Group Ring Groups: Volume 2 Structure Theorems of Unit Groups 9783110411508, 9783110411492

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Eric Jespers, Ángel del Río Group Ring Groups De Gruyter Graduate

Also of interest Group Ring Groups. Volume 1: Orders and Generic Constructions of Units Jespers, del Río, 2015 ISBN 978-3-11-037278-6, e-ISBN (PDF) 978-3-11-037294-6, e-ISBN (EPUB) 978-3-11-038617-2 Abstract Algebra Robinson, 2015 ISBN 978-3-11-034086-0, e-ISBN (PDF) 978-3-11-034087-7, e-ISBN (EPUB) 978-3-11-038560-1

The Elementary Theory of Groups Fine, Gaglione, Rosenberger, Spellman, 2014 ISBN 978-3-11-034199-7, e-ISBN (PDF) 978-3-11-034203-1, e-ISBN (EPUB) 978-3-11-038257-0

Abstract Algebra Carstensen, Fine, Rosenberger, 2011 ISBN 978-3-11-025008-4, e-ISBN (PDF) 978-3-11-025009-1

Journal of Group Theory Christopher W. Parker, John S. Wilson (Editors-in-Chief) ISSN 1433-5883, e-ISSN 1435-4446

Eric Jespers, Ángel del Río

Group Ring Groups

| Volume 2: Structure Theorems of Unit Groups

Mathematics Subject Classification 2010 16S34, 16U60, 20C05, 16H10, 20F05, 20C40, 20E05, 20E34, 16S35, 16G30, 20H10, 20C10, 20H25, 16-02, 11R52, 11R27, 11S45 Authors Eric Jespers Department of Mathematics Vrije Universiteit Brussel Plainlaan 2 1050 Brussel Belgium [email protected] Ángel del Río Departamento de Matemáticas Universidad de Murcia 30100 Murcia Spain [email protected] Acknowledgements The first author has been partially supported by Fonds Wetenschappelijk Onderzoek – Vlaanderen (including project G.015712N) and Onderzoeksraad Vrije Universiteit Brussel. The second author has been partially supported by Ministerio de Economía y Competitividad project MTM2012-35240 and Fondos FEDER. Both authors would also like to thank support by “Instituut ter bevordering van het Wetenschappelijk Onderzoek en de Innovatie van Brussel” (Brusselse hoofdstedelijke regering) for having received a grant “Brains (back) to Brussels” that supported a sabbatical stay of 6 months by the second author at Vrije Universiteit Brussel.

ISBN 978-3-11-041149-2 e-ISBN (PDF) 978-3-11-041150-8 e-ISBN (EPUB) 978-3-11-041275-8 Library of Congress Cataloging-in-Publication Data A CIP catalog record for this book has been applied for at the Library of Congress. Bibliographic information published by the Deutsche Nationalbibliothek The Deutsche Nationalbibliothek lists this publication in the Deutsche Nationalbibliografie; detailed bibliographic data are available on the Internet at http://dnb.dnb.de. © 2016 Walter de Gruyter GmbH, Berlin/Boston Typesetting: PTP-Berlin, Protago-TEX-Production GmbH, Berlin Printing and binding: CPI books GmbH, Leck ♾ Printed on acid-free paper Printed in Germany www.degruyter.com

Preface This is the second Volume on Group Ring Groups, Units and Orders. In Volume 1 the focus is on the generic constructions of the unit group if the integral group ring ℤG of a finite group G. Since ℤG is an order in the semisimple rational group algebra, it was first proved that the unit group of an order in a semisimple finite dimensional rational algebra is finitely presented. In case ℚG does not have exceptional simple components then for most groups G the Bass units together with the bicyclic units generate a subgroup of finite index in U(ℤG). This Volume is devoted to structure theorems of the unit group and determining units of orders in some of the exceptional simple components of ℚG mentioned above. It was shown by Hartley and Pickel that if G is not abelian then the unit group U(ℤG) contains a non-abelian free subgroup, for most groups G. We will give a survey of the main results and techniques in describing explicit units that generate such a group. In the last chapter of the book we determine when U(ℤG) contains a subgroup of finite index that is the direct product of free-by-free groups. This is the best structural result known. Another large part of the book is devoted to the use of geometric methods to compute the unit group of an order in a quaternion algebra, this via discontinuous actions on hyperbolic spaces. We give a detailed account by first including a new recent proof of the classical result of Poincaré on obtaining presentations from fundamental polyhedra for groups of isometries of Riemann manifolds of constant curvature and, in particular, of hyperbolic space. Then we show how to apply this to orders in some quaternion algebras. An outcome of all the above results is that one has now a good, or at least a much better, idea on describing a subgroup of finite index in U(ZG) for all finite groups G, except for those for which ℚG has a non-commutative epimorphic image that is a non-commutative division algebra that is not a totally definite quaternion algebra. Unit groups of orders in such a division algebra are a big unknown and we refer the reader to Kleinert’s book on this topic [137]. Without being encyclopaedic, we have included what we think are all the main results and techniques used to achieve these results. For other topics we refer the reader to the literature; there are several books devoted to the algebraic structure of group rings: A. Bovdi [31], N. Gupta [78], G. Karpilovsky [126–131], G. Lee [147], I. B. N. Passi [169], W. Plesken [179], C. Polcino Milies and S. K. Sehgal [185], D. S. Passman [171, 172], K. Roggenkamp [197], S. K. Sehgal [201, 202, 204], M. Taylor [217] and A. Zalesskii and A. Mihalev [232]. The outline of the book is as follows. Chapter 14 is on explicit constructions of units that generate a free subgroup of the unit group. Chapter 15 contains background on hyperbolic geometry and discrete groups. In Chapter 16 we give a proof of the presentation part of Poincaré’s Theorem on discontinuous groups of isometries of a Riemann manifold of constant curvature.. Chapter 17 contains algorithms to determine a fundamental polyhedron, and thus generators (and presentations), of such a group.

VI | Preface

Chapter 18 contains applications of these algorithms to unit groups of orders in quaternion algebras and integral group rings. Chapter 19 contains a complete classification of when these geometrical methods apply to calculate the unit group of an integral group ring. It turns out that this happens precisely for the finite groups for which the unit group is virtually a direct product of free-by-free groups. We are indebted to O. Broche, M. Caicedo, G. Janssens, A. Kiefer, Á. Pérez Raposo, J. J. Simón and I. Van Gelder for reading parts of the manuscript, to Ann Kiefer for producing some of the 3D pictures of Chapter 18 and to S. Sánchez Pedreño for LATEX assistance. We would like to express our appreciation to our teachers in group rings: D. Passman and S. K. Sehgal. They not only introduced us to this wonderful area of mathematics but they constantly have been very inspiring and stimulating. And most of all they are good friends that have proved such wonderful results. Finally we like to mention that we enjoyed the many hours we collaborated on these topics with many people: V. Bovdi, C. Corrales, A. Dooms, A. Giambruno, J. Z. Gonçalves, E. G. Goodaire, R. Guralnick, A. Herman, M. Hertweck, S. O. Juriaans, E. Kleinert, A. Konovalov, G. Leal, L. Margolis, A. Olivieri, G. Olteanu, A. Pita, C. Polcino Milies, M. M. Parmenter, M. Ruiz and P. Zalesskii.

Contents Volume 1 Preface | V 1 1.1 1.2 1.3 1.4 1.5 1.6

Units in group rings: an introduction | 1 Constructions of units: elementary matrices and bicyclic units | 2 Construction of units: cyclotomic units and Bass units | 5 Examples: unit groups of some orders in number fields | 9 Examples: unit groups of some non-commutative orders | 14 Examples: group rings of groups of small order | 20 Finite rings | 28

2 2.1 2.2 2.3 2.4 2.5 2.6

Representations of algebras | 32 Semisimple algebras | 32 Splitting fields | 42 Characteristic polynomial, trace and norm | 46 Brauer group | 55 Cohomology | 57 Crossed products | 64

3 3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8

Wedderburn decomposition of semisimple group algebras | 73 Representations and characters of finite groups | 74 Some operations with characters | 81 Wedderburn components from character tables | 85 Wedderburn components from monomial characters | 92 Strongly monomial characters | 102 Induction theorems | 112 Brauer-Witt Theorem | 114 Examples | 120

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8

Dedekind domains, valuations and orders | 123 Localization and algebraic integers | 123 Dedekind domains | 130 Finitely generated modules over Dedekind domains | 137 Extensions of Dedekind domains | 142 Valuations | 146 Orders | 152 The discriminant | 157 Brauer group of a number field | 160

VIII | Contents

5 5.1 5.2 5.3 5.4 5.5

The group of units of an order | 170 Lattices in real vector spaces | 170 Hey’s Theorem and Dirichlet’s Unit Theorem | 173 The group of units of an order is finitely generated | 179 The group of units of an order is finitely presented | 194 Subgroups of finite index | 198

6 6.1 6.2

Cyclotomic integers | 205 Cyclotomic fields | 205 Cyclotomic units | 208

7 7.1 7.2 7.3

Central units | 226 The group of central units of an order | 226 Large subgroups of central units: an algorithm | 237 Bass units as generators of large groups of units | 248

8 8.1 8.2 8.3 8.4 8.5

Generic units | 255 Shifted cyclotomic polynomials | 256 The group of generic units | 262 A logarithm function | 265 A basis of generic units for a subgroup of finite index in U(ℤC n ) | 272 Polynomials of small degree defining units | 276

9 9.1 9.2 9.3 9.4 9.5

K-theory | 282 Grothendieck group | 282 The Whitehead group | 284 Stable range condition | 292 Whitehead group and the stable range condition | 295 Applications of K-theory to units | 303

10 10.1 10.2 10.3 10.4

General linear groups of degree 2 | 312 Number theoretical results | 313 Normality of E2 (I1 ≀ I2 ) in SL2 (R, I1 ≀ I2 ) | 314 The factor group SL2 (R, I1 ≀ I2 ) by E2 (I1 ≀ I2 ) | 319 The group E2 (I) is of finite index in SL2 (R) | 326

11 11.1 11.2 11.3 11.4 11.5

Generators of unit groups of group rings | 337 Bass Unit Theorem | 337 Generalized bicyclic units and Bass units I | 341 Bicyclic units and Bass units | 345 Fixed point free groups and Frobenius complements | 348 Group rings of nilpotent groups | 355

Contents | IX

12 12.1 12.2 12.3 12.4 12.5 12.6 12.7

Exceptional simple components | 359 Components of index one | 359 Components of index two | 366 Generalized bicyclic units and Bass units II | 373 Normal closure of the trivial units | 375 Normal complements | 380 Examples: metacyclic groups | 383 Examples with insufficient Bass units and bicyclic units | 390

13 13.1 13.2 13.3 13.4 13.5 13.6 13.7

Idempotents and central units in group rings | 392 Central subgroups and abelian-by-supersolvable groups | 392 Independent units and abelian-by-supersolvable groups | 397 Central subgroups and strongly monomial groups | 400 Independent units and strongly monomial groups | 402 Primitive idempotents and nilpotent groups | 408 Primitive idempotents and strongly monomial groups | 416 Some metacyclic groups | 420

References | 425 Index of Notation | 435 Index | 439

Volume 2 Preface | V 14 14.1 14.2 14.3 14.4 14.5

Free Groups | 1 Ping-Pong Lemma and free groups generated by bicyclic units | 2 Free groups in integral group rings over Hamiltonian groups | 10 Attractors | 14 Free companions | 21 Attractors revisited | 38

15 15.1 15.2 15.3 15.4 15.5

Hyperbolic geometry | 44 Möbius transformations | 44 Riemann varieties of constant curvature | 47 The groups of isometries of ℍ3 and 𝔹3 | 52 Isometric spheres in 𝔹n | 60 Discrete subgroups of PSL2 (ℂ) | 65

X | Contents

16 16.1 16.2

Poincaré’s Theorem | 69 Polyhedra and tessellations | 69 Group presentations | 80

17 17.1 17.2 17.3 17.4 17.5

Fundamental polyhedra | 95 Dirichlet fundamental polyhedra | 95 Ford fundamental polyhedron | 99 Polyhedra of finite volume | 102 An algorithm to compute a fundamental polyhedron | 106 Symmetries on Dirichlet polyhedra | 113

18 18.1 18.2 18.3

Unit groups of orders in quaternion algebras | 116 Algebras of Kleinian type | 116 Bianchi groups | 119 Calculating fundamental polyhedra for orders in division algebras of Kleinian type | 133 Generators of unit groups of group rings with exceptional components | 142

18.4

19 19.1 19.2 19.3 19.4 19.5 19.6 19.7

Virtually free-by-free groups | 151 Free-by-free groups | 151 Orders and free-by-free unit groups | 158 Virtual cohomological dimension and finite groups of Kleinian type | 162 Examples of groups of Kleinian type | 166 Nilpotent groups of Kleinian type | 172 Groups of Kleinian type that are not nilpotent | 185 Structure theorem | 192

References | 195 Index of Notation | 205 Index | 209

14 Free Groups Recall that if G is a nonabelian finite group then U(ℤG) is finite if and only if G is a Hamiltonian 2-group (Theorem 1.5.6). Hence, obviously, if G is either abelian or a Hamiltonian 2-group then U(ℤG) does not contain a nonabelian free group. However, Hartley and Pickel [81] and Sehgal [201, Theorem VI 4.2] proved independently, and almost simultaneously, that in all other cases U(ℤG) contains a nonabelian free group. This result has important consequences on the structure of U(ℤG), mostly these are stated in the negative. For example, U(ℤG) is not solvable unless G is either abelian or a Hamiltonian 2-group. Since the proofs given by Sehgal and Hartley-Pickel are not constructive, this raised the question of constructions of nonabelian free subgroups in U(ℤG). This was obtained by Marciniak and Sehgal [153] for G non-Hamiltonian, using bicyclic units, and by Ferraz [58] for G Hamiltonian, using Bass units. Hence, one has a constructive proof of the existence of nonabelian free groups and, furthermore, once more Bass units and bicyclic units play a relevant role. In Section 14.1 we present the construction of Marciniak and Sehgal and in Section 14.2 we present the construction of Ferraz. The main tool behind most of the results of this chapter is the Ping-Pong Lemma, attributed to Felix Klein who called it “der Process der Ineinanderschiebung” [133]. Two results that can be proved using the Ping-Pong Lemma are a Theorem of Sanov (Proposition 14.1.4) and a Theorem of Tits on free groups generated by diagonalizable non-singular linear maps, which has a relevant role in the proof of the Tits Alternative [218]. The first is essential for the proof of the Marciniak-Sehgal result and the second is behind a result of Gonçalves, Mandel and Shirvani which is the main tool in the proof of the Ferraz result (Lemma 14.2.1). In Section 14.3 we present some methods, due to Gonçalves and Passman [73], which are useful to prove that some pairs of non-singular linear maps generate a nonabelian free group. In particular, a result is obtained that generalizes the Theorem of Tits mentioned above. The main concept behind these techniques is that of attractor. Roughly speaking, an attractor for a linear map T is a subspace which attracts T n (a) when n becomes large. Attractors of diagonalizable linear maps appeared in methods introduced by Tits. Gonçalves and Passman also consider attractors of linear maps of the form 1 + a with a2 = 0. As an application of these methods, we will investigate in Section 14.4 when for a given Bass unit u there is a nonabelian free group of the form ⟨u, v⟩, with u either a Bass unit or a bicyclic unit. There are many results related to the ones presented in this chapter. The proofs of the results presented contain most of the essential tools on this topic. For a complete survey we refer the reader to [70].

2 | 14 Free Groups

14.1 Ping-Pong Lemma and free groups generated by bicyclic units The final result of this section, due to Marciniak-Sehgal [153], is the construction of a nonabelian free group generated by two bicyclic units. The proof presented is a combination of approaches used by Salwa in [199], Gonçalves and Passman in [74], Jespers, del Río and Ruiz in [107] and Gonçalves and del Río in [68]. In particular, the investigation includes an analysis of units of the form 1 + a with a2 = 0, as is the case of bicyclic units. For the sake of brevity, when two elements u and v of a group generate a free group of rank 2 we will simply say that ⟨u, v⟩ is free. Observe that if u and v are elements of a group then ⟨u, v⟩ is free if and only if u and v are of infinite order and ⟨u, v⟩ = ⟨u⟩ ∗ ⟨v⟩, the free product of u and v. If G1 and G2 are subgroups of a given group G then ⟨G1 , G2 ⟩ = G1 ∗ G2 if the identity element 1 of G is not an alternating product of elements belonging to G1 \{1} and G2 \{1}. Checking this condition purely algebraically is usually difficult. The Ping-Pong Lemma is a useful alternative technique. Lemma 14.1.1 (Ping-Pong Lemma). Let G1 and G2 be subgroups of a group G with |G1 | > 2. Assume that G acts on a set P which contains two different non-empty subsets P1 and P2 such that g(P i ) ⊆ P j if 1 ≠ g ∈ G i and {i, j} = {1, 2}. Then, ⟨G1 , G2 ⟩ = G1 ∗ G2 . Proof. Let g1 , . . . , g k be a list of non-trivial elements belonging to G1 and G2 alternatively. We have to prove that g1 ⋅ ⋅ ⋅ g k ≠ 1. We show this by contradiction. So, assume that g1 ⋅ ⋅ ⋅ g k = 1. Without loss of generality, we may suppose that the extreme elements g1 and g k belong to G1 . Indeed, if g1 , g k ∈ G2 then we take g ∈ G1 \ {1} and add g at the beginning of the list and g−1 to the end of the list; if g1 ∈ G2 and g k ∈ G1 then we take g ∈ G1 \ {1, g k } and add g at the beginning of the list and replace g k by g k g−1 ; finally if g1 ∈ G1 and g k ∈ G2 then we take g ∈ G1 \ {1, g1 } and replace g1 by g −1 g1 and add g at the end of the list. The resulting list satisfies the required condition. Then P1 = (g1 ⋅ ⋅ ⋅ g k )(P1 ) ⊆ P2 and if 1 ≠ g ∈ G2 then P2 = (gg1 ⋅ ⋅ ⋅ g k g−1 )(P2 ) ⊆ P1 . Thus P1 = P2 , contradicting the hypothesis. The assumption that one of the two groups G1 or G2 has at least three elements is necessary. Indeed, consider for example the natural action of the symmetric group S2 on {1, 2}. Then G1 = G2 = S2 and P1 = {1} and P2 = {2} satisfy the assumptions of Lemma 14.1.1, except for the fact that both G1 and G2 have order 2. Of course, ⟨G1 , G2 ⟩ = G1 is not a free product of G1 and G2 . In case u and v belong to U(ℤG) we can consider their projections in the Wedderburn components of ℚG. If u1 and v1 are the projections of respectively u and v in one of the Wedderburn components and ⟨u1 , v1 ⟩ is free then obviously ⟨u, v⟩ is free. In fact the converse is true. This is a consequence of the following elementary result.

14.1 Ping-Pong Lemma and bicyclic units

| 3

Proposition 14.1.2. Let G = G1 × ⋅ ⋅ ⋅ × G n be a direct product of groups, and let g = (g1 , . . . , g n ) and h = (h1 , . . . , h n ) be elements of G, with g i , h i ∈ G i for every i. Then ⟨g, h⟩ is free if and only if ⟨g i , h i ⟩ is free for some i. Proof. Without loss of generality, we may assume that n = 2. The sufficiency is clear. To prove the necessity, assume that neither ⟨g1 , h1 ⟩ nor ⟨g2 , h2 ⟩ is free. Then there are non-trivial words w1 and w2 in the free group on two symbols such that a i = −1 w i (g i , h i ) = 1 for i = 1, 2. Let w = w1 w2 w−1 1 w 2 . Clearly, −1 −1 −1 w(g, h) = (a1 w2 (g1 , h1 )a−1 1 w 2 (g 1 , h 1 ), w 1 (g 2 , h 2 )a 2 w 1 (g 2 , h 2 )a 2 ) −1 = (w2 (g1 , h1 )w−1 2 (g 1 , h 1 ), w 1 (g 2 , h 2 )w 1 (g 2 , h 2 ))

= (1, 1) = 1. Hence, if w ≠ 1 then it follows that ⟨g, h⟩ is not free. If, on the other hand, w = 1 then w1 and w2 belong to a cyclic group and therefore there exists 1 ≠ v ∈ ⟨w1 ⟩ ∩ ⟨w2 ⟩. Then v(g, h) = 1. So again ⟨g, h⟩ is not free. Proposition 14.1.2 and its preceding comment suggest to study free groups in matrices over division rings (in particular over fields) that are simple components of rational group algebras ℚG. In this section we are interested in bicyclic units of ℤG. As these are of the form 1 + a with a2 = 0, also their projections in the simple components of ℚG are matrices of this form. Examples of such matrices are elementary matrices. Recall that if r is an element of a ring R then Eij (r) denotes the matrix having r in the (i, j)-th entry and zeros elsewhere and, if i ≠ j then e ij (r) = I + Eij (r). The size of the matrices Eij (r) and e ij (r) usually is clear from the context. Elementary matrices already appeared in the context of free groups in Corollary 1.4.5, which states, in particular, that if n is an integer with |n| ≥ 2 then ⟨e12 (n), e21 (n)⟩ is a free group. We will also need to consider free groups of the form ⟨e12 (z1 ), e21 (z2 )⟩ with z1 and z2 complex numbers. The following lemma provides a method to decide when two pairs of elementary matrices of the form e12 (z1 ) and e21 (z2 ) are conjugate in GL2 (F), for F a field. Lemma 14.1.3. Suppose z1 , z2 , w1 , w2 are non-zero elements in a field F. Then, there exists U ∈ GL2 (F) such that E12 (z1 )U = E12 (w1 ) and E21 (z2 )U = E21 (w2 )) if and only if z1 z2 = w1 w2 . Proof. Let U ∈ GL2 (F) be such that E12 (z1 )U = E12 (w1 ) and E21 (z2 )U = E21 (w2 ). Then E11 (z1 z2 )U = (E12 (z1 )E21 (z2 ))U = E12 (w1 )E21 (w2 ) = E11 (w1 w2 ). Thus z1 z2 is a nonzero eigenvalue of E11 (w1 w2 ) and hence z1 z2 = w1 w2 . −1 Conversely, assume that z1 z2 = w1 w2 . Let a = w1 z−1 1 and thus w 2 = a z 2 . Consider the diagonal matrix D = diag(1, a). Then E12 (z1 )D = E12 (w1 ) as desired.

and

E21 (z2 )D = E21 (w2 ),

4 | 14 Free Groups

As a first application of the Ping-Pong Lemma we present the following generalization of Corollary 1.4.5. Proposition 14.1.4 (Sanov [200]). If z1 and z2 are complex numbers such that |z1 z2 | ≥ 4 then the elementary matrices e12 (z1 ) and e21 (z2 ) generate a free group. Proof. Because of Lemma 14.1.3, without loss of generality, we may assume that |z1 |, |z2 | ≥ 2 (for example replacing both z1 and z2 by a square root of z1 z2 ). Consider the natural action of GL2 (ℂ) on the vector space ℂ2 of column vectors and take P1 = {(x1 , x2 )T ∈ ℂ2 : |x1 | < |x2 |}

and

P2 = {(x1 , x2 )T ∈ ℂ2 : |x1 | > |x2 |}.

If 1 ≠ g ∈ ⟨e12 (z1 )⟩ then g = e12 (nz1 ), with n a non-zero integer. If (x1 , x2 )T ∈ P1 then g ⋅ (x1 , x2 )T = (x1 + nz1 x2 , x2 )T and |x1 + nz1 x2 | ≥ |nz1 | ⋅ |x2 | − |x1 | ≥ 2|x2 | − |x1 | > |x2 |, because |z1 | ≥ 2. This proves that g⋅P1 ⊆ P2 . Similarly, if 1 ≠ g ∈ ⟨e21 (z2 )⟩ then g⋅P2 ⊆ P1 . Hence, by the Ping-Pong Lemma (Lemma 14.1.1), ⟨e12 (z1 ), e12 (z2 )⟩ is free. By Lemma 14.1.3, the freeness of ⟨e12 (z1 ), e21 (z2 )⟩ depends exclusively on the parameter t = z1 z2 . A free point is a complex number t such that ⟨e12 (1), e21 (t)⟩ is free (some other authors replace e12 (1) by e12 (2), which, by Lemma 14.1.3 is simply a change of scale). The study of free points is an active field of research (see for example [13]). For example, Proposition 14.1.4 can be rephrased saying that if |t| ≥ 4 then t is a free point. Let R be a torsion-free ring, i.e. R is torsion-free as a ℤ-module. Let A = ℚ ⊗ℤ R. Since R is torsion-free as ℤ-module, it is flat and hence the natural homomorphism R → A is injective. We consider this homomorphism as inclusion and hence we consider R as a subring of A. If X is a subset of R then ℚ[X] denotes the ℚ-subalgebra of A generated by X. In particular, if X = {r1 , . . . , r k } then ℚ[r1 , . . . , r k ] = {f(r1 , . . . , r k ) : f ∈ ℚ ⟨T1 , . . . , T k ⟩}, where ℚ ⟨T1 , . . . , T k ⟩ denotes the rational free algebra on noncommuting free variables T1 , . . . , T k . Let a ∈ A. We say that a is algebraic if f(a) = 0 for some f ∈ ℚ[X] \ {0}, or equivalently if ℚ[a] is finite dimensional over ℚ. Otherwise we say that a is transcendental. Let ρ denote the regular representation of A over ℚ. Then, by definition, the eigenvalues of a are the complex eigenvalues of ρ(a). If a is algebraic then the eigenvalues of a are the complex roots of Minℚ (a), the minimum polynomial of a over ℚ. If A = A1 × ⋅ ⋅ ⋅ × A k , a product of rational algebras, then λ is an eigenvalue of a = (a1 , . . . , a k ) with a i ∈ A i , if and only if λ is eigenvalue of some a i . If A is semisimple then λ is an eigenvalue of a if and only if there is a simple ℂ ⊗ℚ A-module M such that λ is an eigenvalue of the endomorphism of M given by multiplication by a. In particular, if G is a finite group and a ∈ ℚG then λ is an eigenvalue of a if and only if λ is eigenvalue of ϕ(a) for some irreducible complex representation ϕ of G.

14.1 Ping-Pong Lemma and bicyclic units | 5

We will consider the algebra F = ℚ⟨X, Y | X 2 = Y 2 = 0⟩ = ℚ ⟨T1 , T2 ⟩ /(T12 , T22 ).

Then F = ℚ[XY] ⊕ Yℚ[XY] ⊕ ℚ[XY]X ⊕ Yℚ[XY]X

and ℚ[XY] is a polynomial ring over ℚ in one variable. If a, b ∈ R are such that a2 = b2 = 0 then there is a unique algebra homomorphism f : F → A mapping X to a and Y to b. The image of this homomorphism is ℚ[a, b] = ℚ[ab] + bℚ[ab] + ℚ[ab]a + bℚ[ab]a. Lemma 14.1.5. Let R be a torsion-free ring and let a, b ∈ R with a2 = b2 = 0. (1) ab is transcendental if and only if ℚ[a, b] is isomorphic to the algebra F = ℚ ⟨X, Y | X 2 = Y 2 = 0⟩. (2) ab is algebraic if and only if ℚ[a, b] is finite dimensional. (3) Assume that ab is algebraic and let λ1 , . . . , λ k be the non-zero eigenvalues of ab. Then there is an algebra isomorphism n

ℚ[a, b]/J(ℚ[a, b]) → ℚm × ∏ M2 (ℚ[μ i ]) i=1

with a 󳨃→ (0, . m. ., 0, E12 (1), . . . , E12 (1)), b 󳨃→ (0, . m. ., 0, E21 (μ1 ), . . . , E21 (μ n )) and {μ1 , . . . , μ n } = {λ1 , . . . , λ k }. Proof. Without loss of generality, we may assume that R = ℤ[a, b], and thus A = ℚ[a, b]. We use the standard bar notation for reduction modulo J(A), the Jacobson radical of A. So, A = A/J(A). (1) and (2). Assume that ab is transcendental. We will prove that the monomials (ab)n , b(ab)n , (ab)n a and b(ab)n a with n ≥ 0 are linearly independent over ℚ, or equivalently that the only polynomials f1 , f2 , f3 and f4 in ℚ[X] with f1 (ab) + bf2 (ab) + f3 (ab)a + bf4 (ab)a = 0

(14.1.1)

are the zero polynomials. Clearly, this proves that the algebra homomorphism F → A mapping X to a and Y to b gives an isomorphism F ≅ ℚ[a, b]. Assume f1 , f2 , f3 and f4 are polynomials in ℚ[X] that satisfy (14.1.1). Multiplying both sides of (14.1.1) by ab, we have abf1 (ab)ab = 0. As, by assumption, ab is transcendental we deduce that f1 = 0. Then, multiplying on the left by a and on the right by ab, we have abf2 (ab)ab = 0. Hence f2 = 0. Similarly f3 = 0 and f4 = 0. If ab is algebraic then ℚ[ab] is finite dimensional over ℚ and hence ℚ[a, b] = ℚ[ab] + bℚ[ab] + ℚ[ab]a + bℚ[ab]a is finite dimensional.

6 | 14 Free Groups

(3) Assume that ab is algebraic. Thus, by (2), A is finite dimensional. Then, A is semisimple and the Wedderburn components of A are of the form ℚ[α, β] = M n (D) with α2 = β2 = 0 and D a division algebra. There is no loss of generality in assuming that A = A = ℚ[a, b] = M n (D) with n ≥ 1 and D a division algebra. We have to prove that either n = 1 and D = ℚ or n = 2 and D = ℚ[λ], for λ the unique non-zero eigenvalue of ab and there is an automorphism of A mapping a to E12 (1) and b to E21 (λ). If n = 1 then a = b = 0 and hence A = ℚ. Assume otherwise that n > 1 and let E = Z(D) and let L be a splitting field of A over E. Then A L = L ⊗E A = M m (L) = L[a, b] = S + bS + Sa + bSa with m = Deg(A) and S = L[ab]. Consider A L as the ring of endomorphisms of an m-dimensional vector space over L. Since a2 = 0 we have Im(a) ⊆ ker(a) and hence 2 dimL Im(a) ≤ m. Consequently dimL Im(ab) ≤ dimL Im(a) ≤ ⌊

m ⌋. 2

The Cayley-Hamilton Theorem then implies that ab is the root of a polynomial of dem gree at most ⌊ m 2 ⌋ + 1. So dimL S ≤ ⌊ 2 ⌋ + 1 and thus m2 = dimE A = dimL A L ≤ 4 (⌊

m ⌋ + 1) . 2

2

m Consequently, ( m 2 ) ≤ ⌊ 2 ⌋ + 1 and therefore n Ind(D) = m ≤ 2. Thus, n = 2 and Ind(D) = 1, that is D = E = L. Hence, we may consider A as the ring of endomorphisms a 2-dimensional vector space V over E. As A is not commutative and because it is generated as a ℚ-algebra by a and b, we have ab ≠ ba. Moreover, 0 ≠ Im(a) ⊆ ker(a) ∩ Im(ab) ⊆ Im(ab) ≠ V and therefore Im(a) = ker(a) = Im(ab) = Ew1 for some 0 ≠ w1 ∈ V. Similarly Im(b) = ker(b). We claim that ker(a) ≠ ker(b). Indeed, assume that ker(a) = ker(b) = Ew1 and let w2 ∈ V be such that aw2 = w1 . Then {w1 , w2 } is a basis of V and b(w2 ) = λw1 for some λ ∈ E, and ab(w1 ) = 0 = ba(w1 ) and ab(w2 ) = 0 = ba(w2 ). However, this contradicts with ab ≠ ba. So, ker(b) = Ev2 with v2 ∈ ̸ ker(a). Let v1 = a(v2 ). Then {v1 , v2 } is a basis of V. With respect to this basis, a can be written as E12 (1) and b as E21 (λ), with b(v1 ) = λv2 and λ ∈ U(E). Then ab = E11 (λ), and thus λ is the unique non-zero eigenvalue of ab and A = M2 (E) = ℚ[ab] + ℚ[ab]a + bℚ[ab] + bℚ[ab]a = M2 (ℚ[λ]). Therefore E = ℚ[λ], as desired.

Theorem 14.1.6. Let R be a torsion-free ring and let a, b ∈ R be such that a2 = b2 = 0. Then ⟨1 + a, 1 + b⟩ is free if and only if either ab is transcendental or ab is algebraic and one of the eigenvalues of ab is a free point. Proof. Assume that ab is transcendental. Then, by Lemma 14.1.5 (1), ℚ[a, b] ≅ F = ℚ ⟨X, Y | X 2 = Y 2 = 0⟩ and in fact there is an isomorphism associating a to X and b to Y. Thus there is an algebra homomorphism ℚ[a, b] → M2 (ℚ) mapping a to E12 (1) and b to E21 (4). By Sanov’s Theorem (Proposition 14.1.4), 4 is a free point, i.e. ⟨e12 (1), e12 (4)⟩ is free. Hence, so is ⟨1 + a, 1 + b⟩.

14.1 Ping-Pong Lemma and bicyclic units | 7

Assume that ab is algebraic and let A = ℚ[a, b] and J = J(A), the Jacobson radical of A. By Lemma 14.1.5 (2), A is finite dimensional over ℚ and hence J is nilpotent. As (1 + J n )/(1 + J n+1 ) is central in (1 + J)/(1 + J n+1 ) we deduce that 1 + J is a nilpotent group and hence so is (1 + J) ∩ ⟨1 + a, 1 + b⟩. Thus ⟨1 + a, 1 + b⟩ is free if and only if so is ⟨1 + a, 1 + b⟩. Let λ1 , . . . , λ k be the non-zero eigenvalues of ab. By Lemma 14.1.5 (3), A = A/J ≅ ℚm × ∏ni=1 M2 (ℚ[μ i ]), with {μ1 , . . . , μ n } = {λ1 , . . . , λ k } and this isomorphism associates 1 + a and 1 + b with (1, . . . , 1, e12 (1), . . . , e12 (1)) and (1, . . . , 1, e21 (μ1 ), . . . , e12 (μ k )) respectively. Because of Proposition 14.1.2, ⟨1 + a, 1 + b⟩ is free if and only if some μ i is a free point, that is, some λ i is a free point. Corollary 14.1.7. Let R be a torsion-free ring and let a, b ∈ R be such that a2 = b2 = 0. (1) If m and n are integers then ⟨(1 + a)m , (1 + b)n ⟩ is free if and only if ⟨1+a, 1+mnb⟩ is free. (2) If ab is not nilpotent then ⟨1 + a, 1 + b m ⟩ is free for some positive integer m. (3) ab is nilpotent if and only if ⟨1 + a, 1 + b⟩ is a nilpotent group. Moreover, if ab is algebraic, then ⟨1 + a, 1 + b⟩ is nilpotent if and only if 0 is the only eigenvalue of ab. Proof. (1) This is an obvious consequence of Theorem 14.1.6 because (1 + a)m = 1 + ma and (1 + b)n = 1 + nb. (2) and (3). First, assume that ab is nilpotent and let S be the multiplicative semigroup generated by a and b. Then there is a positive number n such that S n = 0. Hence, 1 + S is a nilpotent subgroup of U(R). So, ⟨1 + a, 1 + b⟩ is a nilpotent group. Second, assume that ab is transcendental. Then ⟨1 + a, 1 + b⟩ is a free group, by Theorem 14.1.6. So (2) holds for m = 1 and ⟨1 + a, 1 + b⟩ is not nilpotent. Third, assume that ab is algebraic and that it has a non-zero eigenvalue λ. Let m be a positive integer such that |mλ| ≥ 4. Then mλ is an eigenvalue of mab and mλ is a free point by Sanov’s Theorem (Proposition 14.1.4). Therefore ⟨1 + a, (1 + b)m = 1 + mb⟩ is a free group, by Theorem 14.1.6, and hence ⟨1 + a, 1 + b⟩ is not nilpotent. Finally, if ab is algebraic and 0 is the only eigenvalue of ab then ab is nilpotent. Let A be a complex algebra. A trace function on A is a ℂ-linear map T : A → ℂ such that T(ab) = T(ba) for a, b ∈ A, T(e) is a positive real number for all non-zero idempotents e ∈ A and T(a) = 0 for every nilpotent element a ∈ A. Example 14.1.8. For every positive real number μ, there is a trace function t μ on M n (ℂ) given by t μ (A) = μ tr(A) and all the trace functions on M n (ℂ) are of this form. Proof. That t μ is a trace function for every μ > 0 is a consequence of the properties of the trace map tr. Conversely, assume that T is a trace function on M n (ℂ). By linearity, T is completely determined by its action on the matrix units E ij = Eij (1). If i ≠ j then T(Eij ) = 0, since E ij is nilpotent. On the other hand, E ii and E jj are conjugate in M n (ℂ) for every i, j. Therefore, if μ = T(E11 ) then T(E ii ) = μ for every i. So, T = t μ .

