with zN(t) ∈ (1− δ, 1) and so, by definition of ∼, wehave zm(t) ∼ zN(t). If k > 0, then the map t �→ kt on (0, ε) induces a “rotation”rotk of C0. In particular, if we use the coordinate θ = log t on C0, then

rotk(θ) = θ + log k

so rotk is an actual rotation. In the next section we will give a characterization ofconjugacy for one-bump functions by means of Mather invariants.

256 F. Matucci

4. Equivalence of the two points of view

In this section we will show the relation between the stair algorithm and thedefinition of Mather invariant. This will provide an alternative proof of Brin andSquier’s conjugacy invariant.

Theorem 4.1 (Brin and Squier, [3]). Let y, z ∈ PL>+(I) be one-bump functions with

y′(0) = z′(0) and y′(1) = z′(1), and let y∞, z∞ : C0 → C1 be the correspondingMather invariants. Then y and z are conjugate if and only if y∞ and z∞ differ byrotations of the domain and range circles:

C0 C1

C0 C1

........................................................................................ ............

y∞

........................................................................................ ............

z∞

....................................................................................................

rotk

....................................................................................................

rot

Proof. Since y′(0) = z′(0) we can pick the fundamental domain for y and z around0 to be the same. Similarly, we can do it around 1 and so it makes sense to talkabout rotations for C0 and C1. We stress that the Mather invariants y∞ and z∞

that we now use depend on the choice of the fundamental domains around 0 and1 to talk about well-defined compositions.

We assume z = g−1yg for some g ∈ PL+(I) and follow the notation ofthe previous section, taking ε, ε′, δ > 0 small enough and N large enough. ThenzN = g−1yNg and the following diagram commutes, where k = g′(0) and � = g′(1):

C0 C1

C0 C1

(ε′, ε) (1− δ, 1)

(ε′, ε) (1 − δ, 1)

............................................................................................................................................................................... ............

z∞............................................................................................

rotk ............................................................................................

rot

..............................................................................y∞

................................................ ............

...........................................................................................................................................................................................

p0

....................................

......................................................................................................

p0

...........................................................................................................................................................................................

p1

...........................................................................................................................................................................................

p1....................................................................................................................................... .........

...zN..............................................................................g

..............................................................................g

....................................................................................................................................... ............

yN

�

To show the converse, choose g0 ∈ PL+(I) that is linear in the initial boxand such that g′0(0) = k and define the map g to be the following pointwise limit

g(t) := limn→∞ yng0z

−n(t).

By the Stair Algorithm (Theorem 2.3) it is clear that g conjugates y−1 to z−1 (andhence y to z). It remains to show that g ∈ PL+(I). By construction, g has onlyfinitely many breakpoints on the interval [0, 1− δ] for a δ > 0 small enough. Since


g conjugates y and z, then g induces a well-defined map gind : C1 → C1 (given byp1gp

−11 ) and we can build a diagram similar to the one of “only if” part of this

theorem

C0 C1

C0 C1

(ε′, ε) (1− δ, 1)

(ε′, ε) (1 − δ, 1)

............................................................................................................................................................................... ............

z∞............................................................................................

rotk ............................................................................................

gind

..............................................................................y∞

................................................ ............

...........................................................................................................................................................................................

p0

....................................

......................................................................................................

p0

...........................................................................................................................................................................................

p1

...........................................................................................................................................................................................

p1....................................................................................................................................... .........

...zN..............................................................................g

..............................................................................g

....................................................................................................................................... ............

yN

for suitable ε, ε′, δ > 0 small enough and an integer N big enough. By hypothesisthe Mather invariants differ by rotations of the domain and range circles, thereforewe have

rot z∞ = y∞ rotk = gindz∞

and so, by cancellation, gind is a rotation by �. To prove that g ∈ PL+(I) we showthat g is linear around 1 in the following claim:

Claim: If g : I → I is a continuous map and p1 is a projection of a neighbourhoodof 1 to C1 such that p1gp−11 is a well-defined map from C1 to C1 and it is a rotationof C1, then g is linear on (1− δ, 1] for a δ > 0 small enough.

Proof of the claim. Let δ > 0 be small enough so that (1− δ, 1] is contained in thedomain of p1 and let t ∈ (1 − δ, 1]. Following the notation from Section 3, sincep1gp

−11 is a rotation by �, we have

g(t) = g(1− (1− t)) = 1− �mr1(1− t).

for some integer r. Thus, for a λ > 0 close enough to 1, we have

g(1− λ(1 − t)) = 1− �mr1λ(1− t) = 1− λ(1 − g(t)).

By the previous equation, the function

h(t) :=1− g(t)1− t

satisfiesh(t) = h(1− λ(1 − t))

for λ > 0 close enough to 1, hence h is locally constant on (1− δ, 1] and thereforeit is constant. Since h is constant, the map g is then linear around 1. �

Remark 4.2. We have slightly abused the notation in the two cube diagrams of theprevious proof: to simplify the exposition, we have not been careful in choosing

258 F. Matucci

the range sets for g that still surject onto C0 and C1 (although it can be madeprecise).

Remark 4.3. The previous proof shows that two functions y, z are conjugate ifand only if the Stair Algorithm builds a linear map in the final linearity box andthis happens if and only if the two Mather invariants differ by rotations of thedomain and the range circles. The Mather invariant thus gives the “obstruction”to finishing the Stair Algorithm at 1.

Remark 4.4. We stress that the definition of Mather invariant and the constructionof the stair algorithm do not really depend upon the set of breakpoints and slopesof the maps y and z. With little work, the two constructions and their equivalencecan be extended to Thompson-Stein groups (see also [6]).

5. Applications: centralizers and generalizations

Given a map f : S1 → S1, a lift of f is a map F : R → R such that F (t + 1) =F (t) + 1 for all t ∈ R and F induces f when passing the domain and the rangeto quotients via the relation α ∼ α+ 1. Given a lift, we talk about a maximal V -interval to refer to an interval [a, b] such that F is linear with slope V on [a, b] anda, b are breakpoints for F . We will give a short proof of the following well-knownresult.

Theorem 5.1. Let z ∈ PL>+(I). Then the centralizer subgroup CPL+(I)(z) = {g ∈

PL+(I) | gz = zg} is isomorphic to the infinite cyclic group.

Proof. Define the following group homomorphism:

ϕz : CPL+(I)(z) −→ (R,+)g �−→ log g′(0).

Lemma 2.2(3) implies that ϕz is injective. By Theorem 4.1 any function g central-izing z induces two rotations rot, rotk such that

rot z∞ = z∞ rotk

where k = g′(0) and � = g′(1). Observe that R(t) = t+log � and Rk(t) = t+log kare lifts of the two rotations rot, rotk. Choose a lift Z : R → R of z∞. The previousequality implies:

Z(t) + log � = R(Z(t)) = Z(Rk(t)) = Z(t+ log k)

which means that the graph of Z can be shifted “diagonally” onto itself. The mapZ is piecewise-linear and, for any positive number r, has finitely many breakpointson the interval [−r, r]. Hence Z has only finitely many maximal Z ′(0)-intervals thatare contained in [−r, r] and so there is only a discrete set of shifts (that is, valuesof log k = ϕz(g)) which maps the graph of Z onto itself, unless Z is a line.

To see that this is not the case, we show that z∞ has breakpoints. Let Nbe a power large enough so that a fundamental domain near 0 is sent near 1 so


that zN induces z∞, then either z−N or z−2N has a breakpoint in the final box[β, 1]2 (this implies immediately that z∞ must have breakpoints). If they were bothlinear, by applying the chain rule on z−2N = z−N ◦ z−N first at β and then at 1,one sees that the slope z−2N on [β, 1] is simultaneously equal to the product ofthe slopes z′(0)−N (z−N)′(β+) and z′(1)−N (z−N )′(1) and this is impossible since(z−N)′(β+) = (z−N)′(1), but z′(0) < z′(1).

We have thus proved that the image of ϕz must be a discrete subgroup of(R,+) and so, by a standard fact, it is isomorphic to Z. �

The Mather invariant approach is also interesting because it lends itself togeneralizations. Let PLdis(R) the group of all orientation-preserving piecewise-linear homeomorphisms of the real line with a discrete set of breakpoints and letEP be the subgroup of PLdis(R) of the functions that are “eventually periodic atinfinity”, that is functions f ∈ PLdis(R) such that there exist numbers Lf , Rf sothat f(t− 1) = f(t) − 1 for t < Lf and f(t+ 1) = f(t) + 1 for t > Rf . It is easyto define the subset EP> and Mather invariant for functions in EP>: we just modout the intervals (−∞, Lf) and (Rf ,∞) by the relation t ∼ f(t) and then take apower of f high enough so that (f−1(Lf ), Lf ) gets carried to a subset of (Rf ,∞).Similarly, one can partially extend the stair algorithm to build conjugators. It isthus interesting to see how much of these techniques can be extended to overgroupscontaining PL+(I) to compute centralizers and, possibly, to study the conjugacyproblem.

Acknowledgment

The author would like to thank Ken Brown, Jose Burillo, Martin Kassabov andan anonymous referee for helpful comments that improved the presentation of thispaper.

References

[1] J.M. Belk and F. Matucci. Dynamics in Thompson’s group F . Submitted.arXiv:math.GR/0710.3633v1.

[2] R. Bieri and R. Strebel. On groups of PL-homeomorphisms of the real line. notes,1985. Math. Sem. der Univ. Frankfurt.

[3] Matthew G. Brin and Craig C. Squier. Presentations, conjugacy, roots, and central-izers in groups of piecewise linear homeomorphisms of the real line. Comm. Algebra,29(10):4557–4596, 2001.

[4] N. Gill and I. Short. Conjugacy, roots, and centralizers in Thompson’s group F .Preprint. arXiv:math.GR/0709.1987v2 .

[5] Victor Guba and Mark Sapir. Diagram groups. Mem. Amer. Math. Soc., 130(620):viii + 117, 1997.

[6] M. Kassabov and F. Matucci. The simultaneous conjugacy problem in groups ofpiecewise linear functions. preprint. arXiv:math.GR/0607167v2.

260 F. Matucci

[7] John N. Mather. Commutators of diffeomorphisms. Comment. Math. Helv., 49:512–528, 1974.

[8] Melanie Stein. Groups of piecewise linear homeomorphisms. Trans. Amer. Math.Soc., 332(2):477–514, 1992.

Francesco MatucciCentre de Recerca MatematicaApartat 50E-08193 Bellaterra, Barcelona, Spaine-mail: [email protected]



Algebraic Geometry over theAdditive Monoid of Natural Numbers:Systems of Coefficient Free Equations

Pavel V. Morar and Artem N. Shevlyakov

Abstract. In the paper we consider homogeneous systems of linear equationsand classify coordinate monoids over the additive monoid of natural numberswhich are defined by such systems. Further, we apply our results to the wideclass of commutative monoids.

Mathematics Subject Classification (2000). 20M14.

Keywords. Universal algebraic geometry, natural numbers, equations, coordi-nate monoids.

1. Introduction

In the paper [1] by E. Daniyarova, A.G. Myasnikov, and V. Remeslennikov thefollowing unification theorems (Theorem A and B in [1]) were formulated andproved.

Theorem 1.1. [No coefficients, [1]] Let B be an equationally Noetherian algebra ina functional language L. Then for a finitely generated algebra C of L the followingconditions are equivalent:1. Th∀(B) ⊆ Th∀(C), i.e., C ∈ ucl(B);2. Th∃(B) ⊇ Th∃(C);3. C embeds into an ultrapower of B;4. C is discriminated by B;5. C is a limit algebra over B;6. C is defined by a complete atomic type in the theory Th∀(B) in L;7. C is the coordinate algebra of an irreducible algebraic set over B defined by

a system of coefficient-free equations.

Theorem 1.2. [With coefficients, [1]] Let A be an algebra in a functional languageLA and B an A-equationally Noetherian A-algebra. Then for a finitely generatedA-algebra C the following conditions are equivalent:

262 P.V. Morar and A.N. Shevlyakov

1. Th∀,A(B) ⊆ Th∀,A(C), i.e., C ∈ uclA(B);2. Th∃,A(B) ⊇ Th∃,A(C);3. C A-embeds into an ultrapower of B;4. C is A-discriminated by B;5. C is a limit algebra over B;6. C is an algebra defined by a complete atomic type in the theory Th∀,A(B)

in LA;7. C is the coordinate algebra of an irreducible algebraic set over B defined by

a system of equations with coefficients in A.

The equivalence of the conditions from these theorems holds for every algebraB. If we deal with a concrete algebra, its coordinate algebra has additional prop-erties. In our paper we consider positive commutative monoids with cancellation(a monoid is called positive if every sum of non-zero elements is non-zero) andwe prove several theorems about the classification of coordinate monoids in thecase of systems of equations with no coefficients. Our theorems supplement andspecialise Theorem 1.1 to a particular case.

Algebraic geometry over torsion-free abelian groups is quite simple and wasinvestigated in [2], where the following facts were proved.

• Every coordinate group over a torsion-free abelian group A of finite rank isisomorphic to a free abelian group of finite rank.

• Each algebraic set over a torsion-free abelian group A of finite rank is irre-ducible.In this paper we show that algebraic geometry over commutative positive

monoids with cancellation is more complicated than over torsion-free abeliangroups.

Here are the main results.

Theorem A. Let N be a commutative positive monoid with cancellation. A finitelygenerated monoid M is an irreducible coordinate monoid over N for a systemof equations with no coefficients iff M is a commutative positive monoid withcancellation.

Theorem B. Let N be a positive commutative monoid with cancellation. Thenevery coordinate monoid over N for a system of coefficient-free equations is irre-ducible. Hence every algebraic set over N is irreducible.

According to the following theorem, proved in our paper, it is only necessaryto prove Theorems A and B in the case N = N, where N is the additive monoidof natural numbers.

Theorem C. Let M be a commutative positive monoid with cancellation and S bea system of equations with no coefficients. Then the coordinate algebra of S overM is equal to the coordinate algebra of S over N, where N is the additive monoidof natural numbers.

Algebraic Geometry over the Additive Monoid of Natural Numbers 263

In the final section we develop the dimension theory of algebraic sets overthe additive monoid of natural numbers N and offer a method of calculation of thedimension of an algebraic set.

TheoremD. Let Y be an algebraic set over the additive monoid of natural numbersN for a system of equations with no coefficients. Then the dimension of Y is equalto the dimension of the subspace ρ(Y ) of the Euclidian space Rn, generated by Y .

In conclusion the authors would like to thank Professor V. Remeslennikovfor his support, attention and constructive criticism.

2. A-monoids

Let A be a monoid. A monoid B is said to be an A-monoid if B contains asubmonoid S such that S and A are isomorphic. More precisely, an A-monoid isa pair (B, λ), where λ : A→ B is embedding. Suppose C is a submonoid of B andC ⊇ λ(A); then a pair (C, λ) is called an A-submonoid of (B, λ).

We say that an A-monoid B is finitely generated over A if B is generated byA and a finite set C.

Let B1 and B2 be A-monoids, λ1 : A→ B1 and λ2 : A→ B2 be embeddings.A homomorphism ϕ : B1 → B2 is called an A-homomorphism if ϕ(λ1(a)) = λ2(a)for each a ∈ A. Denote by HomA(B1, B2) the set of all A-homomorphisms mappingB1 to B2.

Suppose B,C are A-monoids. We say that C is A-separated by B if for eachpair of distinct elements c1, c2 ∈ C there exists an A-homomorphism ϕ : C → Bsuch that ϕ(c1) �= ϕ(c2). We say that C is A-discriminated by B if for each naturalnumber k and for each set c1, . . . , ck consisting of k pairwise distinct elements of C,there exists an A-homomorphism ϕ : C → B such that ϕ(ci) �= ϕ(cj) with i �= j.

Denote by ResA(B) and DisA(B) the classes of all A-monoids which are A-separated and A-discriminated by B respectively.

2.1. Logical preliminaries

Let L = 〈◦(2), 1〉 be the standard language of monoid theory. L contains the binaryfunctional symbol ◦ and the constant symbol 1 with the standard interpretation.We extend the language L to

LA = L ∪ {ca |a ∈ A}.The class of all A-monoids is defined by the following series of LA-axioms:I) The axioms of monoid theory:

Ax′ : ∀x∀y∀z (x ◦ y) ◦ z = x ◦ (y ◦ z);Ax′′ : ∀x x ◦ 1 = 1 ◦ x = x;

II) The axioms which describe the existence of a submonoid which is isomorphicto A:

• ca = 1 for a = 1; • ca1◦a2 = ca1 ◦ ca2 ; • ca1 = ca2 iff a1 = a2.


