Springer Monographs in Mathematics
For further volumes: www.springer.com/series/3733
Tullio CeccheriniSilberstein r Michel Coornaert
Cellular Automata and Groups
Tullio CeccheriniSilberstein Dipartimento di Ingegneria Università del Sannio C.so Garibaldi 107 82100 Benevento Italy
[email protected] Michel Coornaert Institut de Recherche Mathématique Avancée Université de Strasbourg 7 rue RenéDescartes 67084 Strasbourg Cedex France
[email protected] ISSN 14397382 ISBN 9783642140334 eISBN 9783642140341 DOI 10.1007/9783642140341 Springer Heidelberg Dordrecht London New York Library of Congress Control Number: 2010934641 Mathematics Subject Classification (2010): 37B15, 68Q80, 20F65, 43A07, 16S34, 20C07 © SpringerVerlag Berlin Heidelberg 2010 This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilm or in any other way, and storage in data banks. Duplication of this publication or parts thereof is permitted only under the provisions of the German Copyright Law of September 9, 1965, in its current version, and permission for use must always be obtained from Springer. Violations are liable to prosecution under the German Copyright Law. The use of general descriptive names, registered names, trademarks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. Cover design: deblik Printed on acidfree paper Springer is part of Springer Science+Business Media (www.springer.com)
To Katiuscia, Giacomo, and Tommaso To Martine and Nathalie
Preface
Two seemingly unrelated mathematical notions, namely that of an amenable group and that of a cellular automaton, were both introduced by John von Neumann in the ﬁrst half of the last century. Amenability, which originated from the study of the BanachTarski paradox, is a property of groups generalizing both commutativity and ﬁniteness. Nowadays, it plays an important role in many areas of mathematics such as representation theory, harmonic analysis, ergodic theory, geometric group theory, probability theory, and dynamical systems. Von Neumann used cellular automata to serve as theoretical models for selfreproducing machines. About twenty years later, the famous cellular automaton associated with the Game of Life was invented by John Horton Conway and popularized by Martin Gardner. The theory of cellular automata ﬂourished as one of the main branches of computer science. Deep connections with complexity theory and logic emerged from the discovery that some cellular automata are universal Turing machines. A group G is said to be amenable (as a discrete group) if the set of all subsets of G admits a rightinvariant ﬁnitely additive probability measure. All ﬁnite groups, all solvable groups (and therefore all abelian groups), and all ﬁnitely generated groups of subexponential growth are amenable. Von Neumann observed that the class of amenable groups is closed under the operation of taking subgroups and that the free group of rank two F2 is nonamenable. It follows that a group which contains a subgroup isomorphic to F2 is nonamenable. However, there are examples of groups which are nonamenable and contain no subgroups isomorphic to F2 (the ﬁrst examples of such groups were discovered by Alexander Y. Ol’shanskii and by Sergei I. Adyan). Loosely speaking, a general cellular automaton can be described as follows. A conﬁguration is a map from a set called the universe into another set called the alphabet. The elements of the universe are called cells and the elements of the alphabet are called states. A cellular automaton is then a map from the set of all conﬁgurations into itself satisfying the following local property: the state of the image conﬁguration at a given cell only depends on
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the states of the initial conﬁguration on a ﬁnite neighborhood of the given cell. In the classical setting, for instance in the cellular automata constructed by von Neumann and the one associated with Conway’s Game of Life, the alphabet is ﬁnite, the universe is the two dimensional inﬁnite square lattice, and the neighborhood of a cell consists of the cell itself and its eight adjacent cells. By iterating a cellular automaton one gets a discrete dynamical system. Such dynamical systems have proved very useful to model complex systems arising from natural sciences, in particular physics, biology, chemistry, and population dynamics. ∗
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In this book, the universe will always be a group G (in the classical setting the corresponding group was G = Z2 ) and the alphabet may be ﬁnite or inﬁnite. The left multiplication in G induces a natural action of G on the set of conﬁgurations which is called the Gshift and all cellular automata will be required to commute with the shift. It was soon realized that the question whether a given cellular automaton is surjective or not needs a special attention. From the dynamical viewpoint, surjectivity means that each conﬁguration may be reached at any time. The ﬁrst important result in this direction is the celebrated theorem of Moore and Myhill which gives a necessary and suﬃcient condition for the surjectivity of a cellular automaton with ﬁnite alphabet over the group G = Z2 . Edward F. Moore and John R. Myhill proved that such a cellular automaton is surjective if and only if it is preinjective. As the term suggests it, preinjectivity is a weaker notion than injectivity. More precisely, a cellular automaton is said to be preinjective if two conﬁgurations are equal whenever they have the same image and coincide outside a ﬁnite subset of the group. Moore proved the “surjective ⇒ preinjective” part and Myhill proved the converse implication shortly after. One often refers to this result as to the Garden of Eden theorem. This biblical terminology is motivated by the fact that, regarding a cellular automaton as a dynamical system with discrete time, a conﬁguration which is not in the image of the cellular automaton may only appear as an initial conﬁguration, that is, at time t = 0. The surprising connection between amenability and cellular automata was established in 1997 when Antonio Mach`ı, Fabio Scarabotti and the ﬁrst author proved the Garden of Eden theorem for cellular automata with ﬁnite alphabets over amenable groups. At the same time, and completely independently, Misha Gromov, using a notion of spacial entropy, presented a more general form of the Garden of Eden theorem where the universe is an amenable graph with a dense holonomy and cellular automata are called maps of bounded propagation. Mach`ı, Scarabotti and the ﬁrst author also showed that both implications in the Garden of Eden theorem become false if the underlying group contains a subgroup isomorphic to F2 . The question whether the Garden of Eden theorem could be extended beyond the class of
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amenable groups remained open until 2008 when Laurent Bartholdi proved that the Moore implication fails to hold for nonamenable groups. As a consequence, the whole Garden of Eden theorem only holds for amenable groups. This gives a new characterization of amenable groups in terms of cellular automata. Let us mention that, up to now, the validity of the Myhill implication for nonamenable groups is still an open problem. Following Walter H. Gottschalk, a group G is said to be surjunctive if every injective cellular automaton with ﬁnite alphabet over G is surjective. Wayne Lawton proved that all residually ﬁnite groups are surjunctive and that every subgroup of a surjunctive group is surjunctive. Since injectivity implies preinjectivity, an immediate consequence of the Garden of Eden theorem for amenable groups is that every amenable group is surjunctive. Gromov and Benjamin Weiss introduced a class of groups, called soﬁc groups, which includes all residually ﬁnite groups and all amenable groups, and proved that every soﬁc group is surjunctive. Soﬁc groups can be deﬁned in three equivalent ways: in terms of local approximation by ﬁnite symmetric groups equipped with their Hamming distance, in terms of local approximation of their Cayley graphs by ﬁnite labelled graphs, and, ﬁnally, as being the groups that can be embedded into ultraproducts of ﬁnite symmetric groups (this last characterization is due to G´ abor Elek and Endre Szab´ o). The class of soﬁc groups is the largest known class of surjunctive groups. It is not known, up to now, whether all groups are surjunctive (resp. soﬁc) or not. Stimulated by Gromov ideas, we considered cellular automata whose alphabets are vector spaces. In this framework, the space of conﬁgurations has a natural structure of a vector space and cellular automata are required to be linear. An analogue of the Garden of Eden theorem was proved for linear cellular automata with ﬁnite dimensional alphabets over amenable groups. In the proof, the role of entropy, used in the ﬁnite alphabet case, is now played by the mean dimension, a notion introduced by Gromov. Also, examples of linear cellular automata with ﬁnite dimensional alphabets over groups containing F2 showing that the linear version of the Garden of Eden theorem may fail to hold in this case, were provided. It is not known, up to now, if the Garden of Eden theorem for linear cellular automata with ﬁnite dimensional alphabet only holds for amenable groups or not. We also introduced the notion of linear surjunctivity: a group G is said to be Lsurjunctive if every injective linear cellular automaton with ﬁnite dimensional alphabet over G is surjective. We proved that every soﬁc group is Lsurjunctive. Linear cellular automata over a group G with alphabet of ﬁnite dimension d over a ﬁeld K may be represented by d × d matrices with entries in the group ring K[G]. This leads to the following characterization of Lsurjunctivity: a group is Lsurjunctive if and only if it satisﬁes Kaplansky’s conjecture on the stable ﬁniteness of group rings (a ring is said to be stably ﬁnite if onesided invertible ﬁnite dimensional square matrices with coeﬃcients in that ring are in fact twosided invertible). As a corollary, one has that group rings of soﬁc groups are stably ﬁnite, a result previously established by Elek and Szab´ o
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using diﬀerent methods. Moreover, given a group G and a ﬁeld K, the preinjectivity of all nonzero linear cellular automata with alphabet K over G is equivalent to the absence of zerodivisors in K[G]. As a consequence, another important problem on the structure of group rings also formulated by Irving Kaplansky may be expressed in terms of cellular automata. Is every nonzero linear cellular automaton with onedimensional alphabet over a torsionfree group always preinjective? ∗
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The material presented in this book is entirely selfcontained. In fact, its reading only requires some acquaintance with undergraduate general topology and abstract algebra. Each chapter begins with a brief overview of its contents and ends with some historical notes and a list of exercises at various diﬃculty levels. Some additional topics, such as subshifts and cellular automata over subshifts, are treated in these exercises. Hints are provided each time help may be needed. In order to improve accessibility, a few appendices are included to quickly introduce the reader to facts he might be not too familiar with. In the ﬁrst chapter, we give the deﬁnition of a cellular automaton. We present some basic examples and discuss general methods for constructing cellular automata. We equip the set of conﬁgurations with its prodiscrete uniform structure and prove the generalized CurtisHedlund theorem: a necessary and suﬃcient condition for a selfmapping of the conﬁguration space to be a cellular automaton is that it is uniformly continuous and commutes with the shift. Chapter 2 is devoted to residually ﬁnite groups. We give several equivalent characterizations of residual ﬁniteness and prove that the class of residually ﬁnite groups is closed under taking subgroups and projective limits. We establish in particular the theorems, respectively due to Anatoly I. Mal’cev and Gilbert Baumslag, which assert that ﬁnitely generated residually ﬁnite groups are Hopﬁan and that their automorphism group is residually ﬁnite. Surjunctive groups are introduced in Chap. 3. We show that every subgroup of a surjunctive group is surjunctive and that locally residually ﬁnite groups are surjunctive. We also prove a theorem of Gromov which says that limits of surjunctive marked groups are surjunctive. The theory of amenable groups is developed in Chap. 4. The class of amenable groups is closed under taking subgroups, quotients, extensions, and inductive limits. We prove the theorems due to Erling Følner and Alfred Tarski which state the equivalence between amenability, the existence of a Følner net, and the nonexistence of a paradoxical decomposition. The Garden of Eden theorem is established in Chap. 5. It is proved by showing that both surjectivity and preinjectivity of the cellular automaton are equivalent to the fact that the image of the conﬁguration space has maxi
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mal entropy. We give an example of a cellular automaton with ﬁnite alphabet over F2 which is preinjective but not surjective. Following Bartholdi’s construction, we also prove the existence of a surjective but not preinjective cellular automaton with ﬁnite alphabet over any nonamenable group. In Chap. 6 we present the basic elementary notions and results on growth of ﬁnitely generated groups. We prove that ﬁnitely generated nilpotent groups have polynomial growth. We then introduce the Grigorchuk group and show that it is an inﬁnite ﬁnitely generated periodic group of intermediate growth. We show that every ﬁnitely generated group of subexponential growth is amenable. We also establish the KestenDay characterization of amenability which asserts that a group with a ﬁnite (not necessarily symmetric) generating subset is amenable if and only if 0 is in the 2 spectrum of the associated Laplacian. Finally, we consider the notion of quasiisometry for not necessarily countable groups and we show that amenability is a quasiisometry invariant. In Chap. 7 we consider the notion of local embeddability of groups into a class of groups. For the class of ﬁnite groups, this gives the class of LEF groups introduced by Anatoly M. Vershik and Edward I. Gordon. We discuss several stability properties of local embeddability and show that locally embeddable groups are closed in marked groups spaces. The remaining of the chapter is devoted to the class of soﬁc groups. We show that the three deﬁnitions, namely analytic, geometric, and algebraic, we alluded to before, are equivalent. We then prove the GromovWeiss theorem which states that every soﬁc group is surjunctive. The last chapter is devoted to linear cellular automata. We prove the linear version of the Garden of Eden theorem and show that every soﬁc group is Lsurjunctive. We end the chapter with a discussion on the stable ﬁniteness and the zerodivisors conjectures of Kaplansky and their reformulation in terms of linear cellular automata. Appendix A gives a quick overview of a few fundamental notions and results of topology (nets, compactness, product topology, and the Tychonoﬀ product theorem). Appendix B is devoted to Andr´e Weil’s theory of uniform spaces. It includes also a detailed exposition of the HausdorﬀBourbaki uniform structure on subsets of a uniform space. In Appendix C, we establish some basic properties of symmetric groups and prove the simplicity of the alternating groups. The deﬁnition and the construction of free groups are given in Appendix D. The proof of Klein’s pingpong lemma is also included there. In Appendix E we shortly describe the constructions of inductive and projective limits of groups. Appendix F treats topological vector spaces, the weak∗ topology, and the BanachAlaoglu theorem. The proof of the MarkovKakutani ﬁxed point theorem is presented in Appendix G. In the subsequent appendix, of a pure graphtheoretical and combinatorial ﬂavour, we consider bipartite graphs and their matchings. We prove Hall’s marriage theorem and its harem version which plays a key role in the proof of Tarski’s theorem on amenability. The Baire theorem, the open mapping theorem, as well as other
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complements of functional analysis including uniform convexity are treated in Appendix I. The last appendix deals with the notions of ﬁlters and ultraﬁlters. We would like to express our deep gratitude to Dr. Catriona Byrne, Dr. Marina Reizakis and Annika Eling from Springer Verlag and to Donatas Akmanaviˇcius for their constant and kindest help at all stages of the editorial process. Rome and Strasbourg
Tullio CeccheriniSilberstein Michel Coornaert
Contents
1
Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.1 The Conﬁguration Set and the Shift Action . . . . . . . . . . . . . . . . 1.2 The Prodiscrete Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.3 Periodic Conﬁgurations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.4 Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.5 Minimal Memory . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1.6 Cellular Automata over Quotient Groups . . . . . . . . . . . . . . . . . . 1.7 Induction and Restriction of Cellular Automata . . . . . . . . . . . . 1.8 Cellular Automata with Finite Alphabets . . . . . . . . . . . . . . . . . . 1.9 The Prodiscrete Uniform Structure . . . . . . . . . . . . . . . . . . . . . . . 1.10 Invertible Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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2
Residually Finite Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.1 Deﬁnition and First Examples . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.2 Stability Properties of Residually Finite Groups . . . . . . . . . . . . 2.3 Residual Finiteness of Free Groups . . . . . . . . . . . . . . . . . . . . . . . 2.4 Hopﬁan Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.5 Automorphism Groups of Residually Finite Groups . . . . . . . . . 2.6 Examples of Finitely Generated Groups Which Are Not Residually Finite . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2.7 Dynamical Characterization of Residual Finiteness . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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Surjunctive Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.1 Deﬁnition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Stability Properties of Surjunctive Groups . . . . . . . . . . . . . . . . . 3.3 Surjunctivity of Locally Residually Finite Groups . . . . . . . . . . .
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3.4 Marked Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Expansive Actions on Uniform Spaces . . . . . . . . . . . . . . . . . . . . . 3.6 Gromov’s Injectivity Lemma . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.7 Closedness of Marked Surjunctive Groups . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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4
Amenable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.1 Measures and Means . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Properties of the Set of Means . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.3 Measures and Means on Groups . . . . . . . . . . . . . . . . . . . . . . . . . . 4.4 Deﬁnition of Amenability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.5 Stability Properties of Amenable Groups . . . . . . . . . . . . . . . . . . 4.6 Solvable Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.7 The Følner Conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.8 Paradoxical Decompositions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.9 The Theorems of Tarski and Følner . . . . . . . . . . . . . . . . . . . . . . . 4.10 The Fixed Point Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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5
The Garden of Eden Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Garden of Eden Conﬁgurations and Garden of Eden Patterns 5.2 Preinjective Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.3 Statement of the Garden of Eden Theorem . . . . . . . . . . . . . . . . 5.4 Interiors, Closures, and Boundaries . . . . . . . . . . . . . . . . . . . . . . . 5.5 Mutually Erasable Patterns . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.6 Tilings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.7 Entropy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.8 Proof of the Garden of Eden Theorem . . . . . . . . . . . . . . . . . . . . 5.9 Surjunctivity of Locally Residually Amenable Groups . . . . . . . 5.10 A Surjective but Not Preinjective Cellular Automaton over F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.11 A Preinjective but Not Surjective Cellular Automaton over F2 . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.12 A Characterization of Amenability in Terms of Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.13 Garden of Eden Patterns for Life . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
111 111 112 114 115 121 122 125 128 131
Finitely Generated Amenable Groups . . . . . . . . . . . . . . . . . . . . 6.1 The Word Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.2 Labeled Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.3 Cayley Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.4 Growth Functions and Growth Types . . . . . . . . . . . . . . . . . . . . .
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6.5 The Growth Rate . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.6 Growth of Subgroups and Quotients . . . . . . . . . . . . . . . . . . . . . . 6.7 A Finitely Generated Metabelian Group with Exponential Growth . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6.8 Growth of Finitely Generated Nilpotent Groups . . . . . . . . . . . . 6.9 The Grigorchuk Group and Its Growth . . . . . . . . . . . . . . . . . . . . 6.10 The Følner Condition for Finitely Generated Groups . . . . . . . . 6.11 Amenability of Groups of Subexponential Growth . . . . . . . . . . 6.12 The Theorems of Kesten and Day . . . . . . . . . . . . . . . . . . . . . . . . 6.13 QuasiIsometries . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
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Local Embeddability and Soﬁc Groups . . . . . . . . . . . . . . . . . . . . 7.1 Local Embeddability . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2 Local Embeddability and Ultraproducts . . . . . . . . . . . . . . . . . . . 7.3 LEFGroups and LEAGroups . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 The Hamming Metric . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Soﬁc Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.6 Soﬁc Groups and Metric Ultraproducts of Finite Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 A Characterization of Finitely Generated Soﬁc Groups . . . . . . 7.8 Surjunctivity of Soﬁc Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
233 234 243 246 251 254
Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 The Algebra of Linear Cellular Automata . . . . . . . . . . . . . . . . . 8.2 Conﬁgurations with Finite Support . . . . . . . . . . . . . . . . . . . . . . . 8.3 Restriction and Induction of Linear Cellular Automata . . . . . . 8.4 Group Rings and Group Algebras . . . . . . . . . . . . . . . . . . . . . . . . 8.5 Group Ring Representation of Linear Cellular Automata . . . . 8.6 Modules over a Group Ring . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.7 Matrix Representation of Linear Cellular Automata . . . . . . . . . 8.8 The Closed Image Property . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.9 The Garden of Eden Theorem for Linear Cellular Automata . 8.10 Preinjective but not Surjective Linear Cellular Automata . . . 8.11 Surjective but not Preinjective Linear Cellular Automata . . . 8.12 Invertible Linear Cellular Automata . . . . . . . . . . . . . . . . . . . . . . 8.13 Preinjectivity and Surjectivity of the Discrete Laplacian . . . . 8.14 Linear Surjunctivity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.15 Stable Finiteness of Group Algebras . . . . . . . . . . . . . . . . . . . . . . 8.16 ZeroDivisors in Group Algebras and Preinjectivity of OneDimensional Linear Cellular Automata . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
283 284 288 289 291 294 299 301 305 308 314 315 317 321 324 327
260 265 272 275 278
330 335 338
xvi
Contents
A
Nets and the Tychonoﬀ Product Theorem . . . . . . . . . . . . . . . . A.1 Directed Sets . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.2 Nets in Topological Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.3 Initial Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.4 Product Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . A.5 The Tychonoﬀ Product Theorem . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
343 343 343 346 346 347 349
B
Uniform Structures . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.1 Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.2 Uniformly Continuous Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.3 Product of Uniform Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . B.4 The HausdorﬀBourbaki Uniform Structure on Subsets . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
351 351 353 355 356 358
C
Symmetric Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C.1 The Symmetric Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . C.2 Permutations with Finite Support . . . . . . . . . . . . . . . . . . . . . . . . C.3 Conjugacy Classes in Sym0 (X) . . . . . . . . . . . . . . . . . . . . . . . . . . . C.4 The Alternating Group . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
359 359 360 362 363
D
Free Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.1 Concatenation of Words . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.2 Deﬁnition and Construction of Free Groups . . . . . . . . . . . . . . . . D.3 Reduced Forms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.4 Presentations of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . D.5 The Klein PingPong Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . .
367 367 367 373 375 376
E
Inductive Limits and Projective Limits of Groups . . . . . . . . 379 E.1 Inductive Limits of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 379 E.2 Projective Limits of Groups . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 380
F
The BanachAlaoglu Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . F.1 Topological Vector Spaces . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F.2 The Weak∗ Topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . F.3 The BanachAlaoglu Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . .
383 383 384 384
G
The MarkovKakutani Fixed Point Theorem . . . . . . . . . . . . . . G.1 Statement of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . G.2 Proof of the Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
387 387 387 389
H
The Hall Harem Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 H.1 Bipartite Graphs . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 391 H.2 Matchings . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 393
Contents
xvii
H.3 The Hall Marriage Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 394 H.4 The Hall Harem Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 399 Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 401 I
Complements of Functional Analysis . . . . . . . . . . . . . . . . . . . . . . I.1 The Baire Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.2 The Open Mapping Theorem . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.3 Spectra of Linear Maps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . I.4 Uniform Convexity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
403 403 404 406 407
J
Ultraﬁlters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . J.1 Filters and Ultraﬁlters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . J.2 Limits Along Filters . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . Notes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
409 409 412 415
Open Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 417 Comments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 418 References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 421 List of Symbols . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 429 Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 433
Notation
Throughout this book, the following conventions are used: • N is the set of nonnegative integers so that 0 ∈ N; • the notation A ⊂ B means that each element in the set A is also in the set B so that A and B may coincide; • a countable set is a set which admits a bijection onto a subset of N so that ﬁnite sets are countable; • all group actions are left actions; • all rings are assumed to be associative (but not necessarily commutative) with a unity element; • a ﬁeld is a nonzero commutative ring in which each nonzero element is invertible.
xix
Chapter 1
Cellular Automata
In this chapter we introduce the notion of a cellular automaton. We ﬁx a group and an arbitrary set which will be called the alphabet. A conﬁguration is deﬁned as being a map from the group into the alphabet. Thus, a conﬁguration is a way of attaching an element of the alphabet to each element of the group. There is a natural action of the group on the set of conﬁgurations which is called the shift action (see Sect. 1.1). A cellular automaton is a selfmapping of the set of conﬁgurations deﬁned from a system of local rules commuting with the shift (see Deﬁnition 1.4.1). We equip the conﬁguration set with the prodiscrete topology, that is, the topology of pointwise convergence associated with the discrete topology on the alphabet (see Sect. 1.2). It turns out that every cellular automaton is continuous with respect to the prodiscrete topology (Proposition 1.4.8) and commutes with the shift (Proposition 1.4.4). Conversely, when the alphabet is ﬁnite, every continuous selfmapping of the conﬁguration space which commutes with the shift is a cellular automaton (Theorem 1.8.1). Another important fact in the ﬁnite alphabet case is that every bijective cellular automaton is invertible, in the sense that its inverse map is also a cellular automaton (Theorem 1.10.2). We give examples showing that, when the alphabet is inﬁnite, a continuous selfmapping of the conﬁguration space which commutes with the shift may fail to be a cellular automaton and a bijective cellular automaton may fail to be invertible. In Sect. 1.9, we introduce the prodiscrete uniform structure on the conﬁguration space. We show that a selfmapping of the conﬁguration space is a cellular automaton if and only if it is uniformly continuous and commutes with the shift (Theorem 1.9.1).
1.1 The Conﬁguration Set and the Shift Action Let G be a group. For g ∈ G, denote by Lg the left multiplication by g in G, that is, the map Lg : G → G given by T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 1, © SpringerVerlag Berlin Heidelberg 2010
1
2
1 Cellular Automata
Lg (g ) = gg
for all g ∈ G.
Observe that for all g1 , g2 , g ∈ G one has (Lg1 ◦ Lg2 ) (g ) = Lg1 (Lg2 (g )) = Lg1 (g2 g ) = g1 g2 g = Lg1 g2 (g ) which shows that Lg1 ◦ Lg2 = Lg1 g2 .
(1.1)
Let A be a set. Consider the set AG consisting of all maps from G to A: AG = A = {x : G → A}. g∈G
The set A is called the alphabet. The elements of A are called the letters, or the states, or the symbols, or the colors. The group G is called the universe. The set AG is called the set of conﬁgurations. Given an element g ∈ G and a conﬁguration x ∈ AG , we deﬁne the conﬁguration gx ∈ AG by (1.2) gx = x ◦ Lg−1 . Thus one has
gx(g ) = x(g −1 g )
for all g ∈ G.
The map G × AG → AG (g, x) → gx
is a left action of G on AG . Indeed, for all g1 , g2 ∈ G and x ∈ AG , one has g1 (g2 x) = g1 (x ◦ Lg−1 ) = x ◦ Lg−1 ◦ Lg−1 = x ◦ Lg−1 g−1 = x ◦ L(g1 g2 )−1 2
2
1
2
1
= (g1 g2 )x, where the third equality follows from (1.1). Also, denoting by 1G the identity element of G and by IdG : G → G the identity map, one has 1G x = x ◦ L1G = x ◦ IdG = x. This left action of G on AG is called the Gshift on AG . A pattern over the group G and the alphabet A is a map p : Ω → G deﬁned on some ﬁnite subset Ω of G. The set Ω is then called the support of p.
1.3 Periodic Conﬁgurations
3
1.2 The Prodiscrete Topology Let G be a group and let A be a set. We equip each factor A of AG with the discrete topology (all subsets of A are open) and AG with the associated product topology (see Sect. A.4). This topology is called the prodiscrete topology on AG . This is the smallest topology on AG for which the projection map πg : AG → A, given by πg (x) = x(g), is continuous for every g ∈ G (cf. Sect. A.4). The elementary cylinders C(g, a) = πg−1 ({a}) = {x ∈ AG : x(g) = a}
(g ∈ G, a ∈ A)
are both open and closed in AG . A subset U ⊂ AG is open if and only if U can be expressed as a (ﬁnite or inﬁnite) union of ﬁnite intersections of elementary cylinders. For a subset Ω ⊂ G and a conﬁguration x ∈ AG let xΩ ∈ AΩ denote the restriction of x to Ω, that is, the map xΩ : Ω → A deﬁned by xΩ (g) = x(g) for all g ∈ Ω. If x ∈ AG , a neighborhood base of x is given by the sets V (x, Ω) = {y ∈ AG : xΩ = yΩ } = C(g, x(g)), (1.3) g∈Ω
where Ω runs over all ﬁnite subsets of G. Proposition 1.2.1. The space AG is Hausdorﬀ and totally disconnected. Proof. The discrete topology on A is Hausdorﬀ and totally disconnected, and, by Proposition A.4.1 and Proposition A.4.2, a product of Hausdorﬀ (resp. totally disconnected) topological spaces is Hausdorﬀ (resp. totally disconnected). Recall that an action of a group G on a topological space X is said to be continuous if the map ϕg : X → X given by ϕg (x) = gx is continuous on X for each g ∈ G. Proposition 1.2.2. The action of G on AG is continuous. Proof. Let g ∈ G and consider the map ϕg : AG → AG deﬁned by ϕg (x) = gx. The map πh ◦ϕg is equal to πg−1 h and is therefore continuous on AG for every h ∈ G. Consequently, ϕg is continuous (cf. Sect. A.4).
1.3 Periodic Conﬁgurations Let G be a group and let A be a set. Let H be a subgroup of G. A conﬁguration x ∈ AG is called Hperiodic if x is ﬁxed by H, that is, if one has
4
1 Cellular Automata
hx = x
for all h ∈ H.
Let Fix(H) denote the subset of AG consisting of all Hperiodic conﬁgurations. Examples 1.3.1. (a) One has Fix({1G }) = AG . (b) The set Fix(G) consists of all constant conﬁgurations and may be therefore identiﬁed with A. (c) For G = Z and H = nZ, n ≥ 1, the set Fix(H) is the set of sequences x : Z → A which admit n as a (not necessarily minimal) period, that is, such that x(i + n) = x(i) for all i ∈ Z. Proposition 1.3.2. Let H be a subgroup of G. Then the set Fix(H) is closed in AG for the prodiscrete topology. Proof. We have Fix(H) =
{x ∈ AG : hx = x}.
(1.4)
h∈H
The space AG is Hausdorﬀ by Proposition 1.2.1 and the action of G on AG is continuous by Proposition 1.2.2. Thus the set of ﬁxed points of the map x → gx is closed in AG for each g ∈ G. Therefore Fix(H) is closed in AG by (1.4). Consider the set H\G = {Hg : g ∈ G} consisting of all right cosets of H in G and the canonical surjective map ρ : G → H\G g → Hg. Given an element y ∈ AH\G , i.e., a map y : H\G → A, we can form the composite map y ◦ ρ : G → A which is an element of AG . In fact, we have y ◦ ρ ∈ Fix(H) since (h(y ◦ ρ))(g) = y ◦ ρ(h−1 g) = y(ρ(h−1 g)) = y(ρ(g)) = y ◦ ρ(g) for all g ∈ G and h ∈ H. Proposition 1.3.3. Let H be a subgroup of G and let denote by ρ : G → H\G the canonical surjection. Then the map ρ∗ : AH\G → Fix(H) deﬁned by ρ∗ (y) = y ◦ ρ for all y ∈ AH\G is bijective. Proof. If y1 , y2 ∈ AH\G satisfy y1 ◦ρ = y2 ◦ρ, then y1 = y2 since ρ is surjective. Thus ρ∗ is injective. If x ∈ Fix(H), then hx = x for all h ∈ H, that is, x(h−1 g) = x(g)
for all h ∈ H, g ∈ G.
Thus, the conﬁguration x is constant on each right coset of G modulo H, that is, x is in the image of ρ∗ . This shows that ρ∗ is surjective.
1.3 Periodic Conﬁgurations
5
Corollary 1.3.4. If the set A is ﬁnite and H is a subgroup of ﬁnite index of G, then the set Fix(H) is ﬁnite and one has  Fix(H) = A[G:H] , where [G : H] denotes the index of H in G. Example 1.3.5. Let G = Z and H = nZ, where n ≥ 1. If A is ﬁnite of cardinality k, then  Fix(H) = k n . Suppose now that N is a normal subgroup of G, that is, gN = N g for all g ∈ G. Then, there is a natural group structure on G/N = N \G for which the canonical surjection ρ : G → G/N is a homomorphism. Proposition 1.3.6. Let N be a normal subgroup of G. Then Fix(N ) is a Ginvariant subset of AG . Proof. Let x ∈ Fix(N ) and g ∈ G. Given h ∈ N , then there exists h ∈ N such that hg = gh , since N is normal in G. Thus, we have h(gx) = g(h x) = gx which shows that gx ∈ Fix(N ).
Since every element of Fix(N ) is ﬁxed by N , the action of G on Fix(N ) induces an action of G/N on Fix(N ) which satisﬁes ρ(g)x = gx for all g ∈ G and x ∈ Fix(N ). Suppose that a group Γ acts on two sets X and Y . A map ϕ : X → Y is called Γ equivariant if one has ϕ(γx) = γϕ(x) for all γ ∈ Γ and x ∈ X. Proposition 1.3.7. Let N be a normal subgroup of G and let ρ : G → G/N denote the canonical epimorphism. Then the map ρ∗ : AG/N → Fix(N ) deﬁned by ρ∗ (y) = y ◦ ρ for all y ∈ AG/N is a G/N equivariant bijection. Proof. We already know that ρ∗ is bijective (Proposition 1.3.3). Let g ∈ G and y ∈ AG/N . For all g ∈ G, we have ρ(g)ρ∗ (y)(g ) = gρ∗ (y)(g ) = ρ∗ (y)(g −1 g ) = (y ◦ ρ)(g −1 g ) = y(ρ(g −1 g )) = y((ρ(g))−1 ρ(g )) = ρ(g)y(ρ(g )) = ρ∗ (ρ(g)y)(g ). Thus ρ(g)ρ∗ (y) = ρ∗ (ρ(g)y). This shows that ρ∗ is G/N equivariant.
6
1 Cellular Automata
1.4 Cellular Automata Let G be a group and let A be a set. Deﬁnition 1.4.1. A cellular automaton over the group G and the alphabet A is a map τ : AG → AG satisfying the following property: there exist a ﬁnite subset S ⊂ G and a map μ : AS → A such that τ (x)(g) = μ((g −1 x)S )
(1.5)
for all x ∈ AG and g ∈ G, where (g −1 x)S denotes the restriction of the conﬁguration g −1 x to S. Such a set S is called a memory set and μ is called a local deﬁning map for τ . Observe that formula (1.5) says that the value of the conﬁguration τ (x) at an element g ∈ G is the value taken by the local deﬁning map μ at the pattern obtained by restricting to the memory set S the shifted conﬁguration g −1 x. Remark 1.4.2. (a) Equality (1.5) may also be written τ (x)(g) = μ((x ◦ Lg )S )
(1.6)
by (1.2). (b) For g = 1G , formula (1.5) gives us τ (x)(1G ) = μ(xS ).
(1.7)
As the restriction map AG → AS , x → xS , is surjective, this shows that if S is a memory set for the cellular automaton τ , then there is a unique map μ : AS → A which satisﬁes (1.5). Thus one says that this unique μ is the local deﬁning map for τ associated with the memory set S. Examples 1.4.3. (a) The cellular automaton associated with the Game of Life. Consider an inﬁnite twodimensional orthogonal grid of square cells, each of which is in one of two possible states, live or dead. Every cell c interacts with its eight neighboring cells, namely the North, NorthEast, East, SouthEast, South, SouthWest, West and NorthWest cells (see Fig. 1.1). At each step in time, the following rules for the evolution of the states of the cells are applied (in Figs. 1.2–1.5 we label with a “•” a live cell and with a “◦” a dead cell): • (birth): a cell that is dead at time t becomes alive at time t + 1 if and only if three of its neighbors are alive at time t (cf. Fig. 1.2); • (survival): a cell that is alive at time t will remain alive at time t + 1 if and only if it has exactly two or three live neighbors at time t (cf. Fig. 1.3);
1.4 Cellular Automata
7
Fig. 1.1 The cell c and its eight neighboring cells
Fig. 1.2 A cell that is dead at time t becomes alive at time t + 1 if and only if three of its neighbors are alive at time t
Fig. 1.3 A cell that is alive at time t will remain alive at time t + 1 if and only if it has exactly two or three live neighbors at time t
8
1 Cellular Automata
Fig. 1.4 A live cell that has at most one live neighbor at time t will be dead at time t + 1
• (death by loneliness): a live cell that has at most one live neighbor at time t will be dead at time t + 1 (cf. Fig. 1.4); • (death by overcrowding): a cell that is alive at time t and has four or more live neighbors at time t, will be dead at time t + 1 (cf. Fig. 1.5).
Fig. 1.5 A cell that is alive at time t and has four or more live neighbors at time t, will be dead at time t + 1
Let us show that the map which transforms a conﬁguration of cells at time t into the conﬁguration at time t + 1 according to the above rules is indeed a cellular automaton. Consider the group G = Z2 and the ﬁnite set S = {−1, 0, 1}2 ⊂ G. Then there is a onetoone correspondence between the cells in the grid and the elements in G in such a way that the following holds. If c is a given cell, then c+(0, 1) is the neighboring North cell, c+(1, 1) is the neighboring NorthEast cell, and so on; in other words, c and its eight neighboring cells correspond to the group elements c + s with s ∈ S (see Fig. 1.6). Consider the alphabet A = {0, 1}. The state 0 (resp. 1) corresponds to absence (resp. presence) of life. With each conﬁguration of the states of the cells in the grid we associate a map x ∈ AG deﬁned as follows. Given a cell c we set x(c) = 1 (resp. 0) if the cell c is alive (resp. dead).
1.4 Cellular Automata
9
Fig. 1.6 The cell c and its eight neighboring cells c + s, s ∈ S = {−1, 0, 1}2
Consider the map μ : AS → A given by ⎧ ⎧ ⎪ ⎪ ⎨ s∈S y(s) = 3 ⎪ ⎪ ⎨1 if or ⎪ μ(y) = ⎩ ⎪ ⎪ s∈S y(s) = 4 and y((0, 0)) = 1, ⎪ ⎩ 0 otherwise
(1.8)
for all y ∈ AS . A moment of thought tells us that μ just expresses the rules for the Game of Life. The cellular automaton τ : AG → AG with memory set S and local deﬁning map μ is called the cellular automaton associated with the Game of Life. (b) The Discrete Laplacian. Let G = Z and A = R. Consider the map Δ : RZ → RZ deﬁned by Δ(x)(n) = 2x(n) − x(n − 1) − x(n + 1). Then Δ is the cellular automaton over Z with memory set S = {−1, 0, 1} and local deﬁning map μ : RS → R given by μ(y) = 2y(0) − y(−1) − y(1)
for all y ∈ RS .
This may be generalized in the following way. Let G be an arbitrary group and let S be a nonempty ﬁnite subset of G. Let K be a ﬁeld. Consider the G G map ΔS = ΔG S : K → K deﬁned by x(gs). ΔS (x)(g) = Sx(g) − s∈S
10
1 Cellular Automata
Then ΔS is a cellular automaton over G with memory set S ∪ {1G } and local deﬁning map μ : KS∪{1G } → K given by μ(y) = Sy(1G ) − y(s) for all y ∈ KS∪{1G } . s∈S
This cellular automaton is called the discrete Laplacian over K associated with G and S. (c) The Majority action cellular automaton. Let G be a group and let S be a ﬁnite subset of G. Take A = {0, 1} and consider the map τ : AG → AG deﬁned by ⎧ ⎪ x(gs) > S if ⎨1 2 s∈S S τ (x)(g) = 0 if x(gs) < 2 ⎪ s∈S ⎩ S x(g) if s∈S x(gs) = 2 for all x ∈ AG . Then τ is a cellular automaton over G with memory set S ∪ {1G } and local deﬁning map μ : AS∪{1G } → A given by ⎧ ⎪ y(s) > S if ⎨1 2 s∈S S μ(y) = 0 if y(s) < 2 ⎪ s∈S ⎩ S y(1G ) if s∈S y(s) = 2 for all y ∈ AS∪{1G } . The cellular automaton τ is called the majority action cellular automaton associated with G and S (see Figs. 1.7–1.8). The terminology comes from the fact that given x ∈ AG and g ∈ G, the value τ (x)(g) is equal to a ∈ {0, 1} if there is a strict majority of elements of gS at which the conﬁguration x takes the value a, or to x(g) if no such majority exists. (d) Let G be a group, A a set, and f : A → A a map from A into itself. Then the map τ : AG → AG deﬁned by τ (x) = f ◦ x is a cellular automaton with memory set S = {1G } and local deﬁning map μ : AS → A given by μ(y) = f (y(1G )). Note that, if f is the identity map IdA on A, then τ equals the identity map IdAG on AG . (e) Let G be a group, A a set, and s0 an element of G. Let Rs0 : G → G denote the right multiplication by s0 in G, that is, the map Rs0 : G → G deﬁned by Rs0 (g) = gs0 . Then the map τ : AG → AG deﬁned by τ (x) = x ◦ Rs0 is a cellular automaton with memory set S = {s0 } and local deﬁning map μ : AS → A given by μ(y) = y(s0 ). Proposition 1.4.4. Let G be a group and let A be a set. Then every cellular automaton τ : AG → AG is Gequivariant.
1.4 Cellular Automata
11
Fig. 1.7 The local deﬁning map μ for the majority action on Z associated with S = {+1, −1}
Fig. 1.8 The majority action τ on Z associated with S = {+1, −1}
Proof. Let S be a memory set for τ and let μ : AS → A be the associated local deﬁning map. For all g, h ∈ G and x ∈ AG , we have τ (gx)(h) = μ((h−1 gx)S ) = μ(((g −1 h)−1 x)S ) = τ (x)(g −1 h) = gτ (x)(h). Thus τ (gx) = gτ (x).
Corollary 1.4.5. Let τ : AG → AG be a cellular automaton and let H be a subgroup of G. Then one has τ (Fix(H)) ⊂ Fix(H). Proof. Let x ∈ Fix(H). By the previous Proposition, we have, for every h ∈ H, hτ (x) = τ (hx) = τ (x). Thus τ (x) ∈ Fix(H).
The following characterization of cellular automata will be useful in the sequel.
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1 Cellular Automata
Proposition 1.4.6. Let G be a group and let A be a set. Consider a map τ : AG → AG . Let S be a ﬁnite subset of G and let μ : AS → A. Then the following conditions are equivalent: (a) τ is a cellular automaton admitting S as a memory set and μ as the associated local deﬁning map; (b) τ is Gequivariant and one has τ (x)(1G ) = μ(xS ) for every x ∈ AG . Proof. The fact that (a) implies (b) follows from Proposition 1.4.4 and formula (1.7) Conversely, suppose (b). Then, by using the Gequivariance of τ , we get τ (x)(g) = τ (g −1 x)(1G ) = μ((g −1 x)S ) for all x ∈ AG and g ∈ G. Consequently, τ satisﬁes (a).
An important feature of cellular automata is their continuity (with respect to the prodiscrete topology). In the proof of this property, we shall use the following. Lemma 1.4.7. Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton with memory set S and let g ∈ G. Then τ (x)(g) depends only on the restriction of x to gS. Proof. This is an immediate consequence of (1.5) since (g −1 x)(s) = x(gs) for all s ∈ S. Proposition 1.4.8. Let G be a group and let A be a set. Then every cellular automaton τ : AG → AG is continuous. Proof. Let S be a memory set for τ . Let x ∈ AG and let W be a neighborhood of τ (x) in AG . Then we can ﬁnd a ﬁnite subset Ω ⊂ G such that (cf. equation (1.3)) V (τ (x), Ω) ⊂ W. Consider the ﬁnite set ΩS = {gs : g ∈ Ω, s ∈ S}. If y ∈ AG coincide with x on ΩS, then τ (y) and τ (x) coincide on Ω by Lemma 1.4.7. Thus, we have τ (V (x, ΩS)) ⊂ V (τ (x), Ω) ⊂ W. This shows that τ is continuous.
Proposition 1.4.9. Let G be a group and let A be a set. Let σ : AG → AG and τ : AG → AG be cellular automata. Then the composite map σ ◦ τ : AG → AG is a cellular automaton. Moreover, if S (resp. T ) is a memory set for σ (resp. τ ), then ST = {st : s ∈ S, t ∈ T } is a memory set for σ ◦ τ . Proof. It is clear that the map σ ◦ τ is Gequivariant since σ and τ are Gequivariant (by Proposition 1.4.4). Let S (resp. T ) be a memory set for
1.4 Cellular Automata
13
σ (resp. τ ). For every x ∈ AG , we have σ ◦ τ (x)(1G ) = σ(τ (x))(1G ). By Lemma 1.4.7, σ(τ (x))(1G ) depends only on the restriction of τ (x) to S. By using Lemma 1.4.7 again, we deduce that, for every s ∈ S, the element τ (x)(s) depends only on the restriction of x to sT . Therefore, σ ◦ τ (x)(1G ) depends only on the restriction of x to ST . By applying Proposition 1.4.6, we conclude that σ ◦ τ is a cellular automaton admitting ST as a memory set. Remark 1.4.10. With the hypotheses and notation of the previous proposition, denote by μ : AS → A and ν : AT → A the local deﬁning maps for σ and τ , respectively. Then, the local deﬁning map κ : AST → A for σ ◦ τ may be described in the following way. For y ∈ AST and s ∈ S deﬁne ys ∈ AT by setting ys (t) = y(st) for all t ∈ T . Also, denote by y ∈ AS the map deﬁned by y(s) = ν(ys ) for all s ∈ S. We ﬁnally deﬁne the map κ : AST → A by setting (1.9)
κ(y) = μ(y) for all y ∈ AST . Let x ∈ AG , g ∈ G, s ∈ S, and t ∈ T . We then have (s−1 g −1 x)T (t) = s−1 g −1 x(t) = g −1 x(st) = (g −1 x)ST (st)
= (g −1 x)ST s (t). This shows that
(s−1 g −1 x)T = (g −1 x)ST s
and therefore
τ (g −1 x)(s) = ν (s−1 g −1 x)T = ν (g −1 x)ST s = (g −1 x)ST (s). As a consequence,
τ (g −1 x)S = (g −1 x)ST .
(1.10)
Finally, one has (σ ◦ τ )(x)(g) = σ(τ (x))(g)
= μ (g −1 τ (x))S = μ(τ (g −1 x)S )
(1.11)
(by (1.10)) = μ((g −1 x)ST )
(by (1.9)) = κ (g −1 x)ST . Recall that a monoid is a set equipped with an associative binary operation admitting an identity element. Denote by CA(G; A) the set consisting of all
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1 Cellular Automata
cellular automata τ : AG → AG . In Example 1.4.3(d) we have seen that the identity map IdAG : AG → AG is a cellular automaton. Thus we have: Corollary 1.4.11. The set CA(G; A) is a monoid for the composition of maps.
1.5 Minimal Memory Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. Let S be a memory set for τ and let μ : AS → A be the associated deﬁning map. If S is a ﬁnite subset of G such that S ⊂ S , then S is also a memory set for τ and the local deﬁning map associated with S is the map μ : AS → A given by μ = μ◦p, where p : AS → AS is the canonical projection (restriction map). This shows that the memory set of a cellular automaton is not unique in general. However, we shall see that every cellular automaton admits a unique memory set of minimal cardinality. Let us ﬁrst establish the following result. Lemma 1.5.1. Let τ : AG → AG be a cellular automaton. Let S1 and S2 be memory sets for τ . Then S1 ∩ S2 is also a memory set for τ . Proof. Let x ∈ AG . Let us show that τ (x)(1G ) depends only on the restriction of x to S1 ∩ S2 . To see this, consider an element y ∈ AG such that xS1 ∩S2 = yS1 ∩S2 . Let us choose an element z ∈ AG such that zS1 = xS1 and zS2 = yS2 (we may take for instance the conﬁguration z ∈ AG which coincides with x on S1 and with y on G \ S1 ). We have τ (x)(1G ) = τ (z)(1G ) since x and z coincide on S1 , which is a memory set for τ . On the other hand, we have τ (y)(1G ) = τ (z)(1G ) since y and z coincide on S2 , which is also a memory set for τ . It follows that τ (x)(1G ) = τ (y)(1G ). Thus there exists a map μ : AS1 ∩S2 → A such that τ (x)(1G ) = μ(xS1 ∩S2 )
for all x ∈ AG .
As τ is Gequivariant (Proposition 1.4.4), we deduce that S1 ∩S2 is a memory set for τ by using Proposition 1.4.6. Proposition 1.5.2. Let τ : AG → AG be a cellular automaton. Then there exists a unique memory set S0 ⊂ G for τ of minimal cardinality. Moreover, if S is a ﬁnite subset of G, then S is a memory set for τ if and only if S0 ⊂ S. Proof. Let S0 be a memory set for τ of minimal cardinality. As we have seen at the beginning of this section, every ﬁnite subset of G containing S0 is also a memory set for τ . Conversely, let S be a memory set for τ . As S ∩ S0 is a memory set for τ by Lemma 1.5.1, we have S ∩ S0  ≥ S0 . This implies
1.6 Cellular Automata over Quotient Groups
15
S ∩ S0 = S0 , that is, S0 ⊂ S. In particular, S0 is the unique memory set of minimal cardinality. The memory set of minimal cardinality of a cellular automaton is called its minimal memory set. Remark 1.5.3. A map F : AG → AG is constant if there exists a conﬁguration x0 ∈ AG such that F (x) = x0 for all x ∈ AG . By Gequivariance, a cellular automaton τ : AG → AG is constant if and only if there exists a ∈ A such that τ (x)(g) = a for all x ∈ AG and g ∈ G. Observe that a cellular automaton τ : AG → AG is constant if and only if its minimal memory set is the empty set.
1.6 Cellular Automata over Quotient Groups Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. Suppose that N is a normal subgroup of G and let ρ : G → G/N denote the canonical epimorphism. It follows from Proposition 1.3.7 that the map ρ∗ : AG/N → Fix(N ), deﬁned by ρ∗ (y) = y ◦ ρ for all y ∈ AG/N , is a bijection from the set AG/N of conﬁgurations over the group G/N onto the set Fix(N ) ⊂ AG of N periodic conﬁgurations over G. On the other hand, the set Fix(N ) satisﬁes τ (Fix(N )) ⊂ Fix(N ) by Corollary 1.4.5. Thus, we can deﬁne a map τ : AG/N → AG/N by setting τ = (ρ∗ )−1 ◦ τ Fix(N ) ◦ ρ∗ .
(1.12)
In other words, the map τ is obtained by conjugating by ρ∗ the restriction of τ to Fix(N ), so that the diagram ρ∗
AG/N −−−−→ Fix(N ) ⊂ AG ⏐ ⏐ ⏐ ⏐τ  τ
Fix(N ) AG/N −−−∗−→ ρ
Fix(N )
is commutative. Suppose that S ⊂ G is a memory set for τ and that μ : AS → A is the associated local deﬁning map. Consider the ﬁnite subset S = ρ(S) ⊂ G/N and the map μ : AS → A deﬁned by μ = μ ◦ π, where π : AS → AS is the injective map induced by ρ. Proposition 1.6.1. The map τ : AG/N → AG/N is a cellular automaton over the group G/N admitting S as a memory set and μ : AS → A as the associated local deﬁning map.
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1 Cellular Automata
Proof. Let y ∈ AG/N , g ∈ G, and g = ρ(g). We have τ (y)(g) = τ (y ◦ ρ)(g) = μ((g −1 (y ◦ ρ))S ) = μ((g −1 y)S ). Thus τ is a cellular automaton with memory set S and local deﬁning map μ : AS → A. Consider now the map Φ : CA(G; A) → CA(G/N ; A) given by Φ(τ ) = τ , where τ is deﬁned by (1.12). We have the following: Proposition 1.6.2. The map Φ : CA(G; A) → CA(G/N ; A) is a monoid epimorphism. Proof. Let σ : AG/N → AG/N be a cellular automaton over G/N with memory set T ⊂ G/N and local deﬁning map ν : AT → A. Let S ⊂ G be a ﬁnite set such that ρ induces a bijection φ : S → T . Consider the map μ : AS → A deﬁned by μ(y) = ν(y ◦ φ−1 ) for all y ∈ AS . Let τ : AG → AG be the cellular automaton over G with memory set S and local deﬁning map μ. We have μ(z) = (μ ◦ π)(z) = ν(π(z) ◦ φ−1 ) = ν(z) for all z ∈ AS . It follows that μ = ν and τ = σ. This shows that Φ is surjective. The fact that Φ is a monoid morphism immediately follows from (1.12). Examples 1.6.3. Let G be a group, S ⊂ G a ﬁnite subset, and N a normal subgroup of G. Denote by ρ : G → G/N the canonical epimorphism and suppose that ρ induces a bijection between S and S = ρ(S) ⊂ G/N . (a) Consider the discrete laplacian ΔS : RG → RG associated with G and S (cf. Example 1.4.3(b)). Then Φ(ΔS ) : RG/N → RG/N is the discrete laplacian associated with G/N and S. (b) Consider the majority action cellular automaton τ : {0, 1}G → {0, 1}G associated with G and S (cf. Example 1.4.3(c)). Then Φ(τ ) : {0, 1}G/N → {0, 1}G/N is the majority action cellular automaton associated with G/N and S.
1.7 Induction and Restriction of Cellular Automata Let G be a group and let A be a set. Let H be a subgroup of G. Let CA(G, H; A) denote the set consisting of all cellular automata τ : AG → G A admitting a memory set S such that S ⊂ H. Thus, CA(G, H; A) is the
1.7 Induction and Restriction of Cellular Automata
17
subset of CA(G; A) consisting of the cellular automata whose minimal memory set is contained in H. Recall that a subset N of a monoid M is called a submonoid if the identity element 1M is in N and N is stable under the monoid operation (that is, xy ∈ N for all x, y ∈ N ). If N is a submonoid of a monoid M , then the monoid operation induces by restriction a monoid structure on N . Proposition 1.7.1. The set CA(G, H; A) is a submonoid of CA(G; A). Proof. The identity element of CA(G; A) is the identity map IdAG . We have IdAG ∈ CA(G, H; A) since {1G } is a memory set for IdAG and {1G } ⊂ H. Let σ, τ ∈ CA(G, H; A). Let S (resp. T ) be a memory set for σ (resp. τ ) such that S ⊂ H (resp. T ⊂ H). It follows from Proposition 1.4.9 that ST is a memory set for σ ◦ τ . Since ST ⊂ H, this implies that σ ◦ τ ∈ CA(G, H; A). This shows that CA(G, H; A) is a submonoid of CA(G; A). Let τ ∈ CA(G, H; A). Let S be a memory set for τ such that S ⊂ H and let μ : AS → A denote the associated local deﬁning map. Then, the map τH : AH → AH deﬁned by τH (x)(h) = μ((h−1 x)S )
for all x ∈ AH , h ∈ H,
is a cellular automaton over the group H with memory set S and local deﬁning map μ. Observe that if x ∈ AG is such that x H = x, then x)(h) τH (x)(h) = τ (
for all h ∈ H.
(1.13)
This shows in particular that τH does not depend on the choice of the memory set S ⊂ H. One says that τH is the restriction of the cellular automaton τ to H. Conversely, let σ : AH → AH be a cellular automaton with memory set S and local deﬁning map μ : AS → A. Then the map σ G : AG → AG deﬁned by x)(g) = μ((g −1 x )S ) σ G (
for all x ∈ AG , g ∈ G,
is a cellular automaton over G with memory set S and local deﬁning map μ. If S0 is the minimal memory set of σ and μ0 : AS0 → A is the associated local deﬁning map then μ = μ0 ◦ π, where π : AS → AS0 is the restriction map (see Sect. 1.5). Thus, one has x)(g) = μ((g −1 x )S ) = μ0 ◦ π((g −1 x )S ) = μ0 ((g −1 x )S0 ) σ G ( for all x ∈ AG and g ∈ G. This shows in particular that σ G does not depend on the choice of the memory set S ⊂ H. One says that σ G ∈ CA(G, H; A) is the cellular automaton induced by σ ∈ CA(H; A). Proposition 1.7.2. The map τ → τH is a monoid isomorphism from CA(G, H; A) onto CA(H; A) whose inverse is the map σ → σ G .
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1 Cellular Automata
Proof. To simplify notation, denote by α : CA(G, H; A) → CA(H; A) and β : CA(H; A) → CA(G, H; A) the maps deﬁned by α(τ ) = τH and β(σ) = σ G respectively. It is clear from the deﬁnitions given above that β ◦ α and α ◦ β are the identity maps. Therefore, α is bijective with inverse β. It remains to show that α is a monoid homomorphism. ∈ AG extending x. By applying (1.13), we get Let x ∈ AH and let x x)(h) = x (h) = x(h) α(IdAG )(x)(h) = IdAG ( for all h ∈ H. This shows that α(IdAG )(x) = x for all x ∈ AH , that is, α(IdAG ) = IdAH . Let σ, τ ∈ CA(G, H; A). Let x ∈ AH and let x ∈ AG extending x. By applying (1.13) again, we have α(σ ◦ τ )(x)(h) = (σ ◦ τ )( x)(h) = σ(τ ( x))(h)
(1.14)
for all h ∈ H. On the other hand, since τ ( x) extends α(τ )(x), we have α(σ)(α(τ )(x))(h) = σ(τ ( x))(h) that is, (α(σ) ◦ α(τ ))(x)(h) = σ(τ ( x))(h)
(1.15)
for all h ∈ H. From (1.14) and (1.15), we deduce that α(σ ◦ τ )(x) = (α(σ) ◦ α(τ ))(x) for all x ∈ AH , that is, α(σ ◦ τ ) = α(σ) ◦ α(τ ). Let τ ∈ CA(G, H; A). In order to analyze the way τ transforms a conﬁguration x ∈ AG , we now introduce the set G/H = {gH : g ∈ G} consisting of all left cosets of H in G. Since the cosets c ∈ G/H form a partition of G, we have a natural identiﬁcation AG = c∈G/H Ac . With this identiﬁcation, we have x = ( xc )c∈G/H for each x ∈ AG , where x c ∈ Ac denotes the restriction of x to c. Observe now that if c ∈ G/H and g ∈ c, then τ ( x)(g) depends only on x c (this directly follows from Lemma 1.4.7 since if S is a memory set for τ with S ⊂ H, then gS ⊂ c). This implies that τ may be written as a product τc , (1.16) τ= c∈G/H
xc ) = (τ ( x))c for where τc : Ac → Ac is the unique map which satisﬁes τc ( all x ∈ AG . Note that the notation is coherent when c = H, since, in this case, τc = τH : AH → AH is the cellular automaton obtained by restriction of τ to H. Given a coset c ∈ G/H and an element g ∈ c, denote by φg : H → c the bijective map deﬁned by φg (h) = gh for all h ∈ H. Then φg induces a bijective map φ∗g : Ac → AH given by
1.7 Induction and Restriction of Cellular Automata
φ∗g (x) = x ◦ φg
19
(1.17)
for all x ∈ Ac . It turns out that the maps τc and τH are conjugate by φ∗g : Proposition 1.7.3. With the above notation, we have, τc = (φ∗g )−1 ◦ τH ◦ φ∗g .
(1.18)
In other words, the following diagram τ
Ac −−−c−→ ⏐ ⏐ φ∗ g
Ac ⏐ ⏐φ∗
g
AH −−−−→ AH τH
is commutative. Proof. Let x ∈ Ac and let x ∈ AG extending x. For all h ∈ H, we have (φ∗g ◦ τc )(x)(h) = φ∗g (τc (x))(h) = (τc (x) ◦ φg )(h) = τc (x)(gh) = τ ( x)(gh) x)(h) = g −1 τ ( = τ (g −1 x )(h),
where the last equality follows from the Gequivariance of τ (Proposition ∈ AG extends x ◦ φg ∈ AH . 1.4.4). Now observe that the conﬁguration g −1 x Thus, we have (φ∗g ◦ τc )(x)(h) = τH (x ◦ φg )(h) = τH (φ∗g (x))(h) = (τH ◦ φ∗g )(x)(h). This shows that φ∗g ◦ τc = τH ◦ φ∗g , which gives (1.18) since φ∗g is bijective. The following statement will be used in the proof of Proposition 3.2.1: Proposition 1.7.4. Let G be a group and let A be a set. Let H be a subgroup of G and let τ ∈ CA(G, H; A). Let τH : AH → AH denote the cellular automaton obtained by restriction of τ to H. Then the following hold: (i) τ is injective if and only if τH is injective; (ii) τ is surjective if and only if τH is surjective; (iii) τ is bijective if and only if τH is bijective.
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1 Cellular Automata
Proof. It immediately follows from (1.16) that τ is injective (resp. surjective, resp. bijective) if and only if τc is injective (resp. surjective, resp. bijective) for all c ∈ G/H. Now, (1.18) says that, given c ∈ G/H and g ∈ G, the map τc and τH are conjugate by the bijection φg . We deduce that τc is injective (resp. surjective, resp. bijective) if and only if τH is injective (resp. surjective, resp. bijective). Thus, τ is injective (resp. surjective, resp. bijective) if and only if τH is injective (resp. surjective, resp. bijective).
1.8 Cellular Automata with Finite Alphabets Let G be a group and let A be a ﬁnite alphabet. As a product of ﬁnite spaces is compact by Tychonoﬀ theorem (see Corollary A.5.3), it follows that AG is compact. This topological property is very useful in the study of cellular automata over ﬁnite alphabets. In particular, it may be used to prove the following: Theorem 1.8.1 (CurtisHedlund theorem). Let G be a group and let A be a ﬁnite set. Let τ : AG → AG be a map and equip AG with its prodiscrete topology. Then the following conditions are equivalent: (a) the map τ is a cellular automaton; (b) the map τ is continuous and Gequivariant. Proof. The fact that (a) implies (b) directly follows from Proposition 1.4.4 and Proposition 1.4.8 (this implication does not require the ﬁniteness assumption on the alphabet A). Conversely, suppose (b). Let us show that τ is a cellular automaton. As the map ϕ : AG → A deﬁned by ϕ(x) = τ (x)(1G ) is continuous, we can ﬁnd, for each x ∈ AG , a ﬁnite subset Ωx ⊂ G such that if y ∈ AG coincide with x on Ωx , that is, if y ∈ V (x, Ωx ), then τ (y)(1G ) = τ (x)(1G ). The sets V (x, Ωx ) form an open cover of AG . As AG is compact, there is a ﬁnite subset F ⊂ AG such that the sets V (x, Ωx ), x ∈ F , cover AG . Let us set S = ∪x∈F Ωx and suppose that two conﬁgurations y, z ∈ AG coincide on S. Let x0 ∈ F be such that y ∈ V (x0 , Ωx0 ), that is, yΩx0 = x0 Ωx0 . As S ⊃ Ωx0 we have yΩx0 = zΩx0 and therefore τ (y)(1G ) = τ (x0 )(1G ) = τ (z)(1G ). Thus there is a map μ : AS → A such that τ (x)(1G ) = μ(xS ) for all x ∈ AG . As τ is Gequivariant, it follows from Proposition 1.4.6 that τ is a cellular automaton with memory set S and local deﬁning map μ. When the alphabet A is inﬁnite, a continuous and Gequivariant map τ : AG → AG may fail to be a cellular automaton. In other words, the implication (b) ⇒ (a) in Theorem 1.8.1 becomes false if we suppress the ﬁniteness hypothesis on A. This is shown by the following example.
1.8 Cellular Automata with Finite Alphabets
21
Example 1.8.2. Let G be an arbitrary inﬁnite group and take A = G as the alphabet set. To avoid confusion, we denote by g · h the product of two elements g and h in G. Consider the map τ : AG → AG deﬁned by τ (x)(g) = x(g · x(g)) for all x ∈ AG and g ∈ G. Given x ∈ AG and g, h ∈ G we have g(τ (x))(h) = τ (x)(g −1 · h) = x(g −1 · h · x(g −1 · h)) = x(g −1 · h · [gx](h)) = gx(h · [gx](h)) = τ (gx)(h). This shows that g(τ (x)) = τ (gx) for all x ∈ AG and g ∈ G. Therefore, τ is Gequivariant. Moreover, τ is continuous. Indeed, given x ∈ AG and a ﬁnite set K ⊂ G, let us show that there exists a ﬁnite set F ⊂ G such that, if y ∈ AG and y ∈ V (x, F ), then τ (y) ∈ V (τ (x), K). Set F = K ∪ {k · x(k) : k ∈ K}. Then, if y ∈ V (x, F ), then, for all k ∈ K one has τ (x)(k) = x(k · x(k)) = y(k · x(k)) = y(k · y(k)) = τ (y)(k). This shows that τ (y) ∈ V (τ (x), K). Thus, τ is continuous. However, τ is not a cellular automaton. Indeed, ﬁx g0 ∈ G \ {1G } and, for all g ∈ G, consider the conﬁgurations xg and yg in AG deﬁned by ⎧ ⎪ if h = 1G ⎨g xg (h) = g0 if h = g ⎪ ⎩ 1G otherwise
and yg (h) =
g 1G
if h = 1G otherwise
for all h ∈ G. Note that xg G\{g} = yg G\{g} . Let F ⊂ G be a ﬁnite set and choose g ∈ G \ F (this is possible because G is inﬁnite). Then one has xg F = yg F while τ (xg )(1G ) = xg (xg (1G )) = xg (g) = g0 and τ (yg )(1G ) = yg (yg (1G )) = yg (g) = 1G ,
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so that τ (xg )(1G ) = τ (yg )(1G ). It follows that there is no ﬁnite set F ⊂ G such that, for all x ∈ AG , the value of τ (x) at 1G only depends on the values of xF . This shows that τ is not a cellular automaton (cf. Remark 1.4.2(b)).
1.9 The Prodiscrete Uniform Structure Let G be a group and let A be a set. The prodiscrete uniform structure on AG is the product uniform structure obtained by taking the discrete uniform structure on each factor A of AG = g∈G A (see Appendix B for deﬁnition and basic facts about uniform structures). A base of entourages for the prodiscrete uniform structure on AG is given by the sets WΩ ⊂ AG × AG , where WΩ = {(x, y) ∈ AG × AG : xΩ = yΩ }
(1.19)
and Ω runs over all ﬁnite subsets of G. Observe that, using the notation introduced in (1.3), we have V (x, Ω) = {y ∈ AG : (x, y) ∈ WΩ } for all x ∈ AG . The following statement gives a global characterization of cellular automata in terms of the prodiscrete uniform structure and the Gshift on AG . Theorem 1.9.1. Let A be a set and let G be a group. Let τ : AG → AG be a map and equip AG with its prodiscrete uniform structure. Then the following conditions are equivalent: (a) τ is a cellular automaton; (b) τ is uniformly continuous and Gequivariant. Proof. Suppose that τ : AG → AG is a cellular automaton. We already know that τ is Gequivariant by Proposition 1.4.4. Let us show that τ is uniformly continuous. Let S be a memory set for τ . It follows from Lemma 1.4.7 that if two conﬁgurations x, y ∈ AG coincide on gS for some g ∈ G, then τ (x)(g) = τ (y)(g). Consequently, if the conﬁgurations x and y coincide on ΩS = {gs : g ∈ Ω, s ∈ S} for some subset Ω ⊂ G, then τ (x) and τ (y) coincide on Ω. Observe that ΩS is ﬁnite whenever Ω is ﬁnite. Using the notation introduced in (1.19), we deduce that (τ × τ )(WΩS ) ⊂ WΩ for every ﬁnite subset Ω of G. As the sets WΩ , where Ω runs over all ﬁnite subsets of G, form a base of entourages for the prodiscrete uniform structure
1.9 The Prodiscrete Uniform Structure
23
on AG , it follows that τ is uniformly continuous. This shows that (a) implies (b). Conversely, suppose that τ is uniformly continuous and Gequivariant. Let us show that τ is a cellular automaton. Consider the subset Ω = {1G } ⊂ G. Since τ is uniformly continuous, there exists a ﬁnite subset S ⊂ G such that (τ ×τ )(WS ) ⊂ WΩ . This means that τ (x)(1G ) only depends on the restriction of x to S. Thus, there is a map μ : AS → A such that τ (x)(1G ) = μ(xS ) for all x ∈ AG . Using the Gequivariance of τ , we get τ (x)(g) = [g −1 τ (x)](1G ) = τ (g −1 x)(1G ) = μ((g −1 x)S ) for all x ∈ AG and g ∈ G. This shows that τ is a cellular automaton with memory set S and local deﬁning map μ. Consequently, (b) implies (a). Every uniformly continuous map between uniform spaces is continuous with respect to the associated topologies, and the converse is true when the source space is compact (Theorem B.2.3). The topology deﬁned by the prodiscrete uniform structure on AG is the prodiscrete topology (see Example (1) in Sect. B.3). In the case when A is ﬁnite, the prodiscrete topology on AG is compact by Tychonoﬀ theorem (Theorem A.5.2). Thus Theorem 1.9.1 reduces to the CurtisHedlund theorem (Theorem 1.8.1) in this case. Remark 1.9.2. Suppose that G is countable and A is an arbitrary set. Then the prodiscrete uniform structure (and hence the prodiscrete topology) on AG is metrizable. To see this, choose an increasing sequence ∅ = E 0 ⊂ E 1 ⊂ · · · ⊂ En ⊂ · · · of ﬁnite subsets of G such that n≥0 En = G. Then the sets WEn , n ≥ 0, form a base of entourages for the prodiscrete uniform structure on AG . Consider now the metric d on AG deﬁned by 0 if x = y, d(x, y) = − max{n≥0: xEn =yEn } 2 if x = y. for all x, y ∈ AG . Then we have WEn = {(x, y) ∈ AG × AG : d(x, y) < 2−n+1 } for every n ≥ 0. Consequently, d deﬁnes the prodiscrete uniform structure on AG . Let G be a group and let A be a set. Let H be a subgroup of G. Let us equip AH\G with its prodiscrete uniform structure and Fix(H) ⊂ AG with the
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uniform structure induced by the prodiscrete uniform structure on AG . Recall from Proposition 1.3.7 that there is a natural bijection ρ∗ : AH\G → AG deﬁned by ρ∗ (y) = y ◦ ρ, where ρ : G → H\G is the canonical surjection. Proposition 1.9.3. The map ρ∗ : AH\G → Fix(H) is a uniform isomorphism. Proof. For g ∈ G, let πg : AG → A and πg : AH\G → A denote the projection maps given by x → x(g) and y → y(ρ(g)) respectively. Observe that πg ◦ ρ∗ = πg is uniformly continuous for all g ∈ G. This shows that ρ∗ is uniformly continuous. Similarly, the uniform continuity of (ρ∗ )−1 follows from the fact that πg ◦ (ρ∗ )−1 = πg Fix(H) is uniformly continuous for each g ∈ G. Consequently, ρ∗ is a uniform isomorphism.
1.10 Invertible Cellular Automata Let G be a group and let A be a set. One says that a cellular automaton τ : AG → AG is invertible (or reversible) if τ is bijective and the inverse map τ −1 : AG → AG is also a cellular automaton. This is equivalent to the existence of a cellular automaton σ : AG → AG such that τ ◦ σ = σ ◦ τ = IdAG . Thus, the set of invertible cellular automata over the group G and the alphabet A is exactly the group ICA(G; A) consisting of all invertible elements of the monoid CA(G; A). Theorem 1.10.1. Let A be a set and let G be a group. Let τ : AG → AG be a map and equip AG with its prodiscrete uniform structure. Then the following conditions are equivalent: (a) τ is an invertible cellular automaton; (b) τ is a Gequivariant uniform automorphism of AG . Proof. It is clear that the inverse map of a bijective Gequivariant map from AG onto itself is also Gequivariant. Therefore, the equivalence of conditions (a) and (b) follows from the characterization of cellular automata given in Theorem 1.9.1. Bijective cellular automata over ﬁnite alphabets are always invertible: Theorem 1.10.2. Let G be a group and let A be a ﬁnite set. Then every bijective cellular automaton τ : AG → AG is invertible. Proof. Let τ : AG → AG be a bijective cellular automaton. The map τ −1 is Gequivariant since τ is Gequivariant. On the other hand, τ −1 is continuous with respect to the prodiscrete topology by compactness of AG . Consequently, τ −1 is a cellular automaton by Theorem 1.8.1.
1.10 Invertible Cellular Automata
25
The following example shows that Theorem 1.10.2 becomes false if we omit the ﬁniteness hypothesis on the alphabet set A. Example 1.10.3. Let K be a ﬁeld. Let us take as the alphabet set the ring A = K[[t]] of all formal power series in one indeterminate t with coeﬃcients in K. Thus, an element of A is just a sequence a = (ki )i∈N of elements of K written in the form ki ti , a = k0 + k1 t + k2 t2 + k3 t3 + · · · = i∈N
i and the addition and multiplication of two elements a = i∈N ki t and b = i i i respectively given by a+b = (k +k )t and ab = i i i∈N ki t are i∈N i∈N ki t with ki = i1 +i2 =i ki1 ki2 for all i ∈ N. We take G = Z. Thus, a conﬁguration x ∈ AG is a map x : Z → K[[t]]. Consider the map τ : AG → AG deﬁned by τ (x)(n) = x(n) − tx(n + 1) for all x ∈ AG and n ∈ Z. Clearly τ is a cellular automaton admitting S = {0, 1} as a memory set (the local deﬁning map associated with S is the map μ : AS → A deﬁned by μ(x0 , x1 ) = x0 − tx1 for all x0 , x1 ∈ A). Let us show that τ is bijective. Consider the map σ : AG → AG given by σ(x)(n) = x(n) + tx(n + 1) + t2 x(n + 2) + t3 x(n + 3) + · · · for all x ∈ AG and n ∈ Z. Observe that σ(x)(n) ∈ K[[t]] is well deﬁned by the preceding formula. In fact, if we develop x(n) ∈ K[[t]] in the form x(n) = xn,i ti (n ∈ Z, xn,i ∈ K), i∈N
then σ(x)(n) =
i∈N
⎛ ⎞ i ⎝ xn+j,i−j ⎠ ti . j=0
One immediately checks that σ ◦ τ = τ ◦ σ = IdAG . Therefore, τ is bijective with inverse map τ −1 = σ. Let us show that the map σ : AG → AG is not a cellular automaton. Let F be a ﬁnite subset of Z and choose an integer M ≥ 0 such that F ⊂ (−∞, M ]. Consider the conﬁguration y deﬁned by y(n) = 0 if n ≤ M and y(n) = 1 if n ≥ M + 1, and the conﬁguration z deﬁned by z(n) = 0 for all n ∈ Z. Then y and z coincide on F . However, the value at 0 of σ(y) is σ(y)(0) = tM +1 + tM +2 + tM +3 + · · · while the value of σ(z) at 0 is σ(z)(0) = 0. It follows that there is no ﬁnite subset F ⊂ Z such that σ(x)(0) only depends on the restriction of x ∈ AG
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to F . This shows that σ is not a cellular automaton. Consequently, τ is a bijective cellular automaton which is not invertible. In the next proposition we show that invertibility is preserved under the operations of induction and restriction. Proposition 1.10.4. Let G be a group and let A be a set. Let H be a subgroup of G and let τ ∈ CA(G, H; A). Let τH ∈ CA(H; A) denote the cellular automaton obtained by restriction of τ to H. Then the following conditions are equivalent: (a) τ is invertible; (b) τH is invertible. Moreover, if τ is invertible, then τ −1 ∈ CA(G, H; A) and one has (τ −1 )H = (τH )−1 .
(1.20)
Proof. First recall from (1.16) the factorizations AG = Ac and τ = τc , c∈G/H
(1.21)
c∈G/H
where τc : Ac → Ac satisﬁes τc ( xc ) = (τ ( x))c for all x ∈ AG . Suppose that τ is invertible. Denote by σ ∈ CA(G; A) the inverse cellular automaton τ −1 . It follows from (1.21) that the map τc : Ac → Ac is bijective for each c ∈ G/H and that σ= (τc )−1 , (1.22) c∈G/H
where (τc )−1 : Ac → Ac is the inverse map of τc . Let us show that σ ∈ CA(G, H; A). Let S ⊂ G be a memory set for σ. Let x ∈ AG . It follows from −1 xH ). Thus, we have (1.22) that (σ( x))H = (τH ) ( x))H (1G ) = (τH )−1 ( xH )(1G ). σ( x)(1G ) = (σ( H . Arguing as in the proof of This shows that σ( x)(1G ) only depends on x Lemma 1.5.1, we deduce that S ∩ H is a memory set for σ. Indeed, suppose that two conﬁgurations x , y ∈ AG coincide on S ∩ H. Consider the G on S and with y on G \ S. We conﬁguration z ∈ A which coincide with x z )(1G ) since x and z coincide on S. On the other hand, have σ( x)(1G ) = σ( we have σ( y )(1G ) = σ( z )(1G ) since y and z coincide on H. This implies y )(1G ). Thus, there is a map μ : AS∩H → A such that σ( x)(1G ) = σ( xS∩H ) σ( x)(1G ) = μ( for all x ∈ AG . By applying Proposition 1.4.6, it follows that S ∩ H is a memory set for σ. Since S ∩ H ⊂ H, this shows that τ −1 = σ ∈ CA(G, H; A).
Notes
27
Moreover, it follows from (1.22) that (τ −1 )H = σH = (τH )−1 which gives us (1.20). The equivalence (a) ⇔ (b) is then an immediate consequence of Proposition 1.7.2 which tells us that the restriction map CA(G, H; A) → CA(H; A) is a monoid isomorphism.
Notes Cellular automata were introduced by J. von Neumann (see [vNeu2]) who used them to describe theoretical models of selfreproducing machines. He ﬁrst attempted to get such models by means of partial diﬀerential equations in R3 . Later he changed the perspective and tried to use ideas and methods coming from robotics and electrical engineering. Eventually, in 1952, following a suggestion of S. Ulam, his former colleague at the Los Alamos Laboratories, he constructed a cellular automaton over the group Z2 with an alphabet consisting of 29 states. He then outlined the construction of a pattern, containing approximatively 200,000 cells, which would reproduce itself. The details were later ﬁlled in by A.W. Burks in the 1960s [Bur]. The branch of mathematics which is concerned with the study of the dynamical properties of the shift action is known as symbolic dynamics. Many authors trace the birth of symbolic dynamics back to a paper published in 1898 by J. Hadamard [Had] in which words on two letters were used to code geodesics on certain surfaces with negative curvature. However, as it was pointed out by E.M. Coven and Z.W. Nitecki [CovN], Hadamard’s symbolic description of geodesics is purely static and involves only ﬁnite words. According to the authors of [CovN], the beginning of symbolic dynamics should be placed in a paper by G.A. Hedlund [Hed1] published in 1944. Symbolic dynamics has important applications in dynamical systems, especially in the study of hyperbolic dynamical systems for which symbolic codings may be obtained from Markov partitions. One of the ﬁrst examples of such an application was the use of the properties of the ThueMorse sequence (see Exercise 3.41) by M. Morse [Mors] in 1921 to prove the existence of nonperiodic recurrent geodesics on surfaces with negative curvature. A detailed exposition of symbolic dynamics over Z may be found for example in the books by B. Kitchens [Kit], by P. K˚ urka [Kur], and by D. Lind and B. Marcus [LiM]. In the mid1950s, Hedlund studied the socalled shiftcommuting block maps which turn out to be exactly cellular automata over the group Z. The CurtisHedlund theorem (Theorem 1.8.1), also called CurtisHedlundLyndon’s theorem or Hedlund’s theorem, is named after Hedlund [Hed3]
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who proved it in 1969. Its generalization to inﬁnite alphabets, as stated as in Theorem 1.9.1, was proved by the authors in [CeC7]. Cellular automata were intensely studied from the 1960s, both by pure and applied mathematicians, under diﬀerent names such as tessellation automata, parallel maps, cellular spaces, iterative automata, homogeneous structures, universal spaces, and sliding block codes (cf. [LiM, Section 1.1]). In most cases, these researches focused on cellular automata with ﬁnite alphabet over the groups Z or Z2 (see the surveys [BanMS], [BKM], [Kar3], [Wolfr3]). The Game of Life was invented by the British mathematician J.H. Conway. This cellular automaton was described for the ﬁrst time by M. Gardner [Gar1] in the October 1970 issue of the Scientiﬁc American. From a theoretical computer science point of view, it is important because it has the power of a universal Turing machine, that is, anything that can be computed algorithmically can be computed by using the Game of Life. In the 1980s, S. Wolfram [Wolfr1], [Wolfr2] started a systematic study and empirical classiﬁcation of elementary cellular automata, that is, of cellular automata over Z with alphabet A = {0, 1} and memory set S = {−1, 0, 1}. 3 There are 2(2 ) = 256 such elementary cellular automata. Wolfram introduced a naming scheme for them which is nowadays widely used. Each elementary cellular automaton τ : AZ → AZ is uniquely determined by the eight bit sequence μ(111)μ(110)μ(101)μ(100)μ(011)μ(010)μ(001)μ(000) ∈ A8 , where μ : AS → A is the associated local deﬁning map. This bit sequence is the binary expansion of an integer in the interval [0, 255], called the Wolfram number of τ . For example, the majority action τ : AZ → AZ associated with S = {−1, 0, 1} (cf. Example 1.4.3(c)) is an elementary cellular automaton. Its local deﬁning map μ gives μ(111)μ(110)μ(101)μ(100)μ(011)μ(010)μ(001)μ(000) = 11101000 (cf. Fig. 1.7). It follows that the Wolfram number of τ is 232. Let us also mention that the elementary cellular automaton with Wolfram number 110 was recently proved computationally universal by M. Cook. Wolfram introduced an empirical classiﬁcation of elementary cellular automata into four classes according to the behavior of random initial conﬁgurations under iterations. These are known as Wolfram classes and are deﬁned as follows: (W1) Almost all initial conﬁgurations lead to the same uniform ﬁxedpoint conﬁguration, (W2) Almost all initial conﬁgurations lead to a periodic conﬁguration, (W3) Almost all initial conﬁgurations lead to chaos, (W4) Localized structures with complex interactions emerge. The survey paper [BKM] contains other dynamical classiﬁcations of cellular automata over Z with ﬁnite alphabet due to R. Gilman and to M. Hurley and P. K˚ urka.
Exercises
29
Invertible cellular automata are used to model timereversible processes occurring in physics and biology. A group G is called periodic if every element g ∈ G has ﬁnite order. In [CeC11] it is shown that if G is a nonperiodic group, then for every inﬁnite set A there exists a bijective cellular automaton τ : AG → AG which is not invertible (cf. Theorem 1.10.2 and Example 1.10.3). It was shown by Amoroso and Patt in 1972 [Amo] that it is decidable whether a given cellular automaton with ﬁnite alphabet over Z is invertible. This means that there exists an algorithm which establishes, after a ﬁnite number of steps, whether the cellular automaton corresponding to a given local deﬁning map is invertible or not. On the other hand, J. Kari [Kar1], [Kar2], [Kar3] proved that the similar problem for cellular automata with ﬁnite alphabet over Zd , d ≥ 2, is undecidable. Its proof is based on R. Berger’s undecidability result for the tiling problem of Wang tiles.
Exercises 1.1. An action of a group Γ on a topological space X is said to be topologically mixing if for each pair of nonempty subsets U and V of X there exists a ﬁnite subset F ⊂ Γ such that U ∩ γV = ∅ for all γ ∈ Γ \ F . Show that if G is a group and A is a set then the Gshift on AG is topologically mixing for the prodiscrete topology on AG . 1.2. Let G be a group and let A be a set. Let x ∈ AG and let Ω1 and Ω2 be two subsets of G. Show that V (x, Ω1 ∪ Ω2 ) = V (x, Ω1 ) ∩ V (x, Ω2 ) and WΩ1 ∪Ω2 = WΩ1 ∩ WΩ2 (see (1.3) and (1.19) for the deﬁnition of V (x, Ω) and WΩ ). 1.3. Let G be a countable group and let A be a set. Show that the metric d on AG introduced in Remark 1.9.2 is complete. 1.4. Let G be an uncountable group and let A be a set having at least two elements. Prove that the prodiscrete topology on AG is not metrizable. Hint: Prove that this topology does not satisfy the ﬁrst axiom of countability. 1.5. Let G be a group. Let A and B be two sets. Let τA : AG → AG and τB : B G → B G be cellular automata. For x ∈ (A × B)G , let xA ∈ AG and xB ∈ B G be the conﬁgurations deﬁned by x(g) = (xA (g), xB (g)) for all g ∈ G. Show that the map τ : (A × B)G → (A × B)G given by τ (x)(g) = (τA (xA )(g), τB (xB )(g)) for all g ∈ G is a cellular automaton. 1.6. Let G be a group and let S be a ﬁnite subset of G of cardinality k. Let k A be a ﬁnite set of cardinality n. Show that there are exactly nn cellular automata τ : AG → AG admitting S as a memory set.
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1.7. Let G = Z2 and A = {0, 1}. Let τ : AG → AG denote the cellular automaton associated with the Game of Life. Let y ∈ AG be the constant conﬁguration deﬁned by y(g) = 1 for all g ∈ G (all cells are alive). Find a conﬁguration x ∈ AG such that y = τ (x). 1.8. Let A be a set and suppose that G is a trivial group. Show that the monoid CA(G; A) is canonically isomorphic to the monoid consisting of all maps from A to A (with composition of maps as the monoid operation). Also show that the group ICA(G; A) is canonically isomorphic to the symmetric group of A. 1.9. Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. Show that τ admits a memory set which is reduced to a single element if and only if there exist an element s ∈ G and a map f : A → A such that one has τ (x)(g) = f (x(gs)) for all x ∈ AG and g ∈ G. 1.10. Prove that there are exactly 218 cellular automata τ : {0, 1}Z → {0, 1}Z whose minimal memory set is {−1, 0, 1}. 1.11. Let τ ∈ CA(Z2 ; {0, 1}) denote the cellular automaton associated with the Game of Life. Show that the minimal memory set of τ is the set {−1, 0, 1}2 . 1.12. Let G be a group and let A be a set. Let σ, τ ∈ CA(G; A). Let S0 (resp. T0 , resp. C0 ) denote the minimal memory set of σ (resp. τ , resp. σ ◦ τ ). Prove that C0 ⊂ S0 T0 . Give an example showing that this inclusion may be strict. 1.13. Let G be a group and let A be a set. Let H be a subgroup of G and let τ ∈ CA(G, H; A). Show that τ and τH have the same minimal memory set. 1.14. Prove Proposition 1.4.9 by applying Theorem 1.9.1. Hint: Observe that the composite of two uniformly continuous maps is a uniformly continuous map. 1.15. Let G be a group and let A be a set. Let F be a nonempty ﬁnite subset of G and set B = AF . The sets AG and B G are equipped with their prodiscrete uniform structures and with the Gshift action. Show that the map ΦF : AG → B G deﬁned by ΦF (x)(g) = (g −1 x)F for all x ∈ AG and g ∈ G is a Gequivariant uniform embedding. 1.16. Let G be a group, H ⊂ G a subgroup of G, and let A be a set. Let H\G = {Hg : g ∈ G} be the set of all right cosets of H in G and set B = AH\G . The set AG (resp. B H ) is equipped with its prodiscrete uniform structure and with the Gshift (resp. Hshift) action. Let T ⊂ G be a complete set of representatives for the right cosets of H in G so that G = t∈T Ht. Show that the map Ψ = Ψ (H, T ) : AG → B H deﬁned by Ψ (x)(h)(Ht) = x(ht) for all x ∈ AG , h ∈ H and t ∈ T is an Hequivariant uniform isomorphism.
Exercises
31
1.17. Let G be a group and let A be a set. For each s ∈ G, let τs : AG → AG be the cellular automaton deﬁned by τs (x)(g) = x(gs) for all x ∈ AG , g ∈ G (cf. Example 1.4.3(e)). (a) Show that τs ∈ ICA(G; A) for every s ∈ G. (b) Prove that the map Φ : G → ICA(G; A) deﬁned by φ(s) = τs for all s ∈ G is a group homomorphism. (c) Prove that if A has at least two elements, then Φ is injective but not surjective. 1.18. Let G be a group and let A be a set. Prove that the set consisting of all invertible cellular automata τ : AG → AG admitting a memory set which is reduced to a single element is a subgroup of ICA(G; A) isomorphic to the direct product G × Sym(A). 1.19. (cf. [Amo]) Let G = Z and A = {0, 1}. Fix an integer n ≥ 3 and let S = {−1, 0, 1, . . . , n}. Consider the element α ∈ AS (resp. β ∈ AS ) deﬁned by α(−1) = α(n) = 0 and α(k) = 1 for 0 ≤ k ≤ n − 1 (resp. β(−1) = β(0) = β(n) = 0 and β(k) = 1 for 1 ≤ k ≤ n − 1) and the map μ : AS → A deﬁned by μ(α) = 0, μ(β) = 1 and μ(y) = y(0) for y ∈ AS \{α, β}. Let τ : AG → AG be the cellular automaton with memory set S and local deﬁning map μ. (a) Show that S is the minimal memory set of τ . (b) Show that τ is an invertible cellular automaton and that τ −1 = τ . 1.20. Show that the inverse map of the bijective cellular automaton τ : AZ → AZ studied in Example 1.10.3 is discontinuous, with respect to the prodiscrete topology on AZ , at every conﬁguration x ∈ AZ . 1.21. Let G be a group and let A be a set. A subshift of the conﬁguration space AG is a subset X ⊂ AG which is Ginvariant (i.e., such that gx ∈ X for all x ∈ X and g ∈ G) and closed in AG with respect to the prodiscrete topology. (a) Show that ∅ and AG are subshifts of AG . (b) Show that if x ∈ AG then its orbit closure Gx ⊂ AG is a subshift. (c) Show that if (Xi )i∈I is a family of subshifts of AG then i∈I Xi is a subshift of AG . (d) Show that if (Xi )i∈I is a ﬁnite family of subshifts of AG then i∈I Xi is a subshift of AG . (e) Suppose that A is ﬁnite. Show that if X ⊂ AG is a subshift and τ : AG → AG is a cellular automaton then τ (X) is a subshift of AG . Note: This last statement becomes false when A is inﬁnite (see Example 3.3.3). 1.22. Let G be a group and let A and B be two sets. Let f : A → B be a map and consider the map f∗ : AG → B G deﬁned by f∗ (x) = f ◦ x for all x ∈ AG . (a) Show that if A is ﬁnite and X is a subshift of AG then f∗ (X) is a subshift of B G . Hint: Use the compactness of the conﬁguration space AG .
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(b) Let G = Z, A = Z, B = {0, 1} and let f : A → B be deﬁned by f (n) = 0 if n < 0 and f (n) = 1 otherwise. Let X = {xn : n ∈ Z} ⊂ AZ , where xn (m) = n + m for all n, m ∈ Z. In other words, X is the Zorbit Zx0 of the conﬁgurations x0 . Show that X is a subshift of AZ but f ∗ (X) is not a subshift of B Z . 1.23. Let G be a group and let A be a set. Given a set of patterns P ⊂ AΩ , where the union runs over all ﬁnite subsets Ω of G, we set / P for all g ∈ G and all ﬁnite subsets Ω ⊂ G}, XP = {x ∈ AG : (gx)Ω ∈ the restriction of the conﬁguration gx to Ω. where (gx)Ω denotes (a) Let P ⊂ AΩ be a set of patterns. Show that XP is a subshift of AG . G (b) Conversely, Ω show that if X ⊂ A is a subshift, then there exists a subset P ⊂ A such that X = XP . Such a set P is called a deﬁning set of forbidden patterns for X. 1.24. Let G be a group and let A be a set. Given a ﬁnite subset Ω ⊂ G and a ﬁnite subset A ⊂ AΩ we set X(Ω, A) = {x ∈ AG : (gx)Ω ∈ A for all g ∈ G}. (a) Let Ω ⊂ G and A ⊂ AΩ be ﬁnite subsets. Show that X(Ω, A) is a subshift of AG . A subshift X ⊂ AG is said to be of ﬁnite type if there exists a ﬁnite subset Ω ⊂ G and a ﬁnite subset A ⊂ AΩ such that X = X(Ω, A). Such a set A is then called a deﬁning set of admissible patterns for X and the subset Ω is called a memory set for X. (b) Suppose that A is ﬁnite. Show that a subshift X ⊂ AG is of ﬁnite type if and only if it admits a ﬁnite deﬁning set of forbidden patterns. 1.25. Let G be a countable group and let A be a ﬁnite set. Show that there are at most countably many distinct subshifts X ⊂ AG of ﬁnite type. 1.26. Let G be a group and let A be a ﬁnite set. (a) Let τ1 , τ2 : AG → AG be two cellular automata. Show that the set {x ∈ AG : τ1 (x) = τ2 (x)} ⊂ AG is a subshift of ﬁnite type. (b) Deduce from (a) that if τ : AG → AG is a cellular automaton then the set Fix(τ ) = {x ∈ AG : τ (x) = x} ⊂ AG is a subshift of ﬁnite type. (c) Conversely, show that if X ⊂ AG is a subshift of ﬁnite type then there exists a cellular automaton τ : AG → AG such that X = Fix(τ ). 1.27. Suppose that a group Γ acts continuously on a topological space Z. One says that the action of Γ on Z is topologically transitive if for any pair of nonempty open subsets U and V of Z there exists an element γ ∈ Γ such that U ∩ γV = ∅. Let G be a group and let A be a set. A subshift X ⊂ AG is said to be irreducible if for any ﬁnite subset Ω of G and any two elements x1 , x2 ∈ X,
Exercises
33
there exist a conﬁguration x ∈ X and an element g ∈ G such that xΩ = x1 Ω and (gx)Ω = x2 Ω . Suppose that X ⊂ AG is a subshift. Show that the action of G on X induced by the Gshift is topologically transitive if and only if X is irreducible. 1.28. Let G be a group and let A be a set. Let B ⊂ A and consider the subsets X, Y ⊂ AG deﬁned by X = {b : b ∈ B} ⊂ AG , where b denotes the constant conﬁguration given by b(g) = b for all g ∈ G, and Y = {y ∈ AG : y(g) ∈ B for all g ∈ G}. (a) Show that X and Y are subshifts of AG . (b) Show that if G is inﬁnite then Y is irreducible. (c) Suppose that B has at least two distinct elements. Show that X is not irreducible. 1.29. Let G be a group acting continuously on a nonempty complete metric space X whose topology satisﬁes the second axiom of countability (i.e., admitting a countable base of open subsets). Show that the following conditions are equivalent: (i) the action of G on X is topologically transitive; (ii) there is a point x ∈ X whose Gorbit is dense in X; (iii) there is a dense subset D ⊂ X such that the Gorbit of each point x ∈ D is dense in X. Hint: The implications (iii) ⇒ (ii) and (ii) ⇒ (i) are straightforward. To prove (i) ⇒ (iii), consider a sequence (Un )n∈N of nonempty open subsets of X which form a base of the topology and denote by Ωn the set of points x ∈ X whose Gorbit meets Un . Then observe that each Ωn is and open dense subset of X if (i) is satisﬁed and apply Baire’s theorem (Theorem I.1.1, also cf. Remark I.1.2(ii)). 1.30. Let G be a countable group and let A be a countable (e.g. ﬁnite) set. Let X ⊂ AG be a nonempty subshift. Show that the following conditions are equivalent: (i) the subshift X is irreducible; (ii) there is a conﬁguration x ∈ X whose Gorbit is dense in X; (iii) there is a dense subset D ⊂ X such that the Gorbit of each conﬁguration x ∈ D is dense in X. Hint: Use the results of Exercises 1.3 and 1.29. 1.31. Let G be a group and let A be a set. One says that a subshift X ⊂ AG is topologically mixing if the action of G on X induced by the Gshift is topologically mixing (cf. Exercise 1.1). (a) Let X ⊂ AG be a subshift. Show that X is topologically mixing if and only if for any ﬁnite subset Ω of G and any two conﬁgurations x1 , x2 ∈ X, there exists a ﬁnite subset F ⊂ G such that, for all g ∈ G \ F , there exists a conﬁguration x ∈ X satisfying xΩ = x1 Ω and (gx)Ω = x2 Ω . (b) Show that if G is inﬁnite then every topologically mixing subshift X ⊂ AG is irreducible.
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1.32. Let G be a group and let A be a set. Let Δ ⊂ G be a ﬁnite subset. A subshift X ⊂ AG is said to be Δirreducible if it satisﬁes the following condition: if Ω1 and Ω2 are two ﬁnite subsets of G such that Ω1 and Ω2 δ are disjoint for all δ ∈ Δ, then, given any two conﬁgurations x1 , x2 ∈ X, there exists a conﬁguration x ∈ X which satisﬁes xΩ1 = x1 Ω1 and xΩ2 = x2 Ω2 . A subshift X ⊂ AG is said to be strongly irreducible if there exists a ﬁnite subset Δ ⊂ G such that X is Δirreducible. (a) Show that the subshift Y ⊂ AG described in Exercise 1.28 is {1G }irreducible and therefore strongly irreducible. (b) Show that every strongly irreducible subshift is topologically mixing. 1.33. Let G be a group and let A be a set. Let H be a subgroup of G. H the conﬁguration deﬁned by For x ∈ AG and g ∈ G, denote by xH g ∈ A H xg = (gx)H . H G (a) Check that hxH g = xhg for all x ∈ A , h ∈ H and g ∈ G. (b) Let X ⊂ AH be a subshift. Show that the set X (G) deﬁned by X (G) = G {x ∈ AG : xH g ∈ X for all g ∈ G} is a subshift of A . (G) (c) Show that if H = G then the subshift X ⊂ AG is irreducible for H any subshift X ⊂ A . (d) Let X ⊂ AH be a subshift. Show that X (G) ⊂ AG is of ﬁnite type (resp. topologically mixing, resp. strongly irreducible) if and only if X is of ﬁnite type (resp. topologically mixing, resp. strongly irreducible). (e) Suppose that σ : AH → AH is a cellular automaton and let σ G : AG → G A be the induced cellular automaton (cf. Sect. 1.7). Check that one has H G σ G (x)H g = σ(xg ) for all x ∈ A and g ∈ G. H (f) Show that if X ⊂ A is a subshift such that σ(X) ⊂ X then one has σ G (X (G) ) ⊂ X (G) . 1.34. Let G be a group and let A be a set. Let F be a nonempty ﬁnite subset of G and consider the map ΦF : AG → B G deﬁned in Exercise 1.15, where B = AF . Let X ⊂ AG be a subshift and set X [F ] = ΦF (X). (a) Show that X [F ] is a subshift of B G . (b) Show that X is irreducible (resp. topologically mixing, resp. strongly irreducible) if and only if X [F ] is irreducible (resp. topologically mixing, resp. strongly irreducible). (c) Show that if X is of ﬁnite type then X [F ] is of ﬁnite type. 1.35. Let G be a group, H ⊂ G a subgroup of G, and let A be a set. Let also T ⊂ G be a complete set of representatives for the right cosets of H in G and consider the map Ψ : AG → B H deﬁned in Exercise 1.16, where B = AH\G . Let X ⊂ AG be a subshift and set X (H,T ) = Ψ (X). (a) Show that X (H,T ) is a subshift of B H . (b) Show that if X is of ﬁnite type then X (H,T ) is of ﬁnite type. 1.36. Let A be a set. Let A∗ denote the monoid consisting of all words in the alphabet A (cf. Sect. D.1). Recall that any word w ∈ A∗ can be uniquely
Exercises
35
written in the form w = a1 a2 · · · an , where n ≥ 0 and ai ∈ A for 1 ≤ i ≤ n. The integer n is called the length of the word w and it is denoted by (w). In the sequel, we shall identify the word w = a1 a2 · · · an with the pattern p : {1, 2, . . . , n} → A deﬁned by p(i) = ai for 1 ≤ i ≤ n. Given a subshift X ⊂ AZ and an integer n ≥ 0, we denote by Ln (X) ⊂ A∗ the set consisting of all words w ∈ A∗ for which there exists an element x ∈ X such that w = x(1)x(2) · · · x(n). The set L(X) = ∪n∈N Ln (X) is called the language of X. The elements w ∈ L(X) are called the admissible words of X (or, simply, the Xadmissible words). The elements w ∈ A∗ \ L(X) are called the forbidden words of X. (a) Let X and Y be two subshifts of AZ . Show that one has X ⊂ Y (resp. X = Y ) if and only if L(X) ⊂ L(Y ) (resp. L(X) = L(Y )). (b) One says that a word u ∈ A∗ is a subword of a word w ∈ A∗ if there exist v1 , v2 ∈ A∗ such that w = v1 uv2 . Let X ⊂ AZ be a subshift and let L = L(X). Show that L satisﬁes the following conditions: (i) if w ∈ L, then u ∈ L for every subword u of w; (ii) if w ∈ L, then there exist a, a ∈ A such that awa ∈ L. (c) Conversely, show that if a subset L ⊂ A∗ satisﬁes conditions (i) and (ii) in (b), then there exists a unique subshift X ⊂ AZ such that L = L(X). 1.37. Let A be a set and let X ⊂ AZ be a subshift. (a) Show that X is of ﬁnite type if and only if the following holds: there exists an integer n0 ≥ 0 such that if the words u, v, w ∈ A∗ satisfy (v) ≥ n0 and uv, vw ∈ L(X), then one has uvw ∈ L(X). (b) Show that X is irreducible if and only if for every pair of words u and v in L(X), there exists a word w ∈ A∗ such that uwv ∈ L(X). (c) Show that X is topologically mixing if and only if the following holds: for every pair of words u and v in L(X), there exists an integer n0 ≥ 0 such that for every integer n ≥ n0 there exists a word w ∈ A∗ of length (w) = n satisfying uwv ∈ L(X). (d) Show that X is strongly irreducible if and only if the following holds: there exists an integer n0 ≥ 0 such that, for every pair of words u and v in L(X), and for every integer n ≥ n0 , there exists a word w ∈ A∗ of length (w) = n such that uwv ∈ L(X). 1.38. Let A = {0, 1} and let X ⊂ AZ be the set of all x ∈ AZ such that the following holds: if x(n) = 1, x(n + 1) = x(n + 2) = · · · = x(n + k) = 0, x(n + k + 1) = 1, for some n ∈ Z and k ∈ N, then k is even. (a) Show that X is a subshift (it is called the even subshift). (b) Show that X is not of ﬁnite type. (c) Show that X is strongly irreducible (and therefore topologically mixing and irreducible). 1.39. Let A = {0, 1} and consider the subshift of ﬁnite type X ⊂ AZ deﬁned by X = X{11} = {x ∈ AZ : (x(n), x(n + 1)) = (1, 1) for all n ∈ Z}. Show that X is strongly irreducible (and therefore topologically mixing and irreducible). The subshift X is called the golden mean subshift.
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1.40. Let A = {0, 1} and let X ⊂ AZ be the set consisting of the two conﬁgurations x, y ∈ AZ deﬁned by 0 if n is even 1 if n is even and y(n) = x(n) = 1 otherwise 0 otherwise, for all n ∈ Z. Show that X is an irreducible subshift of ﬁnite type which is not topologically mixing (and therefore not strongly irreducible either). 1.41. Let A be a set. Let X ⊂ AZ be a subshift of ﬁnite type. Show that X is topologically mixing if and only if X is strongly irreducible. 1.42. Let A = {0, 1} and let X ⊂ AZ be the subshift with deﬁning set of forbidden words {01k 0h 1 : 1 ≤ h ≤ k, k = 1, 2, . . .}. Show that X is topologically mixing (and therefore irreducible) but not strongly irreducible. 1.43. Cellular automata between subshifts. Let G be a group and let A be a set. The set AG is equipped with its prodiscrete uniform structure and with the Gshift action. Let X, Y ⊂ AG be two subshifts and τ : X → Y a map. Then the following are equivalent: (i) there exists a cellular automaton τ : AG → AG such that τ (x) = τ (x) for all x ∈ X; (ii) τ is Gequivariant and uniformly continuous. One says that τ : X → Y is a cellular automaton if the two equivalent conditions above are satisﬁed. 1.44. Let G be a group and let A be a ﬁnite set. Let τ : X → Y be a map between subshifts X, Y ⊂ AG . Show that the following conditions are equivalent: (i) τ is a cellular automaton, (ii) τ is Gequivariant and continuous (with respect to the topologies induced on X and Y by the prodiscrete topology on AG ). 1.45. Let G be a group and let A be a ﬁnite set. Let τ : AG → AG be a cellular automaton and let X ⊂ AG be an irreducible (resp. topologically mixing, resp. strongly irreducible) subshift. Show that τ (X) is an irreducible (resp. topologically mixing, resp. strongly irreducible) subshift of AG .
Chapter 2
Residually Finite Groups
This chapter is devoted to the study of residually ﬁnite groups, which form a class of groups of special importance in several branches of mathematics. As their name suggests it, residually ﬁnite groups generalize ﬁnite groups. They are deﬁned as being the groups whose elements can be distinguished after taking ﬁnite quotients (see Sect. 2.1). There are many other equivalent deﬁnitions. For example, a group is residually ﬁnite if and only if it can be embedded into the direct product of a family of ﬁnite groups (Corollary 2.2.6). The class of residually ﬁnite groups is closed under taking subgroups and taking projective limits. It contains in particular all ﬁnite groups, all ﬁnitely generated abelian groups, and all free groups (Theorem 2.3.1). Every ﬁnitely generated residually ﬁnite group is Hopﬁan (Theorem 2.4.3) and the automorphism group of a ﬁnitely generated residually ﬁnite group is itself residually ﬁnite (Theorem 2.5.1). Examples of ﬁnitely generated groups which are not residually ﬁnite are presented in Sect. 2.6. The following dynamical characterization of residually ﬁnite groups is given in Sect. 2.7: a group is residually ﬁnite if and only if there is a Hausdorﬀ topological space on which the group acts continuously and faithfully with a dense subset of points with ﬁnite orbit.
2.1 Deﬁnition and First Examples Deﬁnition 2.1.1. A group G is called residually ﬁnite if for each element g ∈ G with g = 1G , there exist a ﬁnite group F and a homomorphism φ : G → F such that φ(g) = 1F . Proposition 2.1.2. Let G be a group. Then the following conditions are equivalent: (a) G is residually ﬁnite; T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 2, © SpringerVerlag Berlin Heidelberg 2010
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(b) for all g, h ∈ G with g = h, there exist a ﬁnite group F and a homomorphism φ : G → F such that φ(g) = φ(h). Proof. The fact that (b) implies (a) is obvious, since (b) gives (a) by taking h = 1G . Conversely, suppose that G is residually ﬁnite. Let g, h ∈ G with g = h. As gh−1 = 1G , there exist a ﬁnite group F and a homomorphism φ : G → F such that φ(gh−1 ) = 1F . Since φ(gh−1 ) = φ(g)(φ(h))−1 , it follows that φ(g) = φ(h). Therefore, (a) implies (b). Proposition 2.1.3. Every ﬁnite group is residually ﬁnite. Proof. Let G be a ﬁnite group and consider g ∈ G such that g = 1G . If φ = IdG : G → G is the identity map, we have φ(g) = g = 1G . Proposition 2.1.4. The additive group Z is residually ﬁnite. Proof. Consider k ∈ Z such that k = 0. Choose an integer m such that k < m. Then the canonical homomorphism φ : Z → Z/mZ (reduction modulo m) satisﬁes φ(k) = 0. A similar argument gives us the following: Proposition 2.1.5. The group GLn (Z) is residually ﬁnite for every n ≥ 1. Proof. Let A = (aij ) ∈ GLn (Z) with A = In = 1GLn (Z) . Choose an integer m such that aij  < m for all i, j. Then the homomorphism φ : GLn (Z) → GLn (Z/mZ) given by reduction modulo m satisﬁes φ(A) = 1GLn (Z/mZ) . We shall now give examples of groups which are not residually ﬁnite. A group G is called divisible if for each g ∈ G and each integer n ≥ 1, there is an element h ∈ G such that hn = g. Example 2.1.6. The additive groups Q, R, and C are divisible. More generally, every Qvector space, with its additive underlying group structure, is divisible. In particular, every ﬁeld of characteristic 0, with its underlying additive group structure, is divisible. Lemma 2.1.7. Let G be a divisible group and let F be a ﬁnite group. Then every homomorphism φ : G → F is trivial. Proof. Set n = F . Let g ∈ G. As G is divisible, we can ﬁnd h ∈ G such that g = hn . If φ : G → F is a homomorphism, then we have φ(g) = φ(hn ) = φ(h)n = 1F . The preceding lemma immediately yields the following result. Proposition 2.1.8. A nontrivial divisible group cannot be residually ﬁnite.
2.1 Deﬁnition and First Examples
39
Example 2.1.9. The additive group underlying a ﬁeld of characteristic 0 is not residually ﬁnite. In particular, the additive group Q is not residually ﬁnite. In order to give another characterization of residually ﬁniteness, we shall use the following: Lemma 2.1.10. Let G be a group and let H be a subgroup of G. Let K = −1 . Then K is a normal subgroup of G contained in H. Moreover, g∈G gHg if H is of ﬁnite index in G, then Kis of ﬁnite index in G. Proof. Since gHg −1 = H for g = 1G , we have K ⊂ H. Let Sym(G/H) denote the group of permutations of G/H. Consider the action of G on G/H given by left multiplication and let ρ : G → Sym(G/H) be the associated homomorphism. For each g ∈ G, the stabilizer of gH is gHg −1 . Consequently, we have K = Ker(ρ), which shows that K is a normal subgroup of G. The group G/K is isomorphic to Im(ρ) ⊂ Sym(G/H). Suppose that H is of ﬁnite index in G. This means that the set G/H is ﬁnite. This implies that the group Sym(G/H) is ﬁnite. We deduce that the group G/K is ﬁnite, that is, K is of ﬁnite index in G. Given a group G, the intersection of all subgroups of ﬁnite index of G is called the residual subgroup (or proﬁnite kernel ) of G. Proposition 2.1.11. Let G be a group and let N denote the residual subgroup of G. Then: (i) N is equal to the intersection of all normal subgroups of ﬁnite index in G; (ii) N is a normal subgroup of G; (iii) G is residually ﬁnite if and only if N = {1G }. Proof. Denote by N the intersection of all normal subgroups of ﬁnite index of G. The inclusion N ⊂ N is trivial. If H is a subgroup of ﬁnite index of G, then K = ∩g∈G gHg −1 is a normal subgroup of ﬁnite index of G contained in H, by Lemma 2.1.10. This implies N ⊂ K ⊂ H. It follows that N ⊂ N . This shows (i). Assertion (ii) follows from (i) since the intersection of a family of normal subgroups of G is a normal subgroup of G. Suppose that G is residually ﬁnite. Let g ∈ G such that g = 1G . Then there exist a ﬁnite group F and a homomorphism φ : G → F such that φ(g) = 1F . Thus g ∈ / Ker(φ). As the group G/Ker(φ) is isomorphic to F , the subgroup Ker(φ) is of ﬁnite index in G. This shows that N = {1G }. Conversely, suppose that N = {1G }. Let g ∈ G such that g = 1G . By (i), we can ﬁnd a normal subgroup of ﬁnite index K ⊂ G such that g ∈ / K. If φ : G → G/K is the canonical homomorphism, we have φ(g) = 1G/K . This shows that G is residually ﬁnite.
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2.2 Stability Properties of Residually Finite Groups Proposition 2.2.1. Every subgroup of a residually ﬁnite group is residually ﬁnite. Proof. Let G be a residually ﬁnite group and let H be a subgroup of G. Let h ∈ H such that h = 1G . Since G is residually ﬁnite, there exist a ﬁnite group F and a homomorphism φ : G → F such that φ(h) = 1F . If φ : H → F is the restriction of φ to H, we have φ (h) = φ(h) = 1F . Consequently, H is residually ﬁnite. Proposition 2.2.2. Let (Gi )i∈I be a family of residually ﬁnite groups. Then their direct product G = i∈I Gi is residually ﬁnite. Proof. Let g = (gi )i∈I ∈ G such that g = 1G . Then there exists i0 ∈ I such that gi0 = 1Gi0 . Since Gi0 is residually ﬁnite, we can ﬁnd a ﬁnite group F and a homomorphism φ : Gi0 → F such that φ(gi0 ) = 1F . Consider the homomorphism φ : G → F deﬁned by φ = π ◦ φ, where π : G → Gi0 is the projection onto Gi0 . We have φ (g) = φ(gi0 ) = 1F . Consequently, G is residually ﬁnite. Corollary 2.2.3. Let (Gi )i∈I be a family of residually ﬁnite groups. Then their direct sum G = i∈I Gi is residually ﬁnite. Proof. This follows immediately from Proposition 2.2.1 and Proposition 2.2.2, since G is the subgroup of the direct product P = i∈I Gi consisting of all g = (gi ) ∈ P for which gi = 1Gi for all but ﬁnitely many i ∈ I. Corollary 2.2.4. Every ﬁnitely generated abelian group is residually ﬁnite. Proof. If G is a ﬁnitely generated abelian group, then there exist an integer r ≥ 0 and a ﬁnite abelian group T such that G is isomorphic to Zr × T . By using Proposition 2.1.3 and Proposition 2.1.4, we deduce that G is residually ﬁnite. Remark 2.2.5. An arbitrary abelian group need not be residually ﬁnite. For example, the additive group Q is not residually ﬁnite (Example 2.1.9). Corollary 2.2.6. Let G be a group. Then the following conditions are equivalent: (a) the group G is residually ﬁnite; (b) there exist a family (Fi )i∈I of ﬁnite groups such that the group G is isomorphic to a subgroup of the direct product group i∈I Fi . Proof. The fact that (b) implies (a) follows from Proposition 2.1.3, Proposition 2.2.2 and Proposition 2.2.1. Conversely, suppose that G is residually ﬁnite. Then, for each g ∈ G \ {1G }, we can ﬁnd a ﬁnite group Fg and a homomorphism φg : G → Fg such that φg (g) = 1Fg . Consider the group
2.2 Stability Properties of Residually Finite Groups
H=
41
Fg .
g∈G\{1G }
The homomorphism ψ : G → H deﬁned by ψ= φg g∈G\{1G }
is injective. Therefore, G is isomorphic to a subgroup of H. This shows that (a) implies (b). The class of residually ﬁnite groups is closed under taking projective limits (see Sect. E.2 for the deﬁnition of the limit of a projective system of groups): Proposition 2.2.7. If a group G is the limit of a projective system of residually ﬁnite groups, then G is residually ﬁnite. Proof. Let (Gi )i∈I be a projective system of residually ﬁnite groups such that of a projective limit (see Appendix E), G is a G = lim Gi . By construction ←− subgroup of the group i∈I Gi . We deduce that G is residually ﬁnite by using Proposition 2.2.2 and Proposition 2.2.1. A group G is called proﬁnite if G is the limit of some projective system of ﬁnite groups. An immediate consequence of Proposition 2.2.7 is the following: Corollary 2.2.8. Every proﬁnite group is residually ﬁnite.
Example 2.2.9. Let p be a prime number. Given integers n ≥ m ≥ 0, let φn,m : Z/pm Z → Z/pn Z denote reduction modulo pn . Then (Z/pn Z, φn,m ) is a projective system of groups over N. The limit of this projective system is called the group of padic integers and is denoted by Zp . Since Zp is the projective limit of ﬁnite groups, it is proﬁnite and hence residually ﬁnite by Corollary 2.2.8. Remark 2.2.10. A group which is the limit of an inductive system of residually ﬁnite groups need not be residually ﬁnite. For instance, we have seen in Example 2.1.9 that the additive group underlying a ﬁeld of characteristic 0 is not residually ﬁnite. However, such a group is the limit of the inductive system formed by its ﬁnitely generated subgroups, which are all residually ﬁnite by Corollary 2.2.4. If P is a property of groups, one says that a group G is virtually P if G contains a subgroup of ﬁnite index which satisﬁes P. Lemma 2.2.11. Let G be a group. Let H be a subgroup of ﬁnite index of G and let K be a subgroup of ﬁnite index of H. Then K is a subgroup of ﬁnite index of G.
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Proof. Let h1 , . . . , hn be a complete set of representatives of the left cosets of G modulo H and let k1 , . . . , kp be a complete set of representatives of the left cosets of H modulo K. Observe that the elements hi kj , 1 ≤ i ≤ n, 1 ≤ j ≤ p, form a complete set of representatives of the left cosets of G modulo K. Therefore [G : K] = np = [G : H][H : K] < ∞. Proposition 2.2.12. Every virtually residually ﬁnite group is residually ﬁnite. Proof. Let G be a group and let H be a subgroup of ﬁnite index of G. By Lemma 2.2.11, the intersection of the subgroups of ﬁnite index of G is contained in the intersection of the subgroups of ﬁnite index of H. Since a group is residually ﬁnite if and only if the intersection of its subgroups of ﬁnite index is reduced to the identity element (Proposition 2.1.11), we deduce that G is residually ﬁnite if H is residually ﬁnite.
2.3 Residual Finiteness of Free Groups The goal of this section is to establish the following result: Theorem 2.3.1. Every free group is residually ﬁnite. To prove this theorem, we shall use the following: Lemma 2.3.2. The subgroup of SL2 (Z) generated by the matrices 12 10 a= and b = 01 21 is a free group of rank 2. Proof. The group GL2 (R) naturally acts on the set of lines of R2 passing through the origin, that is, on the projective line P1 (R). In nonhomogeneous coordinates this action is given by gt =
g11 t + g12 g21 t + g22
for g = (gij )1≤i,j≤2 ∈ GL2 (R) and t ∈ P1 (R) = R ∪ {∞} representing the line of R2 with slope 1/t passing through (0, 0) (see Fig. 2.1). Note that we have ak t = t + 2k
and
bk t =
t 2kt + 1
for all k ∈ Z. Consider the subsets Y and Z of P1 (R) deﬁned by Y = ] − 1, 1[
and
Z = P1 (R) \ [−1, 1].
2.3 Residual Finiteness of Free Groups
43
Fig. 2.1 The action of GL2 (R) on P1 (R). Here, t = 1 and g = a
One immediately checks that, for all k ∈ Z \ {0}, one has ak Y = ]2k − 1, 2k + 1[ ⊂ Z
and
bk Z = ]1/(2k + 1), 1/(2k − 1)[ ⊂ Y.
By applying the Klein PingPong theorem (Theorem D.5.1), we deduce that a and b generate a free group of rank 2. Proof of Theorem 2.3.1. Since GL2 (Z) is residually ﬁnite by Proposition 2.1.5, it follows from Lemma 2.3.2, Corollary D.5.3, and Proposition 2.2.1 that every free group of ﬁnite rank is residually ﬁnite. Consider now an arbitrary set X and let F (X) denote the free group based on X. Let g ∈ F (X) such that g = 1F (X) . Let Y ⊂ X denote the set of elements x ∈ X such that xk appear in the reduced form of g for some k ∈ Z \ {0}. The subgroup F (Y ) ⊂ F (X) generated by Y is a free group with base Y (see Proposition D.2.4). We have g ∈ F (Y ). As the group F (Y ) is free of ﬁnite rank, it is residually ﬁnite by the ﬁrst part of the proof. Thus we can ﬁnd a ﬁnite group H and a homomorphism φ : F (Y ) → H such that φ(g) = 1H . Consider the unique homomorphism π : F (X) → F (Y ) such that π(y) = y for every y ∈ Y and π(x) = 1F (Y ) for every x ∈ X \ Y . Then the homomorphism φ ◦ π : F (X) → H satisﬁes φ ◦ π(g) = φ(π(g)) = φ(g) = 1H . This shows that F (X) is residually ﬁnite.