8 | 14 Free Groups

Example 14.1.9. If G is a group and R is a commutative ring then the trace function of RG is the map T : RG → R associating to each element of RG the coefficient of 1, i.e. T(∑g∈G r g g) = r1 . Let n = |G| and let ρ : RG → M n (R) be the regular representation, given by left multiplication. Then T(x) = 1n tr(ρ(x)) for every x ∈ RG. In particular, if R = ℂ then T can be considered as the restriction of t 1n to ℂG, so that T is a trace function of ℂG. Corollary 14.1.10 (Salwa [199]). Let A be a complex algebra and let T be a trace function on A. If a, b ∈ A are such that a2 = b2 = 0 and |T(ab)| ≥ 2T(1) then ⟨1 + a, 1 + b⟩ is free. Proof. Without loss of generality, we may assume that A = ℂ[a, b]. As T(ab) ≠ 0, ab is not nilpotent. If ab is transcendental over ℂ then the result follows from Theorem 14.1.6. So, assume that ab is algebraic over ℂ. Then A is finite dimensional and hence A/J is semisimple, where J = J(A), the Jacobson radical of A. As J is nilpotent, T(J) = 0 and therefore T induces a trace function in A/J. So, without loss of generality, we may assume that A is semisimple. Let e1 , . . . , e k be all the primitive central idempotents of A. The restriction of T to each Ae i is a trace function T i on Ae i and T(x) = T1 (xe1 ) + ⋅ ⋅ ⋅ + T k (xe k ) for every x ∈ A. Thus, the assumption |T(ab)| ≥ 2T(1) yields that 2(T1 (e1 ) + ⋅ ⋅ ⋅ + T k (e k )) = 2T(1) ≤ |T(ab)| = |T1 (abe1 ) + ⋅ ⋅ ⋅ + T k (abe k )| ≤ |T1 (abe1 )| + ⋅ ⋅ ⋅ + |T k (abe k )|. Hence, 2T i (e i ) ≤ |T i (abe i )| for some i. Therefore, again without loss of generality, we may assume that A is simple, or equivalently, A = M n (ℂ) for some n. Then T = T μ for some μ > 0, by Example 14.1.8. In fact, n ≠ 1 for otherwise abe i = 0. Thus n = 2, by Lemma 14.1.5 (3). As ab is singular, one of its eigenvalues is 0 and, as ab is not nilpotent, it has a non-zero eigenvalue, say λ. Then μ|λ| = |T(ab)| ≥ 2T(1) = 4μ and so |λ| ≥ 4. We conclude that ⟨1 + a, 1 + b⟩ is free, by Proposition 14.1.4 and Theorem 14.1.6. We now are in a position to present the construction of free groups, due to Marciniak and Sehgal. Let G be a finite group and let g, h ∈ G. Recall that the bicyclic units ̃ = 1 + (1 − h)g h ̃ b(g, h)

and

̃ g) = 1 + hg(1 ̃ b(h, − h)

are both trivial if and only if g normalizes ⟨h⟩. Otherwise both are non-trivial. Thus G contains a non-trivial bicyclic unit if and only if G is nonabelian and non-Hamiltonian. ∗ Recall also that ∗ stands for the classical involution on ℤG, i.e. (∑g∈G r g g) = ∑g∈G r g g−1 . Theorem 14.1.11 (Marciniak-Sehgal [153]). Let G be a finite group and let u a non-trivial bicyclic unit of ℤG. Then ⟨u, u∗ ⟩ is free.

14.1 Ping-Pong Lemma and bicyclic units | 9

̃ ≠ 1 with Proof. Let T be the trace map of ℂG (see Example 14.1.9) and let u = b(g, h) ∗ −1 −1 ̃ ̃ ̃ g, h ∈ G. Let a = u − 1 = (1 − h)g h and b = a = hg (1 − h ). Then ba = hg −1 (2 − h − ̃ = h(2 ̃ − z − z−1 )h, ̃ with z ∈ ̸ ⟨h⟩. Therefore T(ab) = T(ba) = 2|h| ≥ 4 = 4T(1). h−1 )g h ∗ Hence ⟨u = 1 + a, u = 1 + b⟩ is free, by Corollary 14.1.10. This proves the result for ̃ ≠ 1. A similar argument deals with u = b(h, ̃ g) ≠ 1. u = b(g, h) ̃ ̃ a bicyclic unit of one type, then b(g, h) ̃ ∗ = b(h −1 , g −1 ), a bicyclic unit of If b = b(g, h), the other type. Therefore, Theorem 14.1.11 provides free groups generated by bicyclic units of different type, provided non-trivial bicyclic units exists. Problem 14.1.3 shows how to obtain more free groups generated by bicyclic units of the same or different types.

Problems 14.1.1. Let T be an indeterminate over a field F. Prove that the algebra F⟨X, Y | X 2 = Y 2 = 0⟩ can be realized as the subalgebra of M2 (F[T]) generated by E12 (T) and E21 (1). 14.1.2. Let a ∈ ℂG and let a1 be the coefficient of 1 in a. Prove that if 0 ≠ a = a2 then a1 is positive rational number, and prove that if a is nilpotent then a1 = 0. ̃ =∑ 14.1.3 ([69]). Let G be a finite group. Recall that if X ⊆ G then X x∈X x ∈ ℤG. For every subgroup H of G, h ∈ H and x ∈ G consider the generalized bicyclic units ̃ = 1 + (1 − h)x H ̃ b(h, x, H)

and

̃ x, h) = 1 + Hx(1 ̃ b(H, − h).

(Note that these units are a special case of the definition given in Section 11.2. For simplicity we introduce a new notation for this exercise.) (1) Prove that if x, y ∈ G and H and K are subgroups of G then ̃ ̃ ̃ Hx K = |H x ∩ K| ⋅ HxK

and

|H x ∩ K|, ̃ ̃ T(y Hx K) = { 0,

if y−1 ∈ HxK otherwise.

(2) Let H ≤ K be subgroups of G and let x ∈ G, h ∈ H and k ∈ K be such that x −1 hx ∈ ̸ K. Prove the following statements. ̃ b(̃ K, k, x−1 )⟩ is free. (a) If x−1 kx ∈ ̸ K then ⟨b(h, x, H), ̃ −1 )⟩ is free. ̃ b(x−1 , xkx−1 , xKx (b) If xkx−1 ∈ ̸ K then ⟨b(h, x, H), (3) Prove that the following conditions are equivalent for x, h ∈ G: ̃ ≠ 1, ̃ b(h, ̃ x−1 )⟩ is free, (a) b(x, h) (d) ⟨b(x, h), ̃ x) ≠ 1, ̃ is free, ̃ x), b(x−1 , h)⟩ (b) b(h, (e) ⟨b(h, ̃ ̃ ̃ b(x−1 , xhx ̃ x), b(xhx −1 )⟩ is free, −1 , x −1 )⟩ is free. (c) ⟨b(x, h), (f) ⟨b(h,

10 | 14 Free Groups

14.2 Free groups in integral group rings over Hamiltonian groups It is shown in Theorem 14.1.11 that if u is a non-trivial bicyclic unit then ⟨u, u∗ ⟩ is a free group. This provides a method to construct free groups in ℤG provided that G is neither abelian nor Hamiltonian (see also Problem 14.1.3). However, if G is Hamiltonian then all the bicyclic units of ℤG are trivial. In this section we prove a result of Ferraz [58] which provides free groups U(ℤG) for G a Hamiltonian group which is not a 2-group. The construction uses Bass units (in Theorem VIII.5.7 in [75] another construction of free groups is given). If G is Hamiltonian then G ≅ Q8 × A × E, with A an abelian group of odd order and E an elementary abelian group, by the DedekindBaer Theorem ([196, 5.3.7]). If A = 1 then U(ℤG) = ±G (Theorem 1.5.6). Otherwise G contains a subgroup isomorphic to Q8 × C p , with p an odd prime and therefore it is enough to consider the case when G = Q8 × C p . The main tool to prove the result of Ferraz is the following application of the Ping-Pong Lemma, due to Gonçalves, Mandel and Shirvani. Lemma 14.2.1 ([72]). Let λ1 , λ2 ∈ ℂ \ {0}. Let A and B be the elements of PSL2 (ℂ) represented by the following respective matrices 1 i ( √2 −1

1 ) −i

and

(

λ1 0

0 ). λ2

󵄨 󵄨 If 󵄨󵄨󵄨󵄨 λλ12 󵄨󵄨󵄨󵄨 < 3 − 2√2 then ⟨A, B⟩ = ⟨A⟩ ∗ ⟨B⟩ = C2 ∗ C∞ . Proof. Note that A2 = I. Let σ = √2 − 1 and λ = λλ12 . Observe that σ2 = 3 − 2√2 and thus the assumption implies that |λ| < σ2 < 1. So B has infinite order. Moreover, σ2 + 2σ − 1 = 0 and hence σ = 1−σ 1+σ . ̂ ̂ by Möbius transformations (see Let ℂ = ℂ ∪ {∞} and consider GL2 (ℂ) acting on ℂ Problem 1.4.2), that is az+b

a ( c

, { { acz+d b ) (z) = { c , { d {∞,

if z ≠ ∞ and cz + d ≠ 0, if c ≠ 0 and z = ∞, otherwise.

In particular A(z) = −

z−i 1 − iz−1 iz + 1 = A(z−1 )−1 = = z+i iz − 1 i − z−1

and B(z) = λz. To prove the result, it is sufficient to verify that the assumptions of the Ping-Pong Lemma (Lemma 14.1.1) hold for G1 = ⟨A⟩, G2 = ⟨B⟩, ̂ : |z| < σ or |z| > σ−1 } P1 = {z ∈ ℂ ̂ : σ < |z| < σ−1 }. P2 = {z ∈ ℂ

and

14.2 Free groups in group rings over Hamiltonian groups

| 11

If z ∈ P2 then |B(z)| = |λz| < σ and B−1 (z) = |λ−1 z| > σ−1 . If n ≥ 1 then |B n (z)| = ≤ |B(z)| < σ and |B−n (z)| = |λ−(n−1) B−1 (z)| ≥ |B−1 (z)| > σ−1 . This proves that g(P2 ) ⊆ P1 , for every g ∈ ⟨B⟩ \ {1}. Suppose R is a real number with 0 < R < σ and thus R < 1. On [−R, R] we define the real valued function f R as follows y . f R (y) = 1 − 4 2 R + 2y + 1 |λ n−1 B(z)|

Observe that for y ∈ [−R, R] there exists x ∈ ℝ with x2 + y2 = R2 and thus R2 + 2y + 1 = x2 + (y + 1)2 ≠ 0, because otherwise x = 0, y = −1 and thus R = 1, a contradiction. Note that if z = x + yi, with x, y ∈ ℂ and |z| = R (and thus y ∈ [−R, R]) then 󵄨󵄨 x + (y − 1)i 󵄨󵄨2 x2 + (y − 1)2 R2 − 2y + 1 y 󵄨 󵄨󵄨 = =1−4 2 = f R (y). |A(z)|2 = 󵄨󵄨󵄨 󵄨 = 2 󵄨󵄨 −y − 1 + xi 󵄨󵄨󵄨 x + (y + 1)2 R2 + 2y + 1 R + 2y + 1 +1 Since f R󸀠 (y) = −4 (R2 R+2y+1) 2 < 0, the function f R (y) is decreasing. Therefore, 2

(

1+R 2 1−R 2 ) = f R (R)2 ≤ |A(z)|2 ≤ f R (−R) = ( ) . 1+R 1−R

If 0 ≠ z ∈ P1 then |z| < σ or |z| > σ−1 . If |z| < σ then the previous can be applied with R = |z|. Hence, it follows that σ=

1−σ 1−R 1+R 1+σ < ≤ |A(z)| ≤ < = σ−1 . 1+σ 1+R 1−R 1−σ

Therefore, A(z) ∈ P2 . If |z| > σ−1 then |z−1 | < σ and therefore, by the previous, A(z−1 ) ∈ P2 . Hence A(z) = (A(z−1 ))−1 ∈ P2 . This proves that A(P1 ) ⊆ P2 and thus indeed the assumptions of the Ping-Pong Lemma hold. Proposition 14.2.2. Let F be a subfield of ℂ not containing i = √−1. Consider the maximal subfields F(i) and F(j) of the Hamilton quaternion algebra ℍ(F) = ( −1,−1 F ) and let τ : F(i) → F(j) be the F-isomorphism mapping i to j. Let σ denote the non-trivial element 󵄨 v 󵄨󵄨 of Gal(F(i)/F) and let 0 ≠ v ∈ F(i). If 󵄨󵄨󵄨 σ(v) 󵄨󵄨 < 3 − 2√2, then ⟨v, τ(v)⟩ is free. Proof. Let u = i + j. Observe that τ(x) = x u for every x ∈ F(i) and that F(i) is a splitting field of ℍ(F) (see Problem 2.2.2). More explicitly, there is an isomorphism ρ : F(i) ⊗F 0 1 ). Then 1 ρ(u) is the matrix ℍ(F) → M2 (F(i)) mapping i to diag(i, −i) and j to ( −1 0 √2 A of Lemma 14.2.1 and ρ(v) = B = diag(v, σ(v)). The assumption of the proposition implies that the assumption of Lemma 14.2.1 holds and therefore ⟨A, B⟩ = ⟨A⟩ ∗ ⟨B⟩ = C2 ∗ C∞ . Thus, ⟨B = ρ(v), B A = ρ(v)ρ(u) ⟩ is free and hence so is ⟨v, τ(v) = v u ⟩, as desired. Let G be a finite group. Assume g ∈ G has order n and let u = u k,m (g), a Bass unit. If ρ is a complex representation of G then ρ(g) is diagonalizable and its eigenvalues are n-th roots of unity. Therefore ρ(u) is diagonalizable and its eigenvalues are of the form u k,m (ξ) = (1 + ξ + ξ 2 + ⋅ ⋅ ⋅ + ξ k−1 )m +

1 − km (1 + ξ + ξ 2 + ⋅ ⋅ ⋅ + ξ n−1 ) = η k (ξ)m , n

12 | 14 Free Groups where ξ is an n-th root of unity and (n, k) = 1. As before, we make the convention η k (1) = 1. So far, any choice of a primitive root of unit has been equally good. However, it will become relevant as we want to apply Proposition 14.2.2 to cyclotomic units v in v ℚ(ζ n , i) and thus we need to control the absolute value of σ(v) . We are ready to address the Ferraz construction of free groups. Let p be an odd prime and set G = Q8 × C p = ⟨a, b, x | a4 = a2 b2 = x p = (a, x) = (b, x) = 1, a b = a−1 ⟩. By Lemma 1.2.1, in order to construct a free group generated by Bass units, one needs Bass units of the type u k,m (g) with 2 ≤ k ≤ |g| − 2. In particular, |g| ∈ ̸ {1, 2, 3, 4, 6}. Thus, for our group, |g| = p, 2p or 4p. However, the elements of G of order p or 2p are central and hence we should take g of order 4p. This suggests to consider subgroups of the form ⟨u k,m (ax), u l,r (bx)⟩ with 2 ≤ k ≤ 4p − 2 and gcd(k, 4p) = gcd(l, 4p) = 1. Note that, in particular, m ≥ 2. Let F = ℚ(ζ p ) and let σ ∈ Gal(F(i)/F) and τ : F(i) → F(j) as in Proposition 14.2.2. Recall from Section 3.8 and Theorem 3.3.6, that ℚG ≅ ℚQ8 ⊗ℚ C p ≅ (4ℚ⊕ℍ(ℚ))⊗ℚ (ℚ⊕ ℚ(ζ p )). Hence, one of the Wedderburn components of FG is the algebra of Hamiltonian quaternions ℍ(F) and the projection ρ : ℚG → ℍ(F) is given by a 󳨃→ i, b 󳨃→ j and x 󳨃→ ζ p . Therefore, if we set v = η k (ζ p i) then we have ρ(u k,m (ax)) = v m ξ6

and

ρ(u k,m (bx)) = τ(v m ).

First assume that p = 3 and set ξ = ζ3 i. Then ξ 4 = ξ 2 −1 and hence (ξ −1)(ξ 5 −1) = − ξ − ξ 5 + 1 = (ξ 4 − ξ 2 ) − ξ − (ξ 3 − ξ) + 1 = −ξ 3 = i. Therefore, for k = 5, we have v=

ξ5 − 1 = −i(ξ 5 − 1)2 = −i(1 − ζ32 i)2 . ξ −1

As ζ3 = ζ32 and ζ3 − ζ32 = √3i, we have 󵄨󵄨 v 󵄨󵄨 |1 − ζ 2 i|2 (1 − ζ 2 i)(1 + ζ3 i) 2 − √3 󵄨󵄨 󵄨󵄨 3 3 = = 7 − 4√3 < 3 − 2√2. = 󵄨= 󵄨󵄨 󵄨󵄨 σ(v) 󵄨󵄨󵄨 |1 + ζ 2 i|2 (1 + ζ 2 i)(1 − ζ3 i) 2 + √3 3 3 By Proposition 14.2.2 we obtain that the group ⟨v, τ(v)⟩ is free and hence so is the group ⟨v2 = ρ(u5,2 (ax), τ(v)2 = ρ(u5,2 (bx))⟩. Thus, for p = 3, ⟨u5,2 (ax), u5,2 (bx)⟩ is free. Second, suppose that p > 3. Take k = 3 and let ω p = 2π p . Consider the function sin(x) x f(x) = 1−2 1+2 sin(x) which is plotted, together with the functions y = sin(x) and y = 2 , π in Figure 14.1 on the interval [0, 2 ]. Observe that ω p belongs to this interval because p ≥ 5.

14.2 Free groups in group rings over Hamiltonian groups

| 13

y = sin(x) y = 2x y = √3 − 2√2 2π 11 2π 29

2π 7

2π 5

2π 13

y = f(x)

y = −√3 − 2√2 Fig. 14.1: Plots of y =

1−2 sin(x) , 1+2 sin(x)

y = sin(x) and y =

x 2

for x ∈ [0, 2π ].

We have 󵄨󵄨 v 󵄨󵄨 󵄨󵄨󵄨 1 + ζ p i − ζ p2 󵄨󵄨󵄨 󵄨󵄨󵄨 ζ p−1 − ζ p + i 󵄨󵄨󵄨 󵄨󵄨󵄨 (1 − 2 sin(ω p ))i 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 = 󵄨󵄨󵄨f(ω )󵄨󵄨󵄨 . 󵄨󵄨 󵄨=󵄨 󵄨= 󵄨=󵄨 󵄨󵄨 σ(v) 󵄨󵄨󵄨 󵄨󵄨󵄨 1 − ζ p i − ζ p2 󵄨󵄨󵄨 󵄨󵄨󵄨 ζ p−1 − ζ p − i 󵄨󵄨󵄨 󵄨󵄨󵄨󵄨 (−1 − 2 sin(ω p ))i 󵄨󵄨󵄨󵄨 󵄨 p 󵄨 󵄨 󵄨 󵄨 󵄨 m m By Proposition 14.2.2, for ⟨v , τ(v )⟩ to be free it is enough to choose m such that |f(ω p )|m < 3 − 2√2. Elementary calculus shows that f is decreasing in the interval 2 π 2 1 √ √ [0, 2π ]. Moreover f ( 2π 29 ) < 0.1589 < 3 − 2 2 and f ( 2 ) = 9 < 3 − 2 2. Since m ≥ 2, 2π 2 m √ we have |f(ω p )| < 3 − 2 2 for every p ≤ 29. However f ( 31 ) > 0.18 > 3 − 2√2 and so for p ≥ 31 we need to take m > 2. To determine such an m we observe that, as 0 ≤ ω p ≤ 2π , we have ω p < 2 sin(ω p ) (see Figure 14.1). Thus 1 − 2 sin(ω p ) 2 2 p − 2π |f(ω p )| = = −1< −1= . (14.2.1) 1 + 2 sin(ω p ) 1 + 2 sin(ω p ) ωp + 1 p + 2π Now, if m ≥ (

1+√2 2π p m

then

m p + 2π 4π 4π 4mπ >1+ ≥ 3 + 2√2. ) = (1 + ) >1+m p − 2π p − 2π p − 2π p

(14.2.2)

So, from (14.2.1) and (14.2.2), we deduce that if m ≥ 1+2π√2 p then |f(ω p )|m < (3 + 2√2)−1 = 3−2√2, as desired. Notice that 2− 2p ≥ 89 > 1+2π√2 and hence the requirement m≥

1+√2 2π p

is satisfied for m = ϕ(4p) = 2(p − 1). This proves the following result.

Theorem 14.2.3 (Ferraz [58]). Let p an odd prime and consider the group G = Q8 × C p = ⟨a, b, x | a4 = a2 b2 = x p = (a, x) = (b, x) = 1, a b = a−1 ⟩. Let k = 5 if p = 3 and k = 3 otherwise. Then ⟨u k,2(p−1) (ax), u k,2(p−1) (bx)⟩ is free. Combining Theorem 14.1.11, Theorem 14.2.3 and the discussion at the beginning of this section we have the following result, which was proved independently by HartleyPickel [81] and Sehgal [201]. Theorem 14.2.4 (Hartley-Pickel, Sehgal). Let G be a finite group. If U(ℤG) is neither abelian nor a Hamiltonian 2-group (i.e., U(ℤG) is not finite and G is not abelian) then U(ℤG) contains a nonabelian free subgroup.

14 | 14 Free Groups

14.3 Attractors In Section 5.5 we mentioned a fundamental result of Tits, the so called Tits Alternative that states that every subgroup of GLn (F), for F a field of characteristic 0, has either a solvable subgroup of finite index or a nonabelian free subgroup [218]. One of the tools introduced by Tits, to prove that some groups generated by diagonalizable non-singular linear maps are free, consists in observing that the eigenspaces of large eigenvalues behave as attractors, i.e. if T is diagonalizable then T n (v) approaches to the eigenspaces with large eigenvalues for most v’s. This technique has been extended by Gonçalves and Passman in [73] to non-necessarily diagonalizable linear maps of finite dimensional vector spaces over a locally compact field. In this section we present this method. For simplicity, we restrict ourself to the case of linear transformations of vector spaces over the complexes. Let V be a finite dimensional vector space over ℂ. A norm in V is a map ‖.‖ : V → ℝ satisfying the following conditions for every a ∈ ℂ and v, w ∈ V: (1) ‖v‖ = 0 if and only if v = 0. (2) ‖av‖ = |a|‖v‖. (3) ‖v + w‖ ≤ ‖v‖ + ‖w‖. Every norm in V defines a metric in V and all the metrics are equivalent and hence define the same topology. In other words, if ‖ ⋅ ‖ and ‖ ⋅ ‖󸀠 are two norms then there are positive real numbers a and b such that a‖v‖󸀠 ≤ ‖v‖ ≤ b‖v‖󸀠 for every v ∈ V. Every bounded closed subset of V is compact and, in particular, the unit sphere 𝕊 = {v ∈ V : ‖v‖ = 1} is compact. If W is another finite dimensional space with a norm (also denoted ‖ ⋅ ‖) and f : V → W is a linear map then we set ‖f‖ = max{‖f(x)‖ : x ∈ 𝕊}. Note that the maximum exists because every linear map is continuous. This defines a norm in the vector space consisting of the linear maps V → W. Moreover, for every x ∈ V, ‖f(x)‖ ≤ ‖f‖ ⋅ ‖x‖ (x ∈ V). (14.3.1) In fact ‖f‖ = max{κ ≥ 0 : ‖f(x)‖ ≤ κ‖x‖, for every x ∈ V}. In particular, if f is invertible then ‖f −1 (x)‖ ≥

‖x‖ . ‖f‖

(14.3.2)

We fix an integer n ≥ 2, set V = ℂn , the complex vector space of n-tuples, and in the rest of the section we fix the norm ‖(a1 , . . . , a n )‖ = max{|a i | : i = 1, . . . , n}.

14.3 Attractors

|

15

A subset of V is called projective if it contains a non-zero element and it is closed under scalar multiplication by non-zero complex numbers. Therefore, except for the possible presence or not of zero, the projective subsets of V are in one-to-one correspondence with the subsets of the projective space of dimension n − 1. Two projective subsets are said to be disjoint if their intersection is contained in {0}. We define the distance d between projective subsets X and Y of V, between non-zero elements v, w ∈ V and between a projective set and a non-zero vector as follows: d(X, Y) = inf{‖x − y‖ : x ∈ X ∩ 𝕊 and y ∈ Y ∩ 𝕊}, d(v, w) = d(ℂv, ℂw) = inf{‖ax − by‖ : a, b ∈ ℂ, ‖ax‖ = ‖by‖ = 1}, d(v, X) = d(X, v) = d(X, ℂv) = inf{d(x, v) : x ∈ X}. Observe that the distance between projective subsets does not satisfy the triangular inequality. For example, if W1 , W2 and W are non-zero subspaces of V such that W = W1 ⊕ W2 then d(W1 , W2 ) > 0 = d(W1 , W) + d(W, W2 ). However, in the following lemma it is proved that the triangular inequality holds for distances between points and between points and projective subsets. Lemma 14.3.1. (1) If W1 and W2 are non-zero subspaces of V then d(W1 , W2 ) = ‖w1 − w2 ‖ for some w1 ∈ W1 ∈ and w2 ∈ W2 . In particular, if W1 ∩ W2 = 0 then d(W1 , W2 ) > 0. (2) The function d defines a metric on the projective space of V with diameter at most 2. (3) If v, w ∈ V are both non-zero then d(v, w) ≤ 2‖v−w‖ ‖v‖ . (4) If X and Y are projective subsets of V and 0 ≠ v ∈ V then d(X, Y) = inf{d(x, y) : 0 ≠ x ∈ X, 0 ≠ y ∈ Y} ≤ d(X, v) + d(v, Y). Proof. (1) This is a consequence of the fact that (x, y) 󳨃→ ‖x − y‖ is continuous and W1 ∩ 𝕊 and W2 ∩ 𝕊 are compact. (2) Let 0 ≠ u, v, w ∈ V. Clearly d(u, v) = d(v, u) and, if u and v correspond to different points of the projective space, then ℂu ∩ ℂv = 0 and hence d(u, v) > 0, by (1). Also, by (1), there are a, b, b󸀠 , c󸀠 ∈ ℂ such that ‖au‖ = ‖bv‖ = ‖b󸀠 v‖ = ‖c󸀠 w‖ = 1, d(u, v) = ‖au − bv‖ and d(v, w) = ‖b󸀠 v − c󸀠 w‖. Clearly, we obtain that |b| = |b󸀠 | and 󸀠 then ‖cw‖ = ‖c󸀠 w‖ = 1 and hence ‖ bb󸀠 (b󸀠 v − c󸀠 w)‖ = ‖b󸀠 v − c󸀠 w‖. Thus, if we set c = bc b󸀠 d(v, w) = ‖bv − cw‖. Therefore, d(u, v) + d(v, w) = ‖au − bv‖ + ‖bv − cw‖ ≥ ‖au − cw‖ ≥ d(u, w). This proves that d defines a metric on the projective space. Finally, d(u, v) = ‖au − bv‖ ≤ ‖au‖ + ‖bv‖ = 2 and so the diameter is at most 2. (3) Let u󸀠 =

v−w , ‖v‖

u󸀠󸀠 =

w ‖v‖ − ‖w‖ ⋅ ‖w‖ ‖v‖

and

u = u󸀠 − u󸀠󸀠 =

v w − . ‖v‖ ‖w‖

16 | 14 Free Groups 󵄩 v 󵄩󵄩 󵄩󵄩 w 󵄩󵄩 󸀠 Then 󵄩󵄩󵄩󵄩 ‖v‖ 󵄩󵄩󵄩 = 󵄩󵄩󵄩 ‖w‖ 󵄩󵄩󵄩 = 1, ‖u ‖ =

‖v−w‖ ‖v‖

and ‖u󸀠󸀠 ‖ =

| ‖v‖−‖w‖ | ‖v‖

d(v, w) ≤ ‖u‖ ≤ ‖u󸀠 ‖ + ‖u󸀠󸀠 ‖ ≤

≤

‖v−w‖ ‖v‖

and therefore

2‖v − w‖ , ‖v‖

as desired. (4) The first equality is obvious. We prove the inequality by contradiction. So, let α = d(X, v), β = d(v, Y) and assume that d(X, Y) = λ > α + β. Put ϵ = λ − (α + β) > 0. Then, there are x ∈ X and y ∈ Y such that d(x, v) < α + 2ϵ and d(y, v) < β + 2ϵ . Therefore λ = d(X, Y) ≤ d(x, y) < α + β + ϵ = λ, a contradiction. For a projective subset X of V and a positive real number ϵ we define B(X, ϵ) = {v ∈ V : 0 ≠ v, d(v, X) ≤ ϵ}. Clearly B(X, ϵ) is a projective subset of V. Lemma 14.3.2. Let W1 , W2 and W be non-zero subspaces of V such that V = W1 ⊕ W2 . Let π1 : V → W1 and π2 : V → W2 denote the projections along this decomposition. Let μ be a positive real number such that 2μ ≤ d(W2 , W). Then, the following inequalities hold for every x ∈ B(W, μ). (1) d(B(W, μ), W2 ) ≥ μ and, in particular, B(W, μ) ∩ W2 = 0 and π1 (x) ≠ 0. (2) 2‖π1 (x)‖ ≥ μ‖x‖ Proof. Let x ∈ B(W, μ). Because of Lemma 14.3.1 (4), Clearly μ + d(x, W2 ) ≥ d(W, x) + d(x, W2 ) ≥ d(W, W2 ) ≥ 2μ. This proves (1). Because of Lemma 14.3.1 (2), the projective space has diameter at most 2. Hence μ ≤ 1. If x ∈ W1 then x = π1 (x) and therefore 2‖π1 (x)‖ = 2‖x‖ > μ‖x‖. Otherwise 0 ≠ π2 (x) ∈ W2 , and hence, by Lemma 14.3.1 (3), μ ≤ d(B(W, μ), W2 ) ≤ d(x, π2 (x)) ≤

2‖x − π2 (x)‖ 2‖π1 (x)‖ = . ‖x‖ ‖x‖

This finishes the proof of (2). Let T be an endomorphism of V, let X and Y be projective subsets of V and let μ > 0. We say that Y is a (T; μ)-attractor of X if for every ϵ > 0, there is m ϵ ≥ 1 such that T m (B(X, μ)) ⊆ B(Y, ϵ) for every m ≥ m ϵ . We first consider attractors for a diagonalizable linear transformation T. The basic idea is that if λ1 and λ2 are eigenvalues of T and v i belong to the λ i -eigenspace of T, for i = 1, 2, then T(v1 + v2 ) = λ1 v1 + λ2 v2 . Thus T pulls in the direction of v1 with a |λ1 | “force” and in the direction of v2 with |λ2 | “force”. If |λ1 | > |λ2 | then T is “pulling” stronger in the direction of the λ1 -eigenspace than in that of the λ2 -eigenspace. Hence the sequence (T n (v)) approaches the λ1 -eigenspace (See Figure 14.2). So the λ-eigenspaces with large |λ| are good candidates for attractors of

14.3 Attractors

|

17

T 5 (v) T 4 (v) T 3 (v) T 2 (v) T(v) v

Fig. 14.2: Evolution of T n (v) for T a linear map with eigenvalues 2 and 1.4. The sequence approaches the eigenspace with larger eigenvalue.

T while the λ-eigenspaces with |λ| small are good candidates for attractors of T −1 . Clearly one should exclude from the parts attracted the subspace generated by the eigenspaces. We start with an easy lemma. Lemma 14.3.3. Let r and s be non-negative real numbers and let T be a diagonalizable linear map of V such that s ≤ |λ| ≤ r for all the eigenvalues λ of T. Then there is k > 0 such that k−1 s m ‖v‖ ≤ ‖T m (v)‖ ≤ kr m ‖v‖ for every m ≥ 1 and v ∈ V. Proof. Let v1 , . . . , v n be a basis of eigenvectors of T and assume that T(v i ) = λ i v i for every i. Consider the norm ‖ ⋅ ‖󸀠 given by ‖α1 v1 + ⋅ ⋅ ⋅ + α n v n ‖󸀠 = max{|α i | : i = 1, . . . , n}, where α1 , . . . , α n ∈ ℂ. As ‖ ⋅ ‖ and ‖‖󸀠̇ are equivalent norms, there are positive real numbers a and b such that a‖v‖󸀠 ≤ ‖v‖ ≤ b‖v‖󸀠 for every v ∈ V. If v = α1 v1 + ⋅ ⋅ ⋅ + α n v n then 󸀠 m ‖T m (v)‖ ≤ b‖T m (v)‖󸀠 = b‖λ1m α1 v1 + . . . λ m n α n v n ‖ ≤ br max{|α i | : i = 1, . . . , n}

= br m ‖v‖󸀠 ≤ Similarly ‖T m (v)‖ ≥

a m b s ‖v‖.

b m r ‖v‖. a

Thus k =

b a

satisfies the desired condition.

Proposition 14.3.4. Let T be a diagonalizable non-singular endomorphism of a finite dimensional complex vector space V. Let r be a positive real number, let W1 be the subspace of V generated by the λ-eigenspaces with |λ| ≥ r and let W2 be the subspace generated by the other eigenspaces. Assume W1 and W2 are non-zero subspaces of V 2) and suppose W is a non-zero subspace of V with W ∩ W2 = 0 and let 0 < μ ≤ d(W,W . 2 Then W1 is a (T, μ)-attractor of W. Proof. Since T is diagonalizable, V = W1 ⊕ W2 and W1 , W2 and W satisfy the assumptions of Lemma 14.3.2. As in this lemma, let π i denote the projection of V onto W i along this decomposition. Let h = ‖π2 ‖. Then ‖π2 (v)‖ ≤ h‖v‖ for every v ∈ V. Let s be the

18 | 14 Free Groups maximum of all the eigenvalues of the restriction of T to W2 . Then 0 ≤ s < r and, by Lemma 14.3.3, there exist positive real numbers k1 and k2 such that ‖T m (a)‖ ≥ k1 r m ‖a‖ and ‖T m (b)‖ ≤ k2 s m ‖b‖ for every m ≥ 1, a ∈ W1 and b ∈ W2 . Let x ∈ B(W, μ) with ‖x‖ = 1. By Lemma 14.3.2 (1), we have x ∈ ̸ W2 and π1 (x) ≠ 0. Thus, both T m (x) and T m (π1 (x)) are non-zero. By Lemma 14.3.1 (3) and Lemma 14.3.2 (2) we have d(T m (x), W1 ) ≤ d(T m (x), T m (π1 (x))) 2‖T m (x − π1 (x))‖ ‖T m (π1 (x))‖ 2‖T m (π2 (x))‖ = ‖T m (π1 (x))‖ 2k2 s m ‖π2 (x)‖ ≤ k1 r m ‖π1 (x)‖ 4k2 h s m ≤ ( ) . k1 μ r

≤

s m 2h As 4k k1 μ is a positive constant and 0 < r < 1, for every ϵ > 0 we have d(T (x), W 1 ) ≤ ϵ for m sufficiently large. Thus W1 is a (T, μ) attractor of W, as desired.

Let T be a diagonalizable non-singular homomorphism of a finite dimensional complex vector space V. We say that V = T+ ⊕ T0 ⊕ T− is a T-decomposition of V if there are 0 < s < r such that T+ (respectively, T− , T0 ) is the sum of the λ-eigenspaces with |λ| ≥ r (respectively, |λ| ≤ s, s < |λ| < r) and both T+ and T− are non-trivial. Observe that the existence of a T-decomposition implies the presence of eigenvalues with different absolute value and in particular of eigenvalues which are not roots of unity. So T has to be of infinite order. Now we show how to apply Proposition 14.3.4 to produce free groups from diagonalizable endomorphisms. Theorem 14.3.5 (Gonçalves-Passman [73]). Let V be a finite dimensional complex vector space and let T1 and T2 be diagonalizable automorphisms of V. For every i = 1, 2, let V = T i+ ⊕ T i0 ⊕ T i− be a T i -decomposition. If the eight intersections T i± ∩ (T j0 ⊕ T j± ) m m with {i, j} = {1, 2} are trivial then ⟨T1 1 , T2 2 ⟩ is free for m1 and m2 sufficiently large. Proof. Let {i, j} = {1, 2}. Let 2μ be the minimum of the eight distances d(T i± , T j0 ⊕T j± ). By Lemma 14.3.1 (1), μ > 0. By Proposition 14.3.4, T i+ is a (T i ; μ)-attractor of T j0 ⊕ T j− . One the other hand, V = T i+ ⊕ T i0 ⊕ T i− is a T i−1 -decomposition. Thus, T i− is a (T i−1 ; μ)attractor of T j+ ⊕ T j0 . Therefore, if we let P k = B(T k+ , μ) ∪ B(T k− , μ) for k = 1, 2 then m T k k (P j ) ⊆ P k for |m k | sufficiently large. Moreover P1 ≠ P2 because T i± ⊆ P i and P i ∩ T j± = 0. The theorem then follows from the Ping-Pong Lemma (Lemma 14.1.1). The proof of Theorem 14.3.5 shows how attractors can be applied in the search of free groups. Namely, given two non-singular endomorphisms T1 and T2 , the basic idea is to find two different projective subsets X1 and X2 such that X1 is a (T1± ; μ)-attractor of X2 and X2 is a (T2± ; μ)-attractor of X1 . Then apply the Ping-Pong Lemma for suitable

14.3 Attractors

|

19

powers of T1 and T2 with X1 and X2 in the roles of the two sides of the “Ping-Pong board”. The assumptions of Theorem 14.3.5 imply that the four subspaces T i± have the same dimension (Problem 14.3.2). Corollary 14.3.6. Let V be a finite dimensional complex vector space and let T1 and T2 be diagonalizable automorphisms of V. For every i = 1, 2, let V = T i+ ⊕ T i0 ⊕ T i− be a T i -decomposition. Let π i± : V → T i± be the projection along this decomposition and consider each π i± as an endomorphism of V. Suppose that the four subspaces T i± have the same dimension, say r. If the eight compositions π i± ∘ π j± with {i, j} = {1, 2} have m m rank r then ⟨T1 1 , T2 2 ⟩ is free for m1 and m2 sufficiently large. Proof. The assumptions imply that if {i, j} = {1, 2} then π iϵ1 restricts to an isomorphism T jϵ2 → T iϵ1 for every ϵ1 , ϵ2 ∈ {+, −}. Thus T jϵ2 ∩ ker(π iϵ1 ) = 0. Moreover, if {ϵ2 , ϵ󸀠2 } = {+, −} then ker(π iϵ󸀠2 ) = T i0 ⊕ T iϵ2 . Therefore, T jϵ1 ∩ (T i0 + T iϵ2 ) = T jϵ1 ∩ ker(π iϵ󸀠2 ) = {0}. Hence T1 and T2 satisfy the hypothesis of Theorem 14.3.5 and the result follows. We now consider attractors for operators of the form T = I + τ with τ2 = 0. Then, clearly, T n (v) = v + nτ(v) for every n ∈ ℤ. Therefore, we can visualize T and T −1 as “pulling” in the direction of Im(τ) (see Figure 14.3). Hence Im(τ) is a good candidate to be an attractor for T. Of course, there is no “pulling” for the elements of ker(τ). More precisely the following property holds. T 4 (v) T 3 (v) T 2 (v) 3τ(v)

T(v) v T −1 (v)

4τ(v)

2τ(v) τ(v)

T −2 (v) T −3 (v) T −4 (v)

−τ(v) −2τ(v)

−3τ(v) −4τ(v) Fig. 14.3: Evolution of T n (v) for a linear map T of the form 1 + τ, with τ 2 = 0 and for v ∉ ker(τ).