Now we give the definitions of two special types of LA-formulas.• A universal sentence is a formula

∀x1 . . .∀xn ϕ(x1, . . . , xn),

where ϕ is a quantifier-free formula.• A quasi-identity is a universal sentence ∀x1 . . .∀xn ϕ(x1, . . . , xn), where ϕ isof the form

∧mi=1 ti(x) = si(x) → r1(x) = r2(x),

and si, ti, r1, r2 are LA-terms.Let B be an A-monoid. Below we give the definitions of two special classes

of A-monoids.• The A-universal closure of B is a set of all A-monoids C such that, for eachuniversal LA-formula ψ, if ψ is true in B then C |= ψ.

• The A-quasivariety of B is a set of all A-monoids C such that, for eachLA-quasi-identity ϕ, if ϕ is true in B then C |= ϕ.Denote by uclA(B) and qvarA(B) the A-universal closure and the A-quasi-

variety of B. It is easy to show that uclA(B) ⊆ qvarA(B).

3. Introduction to algebraic geometry

In this section we give basic definitions and main theorems of algebraic geometryover monoids. For more details see [1].

3.1. Systems of equations

Suppose X = {x1, . . . , xn} is a fixed set of variables.Let TLA(X) be the set of LA-terms with variables from the set X . The

composition of terms on the set TLA(X) is clearly defined. TLA(X) is said to bean absolutely free algebra with the basis X .

An atomic LA-formula t = s, where t, s ∈ TLA(X), is called an equation overA, and a set of equations

S = {ti = si| i ∈ I, ti, si ∈ TLA(X)}is called a system of equations over A.

Only variables, the symbols ◦, 1 and the constants ca (a ∈ A) can occur ina term t ∈ TLA(X); thus the value t(b1, . . . , bn), bi ∈ B is defined for every A-monoid B. Hence we can seek a solution of every equation over A in an arbitraryA-monoid B.

The n-dimensional affine space over B is the set Bn = {(b1, . . . , bn) | bi ∈ B}.A point p = (b1, . . . , bn) ∈ Bn is said to be a solution of an equation t = s

if B |= t = s with the interpretation xi �→ bi, 1 ≤ i ≤ n. Moreover, p is calleda solution of the system of equations S if p is a solution of every equation of S.Denote by VB(S) the set of all solutions of a system S.


A system S is called incompatible over B if VB(S) = ∅, and otherwise S iscompatible.

If the monoid A is trivial, i.e., A = {1}; then the language LA is equal to L,and every equation over A is coefficient-free. In this case the algebraic geometryover B is called coefficients free.

3.2. Algebraic sets

A set Y ⊆ Bn is said to be algebraic over B if there exists a system of equationsS such that Y = VB(S).

An algebraic set Y is called irreducible if there do not exist algebraic setsY1, Y2 such that Y = Y1 ∪ Y2, Y �= Y1, and Y �= Y2. Otherwise, Y is a reducibleset.

Suppose Y ⊆ Bn and Z ⊆ Bm are algebraic sets. A mapping μ : Y → Z iscalled an LA-mapping if there are terms t1, . . . , tm ∈ TLA(X) such that

μ(b1, . . . , bn) = (t1(b1, . . . , bn), . . . , tm(b1, . . . , bn)) ∈ Z.

Two algebraic sets Y, Z are called isomorphic if there are two LA-mappingsμ : Y → Z, η : Z → Y such that η ◦ μ and μ ◦ η are the identity mappings over Yand Z respectively.

Lemma 3.1. Suppose Y, Z are isomorphic sets; then Y is irreducible iff Z is irre-ducible.

Now we define the category ASA(B) of all algebraic sets over B. The set ofobjects of ASA(B) consists of all nonempty algebraic sets over B, and the set ofmorphisms of this category is the set of all LA-mappings of algebraic sets over B.

Note that the empty set is algebraic over an A-monoid B whenever A �= {1}because ∅ is defined by the system 1 = a, where a ∈ A\{1}. If A = {1}, the emptyset is not algebraic since every system S with the variables X = {x1, . . . , xn} iscompatible and has the solution (1, 1, . . . , 1).

In other words, we have the following statement.

Statement 3.2. ∅ ∈ ASA(B) iff A �= {1}.3.3. Radicals

Let Y be an arbitrary set in Bn. The radical of Y is the set of all equations overA which are satisfied by all points of Y . More formally,

RadB(Y ) = {t(x) = s(x) | B |= t(p) = r(p) ∀ p ∈ Y }.It is clear that the radical RadB(∅) contains all equations over A. The radical

of a system S is RadB(S) = RadB(VB(S)). The following proposition describesthe properties of radicals.

Proposition 3.3.

• Suppose Y1 ⊆ Y2 ⊆ Bn; then

RadB(Y1) ⊇ RadB(Y2).


• Suppose Yi ⊆ Bn, i ∈ I; then

RadB

(⋃i∈I

Yi

)=

⋂i∈I

RadB(Yi).

• Each radical defines a unique algebraic set, i.e., if Y1, Y2 ⊆ Bn are algebraicsets, we have

Y1 = Y2 ⇔ RadB(Y1) = RadB(Y2).

Suppose Y ⊆ Bn. Proposition 3.3 states that the radical RadB(Y ) defines analgebraic set Y = VB(RadB(Y )), which is called the algebraic closure of Y . It iseasy to see that the set Y is algebraic whenever Y = Y .

3.4. Coordinate monoids

Let ∼ be an equivalence relation on an A-monoid C. The relation ∼ is calledcongruence if the following condition holds

m1 ◦m2 = m1 ◦ m2 (1)

for all equivalence classes m1, m2, where mi is the equivalence class of an elementmi ∈ C.

It is clear that each radical determines a congruence over TLA(X). The con-gruence generated by RadB(Y ) is denoted by ΘRadB(Y ). Hence for each algebraicset Y there is the factor monoid ΓA(Y ) = TLA(X)/ΘRadB(Y ). The monoid ΓA(Y )consists of the equivalence classes of terms from TLA(X), and the product of twoclasses is defined by formula (1). It is easy to check that for every nonempty al-gebraic set Y the factor monoid ΓA(Y ) is an A-monoid, and for Y = ∅ we haveΓA(Y ) = {1}.

The factor monoid ΓA(Y ) is called the coordinate A-monoid of the algebraicset Y .

Let p = (b1, . . . , bn) be a point of the affine space Bn. There is a mappingϕp : TLA(X)→ B defined by ϕp(t) = t(b1, . . . , bn). This mapping has the followingproperties.

Proposition 3.4. Suppose Y ⊆ Bn is an algebraic set.• If p ∈ Y is a point, then ϕp ∈ HomA(ΓA(Y ), B).• If ϕ ∈ HomA(ΓA(Y ), B), then there exists a point p ∈ Bn such that ϕ = ϕp.

Let us define the category CMA(B). The set of objects of CMA(B) consistsof all coordinate A-monoids corresponding to nonempty algebraic sets over B. Theset of morphisms of this category consists of all A-homomorphisms ϕ : Γ1 → Γ2.

The following proposition establishes relations between ASA(B) and CMA(B).

Proposition 3.5. Two algebraic sets Y1, Y2 are isomorphic whenever ΓA(Y1) andΓA(Y2) are isomorphic.

Remark 3.6. Further we write Y ∈ ASA(B) if Y is an algebraic set over B and wewrite M ∈ CMA(B) if M is a coordinate monoid corresponding to an algebraicset over B.


We shall say that a coordinate A-monoid ΓA(Y ) is irreducible if Y is irre-ducible. It follows from Lemma 3.1 and Proposition 3.5 that this definition is welldefined.

3.5. Equationally Noetherian monoids

An A-monoid B is said to be A-equationally Noetherian if for every infinite systemof equations S that depends on variables x1, . . . , xn, there exists a finite systemS0 ⊆ S such that VB(S) = VB(S0).

If a monoid B is A-equationally Noetherian we need only consider finitesystems of equations. Moreover, the following theorem holds.

Theorem 3.7. Let B be an A-equationally Noetherian monoid. Then every algebraicset over B can be represented as a finite union of irreducible algebraic sets Yi

(irreducible components). Moreover, if Yi Yj for i �= j, then this decompositionis unique up to a permutation of components.

The next theorem was proved in [2] for groups, but its proof can be eas-ily extended to arbitrary algebraic systems of a language without relations. Weformulate this theorem for monoids.

Theorem 3.8. Let B be an A-equationally Noetherian monoid. Then for an arbi-trary A-monoid C the following conditions are equivalent:1. C is a coordinate monoid of an algebraic set over B (C ∈ CMA(B));2. C is A-separated by B (C ∈ ResA(B));3. C ∈ qvarA(B);4. C ∈ SP(B), where S,P are the operators of taking A-submonoids and direct

products respectively.

Remark 3.9. According to Theorem 3.8, each monoid from quasivariety qvarA(B)can be obtained from B as the result of a composition of the operators S,P.Obviously, S,P preserve the Noetherian property, and so if B is A-equationallyNoetherian, then M is A-equationally Noetherian too.

The main problem of algebraic geometry over an A-monoid B is to classify• algebraic sets of the category ASA(B);• radicals of algebraic sets and• coordinate A-monoids of the category CMA(B).Propositions 3.3 and 3.5 establish the equivalence of these three approaches.

In our paper we follow the third approach and Theorems 1.1 and 3.8 give us apowerful tool for the classification of coordinate monoids.

4. Commutative monoids with cancellation

In this section we begin to study algebraic geometry over the additive monoid ofnatural numbers and give some properties of commutative monoids with cancella-tion.


In the sequel we reserve the notation A for a fixed submonoid of N. Thefollowing proposition plays an important role in the paper.

Proposition 4.1. The monoid N is A-equationally Noetherian.

Proof. Let R be the additive monoid of real numbers and let P be a submonoid ofR. It follows from well-known theorems of linear algebra that R is P -equationallyNoetherian, and the proof of this proposition immediately follows from Remark 3.9and the inclusion N ⊂ R. �

Theorem 3.8 states that all coordinate A-monoids of CMA(N) belong toquasivariety qvarA(N). Hence all quasi-identities that are true in N must be truein every A-monoid of CMA(N). In particular, all monoids of CMA(N) must becommutative because the axiom of commutativity

Axcomm : ∀x∀y x ◦ y = y ◦ x ∼ ∀x∀y (1 = 1)→ x ◦ y = y ◦ x (2)

is a quasi-identity.The commutativity of all coordinate monoids allows us to use the traditional

symbols +, 0 instead of ◦, 1, and further we use the language 〈+, 0〉 as L. Therefore,each LA-term is logically equivalent to an LA-term of the form∑

αixi + a, αi ∈ N, a ∈ A. (3)

Remark 4.2. According to formula (3), we can consider elements of the absolutelyfree algebra TLA(X) as elements of the free commutative monoid generated by Aand X = {x1, . . . , xn}.

A commutative monoid M is called a monoid with cancellation if the quasi-identity

Axcanc : ∀x∀y∀z x+ z = y + z → x = y (4)holds in M .

Theorem 4.3 ([5]). Suppose M is a commutative monoid. Then M embeds into theadditive group whenever M is a monoid with cancellation.

All coordinate A-monoids of algebraic sets over N are monoids with cancel-lation because N |= Axcanc. Therefore, each equation t = s over A can be writtenin the form ∑

i∈I

γixi + a =∑j∈J

γjxj + a′, (5)

where a, a′ ∈ A and I ∩ J = ∅.Remark 4.4. It is easy to see that there may be equations over A that can not betransformed to the form ∑

i∈I

γixi + a′′ =∑j∈J

γjxj ,

where one side of the equation does not have constant term.


5. Coefficient free algebraic geometry over N

In this section we classify coordinate monoids of the category CM0(N).

Definition 5.1. A monoid is called positive if for every pair of nontrivial elementsm1,m2 ∈M the sum m1 +m2 is not equal to 0.

We explain the motivation for the term “positive” in Subsection 5.2.

Remark 5.2. It is easy to prove that a monoid M is positive whenever the setM\{0} is a semigroup.

The quasi-identity of the language LAxpos : ∀x∀y x+ y = 0→ x = 0, (6)

describes the property of positiveness.The following important theorems give us a full description of coordinate

monoids of the category CM0(N).

Theorem A. A finitely generated monoid M is an irreducible coordinate monoidof the category CM0(N) iff M is a commutative positive monoid with cancellation(M |= Axcomm,Axcanc,Axpos).

We prove Theorem A in Subection 5.3.

Theorem B. Every coordinate monoid of the category CM0(N) is irreducible.Hence every algebraic set of the category AS0(N) is also irreducible.

Proof. Theorem B is a simple corollary of Theorem A and Theorem 3.8. Indeed,suppose M ∈ CM0(N). It follows from Theorem 3.8 that M ∈ qvar0(N). Theaxioms Axcomm,Axcanc,Axpos are defined as the quasi-identities (2), (4), (6), sowe see that M |= Axcomm,Axcanc,Axpos. It follows from Theorem A that themonoid M is irreducible. �

The next corollary follows from Theorems 1.1, 3.8, A, and B.

Corollary 5.3. The universal class qvar0(N) is equal to the universal class ucl0(N),and the set of formulas of the language L = 〈+, 0〉1. ∀x∀y∀z (x + y) + z = x+ (y + z);2. ∀x x+ 0 = 0 + x = x;3. ∀x∀y x+ y = y + x4. ∀x∀y∀z x+ z = y + z → x = y;5. ∀x∀y x+ y = 0→ x = 0.

is a complete set of axioms of these classes.

In the following subsection we study properties of finitely generated com-mutative positive monoids with cancellation. These properties are necessary forthe proof of Theorem A (Theorem C is proved as a corollary of Theorem A inSubsection 6).


5.1. Properties of finitely generated commutative positive monoidswith cancellation

Let us study the properties of finitely generated monoids that satisfy the axiomsAxcomm, Axcanc, Axpos. The next proposition holds for these monoids.

Proposition 5.4. Suppose M is a finitely generated commutative positive monoidwith cancellation. Then there exists a natural number n such that M is embeddedin the additive group Zn, and also this group can be embedded in the Euclidianspace Rn so that vectors corresponding to elements of the monoid M have integercoordinates.

Proof. According to Theorem 4.3, there exists an abelian group G and an embed-ding ϕ : M → G. It is clear that G is finitely generated.

The group G is isomorphic to the direct sum

G = Z1 ⊕ · · · ⊕ Zn ⊕ t(G),

where Zi Z and t(G) is the torsion part of G.The monoidM is positive, so it is torsion-free and ϕ(M)∩t(G) = ∅. Therefore

ϕ is an embedding M → Zn. �

Recall some well-known facts of theory of convex analysis in the Euclidianspaces.

Let C ⊆ Rn be a set. The intersection of all linear varieties of Rn that containC is called an affine hull of the set C and denoted by aff(C). A relative interiorri(C) of the set C is a subset of C such that every point of ri(C) has a neighborhoodO with O ∩ aff(C) ⊆ C. It follows from the definition that ri(C) is an open set ofaff(C).

A set K ⊆ Rn is called a cone if for every point x ∈ K and for every positivereal number λ we have λx ∈ K.

A set C ⊆ Rn is called convex if for all points x1, x2 ∈ C and for every realnumber 0 � λ � 1 the point λx1 + (1− λ)x2 is in C.

The next two theorems of convex analysis give us important properties ofconvex sets in Rn.

Theorem 5.5 ([6]). If a set K ⊆ Rn of vectors is closed under addition and multi-plication by positive real numbers, then K is a convex cone.

Theorem 5.6 ([6]). Suppose K1,K2 are nonempty convex cones in Rn and ri(K1)∩ri(K2) = ∅. Then there exists a nontrivial vector a such that

(a, x) ≥ 0 for each x ∈ K1,

(a, x) ≤ 0 for each x ∈ K2,

and there exists a point x0 ∈ K1 ∪K2 satisfying strict inequalities.


Let M ⊆ Zn be a finitely generated monoid with a generating set {m1, . . .,mk}. Denote

LR+(M) =

{k∑

i=1

αimi | αi ∈ R, αi ≥ 0

}.

This is the linear hull of generators m1, . . . ,mk in the space Rn such that all thecoefficients in the sum are real and nonnegative.

It is easy to see that the set LR+(M) is closed under addition and multipli-cation by nonnegative real numbers. Therefore, it follows from Theorem 5.5 thatthis set is a convex cone.

For every monoid M ⊆ Zn we can define the monoid

−M = {−m | m ∈M} ⊆ Zn.

Note that M ∩−M = {0} if M is positive.The next two lemmas hold for finitely generated positive submonoids of the

group Zn.