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2 Residually Finite Groups
2.4 Hopﬁan Groups Deﬁnition 2.4.1. A group G is called Hopﬁan if every surjective endomorphism of G is injective. Examples 2.4.2. (a) Every ﬁnite group is Hopﬁan. (b) The additive group Q is Hopﬁan. Indeed, every endomorphism of the group Q is of the form x → ax for some a ∈ Q, and it is clear that such an endomorphism is surjective (resp. injective) if and only if a = 0. (c) Every simple group is Hopﬁan. (we recall that a simple group is a nontrivial group G such that the only normal subgroups of G are {1G } and G). (d) The additive group Q/Z is not Hopﬁan. Indeed, the endomorphism ψ : Q/Z → Q/Z deﬁned by ψ(x) = 2x is surjective but not injective. Theorem 2.4.3. Every ﬁnitely generated residually ﬁnite group is Hopﬁan. Given groups G1 and G2 , we shall denote by Hom(G1 , G2 ) the set of all homomorphisms from G1 to G2 . We begin by establishing the following result. Lemma 2.4.4. Let G be a ﬁnitely generated group and let F be a ﬁnite group. Then the set Hom(G, F ) is ﬁnite. Proof. Let A be a ﬁnite generating subset of G. Let us set n = A and p = F . As A generates G, any homomorphism u : G → F is completely determined by the elements u(a), a ∈ A. Thus the set Hom(G, F ) contains at most pn elements. Proof of Theorem 2.4.3. Let G be a ﬁnitely generated residually ﬁnite group. Suppose that ψ : G → G is a surjective endomorphism of G. Let K be a normal subgroup of ﬁnite index of G and let ρ : G → G/K denote the canonical homomorphism. Consider the map Φ : Hom(G, G/K) → Hom(G, G/K) deﬁned by Φ(u) = u ◦ ψ for all u ∈ Hom(G, G/K). The map Φ is injective since ψ is surjective by our hypothesis. As the set Hom(G, G/K) is ﬁnite by Lemma 2.4.4, we deduce that Φ is also surjective. In particular, there exists a homomorphism u0 ∈ Hom(G, G/K) such that ρ = u0 ◦ ψ. This implies Ker(ψ) ⊂ Ker(ρ) = K. It follows that Ker(ψ) is contained in the intersection of all normal subgroups of ﬁnite index of G. As G is residually ﬁnite, we deduce that Ker(ψ) = {1G } by Proposition 2.1.11(iii). Thus ψ is injective. This shows that G is Hopﬁan. Since every free group is residually ﬁnite by Theorem 2.3.1, an immediate consequence of Theorem 2.4.3 is the following: Corollary 2.4.5. Every free group of ﬁnite rank is Hopﬁan.
2.5 Automorphism Groups of Residually Finite Groups
45
Remarks 2.4.6. (a) Let X be an inﬁnite set and let F (X) denote the free group based on X. Every surjective but non injective map f : X → X extends to a surjective endomorphism ψ : F (X) → F (X) which is not injective. Consequently, the group F (X) is not Hopﬁan. Since F (X) is residually ﬁnite by Theorem 2.3.1, this shows that we cannot suppress the hypothesis that G is ﬁnitely generated in Theorem 2.4.3. (b) An example of a ﬁnitely generated Hopﬁan group which is not residually ﬁnite will be given in Sect. 2.6 (see Proposition 2.6.1).
2.5 Automorphism Groups of Residually Finite Groups Let G be a group. Recall that an automorphism of G is a bijective homomorphism α : G → G. Clearly, the set Aut(G) consisting of all automorphisms of G is a subgroup of the symmetric group of G. The group Aut(G) is called the automorphism group of G. Theorem 2.5.1. Let G be a ﬁnitely generated residually ﬁnite group. Then the group Aut(G) is residually ﬁnite. Let us ﬁrst establish the following result. Lemma 2.5.2. Let G be a group. Let H1 and H2 be subgroups of ﬁnite index of G. Then the subgroup H = H1 ∩ H2 is of ﬁnite index in G. Proof. Two elements in G are left congruent modulo H if and only if they are both left congruent modulo H1 and left congruent modulo H2 . Therefore, there is an injective map from G/H into G/H1 × G/H2 given by gH → (gH1 , gH2 ). As the sets G/H1 and G/H2 are ﬁnite by hypothesis, we deduce that G/H is ﬁnite, that is, H is of ﬁnite index in G. Proof of Theorem 2.5.1. Let α0 ∈ Aut(G) such that α0 = IdG . Then we can ﬁnd an element g0 ∈ G such that α0 (g0 ) = g0 . As G is residually ﬁnite, there exist a ﬁnite group F and a homomorphism φ : G → F satisfying φ(α0 (g0 )) = φ(g0 ). Consider the set H deﬁned by H= Ker(ψ), ψ∈Hom(G,F )
where Hom(G, F ) denotes, as above, the set of all homomorphisms from G to F . Observe that H is a normal subgroup of G since it is the intersection of a family of normal subgroups of G. On the other hand, for every α ∈ Aut(G), one has
46
2 Residually Finite Groups
α(H) = α
Ker(ψ)
ψ∈Hom(G,F )
=
α(Ker(ψ))
ψ∈Hom(G,F )
=
Ker(ψ ◦ α−1 ).
ψ∈Hom(G,F )
As the map from Hom(G, F ) to itself deﬁned by ψ → ψ ◦ α−1 is bijective (with ψ → ψ ◦ α as inverse map), we get α(H) = Ker(ψ) = H. ψ∈Hom(G,F )
Therefore α induces an automorphism α of G/H, given by α(gH) = α(g)H for all g ∈ G. The map α → α is clearly a homomorphism from Aut(G) to Aut(G/H). Let us show that the group Aut(G/H) is ﬁnite and that α0 = 1Aut(G/H) = IdG/H . Observe ﬁrst that the set Hom(G, F ) is ﬁnite by Lemma 2.4.4. As Ker(ψ) is of ﬁnite index in G for every ψ ∈ Hom(G, F ), we deduce that H is of ﬁnite index in G by applying Lemma 2.5.2. This implies that the group Aut(G/H) is ﬁnite. On the other hand, we have α0 (g0 H) = α0 (g0 )H = g0 H since H is a subgroup of Ker(φ) and φ(g0 ) = g0 . Therefore α0 = IdG/H . This shows that the group Aut(G) is residually ﬁnite. Every free group is residually ﬁnite by Theorem 2.3.1. Therefore we deduce from Theorem 2.5.1 the following result. Corollary 2.5.3. The group Aut(Fn ) is residually ﬁnite for every n ≥ 1.
Every ﬁnitely generated abelian group is residually ﬁnite by Corollary 2.2.4. Thus we have: Corollary 2.5.4. The automorphism group of a ﬁnitely generated abelian group is residually ﬁnite. Remark 2.5.5. The automorphism group of a free abelian group of ﬁnite rank n is isomorphic to GLn (Z). Thus, the residual ﬁniteness of GLn (Z) (Proposition 2.1.5) may also be deduced from Corollary 2.5.4. Corollary 2.5.6. Let R be a ring and let M be a left (or right) module over R. Suppose that M is ﬁnitely generated as a Zmodule. Then the automorphism group AutR (M ) of the Rmodule M is residually ﬁnite. Proof. The group AutR (M ) is a subgroup of AutZ (M ). Since AutZ (M ) is residually ﬁnite by Corollary 2.5.4, we deduce that AutR (M ) is residually ﬁnite by applying Proposition 2.2.1.
2.6 Examples of Finitely Generated Groups Which Are Not Residually Finite
47
Corollary 2.5.7. Let R be a ring. Suppose that R is ﬁnitely generated as a Zmodule. Then the group GLn (R) is residually ﬁnite for every n ≥ 1. Proof. This is an immediate consequence of the preceding corollary since the group GLn (R) is isomorphic to AutR (Rn ), where Rn is viewed as a left module over R. Example 2.5.8. The ring of Gaussian integers Z[i] = {a + bi : a, b ∈ Z} is a free abelian Zmodule of rank 2. Thus, the group GLn (Z[i]) is residually ﬁnite for every n ≥ 1 by Corollary 2.5.7. More generally, consider a number ﬁeld K, that is, a ﬁeld extension of Q such that d = dimQ K < ∞. Let A denote the ring of algebraic integers of K (we recall that an element x ∈ K is called an algebraic integer of K if x is a root of a monic polynomial with integral coeﬃcients). It is a standard fact in algebraic number theory that A is a free Zmodule of rank d. Thus, the group GLn (A) is residually ﬁnite for every n ≥ 1 by Corollary 2.5.7.
2.6 Examples of Finitely Generated Groups Which Are Not Residually Finite We have seen in Example 2.1.9 that the additive group Q is not residually ﬁnite. Observe that the group Q is not ﬁnitely generated. In fact, any ﬁnitely generated abelian group is residually ﬁnite by Corollary 2.2.4. The purpose of this section is to give two examples of ﬁnitely generated groups G1 and G2 which are not residually ﬁnite. Our ﬁrst example is a subgroup of the symmetric group Sym(Z) generated by two elements: Proposition 2.6.1. Let G1 denote the subgroup of Sym(Z) generated by the translation T : n → n + 1 and the transposition S = (0 1). Then G1 is a ﬁnitely generated Hopﬁan group which is not residually ﬁnite. Let us ﬁrst establish the following lemmas: Lemma 2.6.2. Let G be an inﬁnite simple group. Then G is not residually ﬁnite. Proof. The only normal subgroup of ﬁnite index of G is G itself. Therefore G is not residually ﬁnite by Proposition 2.1.11(iii). Given a set X, we recall that Sym0 (X) denotes the subgroup of Sym(X) consisting of all permutations of X whose support is ﬁnite (see Appendix C). Lemma 2.6.3. Let X be an inﬁnite set. Then the group Sym0 (X) is not residually ﬁnite.
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2 Residually Finite Groups
Proof. The subgroup Sym+ 0 (X) ⊂ Sym(X) consisting of all permutations of X with ﬁnite support and signature 1 is an inﬁnite simple group by Theorem C.4.3. Therefore Sym+ 0 (X) is not residually ﬁnite by Lemma 2.6.2. We deduce that Sym0 (X) is not residually ﬁnite by using Proposition 2.2.1. Lemma 2.6.4. The group G1 contains Sym0 (Z) as a normal subgroup. Moreover, G1 is the semidirect product of Sym0 (Z) with the inﬁnite cyclic subgroup of G1 generated by T . Proof. For all i ∈ Z, we have T i ST −i = (i i + 1),
(2.1)
which shows that (i i + 1) ∈ G1 . If i < j, we also have (i j + 1) = (j j + 1)(i j)(j j + 1) which, by induction on j, shows that (i j) ∈ G1 for all i, j ∈ Z with i < j. Since the transpositions (i j), with i, j ∈ Z and i < j, generate Sym0 (Z) by Corollary C.2.4, it follows that G1 contains Sym0 (Z). The fact that Sym0 (Z) is normal in Sym(Z) (Proposition C.2.2) implies that Sym0 (Z) is normal in G1 . Let H denote the subgroup of G1 generated by T . It is clear that H ∩ Sym0 (Z) is reduced to the identity map IdZ . On the other hand, by using (2.1), we see that every element g ∈ G1 may be written in the form g = hσ, where h ∈ H and σ ∈ Sym0 (Z). Therefore, G1 is the semidirect product of Sym0 (Z) and H. Proof of Proposition 2.6.1. The group G1 contains Sym0 (Z) as a subgroup by Lemma 2.6.4. As the group Sym0 (Z) is not residually ﬁnite by Lemma 2.6.3, we conclude that G1 is not residually ﬁnite by applying Proposition 2.2.1. It remains to show that G1 is Hopﬁan. We start by observing that there are exactly two surjective homomorphisms from G1 onto Z. Indeed, as G1 is the semidirect product of Sym0 (Z) with the subgroup generated by T (Lemma 2.6.4), every element g ∈ G1 can be uniquely written in the form g = T k σ, where k ∈ Z and σ ∈ Sym0 (Z), and the map u : G1 → Z deﬁned by u(g) = k is a surjective homomorphism. As all elements of Sym0 (Z) have ﬁnite order, it immediately follows that the only homomorphisms from G1 onto Z are u and −u. Now, let φ : G1 → G1 be a surjective homomorphism. Then u◦φ : G1 → Z is a surjective homomorphism so that, by our preceding observation, we have u◦φ = u or −u. As Ker(u) = Sym0 (Z) and φ : G1 → G1 is onto, it follows that Ker(φ) ⊂ Sym0 (Z) and φ(Sym0 (Z)) = Sym0 (Z). The simplicity of Sym+ 0 (Z) + (Z) is either equal to Sym (Theorem C.4.3) implies that Ker(φ) ∩ Sym+ 0 0 (Z) + + or reduced to the identity. We cannot have Ker(φ) ∩ Sym0 (Z) = Sym0 (Z), that is, Sym+ 0 (Z) ⊂ Ker(φ), since this would imply that φ(Sym0 (Z)) is either reduced to the identity or cyclic of order 2, which contradicts φ(Sym0 (Z)) = Sym0 (Z). Consequently, we have Ker(φ) ∩ Sym+ 0 (Z) = {IdZ }. If Ker(φ) is not
2.6 Examples of Finitely Generated Groups Which Are Not Residually Finite
49
reduced to the identity, it follows that Ker(φ) is a normal subgroup of order 2 of Sym0 (Z). But this is impossible since Proposition C.2.3 combined with Proposition C.3.2 implies that, if X is an inﬁnite set, then every nontrivial element in Sym0 (X) has an inﬁnite number of conjugates in Sym0 (X). Thus Ker(φ) must be reduced to the identity, that is, φ is injective. This shows that G1 is Hopﬁan. Our second example of a ﬁnitely generated but not residually ﬁnite group will also show that the class of residually ﬁnite groups is not closed under extensions. In other words, a group G admitting a normal subgroup N such that both N and G/N are residually ﬁnite may fail to be residually ﬁnite. In order to construct it we consider ﬁrst the group H= Hi , i∈Z
where each Hi is a copy of the alternating group Sym+ 5 . Let ψ : Z → Aut(H) be the homomorphism deﬁned by ψ(n) = αn , where α ∈ Aut(H) is the onestep shift given by α(h) = (hi−1 )i∈Z for all h = (hi )i∈Z ∈ H. Proposition 2.6.5. The semidirect product G2 = H ψ Z is a ﬁnitely generated but not residually ﬁnite group. Proof. By deﬁnition of a semidirect product, G2 is the group with underlying set H × Z and group operation given by (h, n)(h , n ) = (hαn (h ), n + n ) for all (h, n), (h , n ) ∈ H × Z. Let t be the element of G2 deﬁned by t = (1H , 1) and identify each element h ∈ H with the element (h, 0) ∈ G. Then H is a normal subgroup of G2 and the quotient group G2 /H is an inﬁnite cyclic group generated by the class of t. One has (h, n) = htn (2.2) and
tn ht−n = αn (h)
(2.3)
for all h ∈ H and n ∈ Z. It follows from (2.2) that G2 is generated by t and the elements of H. In fact, since H is the direct sum of the groups Hi and Hi = ti H0 t−i
(2.4)
by (2.3), we deduce that G2 is generated by t and the elements of H0 . As the group H0 is ﬁnite, this shows that G is ﬁnitely generated.
50
2 Residually Finite Groups
Let us prove now that G2 is not residually ﬁnite. Suppose on the contrary that G2 is residually ﬁnite. Let g be an element in H0 such that g = 1H0 . The residual ﬁniteness of G2 implies the existence of a ﬁnite group F and a homomorphism φ : G2 → F such that φ(g) = 1F . As H0 is a simple group, the restriction of φ to H0 is injective. Therefore, the group φ(H0 ) is isomorphic m m to H0 and hence to Sym+ 5 . Let m = F . We have φ(t ) = φ(t) = 1F . Thus, it follows from (2.4) that φ(Hm ) = φ(H0 ). Since xy = yx for all x ∈ H0 and y ∈ Hm , this implies that φ(H0 ) is abelian. This contradicts the fact that φ(H0 ) is isomorphic to Sym+ 5 , which is not abelian. Remark 2.6.6. Observe that the group H is residually ﬁnite by Corollary 2.2.3. The quotient group G2 /H is also residually ﬁnite since it is isomorphic to Z. This shows that an extension of a residually ﬁnite group by a residually ﬁnite group need not to be residually ﬁnite.
2.7 Dynamical Characterization of Residual Finiteness Let G be a group and let A be a set. Recall that the set AG = {x : G → A} is equipped with the prodiscrete topology and that G acts on AG by the left shift deﬁned by (1.2). Theorem 2.7.1. Let G be a group. Then the following conditions are equivalent: (a) the group G is residually ﬁnite; (b) for every set A, the set of points of AG which have a ﬁnite Gorbit is dense in AG ; (c) there exists a set A having at least two elements such that the set of points of AG which have a ﬁnite Gorbit is dense in AG ; (d) there exists a Hausdorﬀ topological space X equipped with a continuous and faithful action of G such that the set of points of X which have a ﬁnite Gorbit is dense in X. We recall that an action of a group G on a set X is called faithful if 1G is the only element of G ﬁxing all points of X. Lemma 2.7.2. Let G be a group and let A be a set having at least two elements. Then the action of G on AG is faithful. Proof. Let a and b be two distinct elements in A. Consider an element g0 in G such that g0 = 1G . Let x ∈ AG be the conﬁguration deﬁned by x(g) = a if g = 1G and x(g) = b otherwise. We have g0 x = x since g0 x(1G ) = x(g0−1 ) = b and x(1G ) = a. Consequently, the action of G on AG is faithful. Lemma 2.7.3. Let G be a residually ﬁnite group and let Ω be a ﬁnite subset of G. Then there exists a normal subgroup of ﬁnite index K of G such that the restriction of the canonical homomorphism ρ : G → G/K to Ω is injective.
Notes
51
Proof. Consider the ﬁnite subset S = {g −1 h : g, h ∈ Ω and g = h} ⊂ G. Since G is residually ﬁnite, we can ﬁnd, for every s ∈ S, a normal subgroup / Ns . The set K = ∩s∈S Ns is a normal of ﬁnite index Ns ⊂ G such that s ∈ subgroup of ﬁnite index in G by Lemma 2.5.2. Let ρ : G → G/K be the canonical homomorphism. If g and h are distinct elements in Ω, then g −1 h ∈ / K and hence ρ(g) = ρ(h). Proof of Theorem 2.7.1. Suppose that G is residually ﬁnite. Let A be a set and let W be a neighborhood of a point x in AG . Let us show that W contains a conﬁguration with ﬁnite Gorbit. Consider a ﬁnite subset Ω ⊂ G such that V (x, Ω) = {y ∈ AG : yΩ = xΩ } ⊂ W. By Lemma 2.7.3, we can ﬁnd a normal subgroup of ﬁnite index K ⊂ G such that the restriction to Ω of the canonical homomorphism ρ : G → G/K(= K\G) is injective. This implies that the map Φ : AG/K → AΩ deﬁned by Φ(z) = (z ◦ ρ)Ω is surjective. Thus we can ﬁnd an element z0 ∈ AG/K such that the conﬁgurations z0 ◦ ρ and x coincide on Ω, that is, such that z0 ◦ ρ ∈ V (x, Ω). On the other hand, the conﬁguration z0 ◦ ρ is Kperiodic by Proposition 1.3.3 (observe that K\G = G/K as K is normal in G). As K is of ﬁnite index in G, we deduce that the Gorbit of z0 ◦ ρ is ﬁnite. Thus W contains a conﬁguration whose Gorbit is ﬁnite. This shows that (a) implies (b). Implication (b) ⇒ (c) is trivial. The fact that (c) implies (d) follows from Proposition 1.2.1, Proposition 1.2.2, and Lemma 2.7.2. Let us show that (d) implies (a). Suppose that the group G acts continuously and faithfully on a Hausdorﬀ topological space X and let E denote the set of points of X whose Gorbit is ﬁnite. For x ∈ E, let Stab(x) = {g ∈ G : gx = x} denote the stabilizer of x in G. Observe that x ∈ E if and only if Stab(x) is of ﬁnite index in G. If E is dense in X, then ∩x∈E Stab(x) = {1G } since the action of G on X is continuous and faithful. Thus, by Proposition 2.1.11 we have that G is residually ﬁnite.
Notes A survey article on residually ﬁnite groups has been written by W. Magnus [Mag]. A group G is called linear if there exist an integer n ≥ 1 and a ﬁeld K such that G is isomorphic to a subgroup of GLn (K). A theorem of A.I. Mal’cev [Mal1] asserts that every ﬁnitely generated linear group is residually ﬁnite.
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2 Residually Finite Groups
An example of a ﬁnitely presented group which is residually ﬁnite but not linear was recently given by C. Drut¸u and M. Sapir [DrS]. The residual ﬁniteness of free groups was established by F. Levi [Lev]. The proof presented in Sect. 2.3 is based on the fact that the matrices 10 21
and 12 01 generate a free subgroup of GL2 (Z). This last result is due to I.N. Sanov [San]. The reader may ﬁnd other proofs of the residual ﬁniteness of free groups in [RobD, p. 158] and [MaKS, p. 116]. Hopﬁan groups are named after H. Hopf who used topological methods to prove that fundamental groups of closed orientable surfaces are Hopﬁan (see [MaKS, p. 415]) and raised the question of the existence of ﬁnitely generated nonHopﬁan groups. The fact that every ﬁnitely generated residually ﬁnite group is Hopﬁan (Theorem 2.4.3) was discovered by Mal’cev [Mal1]. The ﬁrst example of a ﬁnitely generated nonHopﬁan group was given by B.H. Neumann [Neu2]. Shortly after, a ﬁnitely presented nonHopﬁan group was found by G. Higman [Hig]. The simplest example of a ﬁnitely presented nonHopﬁan group is certainly provided by the BaumslagSolitar group BS(2, 3) = x, y : yx2 y −1 = x3 , that is, the quotient of the free group F2 based on two generators x and y by the smallest normal subgroup of F2 containing the element x−3 yx2 y −1 (see [BaS], [MaKS], [LS]). On the other hand, the BaumslagSolitar group BS(2, 4) = x, y : yx2 y −1 = x4 is an example of a ﬁnitely presented Hopﬁan group which is not residually ﬁnite (the incorrect statement in [BaS] about the residual ﬁniteness of BS(2, 4) was corrected by S. Meskin in [Mes]). The residual ﬁniteness of the automorphism group of a ﬁnitely generated residually ﬁnite group (Theorem 2.5.1) was proved by G. Baumslag in [Bau]. The group G1 of Sect. 2.6 was considered by Mal’cev in [Mal1]. Given groups A and G, the wreath product A G is the semidirect product H ψ G, where H = ⊕g∈G A and ψ : G → Aut(A) is the group homomorphism associated with the action of G on H ⊂ AG induced by the Gshift (see [RobD], [Rot]). Thus, the group G2 of Sect. 2.6 is the wreath product G2 = Sym+ 5 Z. The proof of Proposition 2.6.5 shows that, if A is a ﬁnitely generated nonabelian simple group and G is a ﬁnitely generated inﬁnite group, then the wreath product A G is a ﬁnitely generated group which is not residually ﬁnite. Examples of ﬁnitely generated nonabelian simple groups are provided by the (ﬁnite) groups Sym+ n , where n ≥ 5, or PSLn (K), where n ≥ 2 and K a ﬁnite ﬁeld having at least 4 elements, or one of the famous inﬁnite ﬁnitely presented simple Thompson groups T or V (see [CFP]).
Exercises 2.1. Show that every quotient of a divisible group is a divisible group. 2.2. Show that every torsionfree divisible abelian group G is isomorphic to a direct sum of copies of Q. Hint: Prove that there is a natural Qvector space structure on G.
Exercises
53
2.3. Let G be a group. (a) Show that it is possible to deﬁne a topology on G by taking as open sets the subsets Ω ⊂ G which satisfy the following property: for each g ∈ Ω there is a subgroup of ﬁnite index H ⊂ G such that gH ⊂ Ω. This topology is called the proﬁnite topology on G. (b) Show that G is residually ﬁnite if and only if the proﬁnite topology on G is Hausdorﬀ. 2.4. Show that the class of residually ﬁnite groups is not closed under taking quotients. Hint: Any group is isomorphic to a quotient of a free group (see Corollary D.4.2). 2.5. Let X be an inﬁnite set. Show that the symmetric group Sym(X) is not residually ﬁnite. Hint: Use Cayley’s theorem (Theorem C.1.2) to prove that Sym(X) contains a subgroup isomorphic to Q or use Lemma 2.6.3. 2.6. Let G be a residually ﬁnite group and let A be a ﬁnite set. (a) Show that there exists a canonical injective homomorphism of the group ICA(G; A) (cf. Sect. 1.10) into the group H Sym(Fix(H)), where H runs over all ﬁnite index subgroups of G. Hint. Use the fact that the conﬁgurations with ﬁnite Gorbit are dense in AG (cf. Theorem 2.7.1). (b) Deduce from (a) that ICA(G; A) is residually ﬁnite. 2.7. Let m be an integer such that m ≥ 2. Let G denote the quotient of the free group F2 on two generators x and y by the normal closure of the single element xyx−1 y −m . Thus, G is the group given by the presentation G = a, b : aba−1 = bm , where a and b denote the images in G of x and y by the quotient homomorphism. (a) Show that every element g ∈ G may be (not uniquely) written in the form g = ai bj ak for some i, j, k ∈ Z. (b) Show that there is a unique group homomorphism φ : G → GL2 (Q) which satisﬁes m0 11 φ(a) = and φ(b) = . 0 1 01 (c) Use (a) to show that φ is injective and that the image of φ is the subgroup φ(G) ⊂ GL2 (Q) given by n m r : n ∈ Z, r ∈ Z[1/m] , φ(G) = 0 1 where Z[1/m] denotes the set of all rationals r ∈ Q which can be written in the form r = umv for some u, v ∈ Z. (d) Use (c) to prove that G is torsionfree. (e) Prove that G is residually ﬁnite. Hint: You have to show that, for any element g ∈ G \ {1G }, there exist a ﬁnite group F and a homomorphism ψ : G → F such that ψ(g) = 1F . Write g = ai bj ak , where i, j, k ∈ Z, and
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treat ﬁrst the case i + k = 0 by using determinants. If i + k = 0, choose a prime number p which is not a divisor of m nor a divisor of j, and consider the homomorphism ψ : G → GL2 (Z/pZ) deﬁned by m0 11 , and ψ(b) = ψ(a) = 0 1 01 where n ∈ Z/pZ denotes the class of n ∈ Z modulo p. 2.8. Let m ≥ 2 be an integer. Denote by Z[1/m] the subgroup of the additive group Q consisting of all rationals r which can be written in the form r = kmn for some k, n ∈ Z. Show that the group Z[1/m] is residually ﬁnite. Hint: Observe that Z[1/m] is isomorphic to a subgroup of the group G studied in Exercise 2.7. 2.9. Let K be a ﬁeld of characteristic p > 0. Show that the additive group K is residually ﬁnite. Hint: Use a base of K seen as a Z/pZ vector space and apply Corollary 2.2.3. 2.10. Show that all elements of the additive group Q/Z have ﬁnite order but that Q/Z is not residually ﬁnite. 2.11. Show that the multiplicative group Q∗ of nonzero rational numbers is residually ﬁnite. Hint: Use the unique factorization of elements of Q∗ into powers of prime numbers and apply Corollary 2.2.3. 2.12. Show that the automorphism group Aut(Q) of the additive group Q is isomorphic to the multiplicative group Q∗ . 2.13. Show that the additive group R is neither residually ﬁnite nor Hopﬁan. 2.14. Let p be a prime. Show that the additive group Zp of padic integers is not Hopﬁan. 2.15. Show that the multiplicative group R∗ of nonzero real numbers is neither residually ﬁnite nor Hopﬁan. 2.16. Let F be a free group of ﬁnite rank n. Suppose that S is a ﬁnite generating subset of F of cardinality S ≤ n. Use the fact that F is Hopﬁan (Corollary 2.4.5) to prove that S = n and that S is a base for F . 2.17. (M. Hall [HalM2]) Let G be a ﬁnitely generated group and let n ≥ 1 be an integer. Show that G contains only a ﬁnite number of subgroups of index n. Hint: Use Lemma 2.1.10 and Lemma 2.4.4. 2.18. Show that the group GLn (Z) is ﬁnitely generated. Hint: For 1 ≤ i, j ≤ n, let Eij denote the n × n matrix all of whose entries are 0 except the entry located on the ith row and the jth column which is equal to 1. Use Euclidean division in Z to show that GLn (Z) is generated by the n2 matrices In − 2Eii , In + Eij , where 1 ≤ i, j ≤ n and i = j.
Exercises
55
2.19. Show that GLn (Z) is Hopﬁan. 2.20. Give a direct proof of the fact that the automorphism group of a ﬁnitely generated abelian group G is residually ﬁnite (Corollary 2.5.4). Hint: Show that there is a ﬁnite group F and an integer n ≥ 0 such that Aut(G) is isomorphic to F × GLn (Z). 2.21. Show that the group G1 considered in Sect. 2.6 contains no normal subgroup H such that both H and G/H are residually ﬁnite. 2.22. Show that the group G2 considered in Sect. 2.6 is Hopﬁan. 2.23. A group G is called almost perfect if all nontrivial ﬁnite quotients of G are nonabelian. Show that, if A is a ﬁnitely generated almost perfect nontrivial group and G is a ﬁnitely generated inﬁnite group, then the wreath product group A G is ﬁnitely generated but not residually ﬁnite. Hint: Follow the proof of Proposition 2.6.5. 2.24. Let G be a group. Denote by Nf q the set of all normal subgroups of ﬁnite index of G, partially ordered by reverse inclusion. (a) Show that Nf q is a directed set. (b) For H, K ∈ Nf q with H ⊂ K, let ϕK,H : G/H → G/K denote the canonical homomorphism. Show that the directed set Nf q together with the homomorphisms ϕK,H form a projective system of groups. The limit of this projective system is called the proﬁnite completion of the group G and is denoted by G. and that the (c) Show that there is a canonical homomorphism η : G → G kernel of η is the residual subgroup of G. (d) Prove that G is residually ﬁnite if and only if the canonical homomor is injective. phism η : G → G
Chapter 3
Surjunctive Groups
Surjunctive groups are deﬁned in Sect. 3.1 as being the groups on which all injective cellular automata with ﬁnite alphabet are surjective. In Sect. 3.2 it is shown that every subgroup of a surjunctive group is a surjunctive group and that every locally surjunctive group is surjunctive. Every locally residually ﬁnite group is surjunctive (Corollary 3.3.6). The class of locally residually ﬁnite groups is quite large and includes in particular all ﬁnite groups, all abelian groups, and all free groups (a still wider class of surjunctive groups, namely the class of soﬁc groups, will be described in Chap. 7). In Sect. 3.4, given an arbitrary group Γ , we introduce a natural topology on the set of its quotient groups. In Sect. 3.7, it is shown that the set of surjunctive quotients is closed in the space of all quotients of Γ .
3.1 Deﬁnition A set X is ﬁnite if and only if every injective map f : X → X is surjective. The deﬁnition given below is related to this characterization of ﬁnite sets. Deﬁnition 3.1.1. A group G is said to be surjunctive if it satisﬁes the following condition: if A is a ﬁnite set, then every injective cellular automaton τ : AG → AG is surjective (and hence bijective). Remark 3.1.2. Given a group G and a ﬁnite set A, it follows from Theorem 1.8.1 that a map f : AG → AG is a cellular automaton if and only if f is Gequivariant (with respect to the Gshift) and continuous (with respect to the prodiscrete topology on AG ). Thus, the deﬁnition of a surjunctive group may be reformulated as follows: a group G is surjunctive if and only if, for any ﬁnite set A, every injective Gequivariant continuous map f : AG → AG is surjective. Proposition 3.1.3. Every ﬁnite group is surjunctive. T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 3, © SpringerVerlag Berlin Heidelberg 2010
57
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Proof. If G is a ﬁnite group and A is a ﬁnite set, then the set AG is ﬁnite. Therefore, every injective cellular automaton τ : AG → AG is surjective.
3.2 Stability Properties of Surjunctive Groups Proposition 3.2.1. Every subgroup of a surjunctive group is surjunctive. Proof. Suppose that H is a subgroup of a surjunctive group G. Let A be a ﬁnite set and let τ : AH → AH be an injective cellular automaton over H. Consider the cellular automaton τ G : AG → AG over G obtained from τ by induction (see Sect. 1.7). The fact that τ is injective implies that τ G is injective by Proposition 1.7.4(i). Since G is surjunctive, it follows that τ G is surjective. By applying Proposition 1.7.4(ii), we deduce that τ is surjective. Proposition 3.2.2. Let G be a group. Then the following conditions are equivalent: (a) G is surjunctive; (b) every ﬁnitely generated subgroup of G is surjunctive. Proof. The fact that (a) implies (b) follows from Proposition 3.2.1. Conversely, let G be a group all of whose ﬁnitely generated subgroups are surjunctive. Let A be a ﬁnite set and let τ : AG → AG be an injective cellular automaton with memory set S. Let H denote the subgroup of G generated by S and consider the cellular automaton τH : AH → AH obtained by restriction of τ (see Sect. 1.7). The fact that τ is injective implies that τH is injective by Proposition 1.7.4(i). As H is ﬁnitely generated, it is surjunctive by our hypothesis on G. It follows that τH is surjective. By applying Proposition 1.7.4(ii), we deduce that τ is also surjective. This shows that (b) implies (a). If P is a property of groups (e.g. being ﬁnite, being nilpotent, being solvable, being free, etc.), a group G is called locally P if all ﬁnitely generated subgroups of G satisfy P. With this terminology, Proposition 3.2.2 may be rephrased by saying that the class of surjunctive groups and the class of locally surjunctive groups coincide. Corollary 3.2.3. Every locally ﬁnite group is surjunctive. Proof. This immediately follows from Proposition 3.2.2 since every ﬁnite group is surjunctive by Proposition 3.1.3. Example 3.2.4. Let X be a set and consider the group Sym0 (X) consisting of all permutations of X which have ﬁnite support (see Sect. C.2). The group Sym0 (X) is locally ﬁnite. Indeed, if Σ is a ﬁnite subset of Sym0 (X), then
3.3 Surjunctivity of Locally Residually Finite Groups
59
the subgroup H ⊂ Sym0 (X) generated by Σ is ﬁnite since it is isomorphic to a subgroup of Sym(A), where A denotes the union of the supports of the elements of Σ. Consequently, the group Sym0 (X) is surjunctive. Observe that Sym0 (X) is inﬁnite when X is inﬁnite. This yields our ﬁrst examples of inﬁnite surjunctive groups. Finally, note that it follows from Lemma 2.6.3 that Sym0 (X) is not residually ﬁnite whenever X is inﬁnite.
3.3 Surjunctivity of Locally Residually Finite Groups Theorem 3.3.1. Every residually ﬁnite group is surjunctive. Let us ﬁrst establish an important property of cellular automata with ﬁnite alphabet, namely the fact that they always have a closed image with respect to the prodiscrete topology on the set of conﬁgurations: Lemma 3.3.2. Let G be a group and let A be a ﬁnite set. Let τ : AG → AG be a cellular automaton. Then the set τ (AG ) is closed in AG for the prodiscrete topology. Proof. Since A is ﬁnite, the space AG is compact by Tychonoﬀ theorem (see Corollary A.5.3). As τ is continuous by Proposition 1.4.8, we deduce that τ (AG ) is a compact subset of AG . This implies that τ (AG ) is closed in AG since AG is Hausdorﬀ by Proposition 1.2.1. The following example shows that Lemma 3.3.2 becomes false if we suppress the hypothesis that the alphabet is ﬁnite. Example 3.3.3. Consider the map τ : NZ → NZ given by τ (x)(n) = max(0, x(n) − x(n + 1)) for all x ∈ NZ and n ∈ Z. Clearly, τ is a cellular automaton over the group Z and the alphabet N with memory set {0, 1}. Consider the conﬁguration y : Z → N deﬁned by y(n) = 1 if n ≥ 0 and y(n) = 0 otherwise. Let F be a ﬁnite subset of Z and choose an integer M ≥ 1 such that F ⊂ [−M +1, M −1]. Consider the conﬁguration xM : Z → N deﬁned by xM (n) = max(0, M − n) if n ≥ 0 and xM (n) = M otherwise. Observe that yM = τ (xM ) is such that yM (n) = 1 if 0 ≤ n ≤ M − 1 and yM (n) = 0 otherwise. Therefore, the conﬁgurations yM and y coincide on [−M + 1, M − 1] and hence on F . Thus y is in the closure of τ (NZ ) in NZ . On the other hand, it is clear that y is not in the image of τ . This shows that τ (NZ ) is not closed in NZ for the prodiscrete topology. In the proof of the surjunctivity of residually ﬁnite groups, we shall also use the following result:
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Lemma 3.3.4. Let G be a group. Suppose that G satisﬁes the following property: for each ﬁnite subset Ω ⊂ G, there exist a surjunctive group Γ and a homomorphism φ : G → Γ such that the restriction of φ to Ω is injective. Then G is surjunctive. Proof. Let A be a ﬁnite set and equip AG with its prodiscrete topology. Suppose that τ : AG → AG is an injective cellular automaton. Let us show that τ is surjective, that is, τ (AG ) = AG . Since τ (AG ) is closed in AG by Lemma 3.3.2, it suﬃces to prove that τ (AG ) is dense in AG . Consider a conﬁguration x ∈ AG and a neighborhood W of x in AG . Then there is a ﬁnite subset Ω ⊂ G such that V (x, Ω) = {y ∈ AG : yΩ = xΩ } ⊂ W. By our hypothesis, we may ﬁnd a surjunctive group Γ and a homomorphism φ : G → Γ such that the restriction of φ to Ω is injective. Let K denote the image of φ and let N denote its kernel. By Proposition 1.3.3, there is a bijective map ψ : AK → Fix(N ) given by ψ(z) = z ◦ φ for all z ∈ AK . Moreover, if we identify AK with Fix(N ) via ψ, the restriction of τ to Fix(N ) yields a cellular automaton τ : AK → AK over the group K (see Proposition 1.6.1). Since the group Γ is surjunctive, the group K is surjunctive by Proposition 3.2.1. We deduce that τ is surjective and hence τ (Fix(N )) = Fix(N ). On the other hand, as the restriction of φ to Ω is injective, we may ﬁnd z0 ∈ AK such that ψ(z0 ) = xΩ . We then have ψ(z0 ) ∈ Fix(N ) ∩ V (x, Ω) = τ (Fix(N )) ∩ V (x, Ω). This shows that W meets the image of τ . Thus τ (AG ) is dense in AG .
Proof of Theorem 3.3.1. Let G be a residually ﬁnite group. If Ω is a ﬁnite subset of G, then there exist, By Lemma 2.7.3, a ﬁnite group Γ and a homomorphism φ : G → Γ such that the restriction of φ to Ω is injective. As ﬁnite groups are surjunctive by Proposition 3.1.3, it follows that G satisﬁes the hypothesis of Lemma 3.3.4. Consequently, G is surjunctive. Corollary 3.3.5. Every free group is surjunctive. Proof. Free groups are residually ﬁnite by Theorem 2.3.1.
Corollary 3.3.6. Every locally residually ﬁnite group is surjunctive. Proof. This immediately follows from Theorem 3.3.1 by using Proposition 3.2.2. Corollary 3.3.7. Every abelian group is surjunctive. Proof. This is an immediate consequence of Corollary 3.3.6 since every abelian group is locally residually ﬁnite by Corollary 2.2.4.
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When X is a ﬁnite set, every surjective map f : X → X is injective. Therefore, a surjective cellular automaton with ﬁnite alphabet over a ﬁnite group is necessarily injective. However, a surjective cellular automaton with ﬁnite alphabet τ : AG → AG may fail to be injective when the group G is inﬁnite as shown by the following example. Example 3.3.8. Take A = Z/2Z = {0, 1} and G = Z. Let τ : AZ → AZ be the map deﬁned by τ (x)(n) = x(n) + x(n + 1) for all x ∈ AZ and n ∈ Z. Clearly, τ is a cellular automaton with memory set S = {0, 1} ⊂ Z and local deﬁning map μ : AS → A given by μ(y) = y(0) + y(1) for all y ∈ AS . The cellular automaton τ is surjective. Indeed, given an element y ∈ AG , the conﬁguration x : Z → Z/2Z deﬁned by ⎧ ⎪ if n = 0, ⎨0 x(n) = y(0) + y(1) + · · · + y(n − 1) if n > 0, ⎪ ⎩ y(n) + y(n + 1) + · · · + y(−1) if n < 0, satisﬁes τ (x) = y. However τ is not injective since the two constant conﬁgurations have the same image under τ .
3.4 Marked Groups Let Γ be a group. A Γ quotient is a pair (G, ρ), where G is a group and ρ : Γ → G is a surjective homomorphism. We deﬁne an equivalence relation on the class of all Γ quotients by declaring that two Γ quotients (G1 , ρ1 ) and (G2 , ρ2 ) are equivalent when there exists a group isomorphism φ : G2 → G1 such that the following diagram is commutative: G2 ρ2 φ∼ =
Γ ρ1
G1 that is, such that ρ1 = φ ◦ ρ2 . An equivalence class of Γ quotients is called a Γ marked group. Observe that two Γ quotients (G1 , ρ1 ) and (G2 , ρ2 ) are equivalent if and only if Ker(ρ1 ) = Ker(ρ2 ). Thus, the set of Γ marked groups may be identiﬁed with the set N (Γ ) consisting of all normal subgroups of Γ . Let us identify the set P(Γ ) consisting of all subsets of Γ with the set {0, 1}Γ by means of the bijection from P(Γ ) onto {0, 1}Γ given by A → χA ,
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where χA : Γ → {0, 1} is the characteristic map of A ⊂ Γ . We equip the set P(Γ ) = {0, 1}Γ with its prodiscrete uniform structure and N (Γ ) ⊂ P(Γ ) with the induced uniform structure. Thus, a base of entourages of N (Γ ) is provided by the sets VF = {(N1 , N2 ) ∈ N (Γ ) × N (Γ ) : N1 ∩ F = N2 ∩ F }, where F runs over all ﬁnite subsets of Γ . Intuitively, two normal subgroups of Γ are “close” in N (Γ ) when their intersection with a large ﬁnite subset of Γ coincide. Proposition 3.4.1. Let Γ be a group. Then the space N (Γ ) of Γ marked groups is a totally disconnected compact Hausdorﬀ topological space. Proof. The space P(Γ ) = {0, 1}Γ is totally disconnected and Hausdorﬀ by Proposition 1.2.1. Moreover, P(Γ ) is compact by Corollary A.5.3 since it is a product of ﬁnite spaces. Thus, it suﬃces to show that N (Γ ) is closed in P(Γ ). To see this, observe that a subset E ∈ P(Γ ) is a normal subgroup of Γ if and only if it satisﬁes (1) 1Γ ∈ E; (2) αβ −1 ∈ E for all α, β ∈ E; (3) γαγ −1 ∈ E for all α ∈ E and γ ∈ Γ . Denoting, for each γ ∈ Γ , by πγ : P(Γ ) = {0, 1}Γ → {0, 1} the projection map corresponding to the γfactor, these conditions are equivalent to (1’) π1Γ (E) = 1; (2’) πα (E)πβ (E)(παβ −1 (E) − 1) = 0 for all α, β ∈ Γ ; (3’) πα (E)(πγαγ −1 (E) − 1) = 0 for all α, γ ∈ Γ . As all projection maps are continuous on P(Γ ), this shows that N (Γ ) is closed in P(Γ ). Remark 3.4.2. If Γ is countable then the uniform structure on N (Γ ) is metrizable. Indeed, the uniform structure on P(Γ ) = {0, 1}Γ is metrizable when Γ is countable (cf. Remark 1.9.2). Let P be a property of groups. A group Γ is called residually P if for each element γ ∈ Γ with γ = 1Γ , there exist a group Γ satisfying P and an epimorphism φ : Γ → Γ such that φ(γ) = 1Γ . Observe that every group which satisﬁes P is residually P. Proposition 3.4.3. Let Γ be a group and let P be a property of groups. Suppose that the class of groups which satisfy P is closed under taking ﬁnite products and subgroups. Then the following conditions are equivalent: (a) Γ is residually P; (b) there exists a net (Ni )i∈I in N (Γ ) which converges to {1Γ } such that Γ/Ni satisﬁes P for all i ∈ I.
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Proof. Suppose that Γ is residually P. For all γ ∈ Γ \{1Γ } we can ﬁnd a group Γγ satisfying P and an epimorphism φγ : Γ → Γγ such that φγ (γ) = 1Γγ . Let I ⊂ P(Γ ) be the directed set consisting of all ﬁnite subsets of Γ not containing 1 partially ordered by inclusion. For i ∈ I we denote by φ = Γ i γ∈i φγ : Γ → γ∈i Γγ the homomorphism deﬁned by φi (γ ) = (φγ (γ ))γ∈I for all γ ∈ Γ . Set Ni = ker(φi ) and let us show that the net (Ni )i∈I in N (Γ ) converges to {1Γ }. Given a ﬁnite set F ⊂ Γ set iF = F \ {1Γ }. Let f ∈ F \ {1Γ } and i ∈ I such that iF ⊂ i. Then φi (f ) = 1Γi since φf (f ) = 1Γf . Thus / {1Γ } ∩ F. f∈ / Ni and f ∈
(3.1)
/ F then Ni ∩ F = ∅ = {1Γ } ∩ F . On the other hand, It follows that if 1Γ ∈ if 1Γ ∈ F , from (3.1) we deduce Ni ∩ F = {1Γ } = {1Γ } ∩ F . In either cases that limi Ni = {1Γ }. Finally, by our we have Ni ∩ F = {1Γ } ∩ F . We deduce assumptions on P, we have that γ∈i Γγ and its subgroup φi (Γ ) ∼ = Γ/Ni satisfy P for all i ∈ I. This shows (a) ⇒ (b). Suppose now that (b) holds. Let γ ∈ Γ \ {1Γ }. Consider the ﬁnite set F = {γ}. Then there exists iF ∈ I such that NiF ∩ F = {1Γ } ∩ F and / NiF . In other words, Γ/NiF satisfy P. As {1Γ } ∩ F = ∅ we have that γ ∈ if φF : Γ → Γ/NiF denotes the canonical epimorphism, we have φF (γ) = 1Γ/NiF . We deduce that Γ is residually P. This shows (b) ⇒ (a). Note that in the above proposition, the assumptions on P are not needed for the implication (b) ⇒ (a) Let A be a set. Consider the set AΓ equipped with its prodiscrete uniform structure and the Γ shift action. For each N ∈ N (Γ ), let Fix(N ) = {x ∈ AΓ : γx = x for all γ ∈ N } ⊂ AΓ denote the set of conﬁgurations which are ﬁxed by N . Recall from Proposition 1.3.6 that Fix(N ) is a closed Γ invariant subset of AΓ . Let us equip P(AΓ ) with the HausdorﬀBourbaki uniform structure associated with the prodiscrete uniform structure on AΓ (see Sect. B.4 for the deﬁnition of the HausdorﬀBourbaki uniform structure on the set of subsets of a uniform space). Theorem 3.4.4. Let Γ be a group and let A be a set. Then the map Ψ : N (Γ ) → P(AΓ ) deﬁned by Ψ (N ) = Fix(N ) is uniformly continuous. Moreover, if A contains at least two elements then Ψ is a uniform embedding. Proof. Let N0 ∈ N (Γ ) and let W be an entourage of P(AΓ ). Let us show that there exists an entourage V of N (Γ ) such that Ψ (V [N0 ]) ⊂ W [Ψ (N0 )].
(3.2)
This will prove that Ψ is continuous. By deﬁnition of the HausdorﬀBourbaki uniform structure on P(AΓ ), there is an entourage T of AΓ such that
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T = {(X, Y ) ∈ P(AΓ ) × P(AΓ ) : Y ⊂ T [X] and X ⊂ T [Y ]} ⊂ W.
(3.3)
Since AΓ is endowed with its prodiscrete uniform structure, there is a ﬁnite subset F ⊂ Γ such that U = {(x, y) ∈ AΓ × AΓ : πF (x) = πF (y)} ⊂ T,
(3.4)
where πF : AΓ → AF is the projection map. Consider now the ﬁnite subset E ⊂ Γ deﬁned by E = F · F −1 = {γη −1 : γ, η ∈ F }, and the entourage V of N (Γ ) given by V = {(N1 , N2 ) ∈ N (Γ ) × N (Γ ) : N1 ∩ E = N2 ∩ E}. We claim that V satisﬁes (3.2). To prove our claim, suppose that N ∈ V [N0 ]. Let x ∈ Fix(N ). The fact that N ∩ E = N0 ∩ E implies that if γ and η are elements of F with γ = νη for some ν ∈ N , then ν ∈ N0 and therefore x(γ) = x(η). Denoting by ρ0 : Γ → Γ/N0 the canonical epimorphism, we deduce that we may ﬁnd an element x0 ∈ AΓ/N0 such that x(γ) = x0 ◦ ρ0 (γ) for all γ ∈ F . We have x0 ◦ ρ0 ∈ Fix(N0 ) and (x, x0 ◦ ρ0 ) ∈ U . Since U ⊂ T by (3.4), this shows that Fix(N ) ⊂ T [Fix(N0 )]. Therefore Ψ is continuous. As N (Γ ) is compact by Proposition 3.4.1, we deduce that Ψ is uniformly continuous by applying Theorem B.2.3. Suppose now that A has at least two elements. Let us show that Ψ is injective. Let N1 , N2 ∈ N (Γ ). Fix two elements a, b ∈ A with a = b and consider the map x : Γ → A deﬁned by x(γ) = a if γ ∈ N1 and x(γ) = b otherwise. We clearly have x ∈ Fix(N1 ). Suppose that Ψ (N1 ) = Ψ (N2 ), that is, Fix(N1 ) = Fix(N2 ). Then for all ν ∈ N2 , we have ν −1 x = x since x ∈ Fix(N1 ) = Fix(N2 ), and hence x(ν) = ν −1 x(1Γ ) = x(1Γ ) = a. This implies N2 ⊂ N1 . By symmetry, we also have N1 ⊂ N2 . Therefore N1 = N2 . This shows that Ψ is injective. As N (Γ ) is compact and P(AΓ ) is Hausdorﬀ, we conclude that Ψ is a uniform embedding by applying Proposition B.2.5.
3.5 Expansive Actions on Uniform Spaces Let X be a uniform space and let Γ be a group acting on X. We consider the diagonal action of Γ on X × X deﬁned by γ(x, y) = (γx, γy) for all γ ∈ Γ and x, y ∈ X. One says that the action of Γ on X is uniformly continuous if the orbit map X → X, x → γx, is uniformly continuous for each γ ∈ Γ . This is
3.6 Gromov’s Injectivity Lemma
65
equivalent to saying that γ −1 V is an entourage of X for all entourage V of X and γ ∈ Γ . The action of Γ on X is said to be expansive if there exists an entourage W0 of X such that γ −1 W0 = ΔX , (3.5) γ∈Γ
where ΔX = {(x, x) : x ∈ X} denotes the diagonal in X × X. Equality (3.5) means that if x, y ∈ X satisfy (γx, γy) ∈ W0 for all γ ∈ Γ , then x = y. An entourage W0 satisfying (3.5) is then called an expansivity entourage for the action of Γ on X. Remarks 3.5.1. (a) If a uniform space X admits an expansive uniformly continuous action of a group Γ , then the topology on X must be Hausdorﬀ. Indeed, if W0 is an expansivity entourage for such an action then ΔX is equal to the intersection of the entourages γ −1 W0 , γ ∈ Γ , by (3.5). (b) Suppose that B is a base of the uniform structure on X. Then an action of a group Γ on X is expansive if and only if it admits an expansivity entourage B0 ∈ B. (c) Let (X, d) be a metric space. Then an action of a group Γ on X is expansive if and only if there exists a real number ε0 > 0 with the following property: if x, y ∈ X satisfy d(γx, γy) < ε0 for all γ ∈ Γ , then x = y. Such an ε is called an expansivity constant for the action of Γ on X. Our basic example of a uniformly continuous and expansive action is provided by the following: Proposition 3.5.2. Let G be a group and let A be a set. Then the Gshift on AG is uniformly continuous and expansive with respect to the prodiscrete uniform structure on AG . Proof. Uniform continuity follows from the fact that G acts on AG by permuting coordinates. Expansiveness is due to the fact that the action of G on itself by left multiplication is transitive. Indeed, consider the entourage W0 of AG deﬁned by W0 = {(x, y) ∈ AG × AG : x(1G ) = y(1G )}. Given g ∈ G, we have (x, y) ∈ g −1 W0 if and only if x(g −1 ) = y(g −1 ). Thus −1 W0 is equal to the diagonal in AG × AG . g∈G g
3.6 Gromov’s Injectivity Lemma The following result will be used in the next section to prove that surjunctive groups deﬁne a closed subset of the space of Γ marked groups for any group Γ .
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3 Surjunctive Groups
Theorem 3.6.1 (Gromov’s injectivity lemma). Let X be a uniform space endowed with a uniformly continuous and expansive action of a group Γ . Let f : X → X be a uniformly continuous and Γ equivariant map. Suppose that Y is a subset of X such that the restriction of f to Y is a uniform embedding. Then there exists an entourage V of X satisfying the following property: if Z is a Γ invariant subset of X such that Z ⊂ V [Y ], then the restriction of f to Z is injective. (We recall that the notation Z ⊂ V [Y ] means that for each z ∈ Z there exists y ∈ Y such that (z, y) ∈ V .) Proof. By expansivity of the action of Γ , there is an entourage W0 of X such that γ −1 W0 = ΔX . (3.6) γ∈Γ
It follows from the axioms of a uniform structure that we can ﬁnd a symmetric entourage S of X such that S ◦ S ◦ S ⊂ W0 .
(3.7)
Since the restriction of f to Y is a uniform embedding, we can ﬁnd an entourage T of X such that (f (y1 ), f (y2 )) ∈ T ⇒ (y1 , y2 ) ∈ S
(3.8)
for all y1 , y2 ∈ Y . Let U be a symmetric entourage of X such that U ◦ U ⊂ T.
(3.9)
Since f is uniformly continuous, we can ﬁnd an entourage E of X such that (x1 , x2 ) ∈ E ⇒ (f (x1 ), f (x2 )) ∈ U
(3.10)
for all x1 , x2 ∈ X. Let us show that the entourage V = S ∩ E has the required property. So let Z be a Γ invariant subset of X such that Z ⊂ V [Y ] and let us show that the restriction of f to Z is injective. Let z and z be points in Z such that f (z ) = f (z ). Since f is Γ equivariant, we have (3.11) f (γz ) = f (γz ) for all γ ∈ Γ . As the points γz and γz stay in Z, the fact that Z ⊂ V [Y ] implies that there are points yγ and yγ in Y such that (γz , yγ ) ∈ V and (γz , yγ ) ∈ V . Since V ⊂ E, it follows from (3.10) that (f (γz ), f (yγ )) and (f (γz ), f (yγ )) are in U . As U is symmetric, we also have (f (yγ ), f (γz )) ∈ U . We deduce that (f (yγ ), f (yγ )) ∈ U ◦ U ⊂ T by using (3.9) and (3.11). This implies (yγ , yγ ) ∈ S by (3.8). On the other hand, we also have (γz , yγ ) ∈ S and (yγ , γz ) ∈ S since V ⊂ S and S is symmetric. It follows that
3.7 Closedness of Marked Surjunctive Groups
67
(γz , γz ) ∈ S ◦ S ◦ S ⊂ W0 by (3.7). This gives us (z , z ) ∈
γ −1 W0 ,
γ∈Γ
and hence z = z by (3.6). Thus the restriction of f to Z is injective.
3.7 Closedness of Marked Surjunctive Groups Theorem 3.7.1. Let Γ be a group. Then the set of normal subgroups N ⊂ Γ such that the quotient group Γ/N is surjunctive is closed in N (Γ ). Proof. Let N ∈ N (Γ ) and let (Ni )i∈I be a net in N (Γ ) converging to N . Suppose that the groups Γ/Ni are surjunctive for all i ∈ I. Let us show that the group Γ/N is also surjunctive. Let A be a ﬁnite set and let τ : AΓ/N → AΓ/N be an injective cellular automaton over the group Γ/N . Let S ⊂ Γ/N be a memory set for τ with associated local deﬁning map μ : AS → A. Choose a subset S ⊂ Γ such that the canonical epimorphism ρ : Γ → Γ/N gives a bijection ψ : S → S, and let e π : AS → AS denote the bijective map induced by ψ. Consider the cellular automaton τ : AΓ → AΓ over Γ with memory set S and local deﬁning map e μ
= μ ◦ π −1 : AS → A. To simplify notation, let us set X = AΓ , f = τ , Y = Fix(N ) and Zi = Fix(Ni ). We claim that the hypotheses of Theorem 3.6.1 are satisﬁed by X, f and Y . Indeed, we ﬁrst observe that the action of Γ on X is uniformly continuous and expansive by Proposition 3.5.2. On the other hand, the cellular automaton f : X → X is uniformly continuous and Γ equivariant by Theorem 1.9.1. Finally, the restriction of f to Y is injective since this restriction is conjugate to τ by Proposition 1.6.1. As Y is a closed subset of X and hence compact, it follows from Proposition B.2.5 that the restriction of f to Y is a uniform embedding. By applying Theorem 3.6.1, it follows that there exists an entourage V of X such that if Z is a Γ invariant subset of X with Z ⊂ V [Y ], then the restriction of f to Z is injective. Since the net (Zi )i∈I converges to Y for the HausdorﬀBourbaki topology on P(X) by Theorem 3.4.4, there is an element i0 ∈ I such that Zi ⊂ V [Y ] for all i ≥ i0 . As the sets Zi are Γ invariant by Proposition 1.3.6, it follows that the restriction of f to Zi is injective for all i ≥ i0 . On the other hand, f (Zi ) ⊂ Zi and the restriction of f to Zi is conjugate to a cellular automaton τi : AΓ/Ni → AΓ/Ni over the group Γ/Ni for all i ∈ I by Proposition 1.6.1. As the groups Γ/Ni are surjunctive by our hypotheses, we deduce that f (Zi ) = Zi for all i ≥ i0 . Now, it follows from Proposition B.4.6 that the net (f (Zi ))i∈I converges to f (Y ) in P(X). Thus, the net (Zi )i∈I converges to both Y and f (Y ). As Y and f (Y ) are closed in X (by compactness of Y ), we deduce that
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3 Surjunctive Groups
Y = f (Y ) by applying Proposition B.4.3. This shows that τ is surjective since τ is conjugate to the restriction of f to Y . Consequently, the group Γ/N is surjunctive.
Notes Surjunctive groups were introduced by W. Gottschalk. The ﬁrst results on these groups are due to W. Lawton who proved in particular Proposition 3.2.1, Corollary 3.2.3, Theorem 3.3.1, and Corollary 3.3.7 (see [Got], [Law]). The characterization of inﬁnite sets by the existence of injective but not surjective selfmaps is known as Dedekind’s deﬁnition of inﬁnite sets. Proposition 3.3.6 together with Mal’cev theorem [Mal1] which says that every ﬁnitely generated linear group is residually ﬁnite implies that every linear group is surjunctive. The problem of the existence of a non surjunctive group was raised by Gottschalk in [Got] and remains open up to now. Note that to prove that all groups are surjunctive it would suﬃce to prove that the symmetric group Sym(N) is surjunctive (this immediately follows from Proposition 3.2.1, Proposition 3.2.2, and the fact that every ﬁnitely generated group is countable and hence isomorphic to a subgroup of Sym(N) by Cayley’s theorem). In [CeC11] it is shown that if G is a nonperiodic group, then for every inﬁnite set A there exists a cellular automaton τ : AG → AG whose image τ (AG ) is not closed in AG with respect to the prodiscrete topology (cf. Lemma 3.3.2 and Example 3.3.3). Theorem 3.6.1 is a uniform version of Lemma 4.H” in [Gro5]. The proof presented in this chapter of the surjunctivity of limits of surjunctive groups (Theorem 3.7.1) closely follows Sect. 4 of [Gro5] (see also [CeC10]). There is another proof based on techniques from model theory (see [Gro5] and [GlG]).
Exercises 3.1. Let K be an algebraically closed ﬁeld. Show that every injective polynomial map f : K → K is surjective. 3.2. Show that every injective polynomial map f : R → R is surjective. 3.3. Show that the polynomial map f : Q → Q deﬁned by f (x) = x3 is injective but not surjective. 3.4. Show that every injective holomorphic map f : C → C is surjective. 3.5. Give an example of a real analytic map f : R → R which is injective but not surjective.
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69
3.6. Let K be a ﬁeld and let V be a vector space over K. Show that V is ﬁnitedimensional if and only if every injective endomorphism f : V → V is surjective. 3.7. Let R be a ring and let M be a left (or right) Rmodule. One says that M is Artinian if every descending chain N0 ⊃ N1 ⊃ N2 ⊃ . . . eventually stabilizes (i.e., there is an integer n0 ≥ 0 such that NN = Nn+1 for all n ≥ n0 ). Show that if M is Artinian then every injective endomorphism f : M → M is surjective. Hint: Consider the sequence of submodules deﬁned by Nn = Im(f n ) for n ≥ 0. 3.8. Let U = {z ∈ C : z = 1}. Show that every injective continuous map f : U → U is surjective. 3.9. Let n ≥ 0 be an integer. An ndimensional topological manifold is a nonempty Hausdorﬀ topological space X such that each point in X admits a neighborhood homeomorphic to Rn . Show that if X is a compact ndimensional topological manifold, then every injective continuous map f : X → X is surjective. Hint: Use Brouwer’s invariance of domain to prove that f (X) is open in X. 3.10. Let P be a property of groups. Show that every subgroup of a locally P group is itself locally P. 3.11. Let P be a property of groups. Let G be a group. Show that G is locally P if and only if all its ﬁnitely generated subgroups are locally P. 3.12. Show that the additive group Q is locally cyclic. 3.13. Show that in a locally ﬁnite group every element has ﬁnite order. 3.14. Let G be an abelian group. Show that G is locally ﬁnite if and only if every element of G has ﬁnite order. 3.15. Show that every subgroup and every quotient of a locally ﬁnite group is a locally ﬁnite group. 3.16. Let G be a group. Suppose that G contains a normal subgroup H such that both H and G/H are locally ﬁnite. Show that G is locally ﬁnite. 3.17. Let G be a group which is the limit of an inductive system of locally ﬁnite groups. Show that G is locally ﬁnite. 3.18. Show that the direct sum of any family of locally ﬁnite groups is a locally ﬁnite group. 3.19. Let G be a group. Let S denote the set consisting of all normal locally ﬁnite subgroups of G. (a) Show that if H ∈S and K ∈ S then HK ∈ S. (b) Show that M = H∈S H is a normal locally ﬁnite subgroup of G and that every normal locally ﬁnite subgroup of G is contained in M .
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3.20. Let G be a locally ﬁnite group and let A be a set. Show that every bijective cellular automaton τ : AG → AG is invertible. 3.21. Let G be a locally ﬁnite group and let A be a ﬁnite set. Show that every surjective cellular automaton τ : AG → AG is injective. 3.22. Let G be a locally ﬁnite group and let A be a set. Let τ : AG → AG be a cellular automaton. Show that τ (AG ) is closed in AG with respect to the prodiscrete topology. Hint: First treat the case when G is ﬁnite, then use Sect. 1.7 and Proposition A.4.3. 3.23. Let G be a group and let A be a nonzero abelian group. Suppose that s0 ∈ G is an element of inﬁnite order. Show that the map τ : AG → AG deﬁned by τ (x)(g) = x(gs0 ) − x(g), for all x ∈ AG and g ∈ G, is a cellular automaton which is surjective but not injective. 3.24. Let X be an inﬁnite set. Show that the symmetric group Sym(X) is not locally residually ﬁnite. 3.25. Let G be a group. Suppose that there exists a family (Ni )i∈I of normal subgroups of G satisfying the following properties: (1) for all i and j in I, there exists k in I such that Nk ⊂ Ni ∩ Nj ; (2) i∈I Ni = {1G }; (3) the group G/Ni is surjunctive for each i ∈ I. Show that G is surjunctive. Hint: Apply Lemma 3.3.4. 3.26. It follows from Theorem 3.3.1 that every residually ﬁnite group is surjunctive. The goal of this exercise is to present an alternative proof of this result. Let G be a residually ﬁnite group and let A be a ﬁnite set. Let τ : AG → AG be an injective cellular automaton. Fix a family (Γi )i∈I of subgroups of ﬁnite index of G such that i∈I Γi = {1G } (the existence of such a family follows from the residual ﬁniteness of G). (a) Show that, for each i ∈ I, the set Fix(Γi ) = {x ∈ AG : gx = x for all g ∈ Γi } is ﬁnite. (b) Show that τ(Fix(Γi )) = Fix(Γi ) for all i ∈ I. (c) Prove that i∈I Fix(Γi ) is dense in AG and conclude. 3.27. Let X be a set. Let (Ai )i∈I be a net of subsets of X. (a) Show that Aj ⊂ Aj . i j≥i
i j≥i
(b) Let B be a subset of X. Show that the net (Ai )i∈I converges to B with respect to the prodiscrete topology on P(X) = {0, 1}X if and only if B= Aj = Aj . i j≥i
i j≥i
Exercises
71
3.28. Let G be a group and let A be a set. Let H be a subgroup of G. In Exercise 1.33, we associated with each subshift X ⊂ AH the subshift G H X (G) = {x ∈ AG : xH g ∈ X for all g ∈ G} ⊂ A , where xg = (gx)H for G H H all x ∈ A , h ∈ H and g ∈ G. Let σ : A → A be a cellular automaton and denote by σ G : AG → AG the induced cellular automaton. Show that if X, Y ⊂ AH are two subshifts such that σ(X) ⊂ Y , then the cellular automaton σ G X (G) : X (G) → Y (G) is injective (resp. surjective) if and only if the cellular automaton σX : X → Y is injective (resp. surjective). 3.29. Let G be a group and let A be a ﬁnite set. Given a subshift Z ⊂ AG we denote by Zf the set of all conﬁgurations in Z whose Gorbit is ﬁnite (cf. Example 1.3.1(c)). We say that Z is surjunctive if every injective cellular automaton σ : Z → Z is surjective. Let X ⊂ AG be a subshift such that Xf is dense in X. (a) Let τ : AG → AG be a cellular automaton. Consider the subshift Y = τ (X). Show that Yf is dense in Y . (b) Deduce from (a) that X is surjunctive. 3.30. Let A be a ﬁnite set and let X ⊂ AZ be an irreducible subshift of ﬁnite type. (a) Show that Xf is dense in X. (b) Deduce from (a) and Exercise 3.29(b) that X is surjunctive (compare with its stronger version in Exercise 6.36). (c) Let A = {0, 1} and consider the subshift of ﬁnite type Y ⊂ AZ deﬁned by Y = {x ∈ AZ : (x(n), x(n + 1)) = (0, 1) for all n ∈ Z}. Show that Yf consists of the two constant conﬁgurations x0 and x1 and therefore it is closed but not dense in Y . (d) Show that the subshift Y is surjunctive. (e) Let A = {0, 1} and let Y as in (c). Consider the associated subshift 2 2 Z = Y (Z ) ⊂ AZ deﬁned in Exercise 1.33. Show that Z is an irreducible subshift of ﬁnite type but that Zf is not dense in Z. (f) Show that the subshift Z is surjunctive. 3.31. Show that the golden mean subshift is surjunctive. Hint: Use Exercise 1.39 and Exercise 3.30(b). 3.32. Show that the even subshift is surjunctive. Hint: Use Exercise 3.30. 3.33. A subshift of ﬁnite type which is not surjunctive (cf. [Weiss, Sect. 4]). Let A = {0, 1, 2} and consider the subshift of ﬁnite type X = X(Ω, A) ⊂ AZ with memory set Ω = {0, 1} ⊂ Z and deﬁning set of admissible words A = {00, 01, 11, 12, 22} ⊂ AΩ . (a) Show that X is not irreducible. (b) Consider the cellular automaton σ : AZ → AZ with memory set S = {0, 1} and local deﬁning map μ : AS → A deﬁned by
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3 Surjunctive Groups
μ(y) =
1 y(0)
if y(0)y(1) = 01 otherwise.
Show that σ(X) ⊂ X. (c) Show that the cellular automaton τ = σX : X → X is injective but not surjective. Deduce that X is not surjunctive. 3.34. Let G be a group and let A be a ﬁnite set. Suppose that G contains an element of inﬁnite order and that A contains at least two distinct elements. (a) Show that there exists a subshift X ⊂ AG which is not surjunctive. (b) Show that if, in addition, G is not inﬁnite cyclic then there exists an irreducible subshift X ⊂ AG which is not surjunctive. Hint: Use Exercise 3.33 and Exercise 1.33(c). 3.35. Given a group Γ acting continuously on a topological space Z, a subset M ⊂ Z is called a minimal set if M is a nonempty Γ invariant subset of Z and the Γ orbit of every point m ∈ M is dense in M . Let G be a group and let A be a set. A subshift X ⊂ AG is called minimal if X is a minimal set with respect to the Gshift action on AG . (a) Show that a subshift X ⊂ AG is minimal if and only if X = ∅ and there is no subshift Y in AG such that ∅ = Y X. (b) Show that every minimal subshift X ⊂ AG is irreducible. (c) Show that the ﬁnite minimal subshifts in AG are precisely the ﬁnite Gorbits. (d) Show that if X ⊂ AG is an inﬁnite minimal subshift then the Gorbit of every conﬁguration x ∈ X is inﬁnite. (e) Show that if X and Y are minimal subshifts in AG then either X = Y or X ∩ Y = ∅. 3.36. Let G be a group and let A be a ﬁnite set. Let X, Y ⊂ AG be two subshifts. Show that if X is nonempty and Y is minimal then every cellular automaton τ : X → Y is surjective. 3.37. Let G be a group and let A be a ﬁnite set. Show that every minimal subshift X ⊂ AG is surjunctive. 3.38. Let G be a group and let A be a ﬁnite set. Show that every nonempty subshift X ⊂ AG contains a minimal subshift. Hint: Apply Zorn’s lemma to the set of nonempty subshifts contained in X. 3.39. Let G be a group and let A be a set. A subset R ⊂ G is called syndetic if there exists a ﬁnite subset K ⊂ G such that the set Kg meets R for every g ∈ G. A conﬁguration x ∈ AG is called almost periodic if for every ﬁnite subset Ω ⊂ G the set R(x, Ω) = {g ∈ G : (gx)Ω = xΩ } is syndetic in G. (a) Show that if x ∈ AG is an almost periodic conﬁguration then its orbit closure X = Gx ⊂ AG is a minimal subshift. Hint: Suppose that the orbit closure X = Gx of a conﬁguration x ∈ AG is not minimal. Then there exists
Exercises
73
a nonempty subshift Y ⊂ X with x ∈ / Y . This implies that there is a ﬁnite subset Ω ⊂ G such that xΩ = yΩ for all y ∈ Y . Let K be a ﬁnite subset of G and consider the set Ω = K −1 Ω. Choose an arbitrary conﬁguration y0 ∈ Y . Since y0 ∈ X, there is an element g0 ∈ G such that (g0 x)Ω = y0 Ω . This implies that (kg0 x)Ω = (ky0 )Ω = xΩ for all k ∈ K. Thus Kg0 does not meet R(x, Ω) so that x is not almostperiodic. (b) Suppose that the set A is ﬁnite. Show that if X ⊂ AG is a minimal subshift, then every conﬁguration x ∈ X is almostperiodic. Hint: Observe that if Ω ⊂ G is a ﬁnite subset and x ∈ X, then the open sets gV , where V = {y ∈ AG : yΩ = xΩ } and g runs over G, cover X by minimality and use the compactness of X. (c) Suppose that the set A is ﬁnite. Show that a nonempty subshift X ⊂ AG is minimal if and only if X is irreducible and every conﬁguration x ∈ X is almostperiodic. 3.40. Let A be a set. The language of a conﬁguration x ∈ AZ is the subset L(x) ⊂ A∗ consisting of all words w ∈ A∗ which can be written in the form w = x(n + 1)x(n + 2) . . . x(n + m) for some integers n ∈ Z and m ≥ 0. (a) Let X ⊂ AZ be a subshift and let x ∈ X. Show that L(x) ⊂ L(X) and that one has L(x) = L(X) if and only if the Zorbit of x is dense in X. (b) Show that a nonempty subshift X ⊂ AZ is minimal if and only if one has L(x) = L(X) for all x ∈ X. (c) Show that a conﬁguration x ∈ AZ is almost periodic if and only if it satisﬁes the following condition: for every word u ∈ L(x), there exists an integer n = n(u) ≥ 0 such that u is a subword of any word v ∈ L(x) of length n. 3.41. The ThueMorse sequence. Let A = {0, 1}. The ThueMorse sequence is the map x : N → A deﬁned by
1 if the number of ones in the binary expansion of n is odd x(n) = 0 otherwise. (a) Check that x(0)x(1) . . . x(16) = 0110100110010110. (b) Show that one has x(k + 3m2n ) = x(k) for all k, m, n ∈ N such that 0 ≤ k ≤ 2n . (c) Show that x satisﬁes the recurrence relations ⎧ ⎪ ⎨x(0) = 0 x(2n) = x(n) ⎪ ⎩ x(2n + 1) = 1 − x(n) and that these relations uniquely determine x.
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3 Surjunctive Groups
(d) Consider the unique monoid homomorphism ϕ : A∗ → A∗ which satisﬁes ϕ(0) = 01 and ϕ(1) = 10. Show that ϕ(x(0)x(1) . . . x(2n − 1)) = x(0)x(1) . . . x(2n+1 − 1) for all n ∈ N. (e) Consider the unique monoid isomorphism ι : A∗ → A∗ which satisﬁes ι(0) = 1 and ι(1) = 0. Show that x(0)x(1) . . . x(2n+1 − 1) = x(0)x(1) . . . x(2n − 1)ι(x(0)x(1) . . . x(2n − 1)) for all n ∈ N. Note: The ThueMorse sequence was ﬁrst studied by E. Prouhet in a 1851 paper dealing with problems in number theory. It is an example of an automatic sequence (see [AlS, Sect. 5.1]). 3.42. The Morse subshift. Let A = {0, 1} and let L = {(x(n), x(n + 1), . . . , x(n + m)) : n, m ∈ N} ⊂ A∗ , where x : N → A is the ThueMorse sequence. (a) Show that the set L satisﬁes the conditions (i) and (ii) in Exercise 1.36(b). (b) Let X ⊂ AZ denote the unique subshift whose language is L(X) = L (cf. Exercise 1.36(c)). Show that X is an inﬁnite minimal subshift (it is called the Morse subshift). Hint: Use Exercises 3.39(a), 3.40 and 3.41(b). 3.43. Toeplitz subshifts. Let A be a ﬁnite set. A conﬁguration x ∈ AZ is said to be a Toeplitz conﬁguration if for every n ∈ Z there exists k ≥ 1 such that x(n) = x(n + kr) for all r ∈ Z. In other words, x is a Toeplitz conﬁguration if there exists a partition of Z into arithmetic progressions such that x is constant on each element of the partition. Given a Toeplitz conﬁguration x ∈ AZ and n ∈ Z we set k(x, n) = min{k ≥ 1 : x(n) = x(n + kr) for all r ∈ Z}. One says that a subshift X ⊂ AZ is a Toeplitz subshift if it is the orbit closure of some Toeplitz conﬁguration x ∈ AZ . (a) Let x ∈ AZ be a Toeplitz conﬁguration. Show that the Zorbit of x is ﬁnite if and only if the set {k(x, n) : n ∈ Z} is ﬁnite. (b) Show that every Toeplitz conﬁguration x ∈ AZ is almostperiodic. Hint: Use Exercise 3.40. (c) Show that every Toeplitz subshift X ⊂ AZ is minimal.
Exercises
3.44. Let A be a ﬁnite set and let X ⊂ AZ be a Toeplitz subshift. (a) Show that X is irreducible. Hint: Use Exercises 3.43(d) and 3.35. (b) Show that X is surjunctive. Hint: Use Exercises 3.43(d) and 3.37.
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Chapter 4
Amenable Groups
This chapter is devoted to the class of amenable groups. This is a class of groups which plays an important role in many areas of mathematics such as ergodic theory, harmonic analysis, representation theory, dynamical systems, geometric group theory, probability theory and statistics. As residually ﬁnite groups, amenable groups generalize ﬁnite groups but there are residually ﬁnite groups which are not amenable and there are amenable groups which are not residually ﬁnite. An amenable group is a group whose subsets admit an invariant ﬁnitely additive probability measure (see Sect. 4.4). The class of amenable groups contains in particular all ﬁnite groups, all abelian groups and, more generally, all solvable groups (Theorem 4.6.3). It is closed under the operations of taking subgroups, taking quotients, taking extensions, and taking inductive limits (Sect. 4.5). The notion of a Følner net, roughly speaking, a net of almost invariant ﬁnite subsets of a group, is introduced in Sect. 4.7. Paradoxical decompositions are deﬁned in Sect. 4.8. The FølnerTarski theorem (Theorem 4.9.1) asserts that amenability, existence of a Følner net, and nonexistence of a paradoxical decomposition are three equivalent conditions for groups. Another characterization of amenable groups is given in Sect. 4.10: a group is amenable if and only if every continuous aﬃne action of the group on a nonempty compact convex subset of a Hausdorﬀ topological vector space admits a ﬁxed point (Corollary 4.10.2).
4.1 Measures and Means Let E be a set. We shall denote by P(E) the set of all subsets of E. Deﬁnition 4.1.1. A map μ : P(E) → [0, 1] is called a ﬁnitely additive probability measure on E if it satisﬁes the following properties: (1) μ(E) = 1, (2) μ(A ∪ B) = μ(A) + μ(B) for all A, B ∈ P(E) such that A ∩ B = ∅. T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 4, © SpringerVerlag Berlin Heidelberg 2010
77
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4 Amenable Groups
Example 4.1.2. Let F be a nonempty ﬁnite subset of E. Then the map μF : P(E) → [0, 1] deﬁned by μF (A) =
A ∩ F  F 
for all A ⊂ E, is a ﬁnitely additive probability measure on E. Proposition 4.1.3. Let μ : P(E) → [0, 1] be a ﬁnitely additive probability measure on E. Then one has: (i) μ(∅) = 0; (ii) μ(A ∪ B) = μ(A) + μ(B) − μ(A ∩ B); (iii) μ(A ∪ B) ≤ μ(A) + μ(B); (iv) A ⊂ B ⇒ μ(B \ A) = μ(B) − μ(A); (v) A ⊂ B ⇒ μ(A) ≤ μ(B); for all A, B ∈ P(E). Proof. We have μ(E) = μ(E ∪ ∅) = μ(E) + μ(∅), which implies (i). For all A, B ∈ P(E), we have μ(A ∪ B) = μ(A) + μ(B \ A) and μ(B) = μ(A ∩ B) + μ(B \ A), which gives (ii). Since μ(A ∩ B) ≥ 0, equality (ii) implies (iii). If A ⊂ B, we have μ(B) = μ(B \ A) + μ(A), which implies (iv). Finally, inequality (v) follows from (iv) since μ(B \ A) ≥ 0.
Consider now the real vector space ∞ (E) consisting of all bounded functions x : E → R. Recall that ∞ (E) is a Banach space for the norm · ∞ deﬁned by x ∞ = sup x(a). a∈E
∞
We equip (E) with the partial ordering ≤ given by x ≤ y ⇐⇒ (x(a) ≤ y(a) for all a ∈ E). For each λ ∈ R, we shall also denote by λ the element of ∞ (E) which is identically equal to λ on E. Deﬁnition 4.1.4. A mean on E is a linear map m : ∞ (E) → R such that: (1) m(1) = 1, (2) x ≥ 0 ⇒ m(x) ≥ 0 for all x ∈ ∞ (E). Example 4.1.5. Let S be a countable (ﬁnite or inﬁnite) subset of E and let f : S → R such that: (C1) f(s) > 0 for all s ∈ S; (C2) s∈S f (s) = 1.
4.1 Measures and Means
79
Then the map mf : ∞ (E) → R deﬁned by f (s)x(s) mf (x) = s∈S
is a mean on E. One says that a mean m on E has ﬁnite (resp. countable) support if there exists a ﬁnite (resp. countable) subset S ⊂ E and a map f : S → R satisfying conditions (C1) and (C2) above such that m = mf . Proposition 4.1.6. Let m : ∞ (E) → R be a mean on E. Then one has: (i) m(λ) = λ, (ii) x ≤ y ⇒ m(x) ≤ m(y), (iii) inf E x ≤ m(x) ≤ supE x, (iv) m(x) ≤ x ∞ , for all λ ∈ R and x, y ∈ ∞ (E). Proof. (i) By linearity, we have m(λ) = λm(1) = λ. (ii) If x ≤ y then m(y) − m(x) = m(y − x) ≥ 0 and thus m(x) ≤ m(y). (iii) We have inf E x ≤ x ≤ supE x and therefore inf E x ≤ m(x) ≤ supE x by applying (i) and (ii). (iv) From (iii) we get − x ∞ ≤ m(x) ≤ x ∞ , that is, m(x) ≤ x ∞ .
Consider the topological dual of ∞ (E), that is, the vector space (∞ (E))∗ consisting of all continuous linear maps u : ∞ (E) → R. Recall that (∞ (E))∗ is a Banach space for the operator norm · deﬁned by (4.1) u = sup u(x) x∞ ≤1
for all u ∈ (∞ (E))∗ . Proposition 4.1.7. Let m : ∞ (E) → R be a mean on E. Then m ∈ (∞ (E))∗ and m = 1. Proof. By deﬁnition m is linear. Inequality (iv) in Proposition 4.1.6 shows that m is continuous and satisﬁes m ≤ 1. Since m(1) = 1, we have m = 1.
Let PM(E) (resp. M(E)) denote the set of all ﬁnitely additive probability measures (resp. of all means) on E. We are going to show that there is a natural bijection between the sets PM(E) and M(E). This natural bijection, which is analogous to the Riesz representation in measure theory, may be used to view ﬁnitely additive measures on E as points in the dual space (∞ (E))∗ where classical techniques of functional analysis may be applied. For each subset A ⊂ E, we denote by χA the characteristic map of A, that is, the map χA : E → R deﬁned by χA (x) = 1 if x ∈ A and χA (x) = 0 otherwise.
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4 Amenable Groups
Let m ∈ M(E). Consider the map m : P(E) → R given by m(A) = m(χA ). ∈ [0, 1] by Proposition 4.1.6(ii). Observe that 0 ≤ χA ≤ 1 and therefore m(A) We have m(E) = m(χE ) = m(1) = 1. On the other hand, if A and B are disjoint subsets of E, we have χA∪B = χA + χB and thus m(A ∪ B) = m(A) + m(B) by linearity of m. It follows that m ∈ PM(E). Theorem 4.1.8. The map Φ : M(E) → PM(E) deﬁned by Φ(m) = m is bijective. The proof will be divided in several steps. Denote by E(E) the set of all maps x : E → R which take only ﬁnitely many values. Observe that E(E) is the vector subspace of ∞ (E) spanned by the set of characteristic maps {χA : A ⊂ E}. Lemma 4.1.9. The vector subspace E(E) is dense in ∞ (E). Proof. Let x ∈ ∞ (E) and ε > 0. Set α = inf E x and β = supE x. Choose an integer n ≥ 1 such that (β − α)/n < ε and set λi = α + i(β − α)/n for i = 1, 2, . . . , n. Consider the map y : E → R deﬁned by y(a) = min{λi : x(a) ≤ λi } for all a ∈ E. The map y take its values in the set {λ1 , . . . , λn }. Thus y ∈ E(E). We have x − y ∞ ≤ (β − α)/n < ε by construction. Consequently,
E(E) is dense in ∞ (E). Let μ ∈ PM(E). Let us set, for all x ∈ E(E), μ(x) = μ(x−1 (λ))λ.
(4.2)
λ∈R
Observe that there is only a ﬁnite number of nonzero terms in the righthand side of (4.2) since x ∈ E(E) and μ(∅) = 0. Lemma 4.1.10. Let x ∈ E(E). Suppose that there is a ﬁnite partition (Ai )i∈I of E such that the restriction of x to Ai is constant equal to αi for each i ∈ I. Then one has μ(x) = μ(Ai )αi . i∈I
Proof. Since μ is ﬁnitely additive, we have μ(Ai )αi = μ(Ai ) λ = μ(x−1 (λ))λ = μ(x). i∈I
λ∈R
αi =λ
λ∈R
4.1 Measures and Means
81
Lemma 4.1.11. The map μ : E(E) → R is linear and continuous. Proof. Let x, y ∈ E(E) and ξ, η ∈ R. Denote by V (resp. W ) the set of values taken by x (resp. y). The subsets x−1 (α) ∩ y −1 (β), (α, β) ∈ V × W , form a ﬁnite partition of E. By applying Lemma 4.1.10, we get μ(ξx + ηy) = μ(x−1 (α) ∩ y −1 (β))(ξα + ηβ) (α,β)∈V ×W
⎛
⎞
= ξ⎝
μ(x−1 (α) ∩ y −1 (β))α⎠
(α,β)∈V ×W
⎛
+ ⎝η
⎞ μ(x−1 (α) ∩ y −1 (β))β ⎠
(α,β)∈V ×W
= ξμ(x) + ημ(y). Consequently, μ is linear. Formula (4.2) implies inf E x ≤ μ(x) ≤ supE x since μ(x−1 (λ)) = μ(E) = 1. λ∈R
It follows that μ(x) ≤ x ∞ for all x ∈ E(E). This shows that the linear
map μ is continuous. Lemma 4.1.12. Let X be a normed vector space and let Y be a dense vector subspace of X. Suppose that ϕ : Y → R is a continuous linear map. Then there exists a continuous linear map ϕ : X → R extending ϕ (that is, such that ϕ Y = ϕ). Proof. Let x ∈ X. Since Y is dense in X, we can ﬁnd a sequence (yn )n≥0 of points of Y which converges to x. The sequence (ϕ(yn )) is a Cauchy sequence since ϕ(yp ) − ϕ(yq ) = ϕ(yp − yq ) ≤ ϕ yp − yq for all p, q ≥ 0. As R is complete, this sequence converges. Set ϕ(x) = lim ϕ(yn ). If (yn ) is another sequence of points in Y converging to x, then ϕ(yn ) − ϕ(yn ) ≤ ϕ yn − yn . Thus we have lim ϕ(yn ) = lim ϕ(yn ). This shows that ϕ(x) does not depend of the choice of the sequence (yn ). The linearity of ϕ gives the linearity of ϕ by taking limits. We also get ϕ(x) ≤ ϕ x for all x ∈ X by taking limits. This shows that the linear map ϕ is continuous. If x ∈ Y , we have ϕ(x) = ϕ(x) since we can take as extends ϕ.
(yn ) the constant sequence yn = x in this case. Thus ϕ Proof of Theorem 4.1.8. Let μ ∈ PM(E). By Lemma 4.1.9, Lemma 4.1.11 and Lemma 4.1.12, the map μ : E(E) → R can be extended to a continuous linear map μ : ∞ (E) → R. We have
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4 Amenable Groups
μ (1) = μ(1) = μ(E) = 1. If y ∈ E(E) and y ≥ 0, then μ (y) = μ(y) ≥ 0 by (4.2). Consider now an element x ∈ ∞ (E) such that x ≥ 0. Let (yn )n≥0 be a sequence of elements of E(E) converging to x. Let us set zn = yn . Since zn ∈ E(E) and zn ≥ 0, we have μ (zn ) ≥ 0. On the other hand, the sequence (zn ) converges to x since, by the triangle inequality, x − zn ∞ ≤ x − yn ∞ . It follows that μ (x) = lim μ(zn ) ≥ 0. n→∞
Thus μ ∈ M(E). For every subset A ⊂ E, we have μ (χA ) = μ(χA ) = μ(A). Therefore Φ( μ) = μ. This proves that Φ is surjective. It remains to show that Φ is injective. To see this, consider m1 , m2 ∈ M(E) such that Φ(m1 ) = Φ(m2 ). This means that m1 (χA ) = m2 (χA ) for every subset A ⊂ E. By linearity, this implies that m1 and m2 coincide on E(E). Since E(E) is dense in ∞ (E) by Lemma 4.1.9, we deduce that m1 = m2 by
continuity of m1 and m2 . This shows that Φ is injective.
4.2 Properties of the Set of Means Let E be a set. The topology on (∞ (E))∗ associated with the operator norm · deﬁned in (4.1) is called the strong topology on (∞ (E))∗ . Another topology that is commonly used is the weak∗ topology on (∞ (E))∗ . Recall that the weak∗ topology on (∞ (E))∗ is by deﬁnition the smallest topology for which the evaluation map ψx : (∞ (E))∗ → R u → u(x) is continuous for each x ∈ ∞ (E) (see Sect. F.2). By Proposition 4.1.7, the set M(E) is contained in the unit sphere {u : u = 1} ⊂ (∞ (E))∗ . The following result will play an important role in the sequel. Theorem 4.2.1. The set M(E) is a convex compact subset of (∞ (E))∗ with respect to the weak∗ topology.
4.3 Measures and Means on Groups
83
Proof. Let m1 , m2 ∈ M(E) and t ∈ [0, 1]. Then (tm1 + (1 − t)m2 )(1) = tm1 (1) + (1 − t)m2 (1) = t + (1 − t) = 1 and (tm1 + (1 − t)m2 )(x) = tm1 (x) + (1 − t)m2 (x) ≥ 0 for all x ∈ ∞ (E) such that x ≥ 0. Thus tm1 + (1 − t)m2 ∈ M(E). This shows that M(E) is convex. Equip now (∞ (E))∗ with the weak∗ topology and suppose that (mi )i∈I is a net in M(E) converging to u ∈ (∞ (E))∗ . Then, for every i ∈ I, we have ψ1 (mi ) = mi (1) = 1 and ψx (mi ) = mi (x) ≥ 0 for all x ∈ ∞ (E) such that x ≥ 0. By taking limits, we get u(1) = ψ1 (u) = 1 and u(x) = ψx (u) ≥ 0 for all x ∈ ∞ (E) such that x ≥ 0. Thus u ∈ M(E). This shows that M(E) is closed in (∞ (E))∗ (Proposition A.2.1). As M(E) is contained in the unit ball of (∞ (E))∗ , which is compact for the weak∗ topology by the BanachAlaoglu Theorem (Theorem F.3.1), it follows that M(E) is compact.
4.3 Measures and Means on Groups In this section, we shall see that the set of ﬁnitely additive measures and the set of means carry additional structure when the underlying set is a group. Indeed, in this case, the group naturally acts on both sets. Moreover, there are involutions coming from the operation of taking inverses in the group. More precisely, let G be a group. The group G naturally acts on the left and on the right on each of the sets PM(G) and M(G) in the following way. Firstly, for μ ∈ PM(G) and g ∈ G, we deﬁne the maps gμ : P(G) → [0, 1] and μg : P(G) → [0, 1] by gμ(A) = μ(g −1 A)
and
μg(A) = μ(Ag −1 )
for all A ∈ P(G). One clearly has gμ ∈ PM(G) and μg ∈ PM(G). Moreover, it is straightforward to check that the map (g, μ) → gμ (resp. (μ, g) → μg) deﬁnes a left (resp. right) action of G on PM(G). Note that these two actions commute in the sense that g(μh) = (gμ)h for all g, h ∈ G and μ ∈ PM(G). On the other hand, recall that we introduced in Sect. 1.1 a left action of G on RG (the Gshift) by deﬁning, for all g ∈ G and x ∈ RG , the element gx ∈ RG by gx(g ) = x(g −1 g ) for all g ∈ G. Similarly, we make G act on the right on RG by deﬁning the element xg ∈ RG by xg(g ) = x(g g −1 ) for all g ∈ G. These two actions of G on RG are linear and commute. Moreover, the vector subspace ∞ (G) ⊂ RG is left invariant by both actions. Observe that gx ∞ = xg ∞ = x ∞ for all g ∈ G and x ∈ ∞ (G). Thus the left (resp. right) action of G on ∞ (G) is isometric (and therefore continuous).
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4 Amenable Groups
By duality, we also get a left and a right action of G on (∞ (G))∗ . More precisely, for g ∈ G and u ∈ (∞ (G))∗ , we deﬁne the elements gu and ug by gu(x) = u(g −1 x)
and
ug(x) = u(xg −1 ),
for all x ∈ ∞ (G). We have gu ∈ (∞ (G))∗ and ug ∈ (∞ (G))∗ since the left and right actions of G on ∞ (G) are linear and continuous. Note that the set M(G) is invariant under both actions of G on (∞ (G))∗ . Proposition 4.3.1. The left (resp. right) action of G on M(G) is aﬃne and continuous with respect to the weak∗ topology on M(G). Proof. It is clear that the left (resp. right) action of G on (∞ (G))∗ is linear. On the other hand, these actions are continuous if we equip (∞ (G))∗ with the weak∗ topology. Indeed, if we ﬁx g ∈ G, the map u → gu is continuous on (∞ (G))∗ since, for each x ∈ ∞ (G), the map u → gu(x) is the evaluation map at g −1 x, which is continuous by deﬁnition of the weak∗ topology. Similarly, the map u → ug is continuous. Since M(G) is a convex subset of (∞ (G))∗ , we deduce that the restrictions of both actions to M(G) are aﬃne and continuous.
Given μ ∈ PM(G), we deﬁne the map μ∗ : P(G) → [0, 1] by μ∗ (A) = μ(A−1 ) for all A ∈ P(G). It is clear that μ∗ ∈ PM(G) and that the map μ → μ∗ is an involution of PM(G). For x ∈ ∞ (G), deﬁne x∗ ∈ ∞ (G) by x∗ (g) = x(g −1 ) for all g ∈ G. The map x → x∗ is an isometric involution of ∞ (G). By duality, it gives an isometric involution u → u∗ of (∞ (G))∗ deﬁned by u∗ (x) = u(x∗ )
for all x ∈ ∞ (G).
Note that m∗ ∈ M(G) for all m ∈ M(G). Proposition 4.3.2. Let g ∈ G, x ∈ ∞ (G), μ ∈ PM(G) and u ∈ (∞ (G))∗ . Then one has (i) (gx)∗ = x∗ g −1 ; (ii) (xg)∗ = g −1 x∗ ; (iii) (gμ)∗ = μ∗ g −1 ; (iv) (μg)∗ = g −1 μ∗ ; (v) (gu)∗ = u∗ g −1 ; (vi) (ug)∗ = g −1 u∗ .
4.4 Deﬁnition of Amenability
85
Proof. For every h ∈ G, we have (gx)∗ (h) = gx(h−1 ) = x(g −1 h−1 ) = x∗ (hg) = x∗ g −1 (h), which gives (gx)∗ = x∗ g −1 . The proofs of the other properties are similar.
Proposition 4.3.3. Let g ∈ G and m ∈ M(G). Then one has: (i) gm = g m; (ii) mg = mg; ∗ = m ∗. (iii) m We need the following result. Lemma 4.3.4. Let g ∈ G and A ⊂ G. Then one has: (i) (χA )∗ = χA−1 ; (ii) gχA = χgA . Proof. (i) For h ∈ G one has (χA )∗ (h) = χA (h−1 ) = 1 ⇔ h−1 ∈ A ⇔ h ∈ A−1 ⇔ χA−1 (h) = 1. This shows that (χA )∗ = χA−1 . (ii) For h ∈ G one has (gχA )(h) = χA (g −1 h) = 1 ⇔ g −1 h ∈ A ⇔ h ∈ gA ⇔ χgA (h) = 1. This shows that gχA = χgA .
Proof of Proposition 4.3.3. For every A ∈ P(G), we have, using Lemma 4.3.4(ii), gm(A) = gm(χA ) = m(g −1 χA ) = m(χg−1 A ) = m(g −1 A) = g m(A). Thus gm = g m. Similarly, we get mg = mg. On the other hand, for every A ∈ P(G), using Lemma 4.3.4(i) one obtains ∗ (A) −1 ) = m(χA−1 ) = m((χA )∗ ) = m∗ (χA ) = m m ∗ (A) = m(A ∗ . and this shows that m ∗ = m
Remark 4.3.5. Properties (i) and (ii) in Proposition 4.3.3 say that the bijective map Φ : M(G) → PM(G) given by m → m (see Theorem 4.1.8) is biequivariant.
4.4 Deﬁnition of Amenability Let G be a group. A ﬁnitely additive probability measure μ ∈ PM(G) is called leftinvariant (resp. rightinvariant) if μ is ﬁxed under the left (resp. right) action of G on PM(G), that is, if it satisﬁes gμ = μ (resp. μg = μ) for all g ∈ G. One says that μ is biinvariant if μ is both left and rightinvariant.
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4 Amenable Groups
Similarly, a mean m ∈ M(G) is called leftinvariant (resp. rightinvariant) if m is ﬁxed under the left (resp. right) action of G on M(G). One says that m is biinvariant if m is both left and right invariant. Proposition 4.4.1. Let μ ∈ PM(G). Then μ is leftinvariant (resp. rightinvariant) if and only if μ∗ is rightinvariant (resp. leftinvariant). Proof. This immediately follows from Proposition 4.3.2(ii).
Proposition 4.4.2. Let m ∈ M(G). Then m is leftinvariant (resp. rightinvariant) if and only if m∗ is rightinvariant (resp. leftinvariant). Proof. This immediately follows from Proposition 4.3.2(iii).
Proposition 4.4.3. Let m ∈ M(G). Then m is leftinvariant (resp. rightinvariant, resp. biinvariant) if and only if the associated ﬁnitely additive probability measure m ∈ PM(G) is leftinvariant (resp. rightinvariant, resp. biinvariant). Proof. This immediately follows from assertions (i) and (ii) of Proposition 4.3.3.
Proposition 4.4.4. Let G be a group. Then the following conditions are equivalent: (a) there exists a leftinvariant ﬁnitely additive probability measure μ : P(G) → [0, 1] on G; (b) there exists a rightinvariant ﬁnitely additive probability measure μ : P(G) → [0, 1] on G; (c) there exists a biinvariant ﬁnitely additive probability measure μ : P(G) → [0, 1] on G; (d) there exists a leftinvariant mean m : ∞ (G) → R on G; (e) there exists a rightinvariant mean m : ∞ (G) → R on G; (f) there exists a biinvariant mean m : ∞ (G) → R on G. Proof. Equivalences (a) ⇔ (d), (b) ⇔ (e) and (c) ⇔ (f) immediately follow from Proposition 4.4.3. The equivalence between (a) and (b) follows from Proposition 4.4.1. Implication (f) ⇒ (d) is trivial. Thus it suﬃces to show (d) ⇒ (f) to conclude. Suppose that there exists a leftinvariant mean m : ∞ (G) → R. For each x ∈ ∞ (G), deﬁne the map x : G → R by x (g) = m(xg) for all g ∈ G. By Proposition 4.1.6(iv), we have  x(g) = m(xg) ≤ xg ∞ = x ∞ . Therefore x ∈ ∞ (G) for all x ∈ ∞ (G). Consider now the map M : ∞ (G) → R deﬁned by
4.4 Deﬁnition of Amenability
87
M (x) = m( x) for all x ∈ ∞ (G). Clearly M is a mean on G. Let h ∈ G and x ∈ ∞ (G). For all g ∈ G, we have
hx(g) = m(hxg) = m(xg) = x (g),
=x since m is leftinvariant. We deduce that hx . This implies
= m( M (hx) = m(hx) x) = M (x). Consequently M is leftinvariant. On the other hand, for all g ∈ G, we have
xh(g) = m((xh)g) = m(x(hg)) = x (hg) = h−1 x (g).
= h−1 x It follows that xh . Therefore, by using again the fact that m is leftinvariant, we get
= m(h−1 x M (xh) = m(xh) ) = m( x) = M (x). Therefore the mean M is also rightinvariant. This shows that (d) implies (f).
Deﬁnition 4.4.5. A group G is called amenable if it satisﬁes one of the equivalent conditions of Proposition 4.4.4. Proposition 4.4.6. Every ﬁnite group is amenable. Proof. If G is a ﬁnite group, then the map μ : P(G) → [0, 1] deﬁned by μ(A) =
A G
for all A ∈ P(G)
is a biinvariant ﬁnitely additive probability measure on G. Theorem 4.4.7. The free group on two generators F2 is not amenable.
Proof. Suppose that there exists a leftinvariant ﬁnitely additive probability measure μ : P(F2 ) → [0, 1]. Denote by a and b the canonical generators of F2 . Let A ⊂ F2 be the set of elements of F2 whose reduced form begins by a nonzero (positive or negative) power of a. We have F2 = A ∪ aA and hence μ(F2 ) ≤ μ(A) + μ(aA) = 2μ(A). Since μ(F2 ) = 1, this implies μ(A) ≥
1 . 2
(4.3)
On the other hand, the subsets A, bA and b2 A are pairwise disjoint. Thus one has μ(F2 ) ≥ μ(A) + μ(bA) + μ(b2 A) = 3μ(A). This gives
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4 Amenable Groups
μ(A) ≤
1 3
contradicting (4.3).
4.5 Stability Properties of Amenable Groups Proposition 4.5.1. Every subgroup of an amenable group is amenable. Proof. Let G be an amenable group and let H be a subgroup of G. Let μ : P(G) → [0, 1] be a leftinvariant ﬁnitely additive probability measure on G. Choose a complete set of representatives of the right cosets of G modulo H, that is, a subset R ⊂ G such that, for each g ∈ G, there exists a unique element (h, r) ∈ H × R satisfying g = hr. Let us check that the map μ : P(H) → [0, 1] deﬁned by Ar for all A ∈ P(H) μ (A) = μ r∈R
is a leftinvariant ﬁnitely additive probability measure on H. Firstly, we have Hr = μ(G) = 1. μ (H) = μ r∈R
On the other hand, if A and B are disjoint subsets of H, then μ (A ∪ B) = μ (A ∪ B)r r∈R
=μ
=μ
∪
Ar
r∈R
Ar
+μ
r∈R
=μ (A) + μ (B), r∈R
are disjoint.
Ar
and
Br
r∈R
r∈R
since the sets
r∈R
Br
Br
4.5 Stability Properties of Amenable Groups
89
Finally, for all h ∈ H and A ∈ P(H), we have μ (hA) = μ hAr = μ h Ar = μ Ar = μ (A). r∈R
r∈R
r∈R
This shows that H is amenable. Combining Proposition 4.5.1 with Theorem 4.4.7, we get:
Corollary 4.5.2. If G is a group containing a subgroup isomorphic to the free group on two generators F2 , then G is not amenable.
Examples 4.5.3. (a) Let X be a set having at least two elements. Then the free group F (X) based on X is not amenable. Indeed, the subgroup of F (X) generated by two distinct elements a, b ∈ X is isomorphic to F2 . (b) The group SL(n, Z) is not amenable for n ≥ 2. Indeed, as we haveseen 12 in Lemma 1 0 2.3.2, the subgroup of SL(2, Z) generated by the matrices 0 1 and 2 1 is isomorphic to F2 . Proposition 4.5.4. Every quotient of an amenable group is amenable. Proof. Let G be an amenable group and let H be a normal subgroup of G. Let ρ : G → G/H denote the canonical homomorphism. Consider a leftinvariant ﬁnitely additive probability measure μ : P(G) → [0, 1] on G. Let us show that the map μ : P(G/H) → [0, 1] deﬁned by μ (A) = μ(ρ−1 (A))
for all A ∈ P(G/H)
is a leftinvariant ﬁnitely additive probability measure on G/H. Firstly, we have μ (G/H) = μ((ρ−1 (G/H)) = μ(G) = 1. On the other hand, if A and B are disjoint subsets of G/H, then μ (A ∪ B) = μ(ρ−1 (A ∪ B)) = μ(ρ−1 (A) ∪ ρ−1 (B)) = μ(ρ−1 (A)) + μ(ρ−1 (B)) (A ∪ B) = μ (A) + μ (B). since ρ−1 (A) ∩ ρ−1 (B) = ∅. Thus we have μ Finally, if g ∈ G and A ⊂ G/H, we have (A). μ (ρ(g)A) = μ(ρ−1 (ρ(g)A)) = μ(gρ−1 (A)) = μ(ρ−1 (A)) = μ As ρ is surjective, it follows that μ is leftinvariant with respect to G/H. This shows that G/H is amenable.
In the next proposition we show that the class of amenable groups is closed under the operation of taking extensions of amenable groups by amenable groups.
90
4 Amenable Groups
Proposition 4.5.5. Let G be a group and let H be a normal subgroup of G. Suppose that the groups H and G/H are amenable. Then the group G is amenable. Proof. Since H is amenable, we can ﬁnd a Hleftinvariant mean m0 : ∞ (H) → : G/H → R deﬁned by R. Let x ∈ ∞ (G). Consider the map x x (g) = m0 ((g −1 x)H )
for all g ∈ G,
is where g = gH = Hg ∈ G/H denote the class of g modulo H. The map x well deﬁned. Indeed, if g1 and g2 are elements of G such that g1 = g2 , then g2 = g1 h for some h ∈ H, and hence m0 ((g2−1 x)H ) = m0 ((h−1 g1−1 x)H ) = m0 (h−1 (g1−1 x)H ) = m0 ((g1−1 x)H ) ∈ ∞ (G/H) since, by Proposince m0 is Hleftinvariant. Observe also that x sition 4.1.6(iv),  x(g) = m0 ((g −1 x)H ) ≤ sup (g −1 x)(h) ≤ g −1 x ∞ = x ∞ h∈H
for all g ∈ G. As the group G/H is amenable, there exists a G/Hleftinvariant mean m1 : ∞ (G/H) → R. Let us set, for each x ∈ ∞ (G), x). m(x) = m1 ( Clearly m is a mean on G. Let us show that it is Gleftinvariant. Let g ∈ G and x ∈ ∞ (G). For all g ∈ G, we have g x(g ) = m0 ((g
−1
gx)H ) = m0 (((g −1 g )−1 x)H ) = x (g −1 g ) = x (g −1 g )
x(g ). = g x. As m1 is G/Hleftinvariant, it follows that Thus g x = g g x) = m1 (g x) = m1 ( x) = m(x), m(gx) = m1 ( which shows that m is Gleftinvariant.Therefore G is amenable.
Corollary 4.5.6. Suppose that G1 and G2 are amenable groups. Then the group G = G1 × G2 is amenable. Proof. The set H = {(g1 , 1G2 ) : g1 ∈ G1 } is a normal subgroup of G isomor
phic to G1 with quotient G/H isomorphic to G2 . Remark 4.5.7. It immediately follows from the preceding corollary that every direct product of a ﬁnite number of amenable groups is amenable. However, a direct product of inﬁnitely many amenable groups is not necessarily amenable. For example, it follows from Theorem 2.3.1 and Corollary 2.2.6
4.5 Stability Properties of Amenable Groups
91
that there exists a (countable) family of ﬁnite, and hence amenable, groups (Gi )i∈I such that the group G = i∈I Gi contains a subgroup isomorphic to F2 . Such a group G is not amenable by Corollary 4.5.2. Corollary 4.5.8. Every virtually amenable group is amenable. Proof. Let G be a virtually amenable group. Let H be an amenable subgroup of ﬁnite index of G. By Lemma 2.1.10, the set K = ∩g∈G gHg −1 is a normal subgroup of ﬁnite index of G contained in H. The group K is amenable by Proposition 4.5.1. On the other hand the group G/K is ﬁnite and hence amenable. Consequently, G is amenable by Proposition 4.5.5.
Our next goal is to show that an inductive limit of amenable groups is amenable. In the proof we shall use the following: Lemma 4.5.9. Let G be a group. Suppose that there is a net (mi )i∈I in M(G) such that, for each g ∈ G, the net (gmi − mi )i∈I converges to 0 in (∞ (G))∗ for the weak∗ topology. Then G is amenable. Proof. Since M(G) is compact for the weak∗ topology by Theorem 4.2.1, we may assume, after taking a subnet if necessary, that the net (mi ) converges to a mean m ∈ M(G). Let g ∈ G. Since the left action of G on M(G) is continuous by Proposition 4.3.1, we deduce that, for every x ∈ ∞ (G) the net (gmi − mi )(x) = gmi (x) − mi (x) converges to 0. By taking limits, we get gm(x) − m(x) = 0. Therefore gm = m. This shows that the mean m is leftinvariant. Thus G is amenable.
Proposition 4.5.10. Every group which is the limit of an inductive system of amenable groups is amenable. Proof. Let (Gi )i∈I be an inductive system of amenable groups and set G = lim Gi . Consider the family (Hi )i∈I of subgroups G deﬁned by Hi = hi (Gi ), −→ where hi : Gi → G is the canonical homomorphism. As Hi is amenable by Proposition 4.5.4, we can ﬁnd, for each i ∈ I, a Hi leftinvariant mean m
i : ∞ (Hi ) → R. Consider the family mi : ∞ (G) → R of means on G deﬁned by
i (xHi ) for all x ∈ ∞ (G). mi (x) = m Let g ∈ G. Since G = lim Gi , there exists i0 (g) ∈ I such that g ∈ Hi for all −→ i ≥ i0 (g). For x ∈ ∞ (G) and i ≥ i0 (g), we have
i (xHi ) − m
i (xHi ) = 0, (gmi − mi )(x) = gmi (x) − mi (x) = g m since m
i is Hi leftinvariant. Thus gmi −mi = 0 for all i ≥ i0 (g). By applying Lemma 4.5.9, we deduce that G is amenable.
Every group is the inductive limit of its ﬁnitely generated subgroups. Therefore we have:
92
4 Amenable Groups
Corollary 4.5.11. Every locally amenable group is amenable.
Since every ﬁnite group is amenable, we obtain in particular the following: Corollary 4.5.12. Every locally ﬁnite group is amenable.
Example 4.5.13. Let X be a set. Then the group Sym0 (X) consisting of all permutations of X with ﬁnite support is locally ﬁnite (see Example 3.2.4). Consequently, the group Sym0 (X) is amenable. The groups Sym0 (X), where X is a inﬁnite set, are our ﬁrst examples of inﬁnite amenable groups. Recall from Lemma 2.6.3 that Sym0 (X) is not residually ﬁnite whenever X is inﬁnite. Corollary 4.5.14. Let (Gi )i∈I be a family of amenable groups. Then their direct sum G = i∈I Gi is amenable. Proof. Every ﬁnitely generated subgroup of G is a subgroup of some ﬁnite product of the groups Gi and hence amenable by Corollary 4.5.6 and Proposition 4.5.1. Thus G is locally amenable and therefore amenable by Corollary 4.5.11.
4.6 Solvable Groups Theorem 4.6.1. Every abelian group is amenable. Proof. Let G be an abelian group. Equip (∞ (G))∗ with the weak∗ topology. By Theorem 4.2.1, the set M(G) is a nonempty convex compact subset of (∞ (G))∗ . On the other hand, it follows from Proposition 4.3.1 that the action of G on M(G) is aﬃne and continuous (note that the left and the right actions coincide since G is Abelian). By applying the MarkovKakutani ﬁxedpoint Theorem (Theorem G.1.1), we deduce that G has at least one ﬁxed point in M(G). Such a ﬁxed point is clearly a biinvariant mean on G. This shows that G is amenable.
Let G be a group. Recall the following deﬁnitions. The commutator of two elements h and k in G is the element [h, k] ∈ G deﬁned by [h, k] = hkh−1 k −1 . If H and K are subgroups of G, we denote by [H, K] the subgroup of G generated by all commutators [h, k], where h ∈ H and k ∈ K. Note that [H, K] ⊂ K if K is normal in G. Note also that [H, K] is normal in G if H and K are both normal in G. The subgroup D(G) = [G, G] is called the derived subgroup, or commutator subgroup, of G. The subgroup D(G) is normal in G and the quotient group G/D(G) is abelian. Observe that G is abelian if and only if [G, G] = {1G }. A group is called metabelian if its derived subgroup is abelian. The derived series of a group G is the sequence (Di (G))i≥0 of subgroups of G inductively deﬁned by D0 (G) = G and Di+1 (G) = D(Di (G)) for all i ≥ 0. One has
4.6 Solvable Groups
93
G = D0 (G) ⊃ D1 (G) ⊃ D2 (G) ⊃ . . . with Di+1 (G) normal in Di (G) and Di (G)/Di+1 (G) abelian for all i ≥ 0. The group G is said to be solvable if there is an integer i ≥ 0 such that Di (G) = {1G }. The smallest integer i ≥ 0 such that Di (G) = {1G } is then called the solvability degree of G. Examples 4.6.2. (a) A group is solvable of degree 0 if and only if it is reduced to the identity element. (b) The solvable groups of degree 1 are the nontrivial abelian groups. (c) The solvable groups of degree 2 are the nonabelian metabelian groups. (d) Let K be a ﬁeld. The aﬃne group over K is the subgroup Aﬀ(K) of Sym(K) consisting of all permutations of K which are of the form x → ax + b, where a, b ∈ K and a = 0. The group D1 (Aﬀ(K)) is the group of translations x → x + b, b ∈ K. Since D1 (Aﬀ(K)) is abelian, we have Di (Aﬀ(K)) = {1Aﬀ(K) } for all i ≥ 2. Consequently, the group Aﬀ(K) is solvable of degree 2. + (e) The alternating groups Sym+ 2 and Sym3 are abelian and hence solvable + of degree 1. The group Sym4 is solvable of degree 2. For n ≥ 5, the alternating + + i group Sym+ n is simple (see Remark C.4.4) and therefore D (Symn ) = Symn + for all i ≥ 0. Thus Symn is not solvable for n ≥ 5. (f) The symmetric group Symn is solvable of degree 1 for n = 2, solvable of degree 2 for n = 3, solvable of degree 3 for n = 4, and not solvable for n ≥ 5. Theorem 4.6.3. Every solvable group is amenable. Proof. We proceed by induction on the solvability degree i of the group. For i = 0, the group is reduced to the identity element and there is nothing to prove. Suppose now that the statement is true for solvable groups of degree i for some i ≥ 0. Let G be a solvable group of degree i + 1. Then its derived subgroup D(G) is solvable of degree i and hence amenable by our induction hypothesis. As G/D(G) is abelian and therefore amenable by Theorem 4.6.1, we deduce that G is amenable by applying Proposition 4.5.5.
Remark 4.6.4. The alternating group Sym+ 5 is amenable since it is ﬁnite. However, as mentioned above, Sym+ 5 is not solvable. An example of an inﬁnite amenable group which is not solvable is provided by the group Z × Sym+ 5. Let G be a group. The lower central series of G is the sequence (C i (G))i≥0 of subgroups of G deﬁned by C 0 (G) = G and C i+1 (G) = [C i (G), G] for all i ≥ 0. An easy induction shows that C i (G) is normal in G and that C i+1 (G) ⊂ C i (G) for all i. The group G is said to be nilpotent if there is an integer i ≥ 0 such that C i (G) = {1G }. The smallest integer i ≥ 0 such that C i (G) = {1G } is then called the nilpotency degree of G.
94
4 Amenable Groups
Example 4.6.5. Let R be a nontrivial commutative ring. The Heisenberg group with coeﬃcients in R is the subgroup HR of GL3 (R) consisting of all matrices of the form ⎛ ⎞ 1yz M (x, y, z) = ⎝0 1 x⎠ (x, y, z ∈ R). 001 One easily checks that the center Z(HR ) of HR consists of all matrices of the form M (0, 0, z), z ∈ R, and that Z(HR ) is isomorphic to the additive group (R, +). Moreover, one has D(HR ) = Z(HR ). It follows that HR is nilpotent of degree 2. Proposition 4.6.6. Every nilpotent group is solvable. Proof. An easy induction yields Di (G) ⊂ C i (G) for all i ≥ 0.
Remark 4.6.7. We have seen in Example 4.6.2(d) that the aﬃne group Aﬀ(K) over a ﬁeld K is solvable. However, the group G = Aﬀ(K) is not nilpotent. Indeed, the lower central series satisﬁes C i (G) = D(G) = {1G } for all i ≥ 1. From Theorem 4.6.3 and Proposition 4.6.6, we get: Corollary 4.6.8. Every nilpotent group is amenable.
4.7 The Følner Conditions Proposition 4.7.1. Let G be a group. Then the following conditions are equivalent: (a) for every ﬁnite subset K ⊂ G and every real number ε > 0, there exists a nonempty ﬁnite subset F ⊂ G such that F \ kF  0, there exists a nonempty ﬁnite subset F ⊂ G such that F \ F k 0. We equip J with the partial ordering ≤ deﬁned by (K, ε) ≤ (K , ε ) ⇔ (K ⊂ K and ε ≥ ε ). Note that (J, ≤) is a lattice, that is, a partially ordered set in which any two elements admit a supremum (also called a join) and an inﬁmum (also called a meet). Indeed, one has sup{(K, ε), (K , ε )} = (K ∪ K , min(ε, ε )) and inf{(K, ε), (K , ε )} = (K ∩ K , max(ε, ε )). In particular, J is a directed set. By (a), for each j = (K, ε) ∈ J, there exists a nonempty ﬁnite subset Fj ⊂ G such that Fj \ kFj  0. Set j0 = ({g}, ε0 ). If j = (K, ε) satisﬁes j ≥ j0 , then g ∈ K and ε ≤ ε0 . Thus we have Fj \ gFj  < ε0 , Fj 
(4.10)
for all j ≥ j0 by (4.9). This shows that the net (Fj )j∈J satisﬁes (4.5). Consequently, (a) implies (b). Finally, let us show (b) ⇒ (a). Suppose (b). Let K ⊂ G be a ﬁnite subset and let ε > 0. By (4.5), for every k ∈ K there exist j(k) ∈ J such that Fj \ kFj  0 such that for every nonempty ﬁnite subset F ⊂ G one has F \ k0 F  ≥ ε0 F 
(4.16)
for some k0 = k0 (F ) ∈ K0 . Consider the set K1 = K0 ∪ {1G }. Let F be a nonempty ﬁnite subset of G. Observe that K1 F ⊃ F and K1 F \F = K0 F \F . Thus, we have K1 F  − F  = K1 F \ F  = K0 F \ F  ≥ k0 F \ F  = F \ k0 F  (since F  = k0 F ) ≥ ε0 F  (by (4.16)), which gives K1 F  ≥ (1 + ε0 )F .
102
4 Amenable Groups
Choose n0 ∈ N such that (1 + ε0 )n0 ≥ 2 and set K = K1n0 . Then, we have KF  = K1n0 F  ≥ (1 + ε0 )K1n0 −1 F  ≥ · · · ≥ (1 + ε0 )n0 F  so that KF  ≥ 2F  for every ﬁnite subset F ⊂ G. This shows that (b) implies (c). (c) ⇒ (e). Suppose that G satisﬁes condition (c), that is, there exists a ﬁnite subset K ⊂ G such that KF  ≥ 2F  for every ﬁnite subset F ⊂ G.
(4.17)
Consider the bipartite graph GK (G) = (G, G, E) (see Appendix H), where the set E ⊂ G × G of edges consists of all the pairs (g, h) such that g ∈ G and h ∈ Kg. We claim that GK (G) satisﬁes the Hall 2harem conditions. Indeed, if F is a ﬁnite subset of G, then, using the terminology for bipartite graphs introduced in Sect. H.1, the right and left neighborhoods of F in GK (G) are the sets NR (F ) = KF and NL (F ) = K −1 F respectively. Therefore, we have NR (F ) = KF  ≥ 2F , by applying (4.17). On the other hand, if k ∈ K, then NL (F ) ⊃ k−1 F so that 1 NL (F ) ≥ k−1 F  = F  ≥ F . 2 This proves our claim. Thus, by virtue of the Hall harem Theorem (Theorem H.4.2), we deduce the existence of a perfect (1, 2)matching M for GK (G). In other words, there exists a 2toone surjective map ϕ : G → G such that (ϕ(g), g) ∈ E, that is, g(ϕ(g))−1 ∈ K for all g ∈ G. This shows that (c) implies (e). (e) ⇒ (g). Suppose (e), that is, there exist a 2toone surjective map ϕ : G → G and a ﬁnite set K ⊂ G such that g(ϕ(g))−1 ∈ K for all g ∈ G.