20 | 14 Free Groups Proposition 14.3.7. Let τ be an endomorphism of V = ℂn with τ2 = 0 and let I = Im(τ), K = ker(τ) and T = I + τ. If W a non-zero subspace of V such that W ∩ K = 0 and μ ≤ d(W,K) then I is a (T; μ)-attractor and a (T −1 ; μ)-attractor of W. 2 Proof. Without loss of generality, we may assume that V = K ⊕ W. Let π K : V → K and π W : V → W denote the projections along this decomposition. Let x ∈ B(W, μ). Lemma 14.3.2, applied to W1 = W and W2 = K, yields that d(B(W, μ), K) ≥ μ

and

2‖π W (x)‖ ≥ μ‖x‖.

(14.3.3)

The restriction of τ to W yields an isomorphism σ : W → I. Let s = ‖σ−1 ‖. Applying (14.3.2) to f = σ−1 we have ‖τ(w)‖ ≥ ‖w‖ s for every w ∈ W. Combining this with (14.3.3) we obtain ‖π W (x)‖ μ‖x‖ ‖τ(π W (x))‖ ≥ ≥ . (14.3.4) s 2s We have x + mτ(π W (x)) = x + mτ(x) = T m (x) ≠ 0 and 0 ≠ τ(π W (x)) ∈ I. Then applying Lemma 14.3.1 (3) and (14.3.4) we obtain d(T m (x), I) ≤ d(T m (x), mτ(π W (x))) ≤ Therefore, if |m| ≥ attractor.

4s μϵ

2‖T m (x) − mτ(π W (x))‖ 2‖x‖ 4s = ≤ . ‖mτ(π W (x))‖ |m| ⋅ ‖τ(π W (x))‖ |m|μ

then T m (x) ∈ B(I, ϵ). Thus I is a (T; μ)-attractor and a (T −1 ; μ)-

Theorem 14.3.8 (Gonçalves-Passman [73]). Let V be a finite dimensional complex vector space and let T and S be non-singular endomorphisms of V. Suppose that T = I + τ with τ2 = 0 ≠ τ and S is diagonalizable. Let K = ker(τ), I = Im(τ) and let V = S+ ⊕S0 ⊕S− be an S-decomposition of V. Suppose that the four intersections I ∩ (S0 ⊕ S± ) and K ∩ S± are trivial. Then ⟨T m1 , S m2 ⟩ is free for m1 and m2 sufficiently large. Proof. Let 2μ be the, minimum of the four distances d(I, S0 ⊕ S± ) and d(K, S0 ⊕ S± ). By Lemma 14.3.1 (1), μ > 0. By Proposition 14.3.7, I is a (T; μ)-attractor and a (T −1 ; μ)attractor of S0 ⊕ S± . Moreover, by Proposition 14.3.4, S+ is a (S; μ)-attractor of I and S− is a (S−1 ; μ)-attractor of I. So, taking P = B(I, μ) and Q = B(S+ , μ) ∪ B(S− , μ), we have T m1 (P) ⊆ Q and S m2 (Q) ⊆ P for m1 and m2 sufficiently large. Thus the theorem follows from the Ping-Pong Lemma.

Problems 14.3.1. The original proof of Hartley and Pickel for Corollary 14.2.4 is not constructive and used Sanov’s Theorem (Proposition 14.1.4) and Tits’ version of Theorem 14.3.5. Find such a proof, that is, prove that if G is a finite group which is neither abelian nor a Hamiltonian 2-group then U(ℤG) contains a nonabelian free group using

14.4 Free companions

| 21

only Lemma 4.6.9 (4), Proposition 14.1.4 and Theorem 14.3.5. By Hamiltonian that eigenspaces 14.3.2. Prove that, under the conditions of Theorem 14.3.5, the four subspaces T i± have the same dimension and under the conditions of Theorem 14.3.8, S+ and S− have the same dimension. 14.3.3. Let V be a finite dimensional complex vector space. For a diagonalizable nonsingular endomorphism A of V let A(T) denote the sum of the eigenspaces of T for eigenvalues of maximum absolute value and A󸀠 (T) the sum of the remaining eigenvalues. Let T1 and T2 be diagonalizable non-singular endomorphisms of V. Proposition 3.12 of [218] states that if A(T i ) and A(T i−1 ) are one-dimensional and A(T i ) ∪ m m A(T i−1 ) ⊆ V\(A󸀠 (T j )∪A󸀠 (T j−1 )) for every {i, j} = {1, 2} then ⟨T1 1 , T2 2 ⟩ is free for m1 and m2 sufficiently large. Prove that the Tits Proposition is equivalent to Theorem 14.3.5. Hint: Let T1 and T2 satisfy the assumptions of Theorem 14.3.5 and let k be the common dimension of A(T i ) and A󸀠 (T i−1 ) for each i = 1, 2 (Problem 14.3.2). Consider the endomorphisms induced by T1 and T2 in the exterior product Γ k V. 14.3.4. Let τ1 and τ2 be non-zero endomorphisms of a finite dimensional complex vector space such that τ21 = τ22 = 0. Let T i = I + τ i , K i = ker(τ i ) and I i = Im(τ i ), for i = 1, 2. Prove that if the two intersections I i ∩ K j with {i, j} = {1, 2} are trivial then m m ⟨T1 1 , T2 2 ⟩ is free for m1 and m2 sufficiently large. (Actually find two proofs, one using the methods of Section 14.1 and the other using attractors).

14.4 Free companions Let u be an element of a group Γ. A free companion of u in Γ is an element v ∈ Γ such that ⟨u, v⟩ is nonabelian free. By Theorem 14.1.11 every non-trivial bicyclic unit in ℤG has a free companion in U(ℤG) which is a bicyclic unit. We cannot expect a similar result for Bass units because there are non-trivial torsion Bass units. More precisely, a Bass unit u k,m (g) is torsion if and only if k ≡ ±1 mod |g| (Lemma 1.2.1). Even more, if n ∈ ̸ {1, 2, 3, 4, 6} then the integral group ring of the dihedral group D2n has noncentral non-torsion Bass units but all of them have finite order modulo the center of U(ℤD2n ) (see Problem 5.2.3). Of course, such Bass units cannot have a free companion in U(ℤG). In this section we investigate the existence of free companions of Bass units. The main result in the section is a result of Gonçalves and del Río which states that, under certain mild conditions, if u is a Bass unit of infinite order modulo the center of U(ℤG) then some power of u has a free companion among the powers of Bass units or bicyclic units (Theorem 14.4.13). The strategy for the proof of this result was first introduced by Gonçalves and Passman to prove that if G is a nonabelian group of order non-divisible by 6 then U(ℤG) has a nonabelian free subgroup generated by Bass

22 | 14 Free Groups units [73]. Under the same assumptions, Gonçalves and del Río proved that U(ℤG) has a free subgroup generated by a Bass unit and a bicyclic unit [69]. We start by investigating the representations of Bass units. Let n be an integer greater than 1. In Section 14.2 we have seen that if ρ is a representation of a finite group G then the image under ρ of every Bass unit by ρ is diagonalizable and if g is an element of G of order n then the eigenvalues of ρ(u k,m (g)) are of the form u k,m (ζ na )) = η k (ζ na )m with a = 0, 1, . . . , n (by convention η k (1) = 1). Motivated by Theorem 14.3.4, we 2 now investigate the absolute value of the cyclotomic units η k (ζ na ). Using ζ2n = ζ n and −a a πa ζ2n − ζ2n = 2i sin( n ) we have η k (ζ na ) =

kπa −ak ak ak ζ2n − ζ2n ζ nak − 1 ζ2n a(k−1) sin( n ) = ⋅ = ζ . 2n a −a a ζ na − 1 ζ2n sin( πa ζ2n − ζ2n n )

Therefore

󵄨󵄨 sin( kπa ) 󵄨󵄨 󵄨 n 󵄨󵄨󵄨 |η k (ζ na )| = 󵄨󵄨󵄨󵄨 󵄨󵄨. 󵄨󵄨 sin( πa n ) 󵄨󵄨 This suggests to study the real function f k (x) =

(14.4.1)

sin(kπx) sin(πx)

in the open interval (0, 1). Clearly f k (1 − x) = (−1)k−1 f k (x), and hence it is enough to consider the function on the interval (0, 12 ]. Moreover, we can extend f k continux ously to [0, 1] by setting f k (0) = limx→0 f k (x) = k limx→0 sin(kx) kx sin(x) = k and f k (1) = (−1)k−1 f(0) = (−1)k−1 k. Figure 14.4 represents the plot of f k for k = 2, 7 and 21. In this plot we observe that f k decreases from k to 0 in the interval [0, 1k ] and then remains relatively small in the interval [ 1k , 12 ]. The following lemma shows that this is a general phenomenon for k ≥ 2. 21

1 f21 ( 42 )

7 1 f7 ( 14 )

2

f2 ( 14 ) 1 42

1 14

1 4

Fig. 14.4: Plot of y = f k (x) in the interval [0, 12 ] for k = 2, 7 and 21.

1 2

14.4 Free companions | 23

Lemma 14.4.1. Let k be an integer with k ≥ 2 and let r, x ∈ ℝ with 0 ≤ r ≤ r ≤ x ≤ 12 . Then |f k (x)| ≤ f k (r) > 1.

1 2k

and

Proof. First we prove that f k is strictly decreasing on the interval (0, 1/k]. This is easy in the interval [1/2k, 1/k] because in this interval the numerator is decreasing and the numerator is increasing. To prove that f k is also decreasing on (0, 1/2k] we consider 1 the functions g(x) = tan(πx) and h(x) = tan(kπx)/k for x ∈ (0, 2k ). Observe that f k󸀠 (x) =

kπ cos(kπx) cos(πx) sin2 (πx)

(g(x) − h(x)).

1 , both cos(kπx) are cos(πx) are positive, and hence it is enough to Since 0 < x < 2k show that g(x) < h(x). We have g(0) = h(0) and g 󸀠 (x) = cosπ2 πx < cos2 π(kπx) = h󸀠 (x) for x ∈ (0, 1/2k], since the cosine function is strictly decreasing in the interval (0, π/2]. Hence for each x ∈ (0, 1/2k] we have g(x) < h(x), as desired. Thus, f k indeed is decreasing on (0, 1/k]. Let 0 < r ≤ 1/2k. Using the inequalities y(1 − y2 /6) ≤ sin(y) ≤ y, which hold for each y > 0, we have

sin(kπr) kπr(1 − (kπr)2 /6) π2 ≥ ≥ k (1 − ) sin(πr) πr 24 2 π ≥ 2 (1 − ) > 1.177 > 1. 24

f k (r) =

(14.4.2)

By the previous paragraph f k (r) ≥ f k (x) for every r ≤ x ≤ 1/k. Assume now that 1/k < x ≤ 1/2. As the sine function is increasing on [0, π/2], we have π (π/k)2 π π2 π sin(πx) ≥ sin ( ) ≥ (1 − ) ≥ (1 − ). k k 6 k 24 Combining this with π(1 − π 2 /24)2 > 1.089 > 1 and (14.4.2) we deduce |f k (x)| ≤

1 k π2 ≤ < k (1 − ) ≤ f k (r), 2 |sin(πx)| π(1 − π /24) 24

as desired. The following lemma determines the maximum of {|η k (ζ na )| : a ∈ ℤ}. Observe that η1 (ζ na ) = 1 so the only interesting case holds for k ≢ 1 mod n. Lemma 14.4.2. Let n and k be relatively prime integers and assume that k ≢ ±1 mod n. Let l ∈ ℤ with kl ≡ 1 mod n. Then, for every a ∈ {1, . . . , n − 1}, 1 > |η k (ζ nl )| ≤ |η k (ζ na )| = |η k (ζ n−a )| ≤ |η k (ζ n )| > 1. Thus, if m is a positive integer with k m ≡ 1 mod n then 1 > |u k,m (ζ nl )| ≤ |u k,m (ζ na )| = |u k,m (ζ n−a )| ≤ |u k,m (ζ n )| > 1. Proof. Let a ∈ {1, . . . , n − 1}. The equality |η k (ζ na )| = |η k (ζ n−a )| is clear.

24 | 14 Free Groups As η n+k (ζ na ) = η k (ζ na ), one may assume that 2 ≤ k ≤ n − 2. Moreover, by equation (14.4.1), 󵄨󵄨 sin ( (n−k)πa ) 󵄨󵄨 󵄨󵄨 sin ( kπa ) 󵄨󵄨 󵄨 󵄨󵄨 n n a 󵄨󵄨󵄨 = 󵄨󵄨󵄨 󵄨 |η n−k (ζ na )| = 󵄨󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 πa πa 󵄨󵄨 = |η k (ζ n )|. 󵄨󵄨 sin ( n ) 󵄨󵄨 󵄨󵄨 sin ( n ) 󵄨󵄨 1 Thus one also may assume that 2 ≤ k ≤ 2n . Then an ≥ 1n ≤ 2k . Applying Lemma 14.4.1 1 a with r = n , we deduce that |η k (ζ n )| = |f k (a/n)| ≤ |f k (1/n)| = |η k (ζ n )| > 1. Applying this to the units of the form η l (ζ nka ) and using (1.2.2), we obtain |η k (ζ na )| = |η l (ζ nka )|−1 ≥ |η l (ζ nk )|−1 = |η k (ζ nl )| < 1.

In the following lemma we investigate when u k,m (ζ na ) and u k,m (ζ nb ) have the same absolute value provided n is a prime power. Lemma 14.4.3. Let p be a prime integer and n integer that is a power of p. (1) Suppose that ξ1 +⋅ ⋅ ⋅+ ξ r = ε1 +⋅ ⋅ ⋅+ ε r with r < p and ξ1 , . . . , ξ r , ε1 , . . . , ε r complex n-th roots of unity. Then there is a permutation σ ∈ S r such that ε i = ξ σ(i) for every i. (2) Assume that p ≥ 5. Let 2 ≤ k ≤ n − 2 and m a positive integer with k m ≡ 1 mod n. Let a, b ∈ ℤ. Then (a) |u k,m (ζ na )| = 1 if and only if ka ≡ ±a mod n. In particular, if p ∤ a then |u k,m (ζ na )| ≠ 1. (b) If |u k,m (ζ na )| = |u k,m (ζ nb )| ≠ 1 then a ≡ ±b mod n. (c) If n divides m and a ≡ ±b mod n then u k,m (ζ na ) = u k,m (ζ nb ). Proof. (1) Arguing by contradiction and induction on r we may assume, without loss of generality, that ε i ≠ ξ j for every i, j. Moreover, multiplying with ξ1−1 and reordering the ξ i ’s, we also may assume that ξ i = 1 if and only if 1 ≤ i ≤ k, for some k ≥ 1. Let T = pn Trℚ(ζ n )/ℚ : ℚ(ζ n ) → ℚ. If ξ is an n-th root of unity then p − 1, { { T(ξ) = {−1, { {0;

if ξ = 1; if ξ has order p; otherwise.

So, T(ξ1 ) = ⋅ ⋅ ⋅ = T(ξ k ) = p − 1 and T(ϵ i ) ≤ 0 for every 1 ≤ i ≤ r. By assumption, r < p and thus r − (k + 1) ≤ p − 2. Hence, 0 < k(p − 1) + T(ξ k+1 + ⋅ ⋅ ⋅ + ξ r ) = T(ξ1 + ⋅ ⋅ ⋅ + ξ r ) = T(ε1 + ⋅ ⋅ ⋅ + ε r ) ≤ 0, a contradiction. (2a) Suppose |u k,m (ζ na )| = 1. As n is odd, a ≡ 2a1 mod n for some integer a1 . So, 1= a k

Since

−a k

ζ n 1 −ζ n 1 a −a ζ n 1 −ζ n 1

|u k,m (ζ na )|

󵄨m 󵄨󵄨 2a1 k 󵄨󵄨 a1 k −2a1 k 󵄨󵄨m 󵄨󵄨 󵄨󵄨 ζ2n − ζ2n 󵄨󵄨 ζ n − ζ n−a1 k 󵄨󵄨󵄨 󵄨 󵄨󵄨 . 󵄨 󵄨 = 󵄨󵄨 2a 󵄨 = 󵄨󵄨 a1 󵄨󵄨 ζ 1 − ζ −2a1 󵄨󵄨󵄨 󵄨󵄨 ζ n − ζ n−a1 󵄨󵄨󵄨 󵄨 󵄨 󵄨 2n 󵄨 2n a k

−a1 k

∈ ℝ, it has to be ±1. Hence, ζ n 1 − ζ n

a

−a

= ±(ζ n 1 − ζ n 1 ). So, by part (1),

a1 k ≡ ±a1 mod n and hence ak ≡ ±a mod n. The converse is clear. (2b) Suppose |u k,m (ζ na )| = |u k,m (ζ nb )| ≠ 1. Note that this implies that ζ na ≠ 1 and b ζ n ≠ 1, and by part (2a) also ζ nak ≠ 1 and ζ nbk ≠ 1. Arguing as above, let a1 and b1 be

14.4 Free companions | 25

integers such that a ≡ 2a1 mod n and b ≡ 2b1 mod n. We obtain 󵄨m 󵄨󵄨 a1 k 󵄨󵄨 b1 k −b1 k 󵄨󵄨m 󵄨󵄨 󵄨󵄨 ζ n − ζ n−a1 k 󵄨󵄨󵄨 󵄨 󵄨󵄨 = |u k,m (ζ na )| = |u k,m (ζ nb )| = 󵄨󵄨󵄨 ζ n − ζ n 󵄨󵄨 󵄨󵄨 󵄨󵄨 a1 󵄨󵄨 b1 −a1 󵄨󵄨 −b1 󵄨󵄨 󵄨󵄨 󵄨󵄨 󵄨󵄨 ζ n − ζ n 󵄨󵄨 ζ n − ζ n and

−a1 k

a k

ζn 1 − ζn

−a1

a

ζn 1 − ζn

=ϵ

b k

−b1 k

b

−b1

ζn 1 − ζn

,

ζn 1 − ζn

with ϵ = ±1. Therefore a k+b1

(ζ n 1

−(a1 k+b1 )

+ ζn

a k−b1

) − (ζ n 1

−(a1 k−b1 )

+ ζn

−

−a k b ζ n 1 )(ζ n 1

−

−

−a b k ζ n 1 )(ζ n 1

− ζn

=

a k (ζ n 1

=

a ϵ(ζ n 1

=

b k+a ϵ((ζ n 1 1

=

b k+ϵa1 (ζ n 1

+

+

)

−b ζn 1 ) −b1 k

−(b k+a ) ζn 1 1 )

−(b k+ϵa1 ) ζn 1 )

) b k−a1

− (ζ n 1 −

−(b1 k−a1 )

) + ζn

b k−ϵa1 (ζ n 1 )

+

))

−(b k−ϵa1 ) ζn 1 ).

That is, a k+b1

ζn 1

−a1 k−b1

+ ζn

b k−ϵa1

+ ζn 1

−b1 k+ϵa1

+ ζn

b k+ϵa1

= ζn 1

−b1 k−ϵa1

+ ζn

a k−b1

+ ζn 1

−a1 k+b1

+ ζn

.

By part (1), the four summands on the left side coincide (up to a permutation) with the ±(b −a k) a k+b 2a k four summands of the right side. However, ζ n 1 1 ≠ ζ n 1 1 , as ζ n 1 = ζ nak ≠ 1 and 2b1 ζ n = ζ nb ≠ 1. Therefore, a1 k + b1 ≡ ±(b1 k + ϵa1 )

mod n

a1 k − b1 ≡ ±(b1 k − ϵa1 )

mod n

Adding and subtracting these congruences, we deduce that either a ≡ 2a1 ≡ ±2b1 ≡ ±b mod n, or a1 k ≡ ±a1 mod n and b1 k ≡ ±b1 mod n. In the former case we obtain the desired conclusion. The latter case is impossible as it implies |u k,m (ζ na )| = |u k,m (ζ mb )| = 1. (2c) Observe that η k (ζ n−a ) =

ζ n−ak − 1 ζ n−ak ζ nak − 1 a(1−k) = ζn = −a ⋅ a η k (ζ na ). ζn − 1 ζ n−a − 1 ζn

Hence, if n | m then u k,m (ζ n−a ) = η k (ζ n−a )m = η k (ζ na )m = u k,m (ζ na ). Thus u k,m (ζ na ) = u k,m (ζ nb ) if a ≡ ±b mod n. With the help of Lemma 14.4.3 one can characterize when some Bass units are of finite order modulo the center of U(ℤG). Lemma 14.4.4. Let p be a prime number. Let G be a finite group, g an element of G of prime order p and let u = u k,m (g). Then, u has infinite order modulo the center of U(ℤG) if and only if 2 ≤ k ≤ p − 2 and the conjugacy class of g in G is not contained in {g, g−1 }.

26 | 14 Free Groups Proof. By Lemma 1.2.1, u is of infinite order if and only if 2 ≤ k ≤ p − 2. Thus, throughout the proof, we may assume that u has infinite order. Because of Example 1.5.2, p ≥ 5. Let ξ = ζ p , a primitive p-th root of unity. Then, by Proposition 6.1.2, ℤ[ξ] is the ring of integers of ℚ(ξ) and ℤ[ξ + ξ −1 ] is the ring of integers of ℚ(ξ + ξ −1 ) = ℚ(ξ) ∩ ℝ. By Corollary 5.2.7, U(ℤ[ξ + ξ −1 ]) has finite index in U(ℤ[ξ]). Let n = [U(ℤ[ξ]) : U(ℤ[ξ + ξ −1 ])]. Assume that the conjugacy class of g in G is contained in {g, g−1 }. We claim that u n belongs to the center of U(ℤG). To prove this, one may assume that G = ⟨g, h⟩ with g h = g−1 . Then H = ⟨g, h2 ⟩ is an abelian normal subgroup of index 2 in G and thus it contains G󸀠 . Hence, by Corollary 3.5.13, every (non-linear) irreducible complex character of G is of the form λ G for λ a linear representation of H. So λ(g) = ξ i for some i and thus λ(g h ) = λ(g −1 ) = ξ −i . Let ρ be the representation of G affording the character λ G as given in Problem 3.2.3. Therefore ρ(u) = diag(α, α), with α = u k,m (ξ i ) = η k (ξ i )m and ρ(huh−1 ) = diag(α, α). As n = [U(ℤ[ξ]) : U(ℤ[ξ + ξ −1 ])], it follows that α n ∈ U(ℤ[ξ + ξ −1 ]) ∈ ℝ and therefore ρ(hu n h−1 ) = diag(α n , α n ) = α n I = ρ(u n ). This shows that ρ(u n ) is central in ρ(ℤG), for every irreducible representation ρ of G and hence u n is central in U(ℤG). Otherwise, i.e. if the conjugacy class of G is not contained in {g, g−1 }, there is h ∈ G such that g h ∈ ̸ {g, g−1 }. We consider separately the cases g h ∈ ⟨g⟩ and g h ∈ ̸ ⟨g⟩. Assume first that g h ∈ ̸ ⟨g⟩. Then Supp(u r ) ∩ Supp(hu r h−1 ) ⊆ ⟨g⟩ ∩ ⟨hgh−1 ⟩ = 1, for every r ≥ 1. Since u has infinite order, we deduce that u r ≠ hu r h−1 and hence u has infinite order modulo the center of U(ℤG). Suppose now that g h = g i . Because g h ∈ ̸ {g, g−1 }, we also know that i ≢ ±1 mod p. Let H = ⟨g, h⟩ = ⟨g⟩ ⋊ ⟨h⟩ and K = ⟨g, h t ⟩. Then K is an abelian normal subgroup of index t in H and K has a linear representation λ with λ(g) = ξ . Let ρ = λ H , the representation of H induced by λ as 2 t−1 given in Problem 3.2.3. So, ρ(g) = diag(ξ, ξ i , ξ i , . . . , ξ i ) and for every r ≥ 1 we have 2

t−1

ρ(u r ) = ρ(u k,rm (g)) = diag(u k,rm (ξ), u k,rm (ξ i ), u k,rm (ξ i )), . . . , u k,rm (ξ i ))). By Lemma 14.4.3 (2b), we have |u k,rm (ξ)| ≠ |u k,rm (ξ i )|, since i ≢ ±1 mod p. We deduce that ρ(u r ) is not central in the image of ρ and therefore u r is not central in ℤH. In particular, u has infinite order modulo the center of U(ℤG). The natural classes to look for free companions of a Bass unit (of infinite order modulo the center) are those of the Bass units and the bicyclic units. The following two examples show some limitations. Example 14.4.5. Consider the group G = A ⋊ ⟨b⟩, where A is a p-elementary abelian group, b has order 3 and ⟨b⟩ acts irreducibly on A, that is, no proper non-trivial subgroup of A is invariant under the action of b. In particular, Z(G) ∩ A = 1. Suppose p ≥ 5 and let a ∈ A \ Z(G), k and m positive integers such that 2 ≤ k ≤ p − 2 and k m ≡ 1 mod p and u = u k,m (a). Clearly a b ∈ ̸ {a, a−1 } and hence u has infinite order modulo the center of ℤG, by Lemma 14.4.4. However u does not have a free companion among the Bass units. Indeed, if g ∈ G \ A then g3 ∈ A ∩ Z(G) = 1 and therefore, by Ex-

| 27

14.4 Free companions

ample 1.5.2, every Bass unit based on g has finite order. Consequently, every two Bass units of infinite order commute, as they belong to ℤA. We will see in Proposition 14.4.9 that a power of u has a free companion which is a power of a bicyclic unit. Example 14.4.6 (Following an idea in [59]). Let G = ⟨b⟩p e ⋊⟨a⟩p , a semidirect product, e−1 where a has prime order p ≥ 5, b has order p e , with e ≥ 2, and b a = b1+p . As in Example 14.4.5, there is a Bass unit u = u k,m (a) of infinite order modulo the center of ℤG. We claim that no power of u has a free companion among powers of bicyclic units. Because of symmetry, we only prove the claim for non-trivial bicyclic units of the type v = b(h, g̃ ) = 1 + (1 − g)h g̃ of U(ℤG). Then ⟨g⟩ is not normal in G and therefore it does e−1 not contain G󸀠 = ⟨b p ⟩. Hence ⟨g⟩ ∩ ⟨b⟩ = 1, so that g p = 1. Thus, g = b j a for some j such that e−1 p 1 = (b j a)p = b

j(p e +1) (1+p e−1)

−1

.

p

p e−1 )p

Therefore divides j((1 + − 1). However, (1 + p e−1 )p − 1 is not multiple of e−1 p e+1 , since (1 + p e−1 )p ≡ 1 + p e mod p e+1 . So, p e−1 | j and hence b j ∈ ⟨b p ⟩ ⊆ Z(G). Thus g = za t for some central element z with z p = 1 and 1 ≤ t < p. Let t󸀠 be the 󸀠 󸀠 p−1 multiplicative inverse of t modulo p. Then a = z−t g t . Write u = u k,m (a) = ∑x=0 α x a x 󸀠 p−1 −t x with α x ∈ ℤ. Let w = ∑x=0 α x z . We get p2e−1

󸀠

󸀠

g̃ u = g̃ ∑ α x z−t x g t x = g̃ ∑ α x z−t x

󸀠

x

= g̃ w.

x

As w is central and g and u commute we have u−j v i u j = 1 + iw j u−j (1 − g)h g̃ for every i, j. Using g̃ w j u−j (1 − g) = g̃ u j u−j (1 − g) = 0, it is easy to see that A = ⟨u−j vu j : j ∈ ℤ⟩ is an abelian normal subgroup of ⟨u, v⟩ such that ⟨u, v⟩ /A is cyclic. Thus ⟨u, v⟩ does not contain a nonabelian free group. This proves the claim. However, in Proposition 14.4.7 we prove that some power of u has a free companion which is a Bass unit. Our next aim is to prove the statements given at the end of Examples 14.4.5 and 14.4.6, namely we will present a bicyclic unit which is a free companion of the Bass unit of Example 14.4.6 and a Bass unit which is a free companion of the Bass unit of Example 14.4.6. As a preparation, we introduce the following n × n-matrices 0 ( Pn = (

1 0

1 .. .

..

. 0

(1

1 1 ) (1 ) and W n = ( .. . 1 0) (1

1 ζn ζ n2 .. . ζ nn−1

1 ζ n2 ζ n4 .. . 2(n−1)

ζn

... ... ... .. .

1 ζ nn−1 ζn

...

ζn

Observe that W n is the Vandermonde matrix of the n-th roots of unity and n−1

−ij jk

n−1

∑ ζ n ζ n = ∑ (ζ nk−i )j = { j=0

j=0

n, 0,

if k = i; otherwise.

2(n−1) )

(n−1)2

). )

28 | 14 Free Groups Thus (W n )−1 = 1n W n , where W n = (ζ −ij )i,j≤n , the componentwise conjugate of W n . Moreover, the i-th column of W n is an eigenvector of P n with eigenvalue ζ ni and so (W n )−1 P n W n = diag(1, ζ n , ζ n2 , . . . , ζ nn−1 ).

(14.4.3)

The following proposition provides the Bass unit promised at the end of Example 14.4.6. Proposition 14.4.7 (Gonçalves-Passman [73]). Let p and q be prime integers and let G = ⟨b⟩q e ⋊ ⟨a⟩p be a nonabelian semidirect product of a cyclic group of order q e by a cyclic group of order p. Consider the Bass units u = u k,m (a) and v = u l,n (b). If u and v have infinite order then ⟨u r , v s ⟩ is free for r and s sufficiently large. Proof. Suppose that u and v have infinite order. Then k ≢ ±1 mod p and l ≢ ±1 mod q e , by Lemma 1.2.1. In particular p ≥ 5. Moreover, p | φ(q e ) = q e−1 (q − 1) and hence either p = q or p | q − 1. So, also q ≥ 5. Consider a faithful linear representation ϕ of ⟨b⟩ and let ρ be the representation of G given by ρ(b) = diag(ϕ(b), ϕ(b a ), . . . , ϕ(b a

p−1

) = diag(ξ1 , ξ2 , . . . , ξ p ) and

ρ(a) = P p .

Let T1 = ρ(v) and T2 = ρ(u). To prove the result it is enough to show that there are a T1 decomposition and a T2 -decomposition satisfying the conditions of Corollary 14.3.6. For every i ∈ {0, . . . , p − 1}, let f i = E(i+1)(i+1) (1), the diagonal matrix having 1 at the i + 1 diagonal entry and zeros elsewhere. p−1 As G is nonabelian, b, b a , . . . , b a are all different and hence ξ1 , ξ2 , . . . , ξ p i are different roots of unity of order q e . Moreover, as p is odd, b a ≠ b−1 for every i and hence ξ1 , ξ2 , . . . , ξ p are pairwise non-conjugate. By statements (2a) and (2b) of Lemma 14.4.3, we have 1 ≠ |u l,n (ξ i )|, 1 ≠ |u l,n (ξ j )| and |u l,n (ξ i )| ≠ |u l,n (ξ j )| for every i ≠ j. Moreover T1 = ρ(v) = diag(u l,n (ξ1 ), u l,n (ξ2 ), . . . , u l,n (ξ p )). Therefore V = ℂp has a T1 -decomposition V = T1+ ⊕ T10 ⊕ T1− with dimℂ T1+ = dimℂ T1− = 2. Hence there exist different 0 ≤ i, j, i󸀠 , j󸀠 ≤ p − 1 with |{i, j, i󸀠 , j󸀠 }| = 4 such that if e1+ = f i + f j and e1− = f i󸀠 + f j󸀠 then the projections V → T1± are given by x 󳨃→ e1± x. On the other hand, by (14.4.3) we have W p−1 ρ(a)W p = diag(1, ζ p , . . . , ζ p

p−1

)

and therefore (W p )−1 T2 W p = diag(u k,m (1), u k,m (ζ p ), . . . , u k,m (ζ p

p−1

)).

Let T2+ be the sum of the eigenspaces of ρ(u) with eigenvalue of maximal absolute value, T2− the sum of the eigenspaces of minimum absolute value and T20 the sum of the remaining eigenspaces. By Lemma 14.4.2 and Lemma 14.4.3, dimℂ T2+ =

14.4 Free companions

| 29

dimℂ T2− = 2. Moreover the projections V → T2± along this decomposition are given by x 󳨃→ e2± x, where e2+ = W p (f1 + f p−1 )(W p )−1 , e2− = W p (f l + f p−l )(W p )−1 and kl ≡ 1 mod p. Therefore, to verify the hypotheses of Corollary 14.3.6, we need to check that the eight matrices e i± e j± with {i, j} = {1, 2} have rank 2. Observe that all the matrices e i± are symmetric and therefore (e1± e2± )T = e2± e1± . Thus it is enough to verify the rank condition for the four matrices e1± e2± . By symmetry we only check one of them. For example, the rank of e1+ e2− coincides with the rank of e1+ W p (f l + f p−l ). This matrix has only four non-zero entries which form the minor 󵄨󵄨 li 󵄨󵄨ζ p 󵄨󵄨 󵄨󵄨 lj 󵄨󵄨ζ p

(p−l)i 󵄨󵄨 󵄨󵄨 ζp li+(p−l)j (p−l)i+lj l(i−j) l(j−i) 󵄨 − ζp = ζp − ζp . (p−l)i 󵄨󵄨󵄨 = ζ p ζp 󵄨󵄨 l(j−i)

This minor is non-zero because 2l ≢ 0 mod p and i ≢ j mod p and hence ζ p

≠

l(i−j) ζp .

We need the following lemma. Lemma 14.4.8. Let p be a prime. Let ξ0 , . . . , ξ q−1 be a list of p-th roots of unity such that ξ r ∈ ̸ {ξ s , ξ s } for some 0 ≤ r, s ≤ q − 1. Let k and m be integers with k ≢ ±1 mod p and k m ≡ 1 mod p. Consider the diagonal matrix S = diag(u k,m (ξ0 ), . . . , u k,m (ξ q−1 )) and the matrix ξ0 − ξ1 ξ1 − ξ2 τ=( ... ξ q−1 − ξ0

ξ0 − ξ1 ξ1 − ξ2 ... ξ q−1 − ξ0

... ... ... ...