Lemma 5.7. Suppose M ⊆ Zn is a finitely generated positive monoid. Then

LR+(M) ∩ LR+(−M) = {0}.Proof. Let {m1, . . . ,mk} be a set of generators of M .

Assume the converse, i.e., there exists a nonzero vector y ∈ LR+(M) ∩LR+(−M). We have

y =k∑

i=1

αimi, (αi ∈ R+),

y =k∑

i=1

βi(−mi) (βi ∈ R+).

Therefore,k∑

i=1

(αi + βi)mi =k∑

i=1

γimi = 0

for nonnegative real numbers γ1, . . . , γk, and not all of these numbers equal 0.Without loss of generality it can be assumed that γ1, . . . , γp are positive and

γp+1, . . . , γk are equal to 0.Let us consider the system of linear equations Mγ = 0 with respect to the

variables γ = (γ1, . . . , γp), where the ith column of the matrix M is the vector mi.This system can be rewritten in the form

p∑i=1

γimi = 0. (7)

It is obvious that the vector (γ1, . . . , γp) with positive coordinates is a solutionof the system (7). Since there exists a nontrivial solution, the set of free variablesis nonempty.


Let γ1, . . . , γl be free variables and γl+1, . . . , γp be dependent variables. Sinceall elements of the matrixM are integers and the algorithm of Gaussian eliminationuses only arithmetic operations (addition, subtraction, multiplication, division), wecan represent the variables γl+1, . . . , γp as a linear combination of the free variablesγ1, . . . , γl with rational coefficients.

Now let us prove that there is a nonzero positive rational solution (r1, . . . , rp)of (7). If all coordinates of the vector (γ1, . . . , γp) are rational our proof is finished.Suppose that some of elements of (γ1, . . . , γp) are irrational. We can consider thevariables γl+1, . . . , γp as linear functions with the arguments γ1, . . . , γl. It is obvi-ous that γl+1, . . . , γp are continuous and positive at the point (γ1, . . . , γl). Hencethere is a neighbourhood Oγ of (γ1, . . . , γl) such that the functions γl+1, . . . , γp arepositive at every point x ∈ Oγ .

Let (r1, . . . , rl) be an arbitrary point with rational coordinates from Oγ .The values r1, . . . , rl of the free variables γ1, . . . , γl determine a positive rationalsolution (r1, . . . , rp) of the system (7).

Since (r1, . . . , rp) is a solution of (7), we can substitute (r1, . . . , rp) instead ofthe variables (γ1, . . . , γp). We obtain a nontrivial linear combination with positiverational coefficients

p∑i=1

rimi = 0.

If we multiply the both sides of the last equation by a suitable number, we obtain anontrivial linear combination with natural coefficients. We obtain a contradictionbecause M is a positive monoid.

This contradiction proves that LR+(−M)⋂LR+(M) = {0}. �

Lemma 5.8. Suppose M ⊆ Zn is a finitely generated positive monoid. Then thereexists a vector a ∈ Rn such that for every nontrivial element m ∈ M the scalarmultiplication (a,m) is positive.

Proof. Since the lemma obviously holds for the trivial monoidM = {0}, we assumethat M �= {0}.

Let O0 be a neighbourhood of the zero vector in the space Rn. Since LR+(M),LR+(−M) are convex cones, we have

LR+(M) ∩O0 �= {0}, LR+(−M) ∩O0 �= {0}and

LR+(M) ∪ LR+(−M) ⊂ aff(LR+(M)) and aff(LR+(M)) = aff(LR+(−M)).

We obtain

O0 ∩ aff(LR+(M)) LR+(M), O0 ∩ aff(LR+(−M)) LR+(−M).

From the two previous formulas we obtain

0 /∈ ri(LR+(M)), 0 /∈ ri(LR+(−M)).


Therefore, ri(LR+(M)) ∩ ri(LR+(−M)) = ∅. According to Theorem 5.6, there is avector a ∈ Rn such that (a, x) ≥ 0 for all x ∈ LR+(M) and there exists a point pof LR+(M) such that (a, p) > 0.

Let {m1, . . . ,mk} be a set of generators of M . Since M ⊆ LR+(M), we have(a,m) ≥ 0 for all m ∈ M . If the generators m1, . . . ,mk are orthogonal to a, theneach vector from LR+(M) is also orthogonal to a and there does not exist a vectorx ∈ LR+(M) such that (a, x) > 0. Therefore, there exists a generator mi such thatthe scalar multiplication (a,mi) is positive.

Further we prove by induction on the number of generators ofM the existenceof a vector a′ ∈ Rn such that (a′, x) > 0 for each element x ∈M .

Clearly, if M = 〈m1〉 we may set a′ = m1.Let us prove the step of induction. Suppose M = 〈m1, . . . ,mk〉 ⊆ Zn. Using

the a found above, without loss of generality it may be assumed that m1 = mi, so(a,mi) > 0 and (a,m) ≥ 0 for all m ∈M . The inductive assumption gives b suchthan (b,mi) > 0 for all i ≥ 2.

Since the scalar multiplication is a continuous function, there exists a realnumber ε > 0 such that (a+ εb,m1) > 0 and,

(a+ εb,mi) ≥ ε(b,mi) > 0

for i ≥ 2.We obtain that scalar multiplication (a+ εb,mi) is positive for all generators

mi. Since an element m ∈ M is a linear combination of generators with naturalcoefficients, we have (a+ εb,m) > 0. �

For a monoid M define the set

V>(M) = {a ∈ Rn|(a,m) > 0 for every nonzero m ∈M}.The set V>(M) is open in Rn, and it is not empty if the monoidM ⊆ Zn is finitelygenerated and positive (Lemma 5.8).

Let x1, . . . , xt be pairwise distinct vectors of the space Rn. We say that avector a ∈ Rn distinguishes the vectors x1, . . . , xt if for every i �= j we have(a, xi) �= (a, xj).

Let us prove the next lemma.

Lemma 5.9. Suppose M ⊆ Zn is a finitely generated positive monoid and x1, . . . , xt

are pairwise distinct elements of M . Then there exists a vector a ∈ V>(M) ∩ Znn

that distinguishes x1, . . . , xt.

Proof. The proof is by induction on t.For t = 2 we take an arbitrary vector a′ ∈ V>(M). If (a′, x1) �= (a′, x2), there

is nothing to prove. Suppose (a′, x1) = (a′, x2). Since the set V>(M) is open inRn, there is a neighbourhood Oa′ of a′ such that Oa′ ⊆ V>(M).

The vector a′ belongs to the closure of the set X = {x|(x, x1) �= (x, x2)}.Since X is open, the intersection Oa′ ∩X is nonempty and open in Rn.

Since Oa′∩X is open, there is a vector a′′ ∈ Oa′∩X with rational coordinatessuch that a′′ distinguishes x1 and x2. Let α be a suitable natural number such


that the vector αa′′ has natural coordinates. It is obvious that αa′′ satisfies allconditions of the lemma.

Let us proceed with the induction step. Assume that the lemma holds for allnumbers t′ < t. Suppose x1, . . . , xt is a set of pairwise distinct elements of M . Bythe inductive assumption there are vectors a1, . . . , at ∈ V>(M) ∩ Zn such that ai

distinguishes the elements x1, . . . , xi−1, xi+1, . . . , xt.We put

a = a1 +A1a2 +A1A2a3,

whereAi = max

1�j�t{(ai, xj)} + 1.

Since Ai ∈ N, we have a ∈ V>(M) ∩ Zn.Let us prove that a distinguishes all pairs xi, xj (1 � i �= j � t). Consider all

possible cases, where integer remainders and quotients are denoted by mod and ÷respectively.1) If i > 1, j > 1, we have

(a, xi)modA1 = (a1, xi),

(a, xj)modA1 = (a1, xj).

Since (a1, xi) �= (a1, xj), we obtain (a, xi) �= (a, xj).2) If i = 1, j > 2, we have

((a, x1)modA1A2)÷A1 = (a2, x1),

((a, xj)modA1A2)÷A1 = (a2, xj).

Since (a2, x1) �= (a2, xj), we obtain (a, x1) �= (a, xj).3) If i = 1, j = 2, we have

(a, x1)÷ (A1A2) = (a3, x1),

(a, x2)÷ (A1A2) = (a3, x2).

Since (a3, x1) �= (a3, x2), we obtain (a, x1) �= (a, x2). �

5.2. Ordering of submonoids of Zn

In this subsection we explain the motivation of the term “positive”. We study howfor a given finitely generated positive monoid M ⊆ Znwe can define a monotoneorder ≤ on Zn such that this monoid will be positive with respect to ≤.

An order ≤ on an abelian group G is called monotone if the following condi-tions hold:

• ∀x∀y(x ≤ y) ∨ (y ≤ x)• ∀x, y, z(x ≤ y) ∧ (y ≤ z)→ (x ≤ z)• ∀x, y, z(x ≤ y)→ (x+ z ≤ y + z)

Note that the definition allows the existence of elements g1, g2 ∈ G such thatg1 ≤ g2, g2 ≤ g1, but g1 �= g2.


Proposition 5.10 explains the choice of the term “positive monoid”.

Proposition 5.10. Suppose a finitely generated positive monoid M is embedded inZn. Then we can define a monotone order ≤ on Zn such that 0 ≤ m and m 0for all non-trivial m ∈M .

Proof. Let M ⊆ Zn be a finitely generated positive monoid. Using Lemma 5.8 wemay choose a vector a ∈ Zn such that (a,m) > 0 for all non-trivial elements ofM . For all x1, x2 ∈ Zn we put x1 ≤ x2 whenever (x1, a) ≤ (x2, a). It follows fromthe properties of scalar multiplication that the order ≤ is monotone and 0 ≤ m,m 0 whenever m ∈M\{0}. �

5.3. Proof of Theorem ATheorem A. A finitely generated monoid M is an irreducible coordinate monoidof the category CM0(N) iff M is commutative positive and has cancellation (M |=Axcomm,Axcanc,Axpos).

Proof. Let M be an irreducible coordinate monoid over N (M ∈ CM0(N)). It fol-lows from Theorem 1.1 thatM belongs to the universal closure ucl0(N). Therefore,the universal formulas Axcomm, Axcanc, Axpos must be true in M .

Assume now thatM is a finitely generated commutative positive monoid withcancellation. Let us prove thatM is a coordinate monoid over N. By Theorem 1.1,it is sufficient to show that N discriminates M .

Suppose m1, . . . ,mk are pairwise distinct elements of M . Using Proposi-tion 5.4, we have that M is embedded into Zn. By Lemma 5.9, there is a vectora ∈ Zn ∩ V>(M) distinguishing m1, . . . ,mk.

Consider the mapping ψ(x) = (a, x). Since a ∈ Zn∩V>(M), we have ψ(x) ∈ Nfor all x ∈ M . Moreover, ψ(x + y) = ψ(x) + ψ(y) for all x, y ∈ M . Therefore,ψ : M → N is a homomorphism. Since a distinguishes m1, . . . ,mk, we obtain thatψ(mi) �= ψ(mj) (i �= j) so N discriminates M . �

6. Geometric and universal equivalence

It is easy to show that Theorems A and B proved for N can be extended to thesubclass of commutative monoids, in particular, to the class of direct products Nk.It means that the categories CMA(N) and CMA(Nk) are equal and, moreover, thecategories of irreducible coordinate monoids over Nk and N are also equal. There-fore, N and Nk are called geometrically and universally equivalent. Below we givea definition of the geometric and universal equivalence over monoids and describethe class of monoids that are geometrically and universally equivalent to N.

A-monoids M1,M2 are called geometrically equivalent if RadM1(S) =RadM2(S) for every system S of equations over A. We denote the geometricalequivalence of A-monoids M1,M2 by M1 ≈A M2.

This notion for algebraic systems was introduced by Plotkin [4]. Below weformulate the Plotkin’s problem over monoids.


Plotkin’s problem. Find necessary and sufficient conditions for the geometric equiv-alence of A-monoids M1,M2.

The following theorem solves this problem in the Noetherian case.

Theorem 6.1 ([2], [3]). Suppose A-monoids M1,M2 are A-equationally Noetherian.Then M1 ≈A M2 ⇐⇒ qvarA(M1) = qvarA(M2).

Theorem 6.1 was proved in [2] for groups (Theorem 2) and in [3] for Liealgebras (Lemma 3.47). The proof of this theorem is universal and uses ideas frommodel theory. Therefore, Theorem 6.1 can be proved for monoids with the samereasoning.

A-monoids M1,M2 are called universal equivalent if M1 |= ϕ⇔M2 |= ϕ forevery LA-formula ϕ. Denote byM1 ≈∀ M2 the universal equivalence of A-monoidsM1,M2. Obviously, M1 ≈∀ M2 implies uclA(M1) = uclA(M2).

Note that geometric equivalence does not imply universal equivalence andvice versa.

The main result of this section is Theorem C, which states that a non-trivialmonoid belongs to the quasivariety qvar0(N) if and only if it is geometricallyequivalent to N.

We need the following two lemmas to prove Theorem C. The proof of thefirst simple lemma is omitted.

Lemma 6.2. Suppose M1 and M2 are A-monoids.• qvarA(M1) = qvarA(M2) iff M2 ∈ qvarA(M1) and M1 ∈ qvarA(M2);• uclA(M1) = uclA(M2) iff M2 ∈ uclA(M1) and M1 ∈ uclA(M2).

Lemma 6.3. Suppose M ∈ qvar0(N). Then N ∈ ucl0(M).

Proof. Let m be a nonzero element of M . Since M is positive, the submonoidgenerated by m is isomorphic to N. Suppose ϕ is a universal formula such thatM |= ϕ. It is clear that 〈m〉 N |= ϕ. Therefore we obtain N ∈ ucl0(M). �

TheoremC. A nontrivial monoid M belongs to the quasivariety qvar0(N) iff M ≈0N. In this case all algebraic sets over M are irreducible (qvar0(M) = ucl0(M)),and also M ≈∀ N.

Proof. Suppose M ∈ qvar0(N) and M �= {0}. By Lemma 6.3, we haveN ∈ ucl0(M) ⊆ qvar0(M). Since M ∈ qvar0(N), using Lemma 6.2, we obtainqvar0(N) = qvar0(M). It follows from Theorem 6.1 and Remark 3.9 that M ≈0 N.

Suppose M ≈0 N. It easily follows from Theorem 6.1 and Lemma 6.2 thatM ∈ qvar0(N).

Using Theorem B, 1.1, 3.8, we haveM ∈ qvar0(N) = ucl0(N). By Lemma 6.3,we obtain ucl0(M) = ucl0(N) = qvar0(N) = qvar0(M). Therefore M ≈∀ N and allalgebraic sets over M are irreducible. �

Corollary 6.4. A monoid M �= {0} is geometrically and universally equivalent toN if and only if M is a commutative positive monoid with cancellation.


7. Dimension theory

Since all the algebraic sets of the category AS0(N) are irreducible (Theorem B),we give the definition of dimension an algebraic set only for irreducible algebraicsets.

Let Y be an irreducible algebraic set and Y = Y0 ⊃ Y1 ⊃ · · · ⊃ Ym be aseries of irreducible sets such that this series has the maximal length among allthe series of this form. The length of this series is called the dimension, dimY ,of the set Y . Note that the dimension of every algebraic set is finite because N isequationally Noetherian.

Suppose Y ∈ AS0(N) and Y ⊆ Nn. We embed Y into Rn using the naturalembedding of Nn into Rn. The subspace generated by the image of Y in Rn isdenoted by ρ(Y ).

The main result of this section is Theorem D. Using this theorem we caneasily reduce the calculation of the dimension of an algebraic set to the calculationof the dimension of the corresponding subspace of Rn.

Lemma 7.1. Suppose Y1, Y2 ∈ AS0(N), and ρ(Y1) = ρ(Y2), then Y1 = Y2.

Proof. Assume the converse, i.e., Y1 �= Y2. Then there is a point y ∈ Y1\Y2.Let e1, . . . , em be a basis of ρ(Y1). Then y ∈ Nn is equal to a nontrivial linearcombination of e1, . . . , em

y = α1e1 + · · ·+ αmem.

We can assume that α1 �= 0. Hence the set of vectors y, e2, . . . , em is also a basis ofρ(Y1). Since y /∈ ρ(Y2), we obtain ρ(Y1) �= ρ(Y2), in contradiction to the conditionsof the lemma. �

Theorem D. For every algebraic set Y of the category AS0(N) we have dimY =dim ρ(Y ).