(4.18)
By the axiom of choice, we can ﬁnd maps ψ1 , ψ2 : G → G such that, for every g ∈ G, the elements ψ1 (g) and ψ2 (g) are the two preimages of g for ϕ. Observe that θ1 (g) = ψ1 (g)g −1 and θ2 (g) = ψ2 (g)g −1 belong to K for every g ∈ G by (4.18). For each k ∈ K, deﬁne Ak and Bk by Ak = {g ∈ G : θ1 (g) = k} We have G=
k∈K
and Bk = {g ∈ G : θ2 (g) = k}.
Ak =
k∈K
Bk .
(4.19)
4.10 The Fixed Point Property
103
ψ1 (G) = On the other hand, observe that if g ∈Ak then ψ1 (g) = kg. Thus ψ2 (G), k∈K kAk . Similarly, we have ψ2 (G) = k∈K kBk . As G = ψ1 (G) we deduce that (4.20) kAk kBk . G= k∈K
k∈K
Combining together (4.19) and (4.20), ,we deduce that (K, (Ak )k∈K , (Bk )k∈K ) is a left paradoxical decomposition for G. This shows that (e) implies (g). (g) ⇒ (a). Suppose that G admits a left paradoxical decomposition (K, (Ak )k∈K , (Bk )k∈K ). If μ : P(G) → [0, 1] is a left invariant ﬁnitely additive probability measure on G, then (4.11) gives 1 = μ(G)
=μ =
kAk
k∈K
μ(kAk ) +
k∈K
=
kBk
k∈K
μ(kBk )
k∈K
μ(Ak ) +
k∈K
= μ(
μ(Bk )
k∈K
Ak ) + μ(
k∈K
Bk )
k∈K
= μ(G) + μ(G) = 2, which is clearly absurd. Therefore G is not amenable. This shows that (g) implies (a).
4.10 The Fixed Point Property The following ﬁxed point theorem is a generalization of the MarkovKakutani theorem (Theorem G.1.1). Theorem 4.10.1. Let G be an amenable group acting aﬃnely and continuously on a nonempty convex compact subset C of a Hausdorﬀ topological vector space X. Then G ﬁxes at least one point in C. Proof. Let (Fj )j∈J be a left Følner net for G. Choose an arbitrary point x ∈ C and set xj =
1 hx. Fj  h∈Fj
104
4 Amenable Groups
for each j ∈ J. Observe that xj ∈ C since C is convex. By compactness of C, we can assume, after taking a subnet, that the net (xj )j∈J converges to a point c ∈ C. Let g ∈ G. For every j ∈ J, we have ⎛ ⎞ 1 1 gxj − xj = g ⎝ hx⎠ − hx Fj  Fj  h∈Fj
h∈Fj
1 1 = ghx − hx (since the action of G on C is aﬃne) Fj  Fj  h∈Fj
h∈Fj
1 1 = hx − hx, Fj  Fj  h∈gFj
h∈Fj
which yields, after simpliﬁcation, gxj − xj =
1 Fj 
hx −
h∈gFj \Fj
1 Fj 
hx.
(4.21)
h∈Fj \gFj
Consider the points yj =
1 gFj \ Fj 
hx,
and
zj =
h∈gFj \Fj
1 Fj \ gFj 
hx.
h∈Fj \gFj
Note that yj and zj belong to C by convexity of C. We have Fj \ gFj  = gFj \ Fj  since the sets Fj and gFj have the same cardinality (cf. (4.15)). Setting λj =
gFj \ Fj  Fj \ gFj  = , Fj  Fj 
equality (4.21) gives us gxj − xj = λj yj − λj zj . The net (λj ) converges to 0 since (Fj ) is a left Følner net. As C is compact, it follows that lim λj yj = lim λj zj = 0, j
j
by using Lemma G.2.2. Therefore, we have gc − c = lim(gxj − xj ) = 0. j
This shows that c is ﬁxed by G.
Notes
105
Corollary 4.10.2. Let G be a group. Then the following conditions are equivalent: (a) the group G is amenable; (b) every continuous aﬃne action of G on a nonempty convex compact subset of a Hausdorﬀ topological vector space admits a ﬁxed point; (c) every continuous aﬃne action of G on a nonempty convex compact subset of a Hausdorﬀ locally convex topological vector space admits a ﬁxed point. Proof. Implication (a) ⇒ (b) follows from the preceding theorem. Implication (b) ⇒ (c) is trivial. Finally,(c) ⇒ (a) follows from Theorem 4.2.1, Proposition 4.3.1 and the fact that (∞ (G))∗ is a locally convex Hausdorﬀ topological vector space for the weak∗ topology (see Sect. F.2).
Notes The theory of amenable groups emerged from the study of the axiomatic properties of the Lebesgue integral and the discovery of the BanachTarski paradox at the beginning of the last century (see [Har3], [Pat], [Wag]). The ﬁrst deﬁnition of an amenable group, by the existence of an invariant ﬁnitely additive probability measure, is due to J. von Neumann in [vNeu1]. Von Neumann proved in particular that every abelian group is amenable (Theorem 4.6.1) and that an amenable group cannot contain a subgroup isomorphic to F2 (Corollary 4.5.2). The term amenable was introduced in the 1950s by M.M. Day, who played a central role in the development of the modern theory of amenable groups by using means and applying techniques from functional analysis. Moreover, Day extended the notion of amenability to semigroups, for which one has to distinguish between right amenability and left amenability. The question of the existence of a nonamenable group containing no subgroup isomorphic to F2 , which is called by some authors the von Neumann conjecture or Day’s problem, was answered in the aﬃrmative by A.Yu. Ol’shanskii [Ols] who gave an example of a ﬁnitely generated nonamenable group all of whose proper subgroups are cyclic. Other examples of nonamenable groups with no subgroup isomorphic to F2 were found by S.I. Adyan [Ady] who showed that the free Burnside group B(m, n) is nonamenable for m ≥ 2 and n odd with n ≥ 665. The free Burnside group B(m, n) is the quotient of the free group Fm by the subgroup of Fm generated by all npowers, that is, all elements of the form wn for some w ∈ Fm . The order of every element of B(m, n) divide n. In particular, B(m, n) is a periodic group, that is, a group in which every element has ﬁnite order. It is clear that a periodic group cannot contain a subgroup isomorphic to F2 . A geometric method for constructing ﬁnitely generated nonamenable periodic groups was described by M. Gromov in [Gro3]. Examples of ﬁnitely presented nonamenable groups
106
4 Amenable Groups
which contain no subgroup isomorphic to F2 were given by A.Yu Ol’shanskii and M. Sapir [OlS]. There is also a more general notion of amenability for locally compact groups and actions of locally compact groups (see [Gre], [Pat]). A group G is called polycyclic if it admits a ﬁnite sequence of subgroups {1G } = H0 ⊂ H1 ⊂ H2 ⊂ · · · ⊂ Hn = G such that Hi is normal in Hi+1 and Hi+1 /Hi is a (ﬁnite or inﬁnite) cyclic group for each 0 ≤ i ≤ n − 1. Clearly, every polycyclic group is solvable. It is not hard to prove that every ﬁnitely generated nilpotent group is polycyclic. It was shown by L. Auslander [Aus] that every polycyclic group is isomorphic to a subgroup of SLn (Z) for some integer n ≥ 1. It follows in particular that every polycyclic group is residually ﬁnite. This last result is due to K.A. Hirsch [Hir] (see also [RobD, p. 154]). P. Hall [Hall2] proved that every ﬁnitely generated metabelian group is residually ﬁnite and gave an example of a ﬁnitely generated solvable group of degree 3 which is not residually ﬁnite. The equivalence between Følner conditions and amenability was established by E. Følner in [Føl]. The proof was later simpliﬁed by I. Namioka in [Nam]. The equivalence between amenability and the nonexistence of a paradoxical decomposition is due to A. Tarski (see [Tar1], [Tar2] and [CGH1]). Let G be a group. Given a left (or right) paradoxical decomposition P = (K, (Ak )k∈K , (Bk )k∈K ) of G, the integer number c(P) = m + n, where m = {k ∈ K : Ak = ∅} and n = {k ∈ K : Bk = ∅}, is called the complexity of P. Then the quantity T (G) = inf c(P), where the inﬁmum is taken over all left (or right) paradoxical decompositions P of G, is called the Tarski number of G. One uses the convention that T (G) = ∞ if G admits no paradoxical decompositions, that is, if G is amenable (cf. Theorem 4.9.1). It was proved by B. Jonsson, a student of Tarski in the 1940s, that a group G has Tarski number T (G) = 4 if and only if G contains a subgroup isomorphic to F2 , the free group of rank 2. In [CGH1, CGH2] it was shown that for the free Burnside groups B(m, n) with m ≥ 2 and n ≥ 665 odd one has 6 ≤ T (B(m, n)) ≤ 14. The computations involve spectral analysis and CheegerBuser type isoperimetric inequalities (see Sect. 6.10 and Sect. 6.12) and Adyan’s cogrowth estimates [Ady] for the free Burnside groups B(m, n) (see (6.117) in the Notes for Chap. 6). The extension of the MarkovKakutani ﬁxed point theorem to amenable groups (cf. Theorem 4.10.1) is due to Day [Day2].
Exercises 4.1. Let G be an inﬁnite group and let μ : P(G) → [0, 1] be a left (or right) invariant ﬁnitely additive probability measure on G. Show that every ﬁnite subset A ⊂ G satisﬁes μ(A) = 0.
Exercises
107
4.2. Let X be an inﬁnite set. Prove that the symmetric group Sym(X) is not amenable. Hint: Show that Sym(X) contains a subgroup isomorphic to the free group F2 . 4.3. Show that the ﬁnitely generated and non residually ﬁnite groups G1 and G2 described in Sect. 2.6 are amenable. Hint: Each of these groups is the semidirect product of a locally ﬁnite group and an inﬁnite cyclic group. 4.4. Let G be a group. (a) Suppose that H and K are normal subgroups of G. Prove that HK is a normal subgroup of G and that the groups HK/K and H/(H ∩ K) are isomorphic. (b) Suppose that H and K are normal amenable subgroups of G. Prove that HK is a normal amenable subgroup of G. (c) Show that the set of all normal amenable subgroups of G has a maximal element for inclusion. This maximal element is called the amenable radical of the group G. Hint: Use (b) and Proposition 4.5.10 to show that the union of all normal amenable subgroups of G is a normal amenable subgroup of G. 4.5. Let G be a group. Show that G is metabelian if and only if it contains a normal subgroup N such that the group G/N is abelian. 4.6. Let G be a ﬁnitely generated solvable group. Show that if all elements of G have ﬁnite order then G is ﬁnite. Hint: Use induction on the solvability degree of G. 4.7. Show that every subgroup of a solvable (resp. nilpotent) group is solvable (resp. nilpotent). 4.8. Show that every quotient of a solvable (resp. nilpotent) group is solvable (resp. nilpotent). 4.9. Let G be a group containing a normal subgroup H such that both H and G/H are solvable. Show that G is solvable. 4.10. Show that the direct product of two nilpotent groups is a nilpotent group. 4.11. Show that a semidirect product of two nilpotent groups may fail to be nilpotent. Hint: Take for example the symmetric group Sym3 , which is the semidirect product of two cyclic groups. 4.12. Let G be a group. Show that G is solvable if and only if there is a ﬁnite sequence {1G } = H0 ⊂ H1 ⊂ H2 ⊂ · · · ⊂ Hn = G of subgroups of G such that Hi is normal in Hi+1 and Hi+1 /Hi is abelian for all 0 ≤ i ≤ n − 1.
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4.13. Let G be a group and i ≥ 0. Show that the group C i (G)/C i+1 (G) is contained in the center of G/C i+1 (G). 4.14. Show that every nontrivial nilpotent group has a nontrivial center. 4.15. Let G be a nilpotent group of nilpotency degree d ≥ 1. Show that the quotient group G/C d−1 (G) is nilpotent of nilpotency degree d − 1. 4.16. Let G be a ﬁnite group whose order is a power of a prime number. Show that G is nilpotent. Hint: Prove that the center of G is nontrivial by considering the action of G on itself by conjugation and then proceed by induction. 4.17. Let G be a group. Denote by Nnq (resp. Nsq , resp. Naq ) the set of all normal subgroups N ⊂ G such that the quotient group G/N is nilpotent (resp. solvable, resp. amenable). The sets Nnq , Nsq and Naq are partially ordered by reverse inclusion. (a) Show that if N1 and N2 are two elements of Nnq (resp. Nsq , resp. Naq ), then N1 ∩ N2 is an element of Nnq (resp. Nsq , resp. Naq ). Hint: observe that if ρ1 : G → G/N1 and ρ2 : G → G/N2 are the canonical homomorphisms, then the map ψ : G → G/N1 × G/N2 deﬁned by ψ(g) = (ρ1 (g), ρ2 (g)) is a homomorphism whose kernel is N1 ∩ N2 . (b) Show that Nnq (resp. Nsq , resp. Naq ) gives rise to a projective system of groups in a natural way. The limit of this projective system is called the pronilpotent completion (resp. prosolvable completion, resp. proamenable s , resp. G a ). n (resp. G completion) of the group G and is denoted by G s , (c) Show that there is a canonical homomorphism G → Gn (resp. G → G a ) and that this homomorphism is injective if and only if G is resp. G → G residually nilpotent (resp. residually solvable, resp. residually amenable). 4.18. Recall that a group G is called polycyclic if it admits a ﬁnite sequence of subgroups {1G } = H0 ⊂ H1 ⊂ H2 ⊂ · · · ⊂ Hn = G such that Hi is normal in Hi+1 and Hi+1 /Hi is a (ﬁnite or inﬁnite) cyclic group for each 0 ≤ i ≤ n − 1. (a) Show that every polycyclic group is ﬁnitely generated. (b) Show that every subgroup of a polycyclic group is polycyclic. (c) Deduce from (a) and (b) that every subgroup of a polycyclic group is ﬁnitely generated. 4.19. The lamplighter group is the wreath product L = (Z/2Z) Z. Thus, L is the semidirect product of the groups H = ⊕n∈Z An and Z, where An = Z/2Z for all n ∈ Z and Z acts on H by the Zshift. (a) Show that the group L is metabelian (and therefore solvable) and residually ﬁnite.
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109
(b) Prove that L is ﬁnitely generated. Hint: Show that L is generated by the two elements s and t corresponding respectively to the nontrivial element of A0 and to the canonical generator of Z. (c) Show that L is not polycyclic. Hint: Observe that H is not ﬁnitely generated and use Exercise 4.18(c). 4.20. Show that every ﬁnite solvable group is polycyclic. 4.21. Let m be an integer such that m ≥ 2. Let G be the group given by the presentation G = a, b : aba−1 = bm . Use the results in Exercise 2.7 to prove that the commutator subgroup [G, G] is isomorphic to the additive group Z[1/m] and that the quotient group G/[G, G] is inﬁnite cyclic. 4.22. Let G be a locally ﬁnite group. Let S denote the directed set consisting of all ﬁnitely generated subgroups of G partially ordered by inclusion. Prove that the net (H)H∈S is a Følner net for G. 4.23. Show that the sequence (Fn )n∈N , where Fn consists of all rational numbers of the form k/n! with k ∈ N and k ≤ (n + 1)!, is a Følner sequence for the additive group Q. 4.24. Let G = HZ denote the integral Heisenberg group (cf. Example 4.6.5). For each integer n ≥ 0, deﬁne the subset Fn ⊂ G by ⎧⎛ ⎫ ⎞ ⎨ 1xz ⎬ Fn = ⎝0 1 y ⎠ ∈ G : 1 ≤ x ≤ n, 1 ≤ y ≤ n, 1 ≤ z ≤ n2 . ⎩ ⎭ 001 Show that the sequence (Fn )n≥0 is a Følner sequence for G. 4.25. Let G be a group. Show that G is amenable if and only if the following condition holds: for every ﬁnite subset K ⊂ G and every ε > 0, there exists a ﬁnite subset F ⊂ G such that KF  < (1 + ε)F . 4.26. Let G be a countable amenable group. Show that G admits a left (resp. right) Følner sequence (Fn )n∈N which satisﬁes G = n∈N Fn and Fn ⊂ Fn+1 for all n ∈ N. 4.27. Let (K, (Ak )k∈K , (Bk )k∈K ) be a left (or right) paradoxical decomposition of a group G. Show that K ≥ 3. 4.28. Let G be a group and H ⊂ G a subgroup. Let (K, (Ak )k∈K , (Bk )k∈K ) be a left paradoxical decomposition of H and T ⊂ G a set of representa tives for the right cosets of H in G. For each k ∈ K set A k = t∈T Ak t and Bk = t∈T Bk t. Show that (K, (A k )k∈K , (Bk )k∈K ) is a left paradoxical decomposition of G.
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4.29. Let T (G) ∈ N ∪ {∞} denote the Tarski number of a group G. (a) Show that T (G) ≥ 4 for all groups G. (b) Show that T (G) = 4 if and only if G contains a subgroup isomorphic to F2 . Hint: Use Example 4.8.2 and the Klein PingPong theorem (Theorem D.5.1). (c) Let H be a subgroup of a group G. Show that T (G) ≤ T (H). (d) Let N be a normal subgroup of a group G. Show that T (G) ≤ T (G/N ). (e) Let G be a group. Show that there exists a ﬁnitely generated subgroup H ⊂ G such that T (H) = T (G). (f) Let G be a group. Suppose that all elements of G have ﬁnite order. Show that T (G) ≥ 6.
Chapter 5
The Garden of Eden Theorem
The Garden of Eden Theorem gives a necessary and suﬃcient condition for the surjectivity of a cellular automaton with ﬁnite alphabet over an amenable group. It states that such an automaton is surjective if and only if it is preinjective. As the name suggests it, preinjectivity is a weaker notion than injectivity. It means that any two conﬁgurations which have the same image under the automaton must be equal if they coincide outside a ﬁnite subset of the underlying group (see Sect. 5.2). We shall establish the Garden of Eden theorem in Sect. 5.8 by showing that both the surjectivity and the preinjectivity are equivalent to the maximality of the entropy of the image of the cellular automaton. The entropy of a set of conﬁgurations with respect to a Følner net of an amenable group is deﬁned in Sect. 5.7. Another important tool in the proof of the Garden of Eden theorem is a notion of tiling for groups introduced in Sect. 5.6. The Garden of Eden theorem is used in Sect. 5.9 to prove that every residually amenable (and hence every amenable) group is surjunctive. In Sect. 5.10 and Sect. 5.11, we give simple examples showing that both implications in the Garden of Eden theorem become false over a free group of rank two. In Sect. 5.12 it is shown that a group G is amenable if and only if every surjective cellular automaton with ﬁnite alphabet over G is preinjective. This last result gives a characterization of amenability in terms of cellular automata.
5.1 Garden of Eden Conﬁgurations and Garden of Eden Patterns Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. A conﬁguration y ∈ AG is called a Garden of Eden conﬁguration for τ if y is not in the image of τ . Thus the surjectivity of τ is equivalent to the nonexistence of Garden of Eden conﬁgurations. T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 5, © SpringerVerlag Berlin Heidelberg 2010
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5 The Garden of Eden Theorem
The biblical terminology “Garden of Eden” (a peaceful place where we will never return) comes from the fact that one often regards a cellular automaton τ : AG → AG from a dynamical viewpoint. This means that one thinks of a conﬁguration as evolving with time according to τ : if x ∈ AG is the conﬁguration at time t = 0, 1, 2, . . ., then τ (x) is the conﬁguration at time t + 1. A conﬁguration x ∈ AG \ τ (AG ) is called a Garden of Eden conﬁguration because it may only appear at time t = 0. A pattern p : Ω → A is called a Garden of Eden pattern for τ if there is no conﬁguration x ∈ AG such that τ (x)Ω = p. It follows from this deﬁnition that if p : Ω → G is a Garden of Eden pattern for τ , then any conﬁguration y ∈ AG such that yΩ = p is a Garden of Eden conﬁguration for τ . Thus, the existence of a Garden of Eden pattern implies the existence of Garden of Eden conﬁgurations, that is, the nonsurjectivity of τ . It turns out that the converse is also true when the alphabet set is ﬁnite: Proposition 5.1.1. Let G be a group and let A be a ﬁnite set. Let τ : AG → AG be a cellular automaton. Suppose that τ is not surjective. Then τ admits a Garden of Eden pattern. Proof. We know that the set τ (AG ) is closed in AG for the prodiscrete topology by Lemma 3.3.2. It follows that the set AG \ τ (AG ) is open in AG . Therefore, if y ∈ AG is a Garden of Eden conﬁguration for τ , we may ﬁnd a ﬁnite subset Ω ⊂ G such that V (y, Ω) = {x ∈ AG : xΩ = yΩ } ⊂ AG \ τ (AG ). In other words, every conﬁguration extending yΩ is not in τ (AG ), that is, yΩ is a Garden of Eden pattern for τ .
5.2 Preinjective Maps Let G be a group and let A be a set. Two conﬁgurations x1 , x2 ∈ AG are called almost equal if the set {g ∈ G : x1 (g) = x2 (g)} is ﬁnite. It is clear that being almost equal deﬁnes an equivalence relation on the set AG . Given a subset X ⊂ AG and a set Z, a map f : X → Z is called preinjective if it satisﬁes the following condition: if two conﬁgurations x1 , x2 ∈ X are almost equal and such that f (x1 ) = f (x2 ), then x1 = x2 . It immediately follows from this deﬁnition that the injectivity of f implies its preinjectivity. The converse is trivially true when the group G is ﬁnite. However, a preinjective map f : AG → Z may fail to be injective when G is inﬁnite. Examples 5.2.1. (a) Let us take G = Z and A = Z/3Z. Consider the cellular automaton τ : AG → AG deﬁned by τ (x)(n) = x(n−1)+x(n)+x(n+1) for all
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113
x ∈ AG and n ∈ G. Then τ is preinjective. Indeed, suppose that x1 , x2 ∈ AG are two conﬁgurations such that the set Ω = {n ∈ G : x1 (n) = x2 (n)} is a nonempty ﬁnite subset of Z. Let n0 denote the largest element in Ω. Then τ (x1 )(n0 + 1) = τ (x2 )(n0 + 1) and hence τ (x1 ) = τ (x2 ). This sows that τ is preinjective. However, τ is not injective since the constant conﬁgurations c0 , c1 ∈ AG given by c0 (n) = 0 and c1 (n) = 1 for all n ∈ Z have the same image c0 by τ . (b) Let G = Z2 and A = {0, 1}. Consider the cellular automaton τ : AG → G A associated with the Game of Life (cf. Example 1.4.3(a)). Let x1 ∈ AG be the conﬁguration with no live cells (x1 (g) = 0 for all g ∈ G) and let x2 ∈ AG be the conﬁguration with only one live cell at the origin (x2 (g) = 1 if g = (0, 0) and x2 (g) = 0 otherwise). Then x1 and x2 are almost equal and one has τ (x1 ) = τ (x2 ) = x1 . Therefore τ is not preinjective. (c) Let G be a group and let A = {0, 1}. Let S be a ﬁnite subset of G having at least 3 elements. Let τ : AG → AG be the majority action cellular automaton associated with G and S (cf. Example 1.4.3(c)). Let x0 ∈ AG be the conﬁguration deﬁned by x0 (g) = 0 for all g ∈ G. Let x1 ∈ AG be the conﬁguration deﬁned by x1 (1G ) = 1 and x1 (g) = 0 if g = 1G . One has τ (x0 ) = τ (x1 ) = x0 and {g ∈ G : x0 (g) = x1 (g)} = {1G }. Consequently, τ is not preinjective. The operations of induction and restriction of cellular automata with respect to a subgroup of the underlying group have been introduced in Sect. 1.7. It turns out that preinjectivity, like injectivity and surjectivity (see Proposition 1.7.4), is preserved by these operations. More precisely, we have the following result (which will not be used in the proof of the Garden of Eden theorem given below): Proposition 5.2.2. Let G be a group and let A be a set. Let H be a subgroup of G and let τ ∈ CA(G, H; A). Let τH ∈ CA(H; A) denote the cellular automaton obtained by restriction of τ to H. Then, τ is preinjective if and only if τH is preinjective. Proof. First recall from (1.16) the factorizations AG = Ac and τ = τc , c∈G/H
(5.1)
c∈G/H
xc ) = (τ ( x))c for all x ∈ Ac . where τc : Ac → Ac satisﬁes τc ( Suppose that τ is preinjective. We want to show that τH is preinjective. So let x, y ∈ AH be almost equal conﬁgurations over H such that τH (x) = τH (y). Let us ﬁx an arbitrary element a0 ∈ A and extend x and y to conﬁgurations x and y in AG by setting x(g) if g ∈ H, y(g) if g ∈ H, x (g) = and y(g) = a0 otherwise a0 otherwise
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for all g ∈ G. Note that the conﬁgurations x and y are almost equal since {g ∈ G : x (g) = y(g)} = {h ∈ H : x(h) = y(h)}. By construction, x c = yc for all c ∈ G/H \ {H}, while x H = x and yH = y. From (5.1), we deduce that τ ( x) = τ ( y ). It follows that x = y, by preinjectivity of τ . This implies that x = x H equals yH = y. This shows that τH is preinjective. , y ∈ AG be almost equal Conversely, suppose that τH is preinjective. Let x conﬁgurations over G such that τ ( x) = τ ( y ). For each g ∈ G, consider the g (h) = x (gh) and yg (h) = y(gh) for conﬁgurations x g , yg ∈ AH deﬁned by x all h ∈ H. Observe that the conﬁgurations x g and yg are almost equal since x and y are almost equal. On the other hand, we have x g = φ∗g ( xc ) and yg = ∗ ∗ c H φg ( y c ), where c = gH ∈ G/H and φg : A → A is the bijective map deﬁned x) = τ ( y ), we have τc ( xc ) = τc ( y c ) by φ∗g (u)(h) = u(gh) for all u ∈ Ac . As τ ( xg ) = τH ( yg ) since φ∗g conjugates τc and τH by by (5.1). We deduce that τH ( Proposition 1.18. Consequently, we have x g = yg for all g ∈ G by the pre = y. Therefore, τ is preinjective. injectivity of τH . This implies that x
5.3 Statement of the Garden of Eden Theorem The Garden of Eden theorem gives a necessary and suﬃcient condition for the surjectivity of a cellular automaton with ﬁnite alphabet over an amenable group. Its name comes from the fact that the surjectivity of a cellular automaton is equivalent to the absence of Garden of Eden conﬁgurations (see Sect. 5.1). Theorem 5.3.1 (The Garden of Eden theorem). Let G be an amenable group and let A be a ﬁnite set. Let τ : AG → AG be a cellular automaton. Then one has τ is surjective ⇐⇒ τ is preinjective. The proof of Theorem 5.3.1 will be given in Sect. 5.8 (see Theorem 5.8.1). Let us ﬁrst present some applications of this theorem. Examples 5.3.2. (a) Let G = Z and A = Z/3Z. We have seen in Example 5.2.1(a) that the cellular automaton τ : AG → AG deﬁned by τ (x)(n) = x(n − 1) + x(n) + x(n + 1) is preinjective. Since Z is amenable (cf. Theorem 4.6.1), it follows from the Garden of Eden theorem that τ is surjective. In fact, a direct proof of the surjectivity of τ is not diﬃcult (see Exercise 5.5). (b) Let G = Z2 and A = {0, 1}. Consider the cellular automaton τ : AG → G A associated with the Game of Life. We have seen in Example 5.2.1(b) that τ is not preinjective. The group Z2 is amenable by Theorem 4.6.1. By applying the Garden of Eden theorem, we deduce that τ is not surjective (see Sect. 5.13 for a direct proof).
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(c) Let G be a group and let A = {0, 1}. Let S be a ﬁnite subset of G having at least 3 elements. Let τ : AG → AG be the majority action cellular automaton associated with G and S. We have seen in Example 5.2.1(c) that τ is not preinjective. Thus it follows from the Garden of Eden theorem that if G is amenable then τ is not surjective. We shall see in Sect. 5.10 that if G is the free group F2 , then τ is surjective.
5.4 Interiors, Closures, and Boundaries Let G be a group. Let E and Ω be subsets of G. The Einterior Ω −E and the Eclosure Ω +E of Ω are the subsets of G deﬁned respectively by Ω −E = {g ∈ G : gE ⊂ Ω} Ω
+E
and
= {g ∈ G : gE ∩ Ω = ∅}
(see Figs. 5.1–5.2). Observe that Ω −E =
Ωe−1
(5.2)
Ωe−1 = ΩE −1 .
(5.3)
e∈E
and Ω +E =
e∈E
Fig. 5.1 The Einterior Ω −E of a set Ω. Here, Ω ⊂ R2 , E = {e, f } with e = (2, 1) and f = (−1, −1)
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5 The Garden of Eden Theorem
Fig. 5.2 The Eclosure Ω +E of a set Ω. Here, Ω ⊂ R2 , E = {e, f } with e = (2, 1) and f = (−1, −1)
The Eboundary of Ω is the subset ∂E (Ω) of G deﬁned by ∂E (Ω) = Ω +E \ Ω −E (see Fig. 5.3). Examples 5.4.1. (a) If E = ∅, then Ω −E = G and Ω +E = ∂E (Ω) = ∅. (b) If E = {1G }, then Ω −E = Ω +E = Ω and ∂E (Ω) = ∅. (c) If a ∈ G and E = {1G , a}, then Ω −E = Ω ∩ Ωa−1 , Ω +E = Ω ∪ Ωa−1 , so that ∂E (Ω) = (Ω ∪ Ωa−1 ) \ (Ω ∩ Ωa−1 ) = (Ω \ Ωa−1 ) ∪ (Ωa−1 \ Ω)
(5.4)
is the symmetric diﬀerence between the sets Ω and Ωa−1 (see Fig. 5.4). (d) Let us take G = Z2 and E = {−1, 0, 1}2 . Let a, b, c, d ∈ Z and consider the rectangle Ω = [a, b] × [c, d] = {(x, y) ∈ Z2 : a ≤ x ≤ b, c ≤ y ≤ d}. Then one has Ω −E = [a + 1, b − 1] × [c + 1, d − 1]
and
Ω +E = [a − 1, b + 1] × [c − 1, d + 1]
(see Fig. 5.5). Here are some general properties of the sets Ω −E , Ω +E and ∂E (Ω) which we shall frequently use in the sequel. Proposition 5.4.2. Let G be a group. Let E, E1 , E2 and Ω be subsets of G. Then the following hold: (i) (G \ Ω)−E = G \ Ω +E ; (ii) (G \ Ω)+E = G \ Ω −E ;
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Fig. 5.3 The Eboundary ∂E (Ω) of a set Ω. Here, Ω ⊂ R2 , E = {e, f } with e = (2, 1) and f = (−1, −1)
Fig. 5.4 The Einterior Ω −E = Ω ∩ Ωa−1 , the Eclosure Ω +E = Ω ∪ Ωa−1 , and the Eboundary ∂E (Ω) = (Ω ∪ Ωa−1 ) \ (Ω ∩ Ωa−1 ) of a set Ω when E = {1G , a}
Fig. 5.5 The Einterior Ω −E , the Eclosure Ω +E , and the Eboundary ∂E (Ω) of a rectangle Ω ⊂ Z2 . Here, Ω = [a, b] × [b, c] and E = {−1, 0, −1}2
(iii) if (iv) if (v) if (vi) if
a ∈ E, then Ω −E ⊂ Ωa−1 ⊂ Ω +E ; 1G ∈ E, then Ω −E ⊂ Ω ⊂ Ω +E ; E is nonempty and Ω is ﬁnite, then Ω −E is ﬁnite; E and Ω are both ﬁnite, then Ω +E and ∂E (Ω) are ﬁnite;
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(vii) if E1 ⊂ E2 , then Ω −E2 ⊂ Ω −E1 , Ω +E1 ⊂ Ω +E2 and ∂E1 (Ω) ⊂ ∂E2 (Ω); (viii) if h ∈ G, then h(Ω −E ) = (hΩ)−E , h(Ω +E ) = (hΩ)+E and h(∂E (Ω)) = ∂E (hΩ). Proof. (i) By deﬁnition, we have g ∈ G \ Ω +E if and only if gE does not meet Ω, that is, if and only if g ∈ (G \ Ω)−E . (ii) By replacing Ω by G \ Ω in (i) we get Ω −E = G \ (G \ Ω)+E which gives (ii) after taking complements. (iii) If a ∈ E, then Ω −E ⊂ Ωa−1 by (5.2) and Ωa−1 ⊂ Ω +E by (5.3). (iv) Assertion (iii) gives (iv) by taking a = 1G . (v) If a ∈ E and Ω is ﬁnite, then Ω −E  ≤ Ωa−1  = Ω by (iii). (vi) If E and Ω are both ﬁnite, then Ω +E  ≤ ΩE by (5.3) so that Ω +E is ﬁnite. The set ∂E (Ω) is then also ﬁnite since it is contained in Ω +E . (vii) The ﬁrst two statements follow immediately from (5.2) and (5.3), respectively, and imply that ∂E1 (Ω) = Ω +E1 \ Ω −E1 ⊂ Ω +E2 \ Ω −E1 ⊂ Ω +E2 \ Ω −E2 = ∂E2 (Ω). (viii) By using (5.2) we have h(Ω −E ) = h ∩e∈E Ωe−1 = ∩e∈E hΩe−1 = (hΩ)−E , which gives the ﬁrst statement. Similarly, from (5.3) we get h(Ω +E ) = h(ΩE −1 ) = (hΩ)E −1 = (hΩ)+E . Finally, we have h(∂E (Ω)) = h(Ω +E \ Ω −E ) = hΩ −E − hΩ −E = ∂E (hΩ). Proposition 5.4.3. Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton with memory set S. Let x and x be elements of AG . Suppose that there is a subset Ω of G such that x and x coincide on Ω (resp. on G \ Ω). Then the conﬁgurations τ (x) and τ (x ) coincide on Ω −S (resp. on G \ Ω +S ). Proof. Suppose that x and x coincide on Ω. If g ∈ Ω −S , then gS ⊂ Ω and therefore τ (x)(g) = τ (x )(g) by Lemma 1.4.7. It follows that τ (x) and τ (x ) coincide on Ω −S . Suppose now x and x coincide on G \ Ω. Then τ (x) and τ (x ) coincide on (G \ Ω)−S = G \ Ω +S by the ﬁrst part of the proof and Proposition 5.4.2(i). Proposition 5.4.4. Let G be a group and let (Fj )j∈J be a net of nonempty ﬁnite subsets of G. Then the following conditions are equivalent:
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119
(a) the net (Fj )j∈J is a right Følner net for G; (b) one has lim j
∂E (Fj ) =0 Fj 
for every ﬁnite subset E ⊂ G.
Proof. (b) ⇒ (a). Suppose (b). Let g ∈ G and take E = {1G , g −1 }. By (5.4), we have Fj \ Fj g ⊂ ∂E (Fj ), and hence Fj \ Fj g ≤ ∂E (Fj ). Therefore, property (b) implies lim j
Fj \ Fj g = 0. Fj 
This shows that (Fj ) is a right Følner net for G. (a) ⇒ (b). Let E be a ﬁnite subset of G. By (5.2) and (5.3) we have −1 −1 \ ∂E (Fj ) = Fj a Fj b a∈E
=
Fj a
−1
a∈E
=
Fj a
−1
b∈E
G\
a∈E
=
Fj b
−1
b∈E
(G \ Fj b
−1
)
b∈E
(Fj a−1 \ Fj b−1 ).
a,b∈E
This implies ∂E (Fj ) ≤
Fj a−1 \ Fj b−1 .
(5.5)
a,b∈E
Now observe that, for all a, b ∈ E, we have Fj a−1 \ Fj b−1  = Fj \ Fj b−1 a, since right multiplication by a is bijective on G. Therefore, inequality (5.5) gives us Fj \ Fj g ∂E (Fj ) ≤ E2 max , g∈K Fj  Fj  where K is the ﬁnite subset of G deﬁned by K = {b−1 a : a, b ∈ E}. This shows that (a) implies (b).
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5 The Garden of Eden Theorem
Corollary 5.4.5. Let G be a group. Then the following conditions are equivalent: (a) G is amenable; (b) for every ﬁnite subset E ⊂ G and every real number ε > 0, there exists a nonempty ﬁnite subset F ⊂ G such that ∂E (F ) < ε. F 
(5.6)
Proof. Suppose ﬁrst that G is amenable. It follows from the TarskiFølner theorem (Theorem 4.9.1) and Proposition 4.7.1 that there exists a right ∂ (F ) Følner net (Fj )j∈J in G. By Proposition 5.4.4 we have that limj EFj j = 0 for every ﬁnite subset E ⊂ G. Thus, given ε > 0 and a ﬁnite subset S ⊂ G, ∂ (F ) there exists j0 ∈ J such that EFj j < ε for all j ≥ j0 . Taking F = Fj0 we deduce (5.6). This shows (a) ⇒ (b). Conversely, suppose (b). Let J denote the set of all pairs (E, ε), where E is a ﬁnite subset of G and ε > 0. We equip J with the partial ordering ≤ deﬁned by (E, ε) ≤ (E , ε ) ⇔ (E ⊂ E and ε ≤ ε). Then J is a directed set. By (b), for every j = (E, ε) ∈ J, there exists a nonempty ﬁnite subset Fj ⊂ G such that ∂E (Fj ) < ε. Fj 
(5.7)
Let us show that lim j
∂E (Fj ) = 0 for every ﬁnite subset E ⊂ G. Fj 
(5.8)
Fix a ﬁnite subset E0 ⊂ G and ε0 > 0. Let j ∈ J and suppose that j ≥ j0 , where j0 = (E0 , ε0 ) ∈ J. By virtue of Proposition 5.4.2(vii) we have ∂E0 (Fj ) ⊂ ∂E (Fj ) so that, from (5.7), we deduce that ∂E (Fj ) ∂E0 (Fj ) ≤ < ε ≤ ε0 . Fj  Fj  This shows (5.8). From Proposition 5.4.4 and Proposition 4.7.1 we deduce that G satisﬁes the Følner conditions. Thus G is amenable by virtue of the TarskiFølner theorem (Theorem 4.9.1).
5.5 Mutually Erasable Patterns
121
5.5 Mutually Erasable Patterns In this section, we give a characterization of preinjective cellular automata based on the notion of mutually erasable patterns. This leads to an equivalent deﬁnition of preinjectivity which is frequently used in the literature. However, the material contained in this section will not be used in the proof of the Garden of Eden theorem so that the reader who is only interested in this proof may go directly to the next section. Let G be a group and let A be a set. Let Z be a set and let f : AG → Z be a map. Two distinct patterns p1 , p2 : Ω → Z with the same support Ω ⊂ G are called mutually erasable (with respect to f ) if they satisfy the following condition: if x1 , x2 ∈ AG are conﬁgurations such that x1 Ω = p1 , x2 Ω = p2 and x1 G\Ω = x2 G\Ω , then f (x1 ) = f (x2 ). Example 5.5.1. Let G = Z2 and A = {0, 1}. Consider the cellular automaton τ : AG → AG associated with the Game of Life (see Example 1.4.3(a)). Let Ω = {−1, 0, 1}2 be the 3 × 3 square in Z2 centered at the origin and consider the pattern p1 (resp. p2 ) with support Ω deﬁned by p1 (g) = 0 for all g ∈ Ω (resp. p2 (g) = 1 if g = (0, 0) and p2 (g) = 0 otherwise). Then it is clear that p1 and p2 are mutually erasable patterns for τ (see Fig. 5.6).
Fig. 5.6 Two mutually erasable patterns for the Game of Life (recall that ◦ denotes a dead cell, while • denotes a live cell)
Suppose that p1 and p2 are mutually erasable patterns for a map f : AG → Z. Let Ω denote their common support and consider two conﬁgurations x1 and x2 which coincide outside Ω and such that x1 Ω = p1 and x2 Ω = p2 . Then x1 and x2 are almost equal and f (x1 ) = f (x2 ). On the other hand, x1 = x2 since p1 = p2 and therefore f is not preinjective. This shows that a preinjective map f : AG → Z admits no mutually erasable patterns. It turns out that for cellular automata, the converse is true:
122
5 The Garden of Eden Theorem
Proposition 5.5.2. Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton. Then τ is preinjective if and only if it does not admit mutually erasable patterns. Proof. It remains only to show that if τ is not preinjective then it admits mutually erasable patterns. So let us assume that τ is not preinjective. This means that there exist conﬁgurations x1 , x2 ∈ AG satisfying τ (x1 ) = τ (x2 ) such that Σ = {g ∈ G : x1 (g) = x2 (g)} is a nonempty ﬁnite subset of G. Let S be a memory set for τ such that S = S −1 and 1G ∈ S. Consider the 2 ﬁnite sets S 2 = {s1 s2 : s1 , s2 ∈ S} and Ω = Σ +S . Let us show that the patterns p1 = x1 Ω and p2 = x2 Ω are mutually erasable. First observe that p1 = p2 since Σ ⊂ Ω. Suppose now that y1 , y2 ∈ AG are two conﬁgurations coinciding outside Ω such that y1 Ω = p1 and y2 Ω = p2 . Then y1 and y2 coincide outside Σ since p1 and p2 coincide on Ω \ Σ. This implies τ (y1 )(g) = τ (y2 )(g) for all g ∈ G \ Σ +S
(5.9)
by Proposition 5.4.3. On the other hand, for i = 1, 2, the conﬁgurations yi and xi coincide on Ω by construction. Thus, it follows from Proposition 5.4.3 that τ (yi ) and τ (xi ) coincide on Ω −S . As τ (x1 ) = τ (x2 ), we deduce that the conﬁgurations τ (y1 ) and τ (y2 ) coincide on Ω −S . Now observe that Σ +S ⊂ Ω −S . Indeed, if g ∈ Σ +S , that is, gs0 ∈ Σ for some s0 ∈ S, then gs ∈ Ω for all s ∈ S since gss−1 s0 = gs0 ∈ gsS 2 ∩ Σ. It follows that τ (y1 ) and τ (y2 ) coincide on Σ +S . Combining this with (5.9), we conclude that τ (y1 ) = τ (y2 ). This sows that p1 and p2 are mutually erasable patterns.
5.6 Tilings Let G be a group. Let E and E be subsets of G. A subset T ⊂ G is called an (E, E )tiling of G if the sets gE, g ∈ T are pairwise disjoint and if the sets gE , g ∈ T cover G. In other words, T ⊂ G is an (E, E )tiling if and only if the following conditions are satisﬁed: (T1) g1 E ∩ g2 E = ∅ for all g1 , g2 ∈ T such that g1 = g2 ; (T2) G = g∈T gE . Examples 5.6.1. (a) In the additive group R, the set Z is a ([0, 1[, [0, 1[)tiling and the set [0, 1] is a (2Z, Z)tiling. (b) If G is a group and H is a subgroup of G, then every complete set of representatives for left cosets of G modulo H is an (H, H)tiling. Remark 5.6.2. Let G be a group and let E and E be subsets of G. If T is an (E, E )tiling of G and if E1 and E1 are subsets of G such that E1 ⊂ E and E ⊂ E1 , then it is clear that T is also an (E1 , E1 )tiling of G.
5.6 Tilings
123
The Zorn lemma may be used to prove the existence of (E, E )tilings for any subset E of G and for E ⊂ G “large enough”. More precisely, we have the following: Proposition 5.6.3. Let G be a group and let E be a nonempty subset of G. Let E = {g1 g2−1 : g1 , g2 ∈ E}. Then there is an (E, E )tiling T ⊂ G. Proof. Consider the set S consisting of all subsets S ⊂ G such that the sets (gE)g∈S are pairwise disjoint. Observe that S is not empty since {1G } ∈ S. On the other hand, the set S, partially ordered by inclusion, is inductive.
Indeed, if S is a totally ordered subset of S, then the set M = S∈S S belongs to S and is an upper bound for S . By applying Zorn’s lemma, we deduce that S admits a maximal element T . The sets (gE)g∈T are pairwise disjoint since T ∈ S. On the other hand, consider an arbitrary element h ∈ G. By maximality of T , we can ﬁnd g ∈ T such that the set hE meets gE. This implies h ∈ gE . This shows that the sets (gE )g∈T cover G. Consequently, T is an (E, E )tiling of G. Proposition 5.6.4. Let G be an amenable group and let (Fj )j∈J be a right Følner net for G. Let E and E be ﬁnite subsets of G and suppose that T ⊂ G is an (E, E )tiling of G. Let us set, for each j ∈ J, Tj = T ∩ Fj−E = {g ∈ T : gE ⊂ Fj }. Then there exist a real number α > 0 and an element j0 ∈ J such that Tj  ≥ αFj 
for all j ≥ j0 .
Proof. After possibly replacing E by E ∪ E , we can assume that E ⊂ E . Let us set Tj+ = T ∩ Fj+E = {g ∈ T : gE ∩ Fj = ∅}. As the sets gE , g ∈ Tj+ , cover Fj , we have Fj  ≤ Tj+  · E , which gives Tj+  1 ≥ Fj  E 
(5.10)
for all j ∈ J. Observe now that Tj+ \ Tj = T ∩ Fj+E \ T ∩ Fj−E
= T ∩ (Fj+E \ Fj−E )
⊂ T ∩ (Fj+E \ Fj−E ) ⊂ T ∩ ∂E (Fj ) ⊂ ∂E (Fj ) where the ﬁrst inclusion follows from E ⊂ E . Thus we have ∂E (Fj ) ≥ Tj+  − Tj .
124
5 The Garden of Eden Theorem
Using (5.10), we then deduce Tj+  ∂E (Fj ) 1 ∂E (Fj ) Tj  ≥ − ≥ − . Fj  Fj  Fj  E  Fj  Hence we have
1 Tj  ≥α= Fj  2E 
for j large enough, by Proposition 5.4.4 (see Fig. 5.7).
Fig. 5.7 The set T7 = T ∩ F7−E ⊂ Z2 , where E = {(0, 0), (1, 0)}, E = {0, 1} × {0, 1} and
T = (2Z + 1) × (2Z + 1) ⊂ Z2 is an (E, E )tiling in Z2 . Note that 1 8
=
1 2E 
=α
T7  F7 
=
12 64
=
3 16
≥
2 16
=
5.7 Entropy
125
5.7 Entropy In this section, G is an amenable group, F = (Fj )j∈J is a right Følner net for G, and A is a ﬁnite set. For E ⊂ G, we denote by πE : AG → AE the canonical projection (restriction map). We thus have πE (x) = xE for all x ∈ AG . Deﬁnition 5.7.1. Let X ⊂ AG . The entropy entF (X) of X with respect to the right Følner net F = (Fj )j∈J is deﬁned by entF (X) = lim sup j
log πFj (X) . Fj 
Here are some immediate properties of entropy. Proposition 5.7.2. One has (i) entF (AG ) = log A; (ii) entF (X) ≤ entF (Y ) if X ⊂ Y ⊂ AG ; (iii) entF (X) ≤ log A for all X ⊂ AG . Proof. (i) If X = AG , then, for every j, we have πFj (X) = AFj and therefore log πFj (X) log AFj  Fj  log A = = = log A. Fj  Fj  Fj  Thus we have entF (X) = log A. (ii) If X ⊂ Y , then πFj (X) ⊂ πFj (Y ) and hence πFj (X) ≤ πFj (Y ) for all j. This implies entF (X) ≤ entF (Y ). (iii) This follows immediately from (i) and (ii). An important property of cellular automata is the fact that applying a cellular automaton to a set of conﬁgurations cannot increase the entropy of the set. More precisely, we have the following: Proposition 5.7.3. Let τ : AG → AG be a cellular automaton and let X ⊂ AG . Then one has entF (τ (X)) ≤ entF (X). Proof. Let Y = τ (X). Let S ⊂ G be a memory set for τ . After replacing S by S ∪ {1G }, we can assume that 1G ∈ S. Let Ω be a ﬁnite subset of G. Observe ﬁrst that τ induces a map τΩ : πΩ (X) → πΩ −S (Y ) deﬁned as follows. If u ∈ πΩ (X), then τΩ (u) = (τ (x))Ω −S , where x is an element of X such that xΩ = u. Note that the fact that τΩ (u) does not depend on the choice of such an x follows from Proposition 5.4.3.
126
5 The Garden of Eden Theorem
Clearly τΩ is surjective. Indeed, if v ∈ πΩ −S (Y ), then there exists x ∈ X such that (τ (x))Ω −S = v. Then, setting u = πΩ (x) we have, by construction, τΩ (u) = v. Therefore, we have πΩ −S (Y ) ≤ πΩ (X).
(5.11)
Observe now that Ω −S ⊂ Ω, since 1G ∈ S (cf. Proposition 5.4.2(iv)). Thus −S πΩ (Y ) ⊂ πΩ −S (Y ) × AΩ\Ω . This implies log πΩ (Y ) ≤ log πΩ −S (Y ) × AΩ\Ω
−S

Ω\Ω −S
= log πΩ −S (Y ) + log A  = log πΩ −S (Y ) + Ω \ Ω −S  log A ≤ log πΩ (X) + Ω \ Ω −S  log A, by (5.11). As Ω \ Ω −S ⊂ ∂S (Ω), we deduce that log πΩ (Y ) ≤ log πΩ (X) + ∂S (Ω) log A. By taking Ω = Fj , this gives us log πFj (X) ∂S (Fj ) log πFj (Y ) ≤ + log A. Fj  Fj  Fj  Since lim j
∂S (Fj ) =0 Fj 
by Proposition 5.4.4, we ﬁnally get entF (Y ) = lim sup j
log πFj (Y ) log πFj (X) ≤ lim sup = entF (X). Fj  Fj  j
It follows from Proposition 5.7.2 that the maximal value for the entropy of a subset X ⊂ AG is log A. The following result gives a suﬃcient condition on X which implies that its entropy is strictly less than log A. Proposition 5.7.4. Let X ⊂ AG . Suppose that there exist ﬁnite subsets E and E of G and an (E, E )tiling T ⊂ G such that πgE (X) AgE for all g ∈ T . Then one has entF (X) < log A. Proof. For each j ∈ J, let us deﬁne, as in Proposition 5.6.4 (see Fig. 5.7), the subset Tj ⊂ T by Tj = T ∩ Fj−E = {g ∈ T : gE ⊂ Fj } and set gE Fj∗ = Fj \ g∈Tj
(see Fig. 5.8). By hypothesis, we have πgE (X) ≤ AgE  − 1 = AgE − 1 for all g ∈ T.
(5.12)
5.7 Entropy
127
‘ Fig. 5.8 The set F7∗ = F7 \ g∈T7 gE = F7 \ T7 E ⊂ Z2 , where E and T are as in Fig. 5.7 (note that here F7 is the whole square, while F7∗ ⊂ F7 consists of its •points)
As
∗
πFj (X) ⊂ AFj ×
πgE (X),
g∈Tj
we get ∗
log πFj (X) ≤ log AFj ×
πgE (X)
g∈Tj
= Fj∗  log A +
log πgE (X)
g∈Tj
≤ Fj∗  log A +
log(AgE − 1) (by (5.12))
g∈Tj
=
Fj∗  log A
+
g∈Tj
gE log A +
g∈Tj
= Fj  log A + Tj  log(1 − A−E ),
log(1 − A−gE )
128
since
5 The Garden of Eden Theorem
Fj  = Fj∗  +
gE
gE = E.
and
g∈Tj
By setting c = − log(1 − A−E ) (note that c > 0), this gives us log πFj (X) ≤ Fj  log A − cTj 
for all j ∈ J.
Now, by Proposition 5.6.4, there exist α > 0 and j0 ∈ J such that Tj  ≥ αFj  for all j ≥ j0 . Thus log πFj (X) ≤ log A − cα Fj 
for all j ≥ j0 .
This implies that entF X = lim sup j
log πFj (X) ≤ log A − cα < log A. Fj 
Recall from Sect. 1.1 that G acts on the left on A by the shift (g, x) → gx deﬁned by gx(g ) = x(g −1 g ) for g, g ∈ G and x ∈ AG . G
Corollary 5.7.5. Let X be a Ginvariant subset of AG . Suppose that there exists a ﬁnite subset E ⊂ G such that πE (X) AE . Then one has entF (X) < log A. Proof. Let E = {g1 g2−1 : g1 , g2 ∈ E}. By Proposition 5.6.3, we may ﬁnd an (E, E )tiling T ⊂ G. Since πE (X) AE and X is Ginvariant, we have πgE (X) AgE for all g ∈ G. This implies entF (X) < log A by Proposition 5.7.4.
5.8 Proof of the Garden of Eden Theorem The purpose of this section is to establish the following: Theorem 5.8.1. Let G be an amenable group and let A be a ﬁnite set. Let F = (Fj )j∈J be a right Følner net for G. Let τ : AG → AG be a cellular automaton. Then the following conditions are equivalent: (a) τ is surjective; (b) entF (τ (AG )) = log A; (c) τ is preinjective. Note that this will prove the Garden of Eden theorem (Theorem 5.3.1) since the Garden of Eden theorem asserts the equivalence of conditions (a) and (c) in Theorem 5.8.1.
5.8 Proof of the Garden of Eden Theorem
129
We divide the proof of Theorem 5.8.1 into several lemmas. In these lemmas, it is assumed that the hypotheses of Theorem 5.8.1 are satisﬁed: G is an amenable group, F = (Fj )j∈J is a right Følner net for G, A is a ﬁnite set, and τ : AG → AG is a cellular automaton. Lemma 5.8.2. Suppose that τ is not surjective. Then one has entF (τ (AG )) < log A. Proof. By Proposition 5.1.1, τ admits a Garden of Eden pattern. This means that there is a ﬁnite subset E ⊂ G such that πE (τ (AG )) AE . The set τ (AG ) is Ginvariant since τ is Gequivariant by Proposition 1.4.4. We deduce that entF (τ (AG )) < log A by applying Corollary 5.7.5. Lemma 5.8.3. Suppose that ent(τ (AG )) < log A.
(5.13)
Then τ is not preinjective. Proof. Let S be a memory set for τ such that 1G ∈ S. Let Y = τ (AG ). We have Fj−S ⊂ Fj ⊂ Fj+S by Proposition 5.4.2(iv) and therefore Fj+S \ Fj ⊂ +S
∂S (Fj ). As πF +S (Y ) ⊂ πFj (Y ) × AFj
\Fj
, it follows that
j
log πF +S (Y ) ≤ log πFj (Y ) + Fj+S \ Fj  log A j
≤ log πFj (Y ) + ∂S (Fj ) log A. This implies log πF +S (Y ) j
Fj 
≤
log πFj (Y ) ∂S (Fj ) + log A. Fj  Fj 
As ent(Y ) = lim sup j
by hypothesis, and lim j
(5.14)
log πFj (Y ) < log A Fj 
∂S (Fj ) =0 Fj 
by Proposition 5.4.4, we deduce from inequality (5.14) that there exists j0 ∈ J such that log πF +S (Y ) j0 < log A. (5.15) Fj0  Let us ﬁx an arbitrary element a0 ∈ A and denote by Z the ﬁnite set of conﬁgurations z ∈ AG such that z(g) = a0 for all g ∈ G\Fj0 . Inequality (5.15) gives us
130
5 The Garden of Eden Theorem
πF +S (Y ) < AFj0  = Z. j0
Observe that τ (z1 ) and τ (z2 ) coincide outside Fj +S for all z1 , z2 ∈ Z. Thus 0
τ (Z) = πF
+S j0
(τ (Z)) ≤ πF
+S j0
(Y ) < Z.
This implies that we may ﬁnd distinct conﬁgurations z1 , z2 ∈ Z such that τ (z1 ) and τ (z2 ). Since z1 and z2 coincide outside the ﬁnite set Fj0 , this shows that τ is not preinjective. Lemma 5.8.4. Suppose that τ is not preinjective. Then one has entF (τ (AG )) < log A.
(5.16)
Proof. Since τ is not preinjective, we may ﬁnd two conﬁgurations x1 , x2 ∈ AG satisfying τ (x1 ) = τ (x2 ) such that the set Ω = {g ∈ G : x1 (g) = x2 (g)} is a nonempty ﬁnite subset of G. Observe that, for each h ∈ G, the conﬁgurations hx1 and hx2 satisfy τ (hx1 ) = τ (hx2 ) (since τ is Gequivariant by Proposition 1.4.4) and {g ∈ G : hx1 (g) = hx2 (g)} = hΩ. Let S be a memory set for τ such that 1G ∈ S. Then the set R = {s−1 s : s, s ∈ S} is ﬁnite and we have 1G ∈ R. Let E = Ω +R . By Proposition 5.6.3, we may ﬁnd a ﬁnite subset E ⊂ G and an (E, E )tiling T ⊂ G. Consider the subset Z ⊂ AG consisting of all conﬁgurations z ∈ AG such that zhE = (hx1 )hE
for all h ∈ T.
Observe that, for each h ∈ T , we have πhE (Z) AhE since (hx1 )hE ∈ / πhE (Z). We deduce that entF (Z) < log A by applying Proposition 5.7.4. As entF (τ (Z)) ≤ entF (Z) by Proposition 5.7.3, this implies entF (τ (Z)) < log A.
(5.17)
Thus, to establish inequality (5.16), it suﬃces to prove that τ (AG ) = τ (Z). To see this, consider an arbitrary conﬁguration x ∈ AG and let us show that there is a conﬁguration z ∈ Z such that τ (x) = τ (z). Let T = {h ∈ T : xhE = (hx1 )hE }.
5.9 Surjunctivity of Locally Residually Amenable Groups
131
Let z ∈ AG be the conﬁguration deﬁned by hx2 (g) if there is h ∈ T such that g ∈ hE, z(g) = x(g) otherwise. Notice that the conﬁguration z is obtained from x by modifying the values taken by x only on the subsets of the form hΩ, where h ∈ T , (since, as we have seen above, hx1 and hx2 coincide outside hΩ). By construction, we have z ∈ Z. Let g ∈ G. Let us shows that τ (x)(g) = τ (z)(g). Suppose ﬁrst that gS does not meet any of the sets hΩ, h ∈ T . Then we have zgS = xgS . We deduce that τ (z)(g) = τ (x)(g) by applying Lemma 1.4.7. Suppose now that there is an element h ∈ T such that gS meets hΩ. This means that there exists an element s0 ∈ S such that gs0 ∈ hΩ. For each s ∈ S, we have gss−1 s0 = gs0 ∈ hΩ. As s−1 s0 ∈ R, this implies gs ∈ (hΩ)+R = hΩ +R = hE.
(by Proposition 5.4.2(viii))
We deduce that gS ⊂ hE. Thus we have τ (x)(g) = τ (hx1 )(g) since xhE = hx1 hE . Similarly, by applying Lemma 1.4.7, we get τ (z)(g) = τ (hx2 )(g), since z and hx2 coincide on hE. As τ (hx1 ) = τ (hx2 ), we deduce that τ (x)(g) = τ (z)(g). Thus τ (z) = τ (x). This shows that τ (AG ) = τ (Z) and completes the proof of the lemma. Proof of Theorem 5.8.1. If τ is surjective, then τ (AG ) = AG and hence entF (τ (AG )) = entF (AG ) = log A. Thus (a) implies (b). Since the converse implication follows from Lemma 5.8.2, we deduce that conditions (a) and (b) are equivalent. The fact that (c) implies (b) follows from Lemma 5.8.3 and the converse implication follows from Lemma 5.8.4. Thus, conditions (b) and (c) are also equivalent.
5.9 Surjunctivity of Locally Residually Amenable Groups The notion of a residually ﬁnite group was introduced in Chap. 2. More generally, if P is a property of groups, a group G is called residually P if for each element g ∈ G with g = 1G , there exist a group Γ satisfying P and an epimorphism φ : G → Γ such that φ(g) = 1Γ . Observe that every group which satisﬁes P is residually P.
132
5 The Garden of Eden Theorem
According to the preceding deﬁnition, a group G is called residually amenable if for each element g ∈ G with g = 1G , there exist an amenable group Γ and an epimorphism φ : G → Γ such that φ(g) = 1Γ . Note that, as every subgroup of an amenable group is amenable, it is not necessary to require that the homomorphism φ is surjective in this deﬁnition. Observe also that every subgroup of a residually amenable group is residually amenable and that the fact that every ﬁnite group is amenable (Proposition 4.4.6) implies that every residually ﬁnite group is residually amenable. Theorem 5.9.1. Every residually amenable group is surjunctive. Let us ﬁrst establish the following: Lemma 5.9.2. Let G be a residually amenable group and let Ω be a ﬁnite subset of G. Then there exist an amenable group Γ and a homomorphism ρ : G → Γ such that the restriction of ρ to Ω is injective. Proof. Consider the ﬁnite subset S ⊂ G deﬁned by S = {g −1 h : g, h ∈ Ω and g = h}. Since G is residually amenable, we can ﬁnd, for each s ∈ S, an amenable group Λs and a homomorphism φs : G → Λs such that φs (s) = 1Λs . Let us show that the group Λs Γ = s∈S
and the homomorphism ρ : G → Γ given by ρ= φs s∈S
have the required properties. The fact that the group Γ is amenable follows from Corollary 4.5.6. On the other hand, suppose that g and h are distinct elements of Ω. Then s = g −1 h ∈ S and φs (g) = φs (h) since (φs (g))−1 φs (h) = φs (g −1 h) = φs (s) = 1Λs . This implies ρ(g) = ρ(h). Therefore, the restriction of ρ to Ω is injective. Proof of Theorem 5.9.1. Since every injective cellular automaton is preinjective, the Garden of Eden theorem (Theorem 5.3.1) implies that every amenable group is surjunctive. By applying Lemma 3.3.4 and Lemma 5.9.2, it follows that every residually amenable group is surjunctive. As an immediate consequence of Theorem 5.9.1 and Proposition 3.2.2, we obtain the following: Corollary 5.9.3. Every locally residually amenable group is surjunctive.
5.11 A Preinjective but Not Surjective Cellular Automaton over F2
133
5.10 A Surjective but Not Preinjective Cellular Automaton over F2 The Garden of Eden theorem (Theorem 5.3.1) implies that every surjective cellular automaton with ﬁnite alphabet over an amenable group is necessarily preinjective. In this section, we give an example of a surjective but not preinjective cellular automaton with ﬁnite alphabet over the free group F2 . Let G = F2 be the free group on two generators a and b. Let S = {a, b, a−1 , b−1 } and A = {0, 1}. Consider the cellular automaton τ : AG → AG deﬁned by ⎧ ⎪ if x(gs) > 2, ⎨1 s∈S τ (x)(g) = 0 x(gs) < 2, if ⎪ s∈S ⎩ x(g) if s∈S x(gs) = 2. Thus τ is the majority action automaton associated with G and S (see Example 1.4.3(c)). We have already seen in Example 5.2.1(c) that τ is not preinjective. Let us show that τ is surjective. Let y ∈ AG be an arbitrary conﬁguration. We may construct a conﬁguration x ∈ AG such that y = τ (x) in the following way. Consider the map ψ : G\{1G } → G which associates to each g ∈ G\{1G } the element of G obtained by suppressing the last factor in the reduced form of g. Thus if g = s1 s2 . . . sn with si ∈ S for 1 ≤ i ≤ n and si si+1 = 1G for 1 ≤ i ≤ n − 1, then ψ(g) = s1 s2 . . . sn−1 . Let x ∈ AG be the conﬁguration deﬁned by x(g) = y(ψ(g)) for all g ∈ G \ {1G } and x(1G ) = 0. Then, for each g ∈ G, the conﬁguration x takes the value y(g) at (at least) three of the four elements gs, s ∈ S. it follows that τ (x) = y. Thus τ is surjective. More generally, let G be a group containing a free subgroup H based on two elements a and b. Let τ : {0, 1}G → {0, 1}G be the majority action cellular automaton over G associated with the set S = {a, b, a−1 , b−1 }. We have seen in Example 5.2.1(c) that τ is not preinjective. Consider its restriction τH : {0, 1}H → {0, 1}H . Note that τH is the majority action cellular automaton over H associated with S. We have seen above that τH is surjective. It follows from Proposition 1.7.4(ii) that τ is also surjective. Thus, every group containing a free subgroup of rank two admits a cellular automaton with ﬁnite alphabet which is surjective but not preinjective. This will be extended to all nonamenable groups in Sect. 5.12.
5.11 A Preinjective but Not Surjective Cellular Automaton over F2 It follows from the Garden of Eden theorem (Theorem 5.3.1) that every preinjective cellular automaton with ﬁnite alphabet over an amenable group is
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necessarily surjective. In this section, we give examples of cellular automata with ﬁnite alphabet over the free group F2 which are preinjective but not surjective. Let G = F2 be the free group on two generators a and b. Let H be a nontrivial abelian group. Our alphabet will be the group A = H ×H. We shall use additive notation for the group operations on H and A. Let p1 , p2 : A → A be the group endomorphisms deﬁned respectively by p1 (u) = (h1 , 0) and p2 (u) =(h2 , 0) for all u = (h1 , h2 ) ∈ A. Let us equip the Cartesian product AG = g∈G A with its natural Abelian group structure. Consider the map τ : AG → AG given by τ (x)(g) = p1 (x(ga)) + p2 (x(gb)) + p1 (x(ga−1 )) + p2 (x(gb−1 ))
(5.18)
for all g ∈ G and x ∈ AG . It is clear that τ is a cellular automaton over the group G and the alphabet A with memory set S = {a, b, a−1 , b−1 } and a group endomorphism of AG . Proposition 5.11.1. The cellular automaton τ : AG → AG deﬁned by (5.18) is preinjective but not surjective. Proof. The image of τ is contained in (H × {0})G (H × H)G = AG . Thus τ is not surjective. Let us show that τ is preinjective. Suppose not. Then there exist conﬁgurations x1 , x2 ∈ AG satisfying τ (x1 ) = τ (x2 ) such that the set Ω = {g ∈ G : x1 (g) = x2 (g)} is a nonempty ﬁnite subset of G. The conﬁguration x0 = x1 − x2 ∈ AG satisﬁes τ (x0 ) = τ (x1 ) − τ (x2 ) = 0 and one has Ω = {g ∈ G : x0 (g) = 0}. Consider an element g0 ∈ Ω whose reduced form has maximal length, say n0 . Then x0 (g0 ) is a nonzero element (h0 , k0 ) of A. If h0 = 0, take s0 ∈ {a, a−1 } such that the reduced form of g0 s0 has length n0 + 1. Then, for each s ∈ S \ {s−1 0 }, the length of the reduced form of g0 s0 s is n0 + 2 and hence x(g0 s0 s) = 0. By applying 5.18, we deduce that τ (x0 )(g0 s0 ) = p1 (x0 (g0 )) = (h0 , 0) = 0, which contradicts the fact that τ (x0 ) = 0. If h0 = 0, then k0 = 0 and we proceed similarly by taking s0 ∈ {b, b−1 } such that the reduced form of g0 s0 has length n0 + 1. This gives us τ (x0 )(g0 s0 ) = p2 (x0 (g0 )) = (k0 , 0) = 0, which yields again a contradiction. This shows that τ is preinjective.
If we take for H a ﬁnite abelian group of cardinality H = n ≥ 2 (e.g., the group H = Z/nZ), this gives us a preinjective but not surjective cellular automaton over F2 whose alphabet is ﬁnite of cardinality n2 . From Proposition 1.7.4 we deduce the following:
5.12 A Characterization of Amenability in Terms of Cellular Automata
135
Proposition 5.11.2. Let G be a group containing a free subgroup of rank two. Then there exist a ﬁnite set A and a cellular automaton τ : AG → AG which is preinjective but not surjective.
5.12 A Characterization of Amenability in Terms of Cellular Automata In Sect. 5.10, we gave an example of a cellular automaton with ﬁnite alphabet over the free group F2 which is surjective but not preinjective. In fact, the existence of such an automaton holds for any nonamenable group: Theorem 5.12.1. Let G be a nonamenable group. Then there exists a ﬁnite set A and a cellular automaton τ : AG → AG which is surjective but not preinjective. Proof. Since G is nonamenable, it follows from Theorem 4.9.2 that there exist a 2toone surjective map ϕ : G → G and a ﬁnite subset S ⊂ G such that (5.19) (ϕ(g))−1 g ∈ S for all g ∈ G. Our alphabet will be the Cartesian product A = S × S. Let us ﬁx some total order ≤ on S and an arbitrary element s0 ∈ S. Deﬁne the map μ : AS → A by ⎧ ⎪ ⎨(s , t ) if there exists a unique element (s, t) ∈ S × S with s < t μ(y) = such that y(s) = (s, s ) and y(t) = (t, t ), where s , t ∈ S, ⎪ ⎩ (s0 , s0 ) otherwise, (5.20) for all y ∈ AS . Let us show that the cellular automaton τ : AG → AG with memory set S and local deﬁning map μ has the required properties. We ﬁst observe that S has at least two elements since otherwise (5.19) would imply that ϕ is bijective. Let s1 ∈ S such that s1 = s0 . Consider the conﬁgurations x0 , x1 ∈ AG , where x0 is deﬁned by x0 (g) = (s0 , s0 ) for all g ∈ G, and x1 is deﬁned by x1 (g) = (s0 , s0 ) if g = 1G and x1 (1G ) = (s0 , s1 ). The conﬁgurations x0 and x1 are almost equal since they diﬀer only at 1G . On the other hand, it is clear that x0 and x1 have the same image, namely x0 , by τ . Thus τ is not preinjective. We use the properties of ϕ to prove that τ is surjective. Let x ∈ AG be an arbitrary conﬁguration. Let us show that there is a conﬁguration z ∈ AG such that x = τ (z). We construct z in the following way. Let u : G → S and v : G → S be the maps deﬁned by x(g) = (u(g), v(g)) for all g ∈ G. For each g ∈ G, there are exactly two elements sg , tg ∈ S such that sg < tg and ϕ(gsg ) = ϕ(gtg ) = g. Let us set z(gsg ) = (sg , u(g)) and z(gtg ) = (tg , v(g)).
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Observe that z : G → A is well deﬁned and that the value of z at g ∈ G is either ((ϕ(g))−1 g, u(ϕ(g)) or ((ϕ(g))−1 g, v(ϕ(g)). It immediately follows from the deﬁnition of τ that x = τ (z). This shows that τ is surjective. Combining Theorem 5.12.1 with Theorem 5.3.1, we obtain the following characterization of amenable groups in terms of cellular automata: Corollary 5.12.2. Let G be a group. Then the following conditions are equivalent: (a) G is amenable; (b) every surjective cellular automaton with ﬁnite alphabet over G is preinjective.
5.13 Garden of Eden Patterns for Life Let τ : {0, 1}Z → {0, 1}Z denote the cellular automaton associated with the Game of Life (see Example 1.4.3(a)). As it was observed in Example 5.3.2(b), it follows from the Garden of Eden theorem that τ is not surjective. Thus, Proposition 5.1.1 implies that τ admits Garden of Eden patterns. The purpose of this section is to present a direct proof of the existence of such patterns. This construction will provide a concrete illustration of the ideas underlying the proof that surjectivity implies preinjectivity in the Garden of Eden theorem. Given an integer n ≥ 1, one says that a subset Ω ⊂ Z2 is a square of size n × n in Z2 if there exist p, q ∈ Z such that 2
2
Ω = {p, p + 1, . . . , p + n − 1} × {q, q + 1, . . . , q + n − 1}. Proposition 5.13.1. Let τ : {0, 1}Z → {0, 1}Z be the cellular automaton associated with the Game of Life. Then every square Ω ⊂ Z2 of size n × n with n ≥ 3 × 109 is the support of a Garden of Eden pattern for τ . 2
2
Proof. Let n ≥ 1 be an integer. Consider a square Cn ⊂ Z2 of size 5n × 5n. Let Dn ⊂ Cn be the square of size (5n − 2) × (5n − 2) which is the Sinterior of Dn for S = {−1, 0, 1}2 ⊂ Z2 . Thus Dn is obtained from Cn by removing the 20n − 4 points of Z2 located on the (usual) boundary of Cn . Let us set Xn = {0, 1}Cn and Yn = {0, 1}Dn . The map τ induces a map τn : Xn → Yn deﬁned as follows. If u ∈ Xn , we set τn (u) = (τ (x))Dn , where 2 x ∈ {0, 1}Z satisﬁes xCn = u (the fact that (τ (x))Dn does not depend of the choice of x follows from Proposition 5.4.3 since S is a memory set for τ and Dn is the Sinterior of Cn ). We have 2
Yn  = 2(5n−2) = 225n
2
−20n+4
.
Let us divide the square Cn into n2 squares of size 5 × 5 (see Fig. 5.10).
5.13 Garden of Eden Patterns for Life
137
Fig. 5.9 A square Cn ⊂ Z2 of size 5n × 5n and its Sinterior Dn ⊂ Cn , a square of size (5n − 2) × (5n − 2) where S = {−1, 0, 1}2 ⊂ Z2 ; here n = 3
There are 225 maps from each square of size 5 × 5 to the set {0, 1}. Now observe that if the restriction of an element u ∈ Xn to one of these 5 × 5 squares is identically 0, then we may replace the 0 value at the center of this 5 × 5 square by 1 without changing τn (u). We deduce that 2
τn (Xn ) ≤ (225 − 1)n = (2log2 (2
25
−1) n2
)
= 2(25+log2 (1−2
−25
Therefore we have τn (Xn ) < Yn  if (25 + log2 (1 − 2−25 ))n2 < 25n2 − 20n + 4. This inequality is equivalent to −n2 log2 (1 − 2−25 ) − 20n + 4 > 0, which is veriﬁed if and only if n>
2 (5 + 25 + log2 (1 − 2−25 )), −25 − log2 (1 − 2 )
that is, if and only if n ≥ 465 163 744.
))n2
.
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5 The Garden of Eden Theorem
Fig. 5.10 The square Cn ⊂ Z2 of size 5n × 5n is divided into n2 squares of size 5×5; here, as in Fig. 5.9, n = 3
For such values of n, the map τn is not surjective, i.e., there exists a Garden of Eden pattern whose support is Dn . This shows the existence of a Garden of Eden pattern of size 2 325 818 718 × 2 325 818 718.
Notes According to M. Gardner (see [Gar2, p.230]), it was J. Tuckey who introduced the term “Garden of Eden” in the theory of cellular automata. The ﬁrst contribution to the Garden of Eden theorem goes back to E.F. Moore [Moo] who proved that every surjective cellular automaton with ﬁnite alphabet over Z2 is preinjective. The converse implication was established shortly after by J. Myhill [Myh]. This is the reason why the Garden of Eden theorem for Z2 is often referred to as the MooreMyhill theorem. The next step in the proof of the Garden of Eden theorem was done by A. Mach`ı and F. Mignosi [MaM] who extended it to ﬁnitely generated groups of subexponential growth (see Chap. 6 for the deﬁnition of ﬁnitely generated groups of subexponential growth). Then, the Garden of Eden theorem was proved for
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all ﬁnitely generated amenable groups by A. Mach`ı, F. Scarabotti and the ﬁrst author [CMS1]. The general case may be reduced to the case of ﬁnitely generated groups by considering the restriction of the cellular automaton to the subgroup generated by a memory set and applying Proposition 1.7.4(ii) and Proposition 5.2.2 (see [CeC8]). The proof based on Følner nets which is presented in this chapter is more direct. The term “preinjective” was introduced by M. Gromov in the appendix of [Gro5]. It follows from a result due to D.S. Ornstein and B. Weiss (see [OrW], [Gro6], [Kri]) that the lim sup appearing in the deﬁnition of entropy (Deﬁnition 5.7.1) is in fact a true limit and is independent of the particular choice of the right Følner net F in the group. Versions of the Garden of Eden theorem for cellular automata over certain classes of subshifts (closed invariant subsets of the full shift) may be found in the appendix of [Gro5] and in two papers of F. Fiorenzi [Fio1], [Fio2]. See also [CFS]. The ﬁrst examples of preinjective (resp. surjective) cellular automata which are not surjective (resp. not preinjective) where described by D.E. Muller (unpublished class notes). The underlying group was the modular group G = P SL(2, Z) = (Z/2Z) ∗ (Z/3Z) (note that G contains the free group F2 ). Theorem 5.12.1 is due to L. Bartholdi [Bar]. The computation in Sect. 5.13 is taken from [BCG, Page 828]. The size of the corresponding Garden of Eden pattern is far from being optimal. The ﬁrst explicit example of a Garden of Eden pattern for the cellular automaton associated with Conway’s Game of Life was found by R. Banks in 1971. Banks’ pattern is supported by a rectangle of size 33 × 9 and has 226 alive cells. The smallest known Garden of Eden pattern for Life, found by N. Beluchenko on September 2009, has as support a square 11 × 11 and bears 69 alive cells. It was proved that there exist no Garden of Eden patterns for Life with support contained in a rectangle of size 6 × 5.
Exercises 5.1. Let G be a group and let A be a ﬁnite set. Let τ : AG → AG be a cellular automaton admitting a memory set M ⊂ G such that M  = 1. Show that the following conditions are equivalent: (i) τ is preinjective; (ii) τ is injective; (iii) τ is surjective. 5.2. Let G be a group and let A be a ﬁnite set. Let τ : AG → AG be a nonsurjective cellular automaton. Show that there are uncountably many Garden of Eden conﬁgurations in AG .
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5 The Garden of Eden Theorem
5.3. Life on Z. Let A = {0, 1} and consider the cellular automaton τ : AZ → AZ with memory set S = {−1, 0, 1} and local deﬁning map μ : AS → A given by 1 if s∈S y(s) = 2 μ(y) = 0 otherwise for all y ∈ AS . (a) Show that τ is not preinjective. (b) Deduce from (a) that τ is not surjective either. Hint: Use Theorem 4.6.1 and the Garden of Eden Theorem (cf. Exemple 5.3.2(c)). (c) It follows from Proposition 5.1.1 that τ admits a Garden of Eden pattern. Check that the map p : {0, 1, 2, 3, 4, 5, 6, 7, 8} → A deﬁned by p(0) = p(1) = p(3) = p(5) = p(8) = 1 and p(2) = p(4) = p(6) = p(7) = 0 is a Garden of Eden pattern for τ . 5.4. Let G = Z, A = {0, 1} and let τ : AG → AG be the majority action cellular automaton (cf. Example 1.4.3(c)). We have seen in Example 5.2.1(c) that τ is not preinjective so that, by amenability of the group G (cf. Theorem 4.6.1) and the Garden of Eden Thoerem, τ is not surjective either (cf. Exemple 5.3.2(c)). It follows from Proposition 5.1.1 that τ admits a Garden of Eden pattern. Check that the map p : {1, 2, 3, 4, 5} → A deﬁned by p(1) = p(3) = p(4) = 0 and p(2) = p(5) = 1 is a Garden of Eden pattern for τ . 5.5. Let (A, +) be an abelian group (not necessarily ﬁnite). Let τ : AZ → AZ be the cellular automaton deﬁned by τ (x)(n) = x(n − 1) + x(n) + x(n + 1) for all x ∈ AZ and n ∈ Z. Show that τ is surjective. Hint: Given an arbitrary conﬁguration y ∈ AZ , construct a conﬁguration x ∈ AZ such that x(0) = x(1) = 0A and τ (x) = y. 5.6. Take G = Z and A = Z/2Z = {0, 1}. Let x1 ∈ AG be the conﬁguration deﬁned by x1 (n) = 1 for all n ∈ Z. Let x2 ∈ AG be the conﬁguration deﬁned by x2 (0) = 1 and x2 (n) = 0 for all n ∈ Z \ {0}. Let f : AG → AG be the map deﬁned by f (x2 ) = x1 and f (x) = x for all x ∈ AG \ {x2 }. Let g : AG → AG be the map deﬁned by g(x)(n) = x(n + 1) + x(n) for all n ∈ Z and x ∈ AG . Verify that the maps f and g are both preinjective but that the map g ◦ f is not preinjective. 5.7. Let G be group and let A be a set. Suppose that σ : AG → AG and τ : AG → AG are preinjective cellular automata. Show that the cellular automaton τ ◦ σ is preinjective. 5.8. Let G be a locally ﬁnite group and let A be a set. Show that every preinjective cellular automaton τ : AG → AG is injective. 5.9. Give a direct proof of the Garden of Eden theorem (Theorem 5.3.1) in the case when the group G is locally ﬁnite.
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5.10. Life in a tree. Let G = F4 denote the free group based on four generators a, b, c, d. Let A = {0, 1}. Consider the cellular automaton τ : AG → AG with memory set S = {1G , a, b, c, d, a−1 , b−1 , c−1 , d−1 } and local deﬁning map μ : AS → A given by ⎧ ⎧ ⎪ ⎪ ⎨ s∈S y(s) = 3 ⎪ ⎪ ⎨1 if or ⎪ μ(y) = ⎩ ⎪ ⎪ s∈S y(s) = 4 and y((0, 0)) = 1, ⎪ ⎩ 0 otherwise for all y ∈ AS (cf. Example 1.4.3(a)). Show that τ is surjective but not preinjective. 5.11. Let G be a group. Let E, F and Ω be subsets of G. (a) Show that (Ω −E )−F = Ω −F E and (Ω +E )+F = Ω +F E . (b) Let G = Z, E = {−2, −1, 0, 1, 2}, F = {−1, 0, 1} and Ω = {−10, −9, . . . , −1, 0, 1, . . . , 9, 10}. Check that ∂F (∂E (Ω)) = {−13, −12, −9, −8, 8, 9, 12, 13} and ∂F E (Ω) = {−13, −12, −11, −10, −9, −8, 8, 9, 10, 11, 12, 13}. Deduce that, in general, one has ∂F (∂E (Ω)) = ∂EF (Ω). 5.12. Let G be a group and let E ⊂ G. Determine the sets ∂E (∅) and ∂E (G). 5.13. Let G be a group. Let E and Ω be subsets of G. Show that ∂E (G\Ω) = ∂E (Ω). 5.14. Let G be a group. Let E, Ω1 and Ω2 be subsets of G. Show that ∂E (Ω1 ∪ Ω2 ) ⊂ ∂E (Ω1 ) ∪ ∂E (Ω2 ). Give an example showing that one may have ∂E (Ω1 ∪ Ω2 ) = ∂E (Ω1 ) ∪ ∂E (Ω2 ). 5.15. Let G be a group. Let E, Ω1 and Ω2 be subsets of G. Show that ∂E (Ω1 \ Ω2 ) ⊂ ∂E (Ω1 ) ∪ ∂E (Ω2 ). Give an example showing that one may have ∂E (Ω1 \ Ω2 ) = ∂E (Ω1 ) ∪ ∂E (Ω2 ). 5.16. Let G be a group. Let E and Ω be subsets of G. Show that Ω −gE = Ω −E g −1 , Ω +gE = Ω +E g −1 , and ∂gE (Ω) = ∂E (Ω)g −1 for all g ∈ G. 5.17. Let G be a group. Show that a subset R ⊂ G is syndetic (cf. Exercise 3.39) if and only if there exists a ﬁnite subset S ⊂ G such that the set Ω = R−1 satisﬁes Ω +S = G. 5.18. Let G be a group. Let E and Ω be subsets of G. Show that Ω ⊂ −1 (Ω +E )−E . 5.19. Let G be a group and let A be a set. Let τ : AG → AG be a cellular automaton with memory set M . Let Ω be a subset of G. Show that if two −1 conﬁgurations x1 , x2 ∈ AG coincide on Ω +M then the conﬁgurations τ (x1 ) and τ (x2 ) coincide on Ω.
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5.20. Let (A, +) be a ﬁnite abelian group and let ϕ : A × A → A be a map. Let τ : AZ → AZ be the map deﬁned by τ (x)(n) = x(n)+ϕ(x(n+1), x(n+2)) for all n ∈ Z and x ∈ AZ . (a) Show that τ is a cellular automaton over the group Z and the alphabet A. (b) Show that τ is preinjective. (c) Show that τ is surjective. 5.21. Let G = Z and A = {0, 1}. Verify that, among the 16 cellular automata τ : AZ → AZ with memory set S = {0, 1} ⊂ Z, there are exactly 6 of them which are surjective, 4 of them which are injective, 4 of them which are bijective, and 10 of them which are neither surjective nor injective. Hint: The Garden of Eden Theorem may help. 5.22. Topological entropy. Let G be an amenable group and let F = (Fj )j∈J be a right Følner net for G. If X is a compact topological space equipped with a continuous action of G, the topological entropy hF (X, G), 0 ≤ hF (X, G) ≤ ∞, of X is deﬁned as follows. Suppose that U = (Ui )i∈I is an open cover of X. We denote by N (U) the smallest integer
n ≥ 0 such that there is a ﬁnite subset I0 ⊂ I of cardinality n such that i∈I0 Ui = X (note that there exists such a ﬁnite subset I0 ⊂ I by compactness of X). Given a nonempty subset F ⊂ G, we deﬁne the open cover UF = (Wα )α∈I F indexed by the set I F = {α : F → I} by setting Wα = gUα(g) g∈F
for all α ∈ I F . Finally, we set hF (U) = lim sup j
log N (UFj ) Fj 
and hF (X, G) = sup hF (U), U
where U ranges over all open covers of X. Observe that it is clear from this deﬁnition that if X and Y are two compact topological spaces, each equipped with a continuous action of G, such that there exists a Gequivariant homeomorphism f : X → Y , then one has hF (X, G) = hF (Y, G). (a) Let X be a compact space and let U = (Ui )i∈I and V = (Vk )k∈K be two open covers of X. One says that the open cover V is ﬁner than U if, for each k ∈ K, there exists i ∈ I such that Vk ⊂ Ui . Show that if the open cover V is ﬁner than the open cover U then one has N (U) ≤ N (V). (b) Let X be a compact space and let U = (Ui )i∈I be an open cover of X which forms a partition of X (i.e., such that Ui1 ∩ Ui2 = ∅ for all i1 , i2 ∈ I with i1 = i2 ). Show that N (U) = {i ∈ I : Ui = ∅}.
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(c) Let A be a ﬁnite set and let X ⊂ AG be a subshift. Show that if X is equipped with the action of G induced by the Gshift on AG , then one has hF (X, G) = entF (X). Hint: Consider the open cover T = (Ta )a∈A of X deﬁned by Ta = {x ∈ X : x(1G ) = a} and deduce from (b) that N (TFj ) = πFj (X), where πFj : AG → AFj denotes the projection map. This gives hF (X, G) ≥ hF (T ) = entF (X). To prove hF (X, G) ≤ entF (X), observe that if U = (Ui )i∈I is an arbitrary open cover of X, then there exists a ﬁnite subset Ω ⊂ G such that the open cover TΩ is ﬁner than U. Apply (a) to get N (UFj ) ≤ πFj Ω (X) and ﬁnally use the fact that F is a right Følner net to conclude. Note: It can be shown by using a result due to Ornstein and Weiss (cf. the notes above) that the topological entropy hF (X, G) of a compact space X equipped with a continuous action of an amenable group G is in fact independent of the choice of the right Følner net F. 5.23. Let G be an amenable group and let F = (Fj )j∈J be a right Følner net for G. Let A and B be ﬁnite sets. Suppose that X ⊂ AG and Y ⊂ B G are two subshifts such that there exists a bijective continuous Gequivariant map ϕ : X → Y . Show that entF (X) = entF (Y ). Hint: Use Exercise 5.22. 5.24. Let G be an amenable group, F a right Følner net for G, and A a ﬁnite set. Let X be a subset of AG and let X denote the closure of X in AG for the prodiscrete topology. Show that one has entF (X) = entF (X). 5.25. Let G be an amenable group and let (Fj )j∈J be a right Følner net for G. Let A be a ﬁnite set. Suppose that X (resp. Y ) is a subset of AG having at least two distinct elements. Show that entF (X ∪ Y ) ≤ entF (X) + entF (Y ). 5.26. Let G be an amenable group, A a ﬁnite set, F be a right Følner net for G, and X ⊂ AG a subshift. Let F be a nonempty ﬁnite subset of G and consider the subshift X [F ] ⊂ B G , where B = AF (cf. Exercise 1.34). Show that entF (X [F ] ) = entF (X). 5.27. Let G be a group, A be a set and X ⊂ AG a subshift. Let also H ⊂ G be a subgroup of G and T ⊂ G a complete set of representatives for the right cosets of H in G, and consider the subshift X (H,T ) ⊂ B H , where B = AH\G (cf. Exercise 1.35). Suppose that G and H are isomorphic and denote by φ : G → H an isomorphism. Let F = (Fj )j∈J be a right Følner net in G so that F = (φ(Fj ))j∈J is a right Følner net in G. Show that entF (X (H,T ) ) = entF (X). 5.28. Let G = Z and A = {0, 1}. Set Fn = {0, 1, . . . , n − 1} and recall from Example 4.7.4(b) that F = (Fn )n∈N is a Følner sequence in Z. Show that if X ⊂ AZ is either the even subshift considered in Exercise 1.38 or the golden mean subshift considered in Exercise 1.39 then entF (X) = log ϕ, √ 1+ 5 where ϕ = 2 is the golden number.
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5.29. Entropy of the Morse subshift. Let A = {0, 1} and let x ∈ AN and X ⊂ AZ denote the ThueMorse sequence (cf. Exercises 3.41) and the Morse subshift (cf. Exercise 3.42) respectively. (a) Show that Ln (X) ≤ 8n for all n ≥ 1. Hint: Let n ≥ 1 and denote by k the unique integer such that 2k−1 ≤ n < 2k . Consider the words u = x(0)x(1) · · · x(2k − 1) and v = ι(u) (cf. Exercise 3.41(e)). Observe that any word w ∈ Ln (X) is necessarily a subword of one of the words uv, vv, vu, or uu. Altogether, this gives at most 4 · 2k = 8 · 2k−1 ≤ 8n distinct possibilities for w. (b) Let F = (Fn )n≥1 denote the Følner sequence of Z where Fn = {0, 1, . . . , n − 1}. Deduce from (a) that entF (X) = 0. 5.30. Let A = {0, 1, 2} and set Y = {y ∈ AZ : y(g) ∈ {1, 2} for all g ∈ Z} and X = Y ∪ {x0 }, where x0 ∈ AZ denotes the constant conﬁguration deﬁned by x0 (g) = 0 for all g ∈ Z. (a) Show that X and Y are subshifts of ﬁnite type of AZ and that one has the strict inclusion Y X. (b) Show that X is not irreducible. (c) Check that X and Y have the same entropy entF (X) = entF (Y ) = log 2 with respect to the Følner sequence F = (Fn )n∈N of Z given by Fn = {0, 1, . . . , n − 1}. 5.31. Let G be an amenable group, A a ﬁnite set, X ⊂ AG a strongly irreducible subshift, and Y ⊂ AG a nonempty subshift such that Y X. Let also F = (Fj )j∈J be a right Følner net in G. For a subset Ω ⊂ G we denote by πΩ : AG → AΩ the projection (restriction) map. (a) Show that there exists a ﬁnite subset Ω0 ⊂ G such that πΩ0 (Y ) πΩ0 (X). (b) Let Δ ⊂ G be a ﬁnite subset such that 1G ∈ Δ and X is Δirreducible. Let E = Ω0+Δ and set ξ = πE (X)−1 and E = EE −1 = {ab−1 : a, b ∈ E}. By virtue of Proposition 5.6.3 we can ﬁnd an (E, E )tiling T ⊂ G. For j ∈ J set Tj = {g ∈ T : gE ⊂ Fj }. Show that ξπFj (X) ≤ πFj \gE (X) for all g ∈ Tj . (c) Let p ∈ πΩ0 (X) \ πΩ0 (Y ) (cf. (a)) and let x ∈ X such that xΩ0 = p. For g ∈ G set pg = (gx)gΩ0 ∈ AgΩ0 . Show that pg ∈ πgΩ0 (X) \ πgΩ0 (Y ). F (d) For j ∈ J and g ∈ Tj denote by πgΩj 0 : πFj (X) → πgΩ0 (X) the projection (restriction) map and let pg ∈ πgΩ0 (X) \ πgΩ0 (Y ) (cf. (c)). Using (b) and F the Δirreducibility of X show that πFj (X)\(πgΩj 0 )−1 (pg ) ≤ (1−ξ)πFj (X) for all j ∈ J and g ∈ Tj . (e) For j ∈ J denote by πFj (X)∗ the set of patterns p ∈ πFj (X) such F that πgΩj 0 (p) = pg for all g ∈ Tj . Observe that πFj (X)∗ = πFj (X) \
Fj −1 (pg ) and using the Δirreducibility of X and an inductive arg∈Tj (πgΩ0 ) gument based on (d), show that πFj (X)∗  ≤ (1 − ξ)Tj  πFj (X).
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(f) Observe that πFj (Y ) ⊂ πFj (X)∗ and deduce from (e) that log πFj (Y ) ≤ Tj  log(1 − ξ) + log πFj (X). (g) By Proposition 5.6.4 there exists a real number α > 0 and an element j0 ∈ J such that Tj  ≥ αFj  for all j ≥ j0 . Deduce from (f) that entF (Y ) ≤ α log(1 − ξ) + entF (X). (h) Deduce from (g) that entF (Y ) < entF (X). 5.32. Let G be an amenable group and let A be a ﬁnite set. Let F = (Fj )j∈J be a right Følner net in G. Let X ⊂ AG be a nonempty strongly irreducible subshift. Show that if X is not minimal then one has entF (X) > 0. Hint: Show that, with the notation introduced in Exercise 5.31, one has entF (X) ≥ α log 2. 5.33. Let G be a group and let A be a ﬁnite set. Let Δ be a ﬁnite subset of G and let X ⊂ AG be a Δirreducible subshift. Suppose that (Ωi )i∈I is a (possibly inﬁnite) family of (possibly inﬁnite) subsets of G such that ⎛ ⎞ Ωi+Δ ∩ ⎝ Ωk ⎠ = ∅ for all i ∈ I. k∈I\{i}
Let also (xi )i∈I be a family of conﬁgurations in X. Denote by Pf (G) the set of all ﬁnite subsets of G. For each Λ ∈ Pf (G) let X(Λ) ⊂ X denote the set consisting of all conﬁgurations in X which coincide with xi on Λ ∩ Ωi for all i ∈ I. (a) Show that X(Λ) is closed in X for each Λ ∈ Pf (G). (b) Show that if we ﬁx Λ ∈ Pf (G), then the subsets Ψi = Λ ∩ Ωi are all contained in Λ and satisfy ⎛ ⎞ Ψk ⎠ = ∅ for all i ∈ I. Ψi+Δ ∩ ⎝ k∈I\{i}
(c) Fix Λ ∈ Pf (G) and set IΛ = {i ∈ I : Λ ∩ Ωi = ∅}. Show that IΛ is ﬁnite. Then, applying induction on the cardinality of IΛ , show that, by Δirreducibility of X, one has X(Λ) = ∅. (d) Show that X(Λ1 ) ∩ X(Λ2 ) ∩ · · · ∩ X(Λn ) = X(Λ1 ∪ Λ2 ∪ · · · ∪ Λn ) and deduce that X(Λ1 )∩X(Λ2 )∩· · ·∩X(Λn ) = ∅ for all Λ1 , Λ2 , . . . , Λn ∈ Pf (G). (e) Deduce from (d) that the family (X(Λ))Λ∈Pf (G) has a nonempty intersection. (f) Let x ∈ X be such that x ∈ X(Λ) for each ﬁnite subset Λ ⊂ G (whose existence is guaranteed by (e)). Show that x coincides with xi on Ωi for all i ∈ I. 5.34. Let G be an amenable group and let A be a ﬁnite set. Let F = (Fj )j∈J be a right Følner net in G. Let X ⊂ AG be a (possibly minimal) strongly irreducible subshift containing at least two distinct conﬁgurations x0 and x1 .
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(a) Show that there exists a ﬁnite subset D ⊂ G such that (gx0 )gD = (gx1 )gD for all g ∈ G. (b) Let Δ be a ﬁnite subset of G such that X is Δirreducible and 1G ∈ Δ. Set E = D+Δ . By Proposition 5.6.3 there exists an (E, E )tiling T ⊂ G for some ﬁnite subset E ⊂ G. Consider the subset Z ⊂ X consisting of all the conﬁgurations z ∈ X such that, for all g ∈ T , one has either zgD = (gx0 )gD or zgD = (gx1 )gD . Applying Exercise 5.33 to the family (gD)g∈T show that, given any map ι : T → {0, 1}, there exists a conﬁguration x ∈ X such that xgD = (gxι(g) )gD for all g ∈ T . (c) Deduce that ZF j  ≥ 2Tj  for all j ∈ J, where, Tj = {g ∈ T : gE ⊂ Fj }. (d) By Proposition 5.6.4 there exists a real number α > 0 and an element j0 ∈ J such that Tj  ≥ αFj  for all j ≥ j0 . Deduce from (c) that entF (Z) ≥ α log 2. (e) Deduce that entF (X) > 0. 5.35. Let G be a group and let A be a ﬁnite set. Let X ⊂ AG be a strongly irreducible subshift. Suppose that Δ is a ﬁnite subset of G such that X is Δirreducible. Show that if Ω1 and Ω2 are (possibly inﬁnite) subsets of G such that Ω1+Δ ∩ Ω2 = ∅, then, given any two conﬁgurations x1 and x2 in X, there exists a conﬁguration x ∈ X which coincides with x1 on Ω1 and with x2 on Ω2 . Hint: For each ﬁnite subset Ω ⊂ G, consider the subset X(Ω) ⊂ X consisting of all conﬁgurations in X which coincide with x1 on Ω ∩ Ω1 and with x2 on Ω ∩ Ω2 . Observe that the family formed by the sets X(Ω), where Ω runs over all ﬁnite subsets of G is a family of closed subsets of X having the ﬁnite intersection property and use the compactness of X. 5.36. (cf. Lemma 5.8.3) Let G be an amenable group, F = (Fj )j∈J a right Følner net for G, and A a ﬁnite set. Let τ : X → Y be a cellular automaton, where X, Y ⊂ AG are subshifts such that X is strongly irreducible and entF (Y ) < entF (X). (a) Let S ⊂ G be a memory set for τ . Up to enlarging the subset S if necessary, we can also suppose that 1G ∈ S and that X is Sirreducible. Deduce from the hypothesis entF (Y ) < entF (X) that there exists j0 ∈ J such that πF +S2 (Y ) < πFj0 (X). j0
(b) Fix an arbitrary conﬁguration x0 ∈ X and consider the ﬁnite subset Z ⊂ X consisting of all conﬁgurations z ∈ X which coincide with x0 on G \ . Show that πFj0 (Z) = πFj0 (X). Hint: Use the result of Exercise 5.35(a) Fj+S 0 by taking Ω1 = Fj0 and Ω2 = G \ Fj+S . 0 (c) Use Proposition 5.4.3 to show that τ (z) coincide with τ (x0 ) on G\Fj+S 0 for all z ∈ Z. (d) Deduce from (c) that τ (Z) = πF +S2 (τ (Z)). j0
2
(e) Using the fact that τ (Z) ⊂ Y , deduce from (c), (a) and (b) that there exist conﬁgurations z1 = z2 in Z such that τ (z1 ) = τ (z2 ). (f) Deduce from (e) that τ is not preinjective.
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5.37. Let G be an amenable group, F = (Fj )j∈J a right Følner net for G, and A a ﬁnite set. Let X, Y ⊂ AG be two strongly irreducible subshifts such that entF (X) = entF (Y ). Show that every preinjective cellular automaton τ : X → Y is surjective. Hint: Use the results of Exercise 5.31 and Exercise 5.36. 5.38. Let G be an amenable group and let A be a ﬁnite set. Let X ⊂ AG be a strongly irreducible subshift. Show that every preinjective cellular automaton τ : X → X is surjective. Hint: Use Exercise 5.37 with Y = X. 5.39. Let G be an amenable group and let A be a ﬁnite set. Show that every strongly irreducible subshift X ⊂ AG is surjunctive. Hint: This immediately follows from Exercise 5.38. 5.40. (cf. Lemma 5.8.4) Let G be an amenable group, F = (Fj )j∈J a right Følner net for G, and A a ﬁnite set. Let X ⊂ AG be a strongly irreducible subshift of ﬁnite type and let τ : X → AG be a cellular automaton which is not preinjective. Let x1 , x2 ∈ X such that τ (x1 ) = τ (x2 ) and Ω = {g ∈ G : x1 (g) = x2 (g)} is a nonempty ﬁnite subset of G. Let S be a memory set for both τ and X. Up to enlarging the subset S if necessary, we can also suppose −1 that 1G ∈ S and that X is Sirreducible. Set E = Ω +(S S) = {ωs−1 s : ω ∈ Ω, s, s ∈ S} and E = EE −1 = {ab−1 : a, b, ∈ E}. By Proposition 5.6.3 there exists an (E, E )tiling T ⊂ G. Let Z ⊂ X be the set consisting of all conﬁgurations z in X such that ztE = (tx1 )tE for all t ∈ T . (a) Using the Sirreducibility of X, show that entF (Z) < entF (X). Hint: Use the arguments in Exercise 5.31. (b) From (a) and Proposition 5.7.3 deduce that entF (τ (Z)) ≤ entF (X). (c) Using the fact that S is a memory set for X, show that for every x ∈ X the conﬁguration z deﬁned by tx2 (g) if there is t ∈ T such that g ∈ tE and xtE = tx1 tE , z(g) = x(g) otherwise. satisﬁes z ∈ Z and τ (z) = τ (x) (cf. the end of the proof of Lemma 5.8.4). (d) Deduce from (c) that τ (Z) = τ (X). (e) Deduce from (b) and (d) that entF (τ (X)) < entF (X). 5.41. Let G be an amenable group, F = (Fj )j∈J a right Følner net for G, and A a ﬁnite set. Let X, Y ⊂ AG be two subshifts such that X is strongly irreducible of ﬁnite type and entF (X) = entF (Y ). Show that every surjective cellular automaton τ : X → Y is preinjective. Hint: Use the result of Exercise 5.40. 5.42. The Garden of Eden theorem for strongly irreducible subshifts of ﬁnite type. (cf. [Fio2, Theorem 4.7]). Let G be an amenable group, A a ﬁnite set, X ⊂ AG a strongly irreducible subshift of ﬁnite type, and τ : X → X a cellular automaton. Show that τ is preinjective if and only if it is surjective.
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Hint: Combine together the results from Exercise 5.41 with Y = X and Exercise 5.38. 5.43. Let G be a group. Let F be a nontrivial ﬁnite abelian group and set A = F × F . Suppose that G contains an element h0 ∈ G of inﬁnite order (e.g. G = Z). Denote by H ⊂ G the inﬁnite cyclic subgroup generated by h0 . For each x ∈ AG , let x1 , x2 : G → F be the maps deﬁned by x(g) = (x1 (g), x2 (g)) for all g ∈ G. Consider the subset X ⊂ AG consisting of all conﬁgurations x ∈ AG which satisfy x2 (gh0 ) = x2 (g) for any g ∈ G. In other words, X is formed by all the conﬁgurations x ∈ AG such that x2 is constant on each left coset of H in G. (a) Show that X is a subshift of ﬁnite type. (b) Consider the map τ : X → X deﬁned by τ (x) = (x1 + x2 , 0) for all x = (x1 , x2 ) ∈ X (here 0 denotes the zero conﬁguration, i.e., the constant conﬁguration whose value at each element of G is equal to the identity element of F ). Show that τ is a cellular automaton. (c) Show that the cellular automaton τ deﬁned in (b) is preinjective but not surjective. (d) Show that if H is of inﬁnite index in G (this is the case, for example, when G = Z2 ) then the subshift X is irreducible. 5.44. Let G be a group and let A be a set. Let H be a subgroup of G. Suppose that σ : AH → AH is a cellular automaton and let σ G : AG → AG be the induced cellular automaton (cf. Sect. 1.7). Let X ⊂ AH be a subshift and denote by X (G) ⊂ AG the associated subshift deﬁned in Exercise 1.33. Suppose that σ(X) ⊂ X. Show that the cellular automaton σ G X (G) : X (G) → X (G) is preinjective if and only if the cellular automaton σX : X → X is preinjective. 5.45. Let A = {0, 1}. Let x0 , x1 ∈ AZ denote the two constant conﬁgurations deﬁned by x0 (n) = 0 and x1 (n) = 1 for all n ∈ Z. (a) Show that X = {x0 , x1 } is a subshift of ﬁnite type. (b) Show that the map τ : X → X given by τ (x0 ) = τ (x1 ) = x0 is a cellular automaton which is preinjective but neither surjective nor injective. 5.46. Let A = {0, 1}, G = Z2 , and H = Z × {0} ⊂ G. Consider the subset X = Fix(H) ⊂ AG consisting of all the conﬁgurations x ∈ AG which are ﬁxed by each element of H. (a) Show that X is an irreducible subshift of ﬁnite type of AG . (b) Consider the map τ : X → X deﬁned by τ (x)(g) = 0 for all x ∈ X and g ∈ G. Show that τ is a cellular automaton which is preinjective but neither surjective nor injective. 5.47. ([Fio1, Counterexample 4.27]) Let A = {0, 1, 2} and let X ⊂ AZ be the subshift of ﬁnite type with deﬁning set of forbidden words {01, 02}. (a) Show that X is not irreducible.
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(b) Consider the cellular automaton σ : AZ → AZ with memory set S = {0, 1} and local deﬁning map y(0) if y(1) = 0 μ(y) = 0 otherwise. Show that σ(X) ⊂ X. (c) Show that the cellular automaton σX : X → X is surjective but not preinjective. 5.48. Let A, X ⊂ AZ and σ : AZ → AZ as in Exercise 5.47. Let H = Z×{0} ⊂ 2 Z2 and let X(Z2 ) be the associated irreducible subshift of ﬁnite type in AZ 2 2 2 deﬁned as in Exercise 1.33 and let σ Z : AZ → AZ be the induced cellular 2 2 Z2 automaton. Show that the cellular automaton σ X (Z2 ) : X (Z ) → X (Z ) is surjective but not preinjective. 5.49. ([Fio1, Section 3]) Let A = {0, 1} and let X ⊂ AZ be the even subshift (cf. Exercise 1.38). Let also σ : AZ → AZ be the cellular automaton with memory set S = {0, 1, 2, 3, 4} and local deﬁning map μ : AS → A given by 1 if y(0)y(1)y(2) ∈ {000, 111} or y(0)y(1)y(2)y(3)y(4) = 00100 μ(y) = 0 otherwise. (a) Show that σ(X) ⊂ X. (b) Set τ = σX : X → X. Show that τ is surjective. (c) Show that τ is not preinjective. Hint: Show that the conﬁgurations x1 = · · · 0000(100100)0000 · · · and x2 = · · · 0000(011100)0000 · · · satisfy τ (x1 ) = τ (x2 ) = · · · 1111(100100)1111 · · ·
Chapter 6
Finitely Generated Amenable Groups
This chapter is devoted to the growth and amenability of ﬁnitely generated groups. The choice of a ﬁnite symmetric generating subset for a ﬁnitely generated group deﬁnes a word metric on the group and a labelled graph, which is called a Cayley graph. The associated growth function counts the number of group elements in a ball of radius n with respect to the word metric. We deﬁne a notion of equivalence for such growth functions and observe that the growth functions associated with diﬀerent ﬁnite symmetric generating subsets are in the same equivalence class (Corollary 6.4.5). This equivalence class is called the growth type of the group. The notions of polynomial, subexponential and exponential growth are introduced in Sect. 6.5. In Sect. 6.7 we give an example of a ﬁnitely generated metabelian group with exponential growth. We prove that ﬁnitely generated nilpotent groups have polynomial growth (Theorem 6.8.1). In Sect. 6.9 we consider the Grigorchuk group. It is shown that it is an inﬁnite periodic (Theorem 6.9.8), residually ﬁnite (Corollary 6.9.5) ﬁnitely generated group of intermediate growth (Theorem 6.9.17). In the subsequent section, we show that every ﬁnitely generated group of subexponential growth is amenable (Theorem 6.11.2). In Sect. 6.12 we prove the KestenDay characterization of amenability (Theorem 6.12.9) which asserts that a group with a ﬁnite (not necessarily symmetric) generating subset is amenable if and only if 0 is in the 2 spectrum of the associated Laplacian. Finally, in Sect. 6.13 we consider the notion of quasiisometry for not necessarily countable groups and we show that amenability is a quasiisometry invariant (Theorem 6.13.23).
6.1 The Word Metric Let G be a group. One says that a subset S ⊂ G generates G, or that S is a generating subset of G, if every element g ∈ G can be expressed as a product of elements in T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 6, © SpringerVerlag Berlin Heidelberg 2010
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S ∪ S −1 , that is, for each g ∈ G there exist n ≥ 0, s1 , s2 , . . . , sn ∈ S and ε1 , ε2 , . . . , εn ∈ {1, −1} such that g = sε11 sε22 · · · sεnn .
(6.1)
A subset S ⊂ G is called symmetric if S = S −1 , that is, s−1 ∈ S whenever s ∈ S. Observe that if S ⊂ G is a generating subset of G, then S ∪ S −1 is a symmetric generating subset of G. Recall that G is said to be ﬁnitely generated if it admits a ﬁnite generating subset. If S ⊂ G is a ﬁnite generating subset of G, then S ∪ S −1 is a ﬁnite symmetric generating subset of G. Thus any ﬁnitely generated group admits a ﬁnite symmetric generating subset. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. The Swordlength S (g) = G S (g) of an element g ∈ G is the minimal integer n ≥ 0 such that g can be expressed as a product of n elements in S, that is, S (g) = min{n ≥ 0 : g = s1 s2 · · · sn , si ∈ S, 1 ≤ i ≤ n}.
(6.2)
It immediately follows from the deﬁnition that for g ∈ G one has S (g) = 0 if an only if g = 1G .
(6.3)
Proposition 6.1.1. One has S (g −1 ) = S (g)
(6.4)
S (gh) ≤ S (g) + S (h)
(6.5)
and for all g, h ∈ G. Proof. Let g, h ∈ G. Set m = S (g) and n = S (h). Then there exist s1 , s2 , . . . , sm and t1 , t2 , . . . , tn in S such that g = s1 s2 · · · sm and h = −1 −1 so that S (g −1 ) ≤ m = S (g). t1 t2 · · · tn . We have g −1 = s−1 m · · · s2 s1 −1 Exchanging the roles of g and g we get S (g) ≤ S (g −1 ) and therefore −1 S (g ) = S (g). On the other hand, we have gh = s1 s2 · · · sm t1 t2 · · · tn so that S (gh) ≤ m + n = S (g) + S (h). Consider the map dS = dG S : G × G → N deﬁned by dS (g, h) = S (g −1 h) for all g, h ∈ G. Proposition 6.1.2. The map dS is a metric on G.
(6.6)
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Proof. Let g, h, k ∈ G. It follows from (6.3) that dS (g, h) = 0 if and only if g = h. Moreover, from (6.4) we deduce that dS (g, h) = dS (h, g). Finally, using (6.5) we get dS (g, k) + dS (k, h) = S (g −1 k) + S (k−1 h) ≥ S ((g −1 k)(k−1 h)) = S (g −1 h) = dS (g, h). The metric dS is called the word metric on G associated with the ﬁnite symmetric generating subset S. Proposition 6.1.3. The metric dS is invariant by left multiplication, that is, dS (gg1 , gg2 ) = dS (g1 , g2 ) for all g, g1 , g2 ∈ G. Proof. We have dS (gg1 , gg2 ) = S ((gg1 )−1 gg2 ) = S (g1−1 g −1 gg2 ) = S (g1−1 g2 ) = dS (g1 , g2 ). We may rephrase the previous result by saying that the action of G on itself by left multiplication is an isometric action with respect to the word metric. For g ∈ G and n ∈ N, we denote by BSG (g, n) = {h ∈ G : dS (g, h) ≤ n} the ball of radius n in G centered at the element g ∈ G. When g = 1G we have BSG (1G , n) = {h ∈ G : S (h) ≤ n} and we simply write BSG (n) instead of BSG (1G , n). Also, when there is no ambiguity on the group G, we omit the supscript “G” and we simply write BS (g, n) and BS (n) instead of BSG (g, n) and BSG (n).
6.2 Labeled Graphs Let S be a set. An Slabeled graph is a pair Q = (Q, E), where Q is a set, called the set of vertices, and E is a subset of Q×S ×Q, called the set of edges. The projection map λ : E → S, deﬁned by λ(e) = s for all e = (q, s, q ) ∈ E, is called the labelling map. Also consider the projection maps α, ω : E → Q deﬁned by α(e) = q and ω(e) = q for all e = (q, s, q ) ∈ E. Then, α(e) and ω(e) are called the initial and terminal vertices of the edge e ∈ E.
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Let Q1 = (Q1 , E1 ) and Q2 = (Q2 , E2 ) be two Slabeled graphs. An Slabeled graph homomorphism from Q1 to Q2 is a map φ : Q1 → Q2 such that (φ(q), s, φ(q )) ∈ E2 for all (q, s, q ) ∈ E1 . An Slabeled graph homomorphism φ : Q1 → Q2 which is bijective and such that the inverse map φ−1 : Q2 → Q1 is also an Slabeled graph homomorphism is called a Slabeled graph isomorphism. If such an Slabeled graph isomorphism exists one says that the Slabeled graphs Q1 and Q2 are isomorphic. An Slabeled graph Q = (Q, E) is said to be ﬁnite if the sets Q and E are both ﬁnite. Let Q = (Q, E) be an Slabeled graph. An Slabeled subgraph of Q is an Slabeled graph Q = (Q , E ) such that Q ⊂ Q and E ⊂ E. Let Q ⊂ Q and set E = E ∩ (Q × S × Q ). Then the Slabeled graph Q = (Q , E ) is called the Slabeled subgraph of Q induced on Q . Sometimes, by abuse of language, we shall simply denote by Q the Slabeled graph Q = (Q , E ). If there exist vertices q, q ∈ Q and labels s1 = s2 in S such that (q, s1 , q ), (q, s2 , q ) ∈ E, in other words, if there exist two edges with the same initial and terminal vertices but with diﬀerent labels, then one says that Q has multiple edges. An edge of the form (q, s, q), q ∈ Q, s ∈ S, is called a loop at q. A path in Q is a ﬁnite sequence of edges π = (e1 , e2 , . . . , en ), e1 , e2 , . . . , en ∈ E, such that ω(ei ) = α(ei+1 ) for all i = 1, . . . , n − 1. The integer n is called the length of the path π and it is denoted by (π). The vertices π − = α(e1 ) and π + = ω(en ) are called the initial and terminal vertices of π and one says that π connects π − to π + . The label of a path π = (e1 , e2 , . . . , en ) is deﬁned by λ(π) = (λ(e1 ), λ(e2 ), . . . , λ(en )) ∈ S n . For q ∈ Q, we also admit the empty path starting and ending at q. It has length 0 and its label is the empty word . Let π1 = (e1 , e2 , . . . , en ) and π2 = (e1 , e2 , . . . , em ) be two paths with π1+ = π2− . Then the path π1 π2 = (e1 , e2 , . . . , en , e1 , e2 , . . . , em ) is called the composition of the paths π1 and π2 . Note that (π1 π2 ) = (π1 ) + (π2 ) and that λ(π1 π2 ) = λ(π1 )λ(π2 ). Let q, q ∈ Q. If there is no path connecting q to q we set dQ (q, q ) = ∞, otherwise, we set dQ (q, q ) = min{(π) : π a path connecting q to q }.
(6.7)
A path π connecting q to q with minimal length, that is, such that (π) = dQ (q, q ), is called a geodesic path from q to q . Proposition 6.2.1. Let Q = (Q, E) be a labeled graph. Then, for all q, q , q ∈ Q one has:
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(i) dQ (q, q ) ∈ N ∪ {∞}; (ii) dQ (q, q ) = 0 if and only if q = q ; (iii) dQ (q, q ) < ∞ if and only if there exists a path which connects q to q ; (iv) dQ (q, q ) ≤ dQ (q, q ) + dQ (q , q ) (triangular inequality). Proof. The statements (i), (ii), and (iii) are trivial. Let us prove (iv). If dQ (q, q ) = ∞ or dQ (q , q ) = ∞ there is nothing to prove. Otherwise, let π1 be a geodesic path connecting q to q and π2 a geodesic path connecting q to q . Then π1 π2 connects q to q and therefore dQ (q, q ) ≤ (π1 π2 ) = (π1 ) + (π2 ) = dQ (q, q ) + dQ (q , q ). Note that, in general, for q, q ∈ Q one has dQ (q, q ) = dQ (q , q). For q ∈ Q and r ∈ N we denote by B(q, r) = {q ∈ Q : dQ (q, q ) ≤ r} the ball of radius r centered at q. The labeled graph Q is said to be connected if given any two vertices q, q ∈ Q there exists a path which connects q to q . Equivalently, Q is connected if and only if dQ (q, q ) < ∞ for all q, q ∈ Q. A path π in Q such that π − = π + is said to be closed. The valence (or degree) δ(q) of a vertex q ∈ Q is the cardinality of the set {e ∈ E : α(e) = q}. If δ(q) < ∞ for all q ∈ Q, one says that Q is locally ﬁnite. If all vertices have the same ﬁnite valence k then one says that Q is regular of degree k. Suppose that S is equipped with an involution s → s. Then the labeled graph Q is said to be edgesymmetric if for all e = (q, s, q ) ∈ E one has that the inverse edge e−1 = (q , s, q) also belongs to E. If Q is edgesymmetric and (q, s, q ) ∈ E one says that the vertices q and q are neighbors. Suppose that Q is edgesymmetric. If π = (e1 , e2 , . . . , en ) is a path in Q connecting a vertex q to a vertex q , then π −1 = (en −1 , en−1 −1 , . . . , e2 −1 , e1 −1 ) is a path in Q connecting q to q. The path π −1 is called the inverse of π. Note that (π −1 ) = (π). Corollary 6.2.2. Suppose that Q is edgesymmetric and connected. Then the map dQ : Q × Q → N deﬁned in (6.7) is a metric on the set Q of vertices of Q. Proof. This follows immediately from Proposition 6.2.1 and the fact that dQ (q, q ) = dQ (q , q) for all q, q ∈ Q. The latter immediately follows from the fact that the map π → π −1 yields a lengthpreserving bijection from the set of paths connecting q to q onto the set of paths connecting q to q. The metric dQ is called the graph metric on Q. Let π = (e1 , e2 , . . . , en ) be a path in Q. We associate with π the sequence πQ = (q0 , q1 , . . . , qn ) ∈ Qn+1 deﬁned by qi = α(ei+1 ) for all i = 0, 1, . . . , n and qn = ω(en ). The vertices q0 , q1 , . . . , qn are called the vertices visited by π.