ξ0 − ξ1 ξ1 − ξ2 ). ... ξ q−1 − ξ0

Then ⟨S r , (1 + τ)s ⟩ is free for some r and s sufficiently large. Proof. First of all p ≥ 5 because, k ≢ ±1 mod p. Let M+ and M− be the maximum and minimum of {|u k,m (ξ0 )|, . . . , |u k,m (ξ q−1 )|}. Let X = {0, 1, 2, . . . , q − 1}, X± = {i ∈ X : |u k,m (ξ i )| = M± } and X0 = X \ (X+ ∪ X− ). By statement (2b) of Lemma 14.4.3, any of the two sets {ξ i : i ∈ X± } has cardinality at most 2 and if it has cardinality 2 then its two members are complex conjugate. We consider S and τ as endomorphisms of V = ℂq . Let {e0 , . . . , e q−1 } be the standard basis of V. Let S+ , S− and S0 denote the subspaces of V generated by {e i : i ∈ X+ }, {e i : i ∈ X− } and {e i : i ∈ X0 }, respectively. Then V = S+ ⊕ S0 ⊕ S− is an S-decomposition of V. Let K and I denote the kernel and image of τ. Let W = (S+ ∩ K) ⊕ (S− ∩ K). We use the standard bar notation for reduction modulo W. In particular, if U is a subspace of V then U = (U + W)/W. Without loss of generality, we may replace S by one of its powers. Hence, by (1.2.6), one may assume that p divides m. By statements (2b) and (2c) of Lemma 14.4.3, this implies that if |u k,m (ξ i )| = |u k,m (ξ j )| then u k,m (ξ i ) = u k,m (ξ j ) and hence both S+ and S− are eigenspaces of S with respect to suitable eigenvalues. Thus W = (S+ ∩ K) ⊕ (S− ∩ K) is invariant under the action of S and τ and hence they induce endomorphisms S and

30 | 14 Free Groups τ of V. Moreover V = S+ ⊕ S0 ⊕ S− is an S-decomposition. Observe S± ≠ 0, because K does not contain any e i . Then the dimension of S± is 1. To finish the proof we verify that the assumptions of Theorem 14.3.8 hold for V. The kernel and image of τ are K1 = τ−1 (W) and I respectively. As the dimension of S± is 1 and S(e i ) = u k,m (ξ i )e i ∈ ̸ K for every i ∈ X, we have S± ∩ K1 = 0. Moreover I is 1-dimensional generated by Ψ, with Ψ = (ξ0 −ξ1 , ξ1 −ξ2 , . . . , ξ q−1 −ξ0 ). It only remains to prove that I ∩ (S0 ⊕ S± ) = 0 or equivalently Ψ ∈ ̸ S0 ⊕ S± . Because of symmetry, we prove Ψ ∈ ̸ S0 ⊕ S+ . We prove this by contradiction. So suppose Ψ ∈ S0 ⊕ S+ . Then the projection of Ψ onto S− belongs to K. That is, ∑ (ξ i − ξ i+1 ) = 0, i∈X−

where the subindices are considered modulo q. Now we delete from the equality the zero summands, move to the right side the terms with negative coefficient and add up the equal terms obtaining k

k

n1 λ1 + n2 λ2 = ∑ m i μ i

n1 + n2 = ∑ m i ,

and

i=1

(14.4.4)

i=1

where {λ1 , λ2 } = {ξ i : i ∈ X− , ξ i ≠ ξ i+1 }, {μ1 , . . . , μ k } = {ξ i+1 : i ∈ X− , ξ i ≠ ξ i+1 }, the μ i ’s are pairwise different, n i is the cardinality of {j : λ i = ξ j ≠ ξ j+1 }, for i = 1, 2, and m i is the cardinality of {j ∈ X− : ξ j ≠ ξ j+1 = μ i }. Moreover λ1 and λ2 are complex conjugate and n1 and n2 are non-negative with at least one of them positive. Without loss of generality, we may assume that n1 > 0 and if n2 ≠ 0 then λ1 ≠ λ2 . Let T = Trℚ(ζ p )/ℚ , the Galois trace of ℚ(ζ p )/ℚ. We have T(1) = p − 1 and T(ξ) = −1, for every primitive p-th root of unity ξ . If n2 = 0 then μ i ≠ λ1 for every i and applying T to k

n1 = ∑ m i μ i λ−1 1 i=1

we obtain n1 (p − 1) < 0, yielding a contradiction. Thus n2 ≠ 0 and so λ1 = λ2 ≠ λ2 . Now, applying T on the left equation of (14.4.4), we obtain k

k

− ∑ m i = −n1 − n2 = ∑ m i T(μ i ) i=1

i=1

and hence μ i ≠ 1 for every i. This, together with (14.4.4), and the fact that the set of primitive p-th roots of unity is linearly independent over the rationals, implies that every μ i is either λ1 or λ2 = λ1 . However, by assumption, there is 0 ≤ r ≤ q − 1 such that ξ r ∈ ̸ {λ1 , λ1 }. Let r be the smallest element satisfying this condition. Then r−1 ∈ X− and ξ r−1 ≠ ξ r (recall that we are considering the subindices of the ξ ’s modulo q). Hence μ i = ξ r ∈ ̸ {λ1 , λ1 } for some i, a contradiction. Given a group G and an element a of G, the extended centralizer of a in G is Cen∗G (a) = {g ∈ G : a g ∈ {a, a−1 }}.

14.4 Free companions

| 31

Clearly CenG (a) ⊆ Cen∗G (a) ⊆ N G (⟨a⟩) and [Cen∗G (a) : CenG (a)] ≤ 2. One can now rephrase Lemma 14.4.4 as follows: If a has prime order then a Bass unit u k,m (a) has finite order modulo the center if and only if either it has finite order or Cen∗G (a) = G. The next proposition provides the free companion bicyclic unit promised at the end of Example 14.4.5. Proposition 14.4.9. Let G be a finite group, A an abelian normal subgroup of G and let a ∈ A and g ∈ G. Assume that the order of a is prime, say p, and g ∈ ̸ Cen∗G (a) and ⟨g⟩ ∩ A = 1. Let u = u k,m (a) be a Bass cyclic unit with k ≢ ±1 mod p (in particular, p ≥ 5) and v = b(a, g̃ ). Then ⟨u n , v n ⟩ is free nonabelian for some n. Proof. Without loss of generality, we may assume that G = ⟨a, g⟩ and A = ⟨a x : x ∈ ⟨g⟩⟩ is abelian. As a g ∈ ̸ {a, a−1 }, there is a linear representation ϕ of A such that ϕ(a g ) ∈ ̸ {ϕ(a), ϕ(a−1 )}. Let q denote the order of g (and the index of A in G). For i every i ∈ {0, 1, . . . , q − 1}, let ξ i = ϕ(a g ). Let ρ be the representation G given by ρ(a) = diag(ξ0 , ξ1 , . . . , ξ q−1 ) and ρ(g) = P q . (Notice that, by Problem 3.2.3, the character afforded by ρ is χ G .) Moreover, S = ρ(u) and τ = ρ(v − 1) are as in Lemma 14.4.8. Thus ⟨S r , (1 + τ)s ⟩ is free for r and s sufficiently large and hence so is ⟨u r , v s ⟩. Before proving the main result of this section, we need another result in the flavor of Proposition 14.4.7 and Proposition 14.4.9. Proposition 14.4.10 (Gonçalves-del Río [69]). Let p and q be two different primes. Let ⟨a⟩ be a cyclic group of order p acting faithfully and irreducibly on a non-cyclic elementary abelian q-group H and let G = H ⋊ ⟨a⟩ the corresponding semidirect product. Let ̃ 1 ≠ h ∈ H. If u = u k,m (a) has infinite order modulo the center of U(ℤG) and v = b(a, ha) then ⟨u n , v n ⟩ is a nonabelian free group for some integer n. Proof. As H is non-cyclic, it has a non-trivial linear character χ of H, with χ(h) = 1. i For each i ∈ ℤ, let 𝛾i = χ(h a ). Clearly, we may consider the subindices of the 𝛾’s as elements of ℤ/pℤ. We also label the rows and columns of a p × p matrix by elements of ℤ/pℤ. Accordingly, for 0 ≤ i, j < p, let E i,j denote the matrix having 1 at the (i, j)-entry and zeros elsewhere. i p−1 p−1 Moreover, ∏i=0 h a ∈ Z(G) = 1 and therefore ∏i=0 𝛾i = 1. Using this we may consider the subindices in the following definition as cycle intervals in ℤ/p/ℤ. j−1

𝛾[i,j) = ∏ 𝛾k and 𝛾[j,i) = 𝛾−1 [i,j) = 𝛾j 𝛾j+1 . . . 𝛾p−1 𝛾0 𝛾1 . . . 𝛾i−1 (i ≤ j). k=i i

Observe that H = ⟨h a : i = 0, 1, . . . , p − 1⟩, because the action of ⟨a⟩ on H is i irreducible. Therefore, G = ⟨a, h⟩ and h a ∈ ̸ ker(χ) for some i. Thus, not all the 𝛾i ’s are equal and hence not all the 𝛾[0,i) ’s are equal. Let ρ be the representation of G given by p−1

ρ(h) = diag(𝛾0 , 𝛾1 , . . . , 𝛾p−1 ) = ∑ 𝛾i E i,i i=0

(Notice that the character afforded by ρ is χ G .)

p−1

and

ρ(a) = P p = ∑ E i,i+1 . i=0

32 | 14 Free Groups Let S = ρ(u) and τ = ρ(v − 1), which we consider as endomorphisms of V = ℂp . The strategy of the proof is similar to that of the proof of Lemma 14.4.8, namely we try to verify the hypotheses of Theorem 14.3.8 for S and τ. We will find a similar obstacle as in that proof which we avoid by taking the endomorphism induced by S and τ on a suitable quotient of V. p−1 Observe that ρ(ha) = ∑i=0 𝛾i E i,i+1 and hence p−1

p−1

i=0

i=0

ρ((ha)j ) = ∑ 𝛾i 𝛾i+1 . . . 𝛾i+j−1 E i,i+j = ∑ 𝛾[i,i+j) E i,i+j . Note that the non-zero entries of ρ((ha)j ) appear in different positions for different values of j. So, ̃ = ρ(1 + ha + (ha)2 + ⋅ ⋅ ⋅ + (ha)p−1 ) = ∑ 𝛾[i,j) E i,j = (𝛾[i,j) ). ρ(ha) i,j

̃ As 𝛾[i,0) = 𝛾[i,j) 𝛾[j,0) = 𝛾[i,j) 𝛾−1 [0,j) , all the columns of ρ( ha) are proportional to the first ̃ is 1. Thus the rank of τ = ρ((1 − ha)a ha) ̃ is at column and hence the rank of ρ(ha) most 1. In fact it is 1, because p−1

p−1

i=0

j=0

τ = ( ∑ E i,i − 𝛾i E i,i+1 ) ∑ E j,j+1 (∑ 𝛾[k,l) E k,l ) k,l

= ∑(𝛾[i+1,j) − 𝛾i 𝛾[i+2,j) )E i,j i,j

= ∑(𝛾i+1 − 𝛾i )𝛾[i+2,j) E i,j i,j

= ∑ 𝛾[i+1,j) (1 − 𝛾i 𝛾−1 i+1 )E i,j ≠ 0, i,j

̃ as not all the 𝛾i ’s are equal. Hence the kernel of τ coincides with the nullspace of ρ(ha) ̃ and this is equal to the nullspace of any non-zero row of ρ(ha), for example we can take the first row. In other words, K = ker(τ) = {(x 0 , . . . , x p−1 ) : x0 + 𝛾[0,1) x1 + ⋅ ⋅ ⋅ + 𝛾[0,p−1) x p−1 = 0} and the image of τ is spanned by Ψ = ((𝛾1 − 𝛾0 )𝛾[2,0) , (𝛾2 − 𝛾1 )𝛾[3,0) , . . . , (𝛾0 − 𝛾p−1 )𝛾[1,0) ) . Let v i denote the i-th column of W p . By (14.4.3) we have W p−1 ρ(a)W p = diag(1, ζ p , ζ p2 , . . . , ζ p

p−1

),

−ij

ij

and W p−1 = 1p W p , with W p = (ζ p ) and W p = (ζ p ). Thus v i is an eigenvector of P p = ρ(a) with eigenvalue ζ pi , and an eigenvector of S = ρ(u) with eigenvalue u k,m (ζ pi ). Let β

j

α, β ∈ {0, 1, . . . , p − 1} be such that |u k,m (ζ p )| ≤ |u k,m (ζ p )| ≤ |u k,m (ζ pα )|, for every

14.4 Free companions | 33

j. By Lemma 14.4.2 and statements (2a) and (2b) of Lemma 14.4.3, α, β ≠ 0 and V = S+ ⊕ S0 ⊕ S− is an S-decomposition of V with S+ = ℂv α + ℂv−α , S− = ℂv β + ℂv−β and S0 = ∑i∈{±α,±β} ℂv i . ̸ As K is a hyperplane of the representing space V = ℂp and S+ and S− have dimension 2 we have K∩S+ ≠ 0 ≠ K∩S− . However we can play the same trick as in the proof of Lemma 14.4.8, by assuming that m is a multiple of p, so that, u k,m (ζ pi ) = u k,m (ζ p−i ) ∈ ℝ for every i. Thus S+ and S− are the eigenspaces of S corresponding to the eigenvalues β u k,m (ζ pα ) and u k,m (ζ p ), respectively. Thus W = (K ∩ S+ ) ⊕ (K ∩ S− ) is invariant under the action of S and τ. Hence there are induced endomorphisms S and τ of V = V/W. The kernel of τ is τ−1 (W) and the image is generated by Ψ. Recall that the minimal polynomial of each primitive p-root of unity over ℚ is ij p−1 1 + x + x2 + ⋅ ⋅ ⋅ + x p−1 . Hence ∑j=0 𝛾[0,j) ζ p ≠ 0 for every 1 ≤ i < p, since not all the 𝛾[0,j) ’s are equal. This means that v i ∈ ̸ K for every i. In particular, v α , v β ∈ ̸ K and hence both K ∩ S+ and K ∩ S− have dimension 1. To finish the proof it is enough to check that S and τ satisfy the conditions of Theorem 14.3.8 and for this it only remains to show that Ψ ∈ ̸ S0 ⊕ S± . By symmetry, we prove Ψ ∈ ̸ S0 ⊕ S+ . By means of contradiction assume that Ψ ∈ S0 ⊕ S+ . Then the projection Ψ− of Ψ onto S− , along the S-decomposition, belongs to K. The projection of the column vector x onto S− is 1p W p (E β,β + E−β,−β )W p x. We first calculate W p (E β,β + E−β,−β )W p = (∑ ζ p E i,j )(E β,β + E−β,−β )(∑ ζ p−kl E k,l ) ij

i,j

=

k,l

iβ −βl ∑(ζ p ζ p i,l (i−l)β

= ∑(ζ p

−iβ βl ζ p ζ p )E i,l

+

(l−i)β

+ ζp

)E i,l .

i,l

Therefore pΨ− = W p (E β,β + E−β,−β )W p Ψ −jβ

∑j (ζ p =(

jβ

+ ζ p )(𝛾j+1 − 𝛾j )𝛾[j+2,0)

(1−j)β

+ ζp

(p−1−j)β

+ ζp

∑j (ζ p

(∑j (ζ p

(j−1)β

(i−j)β

So, Ψ− ∈ K if and only if ∑i,j 𝛾[0,i) (ζ p iβ ∑i,j 𝛾[0,j+i) (ζ p

+

−iβ ζ p )(𝛾j+1

)(𝛾j+1 − 𝛾j )𝛾[j+2,0) ) . .. .

(j−p+1)β (j−i)β

+ ζp

)(𝛾j+1 − 𝛾j )𝛾[j+2,0) )

)(𝛾j+1 − 𝛾j )𝛾[j+2,0) = 0, or equivalently

− 𝛾j )𝛾[j+2,0) = 0. This on its turn is equivalent with iβ

∑(∑(𝛾j+1 − 𝛾j )𝛾[j+2,0) (𝛾[0,j+i) + 𝛾[0,j−i) )ζ p = 0 i

j

and also with α i = ∑j (𝛾j+1 − 𝛾j )𝛾[j+2,0) (𝛾[0,j+i) + 𝛾[0,j−i) ) being independent of i.

34 | 14 Free Groups For i = 0 we have α0 = 2 ∑ 𝛾[0,j) (𝛾j+1 − 𝛾j )𝛾[j+2,0) j

= 2(∑ 𝛾[0,j) 𝛾j+1 𝛾[j+2,0) − ∑ 𝛾[0,j) 𝛾j 𝛾[j+2,0) ) j

j

= 2(∑ 𝛾[j+1,j) − ∑ 𝛾[j+2,j+1) ) j

=

j

2(∑ 𝛾−1 j j

− ∑ 𝛾−1 j+1 ) = 0. j

Hence, if Ψ− ∈ K then α i = 0 for every i. In particular, for i = 1 we have 0 = α1 = ∑(𝛾[0,j+1) + 𝛾[0,j−1) )(𝛾j+1 − 𝛾j )𝛾[j+2,0) j

= ∑ (𝛾[0,j+1) 𝛾j+1 𝛾[j+2,0) + 𝛾[0,j−1) 𝛾j+1 𝛾[j+2,0) j

−𝛾[0,j+1) 𝛾j 𝛾[j+2,0) − 𝛾[0,j−1) 𝛾j 𝛾[j+2,0) )

−1 −1 −1 −1 = ∑(1 + 𝛾−1 j−1 𝛾j − 𝛾j 𝛾j+1 − 𝛾j−1 𝛾j+1 ) j −1 −1 −1 −1 = p + ∑ 𝛾−1 j 𝛾j+1 − ∑ 𝛾j 𝛾j+1 − ∑ 𝛾j 𝛾j+2 . j

j

j

If T denotes the trace of the extension ℚ(ζ q )/ℚ then T(ζ qi ) = q − 1 if q divides i and it is −1 otherwise. Therefore applying T in the previous expression we obtain 0 = p(q − 1) + n1 (q − 1) − (p − n1 ) − n2 (q − 1) + (p − n2 ) − n3 (q − 1) + (p − n3 ) = (p + n1 − n2 − n3 )q, where n i is the cardinality of X i for X1 = {j ∈ ℤp : 𝛾j = 𝛾−1 j+1 },

X2 = {j ∈ ℤp : 𝛾j = 𝛾j+1 }

and X3 = {j ∈ ℤp : 𝛾j = 𝛾−1 j+2 }. On the other hand, if j ∈ X2 ∩ X3 then j + 1 ∈ X1 . Hence j 󳨃→ j + 1 is an injective map X2 ∩ X3 → X1 . So p = n2 + n3 − n1 ≤ |X2 | + |X3 | − |X2 ∩ X3 | = |X2 ∪ X3 | and −1 therefore X2 ∪ X3 = ℤp . Thus 𝛾j ∈ {𝛾j+1 , 𝛾−1 j+2 } for every j. If X 2 = 0 then 𝛾j+2 = 𝛾j for every j. Using that p is odd, one deduces that 𝛾j = 𝛾0 = 1, for every j, contradicting the construction. Thus X2 ≠ 0, i.e. 𝛾j0 = 𝛾j0 +1 for some j0 . Now using recursively that −1 𝛾j−1 ∈ {𝛾j , 𝛾−1 j+1 } for every j, we deduce that 𝛾j is either 𝛾j0 or 𝛾j0 for every j. As 𝛾0 = 1, we deduce that 𝛾j = 1 for every j and this yields the final contradiction. Let G be a group and a ∈ G. If Cen∗G (a) = G then a behaves in G as the elements in the cyclic subgroup of index 2 in a dihedral group. This suggests the following terminology, which is handy for our purpose. We say that a is dihedral p-critical in G if it satisfies the following conditions:

14.4 Free companions | 35

(C1) a has order p and Cen∗G (a) ≠ G. (C2) If H is a proper subgroup of G, and a ∈ H, then Cen∗H (a) = H. (C3) If G is a proper quotient of G then Cen∗ (a) = G. (Here, again, we use the standard G bar notation.) The following lemma collects some easy consequences of the definition. Lemma 14.4.11. If a is dihedral p-critical in G then the following properties hold. (1) Every proper subgroup of G containing a is contained in Cen∗G (a). (2) If b ∈ G \ Cen∗G (a) then G = ⟨a, b⟩. (3) If N is a non-trivial normal subgroup of G then G󸀠 ⊆ ⟨a, N⟩. Proof. (1) and (2) are consequences of (C2). By (2), G = ⟨a, b⟩ for some b ∈ G. Let N be a non-trivial normal subgroup of G. By condition (C3), a g ∈ aN ∪ a−1 N for every g ∈ G. Hence (g, a)h = (a gh )−1 a h ∈ ⟨a, N⟩ for every g, h ∈ G. This shows that ⟨a, N⟩ contains M = ⟨(G, a)g : g ∈ G⟩. Since G/M is abelian, because G = ⟨a, b⟩, we deduce that G󸀠 ⊆ ⟨a, N⟩. Recall that the notation ⟨g⟩n represents a cyclic group of order n generated by g. Also, if i and n are coprime integers then o n (i) denotes the order of [i]n in U(ℤn ). Proposition 14.4.12 (Gonçalves-del Río [69]). Let a be a dihedral p-critical element of a finite group G. Then one of the following conditions holds: n−1 (1) G = ⟨b⟩p n ⋊ ⟨a⟩p , with n ≥ 2 and b a = b1+p . (2) G = ⟨a⟩p ⋊ ⟨b⟩q , for q either 4 or an odd prime such that a b = a i and q = o p (i). (3) G = (⟨a⟩p × ⟨z⟩p ) ⋊ ⟨b⟩p , with z ∈ Z(G) and a b = za. (4) G = (⟨a⟩p × ⟨z⟩p ) ⋊ ⟨b⟩2 , with z ∈ Z(G) and a b = za−1 . (5) G = (⟨a⟩p × ⟨a1 ⟩p ) ⋊ ⟨b⟩4 , with a b = a1 and a1b = a−1 . (6) G = A ⋊ ⟨b⟩q , where A is a non-cyclic elementary abelian p-group containing a, with q a prime different from the prime p and the action of ⟨b⟩ on A is faithful and irreducible. (7) G = B ⋊ ⟨a⟩p , where B is an elementary abelian q-group and the action of ⟨a⟩ on B is faithful and irreducible. (8) G is simple. Proof. We set D = Cen∗G (a) and C = CenG (a) and fix an element b ∈ G \ Cen∗G (a). Then G = ⟨a, b⟩, by Lemma 14.4.11 (2). Notice that g2 ∈ C for every g ∈ D and, by Lemma 14.4.11, D is the unique maximal subgroup of G containing a. Case 1: Assume that G󸀠 = ⟨a⟩. Then a b = a i for some i ≢ ±1 mod p. Let t = o p (i). Since U(ℤ/pℤ) is cyclic, the only element of U(ℤ/pℤ) of order 2 is −1. Hence t > 2. If q q q is a prime divisor of t then ⟨a, b q ⟩ is a proper subgroup of G and hence a i = a(b ) = a±1 . Thus i2q ≡ 1 mod p. This shows that t divides 2q for every prime divisor q of t. Hence t is either 4 or an odd prime. If r is a prime divisor of the order of b which is coprime with t then ⟨a, b r ⟩ is again a proper subgroup of G not contained in D,

36 | 14 Free Groups yielding a contradiction. Thus G = ⟨a⟩p ⋊ ⟨b⟩q k and either q = o p (i) is an odd prime or q = 2, k ≥ 2 and o p (i) = 4. In the first case k = 1, because otherwise 1 ≠ b q ∈ Z(G) and G/⟨b q ⟩ = ⟨a⟩p ⋊ ⟨b⟩q k−1 , with a b = a i and i ≢ ±1 mod p, contradicting (C3). In the second case b4 ∈ Z(G) and a similar argument shows that in this case k = 2. Thus G satisfies condition (2). Case 2: Assume that G󸀠 ≠ ⟨a⟩ and Z(G) ≠ 1. By Lemma 14.4.11 (3), G󸀠 is contained in ⟨a, z⟩ = ⟨a⟩ × ⟨z⟩ for every 1 ≠ z ∈ Z(G). Therefore, if H and K are two different minimal subgroups of Z(G) then G󸀠 ⊆ ⟨a, H⟩ ∩ ⟨a, K⟩ = ⟨a⟩, contradicting the assumption G󸀠 ≠ ⟨a⟩. Hence Z(G) has a unique minimal subgroup, or equivalently, Z(G) is cyclic of prime power order, say q k with q prime. From (C1) and (C3) we deduce that a b = a±1 z, with z a central element of order q. In particular, p = q. First, assume that a b = az. Then b−p ab p = z p a = a and hence b p ∈ Z(G). If p b = 1 then G = (⟨a⟩p × ⟨z⟩p ) ⋊ ⟨b⟩p and G satisfies condition (3). Otherwise, b has n−1 order p n with n ≥ 2. Then ⟨b p ⟩ is the only minimal subgroup of Z(G) and replacing n−1 n−1 b by a suitable power if needed, one may assume that z = b p . So b a = b1+p and G satisfies condition (1). 2 Second, assume that a b = a−1 z ≠ az (so p ≠ 2). Then a b = (a−1 z)b = az−1 z = a, p i.e. b2 ∈ Z(G). Hence a b = a−1 z ∈ ̸ {a, a−1 } and therefore G = ⟨a, b p ⟩. If b2 = 1 then G = (⟨a⟩p × ⟨z⟩p ) ⋊ ⟨b⟩2 . Otherwise ⟨z⟩ = Z(G) = ⟨b2 ⟩. Thus b p has order 2 and G = ⟨a, b p ⟩ = (⟨a⟩p × ⟨z⟩p ) ⋊ ⟨b p ⟩2 . In both cases G satisfies condition (4). Case 3: Z(G) = 1 and D is normal in G. In particular, G is not a p-group and a g ∈ D for all g ∈ G. Let A = ⟨a g : g ∈ G⟩. We claim that A is abelian (and thus it is an elementary abelian p-group). To prove this, it is enough to show that if g ∈ G then a g ∈ C. This is clear if p = 2 and otherwise it follows from the fact that (a g )2 ∈ C. This proves the claim. As A is normal in G and a ∈ A, we have G󸀠 ⊆ A ⊆ D by Lemma 14.4.11. Since D is the unique maximal subgroup of G containing a, G/A is a cyclic q-group for some prime q. Furthermore, as Z(G) = 1 and thus G is not a p-group, we have q ≠ p. Then b is a generator of G modulo A. Thus D = ⟨A, b q ⟩ and hence b2q ∈ C. The latter implies that b2q ∈ Z(G) = 1. Therefore G = A ⋊ ⟨b⟩, because A ∩ ⟨b⟩ ⊆ CenG (A)∩CenG (b) = Z(G) = 1. Furthermore, by the definition of A, the abelian group A is a minimal ⟨b⟩-module. By Maschke’s Theorem, the action of ⟨b⟩ on A is irreducible. The action of ⟨b⟩ on A is faithful because its kernel is contained in Z(G) = 1. On the other hand, [G : A] = |b|, it is a power of q and a divisor of 2q. Thus the order of b is either q or 4. If A is cyclic then G satisfies condition (2). If A is non-cyclic and b as prime order then G satisfies condition (6). Finally, if A is non-cyclic and b has 2 order 4 then ⟨a, b2 ⟩ is a proper subgroup of G containing a and thus a b = a±1 . Since 2 b −1 b Z(G) = 1, we get that a = a . Therefore A = ⟨a⟩ × ⟨a1 ⟩, a1 = a and a1b = a−1 and condition (5) holds. Case 4: Z(G) = 1 and D is not normal in G. In particular G󸀠 is not contained in D. Thus ⟨G󸀠 , a⟩ is a subgroup of G containing a and not contained in D. Hence, by Lemma 14.4.11 (1), we have G = ⟨G󸀠 , a⟩ and ⟨a, N⟩ = G for every non-trivial normal

14.4 Free companions

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subgroup N of G. If G is simple then condition (8) holds. Otherwise, for every proper non-trivial normal subgroup N of G we have G = N ⋊ ⟨a⟩ and thus 1 ≠ G󸀠 ⊆ N. Hence G󸀠 , is the unique minimal normal subgroup of G. Then G󸀠 = S1 ×⋅ ⋅ ⋅×S k with S1 , . . . , S k minimal normal subgroups of G󸀠 . We claim that G󸀠 is abelian. Otherwise, S1 , . . . , S k are the only minimal normal subgroups of G󸀠 [76, Proposition 3.2] and hence the action of ⟨a⟩ permutes the S i transitively (and, in particular, k = p). If 1 ≠ s ∈ S1 then 2 p−1 H = ⟨s⟩ × ⟨s a ⟩ × ⟨s a ⟩ × ⋅ ⋅ ⋅ × ⟨s a ⟩ is a proper subgroup of G󸀠 which is invariant under the action of a. Therefore H ⋊ ⟨a⟩ is a proper subgroup of G containing a and not contained in D, contradicting condition (C2). We conclude that G󸀠 is abelian and hence G󸀠 is an elementary abelian q-group, for a prime q different from p and the action of ⟨a⟩ is faithful and irreducible. So condition (7) holds. Finally we are ready to prove the main result of this section. Theorem 14.4.13 (Gonçalves-del Río [69]). Let G be a solvable and finite group, let a be a non-central element of G of prime order and let u = u k,m (a), a Bass unit of ℤG. If u has infinite order modulo the center of U(ℤG) then ℤG contains a Bass unit or a bicyclic unit v such that ⟨u n , v n ⟩ is a nonabelian free group for some integer n. Proof. We give a proof by contradiction. So, assume G and u = u k,m (a) are a counterexample for the theorem with G of minimal order. Then a has prime order, say p, and u has infinite order modulo the center of U(ℤG). By Lemma 14.4.4, a ∈ G \ Cen∗G (a), so that a satisfies condition (C1). Let H be a proper subgroup of G containing a. By the minimality of |G|, u has finite order modulo the center of U(ℤH) and hence a ∈ Cen∗H (a), that is, a satisfies condition (C2). Finally, let G be a proper epimorphic image of G such that Cen∗ (a) ≠ G. By Lemma 14.4.4, u is a Bass unit based on a that is G of infinite order modulo the center of U(ℤG). By the minimality of |G|, ⟨u n , w n ⟩ is a nonabelian free group for some w, which is either a Bass unit or a bicyclic unit in ℤG. Then, by Problem 1.2.6, w = v m with v either a Bass cyclic unit or a bicyclic unit of ℤG and some m. Thus ⟨u nm , v nm ⟩ is free nonabelian, contradicting the assumption. This proves that a satisfies condition (C3). Thus a is dihedral p-critical in G. Therefore, G satisfies one of the conditions of Proposition 14.4.12. As G is solvable and nonabelian, G is not simple. Therefore it satisfies one of the conditions (1)–(7). If it satisfies condition (1) then the assumptions of Proposition 14.4.7 hold and hence there is a Bass unit v such that ⟨u n , v n ⟩ is free for some n. If it satisfies one of the conditions (2)–(6) then the hypothesis of Proposition 14.4.9 hold and if it satisfies condition (7) then the assumptions of Proposition 14.4.10 hold. In all these cases there is a bicyclic unit v such that ⟨u n , v n ⟩ is free for some n. This finishes the proof. It seems that one should be able to drop the solvability hypothesis in Theorem 14.4.13. Observe that the induction argument works in general and the only use of solvability is to exclude simple groups with a dihedral p-critical element with p ≥ 5. Such simple groups have been classified by Gonçalves, Guralnick and del Río in [71]. Namely it is proved that a finite simple group G has a dihedral p-critical element with p ≥ 5 if and

38 | 14 Free Groups only if it is isomorphic to PSL(2, l r ) with l prime and either p = l r = 5 or l ≠ p > 5 and 2r is the multiplicative order of l modulo p. It also has been proved that if l = 2 then every Bass unit of ℤG has a free companion among powers of bicyclic units. However the case where l is odd is still open. This is the only remaining case to deal with, in order to be able to drop the solvability hypothesis in Theorem 14.4.13.

Problems 14.4.1. Let p be an odd positive integer and let G = C p × D8 . Let x, a ∈ G with |x| = p and a of order 2. Prove that every Bass unit of the form u k,m (ax) has finite order modulo the center of U(ℤG). (Hint: See the proof of Lemma 14.4.4.)

14.5 Attractors revisited If A and B are automorphism of a vector space satisfying the conditions of either Theorem 14.3.5 or Theorem 14.3.8 then ⟨A m , B m ⟩ is free for some positive integer m. The proofs of these results however do not give any indication on which m satisfies this condition, or even when m = 1. In this section we present a method to deal with this question under some circumstances. The keystone is yet another application of the Ping-Pong Lemma. Lemma 14.5.1. Let G be a group acting on a set T and let G1 and G2 be subgroups of G. Assume that |G1 | > 2. For i = 1, 2, let U i ⊆ T be such that U i ∩ g(U i ) = 0 for every 1 ≠ g ∈ G i . If U1 ∪ U2 = T then ⟨G1 , G2 ⟩ = G1 ∗ G2 . Proof. Let P i = T \ U j for {i, j} = {1, 2}. Let 1 ≠ g ∈ G i and x ∈ P i . Then x ∈ ̸ U j and hence x ∈ U i . Therefore g(x) ∈ T \ U i = P j . Then the lemma follows from the Ping-Pong Lemma (Lemma 14.1.1). A typical situation where Lemma 14.5.1 can be applied is the case when T is a topological space, G acts by homeomorphisms on T and U i is the interior of a fundamental domain of G i . A fundamental domain for the action of a group G on a topological space T is a subset F of T such that T = ∪g∈G g(F) and F∩g(F) ⊆ ∂(F) for every 1 ≠ g ∈ G (here ∂F denotes the boarder of F). In this case, F 0 ∩ g(F 0 ) = 0 for every 1 ≠ g ∈ G. Hence, if G1 and G2 are subgroups of G and F1 and F2 are fundamental domains of G1 and G2 respectively then U i = F 0i for i = 1, 2 satisfy the first hypothesis of Lemma 14.5.1. We present an application of this idea. In the remainder of the section we consider ̂ = ℂ ∪ {∞} by Möbius transformations. The action of g ∈ GL2 (ℂ) GL2 (ℂ) acting on ℂ ̂ on z ∈ ℂ is denoted g ⋅ z. We start with a lemma that relates eigenvalues with fixed points.

14.5 Attractors revisited

| 39

Lemma 14.5.2. Let g = ( ac db ) ∈ GL2 (ℂ) and z ∈ ℂ. Then, z is a fixed point of the action of g as a Möbius transformation if and only if cz+d is an eigenvalue of the linear transformation g and in this case (z, 1) is an eigenvector of g. In particular, if v = (v1 , v2 ) ∈ ℂ2 with v2 ≠ 0 then v is an eigenvector of g if and only if vv12 is a fixed point of the action of g by Möbius transformation. Proof. Let v = (v1 , v2 ) ∈ ℂ2 with v2 ≠ 0. If v is an eigenvector of g then g⋅

v1 av1 + bv2 v1 = = , v2 cv1 + dv2 v2

That is, vv12 is invariant by the Möbius transformation determined by g. In particular, if v = (z, 1) is an eigenvector of g then g⋅z = z. Conversely, if g⋅z = z then az+b = (cz+d)z and hence a b z z ( ) ( ) = (cz + d) ( ) . c d 1 1 Hence, (z, 1) is an eigenvector of g with eigenvalue cz + d. Suppose that v1 = (v11 , v21 ) and v2 = (v12 , v22 ) are eigenvectors of g with eigenvalues λ1 and λ2 respectively and assume that v1 and v2 are linearly independent over ℂ. v12 Let U = ( vv11 ), so that U −1 gU = diag(λ1 , λ2 ). The points 0 and ∞ are invariants 21 v 22 and by the Möbius transformation defined by diag(λ1 , λ2 ) and therefore U(∞) = vv11 21 U(0) = vv12 are the points invariant by the Möbius transformation defined by g. Assume 22 additionally that |λ1 | > |λ2 |. Then the “disk” ̂ : |λ2 | ≤ |z| ≤ |λ1 |} ∪ {∞} F = {z ∈ ℂ is a fundamental domain for the action of ⟨diag(λ1 , λ2 )⟩ by Möbius transformations and therefore U(F) is a fundamental domain for the action of ⟨g⟩. The following lemma helps us to understand the shape of U(F). In ℂ2 we denote by B(c, R) the open ball with center c ∈ ℂ2 and radius R > 0. Lemma 14.5.3. Let g = ( ac db ) ∈ GL2 (ℂ) and let C be the circle with center 0 and radius R. If |d| = R|c| then g ⋅ C is the line given by the equation 2 Re((R2 ca − db)z) = R2 |a|2 − |b|2 . Otherwise, g ⋅ C is the circle with center Moreover, if |d|2 > R2 |c|2 then g ⋅ B(0, R) = B (

db−R2 ca |d|2 −R2 |c|2

and radius

R|ad−bc| . ||d|2 −R2 |c|2 |

db − R2 ca R|ad − bc| , ) |d|2 − R2 |c|2 ||d|2 − R2 |c|2 |

and, otherwise ̂ \B( g ⋅ B(0, R) = ℂ

R|ad − bc| db − R2 ca , ). |d|2 − R2 |c|2 ||d|2 − R2 |c|2 |

40 | 14 Free Groups 1 d −b ) we get that z ∈ g ⋅ C if and only if |g −1 ⋅ z| = R if Proof. Because g −1 = ad−bc ( −c a 2 2 and only if |dz − b| = R |−cz + a|2 , or equivalently,

|d|2 |z|2 − 2 Re(dbz) + |b|2 = R2 (|c|2 |z|2 − 2 Re(caz) + |a|2 ). This on its turn is equivalent with (|d|2 − R2 |c|2 )|z|2 + 2 Re((R2 ca − db)z) = R2 |a|2 − |b|2 . We claim that either |d| ≠ R|c| or R2 ca ≠ db. Indeed, assume |d| = R|c| and R2 ca = db. As (c, d) ≠ (0, 0), we get that both c and d are non-zero. Moreover |bd| = R2 |ca| = R|ad|, so that |b| = R|a|. Hence 0 ≠ (ad − bc)b = aR2 ca − |b|2 c = 0, a contradiction. If |d| = R|c| then g ⋅ C is the line given by the equation 2 Re((R2 ca − db)z) = R2 |a|2 − |b|2 , as desired. If |d| ≠ R|c| then g ⋅ C is given by the equation |z|2 + 2 Re (

R2 ca − db R2 |a|2 − |b|2 z) = . |d|2 − R2 |c|2 |d|2 − R2 |c|2

or equivalently 󵄨󵄨 󵄨󵄨 2 2 2 2 2 2 2 2 2 󵄨 󵄨󵄨 󵄨󵄨z − db − R ca 󵄨󵄨󵄨 = (R |a| − |b| )(|d| − R |c| ) + |db − R ca| . 󵄨 󵄨󵄨 2 2 2 2 2 2 2 |d| − R |c| 󵄨󵄨󵄨 (|d| − R |c| ) 󵄨󵄨 However, (R2 |a|2 − |b|2 )(|d|2 − R2 |c|2 ) + |db − R2 ca|2 = R2 (|a|2 |d|2 + |b|2 |c|2 ) − 2R2 Re(adbc) = R2 |ad − bc|2 . 2

R|ad−bc| db−R ca Therefore g ⋅ C is the circle with center |d| 2 −R 2 |c|2 and radius ||d|2 −R 2 |c|2 | , as desired. This proves the first part of the statement. The second part now can easily be obtained as well.