Proof. Let Y ∈ AS0(N) andY = Y0 ⊃ · · · ⊃ Ym (8)

be a chain of irreducible algebraic sets. We assume that (8) is a chain of themaximal length. We can construct the chain of subspaces

ρ(Y ) = R0 ⊃ · · · ⊃ Rm, (9)

where Ri = ρ(Yi). By Lemma 7.1, all the sets in (9) are pairwise distinct.Let us show that dimRi − dimRi+1 = 1 for i = 1, . . . ,m − 1. Assume the

converse, i.e., there is a number i such that dimRi − dimRi+1 = k > 1. Sinceρ(Yi) = Ri and ρ(Yi+1) = Ri+1, we have that the sets Ri and Ri+1 are algebraicover R and are defined by a systems Si, Si+1 with natural coefficients. It is clearthat Si ⊂ Si+1 because Yi ⊃ Yi+1. Since k > 1, there exists a system S′ suchthat Si ⊂ S′ ⊂ Si+1. The subspace R′ generated by solutions of S′ satisfies theconditions Ri ⊃ R′ ⊃ Ri+1. The system S′ has natural coefficients, so there is analgebraic set Y ′ over N such that ρ(Y ′) = R′.


It follows from the properties of R′ and Lemma 7.1 that Yi ⊃ Y ′ ⊃ Yi+1,but these inclusions contradict to the condition that the length of the chain (8) ismaximal. We have proved that dimRi − dimRi+1 = 1 for i = 1, . . . ,m− 1, so thechain (9) is maximal.

Since the chains (8),(9) have the same length, we have Y = dim ρ(Y ). �Corollary 7.2. Suppose Y = Y0 ⊃ · · · ⊃ Yn is a series of irreducible algebraic setsof AS0(N), such that the set inclusions are strict and the length of the series cannot be increased. Then dimY = n.

Proof. Given the series Y = Y0 ⊃ · · · ⊃ Yn using the mapping ρ, we can constructthe series ρ(Y ) = ρ(Y0) ⊃ · · · ⊃ ρ(Yn). It is easy to prove that the length of theseries constructed can not be increased. Using the properties of the space Rn weobtain that dim ρ(Y ) = n, and it follows from Theorem D that dimY = n. �

References

[1] E. Daniyarova, A. Miasnikov, V. Remeslennikov, Unification theorems in algebraicgeometry, Algebra and Discrete Mathematics, 1 (2008), 137–164.

[2] A. Myasnikov, V. Remeslennikov, Algebraic geometry over groups II: logical foun-dations, J. Algebra, 234 (2000), 225–276.

[3] E. Daniyarova, Foundations of algebraic geometry over Lie algebras, Herald of OmskUniversity, Combinatorical methods in algebra and logic (2007), 8–39.

[4] B. Plotkin, Varieties of algebras and algebraic varieties. Categories of algebraic va-rieties, Siberian Advances in Math., 7 (1997), 64–97.

[5] The algebraic theory of semigroups, A.H. Clifford and G.B. Preston. American Math-ematical Society, 1961 (volume 1), 1967 (volume 2).

[6] Rockafellar R.T., Convex Analysis, Princeton University Press, Princeton, N.J.,1970.

Pavel V. Morar and Artem N. ShevlyakovOmsk Department of Institute of MathematicsSiberian Branch of the Russian Academy of SciencesPevtsova st. 13Omsk, 644099 Russiae-mail: [email protected]

a [email protected]



Some Graphs Related to Thompson’s Group F

Dmytro Savchuk

Abstract. The Schreier graphs of Thompson’s group F with respect to thestabilizer of 1

2and generators x0 and x1, and of its unitary representation in

L2([0, 1]) induced by the standard action on the interval [0, 1] are explicitlydescribed. The coamenability of the stabilizers of any finite set of dyadicrational numbers is established. The induced subgraph of the right Cayleygraph of the positive monoid of F containing all the vertices of the form xnv,where n ≥ 0 and v is any word over the alphabet {x0, x1}, is constructed. Itis proved that the latter graph is non-amenable.

Mathematics Subject Classification (2000). 20F65.

Keywords. Thompson’s group, amenability, Schreier graphs, Cayley graphs.

Introduction

Thompson’s group F was discovered by Richard Thompson in 1965. A lot offascinating properties of this group were discovered later on, many of which aresurveyed nicely in [CFP96]. It is a finitely presented torsion free group. One ofthe most intriguing open questions about this group is whether F is amenable.Originally this question was asked by Geoghegan in 1979 (see p. 549 of [GS87])and since then dozens of papers were in some extent devoted to it. It was shownin [BS85] that F does not contain a nonabelian free subgroup and in [CFP96]that it is not elementary amenable. So the question of amenability of F is partic-ularly important because F would be an example of a group given by a balancedpresentation (two generators and two relators) of either amenable, but not el-ementary amenable group (the first finitely presented example was constructedby R.Grigorchuk in [Gri98]), or non-amenable group, which does not contain anonabelian free subgroup (the first finitely presented example of this type wasconstructed by Ol’shanskii and Sapir in [OS02]).

The study of the Schreier graphs of F was also partially inspired by thequestion of amenability of F . In particular, if any Schreier graph with respect

The author was supported by NSF grants DMS-0600975 and DMS-0456185.

280 D. Savchuk

to any subgroup is non-amenable the whole group F would be non-amenable.Unfortunately, all Schreier graphs we describe here are amenable which does notgive any information about the amenability of F . But the knowledge about thestructure of Schreier graphs provides some additional information about F itself.

It happens that the described Schreier graph of the action of F on the setof dyadic rational numbers on the interval (0, 1) is closely related to the unitaryrepresentation of F in the space B(L2([0, 1])) of all bounded linear operators onL2([0, 1]). It reflects (modulo a finite part) the dynamics of F on the Haar waveletbasis in L2([0, 1]). We define the Schreier graph of a group action on a Hilbertspace with respect to some basis and make this connection precise.

R. Grigorchuk and S. Stepin in [GS98] reduced the question of amenabil-ity of F to the right amenability of the positive monoid P of F . Moreover, theamenability of F is equivalent to the amenability of the induced subgraph ΓP ofthe Cayley graph ΓF of F with respect to generating set {x0, x1} containing thepositive monoid P . We construct the induced subgraph ΓS of ΓF containing allthe vertices of the form xnv for n ≥ 0, v ∈ {x0, x1}∗ and prove that this graph isnon-amenable. In this construction we use the realization of the elements of thepositive monoid of F as binary rooted forests. The existence of this representationwas originally noted by K. Brown and developed by J. Belk in [Bel04] and Z. Sunicin [Sun07]. It was also used by J. Donelly in [Don07] to construct an equivalentcondition for amenability of F .

The structure of the paper is as follows. In Section 1 the definition and thebasic facts about Thompson’s group are given. Section 2 contains a description ofthe Schreier graph of the action of F on the set of dyadic rational numbers fromthe interval (0, 1). The coamenability of the stabilizers of any finite set of dyadicrational numbers is shown in Section 3. The Schreier graph of the action of F onL2([0, 1]) is constructed in Section 4. The last Section 5 contains a description ofthe subgraph ΓS of ΓP and a proof that ΓS is non-amenable.

I would like to express my warm gratitude to Rostislav Grigorchuk for valu-able comments and bringing my attention to Thompson’s group, and to ZoranSunic, who has pointed to the connection with forest diagrams, which simplifiedthe proofs in the last section. Also I want to thank the referee for constructivecomments and suggestions that were very helpful.

1. Thompson’s group

Definition 1. The Thompson’s group F is the group of all strictly increasing piece-wise linear homeomorphisms from the closed unit interval [0, 1] to itself that aredifferentiable everywhere except at finitely many dyadic rational numbers and suchthat on the intervals of differentiability the derivatives are integer powers of 2. Thegroup operation is superposition of homeomorphisms.

Basic facts about this group can be found in the survey paper [CFP96]. Inparticular, it is proved that F is generated by two homeomorphisms x0 and x1

Some Graphs Related to Thompson’s Group F 281

given by

x0(t) =

⎧⎪⎨⎪⎩t2 , 0 ≤ t ≤ 1

2 ,

t− 14 ,

12 ≤ t ≤ 3

4 ,

2t− 1, 34 ≤ t ≤ 1,

x1(t) =

⎧⎪⎪⎪⎨⎪⎪⎪⎩t, 0 ≤ t ≤ 1

2 ,t2 +

14 ,

12 ≤ t ≤ 3

4 ,

t− 18 ,

34 ≤ t ≤ 7

8 ,

2t− 1, 78 ≤ t ≤ 1.

The graphs of x0 and x1 are displayed in Figure 1.

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Figure 1. Generators of F

Throughout the paper we will follow the following conventions. For any twoelements f , g of F and any x ∈ [0, 1]

(fg)(x) = g(f(x)), fg = gfg−1. (1)

With respect to the generating set {x0, x1} F is finitely presented. But forsome applications it is more convenient to consider an infinite generating set{x0, x1, x2, . . .}, where

xn = (x1)xn−10 .

With respect to this generating set (and with respect to convention (1)) F has anice presentation

F ∼= 〈x0, x1, x2, . . . | xkxn = xn+1xk, 0 ≤ k < n〉. (2)

2. The Schreier graph of the action of F on the set ofdyadic rational numbers

Let G be a group generated by a finite generating set S acting on the set M . TheSchreier graph Γ(G,S,M) of the action of G on M with respect to the generat-ing set S is an oriented labelled graph defined as follows. The set of vertices ofΓ(G,S,M) is M and there is an arrow from x ∈M to y ∈M labelled by s ∈ S ifand only if xs = y.

282 D. Savchuk

For any subgroup H of G, the group G acts on the right cosets in G/H byright multiplication. The corresponding Schreier graph Γ(G,S,G/H) is denotedas Γ(G,S,H) or just Γ(G,H) if the generating set is clear from the context.

Conversely, if G acts onM transitively, then Γ(G,S,M) is canonically isomor-phic to Γ(G,S, StabG(x)) for any x ∈ M , where the vertex y ∈ M in Γ(G,S,M)corresponds to the coset from G/ StabG(x) consisting of all elements of G thatmove x to y.

Consider the subgroup StabF (12 ) of F consisting of all elements of F that fix12 . There is a natural isomorphism ψ : StabF (12 )→ F × F given by

StabF

(12

)( f(t)

ψ�−→(2f( t2

), 2f

( t+ 12

)− 1

)∈ F × F. (3)

This group was studied in [Bur99], where it was shown that it embeds into Fquasi-isometrically.

The Schreier graph Γ(F, {x0, x1}, StabF (12 )) coincides with the Schreier graphof the action of F on the orbit of 12 . Let D be the set of all dyadic rational numbersfrom the interval (0, 1). It is known that F acts transitively on D (which followsalso from the next proposition). Therefore the latter graph coincides with theSchreier graph Γ(F, {x0, x1}, D).

Proposition 1. The Schreier graph Γ(F, {x0, x1}, D) has the following structure(dashed arrows are labelled by x0 and solid arrows by x1)

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Proof. Define the following subsets of D.

An ={ k

2n∣∣ k is odd} ∩ (1

2,34

), n ≥ 3

Bn ={ k

2n∣∣ k is odd} ∩ (3

4,78

), n ≥ 4

Cn = An ∩(12,58

), Dn = An ∩

(58,34

), n ≥ 4


On the graph above, An represents the (n− 3)-rd level of the gray vertices in thebinary tree; Bn is the set of the white vertices between levels n − 4 and n− 3 ofthe tree, which are adjacent to 2 gray vertices; Cn and Dn are the sets of the grayvertices of the (n−3)-rd level having gray and white neighbors above respectively.

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Figure 2. Dynamics of x0 and x1

Now we compute the action of F on this subsets (see Figure 2). We havex−10 (An) = Bn+1, x1(Bn) = Dn, x1(An) = Cn+1, hence (x−10 x1)(An) = Dn+1

and (x−10 x1)(An) ∪ x1(An) = An+1. Therefore, we immediately obtain that themonoid generated by x1 and x−10 x1 is free. Indeed, any relation in this monoidwould create a loop in the Schreier graph contradicting to the fact that all thesets Cn and Dn are disjoint. Hence, the grey part of the graph is a binary tree.Furthermore, for any set A ⊂ R denote αA + β = {αa + β : a ∈ A}. Thenxk0(An) = xk

0x1(An) = 2−k+1(An− 14 ) for k ≥ 1. This corresponds to the rays with

the black vertices sticking out to the right from the gray ones. On the other handsince the actions of x−10 and x−11 on [34 , 1] coincide, for any element f of lengthk ≥ 0 from the monoid generated by x−10 and x−11 we have f(Bn) = 1−2−k(1−Bn).This corresponds to the rays with white vertices. There is one more geodesic linein the graph corresponding to 12 which completes the picture. �

This graph gives alternative proofs of the following well-known facts.

Corollary 1.

(a) Thompson’s group F acts transitively on the set D of all dyadic rationalsfrom the interval (0, 1).

(b) StabF (12 ) acts transitively on the sets of dyadic rationals from the intervals(0, 12 ) and (

12 , 1).

284 D. Savchuk

Proof. Part (a) follows immediately from the structure of the Schreier graphF/ StabF (12 ). Part (b) is a consequence of part (a) and the isomorphism (3). �

Proposition 2. The subgroup StabF (12 ) is a maximal subgroup in F .

Proof. Let f be any element from F \StabF (12 ). Then for any g ∈ F we show thatg ∈ 〈StabF (12 ), f〉. Let g be an arbitrary element in F that does not stabilize 12 .

Denote u = f(12 ) and v = g(12 ). Without loss of generality we may assumeu < 1

2 . Then by transitivity from Corollary 1(b) there exists h ∈ StabF (12 ) suchthat either h(f(12 )) = v or h(f−1(12 )) = v depending on whether v < 1

2 or v >12 . In

any case the element f = fh (or f = f−1h) belongs to 〈StabF (12 ), f〉 and satisfiesf(12 ) = v.

Now for h = gf−1 we have h(12 ) = f−1(g(12 )) = f−1(v) = 12 . Thus h ∈

StabF (12 ) and g = hf ∈ 〈StabF (12 ), f〉. �

Proposition 1 also yields a bound on the length of an element. Namely, ifthe graph of an element f ∈ F passes through the point (a, b) for some dyadicrational numbers a and b, then the length of f with respect to the generating set{x0, x1} is not smaller than the combinatorial distance between a and b in thegraph Γ(F, {x0, x1}, D).

Estimates similar in spirit (also based on the properties of graph of an ele-ment, but in a different realization of F ) were used by J.Burillo in [Bur99] to showthat StabF (12 ) quasi-isometrically embeds into F .

3. Coamenability of stabilizers of several dyadic rationals

In this section we show that for any finite subset {d1, . . . , dn} of dyadic rationalsthe Schreier graphs of F with respect to StabF (d1, . . . , dn) is amenable.

First we recall the definition of an amenable graph.

Definition 2. Given an infinite graph Γ = (V,E) of bounded degree the Cheegerconstant h(Γ) is defined as follows

h(Γ) = infS

|∂S||S| ,

where S runs over all nonempty finite subsets of V , and ∂S, the boundary of S,consists of all vertices of V \ S that have a neighbor in S.

Definition 3. A graph Γ is called amenable if h(Γ) = 0.

Definition 4. A subgroup H of a group G is called coamenable in G if the Schreiergraph Γ(G,H) is amenable.

Note, that coamenability of a subgroup does not depend on the generatingset of G. This follows easily from Gromov’s doubling condition (see Theorem 5 inSection 5).


Proposition 3. Let {d1, . . . , dn} ⊂ D be any finite subset of dyadic rationals. Thenthe subgroup StabF (d1, . . . , dn) of F consisting of all elements stabilizing all thedi’s is coamenable in F .

Proof. First, we describe the structure of the Schreier graph Γ(F, {x0, x1},StabF (d1, . . . , dn)), d1 < d2 < · · · < dn. Analogously to the singleton case thereis a one-to-one correspondence between cosets from F/ StabF (d1, . . . , dn) and allstrictly increasing n-tuples of dyadic rationals. This follows from the fact thatF acts transitively on the latter set (see [CFP96]). There is an edge labelled bys ∈ {x0, x1} from the coset (d′1, . . . , d

′n) to the coset (d′′1 , . . . , d

′′n) if and only if

s(d′i) = d′′i for every i.Geometrically one can interpret this in the following way. Consider a disjoint

union of n copies of Γ(F, {x0, x1}, StabF (12 )) (a layer for each di). Then the coset(d′1, . . . , d

′n) of F/ StabF (d1, . . . , dn) can be represented by the path joining d′i

vertex on the ith layer with d′i+1 vertex on the (i+ 1)th layer (see Figure 3). Theaction of the generators on the set of such paths is induced by the independentactions of the generators on the layers.