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One says that π is simple if qi = qj for all 0 ≤ i = j ≤ n. If π is closed, one says that π is a closed simple path if qi = qj for all 0 ≤ i, j ≤ n such that (i, j) = (0, n). One says that π is proper, or with no back–tracking, if ei+1 = e−1 for i = 1, 2, . . . , n − 1. i An edgesymmetric, connected labeled graph with no loops, no multiple edges and with no closed proper paths is called a tree.
6.3 Cayley Graphs Let G be a ﬁnitely generated group and S a ﬁnite symmetric generating subset of G. The Cayley graph of G with respect to S is the Slabeled graph CS (G) = (Q, E) whose vertices are the group elements, that is, Q = G and the edge set is E = {(g, s, gs) : g ∈ G and s ∈ S}.
Fig. 6.1 An edge in the Cayley graph CS (G) is a triple (g, s, h), where g ∈ G, s ∈ S, and h = gs
Note that, as S is symmetric, the inverse map s → s−1 is an involution on S. Since we have (h, s−1 , g) ∈ E for all (g, s, h) ∈ E, it follows that the Cayley graph CS (G) is edgesymmetric with respect to the inverse map on S. Moreover, the Cayley graph is connected. For, given g, h ∈ G, as S generates G, we can ﬁnd a nonnegative integer n and s1 , s2 , . . . , sn ∈ S such that g −1 h = s1 s2 · · · sn . Then the path π = (e1 , e2 , . . . , en ), where ei = (gs1 s2 · · · si−1 , si , gs1 s2 · · · si−1 si ), i = 1, 2, . . . , n, connects g to h = gs1 s2 · · · sn . Also observe that if 1G ∈ S then CS (G) has a loop at each vertex and that, on the contrary, CS (G) has no loops if 1G ∈ / S. On the other hand, CS (G) has no multiple edges, that is, for all g, h ∈ G, there exists at most one s ∈ S such that (g, s, h) ∈ E, namely s = g −1 h if g −1 h ∈ S. Proposition 6.3.1. Let G be a ﬁnitely generated group. Let S ⊂ G be a ﬁnite symmetric generating subset. Then, the Sdistance of two group elements equals the graph distance of the same elements viewed as vertices in the associated Cayley graph CS (G). In other words, dS (g, h) = dCS (G) (g, h)
(6.8)
for all g, h ∈ G. Proof. Let g, h ∈ G and suppose that π = (e1 , e2 , . . . , en ) is a geodesic path connecting g and h. Let λ(π) = (s1 , s2 , . . . , sn ) be the label of π. It then follows that dS (g, h) = S (g −1 h) = S (s1 s2 · · · sn ) ≤ n = (π) = dCS (G) (g, h).
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Conversely, suppose that dS (g, h) = m. Then we can ﬁnd s1 , s2 , . . . , sm ∈ S such that g −1 h = s1 s2 · · · sm . Then the unique path π = (e1 , e2 , . . . , em ) with (π )− = g and label λ(π ) = s1 s2 · · · sm clearly satisﬁes (π )+ = h, that is, it connects g to h. We deduce that dCS (G) (g, h) ≤ (π ) = m = dS (g, h). Then (6.8) follows. Note that, in particular, two distinct elements g and h in G are neighbors in the graph CS (G) if and only if dS (g, h) = 1. For all g ∈ G, the map s → gs is a bijection from S onto the set gS of all neighbors of g in CS (G). In particular, all vertices g in CS (G) have the same degree δ(g) = S. Summarizing, we have that a Cayley graph is a connected, edgesymmetric and regular labeled graph. Examples 6.3.2. We graphically represent Cayley graphs by connecting two neighboring vertices by a single directed labeled arc e (see Fig. 6.1). Thus one should think of the inverse edge e−1 as the oppositely directed arc with label λ(e)−1 . (a) Let G = Z and take S = {1, −1} as a ﬁnite symmetric generating subset of G. Then the Cayley graph CS (Z) is represented in Fig. 6.2.
Fig. 6.2 The Cayley graph of G = Z for S = {1, −1}
(b) Let G = Z and take S = {1, 0, −1} as a ﬁnite symmetric generating subset of G. Then the Cayley graph CS (Z) is represented in Fig. 6.3. Note that as 0 = 1Z ∈ S, we have a loop at each vertex in CS (Z).
Fig. 6.3 The Cayley graph of G = Z for S = {1, 0, −1}
(c) Let G = Z and S = {2, −2, 3, −3}. Then, the corresponding Cayley graph CS (Z) is represented in Fig. 6.4. (d) Let G = Z×(Z/2Z), where Z/2Z = {0, 1} and take S = {(1, 0), (−1, 0), (0, 1)}. Then the Cayley graph CS (G) is represented by the biinﬁnite ladder as in Fig. 6.5. (e) Let G = D∞ be the inﬁnite dihedral group, that is, the group of isometries of the real line R generated by the reﬂections r : R → R and
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Fig. 6.4 The Cayley graph of G = Z for S = {2, −2, 3, −3}
Fig. 6.5 The Cayley graph of G = Z × (Z/2Z) for S = {(1, 0), (−1, 0), (0, 1)}
s : R → R deﬁned by r(x) = −x s(x) = 1 − x
(symmetry with respect to 0) (symmetry with respect to 1/2)
for all x ∈ R. Note that r2 = s2 = 1G . Taking S = {r, s}, the Cayley graph CS (G) is as in Fig. 6.6.
Fig. 6.6 The Cayley graph of G = D∞ for S = {r, s}
(f) Let G be the inﬁnite dihedral group, as in the previous example. Denote by t : R → R the map deﬁned by t(x) = x + 1 for all x ∈ R (translation). Then we have t = sr and hence s = tr. It follows that the set S = {r, t, t−1 } is also a symmetric generating subset of G and the corresponding Cayley graph CS (G) is as in Fig. 6.7. (g) Let G = Z2 and take S = {(1, 0), (−1, 0), (0, 1), (0, −1)} as a ﬁnite and symmetric generating subset of G. Then, the corresponding Cayley graph CS (Z2 ) is given in Fig. 6.8. (h) Let G = Z2 and take
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159
Fig. 6.7 The Cayley graph of G = D∞ for S = {r, t, t−1 }
Fig. 6.8 The Cayley graph of G = Z2 for S = {(1, 0), (−1, 0), (0, 1), (0, −1)}
S = {(1, 0), (−1, 0), (0, 1), (0, −1), (1, 1), (−1, 1), (1, −1), (−1, −1)}. Then, the corresponding Cayley graph CS (Z2 ) is as in Fig. 6.9. (i) Let G = FX be the free group based on a nonempty ﬁnite set X and take S = X ∪X −1 . Then, the Cayley graph CS (FX ) is a regular tree of degree / S, the Cayley graph CS (FX ) S = 2X (see Fig. 6.10). Indeed, since 1FX ∈ does not have loops. Moreover, if π is a nontrivial proper path in CS (FX ) with label λ(π) = (s1 , s2 , . . . , sn ) ∈ S ∗ , then, the word w = s1 s2 · · · sn ∈ S ∗ is nonempty and, by deﬁnition of properness, it is reduced. Therefore 1FX and h = s1 s2 · · · sn ∈ FX are distinct and we have π + = π − h = π − , that is, π is not closed. This shows that CS (FX ) is a tree.
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Fig. 6.9 The Cayley graph of G = Z2 for S = {(1, 0), (−1, 0), (0, 1), (0, −1), (1, 1), (−1, 1), (1, −1), (−1, −1)}
6.4 Growth Functions and Growth Types Let G be a ﬁnitely generated group and S ⊂ G a ﬁnite and symmetric generating subset of G. The growth function of G relative to S is the function γSG : N → N deﬁned by γSG (n) = BSG (n) = {g ∈ G : S (g) ≤ n}
(6.9)
for all n ∈ N. When there is no ambiguity we omit the supscript “G” and simply denote it by γS . Note that γS (0) = BS (0) = {1G } = 1 and that γS (n) ≤ γS (n + 1) for all n ∈ N. Also, as the map (s1 , s2 , . . . , sn ) → s1 s2 · · · sn is a surjection from (S ∪ {1G })n to BS (n), one has γS (n) ≤ S ∪ {1G }n
(6.10)
for all n ∈ N. Proposition 6.4.1. Let G be a ﬁnitely generated group and let S and S be two ﬁnite symmetric generating subsets of G. Let c = max{S (s) : s ∈ S}.
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Fig. 6.10 The Cayley graph of G = F2 for S = {a, a−1 , b, b−1 }
Then the following holds. (i) S (g) ≤ cS (g) for all g ∈ G; (ii) dS (g, h) ≤ cdS (g, h) for all g, h ∈ G; (iii) BS (n) ⊂ BS (cn) for all n ∈ N; (iv) γS (n) ≤ γS (cn) for all n ∈ N. Proof. (i) Let g ∈ G. Suppose that S (g) = n. Then there exist s1 , s2 , . . . , sn ∈ S such that g = s1 s2 · · · sn . Using (6.5) we get S (g) = S (s1 s2 · · · sn ) ≤
n
S (si ) ≤ cn.
i=1
(ii) By applying (i) we have, for all g, h ∈ G, dS (g, h) = S (g −1 h) ≤ cS (g −1 h) = cdS (g, h).
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Finally, from (ii), we have that dS (g, 1G ) ≤ cn if dS (g, 1G ) ≤ n for all g ∈ G. This gives (iii). Thus γS (n) = BS (n) ≤ BS (cn) = γS (cn) for all n ∈ N.
Two metrics d and d on a set X are said to be Lipschitzequivalent if there exist constants c1 , c2 > 0 such that c1 d(x, y) ≤ d (x, y) ≤ c2 d(x, y) for all x, y ∈ X. Corollary 6.4.2. Let G be a ﬁnitely generated group and let S and S be two ﬁnite symmetric generating subsets of G. Then, the word metrics dS and dS are Lipschitzequivalent. A nondecreasing function γ : N → [0, +∞) is called a growth function. Let γ, γ : N → [0, +∞) be two growth functions. One says that γ dominates γ, and one writes γ γ , if there exists an integer c ≥ 1 such that γ(n) ≤ cγ (cn) for all n ≥ 1. One says that γ and γ are equivalent and one writes γ ∼ γ if γ γ and γ γ. Proposition 6.4.3. We have the following: (i) is reﬂexive and transitive; (ii) ∼ is an equivalence relation; (iii) let γ1 , γ2 , γ1 , γ2 : N → [0, +∞) be growth functions. Suppose that γ1 ∼ γ1 , γ2 ∼ γ2 and that γ1 γ2 . Then γ1 γ2 . Proof. It is clear that is reﬂexive. Let γ1 , γ2 , γ3 : N → [0, +∞) be growth functions. Suppose that γ1 γ2 and that γ2 γ3 . Let c1 and c2 be positive integers such that γ1 (n) ≤ c1 γ2 (c1 n) and γ2 (n) ≤ c2 γ3 (c2 n) for all n ≥ 1. Then, taking c = c1 c2 one has γ1 (n) ≤ c1 γ2 (c1 n) ≤ c1 c2 γ3 (c2 c1 n) = cγ3 (cn) for all n ≥ 1. Thus is also transitive. This shows (i). Property (ii) immediately follows from (i) and the deﬁnition of ∼. Finally, suppose that γ1 , γ2 , γ1 , γ2 satisfy the hypotheses of (iii). Then we have in particular, γ1 γ1 , γ1 γ2 and γ2 γ2 . From (i) we deduce that γ1 γ2 . Let γ : N → [0, +∞) be a growth function. We denote by [γ] the ∼equivalence class of γ. By abuse of notation, we shall also write [γ] ∼ γ(n).
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If γ1 and γ2 are two growth functions, we write [γ1 ] [γ2 ] if γ1 γ2 . This deﬁnition makes sense by virtue of Proposition 6.4.3(iii). Note that, this way, becomes a partial ordering on the set of equivalence classes of growth functions. Examples 6.4.4. (a) Let α and β be nonnegative real numbers. Then nα nβ if and only if α ≤ β, and nα ∼ nβ if and only if α = β. (b) Let γ : N → [0, +∞) be a growth function. Suppose that γ is a polynomial of degree d for some d ≥ 0. Then one has γ(n) ∼ nd . (c) Let a, b ∈ (1, +∞). Then an ∼ bn .
(6.11)
Indeed, suppose for instance that a ≤ b. On the one hand we have an ≤ bn for all n ≥ 1, so that trivially an bn . On the other, setting c = [loga b] + 1 > 1 (here [ · ] denotes the integer part), one has bn = (aloga b )n = a(loga b)n ≤ acn ≤ cacn for all n ≥ 1, so that bn an . We deduce that an ∼ bn . In particular, we have an ∼ exp(n) for all a ∈ (1, +∞). (d) Let d ≥ 0 be an integer. Then nd exp(n) and nd ∼ exp(n). As a consequence if γ : N → [0, +∞) is a growth function such that γ(n) nd , nd then γ exp(n) and γ ∼ exp(n). Indeed, since limn→∞ exp(n) = 0, the d
n sequence ( exp(n) )n≥1 is bounded and we can ﬁnd an integer c ≥ 1 such that nd exp(n)
≤ c for all n ≥ 1. It follows that nd ≤ c exp(n) ≤ c exp(cn) for all n ≥ 1 and therefore nd exp(n). On the other hand, suppose by contradiction that exp(n) nd . Then we can ﬁnd an integer c > 0 such that exp(n) ≤ c(cn)d for all n ≥ 1. But then exp(n) ≤ cd+1 for all n ≥ 1 contradicting the fact that limn→∞ exp(n) = +∞. nd nd d Thus n ∼ exp(n). The remaining statements concerning the growth function γ immediately follow from transitivity of (cf. Proposition 6.4.3(i)) and symmetry of ∼ (cf. Proposition 6.4.3(ii)). From Proposition 6.4.1(iii) and (6.10) we deduce the following: Corollary 6.4.5. Let G be a ﬁnitely generated group and let S and S be two ﬁnite symmetric generating subsets of G. Then, the growth functions associated with S and S are equivalent, that is, γS ∼ γS . Moreover, γS (n) exp(n). Let G be a ﬁnitely generated group. The equivalence class [γS ] of the growth functions associated with the ﬁnite symmetric generating subsets S of G is called the growth type of G and we denote it by γ(G).
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Proposition 6.4.6. Let γ : N → [0, +∞) be a growth function with γ(0) > 0. Then γ ∼ 1 if and only if γ is bounded. Proof. Suppose ﬁrst that γ is bounded. Then we can ﬁnd an integer c ≥ 1 such that γ(n) ≤ c for all n ≥ 1. It follows that γ(n) ≤ c1(n) ≤ c1(cn) for 1 all n ≥ 1. Thus γ 1. On the other hand, setting c = [ γ(0) ] + 1, we have 1(n) = 1 ≤ cγ(0) ≤ cγ(n) ≤ cγ(cn) for all n ≥ 1 so that 1 γ. This shows that γ ∼ 1. Conversely, suppose that γ ∼ 1. Then γ 1 and we can ﬁnd an integer c ≥ 1 such that γ(n) ≤ c1(cn) = c for all n ≥ 1. This shows that γ is bounded. Corollary 6.4.7. Let G be a ﬁnitely generated group. Then γ(G) ∼ 1 if and only if G is ﬁnite. As a consequence, all ﬁnite groups have the same growth type. Proof. Let S be a ﬁnite and symmetric generating subset of G. Suppose that γ(G) ∼ γS (n) ∼ 1. Then, by Proposition 6.4.6 we deduce that γS is bounded, say by an integer c ≥ 1. This shows that G ≤ c and therefore G is ﬁnite. Conversely, if G is ﬁnite, we have γS (n) = BS (n) ≤ G for all n ≥ 1, that is, γS is bounded. From Proposition 6.4.6 we deduce that γ(G) ∼ γS (n) ∼ 1. Proposition 6.4.8. Let G be an inﬁnite ﬁnitely generated group. Then n γ(G). Proof. Let S be a ﬁnite symmetric generating subset of G. Consider the inclusions {1G } = BS (0) ⊂ BS (1) ⊂ BS (2) ⊂ · · · ⊂ BS (n) ⊂ BS (n + 1) ⊂ · · · (6.12) Let us show that if BS (n) = BS (n+1) for some n ∈ N , then BS (n) = BS (m) for all m ≥ n. We proceed by induction on m. Suppose that BS (n) = BS (m) for some m ≥ n + 1. For all g ∈ BS (m + 1) there exist g ∈ BS (m) and s ∈ S such that g = g s. By the inductive hypothesis, g ∈ BS (m − 1) so that g = g s ∈ BS (m − 1)S ⊂ BS (m). Since BS (m) ⊂ BS (m + 1), it follows that BS (m + 1) = BS (m) = BS (n). As a consequence, if BS (n) = BS (n + 1) for some n ∈ N, then we have G = BS (n). Since, by our assumptions, G is inﬁnite, we deduce that all the inclusions in (6.12) are strict. It follows that for all n ∈ N we have n ≤ BS (n) = γS (n). This shows that n γS (n) and therefore n γ(G). Deﬁnition 6.4.9. Let G be a ﬁnitely generated group. One says that G has exponential (resp. subexponential ) growth if γ(G) ∼ exp(n) (resp. γ(G) ∼ exp(n)). One says that G has polynomial growth if there exists an integer d ≥ 0 such that γ(G) nd .
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165
Proposition 6.4.10. Every ﬁnitely generated group of polynomial growth has subexponential growth. Proof. Let G be a ﬁnitely generated group. It follows from Example 6.4.4(d) that if γ(G) nd for some integer d ≥ 0, then γ(G) ∼ exp(n). Examples 6.4.11. (a) Let G = Z. With S = {1, −1} one has that the ball of radius r centered at the element g ∈ G is the interval [g − r, g + r] = {n ∈ Z : g − r ≤ n ≤ g + r}, see Fig. 6.11. We have γS (n) = 2n + 1. It follows that γ(Z) ∼ γS (n) ∼ n. In particular, Z has polynomial growth.
Fig. 6.11 The ball BS (n, r) ⊂ Z with S = {1, −1}; here r = 2
(b) Let G = Z2 and S = {(1, 0), (−1, 0), (0, 1), (0, −1)}. Then the ball of radius r centered at the element g = (n, m) ∈ G is the diagonal square with vertices (n + r, m), (n − r, m), (n, m − r), (n, m + r), see Fig. 6.12. In the 2 language of cellular automata, the ball BSZ (g, 1) is commonly called the von n Neumann neighborhood of g. We have γS (n) = 1 + k=1 4k = 2n2 + 2n + 1. 2 2 2 It follows that γ(Z ) ∼ γS (n) ∼ n . In particular, Z has polynomial growth.
Fig. 6.12 The ball BS (n, r) ⊂ Z2 with S = {(1, 0), (−1, 0), (0, 1), (0, −1)} and r = 2
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6 Finitely Generated Amenable Groups
(c) Let G = Z2 and S = {(1, 0), (−1, 0), (0, 1), (0, −1), (1, 1), (−1, 1), (1, −1), (−1, −1)}. Then the ball of radius r centered at the element g = (n, m) ∈ G is the square [n−r, n+r]×[m−r, m+r], see Fig. 6.13. In the language of cellular automata, 2 the ball BSZ (g, 1) is commonly called the Moore neighborhood of g. We have 2 γSZ (n) = 4n2 + 4n + 1 = (2n + 1)2 = (γSZ (n))2 . Note that this yields again γ(Z2 ) ∼ γS (n) ∼ n2 and the polynomial growth of Z2 (cf. Example 6.4.11(b) above).
Fig. 6.13 The ball BS (n, r) ⊂ Z2 with S = {±(1, 0), ±(0, 1), ±(1, 1), ±(1, −1)} and r = 2
¯ (−1, 0), ¯ (1, ¯1), (−1, ¯1)}. Then the (d) Let G = Z × (Z/2Z) and S = {(1, 0), ball of radius r centered at the element (n, ¯ 0) is represented in Fig. 6.14. We deduce that γS (n) = (2n + 1) + 2(n − 1) + 1 = 4n. Thus γ(Z × (Z/2Z)) ∼ γS (n) ∼ n. In particular, Z × (Z/2Z) has polynomial growth. (e) Let G = D∞ be the inﬁnite dihedral group and S = {r, s}. Then the ball of radius n centered at the element g ∈ G is represented in Fig. 6.15. It follows that γS (n) = 2n + 1. Thus γ(D∞ ) ∼ γS (n) ∼ n. In particular, D∞ has polynomial growth.
6.4 Growth Functions and Growth Types
167
Fig. 6.14 The ball BS ((n, 0), r) ⊂ Z × (Z/2Z) with S = {(1, 0), (−1, 0), (0, 1)}; here r = 2
Fig. 6.15 The ball BS (g, n) ⊂ D∞ with S = {r, s}; here n = 3
(f) Let G = D∞ be the inﬁnite dihedral group and S = {r, t, t−1 }. Then the ball of radius r centered at the element g ∈ G is represented in Fig. 6.16. It follows that γS (n) = (2n + 1) + 2(n − 1) + 1 = 4n. Note that this yields again γ(D∞ ) ∼ γS (n) ∼ n and the polynomial growth of D∞ .
Fig. 6.16 The ball BS (g, n) ⊂ D∞ with S = {r, t, t−1 }; here n = 2
(g) Let G = Fk be the free group of rank k ≥ 2. Let {a1 , a2 , . . . , ak } be a −1 −1 free basis and set S = {a1 , a−1 1 , a2 , a2 , . . . , ak , ak }. Then the ball of radius r centered at the element g ∈ G is the ﬁnite tree rooted at g of depth r (see Fig. 6.17). We have γSFk (n) = 1 + 2k
n−1
(2k − 1)j =
j=0
k(2k − 1)n − 1 . k−1
It follows that γ(Fk ) ∼ γS (n) ∼ (2k − 1)n ∼ exp(n). In particular, Fk has exponential growth.
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6 Finitely Generated Amenable Groups
Fig. 6.17 The ball BS (a, 2) ⊂ F2 with S = {a, a−1 , b, b−1 }
6.5 The Growth Rate Lemma 6.5.1. Let (an )n≥1 be a sequence of positive real numbers such that an+m ≤ an am for all n, m ≥ 1. Then the limit lim
n→∞
exists and equals inf n≥1
√ n
an
√ n a . n
Proof. Fix an integer t ≥ 1 and, for all n ≥ 1 write n = qt+r, with 0 ≤ r < t. √ q/n √ Then an ≤ aqt ar ≤ aqt ar and n an ≤ at n ar . As 0 ≤ r/n < t/n and limn→∞ t/n = 0, we have that limn→∞ r/n = 0 and limn→∞ q/n = 1/t. In √ √ √ q/n 1/n 1/t particular limn→∞ at ar = at = t at . Thus, lim supn→∞ n an ≤ t at . √ √ √ This gives lim supn→∞ n an ≤ inf t≥1 t at ≤ lim inf t→∞ t at completing the proof.
6.5 The Growth Rate
169
Proposition 6.5.2. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. Then, the limit (6.13) λS = lim n γS (n) n→∞
exists and λS ∈ [1, +∞). Proof. From (6.5) we deduce that BS (n + m) ⊆ BS (n)BS (m) and therefore γS (n + m) = BS (n + m) ≤ BS (n)BS (m) ≤ BS (n) · BS (m) = γS (n)γS (m). Thus, the sequence {γS (n)}n≥1 satisﬁes the hypotheses of the previous lemma and λS exists and is ﬁnite. As γS (n) ≥ 1 for all n ∈ N we also have λS ≥ 1. Deﬁnition 6.5.3. The number λS = λG S in (6.13) is called the growth rate of G with respect to S. Proposition 6.5.4. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. Then λS > 1 if and only if G has exponential growth. Proof. Suppose that γ(G) ∼ exp(n). We then have exp(n) γS so that there exists an integer c ≥ 1 such that en ≤ cγS (cn) for all n ≥ 1. This implies √ √ √ 1 < c e = lim cn en ≤ lim cn cγS (cn) = lim cn c · lim cn γS (cn) = λS . n→∞
n→∞
n→∞
n→∞
Conversely, suppose that λS > 1. By Lemma 6.5.1 we have that γS (n) ≥ λnS .
n
γS (n) ≥ λS so
This shows that exp(n) ∼ λnS γS (n). By Corollary 6.4.5 we have γS (n) exp(n) and therefore γ(G) ∼ γS ∼ exp(n). Corollary 6.5.5. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. Then λS = 1 if and only if G has subexponential growth. As an immediate consequence of Proposition 6.5.4 and Proposition 6.4.3(ii) we deduce the following. Corollary 6.5.6. Let G be a ﬁnitely generated group and let S and S be two ﬁnite symmetric generating subsets of G. Then λS = 1 (resp. λS > 1) if and only if λS = 1 (resp. λS > 1).
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6 Finitely Generated Amenable Groups
6.6 Growth of Subgroups and Quotients Proposition 6.6.1. Let G be a group and let N be a normal subgroup of G. Suppose that N and the quotient group G/N are both ﬁnitely generated. Then G is ﬁnitely generated. Proof. Denote by π : G → G/N the quotient homomorphism. Let U ⊂ N and T ⊂ G/N be two ﬁnite symmetric generating subsets of N and G/N respectively. Let S ⊂ G be a ﬁnite symmetric set such that π(S) ⊃ T and S ⊃ U . Let us prove that S generates G. Let g ∈ G. Then there exist h ≥ 0 and t1 , t2 , . . . , th ∈ T such that π(g) = t1 t2 · · · th . Let s1 , s2 , . . . , sh ∈ S be such that π(si ) = ti for i = 1, 2, . . . , h. Setting g = s1 s2 · · · sh we then have π(g ) = π(g). It follows that n = (g )−1 g ∈ ker(π) = N . Therefore, there exist k ≥ 0 and sh+1 , sh+2 , . . . , sh+k ∈ S such that n = sh+1 sh+2 · · · sh+k . It follows that g = g n = s1 s2 · · · sh sh+1 , sh+2 · · · sh+k . This shows that S generates G. Proposition 6.6.2. Let G be a group and let H be a subgroup of ﬁnite index in G. Then, G is ﬁnitely generated if and only if H is ﬁnitely generated. Proof. Let R ⊂ G be a complete set of representatives of the right cosets of H in G such that 1G ∈ R. Note that R = [G : H] < ∞. Suppose ﬁrst that H is ﬁnitely generated. Let S ⊂ H be a ﬁnite symmetric generating subset of H. Given g ∈ G, there exist r ∈ R and h ∈ H such that g = hr. Let s1 , s2 , . . . , sn ∈ S be such that h = s1 s2 · · · sn . Then g = s1 s2 · · · sn r. This shows that the set S ∪ R is a ﬁnite generating subset of G. Suppose now that G is ﬁnitely generated and let S be a ﬁnite symmetric generating subset of G. Let us show that the ﬁnite set S = RSR−1 ∩ H
(6.14)
generates H. Let h ∈ H and write h = s1 s2 · · · sn where si ∈ S. Then, there exists h1 ∈ H and r1 ∈ R such that s1 = h1 r1 . Note that h1 = 1G s1 r1−1 ∈ S . By induction, for all i = 2, 3, . . . , n−1 there exist hi ∈ H and ri ∈ R such that ri−1 si = hi ri . Moreover, hi = ri−1 si ri−1 ∈ S . Thus, setting hn = rn−1 sn we have h = s1 s2 · · · sn −1 = (1G s1 r1−1 )(r1 s2 r2−1 ) · · · (rn−2 sn−1 rn−1 )(rn−1 sn )
= h1 h2 · · · hn−1 hn . −1 −1 Note that hn = h−1 n−1 · · · h2 h1 h ∈ H and, on the other hand, hn = −1 rn−1 sn = rn−1 sn 1G ∈ RSR . Thus also hn ∈ S . This shows that S generates H.
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171
Proposition 6.6.3. Let G be a ﬁnitely generated group and let H be a ﬁnitely generated subgroup of G. Then γ(H) γ(G). Proof. Let SG (resp. SH ) be a ﬁnite symmetric generating subset of G (resp. H). Then the set S = SH ∪ SG is a ﬁnite symmetric generating subset of G. As SH ⊂ S we have BSHH (n) ⊂ BSG (n) and therefore γSHH (n) ≤ γSG (n) for all n ∈ N. Thus, γ(H) γ(G). Corollary 6.6.4. Every ﬁnitely generated group which contains a ﬁnitely generated subgroup of exponential growth has exponential growth. From Example 6.4.11(d) and Corollary 6.6.4 we deduce the following. Corollary 6.6.5. Every ﬁnitely generated group which contains a subgroup isomorphic to the free group F2 has exponential growth. In the following proposition we show that ﬁnitely generated groups and their ﬁnite index subgroups (which are ﬁnitely generated as well, by Proposition 6.6.2) have the same growth type. Proposition 6.6.6. Let G be a ﬁnitely generated group and let H be a ﬁnite index subgroup of G. Then H is ﬁnitely generated and γ(H) = γ(G). Proof. Since [G : H] < ∞, we deduce from Proposition 6.6.2 that H is ﬁnitely generated. It follows from Proposition 6.6.3 that γ(H) γ(G). Let now S be a ﬁnite symmetric generating subset of G and consider the ﬁnite symmetric set S = RSR−1 ∩ H. It follows from the proof of Proposition 6.6.2 that S generates H. Let g ∈ BSG (n) and write g = s1 s2 · · · sn , where s1 , s2 , . . . , sn ∈ S. As in the proof of Proposition 6.6.2 we may ﬁnd r0 = 1G , r1 , . . . , rn ∈ R such that g = s1 s2 · · · sn −1 = (1G s1 r1−1 )(r1 s2 r2−1 ) · · · (rn−2 sn−1 rn−1 )(rn−1 sn rn−1 )rn
= h1 h2 · · · hn−1 hn rn where hi = ri−1 si ri−1 ∈ S . It follows that BSG (n) ⊂ BSH (n)R so that, taking cardinalities, γSG (n) = BSG (n) ≤ BSH (n)R = [G : H]γSH (n) ≤ [G : H]γSH ([G : H]n). This shows that γ(G) γ(H). It follows that γ(G) = γ(H).
Two groups G1 and G2 are called commensurable if there exist ﬁnite index subgroups H1 ⊂ G1 and H2 ⊂ G2 such that H1 and H2 are isomorphic. From the previous proposition we immediately deduce: Corollary 6.6.7. If G1 and G2 are commensurable groups and G2 is ﬁnitely generated, then G1 is ﬁnitely generated and one has γ(G1 ) = γ(G2 ).
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6 Finitely Generated Amenable Groups
Proposition 6.6.8. Let G be a ﬁnitely generated group and let N be a normal subgroup of G. Then the quotient group G/N is ﬁnitely generated and one has γ(G/N ) γ(G). If in addition N is ﬁnite, then γ(G/N ) = γ(G). Proof. Let S be a ﬁnite symmetric generating subset of G and let π : G → G/N denote the canonical quotient homomorphism. Then S = π(S) is a ﬁnite symmetric generating subset of G/N . Thus, for all n ∈ N one has G/N
BS
(n) = π(BSG (n))
(6.15)
and therefore G/N
γS
G/N
(n) = BS
(n) ≤ BSG (n) = γSG (n).
This shows that γ(G/N ) γ(G). Suppose now that N is ﬁnite. From (6.15) we deduce that BSG (n) ⊂ G/N −1 π (BS (n)) and therefore G/N
γSG (n) = BSG (n) ≤ N BS
G/N
(n) = N γS
G/N
(n) ≤ N γS
(N n).
This shows that γ(G) γ(G/N ). It follows that γ(G) = γ(G/N ).
Lemma 6.6.9. Let γ1 , γ2 , γ1 , γ2 : N → [0, +∞) be growth functions. Suppose that γ1 γ1 , γ2 γ2 . Then the products γ1 γ2 , γ1 γ2 : N → [0, ∞) are also growth functions and one has γ1 γ2 γ1 γ2 . Proof. Since the product of nondecreasing functions is also nondecreasing, it is clear that γ1 γ2 and γ1 γ2 are also growth functions. Let c1 and c2 be positive integers such that γ1 (n) ≤ c1 γ1 (c1 n) and γ2 (n) ≤ c2 γ2 (c2 n) for all n ≥ 1. Taking c = c1 c2 we have (γ1 γ2 )(n) = γ1 (n)γ2 (n) ≤ c1 c2 γ1 (c1 n)γ2 (c2 n) ≤ c1 c2 γ1 (c1 c2 n)γ2 (c1 c2 n) = c(γ1 γ2 )(cn) for all n ≥ 1. This shows γ1 γ2 γ1 γ2 .
Given two growth functions γ1 and γ2 we set [γ1 ] · [γ2 ] = [γ1 γ2 ]. This deﬁnition makes sense since if γ1 ∼ γ1 and γ2 ∼ γ2 then γ1 γ2 ∼ γ1 γ2 as it immediately follows from Lemma 6.6.9. Proposition 6.6.10. Let G1 and G2 be two ﬁnitely generated groups. Then the direct product G1 × G2 is also ﬁnitely generated and γ(G1 × G2 ) = γ(G1 )γ(G2 ). Proof. Let S1 and S2 be ﬁnite symmetric generating subsets of G1 and G2 . Then the set
6.7 A Finitely Generated Metabelian Group with Exponential Growth
173
S = (S1 × {1G2 }) ∪ ({1G1 } × S2 ) is a ﬁnite symmetric generating subset of G1 × G2 . Let (g1 , g2 ) ∈ BSG1 ×G2 (n). Then there exist s1,1 , s2,1 , . . . , sk,1 ∈ S1 and s1,2 , s2,2 , . . . , sh,2 ∈ S2 , where h + k ≤ n, such that (g1 , g2 ) = (s1,1 , 1G2 )(s2,1 , 1G2 ) · · · (sk,1 , 1G2 ) · (1G1 , s1,2 )(1G1 , s2,2 ) · · · (1G1 , sh,2 ) = (s1,1 s2,1 · · · sk,1 , s1,2 s2,2 · · · sh,s ). Thus, BSG1 ×G2 (n) ⊂ BSG11 (n) × BSG22 (n) and γSG1 ×G2 (n) ≤ γSG11 (n)γSG22 (n). This shows that γ(G1 × G2 ) γ(G1 )γ(G2 ). On the other hand, if g1 ∈ BSG11 (n) and g2 ∈ BSG22 (n), then (g1 , g2 ) ∈ BSG1 ×G2 (2n) and one has γSG11 (n)γSG22 (n) ≤ γSG1 ×G2 (2n) ≤ 2γSG1 ×G2 (2n). This shows that γ(G1 )γ(G2 ) γ(G1 × G2 ). It follows that γ(G1 × G2 ) = γ(G1 )γ(G2 ). From Example 6.4.11(a) and the previous proposition one immediately deduces the following. Corollary 6.6.11. Let d be a positive integer. Then γ(Zd ) ∼ nd .
Corollary 6.6.12. Every ﬁnitely generated abelian group has polynomial growth. Proof. Let G be a ﬁnitely generated abelian group. Then there exist an integer d ≥ 0 and a ﬁnite group F such that G is isomorphic to the cartesian product Zd × F . It follows from Proposition 6.6.10, Corollary 6.6.11 and Corollary 6.4.7 that γ(G) = γ(Zd )γ(F ) ∼ nd . This shows that G has polynomial growth.
6.7 A Finitely Generated Metabelian Group with Exponential Growth We have seen in Corollary 6.6.12 that every ﬁnitely generated abelian group has polynomial growth. The purpose of the present section is to give an example of a ﬁnitely generated metabelian group with exponential growth. As every metabelian group is solvable and hence amenable (Theorem 4.6.3), this will show in particular that there exist ﬁnitely generated amenable groups, and even ﬁnitely generated solvable groups, whose growth is exponential. Proposition 6.7.1. Let G denote the subgroup of GL2 (Q) generated by the two matrices 20 11 A= and B = . 01 01 Then G is a ﬁnitely generated metabelian group with exponential growth.
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6 Finitely Generated Amenable Groups
Proof. The group G is ﬁnitely generated by deﬁnition. We have G=
k 2 r : k ∈ Z, r ∈ Z[1/2] , 0 1
(6.16)
where Z[1/2] is the subring of Q consisting of all dyadic rationals. Indeed, if we denote by H the righthand side of (6.16), we ﬁrst observe that G ⊂ H since H is clearly a subgroup of GL2 (Q) containing A and B. On the other hand, we also have the inclusion H ⊂ G since any r ∈ Z[1/2] can be written in the form r = 2m n for some m, n ∈ Z, so that
2k r 0 1
= Am B n Ak−m ∈ G.
From (6.16), we deduce that the determinant map yields a surjective homomorphism from G onto an inﬁnite cyclic group whose kernel is isomorphic to the additive group Z[1/2] and is therefore abelian. Since any element in [G, G] has determinant 1, this shows that [G, G] is abelian, that is, that G is metabelian. It remains to show that G has exponential growth. To see this, let us estimate from below the cardinality of the ball BS (3n − 2), where S is the ﬁnite symmetric generating subset of G deﬁned by S = {A, B, A−1 , B −1 } and n ≥ 1 is a ﬁxed integer. Consider the subset E ⊂ G consisting of all matrices of the form 1q M (q) = , 01 where q is an integer such that 0 ≤ q ≤ 2n − 1. Developing q in base 2, we get n q= ui 2i−1 , i=1
where ui ∈ {0, 1} for 1 ≤ i ≤ n. This gives us M (q) = As we have
u u u n−1 un 1 1 1 1 2 2 1 22 3 12 ··· . 01 01 0 1 0 1
1 2i−1 0 1
= Ai−1 BA−(i−1)
for all 1 ≤ i ≤ n, it follows that we can write M (q) in the form M (q) = B u1 AB u2 A−1 A2 B u3 A−2 · · · An−2 B un−1 A−(n−2) An−1 B un A−(n−1) = B u1 AB u2 AB u3 A · · · AB un−1 AB un A−(n−1) .
6.8 Growth of Finitely Generated Nilpotent Groups
175
This shows that S (M (q)) ≤ (2n − 1) + (n − 1) = 3n − 2. We deduce that BS (3n − 2) ≥ E = 2n . It follows that the growth rate of G with respect to S satisﬁes √ √ 3n−2 3 2n = 2 > 1. λS = lim 3n−2 BS (3n − 2) ≥ lim n→∞
n→∞
Thus G has exponential growth.
6.8 Growth of Finitely Generated Nilpotent Groups In this section we prove the following Theorem 6.8.1. Every ﬁnitely generated nilpotent group has polynomial growth. In order to prove this result we need some preliminaries. Lemma 6.8.2. Let G be any group. Let H and K be two normal subgroups of G. Suppose that S ⊆ H and T ⊆ K generate H and K respectively. Then [H, K] is equal to the normal closure in G of the set {[s, t] : s ∈ S, t ∈ T }. Proof. Denote by N the normal closure in G of the set {[s, t] : s ∈ S, t ∈ T } and let us show that N = [H, K]. Since {[s, t] : s ∈ S, t ∈ T } ⊂ [H, K] and [H, K] is a normal subgroup, we have N ⊂ [H, K].
(6.17)
Consider the quotient homomorphism π : G → G/N . For all s ∈ S and t ∈ T we have [π(s), π(t)] = π([s, t]) = 1G/N , that is, π(s) and π(t) commute. It follows that all elements of π(H) commute with all elements of π(K), since S generates H and T generates K. In other words, π([h, k]) = [π(h), π(k)] = 1G/N , for all h ∈ H and k ∈ K. This gives [h, k] ∈ N for all h ∈ H and k ∈ K and therefore [H, K] ⊂ N . From (6.17) we deduce that [H, K] = N . Let G be a group. Recall that the lower central series of G is the sequence (C i (G))i≥0 of normal subgroups of G deﬁned by C 0 (G) = G and C i+1 (G) = [C i (G), G] for all i ≥ 0. Given elements g1 , g2 , . . . , gi ∈ G, i ≥ 3, we inductively set [g1 , g2 , . . . , gi ] = [[g1 , g2 , . . . , gi−1 ], gi ] ∈ C i−1 (G). (i)
If S ⊂ G and i ≥ 2 we then denote by SG the set consisting of all elements of the form
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6 Finitely Generated Amenable Groups
[s1 , s2 , . . . , si ], s1 , s2 , . . . , si ∈ S.
(6.18)
The elements (6.18) are called simple Scommutators of weight i. Note that (i) SG ⊂ C i−1 (G). Lemma 6.8.3. Let G be any group. Let SG ⊂ G be a generating subset of G. Then for all i ≥ 1, the subgroup C i (G) is the normal closure in G of the set (i+1) SG . Proof. Let us prove the statement by induction on i. We have that C 1 (G) = (2) [G, G] is the normal closure in G of SG = {[s1 , s2 ] : s1 , s2 ∈ SG }, as it follows from Lemma 6.8.2 by taking H = K = G and S = T = SG . Thus, the statement holds for i = 1. Suppose by induction that C i−1 (G) is the normal (i) (i+1) closure in G of SG , denote by N the normal closure in G of SG , and (i+1) let us show that C i (G) = N . Since SG ⊂ C i (G) and C i (G) is a normal subgroup, we deduce that N ⊂ C i (G). (6.19) (i)
Denote by π : G → G/N the quotient map. Let w ∈ SG and s ∈ SG . (i+1) Then, by deﬁnition, [w, s] ∈ SG and therefore [w, s] ∈ N . It follows that [π(w), π(s)] = π([w, s]) = 1G/N , that is, π(w) and π(s) commute. As SG generates G, we have that π(SG ) generates G/N and therefore π(w) ∈ Z(G/N ). It follows that π(hwh−1 ) = π(h)π(w)π(h)−1 = π(w) for all h ∈ G, so that π([hwh−1 , s]) = [π(hwh−1 ), π(s)] = [π(w), π(s)] = 1G/N . It follows that [hwh−1 , s] ∈ N.
(6.20)
By the inductive hypothesis, every element in C i−1 (G) can be expressed (i) −1 −1 as a product (h1 w1 h−1 1 )(h2 w2 h2 ) · · · (hm wm hm ) with wk ∈ SG and hk ∈ G, k = 1, 2, . . . , m, for some m ∈ N. By taking H = C i−1 (G), K = G, (i) S = {wh : w ∈ SG , h ∈ G} and T = SG in Lemma 6.8.2, we deduce that i i−1 C (G) = [C (G), G] is the normal closure in G of the set {[hwh−1 , s] : (i) w ∈ SG , h ∈ G, s ∈ SG }. Thus from (6.20) it follows that C i (G) ⊂ N . By using (6.19) this shows that C i (G) = N . Lemma 6.8.4. Let G be a ﬁnitely generated nilpotent group of nilpotency degree d ≥ 1. Then the subgroups C i (G), i = 1, 2, . . . , d − 1 are ﬁnitely generated. Proof. Let SG ⊂ G be a ﬁnite generating subset of G. We ﬁrst show that the quotient groups C i (G)/C i+1 (G) are ﬁnitely generated, i = 0, 1, . . . , d − 1. From Lemma 6.8.3 we have that C i (G) is the normal closure in G of the (i+1) ﬁnite set SG . Therefore, if π : G → G/C i+1 (G) denotes the quotient homomorphism, we have that C i (G)/C i+1 (G) = π(C i (G)) is the normal clo(i+1) (i+1) sure in G/C i+1 (G) of the set π(SG ). But π(SG ) = (π(SG ))(i+1) ⊂ (i+1) Z(C i (G)/C i+1 (G)) and therefore π(SG ) generates C i (G)/C i+1 (G).
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177
Let us now prove the statement by reverse induction starting from i = d−1. It follows from the ﬁrst part of the proof that the subgroup C d−1 (G) ∼ = C d−1 (G)/{1G } = C d−1 (G)/C d (G) is ﬁnitely generated. Suppose by induction that the subgroup C i+1 (G) is ﬁnitely generated for some i ≤ d − 2. From the ﬁrst part of the proof we have that C i (G)/C i+1 (G) is also ﬁnitely generated. Therefore from Proposition 6.6.1 we deduce that C i (G) is ﬁnitely generated. We are now in position to prove the main result of this section. Proof of Theorem 6.8.1. Let G be a ﬁnitely generated nilpotent group of nilpotency degree d. Let us prove the statement by induction on d. If d = 0 then G = {1G } and therefore G has polynomial growth. Suppose now that d ≥ 1 and that all ﬁnitely generated nilpotent groups of nilpotency degree ≤ d − 1 have polynomial growth. We ﬁrst observe that the subgroup H = C 1 (G) is nilpotent of nilpotency degree ≤ d − 1. Indeed, as one immediately checks by induction, we have C i (H) ⊂ C i+1 (G) for all i = 0, 1, . . . , d − 1, so that C d−1 (H) ⊂ C d (G) = {1G }. Moreover, by Lemma 6.8.4, we have that H is ﬁnitely generated. Thus, by the inductive hypothesis we have that H has polynomial growth. Let T ⊂ H be a ﬁnite symmetric generating subset of H. Then there exist integers c1 > 0 and q ≥ 0 such that γTH (n) ≤ c1 (c1 n)q
(6.21)
for all n ≥ 1. Let S = {s1 , s2 , . . . , sk } ⊂ G be a ﬁnite symmetric generating subset of G. Let g ∈ G and suppose that m = G S (g) ≤ n. Then there exist 1 ≤ i1 , i2 , . . . , im ≤ k such that g = si1 si2 · · · sim .
(6.22)
Since g2 g1 = g1 g2 [g2−1 , g1−1 ] and [g2−1 , g1−1 ] ∈ H for all g1 , g2 ∈ G, we can permute the generators in (6.22) and express g in the form g = sj1 sj2 · · · sjm h
(6.23)
where 1 ≤ j1 ≤ j2 ≤ · · · ≤ jm ≤ k and h ∈ H. Let us estimate H T (h). We set (i) L = max{H T (w) : w ∈ S , 2 ≤ i ≤ d}.
(6.24)
As we observed above, exchanging a pair of consecutive generators produces a simple Scommutator of weight two on their right. It is clear that one needs at most n such exchanges to bring sj1 , where j1 = ip1 = min{ip : p = 1, 2, . . . , m}, to the leftmost place. Analogously, one needs at most n such exchanges to bring sj2 , where j2 = min{ip : p = 1, 2, . . . , m; p = p1 }, to the leftmost but one place (in fact on the right of sj1 ). And so on. Altogether,
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there are at most mn ≤ n2 such exchanges. This produces at most n2 simple Scommutators of weight two. But at each step, when moving a generator sjt to the left, we also have to exchange it with all the simple Scommutators on its left that were produced before (namely by sj1 , sj2 , . . . , sjt−1 ). As one easily checks by induction, that this produces, altogether, at most n3 simple Scommutators of weight three, n4 simple Scommutators of weight four, and so on. Continuing this way, since G is nilpotent of nilpotency degree d, all simple Scommutators of weight d+1 are equal to 1G . Thus, the total number of simple Scommutators that are eventually produced in this process is at most n2 + n3 + · · · + nd ≤ dnd . It follows from (6.24) that d H T (h) ≤ Ldn .
(6.25)
On the other hand, we can bound the number of elements in G which are of the form sj1 sj2 · · · sjm , where 1 ≤ j1 ≤ j2 ≤ · · · ≤ jm ≤ k, by c2 nk , where c2 > 0 is a constant independent of n. Indeed, each such group element can be written in the form sn1 1 sn2 2 · · · snk k , where 0 ≤ ni ≤ n for all i = 1, 2, . . . , k. We deduce from (6.21) and (6.25) that γSG (n) ≤ c2 nk c1 (Ldnd )q = Cnδ ≤ C(Cn)δ for all n ≥ 1, where δ = k + qd and C = c1 c2 (Ld)q . Note that C > 0 is a constant independent of n. It follows that G has polynomial growth. From Theorem 6.8.1 and Proposition 6.6.6 we deduce the following: Corollary 6.8.5. Every ﬁnitely generated virtually nilpotent group has polynomial growth.
6.9 The Grigorchuk Group and Its Growth In this section we present the Grigorchuk group and some of its main properties, namely being inﬁnite, periodic, residually ﬁnite and of intermediate growth. Let Σ = {0, 1}. We denote by Σ ∗ = ∪n∈N Σ n the set of all words on the alphabet Σ. Recall that Σ ∗ is a monoid for the word concatenation whose identity element is the empty word (cf. Sect. D.1). Every word w ∈ Σ ∗ may be uniquely written in the form w = σ1 σ2 · · · σn , where σ1 , σ2 , . . . , σn ∈ Σ and n = (w) ∈ N is the length of w. Denote by Sym(Σ ∗ ) the symmetric group on Σ ∗ (cf. Appendix C). We introduce a partial order in Σ ∗ by setting u v, u, v ∈ Σ ∗ , if there exists w ∈ Σ ∗ such that uw = v. We then set Sym(Σ ∗ , ) = {g ∈ Sym(Σ ∗ ) : g(u) g(v) for all u, v ∈ Σ ∗ such that u v}. (6.26)
6.9 The Grigorchuk Group and Its Growth
179
Note that Sym(Σ ∗ , ) is a subgroup of Sym(Σ ∗ ). Moreover, for w ∈ Σ ∗ the length (w) equals the maximum of n ∈ N such that there exists a sequence (wk )0≤k≤n of distinct words in Σ ∗ such that = w0 w1 · · · wn = w. It follows that if g ∈ Sym(Σ ∗ , ) then (g(w)) = (w) for all w ∈ Σ ∗ . Consider the elements a, b, c, d ∈ Sym(Σ ∗ ) deﬁned as follows. We ﬁrst deﬁne a by setting a( ) = (6.27) and a(0w) = 1w,
a(1w) = 0w
(6.28)
for all w ∈ Σ ∗ . Then, for all w ∈ Σ ∗ , we deﬁne b(w), c(w), d(w) by induction on (w). We start by setting b( ) = c( ) = d( ) = .
(6.29)
b(0w) = 0a(w), b(1w) = 1c(w), c(0w) = 0a(w), c(1w) = 1d(w), d(0w) = 0w, d(1w) = 1b(w)
(6.30)
Then we set
for all w ∈ Σ ∗ . By an obvious induction we have a, b, c, d ∈ Sym(Σ ∗ , ).
(6.31)
Example 6.9.1. Let w = 10110 ∈ Σ ∗ . Then we have a(w) = a(10110) = 00110, b(w) = b(10110) = 1c(0110) = 10a(110) = 10010, c(w) = c(10110) = 1d(0110) = 10110, d(w) = d(10110) = 1b(0110) = 10a(110) = 10010. Deﬁnition 6.9.2. The Grigorchuk group is the subgroup G of Sym(Σ ∗ ) generated by the elements a, b, c, d. Observe that by (6.31) we have G ⊂ Sym(Σ ∗ , ).
(6.32)
Proposition 6.9.3. The following relations hold in G. a2 = b2 = c2 = d2 = 1G
(6.33)
bc = cb = d, dc = cd = b, db = bd = c.
(6.34)
and Proof. To prove (6.33), we have to show that g 2 (w) = w
(6.35)
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6 Finitely Generated Amenable Groups
for all w ∈ Σ ∗ and g ∈ {a, b, c, d}. We have a(ε) = ε by (6.27). Moreover from (6.28) we have a2 (0w) = a(1w) = 0w
and a2 (1w) = a(0w) = 1w,
(6.36)
for all w ∈ Σ ∗ . This shows that a2 = 1G . To prove (6.35) for g = b, c and d, we use induction on (w). By virtue of (6.29) this holds when (w) = 0. Suppose that (6.35) holds when (w) = n. From (6.30) and the fact that a2 = 1G we deduce, for all w ∈ Σ ∗ with (w) = n, b2 (0w) = b(0a(w)) = 0a2 (w) = 0w, c2 (0w) = c(0a(w)) = 0a2 (w) = 0w, d2 (0w) = d(0w) = 0w,
b2 (1w) = b(1c(w)) = 1c2 (w) = 1w, c2 (1w) = c(1d(w)) = 1d2 (w) = 1w, d2 (1w) = d(1b(w)) = 1b2 (w) = 1w.
This shows that (6.35) holds for all w ∈ Σ ∗ with (w) = n + 1. This proves (6.35) for g = b, c, d and (6.33) follows. To prove (6.34), we have to show that ij(w) = k(w)
(6.37)
for all w ∈ Σ ∗ and all distinct i, j, k ∈ {b, c, d}. Again, we can use induction on (w). If (w) = 0 then (6.37) follows from (6.29). Suppose by induction that (6.37) holds when (w) = n. From (6.28),(6.30) and (6.33) we deduce, for all w ∈ Σ ∗ with (w) = n, bc(0w) = b(0a(w)) = 0a2 (w) = 0w = d(0w), bc(1w) = b(1d(w)) = 1cd(w) = 1b(w) = d(1w), cd(0w) = c(0w) = 0a(w) = b(0w), cd(1w) = c(1b(w)) = 1db(w) = 1c(w) = b(1w), db(0w) = d(0a(w)) = 0a(w) = c(0w), db(1w) = d(1c(w)) = 1bc(w) = 1d(w) = c(1w). It follows that (6.37) holds for all w ∈ Σ ∗ with (w) = n + 1 whenever (i, j, k) = (b, c, d), (c, d, b), (d, b, c). Thus, by induction bc = d, cd = b and db = c. Finally, using (6.33) we deduce cb = cd2 b = (cd)(db) = bc, dc = db2 c = (db)(bc) = cd and bd = bc2 d = (bc)(cd) = db. This completes the proof of (6.34). It follows from (6.33) that the set S = {a, b, c, d} is a symmetric generating subset of G. We denote by S : G → N the corresponding wordlength function. Every group element g ∈ G can be expressed in the form g = s1 s2 · · · sn
(6.38)
6.9 The Grigorchuk Group and Its Growth
181
with s1 , s2 , . . . , sn ∈ S. We say that the expression (6.38) is a reduced form of g provided that for all i, j = 1, 2, . . . , n − 1 one has that if si = a then si+1 ∈ {b, c, d}, and if sj ∈ {b, c, d} then sj+1 = a. It immediately follows from (6.33) and (6.34) that every group element g ∈ G can be expressed (not necessarily in a unique way) as in (6.38) in reduced form. For all n ∈ N we set Hn = {g ∈ G : g(w) = w for all w ∈ Σ n }.
(6.39)
Proposition 6.9.4. For all n ∈ N, the set Hn is a normal subgroup of G and [G : Hn ] < ∞. (6.40) Moreover, G = H0 ⊃ H1 ⊃ H2 ⊃ · · · ⊃ Hn ⊃ Hn+1 ⊃ · · · .
(6.41)
Proof. Since g(Σ n ) ⊂ Σ n for all g ∈ G (cf. (6.32)) we may consider the (restriction) map θn : G → Sym(Σ n ) deﬁned by θn (g)(w) = g(w) for all g ∈ G and w ∈ Σ n . Clearly, θn is a homomorphism and ker(θn ) = Hn . This shows that Hn is a normal subgroup and that [G : Hn ] = G/Hn  = θn (G) ≤  Sym(Σ n ) < ∞. Finally, let n ∈ N, u ∈ Σ n and g ∈ Hn+1 . Let σ ∈ Σ and set w = uσ ∈ n+1 so that u w. From (6.32) we deduce that g(u) g(w) = uσ and Σ therefore g(u) = u. This shows that g ∈ Hn . Thus Hn ⊃ Hn+1 and (6.41) follows. Corollary 6.9.5. The Grigorchuk group G is residually ﬁnite. Proof. Let h ∈ ∩n∈N Hn . We have h(w) = w for all w ∈ Σ ∗ and therefore h = 1G . This shows that ∩n∈N Hn = {1G }. As [G : Hn ] < ∞ for all n ∈ N we deduce from Proposition 2.1.11 that G is residually ﬁnite. Proposition 6.9.6. We have: (i) the subgroup H1 consists of all group elements g ∈ G which can be expressed in the form (6.38) (not necessarily reduced) with an even number of occurrences of the generator a; (ii) [G : H1 ] = 2; (iii) the group H1 is generated by the elements b, c, d, aba, aca, ada; (iv) the group H1 equals the normal closure in G of the elements b, c and d. Proof. It follows from (6.28) and (6.30) that a generator s ∈ S satisﬁes s(σ) = σ for all σ ∈ Σ if and only if s ∈ {b, c, d}. Thus, g = s1 s2 · · · sn , si ∈ S, belongs to H1 if and only if {i : si = a} is an even number. This also shows that [G : H1 ] = 2. Finally, let g ∈ H1 . Then it can be expressed in one of the following reduced forms:
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g = t0 at1 at2 at3 at4 · · · at2k−1 at2k = t0 (at1 a)t2 (at3 a)t4 · · · (at2k−1 a)t2k , g = at1 at2 at3 at 4 · · · at2k−1 at2k = (at1 a)t2 (at3 a)t4 · · · (at2k−1 a)t2k , g = t0 at1 at2 at3 at4 · · · at2k−1 a = t0 (at1 a)t2 (at3 a)t4 · · · (at2k−1 a), g = at1 at2 at3 at4 · · · at2k−1 a = (at1 a)t2 (at3 a)t4 · · · (at2k−1 a), where t0 , t1 , . . . , t2k+1 ∈ {b, c, d} and k ∈ N (observe that the occurrences of the a’s is 2k). This shows that the set {b, c, d, aba, aca, ada} generates H1 . Now, the normal closure in G of the generators b, c, d is the subgroup H ⊂ G generated by all the conjugates gtg −1 with g ∈ G and t ∈ {b, c, d}. Thus taking g = 1G , a we deduce from (iii) that H1 ⊂ H. On the other hand, since b, c, d ∈ H1 we also have H ⊂ H1 . Thus H = H1 . Let h ∈ H1 . For all w ∈ Σ ∗ there exist w0 , w1 ∈ Σ ∗ , (w0 ) = (w1 ) = (w) such that h(0w) = 0w0 and h(1w) = 1w1 . Denote by h0 , h1 ∈ Sym(Σ ∗ , ) the maps deﬁned by h0 (w) = w0 and h1 (w) = w1 . We thus have h(0w) = 0h0 (w) and h(1w) = 1h1 (w) for all w ∈ Σ ∗ . We denote by φ0 : H1 → Sym(Σ ∗ , ) (resp. φ1 : H → Sym(Σ ∗ , )) the map deﬁned by φ0 (h) = h0 (resp. φ1 (h) = h1 ) and by φ : H1 → Sym(Σ ∗ , ) × Sym(Σ ∗ , ) the product map φ(h) = (h0 , h1 ). From (6.30) we immediately deduce φ(b) = (a, c), φ(c) = (a, d), φ(d) = (1G , b),
φ(aba) = (c, a), φ(aca) = (d, a), φ(ada) = (b, 1G ).
(6.42)
Proposition 6.9.7. We have: (i) the maps φ0 , φ1 : H1 → G are surjective homomorphisms; (ii) the map φ : H1 → G × G is an injective homomorphism. Proof. Let σ ∈ {0, 1}. For all h, h ∈ H and w ∈ Σ ∗ we have σφσ (hh )(w) = hh (σw) = h(σφσ (h )(w)) = σφσ (h)φσ (h )(w). This shows that φσ (hh ) = φσ (h)φσ (h ). It follows that φ0 and φ1 are homomorphisms. From (6.42) and Proposition 6.9.6(iii) we immediately deduce that φ0 (H1 ) = φ1 (H1 ) = G. This shows (i). To show the injectivity of φ, let h ∈ H1 and suppose that φ(h) = 1G×G = (1G , 1G ). We have h( ) = and h(σw) = σhσ (w) = σw for all σ ∈ {0, 1} and w ∈ Σ ∗ . It follows that h = 1G = 1H . This shows (ii). As a consequence of Proposition 6.9.7(ii), we can identify each element h ∈ H1 with its image φ(h) ∈ G × G. Thus, we shall write h = (h0 , h1 ) if h ∈ H1 and φ(h) = (h0 , h1 ). Recall that a group is called periodic if it contains no elements of inﬁnite order.
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183
Theorem 6.9.8. The Grigorchuk group G is an inﬁnite ﬁnitely generated periodic group. Proof. Since H1 is a proper subset of G and φ0 : H1 → G is surjective (cf. Proposition 6.9.7(i)), we deduce that G is inﬁnite. Let now show that G is a 2–group, that is, that the order of every element g ∈ G is a power of 2. The proof is by induction on S (g). By (6.33) the statement is true for S (g) = 1. Let us ﬁrst show that every element g ∈ G \ {1G } is conjugate either to an element in S or to an element g which can be expressed in a reduced form g = at1 at2 · · · atk
(6.43)
with t1 , t2 , . . . , tk ∈ {b, c, d}, k ≥ 1, and such that S (g ) ≤ S (g). Indeed, if g is not conjugate to an element in S, then any element of minimal Slength in the conjugacy class of g necessarily admits one of the following two reduced forms, besides (6.43): t1 at2 · · · atk a and t0 at1 · · · atk with t0 , t1 , . . . , tk ∈ {b, c, d}, t0 = tk , k ≥ 1. Conjugating by a and t0 respectively and replacing tk t0 with t ∈ {b, c, d} according to (6.34), we transform the above two expressions into one of the form (6.43). It is clear that such a process does not increase the word lengths so that S (g ) ≤ S (g). Suppose ﬁrst that g ∈ H1 and (g) > 1. Since the order of any element equals the order of all its conjugates, we may suppose, up to conjugacy, that g admits a reduced expression of the form (6.43) (with k ≥ 2 even, since g ∈ H1 ). Now, the image of each quadruple at2i−1 at2i , i = 1, 2, . . . , k/2, via φ0 or φ1 has wordlength ≤ 2. Thus, S (g0 ), S (g1 ) ≤ S (g)/2 < S (g). By induction, g0 e g1 are 2−elements, and thus g = (g0 , g1 ) is a 2−element as well. Suppose now that g ∈ / H1 . As before, we may suppose, up to conjugacy, that g admits a reduced expression of the form (6.43) (with k odd, since g ∈ H1 ). We distinguish three cases. Case 1. The generator d appears in the expression of g, say d = ti for some 1 ≤ i ≤ k. Then, up to conjugating by the element ti−1 ati−2 a · · · at1 a, we can suppose that d = t1 . Now g 2 = (adat2 )(at3 at4 ) · · · (atk ad)(at2 at3 ) · · · (atk−1 atk ) ∈ H. The image of each quadruple via φ0 (resp. φ1 ) has length 2, with the only exception for those of the form atj ad = (atj a)d (resp. adatj = (ada)tj ) since φ0 (d) = 1G (resp. φ1 (ada) = 1G ). But at least one of these quadruples occurs in g 2 , for instance for j = k (resp. j = 1), so that S (φ0 (g 2 )) ≤ 2k − 1 < 2k = S (g) (resp. S (φ1 (g 2 )) < S (g)). By induction, φ0 (g 2 ) and φ1 (g 2 ) are 2elements so that g 2 = (φ0 (g 2 ), φ1 (g 2 )) and therefore g are 2elements as well. Case 2. Suppose now that d doesn’t occur in the expression of g but c does. As before, up to conjugacy, we may suppose that t1 = c. We then have φ0 (abat ) = ca, φ0 (acat ) = da and φ1 (abat ) = ac, φ1 (acat ) = ad for all t ∈ {b, c}. It follows that S (φ0 (g 2 )) = S (φ1 (g 2 )) = 2k = (g). Observe that φ0 (g 2 ) = da · · · a (resp. φ1 (g 2 ) = a · · · ad) so that, by conjugating by a (resp.
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6 Finitely Generated Amenable Groups
ad) we fall into Case 1. This shows that φ0 (g 2 ) and φ1 (g 2 ) are 2elements. Thus g 2 = (φ0 (g 2 ), φ1 (g 2 )) and therefore g are 2elements. Case 3. Finally, suppose that neither d nor c appear in (6.43) so that necessarily g = (ab)2h+1 for some h ≥ 0. We have g 2 = (ab)4h+2 = ((aba)b)2h+1 ∈ H, so that φ0 (g 2 ) = (ca)2h+1 and φ1 (g 2 ) = (ac)2h+1 . Since S (φ0 (g 2 )) = S (φ1 (g 2 )) = 4h + 2 = S (g), we are in Case 2 and we deduce that φ0 (g 2 ) and φ1 (g 2 ) are 2elements. Thus g 2 = (φ0 (g 2 ), φ1 (g 2 )) and therefore g itself are 2elements. Theorem 6.9.9. The Grigorchuk group G does not have polynomial growth. In order to prove this theorem, let us prove some preliminary results. Lemma 6.9.10. The subgroup of G generated by a and d is dihedral of order 8. Proof. The dihedral group D8 of order 8 has presentation D8 = x, y : x2 , y 2 , (xy)4 . Since a2 = d2 = 1G by (6.33), we are only left to verify that the order of the element ad is 4. We have (ad)2 = (ada)d = (b, 1)(1, b) = (b, b) = 1G , while (ad)4 = ((ad)2 )2 = (b, b)2 = (b2 , b2 ) = (1G , 1G ) = 1G . Recall that two groups G1 and G2 are commensurable if there exist two subgroups of ﬁnite index K1 ⊂ G1 and K2 ⊂ G2 such that K1 and K2 are isomorphic. Lemma 6.9.11. The Grigorchuk group G is commensurable with its own square G × G. Proof. Let us start by showing that the index of φ(H1 ) inside G × G is ﬁnite. Since b, c, d, aba, aca and ada generate H1 (cf. Proposition 6.9.4), we deduce that the elements in (6.42) generate φ(H1 ). Denote by B ⊂ G the normal closure in G of b. The quotient G/B is generated by the images of the generators of G and since cd = b ∈ B, it is generated by the images of a and d. From Lemma 6.9.10 we deduce [G : B] = G/B ≤ 8.
(6.44)
Let g ∈ G and consider the element gbg −1 . Since φ0 (resp. φ1 ) is surjective, there exists an element h ∈ H (resp. h ∈ H) such that φ0 (h) = g (resp. φ1 (h ) = g). It follows that (gbg −1 , 1G ) = φ(hadah−1 ) ∈ φ(H1 ) (resp. (1G , gbg −1 ) = φ(h d(h )−1 ) ∈ φ(H1 )). As g varies in G the elements (gbg −1 , 1) (resp. (1, gbg −1 )) generate the subgroup B0 = B × {1G } ⊂ φ(H1 ) (resp. B1 = {1G } × B ⊂ φ(H1 )). Observe that B0 B1 B and that B0 and B1 are normal subgroups of G × G. Moreover, B0 ∩ B1 = {1G×G }, so that (G×G)/(B0 B1 ) G/B ×G/B. From (6.44) and the fact that B0 B1 ⊂ φ(H1 ) we deduce that [G × G : φ(H1 )] ≤ [G × G : B0 B1 ] = [G : B]2 = 64. This shows that φ(H1 ) has ﬁnite index in G × G.
(6.45)
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185
On the other hand [G : H1 ] = 2 (Proposition 6.9.6(ii)) and since φ is an injective homomorphism (Proposition 6.9.7(ii)), we have that H1 and φ(H1 ) are isomorphic. It follows that G and G × G are commensurable. Proof of Theorem 6.9.9. Since G is inﬁnite (Theorem 6.9.8) we have n γ(G) (cf. Proposition 6.4.8). Since G and G×G are commensurable (Lemma 6.9.11) we deduce from Corollary 6.6.7 that γ(G) = γ(G×G). On the other hand, by Proposition 6.6.10 we have γ(G × G) = γ(G)2 . Using Lemma 6.6.9 it follows that n2 γ(G)2 = γ(G). By induction, we have that n2h γ(G) for all h ∈ N. Thus G cannot have polynomial growth. The remaining of this section is devoted to showing the following: Theorem 6.9.12. The Grigorchuk group G has subexponential growth. To prove this theorem we need some preliminaries. We start with a useful criterion for detecting that certain ﬁnitely generated groups have subexponential growth. Lemma 6.9.13. Let G be a ﬁnitely generated group. Let S ⊂ G be a ﬁnite symmetric generating subset and suppose that there exist an integer M ≥ 2, two constants 0 < k < 1 and K ≥ 0, and an injective homomorphism ψ : H → GM g → (gi )M i=1 where H ⊂ G is a ﬁnite index subgroup of G, such that M
S (gi ) ≤ kS (g) + K
(6.46)
i=1
for all g ∈ H. Then G has subexponential growth. 1
Proof. Let us show that λS = limn→∞ γSG (n) n equals 1. Fix ε > 0. Then there exists an integer n0 ≥ 1 such that γSG (n) < (λS +ε)n for all n ≥ n0 . Since λS ≥ 1, it follows that γSG (n) ≤ γSG (n0 )(λS + ε)n
(6.47)
for all n ∈ N. Let γSH (n) = {h ∈ H : S (h) ≤ n} and ﬁx a system T of left coset representatives of H in G. Set C = maxt∈T S (t). Then, given g ∈ G, there exist unique h ∈ H and t ∈ T such that g = th. Therefore S (h) ≤ S (t) + S (g) ≤ C + (g), so that BSG (n) ⊂ T BSH (n + C). We deduce that γSG (n) ≤ [G : H]γSH (n + C).
(6.48)
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6 Finitely Generated Amenable Groups
On the other hand, by (6.46) we have γSH (n) ≤ γSG (n1 )γSG (n2 ) · · · γSG (nM ) where the sum runs over all M −tuples n1 , n2 , . . . , nM such that kn + K. From (6.47) we then deduce: γSH (n) ≤ γSG (n0 )M (λS + )n1 (λS + )n2 · · · (λS + )nM = γSG (n0 )M (λS + )n1 +n2 +···+nM M ≤ γSG (n0 )(kn + K) (λS + )kn+K .
ni ≤
Using (6.48) we then obtain M (λS + ε)kn+K γSG (n) ≤ [G : H]γSH (n + C) ≤ [G : H] γSG (n0 )(kn + K ) (6.49) where K = K +kC. Taking the nth roots and passing to the limit for n → ∞ in (6.49), the ﬁrst term on the left tends to λS , while the last term on the right tends to (λS + ε)k . Thus λS ≤ (λS + ε)k . Since ε was arbitrary we deduce that λS ≤ λkS . Since by hypothesis k < 1, we have that λS = 1. Finally, from Corollary 6.5.5 we deduce that G has subexponential growth. Lemma 6.9.14. Let n ≥ 1. Then φσ (Hn+1 ) ⊂ Hn for σ = 0, 1. Proof. We proceed by induction on n. We have already noticed that φσ (H1 ) ⊂ G = H0 (Proposition 6.9.7(i)). Suppose that φσ (Hn ) ⊂ Hn−1 and let us show that φσ (Hn+1 ) ⊂ Hn . Let g ∈ Hn+1 and w = σu, with u ∈ Σ n and σ ∈ {0, 1}. Then σu = w = g(w) = σφσ (g)(u), so that φσ (g)(u) = u. This shows that φσ (g) ∈ Hn . It follows that φσ (Hn+1 ) ⊂ Hn . As a consequence of the previous lemma, for all n ≥ 1 and w = σ1 σ2 · · · σn ∈ Σ n the homomorphisms φw : Hn → G deﬁned by φw = φσ1 ◦ φσ2 ◦ · · · ◦ φσn
are well deﬁned. We then deﬁne the homomorphism ψn : Hn → w∈Σ n G by setting ψn (g) = (φw (g))w∈Σ n for all g ∈ Hn . Note that by Proposition 6.9.7(ii) ψn is injective. Thus, identifying Hn with its image ψn (Hn ) ⊂
n w∈Σ n G, we simply write g = (gw )w∈Σ for all g ∈ Hn . In the following, we consider the alphabet Λ = {α, β, γ, δ} and the monoid Λ∗ . The map Λ → S given by α → a, β → b, γ → c and δ → d uniquely extends to a surjective monoid homomorphism π : Λ∗ → G. We say that a
6.9 The Grigorchuk Group and Its Growth
187
word w = λ1 λ2 · · · λn ∈ Λ∗ is reduced if for all i, j = 1, 2, . . . , n − 1 one has λi = α implies λi+1 ∈ {β, γ, δ} and λj ∈ {β, γ, δ} implies λj+1 = α. Consider the following transformation p : Λ2 → Λ∗ deﬁned by setting ⎧ if λ = λ ⎪ ⎪ ⎪ ⎪ if λ = β, λ = γ or λ = γ, λ = β ⎨δ if λ = β, λ = δ or λ = δ, λ = β (6.50) p(λλ ) = γ ⎪ ⎪ = δ or λ = δ, λ = γ β if λ = γ, λ ⎪ ⎪ ⎩ λλ otherwise. Given a word u = λ1 λ2 · · · λn ∈ Λ∗ , and 1 ≤ i ≤ n, we set u(i) = λ1 λ2 · · · λi−1 p(λi λi+1 )λi+2 · · · λn ∈ Λ∗ . Note that u is reduced if and only if u = u(i) for all i = 1, 2, . . . , n − 1. By induction we deﬁne u(i1 ,i2 ,...,ik ) = (u(i1 ,i2 ,...,ik−1 ) )ik for 1 ≤ ik ≤ (u(i1 ,i2 ,...,ik−1 ) ) − 1. Let now w = λ1 λ2 · · · λn ∈ Λ∗ . If w is reduced we set w = w. Otherwise, let i1 be the minimal integer such that w = w(i1 ) . If w(i1 ) is reduced we set w = w(i1 ) . Otherwise, let i2 be the minimal integer such that w(i1 ) = w(i1 ,i2 ) . If w(i1 ,i2 ) is reduced we set w = w(i1 ,i2 ) . And so on. Since (w) > (w(i1 ) ) > (w(i1 ,i2 ) > · · · > (w(i1 ,i2 ,...,ik−1 ) ) > (w(i1 ,i2 ,...,ik ) ) > · · · there exists an integer k such that w(i1 ,i2 ,...,ik ) is reduced. We then set w = w(i1 ,i2 ,...,ik ) . A transformation w(i1 ,i2 ,...,ij−1 ) → w(i1 ,i2 ,...,ij ) is called a leftmost cancellation. It follows that every word in Λ∗ can be transformed into a reduced word by a ﬁnite sequence of leftmost cancellations. We denote by Θ1 ⊂ Λ∗ the subset consisting of all reduced words w on Λ which contain an even number α (w) of occurrences of the letter α. Thus, a word w ∈ Θ1 is an alternate product of terms of the form (αλα) and λ , with λ, λ ∈ {β, γ, δ}. Consider the maps Φ0 : Θ1 → Λ∗ and Φ1 : Θ1 → Λ∗ deﬁned by setting Φσ ((αλ1 α)λ2 · · · ) = Φσ (αλ1 α)Φσ (λ2 ) · · · (resp. Φσ (λ1 (αλ2 α) · · · ) = Φσ (λ1 )Φσ (αλ2 α) · · · for all λ1 , λ2 , . . . ∈ {β, γ, δ} where Φ0 (β) = α, Φ0 (αβα) = γ, Φ1 (β) = γ, Φ1 (αβα) = α, Φ0 (γ) = α, Φ0 (αγα) = δ, Φ1 (γ) = δ, Φ1 (αβα) = α, Φ0 (δ) = , Φ0 (αδα) = β, Φ1 (β) = δ, Φ1 (αβα) = . For n ≥ 1 we inductively deﬁne Θn+1 ⊂ Λ∗ as the subset consisting of all reduced words w in Θn such that the reduced words Φ0 (w) and Φ1 (w) belong to Θn . Also, given w ∈ Θn we recursively set wσ0 σ1 ···σn−1 σ = Φσ (wσ0 σ1 ···σn−1 ) for all σ, σ1 , σ2 , . . . , σn−1 ∈ {0, 1}. Lemma 6.9.15. For all w ∈ Θ1 one has (w0 ) + (w1 ) ≤ (w) + 1.
(6.51)
188
6 Finitely Generated Amenable Groups
Proof. Let w ∈ Θ1 . We distinguish a few cases. Case 1. Suppose that w starts and ends with α so that w = (αλ1 α)λ2 · · · λk−1 (αλk α) with λi ∈ {β, γ, δ}. Note that (w) = 2k + 1. Then wσ = Φσ (w) and Φσ (w) = Φσ (αλ1 α)Φσ (λ2 ) · · · Φσ (λk−1 )Φσ (αλk α) and therefore (wσ ) = (Φσ (w)) ≤ (Φσ (w)) ≤ (Φσ (αλ1 α)) + (Φσ (λ2 )) + · · · + (Φσ (λk−1 )) + (Φσ (αλk α)) (2k + 1) − 1 2 (w) − 1 . = 2 ≤k=
Case 2. Suppose that w starts and ends with letters in {β, γ, δ}, that is, w = λ1 (αλ2 α)λ3 · · · (αλk−1 α)λk with λi ∈ {β, γ, δ}. Note that (w) = 2k − 1. Then wσ = Φσ (w) and Φσ (w) = Φσ (λ1 )Φσ (αλ2 α) · · · Φσ (αλk−1 α)Φσ (λk ) and therefore (wσ ) = (Φσ (w)) ≤ (Φσ (w)) ≤ (Φσ (λ1 )) + (Φσ (αλ2 α)) + · · · + (Φσ (αλk−1 α)) + (Φσ (λk )) (2k − 1) + 1 2 (w) + 1 . = 2 ≤k=
Case 3. Finally, suppose that w starts with α and ends with λ ∈ {β, γ, δ}, or viceversa. To ﬁx ideas, suppose that w = (αλ1 α)λ2 · · · (αλk−1 α)λk with λi ∈ {β, γ, δ}. Passing to the inverse word w−1 one handles the other possibility. Note that (w) = 2k. Then wσ = Φσ (w) and Φσ (w) = Φσ (αλ1 α)Φσ (λ2 ) · · · Φσ (αλk−1 α)Φσ (λk ) and therefore (wσ ) = (Φσ (w)) ≤ (Φσ (w)) ≤ (Φσ (αλ1 α)) + (Φσ (λ2 )) + · · · + (Φσ (αλk−1 α)) + (Φσ (λk )) 2k 2 (w) . = 2 ≤k=
This shows (6.51). Lemma 6.9.16. For all g ∈ H3 one has 1 i,j,k=0
S (gijk ) ≤
5 S (g) + 8. 6
(6.52)
6.9 The Grigorchuk Group and Its Growth
189
Proof. Let g ∈ H3 and let w ∈ Θ3 be such that g = π(w) and S (g) = (w). As a consequence of Lemma 6.9.15 we have 1
(wij ) ≤
i,j=0
1 ((wi ) + 1) i=0
=
1
(6.53) (wi ) + 2
i=0
≤ (w) + 3 and therefore 1
(wijk ) ≤
1
((wij ) + 1)
i,j=0
i,j,k=0
≤
1
(wij ) + 4
i,j=0
(by (6.53)) ≤
1 i=0
(6.54)
(wi ) + 2
+4=
1
(wi ) + 6
i=0
(by (6.51)) ≤ (w) + 1 + 6 = (w) + 7. Now we observe that by deﬁnition of the maps Φ0 and Φ1 , the inequalities in (6.51), (6.53) and (6.54) can be sharpened as follows. (w0 ) + (w1 ) ≤ (w) + 1 − δ (w),
(6.55)
where δ (w) denotes the number of occurrences of the letter δ in the word w. Indeed, every such δ, may appear in w either isolated or in a triplet (αδα). Then, in the ﬁrst case we have Φ0 (δ) = , while in the second one, Φ1 (αδα) = . Similarly, since every letter γ in the word w will give rise via Φ0 or Φ1 to a letter δ in one of w0 and w1 , the above argument shows that, even if some cancellation occurs, (w00 ) + (w01 ) + (w10 ) + (w11 ) ≤ (w) + 3 − γ (w),
(6.56)
where γ (w) denotes the number of occurrences of the letter γ in the word w. Finally, since every letter β in the word w will give rise via Φ0 or Φ1 to a letter γ in one of w0 and w1 , and therefore to a letter δ in one of w00 , w01 , w10 and w11 , the above arguments show that indeed, even if some cancellation occurs,
190
6 Finitely Generated Amenable Groups 1
(wijk ) ≤ (w) + 7 − β (w),
(6.57)
i,j,k=0
where β (w) denotes the number of occurrences of the letter β in the word w. Since (w) = α (w)+β (w)+γ (w)+δ (w) and w is reduced, we necessarily and therefore have β (w) + γ (w) + δ (w) ≥ (w)−1 2 max λ∈{β,γ,δ}
λ (w) >
(w) − 1. 6
Taking into account the inequalities in (6.54) we obtain ⎧ ⎫ 1 1 1 ⎨ ⎬ (wijk ) ≤ min (wi ) + 6, (wij ) + 4 . ⎩ ⎭ i=0
i,j,k=0
(6.58)
(6.59)
i,j=0
Observe that π(wijk ) = gijk so that S (gijk ) ≤ (wijk ) for all i, j, k = 0, 1. Thus, using ﬁrst (6.59), (6.55), (6.56) and (6.57), and then (6.58) we deduce 1 i,j,k=0
1
S (gijk ) ≤
(wijk )
i,j,k=0
≤ min{(w) + 7 − β (w), (w) + 7 − γ (w), (w) + 7 − δ (w)} ≤ (w) + 7 − max{β (w), γ (w), δ (w)} (w) −1 ≤ (w) + 7 − 6 5 ≤ (w) + 8 6 5 = (g) + 8. 6 This shows (6.52).
Proof of Theorem 6.9.12. Let H = H3 . By Proposition 6.9.4 we have [G : H] < ∞. Moreover, by Lemma 6.9.16 we can take ψ = ψ3 , M = 8, k = 5/6 and K = 8, and apply Lemma 6.9.13 to obtain that G has subexponential growth. A ﬁnitely generated group G is said to have intermediate growth if G has subexponential growth but does not have polynomial growth. Note that every ﬁnitely generated group of intermediate growth contains no subgroups isomorphic to F2 , by Corollary 6.6.5. From Theorem 6.9.9 and Theorem 6.9.12 we then get: Theorem 6.9.17. The Grigorchuk group G has intermediate growth.
6.10 The Følner Condition for Finitely Generated Groups
191
6.10 The Følner Condition for Finitely Generated Groups Let G be a group. As right multiplication by elements in G is bijective we have, for all A, B, F ⊂ G, F ﬁnite, and g ∈ G, (A \ B)g = Ag \ Bg,
(6.60)
F \ F g = F g \ F  = F \ F g −1  = F g −1 \ F .
(6.61)
Lemma 6.10.1. Let A, B, C ⊂ G be three ﬁnite sets. Then we have: A \ B ≤ A \ C + C \ B.
(6.62)
Proof. Suppose that x ∈ A \ B, that is, (i) x ∈ A and (ii) x ∈ / B. We distinguish two cases. If x ∈ / C, then, by (i), x ∈ A \ C. If x ∈ C, then, by (ii), x ∈ C \ B. In both cases, x ∈ (A \ C) ∪ (C \ B). It follows that A \ B ⊂ (A \ C) ∪ (C \ B) and (6.62) follows. Let now S ⊂ G be a ﬁnite subset. The isoperimetric constant of G with respect to S is the nonnegative number ιS (G) = inf F
F S \ F  F 
(6.63)
where F runs over all non empty ﬁnite subsets of G. Observe that ιS (G) = ιS∪{1G } (G). Proposition 6.10.2. Let G be a ﬁnitely generated group and let S ⊂ G be a ﬁnite generating subset. Then the following conditions are equivalent: (a) G is amenable; (b) for all ε > 0 there is a ﬁnite subset F = F (ε) ⊂ G such that F \F s < εF 
for all s ∈ S;
(6.64)
(c) ιS (G) = 0. Proof. Suppose (a) and ﬁx ε > 0. By Theorem 4.9.1, G satisﬁes the Følner conditions. For our convenience, we express the Følner conditions as follows. Given any ﬁnite subset K ⊂ G and ε > 0 there exists a ﬁnite subset F = F (K, ε ) ⊂ G such that F \F k < ε F 
for all k ∈ K.
(6.65)
Taking ε = ε, K = S and F = F (S, ε) in 6.65, we then immediately obtain (6.64), and the implication (a) ⇒ (b) follows.
192
6 Finitely Generated Amenable Groups
Suppose (b). Taking g = s in (6.61) we immediately deduce F \F s < εF 
for all s ∈ S ∪ S −1 .
(6.66)
Fix a ﬁnite set K ⊂ G and ε > 0. Since S generates G we can ﬁnd n ∈ N such that K ⊂ (S ∪ S −1 )n . Set ε = ε /n. Let F = F (ε) ⊂ G be a ﬁnite set such that (6.66) holds. Given k ∈ K we can ﬁnd a1 , a2 , . . . , an ∈ S ∪ S −1 such that k = a1 a2 · · · an . Then, recalling (6.62) and (6.60), we have F \ F k = F \ F a1 a2 · · · an  ≤ F \ F an  + F an \ F an−1 an  + F an−1 an \ F an−2 an−1 an  + · · · + F a2 a3 · · · an \ F a1 a2 · · · an  = F \ F an  + F \ F an−1  + F \ F an−2  + · · · + F \ F a1  < nεF  ≤ ε F  for all k ∈ K. This shows that G satisﬁes the Følner conditions (6.65) and therefore G is amenable by Theorem 4.9.1. Thus (b) ⇒ (a). Finally, the equivalence (b) ⇔ (c) immediately follows from (6.61) and the inequalities F s \ F  = F \ F s . F \ F s = F s \ F  ≤ F S \ F  ≤ s ∈S
s ∈S
for any s ∈ S and any ﬁnite subset F ⊂ G.
6.11 Amenability of Groups of Subexponential Growth In this section we show that every ﬁnitely generated group of subexponential growth is amenable. We ﬁrst prove the following: Lemma 6.11.1. Let (an )n≥1 be a sequence of positive real numbers. Then lim inf n→∞
√ an+1 ≤ lim inf n an . n→∞ an
(6.67)
Proof. Set α = lim inf n→∞ aan+1 . If α = 0 there is nothing to prove. Othern ≥β wise, let 0 < β < α. Then, there exists an integer N ≥ 1 such that aan+1 n for all n ≥ N . Thus, for all p = 1, 2 . . . one has aN +p aN +p−1 aN +1 aN +p = · ··· ≥ βp. aN aN +p−1 aN +p−2 aN This gives aN +p ≥ β p aN . It follows that setting n = N + p, one has, for all n≥N
6.12 The Theorems of Kesten and Day
193
an ≥ β n−N aN = β n (β −N aN ). √ Thus, taking the nth roots we obtain n an ≥ β n β −N aN and therefore, lim inf n→∞
√ n an ≥ β.
As (6.68) holds for all β < α, we have lim inf n→∞
(6.68) √ n a n ≥ α = lim inf n→∞
an+1 an .
We are now in position to prove the main result of this section. Theorem 6.11.2. Every ﬁnitely generated group of subexponential growth is amenable. Proof. Let G be a ﬁnitely generated group of subexponential growth. Let S ⊂ G be a ﬁnite symmetric generating subset of G. Then, by virtue of the n previous lemma we have 1 ≤ lim inf n→∞ γSγ(n+1) ≤ lim γ n→∞ S (n) = 1, so S (n) that lim inf n→∞
γS (n+1) γS (n)
= 1. Fix ε > 0 and let n0 ∈ N be such that γS (n0 + 1) < 1 + ε. γS (n0 )
(6.69)
Set F = BS (n0 ) and let us show that F \ F s < εF  for all s ∈ S. Let s ∈ S. Then, as BS (n0 )s ⊂ BS (n0 + 1) for all s ∈ S, we have F \ F s = F s \ F  = BS (n0 )s \ BS (n0 ) ≤ BS (n0 + 1) \ BS (n0 ) = γS (n0 + 1) − γS (n0 ) < εγS (n0 ) = εF , where the last inequality follows from (6.69). From Proposition 6.10.2 we deduce that G is amenable. From Theorem 6.9.12 and Theorem 6.11.2 we deduce the following: Corollary 6.11.3. The Grigorchuk group G is amenable.
6.12 The Theorems of Kesten and Day Let G be a group. Let p ∈ [1, +∞). We consider the real Banach space p G p (G) = x ∈ R : x(g) < ∞ g∈G
194
6 Finitely Generated Amenable Groups
consisting of all psummable real functions on G. For x ∈ p (G), the nonneg 1 p p is called the p norm of x. ative number xp = g∈G x(g) The support of a conﬁguration x ∈ RG is the set {g ∈ G : x(g) = 0}. We denote by R[G] ⊂ RG the vector subspace consisting of all ﬁnitely supported conﬁgurations in RG . Note that R[G] is a dense subspace in p (G). When p = 2, it is possible to endow 2 (G) with a scalar product ·, · deﬁned by setting x(g)y(g) x, y = g∈G
for all x,y ∈ 2 (G). Then, the 2 norm of an element x ∈ 2 (G) is given by x2 = x, x. The space (2 (G), ·, ·) is a real Hilbert space. The p norm of a linear map T : p (G) → p (G) is deﬁned by sup T xp =
T p→p =
p
x∈ (G)
x p ≤1
T xp . (G) xp
sup p
x∈ x=0
Then, T is continuous if and only if T p→p < ∞. We denote by L(p (G)) the space of all continuous linear maps T : p (G) → p (G) and by I : p (G) → p (G) the identity map. Note that, by density of R[G] in p (G), we have T xp x∈R[G] xp
T p→p = sup T xp = sup x∈R[G]
x p ≤1
(6.70)
x=0
for all T ∈ L(p (G)). Let p ∈ [1, +∞) and s ∈ G. For all x ∈ p (G) and g ∈ G we set (Ts(p) x)(g) = x(gs). We then have Ts(p) xpp =
Ts(p) x(g)p
g∈G
=
x(gs)p
g∈G
= =
x(h)p (by setting h = gs)
h∈G xpp (p)
for all x ∈ p (G). We deduce that Ts x ∈ p (G) and that the linear map (p) Ts : p (G) → p (G) has p norm
6.12 The Theorems of Kesten and Day
195
Ts(p) p→p = 1.
(6.71)
(p)
In particular, Ts ∈ L(p (G)). (p) Let now S ⊂ G be a nonempty ﬁnite set. We denote by MS : p (G) → (p) (p) 1 p (G) the map deﬁned by MS = S s∈S Ts . In other words, (p)
(MS x)(g) =
1 x(gs) S s∈S
(p)
for all x ∈ p (G) and g ∈ G. The map MS on p (G) associated with S.
is called the p Markov operator
Proposition 6.12.1. Let G be a group and S ⊂ G a nonempty ﬁnite set. (p) Then the p Markov operator MS : p (G) → p (G) is linear and continuous. Moreover, (p)
MS p→p ≤ 1. (p)
Proof. Since Ts have
(p)
(6.72) (p)
∈ L(p (G)), we deduce that MS
MS p→p =
∈ L(p (G)). Finally, we
1 (p) 1 (p) Ts p→p ≤ Ts p→p = 1 S S s∈S
s∈S
where the last equality follows from (6.71).
Proposition 6.12.2. Let G be a group. Let S ⊂ G be a ﬁnite subset containing 1G . Then, the following conditions are equivalent: (2)
(a) MS 2→2 = 1; (2) (b) given ε > 0 there exists x ∈ R[G] such that x2 = 1 and MS x2 ≥ 1 − ε; (c) given ε > 0 there exists x ∈ R[G] such that x ≥ 0, x2 = 1 and (2) MS x2 ≥ 1 − ε; (d) given ε > 0 there exists x ∈ R[G] such that x ≥ 0, x2 = 1 and x − Ts(2) x2 ≤ ε (2)
for all s ∈ S;
(6.73)
(2)
(e) 1 belongs to the real spectrum σ(MS ) of MS . Proof. The implication (a) ⇒ (b) follows from the deﬁnition of 2 norm and the density of R[G] in 2 (G) (cf. (6.70)). Let now x ∈ R[G]. We have (2) 2 (2) MS x22 ≤ MS x . 2
(6.74)
196
6 Finitely Generated Amenable Groups
Indeed, (2) MS x22
2 1 = Ts(2) x S s∈S 2 2 1 = x(gs) S g∈G s∈S 2 1 ≤ x(gs) S g∈G s∈S 2 1 (2) = Ts x S s∈S 2 2 (2) = MS x . 2
Thus, replacing x by x gives the implication (b) ⇒ (c). Suppose that (d) fails to hold, that is, there exists ε0 > 0 such that for all (2) x ∈ R[G], x ≥ 0, x2 = 1, there exists s0 ∈ S such that x − Ts0 x2 ≥ ε0 . Since 2 (G) is uniformly convex (Lemma I.4.2) there exists δ0 > 0 such that x + T (2) x s0 (6.75) ≤ 1 − δ0 2 2
for all x ∈ R[G] such that x2 = 1. It then follows 1 (2) (2) MS x2 = Ts x S s∈S 2 (2) 2 x + Ts0 x 1 (2) + = T x s S 2 S s∈S\{1G ,s0 } 2 (2) 2 x + Ts0 x 1 ≤ Ts(2) x2 + S 2 S 2
s∈S\{1G ,s0 }
2 1 (1 − δ0 ) + (S − 2) (by (6.75)) ≤ S S 2δ0 =1− S for all x ∈ R[G], such that x2 = 1 and x ≥ 0. This clearly contradicts (c). We have shown (c) ⇒ (d). Let us show (d) ⇒ (e). Suppose that (6.73) holds. Then, for every ε > 0 there exists x ∈ R[G] such that
6.12 The Theorems of Kesten and Day
1 (2) x − MS x2 = x − Ts(2) x S s∈S 2 1 (2) = (x − Ts x) S s∈S
1 ≤ x − Ts(2) x2 S
197
2
s∈S
≤ ε. (2)
Therefore, by Corollary I.2.3, the linear map I − MS ∈ L(2 (G)) is not (2) bijective, that is, 1 ∈ σ(MS ). Finally, the implication (e) ⇒ (a) follow from (I.14) and the fact that (2) MS 2→2 ≤ 1 (Proposition 6.12.1). Lemma 6.12.3. Let S ⊂ G be a nonempty ﬁnite subset. The following conditions are equivalent: (2)
(a) 1 ∈ σ(MS ); (2) (b) 1 ∈ σ(MS∪{1G } ). Proof. If S contains 1G there is nothing to prove. S Suppose that 1G ∈ / S and set α = S+1 . Note that 0 < α < 1 and that (2)
(2)
MS∪{1G } = (1 − α)I + αMS . (2)
(2)
(6.76) (2)
We then have I − MS∪{1G } = (1 − α)(I − MS ) so that I − MS∪{1G } is (2)
bijective if and only if I − MS
(2)
is bijective. In other words, 1 ∈ σ(MS∪{1G } )
(2)
if and only if 1 ∈ σ(MS ).
Before stating and proving the main result of this section we need a little more work. More precisely, we want to show the equivalence between the existence of an almost Sinvariant positive element x ∈ R[G] of norm x2 = 1 (cf. condition (d) in Proposition 6.12.2) and the existence of an almost Sinvariant nonempty ﬁnite set F ⊂ G (cf. condition (b) in Proposition 6.10.2). Let F ⊂ G be a ﬁnite set. Denote, as usual, by χF the characteristic map of F and observe that χF ∈ R[G]. Proposition 6.12.4. Let A, B ⊂ G be two ﬁnite sets. Then χA − χB 1 = χA − χB 22 = A \ B + B \ A.
(6.77)
198
6 Finitely Generated Amenable Groups
Proof. First note that the map χA − χB takes value 1 at each point of A \ B, value −1 at each point of B \ A, and value 0 everywhere else, that is, outside of the set (A \ B) ∪ (B \ A) = (G \ (A ∪ B)) ∪ (A ∩ B). We then have χA (g) − χB (g)2 χA − χB 22 = g∈G
=
χA (g) − χB (g)
g∈G
=
χA (g) − χB (g)
g∈G: χA (g)−χB (g)=1
+
χA (g) − χB (g)
g∈G: χA (g)−χB (g)=−1
=
χA\B (g) +
g∈G
χB\A (g)
g∈G
= A \ B + B \ A. Lemma 6.12.5. Let x, y ∈ 2 (G) such that x2 = y2 = 1. Then, the following holds: (i) x2 ∈ 1 (G) and x2 1 = 1; (ii) x2 − y 2 1 ≤ 2x − y2 ;
1 (iii) suppose that x, y ≥ 0. Then, x − y2 ≤ x2 − y 2 1 2 . Proof. (i) We have x2 1 = g∈G x2 (g) = x22 = 1. (ii) We have x2 − y 2 1 = x2 (g) − y 2 (g) g∈G
=
x(g) − y(g) · x(g) + y(g)
g∈G
= x − y, x + y ≤ x − y2 · x + y2 ≤ x − y2 · (x2 + y2 ) = 2x − y2 , where the ﬁrst inequality follows from CauchySchwarz.
6.12 The Theorems of Kesten and Day
199
(iii) We have x − y22 =
(x(g) − y(g))2
g∈G
=
x(g) − y(g) · x(g) − y(g)
g∈G
≤
x(g) − y(g) · x(g) + y(g)
g∈G
=
x2 (g) − y 2 (g)
g∈G
= x2 − y 2 1 , where the inequality follows from the fact that x, y ≥ 0.
Lemma 6.12.6. Let F ⊂ G be a nonempty ﬁnite set and s ∈ G. Then the following holds: (p)
(i) Ts χF = χF s−1 , for p = 1, 2; (ii) χF 1 = χF 22 = F ; (1) (2) (iii) χF − Ts χF 1 = χF − Ts χF 22 = 2F \ F s. (s)
Proof. For g ∈ G and p = 1, 2, one has that (Tp χF )(g) = χF (gs) equals 1 if and only if gs ∈ F , that is, if and only if g ∈ F s−1 . This shows (i). Moreover, recalling that χF (g) ∈ {0, 1}, for all g ∈ G, on has χF 22 = χF (g)2 = χF (g) = χF 1 = {g ∈ G : χF (g) = 1} = F . g∈G
g∈G
This shows (ii). Finally, (iii) follows from (i), (6.77) (with A = F and B = F s−1 ) and (6.61). Lemma 6.12.7. Let x ∈ R[G] such that x ≥ 0 and x1 = 1. Then there exist an integer n ≥ 1, nonempty ﬁnite subsets Ai ⊂ G and real numbers λi > 0, 1 ≤ i ≤ n, satisfying A1 ⊃ A2 ⊃ · · · ⊃ An and λ1 + λ2 + · · · + λn = 1 such that n χA (6.78) x= λi i . A i i=1 Proof. Let 0 < α1 < α2 < · · · < αn be the values taken by x. For each 1 ≤ i ≤ n, let us set Ai = {g ∈ G : x(g) ≥ αi }.
200
6 Finitely Generated Amenable Groups
Clearly the sets Ai are nonempty ﬁnite subsets of G such that A1 ⊃ A2 ⊃ · · · ⊃ An . On the other hand, we have x = α1 χA1 + (α2 − α1 )χA2 + · · · + (αn − αn−1 )χAn = λ1
χA1 χA χA + λ2 2 + · · · + λ n n , A1  A2  An 
by setting λ1 = α1 A1  and λi = (αi − αi−1 )Ai  for 2 ≤ i ≤ n. Thus λi > 0 for 1 ≤ i ≤ n and n
λi = α1 A1  + (α2 − α1 )A2  + · · · + (αn − αn−1 )An 
i=1
= α1 (A1  − A2 ) + α2 (A2  − A3 ) + · · · + αn An  x(g) = 1. = g∈G
Lemma 6.12.8. With the same notation and hypotheses as in Lemma 6.12.7, we have n 2Ai \ Ai g (6.79) λi x − Tg(1) x1 = Ai  i=1 for every g ∈ G. Proof. Equality (6.78) gives us x−
Tg(1) x
=
n
(1)
λi
χAi − Tg χAi Ai 
λi
χAi − χAi g−1 . Ai 
i=1
(by Lemma 6.12.6(a))
=
n i=1
As we observed before (cf. the proof of Proposition 6.12.4), the map χAi − χAi g−1 takes value 1 at each point of Ai \ Ai g −1 , value −1 at each point of Ai g −1 \ Ai , and value 0 everywhere else. Let us set B= (Ai \ Ai g −1 ) and C = (Ai g −1 \ Ai ). 1≤i≤n
1≤i≤n
Note that the sets B and C are disjoint. Indeed, for all 1 ≤ i, j ≤ n, we have (Ai \ Ai g −1 ) ∩ (Aj \ Aj g −1 ) = ∅ since either Ai ⊂ Aj or Aj ⊂ Ai (which implies Aj g −1 ⊂ Ai g −1 ). It follows that
6.12 The Theorems of Kesten and Day
x − Tg(1) x1 =
201
(x − Tg(1) x)(a)
a∈G
n (χAi − χAi g−1 )(a) λi = Ai  a∈G i=1 n n (χAi − χAi g−1 )(a) (χAi − χAi g−1 )(a) λi λi = + Ai  Ai  i=1 i=1 =
a∈B n i=1
a∈C
λi
Ai \ Ai g Ai 
−1

+
n
−1
λi
i=1
Ai g \ Ai  Ai 
n
2Ai \ Ai g . (by (6.61)) = λi Ai  i=1 Let G be a ﬁnitely generated group and let S be a ﬁnite (not necessarily symmetric) generating subset of G. Consider the combinatorial Laplacian ΔS : RG → RG (cf. Example 1.4.3(b)). For all x ∈ 2 (G) ⊂ RG and g ∈ G we have (ΔS x)(g) = Sx(g) − x(gs) s∈S
= Sx(g) −
(Ts(2) x)(g)
s∈S (2)
= S(I − MS )(x)(g). (2)
This shows that for the map ΔS = ΔS  2 (G) one has (2) (2) ΔS = S I − MS
(6.80)
(2)
and therefore ΔS ∈ L(2 (G)). Let λ ∈ R. From (6.80) we deduce that (2)
(2)
λI − ΔS = λI − S(I − MS ) = −S It follows that (2)
λ ∈ σ(ΔS ) ⇔ and therefore
(2)
λ (2) I − MS . 1− S
λ (2) ∈ σ(MS ) 1− S (2)
0 ∈ σ(ΔS ) ⇔ 1 ∈ σ(MS ).
(6.81)
Theorem 6.12.9 (KestenDay). Let G be a ﬁnitely generated group. Let S ⊂ G be a ﬁnite (not necessarily symmetric) generating subset of G. Then
202
6 Finitely Generated Amenable Groups
the following conditions are equivalent: (a) G is amenable; (2) (b) 0 ∈ σ(ΔS ). Proof. Suppose (a). By Theorem 4.9.1 we have that G satisﬁes the Følner conditions. Fix ε > 0. By Proposition 6.10.2 there exists a ﬁnite subset F ⊂ G 2 such that F \ F s < ε2 F  for all s ∈ S. Set x = √1 χF and note that F 
x2 = 1. We have
1 (2) Ts x2 S s∈S 1 = (x − Ts(2) x)2 S s∈S 1 ≤ x − Ts(2) x2 S s∈S 1 = χF − Ts(2) χF 2 S · F  s∈S 1 1 (by Lemma 6.12.6(iii)) = 2F \ F s 2 S · F  s∈S 1 1 F \ F s 2 = 2 S F  (2)
(I − MS )x2 = x −
s∈S
=
2
F  < ε.
1
1 2
ε2
(2)
From Corollary I.2.3 we deduce that I − MS is not bijective, that is, 1 ∈ (2) (2) σ(MS ). From (6.81) we deduce that 0 ∈ σ(ΔS ). Thus (a) implies (b). To prove the converse implication, ﬁrst observe that, by virtue of Proposition 6.10.2, in order to prove (a), it suﬃces to show that given any ε > 0 there exists a ﬁnite subset F ⊂ G such that F \ F s < εF 
for all s ∈ S.
(6.82) (2)
Fix ε > 0. Suppose (b) and observe that by (6.81) we have 1 ∈ σ(MS ). Also note that, by Lemma 6.12.3, we can suppose that S 1G . Thus, by virtue of the implication (e) ⇒ (c) in Proposition 6.12.2, we can ﬁnd a non(2) ε negative function xε ∈ R[G] such that xε 2 = 1 and xε − Ts xε 2 ≤ 2S for all s ∈ S. Setting x = x2ε , from Lemma 6.12.5 we deduce that x1 = 1 and ε for all s ∈ S. (6.83) x − Ts(1) x1 ≤ S
6.12 The Theorems of Kesten and Day
203
By Lemma 6.12.7, there exist an integer n ≥ 1, nonempty ﬁnite subsets A1 ⊃ A2 ⊃ · · · ⊃ An Ai ⊂ G and real numbers λi > 0, 1 ≤ i ≤ n, nsatisfying χ and λ1 + λ2 + · · · + λn = 1, such that x = i=1 λi AAii . Set Ω = {1, 2, . . . , n} and consider the unique probability measure μ on Ω such that μ({i}) = λi for every i ∈ Ω. Finally, for each g ∈ G, let Ωg denote the subset of Ω deﬁned by
Ai \ Ai g ≥ε . Ωg = i ∈ Ω : Ai  It follows from Lemma 6.12.8 that x − Tg(1) x1 =
λi
i∈Ω
≥
2Ai \ Ai g Ai 
λi
i∈Ωg
≥ 2ε
2Ai \ Ai g Ai 
λi
i∈Ωg
= 2εμ(Ωg ). Therefore, we have (1)
μ(Ωg ) ≤
x − Tg x1 2ε
for all g ∈ G.
By using (6.83), we deduce 1 S
μ(Ωs ) < which implies
μ
Ωs
≤
s∈S
Thus
for all s ∈ S,
μ(Ωs ) < 1.
s∈S
Ωs = Ω.
s∈S
This means that there is some i0 ∈ Ω such that Ai0 \ Ai0 s 0 such that for every h ∈ H there exists g ∈ G such that dH (ϕ(g), h) ≤ δ.
(6.99)
Proof. Suppose that ϕ is a quasiisometric embedding. Let C ⊂ H be a ﬁnite subset such that H = ϕ(H)C and set δ = max{dH (1H , c) : c ∈ C}. Let h ∈ H, then there exist c ∈ C and g ∈ G such that h = ϕ(g)c, that is, ϕ(g)−1 h = c ∈ C. We deduce that dH (ϕ(g), h) ≤ δ and (6.99) follows. This shows (a) ⇒ (b). Conversely, suppose (b) and set C = BSG (1G , δ) ⊂ G. Let h ∈ H. By (6.99) there exists g ∈ G such that dH (ϕ(g), h) ≤ δ. It follows that ϕ(g)−1 h ∈ C, equivalently h ∈ ϕ(g)C. This shows that H = ϕ(G)C. Therefore ϕ is a quasiisometry and (a) follows. Proposition 6.13.14. Let G and H be two ﬁnitely generated groups. Suppose that there is a quasiisometric embedding ϕ : G → H. Then one has γ(G) γ(H). Proof. By Remark 6.13.4, we can assume ϕ(1G ) = 1H . Denote by dG and dH the word metric associated with two ﬁnite symmetric generating subsets SG and SH for G and H respectively. By Proposition 6.13.12, there exist integers α ≥ 1 and β ≥ 0 such that 1 dG (g, g ) − β ≤ dH (ϕ(g), ϕ(g )) ≤ αdG (g, g ) + β α
(6.100)
210
6 Finitely Generated Amenable Groups
for all g, g ∈ G. Note that the left inequality in (6.100) implies that ϕ(g) = ϕ(g ) whenever g, g ∈ G satisfy dG (g, g ) ≥ αβ+1. For n ∈ N, let BSGG (n) ⊂ G (resp. BSHH (n) ⊂ H) denote the ball of radius n centered at 1G (resp. 1H ). Choose a subset En ⊂ BSGG (n) of maximal cardinality such that dG (g, g ) ≥ αβ + 1 for all g, g ∈ En . Setting C = BSGG (αβ + 1), we have BSGG (n) ≤ CEn 
(6.101)
for all n ∈ N . Indeed, the balls of radius C centered at the elements of En cover BSGG (n) by the maximality of En . Now observe that the images by ϕ of the elements of En are all distinct and belong to BSHH (αn + β) by the right inequality in (6.100). This implies that En  ≤ BSHH (αn + β). By using (6.101), we then deduce that BSGG (n) ≤ CBSHH (αn + β) ≤ CBSHH (β)BSHH (αn) ≤ C BSHH (C n) for all n ≥ 1, where C = CαBSHH (β). It follows that γ(G) γ(H).
Corollary 6.13.15. If two ﬁnitely generated groups are quasiisometric, then they have the same growth type. Proof. If G and H are quasiisometric ﬁnitely generated groups, then there exist a quasiisometric embedding from G into H and a quasiisometric embedding from H into G. It follows that γ(G) γ(H) and γ(H) γ(G). Thus we have γ(G) = γ(H). Corollary 6.13.16. Let G and H be two quasiisometric ﬁnitely generated groups. Then G has exponential (resp. subexponential, resp. polynomial, resp. intermediate) growth if and only if H has exponential (resp. subexponential, resp. polynomial, resp. intermediate) growth. An important property of quasiisometric embeddings is the fact that every quasiisometric embedding is uniformly ﬁnitetoone: Proposition 6.13.17. Let G and H be two groups. Let ϕ : G → H be a quasiisometric embedding. Then there exists an integer M ≥ 1 such that ϕ−1 (h) ≤ M for all h ∈ H. Proof. Since ϕ is a quasiisometric embedding, we can ﬁnd a ﬁnite set K ⊂ G such that g1−1 g2 ∈ K whenever g1 , g2 ∈ G satisfy ϕ(g1 )−1 ϕ(g2 ) ∈ {1H }. Let us set M = K. Let h ∈ H. Suppose that g0 ∈ ϕ−1 (h). Then, every g ∈ ϕ−1 (h) satisﬁes ϕ(g0 )−1 ϕ(g) = h−1 h = 1H and therefore g0−1 g ∈ K. Thus, we have ϕ−1 (h) ⊂ g0 K and hence ϕ−1 (h) ≤ g0 K = K = M . Corollary 6.13.18. Let G and H be two groups. Suppose that there exists a quasiisometric embedding ϕ : G → H and that H is ﬁnite. Then G is ﬁnite.
6.13 QuasiIsometries
211
Proof. Taking M as in the preceding proposition, we have G ≤ M H.
Corollary 6.13.19. Let G and H be two groups. Suppose that H is ﬁnite. Then G is quasiisometric to H if and only if G is ﬁnite. Proof. If G is ﬁnite then it is clear from the deﬁnition that any map from G to H is a quasiisometry. On the other hand, if G is quasiisometric to H then G is ﬁnite by Corollary 6.13.18. Proposition 6.13.20. Let G and H be two groups. Suppose that there exists a quasiisometric embedding ϕ : G → H and that H is locally ﬁnite. Then G is locally ﬁnite. Proof. Let K be a ﬁnitely generated subgroup of G and let S ⊂ K be a ﬁnite symmetric generating subset of K. Since ϕ is a quasiisometric embedding, we can ﬁnd a ﬁnite subset F ⊂ H such that ϕ(g1 )−1 ϕ(g2 ) ∈ F whenever g1 , g2 ∈ G satisfy g1−1 g2 ∈ S. Let L denote the subgroup of H generated by {ϕ(1G )} ∪ F . Suppose that k ∈ K. Since S is a symmetric generating subset of K, there exist an integer n ≥ 0 and elements s1 , s2 , . . . , sn ∈ S such that k = s1 s2 · · · sn . Consider the elements k0 , k1 , . . . , kn ∈ K deﬁned by k0 = 1G and ki = ki−1 si for all 1 ≤ i ≤ n. Observe that kn = k and that −1 ϕ(ki−1 )−1 ϕ(ki ) ∈ F , for all 1 ≤ i ≤ n, since ki−1 ki = si ∈ S. Thus, we have ϕ(k) = ϕ(1G )(ϕ(k0 )−1 ϕ(k1 ))(ϕ(k1 )−1 ϕ(k2 )) · · · (ϕ(kn−1 )−1 ϕ(kn )) ∈ L. It follows that ϕ(K) ⊂ L. As H is locally ﬁnite, the subgroup L is ﬁnite. On the other hand, it follows from Proposition 6.13.17 that there exists an integer M ≥ 1 such that ϕ−1 (h) ≤ M for all h ∈ H. We deduce that K ≤ M L < ∞. This shows that G is locally ﬁnite. Corollary 6.13.21. Let G and H be two groups. Suppose that G and H are quasiisometric and that the group G is locally ﬁnite. Then H is locally ﬁnite. Lemma 6.13.22. Let G be a group. Let E, Ω and C be three subsets of G. Then, for every c0 ∈ C one has ∂Ec0 (Ω) = ∂E (Ωc−1 0 ).
(6.102)
and (6.103) ∂EC (Ω) ⊃ ∂Ec0 (Ω). −1 Proof. By (5.3) we have Ω +Ec0 = e∈E Ω(ec0 )−1 = e∈E Ωc−1 = 0 e −1 +E −Ec0 −1 and from (5.2) we deduce that Ω = = (Ωc0 ) e∈E Ω(ec0 ) −1 −1 −1 −E +Ec0 −Ec0 Ωc e = (Ωc ) . It follows that ∂ (Ω) = Ω \ Ω = Ec 0 0 0 e∈E −1 −E −1 +E (Ωc−1 ) \ (Ωc ) = ∂ (Ωc ). Similarly, we have E 0 0 0
212
6 Finitely Generated Amenable Groups
Ω +EC =
Ω(ec)−1 =
e∈E c∈C
Ωc−1 e−1 ⊃
e∈E c∈C
−1 +E Ωc−1 = (Ωc−1 0 e 0 )
e∈E
and Ω −EC =
e∈E c∈C
Ω(ec)−1 =
Ωc−1 e−1 ⊂
e∈E c∈C
−1 +E Ωc−1 = (Ωc−1 . 0 e 0 )
e∈E
+E −E We deduce that ∂EC (Ω) = Ω +EC \ Ω −EC ⊃ (Ωc−1 \ (Ωc−1 = 0 ) 0 ) −1 ∂E (Ωc0 ) = ∂Ec0 (Ω), where the last equality follows from (6.102).
Theorem 6.13.23. Let G and H be two quasiisometric groups. Suppose that H is amenable. Then G is amenable. Proof. Let ϕ : G → H be a quasiisometry and let C ⊂ H be a ﬁnite set such that H = ϕ(G)C = ϕ(G)c. (6.104) c∈C
Let EG ⊂ G be a ﬁnite set and ε > 0. Let us show that there exists a ﬁnite set FG ⊂ G such that ∂EG (FG ) < εFG . (6.105) ⊂ H such Since ϕ is a quasiisometric embedding, we can ﬁnd a ﬁnite set EH that (6.106) g1−1 g2 ∈ EG ⇒ ϕ(g1 )−1 ϕ(g2 ) ∈ EH C. Also, by Proposition 6.13.17 we for all g1 , g2 ∈ G. We then set EH = EH can ﬁnd an integer M ≥ 1 such that
ϕ−1 (F ) ≤ M F 
(6.107)
for all ﬁnite sets F ⊂ H. Since H is amenable, it follows from Corollary 5.4.5 ⊂ H such that that we can ﬁnd a ﬁnite subset FH ) < ∂EH (FH
ε F . M C H
(6.108)
By (6.104) we can ﬁnd c0 ∈ C such that −1 c0 ∩ ϕ(G) = ∅ FH
and
(6.109)
−1 −1 FH c ∩ ϕ(G) ≤ FH c0 ∩ ϕ(G) −1 FH c0
⊂ H and FG = ϕ for all c ∈ C. Set FH = FG = ∅ by (6.109). Then we have ϕ(FG ) = FH ∩ ϕ(G) ⊂ FH
−1
(6.110)
(FH ) ⊂ G. Note that (6.111)
6.13 QuasiIsometries
213
and ϕ(G \ FG ) ⊂ H \ FH .
(6.112)
Moreover,
FH  ≤ C · FG . (6.113) = c∈C (FH ∩ ϕ(G)c) so that Indeed, from (6.104) we deduce that FH FH ≤
FH ∩ ϕ(G)c
c∈C
=
−1 FH c ∩ ϕ(G)
c∈C
≤
−1 FH c0 ∩ ϕ(G) (by (6.110))
c∈C
= C · FH ∩ ϕ(G) ≤ C · FG  (by the equality in (6.111)). Let us show that the set FG ⊂ G has the required property. Suppose that g ∈ ∂EG (FG ). This means that the set gEG meets both FG and G \ FG . Thus, there exist g1 ∈ FG and g2 ∈ G \ FG such that g −1 g1 ∈ EG and g −1 g2 ∈ EG . This implies ϕ(g)−1 ϕ(g1 ) ∈ EH and ϕ(g)−1 ϕ(g2 ) ∈ EH by applying (6.106). As ϕ(g1 ) ∈ FH (by (6.111)) and ϕ(g2 ) ∈ H \ FH (by (6.112)), we deduce meets both FH and H \ FH . In other words, we have that the set ϕ(g)EH (FH ). This shows that ϕ(g) ∈ ∂EH (FH )). ∂EG (FG ) ⊂ ϕ−1 (∂EH
(6.114)
By taking cardinalities, we ﬁnally get (FH )) (by (6.114)) ∂EG (FG ) ≤ ϕ−1 (∂EH (FH ) (by (6.107)) ≤ M ∂EH
c (F = M ∂EH H ) (by (6.102)) 0 ≤ M ∂EH (FH ) (by (6.103))
≤
ε F  (by (6.108)) C H
≤ εFG  (by (6.113)). This shows that FG satisﬁes (6.105). From Corollary 5.4.5 we deduce that G is amenable.