Corollary 14.5.4. Let g ∈ SL2 (ℂ) and assume that g has an eigenvalue λ with |λ| > 1. Suppose that (u, 1) and (v, 1) are eigenvectors of g with eigenvalues λ and λ−1 . Let F be the complement of the union of the open balls B(

|λ|4 u − v |λ|2 |u − v| , ) |λ|4 − 1 |λ|4 − 1

and

B(

|λ|4 v − u |λ|2 |u − v| , ). |λ|4 − 1 |λ|4 − 1

Then F ∪ {u} is a fundamental domain of the action of ⟨g⟩ by Möbius transformations. Proof. Let U = ( 1u 1v ). Then U −1 gU = diag(λ, λ−1 ). Consider the disk ̂ : |λ|−2 ≤ |z| ≤ |λ|2 }. D = {z ∈ ℂ Then D ∪ {∞} is a fundamental domain for the action of ⟨diag(λ, λ−1 )⟩ by Möbius transformations and therefore U ⋅ (D ∪ {∞}) is a fundamental domain for the action of

14.5 Attractors revisited

| 41

⟨g⟩. To prove the result it is sufficient to show that U ⋅ D = F and U ⋅ ∞ = u. The latter is clear. Let C(a, r) denote the circle with center a and radius r. Observe that D is the ̂ \ (C(0, |λ|2 ) ∪ C(0, |λ|−2 )) not containing 0 closure of the connected component of ℂ nor ∞. Hence U ⋅ D is the closure of the connected component of ̂ \ (U ⋅ C(0, |λ|2 ) ∪ U ⋅ C(0, |λ|−2 )) X=ℂ not containing U ⋅ 0 = v nor U ⋅ ∞ = u. By Lemma 14.5.3, ̂ \ (C ( X=ℂ

|λ|4 u − v |λ|2 |u − v| |λ|4 v − u |λ|2 |u − v| , , ) ∪ C( 4 )) . 4 4 |λ| − 1 |λ| − 1 |λ| − 1 |λ|4 − 1

Hence, the three connected components of X are the two open balls B(

|λ|4 u − v |λ|2 |u − v| , ) |λ|4 − 1 |λ|4 − 1

and

B(

|λ|4 v − u |λ|2 |u − v| , ), |λ|4 − 1 |λ|4 − 1

and the complement of the union of their closures. As 󵄨󵄨 4 󵄨󵄨 |u − v| |λ|2 |u − v| 󵄨󵄨 |λ| u − v 󵄨󵄨 󵄨󵄨 󵄨󵄨 = − u < |λ|4 − 1 󵄨󵄨󵄨 |λ|4 − 1 󵄨󵄨󵄨 |λ|4 − 1 and 󵄨󵄨 4 󵄨󵄨 |u − v| |λ|2 |u − v| 󵄨󵄨 |λ| v − u 󵄨󵄨 󵄨󵄨 󵄨 󵄨󵄨 |λ|4 − 1 − v󵄨󵄨󵄨 = |λ|4 − 1 < |λ|4 − 1 , 󵄨 󵄨 the connected component not containing u nor v is precisely the complement of the two closed balls and therefore U ⋅ D = F, as desired. Proposition 14.5.5. Let g1 , g2 ∈ SL2 (ℂ), u1 , v1 , u2 , v2 ∈ ℂ. Assume that (u i , 1) and (v i , 1) are eigenvectors of g i with eigenvalues λ i and λ−1 i respectively and that |λ i | > 1, for every i = 1, 2. Let |λ1 |2 |u1 − v1 | |λ2 |2 |u2 − v2 | μ= + |λ1 |4 − 1 |λ2 |4 − 1 If the four numbers 󵄨󵄨 󵄨 󵄨󵄨 |λ1 |4 u1 − v1 |λ2 |4 u2 − v2 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 , − |λ2 |4 − 1 󵄨󵄨󵄨 󵄨󵄨󵄨 |λ1 |4 − 1 󵄨 󵄨󵄨 󵄨󵄨 |λ1 |4 v1 − u1 |λ2 |4 u2 − v2 󵄨󵄨󵄨 󵄨󵄨 , 󵄨󵄨 − 󵄨󵄨 |λ1 |4 − 1 |λ2 |4 − 1 󵄨󵄨󵄨 󵄨

󵄨 󵄨󵄨 󵄨󵄨 |λ1 |4 u1 − v1 |λ2 |4 v2 − u2 󵄨󵄨󵄨 󵄨󵄨 , 󵄨󵄨 − 󵄨󵄨 |λ1 |4 − 1 |λ2 |4 − 1 󵄨󵄨󵄨 󵄨 󵄨󵄨 󵄨 󵄨󵄨 |λ1 |4 v1 − u1 |λ2 |4 v2 − u2 󵄨󵄨󵄨 󵄨󵄨 󵄨󵄨 , − 󵄨󵄨 |λ1 |4 − 1 |λ2 |4 − 1 󵄨󵄨󵄨 󵄨

are greater than μ then ⟨g1 , g2 ⟩ is a free group of rank 2. Suppose additionally that |λ1 | = |λ2 | and let ρ = |λ1 |2 . If |u1 − v1 | + |u2 − v2 | is smaller than each of the following numbers |ρ(u1 − u2 ) − ρ−1 (v1 − v2 )|,

|ρ(u1 − v2 ) − ρ−1 (v1 − u2 )|,

|ρ(v1 − u2 ) − ρ−1 (u1 − v2 )|,

|ρ(v1 − v2 ) − ρ−1 (u1 − u2 )|

then ⟨g1 , g2 ⟩ is free.

42 | 14 Free Groups Proof. For i = 1, 2, consider the open balls B i,1 = B (

|λ i |4 u i − v i |λ i |2 |u i − v i | , ) |λ i |4 − 1 |λ i |4 − 1

and

B i,2 = B (

|λ i |4 v i − u i |λ i |2 |u i − v i | , ). |λ i |4 − 1 |λ i |4 − 1

Let F i be the complement of B i,1 ∪ B i,2 . By Corollary 14.5.4, F i is a fundamental domain ̂ by Möbius transformations. Observe that μ is the sum of the of the action of ⟨g i ⟩ on ℂ radii of a ball B1i and a ball B2j and the four numbers in the statement of the proposition are the respective distances between the centers of two of such balls. Therefore ̂ and the assumptions mean that B1i ∩ B2j = 0 for every i, j ∈ {1, 2}. Thus F10 ∪ F20 = ℂ hence the proposition follows from Lemma 14.5.1. Assume that g i , u i , v i and λ i are as in Proposition 14.5.5 and suppose that ρ = |λ1 |2 = |λ2 |2 . Then for every n ≥ 1, (u i , 1) and (v i , 1) are eigenvectors of g in with eigenvalues λ ni and λ−n i and suppose that {u 1 , v 1 } ∩ {u 2 , v 2 } = 0. Then for every (x, y) ∈ {u 1 , v 1 } × {u2 , v2 }, the sequence (ρ n (x − y)) diverges and the sequence (ρ−n (x − y)) converges to 0. This shows that for some positive n the four numbers (with ρ replaced by ρ n ) in the last part of the Proposition converge to infinity. Hence to obtain a free group ⟨g1n , g2n ⟩ it is enough to find one n for which they are all greater than |u1 − v1 | + |u2 − v2 |. We close the chapter with an example illustrating Proposition 14.5.5. Example 14.5.6. Let ω = ζ3 = g1 = (

1 ω

−1+√−3 2

and let

1 ) −ω2

and

−ω2 g2 = ( ω

1 ). 1

Then ⟨g12 , g22 ⟩ is a free group of rank 2. − √ √13+1 Proof. Let α = √ √13−1 2 2 i. Observe that α2 = −1 − 2√−3 = 1 + 4ω2 . If z ∈ ℂ then g1 ⋅ z = z if and only if ωz2 + ωz −1 = 0, or equivalently z2 + z − ω2 = 0. 2 √ This on its turn is equivalent with z = −1± 21+4ω = −1±α 2 . Therefore, by Lemma 14.5.2, if −1 + α −1 − α u1 = and v1 = 2 2 then (u1 , 1) and (v1 , 1) are eigenvectors of g1 with respective eigenvalues λ = ωu1 − ω2 =

2 + ω + ωα 2

and

λ−1 = ωv1 − ω2 =

2 + ω − ωα . 2

Similarly, g2 ⋅ z = z if and only if −ω2 z + 1 = ωz2 + z, or equivalently z2 − z − ω2 = 0. Thus, g2 ⋅ z = z if and only if 1+α 1−α and v2 = 2 2 then (u2 , 1) and (v2 , 1) are eigenvectors of g2 with eigenvalues u2 =

ωu2 + 1 =

2 + ω + ωα =λ 2

and

ωv2 + 1 =

2 + ω − ωα = λ−1 . 2

14.5 Attractors revisited

|

43

4 13 ≈ 3.79766, u1 − u2 = v1 − v2 = −1, In this case |u1 − v1 | + |u2 − v2 | = 2|α| = 2√

u1 − v2 = α − 1 and v1 − u2 = −1 − α and ρ = |λ|2 =

3+√13+√6(√13+1) 4

≈ 2.96557. Then, a

|ρ(u1 − u2 ) − ρ−1 (v1 − v2 )| = |ρ(v1 − v2 ) − ρ−1 (u1 − u2 )| ≈ 2.62837 < |u1 − v1 | + |u2 − v2 |. Therefore the hypothesis of Proposition 14.5.5 are not satisfied. However |ρ2 (u1 − u2 ) − ρ−2 (v1 − v2 )| = |ρ2 (v1 − v2 ) − ρ−2 (u1 − u2 )| ≈ 8.68092, |ρ2 (u1 − v2 ) − ρ−2 (v1 − u2 )| ≈ 13.5998,

|ρ2 (v1 − u2 ) − ρ−2 (u1 − v2 )| ≈ 23.1953.

Then ⟨g12 , g22 ⟩ is free by Proposition 14.5.5. The elements g1 and g2 of Example 14.5.6 are the elements X and XY of in the proof of Example 19.1.6. With more advanced methods it is proved there that ⟨g1 , g2 ⟩ is free.

15 Hyperbolic geometry In this chapter we give some classical background on hyperbolic geometry and on discontinuous groups of homeomorphisms of some topological space, mainly a hyperbolic space. This is needed in the later chapters for determining the unit group of orders in some quaternion algebras. We assume the reader is familiar with the geometry of Euclidean and spherical spaces and hence we mostly revise hyperbolic geometry in terms of Euclidean geometry. From the extensive list of standard references on hyperbolic geometry, we mainly made use of [20, 23, 33, 54, 77, 150, 159, 186]. We always use ℝn as model of the n-dimensional Euclidean space with the standard Euclidean norm ‖ ‖. The only n-dimensional spherical space we use is the unit sphere: 𝕊n = {x ∈ ℝn+1 : ‖x‖ = 1}. Further, we use the standard topological notations for a subset Y of a topological space 𝕏: Y is the closure of Y in 𝕏, Y ∘ is the interior of Y and ∂Y is the boundary of Y. Assume that 𝕏 is a metric space with distance function d. If a ∈ 𝕏 and r is a positive real number then B𝕏 (a, r) = {x ∈ 𝕏 : d(x, a) ≤ r}, the closed ball with center a and radius r. In case 𝕏 is clear from the context we simply write B(x, a). The group of isometries of 𝕏, we denote Isom(𝕏). In case 𝕏 is a orientable Riemann manifold then the subgroup of Isom(𝕏) formed by the isometries preserving orientation is denoted Isom+ (𝕏). Recall that a diffeomorphism f of an orientable Riemann manifold preserves the orientation if the differential of f has positive determinant at every point, and reverses the orientation if it is negative at every point. In case f is an isometry, this determinant is 1 for orientation preserving isometries and -1 for orientation reversing isometries. Moreover, Isom+ (𝕏) is a subgroup of index 2 of Isom(𝕏).

15.1 Möbius transformations ̂n = ℝn ∪ {∞}, the one point compactification of Let n be a positive integer and let ℝ n n ℝ . If one identifies ℝ with the hyperplane x n+1 = 0 of ℝn+1 , then the stereographic ̂n → 𝕊n with projection with vertex e n+1 = (0, . . . , 0, 1) defines a bijection μ : ℝ μ(∞) = e n+1 and 1 (15.1.1) μ(x) = (2x, ‖x‖2 − 1). ‖x‖2 + 1 ̂n which makes μ into an isometry is called the chordal metric. The metric in ℝ

15.1 Möbius transformations

|

45

We view Euclidean spheres of positive dimension and affine subspaces of ℝn as ̂n . More precisely, the 0-spheres are the singlemembers of one family of subsets of ℝ ̂n and for each k = 1, . . . , n, we call k-spheres both the k-dimensional ton subsets of ℝ affine subspaces of ℝn and the k-dimensional spheres of ℝn . We assume that ∞ belongs to every affine subspaces of positive dimension but that it does not belong to the Euclidean spheres. ̂n is, by definition, an (n − 1)-sphere. If n ≥ 2 then every hyperA hypersphere of ℝ sphere is determined by an equation of the form a0 ‖x‖2 − 2a ⋅ x + a n+1 = 0,

(15.1.2)

with a0 , a n+1 ∈ ℝ and a ∈ ℝn satisfying ‖a‖2 > a0 a n+1 ,

(15.1.3)

with the convention that ∞ satisfies (15.1.2) if and only if a0 = 0. The Euclidean sphere with center a ∈ ℝn and radius r > 0 we denote by S(a, r) = {x ∈ ℝn : ‖x − a‖ = r}. In particular, 𝕊n = S(0, 1) ⊆ ℝn+1 . ̂n then ℝ ̂n \ S has two connected components. We call If S is a hypersphere of ℝ these the two open half-spaces determined by S. Replacing in (15.1.2) the equality by strict inequalities, we obtain the definition of the two connected components associated to a hypersphere. We make the convention that if a0 > 0 (respectively, a0 < 0) then ∞ belongs to the connected component defined by the inequality > (respectively, 0, then the reflection is defined by ∞, { { { R(x) = {a, { { 2 r {a + ( ‖x−a‖ ) (x − a),

if x = a; if x = ∞; otherwise.

46 | 15 Hyperbolic geometry

If S is an Euclidean hyperplane then S preserves the Euclidean metric. However, if S = S(a, r) and x, y ∈ ̸ {a, ∞} then ‖R(x) − R(y)‖ =

r2 ‖x − y‖ . ‖x − a‖‖y − a‖

(15.1.4)

̂n is a composition of reflections in hyperspheres of A Möbius transformation of ℝ ̂n form a group, denoted GM(ℝ ̂n ). The Möbius transformations of ℝ The following theorem collects the fundamental properties of Möbius transformations. These will be used freely in the following chapters. For proofs we refer to [20]. ̂n . ℝ

̂n ). ̂n and let ϕ ∈ GM(ℝ Theorem 15.1.1. Let R be the reflection in a hypersphere S of ℝ n ̂ (1) ϕ is a homeomorphism of ℝ and it is conformal (i.e. its differential is an orthogonal matrix). ([20, Theorem 3.1.6]) (2) ϕ is orientation preserving if and only if ϕ is a composition of an even number of reflections. Otherwise, it is orientation reversing. The orientation preserving Möbius ̂n ), of index 2 in GM(ℝ ̂n ). transformations form a subgroup, denoted M(ℝ n (3) Every element of Isom(ℝ ) is a composition of reflections in Euclidean hyperplanes. ̂n ). ([20, Theorem 3.1.3]) Thus, Isom(ℝn ) ⊆ GM(ℝ ̂n (i.e., a map of the type x 󳨃→ rx, with r > 0) is the composition (4) Every homothety of ℝ ̂n ). of two reflections in Euclidean spheres centered in 0 and hence it belongs to M(ℝ 󸀠 n n ̂ ̂ (5) ϕ(S) is a hypersphere of ℝ . Moreover, for every hypersphere S of ℝ there is a ̂n such that ψ(S) = S󸀠 . ([20, Theorem 3.2.1 and CorolMöbius transformation ψ of ℝ lary 3.2.4]) (6) If the restriction of ϕ to S is the identity then ϕ is either the identity map or the reflection in S. ([20, Theorem 3.2.4]) (7) ϕ ∘ R ∘ ϕ−1 is the reflection in ϕ(S). ([20, Theorem 3.2.5]) ̂n then y = R(x) if and only if every sphere containing x and y is orthogonal (8) If y ∈ ℝ to S. ([20, Theorem 3.2.6]) (9) If ϕ(0) = 0 and ϕ(𝔹n ) = 𝔹n then there is an orthogonal matrix A with entries in ℝ such that ϕ(x) = Ax, for every x ∈ ℝn . In particular, ϕ preserves Euclidean distance. ([20, Theorem 3.4.1]) (10) Let ψ denote the reflection in the unit sphere S(0, 1) and assume that ϕ is the reflection in an Euclidean sphere S with center a ≠ 0. Then the following conditions are equivalent ([20, Theorem 3.4.2]): (a) S and S(0, 1) are orthogonal. (b) ϕ(0) = ψ(a). (c) ϕ(𝔹n ) = 𝔹n . (11) Let ϕ1 and ϕ2 be reflections in the Euclidean spheres S1 and S2 with centers a1 and a2 respectively. Let B1 and B2 be the interior parts of S1 and S2 . Then the following are equivalent ([20, Problem 3.4.2]): (a) S1 and S2 are orthogonal. (c) ϕ1 (B2 ) = B2 . (b) a1 ≠ a2 and ϕ1 (a2 ) = ϕ2 (a1 ). (d) ϕ2 (B1 ) = B1 .

15.2 Riemann varieties of constant curvature

|

47

Problems ̂n : 15.1.1. Prove the following formula for the chordal metric d on ℝ 2‖x−y‖

, { { 2 2 d(x, y) = { √(‖x‖ +1)(‖y‖ +1) { 2 , { √‖x‖2 +1

if x, y ≠ ∞; if y = ∞.

̂n \S then 15.1.2. Let R be the reflection in a hypersphere S. Prove that if x, R(x) ∈ ℝ R(x) is uniquely determined by the following properties. (1) The Euclidean segment [x, R(x)] is orthogonal to S. (2) If S is an Euclidean hyperplane then the middle point x+R(x) belongs to S. 2 (3) If S is the Euclidean sphere S(a, r) then ‖x − a‖n‖R(x) − a‖ = r2 . In particular, if S is a Euclidean sphere then x and R(x) and the center of S are colinear. 15.1.3. Prove that the Möbius transformations leave invariant the family of k-spheres. ̂n is 2-transitive, i.e. if P1 , P2 , Q1 , Q2 ∈ ℝ ̂n ̂n ) on ℝ 15.1.4. Prove that the action of GM(ℝ ̂n ) such that ϕ(P1 ) = Q1 and with P1 ≠ P2 and Q1 ≠ Q2 then there is ϕ ∈ GM(ℝ ϕ(P2 ) = Q2 .

15.2 Riemann varieties of constant curvature It is well known that a Riemann manifold has constant curvature if and only if it is isometric to an Euclidean space, a sphere or a hyperbolic space. The curvature of the first one is 0, that of the second is positive and the third has negative curvature. In this section, we recall, mostly without proofs, some basic facts on these spaces. We restrict to the background required for our purposes. We will use three models for the hyperbolic n-space. The first model is the Poincaré upper half-space ℍn = {(x1 , . . . , x n ) ∈ ℝn : x n > 0}, with the metric d given by cosh(d(a, b)) = 1 +

‖na − b‖2 , 2a n b n

(15.2.1)

for a = (a1 , . . . , a n ) and b = (b1 , . . . , b n ) in ℍn . In particular, the hyperbolic closed ball Bℍn (a, r) = {x ∈ ℍn : d(x, a) ≤ r} with hyperbolic center a = (a1 , . . . , a n ) and hyperbolic radius r is the Euclidean closed ball given by (x1 − a1 )2 + ⋅ ⋅ ⋅ + (x n−1 − a n−1 )2 + (x n − a n cosh(r))2 ≤ (a n sinh(r))2 .

(15.2.2)

48 | 15 Hyperbolic geometry

In other words, Bℍn (a, r) = Bℝn ((a1 , . . . , a n−1 , a n cosh(r)), a n sinh(r)). Hence, the topology of ℍn is that induced by the Euclidean topology of ℝn and a subset of ℍn is compact if and only if it is closed and bounded in the metric of ℍn , or equivalently, the subset is closed and bounded in ℝn . The second model for the hyperbolic n-space is the open unit ball 𝔹n = {a ∈ ℝn : ‖a‖ < 1} with the metric d given by cosh(d(a, b)) = 1 + 2

‖a − b‖2 . (1 − ‖a‖2 )(1 − ‖b‖2 )

(15.2.3)

The third model for the hyperbolic n-space is the Klein model 𝕂n whose underlying set also is the open unit ball but the distance between two distinct points a and b is given by 1 ‖a − b󸀠 ‖n‖b − a󸀠 ‖ d(a, b) = ln ( ), 2 ‖a − a󸀠 ‖‖b − b󸀠 ‖ where a󸀠 and b󸀠 are the unique two points in the boundary such that a belongs to the Euclidean segment [a󸀠 , b] and b belongs to the Euclidean segment [a, b󸀠 ] (see Figure 15.1).

b󸀠 b a a󸀠

Fig. 15.1: Distance in Klein model: d(a, b) =

1 2

󸀠

󸀠

‖‖b−a ‖ ln( ‖a−b ). ‖a−a󸀠 ‖‖b−b󸀠 ‖

The three models ℍn , 𝔹n and 𝕂n of hyperbolic geometry are isometric. Later, we will present a concrete isometry between ℍn and 𝔹n . To understand the hyperbolic geometry of ℍn and 𝔹n , it is convenient to consider ̂n = ℝn ∪ {∞}. More precisely, ℍn is one of the open halfthese models as subsets of ℝ spaces determined by the hypersphere given by the equation x n = 0 and 𝔹n is one of the open half-spaces determined by 𝕊n . Recall that a geodesic in a Riemann manifold is a curve that locally minimizes length. A maximal geodesic is called a complete geodesic. In this text, we will use geodesic and complete geodesic as synonymous concepts. The geodesics in Euclidean space ℝn are the straight lines. The geodesics in a sphere are the circles with the same

15.2 Riemann varieties of constant curvature

ℍ2

𝔹2

|

49

𝕂2

Fig. 15.2: Geodesics in hyperbolic planes.

center and radius as the sphere. The geodesics in 𝕂n are the intersection of 𝕂n with the straight lines of ℝn . The geodesics in ℍn (respectively, 𝔹n ) are the intersection with ̂n orthogonal to the boundary (Figure 15.2). ℍn (respectively, 𝔹n ) of the 1-spheres of ℝ ̂ n+1 ̃ denote the reflection of ℝ ̂n , let R Given a reflection R in a hypersphere S of ℝ ̃ where if S is defined by an equation as in (15.1.2) with a ∈ ℝn in the hypersphere S, ̂ n+1 : a ‖x‖2 − 2(a, 0) ⋅ x + a and a0 , a n+1 ∈ ℝ satisfying (15.1.3), then S̃ = {x ∈ ℝ 0 n+1 = ̃ = R ̃1 ⋅ ⋅ ⋅ R ̃k , with R1 , . . . , R k 0}. By Theorem 15.1.1 (6), the map ϕ = R1 . . . R k 󳨃→ ϕ ̂ ̃ n+1 ). The map ϕ ̂n ) → GM(ℝ reflections, is a well defined group homomorphism GM(ℝ is called the Poincaré extension of ϕ. Examples 15.2.1. ̃ t) = ̂n ) then ϕ(x, (1) If ϕ is a reflection in the Euclidean hyperplane a ⋅ x = t of GM(ℝ (ϕ(x), t). The same formula holds if ϕ is an Euclidean isometry because it is a composition of reflections in Euclidean hyperplanes. ̂n )) then it is the composition of two re(2) If ϕ is the homothety x 󳨃→ rx (in GM(ℝ ̃ is also the flections in Euclidean spheres centered in 0 and as a consequence ϕ n+1 homothety x 󳨃→ rx (in ℍ ). ̂n ) is an isometry of ℍn+1 . Proposition 15.2.2. The Poincaré extension of ϕ ∈ GM(ℝ Proof. It is enough to prove the result in case ϕ is a reflection. If ϕ is a reflection in ̃ is an Euclidean isometry and preserves the the Euclidean hyperplane a ⋅ x = t then ϕ ̃ is an isometry ̃ last coordinate (i.e., ϕ(x)n+1 = x n+1 ). Using this it is easy to see that ϕ n+1 of ℍ . ̃ is the reflection in the Assume now that R is the reflection in S(a, r). Hence R ̂ n+1 . Let x, y ∈ ℍn+1 . Euclidean sphere S((a, 0), r), considered as hypersphere of ℝ r2 x n+1 ̃ Then ϕ(x)n+1 = ‖x−a‖2 and, using (15.1.4), one easily verifies that 2 ̃ − ϕ(y)‖ ̃ ‖ϕ(x) ‖x − y‖2 = . ̃ n+1 ϕ(y) ̃ n+1 x n+1 y n+1 ϕ(x)

̃ is an isometry of ℍn+1 . Hence, by (15.2.1), ϕ

50 | 15 Hyperbolic geometry It is well known that the group Isom(ℍn+1 ) is generated by the Poincaré extensions of translations, rotations, homotheties and the reflection in 𝕊n (see [159, B.7]). Hence, because of the previous proposition, we obtain the following fact. ̂n ) → Isom(ℍn+1 ), that associates to a Möbius transTheorem 15.2.3. The map GM(ℝ formation its Poincaré extension, is a group isomorphism. In particular, every element ̂n . of Isom(ℍn ) is the restriction to ℍn of a Möbius transformation of ℝ We now present a concrete isometry between ℍn and 𝔹n . Let R1 be the reflection of ℝn in the Euclidean hyperplane x n = 0 and let R2 be the reflection in the Euclidean sphere S(j, √2), where j = (0, . . . , 0, 1) ∈ ℝn . Put η0 = R2 ∘ R1 .

(15.2.4)

If x = (x1 , . . . , x n ) then R1 (x) = x − 2x n j. Hence, η0 (x) = j +

2(R1 (x) − j) 1 = (2x1 , . . . , 2x n−1 , ‖x‖2 − 1) 2 ‖R1 (x) − j‖ ‖R1 (x) − j‖2

and 4 4 + (j ⋅ (R1 (x) − j)) 2 ‖R1 (x) − j‖ ‖R1 (x) − j‖2 4x n =1− ‖R1 (x) − j‖2 4 =1− . ‖x‖2 + 2x n + 1 Using these formulas it easily is verified that η0 restricts to a bijection ‖η0 (x)‖2 = 1 +

η 0 : ℍn → 𝔹n

(15.2.5)

that maps j to 0. Moreover, the restriction of η0 to ℍn is precisely the stereographic projection μ : ℍn → 𝕊n given in (15.1.1). In order to check that η0 : ℍn → 𝔹n is an isometry, we set ‖a − b‖2 2 ‖a − b‖2 and ρ2 (a, b) = ρ1 (a, b) = , 2a n b n (1 − ‖a‖2 )(1 − ‖b‖2 ) for a = (a1 , . . . , a n ) and b = (b1 , . . . , b n ), with ρ1 defined for a n b n ≠ 0 and ρ2 defined for ‖a‖ ≠ 1 and ‖b‖ ≠ 1. Then, 󵄩󵄩 R (a) − j R1 (b) − j 󵄩󵄩󵄩󵄩2 󵄩 1 ‖η0 (a) − η0 (b)‖2 = 4 󵄩󵄩󵄩 − 󵄩 2 ‖R1 (b) − j‖2 󵄩󵄩󵄩 󵄩󵄩 ‖R1 (a) − j‖ = 4(

1 1 R1 (a) − j R1 (b) − j + −2 ⋅ ) ‖R1 (a) − j‖2 ‖R1 (b) − j‖2 ‖R1 (a) − j‖2 ‖R1 (b) − j‖2

‖R1 (a) − j‖2 + ‖R1 (b) − j‖2 − 2(a − (2a n + 1)j) ⋅ (b − (2b n + 1)j) ‖R1 (a) − j‖2 ‖R1 (b) − j‖2 2 ‖a − b‖ =4 ‖R1 (a) − j‖2 ‖R1 (b) − j‖2

=4

15.2 Riemann varieties of constant curvature

|

51

and hence ρ2 (η0 (a), η0 (b)) =

2‖η0 (a) − η0 (b)‖2 (1 − ‖η0 (a)‖2 )(1 − ‖η0 (b)‖2 )

‖a − b‖2 ‖R1 (a) − j‖2 ‖R1 (b) − j‖2 = 4a n 4b n ‖R1 (a) − j‖2 ‖R1 (b) − j‖2 8

= ρ1 (a, b). Using the distance formulas stated in (15.2.1) and (15.2.3), one deduces that η0 : ℍn → 𝔹n indeed is an isometry. Combining this with Theorem 15.2.3 we obtain the following result. Proposition 15.2.4. The map η0 : ℍn → 𝔹n , defined in (15.2.4), is an isometry such that η0 ((0, 0, . . . , 1)) = (0, 0, . . . , 0). The rule 𝛾 󳨃→ η0 𝛾η0 defines a group isomorphism Isom(ℍn ) → Isom(𝔹n ). Every element of Isom(𝔹n ) is the restriction to 𝔹n of a Möbius ̂n . transformation of ℝ ̂n can be made into a model of n-dimensional hyperIn fact, any open half-space of ℝ bolic space. Indeed, if H is an arbitrary open half-space then, by Theorem 15.1.1 (5), there is a Möbius transformation ϕ mapping bijectively ℍn to H and hence one can transfer via ϕ the structure of Riemann manifold from ℍn to H. The antipode of a point x of 𝕊n is denoted by x󸀠 . Every geodesic of 𝕊n containing x also contains x󸀠 . In the following chapter we will consider Riemann manifolds of constant curvature. Recall that these are isometric to either an Euclidean space, a sphere or a hyperbolic space. The curvature of the first one is 0, that of the second is positive and the third has negative curvature. Such an n-dimensional manifold we will simply denoted by 𝕏 and we agree that if 𝕏 is Euclidean then 𝕏 = ℝn ; if it is spherical then 𝕏 = 𝕊n and and if 𝕏 is hyperbolic then it is either ℍn , 𝔹n or 𝕂n . Let x and y be two different points of 𝕏 such that if 𝕏 is spherical then x and y are not antipodal. Then 𝕏 has exactly one geodesic containing x and y. The closed interval in this geodesic and which has x and y as extreme points is called the geodesic segment defined by x and y and is denoted [x, y]. Moreover, we set (x, y) = [x, y] \ {x, y}. In case 𝕏 is one of the three models of hyperbolic space, namely ℍn , 𝔹n or 𝕂n , ̂n intersects ∂𝕏 in two points. We will then the prolongation of each geodesic in ℝ abuse terminology and will say that the geodesic contains these two points. Then, for ̂n there is a unique geodesic in every two distinct points P and Q in the closure of 𝕏 in R 𝕏 containing P and Q and we use also the notation [P, Q] and (P, Q) with the obvious meaning. A subspace of 𝕏 is a subset that contains the geodesic passing through any two of its non-antipodal points and that furthermore is closed under taking antipodes. If 𝕏 = ℝn then the subspaces are the affine varieties. If 𝕏 = 𝕊n then the subspaces are

52 | 15 Hyperbolic geometry the intersection with 𝕊n of the affine subspaces of ℝn+1 containing 0. If 𝕏 is either ℍn or 𝔹n then the subspaces of 𝕏 of dimension k are the intersections with 𝕏 of k̂n orthogonal to the boundary of 𝕏. Finally, the subspaces of 𝕂n are the spheres of ℝ intersections of 𝕂n with subspaces of ℝn . Every subspace S of 𝕏 is a Riemann manifold with the same constant curvature as 𝕏. The codimension of S in 𝕏 is the difference between the dimensions of 𝕏 and S. For every subspace S of 𝕏 of dimension k and every point s ∈ S there is a unique subspace Y of 𝕏 of codimension k, containing s and orthogonal to S. For every non-empty subset U of 𝕏 there is a unique minimal subspace of U containing U, it is called the subspace generated by U. If S is a subspace of 𝕏 of dimension k and x ∈ 𝕏\S then the subspace generated by S∪{x} is of dimension k + 1. Let x and y be different elements of 𝕏. The bisector of x and y is Bis(x, y) = {z ∈ 𝕏 | d(x, z) = d(y, z)}. This is the hyperplane in 𝕏 consisting of the points equidistant from x and y. It also is the unique hyperplane of 𝕏 perpendicular to any geodesic joining x and y and containing one (any) point equidistant from x and y. A subset S of 𝕏 is said to be convex if [x, y] ⊆ S for all distinct non-antipodal x, y ∈ S. Let x ∈ 𝕏 and let U be a subset of 𝕏, such that if 𝕏 is spherical then U does not contain the antipode of x. Then the cone based on U with vertex x is ⋃u∈U [x, u]. In case 𝕏 is one of the three models of the hyperbolic space of dimension n, we extend ̂n . this notion to the case where U ∪ {x} is contained in the closure of 𝕏 in ℝ

Problems 15.2.1. Prove that if d is the metric of 𝔹n then d(0, x) = ln( 1+|x| 1−|x| ).

15.3 The groups of isometries of ℍ3 and 𝔹3 Recall that Isom+ (ℍn ) and Isom+ (𝔹n ) denote the groups consisting of the orientation preserving isometries of ℍn and 𝔹n , respectively. As a reflection in a hyperspace is orientation reversing, the elements of Isom+ (ℍn ) and Isom+ (𝔹n ) are the composition of an even number of reflections in hyperspaces. In this section we give a description of the groups Isom+ (ℍ3 ) and Isom+ (𝔹3 ) in terms of some groups of matrices. We consider ℂ as a subfield of ℍ(ℝ) in the obvious way. Also, we identify ℂ with ℝ2 and ℍ(ℝ) with ℝ4 by representing the elements of ℂ and ℍ(ℝ) in the standard ̂2 and ̂ = ℂ ∪ {∞} is identified with ℝ basis {1, i} and {1, i, j, k} respectively. Thus, ℂ ̂4 . Further, we identify ℍ3 with ℝ⊕ℝi ⊕ℝ+ j, 𝔹3 ̂ = ℍ(ℝ)∪{∞} is identified with ℝ ℍ(ℝ) 2 2 2 2 with {x1 + x2 i + rj : x1 + x2 + r < 1}, ℍ with ℝ⊕ℝ+ i and 𝔹2 with {x1 + x2 i : x21 + x22 < 1}. ̂ ∂𝔹3 = 𝕊2 , ∂ℍ2 = ℝ ̂ and ∂𝔹2 = 𝕊1 . The boundaries of these sets are ∂ℍ3 = ℂ,

15.3 The groups of isometries of ℍ3 and 𝔹3

|

53

̂ by Möbius transformations can be extended to an action The action of GL2 (ℂ) on C ̂ of GL2 (ℍ(ℝ)) on ℍ(ℝ) as follows. If A=(

a c

b ) ∈ GL2 (ℍ(ℝ)) d

̂ then define and z ∈ ℍ(ℝ) −1 {(az + b)(cz + d) , { { A ⋅ z = {ac−1 , { { {∞,

if z ∈ ℍ(ℝ) and cz + d ≠ 0; if c ≠ 0 and z = ∞;

(15.3.1)

otherwise.

̂ It easily is verified that this indeed defines an action of GL2 (ℍ(R)) on ℍ(ℝ). ̂ If A ∈ GL2 (ℍ(ℝ)) and z ∈ ℍ(ℝ) then we set g̃ A (z) = A ⋅ z. The mapping A 󳨃→ g̃ A defines a group homomorphism from GL2 (ℍ(ℝ)) to the group of ̂ bijections of ℍ(ℝ). The kernel of this map is formed by the non-zero central matrices, ̂ i.e. the matrices of the form diag(λ, λ) with λ ∈ ℝ \ {0}. Moreover g̃ A leaves ℂ invarî we denote g A . This defines a group homomorphism from ant and its restriction to ℂ ̂ The kernel of this homomorphism is the set of GL2 (ℂ) to the group of bijections of ℂ. non-zero scalar matrices, i.e. the matrices of the form diag(a, a) with a ∈ ℂ \ {0}. In particular, if λ ∈ ℂ \ {0} and A ∈ GL2 (ℂ) then g λA = g A . This implies that the map A 󳨃→ g A induces a group homomorphism from PSL2 (ℂ) ̂ The image of 𝛾 ∈ PSL2 (ℂ) we denote g𝛾 . Clearly, for to the group of bijections of ℂ. every A ∈ GL2 (ℂ) there is B ∈ SL2 (ℂ) such that g A = g B . Moreover, if A denotes the matrix obtained by taking the conjugates of the entries of A and we extend complex ̂ by setting ∞ = ∞ then conjugation to ℂ ̂ g A (z) = g A (z) (z ∈ ℂ).

(15.3.2)

(Caution: This equality does not hold in general for A ∈ GL2 (ℍ(ℝ)).) If A = ( ac db ) ∈ GL2 (ℂ), with c ≠ 0, then the isometric sphere of A is I A = {z ∈ ℍ(ℝ) : ‖cz + d‖2 = |ad − bc|}, i.e.

d √|ad − bc| I A = S (− , ). c |c|

If 𝛾 ∈ PSL2 (ℂ) is represented by A then we set I𝛾 = I A and we call it the isometric sphere of 𝛾. This is well defined because I λA = I A for λ ∈ U(ℂ).