Figure 3. Cosets in F/ StabF (d1, d2, d3)

Now define

Ei =(

12i+n

,1

2i+n−1 , . . . ,1

2i+1

)∈ F/ StabF (d1, . . . , dn)

and

Sm ={Ei

∣∣ 1 ≤ i ≤ m}.

286 D. Savchuk

Since x1(Ei) = Ei and x0(Ei) = Ei+1 we have that the boundary ∂Sm ={E0, Em+1} and

limm→∞

|∂Sm||Sm|

= limm→∞

2m

= 0.

Thus h(Γ(F, {x0, x1}, StabF (d1, . . . , dn))) = 0 and StabF (d1, . . . , dn) is coa-menable in F . �

The amenability of the action of F on the set of dyadic rational numbersand on the set of the ordered tuples of dyadic rational numbers was also notedindependently by N. Monod and Y. Glasner (private communication).

4. The Schreier graph of the action of F on L2([0, 1])

There is a natural unitary representation of Thompson’s group F in the spaceB(L2([0, 1])) of all bounded linear operators on L2([0, 1]). For g ∈ F and f ∈L2([0, 1]) define

(πgf)(x) =

√dg(x)dx

f(g−1x),

where dg(x)dx denotes the Radon-Nikodym derivative of the measure on [0, 1] induced

by g with respect to Lebesgue measure. In this case it coincides with a regularderivative of a real-valued function g. For the detailed explanation we refer thereader to [BdlHV08, Appendix A.6].

For our purposes it is convenient to consider this action with respect to theorthonormal Haar wavelet basis B = {h(0), h(i)j , i ≥ 0, j = 1 . . . 2i} in L2([0, 1]),where h(0)(x) ≡ 1 and

h(0)1 (x) =

{−1, x < 1

2 ,

1, x ≥ 12 ,

h(i)j (x) =

⎧⎪⎨⎪⎩−2 i

2 , j−12i ≤ x < j−1

2i + 12i+1 ,

2i2 , j−1

2i + 12i+1 ≤ x ≤ j

2i ,

0, x /∈ [ j−12i , j2i ].

This basis has first appeared in 1910 in the paper of Haar [Haa10] and playsan important role in the wavelet theory (see, for example, [Dau92, WS01]).

The convenience of using this basis for us comes from the following fact.Each of the generators x0 and x1 acts on each of the basis functions h

(i)j for i ≥ 3

linearly on the support of h(i)j , so that the image also belongs to B. More precisely,


straightforward computations yield

πx0h(i)j = h

(i+1)j , i ≥ 1, 1 ≤ j ≤ 2i−1,

πx0h(i)j = h

(i)j−2i−2 , i ≥ 2, 2i−1 + 1 ≤ j ≤ 2i−1 + 2i−2,

πx0h(i)j = h

(i−1)j−2i−1 , i ≥ 2, 2i−1 + 2i−2 + 1 ≤ j ≤ 2i,

πx1h(i)j = h

(i)j , i ≥ 1, 1 ≤ j ≤ 2i−1,

πx1h(i)j = h

(i+1)j+2i−1 , i ≥ 2, 2i−1 + 1 ≤ j ≤ 2i−1 + 2i−2,

πx1h(i)j = h

(i)j−2i−3 , i ≥ 3, 2i−1 + 2i−2 + 1 ≤ j ≤ 2i−1 + 2i−2 + 2i−3,

πx1h(i)j = h

(i−1)j−2i−1 , i ≥ 3, 2i−1 + 2i−2 + 2i−3 + 1 ≤ j ≤ 2i. (4)

There is a one-to-one correspondence ψ between B \ {h(0)} and the set ofall dyadic rationals from the interval (0, 1) given by ψ(h(i)j ) = j−1

2i + 12i+1 , that

is, each basis function corresponds to the point of its biggest jump (where thefunction changes the sign).

Below we will use the following simple observation, which can also be usedto derive equalities (4). If a function h(x) ∈ L2([0, 1]) changes its sign at the pointx0 then for any g ∈ F the function (πgh)(x) changes its sign at the point g(x0).This enables us to find the image of h(i)j , i ≥ 3 under action of πxk

, k = 0, 1 in thefollowing easy way:

πxkh(i)j = ψ−1

(xk(ψ(h

(i)j ))

)In other words the following diagram is commutative for k = 0, 1

h(i)j

πxk−−−−→ h(i′)j′

ψ

⏐⏐$ ψ

⏐⏐$j−12i + 1

2i+1 −−−−→xk

j′−12i′ + 1

2i′+1

Now we define the Schreier graph of the action of a group on a Hilbert space.Let H be a Hilbert space with an orthonormal basis {hi, i ≥ 1}. Suppose there isa representation π of a group G = 〈S〉 in the space of all bounded linear operatorsB(H). We denote the image of g ∈ G under π as πg.

Definition 5. The Schreier graph Γ of the action of a group G on a Hilbert space Hwith respect to the basis {hi, i ≥ 1} of H and generating set S ⊂ G is an orientedlabelled graph defined as follows. The set of vertices of Γ is the basis {hi, i ≥ 1}and there is an arrow from hi to hj with label s ∈ S if and only if 〈πs(hi), hj〉 �= 0(in other words the coefficient of πs(hi) at hj in the basis {hi, i ≥ 1} is nonzero).

The argument above shows that the Schreier graph of the Thompson’s groupaction on L2([0, 1]) with respect to the Haar basis and generating set {x0, x1}coincides modulo a finite part with the Schreier graph Γ(F, {x0, x1}, D). In order

288 D. Savchuk

to complete the picture we have to find the images under the action of πx0 andπx1 of those h

(i)j which are not listed in (4).

Again straightforward computations give the following equalities.

πx0h(0) =

(14+√22

)h(0) − 1

4h(0)1 +

(−12+√24

)h(1)1 ,

πx0h(0)1 =

14h(0) +

(−14+√22

)h(0)1 +

(12+√24

)h(1)1 ,

πx0h(1)2 =

(12−√24

)h(0) +

(12+√24

)h(0)1 − 1

2h(1)1 ,

πx1h(0) =

(58+√24

)h(0) +

(−38+√24

)h(0)1 −

√28h(1)2 +

(14−√24

)h(2)3 ,

πx1h(0)1 =

(−38+√24

)h(0) +

(58+√24

)h(0)1 −

√28h(1)2 +

(14−√24

)h(2)3 ,

πx1h(1)1 =

√28h(0) +

√28h(0)1 +

(−14+√22

)h(1)2 +

(12+√24

)h(2)3 ,

πx1h(2)4 =

(−14+√24

)h(0) +

(−14+√24

)h(0)1 +

(12+√24

)h(1)2 − 1

2h(2)3 .

These computations together with Proposition 1 prove the following propo-sition.

Proposition 4. The Schreier graph of Thompson’s group action on L2([0, 1]) withrespect to the Haar basis and the generating set {x0, x1} has the following structure(dashed arrows are labelled by x0 and solid arrows by x1)


5. Parts of the Cayley graph of F

Recall, that the positive monoid P of F is the monoid generated by all generatorsxi, i ≥ 0. As a monoid it has a presentation

P ∼= 〈x0, x1, x2, . . . | xkxn = xn+1xk, 0 ≤ k < n〉,which coincides with the infinite presentation (2) of F . The group F itself can bedefined as a group of left fractions of P (i.e., F = P−1 · P ).

It was shown in [GS98] (see also [Gri90]) that the amenability of F is equiva-lent to the right amenability (with respect to our convention (1)) of P . Moreover,let ΓF be the Cayley graph of F with respect to the generating set {x0, x1} andΓP be the induced subgraph of ΓF containing positive monoid P . The followingproposition is of a folklore type.

Proposition 5. Amenability of F is equivalent to amenability of the graph ΓP .

Proof. Any finite set T in F can be shifted to the positive monoid P , i.e., there issome g ∈ F such that Tg ⊂ P . The boundary ∂P (Tg) of this shifted set in ΓP isnot bigger than the boundary of T in ΓF . Hence, Cheeger constant of ΓP is notbigger than the one of ΓF . Thus, non-amenability of ΓP implies non-amenabilityof F .

Suppose that ΓP is amenable. Then for any ε > 0 there exists a subset T ofP , such that its boundary ∂PT in ΓP satisfies

|∂PT ||T | <

ε

4. (5)

Now we can bound the size of the boundary ∂FT of T in ΓF . We use simpleobservations that for finite sets A and B of the same cardinality |A\B| = |B\A| =12 |AΔB| and that |Tx

−1i ΔT | = |(Tx−1i ΔT )xi| = |TxiΔT |.

We have

∂FT = (Tx0 \ T ) ∪ (Tx1 \ T ) ∪ (Tx−10 \ T ) ∪ (Tx−11 \ T ).Therefore,

|∂FT | ≤ |Tx0 \ T |+ |Tx1 \ T |+ |Tx−10 \ T |+ |Tx−11 \ T |

=12(|Tx0ΔT |+ |Tx1ΔT |+ |Tx−10 ΔT |+ |Tx−11 ΔT |)

= |Tx0ΔT |+ |Tx1ΔT | = 2|Tx0 \ T |+ 2|Tx1 \ T | ≤ 4|∂PT | < ε|T |since Txi \ T ⊂ ∂PT for i = 1, 2 and by (5). This shows that ΓF is also amenablein this case. �

In this section we explicitly construct the induced subgraph ΓS of ΓF con-taining the set of vertices

S = {xnu∣∣ n ≥ 0, u is a word over {x0, x1}}. (6)

We also prove that this graph is non-amenable.

290 D. Savchuk

Since S is included in the positive monoid of F and contains elements fromthe infinite generating set {x0, x1, x2, . . .}, it is natural to use the language of forestdiagrams developed in [Bel04, Sun07] (though the existence of this representationwas originally noted by K.Brown [Bro87]).

First we recall the definition and basic facts about this representation of theelements of F .

A binary forest is an ordered sequence of finite rooted binary trees (some ofwhich may be trivial). The forest is called bounded if it contains only finitely manynontrivial trees.

There is a one-to-one correspondence between the elements of the positivemonoid of F and bounded rooted binary forests. More generally, there is a one-to-one correspondence between elements of F and, so-called, reduced forest diagrams,but for our purposes (and for simplicity) it is enough to consider only the elementsof the positive monoid.

� � � � � � � � ��

There is a natural way to enumerate the leaves of the trees in the forest from leftto right. First we enumerate the leaves of the first tree from left to right, then theleaves of the second tree, etc. Also there is a natural left-to-right order on the setof the roots of the trees in the forest.

The product fg of two rooted binary forests f and g is obtained by stackingthe forest g on the top of f in such a way, that the ith leaf of g is attached to theith root of f.

For example, if g and f have the following diagrams

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�

�


then their product fg is the following rooted binary forest

� � � � � � � �

With this operation the set of all rooted binary forests is isomorphic (see[Bel04, Sun07]) to the positive monoid of Thompson’s group F , where xn corre-sponds to the forest in which all the trees except the (n + 1)st one (which hasnumber n) are trivial and the (n+1)st tree represents a single caret. Below is thepicture of the forest corresponding to x3.

� � � � � �

The multiplication rule for the forests implies the following algorithm forconstruction of the rooted forest corresponding to the element xi1xi2xi3 · · ·xin

of the positive monoid of F . Start from the trivial forest (where all the trees aresingletons) and consequently add the carets at the positions i1, i2,. . . , in (countingfrom 0 the roots of the trees in the forest in previous iteration).

For our main result in this section we need two lemmas.

Lemma 1. Let u be a word from the positive monoid of the form u = xnv, wheren ≥ 2 and v is a word over the alphabet {x0, x1} of length at most n − 2. Thenthis word is not equal in F to any other word of the form xmw, where w is a wordover {x0, x1}.Proof. The forest diagram corresponding to u has a caret c connecting the nthand (n+1)st leaves corresponding to xn and possibly some nontrivial trees to theleft of c.

� � ��

��

�

Figure 4. Forest corresponding to xnv

Indeed, after attaching the caret corresponding to xn all the other carets areattached at positions either 0 or 1. Each of these carets decreases the number of

292 D. Savchuk

trees to the left of caret c by 1. Since originally there were n trees to the left fromc and the length of v is at most n− 2, there must be at least 2 trees to the left ofc in the forest representing u.

Suppose there is another word of the form xmw in the positive monoid of Fwhose corresponding rooted forest coincides with the forest of u. Since there areat least 2 trees to the left of caret c one cannot obtain this caret by applying x0or x1. Therefore it was constructed at the first step with application of xm. Thusxm = xn because this caret connects the nth and (n+ 1)st leaves, which, in turn,implies that v = w in F . But both v and w are the elements of a free submonoidgenerated by x0 and x1, yielding that xnv = xmw as words. �

Lemma 2. Let u be a word from the positive monoid of the form u = xnvx1v′,

where n ≥ 2, v is a word over the alphabet X = {x0, x1} of length n− 2 and v′ isa word over the alphabet X = {x0, x1} of arbitrary length. Then this word is notequal in F to any other word of the form xmw, where w is a word over {x0, x1}.Proof. The rooted forest corresponding to xnv is constructed in Lemma 1 andshown in Figure 4. Note, that there are exactly 2 trees (one of which is showntrivial in Figure 4 but generally both can be nontrivial) to the left of caret c. Atthe next step we apply generator x1, which attaches the new caret d that connectsthe root of the second of these trees to the root of caret c. The resulting forest isshown in Figure 5.

� � ��

��

�

�

Figure 5. Forest corresponding to xnvx1

Next, applying v′ adds some extra carets on top of the picture. The finalrooted forest is shown in Figure 6.

Analogously to Lemma 1 we obtain that if the rooted forest of xmw coincideswith the one of u, the caret c could appear only from the initial application of xm

(since it must be placed before caret d is placed). Hence xn = xm and v = w aswords, because the submonoid generated by x0 and x1 is free. �

Let ΓS be the induced subgraph of the Cayley graph ΓF of F that containsall the vertices of from the set S (recall the definition of S in (6)). As a directcorollary of Lemma 1 and Lemma 2, we can describe explicitly the structure of ΓS

(see Figure 7, where solid edges are labelled by x1 and dashed by x0).


� � ��

��

�

�

Figure 6. Forest corresponding to xnvx1v′

�

��

��

��

�

��

��

��

��

��

Figure 7. Induced subgraph ΓS of the Cayley graph of F

294 D. Savchuk

Proposition 6. The structure of ΓS is as follows(a) ΓS contains the infinite binary tree T corresponding to the free submonoid

generated by x0 and x1;(b) for each n ≥ 2 there is a binary tree Tn in ΓS consisting of n−2 levels which

grows from the vertex xn and does not intersect anything else;(c) Each vertex xnv of the boundary of Tn (i.e., v has length n − 2) has two

neighbors xnvx1 and xnvx0 outside Tn. The first one is the root of an infinitebinary tree which does not intersect anything else. The second one coincideswith the vertex vx0x1 of the binary tree T .

Proposition 7. The graph ΓS is non-amenable.

In order to prove this proposition we will use equivalent to the amenabilitydoubling condition (or Gromov doubling condition) [dlAGCS99].

Theorem A (Gromov’s Doubling Condition). Let X be a connected graph ofbounded degree. Then X is non-amenable if and only if there is some k ≥ 1 suchthat for any finite nonempty subset S ⊂ V (X) we have

|Nk(S)| ≥ 2|S|,where Nk(S) is the set of all vertices v of X such that dX(v, S) ≤ k.

Proof of Proposition 7. In order to use Theorem 5 it is enough to construct twoinjective maps f, g : V (X)→ V (X) with distinct images, that do not move verticesfarther than by distance k.

For any vertex xnv in S put

f(xnv) = xnvx1x0,

g(xnv) = xnvx1x1.

For any vertex xnv of S we have d(xnv, f(xnv)) = 2 and d(xnv, g(xnv)) = 2,so the last condition of Theorem 5 is satisfied.

The relation f(xnv) = f(xmw) implies xnvx1x0 = xmwx1x0 and xnv =xmw. Hence f is an injection. The same is true for g.

Now suppose f(xnv) = g(xmw) or, equivalently,

xnvx1x0 = xmwx1x1 . (7)

The words xnvx1 and xmwx1 represent different vertices in ΓS since otherwisewe would get x0 = x1. According to Proposition 6 the equality (7) is possible onlyin case when xnvx1 is a vertex of the boundary of Tn and xmwx1 is a vertex of T .But by Proposition 6(c) in this case the vertex xnvx1x0 coincides with the vertexvx1x0x1 of T which cannot coincide with xmwx1x1. Indeed, otherwise we get

vx1x0 = xmwx1,

which means that xmwx1 is a vertex of T . The only way a vertex of the form ux1belongs to T is if u is in T . Therefore the last equality may not occur because Tis a tree.