214
6 Finitely Generated Amenable Groups
Notes The idea of looking at a ﬁnitely generated group with a geometer’s eye by investigating the properties of the family consisting of all its word metrics is due to M. Gromov ([Gro1, Gro3, Gro4]) and gave birth to the ﬂourishing branch of mathematics which is commonly known as geometric group theory in the late 1970s. In the 1950s, the notion of growth of a ﬁnitely generated group arose in group theory in relation to volume growth in Riemannian manifolds. This ˇ line of study was initiated by V.A. Efremovich [Efr] and A.S. Svarc [Sva] in the USSR and, slightly later and completely independently, by J. Milnor [Mil1] and J.A. Wolf [Wol] in the USA. In [Mil1] Milnor proved that fundamental groups of closed Riemannian manifolds with negative sectional curvature have exponential growth. Wolf [Wol] proved that a polycyclic group has polynomial growth if it contains a nilpotent subgroup of ﬁnite index and has exponential growth otherwise. Then, Milnor [Mil2] proved that every ﬁnitely generated nonpolycyclic solvable group has exponential growth. Finally, in 1972 H. Bass [Bas] showed that the growth of a nilpotent group G with a ﬁnite symmetric generating subset S is exactly polynomial in the sense that there are positive constants C1 and C2 such that C1 nd ≤ γS (n) ≤ C2 nd , for all n ≥ 1, where d = d(G) ≥ 0 is an integer which can be computed explicitly from the lower central series of G (see [Har1, page 201] for more information on the history and prehistory of these results). Note that the growth estimates of Milnor and Bass imply that a ﬁnitely generated solvable group has either polynomial or exponential growth. It was shown by J. Tits [Tits] (see also [Har1]) that every ﬁnitely generated linear group either is virtually nilpotent or contains a free subgroup of rank two. This last result, which is known as the Tits alternative for linear groups, implies that every ﬁnitely generated linear group has either polynomial growth or exponential growth. The problem of the characterization of ﬁnitely generated groups with polynomial growth remained open until Gromov proved in [Gro2] that a ﬁnitely generated group with polynomial growth contains a nilpotent subgroup of ﬁnite index. It follows from the above mentioned result of Bass and Proposition 6.6.6 that a group of polynomial growth has in fact exactly polynomial growth. Thus, for ﬁnitely generated groups, the notions of polynomial and exactly polynomial growth coincide. The (general) Burnside problem, posed by W. Burnside in 1902, asked whether a ﬁnitely generated periodic group is necessarily ﬁnite. It was answered in the negative in 1964 by E.S. Golod and I.R. Shafarevich [GolS], who gave an example of a ﬁnitely generated inﬁnite pgroup. The Grigorchuk group, also known as the ﬁrst Grigorchuk group, was originally constructed by R. I. Grigorchuk in 1980 [Gri2] as a new example of a ﬁnitely generated inﬁnite periodic group, thus providing another counterexample to the general Burnside problem. In 1984 Grigorchuk [Gri4] proved that this group has intermediate growth (this was announced by Grigorchuk in 1983 [Gri3]),
Notes
215
thus providing a positive answer to the Milnor problem, posed by Milnor in 1968, about the existence of ﬁnitely generated groups of intermediate growth. More precisely, in [Gri4] Grigorchuk proved, among other things, √ that exp( n) γ(G) exp(ns ), where s = log32 (31) ≈ 0.991. The Grigorchuk group also provides the ﬁrst example of an amenable but not elementary amenable group (the class of elementary amenable groups is the smallest class of groups containing all ﬁnite and all abelian groups that is closed under taking subgroups, quotients, extensions, and directed unions), thus answering a question posed by Day in 1957 [Day1]. Among other interesting properties of the Grigorchuk group G, we mention the following (see [CMS2], [Har2], [Gri5], [GriP]): (a) G is not ﬁnitely presented (a recursive set of deﬁning relations for G was found by I.G. Lys¨enok [Lys]), (b) G is just inﬁnite (it is inﬁnite but every proper quotient is ﬁnite), (c) G has solvable word problem, that is, there exists an algorithm that establishes whether, given s1 , s2 , . . . , sn ∈ {a, b, c, d}, one has s1 s2 · · · sn = 1G or not. Originally, the Grigorchuk group was deﬁned as a group of Lebesgue measurepreserving transformations of the unit interval. By representing the elements of Σ ∗ as the vertices of an inﬁnite binary rooted tree, the Grigorchik group may be also realized as a subgroup of the full automorphism group of the tree. Another description of this group was provided by regarding it as a group generated by a ﬁnite automaton (see [GriNS]). Simple random walks on groups were ﬁrst considered by H. Kesten in [Kes1]. Given a ﬁnitely generated group G and a ﬁnite symmetric generating subset S ⊂ G, the simple random walk on G relative to S is the Ginvariant Markov chain with state space G and transition probabilities given by 1 if g −1 h ∈ S p(g, h) = S 0 otherwise. This can be interpreted as follows: a “random walker” on G moves from a group element g with equal probability to one of its S neighbors gs, where s ∈ S. For g, h ∈ G, denote by p(n) (g, h) the probability of reaching h from g after exactly n steps. We then have p(0) (g, h) = δg,h where δg,h is the Kronecker symbol, p(1) (g, h) = p(g, h) and, more generally, p(n) (g, h) = p(n−1) (g, k)p(k, h). k∈G
The quantity p(n) (g, g) does not depend on g ∈ G and is called the return probability after n steps. The number ρ(G, S) = lim sup
n
p(n) (g, g)
(6.115)
n→∞
is called the spectral radius of the simple random walk on G relative to S. Kesten [Kes1, Kes2] proved that one always has
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2 S − 1 ≤ ρ(G, S) ≤ 1 S with equality on the right if and only if G is amenable. Moreover, if S contains no involutions, equality on the left holds if and only if G is a free group and S is the symmetrization of a free base. Theorem 6.12.9 is just an 2 reformulation of the amenability criterion of Kesten’s theorem. Kesten also proved that if S is a ﬁnite symmetric generating subset of a group G and N ⊂ G is a normal subgroup, then, denoting by G = G/N (resp S ⊂ G) the corresponding quotient group (resp. generating subset of G) then ρ(G, S) ≤ ρ(G, S) with equality if and only if N is amenable. Day [Day3] extended Kesten’s amenability criterion to non symmetric random walks. The associated Markov chain is then determined by a probability density whose support generates the group. In this setting, the associated Markov operator on 2 (G) is no more selfadjoint, in general. The key ingredient of this new proof is the uniform convexity (uniform rotundity in Day’s terminology) of Hilbert spaces (cf. Lemma I.4.2) and more generally of p spaces with p > 1 (note that in fact Day considers, more generally, Markov operators on the Banach spaces p (G), for p > 1). For more on this we refer to the paper [KaV] by V.A. Kaimanovich and A.M. Vershik and to W. Woess’ review [Woe1] and monograph [Woe2]. Another important criterion for amenability of ﬁnitely generated groups has been obtained by R.I. Grigorchuk [Gri1]. Let G be a group with m generators. Then G is isomorphic to F/N where F is the free group on m generators and N ⊂ F is a normal subgroup. Let α = α(G; F, N ) be deﬁned by setting α = lim sup n w(n), n→∞
where w(n) equals the number of elements in N at distance at most n from the identity element 1F in the free group F . The nonnegative number α is called the cogrowth of G relative to the presentation G = F ; N . Grigorchuk [Gri1] proved that either α = 1 (this holds if and only if N = {1F }) or √ 2m − 1 ≤ α ≤ 2m − 1. (6.116) He also showed that if ρ = ρ(G, S) is the spectral radius of the simple random walk on G relative to S (the symmetrization of the image of a free base of F under the canonical quotient homomorphism F → G = F/N ), then the following relation holds: ⎧√ √ ⎨ 2m−1 if 1 ≤ α ≤ 2m − 1 m √ (6.117) ρ= √ √ 2m−1 ⎩ 2m−1 √ α if 2m − 1 ≤ α ≤ 2m − 1. + 2m α 2m−1 Then, from (6.117) and Kesten’s criterion he deduced that in (6.116) equality holds on the right if and only if G is amenable. This is called the Grigorchuk
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criterion (or the cogrowth criterion) of amenability. Grigorchuk’s criterion was used by Ol’shanskii [Ols] to show the existence of nonamenable groups without nonabelian free subgroups and by Adyan [Ady] to show that the free Burnside groups B(m, n), with m ≥ 2 generators and exponent n ≥ 665 odd, are nonamenable. The program of the classiﬁcation of all ﬁnitely generated groups up to quasiisometries was posed and initiated by Gromov [Gro4]. The deﬁnition of quasiisometry presented here is modeled after Y. Shalom [Sha].
Exercises 6.1. Let S = {s1 , s2 , . . . , sn } be a ﬁnite subset of Z. Show that S generates Z if and only if gcd(s1 , s2 , . . . , sn ) = 1. 6.2. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. Let x ∈ G and denote by Lx , Rx : G → G the maps deﬁned by Lx (g) = xg and Rx (g) = gx for all g ∈ G. (a) Show that the map g → dS (g, Rx (g)) is constant on G. (b) Show that if x is in the center of G, then Rx is an isometry of (G, dS ). (c) Show that supg∈G dS (g, Lx (g)) < ∞ if and only if the conjugacy class of x in G is ﬁnite. 6.3. Suppose that S and S are two ﬁnite symmetric generating subsets of a group G with S ⊂ S . Show that one has: (i) S (g) ≥ S (g) and dS (g, h) ≥ dS (g, h) for all g, h ∈ G; (ii) BS (n) ⊂ BS (n) and γS (n) ≤ γS (n) for all n ∈ N; (iii) λS ≤ λS . 6.4. Let G1 and G2 be two ﬁnitely generated groups and let G = G1 ×G2 . Let S1 (resp. S2 ) be a ﬁnite symmetric generating subset of G1 (resp. G2 ). Show that S = (S1 × {1G2 }) ∪ ({1G1 } × S2 ) is a ﬁnite symmetric generating subset G1 G2 of G and that one has G S (g) = S1 (g1 ) + S2 (g2 ) for all g = (g1 , g2 ) ∈ G. 6.5. Let S1 and S2 be two sets. For i = 1, 2, let Qi = (Qi , Ei ) be an Si labeled graph. We deﬁne their direct product Q1 × Q2 as the Slabeled graph Q = (Q, E) with: ! ! (1) S = S1 S2 , where denotes the disjoint union; (2) Q = Q1 × Q2 ; (3) E = {((q1 , q2 ), s, (q1 , q2 )) : either q1 = q1 and (q2 , s, q2 ) ∈ E2 , or q2 = q2 and (q1 , s, q1 ) ∈ E1 }. Suppose that S1 (resp. S2 ) is endowed with an involution ι1 : S1 → S1 (resp. ι2 : S2 → S2 ) and that Q1 (resp. Q2 ) is edgesymmetric with respect to such involution. Denote by ι : S → S the map deﬁned by ι(s) = ιi (s) if s ∈ Si , i = 1, 2, and observe that ι is an involution. Show that Q is edgesymmetric with respect to ι.
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6.6. Let G1 and G2 be two ﬁnitely generated groups and let S1 ⊂ G1 and / S1 S2 ⊂ G2 be two ﬁnite and symmetric generating subsets such that 1G1 ∈ and 1G2 ∈ / S2 . Consider the direct product group G = G1 × G2 together with the (ﬁnite and symmetric) generating subset S = (S1 ×{1G2 })∪({1G1 }×S2 ). Denote by CS1 (G1 ), CS2 (G2 ) and CS (G) the corresponding Cayley graphs. If we identify S1 with S1 × {1G2 } (resp. S2 with {1G1 } × S2 ), we may regard CS1 (G1 ) (resp. CS2 (G2 )) as an (S1 ×{1G2 })labeled graph (resp. ({1G1 }×S2 )labeled graph). Show that CS (G) = CS1 (G1 ) × CS2 (G2 ). 6.7. Let G = Z, S = {1, −1} and S = {2, −2, 3, −3}. Find the best possible positive constants C1 and C2 such that C1 S (g) ≤ S (g) ≤ C2 S (g) for all g ∈ G. Hint: Check that C1 = 1/3 and C2 = 2. 6.8. Let G = Zm , where m ≥ 1 is an integer. Consider the ﬁnite and symmetric generating subset S = {±(1, 0, 0, . . . , 0), ±(0, 1, 0, . . . , 0), . . . , ±(0, 0, . . . , 0, 1)} ⊂ Zm . (a) Show that if g = (a1 , a2 , . . . , am ) ∈ Zm then S (g) = a1  + a2  + · · · + am . (b) Let n ∈ N. Set P0 (n) = 1 and, for all integers t ≥ 1 denote by Pt (n) the number of distinct ttuples (a1 , a2 , . . . ,at ) of positive integers such that a1 + a2 + · · · + at ≤ n. Show that Pt (n) = nt for 1 ≤ t ≤ n. Hint: The map (a1 , a2 , . . . , at ) → {a1 , a1 + a2 , . . . , a1 + a2 + · · · + at } establishes a bijection between the set {(a1 , a2 , . . . , at ) ∈ Nt : ai ≥ 1 and a1 + a2 + · · · + at ≤ n} and the set of all subsets of cardinality t of the set {1, 2, . . . , n}. (c) For n, t ∈ N and t ≥ 1 denote by Nt (n) the number of all mtuples m (a1 , a2 , . . . , am ) ∈ Zm with i=1 ai  ≤ n and exactly t many of the ai ’s m m nonzero. Show that γSZ (n) = t=0 Nt (n). (d) Let 0 ≤ t ≤ n and let I be a subset of {1, 2, . . . , n} such that I = t. Show that there are precisely Pt (n) distinct elements g = (a1 , a2 , . . . , am ) ∈ Nm with I = {i : ai > 0} and such that S (g) ≤ n. n (e) Deduce from (d) that there are exactly m t t elements g = (a1 , a2 , . . . , (g) ≤ n. am ) ∈ Nm with {i : ai > 0} = t such that nS . (f) Deduce from (e) that Nt (n) = 2t m t t n m m (g) Deduce from (f) and (c) that γSZ (n) = t=0 2t m t t . 6.9. Suppose that γ, γ : N → [0, +∞) are two growth functions such that γ γ . Show that lim supn→∞ n γ(n) ≤ lim supn→∞ n γ (n). 6.10. Let α = 21 11 ∈ SL2 (Z). Consider the metabelian group G = Z2 α Z, that is, the semidirect product of Z2 by the inﬁnite cyclic subgroup of SL2 (Z) generated by α. Recall that
x G= , z : x, y, z ∈ Z y
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with the multiplication deﬁned by x1 x2 x1 z1 x2 , z1 , z2 = +α , z1 + z2 y1 y2 y1 y2 for all x1 , x2 , y1 , y2 , z1, z2 ∈ Z. for all n ∈ N, where (fk )k∈N is the (a) Show that αn 10 = f2n+1 f2n Fibonacci sequence which is inductively deﬁned by f0 = 0, f1 = 1, and fk = fk−2 + fk−1 for all k ≥ 2. (b) Deduce from (a) that, for any integer n ≥ 1, the set n i−1 1 , 0 : ui ∈ {0, 1} for 1 ≤ i ≤ n ⊂ G ui α A(n) = 0 i=1
has cardinality A(n) = 2n . (c) Consider the subset S ⊂ G deﬁned by S = {a, b, c, a−1 , b−1 , c−1 }, where 0 0 1 a= ,0 ,b = , 0 and c = ,1 . 0 1 0 Show that S is a ﬁnite symmetric generating subset of G and that one has A(n) ⊂ BSG (3n − 2) for all n ≥ 1. (d) Deduce from (b) and (c) that G has exponential growth. 6.11. Let n ≥ 2. Show that GLn (Z) is a ﬁnitely generated group of exponential growth. Hint: Use Exercise 2.18, Lemma 2.3.2 and Corollary 6.6.5. 6.12. Let F2 denote the free group of rank two. Show that the groups GL2 (Z) Use elementary row operations to show that and F2 are commensurable. Hint: the matrices 10 21 and 12 01 generate a ﬁnite index subgroup of GL2 (Z) and apply Lemma 2.3.2. 6.13. Growth of the BaumslagSolitar group BS(1, m). Let m be an integer such that m ≥ 2. Prove that the metabelian group G = a, b : aba−1 = bm studied in Exercises 2.7 and 4.21 has exponential growth. Hint: Use an argument similar to the one used for the case m = 2 in the proof of Proposition 6.7.1. More precisely, take S = {a, b, a−1 , b−1 } and prove that every element of the form g = bk , where 0 ≤ k ≤ mn − 1 and n ≥ 1, has word length S (g) ≤ mn + n − 2 by developing k in base m. 6.14. Aﬃne representation of the BaumslagSolitar group BS(1, m). Let m be an integer such that m ≥ 2. Consider the group G given by the presentation G = a, b : aba−1 = bm (cf. Exercises 2.7, 4.21 and 6.13). (a) Let α, β : R → R be the maps respectively deﬁned by α(x) = mx and β(x) = x + 1 fro all x ∈ R. Show that there is a unique homomorphism ϕ : G → Sym(R) satisfying ϕ(a) = α and ϕ(b) = β.
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(b) Show that ϕ is injective. Hint: Use Exercise 2.7(a). (c) Let n0 be an integer such that mn0 ≥ 3. Consider the elements λ, μ ∈ Sym(R) respectively deﬁned by λ = α−n0 and μ = βλβ −1 . Check that the open intervals i = (−1/2, 1/2) and J = (1/2, 3/2) satisfy λ(J) ⊂ I and μ(I) ⊂ J. (d) Let n ≥ 1 be an integer and let σ1 , σ2 , . . . , σn ∈ {λ, μ}. Prove that σ1 σ2 · · · σn = IdR . Hint: Use (c) and play pingpong as in the proof of Theorem D.5.1. (e) Use (a), (b) and (d) to get another proof of the fact that G has exponential growth. 6.15. Growth of the lamplighter group. Let L = (Z/2Z) Z denote the lamplighter group (cf. Exercise 4.19). Recall that L is the semidirect product of a normal subgroup H = ⊕n∈Z An , where each An is a subgroup of order 2, with an inﬁnite cyclic subgroup N generated by an element t which satisﬁes tat−1 = (an−1 )n∈Z for all a = (an )n∈Z ∈ H. Let s denote the nontrivial element of A0 . (a) Show that S = {s, t, t−1 } is a symmetric generating subset of L. (b) For n ≥ 1, let Bn = ⊕n−1 k=0 Ak . Prove that Bn is a subgroup of H generated by the elements s, tst−1 , t2 st−2 , . . . , tn−1 st−n+1 and that Bn  = 2n . (c) Deduce from (b) that 2n ≤ γSL (3n − 2) for all n ≥ 1. (d) Deduce from (c) that L has exponential growth. 6.16. Growth of the integral Heisenberg group. Let G = HZ denote the Heisenberg group over the ring of integers (cf. Example 4.6.5). Recall that G is the subgroup of SL3 (Z) consisting of all matrices of the form ⎛ ⎞ 1yz M (x, y, z) = ⎝0 1 x⎠ (x, y, z ∈ Z). 001 Let us set A = M (1, 0, 0), B = M (0, 1, 0), C = M (0, 0, 1), and S = {A, A−1 , B, B −1 , C, C −1 }. (a) Verify that M (x, y, z) = Ax B y C z for all x, y, z ∈ Z. (b) Show that S is a ﬁnite symmetric generating subset of G. (c) Show that C x Ay = Ay C x , C x B y = B y C x , and B x Ay = Ay B x C xy for all x, y, z ∈ Z. (d) Deduce from (c) that if P ∈ G satisﬁes S (P ) ≤ n, then there exist x, y, z ∈ Z with x ≤ n, y ≤ n, and z ≤ n2 + n, such that P = Ax B y C z . (e) Deduce from (d) that there exists a constant C1 > 0 such that γS (n) ≤ C1 n4 for all n ≥ 1. (f) Let n, x, y, z be integers such that 0 ≤ x ≤ n, 0 ≤ y ≤ n, and 0 ≤ z ≤ n2 . Show that there exist integers q, r with 0 ≤ q ≤ n and 0 ≤ r ≤ n − 1 such that
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Ax B y C z = Ax−q B n Aq B y−n C r . Hint: Use (c) and Euclidean division of z by n. (g) Deduce from (f) that γS (5n) ≥ (n + 1)2 (n2 + 1) for all n ≥ 0. (h) Deduce from (g) that there exists a constant C2 > 0 such that γS (n) ≥ C2 n4 for all n ≥ 0. (i) Deduce from (e) and (h) that γ(G) ∼ n4 . 6.17. Let G be the Grigorchuk group. (a) Show that G acts transitively on Σ n for all n ∈ N. Hint: Use induction on n. More precisely, let w1 , w2 ∈ Σ n and suppose ﬁrst that w1 and w2 start with the same letter, that is, there exist x ∈ Σ and u1 , u2 ∈ Σ n−1 such that w1 = xu1 and w2 = xu2 . Use induction and Proposition 6.9.7(i). Otherwise, if w1 and w2 do not start with the same letter, observe that w1 and a(w2 ) do start with the same letter and reduce to the previous case. (b) Use (a) to recover the fact that G is inﬁnite (cf. Theorem 6.9.8). 6.18. Let K1 = Z/2Z = {0, 1} and, for n ≥ 2, deﬁne by induction Kn = Kn−1 (Z/2Z). Recall that Kn−1 (Z/2Z) = (Kn−1 )Z/2Z (Z/2Z), so that Kn consists of the elements (f, a) ∈ (Kn−1 )Z/2Z × (Z/2Z) with the multiplication deﬁned by (f1 , a1 )(f2 , a2 ) = (f1 f2a1 , a1 + a2 ), for all f1 , f2 ∈ (Kn−1 )Z/2Z and a1 , a2 ∈ Z/2Z, where f a (a ) = f (a + a ) ∈ Kn−1 for all f ∈ (Kn−1 )Z/2Z and a, a ∈ Z/2Z. The group Kn is called the Kaloujnine 2group of degree n. Set Σ = {0, 1}. For n = 1 and x ∈ Σ, we set 1(x) = 1 − x, and 0(x) = x. For n ≥ 2 let g = (f, a) ∈ Kn and w = xu ∈ Σ n , where x ∈ Σ and u ∈ Σ n−1 . We then set g(w) = a(x)f (x)(u) ∈ Σ n (note that a(x) ∈ Σ, f (x) ∈ Kn−1 , and f (x)(u) ∈ Σ n−1 is deﬁned by induction). (a) Show that the map Kn × Σ n (g, w) → g(w) ∈ Σ n deﬁnes an action of the group Kn on Σ n . (b) Show that this action is faithful. (c) By virtue of (b), we may regard Kn as a subgroup of Sym(Σ n ). Show, by simple counting arguments, that Kn is a Sylow 2subgroup of Sym(Σ n ). (d) Consider the elements g1,n , g2,n , . . . , gn,n ∈ Kn deﬁned by induction as follows. First deﬁne g1,1 = 1 ∈ K1 = Z/2Z and then, for 1 ≤ m ≤ n, set g1,m = (f1,m , 1) ∈ Km , where f1,m : Z/2Z → Km−1 is given by f1,m (a) = 1Km−1 for all a ∈ Z/2Z. Finally, for 2 ≤ k ≤ m ≤ n, set gk,m = (fk,m , 0) ∈ Km , where fk,m : Z/2Z → Km−1 is given by fk,m (0) = gk−1,m−1 and fk,m (1) = (f1,m−1 , 0). Show that g1,n , g2,n , . . . , gn,n generate Kn and verify that gi,n (uxv) = u(1 − x)v ∈ Σ n for all u ∈ Σ i−1 , x ∈ Σ and v ∈ Σ n−i . (e) Deﬁne a map πn : Kn → (Z/2Z)n by induction as follows. π1 : K1 = Z/2Z → Z/2Z is the identity map, while, for n ≥ 2 and (f, a) ∈ Kn we set πn (f, a) = ( b∈(Z/2Z) πn−1 (f (b)), a) ∈ (Z/2Z)n−1 ×(Z/2Z) = (Z/2Z)n for all f ∈ (Kn−1 )Z/2Z and a ∈ Z/2Z. Show that πn is a surjective homomorphism. use induction on n and the fact Hint: To prove that πn is a homomorphism that b∈(Z/2Z) πn−1 (f a (b)) = b∈(Z/2Z) πn−1 (f (b)) for all f ∈ (Kn−1 )Z/2Z
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and a ∈ Z/2Z. To show surjectivity, look at the πn images of the elements g1,n , g2,n , . . . , gn,n ∈ Kn deﬁned in (d). (f) Deduce from (e) that Kn /[Kn , Kn ] (the abelianization of Kn ) is isomorphic to (Z/2Z)n . (g) Consider the map Φn : Σ n → {1, 2, . . . , 2n } given by expansion in base two, that is, Φn (i1 , i2 , . . . , in ) = i1 + 2i2 + 4i3 + · · · + 2n−1 in , for all i1 , i2 , . . . , in ∈ Σ. Verify that, modulo the map Φ3 , one has g1,3 = (1 5)(2 6) × (3 7)(4 8), g2,3 = (1 3)(2 4), and g3,3 = (1 2). 6.19. Let G be the Grigorchuk group. (a) Consider the homomorphism Ψ3 : G → Sym(8) deﬁned by Ψ3 (g) = gΣ 3 (observe that the map Ψ3 is well deﬁned since (g(w)) = (w) for all w ∈ Σ ∗ and g ∈ G). With the notation from Exercise 6.18(f), verify that Ψ3 (a) = (1 5)(2 6)(3 7)(4 8), Ψ3 (b) = (1 3)(2 4)(5 6), Ψ3 (c) = (1 3)(2 4) and Ψ3 (d) = (5 6). (b) Deduce from (a) and Exercise 6.18(e) that Ψ3 (G) = K3 , where K3 ⊂ Sym(8) is the Kaloujnine group (cf. Exercise 6.18). (c) By applying Proposition 6.9.3, deduce from (b) and Exercise 6.18(e) that G/[G, G] (the abelianization of G) is isomorphic to (Z/2Z) × (Z/2Z) × (Z/2Z). 6.20. The word problem for the Grigorchuk group. Let G be the Grigorchuk group. Describe an algorithm which, given any word w ∈ {a, b, c, d}∗ , determines whether w represents the identity element 1G or not. Hint: Given a word w, ﬁrst count the number a (w) of occurrences of the letter a in w. Prove that if a (w) is odd then w does not represent 1G . If a (w) is even, use the maps φ0 , φ1 : H1 → G and apply induction on the length of the word w. 6.21. Let us say that a net (xi )i∈I in a set X converges to inﬁnity if, for /F every ﬁnite subset F ⊂ X, there exists an element i0 ∈ I such that xi ∈ for all i ≥ i0 . Let ϕ : G → H be a map from a group G into a group H. Show that ϕ is a quasiisometric embedding if and only if it satisﬁes the following condition: for any two nets (ui )i∈I and (vi )i∈I in G having the same index set, the net (u−1 i vi )i∈I converges to inﬁnity in G if and only if the net (ϕ(ui )−1 ϕ(vi ))i∈I converges to inﬁnity in H. 6.22. Let G be a group and let E(G) denote the set consisting of all quasiisometries ϕ : G → G. Deﬁne a binary relation ∼ in E(G) by declaring that ϕ1 and ϕ2 ∈ E(G) satisfy ϕ1 ∼ ϕ2 if and only if there exists a ﬁnite subset F ⊂ G such that ϕ1 (g)−1 ϕ2 (g) ∈ F for all g ∈ G. (a) Show that ∼ is an equivalence relation in E(G). (b) Show that if ϕ ∈ E(G) then there exists ψ ∈ E(G) such that ϕ ◦ ψ ∼ IdG and ψ ◦ ϕ ∼ IdG . (b) Show that the composition of maps in E(G) is compatible with ∼ and induces a group structure on the quotient set QI(G) = E(G)/ ∼. (c) Show that if G is a ﬁnite group then the group QI(G) is trivial.
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(d) Show that the group QI(Z) contains a subgroup isomorphic to R×Z/2Z and is therefore uncountable. Hint: Prove that the map ϕλ : Z → Z deﬁned by ϕλ (x) = [λx], where λ ∈ R \ {0} and [α] denotes the integral part of α, that is, the largest integer n such that n ≤ α, is a quasiisometry and that one has fλ1 ∼ fλ2 if and only if λ1 = λ2 . (e) Show that if G and H are quasiisometric groups then the groups QI(G) and QI(H) are isomorphic. 6.23. Let A be a set and let G = (Q, E) be a ﬁnite Alabeled graph. Denote by λ : E → A the labeling map deﬁned by λ(e) = a for every edge e = (q, a, q ) ∈ E. A biinﬁnite path in G is a sequence π = (en )n∈Z of edges en = (qn , an , qn ) ∈ E such that qn = qn+1 for all n ∈ Z. We deﬁne the label of a biinﬁnite path π = (en )n∈Z as being the element λ(π) ∈ AZ given by λ(π)(n) = λ(en ) for all n ∈ Z. (a) Denote by X G the set of the labels λ(π) of all biinﬁnite paths π in G. Show that X G is a subshift of AZ . It is called the subshift deﬁned by the Alabeled graph G. (b) Show that if G is connected then the subshift X G is irreducible. 6.24. Let A = {0, 1} and consider the Alabeled graphs G1 and in G2 in Fig. 6.18. Check that the associated subshifts X G1 and X G2 are the even subshift (cf. Exercise 1.38) and the golden mean subshift (cf. Exercise 1.39) respectively.
Fig. 6.18 The Alabeled graphs G1 and G2
6.25. Let A be a set and let X ⊂ AZ be a subshift of ﬁnite type. Let M be a positive integer such that {1, 2, . . . , M } ⊂ Z is a memory set for X. Consider the Alabeled graph G = G(X, M ) = (Q, E) deﬁned as follows: Q = LM −1 (X) is the set of all Xadmissible words of length M − 1, and the edge set E consists of all triples e = (aw, a, wa ) ∈ Q × A × Q, where a, a ∈ A and w ∈ AM −2 are such that awa ∈ LM (X). (a) Check that X G = X. (b) Show that X is irreducible if and only if G is connected (cf. Exercise 6.23). 6.26. Let A be a set and let X ⊂ AZ be a subshift of ﬁnite type. Let also τ : AZ → AZ be a cellular automaton. Let M be a positive integer such that {1, 2, . . . , M } ⊂ Z is a memory set for both X and τ , and let μ : AM → A denote the corresponding local deﬁning map for τ . Consider the Alabeled
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graph G = G(X, τ, M ) = (Q, E), where Q = LM −1 (X) is the set of all Xadmissible words of length M − 1 and the edge set E consists of all triples e = (aw, μ(awa ), wa ) ∈ Q × A × Q, where a, a ∈ A and w ∈ AM −2 are such that awa ∈ LM (X). Check that X G = τ (X). 6.27. Life on Z. Let A = {0, 1} and consider the cellular automaton τ : AZ → AZ deﬁned in Exercise 5.3. Let G be the Alabeled graph in Fig. 6.19. Check that X G = τ (AG ).
Fig. 6.19 The Alabeled graph G = G(A, τ, 3)
6.28. Let A = {0, 1} and let τ : AZ → AZ be the majority action cellular automaton (cf. Example 1.4.3(c)) associated with the set S = {−1, 0, 1}. Let G be the Alabeled graph in Fig. 6.20. Check that X G = τ (AG ).
Fig. 6.20 The Alabeled graph G = G(A, τ, 3)
6.29. Let A be a set and let G = (Q, E) be a ﬁnite Alabeled graph. (a) Suppose that for each pair (a, a ) ∈ A2 there exists at most one vertex q ∈ Q which is both the terminal vertex of an edge with label a and the
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initial vertex of an edge with label a . Show that the subshift X G ⊂ AZ is of ﬁnite type. Hint: Show that in fact {0, 1} ⊂ Z is a memory set for X G . (b) Show that the hypothesis in (a) is satisﬁed if the labeling map λ : E → A is injective. 6.30. Let A be a set and let G = (Q, E) be a ﬁnite Alabeled graph. We recall that given an edge e = (q, a, q ) ∈ E we denote by α(e) = q ∈ Q (resp. ω(e) = q ∈ Q) the initial (resp. terminal) vertex of e. Consider the Elabeled graph G = (Q , E ), where Q = Q and E = {(α(e), e, ω(e)) ∈ Q × E × Q : e ∈ E}. Note that the labeling map λ : E → E on G is bijective. Identify the set {λ(e) : e ∈ E} ⊂ A formed by the labels of the edges of G with a subset of E in an arbitrary way and consider the cellular automaton τ : E Z → E Z with memory set S = {0} and local deﬁning map μ : E S → E deﬁned by μ(e) = λ(e) for all e ∈ E S = E. Observe that τ (x) ∈ AZ for all x ∈ E Z and show that τ (X G ) = X G . 6.31. Let A be a set and X ⊂ AZ a subshift. Show that the following conditions are equivalent: (i) there exists a ﬁnite Alabeled graph G such that X = X G . (ii) there exists a set B containing A, a subshift Y ⊂ B Z of ﬁnite type and a cellular automaton τ : B Z → B Z such that X = τ (Y ). A subshift X ⊂ AZ is said to be soﬁc if it satisﬁes one of the two above equivalent conditions. Hint: For the implication (i) ⇒ (ii) use Exercise 6.30. For the converse implication, use Exercise 6.25. 6.32. Let A be a set. (a) Show that every subshift X ⊂ AZ of ﬁnite type is soﬁc. (b) Suppose that A has at least two distinct elements. Show that there exists a subshift X ⊂ AZ which is soﬁc but not of ﬁnite type. Hint: The even subshift is soﬁc (cf. Exercise 6.24) but not of ﬁnite type (cf. Exercise 1.38(c)). 6.33. Let A be a set. Let X ⊂ AG be a soﬁc subshift and let τ : AG → AG be a cellular automaton. Show that τ (X) ⊂ AG is a soﬁc subshift. 6.34. A subshift which is not soﬁc (cf. [LiM, Example 3.1.7]). Let A = {0, 1, 2} and consider the subset X ⊂ AZ consisting of all conﬁgurations x ∈ Z such that if x(n) = 0, x(n + 1) = x(n + 2) = · · · = x(n + h) = 1, x(n + h + 1) = x(n + h + 2) = · · · = x(n + h + k) = 2 and x(n + h + k + 1) = 0 for some n ∈ Z and h, k ∈ N, then necessarily h = k. (a) Show that X is a subshift of AZ . It is called the contextfree subshift. (b) Show that X is not soﬁc. Hint: Suppose by contradiction that X = X G for some ﬁnite Alabeled graph G = (Q, E). Let r = Q. Observe that w = 01r+1 2r+1 0 ∈ L(X) so that there exists a path π in G such that λ(π) = w. Let π denote the subpath of π such that λ(π ) = 1r+1 . Since (π ) = r + 1 > r = Q, we can write π = π1 π2 π3 where π2 is a closed path of length s = (π2 ) > 0 (and π1 (resp. π3 ) is a possibly empty path). It follows that / L(X). π = π1 π2 π2 π3 is a path in G and its label is λ(π ) = 01r+1+s 2r+1 0 ∈
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6.35. Let A be a ﬁnite set and let X ⊂ AZ be a soﬁc subshift. A ﬁnite Alabeled graph G = (Q, E) is said to be a minimal presentation of X if (i) X = X G , and (ii) Q ≤ Q  for all Alabeled graphs G = (Q , E ) such that X = X G . Suppose that X is irreducible and that G = (Q, E) is a minimal presentation of X. (a) Show that for every vertex q ∈ Q there exists a word wq ∈ L(X) such that if a path π in G satisﬁes λ(π) = wq , then it passes through q. Hint: By contradiction, if this is not the case for some q ∈ Q then the Alabeled graph G = (Q , E ) where Q = Q\{q} and E = E \{e ∈ E : α(e) = q or ω(e) = q} satisﬁes X G = X and Q  < Q, contradicting the minimality of G. (b) Deduce from (a) that G is connected. Hint: Let q, q ∈ Q. Consider the words wq , wq ∈ L(X) described in (a). Then by irreducibility of X there exists u ∈ L(X) such that wq uwq ∈ L(X). Let π be a path in G such that λ(π) = wq uwq ; then π = π1 π2 π3 where λ(π1 ) = wq , λ(π2 ) = u and λ(π3 ) = wq . By deﬁnition of wq (resp. wq ) the path π1 (resp. π3 ) passes through q (resp. q ), say π1 = π1 π1 with (π1 )+ = q = (π1 )− (resp. π3 = π3 π3 with (π3 )+ = q = (π3 )− ). Then the path π1 π2 π3 connects q to q . (c) Deduce from (b) that for every irreducible soﬁc subshift Y ⊂ AZ there exists a connected ﬁnite Alabeled graph G such that Y = X G . 6.36. Let A be a ﬁnite set and X ⊂ AZ an irreducible soﬁc subshift. (a) Show that the subset Xf consisting of all conﬁgurations in X whose Zorbit is ﬁnite (cf. Example 1.3.1(c)) is dense in X. Hint: Let x ∈ X and let n ∈ N. By virtue of Exercise 6.35, we can ﬁnd a connected Alabeled graph G = (Q, E) such that X = X G . Let π1 be a ﬁnite path in G such that λ(π1 ) = x(0)x(1) · · · x(n − 1) and set q = π1− and q = π1+ . Since G is connected, we can ﬁnd a ﬁnite path π2 in G connecting q to q. Let m = (π2 ). It follows that the path π = π1 π2 is closed and t = (π) = n + m. If w = λ(π) we deduce that wk ∈ L(X) for all k ∈ N. Since X is closed, there exists a conﬁguration y ∈ X such that y(ht)y(ht + 1) · · · y((h + 1)t − 1) = w for all h ∈ Z. In particular, y(0)y(1) · · · y(n − 1) = x(0)x(1) · · · x(n − 1) and y ∈ Xf . (b) Deduce from (a) that X is surjunctive. Hint: Cf. Exercise 3.29. 6.37. Show that the Morse subshift is not soﬁc. Hint: An inﬁnite minimal subshift is irreducible (cf. Exercise 3.35(b)) and contains no conﬁguration whose Zorbit is ﬁnite (cf. Exercise 3.35(d)). On the other hand, by Exercise 6.36(a), every irreducible soﬁc subshift contains an abundance of conﬁgurations with ﬁnite Zorbit. 6.38. Let A be a ﬁnite set and let X ⊂ AZ be an inﬁnite Toeplitz subshift. Show that X is not soﬁc. Hint: The same arguments as for Exercise 6.37 apply. 6.39. Let A be a ﬁnite set. Show that there are at most countably many distinct soﬁc subshifts X ⊂ AZ . Hint: There are at most countably many ﬁnite Alabeled graphs up to isomorphism.
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6.40. Let I be a ﬁnite set. Let B = (bij )i,j∈I be a matrix with real entries (n) bij ≥ 0 for all i, j ∈ I. For an integer n ≥ 1, we denote by B n = (bij )i,j∈I the nth power of B. One says that the matrix B is irreducible if for every i, j ∈ I (n) there exists an integer n = n(i, j) ≥ 1 such that bij > 0. Suppose that B is irreducible. The period per(i) of i ∈ I is the greatest common divisor of the (n) integers n ≥ 1 such that bii > 0. (a) Show that per(i) = per(j) for all i, j ∈ I. The period per(B) of the matrix B is deﬁned as the common value of the numbers per(i), i ∈ I. Hint: (r) (s) Let i, j ∈ I. We can ﬁnd r, s ≥ 1 such that bij > 0 and bji > 0. It follows (r+s)
that bii
(r) (s)
(r+n+s)
≥ bij bji > 0 and bii
(r) (n) (s)
≥ bij bjj bji > 0 for all n ≥ 1 such
(n)
that bjj > 0. It follows from the deﬁnition that per(i) divides both r + s and r + n + s and therefore also divides their diﬀerence n. This shows that per(i) divides per(j). (b) Let n ≥ 1 be an integer. Show that B n is irreducible if and only if n and per(B) are relatively prime. 6.41. Let A be a set and let G = (Q, E) be a ﬁnite Alabeled graph. The adjacency matrix of G is the (Q × Q)matrix BG = (bqq )q,q ∈Q where bqq is the number of edges in E with initial vertex q and terminal vertex q . (a) Show that G is connected if and only if the matrix BG is irreducible. (b) Suppose that G is connected. The period of G is the positive integer deﬁned by per(G) = per(BG ). Show that per(G) is the greatest common divisor of the lengths of all closed paths in G. 6.42. Let A be a set. Given a subshift X ⊂ AZ denote by pern (X) =  Fix(nZ) ∩ X (cf. Example 1.3.1(c)) the number of nZperiodic conﬁgurations in X. Let G be a ﬁnite connected Alabeled graph. (a) Show that for every integer N there exists an integer n ≥ N such that pern (X G ) > 0. (b) The period per(X G ) of X G is deﬁned as the greatest common divisor of all integers n ≥ 1 for which pern (X) > 0. Show that per(X G ) = per(G). 6.43. The N th higher block subshift (cf. Exercise 1.15 and Exercise 1.34). Let A be a set and N a positive integer. Consider the map ΦN : AZ → (AN )Z deﬁned by ΦN (x)(n) = (x(n), x(n + 1), x(n + 2), . . . , x(n + N − 1)) for all x ∈ AZ and n ∈ Z. Similarly, consider the map ϕN : A∗ → (AN )∗ deﬁned as follows: ϕN (w) = ε (the empty word) if (w) < N and ϕ(w) = (a1 , a2 , . . . , aN )(a2 , a3 , . . . , aN +1 ) · · · (am+1 , am+2 , . . . , am+N ) if w = a1 a2 · · · am+N , with m ≥ 0. Let X ⊂ AZ be a subshift and set X [N ] = ΦN (X) ⊂ (AN )Z . Let L ⊂ A∗ be a subset and set L[N ] = ϕN (L) ⊂ (AN )∗ . (a) Show that X [N ] is a subshift of (AN )Z (it is called the N th higher block subshift of X) and that L(X [N ] ) = (L(X))[N ] .
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(b) Show that X is irreducible (resp. topologically mixing, resp. strongly irreducible) if and only if X [N ] is irreducible (resp. topologically mixing, resp. strongly irreducible). (c) Show that if X is of ﬁnite type then X [N ] is of ﬁnite type. (d) Let Fn = {0, 1, 2, . . . , n − 1} ⊂ Z and let F = (Fn )n∈N . Show that entF (X) = entF (X [N ] ). 6.44. Let A be a set and let G = (Q, E) be an Alabeled graph. For N a positive integer, the N higher edge graph G [N ] associated with G is the AN labeled graph (Q[N ] , E [N ] ) deﬁned as follows. For N = 1 one has G [1] = G and, for N ≥ 2, the vertex set Q[N ] is the set of all paths of length N − 1 in G and E [N ] = {(π,λ(e1 )λ(e2 ) · · · λ(eN ), π ) ∈ Q[N ] × AN × Q[N ] : π = (e1 , e2 , . . . , eN −1 ), π = (e2 , e3 , . . . , eN −1 , eN )}. (a) Show that X G = (X G )[N ] . (b) Deduce from (b) that if X ⊂ AZ is a soﬁc subshift, then the subshift [N ] ⊂ (AN )Z is also soﬁc. X [N ]
6.45. The N th higher power subshift (cf. Exercise 1.16 and Exercise 1.35). Let A be a set and N a positive integer. Consider the map ΨN : AZ → (AN )Z deﬁned by ΨN (x)(n) = (x(nN ), x(nN +1), . . . , x((n+1)N −1)) for all x ∈ AZ and n ∈ Z. Similarly, consider the map ψN : A∗ → (AN )∗ deﬁned as follows: ψN (w) = ε if w ∈ A∗ \ (∪n≥1 AnN ) and ψN (w) = (a1 , a2 , . . . , aN )(aN +1 , aN +2 , . . . , a2N ) · · · (a(n−1)N +1 , a(n−1)N +2 , . . . , anN ) if w = a1 a2 · · · anN for some n ≥ 1. Let X ⊂ AZ be a subshift and set X (N ) = ΨN (X) ⊂ (AN )Z . Let L ⊂ A∗ be a subset and set L(N ) = ψN (L) ⊂ (AN )∗ . (a) Show that X (N ) is a subshift of (AN )Z (it is called the N th higher power subshift of X) and that L(X (N ) ) = (L(X))(N ) . (b) Show that if X is of ﬁnite type then X (N ) is of ﬁnite type. (c) Suppose that X is irreducible. Show that X (N ) is irreducible if and only if N and per(X) are relatively prime. (d) Let Fn = {0, 1, 2, . . . , n − 1} ⊂ Z and let F = (Fn )n∈N . Show that entF (X) = entF (X (N ) ). 6.46. Let A be a set and let G = (Q, E) be an Alabeled graph. For N a positive integer, the N higher power graph G (N ) associated with G is the AN labeled graph (Q(N ) , E (N ) ) deﬁned as follows. The vertex set is Q(N ) = Q and the edge set is E (N ) = {(π − , λ(π), π + ) ∈ Q × AN × Q : π a path of length N in G},
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where π − ∈ Q (resp. π + ∈ Q) is the initial (resp. terminal) vertex of the path π. Recall that given a labeled graph H, we denote by BH its adjacency matrix (cf. Exercise 6.40). (a) Show that BG (N ) = (BG )N . (N ) (b) Show that X G = (X G )(N ) . (c) Deduce from (b) that if X ⊂ AZ is a soﬁc subshift, then the subshift (N ) ⊂ (AN )Z is also soﬁc. X 6.47. (cf. [Sca, Lemma]) Let G = (Q, E) be a ﬁnite labeled graph. Suppose that G is connected and let e ∈ E. Show that there exists a positive integer n0 such that if π is any path in G of length n0 , then there exists a path π = (e1 , e2 , . . . , en0 ) of the same length n0 , with the same initial and terminal vertices as π, and such that ei = e for some 1 ≤ i ≤ n0 . Hint: Given a path π = (e1 , e2 , . . . , en ) in G, denote by πQ = (q0 , q1 , . . . , qn ) the associated sequence of visited vertices (cf. Sect. 6.2). We deﬁne a decomposition of π as follows. Let i1 be the largest index such that the vertices q0 , q1 , . . . , qi1 −1 are all distinct. Then qi1 = qj1 for a suitable j1 < i1 and we set r1 = (e1 , e2 , . . . , ej1 ) and c1 = (ej1 +1 , ej1 +2 , . . . , ei1 ). Continuing this way, we obtain a decomposition of the path π = r1 c1 r2 c2 · · · rk ck rk+1 where the c1 , c2 , . . . , ck are closed simple paths and r1 , r2 , . . . , rk+1 are simple (possibly empty) paths. With this notation, for 1 ≤ s ≤ r and a positive integer d, we say that the path πs = r1 c1 r2 c2 rs rs+1 · · · ck rk is obtained from π by collapsing the sth closed simple path cs . Similarly, if π is a closed path such that (π )− = π − and d is a positive integer, we say that the path (π )d π is obtained from π by adding d copies of π at the beginning of π. Now, since G is connected, we can ﬁnd a closed path π with initial (= terminal) vertex π − containing the edge e. If n0 is large enough then in the decomposition π = r1 c1 r2 c2 · · · rk ck rk+1 there exists a cycle c that is repeated many times. If the length of c is , the length of π is m and the cycle c is repeated at least m times, then we may collapse the ﬁrst m copies of c and then add copies of π at the beginning of π to obtain the desired path π . 6.48. Let G = (Q, E) be a ﬁnite labeled graph. We denote by Pn (G) the set of all paths of length n in G and we deﬁne the entropy of G as ent(G) = lim
n→∞
log Pn (G) . n
Show that the above limit exists and is ﬁnite. Hint: Use Lemma 6.5.1. 6.49. (cf. [Sca, Theorem]) Let G = (Q, E) be a ﬁnite connected labeled graph. Let e ∈ E and denote by H = (Q , E ) the labeled subgraph of G were Q = Q and E = E \ {e}. Let n0 be the positive integer given by Exercise 6.47 and set α = Pn0 (G)−1 . (a) Show that P(k−1)n0 (G) ≥ αPkn0 (G) for all k = 1, 2, . . .. (b) We express a path π ∈ Pkn0 (G) as the composition π = π1 π2 · · · πk , where πi ∈ Pn0 (G), 1 ≤ i ≤ k. Also, for i = 1, 2, . . . , k, we denote by ϕi the set
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6 Finitely Generated Amenable Groups
of all paths π ∈ Pkn0 (G) such that πi contains the edge e. Show that ϕ1  ≥ P(k−1)n0 (G) and deduce from (a) that Pkn0 (G)\ϕ1  ≤ (1−α)Pn0 (G). Hint: ˜ ∈ Pkn0 (G) By Exercise 6.47 for any π ˜ ∈ P(k−1)n0 (G) there exits a path π π such that π contains e. i−1 i i (G) = Pkn0 (G)\ t=1 ϕt , C h = {π ∈ Pkn (G) : (c) For 1 ≤ i ≤ k we set Pkn 0 0 h πi contains e} and denote by D the set of pairs (σ1 , σ2 ) ∈ P(i−1)n0 (G) × i P(k−i)n0 (G) such that there exists π = σ1 σσ2 ∈ Pkn (G). Show that C h  ≥ 0 i (G). Hint: Use Exercise 6.47 to show that for every (σ1 , σ2 ) ∈ Dh  ≥ αPkn 0 i D there exists π = σ1 π σ2 ∈ Pkn0 (G) such that π contains e. k (d) Deduce from (c) that Pkn0 (G) \ i=1 ϕi  ≤ (1 − α)k Pkn0 (G). k (e) Observe that Pkn0 (H) ⊂ Pkn0 (G) \ i=1 ϕi and deduce from (d) that ent(H) ≤ log(1−α) + ent(G). n0 (f) Deduce from (e) that ent(H) < ent(G). 6.50. Let A be a ﬁnite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn )n∈N . Let X ⊂ AZ be a subshift and let w ∈ L(X). Set Xw = {x ∈ X : x(n + 1)x(n + 2) · · · x(n + N ) = w for all n ∈ Z}, where N = (w). (a) Show that Xw ⊂ X is a subshift of AZ . (b) Show that (Xw )[N ] = (X [N ] )ΦN (w) (cf. Exercise 6.43). (c) Suppose from now on that X is irreducible of ﬁnite type. Let M be a positive integer such that {1, 2, . . . , M } is a memory set for X, and let G be the Alabeled graph such that X = X(G, M ) (cf. Exercise 6.25). Let H denote the labeled subgraph of G [N ] obtained by removing all edges labeled by ΦN (w). Show that (X [N ] )ΦN (w) = XH . (d) Show that entF (Xw ) < entF (X). Hint: Use Exercises 6.49 and 6.43(d). 6.51. Let A be a ﬁnite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn )n∈N . Let X ⊂ AZ be an irreducible subshift of ﬁnite type. Show that if Y ⊂ AZ is such that Y X, then entF (Y ) < entF (X). Hint: Show that there exists w ∈ L(X) \ L(Y ) such that Y ⊂ Xw ⊂ X. Then apply Exercise 6.50(d) and Proposition 5.7.2(ii). 6.52. (cf. [Hed3], [CovP], and [LiM, Theorem 8.1.16]). Let A be a ﬁnite set. Let F = (Fn )n∈N , where Fn = {0, 1, . . . , n} ⊂ Z. Also let X ⊂ AZ an irreducible subshift of ﬁnite type and τ : X → X a preinjective cellular automaton. (a) Let M be a positive integer such that {1, 2, . . . , M } is a memory set for X. Set Y = τ (X) and consider the labeled graph G = G(X, τ, M ) (cf. Exercise 6.26). Note that for every w ∈ L(Y ) there exists a path π ∈ G such that w = λ(π). Show that, by preinjectivity of τ , for all q, q ∈ Q and w ∈ L(Y ) there exists at most one path π in G with initial (resp. terminal) vertex π − = q (resp. π + = q ) such that w = λ(π). (b) Deduce from (a) that Ln (Y ) ≤ Ln (X) ≤ Ln (Y ) · Q2 . (c) Deduce from (b) that entF (τ (X)) = entF (X).
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6.53. Let A be a ﬁnite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn )n∈N . Let X, Y ⊂ AZ be two irreducible subshifts of ﬁnite type such that entF (X) = entF (Y ). Show that every preinjective cellular automaton τ : X → Y is surjective. Hint: Use the results of Exercise 6.51 and Exercise 6.52. 6.54. Let A be a ﬁnite set. Let X ⊂ AG be an irreducible subshift of ﬁnite type. Show that every preinjective cellular automaton τ : X → X is surjective. Hint: Use Exercise 6.53 with Y = X. 6.55. (cf. [Hed3], [CovP], and [LiM, Theorem 8.1.16]). Let A be a ﬁnite set and X ⊂ AZ an irreducible subshift of ﬁnite type. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn )n∈N . Let τ : X → X be a cellular automaton. Suppose that τ is not preinjective. (a) Show that there exist a positive integer N and two distinct conﬁgurations x1 , x2 ∈ X such that x1 Z\{1,2,...,N } = x2 Z\{1,2,...,N } and τ (x1 ) = τ (x2 ). (b) Up to enlarging N if necessary, we may suppose that {1, 2, . . . , N } is a memory set for both X and τ . Let G = G(X, τ, N ) = (Q, E) (cf. Exercise 6.26). Show that there exist two vertices q, q ∈ Q and two paths π1 , π2 in G with initial (resp. terminal) vertices π1− = π2− = q (resp. π1+ = π2+ = q ) such that λ(π1 ) = x1 {1,2,...,N } and λ(π2 ) = x2 {1,2,...,N } . One says that the two paths π1 , π2 constitute a diamond in G (see Fig. 6.21).
Fig. 6.21 A diamond in an Alabeled graph G
(c) Up to further enlarging N if necessary, we may suppose that N is relatively prime with per(X) (cf. Exercise 6.40 and Exercise 6.42). By Exercise 6.45(c) the N th power graph G (N ) is also irreducible and π1 and π2 are edges in G (N ) with the same initial vertex and same terminal vertex. Let H be the Alabeled subgraph of G (N ) obtained by removing the edge π1 . Deduce from Exercise 6.49 that entF (XH ) < entF (XG (N ) ). (d) Show that XH = Y (N ) , where Y = τ (X). (e) Deduce from (c) and (d) and from Exercise 6.45(d) that entF (τ (X)) < entF (X). 6.56. Let A be a ﬁnite set. Let Fn = {0, 1, . . . , n} ⊂ Z and F = (Fn )n∈N . Let X, Y ⊂ AZ be two subshifts such that X is irreducible of ﬁnite type and entF (X) = entF (Y ). Show that every surjective cellular automaton τ : X → Y is preinjective. Hint: Use the result of Exercise 6.55. 6.57. The Garden of Eden theorem for irreducible subshifts of ﬁnite type over Z [Fio1, Corollary 2.19]. Let A be a ﬁnite set and X ⊂ AZ an irreducible
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subshift of ﬁnite type. Let τ : X → X be a cellular automaton. Show that τ is surjective if and only if it is preinjective. Hint: Combine together the results from Exercise 6.56 with Y = X and Exercise 6.54. 6.58. Let A = {0, 1}. Consider the subset X ⊂ AZ consisting of all conﬁgurations x ∈ AZ such that {n ∈ Z : x(n) = 1} is an interval of Z. Let σ : AZ → AZ be the cellular automaton with memory set S = {0, 1} and local deﬁning map μ : AS → A given by 1 if (y(0), y(1)) = (0, 1), μ(y) = y(0) otherwise. (a) Show that X is a soﬁc subshift. (b) Show that X is neither of ﬁnite type nor irreducible. (c) Check that τ (X) ⊂ X (d) Show that the cellular automaton τ = σX : X → X is injective (and therefore preinjective) but not surjective. (e) Deduce from (d) that X is not surjunctive.
Chapter 7
Local Embeddability and Soﬁc Groups
In this chapter we study the notions of local embeddability and soﬁcity for groups. Roughly speaking, a group is locally embeddable into a given class of groups provided that the multiplicative table of any ﬁnite subset of the group is the same as the multiplicative table of a subset of some group in the class (cf. Deﬁnition 7.1.3). In Sect. 7.1 we discuss several stability properties of local embeddability. Subgroups of locally embeddable groups are locally embeddable (Proposition 7.1.7). Moreover, as the name suggests, local embeddability is a local property, that is, a group is locally embeddable into a class of groups if and only if all its ﬁnitely generated subgroups are locally embeddable into the class (Proposition 7.1.8). When the class is closed under ﬁnite direct products, the class of groups which are locally embeddable in the class is closed under (possibly inﬁnite) direct products (Proposition 7.1.10). We also show that a marked group which is a limit of groups belonging to a given class is locally embeddable into this class (Theorem 7.1.16). Conversely, we prove that if the given class is closed under taking subgroups and the marking group is free, then any marked group which is locally embeddable is a limit of marked groups which are in the class (Theorem 7.1.19). If C is a class of groups which is closed under ﬁnite direct products, then every group which is residually C is locally embeddable into C (Corollary 7.1.14). Conversely, under the hypothesis that C is closed under taking subgroups, every ﬁnitely presented group which is locally embeddable into C is residually C (Corollary 7.1.21). In Sect. 7.2 we present a characterization of local embeddability in terms of ultraproducts: a group is locally embeddable into C if and only if it can be embedded into an ultraproduct of a family of groups in C. Section 7.3 is devoted to LEF and LEAgroups. A group is called LEF (resp. LEA) if it is locally embeddable into the class of ﬁnite (resp. amenable) groups. As the class of ﬁnite (resp. amenable) groups is closed under taking subgroups and taking ﬁnite direct products, all the results obtained in the previous section can be applied. This implies in particular that every locally residually ﬁnite (resp. locally residually amenable) group is LEF (resp. LEA). T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 7, © SpringerVerlag Berlin Heidelberg 2010
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We give an example of a ﬁnitely generated amenable group which is LEF but not residually ﬁnite (Proposition 7.3.9). This group is not ﬁnitely presentable since every ﬁnitely presented LEFgroup is residually ﬁnite. The Hamming metric, which is a biinvariant metric on the symmetric group of a ﬁnite set, is introduced in Sect. 7.4. In Sect. 7.5 we deﬁne the class of soﬁc groups. These groups are the groups admitting local approximations by ﬁnite symmetric groups equipped with their Hamming metric. Subgroups and direct products of soﬁc groups are soﬁc. Every LEAgroup is soﬁc (Corollary 7.5.11). In particular, residually amenable groups, and therefore amenable groups and residually ﬁnite groups, are soﬁc. A group is soﬁc if and only if it can be embedded into an ultraproduct of a family of ﬁnite symmetric groups equipped with their Hamming metrics (Theorem 7.6.6). In Sect. 7.7 we give a characterization of ﬁnitely generated soﬁc groups in terms of their Cayley graphs. More precisely, we show that a ﬁnitely generated group G with a ﬁnite symmetric generating subset S ⊂ G is soﬁc if and only if, for every integer r ≥ 0 and every ε > 0, there exists a ﬁnite Slabeled graph Q such that there is a proportion of at least 1 − ε of vertices q ∈ Q such that the ball of radius r centered at q in Q is isomorphic, as a labeled graph, to a ball of radius r in the Cayley graph of G associated with S (Theorem 7.7.1). The last section of this chapter is devoted to the proof of the surjunctivity of soﬁc groups (Theorem 7.8.1).
7.1 Local Embeddability Deﬁnition 7.1.1. Let G and C be two groups. Given a ﬁnite subset K ⊂ G, a map ϕ : G → C is called a Kalmosthomomorphism of G into C if it satisﬁes the following conditions: (KAH1) ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K; (KAH2) the restriction of ϕ to K is injective. Note that in the preceding deﬁnition, the map ϕ is not required to be a homomorphism nor to be globally injective. Remark 7.1.2. If ϕ : G → C is a Kalmosthomomorphism and ϕ : G → C is a map which coincides with ϕ on K ∪ K 2 , then ϕ is also a Kalmosthomomorphism. Let now C be a class of groups, that is, a collection of groups satisfying the following condition: if C ∈ C and C is a group which is isomorphic to C, then C ∈ C. For example, C might be the class of ﬁnite groups, the class of nilpotent groups, the class of solvable groups, the class of amenable groups, or the class of free groups.
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Deﬁnition 7.1.3. Let C be a class of groups. One says that a group G is locally embeddable into the class C if, for every ﬁnite subset K ⊂ G, there exist a group C ∈ C and a Kalmosthomomorphism ϕ : G → C. Examples 7.1.4. (a) Let C be a class of groups and let G be a group in C. Then G is locally embeddable into C. Indeed, the identity map IdG : G → G is a Kalmosthomomorphism of G into itself for every ﬁnite subset K ⊂ G. (b) The group Z is locally embeddable into the class of ﬁnite groups. Indeed, let K be a ﬁnite subset of Z. Choose an integer n ≥ 0 such that K ⊂ [−n, n]. Then, the quotient homomorphism ϕ : Z → Z/(2n + 1)Z is a Kalmosthomomorphism. Remark 7.1.5. Let G be a group and let C be a class of groups which is closed under taking subgroups. Suppose that G is ﬁnite and it is locally embeddable into C. Then G ∈ C. Indeed, any Galmost homomorphism ϕ : G → C from G into a group C ∈ C is an injective homomorphism. The following characterization of local embeddability will not be used in the sequel but it presents some interest on its own. Proposition 7.1.6. Let G be a group and let C be a class of groups which is closed under taking subgroups. Then the following conditions are equivalent: (a) G is locally embeddable into C; (b) for every ﬁnite subset K ⊂ G, there exist a set L with K ⊂ L ⊂ G and a binary operation : L × L → L such that (L, ) is a group in C and k1 k2 = k1 k2 for all k1 , k2 ∈ K. Proof. Suppose (a). Fix a ﬁnite subset K ⊂ G. If G is ﬁnite, then G ∈ C by Remark 7.1.5. Therefore, condition (b) is satisﬁed in this case since we can take as (L, ) the group G itself. So let us assume that G is inﬁnite. Consider the ﬁnite subset K ⊂ G deﬁned by K = K ∪ K 2 . Since G is locally embeddable into C, we can ﬁnd a group C ∈ C and a K almosthomomorphism ϕ : G → C . Note that, as K is ﬁnite, the subgroup C ⊂ C generated by ϕ (K ) is countable. Since G is inﬁnite, it follows that there exists a surjective map σ : G → C which coincides with ϕ on K . Let us deﬁne a map ψ : C → G as follows. If c ∈ σ(K ), we set ψ(c) = k , where k is the unique element in K such that σ(k ) = c. For c ∈ C \ σ(K ), we take as ψ(c) an arbitrary element in σ −1 (c). Denote by L = ψ(C) ⊂ G the image of ψ. Observe that K ⊂ K ⊂ L and that ψ induces a bijection from C onto L. Let us use this bijection to transport the group structure from C to L. If we denote by the corresponding group operation on L, this means that ψ induces a group isomorphism from C onto (L, ). It follows that we have 1 2 = ψ(σ(1 )σ(2 )) for all 1 , 2 ∈ L. As C is closed under taking subgroups, C and hence (L, ) belong to C. Moreover, for all k1 , k2 ∈ K, we have
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k1 k2 = ψ(σ(k1 )σ(k2 )) = ψ(ϕ (k1 )ϕ (k2 )) = ψ(ϕ (k1 k2 )) = ψ(σ(k1 k2 )). Since K 2 ⊂ K , we deduce that k1 k2 = k1 k2 for all k1 , k2 ∈ K. This shows that condition (b) is satisﬁed. Conversely, suppose (b). Fix a ﬁnite subset K ⊂ G. Set C = (L, ), where L is as in (b). Let ϕ : G → C be any map extending the identity map ι : L → L. Then ϕK = ιK is injective. On the other hand, for all k1 , k2 ∈ K, we have k1 k2 = k1 k2 ∈ L and therefore ϕ(k1 k2 ) = ι(k1 k2 ) = k1 k2 = k1 k2 = ϕ(k1 ) ϕ(k2 ) Thus, ϕ is a Kalmosthomomorphism of G into the group C in C. This shows that G is locally embeddable into C. Proposition 7.1.7. Let C be a class of groups. Every subgroup of a group which is locally embeddable into C is locally embeddable into C. Proof. Let G be a group which is locally embeddable into C and let H be a subgroup of G. Given a ﬁnite subset K ⊂ H, let ϕ : G → C be a Kalmosthomomorphism of G into a group C ∈ C. Then the restriction map ϕH : H → C is a Kalmosthomomorphism of H into C. This shows that H is locally embeddable into C. As the name suggests, local embeddability is a local property for groups: Proposition 7.1.8. Let G be a group and let C be a class of groups. Then G is locally embeddable into C if and only if every ﬁnitely generated subgroup of G is locally embeddable into C. Proof. The necessity of the condition follows from Proposition 7.1.7. Conversely, suppose that every ﬁnitely generated subgroup of G is locally embeddable into C. Let K ⊂ G be a ﬁnite subset and denote by H ⊂ G the subgroup generated by K. Then, as H is locally embeddable into C, there exists a Kalmosthomomorphism ϕ : H → C of H into a group C ∈ C. Extend arbitrarily ϕ to G, for example by setting ϕ(g) = 1C for all g ∈ G \ H. Then ϕ : G → C is a Kalmosthomomorphism of G into C. This shows that G is locally embeddable into C. Let C be a class of groups. Denote by C the class consisting of all groups which are locally embeddable into C. Proposition 7.1.9. Let C be a class of groups. Then C = C.
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Proof. Let G be a group in C. Then, given any ﬁnite subset K ⊂ G there exist a group C ∈ C and a Kalmosthomomorphism ϕ : G → C . Consider the ﬁnite subset K = ϕ(K) ⊂ C and let ϕ : C → C be a K almosthomomorphism of C into a group C ∈ C. Then the composition map Φ = ϕ ◦ ϕ : G → C is a Kalmosthomomorphism. Indeed, for k1 , k2 ∈ K one has Φ(k1 k2 ) = ϕ (ϕ(k1 k2 )) = ϕ (ϕ(k1 )ϕ(k2 )) = ϕ (ϕ(k1 ))ϕ (ϕ(k2 )) = Φ(k1 )Φ(k2 ). Moreover, as ϕK and ϕ K are injective, and K = ϕ(K), one has that ΦK is also injective. It follows that G is locally embeddable into C. This shows that C ⊂ C. The inclusion C ⊂ C follows from the observation in Example 7.1.4(a). This shows that C = C. A class of groups C is said to be closed under ﬁnite direct products if one has G1 × G2 ∈ C whenever G1 ∈ C and G2 ∈ C. Proposition 7.1.10. Let C be a class of groups which is closed under ﬁnite direct products. Let (Gi )i∈I be a family of groups which are locally embeddable into C. Then, their direct product G = i∈I Gi is locally embeddable into C. Proof. For each i ∈ I, let πi : G → Gi denote the projection homomorphism. Fix a ﬁnite subset K ⊂ G. Then there exists a ﬁnite subset J⊂ I such that the projection homomorphism πJ = j∈J πj : G → GJ = j∈J Gj is injective on K. Since the group Gj is locally embeddable into C, we can ﬁnd, for each j ∈ J, a πj (K)almost homomorphism ϕj : Gj → Cj of Gj into a groupCj ∈ C. As the class C is closed under ﬁnite direct products, the group is also in C. Consider the map ϕ : G → C deﬁned by ϕ = ϕJ ◦πJ , C = j∈J Cj where ϕJ = j∈J ϕj : GJ → C. For all k = (ki )i∈I , k = (ki )i∈I ∈ K, we have ϕ(kk ) = ϕJ (πJ (kk )) = ϕJ ((kj kj )j∈J ) = (ϕj (kj kj ))j∈J = (ϕj (kj )ϕj (kj ))j∈J = (ϕj (kj ))j∈J (ϕj (kj ))j∈J = ϕ(k)ϕ(k ). Moreover, ϕK is injective. Indeed, given k = (ki )i∈I , k = (ki )i∈I ∈ K, if ϕ(k) = ϕ(k ) then ϕj (kj ) = ϕj (kj ) for all j ∈ J. By injectivity of ϕj πj (K) , we deduce that kj = kj for all j ∈ J. This implies that k = k since πJ K is injective. This shows that ϕ is a Kalmosthomomorphism of G into C. It follows that G is locally embeddable into C.
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Corollary 7.1.11. Let C be a class of groups which is closed under ﬁnite direct products. Let (Gi )i∈I be a family of groups which are locally embeddable into C. Then their direct sum G = i∈I Gi is locally embeddable into C. Proof. This follows immediately from Proposition 7.1.7 and Proposition 7.1.10, since G is the subgroup of the direct product P = i∈I Gi consisting of all g = (gi )i∈I ∈ P for which gi = 1Gi for all but ﬁnitely many i ∈ I. Corollary 7.1.12. Let C be a class of groups which is closed under ﬁnite direct products. If a group G is the limit of a projective system of groups which are locally embeddable into C, then G is locally embeddable into C. Proof. Let (Gi )i∈I be a projective system of groups which are locally embedof a projective limit (see dable into C such that G = lim Gi . By construction ←− Appendix E), G is a subgroup of the group i∈I Gi . We deduce that G is locally embeddable into C by using Proposition 7.1.10 and Proposition 7.1.7. Recall that if C is a class of groups, then a group G is called residually C if for each element g ∈ G with g = 1G , there exist a group C ∈ C and a surjective homomorphism φ : G → C such that φ(g) = 1C . Proposition 7.1.13. Let C be a class of groups which is closed under ﬁnite direct products. Let G be a group which is residually C. Then, for every ﬁnite subset K ⊂ G, there exist a group C ∈ C and a homomorphism ϕ : G → C whose restriction to K is injective. Proof. Fix a ﬁnite subset K ⊂ G and consider the set L = {hk −1 : h, k ∈ K and h = k}. C and a Since G is residually C, we can ﬁnd, for each g ∈ L, a group Cg ∈ homomorphism φg : G → Cg such that φg (g) = 1Cg . The group C = g∈L Cg is in the class C since L is ﬁnite and C is closed under ﬁnite direct products. Consider the group homomorphism ϕ = g∈L φg : G → C. If h and k are distinct elements in K, then g = hk−1 ∈ L and φg (hk−1 ) = φg (g) = 1Cg . It follows that ϕ(hk−1 ) = 1C and hence ϕ(h) = ϕ(k). Thus, the restriction of ϕ to K is injective. Corollary 7.1.14. Let C be a class of groups which is closed under ﬁnite direct products. Then every group which is residually C is locally embeddable into C. Proof. Let G be a group which is residually C and let K ⊂ G be a ﬁnite subset. By Proposition 7.1.13, there exist a group C ∈ C and a homomorphism ϕ : G → C whose restriction to K is injective. Such a ϕ is a Kalmosthomomorphism. Consequently, G is locally embeddable into C.
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Corollary 7.1.15. Let C be a class of groups which is closed under ﬁnite direct products. Then every group which is locally residually C is locally embeddable into C. Proof. This immediately follows from Corollary 7.1.14 and Proposition 7.1.8. Given a group Γ , let N (Γ ) denote the space of all Γ marked groups (cf. Sect. 3.4). Recall that N (Γ ) may be identiﬁed with the set consisting of all normal subgroups of Γ and that N (Γ ) is a compact Hausdorﬀ space for the topology induced by the prodiscrete topology on P(Γ ) = {0, 1}Γ . Theorem 7.1.16. Let Γ be a group and let C be a class of groups. Let N ∈ N (Γ ). Suppose that there exists a net (Ni )i∈I which converges to N in N (Γ ) such that Γ/Ni ∈ C for all i ∈ I. Then the group Γ/N is locally embeddable into C. Proof. Denote by ρ : Γ → Γ/N and ρi : Γ → Γ/Ni the quotient homomorphisms. Fix a ﬁnite subset K ⊂ Γ/N and let F ⊂ Γ be a ﬁnite symmetric subset such that 1Γ ∈ F and ρ(F ) = K ∪K −1 ∪{1Γ/N }. Since the net (Ni )i∈I converges to N , we can ﬁnd i0 ∈ I such that N ∩ F 4 = N i0 ∩ F 4 . Set C = Γ/Ni0 and deﬁne a map ϕ : Γ/N → C by setting ρi0 (f ) if g = ρ(f ) for some f ∈ F 2 ϕ(g) = 1C otherwise.
(7.1)
(7.2)
Note that ϕ is well deﬁned. Indeed, suppose that g = ρ(f1 ) = ρ(f2 ), for some f1 , f2 ∈ F 2 . Then, from 1Γ/N = ρ(f1 )−1 ρ(f2 ) = ρ((f1 )−1 f2 ), we deduce that (f1 )−1 f2 ∈ ker(ρ) = N. (7.3) As (f1 )−1 f2 ∈ F 4 , we deduce from (7.3) and (7.1) that (f1 )−1 f2 ∈ Ni0 = ker(ρi0 ). Therefore, we have ρi0 (f1 ) = ρi0 (f2 ). Let us check now that ϕ is a Kalmosthomomorphism. Let k1 , k2 ∈ K. Then, there exist f1 , f2 ∈ F such that ρ(f1 ) = k1 and ρ(f2 ) = k2 . Observe that f1 , f2 ∈ F 2 since F ⊂ F 2 . Thus, we get ϕ(k1 ) = ρi0 (f1 ) and ϕ(k2 ) = ρi0 (f2 ) by applying (7.2). On the other hand, we also have f1 f2 ∈ F 2 and k1 k2 = ρ(f1 )ρ(f2 ) = ρ(f1 f2 ). Therefore, by applying again (7.2), we obtain ϕ(k1 k2 ) = ρi0 (f1 f2 ) = ρi0 (f1 )ρi0 (f2 ) = ϕ(k1 )ϕ(k2 ). Moreover, if ϕ(k1 ) = ϕ(k2 ), then ρi0 (f1 ) = ρi0 (f2 ) and hence f1−1 f2 ∈ ker(ρi0 ) = Ni0 . As f1−1 f2 ∈ F 2 ⊂ F 4 , we deduce from (7.1) that f1−1 f2 ∈
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N = ker(ρ). Therefore k1 = ρ(f1 ) = ρ(f2 ) = k2 . This shows that ϕ is injective on K. We have shown that ϕ is a Kalmosthomomorphism of Γ/N into the group C ∈ C. Therefore, Γ/N is locally embeddable into C. Corollary 7.1.17. Let Γ be a group and let C be a class of groups. Then the set of all N ∈ N (Γ ) such that Γ/N is locally embeddable into C is closed (and hence compact) in N (Γ ). Proof. Let (Ni )i∈I be a convergent net in N (Γ ) and suppose that the groups Γ/Ni are locally embeddable into C for all i ∈ I. Let N = limi∈I Ni . By applying Theorem 7.1.16, we deduce that Γ/N is locally embeddable into the class of groups which are locally embeddable into C. It follows from Proposition 7.1.9 that Γ/N is locally embeddable into C. For the next theorem we need some (slightly technical) preliminaries. Let n ≥ 2 be an integer and G, C two groups. Given a ﬁnite subset K ⊂ G, we say that a map ϕ : G → C is an nKalmosthomomorphism if it satisﬁes the following conditions: (nKAH1) ϕ(k1ε1 k2ε2 · · · ktεt ) = ϕ(k1 )ε1 ϕ(k2 )ε2 · · · ϕ(kt )εt for all ki ∈ K ∪ {1G }, εi ∈ {−1, 1}, 1 ≤ i ≤ t ≤ n; (nKAH2) ϕ(1G ) = 1C ; (nKAH3) the restriction of ϕ to (K ∪ K −1 ∪ {1G })n is injective. Lemma 7.1.18. Let G be a group and let C be a class of groups. Let n ≥ 2 be an integer. Then, the following conditions are equivalent: (a) G is locally embeddable into C; (b) for every ﬁnite subset K ⊂ G, there exist a group C ∈ C and an nKalmosthomomorphism ϕ : G → C. Proof. It is clear that any nKalmosthomomorphism is also a Kalmosthomomorphism (take ε1 = ε2 = 1 and t = 2 in (nKAH1), and observe that K ⊂ (K ∪ K −1 ∪ {1G })n ). Thus, (b) implies (a). Conversely, suppose (a). Given a ﬁnite subset K ⊂ G, set K = (K ∪ −1 K ∪ {1G })n . As G is locally embeddable into C, there exists a K almosthomomorphism ϕ : G → C of G into a group C ∈ C. Let us show that ϕ is an nKalmosthomomorphism. As the restriction of ϕ to K is injective, Property (nKAH3) is trivially satisﬁed. On the other hand, we have ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K . For k1 = k2 = 1G , this gives us ϕ(1G ) = ϕ(1G )2 , so that Property (nKAH2) also holds. By taking k1 ∈ K and k2 = k1−1 , we get 1C = ϕ(1G ) = ϕ(k1 k1−1 ) = ϕ(k1 )ϕ(k1−1 ). This implies that ϕ(k1−1 ) = ϕ(k1 )−1 for all k1 ∈ K . Then, Property (nKAH1) immediately follows by induction on t. This shows that ϕ is an nKalmosthomomorphism of G into C. Theorem 7.1.19. Let F be a free group and let C be a class of groups which is closed under taking subgroups. Let N ∈ N (F ) and suppose that the group F/N is locally embeddable into C. Then there exists a net (Ni )i∈I which converges to N in N (F ) such that F/Ni ∈ C for all i ∈ I.
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Proof. Let X ⊂ F be a free base of F and denote by ρ : F → F/N the quotient homomorphism. Let I denote the set of all ﬁnite subsets E of F partially ordered by inclusion. Given E ∈ I, we denote by UE ⊂ X the subset of X consisting of all elements which appear in the reduced form of some element in E. Since E is ﬁnite, UE is also ﬁnite and there exists nE ∈ N such that each w ∈ E may be written in the form w = xε11 xε22 · · · xεt t
(7.4)
for suitable xi ∈ UE , εi ∈ {−1, 1}, i = 1, 2, . . . , t, and 1 ≤ t ≤ nE . Set KE = ρ(BnE ) ⊂ F/N , where BnE denotes the ball of radius nE centered at the identity element 1F in the Cayley graph of the subgroup of F generated by UE . Since F/N is locally embeddable into C it follows from Lemma 7.1.18 that there exists a group CE ∈ C and an nE KE almosthomomorphism ϕE : F/N → CE . As F is a free group with base X, there exists a unique homomorphism ρE : F → CE such that ρE (x) = ϕE (ρ(x)) for all x ∈ X. Setting NE = ker(ρE ) we have NE ∈ N (F ). Moreover, since C is closed under taking subgroups, we have that the quotient group F/NE , being isomorphic to ρE (F ), which is a subgroup of CE , also belongs to C. Let us show that the net (NE )E∈I converges to N . Fix a ﬁnite subset E0 ⊂ F . Let E ∈ I be such that E0 ⊂ E. Let us show that N ∩ E0 = N E ∩ E 0 .
(7.5)
Given w ∈ E0 , we can write w as in (7.4). Using the fact that ϕE is an nE KE almosthomomorphism, we get ρE (w) = ρE (xε11 xε22 · · · xεt t ) = ρE (x1 )ε1 ρE (x2 )ε2 · · · ρE (xt )εt = ϕE (ρ(x1 ))ε1 ϕE (ρ(x2 ))ε2 · · · ϕE (ρ(xt ))εt = ϕE (ρ(x1 )ε1 ρ(x2 )ε2 · · · ρ(xt )εt ) = ϕE (ρ(xε11 xε22 · · · xεt t )) = ϕE (ρ(w)). As ϕE is injective on KE , and ρ(w) ∈ ρ(BE ) = KE , we deduce that ρE (w) = 1F/NE if and only if ρ(w) = 1F/N , equivalently, w ∈ NE if and only if w ∈ N . Thus (7.5) follows. This shows that N = limE NE . From Theorem 7.1.16 (with Γ a free group F ) and Theorem 7.1.19 we immediately deduce the following. Corollary 7.1.20. Let F be a free group and let C be a class of groups which is closed under taking subgroups. Let N ∈ N (F ). Then the following conditions are equivalent. (a) the group F/N is locally embeddable into C;
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(b) there exists a net (Ni )i∈I which converges to N in N (F ) such that F/Ni ∈ C for all i ∈ I. Corollary 7.1.21. Let C be a class of groups which is closed under taking subgroups. Then every ﬁnitely presented group which is locally embeddable into C is residually C. Proof. Let G be a ﬁnitely presented group which is locally embeddable into C. Fix g ∈ G \ {1G }. Since G is ﬁnitely presented, there exists a free group F of ﬁnite rank and a ﬁnite subset R ⊂ F such that G = F/N , where N ∈ N (F ) denotes the normal closure of R in F . By Theorem 7.1.19, there exists a net (Ni )i∈I in N (F ) converging to N such that F/Ni ∈ C for all i ∈ I. Let ρ : F → G denote the quotient homomorphism and choose an element f ∈ F such that ρ(f ) = g. Since the net (Ni )i∈I converges to N , we can ﬁnd i0 ∈ I such that (7.6) N ∩ (R ∪ {f }) = Ni0 ∩ (R ∪ {f }). As R ⊂ N , this implies R ⊂ Ni0 and hence N ⊂ Ni0 . Thus, there is a /N canonical epimorphism φ : G = F/N → C = F/Ni0 . Observe now that f ∈ and therefore f ∈ / Ni0 by (7.6). It follows that φ(g) = 1C . As C ∈ C, this shows that G is residually C. Corollary 7.1.22. Let C be a class of groups which is closed under taking subgroups and under ﬁnite direct products. Then a ﬁnitely presented group is locally embeddable into C if and only if it is residually C. Proof. The fact that every ﬁnitely presented group which is locally embeddable into C is residually C follows from the previous corollary. The converse implication follows from Corollary 7.1.14. We end this section by producing an example showing that Theorem 7.1.19 becomes false if we omit the hypothesis that F is free. Let us ﬁrst establish the following: Proposition 7.1.23. Let C be a class of groups. Let G be a group. Suppose that there exists a net (Ni )i∈I which converges to {1G } in N (G) such that G/Ni ∈ C for all i ∈ I. Then G is residually C. Proof. Let g ∈ G \ {1G }. Let us set F = {1G , g}. Since the net (Ni )i∈I converges to {1G } in N (G), there exists i0 ∈ I such that Ni0 ∩ F = {1G } ∩ F . / Ni0 . Thus, the quotient homomorphism As F ∩{1G } = {1G }, this implies g ∈ φ : G → G/Ni0 satisﬁes φ(g) = 1G/Ni0 . As G/Ni0 ∈ C, this shows that G is residually C. Now, to exhibit the promised example showing the necessity of the freeness hypothesis on F in Theorem 7.1.19, we consider the additive group G = Q of rational numbers and take as C the class of ﬁnite groups. It follows from Example 2.1.9 that Q is not residually ﬁnite. Thus, by the preceding propo
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sition, there is no net (Ni )i∈I converging to {1G } in N (G) such that G/Ni is ﬁnite for all i ∈ I. On the other hand, G is locally residually ﬁnite since every ﬁnitely generated abelian group is residually ﬁnite by Corollary 2.2.4. Therefore, G is locally embeddable into C by Corollary 7.1.15.
7.2 Local Embeddability and Ultraproducts Suppose that we are given a family of groups (Gi )i∈I and a ﬁlter ω (cf. Sect. J.1) on the index set I. Consider the group P = Gi . i∈I
Let α = (αi )i∈I and β = (βi )i∈I be elements of P . We write α ∼ω β if {i ∈ I : αi = βi } ∈ ω. Proposition 7.2.1. One has: (i) α ∼ω α; (ii) α ∼ω β if (iii) if α ∼ω β (iv) if α ∼ω β (v) α ∼ω β if
and and and and
only if β ∼ω α; β ∼ω γ, then α ∼ω γ; γ ∼ω δ, then αγ ∼ω βδ; only if α−1 ∼ω β −1 ;
for all α, β, γ, δ ∈ P . Proof. Let α = (αi )i∈I , β = (βi )i∈I , γ = (γi )i∈I and δ = (δi )i∈I ∈ P . We have α ∼ω α since {i ∈ I : αi = αi } = I belongs to ω (cf. (F6) in Sect. J.1). This shows (i). Also, since {i ∈ I : αi = βi } = {i ∈ I : βi = αi } we deduce (ii). Suppose now that α ∼ω β and β ∼ω γ. We have {i ∈ I : αi = γi } ⊃ {i ∈ I : αi = βi } ∩ {i ∈ I : βi = γi }. Thus (iii) follows from the fact that ω, being a ﬁlter, is closed under ﬁnite intersections and taking supersets (cf. (F2) and (F3) in Sect. J.1). Suppose now that α ∼ω β and γ ∼ω δ, that is, {i ∈ I : αi = βi } and {i ∈ I : γi = δi } both belong to ω. We have {i ∈ I : αi γi = βi δi } ⊃ {i ∈ I : αi = βi } ∩ {i ∈ I : γi = δi }. As ω is closed under ﬁnite intersections and taking supersets, we deduce that {i ∈ I : αi γi = βi δi } ∈ ω, that is, αγ ∼ω βδ. This shows (iv). Finally, from {i ∈ I : αi = βi } = {i ∈ I : αi−1 = βi−1 } we deduce (v). Note that it follows from (i), (ii) and (iii) in Proposition 7.2.1 that ∼ω is an equivalence relation in P . Consider now the subset Nω ⊂ P deﬁned by Nω = {α ∈ P : α ∼ω 1P = (1Gi )i∈I }.
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Proposition 7.2.2. The set Nω is a normal subgroup of P . Proof. This is an easy consequence of the properties of ∼ω stated in Proposition 7.2.1. Indeed, we immediately deduce from Proposition 7.2.1(i) that 1P ∈ Nω . Suppose now that α, β ∈ Nω , that is, α ∼ω 1P and β ∼ω 1P . It follows from Proposition 7.2.1(iv) that αβ ∼ω 1P 1P = 1P , that is, αβ ∈ Nω . = 1P , Similarly, from Proposition 7.2.1(v) we deduce that α−1 ∼ω 1−1 P that is, α−1 ∈ Nω . Thus, Nω is a subgroup of P . Finally, let γ ∈ P . Proposition 7.2.1(iv) implies that γαγ −1 ∼ω γ1P γ −1 = 1P . It follows that γαγ −1 ∈ Nω for all α ∈ Nω and γ ∈ P . This shows that Nω is a normal subgroup of P . Observe that, given α and β in P , one has αNω = βNω ⇐⇒ α ∼ω β.
(7.7)
Indeed, one has αNω = βNω if and only if αβ −1 ∈ Nω , that is, if and only if αβ −1 ∼ω 1P . This is equivalent to α ∼ω β by Proposition 7.2.1(iv). The quotient group Pω = P/Nω is called the reduced product of the family of groups (Gi )i∈I with respect to the ﬁlter ω. In the particular case when ω is an ultraﬁlter, one also says that Pω is the ultraproduct of the family of groups (Gi )i∈I with respect to the ultraﬁlter ω. Theorem 7.2.3. Let C be a class of groups, (Gi )i∈I a family of groups such that Gi ∈ C for all i ∈ I, and ω an ultraﬁlter on the index set I. Then the ultraproduct Pω of the family of groups (Gi )i∈I with respect to the ultraﬁlter ω is locally embeddable into C. Proof. Fix a ﬁnite subset K ⊂ Pω . We want to show that there exist a group C ∈ C and a Kalmosthomomorphism ϕ : Pω → C. Choose a representative of each element g ∈ Pω , that is, an element g = ( gi )i∈I ∈ P such that g = gNω . If h and k are arbitrary elements of K, we ω = ∼ω have hkN h kNω and therefore hk h k by (7.7). It follows that the set i = hi Ih,k = {i ∈ I : hk ki } belongs to ω. Thus, since ω is closed under ﬁnite intersections (cf.(F5) in Sect. J.1), we have that Ih,k IK = h,k∈K
also belongs to ω. On the other hand, if h and k are distinct elements of K, ki } does not belongs to ω. As ω is an ultraﬁlter, the subset {i ∈ I : hi = this implies that Ih,k = I \ {i ∈ I : hi = ki } = {i ∈ I : hi = ki } belongs to ω (cf. (UF) in Sect. J.1). Using again the fact that ω is closed under ﬁnite intersections, we deduce that = Ih,k IK h,k∈K h=k
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also belongs to ω. Since ω is closed under ﬁnite intersections and ∅ ∈ / ω (cf. (F1) in Sect. J.1), we can ﬁnd an index i0 ∈ I such that . i0 ∈ IK ∩ IK
Consider the map ϕ : Pω → Gi0 deﬁned by ϕ(g) = gi0 for all g ∈ Pω . The map ϕ satisﬁes the following properties: i = (1) ϕ(hk) = hk hi 0 ki0 = ϕ(h)ϕ(k) for all h, k ∈ K (because i0 ∈ IK ); 0 (2) ϕ(h) = hi0 = ki0 = ϕ(k) for all h, k ∈ K such that h = k (because ). i0 ∈ IK This shows that ϕ is a Kalmost homomorphism. It follows that Pω is locally embeddable into the class C. Remark 7.2.4. When the ultraﬁlter ω is principal, then the group Pω itself belongs to C. Indeed, in this case, there is an element i0 ∈ I such that ω consists of all subsets of I containing i0 . This implies that α ∼ω β if and only if αi0 = βi0 for all α = (αi )i∈I , β = (βi )i∈I ∈ P . Therefore, Nω consists of all α ∈ P such that αi0 = 1Gi0 . It follows that the group Pω is isomorphic to the group Gi0 and therefore that Pω ∈ C. Theorem 7.2.5. Let C be a class of groups and let G be a group. The following conditions are equivalent: (a) G is locally embeddable into C; (b) there exists a family of groups (Gi )i∈I such that Gi ∈ C for all i ∈ I and an ultraﬁlter ω on I such that G is isomorphic to a subgroup of the ultraproduct Pω of the family (Gi )i∈I with respect to the ultraﬁlter ω. Proof. If G satisﬁes (b), then G is locally embeddable into C since Pω is locally embeddable into C by Theorem 7.2.3 and every subgroup of a group which is locally embeddable into C is itself locally embeddable into C by Proposition 7.1.7. Conversely, suppose that G is locally embeddable into C. Consider the set I consisting of all ﬁnite subsets of G. For each K ∈ I we deﬁne the set IK = {K ∈ I : K ⊂ K }. Observe that IK = ∅ as K ∈ IK . Moreover, the family of nonempty subsets (IK )K∈I is closed under ﬁnite intersections, since IK1 ∩ IK2 = IK1 ∪K2 for all K1 , K2 ∈ I. It follows from Proposition J.1.3 and Theorem J.1.6 that there exists an ultraﬁlter ω on I such that IK ∈ ω for all K ∈ I. As G is locally embeddable into C, we can ﬁnd, for each K ∈ I, a group GK in C and a Kalmosthomomorphism ϕK : G → GK .
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Consider the ultraproduct Pω of the family (GK )K∈I with respect to ω. By into C. Let ϕ : G → P = Theorem 7.2.3, Pω is locally embeddable K∈I GK denote the product map ϕ = K∈I ϕK . Thus, we have ϕ(g) = (ϕK (g))K∈I for all g ∈ G. Let ρ : P → Pω = P/Nω denote the canonical epimorphism. Let us show that the composite map Φ = ρ ◦ ϕ : G → Pω is an injective homomorphism. This will prove that G is isomorphic to a subgroup of Pω . Let g, h ∈ G. Consider the element K0 = {g, h} ∈ I. If K ∈ I satisﬁes K0 ⊂ K, then ϕK (gh) = ϕK (g)ϕK (h) since ϕK is a Kalmosthomomorphism. Thus, the set {K ∈ I : ϕK (gh) = ϕK (g)ϕK (h)} contains IK0 and therefore belongs to ω. This implies that ϕ(h). ϕ(gh) ∼ω ϕ(g) Therefore, we have Φ(gh) = Φ(g)Φ(h) by (7.7). This shows that Φ is a homomorphism. On the other hand, if g = h, we have ϕK (g) = ϕK (h) for all K ∈ I such that K0 ⊂ K. This implies that {K ∈ I : ϕK (g) = / ω, ϕK (h)} ∈ ω. As ω is a ﬁlter, it follows that {K ∈ I : ϕK (g) = ϕK (h)} ∈ that is, ϕ(g) ∼ω ϕ(h). Therefore we have Φ(g) = Φ(h), again by (7.7). Consequently, Φ is injective. This shows that (a) implies (b).
Remark 7.2.6. Suppose that G is a group and that C is a class of groups which is closed under taking subgroups. If G is locally embeddable into C and G ∈ /C then for any family of groups (Gi )i∈I and any ultraﬁlter ω on the index set I satisfying condition (b) in Theorem 7.2.5, we deduce from Remark 7.2.4 that the ultraﬁlter ω is necessarily nonprincipal.
7.3 LEFGroups and LEAGroups Let us rewrite the results obtained in Sect. 7.1 in the particular case when C is either the class of ﬁnite groups or the class of amenable groups. Note that these classes are closed under taking subgroups and taking ﬁnite direct
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products (this is trivial for ﬁnite groups and follows from Proposition 4.5.1 and Corollary 4.5.6 for amenable groups), so that all the results established in Sect. 7.1 apply to these two classes. We shall use the following terminology, which is very popular in the ﬁeld. A group which is locally embeddable into the class of ﬁnite groups is brieﬂy called an LEF group. Similarly, a group which is locally embeddable into the class of amenable groups is called an LEAgroup. Note that every LEFgroup is LEA since the class of ﬁnite groups is contained in the class of amenable groups by Proposition 4.4.6. We deduce from Proposition 7.1.6 the following characterization of LEFgroups. Proposition 7.3.1. Let G be a group. Then the following conditions are equivalent: (a) G is an LEFgroup; (b) for every ﬁnite subset K ⊂ G, there exist a ﬁnite set L with K ⊂ L ⊂ G and a binary operation : L × L → L such that (L, ) is a group and k1 k2 = k1 k2 for all k1 , k2 ∈ K. Analogously, we deduce from Proposition 7.1.6 the following characterization of LEAgroups. Proposition 7.3.2. Let G be a group. Then the following conditions are equivalent: (a) G is an LEAgroup; (b) for every ﬁnite subset K ⊂ G, there exist a set L with K ⊂ L ⊂ G and a binary operation : L × L → L such that (L, ) is an amenable group and k1 k2 = k1 k2 for all k1 , k2 ∈ K. From Theorem 7.2.5 we deduce the following characterization of LEF (resp. LEA) groups in terms of ultraproducts. Corollary 7.3.3. Let G be a group. The following conditions are equivalent: (a) G is an LEF (resp. LEA) group; (b) there exists a family of ﬁnite (resp. amenable) groups (Gi )i∈I and an ultraﬁlter ω on I such that G is isomorphic to a subgroup of the ultraproduct Pω of the family (Gi )i∈I with respect to the ultraﬁlter ω. From Corollary 7.1.15, we get: Proposition 7.3.4. Every locally residually ﬁnite group is LEF and therefore LEA.
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Corollary 7.3.5. All ﬁnite groups, all residually ﬁnite groups, all proﬁnite groups, all free groups, all abelian groups, and all locally ﬁnite groups are LEF and therefore LEA. Proof. In order to complete the proof, it suﬃces to recall that free groups (resp. proﬁnite groups) are residually ﬁnite by Theorem 2.3.1 (resp. Corollary 2.2.8) and that abelian groups are locally residually ﬁnite by Corollary 2.2.4. Similarly, by applying Corollary 7.1.15 to the class of amenable groups, we get: Proposition 7.3.6. Every locally residually amenable group is LEA.
From Proposition 7.1.7, Proposition 7.1.8, Proposition 7.1.10, Corollary 7.1.11, Corollary 7.1.12, Corollary 7.1.17, and Theorem 7.1.19, we get: Proposition 7.3.7. The following assertions hold: (i) every subgroup of an LEFgroup (resp. LEAgroup) is an LEFgroup (resp. LEAgroup); (ii) a group is LEF (resp. LEA) if and only if all its ﬁnitely generated subgroups are LEF (resp. LEA); a family of LEFgroups (resp. LEAgroups). Then their (iii) let (Gi )i∈I be direct product i∈I Gi is an LEFgroup (resp. LEAgroup); be a family of LEFgroups (resp. LEAgroups). Then their (iv) let (Gi )i∈I direct sum i∈I Gi is an LEFgroup (resp. LEAgroup); (v) let (Gi )i∈I be a projective system of LEFgroups (resp. LEAgroups). Then their projective limit G = lim Gi is an LEFgroup (resp. LEA←− group); (vi) Let Γ be a group. Then the set of all N ∈ N (Γ ) such that Γ/N is LEF (resp. LEA) is closed (and hence compact) in N (Γ ). (vii) let F be a free group and let N ∈ N (F ). Then F/N is an LEFgroup (resp. LEAgroup) if and only if there exists a net (Ni )i∈I in N (F ) with F/Ni ﬁnite (resp. amenable) for all i ∈ I such that N = limi Ni . Finally, we deduce from Corollary 7.1.21 the following: Proposition 7.3.8. A ﬁnitely presented group is LEF (resp. LEA) if and only if it is residually ﬁnite (resp. residually amenable). As we have observed at the end of the previous section, the additive group Q is LEF but not residually ﬁnite. On the other hand, it follows from Proposition 7.3.8 that every ﬁnitely presented LEFgroup is residually ﬁnite. The group Q is not ﬁnitely generated. However, there exist ﬁnitely generated LEFgroups which are not residually ﬁnite. An example of such a group is provided by the group G1 introduced in Sect. 2.6:
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Proposition 7.3.9. The group G1 of Sect. 2.6 is a ﬁnitely generated amenable LEFgroup which is not residually ﬁnite. Proof. By construction, the group G1 is ﬁnitely generated since it is deﬁned as being a subgroup of Sym(Z) generated by two elements, namely the transposition (0 1) and the translation n → n + 1. The group G1 is not residually ﬁnite by Proposition 2.6.1. Recall that, denoting by Sym0 (Z) the normal subgroup of Sym(Z) consisting of all permutations of Z with ﬁnite support, the group G1 is the semidirect product of Sym0 (Z) with the inﬁnite cyclic group generated by the translation T : n → n + 1 (see Lemma 2.6.4). Therefore, each element g ∈ G1 can be uniquely written in the form g = T i(g) σ(g), where i(g) ∈ Z and σ(g) ∈ Sym0 (Z). Note that we have i(gh) = i(g) + i(h) and σ(gh) = T −i(h) σ(g)T i(h) σ(h) for all g, h ∈ G1 . To prove that G1 is an LEFgroup, consider a ﬁnite subset K ⊂ G1 . Let us show that there exist a ﬁnite group F and a Kalmosthomomorphism of G1 into F . Let = maxk∈K i(k) and choose an integer r ≥ such that the supports of the elements T −j σ(k)T j ∈ Sym0 (Z) are contained in the interval [−r, r] for all − ≤ j ≤ and k ∈ K. Let us set R = 4r, X = {−R, −R + 1, . . . , −1, 0, 1, . . . , R − 1, R}, and F = Sym(X). Consider the (2R + 1)cycle γ ∈ F given by γ = (−R − R + 1 − R + 2 · · · R − 1 R). For all σ ∈ Sym0 (Z) whose support is contained in X, let σ ∈ F denote the element deﬁned by σ(x) = σ(x) for all x ∈ X. Observe that if σ, σ ∈ Sym0 (Z) have both their support contained in X, then so does σσ and that we have σσ = σσ .
(7.8)
Moreover, for all k ∈ K and j ∈ Z such that − ≤ j ≤ , the supports of σ(k) and T −j σ(k)T j are contained in X and we have T −j σ(k)T j = γ −j σ(k)γ j .
(7.9)
Indeed, suppose ﬁrst that x ∈ [−2r, 2r]. Then x + j ∈ [−3r, 3r] ⊂ X and we have [T −j σ(k)T j ](x) = [T −j σ(k)](x + j) = T −j (σ(k)(x + j)) = σ(k)(x + j) − j, and
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[γ −j σ(k)γ j ](x) = [γ −j σ(k)](x + j)
= γ −j σ(k)(x + j) = γ −j (σ(k)(x + j)) = σ(k)(x + j) − j. Suppose now that x ∈ X \ [−2r, 2r]. Then [T −j σ(k)T j ](x) = x since the support of T −j σ(k)T j is contained in [− + r, r + ] ⊂ [−2r, 2r]. On the other hand, if we denote, for y ∈ Z, by y the unique element in {y + (2R + 1)n : n ∈ Z} ∩ [−R, R], we have x + j ∈ X \ [−r, r]
(7.10)
and + j) [γ −j σ(k)γ j ](x) = [γ −j σ(k)](x
+ j) = γ −j σ(k)(x + j) by (7.10) = γ −j (x = x. This shows (7.9). Deﬁne a map ϕ : G1 → F by setting ϕ(g) = γ i(g) σ(g) if the support of σ(g) is contained in X, and ϕ(g) = 1F otherwise. Let us show that ϕ is a Kalmosthomomorphism. Let k1 , k2 ∈ K. We have ϕ(k1 k2 ) = ϕ[T i(k1 )+i(k2 ) (T −i(k2 ) σ(k1 )T i(k2 ) σ(k2 ))] = γ i(k1 )+i(k2 ) T −i(k2 ) σ(k1 )T i(k2 ) σ(k2 ) (by (7.8)) = γ i(k1 )+i(k2 ) · T −i(k2 ) σ(k1 )T i(k2 ) σ(k2 ) (by (7.9)) = γ i(k1 )+i(k2 ) · γ −i(k2 ) σ(k1 )γ i(k2 ) · σ(k2 ) = γ i(k1 ) σ(k1 )γ i(k2 ) σ(k2 ) = ϕ[T i(k1 ) σ(k1 )]ϕ[T i(k2 ) σ(k2 )] = ϕ(k1 )ϕ(k2 ). Let us show that ϕK is injective. Let k1 , k2 ∈ K and suppose that ϕ(k1 ) = ϕ(k2 ). This implies that γ i(k1 ) σ(k1 ) = γ i(k2 ) σ(k2 ), that is, γ i(k1 )−i(k2 ) = −1
−1
σ(k2 ) · σ(k1 ) . But the support of σ(k2 ) · σ(k1 ) is contained in [−r, r] while the support of γ i , i ∈ Z, is the whole set X if i is not a multiple of 2R + 1. As i(k1 ) − i(k2 ) ≤ 2 < 2R + 1, we deduce that i(k1 ) − i(k2 ) = 0 −1
and σ(k2 ) · σ(k1 ) = 1F . This implies i(k1 ) = i(k2 ) and σ(k1 ) = σ(k2 ), and therefore k1 = k2 . This shows that the restriction of ϕ to K is injective. It
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follows that ϕ is a Kalmost homomorphism and therefore that G1 is an LEF group. Finally, observe that the group Sym0 (Z) is locally ﬁnite and therefore amenable by Corollary 4.5.12. Now, G1 is the semidirect product of the amenable group Sym0 (Z) with an inﬁnite cyclic (and therefore amenable) group so that, by Proposition 4.5.5, it is an amenable group as well. Remarks 7.3.10. (a) From Proposition 7.3.4 and Proposition 7.3.9, we deduce that the class of locally residually ﬁnite groups is strictly contained in the class of LEFgroups. (b) Proposition 7.3.8 and Proposition 7.3.9 imply that the group G1 is not ﬁnitely presentable.
7.4 The Hamming Metric Let G be a group. A metric d on G is called leftinvariant (resp. rightinvariant) if d(hg1 , hg2 ) = d(g1 , g2 ) (resp. d(g1 h, g2 h) = d(g1 , g2 )) for all g1 , g2 , h ∈ G. A metric on G which is both left and rightinvariant is called biinvariant. Note that a leftinvariant (resp. rightinvariant) metric d on G is entirely determined by the map g → d(1G , g), g ∈ G, since d(h, k) = d(1G , h−1 k) (resp. d(h, k) = d(1G , kh−1 )) for all h, k ∈ G. Let now F be a nonempty ﬁnite set and consider the symmetric group Sym(F ). For α ∈ Sym(F ), we denote by Fix(α) the set {x ∈ F : α(x) = x} of ﬁxed points of α. The support of α is the set {x ∈ F : α(x) = x} = F \Fix(α), so that we have {x ∈ F : α(x) = x} = F  −  Fix(α).
(7.11)
Consider the map dF : Sym(F ) × Sym(F ) → R deﬁned by dF (α1 , α2 ) =
{x ∈ F : α1 (x) = α2 (x)} F 
(7.12)
for all α1 , α2 ∈ Sym(F ). Observe that the set {x ∈ F : α1 (x) = α2 (x)} = {x ∈ F : x = α1−1 α2 (x)} is the support of α1−1 α2 , so that (7.11) gives us dF (α1 , α2 ) = 1 −
 Fix(α1−1 α2 ) . F 
(7.13)
Proposition 7.4.1. Let F be a nonempty ﬁnite set. Then dF is a biinvariant metric on Sym(F ). Proof. It is immediate from the deﬁnition that dF (α1 , α2 ) ≥ 0 and dF (α1 , α2 ) = dF (α2 , α1 ) for all α1 , α2 ∈ Sym(F ). Moreover, the equality dF (α1 , α2 ) = 0 holds if and only if α1 (x) = α2 (x) for all x ∈ F , that is, if and only if α1 = α2 .
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Let now α1 , α2 , α3 ∈ Sym(F ). If x ∈ F satisﬁes α1 (x) = α2 (x), then α1 (x) = α3 (x) or α2 (x) = α3 (x). Thus, we have the inclusion {x ∈ F : α1 (x) = α2 (x)} ⊂ {x ∈ F : α1 (x) = α3 (x)} ∪ {x ∈ F : α2 (x) = α3 (x)}. This implies 1 {x ∈ F : α1 (x) = α2 (x)} F  1 {x ∈ F : α1 (x) = α3 (x)} ∪ {x ∈ F : α2 (x) = α3 (x)} ≤ F  1 ({x ∈ F : α1 (x) = α3 (x)} + {x ∈ F : α2 (x) = α3 (x)}) ≤ F  = dF (α1 , α3 ) + dF (α2 , α3 ).
dF (α1 , α2 ) =
This shows that dF also satisﬁes the triangle inequality. Therefore, dF is a metric on Sym(F ). It remains to show that dF is biinvariant. Let α1 , α2 , β ∈ Sym(F ). Since β is bijective, we have {x ∈ F : βα1 (x) = βα2 (x)} = {x ∈ F : α1 (x) = α2 (x)}. This implies that 1 {x ∈ F : βα1 (x) = βα2 (x)} F  1 {x ∈ F : α1 (x) = α2 (x)} = F  = dF (α1 , α2 ).
dF (βα1 , βα2 ) =
Thus, dF is leftinvariant. On the other hand, we have {x ∈ F : α1 β(x) = α2 β(x)} = β −1 ({x ∈ F : α1 (x) = α2 (x)}), which implies 1 {x ∈ F : α1 β(x) = α2 β(x)} F  1 −1 β ({x ∈ F : α1 (x) = α2 (x)}) = F  1 {x ∈ F : α1 (x) = α2 (x)} = F  = dF (α1 , α2 ).
dF (α1 β, α2 β) =
Consequently, dF is also rightinvariant.
Deﬁnition 7.4.2. Let F be a nonempty ﬁnite set. The biinvariant metric dF is called the (normalized) Hamming metric on Sym(F ). Suppose now that m is a positive integer and that F1 , F2 , . . . , Fm are nonempty ﬁnite sets. Consider the Cartesian product F = F1 × F2 × · · · × Fm
7.4 The Hamming Metric
253
m and the natural group homomorphism Φ : i=1 Sym(Fi ) → Sym(F ) deﬁned by Φ(α)(x) = (α1 (x1 ), α2 (x2 ), . . . , αm (xm )) m for all α = (α1 , α2 , . . . , αm ) ∈ i=1 Sym(Fi ) and x = (x1 , x2 , . . . , xm ) ∈ F . Proposition 7.4.3. With the above notation, one has dF (Φ(α), Φ(β)) = 1 −
m
(1 − dFi (αi , βi ))
(7.14)
i=1
for all α = (αi )1≤i≤m and β = (βi )1≤i≤m in
m
Proof. First observe that if α = (αi )1≤i≤m ∈ m Fix(Φ(α)) = i=1 Fix(αi ), and therefore
i=1 Sym(Fi ). m i=1 Sym(Fi ),
then we have
 Fix(α) F  m  Fix(αi ) m = 1 − i=1 i=1 Fi  m  Fix(αi ) =1− Fi  i=1
dF (IdF , Φ(α)) = 1 −
=1−
m
(1 − dFi (IdFi , αi )).
i=1
We deduce that, for all α = (αi )1≤i≤m , β = (βi )1≤i≤m ∈ have dF (Φ(α), Φ(β)) = d(IdF , Φ(α)−1 Φ(β))
m i=1
Sym(Fi ), we
(by leftinvariance of dF )
−1
= dF (IdF , Φ(α β)) m =1− (1 − dFi (IdFi , αi−1 βi ) =1−
i=1 m
(1 − dFi (αi , βi ))
i=1
where the last equality follows from the leftinvariance of dFi .
Corollary 7.4.4. Let m be a positive integer and let F be a nonempty ﬁnite set. Consider the homomorphism Ψ : Sym(F ) → Sym(F m ) deﬁned by Ψ (α)(x1 , x2 , . . . , xm ) = (α(x1 ), α(x2 ), . . . , α(xm ))
(7.15)
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7 Local Embeddability and Soﬁc Groups
for all α ∈ Sym(F ) and x1 , x2 , . . . , xm ∈ F . Then one has dF m (Ψ (α), Ψ (β)) = 1 − (1 − dF (α, β))m for all α, β ∈ Sym(F ).
(7.16)
7.5 Soﬁc Groups Deﬁnition 7.5.1. Let G be a group, K ⊂ G a ﬁnite subset, and ε > 0. Let F be a nonempty ﬁnite set. A map ϕ : G → Sym(F ) is called a (K, ε)almosthomomorphism if it satisﬁes the following conditions: ((K, ε)AH1) for all k1 , k2 ∈ K, one has dF (ϕ(k1 k2 ), ϕ(k1 )ϕ(k2 )) ≤ ε; ((K, ε)AH2) for all k1 , k2 ∈ K, k1 = k2 , one has dF (ϕ(k1 ), ϕ(k2 )) ≥ 1 − ε, where dF denotes the normalized Hamming metric on Sym(F ). Deﬁnition 7.5.2. A group G is called soﬁc if it satisﬁes the following condition: for every ﬁnite subset K ⊂ G and every ε > 0, there exist a nonempty ﬁnite set F and a (K, ε)almosthomomorphism ϕ : G → Sym(F ). Proposition 7.5.3. Every ﬁnite group is soﬁc. Proof. Let G be a ﬁnite group. Consider the map L : G → Sym(G) deﬁned by L(g)(h) = gh for all g, h ∈ G. As L is a homomorphism (cf. the proof of Cayley’s theorem (Theorem C.1.2)), we have dG (L(g1 g2 ), L(g1 )L(g2 )) = 0
(7.17)
for all g1 , g2 ∈ G. Moreover, for all distinct g1 , g2 ∈ G we have L(g1 )(h) = g1 h = g2 h = L(g2 )(h) for all h ∈ G so that dG (L(g1 ), L(g2 )) = 1.
(7.18)
This shows that L is a (K, ε)almosthomomorphism for all K ⊂ G and ε > 0. It follows that G is soﬁc. Proposition 7.5.4. Every subgroup of a soﬁc group is soﬁc. Proof. Let G be a soﬁc group and let H be a subgroup of G. Fix a ﬁnite subset K ⊂ H and ε > 0. As G is soﬁc, there exists a nonempty ﬁnite set F and a (K, ε)almosthomomorphism ϕ : G → Sym(F ). Then the restriction map ϕH : H → Sym(F ) is a (K, ε)almosthomomorphism. This shows that H is soﬁc. Proposition 7.5.5. Every locally soﬁc group is soﬁc.
7.5 Soﬁc Groups
255
Proof. Let G be a locally soﬁc group. Let K ⊂ G be a ﬁnite subset and ε > 0. Denote by H the subgroup of G generated by K. Then, as H is soﬁc, there exist a nonempty ﬁnite set F and a (K, ε)almosthomomorphism ψ : H → Sym(F ). Extend arbitrarily ψ to a map ϕ : G → Sym(F ), for example by setting ϕ(g) = IdF for all g ∈ G \ H. It is clear that ϕ is a (K, ε)almosthomomorphism. This shows that G is soﬁc. From Proposition 7.5.3 and Proposition 7.5.5, we immediately deduce that every locally ﬁnite group is soﬁc. As any locally ﬁnite group is amenable by Corollary 4.5.12, this is actually covered by the following: Proposition 7.5.6. Every amenable group is soﬁc. Proof. Suppose that G is an amenable group. Let K ⊂ G be a ﬁnite subset and ε > 0. Set S = ({1G } ∪ K ∪ K −1 )2 . Since G is amenable, it follows from Theorem 4.9.1 and Proposition 4.7.1(a) that there exists a nonempty ﬁnite subset F ⊂ G such that ε F  (7.19) F \ sF  ≤ S for all s ∈ S. Consider the set E = s∈S sF . Observe that E ⊂ F since 1G ∈ S. In fact, as S = S −1 , we get sE ⊂ F
(7.20)
for all s ∈ S. Moreover, we have F \ E = F \ =
sF 
s∈S
(F \ sF ) (7.21)
s∈S
≤
F \ sF 
s∈S
≤ εF 
by (7.19).
This implies E ≥ (1 − ε)F .
(7.22)
For each g ∈ G, we have F  = gF  and hence F \ gF  = gF \ F . Therefore, we can ﬁnd a bijective map αg : gF \ F → F \ gF . Consider the map ϕ : G → Sym(F ) deﬁned by setting gf if gf ∈ F ϕ(g)(f ) = αg (gf ) otherwise for all g ∈ G and f ∈ F (see Fig. 7.1).
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7 Local Embeddability and Soﬁc Groups
Fig. 7.1 The maps αg : gF \ F → F \ gF and ϕ(g) ∈ Sym(F ). We have ϕ(g)(f1 ) = gf1 , ϕ(g)(f2 ) = αg (gf2 ) and ϕ(g)(f3 ) = αg (gf3 )
Now, suppose that k1 , k2 ∈ K and f ∈ E. Then we have k2 , k1 k2 ∈ S, so that k2 f, k1 k2 f ∈ F by (7.20). This implies ϕ(k1 k2 )(f ) = k1 k2 f and (ϕ(k1 )ϕ(k2 ))(f ) = ϕ(k1 )(ϕ(k2 )(f )) = ϕ(k1 )(k2 f ) = k1 k2 f . Therefore, the permutations ϕ(k1 k2 ) and ϕ(k1 )ϕ(k2 ) coincide on E. From (7.21), we deduce that F \ E ≤ε dF (ϕ(k1 k2 ), ϕ(k1 )ϕ(k2 )) ≤ F  for all k1 , k2 ∈ K. On the other hand, if k1 , k2 ∈ K, k1 = k2 , and f ∈ E, then we have k1 f, k2 f ∈ F so that ϕ(k1 )(f ) = k1 f = k2 f = ϕ(k2 )(f ). By using (7.22), we deduce that E ≥1−ε dF (ϕ(k1 ), ϕ(k2 )) ≥ F  for all k1 , k2 ∈ K with k1 = k2 . Thus, the map ϕ : G → Sym(F ) is a (K, ε)almosthomomorphism. This shows that G is a soﬁc group. Observe that, when G is ﬁnite, the proof of Proposition 7.5.6 reduces to that of Proposition 7.5.3 by taking F = G. Proposition7.5.7. Let (Gi )i∈I be a family of soﬁc groups. Then, their direct product G = i∈I Gi is soﬁc. Proof. For each i ∈ I, let πi : G → Gi denote the projection homomorphism. Fix a ﬁnite subset K ⊂ G and ε > 0. Then there exists a ﬁnite subset J ⊂ I
7.5 Soﬁc Groups
257
such that the projection πJ = j∈J πj : G → GJ = j∈J Gj is injective on K. Choose a constant 0 < η < 1 small enough so that 1 − (1 − η)J ≤ ε
(7.23)
η ≤ ε.
(7.24)
and Since the group Gj is soﬁc for each j ∈ J, we can ﬁnd a nonempty ﬁnite set Fj and a (πj (K), η)almost homomorphism ϕj : Gj → Sym(Fj ). Consider the nonempty ﬁnite set F = j∈J Fj and the map ϕ : G → Sym(F ) deﬁned by ϕ(g)(f ) = (ϕj (gj )(fj ))j∈J for all g = (gi )i∈I ∈ G, and f = (fj )j∈J ∈ F . Then, for all k, k ∈ K, we have, by applying (7.14), 1 − dFj (ϕj (kj kj ), ϕj (kj )ϕj (kj ) dF (ϕ(kk ), ϕ(k)ϕ(k )) = 1 − j∈J
≤ 1 − (1 − η)J ≤ε
(by (7.23)).
On the other hand, if k and k are distinct elements in K, then there exists j0 ∈ J such that kj0 = kj 0 . This implies, again by using (7.14), dF (ϕ(k), ϕ(k )) = 1 −
(1 − dFj (ϕj (kj ), ϕj (kj ))
j∈J
≥ 1 − (1 − dFj0 (ϕj0 (kj0 ), ϕj0 (kj 0 )) ≥1−η ≥1−ε
(by (7.24)).
This shows that ϕ is a (K, ε)almosthomomorphism of G. It follows that G is soﬁc. Corollary7.5.8. Let (Gi )i∈I be a family of soﬁc groups. Then their direct sum G = i∈I Gi is soﬁc. Proof. This follows immediately from Proposition 7.5.4 and Proposition 7.5.7, since G is the subgroup of the direct product P = i∈I Gi consisting of all g = (gi ) ∈ P for which gi = 1Gi for all but ﬁnitely many i ∈ I. Corollary 7.5.9. The limit of a projective system of soﬁc groups is soﬁc. Proof. Let (Gi )i∈I be a projective system of soﬁc groups such that G = lim Gi . By construction of a projective limit (see Appendix E), G is a sub←−
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7 Local Embeddability and Soﬁc Groups
group of the group i∈I Gi . We deduce that G is soﬁc by using Proposition 7.5.7 and Proposition 7.5.4. Proposition 7.5.10. Every group which is locally embeddable into the class of soﬁc groups is soﬁc. Proof. Let G be a group which is locally embeddable into the class of soﬁc groups. Let K ⊂ G be a ﬁnite subset and ε > 0. By deﬁnition of local embeddability, there exists a soﬁc group G and a Kalmosthomomorphism ϕ : G → G . Set K = ϕ(K). By soﬁcity of G , there exists a nonempty ﬁnite set F and a (K , ε)almosthomomorphism ϕ : G → Sym(F ). Let us prove that the composite map Φ = ϕ ◦ ϕ : G → Sym(F ) is a (K, ε)almosthomomorphism. Let k1 , k2 ∈ K. Then we have dF (Φ(k1 k2 ), Φ(k1 )Φ(k2 )) = dF (ϕ (ϕ(k1 k2 )), ϕ (ϕ(k1 ))ϕ (ϕ(k2 ))) = dF (ϕ (ϕ(k1 )ϕ(k2 )), ϕ (ϕ(k1 ))ϕ (ϕ(k2 ))) ≤ε
(as ϕ(k1 ), ϕ(k2 ) ∈ K ).
Finally, let k1 , k2 ∈ K be such that k1 = k2 . Since ϕK is injective, we have ϕ(k1 ) = ϕ(k2 ) and therefore dF (Φ(k1 ), Φ(k2 )) = dF (ϕ (ϕ(k1 )), ϕ (ϕ(k2 ))) ≥ 1 − ε. Thus, Φ is a (K, ε)almosthomomorphism. It follows that G is soﬁc.
Since every amenable group is soﬁc by Proposition 7.5.6, an immediate consequence of Proposition 7.5.10 is the following: Corollary 7.5.11. Every LEAgroup is soﬁc. In particular, every LEFgroup, every locally residually amenable group, every locally residually ﬁnite group, every residually amenable group, and every residually ﬁnite group is soﬁc. As the class of soﬁc groups is closed under direct products by Proposition 7.5.7, it follows from Corollary 7.1.15 that every locally residually soﬁc group is locally embeddable into the class of soﬁc groups. By applying Proposition 7.5.10, we get: Corollary 7.5.12. Every locally residually soﬁc group is soﬁc.
It follows from Proposition 7.5.10 that the class of groups which are locally embeddable into the class of soﬁc groups coincide with the class of soﬁc groups. By applying Corollary 7.1.17, we then deduce the following: Corollary 7.5.13. Let Γ be a group. Then the set of Γ marked groups N ∈ N (Γ ) such that Γ/N is soﬁc is closed (and hence compact) for the prodiscrete topology on N (Γ ) ⊂ P(Γ ) = {0, 1}Γ .