54 | 15 Hyperbolic geometry The name “isometric sphere” comes from the fact that g̃ A acts as an Euclidean isometry on I A . Indeed, let z = x + rj and w = y + sj be elements of I A . Then c|z|2 c̄ + cz d̄ + d z̄ c̄ + |d|2 = c|w|2 c̄ + cw d̄ + d w̄ c̄ + |d|2 = |ad − bc|. Therefore (az + b)(cz + d) − (aw + b)(cw + d) = (az + b)(z̄ c̄ + d)̄ − (aw + b)(w̄ c̄ + d)̄ = a|z|2 c̄ + az d̄ + b z̄ c̄ − a|w|2 c̄ − aw d̄ − b w̄ c̄ = c −1 (ac|z|2 c̄ + acz d̄ + bc z̄ c̄ − ac|w|2 c̄ − acw d̄ − bc w̄ c)̄ = c−1 (ad(w̄ − z)̄ c̄ + bc z̄ c̄ − bc w̄ c)̄ = c−1 (ad − bc)(w̄ − z)̄ c.̄ Thus −1 ̃ ‖g̃ − (aw + b)(cw + d)−1 ‖ A (z) − g A (w)‖ = ‖(az + b)(cz + d)

‖(az + b)(cz + d) − (aw + b)(cw + d)‖ |ad − bc| = ‖w̄ − z‖̄ = ‖z − w‖, =

as desired. Recall that if 𝛾 ∈ PSL2 (ℂ) then the intersection of I𝛾 with ℂ is called the isometric circle of g𝛾 (see Problem 1.4.2). Abusing notation we also denote this as I𝛾 . 3 If A = ( ac db ) ∈ GL2 (ℂ) then the restriction of g̃ A to ℍ takes the following form g̃ A (z + rj) =

(az + b)(cz + d) + acr2 (ad − bc)r + j. |cz + d|2 + |c|2 r2 |cz + d|2 + |c|2 r2

(15.3.3)

3 In particular, if A ∈ SL2 (ℂ) then g̃ A restricts to a bijection of ℍ , which we also denote g̃ A . The following proposition justifies why we use the same notation as for the Poincaré extension.

Proposition 15.3.1. If A ∈ SL2 (ℂ) then the Poincaré extension of g A is g̃ A . Furthermore, ̂2 ) = {g A : A ∈ GL2 (ℂ)} = {g A : A ∈ SL2 (ℂ)}. M(ℝ ̂ If S is given by the equation Proof. Let R be the reflection in a hypersurface S of ℂ. a ⋅ x = t with a = a1 + a2 i ∈ ℂ \ {0} then for z = x1 + x2 i we have R(z) = z − 2 (z ⋅

a t a − ) |a| |a| |a|

= x1 + x2 i − 2 =

(x1 a1 + x2 a2 − t) a |a|2

1 ((x1 + x2 i)(a21 + a22 ) − 2(x1 a21 + x2 a1 a2 + x1 a1 a2 i + x2 a22 i − ta)) a21 + a22

15.3 The groups of isometries of ℍ3 and 𝔹3

= =

|

55

1 ((a22 − a21 )x1 − 2a1 a2 x2 + ((a21 − a22 )x2 − 2a1 a2 x1 )i + 2ta a21 + a22 (a22 − a21 − 2a1 a2 i)z + 2ta a21 + a22

.

If, on the other hand, S = S(a, t) then R(z) = a +

t2 az + t2 − ‖a‖2 . = z−a z−a

In both cases, R(z) = g A (z) for some A ∈ GL2 (ℂ). Because of (15.3.2), it follows that if ϕ is the composition of two reflections then, ϕ(z) = g B (g A (z) = g BA (z), for some ̂2 ) is the composition of an even number of A, B ∈ GL2 (ℂ). As every element ψ of M(ℝ reflections, it follows that ψ is of the form g A for some A ∈ GL2 (ℂ). So, we have shown ̂2 ) ⊆ {g A : A ∈ GL2 (ℂ)} = {g A : A ∈ SL2 (ℂ)}. that M(ℝ To prove the reverse inclusion, let A = ( ac db ) ∈ SL2 (ℂ). If c = 0 then ad = 1 and g A (z) = a2 z + ab. This is the composition of an homothety (multiplication by |a|2 ), a a2 ̂2 rotation (multiplication by |a| 2 ) and a translation (adding ab). Thus g A ∈ M(ℝ ) and, because of Examples 15.2.1, we deduce that the Poincaré extension of g A maps z + rj to g A (z) + |a|2 rj = g A (z) + |d|r 2 j = g̃ A (z + rj), as required in the first part of the statement of the proposition. Assume now that c ≠ 0 and let σ be the reflection in the isometric circle of g A . Then z + dc d 1 d =− + σ(z) = − + c |c|2 󵄨󵄨󵄨z + d 󵄨󵄨󵄨2 c c(cz + d) 󵄨󵄨 c 󵄨󵄨 and hence 1 cσ(z) + d = . cz + d Thus, g A σ(z) =

aσ(z) + b a(cσ(z) + d) + bc − ad a cz + d = = − = αz + β, cσ(z) + d c(cσ(z) + d) c c

for some α, β ∈ ℂ with |α| = 1. Hence, g A σ is an orientation reversing Euclidean isometry and hence it is the composition of an odd number of reflections in hypersurfaces. Thus, g A is the composition of an even number of reflections in hypersurfaces. ̂2 ). This finishes the proof of the reverse inclusion. It remains to Therefore g A ∈ M(ℝ show that also in this case the first part of the statement of the proposition holds. As g A σ is an Euclidean isometry, we get from Example 15.2.1 that the Poincaré extension of g A σ is given by ϕ(z + rj) = g A σ(z) + rj, and as the Poincaré extension of σ is the 1 reflection in the three-dimensional sphere S(− dc , |c| ), it is given by ̃ (z + rj) = − σ

d (z + dc + rj) d 1 z + rj + c d =− + = z1 + r1 j. + 2󵄨 2 󵄨 c |c| 󵄨󵄨z + rj + d 󵄨󵄨 c |cz + d|2 + |c|2 r2 󵄨󵄨 󵄨 c󵄨

56 | 15 Hyperbolic geometry

Using the above expression for g A σ, we obtain (g A σ)(z1 ) = (g A σ)(−

(z + dc ) d + ) c |cz + d|2 + |c|2 r2

=

a (cz + d) − c c(|cz + d|2 + |c|2 r2 )

=

a(|cz + d|2 + |c|2 r2 ) + (bc − ad)(cz + d) c(|cz + d|2 + |c|2 r2 )

=

a|c|2 |z|2 + aczd + a|c|2 r2 + b|c|2 z + bcd c(|cz + d|2 + |c|2 r2 )

=

(az + b)(cz + d) + acr2 . |cz + d|2 + |c|2 r2

Therefore, the Poincaré extension of g A σ maps z + rj to ̃ (z + rj) = ϕ(z1 + r1 j) = (g A σ)(z1 ) + r1 j ϕσ rj (az + b)(cz + d) + acr2 + |cz + d|2 + |c|2 r2 |cz + d|2 + |c|2 r2 = g̃ A (z + rj),

=

as desired. As an immediate application of Theorem 15.2.3 and Proposition 15.3.1 we obtain the desired group isomorphism between PSL2 (ℂ) and Isom+ (ℍ3 ). Theorem 15.3.2. The map 𝛾 󳨃→ g̃𝛾 is a group isomorphism from PSL2 (ℂ) to Isom+ (ℍ3 ). A similar argument shows that 𝛾 󳨃→ g𝛾 induces a group isomorphism PSL2 (ℝ) → Isom+ (ℍ2 ). +

We now turn our attention to Isom

(𝔹3 ).

1 A0 = ( −j

(15.3.4)

Let −j ). 1

If A = ( ac db ) ∈ M2 (ℂ) then A0 AA−1 0 =( with x=

1 (a + d + (b − c)j) 2

and

x y

yk ) xk y=

1 (b + c + (d − a)j). 2

Note that if det(A) = 1 then (a + d)(d − a) − (b − c)(b + c) = −|a|2 + |d|2 − |b|2 + |c|2 ∈ ℝ. k On the other hand, if B = ( xy xy k ) with x = x0 + x1 j, y = y0 + y1 j and x0 , x1 , y0 , y1 ∈ ℂ then x0 − y1 x1 + y0 A−1 ). 0 BA 0 = ( y0 − x1 x0 + y1

15.3 The groups of isometries of ℍ3 and 𝔹3

| 57

This proves that A0 M2 (ℂ) A−1 0 = {(

x y

yk ) : x, y ∈ ℍ(ℝ)} . xk

(15.3.5)

Moreover, det (

x0 − y1 y0 − x1

x1 + y0 ) = ‖x‖2 − ‖y‖2 + 2 Im(x0 y1 − x1 y0 )i x0 + y1

and xy = (x0 + x1 j)(y0 − y1 j) = x0 y0 + x1 y1 + (x1 y0 − x0 y1 )j. It is now readily verified that A0 SL2 (ℂ) A−1 (15.3.6) 0 = SB2 (ℍ(ℝ)), where SB2 (ℍ(ℝ)) = {(

x y

yk ) : x, y ∈ ℍ(ℝ), ‖x‖2 − ‖y‖2 = 1 and xy ∈ ℂ + ℝj} . xk

Let PSB2 (ℍ(ℝ)) = SB2 (ℍ(ℝ))/ ⟨−I⟩ . By (15.3.6) the map Ψ : SL2 (ℂ) → SL2 (ℍ(ℝ)) defined by Ψ(A) = A0 AA−1 0 induces an isomorphism PSL2 (ℂ) → PSB2 (ℍ(ℝ)), which we also denote Ψ. If 𝛾 ∈ PSB2 (ℍ(ℝ)) is represented by A ∈ SB2 (ℍ(ℝ)) then we denote g̃𝛾 = g̃ A . This is obviously well defined. 2 + Recall that the mapping y + rj 󳨃→ j + |y|2 +(r+1) 2 (y − (r + 1)j), with y ∈ ℂ and r ∈ ℝ , defined in (15.2.4), is an isometry η0 : ℍ3 → 𝔹3 . It easily is verified that this mapping is the restriction of g A0 to ℍ3 , with 1 A0 = ( −j

−j ). 1

̂ by setting η0 = ̃ g A0 . Hence We extend η0 to ℍ(ℝ) ∞, { { { η0 (w) = {j, { { −1 {(w − j)(1 − jw) ,

if w = −j; if w = ∞; otherwise.

This implies that Isom(𝔹3 ) ≅ η0 Isom(ℍ3 )η−1 0 . Combining this with Theorem 15.3.2 we obtain the following description of Isom(𝔹3 ). Theorem 15.3.3. The following is a commutative diagram of isomorphisms PSL2 (ℂ) 󳨀→ Isom+ (ℍ3 ) Ψ↓ ↓ PSB2 (ℍ(ℝ)) 󳨀→ Isom+ (𝔹3 )

(15.3.7)

The two horizontal maps are given by 𝛾 󳨃→ g̃𝛾 and the right vertical map is given by + 3 𝛾 ∈ Isom+ (ℍ3 ) 󳨃→ η0 𝛾η−1 0 ∈ Isom (𝔹 ).

58 | 15 Hyperbolic geometry A 2-dimensional version of Theorem 15.3.3 holds with PSL2 (ℂ) and PSB2 (ℍ(ℝ)) replaced by PSL2 (ℝ) and PSB2 (ℂ) = SB2 (ℂ)/(−I), respectively, where SB2 (ℂ) = {(

x y

y ) ∈ GL2 (ℂ) : |x|2 − |y|2 = 1} . x

Thus there is a commutative diagram of isomorphisms PSL2 (ℝ) 󳨀→ Isom+ (ℍ2 ) Ψ↓ ↓ PSB2 (ℂ) 󳨀→ Isom+ (𝔹2 )

(15.3.8)

where the horizontal maps are given by 𝛾 󳨃→ g𝛾 , the left vertical map is induced by conjugation by ( 1i 1i ), and the right vertical map is given by 𝛾 ∈ Isom+ (ℍ2 ) 󳨃→ η0 𝛾η−1 0 ∈ Isom+ (𝔹2 ). We will abuse notation and consider the horizontal isomorphisms of (15.3.7) and (15.3.8) as identifications. Hence, if 𝛾 ∈ PSL2 (ℂ) (or 𝛾 ∈ PSB2 (ℍ(ℝ)) then we simply write g̃𝛾 as 𝛾. Also if 𝛾 ∈ PSL2 (ℝ) or 𝛾 ∈ PSB2 (ℂ) then g𝛾 is simply written as 𝛾. Let ρ : ℍ(ℝ) 󳨀→ M2 (ℂ) be the mapping defined by a −b

x = a + bj 󳨃→ (

b ), a

where a, b ∈ ℂ. Clearly, ρ is a homomorphism of ℝ-algebras. Therefore, the reduced norm of ℍ(ℝ) over ℝ, which we will abbreviate as RNr, is given by RNr(x) = |a|2 + |b|2 = ‖x‖2 . Let A∗ denote the Hermitian transpose of A ∈ M2 (ℂ), i.e. if A = ( ac db ) then A∗ = (

a b

c ). d

A matrix A ∈ GL2 (ℂ) is said to be unitary if AA∗ = I. The special unitary group of degree 2 is SU2 (ℂ) = {A ∈ SL2 (ℂ) : A∗ = A−1 }. Also, if A = ( ac db ) ∈ M2 (ℂ) then the norm ‖A‖ is defined as (|a|2 + |b|2 + |c|2 + |d|2 )1/2 . If 𝛾 ∈ PSL2 (ℂ) is represented by A ∈ SL2 (ℂ) then we define ‖𝛾‖ = ‖A‖. Next we give a description of the restriction to the special unitary group of the isomorphisms of Theorem 15.3.3. In order to do so, it is useful to note that the norm in SL2 (ℂ) and the geometry of ℍ3 are linked by the following formula ‖𝛾‖2 = 2 cosh(d(j, 𝛾(j))),

(15.3.9)

15.3 The groups of isometries of ℍ3 and 𝔹3

| 59

with 𝛾 ∈ PSL2 (ℂ). Indeed, let A = ( ac db ) ∈ SL2 (ℂ) be a representative of 𝛾. Then,

𝛾(j) = g̃A (j) = (aj + b)(cj + d)−1 1 (aj + b)(−cj + d) |c|2 + |d|2 1 = 2 (ac + bd + j). |c| + |d|2 =

As 1 = (ad − bc)(ad − bc) = |a|2 |d|2 + |b|2 |c|2 − adbc − adbc, we obtain from equation (15.2.1) that 2 cosh(d(j, 𝛾(j))) = 2 + (|c|2 + |d|2 )‖j − 𝛾(j)‖2 =2+

‖(|c|2 + |d|2 − 1)j − (ac + bd)‖2 |c|2 + |d|2

=2+

|ac + bd|2 + (|c|2 + |d|2 − 1)2 |c|2 + |d|2

=2+

(ac + bd)(ac + bd) + (|c|2 + |d|2 )2 − 2(|c|2 + |d|2 ) + 1 |c|2 + |d|2

|a|2 |c|2 + |b|2 |d|2 + adbc + adbc + (|c|2 + |d|2 )2 + 1 |c|2 + |d|2 2 2 2 2 |a| |c| + |b| |d| + |a|2 |d|2 + |b|2 |c|2 + (|c|2 + |d|2 )2 = |c|2 + |d|2 =

= |a|2 + |b|2 + |c|2 + |d|2 = ‖A‖2 , as claimed. As cosh is strictly increasing on [0, +∞), we deduce from (15.3.9) that ‖A‖ ≥ 2

(15.3.10)

‖𝛾‖2 = 2 cosh(d(i, 𝛾(i))) (𝛾 ∈ PSL2 (ℝ)).

(15.3.11)

for every A ∈ SL2 (ℂ). A similar argument shows

Proposition 15.3.4. Let A ∈ SL2 (ℂ). The following statements are equivalent: (1) A ∈ SU2 (ℂ), (2) A = ρ(x) for some x ∈ ℍ(ℝ) with RNr(x) = 1, (3) ‖A‖2 = 2, (4) g̃ A (j) = j, (5) g̃ Ψ(A) (0) = 0, (6) Ψ(A) is a diagonal matrix, (7) g̃ Ψ(A) is an Euclidean isometry. In particular, if PSU2 (ℂ) = SU2 (ℂ)/ ⟨−I⟩ and Isom+x (X) denotes the orientation preserving automorphisms fixing x, for X = ℍ3 or 𝔹3 , then the commutative diagram of

60 | 15 Hyperbolic geometry

isomorphisms of Theorem 15.3.3 restricts to a commutative diagram 󳨀→ Isom+j (ℍ3 ) PSU2 (ℂ) ↓ ↓ {diag(x, x k ) : x ∈ ℍ(ℝ), ‖x‖ = 1} / ⟨−I⟩ 󳨀→ Isom+0 (𝔹3 ) Proof. (2) implies (1) and (3). If A = ρ(x) with x = z + wj ∈ ℍ(ℝ) and z, w ∈ ℂ then ρ(x)ρ(x)∗ = (

z −w

w z )( z w

∗

−w ) = (|z|2 + |w|2 )I = RNr(x)I. z

(15.3.12)

Thus, if RNr(x) = 1 then A ∈ SU2 (ℂ) and ‖A‖2 = 2. d −b ) = A −1 = A ∗ = ( a c ). (1) implies (2). Let A = ( ac db ) ∈ SU2 (ℂ). Then ( −c a b d Therefore d = a and b = −c. Thus A = ρ(a + bj) and RNr(a + bj) = 1. (3) implies (2) AA∗ = (

a c

a b )( d b

c |a|2 + |b|2 )=( d ca + db

ac + bd ). |c|2 + |d|2

and |a − d|2 + |b + c|2 = ‖A‖2 − 2. Therefore, if ‖A‖2 = 2 then a = d and b = −c, or equivalently, A = ρ(a + bj). That (3) and (4) are equivalent is a consequence of the equality (15.3.9) That (4) and (5) are equivalent is a consequence of the equalities ̃ g A0 (j) = 0 and g A0 = ̃ g̃ g A0 g̃ Ψ(A) ̃ A. The equivalence of (5) and (6) is obvious. (6) implies (7). If Ψ(A) is diagonal then, for every z ∈ 𝔹3 , we have g̃ Ψ(A) (z) = k −1 ̃ xz(x ) , for some x ∈ ℍ(ℝ) with ‖x‖ = 1. Hence, ‖g̃ Ψ(A) (z)‖ = ‖z‖ and thus g Ψ(A) is an Euclidean isometry. 3 (7) implies (5). If g̃ Ψ(A) is an Euclidean isometry then it leaves 𝔹 invariant and thus it fixes 0.

Problems 15.3.1. Let 𝛾 ∈ PSL2 (ℂ) be such that 𝛾(∞) ≠ ∞. Prove that if z ∈ ℂ then |𝛾󸀠 (z)| > 1 if and only if z belongs to the interior of the isometric circle of 𝛾.

15.4 Isometric spheres in 𝔹n In this section we introduce a notion of isometric sphere for the elements of SB2 (ℍ(ℝ)) not fixing 0, similar to the notion of isometric circle of elements of SL2 (ℂ), not fixing ∞. We start by introducing some notation for some special bisectors in 𝔹n and ℍn .

15.4 Isometric spheres in 𝔹n

| 61

Let 𝕏n be either ℍn or 𝔹n . We consider 𝕏n as a subset of ℝn . By definition, the center C of 𝕏n is j = (0, 0, . . . , 0, 1) if 𝕏n = ℍn and 0 if 𝕏n = 𝔹n . Given 𝛾 ∈ Isom(𝕏n ) ̂n such that with 𝛾(C) ≠ C we let Bis𝕏n (𝛾) denote the unique hypersphere of ℝ Bis𝕏n (𝛾) ∩ 𝕏n = Bis(C, 𝛾(C)). Let Bis>𝕏n (𝛾) and Bisℍ3 (𝛾) = B(P, R) if and only if |c|2 + |d|2 < 1 and otherwise Bisℍ3 (𝛾) = B(P, R) if and only if ‖j − P‖ < ‖𝛾(j) − P‖. This precisely holds for λ > 1, or equivalently |c|2 + |d|2 < 1. Otherwise Bis 1. −1

x y (4) Let x = 12 (d+a+(c−b)j) and y = 12 (−b−c+(a−d)j). Then A0 ( ac db ) A−1 0 = ( y x k ). −1 Moreover, Bis𝔹3 (Ψ(𝛾)), is the isometric sphere of Ψ(𝛾 ) and, by Proposition 15.4.3, it is the sphere with center −y−1 x k and radius ‖y‖−1 . By straightforward calculations we ‖𝛾‖2 −2 obtain ‖y‖2 = 4 and −y−1 x k = ‖𝛾‖42 −2 (2v + (|a|2 + |b|2 − |c|2 − |d|2 )j), as desired. (5) is a direct consequence of (4). k

Let d𝛾 denote the Euclidean distance between the points of intersection of the segment [0, P] with 𝕊2 and S(P𝛾 , R𝛾 ). Clearly, d𝛾 = 1 + R𝛾 − ‖P𝛾 ‖. If 𝛾 ∈ PSU2 (ℂ) then we set d𝛾 = 1. Note that the Euclidean volume of the intersection of 𝔹3 ∩ B(P𝛾 , R𝛾 ) is a function of d𝛾 . Lemma 15.4.5 ([109]). d𝛾 is a strictly decreasing function of ‖𝛾‖2 , for 𝛾 ∈ PSL2 (ℂ). Proof. By (15.3.10) we have ‖𝛾‖2 ∈ [2, +∞). Moreover, if 𝛾 ∈ ̸ PSU2 (ℂ) then ‖𝛾‖2 > 2 and, by Lemma 15.4.4 (5), we have 1

‖𝛾‖2 + 2 2 −( 2 d𝛾 = 1 + ) = f(‖𝛾‖2 ), 1 ‖𝛾‖ − 2 (‖𝛾‖2 − 2) 2 2

where f : (2, ∞) → ℝ is the function defined by f(x) = 1 +

2 − √x + 2 . √x − 2

To prove the result it is now enough to notice that f is strictly decreasing on the interval (2, +∞) (see Figure 15.3). Indeed, one easily verifies that f 󸀠 (x) =

2 − √x + 2 √x + 2√(x − 2)3

which takes negative values for x > 2.

Problems 15.4.1. Let 𝛾 ∈ PSL2 (ℂ) be represented by A = ( ac db ) ∈ SL2 (ℂ). Let Ψ(A) = ( xy with x, y ∈ ℍ(ℝ). Prove that

‖x‖2

=

2+‖𝛾‖2 4 ,

‖y‖2

=

‖𝛾‖2 −2 4

and ‖Ψ(𝛾)‖ = ‖𝛾‖.

yk xk

),

15.5 Discrete subgroups of PSL2 (ℂ) | 65

1

0

2

Fig. 15.3: Plot of y = 1 +

2−√x+2 √x−2

in the interval [2, 7].

15.4.2. Let 𝛾, δ ∈ PSL2 (ℂ) and assume that δ(j) = j. Prove that ‖δ𝛾δ−1 ‖ = ‖𝛾‖. Prove a similar result for elements of PSL2 (ℝ), PSB2 (ℍ(ℝ)) or PSB2 (ℂ). 15.4.3. Prove that if 𝛾 ∈ PSB2 (ℍ(ℝ)) then ‖𝛾‖2 = 2 cosh(d(0, 𝛾(0))), where d denotes the distance in 𝔹3 . 15.4.4. Let 𝕏n be either ℍn or 𝔹n , let C be the center of 𝕏n and let 𝛾1 , 𝛾2 ∈ Isom(𝕏n ) with 𝛾1 (C) ≠ C ≠ 𝛾2 (C). Prove that Bis𝕏n (𝛾1 ) = Bis𝕏n (𝛾2 ) if and only if 𝛾1 (C) = 𝛾2 (C). 15.4.5. Let 𝛾1 , 𝛾2 ∈ PSL2 (ℂ) \ PSU2 (ℂ). Prove that Ψ(𝛾1 ) and Ψ(𝛾2 ) have the same isometric sphere if and only if 𝛾2 𝛾−1 1 ∈ PSU2 (ℂ). 15.4.6. Prove the following properties for 𝛾 = ( ac db ) ∈ SL2 (ℂ) \ SU2 (ℂ): (1) j ∈ Bis 1. Therefore, the only geodesic of ℍ invariant by 𝛾 is x = y = 0 and ℍ3𝛾 = 0. If λ2 ∈ ̸ ℝ then no Euclidean line containing 0 is invariant under 𝛾. Otherwise, λ2 is negative and 𝛾 interchanges the two open half-spaces determined by each Euclidean line containing 0. Hence A and 𝛾 satisfy the conditions of (4). Remark 15.5.3. Let 𝛾 ∈ PSL2 (ℂ) \ {1}. The proof of Proposition 15.5.2 shows that 𝛾 is parabolic if and only if there is t ∈ ℂ such that 𝛾 is conjugate to 𝛾t with 𝛾t (x) = z + t for every z ∈ ℂ. In all the other cases there is a ∈ ℂ \ {0, 1} such that 𝛾 is conjugate to 𝛾a with 𝛾(z) = az for every z ∈ ℂ. If |a| = 1 then 𝛾 is elliptic, if a ∈ ℝ \ {−1} then 𝛾 is hyperbolic and in all the other cases 𝛾 is loxodromic. Lemma 15.5.4. Let 𝛾, δ ∈ PSL2 (ℂ) be such that 𝛾 and δ have exactly one common fixed point and 𝛾 is either hyperbolic or loxodromic. Then ⟨𝛾, δ⟩ is not discrete. Proof. By Remark 15.5.3 one may assume that 𝛾 is represented by a diagonal matrix A = diag(a0 , a−1 0 ) with |a 0 | ≠ 1. Then 𝛾 has two fixed points 0 and ∞. After another conjugation one may assume that the common fixed point of 𝛾 and δ is ∞. So δ(z) is represented by a matrix of the form B = ( 0a ab−1 ) with b ≠ 0. Replacing 𝛾 by 𝛾−1 , 2n

if needed, one may assume that |a0 | < 1. Then A n BA−n = ( a a0−1b ) and therefore 0 a {𝛾n δ𝛾−n : n ∈ ℕ} is an infinite set contained in a bounded subset of SL2 (ℂ). Hence ⟨δ, 𝛾⟩ is not discrete.

68 | 15 Hyperbolic geometry

Problems 15.5.1. Prove that every discrete subgroup of U(ℂ) is either of the form ⟨ζ n ⟩ or of the form ⟨ζ n ⟩ × ⟨a⟩ with a ∈ ℂ and |a| > 1. 15.5.2. Show that a topological group G is discrete if and only if every sequence (g n ) in G which converges to 1 satisfies g n = 1 for almost all n. 15.5.3. Let G be a discontinuous group of a topological space X. Prove that if x ∈ X and (g n ) is a sequence of distinct elements in G then the sequence (g n (x)) cannot converge to any y ∈ X. 15.5.4. Show that a subgroup G of SL2 (ℂ) is discrete if and only if for each positive real number k the set {A ∈ G : ‖A‖ ≤ k} is finite. 15.5.5. Let K be either the rationals or a quadratic imaginary extension of the rationals and assume R is an order in K. Prove that SL2 (R) is a discrete group. 15.5.6. Let G be a subgroup of PSL2 (ℂ). Prove that the following properties are equivalent: (1) G is discrete in PSL2 (ℂ). (2) Ψ(G) is discrete in PSB2 (ℍ(ℝ)). (3) G is a discontinuous subgroup of Isom(ℍ3 ). (4) Ψ(G) is a discontinuous subgroup of Isom(𝔹3 ).

16 Poincaré’s Theorem In this chapter, we give a self-contained proof of the Poincaré Polyhedron Theorem on presentations of discontinuous groups of isometries of a Riemann manifold of constant curvature. In summary, it states that if P is a polyhedron with side pairing transformations satisfying several conditions, then the group G generated by those side pairing transformations is discontinuous, P is a fundamental polyhedron for G and the reflection relations and cycle relations form a complete set of relations for G. Poincaré first published the theorem for dimension two in [180], then one year later also for dimension three in [181]. A lot on this theorem can be found in the literature, see for instance the books [20, 33, 54, 159, 186]. There are also various articles on this theorem, such as [4, 47, 55, 158]. The proofs in [55] and [159] focus mostly on the part of Poincaré’s Theorem which states that if a polyhedron satisfies some conditions then it is a fundamental polyhedron of a discontinuous group. As a consequence the presentation part of the theorem is obtained somehow indirectly and the intuition on the presentation part is hidden in these proofs. In [110], Jespers, Kiefer and del Río gave a self-contained proof of the presentation part of Poincaré’s Theorem. It is this proof that we will present in this chapter.

16.1 Polyhedra and tessellations In this section 𝕏 denotes an n-dimensional Riemann manifold of constant curvature. A hyperplane of 𝕏 is a codimension 1 subspace. If H is a hyperplane of 𝕏 then 𝕏 \ H has two connected components, called the open half-spaces defined by H. If U is one of them then the other open half-space defined by H is denoted U 󸀠 and we have H = ∂U = ∂U = ∂U 󸀠 = ∂U 󸀠 and U = U ∪ H,

∘

U = U,

U 󸀠 = 𝕏 \ U,

U = 𝕏 \ U󸀠.

The sets U and U 󸀠 are called the closed half-spaces defined by H. Moreover, if Z is one of these two closed half-spaces defined by H then the other is denoted by Z 󸀠 . Lemma 16.1.1. Let x ∈ 𝕏 and U ⊆ 𝕏 and assume that one of the following conditions holds. (1) U is an open subset of a hyperplane H of 𝕏 and x ∈ ̸ H. (2) U is an open subset of 𝕏 and if 𝕏 is spherical then U does not contain the antipode of x. Then ∪u∈U (x, u) is an open subset of 𝕏 and it is dense in the cone based on U with vertex x.

70 | 16 Poincaré’s Theorem Proof. (1) Assume that U is an open subset of a hyperplane H and x ∈ ̸ H. First observe that, in the spherical case, H is closed under taking antipodes and therefore it does not contain the antipode of x. Let C = ∪u∈U (x, u). We have to prove that U is open. We first reduce the statement to the Euclidean case. This is clear for the hyperbolic geometry by using the Klein model. To reduce the spherical case to the Euclidean case, consider 𝕊n as a subset of ℝn+1 and let V be the half-space of S n with border H and containing x and let E be the hyperplane of ℝn+1 tangent to the sphere in the point of V whose tangent in 𝕊n is parallel to H. The stereographic projection from the center of the sphere is a bijection V → E mapping the intersections of V with the geodesics of the sphere to the Euclidean geodesics of E. Hence the statement for the Euclidean geometry implies the statement for the hyperbolic and spherical geometries. So we only have to prove the statement for 𝕏 = ℝn . Making use of some affine transformations we may, without loss of generality, assume that x is the origin and H is given by the equation x n = 1. As U is a union of squares of the form (a1 , b1 ) × ⋅ ⋅ ⋅ × (a n−1 , b n−1 ) × {1} it is enough to prove the statement under the assumption that U is one of these squares. Hence, again making use of some linear transformations if needed, we may assume that U = (−1, 1)n−1 × {1}. Then C = {(x1 , . . . , x n ) : 0 < x n < 1 and |x i | < x n for each 1 < i < n}, which is clearly an open subset. (2) Assume now that U is open and, in case 𝕏 is spherical it does not contain the antipode of x. Every u ∈ 𝕏\{x} belongs to some hyperplane H u such that x ∈ ̸ H u . Thus, by part (1), ∪u∈U (x, u) = ∪u∈U ∪v∈U∩H u (x, v) is open and its closure contains U ∪ {x}. So, it is dense in the cone based on U with vertex x. Lemma 16.1.2. If D is an intersection of closed half-spaces in 𝕏 then either D∘ ≠ 0 or D is contained in a hyperplane of 𝕏. Moreover, if D∘ ≠ 0 then D∘ is dense in D. Proof. We may assume that D is not empty. Let S be the set whose elements are the subspaces S of 𝕏 with the property that D ∩ S has a non-empty interior, say V, as a subset of S and such that V is dense in D ∩ S. Clearly, if x ∈ D then {x} ∈ S. So S ≠ 0. Let S be a maximal element of S. It is enough to show that D ⊆ S. Indeed, for if this holds then either S is contained in a hyperplane, and hence so is D, or 𝕏 = S ∈ S and thus D∘ is dense in D. Assume that D ⊈ S and let x ∈ D \ S. Let V be the interior of D ∩ S considered as a subset of S. By definition, V is a non-empty open subset of S. Let T be the subspace generated by S ∪ {x}. Then S is a hyperplane of T. By Lemma 16.1.1, C x = ∪v∈V (x, v) is an open subset of T contained in D and x ∈ ∂C x . As this property holds for every x ∈ (D ∩ T) \ S, we obtain that ∪x∈(D∩T)\S C x is open in T and dense in D ∩ T. Therefore T ∈ S, contradicting the maximality of S. Thus D ⊆ S, as desired. A set H of subsets of 𝕏 is said to be locally finite if each point in 𝕏 has a neighborhood that intersects only finitely many of the sets in H, or equivalently if {H ∈ H : H ∩K ≠ 0} is finite for every compact subset K in 𝕏.

16.1 Polyhedra and tessellations

|

71

A non-empty subset P of 𝕏 is said to be a polyhedron of 𝕏 if P = ∩Z∈H Z for a family H of closed half-spaces such that {∂Z : Z ∈ H} is locally finite. In this case one says that H defines the polyhedron P. For example, a subspace S is a polyhedron because

S = ∩ki=1 H i for some hyperplanes H1 , . . . , H k and then S = ∩ki=1 Z i ∩ Z 󸀠i where Z i and Z 󸀠i are the two closed subspaces containing H i . Let P be a polyhedron and let S be the subspace generated by P. By Lemma 16.1.2, P contains a non-empty open subset of S (and it is dense in S). We then say that P is thick in S (note that S is the unique subspace of 𝕏 in which S is thick). If P is thick in 𝕏 we simply say that P is thick. The dimension (respectively, codimension) of P is by definition the dimension of S (respectively, the codimension of S in 𝕏). The relatively interior of P, denoted P r , is the interior of P in the subspace generated by P. Lemma 16.1.3. Let P be a thick polyhedron and let H be a family of closed half-spaces defining P. Then ∂P = ⋃Z∈H P ∩ ∂Z and the following conditions are equivalent for a closed half-space Z0 of 𝕏: (1) P ≠ ⋂Z∈H\{Z0 } Z. (2) P ⊆ Z0 and P ∩ ∂Z0 is thick in ∂Z0 . Proof. The inclusion ∂P ⊇ ⋃Z∈H P ∩ ∂Z is clear. For the converse inclusion assume that x ∈ P \ ∂Z for every Z ∈ H. As {∂Z : Z ∈ H} is locally finite, any ball of 𝕏 with center x intersects only finitely many ∂Z with Z ∈ H and hence x has a neighborhood not intersecting any ∂Z. This implies that x ∈ P∘ and thus x ∈ ̸ ∂P. Since P is closed it follows that ∂P = ⋃Z∈H P ∩ ∂Z. Let P0 = ∩Z∈H\{Z0 } Z. (1) implies (2). Assume that P ≠ P0 . Clearly Z0 ∈ H and therefore P ⊆ Z0 . Let x ∈ P0 \ P. As P is thick there is a non-empty open subset U of 𝕏 contained in P such that if 𝕏 is spherical then the antipode of x is not in U. Then x ∈ 𝕏 \ Z0 = (Z0󸀠 )∘ and therefore every open segment (x, u) with u ∈ U intersects ∂Z0 . By Lemma 16.1.1, C = ∪u∈U (x, u) is an open subset of 𝕏 contained in P0 and hence C∩∂Z0 is a non-empty open subset of ∂Z0 contained in P. Therefore P ∩ ∂Z0 is thick in ∂Z0 . (2) implies (1). Assume that Z0 satisfies (2). Let x be an element of the interior of P ∩ ∂Z0 in ∂Z0 . It easily is verified that then x ∈ ̸ ∂Z for every Z ∈ H \ {Z0 }. Hence, x ∈ (P0 )∘ . Clearly x ∈ ̸ (Z0 )∘ as x ∈ ∂Z0 . Because P0 ⊆ (Z0 )0 it follows that x ∈ ̸ P∘ . Thus P ≠ P 0 .

Observe that condition (2) of Lemma 16.1.3 does not depend on H, but only depends on P. A closed half-space Z0 of 𝕏 satisfying the equivalent conditions of Lemma 16.1.3 is called an essential half-space of P and ∂Z0 is called an essential hyperplane of P. One says that P is geometrically finite if it has finitely many essential hyperplanes. Lemma 16.1.4. Every thick polyhedron of 𝕏 is the intersection of its essential halfspaces and, in particular, ∂P is the union of the intersection of P with the essential hyperplanes of P.