Thus by Theorem 5 the graph ΓS is non-amenable. �


References

[BdlHV08] Bachir Bekka, Pierre de la Harpe, and Alain Valette. Kazhdan’s property(T ), volume 11 of New Mathematical Monographs. Cambridge UniversityPress, Cambridge, 2008.

[Bel04] James M. Belk. Thompson’s group F . PhD thesis, Cornell University, 2004.

[Bro87] Kenneth S. Brown. Finiteness properties of groups. In Proceedings of theNorthwestern conference on cohomology of groups (Evanston, Ill., 1985),volume 44, pages 45–75, 1987.

[BS85] Matthew G. Brin and Craig C. Squier. Groups of piecewise linear homeo-morphisms of the real line. Invent. Math., 79(3):485–498, 1985.

[Bur99] Jose Burillo. Quasi-isometrically embedded subgroups of Thompson’s groupF . J. Algebra, 212(1):65–78, 1999.

[CFP96] J.W. Cannon, W.J. Floyd, and W.R. Parry. Introductory notes on RichardThompson’s groups. Enseign. Math. (2), 42(3-4):215–256, 1996.

[Dau92] Ingrid Daubechies. Ten lectures on wavelets, volume 61 of CBMS-NSF Re-gional Conference Series in Applied Mathematics. Society for Industrial andApplied Mathematics (SIAM), Philadelphia, PA, 1992.

[dlAGCS99] P. de lya Arp, R.I. Grigorchuk, and T. Chekerini-Sil′berstaın. Amenabil-ity and paradoxical decompositions for pseudogroups and discrete metricspaces. Tr. Mat. Inst. Steklova, 224 (Algebra. Topol. Differ. Uravn. i ikhPrilozh.):68–111, 1999.

[Don07] John Donnelly. Ruinous subsets of Richard Thompson’s group F . J. PureAppl. Algebra, 208(2):733–737, 2007.

[Gri90] R.I. Grigorchuk. Growth and amenability of a semigroup and its group ofquotients. In Proceedings of the International Symposium on the SemigroupTheory and its Related Fields (Kyoto, 1990), pages 103–108, Matsue, 1990.Shimane Univ.

[Gri98] R.I. Grigorchuk. An example of a finitely presented amenable group thatdoes not belong to the class EG. Mat. Sb., 189(1):79–100, 1998.

[GS87] S.M. Gersten and John R. Stallings, editors. Combinatorial group theory andtopology, volume 111 of Annals of Mathematics Studies. Princeton UniversityPress, Princeton, NJ, 1987. Papers from the conference held in Alta, Utah,July 15–18, 1984.

[GS98] R.I. Grigorchuk and A.M. Stepin. On the amenability of cancellation semi-groups. Vestnik Moskov. Univ. Ser. I Mat. Mekh., (3):12–16, 73, 1998.

[Haa10] Alfred Haar. Zur Theorie der orthogonalen Funktionensysteme.Math. Ann.,69(3):331–371, 1910.

[OS02] Alexander Yu. Ol′shanskii and Mark V. Sapir. Non-amenable finitely pre-sented torsion-by-cyclic groups. Publ. Math. Inst. Hautes Etudes Sci.,(96):43–169 (2003), 2002.

[Sun07] Zoran Sunic. Tamari lattices, forests and Thompson monoids. European J.Combin., 28(4):1216–1238, 2007.

296 D. Savchuk

[WS01] Gilbert G. Walter and Xiaoping Shen. Wavelets and other orthogonal sys-tems. Studies in Advanced Mathematics. Chapman & Hall/CRC, Boca Ra-ton, FL, 2001.

Dmytro SavchukDepartment of MathematicsTexas A&M UniversityCollege Station, TX 77843-3368, USAe-mail: [email protected]



Generating Tuples of Virtually Free Groups

Richard Weidmann

Abstract. We give a complete description of all generating tuples of a virtuallyfree group, i.e., we give a parametrization of Epi(Fn,Γ) where n ∈ N and Γis a virtually free group.

Mathematics Subject Classification (2000). 20F05, 20F67.

Keywords. Nielsen equivalence, virtually free groups.

1. Introduction

For a free group of rank k ≤ n it is the fundamental result of J. Nielsen [N] that saysthat for any two epimorphisms φ1, φ2 : Fn → Fk there exists an automorphismα ∈ AutFn such that φ1 = φ2 ◦ α. This is equivalent to saying that any twogenerating n-tuples of Fk are Nielsen equivalent.

The analogous statement is true for free Abelian groups and some finitegroups. By the proof of Grushko’s theorem the property of having unique Nielsenclasses of generating tuples is further closed under free products. Uniqueness of theNielsen class of minimal generating tuples of surface groups has been establishedby Zieschang [Z].

While Nielsen classes are often not unique there are still a number of situ-ations where the finiteness of the number of Nielsen classes can be established.We restrict our discussion to negatively curved groups. Delzant [Del] proves thattorsion-free hyperbolic groups have only finitely many Nielsen classes of generat-ing pairs. This has been generalized to tuples of arbitrary size provided that thegroup is locally quasiconvex [KW1] or Kleinian [KW2]. It has further recently beenshown that torsion-free (non-hyperbolic) Kleinian groups can have infinitely manyNielsen classes of generating pairs [HW].

It turns out that the torsion-freeness assumption cannot be dropped in theabove statements. Indeed there are 2-generated virtually free groups that haveinfinitely many Nielsen equivalence classes of generating pairs, see Section 6. Themain purpose of this article is to show that there is nevertheless a finite descrip-tion of the set of all generating tuples of a given virtually free group. This gives

298 R. Weidmann

an equivalence relation that is coarser but still related to Nielsen equivalence. Thepicture formally resembles the Makanin-Razborov diagrams describing homomor-phisms from a finitely presented group to a free group, see [R], [KM], [S].

Theorem 1.1. For any finitely generated virtually free group Γ and n ≥ rankΓthere exists a finite directed rooted tree T with root v0 satisfying1. Any vertex v ∈ V T is labelled by a virtually free group Gv and Gv0 = Fn

2. Any valence 1 vertex different from v0 is labeled by Γ3. Any edge e ∈ ET is labeled by an epimorphism πe : Gα(e) → Gω(e) where

α(e) is the initial and ω(e) is the terminal vertex of the edge e.such that for any epimorphism φ : Fn → Γ there exists a reduced directed pathe1, . . . , ek from v0 to some valence 1 vertex ω(ek) such that

φ = αk ◦ πek◦ αk−1 ◦ · · · ◦ α1 ◦ πe1 ◦ α0

where α0 ∈ AutFn and αi ∈ AutGω(ei) for 1 ≤ i ≤ k.

It follows from the proof, see also Section 5, that the description is effective,i.e., that the tree T and its labels can be effectively computed.

The paper is organized as follows. In Section 2 we briefly discuss virtuallyfree groups. In Section 3 we discuss the simple concept of the weakly reducedcore of a graph of groups before giving the proof of Theorem 1.1 in Section 4.We then briefly discuss a solution to the rank problem for virtually free groups inChapter 5. We conclude by showing that virtually free groups can have infinitelymany T -systems which is equivalent to saying that there are examples of virtuallyfree groups for which the tree T in Theorem 1.1 cannot be chosen such that anyvertex is in distance one from the root.

I would like to thank Bernhard Muhlherr for pointing out that SL2(p) pro-vides examples for Lemma 6.3.

2. Virtually free groups

A group is called virtually free if it contains a free subgroup of finite index. Thereare many characterizations of virtually free groups; the following is due to Karrass,Pietrowski and Solitar [KPS]. Their proof relies heavily on Stallings’ theorem onends [St].

Theorem 2.1. A finitely generated group is virtually free iff it is the fundamentalgroup of a finite graph of finite groups.

Note that free groups are residually finite and that a group G with finiteindex subgroup H is residually finite iff H is residually finite. As residually finitegroups are further hopfian by Malcev’s theorem [M] we have the following:

Lemma 2.2. Finitely generated virtually free groups are residually finite and there-fore hopfian.

Generating Tuples of Virtually Free Groups 299

3. The reduced core of a graph of groups

For the remainder of this article all graphs of groups are assumed to have finiteunderlying graph. We will follow the notational conventions from [KMW].

Recall that a graph of groups A is called minimal if the corresponding Bass-Serre tree contains no proper π1(A)-invariant subtree; this is equivalent to sayingthat A contains no valence 1 vertex v such that the boundary monomorphismαe : Ae → Aα(e) = Av is surjective where e is the unique edge with α(e) = v.

We further call a graph of groups with a basepoint (A, v0) minimal if Acontains no valence 1 vertex v �= v0 such that the boundary monomorphism αe :Ae → Aα(e) = Av is surjective where e is the unique edge with α(e) = v.

The pair (A, v0) is further called weakly reduced if it contains no valence 2vertex v �= v0 such that such that both αe : Ae → Aα(e) = Av and αf : Af →Aα(f) = Av are surjective where e and f are the two edges with α(e) = α(f) = v.Note a graph of groups that is not weakly reduced can be obtained from anothergraph of groups with fewer edges by performing edge subdivisions.

We can now assign to any pair (A, v0) its weakly reduced core rcore(A, v0)in the obvious way by first dropping all subdivisions at vertices different from v0to make it weakly reduced and then dropping all inessential valence 1 verticesdifferent from v0. For any pair (A, v0) we denote the obvious isomorphism

π1(rcore(A, v0), v0)→ π1(A, v0)

by θA,v0 .The following result is essentially due to Linnell [L], it bounds the complexity

of the weakly reduced core provided there is a uniform bound on the edge groups.The proof that gives the below bound is due to Dunwoody [D]; see [W] for astatement where the following can be immediately extracted from.

Theorem 3.1 (Linnell). Let (A, v0) be a graph of groups with base point such thatall edge groups are of order at most C and that π1(A) is r-generated.

Then rcore(A, v0) has at most C · r · 3 edges.

4. The proof of the theorem

Throughout this section we assume that Γ is a fixed finitely generated virtuallyfree group and that n ≥ rank Γ. We choose a reduced minimal finite graph offinite groups A and a base vertex v0 ∈ V A such that Γ = π1(A, v0). Recall that anA-graph (B, u0) encodes a morphism from some graph of groups B to A such thatu0 is mapped to v0 and the morphism is injective on vertex groups; see [KMW]for details. Fn is assumed to be the free group in x1, . . . , xn.

Lemma 4.1. There exists a finite set of finite graphs of finite groups with basevertices (A1, w1), . . . , (Ak, wk) such that the following hold:

If (B, u0) is an A-graph such that π1(B, u0) is n-generated then rcore(B, u0)is isomorphic to some (Ai, wi).

300 R. Weidmann

Proof. This follows immediately from Theorem 3.1 and the uniform bound on theorders of all vertex groups of A (and therefore B). �

Note that one pair in this list, say (A1, w1), must be the graph with a singlevertex w1 and n loop edges e1, . . . , en (and all vertex and edge groups trivial).Here we can choose an isomorphism

β0 : Fn → π1(A1, w1)

mapping the basis element xi to the homotopy class represented by ei. Therefurther is a pair, say (Ak, wk), that is isomorphic to (A, v0). We fix an isomorphismbetween (Ak, wk) and (A, v0) with induced isomorphism

βt : π1(Ak, wk)→ π1(A, v0) = Γ.

Note that both β0 and βt are only unique up to an automorphism induced by anautomorphism of (A1, w1) and (Ak, wk), respectively.

The proof of Theorem 1.1 now follows immediately from the following twolemmata. The first lemma shows that we can find a tree T of finite diameter thathas the properties demanded in Theorem 1.1, the second one implies that we canchoose the tree to be locally finite. Together they clearly imply the assertion ofTheorem 1.1.

Lemma 4.2. For any epimorphism φ : Fn → Γ there exists a sequence

(A1, w1) = (Ai1 , wi1 ), . . . , (Aim , wim) = (Ak, wk)

of pairwise distinct pairs (from Lemma 4.1) and epimorphisms

πj : π1(Aij , wij )→ π1(Aij+1 , wij+1 )

for 1 ≤ j ≤ m− 1 such that

φ = βt ◦ αm ◦ πm−1 ◦ · · · ◦ α2 ◦ π1 ◦ α1 ◦ β0where αj ∈ Aut

(π1(Aij , wij )

)for 1 ≤ j ≤ m.

Proof. Let γi = φ(xi) for 1 ≤ i ≤ n, i.e., the epimorphism φ : Fn → Γ is givenby xi �→ γi. We represent the γi by not necessarily reduced non-trivial A-paths pibased at vk. Thus [pi] = γi for 1 ≤ i ≤ n, here [pi] is the homotopy class of pi. Wedefine (B0, u0) to be the S-wedge corresponding to S = (p1, . . . , pn).

Thus B0 is the A-graph that is a wedge of n circles (possibly subdivided intomany edges) with trivial edge and vertex groups such that the loops are labeledby the pi. Note that rcore(B0, u0) is isomorphic to (A1, w1). Clearly π1(B0, u0) isisomorphic to Fn and we can choose an isomorphism χ : Fn → π1(B0, u0) suchthat

φ = ν0 ◦ χwhere ν0 : π1(B, u0) → π1(A, v0) = Γ is the homomorphism induced by the A-graph (B0, u0).

Let now (B0, u0), . . . , (Bl, ul) be a sequence of A-graphs such that (Bi+1, ui+1)is obtained from (Bi, ui) by a Stallings fold and that (Bl, ul) = (A, v0). Such asequence exists as all edge groups of A are finite; this even implies that any folding


sequence terminates and has this property. Note, that all π1(Bi, ui) are finitelygenerated virtually free groups as they are fundamental groups of finite graphs offinite groups.

For i = 1, . . . , l the fold induces an epimorphism

ψi : π1(Bi−1, ui−1)→ π1(Bi, ui)

whose product recovers ν0, i.e., we have

ν0 = ψl ◦ · · · ◦ ψ2 ◦ ψ1.

For each i choose ji such that rcore(Bi, ui) is isomorphic to (Aji , wji). Furtherchoose an isomorphism γi : π1(rcore(Bi, ui), ui)→ π1(Aji , wji ). It follows that foreach i we have an isomorphism

ηi := γi ◦ θ−1Bi,ui: π1(Bi, ui)→ π1(Aji , wji).

Thus we have epimorphisms

ψi := ηi ◦ ψi ◦ η−1i−1 : π1(Aji−1 , wji−1 )→ π1(Aji , wji)

and we getφ = η−1l ◦ ψl ◦ · · · ◦ ψ2 ◦ ψ1 ◦ η1 ◦ χ.

Now as η−1l = βt ◦ α for some α ∈ Aut(π1(Ak, wk)) and η1 ◦ χ = α′ ◦ β0 forsome α′ ∈ Aut(π1(A1, w1)) we have

φ = βt ◦ α ◦ ψl ◦ · · · ◦ ψ2 ◦ ψ1 ◦ α′ ◦ β0.

If now ji = jk for some i < k then

ψk ◦ · · · ◦ ψi+1 : π1(Aij , wij )→ π1(Aik , wik)

is a surjective endomorphism and therefore an automorphism because of the Hopfproperty (Theorem 2.1).

The assertion of the lemma now follows by replacing maximal subsequenceof type ψk ◦ · · · ◦ ψi+1 that connect A-graphs with isomorphic weakly reducedcores by the corresponding automorphisms and relabel the remaining ψi as πj forappropriate i and j. �

Lemma 4.3. For each i and j there exist up to precomposition with someα1 ∈ Aut π1(Ai, wi) and postcomposition with some α2 ∈ Aut π1(Aj , wj) onlyfinitely many epimorphisms π1(Ai, wi)→ π1(Aj , wj) that can occur as some ψk inthe proof of Lemma 4.1.

Proof. Note first that a fold applied to one of the Bi induces a fold on the Aji thatis preceded by at most two subdivisions to make the fold simplicial on the level ofthe Ai, this is discussed in much detail in [D]. The claim of the lemma is now animmediate consequence of the finiteness of the underlying graphs and the uniformbound on the orders of all edge groups and vertex groups involved. �

302 R. Weidmann

5. The rank problem

Using the proof of Theorem 1.1 it is fairly simple to extract a solution to the rankproblem for virtually free groups, i.e., to describe an algorithm that computesthe minimal number of generators of a given virtually free group Γ. The discussionbelow in fact implies that the tree described in Theorem 1.1 can be algorithmicallyconstructed.