7.5 Soﬁc Groups
259
We end this section by showing that extensions of soﬁc groups by amenable groups are soﬁc. This is a generalization of Proposition 7.5.6 since every amenable group is an extension of the trivial group by itself. Proposition 7.5.14. Let G be a group. Suppose that G contains a normal subgroup N such that N is soﬁc and G/N is amenable. Then G is soﬁc. Proof. Let K ⊂ G be a ﬁnite subset and 0 < ε < 1. Denote by g the image of an element g ∈ G under the canonical epimorphism of G onto G/N . Fix a set T ⊂ G of representatives for the cosets of N in G and denote by σ : G/N → T the map which associates with each element in G/N its representative in T . √ Note that σ(g)−1 g ∈ N for all g ∈ G. Also set ε = 1 − 1 − ε, so that 0 < ε < ε and (1 − ε )2 = 1 − ε. Since G/N is amenable and hence soﬁc by Proposition 7.5.6, there exist a nonempty ﬁnite set F1 and a (K, ε )almosthomomorphism ϕ1 : G/N → Sym(F1 ). In fact, in the proof of Proposition 7.5.6 it is shown that we can take F1 ⊂ G/N such that there exists a subset E1 ⊂ F1 with E1  ≥ (1 − ε )F1  satisfying ϕ1 (k)(f1 ) = kf1 ∈ F1 and ϕ1 (hk)(f1 ) = hkf1 ∈ F1 for all h, k ∈ K and f1 ∈ E1 . Set M = N ∩ (σ(F1 )−1 · K · σ(F1 )) ⊂ N . As N is soﬁc, we can ﬁnd a ﬁnite set F2 and an (M, ε )almosthomomorphism ϕ2 : N → Sym(F2 ). Thus, for all m, m ∈ M we can ﬁnd a set E2 ⊂ F2 such that E2  ≥ (1 − ε)F2  and ϕ2 (mm )(f2 ) = ϕ2 (m)(ϕ2 (m )(f2 )) for all f2 ∈ E2 .
(7.25)
Set F = F1 × F2 and E = E1 × E2 and observe that E = E1  · E2  ≥ (1 − ε )2 F1  · F2  = (1 − ε)F .
(7.26)
Consider the map Φ : G → Sym(F ) deﬁned by setting Φ(g)(f1 , f2 ) = (ϕ1 (g)(f1 ), ϕ2 (σ(gf1 )−1 gσ(f1 ))(f2 )) for all g ∈ G and (f1 , f2 ) ∈ F . Let us show that Φ is a (K, ε)almosthomomorphism. Let h, k ∈ K and (f1 , f2 ) ∈ E. Recall that the elements kf1 = ϕ1 (k)(f1 ) and hkf1 = ϕ1 (hk)(f1 ) = ϕ1 (h)(ϕ1 (k)(f1 )) both belong to F1 for all f1 ∈ E1 . It follows that Φ(h)Φ(k)(f1 , f2 ) = Φ(h)(ϕ1 (k)(f1 ), ϕ2 (σ(kf1 )−1 kσ(f1 ))(f2 )) = (ϕ1 (h)(ϕ1 (k)(f1 )), ϕ2 (σ(hϕ1 (k)(f1 ))−1 hσ(ϕ1 (k)(f1 ))) (ϕ2 (σ(kf1 )−1 kσ(f1 ))(f2 ))) = (ϕ1 (hk)(f1 ), ϕ2 (σ(hkf1 )−1 hσ(kf1 )) (ϕ2 (σ(kf1 )−1 kσ(f1 ))(f2 ))) =∗ (ϕ1 (hk)f1 , ϕ2 (σ(hkf1 )−1 hkσ(f1 ))(f2 )) = Φ(hk)(f1 , f2 )
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7 Local Embeddability and Soﬁc Groups
where =∗ follows from (7.25) since m = σ(hkf1 )−1 hσ(kf1 ) and m = σ(kf1 )−1 kσ(f1 ) both belong to (σ(F1 )−1 Kσ(F1 )) ∩ N = M . It follows from (7.26) that dF (Φ(hk), Φ(h)Φ(k)) ≤ ε. Let now h, k ∈ K be such that h = k. We distinguish two cases. If h = k then, as ϕ1 is a (K, ε)almosthomomorphism, we have dF1 (ϕ1 (h), ϕ1 (k)) ≥ 1 − ε. It follows that there exists a subset B ⊂ F1 such that B ≥ (1 − ε)F1  such that ϕ1 (h)(b) = ϕ1 (k)(b) for all b ∈ B. Setting B = B × F2 we have that
and
B  = B · F2  ≥ (1 − ε)F1  · F2  = (1 − ε)F 
(7.27)
Φ(h)(b ) = Φ(k)(b ) for all b ∈ B .
(7.28)
Suppose now that h = k. As ϕ2 is an (M, ε)almosthomomorphism one has dF2 (ϕ2 (m), ϕ2 (m )) > 1 − ε for all m, m ∈ M such that m = m . It follows that for all distinct m, m ∈ M there exists a subset D ⊂ F2 such that D ≥ (1 − ε)F2  and ϕ2 (m)(d) = ϕ2 (m )(d) for all d ∈ D. From h = k we deduce that for all f1 ∈ F1 one has that if m = σ(hf1 )−1 × hσ(f1 ) and m = σ(kf1 )−1 kσ(f1 ), then m, m ∈ M and m = σ(hf1 )−1 hσ(f1 ) = σ(kf1 )−1 hσ(f1 ) = σ(kf1 )−1 kσ(f1 ) = m . Thus ϕ2 (σ(hf1 )−1 hσ(f1 ))(d) = ϕ2 (σ(kf1 )−1 kσ(f1 ))(d) for all d ∈ D. Set D = D × F2 so that D  = D · F2  ≥ (1 − ε)F1  · F2  = (1 − ε)F .
(7.29)
It follows that for all d ∈ D one has Φ(h)(d ) = Φ(k)(d ) for all d ∈ D .
(7.30)
From (7.28) and (7.30), and taking into account (7.27) and (7.29) respectively, we deduce that in either cases dF (Φ(h), Φ(k)) ≥ 1 − ε. This shows that Φ : G → Sym(F ) is a (K, ε)almosthomomorphism. Thus G is soﬁc.
7.6 Soﬁc Groups and Metric Ultraproducts of Finite Symmetric Groups Suppose that we are given a triple T = (I, ω, F) consisting of the following data: a set I, an ultraﬁlter ω on I, and a family F = (Fi )i∈I of nonempty ﬁnite sets indexed by I. Our ﬁrst goal in this section is to associate with such a triple T a soﬁc group GT . We start by forming the direct product group PT = Sym(Fi ). i∈I
7.6 Soﬁc Groups and Metric Ultraproducts of Finite Symmetric Groups
261
Let α = (αi )i∈I and β = (βi )i∈I be elements of PT . Since 0 ≤ dFi (αi , βi ) ≤ 1 for all i ∈ I, it follows from Corollary J.2.6 that the Hamming distances dFi (αi , βi ) have a limit δω (α, β) = lim dFi (αi , βi ) ∈ [0, 1] i→ω
along the ultraﬁlter ω. Proposition 7.6.1. One has: (i) δω (α, α) = 0; (ii) δω (β, α) = δω (α, β); (iii) δω (α, β) ≤ δω (α, γ) + δω (γ, β); (iv) δω (γα, γβ) = δω (α, β); (v) δω (αγ, βγ) = δω (α, β). for all α, β, γ ∈ PT . Proof. Let α = (αi )i∈I , β = (βi )i∈I , γ = (γi )i∈I ∈ PT . For each i ∈ I, we have dFi (αi , αi ) = 0, dFi (βi , αi ) = dFi (αi , βi ), dFi (αi , βi ) ≤ dFi (αi , γi ) + dFi (γi , βi ), dFi (γi αi , γi βi ) = dFi (αi , βi ), and dFi (αi γi , βi γi ) = dFi (αi , βi ) since dFi is a biinvariant metric on Sym(Fi ). This gives us properties (i), (ii), (iii), (iv), and (v) for δω by taking limits along ω (cf. Corollary J.2.10). Consider now the subset NT ⊂ PT deﬁned by NT = {α ∈ PT : δω (1PT , α) = 0}. Proposition 7.6.2. The set NT is a normal subgroup of PT . Proof. This is an easy consequence of the properties of δω stated in Proposition 7.6.1. Indeed, we deduce from Property (i) that δω (1PT , 1PT ) = 0, that is, 1PT ∈ NT . On the other hand, by using successively (iv), (iii), and (ii), we get δω (1PT , α−1 β) = δω (α, β) ≤ δω (α, 1PT )+δω (1PT , β) = δω (1PT , α)+δω (1PT , β) for all α, β ∈ PT . This implies that α−1 β ∈ NT if α, β ∈ NT . Thus, NT is a subgroup of PT . Finally, Properties (iv) and (v) imply that δω (1PT , γαγ −1 ) = δω (γ −1 , αγ −1 ) = δω (γ −1 γ, α) = δω (1PT , α) for all α, γ ∈ PT . It follows that γαγ −1 ∈ NT for all α ∈ NT and γ ∈ PT . This shows that NT is a normal subgroup of PT . Observe that, given α = (αi )i∈I and β = (βi )i∈I in PT , one has αNT = βNT
⇐⇒ δω (α, β) = 0.
(7.31)
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7 Local Embeddability and Soﬁc Groups
Indeed, one has αNT = βNT if and only if α−1 β ∈ NT , that is, if and only if δω (1PT , α−1 β) = 0. This is equivalent to δω (α, β) = 0 by the leftinvariance of δω . Theorem 7.6.3. The group GT = PT /NT is soﬁc. Proof. Fix a ﬁnite subset K ⊂ GT and ε > 0. We want to show that there exist a nonempty ﬁnite set F and a (K, ε)almosthomomorphism ϕ : GT → Sym(F ). Choose a representative of each element g ∈ GT , that is, an element g = ( gi )i∈I ∈ PT such that g = gNT . We ﬁrst introduce some constants that will be used in the proof. If h and h, k) > 0 by (7.31). Let us set k are distinct elements in K, then δω ( η = min
h,k∈K h=k
δω ( h, k) . 2
(7.32)
Note that 0 < η ≤ 1/2. Now choose an integer m ≥ log ε/ log(1 − η), so that 1 − (1 − η)m ≥ 1 − ε.
(7.33)
Finally, choose a real number ξ with 0 < ξ < 1 suﬃciently small to make 1 − (1 − ξ)m ≤ ε.
(7.34)
T = If h and k are arbitrary elements of K, we have hkN h kNT and therefore δω (hk, hk) = 0 by (7.31). It follows that the set i, A(h, k) = {i ∈ I : dFi (hk hi ki ) ≤ ξ}
(7.35)
belongs to ω. On the other hand, if h and k are distinct elements of K, we have δω ( h, k) ≥ 2η by (7.32). As η > 0, this implies that the set C(h, k) = {i ∈ I : dFi ( hi , ki ) ≥ η}
(7.36)
belongs to ω. As any ﬁnite intersection of elements of ω is in ω and therefore nonempty, we deduce that there exists an index j ∈ I such that ⎞ ⎛ ⎞ ⎛ ⎜ ⎟ A(h, k)⎠ ⎜ C(h, k)⎟ j∈⎝ ⎠. ⎝ h,k∈K
h,k∈K h=k
Consider the map ψ : GT → Sym(Fj ) deﬁned by ψ(g) = gj for all g ∈ GT . It immediately follows from (7.35) and (7.36) that the map ψ satisﬁes the following properties:
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263
(1) dFj (ψ(hk), ψ(h)ψ(k)) ≤ ξ for all h, k ∈ K; (2) dFj (ψ(h), ψ(k)) ≥ η for all h, k ∈ K such that h = k. Consider now the Cartesian product F = Fj × Fj × · · · × Fj m times
and the homomorphism Ψ : Sym(Fj ) → Sym(F ) deﬁned by Ψ (σ)(x1 , x2 , . . . , xm ) = (σ(x1 ), σ(x2 ), . . . , σ(xm )) for all σ ∈ Sym(Fj ) and (x1 , x2 , . . . , xm ) ∈ F . By Corollary 7.4.4 we have dF (Ψ (σ), Ψ (σ )) = 1 − (1 − dFj (σ, σ ))m for all σ, σ ∈ Sym(Fj ). It follows that the composite map ϕ = Ψ ◦ ψ : G → Sym(F ) satisﬁes the following properties: (1’) dF (ϕ(hk), ϕ(h)ϕ(k)) ≤ 1 − (1 − ξ)m for all h, k ∈ K; (2’) dF (ϕ(h), ϕ(k)) ≥ 1 − (1 − η)m for all h, k ∈ K such that h = k. As 1−(1−ξ)m ≤ ε by (7.34) and 1−(1−η)m ≥ 1−ε by (7.33), we deduce that ϕ is a (K, ε)almosthomomorphism. This shows that GT is a soﬁc group. Remark 7.6.4. When the ultraﬁlter ω is principal, then the group GT is ﬁnite. Indeed, in this case, there is an element i0 ∈ I such that ω consists of all subsets of I containing i0 . This implies that δω (α, β) = dFi0 (αi0 , βi0 ) for all α = (αi )i∈I , β = (βi )i∈I ∈ PT . Therefore, NT consists of all α ∈ PT such that αi0 = 1Sym(Fi0 ) . It follows that the group GT is isomorphic to the group Sym(Fi0 ). Remark 7.6.5. Let α, α , β, β ∈ PT such that αNT = α NT and βNT = β NT . By applying Proposition 7.6.1(iii) and (7.31), we get dω (α, β) ≤ dω (α, α ) + dω (α , β ) + dω (β , β) = dω (α , β ). By exchanging the roles of α and α and of β and β , we obtain dω (α , β ) ≤ dω (α, β). It follows that dω (α, β) = dω (α , β ). Therefore, if g, h ∈ GT and α, β ∈ PT are such that g = αNT and h = βNT , the quantity Δω (g, h) = δω (α, β)
(7.37)
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is well deﬁned. Moreover, the map Δω : GT × GT → [0, 1] given by (7.37) is a biinvariant metric on GT . This follows immediately from Proposition 7.6.1 taking into account that NT is a normal subgroup. Theorem 7.6.6. Let G be a group. The following conditions are equivalent: (a) G is soﬁc; (b) there exists a triple T = (I, ω, F), where I is a set, ω is an ultraﬁlter on I, and F = (Fi )i∈I is a family of nonempty ﬁnite sets indexed by I, such that G is isomorphic to a subgroup of the group GT = PT /NT . Proof. If G satisﬁes (b), then G is soﬁc since GT is soﬁc by Theorem 7.6.3 and every subgroup of a soﬁc group is itself soﬁc by Proposition 7.5.4. Conversely, suppose that G is soﬁc. Consider the set I consisting of all pairs (K, ε), where K is a ﬁnite subset of G and ε > 0. We partially order the set I by setting (K, ε) (K , ε ) if K ⊂ K and ε ≤ ε. For each i = (K, ε) ∈ I we deﬁne the set Ii = {j ∈ I : i j} ⊂ I. Observe that Ii = ∅ as i ∈ Ii . Moreover, the family of nonempty subsets {Ii }i∈I is closed under ﬁnite intersections, since I(K1 ,ε1 ) ∩ I(K2 ,ε2 ) = I(K1 ∪K2 ,min{ε1 ,ε2 }) for all (K1 , ε1 ), (K2 , ε2 ) ∈ I. It follows from Proposition J.1.3 and Theorem J.1.6 that there exists an ultraﬁlter ω on I such that I(K,ε) ∈ ω for all (K, ε) ∈ I. As G is soﬁc, we can ﬁnd, for each i = (K, ε) ∈ I, a nonempty ﬁnite set Fi and a (K, ε)almosthomomorphism ϕi : G → Sym(Fi ). Consider the triple T = (I, ω, F), where F = (Fi )i∈I , and the associated soﬁc group : G → PT denote the product map ϕ = i∈I ϕi . Thus, GT = PT /NT . Let ϕ we have ϕ(g) = (ϕi (g))i∈I for all g ∈ G. Let ρ : PT → GT = PT /NT denote the canonical epimorphism. Let us show that the composite map Φ = ρ ◦ ϕ : G → GT is an injective homomorphism. This will prove that G is isomorphic to a subgroup of GT . Let g, h ∈ G and let η > 0. Consider the element i0 = ({g, h}, η) ∈ I. If i = (K, ε) ∈ I satisﬁes i0 i, then dFi (ϕi (gh), ϕi (g)ϕi (h)) ≤ ε ≤ η since ϕi is a (K, ε)almosthomomorphism. Thus, the set {i ∈ I : dFi (ϕi (gh), ϕi (g)ϕi (h)) ≤ η} contains Ii0 and therefore belongs to ω. This implies that
7.7 A Characterization of Finitely Generated Soﬁc Groups
265
δω (ϕ(gh), ϕ(g) ϕ(h)) = lim dFi (ϕi (gh), ϕi (g)ϕi (h)) = 0. i→ω
Therefore, we have Φ(gh) = Φ(g)Φ(h) by (7.31). This shows that Φ is a homomorphism. On the other hand, if g = h, we have dFi (ϕi (g), ϕi (h)) ≥ 1 − ε ≥ 1 − η for all i ∈ I such that i0 i. This implies that ϕ(h)) = lim dFi (ϕi (g), ϕi (h)) = 1. δω (ϕ(g), i→ω
(7.38)
Therefore, we have Φ(g) = Φ(h) by (7.31). Consequently, Φ is injective. This shows that (a) implies (b). Remarks 7.6.7. (a) If G is an inﬁnite soﬁc group and T = (I, ω, F ) is a triple satisfying condition (b) in Theorem 7.6.6, then the ultraﬁlter ω is necessarily nonprincipal by Remark 7.6.4. (b) Let G be a soﬁc group and let Φ : G → GT be as in the proof of Theorem 7.6.6. Using the notation from Remark 7.6.5, we deduce from (7.38) that Δω (Φ(g), Φ(h)) = 1 for all g, h ∈ G such that g = h. It follows that the restriction of the biinvariant metric Δω to the subgroup Φ(G) ⊂ GT is the discrete metric.
7.7 A Characterization of Finitely Generated Soﬁc Groups In this section we give a geometric characterization of ﬁnitely generated soﬁc groups in terms of a ﬁniteness condition on their Cayley graphs. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. Given r ∈ N, we denote by BS (r) the ball of radius r centered at the vertex corresponding to the identity element 1G of G in the Cayley graph CS (G) of G with respect to S, with the induced Slabeled graph structure (cf. Sects. 6.1, 6.2 and 6.3). Let also Q = (Q, E) be an Slabeled graph. Given q ∈ Q and r ∈ N, we denote by B(q, r) the ball of radius r centered at q with the induced Slabeled graph structure. Given r ∈ N we denote by Q(r) the set of all q ∈ Q such that there exists an Slabeled graph isomorphism ψq,r : BS (r) → B(q, r)
(7.39)
ψq,r (1G ) = q.
(7.40)
satisfying
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Observe that if such a map ψq,r exists it is unique. We have the inclusions Q = Q(0) ⊃ Q(1) ⊃ Q(2) ⊃ · · · ⊃ Q(r) ⊃ Q(r + 1) ⊃ · · ·
(7.41)
Fig. 7.2 The inclusions Q ⊃ Q(1) ⊃ Q(2) ⊃ Q(3)
Note also that since CS (G) (and therefore the induced Slabeled subgraph BS (r)) is edgesymmetric (with respect to the involution s → s−1 on S (cf. Sect. 6.3)) and ψq,r is an Slabeled graph isomorphism, then B(q, r) = ψq,r (BS (r)) is edgesymmetric as well. Theorem 7.7.1. Let G be a ﬁnitely generated group and let S be a ﬁnite symmetric generating subset of G. The following conditions are equivalent: (a) the group G is soﬁc; (b) for all ε > 0 and r ∈ N, there exists a ﬁnite Slabeled graph Q = (Q, E) such that Q(r) ≥ (1 − ε)Q, (7.42) where Q(r) ⊂ Q denotes the set consisting of all vertices q ∈ Q for which there exists an Slabeled graph isomorphism ψq,r : BS (r) → B(q, r) from the ball BS (r) in the Cayley graph CS (G) of G with respect to S onto the ball B(q, r) in Q satisfying ψq,r (1G ) = q. Before starting the proof of the theorem we present some preliminary results. Lemma 7.7.2. Let Q = (Q, E) be an Slabeled graph and r0 , i ∈ N. Suppose that q0 ∈ Q((i + 1)r0 ). Then B(q0 , r0 ) ⊂ Q(ir0 ). Proof. Let q ∈ B(q, r0 ) and let us show that q ∈ Q(ir0 ). It follows from the triangle inequality that the ball B(q , ir0 ) is entirely contained in the ball B(q0 , (i + 1)r0 ). Moreover, since ψq0 ,(i+1)r0 is isometric, setting g =
7.7 A Characterization of Finitely Generated Soﬁc Groups
267
ψq−1 (q ), we have g ∈ BS (r0 ) so that gh ∈ B((i+1)r0 ) for all h ∈ B(ir0 ). 0 ,(i+1)r0 It follows that the map ψq ,ir0 : BS (ir0 ) → B(q , ir0 )
(7.43)
deﬁned by ψq ,ir0 (h) = ψq0 ,(i+1)r0 (gh) for all h ∈ BS (ir0 ) yields an Slabeled graph isomorphism satisfying ψq ,ir0 (1G ) = ψq0 ,(i+1)r0 (g) = q . This shows that q ∈ Q(ir0 ). We deduce that B(q0 , r0 ) ⊂ Q(ir0 ). Lemma 7.7.3. Let Q = (Q, E) be an Slabeled graph and r0 ∈ N. Let q1 , q2 ∈ Q(2r0 ) such that q1 = q2 and g ∈ BS (r0 ). Then we have ψq1 ,2r0 (g) = ψq2 ,2r0 (g).
(7.44)
Proof. If g = 1G we have ψq1 ,2r0 (g) = ψq1 ,2r0 (1G ) = q1 = q2 = ψq2 ,2r0 (1G ) = ψq2 ,2r0 (g). Suppose now that g = 1G . Suppose by contradiction that ψq1 ,2r0 (g) = ψq2 ,2r0 (g) = q0 . Since ψq1 ,2r0 is isometric we have q0 ∈ B(q1 , r0 ). It follows from Lemma 7.7.2 that q0 ∈ Q(r0 ). As g ∈ BS (r0 ), we can ﬁnd 1 ≤ r ≤ r0 and s1 , s2 , . . . , sr ∈ S such that g = s1 s2 · · · sr . Consider the path π = ((1G , s1 , s1 ), (s1 , s2 , s1 s2 ), . . . , (s1 s2 · · · sr −1 , sr , g)) and observe that it is contained in BS (r0 ). Now, π is mapped by ψq1 ,2r0 and ψq2 ,2r0 into two paths π1 and π2 in Q with initial vertices π1− = q1 , π2− = q2 and same terminal vertex π1+ = ψq1 ,2r0 (g) = q0 = ψq2 ,2r0 (g) = π2+ . Note that since ψq1 ,2r0 and ψq2 ,2r0 are isometric π1 and π2 are both contained in B(q0 , r0 ). Moreover, since ψq1 ,2r0 and ψq2 ,2r0 are labelpreserving, π1 and π2 have the same label s1 s2 · · · sr . The inverse images of π1 and π2 under the Slabel preserving graph isomorphism ψq0 ,r0 : BS (r0 ) → B(q0 , r0 ) are both equal to π. Indeed in a Cayley graph there exists a unique path which ends at a given vertex and with a given label. It follows that π1 = π2 and therefore q1 = π1− = π2− = q2 . This contradicts our assumptions. We deduce that ψq1 ,2r0 (g) = ψq2 ,2r0 (g). Lemma 7.7.4. Let Q = (Q, E) be an Slabeled graph and r0 ∈ N. Let h, k ∈ BS (r0 ) and q0 ∈ Q(2r0 ). We have ψq0 ,2r0 (h) ∈ Q(r0 )
(7.45)
ψq0 ,2r0 (hk) = ψψq0 ,2r0 (h),r0 (k),
(7.46)
and where ψq,r , q ∈ Q(r), r ∈ N, is as in (7.39) and (7.40). Proof. Since ψq0 ,2r0 is isometric we have ψq0 ,2r0 (h) ∈ B(q0 , r0 ). Then (7.45) follows from Lemma 7.7.2. Let us show (7.46). First note that (7.46) makes sense by virtue of (7.45). If k = 1G (7.46) follows from ψq0 ,2r0 (hk) =
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ψq0 ,2r0 (h) = ψψq0 ,2r0 (h),r0 (1G ) = ψψq0 ,2r0 (h),r0 (k). Now suppose that k = 1G . Then we can ﬁnd 1 ≤ r ≤ r0 and s1 , s2 , . . . , sr ∈ S such that k = s1 s2 · · · sr . Consider the path π1 = ((h, s1 , hs1 ), (hs1 , s2 , hs1 s2 ), . . . , (hs1 s2 · · · sr −1 , sr , hk)) and observe that it is contained in BS (2r0 ), since h, k ∈ BS (r0 ). The path π1 is mapped by ψq0 ,2r0 into the path π 1 = ((ψq0 ,2r0 (h),s1 , ψq0 ,2r0 (hs1 )), (ψq0 ,2r0 (hs1 ), s2 , ψq0 ,2r0 (hs1 s2 )), . . . . . . , (ψq0 ,2r0 (hs1 s2 · · · sr −1 ), sr , ψq0 ,2r0 (hk))). As ψq0 ,2r0 is isometric, we have that q = ψq0 ,2r0 (h) belongs to B(q0 , r0 ) and since q0 ∈ Q(2r0 ) we deduce from Lemma 7.7.2 that q ∈ Q(r0 ). Consider the inverse image of the path π 1 under the Slabeled graph isomorphism ψq ,r0 . −1 −1 − Since ψq−1 ,r ((π 1 ) ) = ψq ,r (ψq0 ,2r0 (h)) = ψq ,r (q ) = 1G and ψq ,r0 preserves 0 0 0 the label, this inverse image is necessarily equal to the path π2 = ((1G , s1 , s1 ), (s1 , s2 , s1 s2 ), . . . , (s1 s2 · · · sr −1 , sr , k)), since in a Cayley graph there exists a unique path which starts at a given ver−1 + tex and with a given label. It follows that ψq−1 ,r (ψq0 ,2r0 (hk)) = ψq ,r (π 1 ) = 0 0 π2+ = k, that is, ψq0 ,2r0 (hk) = ψq ,r0 (k). Since q = ψq0 ,2r0 (h), we deduce (7.46). We are now in position to prove Theorem 7.7.1. Proof of Theorem 7.7.1. Suppose that G is soﬁc. Fix ε > 0 and r ∈ N. Set K = BS (2r + 1) and ε = ε(1 + BS (r) · S + BS (r)2 )−1 .
(7.47)
Since G is soﬁc, there exists a nonempty ﬁnite set F and a (K, ε )almosthomomorphism ϕ : G → Sym(F ). We construct an Slabeled graph Q = (Q, E) as follows. We take as vertex set Q = F . Then, as set of edges we take the set E ⊂ Q × S × Q consisting of all the triples (q, s, ϕ(s−1 )(q)), were q ∈ Q and s ∈ S. Note that Q may have loops and multiple edges and that Q is not necessarily edgesymmetric with respect to the involution s → s−1 on S. Observe however that, if q ∈ Q and s ∈ S are ﬁxed, then there exists a unique edge in Q with initial vertex q and label s. For each q ∈ Q denote by ψq : G → Q the map deﬁned by setting ψq (g) = ϕ(g −1 )(q) for all g ∈ G. Denote by Q0 the subset of Q consisting of all q ∈ Q satisfying the following conditions: (∗) ψq (1G ) = q, (∗∗) ψq (gs) = ψψq (g) (s) for all g ∈ BS (r) and s ∈ S, (∗ ∗ ∗) ψq (g) = ψq (h) for all g, h ∈ BS (r) with g = h.
7.7 A Characterization of Finitely Generated Soﬁc Groups
269
Suppose that q ∈ Q0 . Let g ∈ BS (r). If g = 1G we have ψq (g) = ψq (1G ) = q ∈ B(q, r), by (∗). If g = 1G , then there exist 1 ≤ r ≤ r and s1 , s2 , . . . , sr ∈ S such that g = s1 s2 · · · sr . Consider the sequence of edges e1 = (q, s1 , ϕ(s−1 1 )(q)) = (q, s1 , ψq (s1 )), e2 = (ψq (s1 ), s2 , ϕ(s−1 2 )ψq (s1 )) = (ψq (s1 ), s2 , ψψq (s1 ) (s2 )) = (ψq (s1 ), s2 , ψq (s1 s2 )) (by (∗∗)), ······ er = (ψq (s1 s2 · · · sr −1 ), sr , ϕ(s−1 r )(ψq (s1 s2 · · · sr −1 ))) = (ψq (s1 s2 · · · sr −1 ), sr , ψψq (s1 s2 ···sr −1 ) (sr )) = (ψq (s1 s2 · · · sr −1 ), sr , ψq (s1 s2 · · · sr −1 sr )) (by (∗∗)) = (ψq (s1 s2 · · · sr −1 ), sr , ψq (g)). The path π = (e1 , e2 , . . . , er ) connects q to ψq (g) and has length (π) ≤ r. This shows that the graph distance from q to ψq (g) in Q does not exceed r so that ψq (BS (r)) ⊂ B(q, r). Conversely, let q ∈ B(q, r). If q = q then by (∗) we have q = q = ψq (1G ) ∈ ψq (BS (r)). If q = q then there exist 1 ≤ r ≤ r and a sequence of edges (q, s1 , q1 ), (q1 , s2 , q2 ), . . . , (qr −1 , sr , q ) ∈ E. Using (∗∗) as above, we get q = ψq (g), where g = s1 s2 · · · sr ∈ BS (r). This shows that B(q, r) ⊂ ψq (BS (r)). Thus B(q, r) = ψq (BS (r)). Moreover, by condition (∗ ∗ ∗), the map ψq BS (r) is injective. Finally, from (∗∗) we deduce that if g ∈ BS (r) and s ∈ S then we have (ψq (g), s, ψq (gs)) = (ψq (g), s, ψψq (g) (s)) = (ψq (g), s, ϕ(s−1 )ψq (g)) ∈ E. Thus, the map ψq,r = ψq BS (r) : BS (r) → B(q, r) is an Slabeled graph isomorphism such that ψq,r (1G ) = ψq (1G ) = q, where the last equality follows from (∗). We deduce that Q0 ⊂ Q(r). Let’s now estimate from below the cardinality of Q0 . We denote by dQ the normalized Hamming metric on Sym(Q). In order to estimate the cardinality of the set of q ∈ Q for which condition (∗) is satisﬁed, let us ﬁrst observe that dQ (IdQ , ϕ(1G )) ≤ ε .
(7.48)
Indeed, since ϕ is a (K, ε )almosthomomorphism, taking k1 = k2 = 1G in property (i) of Deﬁnition 7.5.1, we deduce that dQ (ϕ(1G ), ϕ(1G )ϕ(1G )) ≤ ε which, by leftinvariance of dQ , implies (7.48). From (7.48) we deduce the
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existence of a subset Q ⊂ Q of cardinality Q  ≤ ε Q such that ϕ(1G )(q) = q
(7.49)
for all q ∈ Q \ Q . It follows that ψq (1G ) = ϕ(1−1 G )(q) = ϕ(1G )(q) = q, for all q ∈ Q \ Q so that condition (∗) is satisﬁed in Q \ Q . Let now estimate the cardinality of the set of q ∈ Q for which condition (∗∗) is satisﬁed. Let g ∈ BS (r) and s ∈ S. As ϕ is a (K, ε )almosthomomorphism we have dQ (ϕ(s−1 g −1 ), ϕ(s−1 )ϕ(g −1 )) ≤ ε so that there exists a subset Q (g, s) ⊂ Q with Q (g, s) ≤ ε Q such that ψq (gs) = ϕ((gs)−1 )(q) = ϕ(s−1 g −1 )(q) = ϕ(s−1 )ϕ(g −1 )(q) = ϕ(s−1 )(ψq (g)) = ψψq (g) (s) for all q ∈ Q \ Q (g, s). Setting Q =
Q (g, s)
g∈BS (r) s∈S
we have Q  ≤ BS (r) · Sε Q and condition (∗∗) holds for all q ∈ Q \ Q . Fix now two distinct elements g, h ∈ BS (r). Since ϕ is a (K, ε )almosthomomorphism and g −1 , h−1 ∈ K, by property (ii) of Deﬁnition 7.5.1 we deduce that dQ (ϕ(g −1 ), ϕ(h−1 )) ≥ 1 − ε . Thus we can ﬁnd a subset Q (g, h) ⊂ Q of cardinality Q (g, h) ≤ ε Q such that ψq (g) = ϕ(g −1 )(q) = ϕ(h−1 )(q) = ψq (h)
(7.50)
for all q ∈ Q \ Q (g, h). Let now g = h vary in BS (r) and set
Q = Q (g, h). g,h∈BS (r) g=h
Observe that Q  ≤ BS (r)2 ε Q. It follows from (7.50) that condition (∗ ∗ ∗) holds for all q ∈ Q \ Q . In conclusion, conditions (∗), (∗∗) and (∗ ∗ ∗) are satisﬁed for all q ∈ Q outside of Q ∪ Q ∪ Q . We have Q ∪ Q ∪ Q  ≤ (1 + BS (r) · S + BS (r)2 )ε Q = εQ,
7.7 A Characterization of Finitely Generated Soﬁc Groups
271
where the equality follows from (7.47). We deduce that Q(r) ≥ Q0  ≥ (1 − ε)Q. Thus G satisﬁes condition (b). This shows (a) ⇒ (b). Conversely, suppose (b). Fix a ﬁnite set K ⊂ G and ε > 0. Let r0 ∈ N be such that K ∪ K 2 ⊂ BS (r0 ). Let Q = (Q, E) be the ﬁnite Slabeled graph given by condition (b) corresponding to r = 2r0 and ε. Let g ∈ BS (r0 ). Since the map from Q(2r0 ) into Q deﬁned by q → ψq,2r0 (g) is injective (by Lemma 7.44), we have that {ψq,2r0 (g) : q ∈ Q(2r0 )} = Q(2r0 )
(7.51)
Q \ {ψq,2r0 (g) : q ∈ Q(2r0 )} = Q \ Q(2r0 ).
(7.52)
and therefore
As a consequence, there exists a bijection αg : Q \ Q(2r0 ) → Q \ {ψq,2r0 (g) : q ∈ Q(2r0 )}. Since ψq,2r0 (1G ) = q for all q ∈ Q(2r0 ), we have {ψq,2r0 (1G ) : q ∈ Q(2r0 )} = Q(2r0 ) and therefore we can take α1G = IdQ\Q(2r0 ) .
(7.53)
Consider now the map ϕ : G → Sym(Q) deﬁned by ⎧ −1 ⎪ ⎨ψq,2r0 (g ) ϕ(g)(q) = αg−1 (q) ⎪ ⎩ q
if g ∈ BS (r0 ) and q ∈ Q(2r0 ); if g ∈ BS (r0 ) and q ∈ Q \ Q(2r0 );
(7.54)
otherwise.
Note that ϕ(g) ∈ Sym(Q) for all g ∈ G, by construction. Let us show that the map ϕ : G → Sym(Q) is a (K, ε)almosthomomorphism. Let k1 , k2 ∈ K ⊂ BS (r0 ) and q ∈ Q(2r0 ). We have ϕ(k1 k2 )(q) = ψq,2r0 (k2−1 k1−1 ) = ψψq,2r
−1 0 (k2 ),r0
(k1−1 ) (by (7.46))
= ϕ(k1 )(ψq,2r0 (k2−1 )) = [ϕ(k1 )ϕ(k2 )](q). This shows that on Q(2r0 ) we have ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ). As Q(2r0 ) = Q(r) ≥ (1 − ε)Q
(7.55)
we deduce that dQ (ϕ(k1 k2 ), ϕ(k1 )ϕ(k2 )) ≤ ε. Finally, suppose that k1 = k2 . We have ϕ(k1 )(q) = ψq,2r0 (k1−1 ) = ψq,2r0 (k2−1 ) = ϕ(k2 )(q), since k1−1 , k2−1 ∈ BS (2r0 ) and ψq,2r0 is injective.
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7 Local Embeddability and Soﬁc Groups
From (7.55) we deduce that dQ (ϕ(k1 ), ϕ(k2 )) ≥ 1 − ε. It follows that ϕ is a (K, ε)almosthomomorphism. Therefore G is soﬁc. This shows that (b) ⇒ (a).
7.8 Surjunctivity of Soﬁc Groups In this section we prove that soﬁc groups are surjunctive. Note that this result covers the fact that locally residually amenable groups are surjunctive, which had been previously established in Corollary 5.9.3. Indeed, all locally residually amenable groups are soﬁc by Corollary 7.5.11. Theorem 7.8.1 (GromovWeiss). Every soﬁc group is surjunctive. Let us ﬁrst establish the following: Lemma 7.8.2. Let G be a group, A a ﬁnite set, and equip AG with the prodiscrete topology. Let X ⊂ AG be a closed Ginvariant subset and let f : X → AG be a continuous Gequivariant map. Then there exists a cellular automaton τ : AG → AG such that f = τ X . Proof. From the continuity of f we deduce the existence of a ﬁnite set S ⊂ G such that if two conﬁgurations y, z ∈ X coincide on S then f (y)(1G ) = f (z)(1G ). Let a0 ∈ A and consider the map μ : AS → A deﬁned by setting f (x)(1G ) if there exists x ∈ X such that xS = u μ(u) = otherwise a0 for all u ∈ AS . Then μ is well deﬁned and if we denote by τ : AG → AG the cellular automaton with memory set S and local deﬁning map μ, by G equivariance of f we clearly have τ X = f . Proof of Theorem 7.8.1. Let G be a soﬁc group. Let A be a ﬁnite set of cardinality A ≥ 2 and let τ : AG → AG be an injective cellular automaton. We want to show that τ is surjective. Every subgroup of a soﬁc group is soﬁc by Proposition 7.5.4. On the other hand it follows from Proposition 3.2.2 that a group is surjunctive if all its ﬁnitely generated subgroups are surjunctive. Thus we can assume that G is ﬁnitely generated. Let then S ⊂ G be a ﬁnite symmetric generating subset of G. As usual, for r ∈ N, we denote by BS (r) ⊂ G the ball of radius r centered at 1G in the Cayley graph of G with respect to S. We set Y = τ (AG ). Observe that Y is Ginvariant and, by Lemma 3.3.2, it is closed in AG . The inverse map τ −1 : Y → AG is Gequivariant and, by compactness of G A , it is also continuous. By Lemma 7.8.2, there exists a cellular automaton σ : AG → AG such that σY = τ −1 : Y → AG . Choose r0 large enough so
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that the ball BS (r0 ) is a memory set for both τ and σ. Let μ : ABS (r0 ) → A and ν : ABS (r0 ) → A denote the corresponding local deﬁning maps for τ and σ respectively. We proceed by contradiction. Suppose that τ is not surjective, that is, Y AG . Then, since Y is closed in AG , there exists a ﬁnite subset Ω ⊂ G such that πΩ (Y ) AΩ , where, for a subset E ⊂ G, we denote by πE : AG → AE the projection map. It is not restrictive, up to taking a larger r0 , again if necessary, to suppose that Ω ⊂ BS (r0 ). Thus, πBS (r0 ) (Y ) ABS (r0 ) . Fix ε > 0 such that ε 0 for ! all a ∈ A and a∈A pa = 1) and a countable group G, then, denoting by
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μp = g∈G p the corresponding probability measure on the product space shift. The associated entropy AG , the triple (G, AG , μp ) is called a Bernoulli ! is the nonnegative number H(p) = − a∈A pa log pa . Two Bernoulli shifts (G, AG , μp ) and (G, B G , μq ) are said to be isomorphic (or measurably conjugate) if there exist two Ginvariant subsets X ⊂ AG and Y ⊂ B G such that μp (AG \ X) = μq (B G \ Y ) = 0 and a Gequivariant bijective measurable map θ : X → Y with measurable inverse θ−1 : Y → X such that θ∗ μp = μq . Bowen [Bow, Theorem 1.1] proved that if G is a countable soﬁc group, then two isomorphic Bernoulli shifts (G, AG , μp ) and (G, B G , μq ) have the same entropy, i.e. H(p) = H(q). This result had been established when G = Z by N. Kolmogorov [Ko1, Ko2] in 1958–59 and then extended to countable amenable groups by D. Ornstein and B. Weiss [OrW] in 1987. L. Glebsky and L.M. Rivera [GlR] introduced the concept of a weaklysoﬁc group. This is a natural extension of the deﬁnition of soﬁcity where the Hamming metric on symmetric groups is replaced by general biinvariant metrics on ﬁnite groups. Glebsky and Rivera showed that the existence of a nonweaklysoﬁc group is equivalent to a conjecture on the closure in the proﬁnite topology of products of conjugacy classes in free groups of ﬁnite rank.
Exercises 7.1. Let G and C be two groups. Suppose that K is a ﬁnite symmetric subset of G such that 1G ∈ K. Show that a map ϕ : G → C is a Kalmost homomorphism if and only if it satisﬁes ϕ(k1 k2 ) = ϕ(k1 )ϕ(k2 ) for all k1 , k2 ∈ K and ϕ(k) = 1C for all k ∈ K \ {1G }. 7.2. Show that every group which is locally embeddable into the class of abelian groups is itself abelian. 7.3. Show that every group which is locally embeddable into the class of metabelian groups is itself metabelian. 7.4. Show that every abelian group is locally embeddable into the class of ﬁnite cyclic groups. 7.5. Let C be a class of groups. Suppose that a group G contains " a family (Hi )i∈I of subgroups satisfying the following properties: (1) G = i∈I Hi ; (2) For all i, j ∈ I there exists k ∈ I such that Hi ∪ Hj ⊂ Hk ; (3) Hi is locally embeddable into C for all i ∈ I. Show that G is locally embeddable into C. 7.6. Let C be a class of groups which is closed under ﬁnite direct products. Show that every group which is residually locally embeddable into C is locally embeddable into C.
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7.7. Let C be a class of groups which is closed under taking subgroups and taking ﬁnite direct products. Let G be a group which is residually C. Show that there exists a net (Ni )i∈I which converges to {1G } in N (G) such that G/Ni ∈ C fro all i ∈ I. Hint: Take as I the set of ﬁnite subsets of G partially ordered by inclusion and use Proposition 7.1.13. 7.8. Show that if G is a ﬁnitely presented inﬁnite simple group then G is not LEF. 7.9. Show that if G is a ﬁnitely presented nonamenable simple group then G is not LEA. 7.10. (cf. [VeG]) Let G1 and G2 be two groups. Recall that, by Proposition 7.3.1, Gi (i = 1, 2) is an LEFgroup if and only if the following holds: (∗) for every ﬁnite subset Ki ⊂ Gi there exist a ﬁnite set Li such that Ki ⊂ Li ⊂ Gi and a binary operation i : Li × Li → Li such that (Li , i ) is a group and ki ki = ki ki for all ki , ki ∈ Ki . Suppose that G1 and G2 are LEFgroups. An action of G2 on G1 by group automorphisms, i.e., a group homomorphism ϕ : G2 → Aut(G1 ), is said to be equivariantly approximable, if for all ﬁnite subsets K1 ⊂ G1 and K2 ⊂ G2 there exist ﬁnite groups (L1 , 1 ) and (L2 , 2 ) as in (∗) and an action of L2 on L1 by group automorphisms ψ : L2 → Aut(L1 ) such that if k1 ∈ K1 , k2 ∈ K2 and ϕ(k2 )(k1 ) ∈ K1 then ϕ(k2 )(k1 ) = ψ(k2 )(k1 ). Show that if G1 and G2 are LEFgroups and ϕ : G2 → Aut(G1 ) is an equivariantly approximable action, then the semidirect product G1 ϕ G2 is an LEFgroup. 7.11. Use the previous exercise to show that the group G1 of Sect. 2.6 is an LEFgroup (cf. Proposition 7.3.9). 7.12. Show that every residually free group is torsionfree. 7.13. A group G is called fully residually free if for any ﬁnite subset K ⊂ G, there exist a free group F and a group homomorphism φ : G → F whose restriction to K is injective. (a) Show that every fully residually free group is residually free. (b) Show that every fully residually free group is locally embeddable into the class of free groups. (c) A group G is called commutativetransitive if whenever a, b, c ∈ G\{1G } satisfy ab = ba and bc = cb, then ac = ca. Prove that every group which is locally embeddable into the class of free groups is commutativetransitive. Hint: First prove that if F is a free group and a, b ∈ F satisfy ab = ba, then there exist an element x ∈ F and integers m, n ∈ Z such that a = xm and b = xn . (d) Show that if F is a nonabelian free group, then the group F × Z is residually free but not locally embeddable into the class of free groups.
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7.14. Give an example of a ﬁnitely generated LEFgroup which is neither residually ﬁnite nor amenable. Hint: Take for instance the group G = G1 ×F2 , where G1 is the group introduced in Sect. 2.6 and F2 denotes a free group of rank two. 7.15. Let F be a nonempty ﬁnite set. Denote by GL(RF ) the automorphism group of the real vector space RF = {x : F → R}. For α ∈ Sym(F ), deﬁne λ(α) : RF → RF by λ(α)(x) = x ◦ α−1 for all x ∈ RF . (a) Show that λ(α) ∈ GL(RF ) for all α ∈ Sym(F ). (b) Show that the map λ : Sym(F ) → GL(RF ) is an injective group homomorphism. (c) Show that the Hamming metric dF on Sym(F ) satisﬁes dF (α, β) =
1 Tr(λ(α−1 β)) F 
for all α, β ∈ Sym(F ), where Tr(·) denotes the trace. 7.16. Let H be a real or complex Hilbert space of ﬁnite dimension n ≥ 1. Let L(H) denote the vector space consisting of all linear maps u : H → H. If u ∈ L(H), we denote by Tr(u) the trace of u and by u∗ its adjoint. For u, v ∈ L(H), we set 1 u, vHS = Tr(u ◦ v ∗ ). n (a) Show that ·, ·HS is a scalar product on L(H). Let · HS denote the associated norm. (b) Let U(H) = {u ∈ L(H) : u ◦ u∗ = IdH }. Show that U(H) is a group for the composition of maps. (c) Show that the map dHS : U(H) × U(H) → R deﬁned by dHS (u, v) = u − vHS for all u, v ∈ U(H) is a biinvariant metric on U(H). (The group U(H) is called the unitary group of H and dHS is called the normalized HilbertSchmidt metric on U(H).) 7.17. Let G be a group, K ⊂ G a ﬁnite subset, C a ﬁnite group, and ϕ : G → C a Kalmosthomomorphism. Denote by L : C → Sym(C) the Cayley homomorphism, that is, the map deﬁned by L(g)(h) = gh for all g, h ∈ C and set Φ = L ◦ ϕ : G → Sym(C). Show that Φ is a (K, ε)almosthomomorphism for all ε > 0. 7.18. Let G be a group and let K be a ﬁnite subset of G. Let F be a nonempty ﬁnite set. Show that if 0 < ε < 2/F  then every (K, ε)almosthomomorphism ϕ : G → Sym(F ) is a Kalmosthomomorphism of G into the group Sym(F ). 7.19. Suppose that a group G contains " a family (Hi )i∈I of subgroups satisfying the following properties: (1) G = i∈I Hi ; (2) For all i, j ∈ I there exists k ∈ I such that Hi ∪ Hj ⊂ Hk ; (3) Hi is soﬁc for all i ∈ I. Show that G is soﬁc.
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7.20. Show that every virtually soﬁc group is soﬁc. Hint: Use Proposition 7.5.14. 7.21. By using Exercise 6.5 and Exercise 6.6, give a direct proof of the fact that the direct product of ﬁnitely many groups which satisfy condition (b) in Theorem 7.7.1 also satisﬁes it. 7.22. Let G be a ﬁnitely generated group and suppose that S and S are two ﬁnite symmetric generating subsets of G. Show that the pair (G, S) satisﬁes condition (b) in Theorem 7.7.1 if and only if (G, S ) does. 7.23. Give a direct proof of the fact that every ﬁnitely generated residually ﬁnite group satisﬁes the condition (b) in Theorem 7.7.1. 7.24. Give a direct proof of the fact that every ﬁnitely generated amenable group satisﬁes the condition (b) in Theorem 7.7.1.
Chapter 8
Linear Cellular Automata
In this chapter we study linear cellular automata, namely cellular automata whose alphabet is a vector space and which are linear with respect to the induced vector space structure on the set of conﬁgurations. If the alphabet vector space and the underlying group are ﬁxed, the set of linear cellular automata is a subalgebra of the endomorphism algebra of the conﬁguration space (Proposition 8.1.4). An important property of linear cellular automata is that the image of a ﬁnitely supported conﬁguration by a linear cellular automaton also has ﬁnite support (Proposition 8.2.3). Moreover, a linear cellular automaton is entirely determined by its restriction to the space of ﬁnitelysupported conﬁgurations (Proposition 8.2.4) and it is preinjective if and only if this restriction is injective (Proposition 8.2.5). The algebra of linear cellular automata is naturally isomorphic to the group algebra of the underlying group with coeﬃcients in the endomorphism algebra of the alphabet vector space (Theorem 8.5.2). Linear cellular automata may be also regarded as endomorphisms of the space of ﬁnitelysupported conﬁgurations, viewed as a module over the group algebra of the underlying group with coeﬃcients in the ground ﬁeld (Proposition 8.7.5). This representation of linear cellular automata is always onetoone and, when the alphabet vector space is ﬁnitedimensional, it is also onto (Theorem 8.7.6). The image of a linear cellular automaton is closed in the space of conﬁgurations for the prodiscrete topology, provided that the alphabet is ﬁnite dimensional (Theorem 8.8.1). We exhibit an example showing that if one drops the ﬁnite dimensionality of the alphabet, then the image of a linear cellular automaton may fail to be closed. In Sect. 8.9 we prove a linear version of the Garden of Eden theorem. For the proof, we introduce the mean dimension of a vector subspace of the conﬁguration space. We show that for a linear cellular automaton with ﬁnitedimensional alphabet, both preinjectivity and surjectivity are equivalent to the maximality of the mean dimension of the image of the cellular automaton (Theorem 8.9.6). We exhibit two examples of linear cellular automata with ﬁnitedimensional alphabet over the free group of rank two, one which is preinjective but not surjective, and one which is surjective but T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 8, © SpringerVerlag Berlin Heidelberg 2010
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not preinjective. This shows that the linear version of the Garden of Eden theorem fails to hold for groups containing nonabelian free subgroups (see Sects. 8.10 and 8.11). Provided the alphabet is ﬁnite dimensional, the inverse of every bijective linear cellular automaton is also a linear cellular automaton (Corollary 8.12.2). In Sect. 8.13 we study the preinjectivity and surjectivity of the discrete Laplacian over the real numbers and prove a Garden of Eden type theorem (Theorem 8.13.2) for such linear cellular automata with no amenability assumptions on the underlying group. As an application, we deduce a characterization of locally ﬁnite groups in terms of real linear cellular automata (Corollary 8.13.4). In Sect. 8.14 we deﬁne linear surjunctivity and prove that all soﬁc groups are linearly surjunctive (Theorem 8.14.4). The notion of stable ﬁniteness for rings is introduced in Sect. 8.15. A stably ﬁnite ring is a ring for which onesided invertible square matrices are also twosided invertible. It is shown that linear surjunctivity is equivalent to stable ﬁniteness of the associated group algebra (Corollary 8.15.6). As a consequence, we deduce that group algebras of soﬁc groups are stably ﬁnite for any ground ﬁeld (Corollary 8.15.8). In the last section, we prove that the absence of zerodivisors in the group algebra of an arbitrary group is equivalent to the fact that every nonidenticallyzero linear cellular automaton with onedimensional alphabet is preinjective (Corollary 8.16.12). We recall that in this book all rings are assumed to be associative (but not necessarily commutative) with a unity element, and that a ﬁeld is a nonzero commutative ring in which each nonzero element is invertible.
8.1 The Algebra of Linear Cellular Automata Let G be a group and let V be a vector space over a ﬁeld K. The set V G consisting of all conﬁgurations x : G → V over the group G and the alphabet V has a natural structure of vector space over K in which addition and scalar multiplication are given by (x + x )(g) = x(g) + x (g)
and
(kx)(g) = kx(g)
for all x, x ∈ V G , k ∈ K, and g ∈ G. With the prodiscrete topology, V G becomes a topological vector space (cf. Sect. F.1). The Gshift (see Sect. 1.1) is then Klinear and continuous, that is, for each g ∈ G, the map x → gx is a continuous endomorphism of V G . A linear cellular automaton over the group G and the alphabet V is a cellular automaton τ : V G → V G which is Klinear, i.e., which satisﬁes τ (x + x ) = τ (x) + τ (x ) for all x, x ∈ V G and k ∈ K.
and τ (kx) = kτ (x)
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Proposition 8.1.1. Let G be a group and let V be a vector space over a ﬁeld K. Let τ : V G → V G be a cellular automaton with memory set S ⊂ G and local deﬁning map μ : V S → V . Then τ is linear if and only if μ is Klinear. Proof. Suppose ﬁrst that τ is linear. Let y, y ∈ V S and k ∈ K. Denote by x and x two conﬁgurations in V G extending y and y respectively, i.e., such that xS = y and x S = y . We then have μ(y + y ) = τ (x + x )(1G ) = (τ (x) + τ (x ))(1G ) = τ (x)(1G ) + τ (x )(1G ) = μ(y) + μ(y ). Similarly, as (kx)S = ky, we have μ(ky) = τ (kx)(1G ) = kτ (x)(1G ) = kμ(y). This shows that μ is Klinear. Conversely, suppose that μ is Klinear. Then, for all x, x ∈ V G , k ∈ K and g ∈ G we have τ (x + x )(g) = μ((g −1 (x + x ))S ) = μ((g −1 x)S + (g −1 x )S ) = μ((g −1 x)S ) + μ((g −1 x )S ) = τ (x)(g) + τ (x )(g) = (τ (x) + τ (x ))(g) and τ (kx)(g) = μ((g −1 (kx))S ) = μ(k(g −1 x)S ) = kμ((g −1 x)S ) = kτ (x)(g). This shows that τ (x + x ) = τ (x) + τ (x ) and τ (kx) = kτ (x). It follows that τ is linear. The following result is a linear analogue of the CurtisHedlund theorem (Theorem 1.8.1). Theorem 8.1.2. Let G be a group and let V be a vector space over a ﬁeld K. Let τ : V G → V G be a Gequivariant and Klinear map. Then the following conditions are equivalent: (a) the map τ is a linear cellular automaton; (b) the map τ is uniformly continuous (with respect to the prodiscrete uniform structure on V G ); (c) the map τ is continuous (with respect to the prodiscrete topology on V G ); (d) the map τ is continuous (with respect to the prodiscrete topology on V G ) at the constant conﬁguration x = 0.
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Proof. The implication (a) ⇒ (b) immediately follows from Theorem 1.9.1. Since the topology associated with the prodiscrete uniform structure is the prodiscrete topology (cf. Example B.1.4(a)) and every uniformly continuous map is continuous (cf. Proposition B.2.2), we also have (b) ⇒ (c). The implication (c) ⇒ (d) is trivial. Therefore, we are only left to show that (d) ⇒ (a). Suppose that τ is continuous at 0. Then, the map V G → V deﬁned by x → τ (x)(1G ) is continuous at 0 since the projection maps V G → V deﬁned by x → x(g) are continuous (for the prodiscrete topology) for all g ∈ G and the composition of continuous maps is continuous. We deduce that there exists a ﬁnite subset M ⊂ G such that if x ∈ V G satisﬁes x(m) = 0 for all m ∈ M , then τ (x)(1G ) = 0. By linearity, we have that if two conﬁgurations x and y coincide on M then τ (x)(1G ) = τ (y)(1G ). Thus there exists a linear map μ : V M → V such that τ (x)(1G ) = μ(xM ) for all x ∈ V G . As τ is Gequivariant, we deduce that τ (x)(g) = τ (g −1 x)(1G ) = μ((g −1 x)M ) for all x ∈ V G and g ∈ G. This shows that τ is the (linear) cellular automaton with memory set M and local deﬁning map μ. Thus (d) implies (a). Examples 8.1.3. (a) Let G be a group, let S be a nonempty ﬁnite subset of G, and let K be a ﬁeld. The discrete Laplacian ΔS : KG → KG (cf. Example 1.4.3(b)) is a linear cellular automaton. (b) Let G be a group, V a vector space over a ﬁeld K, and f ∈ EndK (V ). Then the map τ : V G → V G deﬁned by τ (x) = f ◦ x for all x ∈ V G is a linear cellular automaton (cf. Example 1.4.3(d)). (c) Let G be a group, V a vector space over a ﬁeld K, and s0 an element of G. Let Rs0 : G → G be the right multiplication by s0 in G, that is, the map deﬁned by Rs0 (g) = gs0 for all g ∈ G. Then the map τ : V G → V G deﬁned by τ (x) = x ◦ Rs0 is a linear cellular automaton (cf. Example 1.4.3(e)). (d) Let G = Z and K be a ﬁeld. Consider the vector space V = K[t] of all polynomials in the indeterminate t with coeﬃcients in K. A conﬁguration x ∈ V G may be viewed as a sequence x = (xn )n∈Z , where xn = xn (t) is a polynomial for all n ∈ Z. Let S = {0, 1} and consider the Klinear map μ : V S → V deﬁned by μ(p, q) = p − tq for all (p, q) ∈ V S = V × V , where q ∈ V denotes the derivative of the polynomial q. The linear cellular automaton τ : V Z → V Z with memory set S and local deﬁning map μ is then given by τ (x) = y, where yn = xn − txn+1 ∈ V , n ∈ Z, for all x = (xn )n∈Z ∈ V Z. We recall that an algebra over a ﬁeld K (or a Kalgebra) is a vector space A over K endowed with a product A × A → A such that A is a ring with respect to the sum and the product and such that the following associative law holds for the product and the multiplication by scalars: (ha)(kb) = (hk)(ab) for all h, k ∈ K and a, b ∈ A.
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A subset B of a Kalgebra A is called a subalgebra of A if B is both a vector subspace and a subring of A. If V is a vector space over a ﬁeld K, the set EndK (V ) consisting of all endomorphisms of the vector space V has a natural structure of a Kalgebra for which (f + f )(v) = f (v) + f (v), (f f )(v) = (f ◦ f )(v) = f (f (v)) and (kf )(v) = kf (v)
for all f, f ∈ EndK (V ), k ∈ K, and v ∈ V . The identity map IdV is the unity element of EndK (V ). Given a group G and a vector space V over a ﬁeld K, we denote by LCA(G; V ) the set of all linear cellular automata over the group G and the alphabet V . It immediately follows from the deﬁnition of a linear cellular automaton that LCA(G; V ) ⊂ EndK (V G ). Proposition 8.1.4. Let G be a group and let V be a vector space over a ﬁeld K. Then, LCA(G; V ) is a subalgebra of EndK (V G ). Proof. Let τ1 , τ2 ∈ LCA(G; V ). Let S1 and S2 be memory sets for τ1 and τ2 . Then, the set S = S1 ∪ S2 is also a memory set for τ1 and τ2 (cf. Sect. 1.5). Let μ1 : V S → V and μ2 : V S → V be the corresponding local deﬁning maps and set μ = μ1 + μ2 . For all x ∈ V G and g ∈ G we have (τ1 + τ2 )(x)(g) = τ1 (x)(g) + τ2 (x)(g) = μ1 (g −1 xS ) + μ2 (g −1 xS ) = μ(g −1 xS ). This shows that τ1 + τ2 is a cellular automaton with memory set S and local deﬁning map μ. Since the map τ1 + τ2 is Klinear, we deduce that τ1 + τ2 ∈ LCA(G; V ). On the other hand, let k ∈ K and let τ ∈ LCA(G; V ) with memory set S and local deﬁning map μ : V S → V . Then, for all x ∈ V G and g ∈ G, we have (kτ )(x)(g) = kτ (x)(g) = kμ(g −1 xS ) = (kμ)(g −1 xS ). Therefore the Klinear map kτ is a cellular automaton with memory set S and local deﬁning map kμ. It follows that kτ ∈ LCA(G; V ). We clearly have IdV G ∈ LCA(G; V ) (cf. Example 1.4.3(d)). Finally, it follows from Proposition 1.4.9 that if τ1 , τ2 ∈ LCA(G; V ) then the Klinear map τ1 τ2 = τ1 ◦ τ2 is also a cellular automaton and hence τ1 τ2 ∈ LCA(G; V ). This shows that LCA(G; V ) is a subalgebra of EndK (V G ).
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8.2 Conﬁgurations with Finite Support Let G be a group and let V be a vector space over a ﬁeld K. The support of a conﬁguration x ∈ V G is the set {g ∈ G : x(g) = 0V }. We denote by V [G] the subset of V G consisting of all conﬁgurations with ﬁnite support. Proposition 8.2.1. Let G be a group and let V be a vector space over a ﬁeld K. Then the set V [G] is a vector subspace of V G . Moreover, V [G] is dense in V G for the prodiscrete topology. Proof. If k1 , k2 ∈ K and x1 , x2 ∈ V G , then the support of k1 x1 + k2 x2 is contained in the union of the support of x1 and the support of x2 . Therefore, if x1 and x2 have ﬁnite support, so does k1 x1 + k2 x2 . Consequently, V [G] is a vector subspace of V G . Let x ∈ V G and let W ⊂ V G be a neighborhood of x for the prodiscrete topology. By deﬁnition of the prodiscrete topology, there exists a ﬁnite subset Ω ⊂ G such that W contains all conﬁgurations which coincide with x on Ω. It follows that the conﬁguration y ∈ V [G] which coincides with x on Ω and is identically zero outside of Ω is in W . This shows that V [G] is dense in V G . Proposition 8.2.2. Let G be a group and let V be a vector space over a ﬁeld K. Let x, x ∈ V G . Then the conﬁgurations x and x are almost equal if and only if x − x ∈ V [G]. Proof. By deﬁnition, x and x are almost equal if and only if the set {g ∈ G : x(g) = x (g)} is ﬁnite. This is equivalent to x − x ∈ V [G] since {g ∈ G : x(g) = x (g)} = {g ∈ G : (x − x )(g) = 0V }. Proposition 8.2.3. Let G be a group and let V be a vector space over a ﬁeld K. Let τ ∈ LCA(G; V ). Then one has τ (V [G]) ⊂ V [G]. Proof. Denote by S ⊂ G a memory set for τ and let μ : V S → V be the corresponding local deﬁning map. Let x ∈ V [G] and let T ⊂ G denote the support of x. For all g ∈ G, we have τ (x)(g) = μ((g −1 x)S ). As μ is linear (Proposition 8.1.1) and the support of g −1 x is g −1 T , we deduce that τ (x)(g) = 0 if g −1 T ∩ S = ∅. It follows that the support of τ (x) is contained in the ﬁnite set T S −1 ⊂ G. This shows that τ (x) ∈ V [G]. Observe that if τ ∈ LCA(G; V ), then the restriction map τ V [G] : V [G] → V [G] is Klinear, that is, τ V [G] ∈ EndK (V [G]).
8.3 Restriction and Induction of Linear Cellular Automata
289
Let A and B be two algebras over a ﬁeld K. A map F : A → B is called a Kalgebra homomorphism if F is both a vector space homomorphism (i.e., a Klinear map) and a ring homomorphism. This is equivalent to the fact that F satisﬁes F (a + a ) = F (a) + F (a ), F (aa ) = F (a)F (a ), F (ka) = kF (a) for all a, a ∈ A and k ∈ K, and F (1A ) = 1B . Proposition 8.2.4. Let G be a group and let V be a vector space over a ﬁeld K. Then the map Λ : LCA(G; V ) → EndK (V [G]) deﬁned by Λ(τ ) = τ V [G] , where τ V [G] : V [G] → V [G] is the restriction of τ to V [G], is an injective Kalgebra homomorphism. Proof. The fact that Λ is an algebra homomorphism immediately follows from the deﬁnition of the algebra operations on LCA(G; V ) and EndK (V [G]). Suppose that τ ∈ LCA(G; V ) satisﬁes Λ(τ ) = 0. This means that τ (y) = 0 for all y ∈ V [G]. As τ : V G → V G is continuous by Proposition 1.4.8 and V [G] is dense in V G by Proposition 8.2.1 for the prodiscrete topology, we deduce that τ (x) = 0 for all x ∈ V G , that is, τ = 0. This shows that Λ is injective. Proposition 8.2.5. Let G be a group and let V be a vector space over a ﬁeld K. Let τ ∈ LCA(G; V ). Then the following conditions are equivalent: (a) τ is preinjective; (b) τ V [G] : V [G] → V [G] is injective. Proof. Suppose that τ is preinjective. Let x ∈ ker(τ V [G] ). Then x ∈ V [G] and τ (x) = τ V [G] (x) = 0. As the trivial conﬁguration 0 and the conﬁguration x are almost equal and τ (0) = 0 by linearity of τ , the preinjectivity of τ implies that x = 0. This shows that τ V [G] is injective. Conversely, suppose that τ V [G] is injective. Let x, x ∈ V G be two conﬁgurations which are almost equal and such that τ (x) = τ (x ). Then x−x ∈ V [G] by Proposition 8.2.2 and we have τ V [G] (x−x ) = τ (x−x ) = τ (x)−τ (x ) = 0. As τ V [G] is injective, this implies x − x = 0, that is, x = x . Therefore, τ is preinjective.
8.3 Restriction and Induction of Linear Cellular Automata In this section, we show that the operations of restriction and induction for cellular automata that were introduced in Sect. 1.7 preserve linearity. Let G be a group and let V be a vector space over a ﬁeld K. Let H be a subgroup of G. We denote by LCA(G, H; V ) = LCA(G; V ) ∩ CA(G, H; V ) the set of all linear cellular automata τ : V G → V G admitting a memory set S such that S ⊂ H. Recall from Sect. 1.7, that, given a cellular automaton τ : V G → V G with memory set S ⊂ H, we denote by τH : V H → V H
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8 Linear Cellular Automata
the restriction cellular automaton. Similarly, given a cellular automaton σ : V H → V H , we denote by σ G : V G → V G the induced cellular automaton. Proposition 8.3.1. Let τ ∈ CA(G, H; V ). Then, τ ∈ LCA(G, H; V ) if and only if τH ∈ LCA(H; V ). Proof. It follows immediately from the deﬁnitions of restriction and induction that a cellular automaton τ ∈ CA(G, H; V ) is linear if and only if τH ∈ CA(H; V ) is linear. Let A and B be two algebras over a ﬁeld K. A map F : A → B is called a Kalgebra isomorphism if F is a bijective Kalgebra homomorphism. It is clear that if F : A → B is a Kalgebra isomorphism then its inverse map F −1 : B → A is also a Kalgebra isomorphism. Proposition 8.3.2. The set LCA(G, H; V ) is a subalgebra of LCA(G; V ). Moreover, the map τ → τH is a Kalgebra isomorphism from LCA(G, H; V ) onto LCA(H; V ) whose inverse is the map σ → σ G . Proof. Let τ1 , τ2 ∈ LCA(G, H; V ) with memory sets S1 , S2 ⊂ H respectively. Then the linear cellular automaton τ1 + τ2 admits S1 ∪ S2 as a memory set. As S1 ∪ S2 ⊂ H, we have τ1 + τ2 ∈ LCA(G, H; V ). If k ∈ K and τ ∈ LCA(G, H; V ), with memory set S ⊂ H, then S is also a memory set for kτ and therefore kτ ∈ LCA(G, H; V ). This shows that LCA(G, H; V ) is a vector subspace of LCA(G; V ). Since LCA(G, H; V ) is a submonoid of LCA(G; V ) by Proposition 1.7.1, we deduce that LCA(G, H; V ) is a subalgebra of LCA(G; V ). To simplify notation, denote by Φ : LCA(G, H; V ) → LCA(H; V ) and Ψ : LCA(H; V ) → LCA(G, H; V ) the maps deﬁned by Φ(τ ) = τH and Ψ (σ) = σ G respectively. It is clear from the deﬁnitions that Ψ ◦ Φ : LCA(G, H; V ) → LCA(G, H; V ) and Φ ◦ Ψ : LCA(H; V ) → LCA(H; V ) are the identity maps. Therefore, Φ is bijective with inverse Ψ . It remains to show that Φ is a Kalgebra homomorphism. ∈ VG Let τ1 , τ2 ∈ LCA(G, H; V ) and k1 , k2 ∈ K. Let x ∈ V H and let x extending x. By applying (1.13), we have x)(h) = k1 τ1 ( x)(h) + k2 τ2 ( x)(h) Φ(k1 τ1 + k2 τ2 )(x)(h) = (k1 τ1 + k2 τ2 )( for all h ∈ H. We deduce that Φ(k1 τ1 + k2 τ2 )(x) = (k1 Φ(τ1 ) + k2 Φ(τ2 ))(x) for all x ∈ V H , that is, Φ(k1 τ1 + k2 τ2 ) = k1 Φ(τ1 ) + k2 Φ(τ2 ). This shows that Φ is Klinear. Finally, it follows from Proposition 1.7.2 that Φ(IdV G ) = IdV H and Φ(τ1 τ2 ) = Φ(τ1 )Φ(τ2 ) for all τ1 , τ2 ∈ LCA(G, H; V ). We have shown that Φ is a Kalgebra isomorphism.
8.4 Group Rings and Group Algebras
291
8.4 Group Rings and Group Algebras Let G be a group and let R be a ring. We denote by 0R (resp. 1R ) the zero (resp. unity) element of R. We regard R as a left Rmodule over itself. Then RG = {α : G → R} has a natural structure of left Rmodule with addition and scalar multiplication given by (α + β)(g) = α(g) + β(g)
and (rα)(g) = rα(g)
for all α, β ∈ RG , r ∈ R, and g ∈ G. Deﬁne the support of an element α ∈ RG as being the set {g ∈ G : α(g) = 0R }. Let R[G] denote the set consisting of all elements α ∈ RG which have ﬁnite support. Then R[G] is a free submodule of RG . We have R⊂ R = RG , R[G] = g∈G
g∈G
so that the elements δg : G → R, g ∈ G, deﬁned by 1R if h = g δg (h) = 0R if h = g
(8.1)
freely generate R[G] as a left Rmodule. Note that the decomposition of an element α ∈ R[G] in this basis is simply given by the formula α= α(g)δg . g∈G
Let now α and β be two elements of R[G] and denote their supports by S and T . The convolution product of α and β is the element αβ ∈ RG deﬁned by (αβ)(g) = α(h)β(h−1 g) = α(h)β(h−1 g) (8.2) h∈G
h∈S
for all g ∈ G. Note that αβ ∈ R[G] as the support of αβ is contained in ST = {st : s ∈ S, t ∈ T }. By using the change of variables h1 = h and h2 = h−1 g, the convolution product (8.2) may be also expressed as follows: (αβ)(g) = α(h1 )β(h2 ). (8.3) h1 ,h2 ∈G h1 h2 =g
Proposition 8.4.1. Let G be a group and let R be a ring. Then the addition and the convolution product gives a ring structure to R[G]. Proof. We know that (R[G], +) is an abelian group. Let α, β, γ ∈ R[G] and g ∈ G. We have
292
[(αβ)γ](g) =
8 Linear Cellular Automata
(αβ)(h)γ(h−1 g)
h∈G
=
α(k)β(k−1 h)γ(h−1 g)
h∈G k∈G
(by setting s = k
−1
h) =
α(k)β(s)γ(s−1 k −1 g)
s∈G k∈G
=
α(k)(
=
β(s)γ(s−1 k −1 g))
s∈G
k∈G
α(k)(βγ)(k−1 g)
k∈G
= [α(βγ)](g). This shows that (αβ)γ = α(βγ). Thus, the convolution product is associative. On the other hand, given α ∈ R[G] we have, for all g ∈ G, δ1G (h)α(h−1 g) = α(g) = α(k)δ1G (k−1 g) = [αδ1G ](g). [δ1G α](g) = h∈G
k∈G
Thus, δ1G α = αδ1G = α for all α ∈ R[G]. This shows that δ1G is an identity element for the convolution product. Finally, (α + β)(h)γ(h−1 g) [(α + β)γ](g) = h∈G
=
[α(h) + β(h)]γ(h−1 g)
h∈G
=
α(h)γ(h−1 g) +
h∈G
β(k)γ(k−1 g)
k∈G
= (αγ)(g) + (βγ)(g). Thus, we have (α + β)γ = αγ + βγ. Similarly, one shows that α(β + γ) = αβ +αγ. It follows that the distributive laws also hold in R[G]. Consequently, R[G] is a ring with unity element 1R[G] = δ1G . The ring R[G] is called the group ring of G with coeﬃcients in R. Given a ring R, denote by U (R) the multiplicative group consisting of all invertible elements in R. Proposition 8.4.2. Let G be a group and let R be a ring. Then one has δg ∈ U(R[G]) for all g ∈ G. Moreover the map φ : G → U(R[G]) given by φ(g) = δg is a group homomorphism. Proof. Let g1 , g2 , g ∈ G. Then (δg1 δg2 )(g) = h∈G δg1 (h)δg2 (h−1 g) equals 1 if g1−1 g = g2 , that is, if g = g1 g2 , and equals 0 otherwise. This shows that δ g1 δg2 = δg1 g2 .
(8.4)
8.4 Group Rings and Group Algebras
293
We deduce that δg δg−1 = δgg−1 = δ1G = 1R[G] and, similarly, δg−1 δg = 1R[G] . This shows that δg belongs to U(R[G]). The fact that φ is a group homomorphism follows from (8.4). Let G be a group and let R be a ring. For α ∈ R[G] deﬁne α∗ : G → R by setting α∗ (g) = α(g −1 ) for all g ∈ G. If S is the support of α, then the support of α∗ is S −1 . Thus one has α∗ ∈ R[G]. Proposition 8.4.3. Let G be a group and let R be a ring. Let α, β ∈ R[G]. Then one has (i) (1R[G] )∗ = 1R[G] , (ii) (α∗ )∗ = α, (iii) (α + β)∗ = α∗ + β ∗ . Moreover, if the ring R is commutative, then one has (iv) (αβ)∗ = β ∗ α∗ . Proof. (i) We have (1R[G] )∗ = 1R[G] since the support of 1R[G] = δ1G is {1G }. (ii) For all g ∈ G, we have (α∗ )∗ (g) = α((g −1 )−1 ) = α(g). Therefore (α∗ )∗ = α. (iii) For all g ∈ G, we have (α+β)∗ (g) = (α+β)(g −1 ) = α(g −1 )+β(g −1 ) = α∗ (g)+β ∗ (g) = (α∗ +β ∗ )(g). Therefore (α + β)∗ = α∗ + β ∗ . (iv) Suppose that the ring R is commutative. For all g ∈ G, we have (αβ)∗ (g) = (αβ)(g −1 ) = α(h)β(h−1 g −1 ) h∈G
=
β(h−1 g −1 )α(h)
(since R is commutative)
h∈G
=
β ∗ (gh)α∗ (h−1 )
h∈G
=
β ∗ (k)α∗ (k−1 g)
k∈G
= (β ∗ α∗ )(g). Therefore (αβ)∗ = β ∗ α∗ .
If R = (R, +, ·) is a ring, its opposite ring is the ring Rop = (R, +, ∗) having the same underlying set and addition as R and with multiplication ∗ deﬁned by r ∗ s = sr for all r, s ∈ R.
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8 Linear Cellular Automata
From Proposition 8.4.3, we deduce that, if G is a group and R is a commutative ring, then the map α → α∗ is a ring isomorphism between the ring R[G] and its opposite ring (R[G])op . Thus we have Corollary 8.4.4. Let G be a group and let R be a commutative ring. Then the ring R[G] is isomorphic to its opposite ring (R[G])op . Remarks 8.4.5. Let G be a group and let R be a ring. (a) It is immediate to verify that the map R → R[G] deﬁned by r → rδ1G is an injective ring homomorphism. This is often used to regard R as a subring of R[G]. (b) Observe that (rα)β = r(αβ) for all r ∈ R and α, β ∈ R[G]. Therefore, we can use the notation rαβ = (rα)β = r(αβ). If the ring R is commutative, then one has the additional property (r1 α)(r2 β) = r1 r2 αβ for all r1 , r2 ∈ R and α, β ∈ R[G]. (c) Suppose that the group G is abelian and that the ring R is commutative. Then the ring R[G] is commutative. Indeed, it immediately follows from (8.3) that we have αβ = βα for all α, β ∈ R[G] under these hypotheses. Suppose now that we are given an algebra A over a ﬁeld K. Then A[G] is both a Kvector space and a ring; Moreover, we have (k1 α)(k2 β) = k1 k2 αβ for all k1 , k2 ∈ K and α, β ∈ A[G]. Therefore, A[G] is an algebra over K. The Kalgebra A[G] is called the group algebra of G with coeﬃcients in the Kalgebra A. In the particular case A = K, this gives the Kalgebra K[G].
8.5 Group Ring Representation of Linear Cellular Automata Let G be a group and let V be a vector space over a ﬁeld K. Consider the Kalgebra EndK (V )[G], that is, the group algebra of G with coeﬃcients in the Kalgebra EndK (V ) (see Sect. 8.4). For each α ∈ EndK (V )[G], we deﬁne a map τα : V G → V G by setting τα (x)(g) = α(h)(x(gh)) (8.5) h∈G
for all x ∈ V G and g ∈ G. Observe that α(h) ∈ EndK (V ) and x(gh) ∈ V for all x ∈ V G and g, h ∈ G. Note also that there is only a ﬁnite number of nonzero terms in the sum appearing in the right hand side of (8.5) since α has ﬁnite support. It follows that τα is well deﬁned. For each v ∈ V , deﬁne the conﬁguration cv ∈ V [G] by v if g = 1G , (8.6) cv (g) = 0 otherwise.
8.5 Group Ring Representation of Linear Cellular Automata
295
Note that the map from V to V [G] given by v → cv is injective and Klinear. Proposition 8.5.1. Let α ∈ EndK (V )[G]. Then one has: (i) τα ∈ LCA(G; V ); (ii) α(g)(v) = τα (cv )(g) for all v ∈ V and g ∈ G; (iii) the support of α is the minimal memory set of τα . Proof. Let S ⊂ G denote the support of α. Consider the map μ : V S → V given by μ(y) = α(s)(y(s)) s∈S
for all y ∈ V . Then, for all x ∈ V S
and g ∈ G, we have τα (x)(g) = α(h)(x(gh)) G
h∈G
=
α(h)(g −1 x(h)) (8.7)
h∈G
=
α(s)(g
−1
x(s))
s∈S
= μ((g −1 x)S ). Thus τα is the cellular automaton with memory set S and local deﬁning map μ. It is clear from (8.5) that τα is a Klinear map. Thus, we have τα ∈ LCA(G; V ). This shows (i). Let v ∈ V and let cv ∈ V [G] as in (8.6). By applying (8.5), we get, for all g ∈ G, α(h)(cv (g −1 h)) = α(g)(v). τα (cv )(g −1 ) = h∈G
This gives us (ii). Finally, let S0 ⊂ G denote the minimal memory set of τα . We have seen in the proof of (i) that S is a memory set for τα . Therefore, we have S0 ⊂ S. On the other hand, we deduce from (ii) that, for all v ∈ V and g ∈ G, we have α(g)(v) = τα (cv )(g) = μ0 ((g −1 cv )S0 ), where μ0 : V S0 → V denote the local deﬁning map for τα associated with S0 . Since the conﬁguration g −1 cv is identically zero on G \ {g}, it follows that α(g) = 0 if g ∈ / S0 . This implies S ⊂ S0 . Thus, we have S = S0 . This shows (iii). Theorem 8.5.2. Let G be a group and let V be a vector space over a ﬁeld K. Then the map Ψ : EndK (V )[G] → LCA(G; V ) deﬁned by α → τα is a Kalgebra isomorphism. Proof. Let α, β ∈ EndK (V )[G] and k ∈ K. By applying (8.5), we get
296
τα+β (x)(g) =
8 Linear Cellular Automata
[(α + β)(h)](x(gh))
h∈G
=
(α(h) + β(h))(x(gh))
h∈G
=
[α(h)(x(gh)) + β(h)(x(gh))]
h∈G
=
α(h)(x(gh)) +
h∈G
β(h)(x(gh))
h∈G
= τα (x)(g) + τβ (x)(g) and, similarly, τkα (x)(g) =
[(kα)(h)](x(gh))
h∈G
=
k[α(h)](x(gh))
h∈G
=k
[α(h)](x(gh))
h∈G
= kτα (x)(g) for all x ∈ V G and g ∈ G. Thus τα+β = τα + τβ and τkα = kτα . This shows that Ψ is a Klinear map. Let us show that Ψ is a ring homomorphism. Let α, β ∈ EndK (V )[G]. Let x ∈ V G and set y = τβ (x). For all g, h ∈ G, we have β(t)(x(ght)). (8.8) y(gh) = t∈G
It follows that τα (τβ (x))(g) = τα (y)(g) = α(h)(y(gh)) h∈G
(by (8.8)) =
α(h)
t∈G
h∈G
(by setting z = ht) =
α(h)
h∈G
=
z∈G
=
β(t)(x(ght))
β(h−1 z)(x(gz))
z∈G
α(h)β(h−1 z) (x(gz))
h∈G
[αβ](z)(x(gz))
z∈G
= ταβ (x)(g).
8.5 Group Ring Representation of Linear Cellular Automata
297
Thus we have ταβ = τα ◦ τβ , that is, Ψ (αβ) = Ψ (α)Ψ (β). Observe that 1EndK (V )[G] = δ1G , where δ1G : G → EndK (V ) is given by δ1G (h) = IdV if h = 1G and δ1G (h) = 0 otherwise. Thus, if x ∈ V G we have Ψ (1EndK (V )[G] )(x)(g) = Ψ (δ1G )(x)(g) = δ1G (h)(x(gh)) h∈G
= x(g) for all g ∈ G. It follows that Ψ (1EndK (V )[G] ) = IdV G . This proves that Ψ is a ring homomorphism. Let us show now that Ψ is injective. Let α ∈ ker(Ψ ). Then, τα (x) = 0
(8.9)
for all x ∈ V G . Taking x = cv in (8.9), where, for v ∈ V , the element cv ∈ V [G] is as in (8.6), we deduce from Proposition 8.5.1(ii) that α(g)(v) = 0 for all v ∈ V and g ∈ G. Thus, we have α(g) = 0 for all g ∈ G and therefore α = 0. It follows that ker(Ψ ) = {0}, that is, Ψ is injective. Let us show that Ψ is surjective. Suppose that τ ∈ LCA(G; V ) has memory set S and local deﬁning map μ : V S → V . As μ is Klinear (cf. Proposition 8.1.1), there exist Klinear maps αs : V → V , s ∈ S, such that μ(y) = s∈S αs (y(s)) for all y ∈ V S . For all x ∈ V G and g ∈ G, we have τ (x)(g) = μ((g −1 x)S ) =
αs g −1 x(s) = αs (x(gs)) .
s∈S
s∈S
This shows that τ = τα , where α ∈ EndK (V )[G] is deﬁned by αs if h = s ∈ S, α(h) = 0 otherwise.
Thus Ψ is surjective.
If the vector space V is onedimensional over the ﬁeld K, then each endomorphism of V is of the form v → kv, for some k ∈ K. It follows that the Kalgebra EndK (V ) is canonically isomorphic to K in this case. Thus, we get: Corollary 8.5.3. Let G be a group and let V be a onedimensional vector space over a ﬁeld K. Then the map Ψ : K[G] → LCA(G; V ) deﬁned by Ψ (α)(x)(g) = α(h)x(gh) h∈G
for all α ∈ K[G], x ∈ V G , and g ∈ G, is a Kalgebra isomorphism.
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8 Linear Cellular Automata
When G is an abelian group, then the Kalgebra K[G] is commutative for any ﬁeld K by Remark 8.4.5(c). Thus, as an immediate consequence of the preceding corollary, we get: Corollary 8.5.4. Let G be an abelian group and let V be a onedimensional vector space over a ﬁeld K. Then the Kalgebra LCA(G; V ) is commutative. Examples 8.5.5. (a) Let G be a group, S a nonempty ﬁnite subset of G, and let K be a ﬁeld. Consider the element δs ∈ K[G], α = Sδ1G − s∈S
Then, taking V = K, we have, for all x ∈ KG and g ∈ G, Ψ (α)(x)(g) = α(h)x(gh) = Sx(g) − x(gs). s∈S
h∈G
It follows that Ψ (α) is the discrete Laplacian ΔS : KG → KG associated with S (cf. Example 8.1.3(a)). (b) Let G be a group, V a vector space over a ﬁeld K and f ∈ EndK (V ). Consider the element α ∈ EndK (V )[G] deﬁned by α = f δ1G . Let x ∈ V G and g ∈ G. We have α(h)(x(gh)) = f (x(g)). Ψ (α)(x)(g) = h∈G
It follows that Ψ (α) is the linear cellular automaton τ ∈ LCA(G; V ) deﬁned by τ (x) = f ◦ x for all x ∈ V G (cf. Example 8.1.3(b)). (c) Let G be a group, V a vector space over a ﬁeld K, and s0 an element of G. Consider the group algebra element α ∈ EndK (V )[G] deﬁned by α = δs0 . For all x ∈ V G and g ∈ G, we have Ψ (α)(x)(g) = α(h)(x(gh)) = x(gs0 ). h∈G
It follows that Ψ (α) is the linear cellular automaton τ ∈ LCA(G; V ) deﬁned by τ (x) = x ◦ Rs0 , where Rs0 : G → G is the right multiplication by s0 (cf. Example 8.1.3(c)). (d) Let G = Z, let K be a ﬁeld, and let V = K[t] be the vector space consisting of all polynomials in the indeterminate t with coeﬃcients in K. Denote by D and U the elements in EndK (V ) deﬁned respectively by D(p) = p and U (p) = tp, for all p ∈ V . Consider the element α = δ0 − (U ◦ D)δ1 ∈ EndK (V )[Z].
8.6 Modules over a Group Ring
299
For all x = (xn )n∈Z ∈ V Z , we have Ψ (α)(x) = (yn )n∈Z ∈ V Z , where yn = α(m)(xn+m ) = xn − txn+1 m∈Z
for all n ∈ Z. It follows that Ψ (α) is the linear cellular automaton τ ∈ LCA(Z; K[t]) described in Example 8.1.3(d).
8.6 Modules over a Group Ring Let R be a ring and let M be a left Rmodule. We denote by EndR (M ) the endomorphism ring of M . Recall that EndR (M ) is the set consisting of all maps f : M → M satisfying f (x + x ) = f (x) + f (x ) and
f (rx) = rf (x)
for all r ∈ R and x, x ∈ M . The ring operations in EndR (M ) are given by (f + f )(x) = f (x) + f (x)
and
(f f )(x) = (f ◦ f )(x) = f (f (x))
for all f, f ∈ EndR (M ) and x ∈ M , and the unity element of EndR (M ) is the identity map IdM . Suppose that there is a group G which acts on M by endomorphisms. We then deﬁne a structure of left R[G]module on M as follows. For α ∈ R[G] and x ∈ M , deﬁne the element αx ∈ M by α(g)gx. (8.10) αx = g∈G
Note that this deﬁnition makes sense. Indeed, one has α(g) ∈ R and gx ∈ M for each g ∈ G. On the other hand, there is only ﬁnitely many nonzero terms in the right hand side of (8.10) since the support of α is ﬁnite. Proposition 8.6.1. One has: (i) 1R[G] x = x, (ii) α(x + x ) = αx + αx , (iii) (α + β)x = αx + βx, (iv) α(βx) = (αβ)x for all α, β ∈ R[G] and x, x ∈ M . Proof. (i) We have 1R[G] x = δ1G x = x.
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8 Linear Cellular Automata
(ii) We have α(x + x ) =
α(g)g(x + x ) =
g∈G
=
α(g)gx +
g∈G
(α(g)gx + α(g)gx )
g∈G
α(g)gx = αx + αx .
g∈G
(iii) We have (α + β)x =
(α + β)(g)gx =
g∈G
=
α(g)gx +
g∈G
(α(g) + β(g))gx
g∈G
β(g)gx = αx + βx.
g∈G
(iv) We have, by using (8.3), α(βx) = α(h1 )h1 (βx) h1 ∈G
=
α(h1 )h1
h1 ∈G
=
h2 ∈G
α(h1 )β(h2 )h1 h2 x
h1 ∈G h2 ∈G
=
g∈G
=
β(h2 )h2 x
α(h1 )β(h2 ) gx
h1 ,h2 ∈G h1 h2 =g
(αβ)(g)gx
g∈G
= (αβ)x. It follows from Proposition 8.6.1 that the addition on M and the multiplication (α, x) → αx deﬁned by (8.10) gives us a left R[G]module structure on M . Observe that this R[G]module structure extends the Rmodule structure on M if we regard R as a subring of R[G] via the map r → rδ1G (cf. Remark 8.4.5(a)). Proposition 8.6.2. Let f : M → M be a map. Then the following conditions are equivalent: (a) f ∈ EndR[G] (M ); (b) f ∈ EndR (M ) and f is Gequivariant. Proof. Suppose that f satisﬁes (b). Then f (x + x ) = f (x) + f (x ) for all x, x ∈ M since f ∈ EndR (M ). On the other hand, if α ∈ R[G] and x ∈ M ,
8.7 Matrix Representation of Linear Cellular Automata
301
we have, by using the Gequivariance and the Rlinearity of f ,
α(g)gx = α(g)f (gx) = α(g)gf (x) = αf (x). f (αx) = f g∈G
g∈G
g∈G
It follows that f ∈ EndR[G] (M ). This shows that (b) implies (a). Conversely, suppose that f satisﬁes (a). Let x, x ∈ M and r ∈ R. Then we have f (x + x ) = f (x) + f (x ), f (rx) = f (rδ1G x) = rδ1G f (x) = rf (x), and f (gx) = f (δg x) = δg f (x) = gf (x) since f ∈ EndR[G] (M ). This shows that f ∈ EndR (M ) and that f is Gequivariant. Thus, (a) implies (b). Remark 8.6.3. Conversely, suppose that we are given a left R[G]module M . Then, by applying the above construction, we recover the initial R[G]module structure on M if we start from the Rmodule structure on M obtained by restricting the scalars and the Rlinear action of G on M deﬁned by gx = δg x for all g ∈ G and x ∈ M .
8.7 Matrix Representation of Linear Cellular Automata Let G be a group and let V be a vector space over a ﬁeld K. Since the Gshift action on V G is Klinear, it induces a structure of left K[G]module on V G which extends the Kvector space structure on V G (see Sect. 8.6). Note that by applying (8.10) we get, for all α ∈ K[G] and x ∈ V G , αx = α(h)hx, (8.11) h∈G
that is, (αx)(g) =
α(h)x(h−1 g)
h∈G
for all g ∈ G. Thus, αx may be regarded as the “convolution product” of α and x (compare with (8.2)). Let τ : V G → V G be a linear cellular automaton. Then τ is Klinear by deﬁnition. On the other hand, τ is Gequivariant by Proposition 1.4.4. Thus, it follows from Proposition 8.6.2 that τ is an endomorphism of the K[G]module V G . Consequently, we have LCA(G; V ) ⊂ EndK[G] (V G ). It is clear that EndK[G] (V G ) is a subalgebra of the Kalgebra EndK (V G ). Since LCA(G; V ) is a subalgebra of EndK (V G ) by Proposition 8.1.4, we get: Proposition 8.7.1. The set LCA(G; V ) is a subalgebra of the Kalgebra EndK[G] (V G ). Remark 8.7.2. When G is a ﬁnite group, we have LCA(G; V ) = EndK[G] (V G ). Indeed, in this case, every u ∈ EndK[G] (V G ) is a linear cellular automaton
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8 Linear Cellular Automata
with memory set G and local deﬁning map μ : V G → V given by μ(y) = u(y)(1G ), since u(x)(g) = g −1 u(x)(1G ) = u(g −1 x)(1G ) for all x ∈ V G and g ∈ G. Consider now the vector subspace V [G] ⊂ V G consisting of all conﬁgurations with ﬁnite support. Observe that if x ∈ V G has support Ω ⊂ G and g ∈ G, then the support of the conﬁguration gx is gΩ. Therefore, V [G] is a submodule of the K[G]module V G . Recall that, for v ∈ V , the conﬁguration cv ∈ V [G] is deﬁned by v if g = 1G , cv (g) = 0 otherwise. Note that if g ∈ G and v ∈ V , then gcv = δg cv is the conﬁguration which takes the value v at g and is identically 0 on G \ {g}. It follows that every x ∈ V [G] can be written as x= gcx(g) . (8.12) g∈G
Proposition 8.7.3. Suppose that (ei )i∈I is a basis of the Kvector space V . Then the family of conﬁgurations (cei )i∈I is a free basis for the left K[G]module V [G]. Proof. Let x ∈ V [G]. As (ei )i∈I is a Kbasis for V , we can ﬁnd elements αi ∈ K[G], i ∈ I, such that x(g) = αi (g)ei i∈I
for all g ∈ G. This gives us x=
gcx(g)
g∈G
=
g αi (g)cei
g∈G
= = =
i∈I
g∈G
i∈I
i∈I
g∈G
i∈I
αi (g)gcei
αi (g)gcei
αi cei ,
8.7 Matrix Representation of Linear Cellular Automata
303
where the last equality follows from (8.11). If x = 0 ∈ V [G], then x(g) = 0 for all g ∈ G and therefore αi = 0 for all i ∈ I. This shows that (cei )i∈I is a basis for the left K[G]module V [G]. As every vector space admits a basis, we deduce the following Corollary 8.7.4. The left K[G]module V [G] is free.
If τ : V → V is a linear cellular automaton, we have τ (V [G]) ⊂ V [G] by Proposition 8.2.3. Moreover, it follows from Proposition 8.2.4 that the map Λ : LCA(G; V ) → EndK (V [G]) which associates with each τ ∈ LCA(G; V ) its restriction τ V [G] : V [G] → V [G] is an injective homomorphism of Kalgebras. As we have τ V [G] ∈ EndK[G] (V [G]) ⊂ EndK (V [G]) for all τ ∈ LCA(G; V ), we get: G
G
Proposition 8.7.5. Let G be a group and let V be a vector space over a ﬁeld K. Then the map Φ : LCA(G; V ) → EndK[G] (V [G]) deﬁned by Φ(τ ) = τ V [G] , where τ V [G] : V [G] → V [G] is the restriction of τ to V [G], is an injective Kalgebra homomorphism. When the alphabet is ﬁnitedimensional, we have the following: Theorem 8.7.6. Let G be a group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Then the map Φ : LCA(G; V ) → EndK[G] (V [G]) deﬁned by Φ(τ ) = τ V [G] , where τ V [G] : V [G] → V [G] is the restriction of τ to V [G], is a Kalgebra isomorphism. Proof. By the preceding proposition, it suﬃces to show that Φ is surjective. Let u ∈ EndK[G] (V [G]). Consider an element x ∈ V [G]. We have x=
gcx(g)
g∈G
by (8.12). This implies u(x) = u
gcx(g)
g∈G
=
gu(cx(g) )
g∈G
Since u is K[G]linear. We deduce that u(x)(1G ) = u(cx(g) )(g −1 ).
(8.13)
g∈G
Suppose now that dimK (V ) = d and let e1 , e2 , . . . , ed be a Kbasis for V . Denote by Ti ⊂ G the support
of u(cei ), 1 ≤ i ≤ d, and consider the ﬁnite subset S ⊂ G deﬁned by S = 1≤i≤d Ti−1 . d If v ∈ V , we can write v = i=1 ki ei with ki ∈ K, 1 ≤ i ≤ d. We then d have cv = i=1 ki cei and hence
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8 Linear Cellular Automata
u(cv ) = u
d
ki cei
i=1
=
d
ki u(cei ).
i=1
We deduce that u(cv )(g −1 ) = 0 if g ∈ / S. Thus, using (8.13), we get u(x)(1G ) = u(cx(g) )(g −1 ). g∈S
This shows that u(x)(1G ) only depends on the restriction of x to S. More precisely, there is a Klinear map μ : V S → V such that u(x)(1G ) = μ(xS )
for all x ∈ V [G].
Using the K[G]linearity of u, we get u(x)(g) = (g −1 u(x))(1G ) = u(g −1 x)(1G ) = μ((g −1 x)S ) for all x ∈ V [G] and g ∈ G. Therefore u is the restriction to V [G] of the linear cellular automaton τ : V G → V G with memory set S and local deﬁning map μ. This shows that Φ is surjective. Remarks 8.7.7. (a) When G is a ﬁnite group and V is a (not necessarily ﬁnitedimensional) vector space over a ﬁeld K, one has V [G] = V G and Φ : LCA(G; V ) → EndK[G] (V [G]) = EndK[G] (V G ) is the identity map by Remark 8.7.2. (b) Suppose now that G is an inﬁnite group and that V is an inﬁnitedimensional vector space over a ﬁeld K. Then the map Φ : LCA(G; V ) → EndK[G] (V [G]) is not surjective. To see this, take a Kbasis (ei )i∈I for V . As G and I are inﬁnite, we can ﬁnd a family (Ti )i∈I of ﬁnite subsets of G such that i∈I Ti is inﬁnite. Choose, for each i ∈ I, a conﬁguration zi ∈ V [G] whose support is Ti . It follows from Proposition 8.7.3 that V [G] is a free left K[G]module admitting the family (cei )i∈I as a basis. Therefore, there is an element u ∈ EndK[G] (V [G]) such that u(cei ) = zi for all i ∈ I. Let us show that u is not in the image of Φ. We proceed by contradiction. Suppose that u is in the image of Φ. This means that there is a linear cellular automaton τ : V G → V G such that u = τ V [G] . If S is a memory set for τ , this implies that, for x ∈ V [G], the value of u(x)(1G ) depends only on the restriction of x
to S. As S is ﬁnite and i∈I Ti−1 is inﬁnite, we can ﬁnd elements g0 ∈ G and / S and g0 ∈ Ti−1 . Consider the conﬁguration x0 ∈ V [G] i0 ∈ I such that g0 ∈ 0 which takes the value ei0 at g0 and is identically 0 on G \ {g0 }. Observe that x0 = g0 cei0 , so that u(x0 ) = u(g0 cei0 ) = g0 u(cei0 ) = g0 zi0 .
8.8 The Closed Image Property
Thus, we have
305
u(x0 )(1G ) = (g0 zi0 )(1G ) = zi0 (g0−1 ).
We deduce that u(x0 )(1G ) = 0 as g0−1 is in the support Ti0 of zi0 . This gives us a contradiction since x0 coincides with the 0 conﬁguration on S. This shows that Φ is not surjective. (c) By combining the two preceding remarks with Theorem 8.7.6, we deduce that the map Φ : LCA(G; V ) → EndK[G] (V [G]) is surjective if and only if G is ﬁnite or V is ﬁnitedimensional. Let us recall the following basic facts from linear algebra. Let R be a ring and let d ≥ 1 be an integer. We denote by Matd (R) the ring consisting of all d × d matrices A = (aij )1≤i,j≤d with entries aij ∈ R. Recall that the addition and the multiplication in Matd (R) are given by A + B = (aij + bij )1≤i,j≤d
and AB =
d k=1
aik bkj 1≤i,j≤d
for all A = (aij )1≤i,j≤d , B = (bij )1≤i,j≤d ∈ Matd (R). Suppose now that M is a free left Rmodule with basis e1 , e2 , . . . , ed . If u ∈ EndR (M ), then we d have u(ei ) = j=1 aij ej , where aij ∈ R for 1 ≤ i, j ≤ d. Then the map u → (aij )1≤i,j≤d is a ring isomorphism from EndR (M ) onto Matd (Rop ), where Rop denotes the opposite ring of R. Moreover, when R is an algebra over a ﬁeld K, it is an isomorphism of Kalgebras. Let G be a group and let V be a vector space of ﬁnite dimension d ≥ 1 over a ﬁeld K. By Proposition 8.7.3, the left K[G]module V [G] admits a free basis of cardinality d. As the Kalgebras K[G] and (K[G])op are isomorphic by Corollary 8.4.4, we deduce from Theorem 8.7.6 the following: Corollary 8.7.8. Let G be a group and let V be a vector space over a ﬁeld K of ﬁnite dimension dimK (V ) = d ≥ 1. Then the Kalgebras LCA(G; V ) and Matd (K[G]) are isomorphic. Remark 8.7.9. For d = 1, Corollary 8.7.8 tells us that if G is a group and V is a onedimensional vector space over a ﬁeld K, then the Kalgebras LCA(G; V ) and K[G] are isomorphic. Note that this last result also follows from Corollary 8.5.3.
8.8 The Closed Image Property As we have seen in Lemma 3.3.2, the image of a cellular automaton τ : AG → AG is always closed in AG for the prodiscrete topology if the alphabet A is ﬁnite. When A is inﬁnite, the image of τ may fail to be closed in AG (cf. Example 8.8.3). However, it turns out that if A = V is a ﬁnitedimensional
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8 Linear Cellular Automata
vector space over a ﬁeld K and τ : V G → V G is a linear cellular automaton, then the image of τ is closed in V G even when the ﬁeld K is inﬁnite. Theorem 8.8.1. Let G be a group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Let τ : V G → V G be a linear cellular automaton. Then τ (V G ) is closed in V G for the prodiscrete topology. Proof. We split the proof into two steps. Suppose ﬁrst that G is countable. Then we can ﬁnd a sequence (An )n∈N of ﬁnite subsets of G such that G =
n∈N An and An ⊂ An+1 for all n ∈ N. Let S be a memory set for τ
and let Bn = A−S denote the Sinterior of A (cf. Sect. 5.4). Note that G = n n n∈N Bn and Bn ⊂ Bn+1 for all n ∈ N. It follows from Proposition 5.4.3 that if x and x are elements in V G such that x and x coincide on An then the conﬁgurations τ (x) and τ (x ) coincide n ∈ V G a conﬁguration on Bn . Therefore, given xn ∈ V An and denoting by x extending xn , the pattern xn )Bn ∈ V Bn yn = τ ( does not depend on the particular choice of the extension x n . Thus we can deﬁne a map τn : V An → V Bn by setting τn (xn ) = yn for all xn ∈ V An . It is clear that τn is Klinear. Let y ∈ V G and suppose that y is in the closure of τ (V G ). Then, for all n ∈ N there exists zn ∈ V G such that yBn = τ (zn )Bn .
(8.14)
Consider, for each n ∈ N, the aﬃne subspace Ln ⊂ V An deﬁned by Ln = τn−1 (yBn ). We have Ln = ∅ for all n by (8.14). For n ≤ m, the restriction map V Am → V An induces an aﬃne map πn,m : Lm → Ln . Consider, for all n ≤ m, the aﬃne subspace Kn,m ⊂ Ln deﬁned by Kn,m = πn,m (Lm ). We have Kn,m ⊂ Kn,m for all n ≤ m ≤ m since πn,m = πn,m ◦ πm,m . As the sequence Kn,m (m = n, n + 1, . . . ) is a decreasing sequence of ﬁnitedimensional aﬃne subspaces, it stabilizes, i.e., for each n ∈ N there exist a nonempty aﬃne subspace Jn ⊂ Ln and an integer m0 = m0 (n) ≥ n such that Kn,m = Jn if m0 ≤ m. For all n ≤ n ≤ m, we have πn,n (Kn ,m ) ⊂ Kn,m since πn,n ◦ πn ,m = πn,m . Therefore, πn,n induces by restriction an aﬃne map ρn,n : Jn → Jn for all n ≤ n . We claim that ρn,n is surjective. To see this, let u ∈ Jn . Let us choose m large enough so that Jn = Kn,m and Jn = Kn ,m . Then we can ﬁnd v ∈ Lm such that u = πn,m (v). We have u = ρn,n (w), where w = πn ,m (v) ∈ Kn ,m = Jn . This proves the claim. Now, using the surjectivity of ρn,n+1 for all n, we construct by induction a sequence of elements xn ∈ Jn , n ∈ N, as follows. We start by choosing an arbitrary element x0 ∈ J0 . Then, assuming xn has been constructed, we take as xn+1 an arbitrary element in ρ−1 n,n+1 (xn ). Since xn+1 coincides with xn on An , there exists x ∈ V G such that xAn = xn for all n. We have
8.8 The Closed Image Property
307
τ (x)Bn = τn (xn ) = yn = yBn for all n. Since G = ∪n∈N Bn , we deduce that τ (x) = y. This ends the proof in the case when G is a countable group. We now drop the countability assumption on G and prove the theorem in the general case. Let S ⊂ G be a memory set for τ and denote by H ⊂ G the subgroup of G generated by S. Then H is countable since it is ﬁnitely generated. Consider the restriction cellular automaton τH : V H → V H and observe that, by the ﬁrst step of the proof, τH (V H ) is closed in V H
(8.15)
for the prodiscrete topology. With the notation from Sect. 1.7 we have, by virtue of (1.16), that τ (V G ) = τc (V c ). (8.16) c∈G/H
Also, it follows from (1.18) that, for all c ∈ G/H and g ∈ c, τc (V c ) = (φ∗g )−1 τH (V H ) .
(8.17)
As the map φ∗g : V c → V H is a uniform isomorphism and therefore a homeomorphism, it follows from (8.15) and (8.17) that τc (V c ) is closed in V c for all c ∈ G/H. As the product of closed subspaces is closed in the product topology (cf. Proposition A.4.3), we deduce from (8.16) that τ (V G ) is closed in V G . Corollary 8.8.2. Let G be a group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Let τ : V G → V G be a linear cellular automaton. Suppose that every conﬁguration with ﬁnite support y ∈ V [G] lies in the image of τ . Then τ is surjective. Proof. By our hypothesis, we have V [G] ⊂ τ (V G ) ⊂ V G . As V [G] is dense in V G by Proposition 8.2.1 and τ (V G ) is closed in V G by Theorem 8.8.1, we deduce that τ (V G ) = V G . Thus, τ is surjective. In the example below we show that if we drop the ﬁnite dimensionality of the alphabet vector space V , then the image of a linear cellular automaton may fail to be closed in V G . Example 8.8.3. Let G = Z and let K be a ﬁeld. Consider the Kalgebra V = K[t] consisting of all polynomials in the indeterminate t with coeﬃcients in K. A conﬁguration in V Z is therefore a sequence x = (xn )n∈Z , where xn = xn (t) is a polynomial for all n ∈ Z. Consider the Klinear map τ : V Z → V Z deﬁned by setting τ (x) = y where y = (yn )n∈Z is given by yn = xn+1 − txn for all n ∈ Z. Then τ is a linear cellular automaton with memory set {0, 1} and local deﬁning map μ : V {0,1} → V given by μ(x0 , x1 ) = x1 − tx0 .
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8 Linear Cellular Automata
Consider the conﬁguration z = (zn )n∈Z , where zn = 1 for all n ∈ Z. Let Ω be a ﬁnite subset of Z and choose an integer M ∈ Z such that Ω ⊂ [M, ∞). Consider the conﬁguration x = (xn )n∈Z deﬁned by 1 + t + · · · + tn−M if n ≥ M, xn = 0 if n < M. Observe that xn+1 = txn +1 for all n ≥ M , so that the conﬁguration y = τ (x) coincides with z on [M, ∞) and hence on Ω. Thus z is in the closure of τ (V Z ) in V Z . On the other hand, the conﬁguration z is not in the image of τ . Indeed, z = τ (x) for some x ∈ V Z would imply xn+1 = txn + 1 and hence deg(xn+1 ) > deg(xn ) for all n ∈ Z, which is clearly impossible. This shows that τ (V Z ) is not closed in V Z for the prodiscrete topology. Observe that every y ∈ V [Z] is in the image of τ . Indeed, if y = (yn )n∈Z ∈ V [Z] has support contained in [M, ∞) for some M ∈ Z, we can construct x = (xn )n∈Z ∈ V Z with τ (x) = y inductively by setting xn = 0 for all n ≤ M and xn+1 = txn + yn for all n ≥ M . This shows that we cannot omit the hypothesis saying that V is ﬁnitedimensional in Corollary 8.8.2.
8.9 The Garden of Eden Theorem for Linear Cellular Automata In this section, we present a linear version of Theorem 5.8.1. We ﬁrst introduce the notion of mean dimension, which plays here the role which was played by the entropy in the ﬁnite alphabet case, and we present some of its basic properties. In the deﬁnition of mean dimension, the dimension of ﬁnitedimensional vector spaces replaces the cardinality of ﬁnite sets. The proof of the properties of the mean dimension and the proof of the linear version of the Garden of Eden Theorem follow the same lines as their ﬁnite alphabet counterparts (see Sects. 5.7 and 5.8). From now on, in this section, G is an amenable group, F = (Fj )j∈J a right Følner net for G, and V a ﬁnitedimensional vector space over a ﬁeld K. For E ⊂ G, we denote by πE : V G → V E the canonical projection (restriction map). We thus have πE (x) = xE for all x ∈ V G . Note that πE is Klinear so that, if X is a vector subspace of V G , then πE (X) is a vector subspace of V E . Deﬁnition 8.9.1. Let X be a vector subspace of V G . The mean dimension mdimF (X) of X with respect to the right Følner net F = (Fj )j∈J is deﬁned by dim(πFj (X)) mdimF (X) = lim sup , (8.18) Fj  j
8.9 The Garden of Eden Theorem for Linear Cellular Automata
309
where dim(πFj (X)) denotes the dimension of the Kvector subspace πFj (X) ⊂ V Fj . Remark 8.9.2. In the particular case when K is a ﬁnite ﬁeld, every ﬁnitedimensional vector space W over K is ﬁnite of cardinality W  = Kdim(W ) . Therefore, in this case, we have mdimF (X) =
1 entF (X) log K
for every vector subspace X ⊂ V G . Here are some immediate properties of mean dimension. Proposition 8.9.3. One has: (i) mdimF (V G ) = dim(V ); (ii) mdimF (X) ≤ mdimF (Y ) if X ⊂ Y are vector subspaces of V G ; (iii) mdimF (X) ≤ dim V for all vector subspaces X ⊂ V G . Proof. (i) If X = V G , then, for every j, we have πFj (X) = V Fj and therefore dim(πFj (X)) dim(V Fj ) Fj  dim(V ) = = = dim(V ). Fj  Fj  Fj  This implies that mdimF (X) = dim(V ). (ii) If X ⊂ Y are vector subspaces of V G , then πFj (X) ⊂ πFj (Y ) are vector subspaces of V Fj and hence dim(πFj (X)) ≤ dim(πFj (Y )) for all j. This implies mdimF (X) ≤ mdimF (Y ). (iii) This follows immediately from (i) and (ii). An important property of linear cellular automata is the fact that applying a linear cellular automaton to a vector subspace of conﬁgurations cannot increase the mean dimension of the subspace. More precisely, we have the following: Proposition 8.9.4. Let τ : V G → V G be a linear cellular automaton and let X be a vector subspace of V G . Then one has mdimF (τ (X)) ≤ mdimF (X). Proof. Let us set Y = τ (X) and observe that Y is a vector subspace of V G , by linearity of τ . Let S ⊂ G be a memory set for τ . After replacing S by S ∪ {1G }, we can assume that 1G ∈ S. Let Ω be a ﬁnite subset of G. Observe ﬁrst that τ induces a map τΩ : πΩ (X) → πΩ −S (Y ) deﬁned as follows. If u ∈ πΩ (X), then
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8 Linear Cellular Automata
τΩ (u) = (τ (x))Ω −S , where x is an element of X such that xΩ = u. Note that the fact that τΩ (u) does not depend on the choice of such an x follows from Proposition 5.4.3. Clearly τΩ is surjective. Indeed, if v ∈ πΩ −S (Y ), then there exists x ∈ X such that (τ (x))Ω −S = v. Then, setting u = πΩ (x) we have, by construction, τΩ (u) = v. Since τΩ is Klinear and surjective, we have dim(πΩ −S (Y )) ≤ dim(πΩ (X)).
(8.19)
Observe now that Ω −S ⊂ Ω, since 1G ∈ S (cf. Proposition 5.4.2(iv)). Thus −S πΩ (Y ) is a vector subspace of πΩ −S (Y ) × V Ω\Ω . This implies dim(πΩ (Y )) ≤ dim(πΩ −S (Y ) × V Ω\Ω
−S
)
= dim(πΩ −S (Y )) + dim(V Ω\Ω = dim(πΩ −S (Y )) + Ω \ Ω ≤ dim(πΩ (X)) + Ω \ Ω
−S
−S
−S
)
 dim(V )
 dim(V ),
by (8.19). As Ω \ Ω −S ⊂ ∂S (Ω), we deduce that dim(πΩ (Y )) ≤ dim(πΩ (X)) + ∂S (Ω) dim(V ). By taking Ω = Fj , this gives us dim(πFj (Y )) dim(πFj (X)) ∂S (Fj ) ≤ + dim(V ). Fj  Fj  Fj  Since lim j
∂S (Fj ) =0 Fj 
by Proposition 5.4.4, we ﬁnally get mdimF (Y ) = lim sup j
dim(πFj (Y )) dim(πFj (X)) ≤ lim sup = mdimF (X). Fj  Fj  j
By Proposition 8.9.3, the maximal value for the mean dimension of a vector subspace X ⊂ V G is dim(V ). The following result gives a suﬃcient condition on X which guarantees that its mean dimension is strictly less than dim(V ). Proposition 8.9.5. Let X be a Ginvariant vector subspace of V G . Suppose that there exists a ﬁnite subset E ⊂ G such that πE (X) V E . Then one has mdimF (X) < dim(V ). Proof. Let E = {g1 g2−1 : g1 , g2 ∈ E}. By Proposition 5.6.3, we may ﬁnd an (E, E )tiling T ⊂ G.
8.9 The Garden of Eden Theorem for Linear Cellular Automata
311
For each j ∈ J, let us deﬁne, as in Proposition 5.6.4, the subset Tj ⊂ T by Tj = T ∩ Fj−E = {g ∈ T : gE ⊂ Fj } and set
Fj∗ = Fj \
gE.
g∈Tj
Since πE (X) V E and X is Ginvariant, we have πgE (X) V gE for all g ∈ G. We thus have dim(πgE (X)) ≤ dim(V gE ) − 1 = gE dim(V ) − 1 for all g ∈ T. As
∗
πFj (X) ⊂ V Fj ×
(8.20)
πgE (X),
g∈Tj
we get ∗
dim(πFj (X)) ≤ dim(V Fj ×
πgE (X))
g∈Tj
= Fj∗  dim(V ) +
dim(πgE (X))
g∈Tj
≤ Fj∗  dim(V ) +
(gE dim(V ) − 1) (by (8.20))
g∈Tj
=
Fj∗  + gE dim(V ) − Tj  g∈Tj
= Fj  dim(V ) − Tj ,
since
Fj  = Fj∗  +
gE
and
gE = E.
g∈Tj
Now, by Proposition 5.6.4, there exist α > 0 and j0 ∈ J such that Tj  ≥ αFj  for all j ≥ j0 . Thus dim(πFj (X)) ≤ dim(V ) − α Fj 
for all j ≥ j0 .
This implies that mdimF (X) = lim sup j
dim(πFj (X)) ≤ dim(V ) − α < dim(V ). Fj 
312
8 Linear Cellular Automata
We are now in position to state the linear version of the Garden of Eden theorem. Theorem 8.9.6. Let G be an amenable group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Let F = (Fj )j∈J be a right Følner net for G. Let τ : V G → V G be a linear cellular automaton. Then the following conditions are equivalent: (a) τ is surjective; (b) mdimF (τ (V G )) = dim(V ); (c) τ is preinjective. Since every injective cellular automaton is preinjective, we immediately deduce the following: Corollary 8.9.7. Let G be an amenable group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Then every injective linear cellular automaton τ : V G → V G is surjective. We divide the proof of Theorem 8.9.6 into several lemmas. Lemma 8.9.8. Let τ : V G → V G be a linear cellular automaton. Suppose that τ is not surjective. Then mdimF (τ (V G )) < dim(V ). Proof. Let X = τ (V G ) and choose a conﬁguration y ∈ V G \ X. Since V G \ X is an open subset of V G for the prodiscrete topology by Theorem 8.8.1, we / πΩ (X). Therefore we have can ﬁnd a ﬁnite subset Ω ⊂ G such that πΩ (y) ∈ πΩ (X) V Ω . Observe that X is a Ginvariant vector subspace of V G , as τ is Gequivariant and linear. By applying Proposition 8.9.5, we deduce that mdimF (X) < dim(V ). Lemma 8.9.9. Let τ : V G → V G be a linear cellular automaton. Suppose that (8.21) mdimF (τ (V G )) < dim(V ). Then τ is not preinjective. Proof. Let us set X = τ (V G ). Let S be a memory set for τ such that 1G ∈ S. +S As πF +S (X) is a vector subspace of πFj (X) × V Fj \Fj , we have j
dim(πF +S (X)) ≤ dim(πFj (X)) + Fj+S \ Fj  dim(V ) j
≤ dim(πFj (X)) + ∂S (Fj ) dim(V ). We thus have dim(πF +S (X)) j
Fj 
≤
dim(πFj (X)) ∂S (Fj ) + dim(V ). Fj  Fj 
8.9 The Garden of Eden Theorem for Linear Cellular Automata
313
It follows from (8.21) and Proposition 5.4.4 that we can ﬁnd j0 ∈ J such that dim(πF +S (X)) < Fj0  dim(V ). Let Z denote the (ﬁnitedimensional) j0
vector subspace of V G consisting of all conﬁgurations whose support is contained in Fj0 . Observe that τ (x) vanishes outside of Fj+S for every x ∈ Z by 0 Proposition 5.4.3. Thus we have dim(τ (Z)) = dim(πF +S (τ (Z))) j0
≤ dim(πF +S (X)) < Fj0  dim(V ) = dim(Z). j0
This implies that the restriction of τ to Z is not injective. As all conﬁgurations in Z have ﬁnite support, we deduce that τ is not preinjective (cf. Proposition 8.2.5). Lemma 8.9.10. Let τ : V G → V G be a linear cellular automaton. Suppose that τ is not preinjective. Then mdimF (τ (V G )) < dim(V ). Proof. Since the linear cellular automaton τ is not preinjective, we can ﬁnd, by Proposition 8.2.5, an element x0 ∈ V G with nonempty ﬁnite support Ω ⊂ G such that τ (x0 ) = 0. Let S be a memory set for τ such that 1G ∈ S 2 and S = S −1 . Let E = Ω +S . By Proposition 5.6.3, we can ﬁnd a ﬁnite subset F ⊂ G such that G contains a (E, F )tiling T . Note that for each g ∈ G, the support of gx0 is gΩ ⊂ gE. Let us choose, for each g ∈ T , a hyperplane Hg ⊂ V gΩ which does not contain the restriction to gΩ of gx0 . Consider the vector subspace X ⊂ V G consisting of all x ∈ V G such that the restriction of x to gΩ belongs to Hg for each g ∈ T . We claim that τ (V G ) = τ (X). Indeed, let z ∈ V G . Then, for each g ∈ T , there exists a scalar kg ∈ K such that the restriction to gΩ of z + kg (gx0 ) belongs to Hg . Let z ∈ V G be (z + kg (gx0 )) for each g ∈ T and z = z outside such that πgΩ (z ) = πgΩ of g∈T gΩ. We have z ∈ X by construction. On the other hand, since z and z coincide outside g∈T gΩ, we have τ (z ) = τ (z) outside g∈T gΩ +S . 2
Now, if h ∈ gΩ +S for some g ∈ T , then hS ⊂ gΩ +S = gE and therefore τ (z )(h) = τ (z + kg (gx0 ))(h) = τ (z)(h) since gx0 lies in the kernel of τ . Thus τ (z) = τ (z ) and the claim follows. We deduce that mdimF (τ (V G )) = mdimF (τ (X)) ≤ mdimF (X) < dim(V ) where the ﬁrst inequality follows from Proposition 8.9.4 and the second one from Proposition 8.9.5. Proof of Theorem 8.9.6. Condition (a) implies (b) since we have mdimF (V G ) = dim(V ) by Proposition 8.9.3(i). The converse implication follows from Lemma 8.9.8. On the other hand, condition (c) implies (b) by Lemma 8.9.9. Finally, (b) implies (c) by Lemma 8.9.10.