72 | 16 Poincaré’s Theorem Proof. Let P be a thick polyhedron of 𝕏 and let H be a set of closed half-spaces defining P. Let H1 be the set of essential closed half-spaces of P and let P1 = ∩Z∈H1 Z. As H1 ⊆ H we have that P ⊆ P 1 . Assume that this inclusion is strict and take x ∈ P1 \ P. Let Z1 , . . . , Z k be the elements of H whose boundaries contain x. Then, there is a ball U of 𝕏 centered in x such that U ∩ ∂Z = 0 for every Z ∈ H \ {Z1 , . . . , Z k }. Let l be a nonnegative integer such that l ≤ k and Z i is an essential half-space of P if and only if i ≤ l. Then P = ∩Z∈Γ\{Z l+1 ,...,Z k } Z and hence x ∈ ∩li=1 Z i ∩ U = P1 ∩ U = ∩Z∈Γ\{Z l+1 ,...,Z k } Z ∩ U = P ∩ U, a contradiction. The last part of the statement of the lemma follows from Lemma 16.1.3. Lemma 16.1.5. Let Z1 , Z2 and Z3 be closed half-spaces of 𝕏 such that ∂Z1 ∩ ∂Z2 ∩ ∂Z3 has codimension 2 and Z1 ∩ Z2 ∩ Z3 is thick. Then Z1 ∩ Z2 ∩ Z3 = Z i ∩ Z j for some i, j ∈ {1, 2, 3}. Proof. We may assume that Z1 , Z2 and Z3 are pairwise different. Then ∂Z1 , ∂Z2 and ∂Z3 are pairwise different for otherwise Z1 ∩ Z2 ∩ Z3 is not thick. We first prove the result for 𝕏 = ℝn , the Euclidean space. Then each ∂Z i is an Euclidean hyperplane in ℝn and ∂Z1 ∩ ∂Z2 ∩ ∂Z3 is a codimension 2 affine subspace of ℝn . Applying some Euclidean transformation if needed, we may assume that Z1 = {(x1 , x2 , . . . , x n ) ∈ ℝn : x1 ≥ 0} and Z2 = {(x1 , x2 , . . . , x n ) ∈ ℝn : x2 ≥ 0}. Then, ∂Z1 ∩∂Z2 ∩∂Z3 = {(x1 , x2 , . . . , x n ) ∈ ℝn : x1 = x2 = 0} and ∂Z3 = {(x1 , x2 , . . . , x n ) ∈ ℝn : a1 x1 + a2 x2 = 0} with a1 a2 ≠ 0 and a1 > 0. Assume a2 > 0. If Z3 = {(x1 , x2 , . . . , x n ) ∈ ℝn : a1 x1 + a2 x2 ≤ 0} then Z1 ∩ Z2 ∩ Z3 ⊆ ∂Z1 ∩ ∂Z2 contradicting the thickness of Z1 ∩ Z2 ∩ Z3 . So, Z3 = {(x1 , x2 , . . . , x n ) ∈ ℝn : a1 x1 + a2 x2 ≥ 0} and Z1 ∩ Z2 = Z1 ∩ Z2 ∩ Z3 . To finish the proof for 𝕏 = ℝn , it remains to deal with a2 < 0. If Z3 = {(x1 , x2 , . . . , x n ) ∈ ℝn : a1 x1 + a2 x2 ≤ 0} then Z1 ∩ Z3 = Z1 ∩ Z2 ∩ Z3 . Otherwise, Z3 = {(x1 , x2 , . . . , x n ) ∈ ℝn : a1 x1 + a2 x2 ≥ 0} and thus Z2 ∩ Z3 = Z1 ∩ Z2 ∩ Z3 . This finishes the proof in the Euclidean case. In case 𝕏 = 𝕊n ⊆ ℝn+1 each Z i = 𝕊n ∩ Y i with Y i a closed half-space of ℝn+1 such that 0 ∈ ∂Y i . As Z1 ∩ Z2 ∩ Z3 is thick in 𝕊n and Y1 ∩ Y2 ∩ Y3 contains the Euclidean cone with center 0 and base Z1 ∩ Z2 ∩ Z3 , we deduce that Y1 ∩ Y2 ∩ Y3 is thick in ℝn+1 . Then, from the Euclidean case we deduce that Y1 ∩ Y2 ∩ Y3 = Y i ∩ Y j for some i, j ∈ {1, 2, 3} and hence Z1 ∩ Z2 ∩ Z3 = Z i ∩ Z j . To prove the result in the hyperbolic case we use the Klein model 𝕂n seen as subset of ℝn . Then the hyperplanes are the intersection of Euclidean hyperplanes with 𝕂n and the result follows again from the Euclidean case. Lemma 16.1.6. Let H be a countable set of proper subspaces of 𝕏. Then (1) 𝕏 ≠ ⋃S∈H S. (2) If each S ∈ H has codimension at least 2 then for x, y ∈ 𝕏 \ ∪S∈Γ S there is z ∈ 𝕏 \ {x, y, x󸀠 , y󸀠 } such that ([x, z] ∪ [z, y]) ∩ ∪S∈Γ S = 0. In particular, 𝕏 \ ∪S∈H S is path connected.

16.1 Polyhedra and tessellations

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73

Proof. (1) Without loss of generality, we may assume that each element of H is a hyperplane. We argue by induction on the dimension n of 𝕏. The statement is clear for dimension 1 because 𝕏 is uncountable while for n = 1 every element of H has cardinality 1. Since the number of hyperplanes is uncountable, 𝕏 has a hyperplane H with H ∈ ̸ H. The induction hypothesis implies that H ≠ ⋃S∈H H ∩ S and therefore 𝕏 ≠ ⋃S∈H S. (2) For S ∈ H and x ∈ 𝕏 let S x denote the subspace of 𝕏 generated by S ∪ {x}. Because of the assumption, each S x is a proper subspace of 𝕏. Assume x, y ∈ 𝕏 \ ⋃S∈H S. By (1), there exists z ∈ 𝕏\⋃S∈H (S x ∪ S y ). If u ∈ (x, z)∩ S for some S ∈ H then x and u are different and non-antipodal points in S x and hence the geodesic containing both x and u is contained in S x , contradicting the fact that z ∈ ̸ S x . Therefore, the concatenation of the segments [x, z] and [z, y] is a path joining x and y contained 𝕏 \ ⋃S∈H S. Hence, (2) follows. A tessellation of 𝕏 is a set T consisting of thick polyhedra of 𝕏 such that the following conditions are satisfied: (1) 𝕏 = ∪P∈T P, and (2) P0 ∩ Q0 = 0 for every two different members P and Q of T . If only the second condition is satisfied then we call T a partial tessellation of 𝕏. The members of a partial tessellation are called tiles. It is easy to see that P ∩ Q0 = 0 for any two distinct tiles P and Q. In particular, P ∩ Q = ∂P ∩ ∂Q. All the tessellations and partial tessellations that will show up will be locally finite, that is every compact subset intersects only finitely many tiles. One readily verifies that every locally finite partial tessellation has to be countable (see Problem 16.1.1). If P is a tile of a tessellation T then ∂P = ∪Q∈(T\{P}) P ∩ Q. This is not necessarily true if T is a partial tessellation. Definition 16.1.7. A cell C of a partial tessellation T of 𝕏 is a non-empty intersection of tiles satisfying the following property: if P ∈ T then either C ⊆ P or C r ∩ P = 0. Clearly every cell of a locally finite partial tessellation T is a polyhedron and it is contained in only finitely many tiles of T . As the intersection of two different tiles is contained in the boundary of both, the cells of codimension 0 are precisely the tiles and hence the codimension of the intersection of two different tiles is at least 1. By definition, a side of T is a cell of codimension 1 and an edge of T is a cell of codimension 2. If T is a tile of T and C is a cell (respectively, side, edge) of T contained in T then we say that C is a cell (respectively, side, edge) of T in T . In case the tessellation is clear from the context we simply say that C is a cell, side or edge of T. Lemma 16.1.8. Let T be a locally finite tessellation of 𝕏. If x ∈ 𝕏 and C = ⋂T∈T,x∈T T then C is a cell of T and x ∈ C r . Proof. Let x ∈ 𝕏 and let T1 , . . . , T k be the tiles of T containing x. Hence, C = T1 ∩ ⋅ ⋅ ⋅ ∩ T k and T1 , . . . , T k are the only tiles containing C.

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To prove that C is a cell we need to show that if Q is a tile different from any T i then Q ∩ C r = 0. We first consider the case where 𝕏 is spherical and x󸀠 ∈ C. In this case we prove that 𝕏 = T1 ∪ ⋅ ⋅ ⋅ ∪ T k , which of course implies the desired statement. Let U be an open convex neighborhood of x such that U ∩ T = 0 for every T ∈ T \ {T1 , . . . , T k }. Let z ∈ 𝕏. If z = x or x󸀠 then, by assumption, x ∈ T i for some (all) i. Otherwise the geodesic containing x󸀠 and z also contains x and therefore it intersects U \ {x}. In fact there exist non-antipodal elements x1 and x2 in U that both belong to the geodesic containing x and z and are such that x ∈ (x1 , x2 ) and z ∈ ̸ (x1 , x2 ). Then z belongs to either [x󸀠 , x1 ] or [x󸀠 , x2 ]. By symmetry we may assume that z ∈ [x󸀠 , x1 ]. Moreover, as x1 ∈ U ⊆ T1 ∪ ⋅ ⋅ ⋅ ∪ T k , there is i = 1, . . . , k with x1 ∈ T i . Then z ∈ T i , as desired. So we may assume that either 𝕏 is not spherical or x󸀠 ∈ ̸ C and we argue by contradiction. Thus, suppose that there exist Q ∈ T \ {T1 , . . . , T k } and y ∈ Q ∩ C r . In particular, y ≠ x and if 𝕏 is spherical then y ≠ x󸀠 . Let U be an open convex neighborhood of x such that U ∩ T = 0 for every T ∈ T \ {T1 , . . . , T k } and y ∈ ̸ U. Hence, we can take the geodesic g containing x and y and take a point y1 ∈ C r such that y ∈ (x, y1 ). This point exists because g is contained in the subspace generated by C and hence y is an interior point of g ∩ C. By Lemma 16.1.1, W = ∪u∈U (y1 , u) is an open subset of 𝕏. Since y ∈ W ∩ Q and Q is thick, we get that W contains a point z ∈ Q∘ . Let u ∈ U be such that z ∈ (y1 , u). As U ⊆ T1 ∪ ⋅ ⋅ ⋅ ∪ T k , u ∈ T i for some i, (y1 , u) ⊆ T i and therefore z ∈ T i ∩ Q∘ . However, this contradicts with the fact that T i and Q are different tiles of the tessellation T . So, in this case, C indeed is a cell. To prove the second part, assume that x ∈ ̸ C r and let L be the subspace generated by C. Clearly, the dimension of L is positive and k > 1. Therefore C ⊆ ∂T i for every i. Consider C as a thick polyhedron of L. As x ∈ ̸ C r , it follows from Lemma 16.1.3, that x belongs to one of the essential hyperplanes of C, as thick polyhedra of L. Fix y ∈ C r and an open interval (y, z) containing x. Then (y, x) ⊆ C r and (x, z)∩C = 0. Therefore z ∈ ̸ T i for some i. Renumbering the T i ’s and replacing (x, z) by a smaller interval if necessary, one may assume that (x, z) ∩ T1 = 0. We claim that H ∩ [y, z] = {x} (equivalently y ∈ ̸ H) for some essential hyperplane H of T1 . Otherwise y belongs to all the essential hyperplanes of T1 containing x. Then (x, z) is contained in all these essential hyperplanes. If V is an open neighborhood of x only intersecting the essential hyperplanes of T1 containing x then (x, z) ∩ V is a non-empty subset contained in T1 , contradicting the construction. This proves the claim. Note that x ∈ U ∩ H ∩ T1 . Hence, U ∩ H ∩ T1 is a non-empty open subset of H ∩ T1 . So, by the second part of Lemma 16.1.2, there exists w ∈ (U ∩ H ∩ T1 ) ∩ (H ∩ T1 )r . Hence, H is the only essential hyperplane of T1 containing w. We claim that (w, y) ⊆ T1∘ . Indeed, for suppose the contrary, then there exists u ∈ (w, y) ∩ ∂T1 . So u ∈ H1 for some essential hyperplane H1 of T1 . If H1 ≠ H and Z1 is the closed half-space of 𝕏 with ∂Z1 = H1 and T1 ⊆ Z1 then w, y ∈ Z1∘ . Then u ∈ (w, y) ⊆ Z1∘ , a contradiction. So, H1 = H and u ∈ (w, y) ∩ H. Now y ∈ ̸ H and w ∈ T1 . Hence a reasoning as above (interchanging the role of w and y and replacing H1 by H) yields that (w, y) ∩ H = 0, a contradiction.

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Because of the claim and since w ∈ U and U is open in 𝕏, there exists z1 ∈ U such that w ∈ (y, z1 ). As w ∈ H and y ∈ ̸ H, we have z1 ∈ ̸ T1 . However U ⊆ T1 ∪ ⋅ ⋅ ⋅ ∪ T k and hence z1 ∈ T i for some i ≥ 2. Then [y, z1 ] ⊆ T i and we conclude that 0 ≠ (y, w) ⊆ T1∘ ∩T i with i > 1, a contradiction. The cell of T formed by the intersection of the tiles containing x is the smallest cell containing x and we call it the cell of T generated by x. By Lemma 16.1.8, the relative interiors of the cells of T form a partition of 𝕏. The cell generated by x is the unique cell of T whose relative interior contains x. Lemma 16.1.9. Let T be a locally finite partial tessellation of 𝕏. Let T1 and T2 be two tiles of T and let S = T1 ∩ T2 . Assume S has codimension 1 and let H be the hyperplane generated by S. The following properties hold: (1) H is an essential hyperplane of both T1 and T2 . (2) T1 and T2 are contained in different closed half-spaces defined by H. (3) S r ∩ T = 0 for every tile T different from both T1 and T2 . (4) S is a side of T . Proof. Let x ∈ S r and let U be an open neighborhood of x such that U ∩ H ⊆ S. As x ∈ ∂T1 , from Lemma 16.1.4 we obtain that x ∈ H1 , for some essential hyperplane H1 of T1 . If H1 ≠ H then H ∩ U intersects non-trivially the two open half-spaces defined by H1 , contradicting the fact that T1 does not intersect one of these open half-spaces. Therefore H = H1 . Hence (1) follows. It also proves that H is the only essential hyperplane of T1 containing x and, by symmetry, it also is the only essential hyperplane of T2 containing x. So there is an open ball B of 𝕏 centered in x and not intersecting any essential hyperplane of T1 or T2 different from H. Let Z be the closed half-space with boundary H and containing T1 . Then B ∩ Z ∘ is one of the two non-empty connected components of B \ H and it is contained T1∘ . Since T1∘ ∩ T2 = 0 it follows that T2 ⊆ Z 󸀠 . This proves (2). The same argument then shows that B∩Z 󸀠 ⊆ T2 . Therefore B ⊆ T1 ∪T2 . If T is any tile such that x ∈ T then B contains a point in T ∘ . So T ∘ ∩ (T1 ∪ T2 ) ≠ 0 and therefore T = T1 or T = T2 . Hence (3) follows. Clearly (4) is a consequence of (3). The following proposition follows at once from Lemma 16.1.9. Proposition 16.1.10. Every side of a locally finite partial tessellation T is contained in exactly two tiles and it is the intersection of these tiles. Lemma 16.1.11. Let T be a locally finite partial tessellation of 𝕏, let H be a hyperplane of 𝕏 and let Z be a closed half-space defined by H. Then TZ = {T ∩ H : T ∈ T and Z is an essential closed half-space of T}

is a locally finite partial tessellation of H. Proof. Let T1 and T2 be different elements of TZ . If (T1 ∩ H)r ∩(T2 ∩ H)r ≠ 0 then T1 ∩ T2 is a side of T and H is the subspace generated by this side. Then, by Lemma 16.1.9, T1

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and T2 are in different closed half-spaces defined by H, a contradiction. This proves that TZ is a partial tessellation. As T is locally finite, so is TZ . Proposition 16.1.12. If T is a tile of a locally finite tessellation T then ∂T is the union of the sides of T in T . Proof. Let x ∈ ∂T and let T1 , . . . , T k be the tiles containing x and that are different from T. Let U be an open neighborhood of x such that T, T1 , . . . , T k are the only tiles intersecting U. By Lemma 16.1.4, there exists an essential hyperplane H of T in 𝕏 such that x ∈ H. Then H ∩ T is a thick polyhedron of H. Therefore, by Lemma 16.1.2, U ∩ (H ∩ T)r is a non-empty open subset of H. Because T is a tessellation and T ∩ H ⊆ ∂T, we get that U ∩ T ∩ H ⊆ ∪ki=1 T ∩ T i . So U ∩(H ∩ T)r is non-empty open subset of H contained in ∪ki=1 T ∩ T i ∩ H. Hence, T ∩ T i ∩ H is thick in H for some i. So, by Lemma 16.1.9, T ∩ T i is a side of T. Therefore x belongs to a side of T. This proves one of the inclusions of the statement. The other one is obvious. Proposition 16.1.13. Let T be a tile of a locally finite tessellation T of 𝕏. If E is an edge of T in T then E is contained in exactly two sides of T in T . Proof. Let E be an edge of T in T . First we prove by contradiction that E cannot be contained in three different sides of T in T . So, assume that E is contained in three distinct sides, say S1 , S2 and S3 of T in T . Because of Proposition 16.1.10, S i = T ∩ T i with T i a tile different from T. Let H i denote the hyperplane generated by S i . Because of Lemma 16.1.9, each H i is an essential hyperplane of T. Let Z i denote the closed halfspace defined by H i such that T ⊆ Z i . By Lemma 16.1.9, T i ⊆ Z 󸀠i . Furthermore, by Lemma 16.1.5, we may assume that Z1 ∩ Z2 ∩ Z3 = Z2 ∩ Z3 . As each H i is an essential hyperplane of T we deduce that H1 ∈ {H2 , H3 } and hence, we may assume that H1 = H2 . Thus Z1 = Z2 . If H3 = H1 then Z1 = Z3 and S1 , S2 and S3 are tiles of the partial tessellation TZ1 of H1 defined as in Lemma 16.1.11. Since E has codimension 1 in H1 and because it is contained in each S i we get that S i ∩ S j is a side of TZ1 for every i ≠ j. Consequently, S1 , S2 and S3 are distinct tiles of TZ1 containing points in the relative interior of E and hence in the relative interior of a side of TZ1 . This contradicts with Lemma 16.1.9. So H3 ≠ H1 and thus H1 ∩ H3 is the subspace generated by E. Moreover, S1 and S2 are tiles of TZ1 and S1 ∩ S2 is a side of TZ1 . By Lemma 16.1.9, S1 and S2 are in different closed half-spaces of H1 defined by H1 ∩ H3 . These closed half-spaces are H1 ∩ Z3 and H1 ∩ Z3󸀠 . By symmetry we also may assume that S2 ⊆ Z3󸀠 . Hence, T ∩ T2 = S2 ⊆ Z1 ∩ Z3 ∩ Z1󸀠 ∩ Z3󸀠 ⊆ H1 ∩ H3 , in contradiction with the fact that S2 has codimension 1. It remains to prove that E is contained in two different sides of T. Let x ∈ E r . Then x ∈ ∂T and therefore x ∈ S for some side S of T by Proposition 16.1.12. Hence, by the definition of a cell, E ⊆ S. By Lemma 16.1.9, S = T ∩ T1 with T1 a tile of T different from T. Let H be the hyperplane of 𝕏 generated by S. By Lemma 16.1.9, H is an essential hyperplane of both T and T1 . Furthermore, T and T1 are included in different closed half-spaces defined by H. Note that, because of Lemma 16.1.4, a point y ∈ S which

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is not in any essential hyperplane of T or T1 different from H has a neighborhood contained in T ∪ T1 . We claim that there is a hyperplane H1 of 𝕏 different from H such that H1 is an essential hyperplane of either T or T1 and it intersects E r non-trivially. Indeed, for otherwise, for every y ∈ E r there is a neighborhood U y of y in 𝕏 such that U y ⊆ T ∪ T1 . Then ∪y∈E r U y does not intersect any tile of T different from both T and T1 . In particular, the only tiles intersecting E r are T and T1 . So T and T1 are the only tiles containing E and hence E = S, a contradiction. This proves the claim. So let H1 be a hyperplane different from H such that E r ∩ H1 ≠ 0 and H1 is essential hyperplane of either T or T1 . We claim that E r ⊆ H1 . Otherwise E has positive dimension and the subspace L generated by E is not contained in H1 . Hence H1 ∩ L is a hyperplane of L, since 0 ≠ L∩H1 ≠ L. As E r is an open subset of L of dimension at least 1, it has points in the two open half-spaces of L defined by H1 ∩ L. This implies that E r has points in the two open half-spaces of 𝕏 defined by H1 . This contradicts with the facts that E r ⊆ T ∩ T1 and either T or T1 is contained in one closed half-space defined by H1 , because it is an essential hyperplane of either T or T1 . So we have proved that E r ⊆ H1 . Lemma 16.1.5 therefore yields that if H1 is essential in T (respectively, T1 ) then H and H1 are the only essential hyperplanes of T (respectively, T1 ) intersecting E r non-trivially. Let x ∈ E r and let T, T1 , R1 , . . . , R k be the different tiles of T containing x. Then E ⊆ T ∩ T1 ∩ (∩ki=1 R i ), by the definition of a cell. Assume that H1 is an essential hyperplane of T. Consider an open ball centered at x such that U ∩ R = 0 for every tile R ∈ ̸ {T, T1 , R1 , . . . , R k } and U ∩ H2 = 0 for every essential hyperplane H2 of T with H2 ∈ ̸ {H, H1 }. Let V be the open half-space of 𝕏 defined by H and containing T ∘ . Then V ∩ U ∩ H1 is a non-empty open subset of H1 contained in the boundary of T and hence it is also contained in ∪ki=1 T ∩ R i , because V ∩ T1 = 0. Thus T ∩ R i has codimension 1 for some i and hence it is a side of T containing E. Finally assume that H1 is an essential hyperplane of T1 and not of T. In this case we consider an open ball U in 𝕏 with center x such that U ∩ R = 0 for every tile R ∈ ̸ {R1 , . . . , R k } and U ∩ H2 = 0 for every essential hyperplane H2 of T with H2 ≠ H. Moreover, let Z1 be the open half-space defined by H1 not containing T1 . Then Z1 ∩ U ∩ H is a non-empty open subset of H contained in the boundary of T. Hence Z1 ∩ U ∩ H ⊆ ∪ki=1 T ∩ R i and therefore T ∩ R i is a side of T containing E, for some i. In both cases E is contained in two different sides of T containing E, namely S and T ∩ R i , as desired. Note that if the dimension of 𝕏 is at least 3 then an edge of a tile T in T is not necessarily the intersection of two sides of T, although it is contained in exactly two distinct sides. Moreover, even if an edge E is the intersection of two sides of a tile, it could be properly contained in the intersection of two sides of another tile (see Figure 16.1).

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F

D C

E B A

Fig. 16.1: The intersection of the sides C ∩ D and D ∩ F of D is the union of the edges A ∩ D (red, fat and continuous) and B ∩ D (blue, fat and dashed). None of these two edges is the intersection of two sides of D while both are the intersection of two sides of C.

Let E, S and T be respectively an edge, a side and a tile of T with E ⊂ S ⊂ T. We define recursively two sequences, one of tiles (T0 , T1 , . . . ) and another of sides (S1 , S2 , . . . ) by setting T0 = T, S1 = S, T i = tile containing S i and different from T i−1 , and

(16.1.1)

S i+1 = side of T i containing E and different from S i . This is well defined by Proposition 16.1.10 and Proposition 16.1.13 and we have for i ≥ 1: T i−1 and T i are the only tiles containing S i

and

S i and S i+1 are the only sides of T i containing E. For example, assume that Figure 16.1 represents part of a tessellation of ℝ3 and take T = A, S = A ∩ C and E = A ∩ C, the red, fat, continuous segment. Then, the sequence of tiles is periodic of period 5 starting with (A, C, D, F, E). If T = B, S = B ∩ C and E is the blue, fat, dashed segment then again the sequence of tiles is periodic of period 5 starting with (B, C, D, F, E). If one considers the edge E = A ∩ B ∩ C then, with an appropriate side and tile selection, we obtain a sequence of tiles of period 3 starting with (A, B, C). We will show that this behavior is general. As every edge is contained in finitely many tiles, the sequences only have finitely many different elements. Moreover, if k is minimum such that T k = T m for some m > k then k = 0. Indeed, S k and S k+1 are the only sides of T k containing E and the same happens for S m and S m+1 . Therefore either S k = S m and S k+1 = S m+1 or S k = S m+1 and S k+1 = S m . In the former case T m−1 contains S m = S k = T k−1 ∩T k and it is different from T m = T k . Therefore, if k ≠ 0 then T m−1 = T k−1 . In the latter case, if k ≠ 0 then T m+1 contains S m+1 = S k = T k−1 ∩ T k and it is different form T m = T k . Hence, T m+1 = T k−1 . In both cases we obtain a contradiction with the minimality of k.

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Let m be the minimum positive integer with T0 = T m , then (T0 , T1 , . . . , T m )

(16.1.2)

is called an edge loop of E. This definition depends on the choice of the tile T and the side S. However, if S is replaced by another side S󸀠 containing E and included in T (there is only one option by Proposition 16.1.13) then the edge loop obtained is (T m , T m−1 , . . . , T2 , T1 , T0 ). If we replace T by one of the tiles T i and S by one of the sided of T i containing E, i.e. either S i+1 or S i then the edge loop is either (T i , T i+1 , . . . , T m , T1 , . . . , T i−1 , T i ) or (T i , T i−1 , . . . , T1 , T m , . . . , T i+1 , T i ). The next lemma shows that there are no other alternatives. Lemma 16.1.14. If E is an edge and (T0 , T1 , . . . , T m ) is an edge loop of E then T1 , . . . , T m are precisely the tiles containing E. Proof. Note that, as stated before, there are only finitely many tiles containing E. We can order them in such a way that T0 , . . . , T n are all these tiles and T0 , . . . T m are the tiles forming the edge loop. Let z ∈ E r . Then z ∈ T i for every 0 ≤ i ≤ n and there exists λ > 0 such that the ball B = B(z, λ) intersects a tile T if and only T = T i for some 0 ≤ i ≤ n. Moreover B = B1 ∪ B2 with

m

B1 = ⋃ B ∩ T i i=0

n

and

B2 = ⋃ B ∩ T i . i=m+1

We prove the result by contradiction. So suppose that n > m. Hence, both B1 and B2 are non-empty closed sets. Moreover B1 ∩ B2 ⊆ ⋃0≤i≤m,m+1≤j≤n (T i ∩ T j ). We claim that T i ∩ T j is of codimension 1 for at least one 0 ≤ i ≤ m and one m + 1 ≤ j ≤ n. Otherwise set B󸀠1 = B1 \(B1 ∩ B2 ), B󸀠2 = B2 \(B1 ∩ B2 ) and B󸀠 = B \(B1 ∩ B2 ) = B󸀠1 ∪ B󸀠2 . Clearly, B󸀠1 and B󸀠2 are disjoint. Moreover, T i is thick and T i0 is dense in T i . Hence B ∩ T i contains an open subset of 𝕏. Thus dim(B i ) = dim(𝕏). However dim(T i ∩ T j ) < dim(𝕏) for each i ≠ j and hence dim(B1 ∩ B2 ) < dim(𝕏). Thus B󸀠i ≠ 0. Moreover, as T i ∩ T j has codimension at least 2 for every 0 ≤ i ≤ n and m + 1 ≤ j ≤ n, by Lemma 16.1.6, B󸀠 is path-connected and hence connected. Thus, B󸀠 is a connected subspace as the intersection of two disjoint closed subspaces, which is a contradiction. Hence there exists 0 ≤ i ≤ m and m + 1 ≤ j ≤ n such that T i ∩ T j is of codimension 1, and hence it is a side containing E by Proposition 16.1.10. Denote this side by S∗ . By the definition of an edge loop, S i = T i−1 ∩ T i and S i+1 = T i ∩ T i+1 are two different sides contained in T i and containing E (indices are interpreted modulo m.) Moreover T i−1 ≠ T j and T i+1 ≠ T j and hence E is contained in three different sides, which contradicts Lemma 16.1.13.

Problems 16.1.1. Prove that every locally finite tessellation is countable.

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16.2 Group presentations In this section 𝕏 is a Euclidean, spherical or hyperbolic space and G is a discontinuous group of isometries of 𝕏. The action of G on 𝕏 induces a dimension preserving action on the set of subspaces of 𝕏. In particular, G acts on the set of polyhedra of 𝕏. This action can be used to obtain a presentation of G from a fundamental polyhedron of G. As mentioned in the introduction of the chapter, this method was introduced by Poincaré in [180] and [181] for two and three-dimensional hyperbolic spaces. Poincaré also determined conditions on a polyhedron in 𝕏 to be the fundamental polyhedron of a discontinuous group of isometries of 𝕏. In this section we give the proof presented in [110] of the presentation part of Poncaré’s proof. Definition 16.2.1. A fundamental polyhedron of G is a polyhedron P of 𝕏 such that TP = {𝛾(P) : 𝛾 ∈ G} is a tessellation of 𝕏. The polyhedron P is said to be locally finite if TP is locally finite. Observe that if P is a fundamental polyhedron of G then (1) 𝕏 = ∪𝛾∈G 𝛾(P) and (2) P∘ ∩ 𝛾(P)∘ = 0 for every 1 ≠ 𝛾 ∈ G. Conversely, assume that P satisfies (1) and (2). Then α(P)∘ ∩ β(P)∘ = 0 for any distinct elements α and β of G. Moreover, if P were not thick in 𝕏 then it follows from Lemma 16.1.2 that it is contained in a hyperplane. As G is countable, we get that ∪𝛾∈G 𝛾(P) is contained in a countable union of hyperplanes, in contradiction with Lemma 16.1.6. Hence P is thick and hence 𝛾(P) is thick for every 𝛾 ∈ G. Thus P is a fundamental polyhedron of G if and only if (1) and (2) hold. Throughout this section P is a locally finite fundamental polyhedron of G and T = TP . When we refer to cells, tiles, sides or edges it is always with respect to T . Since every cell is contained in only finitely many tiles, the stabilizer of one cell is finite. If S is a side of P then, by Proposition 16.1.10, there is a unique g ∈ G\{1} such that S = P∩g(P) and S r ∩ h(P) = 0 for every h ∈ G \ {1, g}. We denote this g as 𝛾S . It is called a (side) pairing transformation. So, S = P ∩ 𝛾S (P). If g is a pairing transformation then P ∩ g−1 (P) = g−1 (P ∩ g(P)) also is a side of P and hence g −1 is a pairing transformation as well. In this case one denotes S g = P ∩ g(P) and one says then that S g and S g−1 are paired sides. If S is a side then the side paired with S is denoted S󸀠 . So, S󸀠 = 𝛾−1 S (S). If E is an edge of P then, by Proposition 16.1.13, it is contained in exactly two sides, say S g and S g1 . Because G permutes edges, g −1 (E) is an edge of the tessellation and it is contained in S g−1 . Therefore g−1 (E) and g1−1 (E) are edges of P.

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Some relations amongst the side pairing transformations can be deduced. A first type of relations is easily obtained. Indeed, if S1 and S2 are two paired sides then 𝛾S1 = 𝛾−1 S2 . Such a relation is called a pairing relation. In case S is a side paired with itself then the pairing relation takes the form 𝛾2S = 1 and such a relation usually is called a reflection relation. To define the second type of relations, we introduce the following definition. Definition 16.2.2. A loop of G with respect to P (or simply a loop of G, if the polyhedron is clear from the context) is a finite ordered list (g0 , g1 , . . . , g n ) of elements of G such that g0 = g n and g i−1 (P) ∩ g i (P) is a side for each 1 ≤ i ≤ n (equivalently, each g −1 i−1 g i is a pairing transformation). Example 16.2.3. (1) If g, h ∈ G and g −1 h is a pairing transformation then (g, h, g) is a loop of G. (2) If (T0 = g0 (P), T1 = g1 (P), . . . , T m = g m (P)) is an edge loop then (g0 , g1 , . . . , g m ) is a loop of G. (3) If 𝛾S1 ⋅ ⋅ ⋅ 𝛾S m = 1 for sides S1 , . . . , S n of P then (1, 𝛾S1 , 𝛾S1 𝛾S2 , . . . , 𝛾S1 ⋅ ⋅ ⋅ 𝛾S n−1 , 𝛾S1 ⋅ ⋅ ⋅ 𝛾S m = 1) is a loop. Conversely, if (g0 , g1 , . . . , g m ) is a loop of G then S i = P ∩ g −1 i−1 g i (P) is a side of P and g−1 g = 𝛾 for every i ∈ {1, . . . , m}. Furthermore, 𝛾 ⋅ ⋅ ⋅ 𝛾S m = 1 . i S S i 1 i−1 By Example 16.2.3 (3), relations amongst pairing transformations are completely determined by loops. In case a loop (g0 , . . . , g m ) is determined by an edge loop, as in Example 16.2.3 (2), then the resulting relation 𝛾S1 ⋅ ⋅ ⋅ 𝛾S m = 1 is called an edge loop relation. Note that the pairing relations are the relations corresponding to the loops (g, h, g) for g −1 h a pairing transformation. We give an alternative interpretation of the edge loop relations. Let E be an edge of P and choose one of the two sides S of P containing E. Recursively one obtains a sequence E1 , E2 , . . . of edges of P and a sequence of sides S1 , S2 , . . . , with E i ⊆ S i for each i, and which is uniquely determined by the following rules: E1 = E,

S1 = S,

E n+1 = 𝛾−1 S n (E n ) and

S n+1 and S󸀠n = 𝛾−1 S n (S n ) are the two sides of P containing E n+1 .

(16.2.1)

Let g n = 𝛾S1 ⋅ ⋅ ⋅ 𝛾S n for every n ≥ 0 (in particular, we agree that g0 = 1). Observe that g −1 n (E) = E n+1 ⊆ P. In particular E ⊆ g n (P) and hence E is contained in the tiles T0 = g0 (P) = P,

T1 = g1 (P),

T2 = g2 (P), . . .

󸀠 and T n−1 ∩ T n is a side of T for every n ≥ 1. Moreover g−1 n (T n−1 ∩ T n ) = S n ≠ S n+1 = −1 g n (T n ∩ T n+1 ). Therefore T n−1 ∩ T n and T n ∩ T n+1 are the two sides of T n containing E. This proves that T0 = P, T1 , T2 , . . . is a sequence of tiles as defined in (16.1.1). We know this is a periodic sequence and if it has period m then (T0 , T1 , . . . , T m ) is the edge loop defined by ES and P and (g0 = 1, g1 , . . . , g m ) is the loop of G associated to

82 | 16 Poincaré’s Theorem

this edge loop. As T i determines g i , the sequence g0 , g1 , . . . also is periodic of period −1 m. As E i = g −1 i (E) and S i = P ∩ g i−1 g i (P), the sequence of pairs (E i , S i ) also is periodic, m say of period k and let t = k . Then t is a positive integer and the edge loop relation associated to the loop (g0 , g1 , . . . , g m ) takes the form 1 = g m = (𝛾S1 ⋅ ⋅ ⋅ 𝛾S k )t . This usually is called a cycle relation. This is the second type of relations we need for the Poincaré result. Observe that cycle relation and edge loop relation are synonymous concepts. Since m is the minimum integer so that g m = g0 = 1 and g kt = (𝛾S1 ⋅ ⋅ ⋅ 𝛾S k )t , we deduce that t is the order of 𝛾S1 ⋅ ⋅ ⋅ 𝛾S k . An alternative way to see that 𝛾S1 . . . 𝛾S k has finite order is by observing that it stabilizes the edge E and the stabilizer of every cell is finite. Some of the cycle relations are redundant. For example, if S and R are the two sides of P containing the edge E and if (T0 = P, T1 , T2 , . . . , T m−1 , T m = P) is the edge loop obtained by applying the above procedure to E and S then the edge loop obtained by applying the procedure to E and R is (T0 = P, T m−1 , . . . , T2 , T1 , T0 = P). These two −1 t loops give rise to equivalent cycle relations: (𝛾S1 . . . 𝛾S k )t = 1 and (𝛾−1 S k . . . 𝛾S 1 ) = 1. This is because if the period of the list (E i , S i ) obtained from E and S is k then 󸀠 𝛾−1 S k−1 (S k−1 ) = R and hence the list of pairs of edges and sides starting with (E, R) is (E, R), (E k−1 , S󸀠k−1 ), . . . , (E1 , S󸀠1 ), . . . . On the other hand, if we replace E by one of the edges E i then the sequence of pairs of edges and sides obtained is a shift of the list obtained with E and S or R. Then, the cycle relation obtained with E i is a conjugate of the cycle relation associated with E. The edges in the list E1 , . . . , E k form a cycle of edges of P. Clearly, the nonequivalent cycles of edges of P define a partition of the edges of P. Example 16.2.4. Let n ≥ 3 and let D2n be the group of isometries of the Euclidean plane fixing a regular polygon with n sides centered in the origin. Then, the acute wedge P between the two half-lines S1 = {(x, 0) : x > 0} and S2 = {(x, x tan( πn )) : x > 0} is a fundamental polyhedron of D2n . Let g i be the reflection in the line containing S i . Then S i = P∩g i (P) and g2i = 1. So S1 and S2 are the two sides of P, as fundamental polyhedron of G and the pairing relations are the reflection relations S21 = S22 = 1. The only edge is the vertex consisting of the single point (0, 0). The sequence of edges and sides starting with E and S1 is periodic of period 2 starting with (E, S1 ), (E, S2 ). Clearly, R = S1 S2 has finite order and in fact it has order n because it is the rotation n around (0, 0) of angle 2π n . Therefore the only cycle relation is (S 1 S 2 ) = 1. We are ready to state the main result of this section. Theorem 16.2.5 (Poincaré’s Theorem). Let 𝕏 be either an Euclidean, hyperbolic or spherical space. Let P be a locally finite fundamental polyhedron for a discontinuous group of isometries of 𝕏. The pairing transformations generate G, that is G = ⟨𝛾 ∈ G : P ∩ 𝛾(P) is a side of P⟩ , and the pairing and cycle relations form a complete set of relations for G.