Indeed first write Γ as the fundamental group of a finite graph of finitegroups, such a decomposition can be found by enumerating presentations of Γby enumerating Tietz transformations until there is a presentation from which onecan read off a presentation of Γ as a fundamental group of a finite graph of finitegroups A. It is then easy to transform A into a minimal and reduced graph ofgroups.

Note that there are only finitely many finite graphs of minimal weakly reducedgraphs of groups of rank k provided there is a uniform bound on the orders of vertexgroups. These can all be constructed by enumerating folding sequences startingwith the wedge of k circles as in the proof of Theorem 1.1. One can then check foreach k whether A is in the list.

6. Nielsen equivalence and T-systems

In this section we show that the tree T of Theorem 1.1 cannot always be chosensuch that any vertex is in distance 1 from the base vertex v0.

Recall that two generating n-tuples of a group Γ are called Nielsen-equivalentif they are the images of two bases of Fn under a fixed epimorphism φ. As any twobases of Fn are related by an automorphism of Fn it follows that Γ has only finitelymany Nielsen-equivalence classes of generating n-tuples if there exist finitely manyepimorphisms φ1, . . . , φm : Fn → Γ such that for any epimorphism φ : Fn → Γthere exists some α ∈ AutFn and some i such that φ = φi ◦ α.

A coarser equivalence relation is that of Tn-systems. Two generating n-tuplesT1 and T2 of Γ are said to lie in the same Tn-system if there exists some β ∈ Aut Γsuch that T1 and β(T2) are Nielsen-equivalent. Thus a group Γ has only finitelymany Tn systems if there exist finitely many epimorphisms φ1, . . . , φm : Fn → Γsuch that for any epimorphism φ : Fn → Γ there exists some α ∈ AutFn and someβ ∈ AutΓ and some i such that φ = β ◦ φi ◦ α.

Thus the tree T in Theorem 1.1 can be chosen such that any vertex is indistance 1 from v0 iff Γ has only finitely many Tn-systems. We will now show thatfor virtually free groups there can be infinitely many T2-systems. As all inner au-tomorphisms lift to the free group it follows that a group that has infinitely manyNielsen equivalence classes of generating n-tuples and has finite outer automor-phism group must have infinitely many T2-systems. Thus the following theoremproves the claim.

Theorem 6.1. There exists a virtually free group G with |Out(G)| <∞ with a se-quence (gi, hi), i ∈ N, of generating pairs that are pairwise non-Nielsen equivalent.


We first record two lemmata. The first one is due to Nielsen [N0] and is areformulation of the well-known fact that any automorphism of the free groupF (a, b) maps [a, b] to a conjugate of [a, b] or [a, b]−1. It gives us a simple way todistinguish Nielsen equivalence classes of generating pairs.

Lemma 6.2. Let G be a group and g1, g2, h1, h2 ∈ G such that (g1, h1) ∼ (g2, h2).Then [g1, h1] is conjugate to [g2, h2] or [g2, h2]−1. �

The next lemma postulates the existence of a finite group with specific prop-erties.

Lemma 6.3. There exists a finite group H generated by conjugate elements x andy and such that there exists an element z ∈ H − 〈x〉 such that [x, z] = 1.

Proof. An example is SL2(p) with odd p. SL2(p) is generated by the two matrices

x =(1 10 1

)and y =

(1 01 1

)which are conjugate by

(0 11 0

). The group

has further a non-trivial center which has trivial intersection with 〈x〉. This provesthe claim. �

Let now H be a group as in Lemma 6.3 and p be the order of x (and y).Choose w ∈ H such that wxw−1 = y. Choose further q such that p and q arecoprime and let H = H⊕Zq, write Zq as additive group generated by 1. Note thatthe subgroup 〈(x, 0), (idH , 1)〉 is generated by (x, 1) and is cyclic of order pq. LetK = 〈k | k2pq〉 be the cyclic group of order 2pq. We construct a graph of groups Aas follows:1. The underlying graph A has 2 vertices v1 and v2 with vertex group Av1 = K

and v2 = H.2. There exists an edge e with edge group Ae = 〈a | apq〉, α(e) = v1, ω(e) = v2,

αe(a) = k2 and ωe(a) = (x, 1).3. There exists a loop edge f with α(e) = ω(e) = v2, edge group Af = H and

αe(h) = ωe(h) = (h, 0) for all h ∈ H = Af .Let now G be the fundamental group of A. As all vertex groups of A are

finite it follows that G is virtually free. Clearly G has the presentation

〈x, y, u, k, t |uq, [t, x], [t, y], [u, x], [u, y], k2 = xu,R〉where R is a set of defining relators for H with respect to the generators x andy, t is the stable letter of the HNN-extension corresponding to the edge f and ucorresponds to (idH , 1). Note that both u = k2p and x = k2q; thus G is generatedby k, y and t.

We now put gi = k and hi = (kz)itw−1 for all i ≥ 0 where z is the elementpostulated in Lemma 6.3. Together with the remark preceding Theorem 6.1 theproof of Theorem 6.1 is clearly an immediate consequence of the following threesimple lemmata.

Lemma 6.4. G = 〈gi, hi〉 for any i ≥ 0.

304 R. Weidmann

Proof. Note first that

h−1i g2qi hi = wt−1(kz)−ik2q(kz)itw−1 = wt−1(kz)−ix(kz)itw−1 = wxw−1 = y

as t, z and k commute with k2q = x. Thus k, y ∈ 〈gi, hi〉. As further hi = (kz)itw−1

and w ∈ 〈x, y〉 ⊂ 〈k, y〉 this implies that t ∈ 〈gi, hi〉. It follows that y, k, t ∈ 〈gi, hi〉which proves the assertion of the lemma. �

Lemma 6.5. (gi, hi) and (gi, hi) are Nielsen equivalent iff i = j.

Proof. Let i �= j. In the light of Lemma 6.2 we only need to show that [gi, hi]is neither conjugate to [gj , hj ] nor to [gj , hj ]−1. To do so it suffices to show that[gi, hi] and [gj , hj] act with different translation length on the Bass-Serre tree of Aor equivalently that the cyclically reduced reduced forms are of different lengths.This is obvious as [gi, hi] = k · (kz)itw−1 · k−1 ·wt−1(z−1k−1)i can be representedby the cyclically reduced A-path

k(k,e,z,e−1)i−1,k,e,z,f,w−1,e−1,k−1,e,w,f−1,(z−1,e−1,k−1,e)i−1,z−1,e−1,k−1

which is of length 4i+ 4. �

Lemma 6.6. The group G has finite outer automorphism group.

Proof. By construction G is the fundamental group of a graph of finite groups withtwo edge groups Ae and Af such that neither is conjugate into the other. Thisguarantees the finiteness of the outer automorphism group, see [P]. �

References

[D] M.J. Dunwoody Folding sequences Geometry & Topology Monographs. Volume1: The Epstein Birthday Schrift, 139–158.

[Del] T. Delzant, Sous-groupes a deux generateurs des groupes hyperboliques. E. Ghys,A. Haefliger and A. erjovski (ed.) et al., Group theory from a geometrical view-point. Singapore: World Scientific., 1991, 177–189.

[HW] M. Heusener and R. Weidmann, Generating Pairs of 2-Bridge knot groups, toappear in Geom. Ded.

[KW1] I. Kapovich and R. Weidmann, Freely indecomposable groups acting on hyperbolicspaces, IJAC 14 (2004), 115–171.

[KW2] I. Kapovich and R. Weidmann, Kleinian groups and the rank problem, Geometryand Topology 9 (2005), 375–402.

[KMW] I. Kapovich, A. Myasnikov and R. Weidmann. A-graphs, foldings and the inducedsplittings IJAC 15 no.1, 2005, 95–128.

[KM] O. Kharlampovich and M. Myasnikov, Irreducible Affine Varieties over a freegroup, J. Algebra 200, 1998, 517–570.

[KPS] A. Karrass, A. Pietrowski and D. Solitar, Finite and infinite cyclic extensions offree groups. Collection of articles dedicated to the memory of Hanna Neumann,IV. J. Austral. Math. Soc. 16, 1973, 458–466.

[L] P.A. Linnell, On accessibility of groups J. Pure Appl. Algebra 30, 1983, 39–46.


[M] A.I. Malcev, On isomorphic representations of infinite groups by matrices Mat.Sb. 8, 1940, 405–422.

[N0] J. Nielsen, Die Isomorphismen der allgemeinen, unendlichen Gruppe mit zweiErzeugenden, Math. Ann. 78, 1917, 385–397.

[N] J. Nielsen, Om Regning med ikke kommutative Faktorer og dens Anvendlese iGruppenteorien, Mat. Tidskrift B, 1921, 77–94.

[P] M.R. Pettet, Virtually free groups with finitely many outer automorphisms,Trans. AMS 349, no. 10, 1997, 4565–4587.

[R] A.A. Razborov, On systems of Equations in a Free Group, PhD thesis, SteklovMath. Inst., 1987.

[Sc] G.P. Scott The geometries of 3-manifolds Bull. Lond. Math. Soc. 15, 1983, 401–487.

[S] Z. Sela, Diophantine geometry over groups I. Makanin Razborov Diagrams. Publ.Math. IHES 93, 2001, 31–105.

[St] J. Stallings, Groups of cohomological dimension one. Applications of CategoricalAlgebra (Proc. Sympos. Pure Math., Vol. XVIII, New York, 1968), 1970, 124–128.

[W] R. Weidmann, On accessibility of finitely generated groups, to appear in Q. J.Math.

[Z] H. Zieschang, Uber die Nielsensche Kurzungsmethode in freien Produkten mitAmalgam, Inv. Math. 10, 4–37, 1970.

Richard WeidmannDepartment of MathematicsHeriot Watt-UniversityRiccartonEdinburgh, EH14 4AS, UKe-mail: [email protected]



Limits of Thompson’s Group F

Roland Zarzycki

Abstract. Let F be the Thompson’s group 〈x0, x1|[x0x−11 , x−i

0 x1xi0], i = 1, 2〉.

Let Gn = 〈y1, . . . , ym, x0, x1|[x0x−11 , x−i

0 x1xi0], y−1

j gj,n(x0, x1), i = 1, 2, j ≤m〉, where gj,n(x0, x1) ∈ F , n ∈ N, be a family of groups isomorphic to Fand marked by m+2 elements. If the sequence (Gn)n<ω is convergent in thespace of marked groups and G is the corresponding limit we say that G is anF -limit group. The paper is devoted to a description of F -limit groups.

Mathematics Subject Classification (2000). 20E06, 20E18, 20F69.

Keywords. Thompson’s Group F , limit groups, HNN-extensions, free prod-ucts, group laws.

1. Preliminaries

The notion of limit group was introduced by Z. Sela in his work on characterizationof elementary equivalence of free groups [12]. This approach has been extended inthe paper of C. Champetier and V. Guirardel [7], where the authors look at limitgroups as limits of convergent sequences in a space of marked groups. They havegiven a description of Sela’s limit groups in these terms (with respect to the classof free groups). This approach has been also applied by L. Guyot and Y. Stalder[10] to the class of Baumslag-Solitar groups.

Thompson’s group F has remained one of the most interesting objects ingeometric group theory. We study F -limit groups. We show in this paper, thatamong F -limit groups there are no free products of F with any non-trivial group.Moreover, we prove that among F -limit groups there are no HNN-extensions overcyclic subgroups.

In the remaining part of the section we recollect some useful definitions andfacts concerning limit groups and Thompson’s group F . In Section 2 we present re-sults concerning free products and in Section 3 results concerning HNN-extensions.

A marked group (G,S) is a group G with a distinguished set of generatorsS = (s1, s2, . . . , sn). For fixed n, let Gn be the set of all n-generated groups markedby n generators (up to isomorphism of marked groups). Following [7] we put certain

308 R. Zarzycki

metric on Gn. We will say that two marked groups (G,S), (G′, S′) ∈ Gn are atdistance less or equal to e−R if they have exactly the same relations of length atmost R. The set Gn equipped with this metric is a compact space [7]. Limit groupsare simply limits of convergent sequences in this metric space.

Definition 1.1. Let G be an n-generated group. A marked group in Gn is a G-limitgroup if it is a limit of marked groups each isomorphic to G.

To introduce the Thompson’s group F we will follow [5].

Definition 1.2. Thompson’s group F is the group given by the following infinitegroup presentation:

〈x0, x1, x2, . . . |xjxi = xixj+1(i < j)〉In fact F is finitely presented:

F = 〈x0, x1|[x0x−11 , x−i0 x1x

i0], i = 1, 2〉.

Every non-trivial element of F can be uniquely expressed in the normal form:

xb00 x

b11 x

b22 . . . xbn

n x−ann . . . x−a2

2 x−a11 x−a0

0 ,

where n, a0, . . . , an, b0, . . . , bn are non-negative integers such that:i) exactly one of an and bn is nonzero;ii) if ak > 0 and bk > 0 for some integer k with 0 ≤ k < n, then ak+1 > 0 orbk+1 > 0.

We study properties of F -limit groups. For this purpose let us considera sequence, (gi,n)n<ω , 1 ≤ i ≤ t, of elements taken from the group F andthe corresponding sequence of limit groups marked by t + 2 elements, Gn =(F, (x0, x1, g1,n, . . . , gt,n)), n ∈ N, where x0 and x1 are the standard generatorsof F . Assuming that such a sequence is convergent in the space of groups markedby t+ 2 elements, denote by G = (〈x0, x1, g1, . . . , gt|RF ∪RG〉, (x0, x1, g1, . . . , gt))the limit group formed in that manner; here x0, x1 are “limits” of constant se-quences (x0)n<ω and (x0)n<ω, gi is the “limit” of (gi,n)n<ω for 1 ≤ i ≤ t, RF andRG refer respectively to the set of standard relations taken from F and the set(possibly infinite) of new relations.

It has been shown in [7] that in the case of free groups some standard con-structions can be obtained as limits of free groups. For example, it is possible toget Zk as a limit of Z and Fk as a limit of F2. On the other hand, the directproduct of F2 and Z can not be obtained as a limit group. HNN-extensions oftenoccur in the class of limit groups (with respect to free groups). For example, thefollowing groups are the limits of convergent sequences in the space of free groupsmarked by three elements: the free group of rank 3, the free abelian group of rank3 or an HNN-extension over a cyclic subgroup of the free group of rank 2 ([6]). Allnon-exceptional surface groups form another broad class of interesting examples([2], [3]).

In the case of Thompson’s group the situation is not so clear. Since thecentrum of F is trivial it is surely not possible to obtain any direct product with

Limits of Thompson’s Group F 309

the whole group as an F -limit group. In 1985 Brin and Squier [4] showed thatThompson’s group F does not satisfy any law (also see Abert’s paper [1] for ashorter proof). However, in this paper we show that there are certain non-trivialwords with constants over F (which will be called later laws with constants), whichare equal to the identity for each evaluation in F . This implies that no free productof F with any non-trivial group is admissible as limit group with respect to F (seeSection 2). Moreover, we prove that HNN-extensions over a cyclic subgroup arenot admissible as limit groups with respect to F (see Section 3).

There are many geometric interpretations of F , but here we will use thefollowing one. Consider the set of all strictly increasing continuous piecewise-linearfunctions from the closed unit interval onto itself. Then the group F is realizedby the set of all such functions, which are differentiable except at finitely manydyadic rational numbers and such that all slopes (derivatives) are integer powersof 2. The corresponding group operation is just the composition. For the furtherreference it will be useful to give an explicit form of the generators x0, x1, . . . interms of piecewise-linear functions:

xn(t) =

⎧⎪⎪⎪⎪⎪⎪⎨⎪⎪⎪⎪⎪⎪⎩

t , t ∈ [0, 2n−12n ]

12 t+

2n−12n+1 , t ∈ [ 2n−1

2n , 2n+1−12n+1 ]

t− 12n+2 , t ∈ [ 2n+1−1

2n+1 , 2n+2−12n+2 ]

2t− 1 , t ∈ [ 2n+2−12n+2 , 1]

for n = 0, 1, . . ..For any diadic subinterval [a, b] ⊂ [0, 1], let us consider the set, of elements

in F , which are trivial on its complement, and denote it by F[a,b]. We know thatit forms a subgroup of F , which is isomorphic to the whole group. Let us denoteits standard infinite set of generators by x[a,b],0, x[a,b],1, x[a,b],2, . . ..

Let us consider an arbitrary element g in F and treat it as a piecewise-linear homeomorphism of the interval [0, 1]. Let supp(g) be the set {x ∈ [0, 1] :g(x) �= x} and supp(g) the topological closure of supp(g). We will call each pointfrom the set Pg = (supp(g) \ supp(g)) ∩ Z[12 ] a dividing point of g. This set isobviously finite and thus we get a finite subdivision of [0, 1] of the form [0 =p0, p1], [p1, p2], . . . , [pn−1, pn = 1] for some natural n. It is easy to see that g can bepresented as g = g1g2 . . . gn, where gi ∈ F[pi−1,pi] for each i. Since g can act triviallyon some of these subintervals, some of the elements g1, . . . , gn may be trivial. Wecall the set of all non-trivial elements from {g1, . . . , gn} the defragmentation of g.