314
8 Linear Cellular Automata
We end this section by showing that Corollary 8.9.7 as well as both implications (a) ⇒ (c) and (c) ⇒ (a) in Theorem 8.9.6 fail to hold when the vector space V is inﬁnitedimensional. Example 8.9.11. Let G be any group and let V be an inﬁnitedimensional vector space over a ﬁeld K. Let us choose a basis subset B for V . Every map ϕ : B → B uniquely extends to a Klinear map ϕ : V → V . The product G : V G → V G is a linear cellular automaton with memory set map τϕ = ϕ Since B is inﬁnite, we can ﬁnd a map S = {1G } and local deﬁning map ϕ. ϕ1 : B → B which is surjective but not injective and a map ϕ2 : B → B which is injective but not surjective. Consider ﬁrst the cellular automaton τ1 = τϕ1 . Let us show that τ1 is surjective but not preinjective. Observe that ϕ 1 is surjective. As a product of surjective maps is a surjective map, it follows that τ1 is surjective. On the 1 ) \ {0}. Consider the other hand, as ϕ 1 is not injective, there exists v ∈ ker(ϕ conﬁguration x ∈ V G deﬁned by x(1G ) = v and x(g) = 0 for all g ∈ G \ {1G }. Then x ∈ V [G] ∩ ker(τ1 ) and x = 0. This shows that τ1 is not preinjective. Consider now the cellular automaton τ2 = τϕ2 . Let us show that τ2 is injective (and therefore preinjective) but not surjective. Suppose that x ∈ 2 (x(g)) for all g ∈ G. As ϕ 2 is injective, ker(τ2 ), that is, 0 = τ2 (x)(g) = ϕ we deduce that x(g) = 0 for all g ∈ G, in other words, x = 0. This shows that τ2 is injective. On the other hand, as ϕ 2 is not surjective, there exists v ∈ V \ϕ 2 (V ). Consider the constant conﬁguration x ∈ V G where x(g) = v for all g ∈ G. It is clear that x ∈ V G \ τ2 (V G ). This shows that τ2 is not surjective.
8.10 Preinjective but not Surjective Linear Cellular Automata In this section, we give examples of linear cellular automata with ﬁnitedimensional alphabet which are preinjective but not surjective. We recall that the underlying groups for such automata cannot be amenable by Theorem 8.9.6. Proposition 8.10.1. Let G = F2 be the free group of rank two and let V be a twodimensional vector space over a ﬁeld K. Then there exists a linear cellular automaton τ : V G → V G which is preinjective but not surjective. Proof. We may assume V = K2 . Let a and b denote the canonical generators of G = F2 . Let p1 and p2 be the elements of EndK (V ) deﬁned respectively by p1 (v) = (k1 , 0) and p2 (v) = (k2 , 0) for all v = (k1 , k2 ) ∈ V . Consider the linear cellular automaton τ : V G → V G given by τ (x)(g) = p1 (x(ga)) + p2 (x(gb)) + p1 (x(ga−1 )) + p2 (x(gb−1 ))
8.11 Surjective but not Preinjective Linear Cellular Automata
315
for all x ∈ V G and g ∈ G. This cellular automaton is the one described in Sect. 5.11 for H = K. By Proposition 5.11.1, τ is preinjective but not surjective. If H is a subgroup of a group G and τ : V H → V H is a linear cellular automaton over H which is preinjective and not surjective, then the induced cellular automaton τ G : V G → V G is also linear, preinjective and not surjective (see Proposition 1.7.4 and Proposition 5.2.2). Therefore, an immediate consequence of Proposition 8.10.1 is the following: Corollary 8.10.2. Let G be a group containing a free subgroup of rank two and let V be a twodimensional vector space over a ﬁeld K. Then there exists a linear cellular automaton τ : V G → V G which is preinjective but not surjective. This shows that implication (c) ⇒ (a) in Theorem 8.9.6 becomes false if G is a group containing a free subgroup of rank two.
8.11 Surjective but not Preinjective Linear Cellular Automata We now present examples of linear cellular automata with ﬁnitedimensional alphabet which are surjective but not preinjective. We recall that, by Theorem 8.9.6, the underlying groups for such examples are necessarily nonamenable. Proposition 8.11.1. Let G = F2 be the free group of rank two and let V be a twodimensional vector space over a ﬁeld K. Then there exists a linear cellular automaton τ : V G → V G which is surjective but not preinjective. Proof. Let a, b denote the canonical generators of G. We can assume that V = K2 . Consider the elements q1 , q2 ∈ EndK (V ) respectively deﬁned by q1 (v) = (k1 , 0) and q2 (v) = (0, k1 ) for all v = (k1 , k2 ) ∈ V . Let τ : V G → V G be the linear cellular automaton given by τ (x)(g) = q1 (x(ga)) + q1 (x(ga−1 )) + q2 (x(gb)) + q2 (x(gb−1 )) for all x ∈ V G , g ∈ G. A memory set for τ is the set S = {a, b, a−1 , b−1 }. Let k0 be any nonzero element in K. Consider the conﬁguration x0 ∈ V G deﬁned by (0, k0 ) if g = 1G x0 (g) = (0, 0) otherwise. Then, x0 is almost equal to 0 but x0 = 0. As τ (x0 ) = 0, we deduce that τ is not preinjective.
316
8 Linear Cellular Automata
However, τ is surjective. To see this, let z = (z1 , z2 ) ∈ KG × KG = V G . Let us show that there exists x ∈ V G such that τ (x) = z. We deﬁne x(g) by induction on the graph distance (cf. Sect. 6.2), which we denote by g, of g ∈ G from 1G in the Cayley graph of G. We ﬁrst set x(1G ) = (0, 0). Then, for s ∈ S we set ⎧ ⎨ (z1 (1G ), 0) if s = a x(s) = (z2 (1G ), 0) if s = b (8.22) ⎩ (0, 0) otherwise. Suppose that x(g) has been deﬁned for all g ∈ G with g ≤ n, for some n ≥ 1. For g ∈ G with g = n, let g ∈ G and s ∈ S be the unique elements such that g  = n − 1 and g = g s . Then, for s ∈ S with s s = 1G , we set ⎧ (z1 (g) − x1 (g ), 0) if s ∈ {a, a−1 } and s = s ⎪ ⎪ ⎪ ⎪ if s ∈ {a, a−1 } and s = b ⎨ (z2 (g), 0) if s ∈ {b, b−1 } and s = a x(gs) = (z1 (g), 0) (8.23) ⎪ −1 ⎪ (z (g) − x (g ), 0) if s ∈ {b, b } and s = s ⎪ 2 2 ⎪ ⎩ (0, 0) otherwise. Let us check that τ (x) = z. We have τ (x)(1G ) = q1 (x(a)) + q1 (x(a−1 )) + q2 (x(b)) + q2 (x(b−1 )) (by (8.22)) = q1 (z1 (1G ), 0) + q1 (0, 0) + q2 (z2 (1G ), 0) + q2 (0, 0) = (z1 (1G ), 0) + (0, 0) + (0, z2 (1G )) + (0, 0) = (z1 (1G ), z2 (1G )) = z(1G ). Let now g = g s ∈ G with g = g  + 1 > 2. Suppose, for instance, that s = a. We then have: τ (x)(g) = τ (x)(g a) (by (8.23)) = q1 (x(g a2 )) + q1 (x(g )) + q2 (x(g ab)) + q2 (x(g ab−1 )) = q1 (z1 (g) − x1 (g ), 0) + q1 (x1 (g ), x2 (g )) + q2 (z2 (g), 0) + q2 (0, 0) = (z1 (g) − x1 (g ), 0) + (x1 (g ), 0) + (0, z2 (g)) + (0, 0) = (z1 (g), z2 (g)) = z(g). The cases when s = a−1 , b, b−1 are similar. It follows that τ (x) = z. This shows that τ is surjective. If H is a subgroup of a group G, and τ : V H → V H is a linear cellular automaton over H which is surjective and not preinjective, then the induced cellular automaton τ G : V G → V G is also linear, surjective and not
8.12 Invertible Linear Cellular Automata
317
preinjective (see Proposition 1.7.4 and Proposition 5.2.2). Therefore, an immediate consequence of Proposition 8.11.1 is the following: Corollary 8.11.2. Let G be a group containing a free subgroup of rank two and let V be a twodimensional vector space over a ﬁeld K. Then there exists a linear cellular automaton τ : V G → V G which is surjective but not preinjective. As a consequence, we deduce that implication (a) ⇒ (c) in Theorem 8.9.6 becomes false if the group G contains a free subgroup of rank two.
8.12 Invertible Linear Cellular Automata Theorem 8.12.1. Let G be a group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Let τ : V G → V G be an injective linear cellular automaton. Then there exists a linear cellular automaton σ : V G → V G such that σ ◦ τ = IdV G . Proof. We split the proof into two steps. Suppose ﬁrst that G is countable. Since τ is injective, it induces a bijective map τ : V G → Y where Y = τ (V G ) is the image of τ . The fact that τ is Klinear and Gequivariant implies that Y is a Ginvariant vector subspace of V G and that the inverse map τ−1 : Y → V G is Klinear and Gequivariant. Let us show that the following local property is satisﬁed by τ−1 : there exists a ﬁnite subset T ⊂ G such that (∗) for y ∈ Y , the element τ−1 (y)(1G ) only depends on the restriction of y to T . Let us assume by contradiction that there exists no such T . Let S be a memory set for τ such that 1G ∈ S. Since G is countable, we
can ﬁnd a sequence (An )n∈N of ﬁnite subsets of G such that G = n∈N An , −S S ⊂ A0 and An ⊂ An+1 for all n ∈ N.
Let Bn = An denote the Sinterior of An (cf. Sect. 5.4). Note that G = n∈N Bn and Bn ⊂ Bn+1 for all n ∈ N. Since there exists no ﬁnite subset T ⊂ G satisfying condition (∗), we can ﬁnd, for each n ∈ N, two conﬁgurations yn , yn ∈ Y such that yn Bn = yn Bn and τ−1 (yn )(1G ) = τ−1 (yn )(1G ). By linearity of τ−1 , the conﬁguration yn = yn − yn ∈ Y satisﬁes yn Bn = 0 and τ−1 (yn )(1G ) = 0. It follows from Proposition 5.4.3 that if x and x are elements in V G such that x and x coincide on An then the conﬁgurations τ (x) and τ (x ) coincide n ∈ V G a conﬁguration on Bn . Therefore, given xn ∈ V An and denoting by x extending xn , the pattern xn )Bn ∈ V Bn un = τ ( does not depend on the particular choice of the extension x n of xn . Thus we can deﬁne a map τn : V An → V Bn by setting τn (xn ) = un . It is clear that τn is Klinear.
318
8 Linear Cellular Automata
Consider, for each n ∈ N, the vector subspace Ln ⊂ V An deﬁned by Ln = Ker(τn ). where for n ≤ m, the restriction map V Am → V An induces a Klinear map πn,m : Lm → Ln . Indeed, if u ∈ Lm , then we have uAn ∈ Ln since τm (u) = 0 and therefore τn (uAn ) = (τm (u))Bn = 0. Consider now, for all n ≤ m, the vector subspace Kn,m ⊂ Ln deﬁned by Kn,m = πn,m (Lm ). We have Kn,m ⊂ Kn,m for all n ≤ m ≤ m since πn,m = πn,m ◦πm,m . Therefore, if we ﬁx n, the sequence Kn,m , where m = n, n+1, . . ., is a decreasing sequence of vector subspaces of Ln . As Ln ⊂ V An is ﬁnitedimensional, this sequence stabilizes, i.e., there exist a vector subspace Jn ⊂ Ln and an integer kn ≥ n such that Kn,m = Jn for all m ≥ kn . For all n ≤ n ≤ m, we have πn,n (Kn ,m ) ⊂ Kn,m since πn,n ◦ πn ,m = πn,m . Therefore, πn,n induces by restriction a linear map ρn,n : Jn → Jn for all n ≤ n . We claim that ρn,n is surjective. To see this, let u ∈ Jn . Let us choose m large enough so that Jn = Kn,m and Jn = Kn ,m . Then we can ﬁnd v ∈ Lm such that u = πn,m (v). We have u = ρn,n (w), where w = πn ,m (v) ∈ Kn ,m = Jn . This proves the claim. Now, using the surjectivity of ρn,n+1 for all n, we construct by induction a sequence of elements zn ∈ Jn , n ∈ N, as follows. We start by taking as z0 the restriction of xk0 to A0 . Observe that xk0 ∈ Lk0 and hence z0 = π0,k0 (xk0 ) ∈ J0 . Then, assuming that zn has been constructed, we take as zn+1 an arbitrary element in ρ−1 n,n+1 (zn ). Since zn+1 coincides with zn on An , there exists a unique element z ∈ V G such that zAn = zn for all n. We have z ∈ Y , since zn ∈ πAn (Y ) for all n and X is closed in V G . As z(1G ) = z0 (1G ) = xk0 (1G ) = 0, we have z = 0. On the other hand, τ (z) = 0 since τ (z)Bn = τn (zn ) = 0 for all n by construction. This contradicts the injectivity of τ . Thus, there exists a ﬁnite subset T ⊂ G satisfying (∗). z )(1G ), where Consider the linear map ν : V T → V deﬁned by ν(z) = τ−1 ( z ∈ V G is any conﬁguration extending the pattern z ∈ V T . Then the linear cellular automaton σ : V G → V G with memory set T and local deﬁning map ν clearly satisﬁes σ ◦ τ = IdV G . Indeed, if x ∈ V G , then denoting by y = τ (x) ∈ Y its image by τ , we have x = τ−1 (y) and (σ ◦ τ )(x)(1G ) = σ(y)(1G ) = ν(yT ) = τ−1 (y)(1G ) = x(1G ) showing that (σ ◦ τ )(x) = x, by Gequivariance of σ ◦ τ . It follows that σ ◦ τ = IdV G and this proves the statement when G is a countable group. We now drop the countability assumption on G and prove the theorem in the general case. Let S ⊂ G be a memory set for τ and denote by μ : V S → V the corresponding local deﬁning map. Let H ⊂ G be the subgroup generated by S. Note that H is countable. Consider the set G/H of all left cosets of H in G. For c ∈ G/H denote by
8.12 Invertible Linear Cellular Automata
319
πc : V G → V c =
V
g∈c
the projection map. We have V G = c∈G/H V c and, for every x ∈ V G we write x = (xc )c∈G/H , where xc = πc (x) ∈ V c . For c ∈ G/H and g ∈ c denote by φg : H → c the linear map deﬁned by φg (h) = gh for all h ∈ H. Consider the map φ∗g : V c → V H deﬁned by φ∗g (z) = z ◦ φg . Then, if x = (xc )c∈G/H ∈ V G and c ∈ G/H, we have (φg ∗ (xc ))(h) = xc (gh) = x(gh) = (g −1 x)(h) = (g −1 x)H (h) for all h ∈ H, that is,
φ∗g (xc ) = (g −1 x)H .
(8.24)
For c ∈ G/H, deﬁne the map τc : V c → V c by setting τc (z)(g) = μ((φ∗g (z))S ) for all z ∈ V c and g ∈ c. Note that τH : V H → V H is the restriction of τ to the subgroup H. We then have τ= τc . (8.25) c∈G/H
From (8.25) we immediately deduce that Y = c∈G/H Yc , where Yc = πc (Y ) = τc (V c ), and that τc is injective for all c ∈ G/H. By the ﬁrst part of the present proof, there exists a linear cellular automaton σH : V H → V H such that σH ◦ τH = IdV H .
(8.26)
Let T ⊂ H be a memory set for σH and let ν : V T → V be the corresponding local deﬁning map. Consider the linear cellular automaton σ : V G → V G deﬁned by setting σ(y)(g) = ν((g −1 y)T ) for all y ∈ V G and g ∈ G. Note that σ = (σH )G is the induced cellular automaton of σH from H to G, so that, if σc : V c → V c is the map deﬁned by setting σc (z)(g) = ν((φ∗g (z))T ) for all z ∈ V c and g ∈ c then σ=
σc .
c∈G/H
Given c ∈ G/H, z ∈ V c and g ∈ c, we have
(8.27)
320
8 Linear Cellular Automata
(σc ◦ τc )(z)(g) = σc (τc (z))(g) = [φ∗g σc (τc (z))](1G ) (by (8.24)) = σH (φ∗g (τc (z))(1G ) (again by (8.24)) = σH (τH (φ∗g (z)))(1G ) (by (8.26)) = (φ∗g (z))(1G ) = z(g). This shows that (σc ◦ τc )(z) = z for all z ∈ V c . We deduce that σc ◦ τc = IdV c for all c ∈ C. It follows from (8.25) and (8.27) that σ ◦ τ = IdV G . We recall that given a group G and a set A, a cellular automaton τ : AG → A is said to be invertible if τ is bijective and the inverse map τ −1 : AG → AG is also a cellular automaton. When A is a ﬁnite set, every bijective cellular automaton τ : AG → AG is invertible by Theorem 1.10.2. The following is a linear analogue. G
Corollary 8.12.2. Let G be a group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Then every bijective linear cellular automaton τ : V G → V G is invertible. Proof. Let τ : V G → V G be a bijective linear cellular automaton. It follows from Theorem 8.12.1 that there exists a linear cellular automaton σ : V G → V G such that σ◦τ = IdV G . This implies that τ −1 = σ is a cellular automaton. Thus τ is an invertible cellular automaton. Remark 8.12.3. Let G = Z and let K be a ﬁeld. Consider the inﬁnite dimensional Kvector space V = K[[t]] consisting of all formal power series in one indeterminate t with coeﬃcients in K. Thus, an element of V is just a sequence v = (ki )i∈N of elements of K written in the form v = k0 + k1 t + k2 t2 + k3 t3 + · · · = ki ti . i∈N
Consider the map τ : V Z → V Z deﬁned by τ (x)(n) = x(n) − tx(n + 1) for all x ∈ V Z , n ∈ Z. In Example 1.10.3 we have showed that τ is a bijective cellular automaton whose inverse map τ −1 : V Z → V Z is not a cellular automaton. It is clear that τ is Klinear. This shows that Corollary 8.12.2 becomes false if we omit the ﬁnitedimensionality of the alphabet V . Let G be a group and let V be a vector space over a ﬁeld K. We have seen in Sect. 1.10 that the set ICA(G; V ) consisting of all invertible cellular automata τ : V G → V G is a group for the composition of maps. The subset of ICA(G; V ) consisting of all invertible linear cellular automata is a subgroup
8.13 Preinjectivity and Surjectivity of the Discrete Laplacian
321
of ICA(G; V ) since it is the intersection of ICA(G; V ) with the automorphism group of the Kvector space V G . Given a ring R and an integer d ≥ 1, we denote by GLd (R) the group of invertible elements of the matrix ring Matd (R). An immediate consequence of Corollary 8.12.2 and Corollary 8.7.8 is the following Corollary 8.12.4. Let G be a group and let V be a vector space over a ﬁeld K of ﬁnite dimension dimK (V ) = d ≥ 1. Then the set consisting of all bijective linear cellular automata τ : V G → V G is a subgroup of ICA(G; V ) isomorphic to GLd (K[G]).
8.13 Preinjectivity and Surjectivity of the Discrete Laplacian Let G be a group and let K be a ﬁeld. Given a nonempty ﬁnite subset S of G, we recall that the discrete Laplacian associated with G and S is the linear cellular automaton ΔS : KG → KG (with memory set S ∪ {1G }) deﬁned by ΔS (f )(g) = Sf (g) − f (gs) s∈S
for all f ∈ KG and g ∈ G, where S denotes the cardinality of S (cf. Example 1.4.3(b) and Example 8.1.3(a)). Let us observe that ΔS is never injective since all constant maps f : G → K are in the kernel of ΔS . Proposition 8.13.1. Let G be a group and let K be a ﬁeld. Let S ⊂ G be a nonempty ﬁnite subset and suppose that the subgroup H ⊂ G generated by S is ﬁnite. Then ΔS is neither preinjective nor surjective. Proof. Let (ΔS )H : KH → KH denote the restriction cellular automaton of ΔS to H. Observe that (ΔS )H is the discrete Laplacian on KH associated with H and S. As we have seen above, (ΔS )H is not injective. As H is ﬁnite, it follows that (ΔS )H is not preinjective. On the other hand, the noninjectivity of (ΔS )H implies its nonsurjectivity since (ΔS )H is Klinear and KH is ﬁnitedimensional. By applying Proposition 5.2.2 (resp. Proposition 1.7.4), we deduce that ΔS is not preinjective (resp. not surjective). In the remaining of this section, we assume that the ﬁeld K is the ﬁeld R of real numbers. The following result is a Garden of Eden type theorem for the discrete laplacian. Note that there is no amenability hypothesis for the underlying group.
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Theorem 8.13.2. Let G be a group and let S be a nonempty ﬁnite subset of G. Let ΔS : RG → RG denote the associated discrete real laplacian. Then the following conditions are equivalent: (a) ΔS is surjective; (b) the subgroup of G generated by S is inﬁnite; (c) ΔS is preinjective. Let us ﬁrst establish the following simple fact. Recall that given a group G, the subsemigroup generated by a subset S ⊂ G is the smallest subset P of G containing S which is closed under the group operation, i.e., such that p1 p2 ∈ P for all p1 , p2 ∈ P . Clearly, P consists of all elements of the form p = s1 s2 · · · sn where n ≥ 1 and si ∈ S for 1 ≤ i ≤ n. Lemma 8.13.3. Let G be a group and let S be a subset of G. Then the following conditions are equivalent: (a) the subsemigroup of G generated by S is inﬁnite; (b) the subgroup of G generated by S is inﬁnite. Proof. Let P (resp. H) denote the subsemigroup (resp. subgroup) of G generated by S. The implication (a) ⇒ (b) is obvious since P ⊂ H. Suppose that S is nonempty and that P is ﬁnite. Then, for each s ∈ S, the set {sn : n ≥ 1} ⊂ P is also ﬁnite. This implies that every element of S has ﬁnite order. Therefore, for each s ∈ S, we can ﬁnd an integer n ≥ 2 such that sn = 1G . It follows that s−1 = sn−1 ∈ S. Consequently, we have P −1 = P and 1G ∈ P . This implies H = P , so that H is ﬁnite. This shows (b) ⇒ (a). Proof of Theorem 8.13.2. The implication (a) ⇒ (b) immediately follows from Proposition 8.13.1. Suppose that the subgroup generated by S is inﬁnite. Let f be an element of R[G] such that ΔS (f ) = 0. Let us show that f = 0. Let g0 ∈ G such that f (g0 ) = maxg∈G f (g). Note that such a g0 exists since f takes only ﬁnitely many values. As 0 = ΔS (f )(g0 ) = Sf (g0 ) − f (g0 s), s∈S
we get f (g0 ) ≤
1 f (g0 s) S
(8.28)
s∈S
by applying the triangle inequality. We deduce from (8.28) and the deﬁnition of g0 that f (g0 s) = f (g0 ) for all s ∈ S. By iterating the previous argument, we get f (g0 p) = f (g0 ) for all p ∈ P , where P denotes the subsemigroup of G generated by S. As P is inﬁnite (cf. Lemma 8.13.3), this implies that f (h) = f (g0 ) = maxg∈G f (g) for inﬁnitely many h ∈ G. Since f has ﬁnite
8.13 Preinjectivity and Surjectivity of the Discrete Laplacian
323
support, it follows that f = 0. Therefore, ΔS is preinjective. This proves the implication (b) ⇒ (c). To complete the proof, it suﬃces to prove that (c) implies (a). Suppose ﬁrst that G is an amenable group. It follows from the implication (c) ⇒ (a) in Theorem 8.9.6 that if ΔS is preinjective, then it is surjective. This shows the implication (c) ⇒ (a) for G amenable. Suppose now that G is nonamenable (and hence inﬁnite) and that S is a generating subset for G. This implies that G is countable. Consider the Hilbert space 2 (G) ⊂ RG consisting of all squaresummable real functions on (2) G and the continuous linear map ΔS : 2 (G) → 2 (G) obtained by restriction 2 of ΔS to (G) (cf. Sect. 6.12). By the KestenDay amenability criterion (Theorem 6.12.9), the nona(2) menability of G implies that 0 does not belong to the spectrum σ(ΔS ) (2) (2) of ΔS . Thus, ΔS is bijective and hence (2)
R[G] ⊂ 2 (G) = ΔS (2 (G)) = ΔS (2 (G)) ⊂ ΔS (RG ). By applying Corollary 8.8.2, we deduce that ΔS (RG ) = RG . This shows that ΔS is surjective in the case when G is nonamenable and S generates G. Finally, suppose now that G is an arbitrary nonamenable group and that ΔS is preinjective. Denote by H the subgroup of G generated by S. Then the restriction cellular automaton (ΔS )H : RH → RH , which is the discrete laplacian associated with H and S, is preinjective by Proposition 5.2.2. The subgroup H may be amenable or not. However, it follows from the two preceding cases that (ΔS )H is surjective. By applying Proposition 1.7.4, we deduce that ΔS is surjective as well. This completes the proof that (a) implies (c). As a consequence of Theorem 8.13.2, we obtain the following characterization of locally ﬁnite groups in terms of real linear cellular automata: Corollary 8.13.4. Let G be a group and let V be a real vector space of ﬁnite dimension d ≥ 1. Then the following conditions are equivalent: (a) G is locally ﬁnite; (b) every surjective linear cellular automaton τ : V G → V G is injective. Proof. Suppose (a). Let τ : V G → V G be a surjective linear cellular automaton with memory set S ⊂ G. As G is locally ﬁnite, the subgroup H generated by S is ﬁnite. Consider the linear cellular automaton τH : V H → V H obtained from τ by restriction. Observe that τH is surjective by Proposition 1.7.4(ii). Since V H is ﬁnitedimensional, it follows that τH is injective. By applying Proposition 1.7.4(i), we deduce that τ is also injective. This shows that (a) implies (b). Now, suppose that G is not locally ﬁnite. Let us show that there exists a linear cellular automaton τ : V G → V G which is surjective but not injective.
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We can assume that V = Rd . Since G is not locally ﬁnite, we can ﬁnd a ﬁnite subset S ⊂ G such that the subgroup H generated by S is inﬁnite. By Theorem 8.13.2, the discrete Laplacian ΔS : RG → RG is surjective. As mentioned above ΔS is not injective since all constant conﬁgurations are in its kernel. Consider now the product map τ = (ΔS )d : (Rd )G → (Rd )G , where we use the natural identiﬁcation (Rd )G = (RG )d . Clearly, τ is a linear cellular automaton admitting S ∪ {1G } as a memory set. On the other hand, τ is surjective but not injective since any product of surjective (resp. noninjective) maps is a surjective (resp. noninjective) map. This shows that (b) implies (a).
8.14 Linear Surjunctivity In analogy with the ﬁnite alphabet case, we introduce the following deﬁnition. Deﬁnition 8.14.1. A group G is said to be Lsurjunctive if, for any ﬁeld K and any ﬁnitedimensional vector space V over K, every injective linear cellular automaton τ : V G → V G is surjective. Proposition 8.14.2. Every subgroup of an Lsurjunctive group is Lsurjunctive. Proof. Suppose that H is a subgroup of a Lsurjunctive group G. Let V be a ﬁnitedimensional vector space over a ﬁeld K and let τ : V H → V H be an injective linear cellular automaton over H. Consider the cellular automaton τ G : V G → V G over G obtained from τ by induction (see Sect. 1.7). The fact that τ is injective implies that τ G is injective by Proposition 1.7.4(i). Also, τ G is linear by Proposition 8.3.1. Since G is Lsurjunctive, it follows that τ G is surjective. By applying Proposition 1.7.4(ii), we deduce that τ is surjective. This shows that H is Lsurjunctive. Proposition 8.14.3. Let G be a group. Then the following conditions are equivalent: (a) G is Lsurjunctive; (b) every ﬁnitely generated subgroup of G is Lsurjunctive. Proof. The fact that (a) implies (b) follows from Proposition 8.14.2. Conversely, let G be a group all of whose ﬁnitely generated subgroups are Lsurjunctive. Let V be a ﬁnite dimensional vector space over a ﬁeld K and let τ : V G → V G be an injective linear cellular automaton with memory set S. Let H denote the subgroup of G generated by S and consider the linear cellular automaton τH : V H → V H obtained by restriction of τ (see Sect. 1.7 and Proposition 8.3.1). The fact that τ is injective implies that τH is injective by Proposition 1.7.4(i). As H is ﬁnitely generated, it is Lsurjunctive by
8.14 Linear Surjunctivity
325
our hypothesis on G. It follows that τH is surjective. By applying Proposition 1.7.4(ii), we deduce that τ is also surjective. This shows that (b) implies (a). Note that the preceding proposition may be stated by saying that a group is Lsurjunctive if and only if it is locally Lsurjunctive. Every ﬁnite group is obviously Lsurjunctive. Indeed, if G is a ﬁnite group and V is a ﬁnitedimensional vector space, then the vector space V G is also ﬁnitedimensional and therefore every injective endomorphism of V G is surjective. More generally, it follows from Corollary 8.9.7 that every amenable group is Lsurjunctive. In fact, we have the following result, which is a linear analogue of Theorem 7.8.1, Theorem 8.14.4. Every soﬁc group is Lsurjunctive. Proof. Let G be a soﬁc group. Let V be a ﬁnitedimensional vector space over a ﬁeld K of dimension dimK (V ) = d ≥ 1 and let τ : V G → V G be an injective linear cellular automaton. We want to show that τ is surjective. Every subgroup of a soﬁc group is soﬁc by Proposition 7.5.4. On the other hand, it follows from Proposition 3.2.2 that a group is Lsurjunctive if all its ﬁnitely generated subgroups are Lsurjunctive. Thus we can assume that G is ﬁnitely generated. Let then S ⊂ G be a ﬁnite symmetric generating subset of G. As usual, for r ∈ N, we denote by BS (r) ⊂ G the ball of radius r centered at 1G in the Cayley graph associated with (G, S). We set Y = τ (V G ). Observe that Y is a Ginvariant vector subspace of V G . On the other hand, it follows from Theorem 8.8.1 that Y is closed in V G with respect to the prodiscrete topology. By Theorem 8.12.1, there exists a linear cellular automaton σ : V G → V G such that σ ◦ τ = IdV G . Choose r0 large enough so that the ball BS (r0 ) is a memory set for both τ and σ. Let μ : V BS (r0 ) → V and ν : V BS (r0 ) → V denote the corresponding local deﬁning maps for τ and σ respectively. We proceed by contradiction. Suppose that τ is not surjective, that is, Y V G . Then, since Y is closed in V G , there exists a ﬁnite subset Ω ⊂ G such that Y Ω V Ω . It is not restrictive, up to taking a larger r0 , again if necessary, to suppose that Ω ⊂ BS (r0 ). Thus, Y BS (r0 ) V BS (r0 ) . Let ε > 0 be such that ε
1 − (1 − ε)−1 < 1 +
1 dB(2r0 )+1
1 . dB(2r0 )
(8.29) which yields (8.30)
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Since G is soﬁc, we can ﬁnd a ﬁnite Slabeled graph (Q, E, λ) such that Q(3r0 ) ≥ (1 − ε)Q,
(8.31)
where we recall that Q(r), r ∈ N, denotes the set of all q ∈ Q such that there exists an Slabeled graph isomorphism ψq,r : BS (r) → B(q, r) satisfying ψq,r (1G ) = q (cf. Theorem 7.7.1). Note the inclusions Q(r0 ) ⊃ Q(2r0 ) ⊃ · · · ⊃ Q(ir0 ) ⊃ Q((i + 1)r0 ) ⊃ · · · . (cf. (7.41); see also Fig. 7.2). Also recall from Lemma 7.7.2 that B(q, r0 ) ⊂ Q(ir0 ) for all Q((i + 1)r0 ) and i ≥ 0. For each integer i ≥ 1, we deﬁne the map μi : V Q(ir0 ) → V Q((i+1)r0 ) by setting, for all u ∈ V Q(ir0 ) and q ∈ Q((i + 1)r0 ), μi (u)(q) = μ uB(q,r0 ) ◦ ψq,r0 (1G ), where ψq,2r0 is the unique isomorphism of Slabeled graphs from BS (r0 ) ⊂ G to B(q, r0 ) ⊂ Q sending 1G to q (cf. 7.39 and 7.40). Similarly, we deﬁne the map νi : V Q(ir0 ) → V Q((i+1)r0 ) by setting, for all u ∈ V Q(ir0 ) and q ∈ Q((i + 1)r0 ), νi (u)(q) = ν uB(q,r0 ) ◦ ψq,r0 (1G ). From the fact that τ −1 ◦ τ is the identity map on V G , we deduce that the composite νi+1 ◦μi : V Q(ir0 ) → V Q((i+2)r0 ) is the identity on V Q((i+2)r0 ) . More precisely, denoting by ρi : V Q(ir0 ) → V Q((i+2)r0 ) the restriction map, we have that νi+1 ◦ μi = ρi for all i ≥ 1. In particular, we have ν2 ◦ μ1 = ρ1 . Thus, setting Z = μ1 (V Q(r0 ) ) ⊂ V Q(2r0 ) , we deduce that ν2 (Z) = ρ1 (V Q(r0 ) ) = V Q(3r0 ) . It follows that (8.32) dim(Z) ≥ dQ(3r0 ). Let Q ⊂ Q(3r0 ) be as in (7.59) and set Q = q ∈Q B(q , r0 ). Note that Q ⊆ Q(2r0 ) so that Q(2r0 ) = Q  · BS (r0 ) + Q(2r0 ) \ Q .
(8.33)
Now observe that, for all q ∈ Q(2r0 ), we have a natural isomorphism of vector spaces ZB(q,r0 ) → Y BS (r0 ) given by u → u◦ψq,r0 , where ψq,r0 denotes as above the unique isomorphism of Slabeled graphs from BS (r0 ) to B(q, r0 ) such that ψq,r0 (1G ) = q. Since Y BS (r0 ) V BS (r0 ) , this implies that dim ZB(q,r0 ) = dim(Y BS (r0 ) ) ≤ d · BS (r0 ) − 1, (8.34) for all q ∈ Q . Thus we have
8.15 Stable Finiteness of Group Algebras
dim(Z) ≤ dim ZQ + dim ZQ(2r0 )\Q
327
≤ Q  · (d · BS (r0 ) − 1) + d · Q(2r0 ) \ Q  Q  = d Q(2r0 ) − d where the last equality follows from (8.33). Comparing this with (8.32) we obtain Q  . Q(3r0 ) ≤ Q(2r0 ) − d Thus, Q ≥ Q(2r0 ) ≥ Q(3r0 ) +
Q  d
Q(3r0 ) by (7.59), dB(2r0 ) 1 = Q(3r0 ) 1 + dB(2r0 )
≥ Q(3r0 ) +
> Q(3r0 )(1 − ε)−1 where the last inequality follows from (8.30). This yields Q(3r0 ) < (1 − ε)Q which contradicts (8.31). This shows that τ (V G ) = Y = V G , that is, τ is surjective. It follows that the group G is Lsurjunctive.
8.15 Stable Finiteness of Group Algebras Let R be a ring and denote by 1R its unity element. If a, b ∈ R satisfy ab = 1R , then one says that b is a rightinverse of a and that a is a leftinverse of b. An element a ∈ R is said to be rightinvertible (resp. leftinvertible) if it admits a rightinverse (resp. a leftinverse). Every invertible element in R is both rightinvertible and leftinvertible. Conversely, suppose that an element a ∈ R admits a rightinverse b and a leftinverse b . Then, we have ab = 1R and b a = 1R , so that b = (b a)b = b (ab ) = b . This shows that a is invertible with inverse a−1 = b = b . Thus, if an element is both rightinvertible and leftinvertible, then it is invertible. One says that the ring R is directly ﬁnite if every rightinvertible element (or, equivalently, every leftinvertible element) is invertible. This is equivalent to saying that if any two elements a, b ∈ R satisfy ab = 1R , then they also satisfy ba = 1R .
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The ring R is said to be stably ﬁnite if the matrix ring Matd (R) is directly ﬁnite for any d ≥ 1. Observe that every stably ﬁnite ring R is directly ﬁnite since Mat1 (R) = R. Proposition 8.15.1. Every ﬁnite ring is stably ﬁnite. Proof. Let R be a ﬁnite ring. Suppose that a, b ∈ R satisfy ab = 1R . Consider the map f : R → R deﬁned by f (r) = ar. We have f (br) = a(br) = (ab)r = 1R r = r for all r ∈ R. Therefore, the map f is surjective. As R is ﬁnite, this implies that f is also injective. Since f (ba) = a(ba) = (ab)a = a = f (1R ), we deduce that ba = 1R . This shows that every ﬁnite ring is directly ﬁnite. If the ring R is ﬁnite, then the ring Matd (R) is also ﬁnite for every d ≥ 1. Consequently, every ﬁnite ring is stably ﬁnite. Proposition 8.15.2. Every commutative ring is stably ﬁnite. Proof. Let d ≥ 1 and suppose that a ∈ Matd (R) is rightinvertible. Then there exists b ∈ Matd (R) such that ab = Id . This implies that det(a) det(b) = det(ab) = det(Id ) = 1R . It follows that det(a) is an invertible element in R. Therefore a is invertible in Matd (R). This shows that Matd (R) is directly ﬁnite for any d ≥ 1. Consequently, R is stably ﬁnite. Let us give an example of a ring which is not directly ﬁnite. Example 8.15.3. Let R be a nonzero ring and consider the free left Rmodule M = ⊕n∈N R. Every element in M can be represented in the form m = (mn )n∈N where mn ∈ R for all n ∈ N and mn = 0R for all but ﬁnitely many n ∈ N. Consider the maps a : M → M and b : M → M deﬁned by setting a(m) = m (resp. b(m) = m ) where mn = mn−1 for n ≥ 1 and m0 = 0R (resp. mn = mn+1 for all n ∈ N). Then a and b are obviously Rlinear, in other words, a, b ∈ EndR (M ) and ab = IdM = 1EndR (M ) . Consider the element m ∈ M such that m0 = 1R and mn = 0R for all n ≥ 1. As a(m) = 0 we have ba(m) = 0. This shows that ba = 1EndR (M ) . It follows that the ring EndR (M ) is not directly ﬁnite. Let G be a group and let V be a vector space over a ﬁeld K. Recall that the set LCA(G; V ) consisting of all linear cellular automata τ : V G → V G has a natural structure of Kalgebra in which the multiplication is given by the composition of maps. Proposition 8.15.4. Let G be a group and let V be a vector space over a ﬁeld K. Let τ : V G → V G be a linear cellular automaton. Then the following hold: (i) if τ is leftinvertible in LCA(G; V ), then τ is injective;
8.15 Stable Finiteness of Group Algebras
329
(ii) if τ is rightinvertible in LCA(G; V ), then τ is surjective; (iii) if τ is invertible in LCA(G; V ), then τ is bijective. If, in addition, the vector space V is ﬁnitedimensional, then: (iv) τ is leftinvertible in LCA(G; V ) if and only if τ is injective; (v) τ is invertible in LCA(G; V ) if and only if τ is bijective. Proof. (i) Suppose that τ ∈ LCA(G; V ) is leftinvertible. This means that there exists σ ∈ LCA(G; V ) such that σ ◦ τ = IdV G . As IdV G is injective, this implies that τ is injective. (ii) Suppose that τ ∈ LCA(G; V ) is rightinvertible. This means that there exists σ ∈ LCA(G; V ) such that τ ◦ σ = IdV G . As IdV G is surjective, this implies that τ is surjective. (iii) This immediately follows from (i) and (ii). (iv) This immediately follows from (i) and Theorem 8.12.1. (v) This immediately follows from (iii) and Corollary 8.12.2. Remarks 8.15.5. (a) If V is inﬁnitedimensional, it may happen that a bijective linear cellular automaton τ : V G → V G is not leftinvertible in LCA(G; V ). Therefore, the converses of assertions (i) and (iii) in Proposition 8.15.4 are false if we do not add the hypothesis that V is ﬁnitedimensional. To see this, consider the bijective linear cellular automaton τ : V Z → V Z described in Remark 8.12.3. Then, there is no σ ∈ LCA(G; V ) such that σ ◦ τ = IdV G since otherwise the inverse map of τ would coincide with σ, which is impossible as τ is not an invertible cellular automaton. (b) The converse of assertion (ii) in Proposition 8.15.4 does not hold even under the additional hypothesis that V is ﬁnitedimensional. For instance, take an arbitrary ﬁeld K and consider the linear cellular automaton τ : KZ → KZ deﬁned by τ (x)(n) = x(n + 1) − x(n) for all x ∈ KZ and n ∈ Z. Observe that τ is surjective. Indeed, given y ∈ KZ , the conﬁguration x ∈ KZ deﬁned by ⎧ ⎪ if n = 0, ⎨0 x(n) = y(0) + y(1) + · · · + y(n − 1) if n > 0, ⎪ ⎩ y(n) + y(n + 1) + · · · + y(−1) if n < 0, clearly satisﬁes τ (x) = y. However, τ is not rightinvertible in LCA(Z; K). To see this, observe that, as the Kalgebra LCA(Z; K) is commutative by Corollary 8.5.4, the rightinvertibility of τ would imply the bijectivity of τ by Proposition 8.15.4(iii). But τ is not injective as all constant conﬁgurations are mapped to 0. Corollary 8.15.6. Let G be a group and let K be a ﬁeld. Let V be a vector space over K of ﬁnite dimension dimK (V ) = d ≥ 1. Then the following conditions are equivalent: (a) every injective linear cellular automaton τ : V G → V G is surjective; (b) the Kalgebra LCA(G; V ) is directly ﬁnite;
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(c) the Kalgebra Matd (K[G]) is directly ﬁnite. Proof. The equivalence between conditions (b) and (c) follow from the fact that the Kalgebras LCA(G; V ) and Matd (K[G]) are isomorphic by Corollary 8.7.8. Suppose (a). Let τ ∈ LCA(G; V ) be a leftinvertible element in LCA(G; V ). Then τ is injective by Proposition 8.15.4(i). Condition (a) implies then that τ is surjective and hence bijective. By applying Proposition 8.15.4(v), we deduce that τ is invertible in LCA(G; V ). This shows that (a) implies (b). Conversely, suppose (b). Let τ : V G → V G be an injective linear cellular automaton. Then τ is leftinvertible in LCA(G; V ) by Proposition 8.15.4(iv). It follows from condition (b) that τ is invertible in LCA(G; V ). By using Proposition 8.15.4(ii), we deduce that τ is surjective. This shows that (b) implies (a). Corollary 8.15.7. Let G be a group. Then the following conditions are equivalent: (a) the group G is Lsurjunctive; (b) for any ﬁeld K, the group algebra K[G] is stably ﬁnite.
From Theorem 8.14.4 and Corollary 8.15.7, we deduce the following: Corollary 8.15.8. Let G be a soﬁc group and let K be a ﬁeld. Then the group algebra K[G] is stably ﬁnite.
8.16 ZeroDivisors in Group Algebras and Preinjectivity of OneDimensional Linear Cellular Automata Let R be a ring. A nonzero element a ∈ R is said to be a left zerodivisor (resp. a right zerodivisor ) if there exists a nonzero element b in R such that ab = 0 (resp. ba = 0). Note that the existence of a left zerodivisor in R is equivalent to the existence of a right zerodivisor. One says that the ring R has no zerodivisors if R admits no left (or, equivalently, no right) zerodivisors. Examples 8.16.1. (a) Let R = Z/nZ, n ≥ 2, be the (commutative) ring of integers modulo n. Then R has no zerodivisors if and only if n is a prime number. (b) Let G be a group and let R be a nonzero ring. Suppose that G contains an element g0 of ﬁnite order n ≥ 2 and let H = {1G , g0 , g02 , . . . , g0n−1 } denote the subgroup of G generated by g0 . Consider the elements α, β ∈ R[G] deﬁned respectively by
8.16 ZeroDivisors in Group Algebras
⎧ ⎪ ⎨1 α(g) = −1 ⎪ ⎩ 0
if g = 1G , if g = g0 , if g ∈ / {1G , g0 }.
331
1 and β(g) = 0
if g ∈ H, otherwise.
One easily checks that α = 0, β = 0, and αβ = βα = 0. Thus, α and β are both left and right zerodivisors in R[G]. Let R be a ring. An element x ∈ R is called an idempotent if it satisﬁes 1. x2 = x. An idempotent x ∈ R is said to be proper if x = 0 and x = Proposition 8.16.2. Let R be a ring. Then the following hold: (i) if R has no zerodivisors, then R has no proper idempotents; (ii) if R has no proper idempotents, then R is directly ﬁnite. Proof. (i) Every idempotent x ∈ R satisﬁes x(x − 1) = x2 − x = 0. If R has no zerodivisors, this implies x = 0 or x = 1. (ii) Suppose that R has no proper idempotents. Let a, b ∈ R such that ab = 1. Then we have (ba)2 = b(ab)a = ba so that ba is an idempotent. Therefore, ba = 0 or ba = 1. If ba = 0, then a = (ab)a = a(ba) = 0 so that 0 = 1 and R is reduced to 0. Therefore, we have ba = 1 in all cases. This shows that R is directly ﬁnite. Remarks 8.16.3. (a) A ring without proper idempotents may admit zerodivisors. For example, the ring Z/8Z has no proper idempotents. However, the classes of 2, 4, and 6 are zerodivisors in Z/8Z. (b) A directly ﬁnite ring may admit proper idempotents. For example, take a nonzero commutative ring R. Then the ring 2 (R) is directly ﬁ Mat nite by Proposition 8.15.2. However, the matrices 10 00 and 00 01 are proper idempotents in Mat2 (R). A group G is called a uniqueproduct group if, given any two nonempty ﬁnite subsets A, B ⊂ G, there exists an element g ∈ G which can be uniquely expressed as a product g = ab with a ∈ A and b ∈ B. Remark 8.16.4. A uniqueproduct group is necessarily torsionfree. Indeed, suppose that G is a group containing an element g0 of order n ≥ 2. Take A = B = {1G , g0 , g02 , . . . , g0n−1 }. Then AB = A and there is no g ∈ AB which can be uniquely written in the form g = ab with a ∈ A and b ∈ B. Recall that a total ordering on a set X is a binary relation ≤ on X which is reﬂexive, antisymmetric, transitive, and such that one has x ≤ y or y ≤ x for all x, y ∈ X. A group G is called orderable if it admits a leftinvariant total ordering, that is, a total ordering ≤ such that g1 ≤ g2 implies gg1 ≤ gg2 for all g, g1 , g2 ∈ G.
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Examples 8.16.5. (a) Every subgroup of an orderable group is orderable. Indeed, if H is a subgroup of a group G and ≤ is a leftinvariant total ordering on G, then the restriction of ≤ to H is a leftinvariant total ordering on H. (b) The additive groups Z, Q, and R are orderable since the usual ordering on R is translationinvariant. (c) The direct product of two orderable groups is an orderable group. Indeed, let G1 and G2 be two groups and suppose that ≤1 and ≤2 are leftinvariant total orderings on G1 and G2 respectively. Then the lexicographic ordering ≤ on G1 × G2 deﬁned by ⎧ ⎪ ⎨g1 ≤1 g2 (g1 , g2 ) ≤ (h1 , h2 ) ⇐⇒ or ⎪ ⎩ g1 = g2 and h1 ≤2 h2 is a leftinvariant total ordering on G1 × G2 . (d) More generally, the direct product of any family (ﬁnite or inﬁnite) of orderable groups is an orderable group. Indeed, let (Gi )i∈I be a family of orderable groups. Let ≤i be a leftinvariant total ordering on Gi for each i ∈ I. Let us ﬁx a wellordering on I, i.e., a total ordering such that every nonempty subset of I admits a minimal element (the fact that any set can be wellordered is a basic fact in set theory which may be deduced from the Axiom of Choice). Then the lexicographic ordering on G = i∈I Gi , which is deﬁned by setting g ≤ h for g = (gi ), h = (hi ) ∈ G if and only if g = h or gi0 < hi0 where i0 = min{i ∈ I : gi = hi }, is a leftinvariant total ordering on G. As ⊕i∈I Gi is a subgroup of i∈I Gi , it follows that the direct sum of any family of orderable groups is an orderable group. Since a free abelian group is isomorphic to a direct sum of copies of Z, we deduce in particular that every free abelian group is orderable. (e) In fact, every torsionfree abelian group is orderable. Indeed, if G is a torsionfree abelian group then G embeds in the Qvector space G ⊗Z Q via the map g → g ⊗ 1. On the other hand, the additive group underlying a Qvector space V is isomorphic to a direct sum of copies of Q (since V admits a Qbasis) and hence orderable. (f) Suppose that G is a group containing a normal subgroup N such that both N and G/N are orderable groups. Then the group G is orderable. Indeed, let ≤1 (resp. ≤2 ) be a leftinvariant total ordering on N (resp. G/N ). Then one easily checks that the binary relation ≤ on G deﬁned by setting ⎧ ⎪ ⎨ρ(g) ≤2 ρ(h) g ≤ h ⇐⇒ or ⎪ ⎩ ρ(g) = ρ(h) and 1G ≤1 g −1 h is a leftinvariant total ordering on G. (g) Let R be any ring. Consider the Heisenberg group
8.16 ZeroDivisors in Group Algebras
⎧ ⎛ ⎞ 1yz ⎨ HR = M (x, y, z) = ⎝0 1 x⎠ ⎩ 001
333
:
x, y, z ∈ R
⎫ ⎬ ⎭
.
(cf. Example 4.6.5). The kernel of the group homomorphism of HR onto R2 given by M (x, y, z) → (x, y) is the normal subgroup ⎧⎛ ⎫ ⎞ ⎨ 10z ⎬ N = ⎝0 1 0⎠ : z ∈ R . ⎩ ⎭ 001 As the groups N and HR /N are both abelian, it follows from Examples (e) and (f) above that HR is orderable. Given a set X equipped with a total ordering ≤, we denote by Sym(X, ≤) the group of orderpreserving permutations of X, that is, the subgroup of Sym(X) consisting of all bijective maps f : X → X such that x ≤ y implies f (x) ≤ f (y) for all x, y ∈ X. Proposition 8.16.6. Let G be a group. Then the following conditions are equivalent: (a) G is orderable; (b) there exists a set X equipped with a total ordering ≤ such that G is isomorphic to a subgroup of Sym(X, ≤). Proof. Suppose that ≤ is a left invariant total ordering on G. Then the action of G on itself given by left multiplication is orderpreserving. Thus G is isomorphic to a subgroup of Sym(G, ≤). This shows that (a) implies (b). Conversely, suppose that X is a set equipped with a total ordering ≤. Choose a well ordering ≤W on X. Then we can deﬁne a lexicographic ordering ≤L on Sym(X, ≤) by setting f ≤L g for f, g ∈ Sym(X, ≤) if and only if f = g or f (x0 ) ≤ g(x0 ) where x0 = min{x ∈ X : f (x) = g(x)}. Clearly ≤L is a left invariant total ordering on Sym(X, ≤). It follows that Sym(X, ≤) is an orderable group. As any subgroup of an orderable group is orderable, we conclude that (b) implies (a). Corollary 8.16.7. The group Homeo+ (R) of orientationpreserving homeomorphisms of R is orderable. Proposition 8.16.8. Every orderable group is a uniqueproduct group. Proof. Let G be an orderable group and let ≤ be a leftinvariant total ordering on G. Let A and B be nonempty ﬁnite subsets of G. Set bm = min B and let am ∈ A be the unique element such that am bm = min Abm . Consider the element g = am bm . For all a ∈ A and b ∈ B, we have am bm ≤ abm ≤ ab. Thus, if g = ab for some a ∈ A and b ∈ B, then am bm = abm = ab, so that we get am = a and bm = b after right and left cancellation. This shows that G is a uniqueproduct group.
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Proposition 8.16.9. Let G be a uniqueproduct group and let R be a ring with no zerodivisors. Then the group ring R[G] has no zerodivisors. Proof. Let α and β be nonzero elements in R[G], and denote by A and B their supports. As G is a uniqueproduct group, there is an element g ∈ AB such that there exists a unique element (a, b) ∈ A × B such that g = ab. Then we have α(h1 )β(h2 ) = α(a)β(b). (αβ)(g) = h1 ∈A,h2 ∈B h1 h2 =g
As R has no zerodivisors, this implies (αβ)(g) = 0. Thus, we have αβ = 0. This shows that R[G] has no zerodivisors. Corollary 8.16.10. Let G be a uniqueproduct group and let K be a ﬁeld. Then the group algebra K[G] has no zerodivisors. Let G be a group and let K be a ﬁeld. We have seen in Corollary 8.5.3 that the map Ψ : K[G] → LCA(G; K) deﬁned by Ψ (α)(x)(g) = α(h)x(gh) h∈G
for all α ∈ K[G], x ∈ KG , and g ∈ G, is a Kalgebra isomorphism. Proposition 8.16.11. Let G be a group and let K be a ﬁeld. Let α be a nonzero element in K[G]. Then the following conditions are equivalent: (a) α is not a left zerodivisor in K[G]; (b) the linear cellular automaton Ψ (α) : KG → KG is preinjective. Proof. By Proposition 8.2.5, the linear cellular automaton Ψ (α) is not preinjective if and only if there exists a nonzero element β ∈ K[G] such that Ψ (α)(β) = 0. As Ψ (α)(β)(g) = α(h)β(gh) = α(h)β ∗ (h−1 g −1 ) = (αβ ∗ )(g −1 ), h∈G
h∈G
for all β ∈ K[G] and g ∈ G, we deduce that Ψ (α) is not preinjective if and only if there exists a nonzero element β ∈ K[G] such that αβ ∗ = 0, that is, if and only if α is a left zerodivisor in K[G]. This shows that conditions (a) and (b) are equivalent. Corollary 8.16.12. Let G be a group and let K be a ﬁeld. Then the following conditions are equivalent: (a) the group algebra K[G] has no zerodivisors; (b) every nonidenticallyzero linear cellular automaton τ : KG → KG is preinjective.
Notes
335
From Corollary 8.16.12 and Theorem 8.9.6 we deduce the following. Corollary 8.16.13. Let G be an amenable group and let K be a ﬁeld such that K[G] has no zerodivisors. Then every nonidenticallyzero linear cellular automaton τ : KG → KG is surjective. Observe that it follows from Theorem 4.6.1, Example 8.16.5(e), Proposition 8.16.8, and Corollary 8.16.10 that torsionfree abelian groups satisfy the hypotheses of Corollary 8.16.13 for any ﬁeld K. This implies in particular that if G is a torsionfree abelian group and S is a nonempty subset of G which is not reduced to the identity element, then the discrete Laplacian ΔS : KG → KG is surjective for any ﬁeld K.
Notes In the literature, the term linear is used by some authors with a diﬀerent meaning, namely to designate a cellular automaton τ : AG → AG for which the alphabet A is a ﬁnite abelian group and τ is a group endomorphism of AG . Such cellular automata are also called additive cellular automata. Linear cellular automata with vector spaces as alphabets were considered by the authors in a series of papers starting with [CeC1]. The fact that the algebra of linear cellular automata over a group G whose alphabet is a ddimensional vector space over a ﬁeld K is isomorphic to the algebra of d × d matrices with coeﬃcients in K[G] (Corollary 8.7.8) was proved in Sect. 6 of [CeC1] for d = 1 and in Sect. 4 of [CeC2] for all d ≥ 1. The representations of linear cellular automata both as elements in EndK (V )[G] and as elements in EndK[G] (V [G]) were given in Sect. 4 of [CeC5]. Recall that a Laurent polynomial over a ﬁeld K is a polynomial in the variable t and its inverse t−1 with coeﬃcients in K. The Kalgebra of Laurent polynomials is thus denoted by K[t, t−1 ]. Also, a Laurent polynomial matrix is a matrix whose entries are Laurent polynomials. For d ≥ 1, denote by Matd (K[t, t−1 ]) the Kalgebra of d × d Laurent polynomial matrices over K. When G = Z, there are canonical isomorphisms of Kalgebras K[Z] ∼ = Matd (K[t, t−1 ]). In [LiM, Sect. 1.6] = K(t, t−1 ] and LCA(Z; Kd ) ∼ d cellular automata τ ∈ LCA(Z; K ) are called (d × d) convolutional encoders. The notion of mean dimension for vector subspaces of V G , where G is an amenable group and V is a ﬁnitedimensional vector space, was introduced by Gromov in [Gro5] and [Gro6]. Mean dimension was used by Elek [Ele] to prove that, given any amenable group G and any ﬁeld K, there exists a nontrivial homomorphism from the Grothendieck group of ﬁnitely generated modules over the group algebra K[G] into the additive group of real numbers. As for entropy, it can be shown, as an application of the OrnsteinWeiss convergence
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theorem (see [OrW], [LiW], [Gro6], [Ele], [Kri]) that the lim sup in (8.18) is in fact a true limit and does not depend on the Følner net F . The fact that the image of a linear cellular automaton with ﬁnite dimensional alphabet is closed (Theorem 8.8.1) was proved in Sect. 3 of [CeC1]. The linear version of the Garden of Eden theorem (Theorem 8.9.6) was ﬁrst proved for countable amenable groups in [CeC1] and then extended to all amenable groups in [CeC8] using induction and restriction for linear cellular automata. The linear version of the Garden of Eden theorem was generalized to Rlinear cellular automata with coeﬃcients in semisimple left Rmodules of ﬁnite length over a ring R and over amenable groups in [CeC4]. Invertibility of linear cellular automata with ﬁnite dimensional alphabet V was proved in Sect. 3 of [CeC2] for bijective linear cellular automata τ : X → Y between closed linear subshifts X, Y ⊂ V G over countable groups and in [CeC8] for bijective linear cellular automata τ : V G → V G over any group G. In [CeC11] it is shown that if G is a nonperiodic group, then for every inﬁnitedimensional vector space V over a ﬁeld K there exist a bijective cellular automaton τ : V G → V G which is not invertible (cf. Theorem 8.12.1 and Remark 8.12.3) and a cellular automaton τ : V G → V G whose image τ (V G ) is not closed in V G with respect to the prodiscrete topology (cf. Theorem 8.8.1 and Example 8.8.3). Theorem 8.13.2 and Corollary 8.13.4 were proved in [CeC6] and [CeC9]. Directly ﬁnite rings are sometimes called Dedekind ﬁnite rings or von Neumann ﬁnite rings. In [Coh], P.M. Cohn constructed, for each integer d ≥ 1, a ring R such that Matd (R) is directly ﬁnite but Matd+1 (R) is not directly ﬁnite. I. Kaplansky [Kap2, p. 122], [Kap3, Problem 23] observed that techniques from the theory of operator algebras could be used to prove that, for any group G and any ﬁeld K of characteristic 0, the group algebra K[G] is stably ﬁnite and asked whether this property remains true for ﬁelds of characteristic p > 0. The stable ﬁniteness of K[G] in arbitrary characteristic was established for freebyabelian groups G by P. Ara, K.C. O’Meara and F. Perera in [AOP]. This was extended to all soﬁc groups by Elek and Szab´ o [ES1], using the notion of von Neumann dimension for continuous regular rings. The proof of Elek and Szab´ o’s result via linear cellular automata which is presented in this chapter (cf. Corollary 8.15.8) was given in [CeC2]. The notion of Lsurjunctivity was also introduced in [CeC2] and the equivalence between Lsurjunctivity and stable ﬁniteness was established in Corollary 4.3 therein. In [CeC3], the authors proved that if R is a ring and G is a residually ﬁnite group, then every injective Rlinear cellular automaton over G whose alphabet is an Artinian left Rmodule is surjunctive. In [CeC5], it was shown that if R is a ring, then Rlinear cellular automata with coeﬃcients in left Rmodules of ﬁnite length (thus a stronger condition than being Artinian) over soﬁc groups (thus a weaker condition than being residually ﬁnite) are surjunctive. This last result was used to show in [CeC5] that the group ring R[G] is stably ﬁnite whenever R is a left (or right) Artinian ring and G is a
Notes
337
soﬁc group. This yields an extension of the Elek and Szab´ o stable ﬁniteness result since any division ring is Artinian. Uniqueproduct groups were introduced by W. Rudin and H. Schneider in [RuS] under the name of Ωgroups. The question of the existence of a torsionfree group which is not uniqueproduct was raised by Rudin and Schneider [RuS, p. 592]. This question was answered in the aﬃrmative by E. Rips and Y. Segev [RiS] (see also [Pro]). More information on orderable groups may be found for example in [BoR], [Pas], and [Gla]. A group G is said to be locally indicable if any nontrivial ﬁnitely generated subgroup of G admits an inﬁnite cyclic quotient. Every locally indicable group is orderable (see for example [BoR, Theorem 7.3.1] or [Gla, Lemma 6.9.1]). This implies in particular that locally nilpotent torsionfree groups, free groups, and fundamental groups of surfaces not homeomorphic to the real projective plane are all orderable groups. It was observed by G.M. Bergman [Ber] that the universal covering group SL 2 (R) of SL2 (R) is orderable but not locally indi cable. The group SL 2 (R) is orderable since it has a natural faithful action by orientationpreserving homeomorphisms of the real line but it is not locally indicable since it contains nontrivial ﬁnitely generated perfect groups. By a recent result due to D.W. Morris [Morr, Theorem B], every amenable orderable group is locally indicable. The orderability of the braid groups Bn was established independently by P. Dehornoy [Deh] and W. Thurston. It follows from a result of E.A. Gorin and V.Ja. Lin [GoL] that the group Bn is not locally indicable for n ≥ 5. A group is called biorderable if it admits a total ordering which is both left and right invariant. All free groups and all locally nilpotent torsionfree groups are biorderable. For n ≥ 3, the braid group Bn is not biorderable. However, the pure braid groups Pn (i.e., the kernel of the natural epimorphism of Bn onto Symn ) is biorderable for all n. A theorem due to A.I. Mal’cev [Mal2] and B.H. Neumann [Neu1] says that, given a group G equipped with a total ordering which is both left and right invariant and a ﬁeld K, the vector subspace of KG consisting of all maps x : G → K whose support is a wellordered subset of G is a division Kalgebra for the convolution product (see [Pas, Theorem 2.11 in Chap. 13]). When G = Z, this division algebra is the ﬁeld of Laurent series with coeﬃcients in K. The Mal’cevNeumann theorem implies in particular that, for any biorderable group G and any ﬁeld K, the group algebra K[G] embeds in a division Kalgebra and is therefore stably ﬁnite. A famous conjecture attributed to Kaplansky is the zerodivisor conjecture which states that if G is a torsionfree group then the group algebra K[G] has no zerodivisors for any ﬁeld K (see [Kap1], [Kap2, p. 122], [Pas, Chap. 13]). The observation that uniqueproduct groups (and hence orderable groups) satisfy the Kaplansky zerodivisor conjecture (see Corollary 8.16.10) was made by Rudin and Schneider [RuS, Theorem 3.2]. The class of elementary amenable groups is the smallest class of groups containing all ﬁnite
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and all abelian groups that is closed under taking subgroups, quotients, extensions, and directed unions. It is known (see [KLM, Theorem 1.4]) that torsionfree elementary amenable groups satisfy the Kaplansky zerodivisors conjecture. This implies in particular that all torsionfree virtually solvable groups satisfy the Kaplansky zerodivisors conjecture. According to Corollary 8.16.12, the zerodivisor conjecture is equivalent to saying that if G is a torsionfree group and K is a ﬁeld then every nonidenticallyzero linear cellular automaton τ : KG → KG is preinjective. This reformulation of the Kaplansky zerodivisor conjecture in terms of linear cellular automata was given in Sect. 6 of [CeC1].
Exercises 8.1. Let G be a group and let R be a nonzero ring. Show that the ring R[G] is commutative if and only if both G and R are commutative. 8.2. Recall that the center of a ring A is the subring B of A consisting of the elements x ∈ A which satisfy ax = xa for all a ∈ A. Let G be a group and let R be a commutative ring. Let α ∈ R[G]. Show that α is in the center of R[G] if and only if α is constant on each conjugacy class of G. 8.3. One says that a group G has the ICCproperty if every conjugacy class of G except {1G } is inﬁnite. (a) Show that if a group G has the ICCproperty then the center of G is reduced to the identity element. (b) Show that if a group G has the ICCproperty then every normal subgroup of G which is not reduced to the identity element is inﬁnite. (c) Let X be an inﬁnite set. Show that the group Sym0 (X), which consists of all permutations of X with ﬁnite support, has the ICCproperty. (d) Show that every nonabelian free group has the ICCproperty. (e) Let G be a group with the ICCproperty and let R be a commutative ring. Show that the center of R[G] consists of the maps x : G → R which satisfy x(g) = 0R for all g ∈ G \ {1G }. Hint: Use the result of Exercise 8.2. (f) Let G be a group and let R be a nonzero commutative ring. Show that G has the ICCproperty if and only if the center of R[G] is isomorphic to the ring R. 8.4. Let G be a group, K a ﬁeld, and d ≥ 1 an integer. Show that the Kalgebras Matd (K)[G] and Matd (K[G]) are isomorphic. 8.5. Show that Theorem 8.13.2 remains valid if the ﬁeld R is replaced by the ﬁeld C of complex numbers. Hint: Write every f ∈ CG in the form f = f0 +if1 , where f0 , f1 ∈ RG .
Exercises
339
8.6. Let G be a group and let S ⊂ G be a nonempty ﬁnite subset. Denote by ·, · the standard scalar product on R[G] ⊂ 2 (G) and let · denote the associated norm. (a) Show that for all x ∈ R[G] and λ ∈ R, one has x, ΔS (x) + λx =
1 x(g) − x(gs)2 + λx2 . 2 g∈G s∈S
(b) Show that if the subgroup generated by S is inﬁnite then for any non–empty ﬁnite subset F ⊂ G there exists s ∈ S such that F s ⊂ F . (c) Deduce from (a) and (b) that if λ ≥ 0 and the subgroup generated by S is inﬁnite, then the linear cellular automaton ΔS + λ IdRG : RG → RG is preinjective. 8.7. Showthat if (Ri )i∈I is a family of directly ﬁnite rings then the product ring P = i∈I Ri is directly ﬁnite. 8.8. Show that every subring of a directly ﬁnite (resp. stably ﬁnite) ring is directly ﬁnite (resp. stably ﬁnite). 8.9. Let K be a ﬁeld and let V be an inﬁnitedimensional vector space over K. Show that the Kalgebra EndK (V ) is not directly ﬁnite. 8.10. Let R be a ring and let M be a left Rmodule. One says that the module M is Hopﬁan if every surjective endomorphism of M is injective. Show that if M is Hopﬁan, then the ring EndR (M ) is directly ﬁnite. 8.11. Let R be a ring and let M be a left Rmodule. One says that the module M is Noetherian if its submodules satisfy the ascending chain condition, i.e., every increasing sequence N 1 ⊂ N2 ⊂ . . . of submodules of M stabilizes (there is an integer i0 ≥ 1 such that Ni = Ni0 for all i ≥ i0 ). Show that if M is Noetherian, then M is Hopﬁan. Hint: Suppose that f is a surjective endomorphism of M and consider the sequence of submodules Ni = Ker(f i ), i ≥ 1. 8.12. Let R be a ring and let P be a left Rmodule. One says that the module P is projective if for every homomorphism f : P → M and every surjective → M of left Rmodules, there exists a homomorphism homomorphism g : M such that f = g ◦ h. Show that if the module P is projective, then h: P → M P is Hopﬁan if and only if the ring EndR (P ) is directly ﬁnite. 8.13. Let R be a ring and let d ≥ 1 be an integer. Equip Rd with its natural structure of left Rmodule. Show that Rd is Hopﬁan if and only if the ring Matd (R) is directly ﬁnite.
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8.14. One says that a ring R is left Noetherian if R is Noetherian as a left module over itself. (a) Let R be a left Noetherian ring. Show by induction that Rd is Noetherian as a left Rmodule for each integer d ≥ 1. (b) Show that every left Noetherian ring is stably ﬁnite. Hint: Use Exercises 8.11 and 8.13. 8.15. One says that a ring R has the unique rank property if it satisﬁes the following condition: if m and n are positive integers such that Rm and Rn are isomorphic as left Rmodules, then one has m = n. Show that every stably ﬁnite ring has the unique rank property. 8.16. Let K be a ﬁeld and let V be an inﬁnitedimensional vector space over K. Show that the ring R = EndK (V ) does not have the unique rank property. Hint: Observe that the vector spaces V and V ⊕ V are isomorphic and then prove that the set consisting of all Klinear maps f : V ⊕ V → V , with its natural structure of left Rmodule, is isomorphic to both R and R2 . 8.17. Show that every division ring is stably ﬁnite. 8.18. A ring R is said to be unitregular if for any a ∈ R there exists an invertible element u ∈ R such that a = aua. (a) Show that every division ring is unitregular. (b) Show that if K is a ﬁeld then the ring Matd (K) is unitregular for every d ≥ 1. Hint: Prove that if A ∈ Matd (K) has rank r then there exist invertible matrices U, V ∈ GLd (K) such that A = U Dr V , where Dr ∈ Matd (K) is the diagonal matrix deﬁned by δij = 1 if 1 ≤ i = j ≤ r and δij = 0 otherwise, and then observe that A = AXA, where X = (U V )−1 . (c) Show that every unitregular ring is directly ﬁnite. (d) Prove that the ring Z is not unitregular. (e) A ring R is called a Boolean ring if a2 = a for all a ∈ R. Show that every Boolean ring is commutative and unitregular. 8.19. One says that a ring R is a right Ore ring if R has no zerodivisors and if for any pair a, b of nonzero elements in R there exist nonzero elements u, v ∈ R such that au = bv. Left Ore rings are deﬁned similarly. Show that any right (or left) Ore ring is directly ﬁnite. Hint: Prove that ab = 1R implies ba = 1R by a direct argument, or show that R can be embedded as a subring of a division ring by adapting the construction of the ﬁeld of fractions of an integral domain. 8.20. Let K be a ﬁeld. Show that every ﬁnitedimensional Kalgebra is stably ﬁnite. 8.21. Let G be an amenable group and let F be a Følner net for G. Let V be a ﬁnitedimensional vector space. Suppose that X (resp. Y ) is a vector subspace of V G . Show that mdimF (X ∪ Y ) ≤ mdimF (X) + mdimF (Y ).
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8.22. Use Corollary 8.15.7 and the result of Exercise 8.20 to recover the fact that every ﬁnite group is Lsurjunctive. 8.23. Use Proposition 8.15.2 and Corollary 8.15.7 to recover the fact that every abelian group is Lsurjunctive. 8.24. It follows from Theorem 8.14.4 that every residually ﬁnite group is Lsurjunctive. The goal of this exercise is to present an alternative proof of this result. Let G be a residually ﬁnite group and let V be a ﬁnitedimensional vector space over a ﬁeld K. Let τ : V G → V G be an injective linear cellular automaton. Fix a family (Γi )i∈I of subgroups of ﬁnite index of G such that i∈I Γi = {1G } (the existence of such a family follows from the residual ﬁniteness of G). (a) Show that, for each i ∈ I, the set Fix(Γi ) = {x ∈ V G : gx = x for all g ∈ Γi } is a ﬁnitedimensional vector subspace of V G . (b) Show that τ (Fix(Γi )) = Fix(Γi ) for all i ∈ I. (c) Prove that i∈I Fix(Γi ) is dense in V G and conclude. 8.25. Let R be a ring and let x ∈ R. Show that x is an idempotent if and only if 1R − x is an idempotent. 8.26. Show that any subgroup of a uniqueproduct group is a uniqueproduct group. 8.27. Let G be a group. Suppose that G contains a normal subgroup N such that N and G/N are both uniqueproduct groups. Show that G is a uniqueproduct group. Hint: See for example [RuS, Theorem 6.1]. 8.28. Show that every residually orderable group is orderable. 8.29. Show that the limit of a projective system of orderable groups is orderable. 8.30. A group G is called biorderable if it admits a total ordering ≤ which is both left and right invariant, i.e., such that g1 ≤ g2 implies gg1 ≤ gg2 and g1 g ≤ g2 g for all g, g1 , g2 ∈ G. Let G be a biorderable group. Suppose that an element g ∈ G satisﬁes the following property: there exist an integer n ≥ 1 and elements h1 , h2 , · · · , hn ∈ G such that −1 −1 h1 gh−1 1 h2 gh2 · · · hn ghn = 1G .
Show that g = 1G . 8.31. Let G be a biorderable group. Show that if g, h ∈ G are such that g n = hn for some integer n ≥ 1, then g = h. 8.32. The Klein bottle group is the group K given by the presentation K = x, y; xyx−1 y. Thus, K is the quotient group K = F/N , where F is the free
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group based on x and y, and N is the normal closure of xyx−1 y in F . Let ρ : F → K denote the quotient homomorphism. (a) Let H denote the subgroup of K generated by ρ(y). Show that H is normal in K and that H and K/H are both inﬁnite cyclic. (b) Show that K is an orderable group. (c) Show that the group K is not biorderable. 8.33. Let G be an amenable group, F a right Følner net for G, and V a ﬁnitedimensional vector space over some ﬁeld K. Let X be a vector subspace of V G and let X denote the closure of X in V G for the prodiscrete topology. Show that X is a vector subspace of V G and that one has mdimF (X) = mdimF (X).
Appendix A
Nets and the Tychonoﬀ Product Theorem
A.1 Directed Sets Recall that a partially ordered set is a set I equipped with a binary relation ≤ which is both reﬂexive (i ≤ i for all i ∈ I) and transitive (i ≤ j and j ≤ k implies i ≤ k for all i, j, k ∈ I). A directed set is a partially ordered set I which satisﬁes the following condition: for all i, j ∈ I, there exists an element k ∈ I such that i ≤ k and j ≤ k. Examples A.1.1. (a) The set Z equipped with the relation ≤ deﬁned by i ≤ j ⇐⇒ i divides j is a directed set. (b) If E is an arbitrary set, then the set P(E) of all subsets of E is a directed set for inclusion. (c)) If X is a topological space and x is a point of X, then the set of neighborhoods of x, equipped with the relation ≤ deﬁned by V ≤ W ⇐⇒ W ⊂ V, is a directed set. Indeed, if V and W are neighborhoods of x, then V ∩ W is a neighborhood of x satisfying V ≤ V ∩ W and W ≤ V ∩ W .
A.2 Nets in Topological Spaces Let X be a set. A net of points of X (or net in X) is a family (xi )i∈I of points of X indexed by some directed set I.
T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 9, © SpringerVerlag Berlin Heidelberg 2010
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Let (xi )i∈I and (yj )j∈J be nets in a set X indexed by directed sets I and J respectively. One says that the net (yj ) is a subnet of the net (xi ) if there is a map ϕ : J → I which satisﬁes the following conditions: (SN1) yj = xϕ(j) for all j ∈ J; (SN2) for each i ∈ I, there exists j ∈ J such that if k ∈ J and j ≤ k then i ≤ ϕ(k). Let X be a topological space. Let (xi )i∈I be a net in X and let a ∈. One says that the net (xi ) converges to a, or that a is a limit point of the net (xi ), if the following condition is satisﬁed: for each neighborhood V of a in X, there exists an element i0 ∈ I such that xi ∈ V for all i ≥ i0 . One says that the net (xi ) is convergent if there exists a point a in X such that (xi ) converges to a. It is clear that if the net (xi ) converges to a, then every subnet of (xi ) also converges to a. Proposition A.2.1. Let X be a topological space, Y ⊂ X and a ∈ X. Let Y denote the closure of Y in X. Then the following conditions are equivalent: (a) a ∈ Y ; (b) there exists a net (yi )i∈I of points of Y which converges to a in X. Proof. Suppose that (yi )i∈I is a net of points of Y which converges to a. Then, for each neighborhood V of a, there is an element i0 ∈ I such that yi ∈ V for all i ≥ i0 . Thus, every neighborhood of a meets Y . This shows that a ∈ Y . Conversely, suppose that a ∈ Y . Let I denote the directed set consisting of all neighborhoods of a in X partially ordered by reverse inclusion, that is, V ≤ W ⇐⇒ W ⊂ V. Since a is in the closure of Y , we can ﬁnd for each neighborhood V ∈ I a point yV in Y such that yV ∈ V . It is clear that the net (yV )V ∈I converges to a.
Proposition A.2.2. A topological space X is Hausdorﬀ if and only if every convergent net in X admits a unique limit. Proof. Suppose that X is Hausdorﬀ. Consider a net (xi )i∈I in X which converges to some point a ∈ X. Let b be a point in X with a = b. Since X is Hausdorﬀ, there exist a neighborhood V of a and a neighborhood W of b such that V ∩ W . For i large enough, the point xi is in V and therefore not in W . Therefore the net (xi )i∈I does not converge to b. This shows that every convergent net in X has a unique limit. Suppose now that X is not Hausdorﬀ. Then there exist distinct points a and b in X such that each neighborhood of a meets each neighborhood of b. Consider the set I consisting of all pairs (V, W ), where V is a neighborhood of
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a and W is a neighborhood of b. partially ordered by declaring that (V , W ) ≤ (V, W ) if and only if V ⊂ V and W ⊂ W . Clearly I is a directed set. If we choose, for each i = (V, W ) ∈ I a point xi ∈ V ∩ W , then the net (xi )i∈I admits both points a and b as limits. This proves the converse implication.
Let (xi )i∈I be a net in a topological space X. One says that a point a ∈ X is a cluster point of the net (xi ) if it satisﬁes the following condition: for each neighborhood V of a in X and each i ∈ I, there exists an element j ∈ I such that i ≤ j and xj ∈ V . Proposition A.2.3. Let X be a topological space, (xi )i∈I a net in X, and a ∈ X. Then the following conditions are equivalent: (a) the point a is a cluster point of the net (xi ); (b) the net (xi ) admits a subnet converging to a. Proof. Suppose that (yj )j∈J is a subnet of the net (xi )i∈I converging to a. Let ϕ : J → I be a map satisfying conditions (SN1) and (SN2) above. Consider a neighborhood V of a and an element i0 ∈ I. By (SN2), we may ﬁnd j0 ∈ J such that i0 ≤ ϕ(k) for all k ∈ J satisfying j0 ≤ k. Since the net (yj ) converges to a, there exists k0 ∈ J such that j0 ≤ k0 and yk0 ∈ V . Then we have i0 ≤ ϕ(k0 ) and xϕ(k0 ) = yk0 ∈ V . This shows that a is a cluster point for the net (xi ). Thus (b) implies (a). Conversely, suppose that a is a cluster point for the net (xi ). Denote by Na the set of all neighborhoods of a, partially ordered by reverse inclusion. Let J be the subset of the Cartesian product I × Na consisting of all pairs (i, V ) ∈ I × Na such that xi ∈ V . The fact that a is a cluster point of the net (xi ) implies that J is a directed set for the partial ordering ≤ deﬁned by def
(i, V ) ≤ (i , V ) ⇐⇒ (i ≤ i and V ≤ V ). Consider the non decreasing map ϕ : J → I given by ϕ((i, V )) = i. If we set yj = xϕ(j) for all j ∈ J, it is clear that ϕ satisﬁes conditions (SN1) and (SN2) above and that the net (yj )j∈J converges to a. This shows that (a) implies (b).
Note that if f : X → Y is a continuous map between topological spaces and a ∈ X is a limit (resp. cluster) point of the net (xi ), then f (a) is a limit (resp. cluster) point of the net (f (xi ). This immediately follows from the deﬁnition and the fact that f −1 (W ) is a neighborhood of a for each neighborhood W of f (a).
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A.3 Initial Topology Let X be a set and let (Yλ )λ∈Λ be a family of topological spaces indexed by an arbitrary set Λ. Suppose that we are given, for each λ ∈ Λ, a map fλ : X → Yλ . Then one constructs a topology on X in the following way. Let F denote the set of all subsets of X of the form fλ−1 (Uλ ), where λ ∈ Λ and Uλ is an open subset of Yλ . Let B be the set of all subsets of X which may be written as a ﬁnite intersection of elements of F. Finally, let T denote the set of all subsets of X which may be written as a (ﬁnite or inﬁnite) union of elements of B. It is straightforward to verify that the set T is the set of open sets of a topology on X which admits B as a base, and that this topology is the smallest topology on X for which all maps fλ : X → Yλ are continuous. The topology on X whose open sets are the elements of T is called the initial topology on X associated with the topological spaces (Yλ )λ∈Λ and the maps (fλ )λ∈Λ . If Z is a topological space and g : Z → X is a map, then g is continuous with respect to the initial topology on X if and only if all composite maps fλ ◦g : Z → Yλ , λ ∈ Λ, are continuous. If (xi )i∈I is a net in X and a is a point in X, then the net (xi )i∈I converges to a if and only if the net (fλ (xi ))i∈I converges to fλ (a) for every λ ∈ Λ. Similarly, a is a cluster point of the net (xi )i∈I if and only if fλ (a) is a cluster point of the net (fλ (xi ))i∈I for every λ ∈ Λ.
A.4 Product Topology Let (Xλ )λ∈Λ be a family of topological spaces indexed by a set Λ. The initial topology on the cartesian product X = λ∈Λ Xλ associated with the projection maps πλ : X → Xλ is called the product topology on X. Abase for the product topology on X consists of all subsets of the form V = λ∈Λ Uλ , where Uλ is an open subset of Xλ for each λ ∈ Λ and Uλ = Xλ for all but ﬁnitely many λ ∈ Λ. In the case when each Xλ is endowed with the discrete topology, the product topology on X = λ∈Λ Xλ is called the prodiscrete topology. Proposition A.4.1. Let (Xλ )λ∈Λ be a family of Hausdorﬀ topological spaces. Then X = λ∈Λ Xλ is Hausdorﬀ for the product topology. Proof. Let x = (xλ ) and y = (yλ ) be distinct points of X. Then there exists λ0 ∈ Λ such that xλ0 = yλ0 . Since Xλ0 is Hausdorﬀ, we may ﬁnd disjoint open subsets U and V of Xλ0 containing xλ0 and yλ0 respectively. The pullbacks of U and V by the projection map πλ0 : X → Xλ0 are disjoint open subsets of X containing x and y respectively. Therefore X is Hausdorﬀ.
Recall that a topological space X is called totally disconnected if any nonempty connected subset of X is reduced to a single point.
A.5 The Tychonoﬀ Product Theorem
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Proposition A.4.2. Let (X λ )λ∈Λ be a family of totally disconnected topological spaces. Then X = λ∈Λ Xλ is totally disconnected for the product topology. Proof. Let C be a nonempty connected subset of X. Then, for each λ ∈ Λ, the image of C by the projection map πλ : X → Xλ is a nonempty connected subset of Xλ . Since Xλ is totally disconnected, the set πλ (C) is reduced to a single point for every λ ∈ Λ. This implies that C is reduced to a single point. Therefore, X is totally disconnected.
Suppose Proposition A.4.3. Let (Xλ )λ∈Λ be a family of topological spaces. of X for each λ ∈ Λ. Then F = that Fλ is a closed subset λ λ∈Λ Fλ is a closed subset of X = λ∈Λ Xλ for the product topology. Proof. We have F =
πλ−1 (Fλ ).
λ∈Λ
Thus F is closed in X as it is the intersection of a family of closed subsets of X.
A.5 The Tychonoﬀ Product Theorem Recall that a topological space X is called compact if every open cover of X admits a ﬁnite subcover. This means that if (Uα )α∈A is a family of open subsets of X with X = α∈A Uα , then there exists a ﬁnite subset B ⊂ A such that X = α∈B Uα . By taking complements, one sees that the compactness of X is equivalent to the fact that every family (F α )α∈A of closed subsets of X with the ﬁnite intersection property,, that is, α∈B Fα = ∅ for every ﬁnite subset B ⊂ A, has a nonempty intersection. It is well known that a metric space X is compact if and only if every sequence in X admits a convergent subsequence. There is an analogous characterization of compactness for general topological spaces using nets: Theorem A.5.1. Let X be a topological space. Then the following conditions are equivalent: (a) X is compact; (b) every net in X admits a cluster point; (c) every net in X admits a convergent subnet. Proof. The equivalence between conditions (b) and (c) follows from Proposition A.2.3. Thus, it suﬃces to prove that conditions (a) and (b) are equivalent. Suppose ﬁrst that X is compact. Let (xi )i∈I be a net in X. For each i ∈ I, deﬁne the subset Yi ⊂ X by
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Yi = {xj : j ∈ I and i ≤ j}. Let J be a ﬁnite subset of I. Since I is a directed set, we may ﬁnd an element k∈I such that i ≤ k for all i ∈ J. This implies xk ∈ Yi for all i ∈ J and hence i∈J Yi ⊃ i∈J Yi = ∅. It follows that the family of closed sets (Yi )i∈I has By compactness of X, we deduce that the ﬁnite intersection property. Y = ∅. Let a be a point in i i∈I i∈I Yi . Then, any neighborhood of a meets Yi for all i ∈ I. This means that a is a cluster point of the net (xi ). Therefore, (a) implies (b). Conversely, suppose that every net in X admits a cluster point. Let (Fα )α∈A be a family of closed subsets of X with the ﬁnite intersection property. Consider the directed set E consisting of all ﬁnite subsets of A partially ordered by inclusion. Choose, for each E ∈ E, an element xE ∈ α∈E Fα . By our hypothesis, the net (xE )E∈E admits a cluster point a ∈ X. Let α0 ∈ A. If V is a neighborhood of a, then there exists a ﬁnite set E0 ⊂ A such that {α0 } ⊂ E0 and xE0 ∈ V . Since xE0 ∈ α∈E0 Fα ⊂ Fα0 , it follows that that any neighborhood of a meets Fα0 . Since the set Fα0 is closed, we deduce an arbitrary element in A, we conclude that a ∈ a ∈ Fα0 . As α0 was α∈A Fα . This shows that i∈I Fi = ∅. Consequently, the space X is compact. This proves that (b) implies (a).
Theorem A.5.2 (Tychonoﬀ theorem). Let (Xλ )λ∈Λ be a family of com pact topological spaces. Then X = λ∈Λ Xλ is compact for the product topology. Proof. By Theorem A.5.1, it suﬃces to prove that every net in X admits a cluster point. Let (xi ) be net in X. We shall prove that (xi ) admits a cluster point by applying Zorn’s Lemma.Let us ﬁrst introduce some notation. Given a subset A ⊂ Λ, we set X(A) = λ∈A Xλ and equip X(A) with the product B topology. If A and B are subsets of Λ such that A ⊂ B, we denote by πA the projection map X(B) → X(A). Consider the set E consisting of all pairs (A, a), where A is a subset of Λ and a ∈ X(A) is a cluster point of the net Λ (xi ))i∈I . We partially order E by declaring that two elements (A, a) and (πA B (B, b) satisfy (A, a) ≤ (B, b) if and only if A ⊂ B and πA (b) = a. The set E is not empty since it contains the pair (A, a), where A = ∅ and a is the unique element of X(∅). On the other hand, E isinductive. Indeed, suppose that F is a totally ordered subset of E. Let B = (A,a)∈F A and consider the unique B element b ∈ X(B) such that πA (b) = a for all (A, a) ∈ F. Clearly (B, b) ∈ E and (B, b) is an upper bound for F. By applying Zorn’s Lemma, we deduce that E admits a maximal element (M, m). To prove that the net (xi ) admits a cluster point, it suﬃces to show that M = Λ. Suppose not and choose an element λ0 ∈ Λ\M . Since the space Xλ0 is compact, the net (πλ0 (xi )) admits a cluster point a0 ∈ Xλ0 . Let us set M = M ∪ {λ0 } and consider the element M M (m ) = m and π{λ (m ) = a0 . Clearly m is a m ∈ X(M ) deﬁned by πM 0} Λ cluster point of the net (πM (xi )) and (M, m) ≤ (M , m ). This contradicts the maximality of (M, m) and completes the proof.
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It follows from the deﬁnition of compactness that if a topological space X has only ﬁnitely many open subsets, then X is compact. In particular, every ﬁnite topological space is compact. Therefore, an immediate consequence of Tychonoﬀ theorem is the following: Corollary A.5.3 (Tychonoﬀtheorem). Let (Xλ )λ∈Λ be a family of ﬁnite topological spaces. Then X = λ∈Λ Xλ is compact for the product topology.
Notes The original Tychonoﬀ theorem was only stated for product of compact intervals. The proof of the general Tychonoﬀ theorem we have presented here is based on the one given by P. Chernoﬀ in [Che]. Three other proofs may be found in Kelley’s book [Kel]: a proof using Alexander’s subbase theorem, Bourbaki’s proof [Bou] using ultraﬁlters, and a proof based on universal nets. There is also another proof using nonstandard analysis in the book of A. Robinson [RobA].
Appendix B
Uniform Structures
B.1 Uniform Spaces Let X be a set. We shall use the following notation. We denote by ΔX the diagonal in X × X, that is, ΔX = {(x, x) : x ∈ X} ⊂ X × X. Suppose that R is a subset of X × X (in other words, R is a binary relation on X). For y ∈ X, we deﬁne the set R[y] ⊂ X by R[y] = {x ∈ X : (x, y) ∈ R}. −1
The inverse R ⊂ X × X of R is deﬁned by −1
R = {(x, y) : (y, x) ∈ R}. −1
One says that R is symmetric if it satisﬁes R = R. If R and S are subsets of X × X, we deﬁne their composite R ◦ S ⊂ X × X by R ◦ S = {(x, y) : there exists z ∈ X such that (x, z) ∈ R and (z, y) ∈ S}. Deﬁnition B.1.1. Let X be a set. A uniform structure on X is a non–empty set U of subsets of X × X satisfying the following conditions: (UN1) if V ∈ U, then ΔX ⊂ V ; (UN2) if V ∈ U and V ⊂ V ⊂ X × X, then V ∈ U; (UN3) if V ∈ U and W ∈ U, then V ∩ W ∈ U; −1
(UN4) if V ∈ U, then V ∈ U; (UN5) if V ∈ U, then there exists W ∈ U such that W ◦ W ⊂ V . T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 10, © SpringerVerlag Berlin Heidelberg 2010
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A set X equipped with a uniform structure U is called a uniform space and the elements of U are called the entourages of X (see Fig. B.1).
Fig. B.1 An entourage in a uniform space X
Examples B.1.2. (a) Let X be a set. Then U = {X ×X} is a uniform structure on X. This uniform structure is called the trivial uniform structure on X. It is the smallest uniform structure on X. (b) The discrete uniform structure on a set X is the uniform structure whose entourages consist of all subsets of X × X containing ΔX . This is the largest uniform structure on X. It follows from (UN2) that the discrete uniform structure on X is the only uniform structure on X admitting the diagonal ΔX ⊂ X × X as an entourage. (c) Suppose that d is a metric on X. For each ε > 0 let Vε ⊂ X × X denote the set of pairs (x, y) such that d(x, y) < ε. Let U be the set of all subsets W ⊂ X × X such that one can ﬁnd ε > 0 for which Vε ⊂ W . Then U is a uniform structure on X which is called the uniform structure associated with the metric d. A uniform structure U on a set X is said to be metrizable if U is the uniform structure on X associated with some metric on X. Example B.1.3. If d is the discrete metric on a set X, that is, the metric given by d(x, y) = 0 if x = y and d(x, y) = 1 otherwise, then the uniform structure deﬁned by d is the discrete uniform structure on X. Thus the discrete uniform structure on X is metrizable. Let X be a uniform space. One easily veriﬁes that is possible to deﬁne a topology on X by taking as open sets the subsets Ω ⊂ X which satisfy the following property: for each x ∈ Ω, there exists an entourage V ⊂ X × X such that V [x] = Ω. One says that this topology is the topology associated with the uniform structure on X. A subset N ⊂ X is a neighborhood of a point x ∈ X for this topology if and only if there exists an entourage V such
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that N = V [x]. This topology is Hausdorﬀ if and only if the intersection of the entourages of X coincides with the diagonal ΔX ⊂ X × X. Examples B.1.4. (a) The topology associated with the discrete uniform structure on a set X is the discrete topology on X (every subset is open). (b) If U is the uniform structure associated with a metric d on a set X, then the topology deﬁned by U coincides with the topology deﬁned by d. Let U be a uniform structure on a set X. If Y is a subset of X, then UY = {V ∩ (Y × Y ) : V ∈ U} is a uniform structure on Y , which is said to be induced by U. The topology on Y associated with UY is the topology induced by the topology on X associated with U. A subset B ⊂ U is called a base of U if for each W ∈ U there exists V ∈ B such that V ⊂ W . Example B.1.5. If d is a metric on X, then B = {Vε : ε > 0}, where Vε = {(x, y) ∈ X × X : d(x, y) < ε}, is a base for the uniform structure on X associated with d. The proof of the following statement is straightforward. Proposition B.1.6. Let X be a set and let B be a nonempty set of subsets of X × X. Then B is a base for some (necessarily unique) uniform structure on X if and only if it satisﬁes the following properties: (BU1) if V ∈ B, then ΔX ⊂ V ; (BU2) if V ∈ B and W ∈ B, then there exists U ∈ B such that U ⊂ V ∩ W ; −1
(BU3) if V ∈ B, then there exists W ∈ B such that W ⊂ V ; (BU4) if V ∈ B, then there exists W ∈ B such that W ◦ W ⊂ V .
B.2 Uniformly Continuous Maps Let X and Y be uniform spaces. A map f : X → Y is called uniformly continuous if it satisﬁes the following condition: for each entourage W of Y , there exists an entourage V of X such that (f ×f )(V ) ⊂ W . Here f ×f denotes the map from X × X into Y × Y deﬁned by (f × f )(x1 , x2 ) = (f (x1 ), f (x2 )) for all (x1 , x2 ) ∈ X × X. If B (resp. B ) is a base of the uniform structure on X (resp. Y ), then a map f : X → Y is uniformly continuous if and only if it satisﬁes the following condition: for each W ∈ B , there exists V ∈ B such that (f × f )(V ) ⊂ W . Note that this condition is equivalent to the fact that (f × f )−1 (W ) is an entourage of Y for each entourage W of X.
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Example B.2.1. Let (X, dX ) and (Y, dY ) be metric spaces. Then a map f : X → Y is uniformly continuous if and only if it satisﬁes the following condition: for each ε > 0, there exists δ > 0 such that dX (x1 , x2 ) < δ implies dY (f (x1 ), f (x2 )) < ε. Proposition B.2.2. Let X and Y be uniform spaces. Then every uniformly continuous map f : X → Y is continuous (with respect to the topologies on X and Y associated with the uniform structures). Proof. Suppose that f : X → Y is uniformly continuous. Let x ∈ X and let N ⊂ Y be a neighborhood of f (x). Then there exists an entourage W of Y such that W [f (x)] = N . Since f is uniformly continuous, the set V = (f × f )−1 (W ) is an entourage of X. The set V [x] is a neighborhood of x and satisﬁes f (V [x]) ⊂ W [f (x)] = N . This shows that f is continuous. A continuous map between uniform spaces may fail to be uniformly continuous. For example, the map x → x2 is not uniformly continuous on R (equipped with the uniform structure associated with its usual metric). However, this is true when the source space is compact: Theorem B.2.3. Let X and Y be uniform spaces and suppose that X is compact. Then every continuous map f : X → Y is uniformly continuous. Let us ﬁrst establish the following: Lemma B.2.4 (Lebesgue lemma). Let (Ωi )i∈I be an open cover of a compact uniform space X. Then there exists an entourage Λ of X satisfying the following property: for each x ∈ X, there exists an index i ∈ I such that Λ[x] ⊂ Ωi . Proof. Let us choose, for each x ∈ X, an index i(x) ∈ I such that x ∈ Ωi(x) . Since Ωi(x) is a neighborhood of x, there is an entourage Vx such that Vx [x] = Ωi(x) . By (UN5), we may ﬁnd an entourage Wx such that Wx ◦ Wx ⊂ Vx . The set Wx [x] is a neighborhood of x for each x ∈ X. By compactness of X, there exists a ﬁnite subset A ⊂ X such that X = a∈A Wa [a]. Let us show that the entourage Wa Λ= a∈A
has the required property. Let x ∈ X. Choose a point a ∈ A such that x ∈ Wa [a]. Suppose that y ∈ Λ[x]. Since (x, a) ∈ Wa and (y, x) ∈ Λ ⊂ Wa , we have (y, a) ∈ Wa ◦ Wa ⊂ Va . Thus y ∈ Va [a]. Since Va [a] ⊂ Ωi(a) , this shows that Λ[x] ⊂ Ωi(a) . Proof of Theorem B.2.3. Let f : X → Y be a continuous map and let W be an entourage of Y . By (UN3), (UN4), and (UN5), we may ﬁnd a symmetric entourage S of Y such that S ◦ S ⊂ W . Since f is continuous, there exists, for each x ∈ X, an open neighborhood Ωx of X such that f (Ωx ) ⊂ S[f (x)].
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355
By Lemma B.2.4, we may ﬁnd an entourage Λ of X such that, for each y ∈ X, there exists x ∈ X such that Λ[y] ⊂ Ωx . Suppose that (x1 , x2 ) ∈ Λ. Choose a ∈ X such that Λ[x2 ] ⊂ Ωa . Since x1 and x2 are in Λ[x2 ], we deduce that the points f (x1 ) and f (x2 ) are in S[f (a)]. It follows that (f (x1 ), f (a)) and (f (a), f (x2 )) are in S, and hence (f (x1 ), f (x2 )) ∈ S ◦ S ⊂ W . Thus (f × f )(Λ) ⊂ W . This shows that f is uniformly continuous. Let X and Y be uniform spaces. One says that a map f : X → Y is a uniform isomorphism if f is bijective and both f and f −1 are uniformly continuous. One says that a map f : X → Y is a uniform embedding if f is injective and induces a uniform isomorphism between X and f (X) ⊂ Y . Proposition B.2.5. Let X and Y be uniform spaces with X compact and Y Hausdorﬀ. Suppose that f : X → Y is a continuous injective map. Then f is a uniform embedding. Proof. As X is compact and Y is Hausdorﬀ, f induces a homeomorphism from X onto f (X). This homeomorphism is a uniform isomorphism by Theorem B.2.3.
B.3 Product of Uniform Spaces Let X be a set. Suppose that we are given a family (Xλ )λ∈Λ of uniform spaces and a family (fλ )λ∈Λ of maps fλ : X → Xλ . Then the initial uniform structure associated with these data is the smallest uniform structure on X such that all maps fλ : X → Xλ , λ ∈ Λ, are uniformly continuous. In the particular case when X = λ∈Λ Xλ and fλ : X → Xλ is the projection map, the associated initial uniform structure on X is called the product uniform structure. A base of entourages for the product uniform structure on X is obtained by taking all subsets of X × X which are of the form Vλ ⊂ Xλ × Xλ λ∈Λ
λ∈Λ
=
Xλ
λ∈Λ
×
Xλ
λ∈Λ
= X × X, where Vλ ⊂ Xλ × Xλ is an entourage of Xλ and Vλ = Xλ × Xλ for all but ﬁnitely many λ ∈ Λ. When each Xλ is endowed with the discrete uniform structure, the product uniform structure on X = λ∈Λ Xλ is called the prodiscrete uniform structure.
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B.4 The HausdorﬀBourbaki Uniform Structure on Subsets Let X be a uniform space with uniform structure U. In this section, we construct a uniform structure on the set P(X) of all subsets of X. We shall use the following notation. Suppose that R is a subset of X × X. Given a subset Y ⊂ X, we set R[y] = {x ∈ X : (x, y) ∈ R for some y ∈ Y }, (B.1) R[Y ] = y∈Y
⊂ P(X) × P(X) by and we deﬁne the subset R = {(Y, Z) ∈ P(X) × P(X) : Y ⊂ R[Z] and Z ⊂ R[Y ]} . R
(B.2)
Proposition B.4.1. The set {V : V ∈ U} is a base for a uniform structure on P(X). Proof. Let us check that the conditions of Proposition B.1.6 are satisﬁed. Property (BU1) follows from the fact that we have Y ⊂ V [Y ] for all Y ∈ P(X) and V ∈ U since ΔX ⊂ V . Then observe that (V ∩ W )[Y ] = {x ∈ X : (x, y) ∈ V ∩ W for some y ∈ Y } ⊂ {x ∈ X : (x, yV ) ∈ V and (x, yW ) ∈ W for some yV , yW ∈ Y } = V [Y ] ∩ W [Y ],
for all for all V, W ∈ U and Y ⊂ X. Therefore we have V ∩ W ⊂ V ∩ W V, W ∈ U, so that (BU2) is satisﬁed. Property (BU3) follows from the fact that the set V is a symmetric subset of P(X)×P(X) for each V ∈ U. Finally, let us verify (BU4). Let V ∈ U and take W ∈ U such that W ◦ W ⊂ V . We
◦W
⊂ V . To see this, let (Y, Z) ∈ W
◦W
. This means that claim that W
. In particular there exists T ∈ P(X) such that (Y, T ) ∈ W and (T, Z) ∈ W we have Y ⊂ W [T ] and T ⊂ W [Z]. Thus, given y ∈ Y , there exist t ∈ T and z ∈ Z such that (y, t) ∈ W and (t, z) ∈ W . We have (y, z) ∈ W ◦W ⊂ V . This shows that Y ⊂ V [Z]. Similarly, we get Z ⊂ V [Y ] by using Z ⊂ W [T ] and T ⊂ W [Y ]. We deduce that (Y, Z) ∈ V . This proves the claim. Consequently, (BU4) is satisﬁed. The uniform structure on P(X) admitting {V : V ∈ U} as a base is called the HausdorﬀBourbaki uniform structure on P(X) and the topology associated with this uniform structure is called the HausdorﬀBourbaki topology on P(X). Remarks B.4.2. (a) The empty set is an isolated point in P(X). (b) One easily checks that the map i : X → P(X) deﬁned by i(x) = {x} is a uniform embedding.
B.4 The HausdorﬀBourbaki Uniform Structure on Subsets
357
Proposition B.4.3. Let X be a uniform space. Let Y and Z be closed subsets of X. Suppose that there is a net (Ti )i∈I of subsets of X which converges to both Y and Z with respect to the HausdorﬀBourbaki topology on P(X). Then one has Y = Z. Proof. Let y ∈ Y and let Ω be a neighborhood of y in X. Then there is a symmetric entourage V of X such that V [y] ⊂ Ω. Choose an entourage W of X such that W ◦ W ⊂ V . Since the net (Ti )i∈I converges to both Y and Z, we can ﬁnd an element i0 ∈ I such that ⊂ W [Ti0 ] and Ti0 ⊂ W [Z]. Thus, there exist t ∈ Ti0 and z ∈ Z such that (y, t) ∈ W and (t, z) ∈ W . This implies (y, z) ∈ W ◦ W ⊂ V . As V is symmetric, it follows that (z, y) ∈ V and hence z ∈ V [y] ⊂ Ω. This shows that y is in the closure of Z. Since Z is closed in X, we deduce that Y ⊂ Z. By symmetry, we also have Z ⊂ Y . Consequently, Y = Z. By using Proposition A.2.2, we immediately deduce from Proposition B.4.3 the following: Corollary B.4.4. Let X be a uniform space. Then the topology induced by the HausdorﬀBourbaki topology on the set of closed subsets of X is Hausdorﬀ. Remark B.4.5. Suppose that (X, d) is a metric space and let Cb (X) denote the set consisting of all closed bounded subsets of X. For x ∈ X and r > 0, denote by B(x, r) the open ball of radius r centered at x. Then it is not diﬃcult to verify that the map δ : Cb (X) × Cb (X) → R deﬁned by δ(Y, Z) = inf{r > 0 : Z ⊂ B(y, r) and Y ⊂ B(z, r)} y∈Y
z∈Z
is a metric on Cb (X) and that the uniform structure associated with δ is the uniform structure induced by the HausdorﬀBourbaki structure on the set of subsets of X. The metric δ is called the Hausdorﬀ metric on Cb (X). Proposition B.4.6. Let X and Y be uniform spaces and let f : X → Y be a uniformly continuous map. Then the map f∗ : P(X) → P(Y ) which sends each subset A ⊂ X to its image f (A) ⊂ Y is uniformly continuous with respect to the HausdorﬀBourbaki uniform structures on P(X) and P(Y ). Proof. Let W be an entourage of Y and let
= {(B1 , B2 ) ∈ P(Y ) × P(Y ) : B2 ⊂ W [B1 ] and B1 ⊂ W [B2 ]} W be the associated entourage of P(Y ). Since f is uniformly continuous, there is an entourage V of X such that (f × f )(V ) ⊂ W.
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Suppose that (A1 , A2 ) ∈ V , that is, A2 ⊂ V [A1 ] and A1 ⊂ V [A2 ]. If a1 ∈ A1 , then there exists a2 ∈ A2 such that (a1 , a2 ) ∈ V and hence (f (a1 ), f (a2 )) ∈ W . Therefore, we have f∗ (A1 ) ⊂ W [f∗ (A2 )]. Similarly, we get f∗ (A2 ) ⊂
. This shows that (f∗ × W [f∗ (A1 )]. It follows that (f∗ (A1 ), f∗ (A2 )) ∈ W
. Consequently, f∗ is uniformly continuous. f∗ )(V ) ⊂ W
Notes Uniform structures were introduced by Andr´e Weil [Weil]. The reader is referred to [Bou, Ch. 2], [Kel, Ch. 6], and [Jam] for a detailed exposition of the general theory of uniform spaces. The HausdorﬀBourbaki uniform structure on the set of subsets of a uniform space was introduced in exercises by Bourbaki (see [Bou, ch. II exerc. 5 p. 34 and exerc. 6 p. 36]).
Appendix C
Symmetric Groups
C.1 The Symmetric Group Let X be a set. A permutation of X is a bijective map σ : X → X. Let Sym(X) denote the set of all permutations of X. We equip Sym(X) with a group structure by deﬁning the product σ1 σ2 of two elements σ1 , σ2 ∈ Sym(X) as the composite map σ1 ◦ σ2 . The associative property follows from the associativity of the composition of maps. The identity map IdX : X → X is the identity element and the inverse of σ ∈ Sym(X) is the inverse map σ −1 . The group Sym(X) is called the symmetric group on X. Remark C.1.1. Suppose that f : X → Y is a bijection from a set X onto a set Y . Then the map f∗ : Sym(X) → Sym(Y ) deﬁned by f∗ (σ) = f ◦ σ ◦ f −1 is a group isomorphism. As a consequence, symmetric groups on equipotent sets are isomorphic. Theorem C.1.2 (Cayley’s theorem). Every group G is isomorphic to a subgroup of Sym(G). Proof. Let G be a group. Given g ∈ G denote by Lg : G → G the left multiplication by g, that is, the map deﬁned by Lg (h) = gh for all h ∈ G. Observe that Lg ∈ Sym(G). Indeed, Lg is bijective since, given h, h ∈ G, we have Lg (h) = h if and only if h = g −1 h . Let us show that the map L : G → Sym(G) g → Lg is a group homomorphism. Given g, g , h ∈ G we have Lgg (h) = gg h = Lg (g h) = Lg (Lg (h)) which shows that Lgg = Lg Lg . Moreover, L is injective. Indeed, if g ∈ ker(L), that is, Lg = IdG , we have g = g · 1G = Lg (1G ) = 1G . This shows that ker(L) = {1G }, and therefore L is injective. It follows that G is isomorphic to L(G) ⊂ Sym(G). T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 11, © SpringerVerlag Berlin Heidelberg 2010
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C.2 Permutations with Finite Support Let X be a set. The support of a permutation σ ∈ Sym(X) is the set S(σ) ⊂ X consisting of all x ∈ X such that σ(x) = x. Proposition C.2.1. Let σ, τ ∈ Sym(X). Then (i) σ(S(σ)) = S(σ); (ii) S(σ) = S(σ −1 ); (iii) S(στ ) ⊂ S(σ) ∪ S(τ ); (iv) if S(σ) ∩ S(τ ) = ∅, then στ = τ σ; (v) S(τ στ −1 ) = τ (S(σ)). Proof. (i) Let x ∈ X. Then, x ∈ S(σ), that is, σ(x) = x, if and only if σ(σ(x)) = σ(x), that is, σ(x) ∈ S(σ). (ii) This follows from the fact that σ(x) = x if and only if x = σ −1 (x). (iii) Suppose that x ∈ X \ (S(σ) ∪ S(τ )). Then σ(x) = x = τ (x) and therefore (στ )(x) = σ(τ (x)) = σ(x) = x. It follows that x ∈ / S(στ ). (iv) Let x ∈ X. Suppose ﬁrst that x ∈ X \ (S(σ) ∪ S(τ )). Then, by (iii), x∈ / S(στ ) and x ∈ / S(τ σ), so that σ(τ (x)) = x = τ (σ(x)). Suppose now that x is in the support of one of the two permutations, say x ∈ S(σ). It then follows from our assumptions that x ∈ / S(τ ) and therefore τ (x) = x. Also, by (i), σ(x) ∈ S(σ) and therefore, again by our assumptions, σ(x) ∈ / S(τ ), so that τ (σ(x)) = σ(x). We thus have τ (σ(x)) = σ(x) = σ(τ (x)). It follows that στ = τ σ. (v) Let x ∈ X. Then, x ∈ S(σ), that is σ(x) = x, if and only if, (τ στ −1 )(τ (x)) = τ (σ(x)) is not equal to τ (x). Thus, x ∈ S(σ) if and only if τ (x) ∈ S(τ στ −1 ). Let Sym0 (X) denote the subset of Sym(X) consisting of all permutations of X with ﬁnite support. Proposition C.2.2. Let X be a set. Then the set Sym0 (X) is a normal subgroup of Sym(X). Proof. The support of the identity map IdX is the empty set and therefore IdX ∈ Sym0 (X). By Proposition C.2.1(ii), if σ ∈ Sym0 (X) then σ −1 ∈ Sym0 (X). On the other hand, by Proposition C.2.1(iii), the set Sym0 (X) is closed under multiplication. Thus Sym0 (X) is a subgroup of Sym(X). Finally, from Proposition C.2.1(v) we deduce that if σ ∈ Sym0 (X) then τ στ −1 ∈ Sym0 (X) for all τ ∈ Sym(X). It follows that Sym0 (X) is a normal subgroup of Sym(X). Let r ≥ 2 be an integer and let x1 , x2 , . . . , xr be distinct elements in X. We denote by γ = (x1 x2 · · · xr )
C.2 Permutations with Finite Support
361
the permutation of X that maps x1 to x2 , x2 to x3 , . . . , xr−1 to xr , xr to x1 , and maps each element of X \ {x1 , x2 , . . . , xr } to itself. Thus, the support of γ is the set {x1 , x2 , . . . , xr }. One says that γ is a cycle of length r, or an rcycle. A 2cycle is called a transposition. Observe that γ = (x γ(x) γ 2 (x) · · · γ r−1 (x)) for all x ∈ {x1 , x2 , . . . , xr } and that the inverse of γ is the rcycle γ −1 = (xr xr−1 · · · x2 x1 ). Proposition C.2.3. Let X be a set and let σ ∈ Sym0 (X). Then there exists an integer n ≥ 0 and cycles γ1 , γ2 , . . . , γn with pairwise disjoint supports such that σ = γ1 γ 2 · · · γ n . (C.1) Moreover, such a factorization is unique up to a permutation of the factors. Proof. If σ = IdX then n = 0 and there is nothing to prove. Suppose now that σ = IdX . Let S = S(σ) ⊂ X be the support of σ. We introduce an equivalence relation on S by setting ∼ y ifand only if x there exists k ∈ Z such that y = σ k (x). Let S = X1 X2 · · · Xn be the partition of S into the equivalence classes of ∼ and let us set ri = Xi  for 1 ≤ i ≤ n. Note that ri ≥ 2 for all i. For each i = 1, 2, . . . , n, choose a representative xi ∈ Xi . Observe that the elements xi , σ(xi ), σ 2 (xi ), . . . , σ ri −1 (xi ) are all distinct since otherwise the class of xi would have less that ri elements. Consider the cycle γi = (xi σ(xi ) σ 2 (xi ) · · · σ ri −1 (xi )). The support of γi is Xi . Thus, the cycles γ1 , γ2 , . . . , γn have pairwise disjoint supports. Clearly σ = γ1 γ2 · · · γn . Suppose now that σ = δ1 δ2 · · · δs , where δ1 , δ2 , . . . , δs are cycles with pairwise disjoint supports. The supports of the cycles δi are the equivalence classes of ∼. We deduce that s = n. Moreover, up to a permutation of the factors, we may suppose that the support of γi equals the support of δi for all i = 1, 2, . . . , n. We have γi = (xi γi (xi ) γi2 (xi ) · · · γiri −1 (xi )) = (xi σ(xi ) σ 2 (xi ) · · · σ ri −1 (xi )) = (xi δi (xi ) δi2 (xi ) · · · δiri −1 (xi )) = δi and this completes the proof.
Corollary C.2.4. Every permutation in Sym0 (X) can be expressed as a product of transpositions.
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C Symmetric Groups
Proof. First observe that every cycle is a product of transpositions. Indeed, for all distinct x1 , x2 , . . . , xr ∈ X, we have (x1 x2 · · · xr ) = (x1 xr )(x1 xr−1 ) · · · (x1 x3 )(x1 x2 ).
(C.2)
By applying Proposition C.2.3 we deduce that every σ ∈ Sym0 (X) is a product of transpositions.
C.3 Conjugacy Classes in Sym0 (X) Let G be a group and let H ⊂ G be a subgroup. We recall that two elements h and h in H are said to be conjugate in G (resp. in H) if there exists an element g ∈ G (resp. g ∈ H) such that h = ghg −1 . Clearly conjugacy in G (resp. in H) deﬁnes an equivalence relation on H. Proposition C.3.1. Let X be a set. Let γ ∈ Sym0 (X) be a cycle of length r and let σ ∈ Sym(X). Then σγσ −1 is also a cycle of length r. More precisely, if γ = (x1 x2 · · · xr ), then σγσ −1 equals the cycle (σ(x1 ) σ(x2 ) · · · σ(xr )).
(C.3)
Proof. First observe that, by Proposition C.2.1(v) the support of σγσ −1 is the set {σ(x1 ), σ(x2 ), . . . , σ(xn )}. Given 1 ≤ i ≤ r, we have (σγσ −1 )(σ(xi )) = (σγ)(xi ) = σ(xi+1 ), where r + 1 = 1. It follows that σγσ −1 = (σ(x1 ) σ(x2 ) · · · σ(xr )).
Let σ ∈ Sym0 (X). The type of σ is the sequence t(σ) = (tr )r≥2 where tr is the number of cycles of length r in the factorization of σ as a product of cycles with pairwise disjoint supports (cf. Proposition C.2.3). Proposition C.3.2. Let X be a set and let σ and σ in Sym0 (X). Then the following conditions are equivalent: (a) σ and σ are conjugate in Sym0 (X); (b) σ and σ are conjugate in Sym(X); (c) σ and σ have the same type. Proof. The implication (a) ⇒ (b) is obvious. Suppose (b). let σ = γ 1 γ2 · · · γ r
(C.4)
be the factorization of σ as a product of cycles with disjoint supports and let α ∈ Sym(X) be a permutation such that σ = ασα−1 . From (C.4) we deduce that σ = (αγ1 α−1 )(αγ2 α−1 ) · · · (αγr α−1 ). It follows from Proposition C.3.1 that t(σ ) = t(σ). This shows (b) ⇒ (c).
C.4 The Alternating Group
363
Finally, suppose that σ = (x1 x2 · · · xr1 )(y1 y2 · · · yr2 ) · · · (z1 z2 · · · zr ) and
σ = (x1 x2 · · · xr1 )(y1 y2 · · · yr 2 ) · · · (z1 z2 · · · zr )
are two permutations of the same type. Consider a permutation α, with support the union of the supports of σ and σ , which maps xi to xi for all i = 1, 2, . . . , r1 , yj to yj , for all j = 1, 2, . . . , r2 , . . ., and zk to zk for all k = 1, 2, . . . , r . Note that α ∈ Sym0 (X). Then (cf. the proof of Proposition C.3.1) ασα−1 = α . It follows that σ and σ are conjugate in Sym0 (X).