Fact 1.3 (Corollary 15.36 in [8], Proposition 3.2 in [11]). The centralizer of anyelement g ∈ F is the direct product of finitely many cyclic groups and finitely manygroups isomorphic to F .

Moreover if the element g ∈ F has the defragmentation g = g1 . . . gn, thensome roots of the elements g1, . . . , gn are the generators of cyclic components ofthe decomposition of the centralizer above. The components of this decomposition

310 R. Zarzycki

which are isomorphic to F are just the groups of the form F[a,b], where [a, b]is one of the subintervals [pi−1, pi] ⊂ [0, 1], which are stabilized pointwise by g.Generally, if we interpret the elements of F as functions, the relations occurringin the presentation of F , [x0x−11 , x−i

0 x1xi0] for i = 1, 2, have to assure, that two

functions, which have mutually disjoint supports except of finitely many points,commute. In particular, these relations imply analogous relations for different i >2. According to the fact that x−i

0 x1xi0 = xi+1, we conclude that all the relations

of the form [x0x−11 , xM ], M > 1, hold in Thompson’s group F . We often refer tothese geometrical observations.

I am grateful to the referee for his helpful remarks.

2. Free products

Brin and Squier have shown in [4] that the Thompson’s group F does not satisfyany group law. In this section we show how to construct words with constantsfrom F , which are equal to the identity for any substitution in F .

Definition 2.1. Let w(y1, . . . , yt) be a non-trivial word over F , reduced in the groupFt ∗F and containing at least one variable. We will call w a law with constants inF if for any g = (g1, . . . , gt) ∈ F t, the value w(g) is equal to 1F .

The following proposition gives a construction of certain laws with constants in F .

Proposition 2.2. Consider the standard action of Thompson’s group F on [0, 1].Suppose we are given four pairwise disjoint closed diadic subintervals Ii = [pi, qi] ⊂[0, 1], 1 ≤ i ≤ 4, and assume that p1 < p2 < p3 < p4. Then for any non-trivialh1 ∈ FI1 , h2 ∈ FI2 , h3 ∈ FI3 and h4 ∈ FI4 , the word w obtained from

[y−1h−11 yh−14 y−1h1yh4, y−1h−12 yh−13 y−1h2yh3]

by reduction in Z ∗ F (we treat the variable y as a generator of Z) is a law withconstants in F .

Proof. We will use the following notation: w14 = y−1h−11 yh−14 y−1h1yh4 and w23 =y−1h−12 yh−13 y−1h2yh3. It is easy to see that w cannot be reduced to a constant.We claim that

for any any g ∈ F satisfying g(q1) < p4 and g(p4) > q1 the word w14(g)is equal to the identity.

To show this we consider the action of w14(g) on each point from [0, 1]. Assume,that t ∈ [0, g−1(q1)). Since t /∈ supp(h4) we have:

w14(g)(t) = g−1h−11 gh−14 g−1h1g(h4(t)) = g−1h−11 gh−14 g−1(h1(g(t))).

By g−1(h1(g(t))) < g−1(h1(q1)) = g−1(q1) < p4 we see h−14 (g−1(h1(g(t)))) =g−1(h1(g(t))). Thus:

w14(g)(t) = g−1h−11 gg−1(h1(g(t))) = t.


If t ∈ [g−1(q1), 1] then since h±14 (t) ≥ min(t, p4), we have g(h±14 (t)) ≥ q1 andhence:

w14(g)(t) = g−1h−11 gh−14 g−1h1g(h4(t))

= g−1h−11 gh−14 g−1g(h4(t)) = g−1h−11 g(t) = t.

It follows that for g such that g(q1) < p4 and g(p4) > q1, w14(g) = 1F and hencew(g) = [w14(g), w23(g)] = [1F , w23(g)] = 1F . Thus we are left with the case wheng(q1) ≥ p4 (the proof of the case g(p4) ≤ q1 uses the same argument). Now we willprove that

for any g ∈ F satisfying g(q1) ≥ p4 the word w23(g) is equal to theidentity.

Assume that t ∈ [0, g−1(p2)]. Since g(q1) ≥ p4, we have q1 ≥ g−1(p4) > g−1(p2) ≥t. Thus:

w23(g)(t) = g−1h−12 gh−13 g−1h2gh3(t)

= g−1h−12 gh−13 g−1h2g(t) = g−1h−12 gh−13 g−1g(t) = t.

Now assume that t ∈ (g−1(p2), g−1(q2)). Then since again q1 ≥ g−1(p4) >g−1(q2) ≥ t we obtain:

w23(g)(t) = g−1h−12 gh−13 g−1h2gh3(t) = g−1h−12 gh−13 g−1h2g(t).

Since h2(g(t)) ∈ (p2, q2) we have g−1(h2(g(t))) < g−1(q2) < q1 and:

w23(g)(t) = g−1h−12 gh−13 (g−1h2g(t)) = g−1h−12 g(g−1h2g(t)) = t.

Assume that t ∈ [g−1(q2), g−1(p3)]. Since we still have g−1(p3) < q1, we see that:

w23(g)(t) = g−1h−12 gh−13 g−1h2gh3(t)

= g−1h−12 gh−13 g−1h2g(t) = g−1h−12 gh−13 g−1g(t) = t.

Let t ∈ (g−1(p3), g−1(q3)). Then since g(p3) > q2 and h3(t) �= t ⇒ h3(t) > p3:

w23(g)(t) = g−1h−12 gh−13 g−1h2gh3(t)

= g−1h−12 gh−13 g−1gh3(t) = g−1h−12 g(t) = t.

Finally assume t ∈ [g−1(q3), 1] (and then g(t) > q2). Similarly as above:

w23(g)(t) = g−1h−12 gh−13 g−1h2gh3(t)

= g−1h−12 gh−13 g−1g(h3(t)) = g−1h−12 g(t) = t.

Now we see that for g such that g(q1) ≥ p4, we have w23(g) = 1F and hencew(g) = [w14(g), w23(g)] = [w14(g), 1F ] = 1F . The proof is finished. �

We now apply the construction from Proposition 2.2 to limits of Thompson’sgroup F .

Theorem 2.3. Suppose we are given a convergent sequence of marked groups((Gn, (x0, x1, gn,1, . . . , gn,s)))n<ω, where Gn = F , (gn,1, . . . , gn,s) ∈ F , n ∈ N,and denote by G its limit. Then G �= F ∗G for any non-trivial G.

312 R. Zarzycki

Before the proof we formulate a general statement, which exposes the mainpoint of our argument.

Proposition 2.4. Let H = 〈h1, . . . , hm〉 be a finitely generated torsion-free group,which satisfies a one variable law with constants and does not satisfy any lawwithout constants. Let G be the limit of a convergent sequence of marked groups((Gn, (h1, . . . , hm, gn,1, . . . , gn,t)))n<ω, where (gn,1, . . . , gn,t) ∈ H, Gn = H, n ∈ N.Then G �= H ∗K for any non-trivial K < G.

Proof. It is clear that G is torsion-free. To obtain a contradiction suppose thatG = H ∗ K, K �= {1}, and G is marked by a tuple (h1, . . . , hm, f1, . . . , ft). Letf = u(h, f) be an element of K \ {1} and let w(y) be a law with constants in H .Obviously w(u(h, gn,1, . . . , gn,t)) = 1H for all n < ω. It follows from the definitionof an H-limit group that w(u(h, f)) = 1G. Since w was chosen to be non-trivial,with constants from H and |f | =∞, we obtain a contradiction with the fact thatG is the free product of H and K. �Proof of Theorem 2.3. It follows directly from Proposition 2.2, that there is someword w(y), which is a law with constants in F , and hence we just apply Proposition2.4 for H = F , h1 = x0 and h2 = x1. �

3. HNN-extensions

Now we proceed to discuss the case of HNN-extensions. For this purpose we con-sider a sequence of groups marked by three elements, (Gn)n<ω, and the corre-sponding limit group G = (〈x0, x1, g|RF ∪RG〉, (x0, x1, g)). The following theoremis the main result of the section.

Theorem 3.1. Let (Gn)n<ω be a convergent sequence of groups, where Gn =(F, (x0, x1, gn)), and let G = (〈x0, x1, g|RF ∪ RG〉, (x0, x1, g)) be its limit. ThenG is not an HNN-extension of Thompson’s group F of the following form

〈x0, x1, g|RF , ghg−1 = h′〉 for some h, h′ ∈ F.

In what follows we will need two easy technical lemmas:

Lemma 3.2. Suppose g ∈ F and let xa00 x

a11 . . . xan

n x−bnn . . . x−b1

1 x−b00 be its normal

form. There is M ∈ N such that for all m > M :

g−1xmg = xm+t or gxmg−1 = xm+t,

where t =|∑ni=0(ai − bi) |.

Proof. Consider the case when∑n

i=0(ai − bi) ≥ 0. Then for sufficiently large m:

xb00 x

b11 . . . xbn

n x−ann . . . x−a1

1 x−a00 xmx

a00 x

a11 . . . xan

n x−bnn . . . x−b1

1 x−b00

= xb00 x

b11 . . . xbn

n xm+∑n

i=0 aix−bnn . . . x−b1

1 x−b00 = xm+

∑ni=0(ai−bi).

In the case when∑n

i=0(ai − bi) < 0 we consider the symmetric conjugation andapply the same argument. �


Lemma 3.3. Under the assumptions of Lemma 3.2 the numbers M and t definedin that lemma, additionally satisfy the property that for all m > M and k > 0:

g−kxmgk = xm+kt or gkxmg

−k = xm+kt.

Proof. If g−1xmg = xm+t holds (one of possible conclusions of Lemma 3.2) thenM ≤ m + t and applying Lemma 3.2 k times we obtain the result. The casegxmg

−1 = gm+t is similar. �Proof of Theorem 3.1. First we prove the theorem in the case of centralized HNN-extensions.

Suppose that h = h′ �= 1 in the formulation, i.e., the limit group has a relationof the form ghg−1 = h and denote by H the corresponding HNN-extension ofThompson’s group 〈x0, x1, g|RF , ghg

−1 = h〉. Assume that ghg−1 = h is satisfiedin G. From the definition of a limit group it follows, that gnhg−1n =F h for almostall n. Denote by C(h) the centralizer of h and by C1⊕ · · · ⊕Cm its decompositiontaken from Fact 1.3. As almost all gn commute with h, almost all gn have adecomposition of the form gn = gn,1 . . . gn,m, where gn,i ∈ Ci.

As h �= 1, at least one of the factors C1⊕· · ·⊕Cm is isomorphic to Z, say Ci0 .Denote by [a, b] the support of elements taken from the subgroup Ci0 . It followsfrom the construction of this decomposition, that h can only fix finitely manypoints in [a, b].

Let us consider the sequence (gn,i0)n<ω . Without loss of generality we maysuppose that (gn,i0)ω<∞ consists of powers of some element of F (which is agenerator of Ci0). Consider the case when it has infinitely many occurrencesof the same element. If g′ occurs infinitely many times in this sequence, theninfinitely many gn(g′)−1 commute with x[a,b],0, x[a,b],1. That gives us a subse-quence (gkn,i0)n<ω for which the relation [gkn(g′)−1, f ] holds for all n and for allf ∈ 〈x[a,b],0, x[a,b],1〉. As 〈x[a,b],0, x[a,b],1〉 is isomorphic to Thompson’s group F , wewill find a word of the form g′y−1f−1y(g′)−1f with f ∈ 〈x[a,b],0, x[a,b],1〉, whichis trivial for y = limn→∞ gkn in the limit group corresponding to the sequence(gkn,i0)n<ω and non-trivial for y = g in the group H . Indeed, it follows fromBritton’s Lemma on irreducible words in an HNN-extension ([9], page 181), thatthe considered word can be reduced in H only if f−1 lies in the cyclic subgroupgenerated by h. But f ∈ 〈x[a,b],0, x[a,b],1〉 can be easily chosen outside 〈h〉.

Let us now assume that the sequence (gn,i0)n<ω is not stabilizing. By thediscussion from the end of Section 1 we see that

for all m > 1, [x[a,b],0x−1[a,b],1, x[a,b],m] = 1. (†)

On the other hand any gn,i0 is a power of some fixed element from Ci0 . Thus wesee by Lemma 3.3 that for M found for the generator of Ci0 as in Lemma 3.2:

(∀n) g−1n,i0x[a,b],Mgn,i0 = x[a,b],M+tn

or(∀n) gn,i0x[a,b],Mg−1n,i0

= x[a,b],M+tn, tn ≥ 0.

314 R. Zarzycki

Substituting x[a,b],M+tninto (†) instead of x[a,b],m we have:

[x[a,b],0x−1[a,b],1, g

−1n,i0

x[a,b],Mgn,i0 ] = 1or

[x[a,b],0x−1[a,b],1, gn,i0x[a,b],Mg−1n,i0

] = 1.

Thus we see that one of the following relations holds for all n:

[x[a,b],0x−1[a,b],1, g−1n x[a,b],Mgn] = 1

or[x[a,b],0x−1[a,b],1, gnx[a,b],Mg−1n ] = 1.

Suppose that the first relation holds for all n’s. Consider the corresponding wordin the group H :

w = x[a,b],1x−1[a,b],0g

−1x−1[a,b],Mgx[a,b],0x−1[a,b],1g

−1x[a,b],Mg.

We claim that w �= 1 in this HNN-extension. Once again, it follows from Brit-ton’s Lemma on irreducible words in an HNN-extension, that we can reducew if x[a,b],M is a power of h or x[a,b],0x

−1[a,b],1 is a power of h. We know that

x[a,b],0, x[a,b],1, . . . , x[a,b],m, . . . generate the group F[a,b], which is isomorphic toF . From the properties of F we know that for different m,m′ > M , x[a,b],m andx[a,b],m′ do not have a common root. Thus, possibly increasing the number M , wecan assume that x[a,b],M is not a power of h. If x[a,b],0x−1[a,b],1 = hd for some integerd, then hd fixes pointwise the segment [ 14a+

34b, b] ⊂ [a, b]. Hence h also fixes some

final subinterval of [a,b]. This gives a contradiction as h was chosen to fix onlyfinitely many points in [a, b]. This finishes the case of centralized HNN-extensions.

Generally, let us consider the situation, where in the limit group we have onerelation of the form ghg−1 = h′ for some h, h′ ∈ F . By the construction of limitgroups h′ = hf for some element f ∈ F . Indeed, if h and h′ are not conjugatedin F , then there is no sequence (gn)n<ω in F with gnhg

−1n = h′ for almost all n.

We now apply the argument above: let (fgn)n<ω be a sequence convergent to theelement fg. It obviously commutes with h, so we can repeat step by step the proofabove. That completes the proof. �

References

[1] M. Abert, Group laws and free subgroups in topological groups, Bull. London Math.Soc. 37 (2005), 525–534.

[2] B. Baumslag, Residually free groups, Proc. London Math. Soc. (3), 17:402–418, 1967.

[3] G. Baumslag, On generalised free products, Math. Z., 78:423–438, 1962.

[4] M.G. Brin, C.C. Squier, Groups of piecewise linear homeomorphisms of the real line,Invent. Math. 79 (1985), 485–498.

[5] J.W. Cannon, W.J. Floyd, W.R. Parry, Introductory notes on Richard Thompson’sgroups, Enseign. Math. (2) 42 (1996), 215–256.


[6] B. Fine, A.M. Gaglione, A. Myasnikov, G. Rosenberger, D. Spellman, A classificationof fully residually free groups of rank three or less, J. Algebra, 200 (2):571–605, 1998.

[7] C. Champetier, V. Guirardel, Limit groups as limits of free groups: compactifyingthe set of free groups, Israel J. Math. 146 (2005), 1–76.

[8] V. Guba, M. Sapir, Diagram groups, Memoirs of the American Mathematical Society,Volume 130, Number 620, November 1997.

[9] R. Lyndon, P. Schupp, Combinatorial group theory, Springer, Berlin 1977.

[10] L. Guyot, Y. Stalder, Limits of Baumslag-Solitar groups and other families of markedgroups with parameters, eprint arXiv:math/0507236.

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Roland ZarzyckiInstitute of PhilosophyUniversity of Wroclawul. Koszarowa 351-149 Wroclaw, Polande-mail: [email protected]

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