C.4 The Alternating Group Proposition C.4.1. Let X be a set and let σ ∈ Sym0 (X). Suppose that σ can be expressed as a product of n transpositions. Then the parity of n only depends on σ. Proof. Suppose that σ can be expressed both as a product of an even and as a product of an odd number of transpositions. Then, the same holds for σ −1 . It follows that choosing an even writing for σ and an odd one for σ −1 , we can write the identity element IdX = σσ −1 as a product of an odd number of transpositions, say (C.5) IdX = τ1 τ2 · · · τ2m+1 . Let x be an element in X appearing in the support of one of the transpositions τi in (C.5). As transpositions with disjoint support commute (cf. Proposition C.2.1(iv)) and (y z)(x z) = (x y)(y z) for all distinct elements y, z in X \ {x}, we can move all transpositions of the form (x y) to the left in (C.5). In other words, we can write the identity IdX as a product of 2m + 1 transpositions · · · τ2m+1 IdX = τ1 τ2 · · · τr τr+1
(C.6)
where 1 ≤ r ≤ 2m + 1 and x belongs to the support of τi if and only if 1 ≤ i ≤ r. Let τr = (x y) and observe that it cannot appear only once in the product (C.6), otherwise the element x would be mapped onto y, while it has to remain ﬁxed, since that product is the identity. It follows that there exists 1 ≤ j ≤ r − 1 such that τj = τr and τi = τr for all i = j + 1, j + 2, . . . , r − 1. Now, for all j + 1 ≤ i ≤ r − 1, if τi = (x z), we have τi τr = (x z)(x y) = (x y)(y z) = τr (y z). Thus, we can move to the left the transposition τr next to τj and cancel them out, without changing the value of the product. Repeating this operation, we reduce every time by 2 the
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C Symmetric Groups
number of transpositions in (C.6). Eventually, we reach a single transposition. This clearly yields a contradiction. Consider the map ε : Sym0 (X) → {−1, 1}
(C.7)
deﬁned by setting ε(σ) = 1 if σ is a product of an even number of permutations and ε(σ) = −1 otherwise. Note that ε is well deﬁned by virtue of Proposition C.4.1. It is obvious that ε is a group homomorphism. Moreover, ε is surjective if X has at least two elements. The normal subgroup ker(ε) ⊂ Sym0 (X) is called the alternating group on X and it is denoted by Sym+ 0 (X). Proposition C.4.2. Let X be a set. An rcycle is in Sym+ 0 (X) if and only if r is odd. In particular, every 3cycle belongs to Sym+ 0 (X). Proof. We have seen (cf. (C.2)) that and rcycle is a product of r − 1 transpositions. Recall that a nontrivial group G is simple if the only normal subgroups of G are the trivial subgroup {1G } and G itself. Theorem C.4.3. Let X be a set having at least ﬁve distinct elements. Then the group Sym+ 0 (X) is simple. Proof. Every element of Sym+ 0 (X) is a product of permutations of the form (s t)(u v) or (s t)(s u), where s, t, u and v are distinct elements of X. Since (s t)(u v) = (s u t)(s u v) and (s t)(s u) = (s u t), it follows that Sym+ 0 (X) is generated by the set of all 3cycles. Let x and y be two distinct elements in X. Clearly, any 3cycle is of one of the forms (x y s), (x s y), (x s t), (y t u), or (s t u), where s, t, u are distinct elements in X \ {x, y}. We have (x s y) = (x y s)2 , (x s t) = (x y t)(x s y) = (x y t)(x y s)2 , (y s t) = (x t y)(x y s) = (x y t)2 (x y s), (s t u) = (x s y)(x y u)(x t y)(x y s) = (x y s)2 (x y u)(x y t)2 (x y s). This shows that Sym+ 0 (X) is generated by the 3cycles (x y z), where z ∈ X \ {x, y}. Let now N ⊂ Sym+ 0 (X) be a nontrivial normal subgroup. Let us show that N = Sym+ 0 (X). We distinguish a few cases (corresponding to the diﬀerent possible cycle structures of a nontrivial element in N ). Case 1. N contains a 3cycle (x y s). Then, for any z ∈ X \ {x, y, s} we have that the 3cycle
C.4 The Alternating Group
365
(x y z) = (x y)(s z)(x y s)2 (s z)(x y) = [(x y)(s z)]−1 (x y z)2 [(s z)(x y)] belongs to N . From the preceding part of the proof, we deduce that N = Sym+ 0 (x). Case 2. N contains an element σ whose factorization as product of cycles with disjoint supports contains a cycle (x1 x2 · · · xr ) of length r ≥ 4. Write σ = (x1 x2 · · · xr )ρ, where ρ ∈ Sym0 (X) is the product of the remaining cycles. Consider the cycle γ = (x1 x2 x3 ). Then, σ(γσ −1 γ −1 ) ∈ N , as N is a normal subgroup. By Proposition C.3.1, we have σ(γσ −1 γ −1 ) = (σγσ −1 )γ −1 = (x2 x3 x4 )(x3 x2 x1 ) = (x1 x4 x2 ). Thus, N contains a 3cycle and, by Case 1, N = Sym+ 0 (X). Case 3. N contains an element σ whose factorization as a product of cycles with disjoint supports contains at least two cycles (x1 x2 x3 ) and (x4 x5 x6 ) of length 3. Write σ = (x1 x2 x3 )(x4 x5 x6 )ρ, where ρ ∈ Sym0 (X) is the product of the remaining cycles. Consider the cycle γ = (x1 x2 x4 ). Then, as above, σγσ −1 γ −1 ∈ N . But, again by Proposition C.3.1, we have σγσ −1 γ −1 = (x2 x3 x5 )(x4 x2 x1 ) = (x1 x4 x3 x5 x2 ). Thus, N contains a 5cycle and by Case 2, N = Sym+ 0 (X). Case 4. N contains an element σ which factorizes as a product of one single 3cycle (x1 x2 x3 ) and transpositions with disjoint supports. Write σ = (x1 x2 x3 )ρ, where ρ ∈ Sym0 (X) is the product of the transpositions. Note that ρ2 = IdX . Then σ 2 ∈ N and σ 2 = (x1 x2 x3 )ρ(x1 x2 x3 )ρ = (x1 x2 x3 )2 ρ2 = (x1 x3 x2 ). Thus, again as in Case 1, N = Sym+ 0 (X). Case 5. N contains an element σ which is the product of an (even) number of transpositions with disjoint supports. We can write σ = (x1 x2 )(x3 x4 )ρ where ρ ∈ Sym0 (X) satisﬁes ρ2 = IdX . Consider the cycle γ = (x1 x2 x3 ). Then, the element π = σγσ −1 γ −1 belongs to N . By Proposition C.3.1, π = σγσ −1 γ −1 = (x2 x1 x4 )(x1 x3 x2 ) = (x1 x3 )(x2 x4 ). Since X has at least ﬁve distinct elements, there exists an element y ∈ X \ {x1 , x2 , x3 , x4 }. Set δ = (x1 x3 y). Then, π(δπ −1 δ −1 ) ∈ N . But, one more time by Proposition C.3.1, π(δπ −1 δ −1 ) = (πδπ −1 )δ −1 = (x3 x1 y)(y x3 x1 ) = (x1 x3 y). We are again in Case 1, and therefore N = Sym+ 0 (X). + In all cases, N = Sym+ 0 (X), and this shows that Sym0 (X) is simple.
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C Symmetric Groups
Note that if X is a ﬁnite set, then Sym0 (X) = Sym(X). Given an integer n ≥ 1, we denote by Symn the symmetric group of the set {1, 2, . . . , n}. The group Symn is called the symmetric group of degree n. By Remark C.1.1, if X is a ﬁnite set with X = n, we have Symn ∼ = Sym(X). The subgroup Sym+ 0 ({1, 2, . . . , n}) is called the alternating group of degree n and it is denoted by Sym+ n. + Remark C.4.4. The groups Sym+ 1 (= Sym1 ) and Sym2 are trivial groups. + The group Sym3 = {Id{1,2,3} , (1 2 3), (1 3 2)} is cyclic of order 3 and therefore it is a simple group. On the other hand, the subgroup
K = {Id{1,2,3,4} , (1 2)(3 4), (1 3)(2 4), (1 4)(2 3)} ⊂ Sym+ 4 + has index two in Sym+ 4 . It follows that K is a normal subgroup of Sym4 . + Therefore, Sym4 is not a simple group. From Theorem C.4.3 we deduce that the group Sym+ n is simple for all n ≥ 5.
Appendix D
Free Groups
D.1 Concatenation of Words Let A be a set. A word on the alphabet set A is an element of the set An , A∗ = n∈N
where An is the Cartesian product of A with itself n times, that is, the set consisting of all ntuples (a1 , a2 , . . . , an ) with ak ∈ A for 1 ≤ k ≤ n. The unique element of A0 is denoted by and is called the empty word . The concatenation of two words w = (a1 , a2 , . . . , am ) ∈ Am and w = (a1 , a2 , . . . , an ) ∈ An is the word ww ∈ Am+n deﬁned by ww = (a1 , a2 , . . . , am , a1 , a2 , . . . , an ). We have w = w = w and (ww )w = w(w w ) for all w, w , w ∈ A∗ . Thus, A∗ is a monoid for the concatenation product whose identity element is the empty word . Observe that each word w = (a1 , a2 , . . . , an ) ∈ An may be uniquely written as a product of elements of A = A1 , namely w = a1 a2 · · · an .
D.2 Deﬁnition and Construction of Free Groups Deﬁnition D.2.1. A based free group is a triple (F, X, i), where F is a group, X is a set, and i : X → F is a map from X to F satisfying the following universal property: for every group G and any map f : X → G, there exists a unique homomorphism φ : F → G such that f = φ ◦ i.
T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 12, © SpringerVerlag Berlin Heidelberg 2010
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D Free Groups
F i
X
φ
f
G
A group F is called free if there exist a set X and a map i : X → F such that the triple (F, X, i) is a based free group. One then says that (X, i) is a free base for F and that F is a free group based on (X, i). Remarks D.2.2. (a) If (F, X, i) is a based free group and α : Y → X is a bijective map from a set Y onto X, then the triple (F, Y, i ◦ α) is also a based free group. Indeed, if f : Y → G is a map from Y into a group G, then there is a unique homomorphism φ : F → G such that f = φ ◦ i ◦ α, namely the unique homomorphism φ : F → G satisfying f ◦ α−1 = φ ◦ i. (b) If (F, X, i) is a based free group and ψ : F → F is an isomorphism from F onto a group F , then the triple (F , X, ψ ◦ i) is a based free group. Indeed, if f : X → G is a map from X into a group G, then there exists a unique homomorphism φ : F → G such that f = φ ◦ ψ ◦ i, namely the homomorphism given by φ = φ ◦ ψ −1 , where φ : F → G is the unique homomorphism satisfying f = φ ◦ i. Proposition D.2.3. Let (F, X, i) be a based free group. Then the following hold: (i) the map i is injective; (ii) the set i(X) generates the group F ; (iii) the triple (F, X , i ), where X = i(X) and i : X → F is the inclusion map, is also a based free group. Proof. (i) Let x1 and x2 be two distinct elements in X. Consider the map f : X → Z/2Z deﬁned by f (x) = 0 if x = x2 and f (x2 ) = 1. Since (F, X, i) is a based free group, there exists a homomorphism φ : F → Z/2Z such that f = φ ◦ i. As f (x1 ) = f (x2 ), this implies i(x1 ) = i(x2 ). Therefore, the map i is injective. (ii) Denote by H the subgroup of F generated by i(X). Consider the map i∗ : X → H deﬁned by i∗ (x) = i(x) for all x ∈ X. Since (F, X, i) is a based free group, there exists a homomorphism φ : F → H such that i∗ = φ ◦ i. Consider now the inclusion map ρ : H → F . The homomorphisms IdF and ρ ◦ φ satisfy IdF ◦i = ρ ◦ φ ◦ i. By uniqueness, we get IdF = ρ ◦ φ. This implies that ρ is surjective, that is, H = F . Therefore, i(X) generates F . (iii) The fact that (F, X , i ) is a based free group immediately follows from Remark D.2.2(a) since i = i ◦ j −1 where j : X → X is the bijective map deﬁned by j(x) = i(x) for all x ∈ X. From Proposition D.2.3(iii), we deduce that if F is a free group then there exists a subset X ⊂ F such that the triple (F, X, i), where i : X → F is the inclusion map, is a based free group. Such a subset X ⊂ F is then called a free base subset, or simply a base, for F .
D.2 Deﬁnition and Construction of Free Groups
369
Proposition D.2.4. Let (F, X, i) be a based free group and let Y ⊂ X. Let K denote the subgroup of F generated by Y and let j : Y → K be the map deﬁned by j(y) = i(y) for all y ∈ Y . Then (K, Y, j) is a based free group. Proof. Let G be a group and let f : Y → G be a map. Let us show that there exists a unique homomorphism φ : K → G satisfying f = φ ◦ j. Uniqueness follows from the fact that j(Y ) = i(Y ) generates K. Choose a map f : X → G extending f . As (F, X, i) is a based free group, there exists a homomorphism φ : F → G such thatf = φ ◦ i. Then φ = φ K : K → G satisﬁes f = φ ◦ j. This proves that (K, Y, j) is a based free group. In the case when i is an inclusion map, this gives us the following: Corollary D.2.5. Let F be a free group with base X ⊂ F . Let Y ⊂ X and let K denote the subgroup of F generated by Y . Then K is a free group with base Y . Proposition D.2.6. Let (F1 , X1 , i1 ) and (F2 , X2 , i2 ) be two based free groups. Suppose that there is a bijective map u : X1 → X2 . Then there exists a unique group isomorphism ϕ : F1 → F2 satisfying ϕ ◦ i1 = i2 ◦ u. ϕ
F1 −−−−→ ⏐ i1 ⏐
F2 ⏐i ⏐2
X1 −−−−→ X2 u
Proof. Since (F1 , X1 , i1 ) is a based free group, there exists a unique homomorphism ϕ : F1 → F2 such that i2 ◦ u = ϕ ◦ i1 . It suﬃces to show that ϕ is bijective. To see this, we now use the fact that (F2 , X2 , i2 ) is a based free group. This implies that there exists a homomorphism ϕ : F2 → F1 such that i1 ◦u−1 = ϕ ◦i2 . The maps IdF1 and ϕ ◦ϕ are endomorphisms of F1 satisfying IdF1 ◦i1 = i1 and (ϕ ◦ ϕ) ◦ i1 = i1 . Since (F1 , X1 , i1 ) is a based free group, it follows that IdF1 = ϕ ◦ ϕ by uniqueness. Similarly, we get ϕ ◦ ϕ = IdF2 . This shows that ϕ is bijective. Theorem D.2.7. Let X be a set. Then there exist a group F and a map i : X → F such that the triple (F, X, i) is a based free group. Proof. Let X be a disjoint copy of X, that is, a set X such that X ∩ X = ∅ together with a bijective map γ : X → X . Let A = X ∪ X . For each a ∈ A, deﬁne the element a ∈ A by setting γ(a) if a ∈ X, a= −1 γ (a) if a ∈ X . Observe that the map a → a is an involution of A, that is, it satisﬁes a=a for all a ∈ A.
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D Free Groups
Consider now the set A∗ consisting of all words on the alphabet set A (see Sect. D.1). Recall that A∗ is a monoid for the concatenation product whose identity element is the empty word . We say that a word w ∈ A∗ may be obtained from a word w ∈ A∗ by an elementary reduction if there exist an element a ∈ A and words u, v ∈ A∗ such that w = uv and w = ua av. Given words w, w ∈ A∗ , we write w ∼ w if either w may be obtained from w by an elementary reduction or w may be obtained from w by an elementary reduction. Finally, we deﬁne a relation ≡ on A∗ by writing w ≡ w for w, w ∈ A∗ if and only if there exist an integer n ≥ 1 and a sequence of words w1 , w2 , . . . , wn ∈ A∗ with w1 = w and wn = w such that wi ∼ wi+1 for each i = 1, 2, . . . , n − 1. It is immediate to check that ≡ is an equivalence relation on A∗ . Denote by [w] the equivalence class of an element w ∈ A∗ and consider the quotient set F = A∗ / ≡, that is, the set consisting of all equivalence classes [w], w ∈ A∗ . Observe that if u, u , v ∈ A∗ and u ∼ u then uv ∼ u v and vu ∼ vu . It follows from this observation that if the words u, v, u , v ∈ A∗ satisfy u ≡ u and v ≡ v then uv ≡ u v , so that we can deﬁne the product of [u] and [v] in F by setting [u][v] = [uv]. Let us show that this product gives a group structure on F . The associativity immediately follows from the associativity of the concatenation product in A∗ . Indeed, for all u, v, w ∈ A∗ , we have ([u][v])[w] = [uv][w] = [(uv)w] = [u(vw)] = [u][vw] = [u]([v][w]). On the other hand, for all w ∈ A∗ , we have [][w] = [w] = [w] and [w][] = [w] = [w] which shows that [] is an identity element. Finally, let us show that every element in F admits an inverse. For w = a1 a2 · · · an ∈ A∗ , where n ≥ 0 and ∈ A∗ by ak ∈ A for 1 ≤ k ≤ n, deﬁne the word w w = a na n−1 · · · a1 . Observe that ww = a1 a2 · · · an a n · · · a2 a1 ∼ a1 · · · an−1 a n−1 · · · a1 .. . ∼ a1 a2 a2 a1 ∼ a1 a1 ∼ .
D.2 Deﬁnition and Construction of Free Groups
371
Thus, we have ww ≡ . This gives us [w][w] = [ww] = []. Similarly, we get [w][w] = []. It follows that [w] is an inverse of [w] for the product operation in F . This proves that F is a group. Consider the map i : X → F deﬁned by i(x) = [x] for all x ∈ X. If w = a1 a2 · · · an ∈ A∗ , where ak ∈ A for 1 ≤ k ≤ n, then [w] = [a1 a2 · · · an ] = [a1 ][a2 ] · · · [an ]. a)−1 if a ∈ X , we deduce that Since [a] = i(a) if a ∈ X and [a] = [ a]−1 = i( i(X) generates the group F . Let us show that the triple (F, X, i) is a based free group. Suppose that f : X → G is a map from X into a group G. We have to prove that there exists a unique homomorphism φ : F → G such that f = φ ◦ i. Uniqueness follows from the fact that i(X) generates F . To construct φ, we ﬁrst extend f to a map g : A → G by setting f (a) if a ∈ X, g(a) = −1 if a ∈ X . f ( a) Observe that g( a) = g(a)−1 for all a ∈ A. Deﬁne now a map Φ : A∗ → G by setting Φ(w) = g(a1 )g(a2 ) · · · g(an ) for all w = a1 a2 · · · an ∈ A∗ . Note that we have Φ(ww ) = Φ(w)Φ(w )
(D.1)
for all w, w ∈ A∗ . Moreover, if w may be obtained from w by an elementary av for some a ∈ A and u, v ∈ A∗ , then reduction, that is, w = uv and w = ua a)Φ(v) = Φ(u)g(a)g(a)−1 Φ(v) = Φ(u)Φ(v) = Φ(w). Φ(w ) = Φ(u)g(a)g( It follows that Φ(w) = Φ(w ) whenever w, w ∈ A∗ satisfy w ∼ w . By induction, we deduce that Φ(w) = Φ(w ) for all w, w ∈ A∗ such that w ≡ w . Thus, we can deﬁne a map φ : F → G by setting φ([w]) = Φ(w) for all w ∈ A∗ . By using (D.1), we get φ([w][w ]) = φ([ww ]) = Φ(ww ) = Φ(w)Φ(w ) = φ([w])φ([w ]). Therefore, φ is a group homomorphism. On the other hand, for all x ∈ X, we have φ ◦ i(x) = φ([x]) = g(x) = f (x), which shows that φ ◦ i = f . Consequently, the triple (F, X, i) is a based free group. Given an arbitrary set X, it follows from Theorem D.2.7 that there exist a based free group (F, X, i). Then one often says that F = F (X) is the free
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D Free Groups
group based on X (this is a minor abuse of language since if (F , X, i ) is another based free group then there is a unique isomorphism ϕ : F → F such that ϕ ◦ i = i , by Proposition D.2.6). For k ∈ N, the free group based on {1, 2, . . . , k} is denoted by Fk . We recall that one says that two sets X1 and X2 are equipotent, or that they have the same cardinality, if there exists a bijective map u : X1 → X2 . Theorem D.2.8. Let F1 and F2 be two free groups based on X1 ⊂ F1 and X2 ⊂ F2 . Then the following conditions are equivalent: (a) the groups F1 and F2 are isomorphic; (b) the sets X1 and X2 are equipotent. For the proof, we shall need a few preliminary results. We use the following notation. If G1 and G2 are groups, we denote by Hom(G1 , G2 ) the set consisting of all homomorphisms φ : G1 → G2 . We denote by P(X) the set of all subsets of a set X. Lemma D.2.9. Let F be a free group based on X ⊂ F . Then the sets Hom(F, Z/2Z) and P(X) are equipotent. Proof. Since F is free with base X, each map f : X → Z/2Z can be uniquely extended to a homomorphism φ : F → Z/2Z. Therefore, the restriction map yields a bijection from Hom(F, Z/2Z) onto the set of maps from X to Z/2Z. Consequently, the sets Hom(F, Z/2Z) and P(X) are equipotent. Lemma D.2.10. Let F be a free group based on X ⊂ F . Suppose that the set X is inﬁnite. Then the sets X and F are equipotent. Proof. Let A = X ∪ X −1 . Since X generates F , the map ρ : A∗ → F deﬁned by ρ(a1 , a2 , . . . , an ) = a1 a2 . . . an is surjective. As X is inﬁnite, the sets X, A, and A∗ are all equipotent. It follows that there is a surjective map from X onto F and therefore an injective map from F into X. By applying the CantorBernstein theorem (cf. Corollary H.3.5), we deduce that X and F are equipotent. Proof of Theorem D.2.8. The fact that (b) implies (a) immediately follows from Proposition D.2.6. To prove the converse implication, suppose that there exists an isomorphism α : F1 → F2 . Then α induces a bijective map between the sets Hom(F1 , Z/2Z) and Hom(F2 , Z/2Z). It follows that P(X1 ) and P(X2 ) are equipotent by Lemma D.2.9. If X1 is inﬁnite, this implies that X2 is also inﬁnite, and we conclude that X1 and X2 are equipotent by applying Lemma D.2.10. On the other hand, if X1 is ﬁnite, the fact that P(X1 ) and P(X2 ) are equipotent implies that X2 is also ﬁnite and that 2X1  = 2X2  . This gives us X1  = X2  and we conclude that X1 and X2 are also equipotent in this case.
D.3 Reduced Forms
373
Corollary D.2.11. Let F be a free group and let X1 , X2 ⊂ F be two bases of F . Then X1 and X2 are equipotent. Let F be a free group. The cardinality of a base X ⊂ F depends only on F by Corollary D.2.11. This cardinality is called the rank of F . Two free groups are isomorphic if and only if they have the same rank by Theorem D.2.8. A group is free of ﬁnite rank k ∈ N if and only if it is isomorphic to Fk .
D.3 Reduced Forms In order to state the ﬁrst result of this section, we use the notation introduced in the proof of Theorem D.2.7. A word w ∈ A∗ is said to be reduced if it contains no subword of the form a a with a ∈ A, that is, if there is no word w ∈ A∗ which can be obtained from w by applying an elementary reduction. Note that the empty word is reduced and that every subword of a reduced word is itself reduced. Theorem D.3.1. Every equivalence class for ≡ contains a unique reduced word. Proof. It is clear that any word w ∈ A∗ can be transformed into a reduced word by applying a suitable ﬁnite sequence of elementary reductions. This shows the existence of a reduced word in any equivalence class for ≡. Let us prove uniqueness. Consider the subset R ⊂ A∗ consisting of all reduced words. For a ∈ A and r ∈ R, deﬁne the word αa (r) by w if r = aw for some w ∈ A∗ , αa (r) = ar otherwise. Note that the word αa (r) is always reduced. Thus, we get a map αa : R → R deﬁned for each a ∈ A. Let us show that αa ◦ αea = αea ◦ αa = IdR .
(D.2)
Let a ∈ A and consider an arbitrary element r ∈ R. If r = aw for some w ∈ A∗ , then αea (r) = w and hence αa (αea (r)) = αa (w) = aw = r (observe ar that w cannot start by a since r is reduced). Otherwise, we have αea (r) = ar) = r. It follows that αa ◦ αAe = IdR . By and therefore αa (αea (r)) = αa ( replacing a by a in this equality, we get αea ◦ αa = IdR since · is an involution on A. This proves (D.2). From (D.2), we deduce that αa ∈ Sym(R) for all a ∈ A. By setting ρ(w) = αa1 ◦ αa2 ◦ · · · ◦ αan
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for every word w = a1 a2 · · · an ∈ A∗ , we get a monoid homomorphism ρ : A∗ → Sym(R). Note that ρ(r)() = r
for all r ∈ R.
(D.3)
On the other hand, it immediately follows from (D.2) that ρ(w) = ρ(w ) whenever w, w ∈ A∗ satisfy w ≡ w . Thus, if r1 an r2 are reduced words in the same equivalence class for ≡, we have ρ(r1 ) = ρ(r2 ) and therefore r1 = r2 by applying (D.3). Corollary D.3.2. Let (F, X, i) be a based free group. Then every element f ∈ F can be uniquely written in the form f = i(x1 )h1 i(x2 )h2 · · · i(xn )hn
(D.4)
with n ≥ 0, xi ∈ X and hi ∈ Z \ {0} for 1 ≤ i ≤ n, and xi = xi+1 for 1 ≤ i ≤ n − 1. Deﬁnition D.3.3. The expression (D.4) is called the reduced form of the element f in the based free group (F, X, i). Proof of Corollary D.3.2. We can assume that (F, X, i) is the based free group constructed in the proof of Theorem D.2.7. By construction, the element f ∈ F is an equivalence class for ≡. This equivalence class contains a unique reduced word r by Theorem D.3.1. The word r can be uniquely written in the form r = ak11 ak22 · · · aknn , where n ≥ 0, ai ∈ A and ki ∈ N \ {0} for 1 ≤ i ≤ n, and ai = ai+1 for 1 ≤ i ≤ n − 1. This gives us an expression of the form (D.4) with xi = ai and hi = ki if ai ∈ X, and xi = ai and hi = −ki otherwise. Uniqueness of such an expression for f follows from the uniqueness of the reduced word r ∈ f . Corollary D.3.4. Let G be a group. Let U ⊂ G be a subset such that uk11 uk22 · · · uknn = 1G ,
(D.5)
ui+1 for for all u1 , u2 , . . . , un ∈ U and k1 , k2 , . . . , kn ∈ Z \ {0} with ui = 1 ≤ i ≤ n − 1 and n ≥ 1. Then the subgroup of G generated by U is free with base U . Proof. Denote by H the subgroup of G generated by U and by ι : U → H the inclusion map. Let (F, U, i) be a based free group (cf. Theorem D.2.7). Then there exists a unique homomorphism φ : F → H such that ι = φ ◦ i. Note that φ is surjective since φ(i(U )) = ι(U ) = U generates H. Let us show that φ is also injective. Consider an element f ∈ F written in reduced form, that is, in the form f = i(u1 )h1 i(u2 )h2 · · · i(un )hn with n ≥ 0, ui ∈ U and hi ∈ Z \ {0} for 1 ≤ i ≤ n, and ui = ui+1 for 1 ≤ i ≤ n − 1. Then we
D.4 Presentations of Groups
375
have φ(f ) = uh1 1 uh2 2 · · · uhnn since φ(i(ui )) = ι(ui ) = ui for 1 ≤ i ≤ n. It follows from (D.5) that φ(f ) = 1H if and only if n = 0, that is, if and only if f = 1F . This shows that φ is an isomorphism. It follows from Remark D.2.2 that (H, U, ι) is a based free group.
D.4 Presentations of Groups Proposition D.4.1. Let G be a group and let S be a generating subset of G. Let F denote the free group based on S. Then, the group G is isomorphic to a quotient of F . Proof. Let i : S → F and f : S → G denote the inclusion maps. Since F is free with base S, there exists a homomorphism φ : F → G such that f = φ ◦ i. This implies that S is contained in the image of φ. Since S generates G, we deduce that φ is surjective. Therefore, the group G is isomorphic to the quotient group F/ Ker(φ). As every group admits a generating subset (e.g., the group itself), we deduce the following: Corollary D.4.2. Every group is isomorphic to a quotient of a free group. A group is said to be ﬁnitely generated if it admits a ﬁnite generating subset. Proposition D.4.1 gives us: Corollary D.4.3. Every ﬁnitely generated group is isomorphic to a quotient of a free group of ﬁnite rank. Let A be a subset of a group G. The intersection of all normal subgroups of G containing A is a normal subgroup of G which is called the normal closure of A in G. Let G be a group. By Corollary D.4.2, there exist a free group F and an epimorphism φ : F → G. Let X be a free base for F . If R is a subset of F whose normal closure is the kernel of φ, then one says that G admits the presentation G = X; R. (D.6) Note that as X generates F (cf. Proposition D.2.3(ii)), we have that φ(X) generates G. The elements x ∈ X (rather than the φ(x) ∈ G, x ∈ X) are called, by abuse of language, the generators of the presentation (D.6). The elements r ∈ R are called the relators of the presentation (D.6). For every r ∈ R let ur , vr be elements in F such that r = ur (vr )−1 . Then (D.6) is often expressed as G = X : r = 1, r ∈ R or G = X : ur = vr , r ∈ R. Note that G admits a presentation (D.6) with X ﬁnite if and only if G is ﬁnitely generated (cf. Corollary D.4.3).
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If a group G admits a presentation (D.6) with both X and R ﬁnite, then G is said to be ﬁnitely presented.
D.5 The Klein PingPong Theorem The following theorem is often used to prove that a group is free: Theorem D.5.1. Let G be a group. Let X be a generating subset of G having at least two distinct elements. Suppose that G acts on a set E and that there is a family (Ax )x∈X of nonempty pairwise disjoint subsets of E such that ⎛ ⎞ xk ⎝ Ay ⎠ ⊂ Ax for all x ∈ X and k ∈ Z \ {0}. (D.7) y∈X\{x}
Then G is a free group with base X. Proof. Consider an element g ∈ G written as a nontrivial reduced word on the generating subset X, that is, in the form g = xk11 xk22 . . . xknn , where n ≥ 1, xi ∈ X and ki ∈ Z \ {0} for 1 ≤ i ≤ n, and xi = xi+1 for 1 ≤ i ≤ n − 1. By Corollary D.3.4, we have to show that g = 1G . Suppose ﬁrst that either X contains at least three distinct elements or X has exactly two elements and x1 = xn . In this case, we can ﬁnd an element y ∈ X such that y = x1 and y = xn . By successive applications of D.7, we get k
k
k
k
n−2 n−1 kn xn−1 xn Ay gAy = xk11 xk22 . . . xn−2 n−2 n−1 ⊂ xk11 xk22 . . . xn−2 xn−1 Axn
k
n−2 ⊂ xk11 xk22 . . . xn−2 Axn−1
⊂ xk11 xk22 . . . Axn−2 ... ⊂ xk11 xk22 Ax3 ⊂ xk11 Ax2 ⊂ Ax1 . As the sets Ay and Ax1 are disjoint and Ay = ∅ by our hypotheses, we deduce that g = 1G .
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It remains to treat the case when X has exactly two elements and x1 = xn . Then kn −k1 1 1 k2 k3 = x2k xk11 gx−k 1 1 x2 x3 . . . xn x1 1 1 is a reduced form of xk11 gx−k . We have xk11 gx−k = 1G by the ﬁrst case. We 1 1 deduce that we also have g = 1G in this case.
Remark D.5.2. The proof of Theorem D.5.1 shows that the hypotheses on the family (Ax )x∈X may be relaxed: in fact, it suﬃces that the subsets Ax ⊂ X, x ∈ X, satisfy D.7 and Ay ⊂ Ax for all distinct elements x, y ∈ X. Corollary D.5.3. Let F be a free group of rank 2 and let n ≥ 2 be an integer. Then F contains a free subgroup of rank n. Proof. Suppose that F is based on the elements a and b. Let G be the subgroup of F generated by the subset X = {ai ba−i : 0 ≤ i ≤ n − 1}. Consider the action of G on F given by left multiplication. Deﬁne, for each x = ai ba−i ∈ X, the subset Ax ⊂ F as being the set of elements of F whose reduced form starts by ai bk for some k ∈ Z \ {0}. The subsets Ax , x ∈ X, clearly satisfy the hypotheses of Theorem D.5.1. Therefore G is free with base X.
Appendix E
Inductive Limits and Projective Limits of Groups
E.1 Inductive Limits of Groups Let I be a directed set. An inductive system of groups over I consists of the following data: (1) a family of groups (Gi )i∈I indexed by I, (2) for each pair i, j ∈ I such that i ≤ j, a homomorphism ψji : Gi → Gj satisfying the following conditions: ψii = IdGi (identity map on Gi ) for all i ∈ I, ψkj ◦ ψji = ψki for all i, j, k ∈ I such that i ≤ j ≤ k. Then one speaks of the inductive system (Gi , ψji ) or simply of the inductive system (Gi ) if the homomorphisms ψji are understood. of groups over I. Consider the relation Let (Gi , ψji ) be an inductive system ∼ on the disjoint union E = i∈I Gi deﬁned as follows. If xi ∈ Gi and xj ∈ Gj are elements of E, then xi ∼ xj if and only if there is an element k ∈ I such that i ≤ k, j ≤ k and ψki (xi ) = ψkj (xj ). It easy to check that ∼ is a equivalence relation on the set E. Let G = E/ ∼ be the set of equivalence classes of ∼. For xi ∈ Gi , let [xi ] ∈ G denote the class of xi . One deﬁnes a binary operation on G in the following way. Given xi ∈ Gi and xj ∈ Gj , one deﬁnes the class [xi ][xj ] by setting [xi ][xj ] = [ψki (xi )ψkj (xj )], where k ∈ I is such that i ≤ k and j ≤ k. One checks that [xi ][xj ] depends neither on the choice of the representatives xi and xj , nor on the choice of k. Moreover, this operation gives a group structure on G. The group G is called the inductive limit (or the direct limit) of the inductive system (Gi ) and one writes G = lim Gi . For each i ∈ I, there is a canonical homomorphism −→ hi : Gi → G deﬁned by hi (xi ) =[xi ]. Note that it immediately follows from the construction of G that G = i∈I hi (Gi ). Moreover, one has T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 13, © SpringerVerlag Berlin Heidelberg 2010
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hj ◦ ψji = hi for all i, j ∈ I such that i ≤ j. Examples E.1.1. (a) Let (Gi )i∈I be a family of groups and, for i, j ∈ I such that i ≤ j, let ψji : Gi → Gj be the trivial homomorphism, namely, ψji (g) = 1Gj for all g ∈ Gi . Then, (Gi , ψji ) is an inductive system whose inductive limit is a trivial group. (b) Let G be a group and denote by H the setof all ﬁnitely generated subgroups of G. Then (H, ⊂) is a directed set and H∈H H = G. Moreover, the set H together with the inclusion maps ψK,H : H → K, for all H, K ∈ H with H ⊂ K, is an inductive system whose limit is canonically isomorphic to G.
E.2 Projective Limits of Groups Let I be a directed set. A projective system of groups over I consists of the following data: (1) a family of groups (Gi )i∈I indexed by I, (2) for each pair i, j ∈ I such that i ≤ j, a homomorphism ϕij : Gj → Gi satisfying the following conditions: ϕii = IdGi (identity map on Gi ) for all i ∈ I, ϕij ◦ ϕjk = ϕik for all i, j, k ∈ I such that i ≤ j ≤ k. Then one speaks of the projective system (Gi , ϕij ) or simply of the projective system (Gi ) if the homomorphisms ϕij are understood. Let (Gi , ϕij ) be a projective system of groups over I. Let P = i∈I Gi denote the direct product of the groups Gi . One immediately checks that G = {(xi ) ∈ P : ϕij (xj ) = xi for all i, j ∈ I such that i ≤ j} is a subgroup of P . The group G is called the projective limit of the projective system (Gi ), and one writes G = lim Gi . For each i ∈ I, there is a canonical ←− homomorphism fi : G → Gi obtained by restriction of the projection map πi : P → Gi . One has ϕij ◦ fj = fi for all i, j ∈ I such that i ≤ j. Examples E.2.1. (a) Let (Gi )i∈I be a family of groups and, for i, j ∈ I such that i ≤ j, let ϕij : Gj → Gi be the trivial homomorphism. Then, (Gi , ϕij ) is a projective system whose projective limit is a trivial group. (b) Let G be a group. Denote by N the set of all normal subgroups of G. Then (N , ⊃) is a directed set. The family of groups (G/H)H∈N , when
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equipped with the canonical quotient homomorphisms ϕH,K : G/K → G/H for all H, K ∈ N with H ⊃ K, gives rise to a projective system whose projective limit is canonically isomorphic to G.
Appendix F
The BanachAlaoglu Theorem
All vector spaces considered in this appendix are vector spaces over the ﬁeld R of real numbers.
F.1 Topological Vector Spaces A topological vector space is a real vector space X endowed with a topology such that the maps X ×X →X (x, y) → x + y and R×X →X (λ, x) → λx are continuous. Example F.1.1. Let · be a norm on a real vector space X and let d denote the metric on X deﬁned by d(x, y) = x − y for all x, y ∈ X. Then the topology deﬁned by d yields a structure of topological vector space on X. Recall that a subset C of a real vector space X is said to be convex if (1 − λ)x + λy ∈ C for all λ ∈ [0, 1] and x, y ∈ C. A topological real vector space X is said to be locally convex if there is a base of neighborhoods of 0 consisting of convex subsets of X.
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F.2 The Weak∗ Topology Let X be a real vector space equipped with a norm · . Recall that a linear map u : X → R is continuous if and only if u is bounded on the unit ball B(X) = {x ∈ X : x ≤ 1}, that is, if and only if there is a constant C ≥ 0 such that u(x) ≤ C for all x ∈ B(X). The topological dual of X is the vector space X ∗ consisting of all continuous linear maps u : X → R. The operator norm on X ∗ is the norm deﬁned by u =
sup u(x). x∈B(X)
The topology deﬁned by the operator norm is called the strong topology on X ∗ . Given x ∈ X, let ψx : X ∗ → R denote the evaluation map u → u(x) at x. The weak∗ topology on X ∗ is the initial topology associated with the family of all evaluation maps ψx : X ∗ → R, x ∈ X. Thus, the weak∗ topology is the smallest topology on X ∗ for which all evaluation maps ψx are continuous. Observe that every subset of X ∗ which is open for the weak∗ topology is also open for the strong topology since all the evaluation maps are continuous for the strong topology. It follows that convergence with respect to the strong topology implies convergence with respect to the weak∗ topology. The weak∗ topology provides a topological vector space structure on X ∗ . A base of open neighborhoods of 0 for the weak∗ topology is given by all subsets of the form V (F, ε) = {u ∈ X ∗ : u(x) < ε for all x ∈ F }, where F is a ﬁnite subset of X and ε > 0. Since all the sets V (F, ε) are convex, the topological vector space structure associated with the weak∗ topology on X ∗ is locally convex. The weak∗ topology on X ∗ is Hausdorﬀ. Indeed, if u1 and u2 are two distinct elements in X ∗ , then there exists x ∈ X such that u1 (x) = u2 (x). If U1 and U2 are disjoint open subsets of R containing u1 (x) and u2 (x) respectively, then ψx−1 (U1 ) and ψx−1 (U2 ) are disjoint open subsets of X ∗ containing u1 and u2 respectively.
F.3 The BanachAlaoglu Theorem In general, the unit ball in X ∗ is not compact for the strong topology (this follows from the fact that the unit ball in a normed space is never compact unless the space is ﬁnitedimensional). However, this ball is always compact for the weak∗ topology. This result, which is known as the BanachAlaoglu
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theorem is one of the central results of classical functional analysis and may be easily deduced from the Tychonoﬀ theorem. Theorem F.3.1 (BanachAlaoglu theorem). Let X be a real normed vector space and let · denote the operator norm on its topological dual X ∗ . Then the unit ball B(X ∗ ) = {u ∈ X ∗ : u ≤ 1} is compact for the weak∗ topology on X ∗ . Proof. Observe ﬁrst that X ∗ is a vector subspace of the vector space RX consisting of all realvalued functions on X. On the other hand, setting Ix = [−x, x] ⊂ R for each x ∈ X. We haveB(X ∗ ) ⊂ x∈X Ix by deﬁnition of the operator norm. Let us equip RX = x∈X R with the product topology. Then it is clear that the topology induced on x∈X Ix is the product topology and that the topology induced on X ∗ is the weak∗ topology. Let f be an element of RX which is the limit of a net (ui ) of elements of B(x∗ ). For all x, y ∈ X and λ ∈ R, we have ui (λx) = λui (x), ui (x + y) = ui (x) + ui (y) and ui (x) ≤ x since ui ∈ B(X ∗ ). By taking limits, we get f (λx) = λf (x), ∗ f (x + y) = f (x) + f (y) and f (x) ≤ x. This shows that f ∈ B(X ). ∗ X Consequently, B(X ) is closed in R . Since x∈X Ix is compact by Tychonoﬀ theorem (Theorem A.5.2), we deduce that B(X ∗ ) is compact.
Appendix G
The MarkovKakutani Fixed Point Theorem
All vector spaces considered in this appendix are vector spaces over the ﬁeld R of real numbers.
G.1 Statement of the Theorem Let C be a convex subset of a real vector space X. A map f : C → C is called aﬃne if f ((1 − λ)x + λy) = (1 − λ)f (x) + λf (y) for all λ ∈ [0, 1] and x, y ∈ C. Theorem G.1.1 (MarkovKakutani). Let K be a nonempty convex compact subset of a Hausdorﬀ topological vector space X. Let F be a set of continuous aﬃne maps f : K → K. Suppose that all elements of F commute, that is, f1 ◦ f2 = f2 ◦ f1 for all f1 , f2 ∈ F. Then there exists a point in K which is ﬁxed by all the elements of F.
G.2 Proof of the Theorem In the proof of the MarkovKakutani theorem, we shall use the following lemmas. Lemma G.2.1. Let K be a compact subset of a topological vector space X and let V be a neighborhood of 0 in X. Then there exists a real number α > 0 such that λK ⊂ V for every real number λ such that λ < α. Proof. Since the multiplication by a scalar R × X → X is continuous, we can ﬁnd, for each x ∈ X, a real number αx and an open neighborhood Ωx ⊂ X of x such that T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 15, © SpringerVerlag Berlin Heidelberg 2010
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λ < αx ⇒ λx Ωx ⊂ V.
(G.1)
The sets Ωx , x ∈ K, form an open cover of K. As K is compact, there is a ﬁnite subset F ⊂ K such that Ωx . (G.2) K⊂ x∈F
If we take α = minx∈F αx , then α > 0 and λ < α ⇒ λK ⊂ V
by (G.1) and (G.2).
Lemma G.2.2. Let K be a compact subset of a topological vector space X. Let (xi )i∈I be a net of points in K and let (λi )i∈I be a net of real numbers converging to 0 in R. Then the net (λi xi )i∈I converges to 0 in X. Proof. Let V be a neighborhood of 0 in X. By Lemma G.2.1, we can ﬁnd α > 0 such that λK ⊂ V for every λ such that λ < α. As the net (λi ) converges to 0, there exists i0 ∈ I such that i ≥ i0 implies λi  < α. Thus we have λi xi ∈ V for all i ≥ i0 . This shows that the net (λi xi ) converges to 0. The following lemma is the theorem of MarkovKakutani in the particular case when the set F is reduced to a single element. Lemma G.2.3. Let K be a nonempty convex compact subset of a Hausdorﬀ topological vector space X and let f : K → K be an aﬃne continuous map. Then f has a ﬁxed point in K. Proof. Let us set C = {y − f (y) : y ∈ K}. The fact that f admits a ﬁxed point in K is equivalent to the fact that 0 ∈ C. Choose an arbitrary point x ∈ K and consider the sequence (xn )n≥1 of points of X deﬁned by xn =
n−1 1 k (f (x) − f k+1 (x)). n k=0
We have f k (x) − f k+1 (x) = f k (x) − f (f k (x)) ∈ C for 0 ≤ k ≤ n − 1. On the other hand, the set C is convex, since K is convex and f is aﬃne. Thus xn ∈ C for every n ≥ 1. As 1 1 xn = x − f n (x). n n and f n (x) ∈ K for every n ≥ 1, it follows from Lemma G.2.2 that the sequence (xn )n≥1 converges to 0. The set C is compact since it is the image of the compact set K by the continuous map y → y − f (y). As every compact subset of a Hausdorﬀ space is closed, we deduce that C is closed in X. Thus 0 ∈ C.
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Proof of Theorem G.1.1. Let f ∈ F and consider the set Fix(f ) = {x ∈ K : f (x) = x} of its ﬁxed points. The set Fix(f ) is not empty by Lemma G.2.3 and it is compact since it is a closed subset of the compact set K. On the other hand, Fix(f ) is convex since K is convex and f is aﬃne. If g ∈ F and x ∈ Fix(f ), then the fact that f and g commute implies that g(x) ∈ Fix(f ) since f (g(x)) = g(f (x)) = g(x). Therefore we can apply Lemma G.2.3 to the restriction of g to Fix(f ). It follows that g ﬁxes a point in Fix(f ), that is, Fix(f ) ∩ Fix(g) = ∅. By induction on n, we get Fix(f1 ) ∩ Fix(f2 ) ∩ · · · ∩ Fix(fn ) = ∅ for all f1 , f2 , . . . , fn ∈ F. Since K is compact, from the ﬁnite intersection property (see Sect. A.5) we deduce that Fix(f ) = ∅. f ∈F
This shows that there is a point in K which is ﬁxed by all elements of F .
Notes The proof presented here is due to S. Kakutani [Kak] (see [Jac]). In the proof of A. Markov [Mar], the local convexity of X is needed.
Appendix H
The Hall Harem Theorem
H.1 Bipartite Graphs A bipartite graph is a triple G = (X, Y, E), where X and Y are arbitrary sets, and E is a subset of the Cartesian product X × Y . The set X (resp. Y ) is called the set of left (resp. right) vertices and E is called the set of edges of the bipartite graph G (see Fig. H.1).
Fig. H.1 The bipartite graph G = (X, Y, E) with X = {x1 , x2 , x3 , x4 , x5 }, Y = {y1 , y2 , y3 , y4 } and E = {(x1 , y1 ), (x2 , y1 ), (x2 , y2 ), (x2 , y3 ), (x3 , y4 ), (x5 , y3 ), (x5 , y4 )}
A bipartite subgraph of a bipartite graph G = (X, Y, E) is a bipartite graph G = (X , Y , E ) with X ⊂ X, Y ⊂ Y and E ⊂ E (see Fig. H.2). Let G = (X, Y, E) be a bipartite graph. Two edges (x, y), (x , y ) ∈ E are said to be adjacent if x = x or y = y . Given a vertex x ∈ X (resp. y ∈ Y ) the rightneighborhood of x (resp. the leftneighborhood of y) is the subset NR (x) ⊂ Y (resp. NL (y) ⊂ X) deﬁned by (see Figs. H.3–H.4):
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Fig. H.2 The bipartite subgraph G = (X , Y , E ) of the bipartite graph G in Fig. H.1 with X = {x1 , x2 , x4 , x5 }, Y = {y1 , y2 , y3 } and E = {(x2 , y1 ), (x2 , y2 ), (x2 , y3 ), (x5 , y3 )}
Fig. H.3 The rightneighborhood NR (x2 ) ⊂ Y of the vertex x2 ∈ X in the bipartite graph G of Fig. H.1
NR (x) = NRG (x) = {y ∈ Y : (x, y) ∈ E} (resp. NL (y) = NLG (y) = {x ∈ X : (x, y) ∈ E}). For subsets A ⊂ X and B ⊂ Y , we deﬁne the rightneighborhood NR (A) of A and the leftneighborhood NL (B) of B by NR (A) = NRG (A) = NR (a) and NL (B) = NLG (B) = NL (b). a∈A
b∈B
One says that the bipartite graph G = (X, Y, E) is ﬁnite if the sets X and Y are ﬁnite. One says that G is locally ﬁnite if the sets NR (x) and NR (y) are ﬁnite for all x ∈ X and y ∈ Y . Note that if G is locally ﬁnite then the sets NR (A) and NL (B) are ﬁnite for all ﬁnite subsets A ⊂ X and B ⊂ Y .
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Fig. H.4 The leftneighborhood NL (y1 ) ⊂ X of the vertex y1 ∈ Y in the bipartite graph G of Fig. H.1
H.2 Matchings Let G = (X, Y, E) be a bipartite graph. A matching in G is a subset M ⊂ E of pairwise nonadjacent edges. In other words, a subset M ⊂ E is a matching if and only if both projection maps p : M → X and q : M → Y are injective. A matching M is called leftperfect (resp. rightperfect) if for each x ∈ X (resp. y ∈ Y ), there exists y ∈ Y (resp. x ∈ X) such that (x, y) ∈ M (see Fig. H.5). Thus, a matching M is leftperfect (resp. rightperfect) if and only if the projection map p : M → X (resp. q : M → Y ) is surjective (and therefore bijective). A matching M is called perfect if it is both leftperfect and rightperfect.
Fig. H.5 A rightperfect matching M ⊂ E in the bipartite graph G of Fig. H.1. Note that there is no leftperfect matching (and therefore no perfect matching) in G since X > Y 
Remarks H.2.1. (a) A subset M ⊂ E is a leftperfect (resp. rightperfect) matching if and only if there is an injective map ϕ : X → Y (resp. an injective
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map ψ : Y → X) such that M = {(x, ϕ(x)) : x ∈ X} (resp. M = {(ψ(y), y) : y ∈ Y }). (b) Similarly, a subset M ⊂ E is a perfect matching if and only if there is a bijective map ϕ : X → Y such that M = {(x, ϕ(x)) : x ∈ X}. Given a bipartite graph G = (X, Y, E), one often regards the set X as a set of boys and Y as a set of girls. One interprets (x, y) ∈ E as the condition that x and y know each other. In this context, a matching M ⊂ E is a process of getting boys and girls that know each other married (no polygamy is allowed here). The matching is leftperfect (resp. rightperfect) if and only if every boy (resp. girl) gets married. Finally, M is perfect if and only if there remain no singles.
H.3 The Hall Marriage Theorem We use the notation  ·  to denote cardinality of sets. Deﬁnition H.3.1 (Hall conditions). Let G = (X, Y, E) be a bipartite graph. One says that G satisﬁes the left (resp. right) Hall condition if NR (A) ≥ A
(H.1)
(resp. NL (B) ≥ B) for every ﬁnite subset A ⊂ X (resp. for every ﬁnite subset B ⊂ Y ). One says that G satisﬁes the Hall marriage conditions if G satisﬁes both the left and the right Hall conditions. Theorem H.3.2. Let G = (X, Y, E) be a locally ﬁnite bipartite graph. Then the following conditions are equivalent. (a) G satisﬁes the left (resp. right) Hall condition; (b) G admits a left (resp. right) perfect matching. Proof. It is obvious that (b) implies (a). Let us prove that (a) implies (b). By symmetry, it suﬃces to show that if G satisﬁes the left Hall condition then it admits a left perfect matching. We ﬁst treat the case when the set X is ﬁnite by induction on n = X. In the case X = 1, the statement is trivially satisﬁed. Suppose that we have proved the statement whenever X ≤ n − 1 and let us prove it for X = n. We distinguish two cases. Case (i): Suppose that NR (A) ≥ A + 1
(H.2)
for all nonempty proper subsets A ⊂ X. Then ﬁx x0 ∈ X and y0 ∈ Y such that (x0 , y0 ) ∈ E. Consider the bipartite subgraph G = (X , Y , E ),
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where X = X \ {x0 }, Y = Y \ {y0 }, and E = E ∩ (X × Y ). Then, for every nonempty subset A ⊂ X , we have NR (A ) ≥ A  + 1 by (H.2), and hence NRG (A ) ≥ A  since NRG (A ) = NR (A ) \ {y0 }. As X  = n − 1, it follows from our induction hypothesis that G admits a leftperfect matching M ⊂ E . Then M = M ∪ {(x0 , y0 )} is a leftperfect matching for G. Case (ii): Suppose that we are not in Case (i). This means that there exists a nonempty proper subset X ⊂ X such that NR (X ) = X .
(H.3)
Then the bipartite subgraph G = (X , Y , E ), where Y = NR (X ) and E = E ∩ (X × Y ) clearly satisﬁes the left Hall condition. As X  ≤ n − 1, there exists, by our induction hypothesis, a leftperfect matching M ⊂ E for G . Consider now the bipartite subgraph G = (X , Y , E ), where X = X \ X , Y = Y \ Y , and E = E ∩ (X × Y ). We claim that the left Hall condition also holds for G . Otherwise, there would be some subset A ⊂ X such that,
NRG (A ) < A ,
(H.4)
and then the left Hall condition for G would be violated by A = X ∪ A ⊂ X since NR (A) = NR (X ∪ A ) = NR (X ) ∪ NR (A )
= NR (X ) ∪ NRG (A )
≤ NR (X ) + NRG (A ) < X  + A  (by (H.3) and (H.4)) = X ∪ A  = A. Therefore, as X  < X = n, induction applies again yielding a leftperfect matching M ⊂ E for G . It then follows that M = M ∪ M is a leftperfect matching for G. This completes the proof that (a) implies (b) in the case when X is ﬁnite. To treat the general case, we shall apply the Tychonoﬀ product theorem. Suppose that G = (X, Y, E) is a (possibly inﬁnite) locally ﬁnite bipartite graph satisfying the left Hall condition, that is, NR (A) ≥ Afor every ﬁnite subset A ⊂ X. Let us equip the Cartesian product K = x∈X NR (x) with its prodiscrete topology, that is, with the product topology obtained by taking the discrete topology on each factor NR (x). As each set NR (x) is ﬁnite, the space K is compact by the Tychonoﬀ product theorem (Corollary A.5.3).
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For each x ∈ X, let πx : K → NR (x) denote the projection map. Let F be the set consisting of all nonempty ﬁnite subsets of X. Consider, for each F ∈ F, the set C(F ) ⊂ K consisting of all z ∈ K which satisfy πx1 (z) = πx2 (z) for all distinct elements x1 and x2 in F . It follows from this deﬁnition and the continuity of the projection maps πx , that C(F ) is an intersection of closed subsets of X and hence closed in K. On the other hand, C(F ) is not empty. Indeed, choose, for each x ∈ X, an element ψ(x) ∈ NR (x) (observe that NR (x) is not empty by the left Hall condition). Also set G = (X , Y , E ) where X = F , Y = NR (F ) and E = E ∩ (X × Y ). Then, the left Hall condition for G implies the left Hall condition for the ﬁnite bipartite graph G and therefore, by Theorem H.3.2, there exists a perfect matching for G . By Remark H.2.1, there exists an injective mapping ϕ : F = X → Y = NR (F ). Then, the element (Φ(x))x∈X ∈ K, deﬁned by Φ(x) = ϕ(x) if x ∈ F and Φ(x) = ψ(x) if x ∈ X \ F , clearly belongs to C(F ). Since C(F1 ) ∩ C(F2 ) ∩ · · · ∩ C(Fn ) ⊃ C(F1 ∪ F2 ∪ · · · ∪ Fn ) for all F1 , F2 , . . . , Fn ∈ F, we deduce that the family {C(F ) : F ∈ F} of subsets of K hasthe ﬁnite intersection property. By compactness of K, there is a point z0 ∈ F ∈F F . Then M = {(x, πx (z0 )) : x ∈ X} is a leftperfect matching for G. Remark H.3.3. The hypothesis of local ﬁniteness for the bipartite graph can not be removed from the statement of the previous theorem. To see this, consider the bipartite graph G = (X, Y, E), where X = Y = N and E = {(0, n + 1) : n ∈ N} ∪ {(n + 1, n) : n ∈ N}. Observe that G is not locally ﬁnite as NR (0) = ∞. Also, it satisﬁes the left Hall condition. Indeed, given a ﬁnite subset A ⊂ X, the set NR (A) is inﬁnite if 0 ∈ A, while NR (A) = {n − 1 : n ∈ A} = A if 0 ∈ / A. However, G admits no left perfect matching. Indeed, for any matching M ⊂ E, either 0 ∈ X remains unmatched, or, if (0, n) ∈ M for some n ∈ N, then n + 1 ∈ X remains unmatched (see Fig. H.6). Theorem H.3.4. Let G = (X, Y, E) be a bipartite graph. Suppose that G admits both a left perfect matching and a right perfect matching. Then G admits a perfect matching. Proof. Let MX (resp. MY ) be a left perfect (resp. right perfect) matching for G. Consider the equivalence relation in M = MX ∪ MY deﬁned by declaring two edges e and e in relation if there exists a ﬁnite sequence e = e0 , e1 , . . . , en = e in M such that ei and ei+1 are adjacent for all 0 ≤ i ≤ n − 1. Then, each equivalence class consists of either (see Fig. H.7): – a single edge, or – a cycle of even length 2n ≥ 4, or – an inﬁnite chain which can be biinﬁnite, or inﬁnite only in one direction.
H.3 The Hall Marriage Theorem
397
Fig. H.6 The bipartite graph G = (X, Y, E), where X = Y = N and E = {(0, n + 1) : n ∈ N} ∪ {(n + 1, n) : n ∈ N}
Note that an equivalence class is reduced to a single edge (x, y) if and only if (x, y) ∈ MX ∩ MY . On the other hand, an equivalence class is a cycle of length 2n if and only if it is of the form C ={(x1 , y1 ), (x2 , y2 ), . . . , (xn , yn )} ∪ {(x1 , y2 ), (x2 , y3 ), . . . , (xn−1 , yn ), (xn , y1 )} with xi ∈ X, all distinct, and yj ∈ Y , all distinct, 1 ≤ i, j ≤ n. In this case, we then set M (C) = {(x1 , y1 ), (x2 , y2 ), . . . , (xn , yn )}. Also, a biinﬁnite chain is an equivalence class of the form C = {(xn , yn ) : n ∈ Z} ∪ {(xn , yn+1 ) : n ∈ Z} with xi ∈ X, all distinct, and yj ∈ Y , all distinct, i, j ∈ Z. In this case, we then set M (C) = {(xn , yn ) : n ∈ Z}.
398
H The Hall Harem Theorem
Fig. H.7 In the bipartite graph G = (X, Y, E), where X = {x0 , x1 , x2 , x3 , x4 , x5 }, Y = {y0 , y1 , y2 , y3 , y4 , y5 } and E = {(xi , yi ) : i = 0, 1, . . . , 5} ∪ {(xi , yi+1 ) : i = 1, 2, . . . , 4} ∪ {(x5 , y1 )}, we have a leftperfect matching MX = {(xi , yi ) : i = 0, 1, . . . , 5} (which is indeed perfect) and a rightperfect matching MY = {(x0 , y0 )} ∪ {(xi , yi+1 ) : i = 1, 2, . . . , 4} ∪ {(x5 , y1 )} (which is also perfect). There are two equivalence classes in M = MX ∪ MY , namely, MX ∩ MY = {(x0 , y0 )}, which consists of a single edge, and C = {(xi , yi ) : i = 1, 2 . . . , 5} ∪ {(xi , yi+1 ) : i = 1, 2, . . . , 4} ∪ {(x5 , y1 )}, which is a cycle of length 10
Finally, if the equivalence class is an inﬁnite chain, which is inﬁnite only in one direction, then it is either of the form C = {(xn , yn ) : n ∈ N} ∪ {(xn , yn+1 ) : n ∈ N} or C = {(xn , yn ) : n ∈ N} ∪ {(xn+1 , yn ) : n ∈ N} with xi ∈ X, all distinct, and yj ∈ Y , all distinct, i, j ∈ N. In both cases, we then set M (C) = {(xn , yn ) : n ∈ N}. It is then clear that the set M ⊂ E deﬁned by M (C), M= C
where C runs over all equivalence classes in M , is the required perfect matching for G. Corollary H.3.5 (Cantor–Bernstein Theorem). Let X and Y be two sets. Suppose that there exist injective maps f : X → Y and g : Y → X. Then there exists a bijective map h : X → Y . Proof. Consider the bipartite graph G = (X, Y, E), where E = {(x, f (x)) : x ∈ X} ∪ {(g(y), y) : y ∈ Y }. Now, MX = {(x, f (x)) : x ∈ X} and
H.4 The Hall Harem Theorem
399
MY = {(g(y), y) : y ∈ Y } are left perfect and right perfect matchings in G, respectively (cf. Remark H.2.1(a)). By the previous theorem, there exists a perfect matching M ⊂ E for G. Then M = {(x, h(x)) : x ∈ X}, where h : X → Y is bijective (cf. Remark H.2.1(b)). Theorem H.3.6 (The Hall marriage Theorem). Let G = (X, Y, E) be a locally ﬁnite bipartite graph. Then the following conditions are equivalent: (a) G satisﬁes the Hallmarriage conditions; (b) G admits a perfect matching. Proof. The left (resp. right) Hall condition implies, by Theorem H.3.2, the existence of a left (resp. right) perfect matching for G. Then, Theorem H.3.4 guarantees the existence of a perfect matching for G. The converse implication is trivial.
H.4 The Hall Harem Theorem Let G = (X, Y, E) be a bipartite graph and let k ≥ 1 be an integer. A subset M ⊂ E is called a perfect (1, k)matching if it satisﬁes the following conditions: (1) for each x ∈ X, there are exactly k elements in y ∈ Y such that (x, y) ∈ M , (2) for each y ∈ Y , there is a unique element x ∈ X such that (x, y) ∈ M (see Fig. H.8). Thus, a subset M ⊂ E is a perfect (1, k)matching if and only if there exists a ktoone surjective map ψ : Y → X such that M = {(ψ(y), y) : y ∈ Y } (recall that a surjective map f : S → T from a set S onto a set T is said to be ktoone if each element in T has exactly k preimages in S). Note that when k = 1, a perfect (1, k)matching is the same thing as a perfect matching. In the language of boys, girls, and marriages, a perfect (1, k)matching is a process for marrying each boy with exactly k girls (among the girls he knows) in such a way that each girl is married with exactly one boy (among the boys she knows). The girls that are married with a given boy constitute his harem. Deﬁnition H.4.1. Let G = (X, Y, E) be a locally ﬁnite bipartite graph and let k ≥ 1 be an integer. One says that G satisﬁes the Hall kharem conditions if NR (A) ≥ kA and (H.5) NL (B) ≥ k1 B for all ﬁnite subsets A ⊂ X and B ⊂ Y . Theorem H.4.2 (The Hall harem Theorem). Let G = (X, Y, E) be a locally ﬁnite bipartite graph and let k ≥ 1 be an integer. Then, the following conditions are equivalent. (a) G satisﬁes the Hall kharem conditions; (b) G admits a perfect (1, k)matching.
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H The Hall Harem Theorem
Fig. H.8 The bipartite graph G = (X, Y, E), where X = {x1 , x2 }, Y = {y1 , y2 , y3 , y4 } and E = {(x1 , yi ) : i = 1, 2, 3, 5} ∪ {(x2 , yj ) : j = 1, 2, 4, 5, 6}, with a perfect (1, 3)matching M = {(x1 , yi ) : i = 1, 2, 3} ∪ {(x2 , yj ) : j = 4, 5, 6} and the harems H(x1 ) and H(x2 ) of x1 and x2 respectively
Proof. The fact that (b) implies (a) is trivial. Let us prove that (a) implies (b). Suppose that the Hall kharem conditions are satisﬁed by G. Let X1 , X2 , . . . , Xk be disjoint copies of X and let φi : X → Xi , i = 1, 2, . . . , k, denote the copy maps. It will be helpful to think of φi (x) as the clone of x ∈ X in Xi . k Consider the new bipartite graph G = (X , Y , E ), where X = i=1 Xi , Y = Y , and E = {(φi (x), y) : (x, y) ∈ E, i = 1, 2, . . . , k} ⊂ X × Y . that Let A be a ﬁnite subset of X and denote by A the set of x ∈ X such some clone of x belongs to A . Observe that A  ≤ kA  and NRG (A ) = NRG (A ). Thus, using (a), we get
NRG (A ) = NRG (A ) ≥ kA  ≥ A .
(H.6)
On the other hand, if B is a ﬁnite subset of Y = Y , then NLG (B ) is the set consisting of all clones of elements of NLG (B ) so that 1 G G B  = B  (H.7) NL (B ) = kNL (B ) ≥ k k by (a). Inequalities H.6 and H.7 say that G satisﬁes the Hall marriage conditions. Therefore, G admits a perfect matching M ⊂ E by Theorem H.3.6. Then the set M , consisting of all pairs (x, y) ∈ E such that (x , y) ∈ M for some clone x of x, is clearly a perfect (1, k)matching for G.
Notes
401
Notes The Hall marriage theorem was ﬁrst established for ﬁnite bipartite graphs by P. Hall [Hall1] and then extended to inﬁnite locally ﬁnite bipartite graphs by M. Hall [HallM1]. The proof of Theorem H.3.2 which is given in this appendix is based on [Halm].
Appendix I
Complements of Functional Analysis
All vector spaces considered in this appendix are vector spaces over the ﬁeld R of real numbers.
I.1 The Baire Theorem Let (X, d) be a metric space. Given x ∈ X and r > 0, we denote by BX (x, r) = {y ∈ X : d(x, y) ≤ r} the closed ball of radius r centered at x and by OB X (x, r) = {y ∈ X : d(x, y) < r} the open ball of radius r centered at x. For a subset A ⊂ X we denote by A (resp. Int A) the closure (resp. the interior) of A. Theorem I.1.1 (Baire’s Theorem). Let (X, d) be a complete metric space. Let (Xn )n≥1 be a sequence of closed subsets such that Int Xn = ∅ for all n ≥ 1. Then
⎛ Int ⎝
(I.1)
⎞ Xn ⎠ = ∅.
(I.2)
n≥1
Proof. Let us set An = X \ Xn , so that An is an open dense subset of X. Let us show that n≥1 An is dense in X. Let x0 ∈ X and r0 > 0. We have to show that ⎞ ⎛ An ⎠ = ∅. BX (x0 , r0 ) ⎝ (I.3) n≥1
T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 17, © SpringerVerlag Berlin Heidelberg 2010
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As A1 is dense and open we can ﬁnd x1 ∈ BX (x0 , r0 ) A1 and r1 > 0 such that BX (x1 , r1 ) ⊂ OB X (x0 , r0 ) A1 0 < r1
0 there exists r > 0 such that
I.2 The Open Mapping Theorem
405
T (OB X (0, r )) ⊃ OB Y (0, r ). Finally, since T (OB X (0, r )) = r T (OB X (0, 1)) and OB Y (0, r ) = r OB Y (0, 1), we are only left to show that there exists r > 0 such that T (OB X (0, 1)) ⊃ OB Y (0, r). (I.6) For all n ≥ 1 set Yn = nT (OB X (0, 1)) = T (OB X (0, n)). As T is surjective we have n≥1 Yn = Y . It follows from Theorem I.1.1 that there exists n0 ≥ 1 such that Int Yn0 = ∅. Since Int(n0 T (OB X (0, 1))) = n0 Int(T (OB X (0, 1))), we deduce that Int(T (OB X (0, 1))) = ∅. Thus we can ﬁnd r > 0 and y ∈ Y such that OB Y (y, 4r) ⊂ T (OB X (0, 1)). (I.7) It follows that y ∈ T (OB X (0, 1)) and, by symmetry, − y ∈ T (OB X (0, 1)).
(I.8)
Summing (I.7) and (I.8) we obtain 2OB Y (0, 2r) = OB Y (0, 4r) = OB Y (y, 4r) − y ⊂ T (OB X (0, 1)) + T (OB X (0, 1)) ⊂ 2T (OB X (0, 1)). We deduce that OB Y (0, 2r) ⊂ T (OB X (0, 1)) so that, for all n ≥ 0, r OB Y 0, n ⊂ T (OB X (0, 1/2n+1 )). 2
(I.9)
Let y0 ∈ OB Y (0, r) and let us show that there exists x ∈ OB X (0, 1) such that y0 = T (x ). We deduce from (I.9) (with n = 0) that there exists x0 ∈ OB X (0, 12 ) such that y0 − T (x0 ) < 2r . Set y1 = y0 − T (x0 ) and observe that y1 ∈ OB Y (0, 2r ). Continuing this way, we ﬁnd a sequence (yn )n≥1 in Y and a sequence (xn )n≥1 in X such that r yn+1 = yn − T (xn ) ∈ OB Y 0, n+1 , (I.10) 2 1 (I.11) xn ∈ OB X 0, n+1 2 and yn − T (xn )
0 there exists x ∈ X such that x = 1 and T (x) < ε. Then T is not bijective. Proof. Suppose by contradiction that T is bijective. Then, by Corollary I.2.2, the inverse linear map T −1 : Y → X is continuous. Thus we can ﬁnd a constant M > 0 such that T −1 (y) ≤ M y for all y ∈ Y . Since T is bijective, this is equivalent to x ≤ M T (x) for all x ∈ X. This clearly contradicts the hypotheses. Thus, T is not bijective.
I.3 Spectra of Linear Maps Let (X, · ) be a Banach space. We denote by L(X) the space of all continuous linear maps T : X → X endowed with the norm T = sup x∈X x=0
T (x) . x
We denote by IdX : X → X the identity map. Deﬁnition I.3.1. Let T ∈ L(X). The set σ(T ) = {λ ∈ R : (T − λ IdX ) is not bijective}
(I.13)
is called the real spectrum of T . Proposition I.3.2. Let T ∈ L(X). Then the spectrum σ(T ) is a compact set and σ(T ) ⊂ [− T , T ]. (I.14) Proof. Let λ ∈ R and suppose that λ > T . For y ∈ X the equation (T − λ IdX )(x) = y
(I.15)
I.4 Uniform Convexity
407
admits a unique solution x ∈ X. Indeed (I.15) is equivalent to x=
1 (T (x) − y). λ
(I.16)
1 Moreover, λ1 (T (x1 )−y)− λ1 (T (x2 )−y) ≤ λ T · x1 −x2 for all x1 , x2 ∈ X 1 and λ T < 1. It then follows from the Banach ﬁxed point theorem that there exists a unique x ∈ X satisfying (I.16). It follows that T − λ IdX is bijective and therefore λ ∈ / σ(T ). This shows (I.14). Let us show that R \ σ(T ) is open. Let λ0 ∈ R \ σ(T ) so that T − λ0 IdX is bijective. By Corollary I.2.2, we have that the inverse map (T − λ0 IdX )−1 is also continuous so that 0 = (T − λ0 IdX )−1 < ∞. Let λ ∈ R such that λ − λ0  < (T −λ0 1IdX )−1 . For y ∈ X we have that the linear equation (I.15) can be written as T (x) − λ0 x = y + (λ − λ0 )x, that is,
x = (T − λ0 IdX )−1 [y + (λ − λ0 )x].
(I.17)
By applying again the Banach ﬁxed point theorem, we deduce that there exists a unique x ∈ X satisfying (I.17). This shows that T −λ IdX is bijective, that is, λ ∈ R\σ(T ). It follows that R\σ(T ) is open. Therefore σ(T ) is closed.
I.4 Uniform Convexity Deﬁnition I.4.1. A normed space (X, · ) is said to be uniformly convex if, for every ε > 0, there exists δ > 0 such that
x + y
ε implies 2 for all x, y ∈ X with x , y ≤ 1. Let Z be a nonempty set not reduced to a single point. Then the Banach space 1 (Z) is not uniformly convex. For instance, if z, z ∈ Z are distinct, setting x = δz and y = δz , one has x 1 = y 1 = 1, x − y 1 = 2, but x+y 2 1 = 1. On the other hand, we have the following: Proposition I.4.2. Let X be a vector space equipped with a scalar product
·, · and denote by · the associated norm. Then the normed space (X, · ) is uniformly convex. In particular, every Hilbert space is uniformly convex. Proof. Let x, y ∈ X such that x , y ≤ 1. Then, we have
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x + y 2 1 1 2
2 + 4 x − y = 4 ( x + y, x + y + x − y, x − y) 1 = ( x, x + y, y + 2 x, y + x, x + y, y − 2 x, y) 4 1 2 x 2 + 2 y 2 = 4 ≤ 1.
Therefore,
x + y 2 1 2
(I.18)
2 ≤ 1 − 4 x − y . √ Let now ε > 0 and set δ = 1 − 12 4 − ε2 . Suppose that x − y > ε. This 2 implies 1 − 14 x − y 2 < 1 − ε4 = (1 − δ)2 . From (I.18) we then deduce that x+y 2 < 1 − δ. This shows that X is uniformly convex.
Appendix J
Ultraﬁlters
J.1 Filters and Ultraﬁlters Let X be a set. We denote by P(X) the set of all subsets of X. Deﬁnition J.1.1. A ﬁlter on X is a nonempty set F ⊂ P(X) satisfying the following conditions: (F1) ∅ ∈ / F; (F2) if A ∈ F and A ⊂ B ⊂ X, then B ∈ F; (F3) if A ∈ F and B ∈ F, then A ∩ B ∈ F. Examples J.1.2. (a) Let X be a set and let A0 be a nonempty subset of X. Then the set {A ∈ P(X) : A0 ⊂ A} is a ﬁlter on X. A ﬁlter F on X is said to be a principal ﬁlter if there exists a nonempty subset A0 of X such that F = {A ∈ P(X) : A0 ⊂ A}. One then says that F is the principal ﬁlter based on A0 . (b) Let X be a set and Ω ⊂ P(X). Suppose that ∅ ∈ / Ω and that, given A1 , A2 ∈ Ω, there exists A ∈ Ω such that A ⊂ A1 ∩ A2 . Then the set F(Ω) = {A ∈ P(X) : there exists A ∈ Ω such that A ⊂ A}
(J.1)
is a ﬁlter on X and one has Ω ⊂ F(Ω). The ﬁlter F(Ω) is called the ﬁlter generated by Ω. (c) Let (I, ≤) be a nonempty directed set. A subset A ⊂ I is called residual in I if there exists i ∈ I such that A ⊃ {j ∈ I : i ≤ j}. Clearly, the set Fr (I) of all residual subsets of I is a ﬁlter on I. It is the ﬁlter generated by the sets {j ∈ I : i ≤ j}, i ∈ I. The ﬁlter Fr (I) is called the residual ﬁlter on I. (d) Let X be an inﬁnite set. Then the set F = {A ∈ P(X) : X \A is ﬁnite} is a ﬁlter. It is called the Fr´echet ﬁlter on X. If we consider the directed set (N, ≤), a subset A ⊂ N is residual if and only if its complement N \ A is ﬁnite. Therefore, the residual ﬁlter on N equals the Fr´echet ﬁlter on N. T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341 18, © SpringerVerlag Berlin Heidelberg 2010
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(e) Let X be a topological space and let x be a point in X. Then, the set Fx of all neighborhoods of x is a ﬁlter on X. (f) Let X be a nonempty uniform space (cf. Appendix B). Then the set U ⊂ P(X × X) of all entourages of X is a ﬁlter on X × X. Note that for a ﬁlter F ⊂ P(X) the following also holds true: (F4) if A ⊂ X, then A and X \ A cannotboth belong to F; n (F5) if A1 , A2 , . . . , An ∈ F, n ≥ 1, then i=1 Ai ∈ F; (F6) X ∈ F. Indeed, (F4) follows from (F1) and (F3) with B = X \ A. From (F3) we deduce by induction condition (F5). Finally, (F6) follows from the fact that F = ∅ and (F2). Proposition J.1.3. Let X be a set and Ω0 ⊂ P(X). Then, there exists a ﬁlter on X containing Ω0 if an only if Ω0 has the ﬁnite intersection property, that is, every ﬁnite family of elements in Ω0 has nonempty intersection. Proof. Let Ω0 ⊂ P(X) be a set and suppose that there exists a ﬁlter F such that Ω0 ⊂ F. By (F1) and (F5) we have that Ω0 has the ﬁnite intersection property. Conversely, suppose that Ω0 has the ﬁnite intersection property. Consider the set Ω consisting of all ﬁnite intersections of elements of Ω0 . Then, Ω satisﬁes the conditions in Example J.1.2(b), and therefore the ﬁlter
generated by Ω is a ﬁlter containing Ω0 . Deﬁnition J.1.4. A ﬁlter ω ⊂ P(X) is called an ultraﬁlter on X if it satisﬁes the condition (UF) if A ⊂ X, then A ∈ ω or (X \ A) ∈ ω. Example J.1.5. Let X be a set and x ∈ X. Then the principal ﬁlter based on {x} is an ultraﬁlter. An ultraﬁlter ω on X is called principal if there exists x ∈ X such that ω is the principal ﬁlter based on {x}. Note that a principal ﬁlter is an ultraﬁlter if and only if it is based on a singleton. If X is ﬁnite, then every ultraﬁlter is, clearly, principal. An ultraﬁlter which is not principal is called nonprincipal (or free). Theorem J.1.6. Let X be a nonempty set. Let F0 be a ﬁlter on X. Then there exists an ultraﬁlter ω on X containing F0 . Let us ﬁrst prove the following: Lemma J.1.7. Let X be a set. Let F be a ﬁlter on X. Suppose there exists a set A0 ∈ P(X) such that A0 ∈ / F and (X \ A0 ) ∈ / F. Then there exists a ﬁlter F on X such that
J.1 Filters and Ultraﬁlters
411
(1) F ⊂ F ; (2) A0 ∈ F . Proof. Consider the principal ﬁlter F0 based on A0 and set F = {B ∈ P(X) : B ⊃ A ∩ A for some A ∈ F and A ∈ F0 }.
(J.2)
Let us show that F satisﬁes the required conditions. First of all, taking A = X we have A = A ∩ X ∈ F for all A ∈ F. This shows (1). On the other hand, taking A = X and A = A0 we have A0 = X ∩ A0 ∈ F, and this shows (2). We are only left to show that F is a ﬁlter. Let B ∈ F and denote by A ∈ F and A ∈ F0 two sets such that A ∩ A ⊂ B. Let us ﬁrst show that A ∩ A = ∅. Suppose the contrary. Then A ⊂ (X \ A ) ⊂ (X \ A0 ). As A belongs to the ﬁlter F, it follows from (F2) that (X \ A0 ) ∈ F, contradicting our assumptions. It follows that A ∩ A = ∅ and therefore B = ∅. This shows (F1). Suppose now that B ∈ P(X) contains B. Then (A ∩ A ) ⊂ B so that B ∈ F . This shows (F2). Finally, let B1 , B2 ∈ F . For i = 1, 2 denote by Ai ∈ F and Ai ∈ F0 two sets such that Ai ∩ Ai ⊂ Bi . We have B1 ∩ B2 ⊃ (A1 ∩ A1 ) (A2 ∩ A2 ) = (A1 ∩ A2 ) (A1 ∩ A2 ). But A1 ∩ A2 belongs to the ﬁlter F, and A1 ∩ A2 ∈ F0 as both A1 and A2 contain A0 . It follows that B1 ∩ B2 ∈ F . This shows (F3). It follows that
F is a ﬁlter. Proof of Theorem J.1.6. Consider the set Φ0 consisting of all ﬁlters on X containing F0 . This is a nonempty set, partially orderedby inclusion. Let Φ be a totally ordered subset of Φ0 . We claim that F = F ∈Φ F is an upper bound for Φ. We only have to show that F belongs to Φ0 . As ∅ ∈ / F for all This shows that F satisﬁes condition (F1). Let F ∈ Φ we also have ∅ ∈ / F. now A ∈ F and B ⊂ X be such that A ⊂ B. Then there exists F ∈ Φ such that A ∈ F. As F is a ﬁlter, we have B ∈ F, by (F2). It then follows that This shows that F satisﬁes condition (F2). Finally, suppose that B ∈ F. Then there exist FA , FB ∈ Φ such that A ∈ FA and B ∈ FB . As Φ A, B ∈ F. is totally ordered, up to exchanging A and B we can suppose that FA ⊂ FB . We then have A, B ∈ FB and therefore A ∩ B ∈ FB , by (F3). It follows that Therefore F satisﬁes condition (F3) as well. This shows that Φ0 A ∩ B ∈ F. is inductive. By Zorn’s lemma, Φ0 contains a maximal element ω. Let us show that ω is an ultraﬁlter on X. Let A0 ⊂ X. Suppose that A0 , (X \ A0 ) ∈ / ω. Then, by Lemma J.1.7, there exists a ﬁlter ω containing A0 and ω. As ω is in Φ0 and properly contains ω, this contradicts the maximality of ω. It follows that either A0 or X \ A0 belongs to ω. This shows that ω satisﬁes condition (UF), and therefore it is an ultraﬁlter.
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Corollary J.1.8. Let X be a set. Let F be a ﬁlter on X. Then the following conditions are equivalent. (a) F is an ultraﬁlter; (b) F is a maximal ﬁlter, that is, if F is a ﬁlter containing F, then F = F. Proof. Suppose (a) and let F be a ﬁlter properly containing F. Let A ∈ F \ F. As F is an ultraﬁlter, (X \ A) ∈ F ⊂ F . Thus, both A and (X \ A) belong to F contradicting (F4). This shows that F = F and the implication (a) ⇒ (b) follows. Suppose now that F is a maximal ﬁlter. By Theorem J.1.6 there exists an ultraﬁlter ω containing F. By maximality we have F = ω, that is, F is an ultraﬁlter. This shows (b) ⇒ (a).
J.2 Limits Along Filters Deﬁnition J.2.1. Let X be a topological space. A ﬁlter F on X is said to be convergent if there exists a point x0 ∈ X such that all neighborhoods of x0 belong to F. One then says that x0 is a limit of F and that F converges to x0 . Remark J.2.2. A topological space X is Hausdorﬀ if and only if every convergent ﬁlter on X has a unique limit point in X. Indeed, suppose that X is Hausdorﬀ and let F be a ﬁlter converging to two distinct points x, y ∈ X. Let U ∈ Fx ⊂ F and V ∈ Fy ⊂ F be such that U ∩ V = ∅. As U, V ∈ F we also have U ∩ V ∈ F and this contradicts (F1). Conversely, suppose that X is not Hausdorﬀ. Then there exist two distinct points x, y ∈ X such that for all U ∈ Fx and V ∈ Fy one has U ∩ V = ∅. It follows that the set Ω = Fx ∪ Fy ⊂ P(X) has the ﬁnite intersection property. Thus, by Proposition J.1.3, there exists a ﬁlter F containing both Fx and Fy . It follows that F converges to both x and y. Theorem J.2.3. Let X be a topological space. Then the following conditions are equivalent: (a) X is compact; (b) every ultraﬁlter on X is convergent. Proof. Suppose (a) and let ω be an ultraﬁlter on X. Let A1 , A2 , . . . , An , n elements in ω and denote by A1 , A2 , . . . , An their closures. As n≥ 1, be n A ⊃ i=1 i i=1 Ai = ∅ (by (F1) and (F5)), we have that the family (A)A∈ω in X has the ﬁnite intersection property andtherefore, by compactness of X, it has a nonempty intersection. Let x ∈ A∈ω A. As x ∈ A for all A ∈ ω, it follows that for every V ∈ Fx we have A ∩ V = ∅. Consider the set Ω = {A ∩ V : A ∈ ω, V ∈ Fx } ⊂ P(X). Observe that Ω has the ﬁnite intersection property and denote by F(Ω) the ﬁlter generated by Ω
J.2 Limits Along Filters
413
(cf. Example J.1.2(b)). We have ω ⊂ Ω ⊂ F(Ω) and, as ω is an ultraﬁlter, it follows from Corollary J.1.8 that ω = F(Ω). As Fx ⊂ Ω, we deduce that Fx ⊂ ω, that is, ω converges to x. Conversely, suppose (b). Let (Ci )i∈I be a family of closed subsets of X with the ﬁnite intersection property. To prove that X is compact we have to show that Ci = ∅. (J.3) i∈I
By Proposition J.1.3 there exists a ﬁlter F such that Ci ∈ F for all i ∈ I. By Theorem J.1.6 there exists an ultraﬁlter ω such that F ⊂ ω. By our assumptions, there exists x ∈ X such that ω converges to x, equivalently, have Ci ∩ V = ∅ for all V ∈ Fx . Fx ⊂ ω. Let i ∈ I. by (F1) and (F3), we As Ci is closed, we have x ∈ Ci . Thus x ∈ i∈I Ci , and (J.3) follows.
Deﬁnition J.2.4. Let X be a set, Y a topological space, y0 a point of Y , f : X → Y a map and F a ﬁlter on X. One says that y0 is a limit of f along F (or that f (x) converges to y0 along F), and one writes f (x) −→ y0 , x→F
if f −1 (V ) = {x ∈ X : f (x) ∈ V } belongs to F for all neighborhoods V of y0 . If such a limit point y0 is unique one writes lim f (x) = y0 .
x→F
Examples J.2.5. (a) Let X and Y be two topological spaces, f : X → Y a map, x0 ∈ X and y0 ∈ Y . One has that f (x) converges to y0 in Y for x tending to x0 if and only if f (x) −→ y0 . x→Fx0
(b) Let X be a topological space and f : N → X a map. The sequence (f (n))n∈N converges to x ∈ X, if and only if x is a limit of f along the Fr´echet ﬁlter on N. (c) Let (I, ≤) be a directed set and let F be the residual ﬁlter on I. Let Y be a topological space and (yi )i∈I a net in Y . Then (yi )i∈I converges to a point y0 ∈ Y if and only if yi −→ y0 . Note that (b) is a particular case of i→F
the present example. Corollary J.2.6. Let X be a set, Y a compact topological space, f : X → Y a map, and ω an ultraﬁlter on X. Then there exists y0 ∈ Y such that f (x) −→ y0 . Moreover, if X is Hausdorﬀ such an y0 is unique. x→F
Proof. The set f (ω) = {f (A) : A ∈ ω} ⊂ P(Y ) has the ﬁnite intersection property since f (A)∩f (B) ⊃ f (A∩B) = ∅ for all A, B ∈ ω. Let F denote the ﬁlter generated by f (ω). By Theorem J.1.6, there exists an ultraﬁlter ω on Y which contains F. As Y is compact, by Theorem J.2.3 there exists y0 ∈ Y such that ω converges to y0 . Let us show that if V is a neighborhood of y0 , then
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f −1 (V ) belongs to ω. Suppose the contrary. As ω is an ultraﬁlter, by (UF) / V } ∈ ω. Setting U = f (X \ f −1 (V )), we have X \ f −1 (V ) = {x ∈ X : f (x) ∈ we have U ∈ f (ω) ⊂ F ⊂ ω . But V ∈ Fy0 ⊂ ω and, by construction, U ∩ V = ∅. As U ∩ V ∈ ω , this contradicts (F1). This shows that y0 is a limit of f along ω. If X is Hausdorﬀ, then uniqueness of y0 follows from Remark J.2.2.
Proposition J.2.7. Let X be a set and let F be a ﬁlter on X. Let Y , Z be two topological spaces and f : X → Y and g : X → Z be two maps. Equip Y × Z with the product topology and let F : X → Y × Z be the map deﬁned by F (x) = (f (x), g(x)) for all x ∈ X. Suppose that there exist y0 ∈ Y and z0 ∈ Z such that f (x) −→ y0 and g(x) −→ z0 . Then F (x) −→ (y0 , z0 ). x→F
x→F
x→F
Proof. Let W ⊂ Y × Z be a neighborhood of the point (y0 , z0 ). By deﬁnition of the product topology, there exist neighborhoods U ⊂ Y of y0 and V ⊂ Z of z0 such that U × V ⊂ W . As f (x) −→ y0 , we have f −1 (U ) ∈ F. Similarly, as x→F
g(x) −→ z0 , we have g −1 (V ) ∈ F. It follows that f −1 (U ) ∩ g −1 (V ) belongs x→F
to F and as F −1 (W ) ⊃ F −1 (U × V ) = f −1 (U ) ∩ g −1 (V ), we deduce from
(F2) that F −1 (W ) ∈ F. This shows that F (x) −→ (y0 , z0 ). x→F
Proposition J.2.8. Let X be a set and let F be a ﬁlter on X. Let Y , Z be two topological spaces and f : X → Y and g : Y → Z be two maps. Suppose that there exists y0 ∈ Y such that f (x) −→ y0 and that g is continuous at y0 . Then (g ◦ f )(x) −→ g(y0 ).
x→F
x→F
Proof. Let W ⊂ Z be a neighborhood of g(y0 ). By continuity of g there exists a neighborhood V ⊂ Y of y0 such that g −1 (W ) ⊃ V . By deﬁnition of limit, we have f −1 (V ) ∈ F. As (g ◦ f )−1 (W ) = f −1 (g −1 (W )) ⊃ f −1 (V ), it follows
from (F2) that (g ◦ f )−1 (W ) ∈ F. This shows that (g ◦ f )(x) −→ g(y0 ). x→F
Proposition J.2.9. Let X be a set and let F be a ﬁlter on X. Let f1 , f2 : X → R be two maps such that f1 ≤ f2 (i.e. f1 (x) ≤ f2 (x) for all x ∈ X). Suppose that there exists y1 ∈ R (resp. y2 ∈ R) such that y1 = limx→F f1 (x) (resp. y2 = limx→F f2 (x)). Then y1 ≤ y2 . 2 and let V1 = Proof. Suppose, by contradiction, that y1 > y2 . Set r = y1 −y 3 (y1 −r, y1 +r) and V2 = (y2 −r, y2 +r). Note that V1 ∩V2 = ∅. By deﬁnition of limit we have f1−1 (V1 ) ∈ F and f2−1 (V2 ) ∈ F. As F is a ﬁlter we deduce that f1−1 (V1 ) ∩ f2−1 (V2 ) ∈ F. On the other hand we have f1−1 (V1 ) ∩ f2−1 (V2 ) = ∅
since f1 ≤ f2 . This contradicts (F2). It follows that y1 ≤ y2 .
Recall that given a set X, we denote by ∞ (X) the Banach space consisting of all bounded real maps f : X → R equipped with the norm f ∞ = sup{f (x) : x ∈ X}. Moreover, a linear map m : ∞ (X) → R satisfying m(1) = 1 and m(x) ≥ 0 for all x ∈ ∞ (E) such that x ≥ 0, is called a mean on X (cf. Deﬁnition 4.1.4).
Notes
415
Corollary J.2.10. Let X be a set and let ω be an ultraﬁlter on X. For every f ∈ ∞ (X) there exists a unique y0 ∈ R such that f (x) −→ y0 . Moreover, x→ω
the map mω : ∞ (X) → R deﬁned by mω (f ) = limx→ω f (x) is a mean on X, in particular mω is continuous and mω = 1. Proof. Let f ∈ ∞ (X). Using the fact that f (x) ∈ Y = [−f ∞ , f ∞ ] for all x ∈ X and that Y is compact Hausdorﬀ, we deduce from Corollary J.2.6 that there exists a unique y0 ∈ R such that f (x) −→ y0 . x→ω Let us show that the map mω is linear. Let a ∈ R. Consider the continuous map g : R → R deﬁned by g(y) = ay for all y ∈ R. Then af = g ◦ f and from Proposition J.2.8 we deduce mω (af ) = lim (af )(x) = a lim f (x) = amω (f ). x→ω
i→ω
Let now f, g ∈ ∞ (X). Consider the map F : X → R2 deﬁned by setting F (x) = (f (x), g(x)) for all x ∈ X and the continuous map G : R2 → R deﬁned by G(y1 , y2 ) = y1 + y2 . Then f + g = G ◦ F and from Proposition J.2.7 and Proposition J.2.8 we deduce mω (f + g) = lim (f + g)(x) = lim f (x) + lim g(x) = mω (f ) + mω (g). x→ω
x→ω
x→ω
This shows that mω is linear. Let now f (x) = 1 for all x ∈ X. For every neighborhood V of 1 ∈ R we have f −1 (V ) = {x ∈ X : f (x) ∈ V } = X ∈ ω, by (F6). This shows that mω (1) = mω (f ) = limx→ω f (x) = 1. Finally, it follows immediately from Proposition J.2.9 that if f ≥ 0 then mω (f ) = limx→ω f (x) ≥ 0. We have shown that mω is a mean. The last properties of mω follow from Proposition 4.1.7.
Notes The deﬁnition of ﬁlter is due to H. Cartan (1937). The full treatment of convergence along ﬁlters is given in Bourbaki [Bou] as an alternative to the similar notion of a net developed in 1922 by E. H. Moore and H. L. Smith. Given a Hausdorﬀ topological space X the set βX of all ultraﬁlters on X can be given the structure of a compact Hausdorﬀ space, called the Stoneˇ Cech compactiﬁcation of X. This is the largest compact Hausdorﬀ space “generated” by X, in the sense that any map from X to a compact Hausdorﬀ space factors through βX in a unique way. The elements x of X correspond to the principal ultraﬁlters ω({x}) on X. Let now X be any set. With every ultraﬁlter ω on X one associates the {0, 1}valued ﬁnitely additive probability measure μω on X deﬁned
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by μω (A) = 1 if A ∈ ω and μω (A) = 0 if A ∈ P(X) \ ω. Conversely, given a {0, 1}valued ﬁnitely additive probability measure μ on X, the set ωμ = {A ∈ P(X) : μ(A) = 1} is an ultraﬁlter on X. This establishes a onetoone correspondence between the ultraﬁlters ω on X and the {0, 1}valued ﬁnitely additive probability measures on X. In Sect. 4.1 we considered the set MP(X) (resp. M(X)) of all ﬁnitely additive probability measures (resp. of all means) on X and we showed that there exists a natural bijective map Φ : MP(X) → M(X). Then one has Φ−1 (μω ) = mω for all ultraﬁlters ω on X.
Open Problems
In the list below we collect some open problems related to the topics treated in this book. (OP1) Let G be an amenable periodic group which is not locally ﬁnite. Does there exist a ﬁnite set A and a cellular automaton τ : AG → AG which is surjective but not injective? (OP2) Let G be a periodic group which is not locally ﬁnite and let A be an inﬁnite set. Does there exist a bijective cellular automaton τ : AG → AG which is not invertible? (OP3) Let G be a periodic group which is not locally ﬁnite and let V be an inﬁnitedimensional vector space over a ﬁeld K. Does there exist a bijective linear cellular automaton τ : V G → V G which is not invertible? (OP4) Is every Gromovhyperbolic group residually ﬁnite (resp. residually amenable, resp. soﬁc, resp. surjunctive)? (OP5) (Gottschalk’s conjecture) Is every group surjunctive? (OP6) Let G be a periodic group which is not locally ﬁnite and let A be an inﬁnite set. Does there exist a cellular automaton τ : AG → AG whose image τ (AG ) is not closed in AG with respect to the prodiscrete topology? (OP7) Let G be a periodic group which is not locally ﬁnite and let V be an inﬁnitedimensional vector space over a ﬁeld K. Does there exist a linear cellular automaton τ : V G → V G whose image τ (V G ) is not closed in V G with respect to the prodiscrete topology? (OP8) Let G and H be two quasiisometric groups. Suppose that G is surjunctive. Is it true that H is surjunctive? (OP9) Let G be a nonamenable group. Does there exist a ﬁnite set A and a cellular automaton τ : AG → AG which is preinjective but not surjective? (OP10) Does there exist a nonsoﬁc group? (OP11) Does there exist a surjunctive group which is nonsoﬁc? T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341, © SpringerVerlag Berlin Heidelberg 2010
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(OP12) Let G and H be two quasiisometric groups. Suppose that G is soﬁc. Is it true that H is soﬁc? (OP13) Let G be a nonamenable group and let K be a ﬁeld. Does there exist a ﬁnitedimensional Kvector space V and a linear cellular automaton τ : V G → V G which is preinjective but not surjective? (OP14) Let G be a nonamenable group and let K be a ﬁeld. Does there exist a ﬁnitedimensional Kvector space V and a linear cellular automaton τ : V G → V G which is surjective but not preinjective? (OP15) (Kaplanski’s stable ﬁniteness conjecture) Is the group algebra K[G] stably ﬁnite for any group G and any ﬁeld K? Equivalently, is every group Lsurjunctive, that is, is it true that, for any group G, any ﬁeld K, and any ﬁnitedimensional Kvector space V , every injective linear cellular a automaton τ : V G → V G is surjective? (OP16) (Kaplanski’s zerodivisors conjecture) Is it true that the group algebra K[G] has no zerodivisors for any torsionfree group G and any ﬁeld K? Equivalently, is it true that, for any torsionfree group G and any ﬁeld K, every nonidenticallyzero linear cellular automaton τ : KG → KG is preinjective? (OP17) Is every uniqueproduct group orderable?
Comments (OP1) The answer to this question is aﬃrmative if G is nonperiodic, i.e., it contains an element of inﬁnite order (see Exercise 3.23), or if G is nonamenable (Theorem 5.12.1). On the other hand, if G is a locally ﬁnite group and A is a ﬁnite set, then every surjective cellular automaton τ : AG → AG is injective (see Exercise 3.21). An example of an amenable periodic group which is not locally ﬁnite is provided by the Grigorchuck group described in Sect. 6.9. (OP2) The answer is aﬃrmative if G is not periodic (cf. [CeC11, Corollary 1.2]). On the other hand, if G is locally ﬁnite and A is an arbitrary set, then every bijective cellular automaton τ : AG → AG is invertible (cf. Exercise 3.20 or [CeC11, Proposition 4.1]). (OP3) The answer is aﬃrmative if G is not periodic (cf. [CeC11, Theorem 1.1]). On the other hand, if G is locally ﬁnite and V is an arbitrary vector space, then every bijective linear cellular automaton τ : V G → V G is invertible (cf. [CeC11, Proposition 4.1]). (OP5) Every soﬁc group is surjunctive (cf. Theorem 7.8.1). (OP6) When A is a ﬁnite set and G is an arbitrary group, it follows from Lemma 3.3.2 that the image of every cellular automaton τ : AG → AG is closed in AG . When A is an inﬁnite set and G is a nonperiodic group, it is shown in [CeC11, Corollary 1.4] that there exists a cellular automaton τ : AG → AG whose image is not closed in AG . On the other hand, when G
Comments
419
is locally ﬁnite, then, for any set A, the image of every cellular automaton τ : AG → AG is closed in AG (cf. Exercise 3.22 or [CeC11, Proposition 4.1]). (OP6) When V is a ﬁnitedimensional vector space over a ﬁeld K and G is an arbitrary group, it follows from Theorem 8.8.1 that the image of every linear cellular automaton τ : V G → V G is closed in V G . When V is an inﬁnitedimensional vector space and G is a nonperiodic group, it is shown in [CeC11, Theorem 1.3] that there exists a linear cellular automaton τ : V G → V G whose image is not closed in V G . On the other hand, when G is locally ﬁnite, then, for any vector space V , the image of every linear cellular automaton τ : V G → V G is closed in V G (cf. [CeC11, Proposition 4.1]). (OP9) The answer is aﬃrmative if G contains a nonabelian free subgroup (cf. Proposition 5.11.1). (OP13) The answer is aﬃrmative if G contains a nonabelian free subgroup (cf. Corollary 8.10.2). (OP14) The answer is aﬃrmative if G contains a nonabelian free subgroup (cf. Corollary 8.11.2). (OP15) See the discussion in the notes at the end of Chap. 8. (OP16) See the discussion in the notes at the end of Chap. 8.
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List of Symbols
Symbol ∼ T p→p 0R , 0 1G 1R , 1 γSG , γS ΔG S , ΔS (2)
ΔS
∂E (Ω) ιS (G) λG S , λS λ(e) λ(π) π−
Deﬁnition Page the empty word 367 the dominance relation in the set of growth functions γ : N → [0, +∞) 162 the equivalence relation in the set of growth functions γ : N → [0, +∞) 162 the p norm of a linear map 194 T : p (E) → p (E) the zero element of the ring R 291 the identity element of the group G 2 the unity element of the ring R 291 the growth function of the group G relative to the ﬁnite symmetric generating subset S⊂G 160 the discrete laplacian on the group G associated with the subset S ⊂ G 9 the restriction of ΔS to the Hilbert space 201 2 (G) the Eboundary of the subset Ω ⊂ G 116 the isoperimetric constant of the group G with respect to the ﬁnite symmetric generating subset S ⊂ G 191 the growth rate of the group G with respect to the ﬁnite symmetric generating subset S⊂G 169 the label of the edge e ∈ E in a labeled graph G = (Q, E) 153 the label of the path π in a labeled graph G = (Q, E) 154, 223 the initial vertex of the path π in an Slabeled graph 154
T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341, © SpringerVerlag Berlin Heidelberg 2010
429
430
Symbol π+ σ(T ) ψq,r Ω −E Ω +E A∗ AG B(q, n) BSG (g, n), BS (g, n) BSG (n), BS (n) CA(G; A) CA(G, H; A)
(C i (G))i≥0 CS (G)
dF dG S , dS dQ D(G) (Di (G))i≥0 entF (X) F (X) Fn Fix(α) Fix(H) G = X; R
List of Symbols
Deﬁnition the terminal vertex of the path π in an Slabeled graph the real spectrum of T ∈ L(X) the Slabeled graph isomorphism from BS (r) onto B(q, r) such that ψq,r (1G ) = q the Einterior of the subset Ω ⊂ G the Eclosure of the subset Ω ⊂ G the monoid consisting of all words on the alphabet A the set of all conﬁgurations x : G → A the ball of radius n in an Slabeled graph Q = (Q, E) centered at the vertex q ∈ Q the ball of radius n in G centered at the element g ∈ G with respect to the word metric the ball of radius n in G centered at the identity element 1G ∈ G with respect to the word metric the monoid consisting of all cellular automata τ : AG → AG the submonoid of CA(G; A) consisting of all cellular automata τ : AG → AG admitting a memory set S such that S ⊂ H the lower central series of the group G the Cayley graph of the group G with respect to the ﬁnite symmetric generating subset S ⊂ G the normalized Hamming distance on Sym(F ) the word metric on G with respect to the ﬁnite symmetric generating subset S ⊂ G the graph metric in the edgesymmetric Slabeled graph Q the derived subgroup of the group G the derived series of the group G the entropy of the subset X ⊂ AG with respect to the right Følner net F the free group based on the set X the free group of rank n the set of ﬁxed points of the permutation α the set of conﬁgurations x ∈ AG ﬁxed by H the presentation of the group G given by the generating subset X and the set of relators R
Page 154 406 265 115 115 367 2 265
153
153 13
16 93
156 251 152 155 92 92 125 371 372 251 4
375
List of Symbols
Symbol gx G = (X, Y, E) HR ICA(G; A) IdX (w) p (E) ∞ (E) G S (g), S (g) LCA(G; V ) LCA(G, H; V )
L(X) Ln (X) L(X) (p) MS
Matd (R) mdimF (X) M(E) PM(E) N (Γ ) NL (B) ⊂ X NL (y) ⊂ X
431
Deﬁnition the conﬁguration deﬁned by gx(h) = x(g −1 h) the bipartite graph with left (resp. right) vertex set X (resp. Y ) and set of edges E the Heisenberg group with coeﬃcients in the ring R the group consisting of all invertible cellular automata τ : AG → AG the identity map on the set X the length of the word w ∈ A∗ the Banach space of all psummable functions x : E → R the Banach space of all bounded functions x: E → R the wordlength of the element g ∈ G with respect to the ﬁnite symmetric generating subset S ⊂ G the algebra of all linear cellular automata τ: VG →VG the subalgebra of LCA(G; V ) consisting of all linear cellular automata τ : V G → V G admitting a memory set S such that S ⊂ H the language associated with the subshift X the set of admissible words of length n of the subshift X the space of all continous endomorphisms of the Banach space X the p Markov operator associated with the ﬁnite subset S ⊂ G the ring consisting of all d × d matrices with entries in the ring R the mean dimension of the vector subspace X ⊂ V G with respect to the right Følner net F the set of all means on the set E the set of all ﬁnitely additive probability measures on the set E the space of all normal subgroups of the group Γ or, equivalently, the space of all Γ marked groups the leftneighborhood of the subset B ⊂ Y in the bipartite graph (X, Y, E) the leftneighborhood of the vertex y ∈ Y in the bipartite graph (X, Y, E)
Page 2 391 94 24 2 35 193 78
152 287
289 35 35 406 195 305
308 79 79
61 392 391
432
Symbol NR (A) ⊂ Y NR (x) ⊂ Y P(E) per(B) per(G) per(X) pern (X) Q(r)
Q = (Q, E) R[G] Rop Sym(X) Sym0 (X) Sym+ 0 (X) Symn Sym+ n Sym(X, ) U (R) V [G] xΩ X∗ X(A) Xf XP XG Z(G)
List of Symbols
Deﬁnition Page the rightneighborhood of the subset A ⊂ X in the bipartite graph (X, Y, E) 392 the rightneighborhood of the vertex x ∈ X in the bipartite graph (X, Y, E) 391 the set of all subsets of the set E 77 the period of the matrix B 227 the period of the labeled graph G 227 the period of the irreducible soﬁc subshift X 227 the number of nZperiodic conﬁgurations in the subshift X 227 the set of all vertices of the Slabeled graph Q = (Q, E) for which there exists an Slabeled graph isomorphism ψq,r : BS (r) → B(q, r) satisfying ψq,r (1G ) = q 265 the Slabeled graph with vertex set Q and edge set E ⊂ Q × S × Q 153 the group ring of the group G with coeﬃcients in the ring R 292 the opposite ring of the ring R 293 the symmetric group of the set X 359 the subgroup of Sym(X) consisting of all permutations with ﬁnite support 360 the alternating group on X 364 the symmetric group of degree n 366 the alternating group of degree n 366 the subgroup of Sym(X) that preserve the partial order of the set X 179, 333 the multiplicative group consisting of all invertible elements in the ring R, 292 the vector subspace of V G consisting of all conﬁgurations x : V → G with ﬁnite support 288 the restriction of the conﬁguration x ∈ AG to the subset Ω ⊂ G 3 the topological dual of the real normed space X 384 the subshift of ﬁnite type deﬁned by the set of admissible patterns A 32 the set of all conﬁgurations in the subshift X whose Gorbit is ﬁnite 71 the subshift deﬁned by the set of forbidden patterns P 32 the subshift deﬁned by the labeled graph G 223 the center of the group G 94
Index
Δirreducible subshift, 34 action continuous —, 3 equivariantly approximable —, 279 expansive —, 65 faithful —, 50 topologically mixing —, 29 topologically transitive —, 32 uniformly continuous —, 64 additive cellular automaton, 335 adjacency matrix of a labeled graph, 227 admissible pattern, 32 admissible word, 35 aﬃne — group, 93 — map, 387 algebra, 286 — homomorphism, 289 — isomorphism, 290 almost — homomorphism, 234, 254 — equal conﬁgurations, 112 — perfect group, 55 — periodic conﬁguration, 72 alphabet, 2 alternating group, 364 — of rank n, 366 amenable — group, 87 elementary — group, 215, 338 Artinian module, 69 automaton additive cellular —, 335 cellular —, 6 linear cellular —, 284 automorphism group, 45
backtracking, 156 Baire theorem, 403 BanachAlaoglu theorem, 385 base — of a uniform structure, 353 free —, 368 based free group, 367 biinvariant metric, 251 biorderable group, 341 bipartite — graph, 391 — subgraph, 391 ﬁnite — graph, 392 locally ﬁnite — graph, 392 Boolean ring, 340 boundary, 116 Burnside problem, 214 CantorBernstein theorem, 398 Cayley graph, 156 cellular automaton, 6 additive —, 335 induced —, 17 invertible —, 24 linear —, 284, 335 reversible —, 24 characteristic map, 79 closed path, 155 closure, 115 cluster point of a net, 345 color, 2 commensurable groups, 171 commutativetransitive group, 279 commutator — of two group elements, 92 — subgroup, 92 simple —, 176
T. CeccheriniSilberstein, M. Coornaert, Cellular Automata and Groups, Springer Monographs in Mathematics, DOI 10.1007/9783642140341, © SpringerVerlag Berlin Heidelberg 2010
433
434 compact topological space, 347 completion proamenable —, 108 pronilpotent —, 108 prosolvable —, 108 complexity of a paradoxical decomposition, 106 composition of paths, 154 concatenation, 367 conﬁguration, 2 Hperiodic —, 3 almost periodic —, 72 Garden of Eden —, 111 language of a —, 73 Toeplitz —, 74 conjugate elements, 362 connected labeled graph, 155 Connes embedding conjecture, 277 contextfree subshift, 225 convergent net, 344 convex subset, 383 convolution product, 291 convolutional encoders, 335 CurtisHedlund theorem, 20 cycle, 361 Day’s problem, 105 Dedekind ﬁnite ring, 336 degree — of a graph, 155 — of a symmetric group, 366 — of a vertex, 155 — of an alternating group, 366 derived — series, 92 — subgroup, 92 directed set, 343 directly ﬁnite ring, 327 discrete uniform structure, 352 divisible group, 38 dominance of growth functions, 162 edge — symmetric labeled graph, 155 — of a bipartite graph, 391 inverse —, 155 of a labeled graph, 153 elementary — amenable group, 215, 338 — reduction, 370 empty — path, 154 — word, 367
Index entourage, 352 entropy topological —, 142 equipotent sets, 372 equivalence of growth functions, 162 equivariant map, 5 equivariantly approximable action, 279 even subshift, 35 expansive action, 65 expansivity constant, 65 entourage, 65 exponential growth, 164 Følner — conditions, 96 — theorem, 99 left — net, 96 left — sequence, 96 right — net, 96 right — sequence, 96 faithful action, 50 Fibonacci sequence, 219 ﬁeld, 284 ﬁlter, 409 — generated, 409 convergent —, 412 Fr´ echet —, 409 limit of a —, 412 principal —, 409 residual —, 409 ultra—, 410 ﬁnite intersection property, 347 ﬁnitely — additive probability measure, 77 — generated group, 152, 375 — presented group, 376 biinvariant — additive probability measure, 85 leftinvariant — additive probability measure, 85 rightinvariant — additive probability measure, 85 forbidden — pattern, 32 — word, 35 Fr´ echet ﬁlter, 409 free — base, 368 — base subset, 368 — group, 368 — group of rank k, 372 — ultraﬁlter, 410 based — group, 367
Index rank of a — group, 373 fully residually free group, 279 Garden of Eden — conﬁguration, 111 — pattern, 112 — theorem, 114, 128 — theorem for linear cellular automata, 312 generating subset, 151 generator of a presentation, 375 golden mean subshift, 35 graph — metric, 155 bipartite —, 391 Cayley —, 156 degree of a regular labeled —, 155 ﬁnite labeled —, 154 labeled —, 153 loop in a labeled —, 154 regular labeled —, 155 tree, 156 Grigorchuk group, 179 abelianization of the —, 222 group — algebra, 294 — of padic integers, 41 — of intermediate growth, 190 — ring, 292 aﬃne —, 93 almost perfect —, 55 alternating —, 364 alternating — of rank n, 366 amenable —, 87 automorphism —, 45 biorderable —, 341 Cayley graph of a —, 156 commutativetransitive —, 279 divisible —, 38 elementary amenable —, 215, 338 ﬁnitely generated —, 152, 375 ﬁnitely presented —, 376 free —, 368 free — of rank k, 372 fully residually free —, 279 Grigorchuk —, 179 Heisenberg —, 94 Hopﬁan —, 44 hyperlinear —, 277 Kaloujnine —, 221 Klein bottle —, 341 Lsurjunctive —, 324 lamplighter —, 108 LEA —, 247
435 LEF —, 247 linear —, 51 locally P —, 58 locally indicable —, 337 marked —, 61 metabelian —, 92 nilpotent —, 93 orderable —, 331 periodic —, 29, 105 polycyclic —, 106, 108 presentation of a —, 375 proﬁnite —, 41 residually C —, 238 residually P —, 62, 131 residually amenable —, 132 residually ﬁnite —, 37 simple —, 44 soﬁc —, 254 solvable —, 93 surjunctive —, 57 symmetric —, 359 symmetric — of rank n, 366 uniqueproduct —, 331 virtually P —, 41 growth — function, 160, 162 — rate, 169 — type of a group, 163 equivalence class of — functions, 163 equivalence of — functions, 162 exponential —, 164 intermediate —, 190 polynomial —, 164 subexponential —, 164 Hall — kharem conditions, 399 — condition, 394 — harem theorem, 399 — marriage theorem, 399 Hamming metric, 252 Hausdorﬀ metric, 357 HausdorﬀBourbaki — topology, 356 uniform structure, 356 Heisenberg group, 94 homomorphism almost —, 234, 254 labeled graph —, 154 Hopﬁan — group, 44 — module, 339 hyperlinear group, 277
436 ICCproperty, 338 idempotent, 331 proper —, 331 induced — cellular automaton, 17 — labeled subgraph, 154 inductive — limit, 379 — system of groups, 379 initial — topology, 346 — uniform structure, 355 interior, 115 intermediate growth, 190 inverse — edge, 155 — path, 155 invertible cellular automaton, 24 irreducible — matrix, 227 — subshift, 32 isomorphism labeled graph —, 154 isoperimetric constant, 191 Kaloujnine group, 221 abelianization of the —, 222 KestenDay theorem, 201 Klein bottle group, 341 Klein PingPong theorem, 376 Lsurjunctive group, 324 labeled graph, 153 — homomorphism, 154 — isomorphism, 154 adjacency matrix of a —, 227 connected —, 155 edgesymmetric —, 155 ﬁnite —, 154 locally ﬁnite —, 155 path in a —, 154 subgraph, 154 subshift deﬁned by a —, 223 labelling map, 153 lamplighter group, 108 language — of a conﬁguration, 73 — of a subshift over Z, 35 Laplacian, 10 lattice, 95 LEAgroup, 247 LEFgroup, 247 leftinvariant
Index — ﬁnitely additive probability measure, 85 — metric, 251 length — of a cycle, 361 — of a word, 35, 152 letter, 2 limit — along an ultraﬁlter, 413 — of a ﬁlter, 412 — point of a net, 344 inductive —, 379 projective —, 380 linear — cellular automaton, 284, 335 — group, 51 Lipschitzequivalence, 162 local deﬁning map, 6 locally — P group, 58 — convex topological vector space, 383 — embeddable, 235 — ﬁnite bipartite graph, 392 — ﬁnite labeled graph, 155 — indicable group, 337 loop, 154 lower central series, 93 majority action, 10 marked group, 61 Markov operator, 195 MarkovKakutani theorem, 387 matching, 393 leftperfect —, 393 perfect —, 393 rightperfect—, 393 mean, 78 — dimension, 308 biinvariant —, 86 leftinvariant —, 86 rightinvariant —, 86 memory set, 6 minimal —, 15 metabelian group, 92 metric biinvariant —, 251 graph —, 155 Hamming —, 252 word —, 153 metrizable uniform structure, 352 Milnor problem, 215 minimal — memory set, 15 — set, 72
Index — subshift, 72 module Artinian —, 69 Hopﬁan —, 339 Noetherian —, 339 projective —, 339 monoid, 13 Moore neighborhood, 166 Morse subshift, 74 entropy of the —, 144 neighbor, 155 net, 343 nilpotency degree, 93 nilpotent group, 93 Noetherian — module, 339 — ring, 340 nonprincipal ultraﬁlter, 410 normal closure, 375 Open mapping theorem, 404 operator norm, 384 opposite ring, 293 orderable group, 331 Ore ring, 340 paradoxical decomposition left —, 98 right —, 98 partially ordered set, 343 path — in a labeled graph, 154 closed —, 155 closed simple —, 156 composition, 154 empty —, 154 inverse —, 155 label of a —, 154 proper —, 156 simple —, 156 pattern, 2 admissible —, 32 forbidden —, 32 periodic group, 29, 105 permutation, 359 support of a —, 360 polycyclic group, 106, 108 polynomial, 164 preinjective map, 112 presentation — of a group, 375 generator of a —, 375 relator of a —, 375
437 principal — ﬁlter, 409 — ultraﬁlter, 410 proamenable completion, 108 prodiscrete — topology, 3, 346 — uniform structure, 22, 355 product — topology, 346 — uniform structure, 355 proﬁnite — completion, 55 — group, 41 — kernel, 39 — topology, 53 projective — limit, 380 — module, 339 — system of groups, 380 pronilpotent completion, 108 proper — idempotent, 331 — path, 156 prosolvable completion, 108 quasiisometric — embedding, 204 — groups, 206 quasiisometry, 204 rank of a free group, 373 reduced — form, 374 — product, 244 — word, 373 regular labeled graph, 155 relator of a presentation, 375 residual — ﬁlter, 409 — set, 409 — subgroup, 39 residually — C group, 238 — P group, 62, 131 — amenable group, 132 — ﬁnite group, 37 restriction, 17 reversible cellular automaton, 24 rightinvariant — ﬁnitely additive probability measure, 85 — metric, 251 ring Boolean —, 340 Dedekind ﬁnite —, 336
438 directly ﬁnite —, 327 group —, 292 Noetherian —, 340 opposite —, 293 Ore —, 340 stably ﬁnite —, 328 unitregular—, 340 von Neumann ﬁnite —, 336 set directed —, 343 partially ordered —, 343 shift, 2 simple — group, 44 — path, 156 closed — path, 156 soﬁc — subshift, 225 — group, 254 solvable group, 93 spectrum real —, 406 stably ﬁnite ring, 328 state, 2 strong topology, 82, 384 strongly irreducible subshift, 34 subalgebra, 287 subexponential growth, 164 subgraph induced labeled —, 154 labeled —, 154 submonoid, 17 subnet, 344 subsemigroup, 322 subshift, 31 N power —, 228 N th higher block —, 227 Δirreducible —, 34 — deﬁned by a labeled graph, 223 — of ﬁnite type, 32 contextfree —, 225 even —, 35 golden mean —, 35 irreducible —, 32 language of a — over Z, 35 minimal —, 72 Morse —, 74, 144 soﬁc —, 225 strongly irreducible —, 34 surjunctive —, 71 Toeplitz —, 74 topologically mixing —, 33 subword, 35
Index support — of a conﬁguration, 288 — of a pattern, 2 — of a permutation, 360 surjunctive — group, 57 —subshift, 71 symbol, 2 symmetric — group, 359 — subset, 152 syndetic subset, 72 Tarski — alternative, 99 — number of a group, 106 TarskiFølner theorem, 99 theorem Baire —, 403 BanachAlaoglu —, 385 CantorBernstein —, 398 CurtisHedlund —, 20 Garden of Eden —, 114, 128 Garden of Eden — for linear cellular automata, 312 GromovWeiss —, 272 Hall harem —, 399 Hall marriage —, 399 KestenDay —, 201 Klein PingPong —, 376 MarkovKakutani —, 387 open mapping —, 404 TarskiFølner —, 99 Tychonoﬀ —, 348 ThueMorse sequence, 73 tiling, 122 Toeplitz — conﬁguration, 74 — subshift, 74 topological — dual, 384 — entropy, 142 — manifold, 69 — vector space, 383 topologically mixing — action, 29 — subshift, 33 topologically transitive action, 32 topology HausdorﬀBourbaki —, 356 initial —, 346 prodiscrete —, 3, 346 product —, 346 proﬁnite —, 53
Index strong —, 384 weak∗ —, 384 total ordering, 331 totally disconnected topological space, 346 transposition, 361 tree, 156 trivial uniform structure, 352 Tychonoﬀ theorem, 348 ultraﬁlter, 410 free —, 410 limit along an — , 413 nonprincipal —, 410 principal —, 410 ultraproduct, 244 uniform — convexity, 407 — embedding, 355 — isomorphism, 355 — structure, 351 HausdorﬀBourbaki — structure, 356 induced — structure, 353 prodiscrete — structure, 22, 355 uniformly continuous — action, 64 — map, 353 unique — product group, 331 — rank property, 340 unitregular ring, 340 universe, 2
439 valence of a vertex, 155 vertex — of a bipartite graph, 391 — of a labeled graph, 153 degree of a —, 155 neighbor, 155 valence of a —, 155 virtually P group, 41 von Neumann — conjecture, 105 — ﬁnite ring, 336 — neighborhood, 165 weak∗ topology, 384 word, 367 — length, 152 — metric, 153 admissible —, 35 empty —, 367 forbidden —, 35 length of a —, 35 reduced —, 373 subword of a —, 35 wreath product, 52 zerodivisor, 330 — conjecture, 337 left —, 330 right —, 330