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83

We are now ready to prove that the pairing transformations generate G. The proof that the pairings and cycle relations form a complete set of relations requires much more work and will be postponed until the end of the section. Proof of Theorem 16.2.5 (Generators). Let T be the tessellation of 𝕏 formed by the polyhedra g(P) with g ∈ G. As G is countable, T is countable and, as every nonempty cell is the intersection of finitely many tiles (those containing a relative interior point of the cell), the number of cells of T is countable. Let Y be the complement in 𝕏 of the union of the cells of codimension at least 2. Then, by Lemma 16.1.6, Y is path-connected. Let g ∈ G and let x ∈ P∘ . So g(x) ∈ g(P)∘ and x, g(x) ∈ Y. Because Y is path connected, there exists a continuous function α : [0, 1] → Y with α(0) = x and α(1) = g(x). The path α can be very odd and we need to choose a path that “travels smoothly” through the tiles of T . Therefore, we need to perform some “deformations” on α. To do so, we let P denote the finite set consisting of all the tiles that intersect the compact set α([0, 1]) and define L = { l ≥ 1 : There is a continuous function β : [0, 1] → Y, a sequence 0 = a0 < a1 < ⋅ ⋅ ⋅ < a l ≤ 1 and different tiles T1 , . . . , T l such that β(a0 ) = x, β((a l , 1]) ∩ (T1 ∪ ⋅ ⋅ ⋅ T l ) = 0, α |(a l ,1] = β |(a1 ,1] , β(a i ) = α(a i ), and β([a i−1 , a i ]) ⊆ T i for every 1 ≤ i ≤ l }. We first prove that 1 ∈ L. To show this, put a1 = max(α−1 (P)), T1 = P and define the continuous function β : [0, 1] → Y as follows. The restriction of β to [0, a1 ] is such that its image runs through the geodesic [x, α(a1 )] from x to α(a1 ). The restriction of β to [a1 , 1] is the same function as the restriction of α to [a1 , 1]. Clearly β, 0 < a0 < a1 ≤ 1 and T1 satisfy the conditions of the definition of L. Observe that if l ∈ L then l ≤ |P| and hence L is bounded. Let l be maximal in L and let β, 0 = a0 < a1 < ⋅ ⋅ ⋅ < a l ≤ 1 and T1 , . . . , T l satisfy the conditions of the definition of L. We claim that a l = 1. Indeed, if a l < 1 we get that β(a l ) ∈ ∂T l and thus, by Lemma 16.1.12, β(a l ) ∈ S, for some side S of T l . Since β(a l ) ∈ Y (and thus β(a l ) is not in a cell of codimension at least 2) we get that S is the cell of T generated by β(a l ). Hence, β(a l ) ∈ S r . By Proposition 16.1.10, S = T l ∩T for some tile T of T with T ≠ T l . Furthermore, T and T l are the only tiles of T containing S. So, β(a l ) does not belong to any other tile of T . Then, T l ∪T contains a neighborhood of β(a l ) = α(a l ). Because β((a l , 1)) ∩ (T1 ∪ ⋅ ⋅ ⋅ ∪ T l ) = α((a l , 1]) ∩ (T1 ∪ ⋅ ⋅ ⋅ T l ) = 0, it follows that β((a l , b)) ⊆ T for some a l < b ≤ 1. Hence T ≠ T i for every i ∈ {1, . . . , l}. Let a l+1 = max(α−1 (T)). So, a l+1 > a l . Put T l+1 = T. One can now define a continuous function β󸀠 : [0, 1] → Y as follows. On [a0 , a l ] ∪ (a l+1 , 1] the function β󸀠 agrees with β. On [a l , a l+1 ] the function β󸀠 is such that its image runs through the geodesic [β(a l ), β(a l+1 )] from β(a l ) to β(a l+1 ) in case β(a l ) and β(a l+1 ) are not antipodal and otherwise one chooses x ∈ T ∘ and then one defines on [a l , a l+1 ] the function β󸀠 such that its image runs first through the geodesic [β(a l ), x] and then through the geodesic [x, β(a l+1 )]. So β󸀠 ([a l , a l+1 ]) ⊆ T l+1 . Clearly, β󸀠 ([a i−1 , a i ]) = β([a i−1 , a i ]) ⊆ T i for 1 ≤

84 | 16 Poincaré’s Theorem i ≤ l and β󸀠 ((a l+1 , 1]) ∩ (T1 ∪ ⋅ ⋅ ⋅ T l+1 ) = 0. So β󸀠 , a0 < a1 < ⋅ ⋅ ⋅ < a l < a l+1 ≤ 1 and T1 , . . . , T l , T l+1 satisfy the conditions of the definition of L and this contradicts with the maximality of l. So, indeed a l = 1 and thus there exists a continuous function α : [0, 1] → Y and a sequence 0 = a0 < a1 < . . . < a l = 1 such that α(a0 ) = x, α(a l ) = g(x) and α([a i−1 , a i ]) ⊆ T i with T i tile of T for each 1 ≤ 1 ≤ l. Clearly T0 = P. Write T i = g i (P) with g i ∈ G. Then, g1 = 1, g l = g and α(a i ) ∈ g i (P) ∩ g i+1 (P) for every i ∈ {1, 2, . . . , l − 1}. Since α(a i ) is not in any cell of codimension greater than 1, each g i (P) ∩ g i+1 (P) is a side of g i (P) and therefore P ∩ g−1 i g i+1 (P) is a side of P. g is a pairing transformation for each 1 ≤ i < l. Finally g = g l = Hence 𝛾i = g−1 i+1 i (g1−1 g2 )(g2−1 g3 ) ⋅ ⋅ ⋅ (g −1 g ) = 𝛾 ⋅ ⋅ ⋅ 𝛾 . Hence, g belongs to the subgroup of G gener1 l−1 l−1 l ated by the pairing transformations. Because g is an arbitrary element of G, the result follows. Let Δ denote the group given by the presentation of Theorem 16.2.5. More precisely, Δ = F/N, where F is the free group with basis the symbols [g], one for each pairing transformation g, and N is the normal closure of the subgroup of F generated by the set X consisting of the pairing and cycle relations, i.e. X is formed by the products [𝛾S ][𝛾S󸀠 ] with S and S󸀠 paired sides of P (pairing relations) and the elements of the form ([𝛾S1 ] . . . [𝛾S k ])t , where (E1 = E, S1 = S, E2 , S2 , . . . ) is the list defined by (16.2.1) for an edge E and a side S containing E, k is the period of the list and t is the order of 𝛾S1 . . . 𝛾S k (cycle relations). Recall that the edge loop relations and the cycle relations are synonymous concepts, so we may replace the cycle relations by the edge loop relations [g0−1 g1 ][g1−1 g2 ] ⋅ ⋅ ⋅ [g −1 n−1 g n ], where (g0 , g1 , . . . , g n−1 , g n ) is an edge loop. Abusing notation, we will consider the symbols [g], with g a pairing transformation, as elements of Δ. Hence, [𝛾S ][𝛾S󸀠 ] = 1, for every side S of P (so [𝛾S ]−1 = [𝛾−1 S ]) and [g0−1 g1 ][g1−1 g2 ] ⋅ ⋅ ⋅ [g −1 n−1 g n ] = 1, for every edge loop (g 0 , g 1 , . . . , g n ) of P. Let g, h ∈ G and let C be a cell of T of codimension m at most 2 that is contained in g(P) ∩ h(P). We define κ C (g, h) ∈ Δ as follows. – If m = 0 then κ C (g, h) = 1. – If m = 1 then κ C (g, h) = [g −1 h]. – If m = 2 then C is an edge contained in g(P) ∩ h(P) and thus, by Lemma 16.1.14, g and h belong to the edge loop of C. Up to a cyclic permutation, we can write the

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85

edge loop of E as (g = k0 , . . . , k t = h, k t+1 , . . . , k m = g) (or the equivalent edge loop (g = k m , k m−1 , . . . , k t = h, k t−1 , . . . , k1 , k0 = g)) and we set −1 −1 −1 −1 κ C (g, h) = [k−1 0 k 1 ][k 1 k 2 ] ⋅ ⋅ ⋅ [k t−1 k t ] = [k m k m−1 ] ⋅ ⋅ ⋅ [k t+1 k t ].

Observe that κ C (g, g) = 1 in the three cases. Also, if m = 0 then g = h. Furthermore, if g(P) ∩ h(P) is a side then either C = g(P) ∩ h(P) or C is an edge and g and h are consecutive elements of the edge loop of C; in both cases, κ C (g, h) = [g −1 h]. Lemma 16.2.6. Let g, h ∈ G and let C be a cell of T of codimension m ≤ 2 that is contained in g(P) ∩ h(P). The following properties hold. (1) κ C (g, h) = κ C (h, g)−1 . (2) If D is cell of T contained in C and of codimension at most 2 then κ D (g, h) = κ C (g, h). (3) If g1 , . . . , g n ∈ G and C ⊆ ⋂ni=1 g i (P) then κ C (g1 , g n ) = κ C (g1 , g2 ) ⋅ ⋅ ⋅ κ C (g n−1 , g n ). Proof. (1) If m = 0, then g = h and there is nothing to prove. If m = 1 then g −1 h = 𝛾S g−1 h and h−1 g = 𝛾S h−1 g = 𝛾S󸀠 −1 and hence κ C (g, h)κ C (h, g) = [𝛾S g−1 h ][𝛾S󸀠 −1 ] = 1, a pairing g

h

g

h

relation. Finally if m = 2, then we can write the edge loop of C as (g = k0 , . . . , k t = h, k t+1 , . . . , k m = g) and thus −1 −1 κ C (g, h) = [k−1 0 k 1 ][k 1 k 2 ] ⋅ ⋅ ⋅ [k t−1 k t ], −1 −1 κ C (h, g) = [k−1 t k t+1 ][k t+1 k t+2 ] ⋅ ⋅ ⋅ [k m−1 k m ].

It is now easy to see that κ C (g, h)κ C (h, g) = 1 and hence the result follows. (2) If C = D then there is nothing to prove. So assume that C ≠ D. If C is a side then D is an edge and g and h are two consecutive elements of the edge loop of D. Then κ D (g, h) = [g−1 h] = κ C (g, h). Otherwise, C is a tile and hence g = h. Thus τ D (g, h) = 1 = τ C (g, h). (3) By induction it is enough to prove the statement for n = 3. So assume n = 3. If either g1 = g2 or g2 = g3 then the desired equality is obvious. So assume that g1 ≠ g2 and g2 ≠ g3 . If C is an edge then, up to a cyclic permutation, possibly reversing the order and making use of Lemma 16.1.14, the edge loop of C is of the form (g1 = k0 , . . . , g2 = k t , . . . , g3 = k l , . . . , k m ). Then, −1 −1 κ C (g1 , g3 ) = [k−1 0 k 1 ][k 1 k 2 ] ⋅ ⋅ ⋅ [k l−1 k l ] −1 −1 −1 −1 = ([k−1 0 k 1 ][k 1 k 2 ] ⋅ ⋅ ⋅ [k t−1 k t ]) ([k t k t+1 ] ⋅ ⋅ ⋅ [k l−1 k l ])

= κ C (g1 , g2 )κ C (g2 , g3 ) Otherwise, S = g1−1 (C) is a side of P, 𝛾S = g1−1 g2 , 𝛾S󸀠 = g2−1 g1 and g1 = g3 . Then, κ C (g1 , g3 ) = 1 = [𝛾S ][𝛾S󸀠 ] = κ C (g1 , g2 )κ C (g2 , g3 ).

86 | 16 Poincaré’s Theorem We denote by 𝕏T the complement in 𝕏 of the union of the cells of T of codimension at least three. By Lemma 16.1.8, every element of 𝕏T is either in the interior of a tile or in the relative interior of a side or an edge. The first ones are those that belong to exactly one tile, the elements of the relative interior of one side belong to exactly two tiles (Lemma 16.1.9) and the remaining elements belong to at least three tiles. If x ∈ 𝕏T , g, h ∈ G and x ∈ C ⊆ g(P) ∩ h(P) for some cell C then the codimension of C is at most 2 and we define κ x (g, h) = κ C (g, h). This is well defined because if D is another cell containing x and contained in g(P) ∩ h(P) with κ C (g, h) ≠ κ D (g, h) then g ≠ h and C ≠ D. Hence neither C nor D is a tile and either C or D is a side. Therefore, g(P) ∩ h(P) is a side and hence, by Lemma 16.2.6 (3), κ C (g, h) = [g −1 h] = κ D (g, h), a contradiction. This proves that indeed κ x (g, h) is well defined. By Lemma 16.2.6 we have κ x (g, h) = κ x (h, g)−1 and if x ∈ ∩ni=1 g i (P) with g1 , . . . , g n ∈ G then κ x (g1 , g n ) = κ x (g1 , g2 ) ⋅ ⋅ ⋅ κ x (g n−1 , g n )

(16.2.2)

Lemma 16.2.7. Let α : [0, 1] → 𝕏T be a continuous function and let 0 ≤ a < b < c ≤ 1 and g, h, k ∈ G be such that α(a) ∈ g(P), α((a, c)) ⊆ C r for a cell C of k(P) and α((a, b)) ⊆ D r for a cell D of h(P). Then κ α(a) (g, k) = κ α(a) (g, h) κ α(b) (h, k). Proof. First of all, observe that α(a), α(b) ∈ C ∩ D, α(a) ∈ g(P) ∩ k(P) ∩ h(P), α(b) ∈ h(P) ∩ k(P), C ⊆ h(P) and D ⊆ k(P), because α is continuous and every cell is the closure of its relative interior. The desired equality is clear if h = k. So, assume that h ≠ k. Then C and D are not tiles because they are included in h(P) ∩ k(P). If C ⊆ g(P) then κ α(a) (g, k) = κ C (g, k) = κ C (g, h) κ C (h, k) = κ α(a) (g, h) κ α(b) (h, k), by Lemma 16.2.6 (3). Assume that C ⊈ g(P). In particular g, h and k are pairwise different and α(a) ∈ ̸ C r . Hence C is a side, because Lemma 16.1.8 implies that 𝕏T ∩ E = E r for every edge E. Again by Lemma 16.1.8, we obtain that α(a) belongs to the relative interior of a cell E of g(P) and E is properly contained in C. Therefore E is an edge and h and k appears consecutively in an edge loop of E, i.e. after a cyclic permutation or a reversing in the ordering, an edge loop of E takes the form (k1 = g, . . . , k i = h, k i+1 = k, . . . , k m ). −1 −1 Then, κ α(a) (g, k) = κ E (g, k) = ([k−1 1 k 2 ] ⋅ ⋅ ⋅ [k i−1 k i ]) [k i k i+1 ] = κ E (g, h)κ C (h, k) = κ α(a) (g, h)κ α(b) (h, k), as desired. Definition 16.2.8. Let α : [a, b] → 𝕏T be a continuous function on a compact interval [a, b]. An α-adapted list is an ordered list L = (a0 , g1 , a1 , g2 , . . . , g n , a n ) such that a = a0 < a1 < . . . < a n = b and α(a i−1 , a i ) is contained in the relative interior of a cell of g i (P) for all 1 ≤ i ≤ n. Given an α-adapted list L, we define Φ(L) = κ α(a1 ) (g1 , g2 ) κ α(a2 ) (g2 , g3 ) ⋅ ⋅ ⋅ κ α(a n−1 ) (g n−1 , g n ), unless n = 1, where we set Φ(L) = 1.

16.2 Group presentations | 87

Observe that if i ∈ {1, . . . , n} then α(a i ) belongs to the boundaries of both α((a i−1 , a i )) and α((a i , a i+1 )). Hence, α(a i ) ∈ g i (P) ∩ g i+1 (P) and thus we get that κ α(a i ) (g i , g i+1 ) is well defined. Lemma 16.2.9. Let α : [a, b] → 𝕏T be a continuous function such that both α(a) and α(b) belong to the interior of some tile. If L and L󸀠 are α-adapted lists then Φ(L) = Φ(L󸀠 ). Proof. Without loss of generality, we may assume that [a, b] = [0, 1]. Let L = (a0 , g1 , a1 , . . . , g n , a n ) and L󸀠 = (a󸀠0 , g1󸀠 , a󸀠1 , . . . , g󸀠m , a󸀠m ). First observe that g1 (P) = g1󸀠 (P) is the only tile containing α(0). Thus g1 = g1󸀠 . If n = 1 then α([0, 1)) ⊆ g1 (P)∘ and therefore g󸀠i = g1 for every i ∈ {1, . . . , m}. Thus Φ(L) = 1 = Φ(L󸀠 ). Similarly, if m = 1 then Φ(L) = 1 = Φ(L󸀠 ). In the remainder of the proof we assume that n, m > 1. We construct an α-adapted list D󸀠α containing L󸀠 and all the a i ’s as follows. For every 1 ≤ i ≤ i󸀠 < n and j ∈ {1, . . . , m − 1} such that a i−1 ≤ a󸀠j−1 < a i ≤ a i󸀠 < a󸀠j ≤ a i󸀠 +1 , we insert in L󸀠 the sublist (g i , a i , . . . , g i󸀠 , a i󸀠 ) between a󸀠j−1 and g󸀠j . Similarly, we construct another list Dα containing L and all the a󸀠i ’s. We can consider the transition from L󸀠 to D󸀠α (or from L to Dα ) as the result of inserting finitely many pairs (g i , a i ). On the other hand, we can consider the transition from Dα to D󸀠α as the result of replacing finitely many group elements. Therefore, to prove the lemma, it is enough to deal with the following two cases: (1) L󸀠 is obtained by inserting in L one pair (h, b) between a i−1 and g i for a i−1 < b < a i and h ∈ G; (2) L󸀠 is obtained by replacing in L one g i by h, and in both cases α((a i−1 , a i )) is contained in the relative interior of a cell contained in g i (P) ∩ h(P) and h ≠ g i . (1) Assume L󸀠 is obtained by inserting one pair (h, b) between a i−1 and g i . We consider separately the cases when i > 1 or i = 1. If i > 1 then a = a i−1 , b, c = a i , g = g i−1 , h and k = g i satisfy the hypothesis of Lemma 16.2.7 and therefore Φ(L) = κ α(a1 ) (g1 , g2 ) ⋅ ⋅ ⋅ κ α(a n−1 ) (g n−1 , g n ) = κ α(1) (g1 , g2 ) ⋅ ⋅ ⋅ κ α(a i−2 ) (g i−2 , g i−1 )κ α(a i−1 ) (g i−1 , h)κ α(b) (h, g i ) κ α(a i ) (g i , g i+1 ) . . . κ α(a n−1 ) (g n−1 , g n ) = Φ(L󸀠 ). If i = 1 then h = g1 because α(a0 ) is in the relative interior of a tile. Thus κ α(b) (h, g1 ) = 1 and hence Φ(L) = κ α(a1 ) (g1 , g2 ) ⋅ ⋅ ⋅ κ α(a n−1 ) (g n−1 , g n ) = κ α(b) (h, g1 )κ α(a1 ) (g1 , g2 ) ⋅ ⋅ ⋅ κ α(a n−1 ) (g n−1 , g n ) = Φ(L󸀠 ). (2) Assume L󸀠 is obtained by replacing g i with h ≠ g i in L. By definition, α((a i−1 , a i )) ⊆ E r for some cell E contained in g i (P) ∩ h(P). Since g i ≠ h, clearly E is either an edge or a side. As both α(0) and α(1) belong to the interior of tiles we

88 | 16 Poincaré’s Theorem have i ≠ 1, n. To prove Φ(L) = Φ(L󸀠 ), it is enough to show κ α(a i−1 ) (g i−1 , h)κ α(a i ) (h, g i+1 ) = κ α(a i−1 ) (g i−1 , g i )κ α(a i ) (g i , g i+1 ). Let D1 and D2 be the cells generated by α(a i−1 ) and α(a i ) respectively. Then D1 , D2 ⊆ E. If E is an edge then D1 = D2 = E and hence κ α(a i−1 ) (g i−1 , h)κ α(a i ) (h, g i+1 ) = κ E (g i−1 , h)κ E (h, g i+1 ) = κ E (g i−1 , g i+1 ) = κ E (g i−1 , g i )κ E (g i , g i+1 ) = κ α(a i−1 ) (g i−1 , g i )κ α(a i ) (g i , g i+1 ), by Lemma 16.2.6 (3). Otherwise, E is a side. If D1 is a side then D1 = E and g i−1 ∈ {g i , h}. Thus κ α(a i−1 ) (g i−1 , h)κ α(a i ) (h, g i+1 ) = κ E (g i−1 , h)κ D2 (h, g i+1 ) = κ D2 (g i−1 , h)κ D2 (h, g i+1 ) = κ D2 (g i−1 , g i+1 ) = κ D2 (g i−1 , g i )κ D2 (g i , g i+1 ) = κ E (g i−1 , g i )κ D2 (g i , g i+1 ) = κ α(a i−1 ) (g i−1 , g i )κ α(a i ) (g i , g i+1 ), by statements (2) and (3) of Lemma 16.2.6. The case where E = D2 is proved similarly. Finally, assume that E is a side and D1 ≠ E and E ≠ D2 . Then D1 and D2 are edges and g i and h are consecutive members of the edge loops of D1 and D2 . After some reordering the edge loop of D1 takes the form (k1 = g i−1 , k2 , . . . , k t = g i , k t+1 = h, . . . , k m ) and the edge loop of D2 takes the form (h1 = g i , h2 = h, . . . , h l = g i+1 , . . . , h n ). Then −1 −1 [k−1 t k t+1 ] = [g i h] = [h 1 h 2 ] and −1 −1 −1 −1 κ α(a i−1 ) (g i−1 , h) κ α(a i ) (h, g i+1 ) = ([k−1 1 k 2 ] . . . [k t−1 k t ][k t k t+1 ]) ([h 2 h 3 ] . . . [h l−1 h l ]) −1 −1 −1 −1 = ([k−1 1 k 2 ] . . . [k t−1 k t ]) ([h 1 h 2 ][h 2 h 3 ] . . . [h l−1 h l ])

= κ α(a i−1 ) (g i−1 , g i ) κ α(a i ) (g i , g i+1 ). This finishes the proof. A parametrization of a geodesic segment [x, y] in 𝕏 is a surjective continuous function α : [a, b] → [x, y] with α(a) = x, α(b) = y and such that the map t → d(x, α(t)) is not decreasing (recall that d denotes the distance function on 𝕏). For such a map, if T is a polyhedron then α−1 (T) is a closed interval (maybe empty or of 0 length). Indeed, it is closed because so is P and α is continuous. To prove that α−1 (T) is an interval, let a ≤ t0 < t1 ≤ b, with α(t0 ), α(t1 ) ∈ T then α([t0 , t1 ]) is the geodesic segment [α(t0 ), α(t1 )] and hence it is contained in T because T is convex. Therefore [t0 , t1 ] ⊆ α−1 (T). This proves that α−1 (T) indeed is an interval.

16.2 Group presentations | 89

By C([a, b], 𝕏T )

we denote the set consisting of the continuous functions α : [a, b] → 𝕏T for which there is a finite ascending sequence a = b0 < b1 ⋅ ⋅ ⋅ < b m = b such that, for every i ∈ {1, . . . , m}, the restriction of α to [b i−1 , b i ] is a parametrization of a geodesic segment. Lemma 16.2.10. If α ∈ C([a, b], 𝕏T ) then there is an α-adapted list. Proof. Without loss of generality, we may assume that [a, b] = [0, 1]. By restricting to the intervals [b i−1 , b i ], one may assume without loss of generality that α is a parametrization of a geodesic segment. Then, for every cell C, α−1 (C) is a closed interval of [0, 1] (maybe empty or of length 0) and the image of α intersects finitely many tiles. We claim that for every t ∈ [0, 1) there is a cell C and an ϵ > 0 such that α((t, t + ϵ)) ⊆ C r and for every t ∈ (0, 1] there is a cell D and an ϵ > 0 such that α((t − ϵ, t)) ⊆ D r . By symmetry, we only prove the first statement. So, fix t ∈ [0, 1) and assume that T1 , . . . , T k are the only tiles containing α(t). For every i ∈ {1, . . . , k}, let ϵ i = max{ϵ ≥ 0 : ϵ ≤ 1 − t, α(t + ϵ) ∈ T i }. Let U be a neighborhood of α(t) not intersecting any tile different from every T1 , . . . , T k . Then there is an ϵ > 0 such that α([t, t + ϵ]) ⊆ U and therefore α(t + ϵ) ∈ T i for some i. Hence, ϵ i > 0 for some i ∈ {1, . . . , k}. Renumbering if necessary, we may assume that m is a positive integer such that m ≤ k and ϵ i > 0 if and only if i ≤ m. Let ϵ = min{ϵ1 , . . . , ϵ m }. Then α((t, t + ϵ)) ⊆ T1 ∩ ⋅ ⋅ ⋅ ∩ T m and T1 , . . . , T m are the unique tiles containing an element of α((t, t + ϵ)). This implies that T1 ∩⋅ ⋅ ⋅∩ T m is a cell, say C, and α((t, t + ϵ)) is contained in the relative interior of C by Lemma 16.1.8. This proves the claim. Let c be the supremum of all t ∈ [0, 1] with the property that there exists an α t adapted list where α t denotes the restriction of α to [0, t]. By the claim c > 0. It remains to be shown that c = 1. Otherwise, by the claim, there exists ϵ > 0 such that α((c−ϵ, c)) is contained in the relative interior of a cell C and α((c, c + ϵ)) is contained in the relative interior of a cell D. Let h, k ∈ G be such that C ⊆ h(P) and D ⊆ k(P). If (a0 = 0, g1 , a1 , . . . , g n , a n = c − ϵ) is an α c−ϵ -adapted list then (a0 = 0, g1 , a1 , . . . , g n , c − ϵ, h, c, k, c + ϵ) is an α c+ϵ -adapted list, contradicting the maximality of c. Let x, y ∈ 𝕏 and let Cx,y ([a, b], 𝕏T ) = {α ∈ C([a, b], 𝕏T ) : α(a) = x, α(b) = y.}

On Cx,y ([a, b], 𝕏T ) we consider the metric d defined as follows. If α, β ∈ C([a, b], 𝕏T ) then d(α, β) = max(d(α(c), β(c)) : c ∈ [a, b]}. Assume also that both x and y belong to the interior of some tile of T . By Lemma 16.2.9 and Lemma 16.2.10 there is a well defined map Φ:Γ→Δ given by Φ(α) = Φ(L),

90 | 16 Poincaré’s Theorem for L an α-adapted list. The next aim is proving that the map Φ : Cx,y ([0, 1], 𝕏T ) → Δ is constant. To do so, we first prove a strong simply connected property on 𝕏T with respect to the elements of Cx,y ([a, b], 𝕏T ). Recall that if α, β ∈ Cx,y ([a, b], 𝕏T ) then a homotopy from α to β is a continuous function H : [a, b] × [a, b] → 𝕏T such that H(a, t) = α(t), H(b, t) = β(t), H(t, 0) = x and H(t, 1) = y for every t ∈ [0, 1]. We say that α and β are strongly homotopic if there is an homotopy from α to β such that H(t, −) ∈ Cx,y ([a, b], 𝕏T ) for every t ∈ [a, b]. Clearly, this defines an equivalence relation on Cx,y ([a, b], 𝕏T ). Moreover, this equivalence relation is preserved by concatenation. More precisely, if α ∈ Cx,y ([a, b], 𝕏T ) and β ∈ Cy,z ([a, b], 𝕏T ) then the concatenation of α and β is the function α ⊕ β : [a, b] → 𝕏T defined by y−x

α(x + 2(t − a)( b−a )), { { (α ⊕ β)(t) = { { a+b y−z {β(y + 2(t − 2 )( b−a )),

if a ≤ t ≤ if

a+b 2

a+b 2 ;

≤ t ≤ b.

If α i , β i ∈ Cx,y ([a, b], 𝕏T ) are so that α i and β i are strongly homotopic for i ∈ {1, 2} then α1 ⊕ β1 and α2 ⊕ β2 are strongly homotopic. Lemma 16.2.11. All the elements of Cx,y ([a, b], 𝕏T ) are strongly homotopic. Proof. Let V and W be two subspaces of 𝕏 of dimension m and n respectively and let ⟨V, W⟩ denote the smallest subspace of 𝕏 containing V and W. If V ∩ W ≠ 0 then the dimension of ⟨V, W⟩ is at most m + n. Otherwise the dimension of ⟨V, W⟩ is at most m + n + 1. Indeed, let w ∈ W and let V1 be the smallest subspace of 𝕏 containing V and w. Then V1 has dimension m + 1, ⟨V, W⟩ = ⟨V1 , W⟩ and the dimension of this space is at most m + n + 1, because V1 ∩ W ≠ 0. Let L be a geodesic line of 𝕏 and let V be a geodesic subspace of 𝕏 of codimension at least 3. Then, by the previous, ⟨L, V⟩ has positive codimension. Therefore, if {L i : i ∈ I} is a countable family of geodesic lines and {V j : j ∈ J} is a countable family of subspaces of codimension at least 3 then, by Lemma 16.1.6, ∪i∈I,j∈J ⟨L i , V j ⟩ is a proper subset of 𝕏. Using this for the case when {V j : j ∈ J} is the family of subspaces generated by the cells of tiles of T of codimension at least 3, we deduce that for every countable family S = {[x i , y i ] : i ∈ I} of geodesic segments contained in 𝕏T , there exists w ∈ 𝕏T \ ∪i∈I,j∈J ⟨L i , V j ⟩, where L i denotes the geodesic line containing [x i , y i ]. This implies that the intersection of ⟨L i , V j ⟩ with the geodesic plane containing both L i and w is contained in L i . Thus, the cone of [x i , y i ] with vertex w does not intersect any V j and hence it is contained in 𝕏T . (Observe that if 𝕏 is spherical then the antipode of w is not in any [x i , y i ] because w ∈ ̸ L i .) Let x󸀠 , y󸀠 ∈ 𝕏 with x󸀠 ≠ y󸀠 and let ρ = d(x󸀠 , y󸀠 ). Let α0 : [0, 1] → [x󸀠 , y󸀠 ] be the parametrization of a segment [x󸀠 , y󸀠 ] of constant speed, that is α0 is the inverse of the 󸀠 map z ∈ [x󸀠 , y󸀠 ] 󳨃→ d(xρ ,z) ∈ [0, 1]. Assume α is an arbitrary parametrization of the segment [x󸀠 , y󸀠 ]. Consider the function H : [0, 1] × [0, 1] → [x󸀠 , y󸀠 ] defined by H(s, t) = α0 (

d(x󸀠 , α(t)) d(x󸀠 , α(t)) + s (t − )) . ρ ρ

16.2 Group presentations | 91

As both the distance function d and α are continuous, H is a continuous function. Thus H(0, t) = α(t) and H(1, t) = α0 (t). Furthermore H(s, 0) = α0 (0) = x󸀠 and H(s, 1) = α0 (1) = y󸀠 . Moreover, as α is a parametrization, the function t 󳨃→ d(x󸀠 , α(t)) is non-decreasing. Therefore, for every s ∈ [0, 1], the function 󸀠 󸀠 󸀠 ,α(t)) t 󳨃→ d(x ρ,α(t)) + s (t − d(x ρ,α(t)) ) = st + (1−s)d(x is non-decreasing. Hence ρ t 󳨃→ d(x󸀠 , H(s, t)) = d (x󸀠 , α0 (

d(x󸀠 , α(t)) d(x󸀠 , α(t)) + s (t − ))) ρ ρ

is non-decreasing too. Therefore H(s, −) is a parametrization of [x󸀠 , y󸀠 ], for every s ∈ [0, 1]. As the image of H is [x󸀠 , y󸀠 ], we have showed that H is a strong homotopy between α and α0 . Without loss of generality, we may assume that [a, b] = [0, 1]. Let α, β ∈ Cx,y ([0, 1], 𝕏T ). We need to show that α and β are strongly homotopic. Clearly, there is an ascending finite list 0 = a0 < a1 < . . . a n = 1 such that the restrictions to [a i−1 , a i ] of α and β are both parametrizations of segments. By the previous paragraph, we may assume without loss of generality that the restriction to each segment [a i−1 , a i ] of α and β has constant speed. We now argue by induction on n. If n = 1 then α = β (because of the constant speed) and hence there is nothing to prove. Let x i = α(a i ) and y i = β(a i ) for i ∈ {0, 1, . . . , n}. By the discussion in the second paragraph of the proof, there exists w ∈ 𝕏T such that all the cones determined by the segments [x i−1 , x i ] and [y i−1 , y i ] and centered in w are contained in 𝕏T . Let α i denote the restriction of α to [a i−1 , a i ] and let β i denote the restriction of β to [a i−1 , a i ]. Further, let α󸀠 : [0, 1] → 𝕏T be such that it agrees with α on [0, a n−2 ], the restriction of α󸀠 to [a n−2 , a n−1 ] is a parametrization of the interval [x n−2 , w] of constant speed and the restriction of α󸀠 to [a n−1 , a n ] is a parametrization of [w, y] of constant speed. Similarly, let β󸀠 : [0, 1] → 𝕏T agree with β on [0, a n−2 ], its restriction to [a n−2 , a n−1 ] is a parametrization of the geodesic interval [y n−2 , w] of constant speed and the restriction of β󸀠 to [a n−1 , 1] is a parametrization of [w, y] of constant speed. By the induction hypothesis, the restrictions of α󸀠 and β󸀠 to [0, a n−1 ] are strongly homotopic. Furthermore α󸀠 and β󸀠 coincide on [a n−1 , 1] and hence α󸀠 and β󸀠 are strongly homotopic. It remains to prove that α and α󸀠 are strongly homotopic, and that so are β and β󸀠 . For this it is enough to prove that the restrictions of α and α󸀠 (respectively, β and β󸀠 ) to [a n−2 , a n ] are strongly homotopic. This reduces the problem to the case where n = 2 and the two geodesic triangles xα(a1 )β(a1 ) and α(a1 )β(a1 )y are contained in 𝕏T . Let 𝛾 : [0, 1] → [α(a1 ), β(a1 )] be a parametrization of [α(a1 ), β(a1 )] of constant speed. For every s ∈ [0, a1 ] let t → H(s, t) be the parametrization of [x, 𝛾(s)] of constant speed and, for s ∈ [a1 , 1], let t → H(s, t) be the parametrization of [𝛾(s), y] of constant speed. In other words, if 0 ≤ t ≤ a1 then H(s, t) belongs to the geodesic segment [x, 𝛾(s)] and t (16.2.3) d(x, H(s, t)) = d(x, 𝛾(s)) . a1

92 | 16 Poincaré’s Theorem On the other hand, if a1 ≤ t ≤ 1 then H(s, t) belongs to the geodesic segment [𝛾(s), b] and t − a1 d(𝛾(s), H(s, t)) = d(𝛾(s), y) . (16.2.4) 1 − a1 Clearly H(s, −) ∈ Cx,y ([0, 1], 𝕏T ) for every s ∈ [0, 1], H(0, −) = α and H(1, −) = β. Finally, it is easy to see that the function H is continuous. Lemma 16.2.12. If both x and y belong to the interior of some tile then Φ : Cx,y ([a, b], 𝕏T ) → Δ is a constant mapping. Proof. Again, without loss of generality, we may assume that [a, b] = [0, 1]. We claim that it is sufficient to show that Φ is locally constant. Indeed, assume this is the case and let α, β ∈ Cx,y ([0, 1], 𝕏T ). By Lemma 16.2.11 there is a strong homotopy H from α to β. Let c denote the supremum of the s ∈ [0, 1] for which Φ(H(s, −)) = Φ(α). Since, by assumption, Φ is constant in a neighborhood of H(x, −), it easily follows that c = 1 and thus Φ(α) = Φ(β). To prove that Φ is locally constant, we show that for every α ∈ Cx,y ([0, 1], 𝕏T ) and for every α-adapted list L = (a0 , g1 , a1 , . . . , g n , a n ) (which exists because of Lemma 16.2.10), there is a positive real number δ such that for every β ∈ Cx,y ([0, 1], 𝕏T ) with d(α, β) < δ there is a β-adapted list D = (b0 , h1 , b1 , . . . , h m , b m ) and an increasing sequence of integers j0 = 0 < j1 < j2 < ⋅ ⋅ ⋅ < j n−1 < j n = m such that L󸀠 = (a0 , h j1 , a1 , h j2 , a2 , . . . , h j i−1 , a i−1 , h j n , a n )

is an α-adapted list and, for every i ∈ {0, 1, . . . , n}, Φ((L󸀠 )i ) = Φ((D)i ), where (L󸀠 )i = (a0 , h1 , a1 , . . . , h i , a i ) and (D)i = (b0 , h1 , b1 , . . . , h j i , b j i ). In particular, by Lemma 16.2.9, we have Φ(α) = Φ(L) = Φ(L󸀠 ) = Φ((L󸀠 )n ) = Φ((D)n ) = Φ(D) = Φ(β), as desired. Since T is locally finite, there is δ1 > 0 such that for every i ∈ {1, . . . , n − 1} and every g ∈ G, if B(α(a i ), 2δ1 )) ∩ g(P) ≠ 0 then α(a i ) ∈ g(P). Since α is continuous i−1 : i ∈ {1, . . . , n − 1}} such that, for every i ∈ {0, 1, . . . , n}, there is ϵ < min { a i −a 2 d(α(t), α(a i )) < δ1 for every t with |t − a i | < ϵ. For every i ∈ {1, . . . , n −1}, let a󸀠i = a i − ϵ 󸀠󸀠 󸀠󸀠 󸀠 󸀠 and a󸀠󸀠 i = a i + ϵ. We also set a n = 1 and a 0 = 0. Observe that a i−1 ≤ a i for every 󸀠 i ∈ {1, . . . , n}. Each α([a󸀠󸀠 i , a i+1 ]) is compact and it is contained in the relative interior of a cell C i contained in g i (P). Using again that T is locally finite we obtain a positive 󸀠 number δ2 such that d(α(t), g(P)) > δ2 for every t ∈ [a󸀠󸀠 i , a i+1 ] and every g ∈ G with C i ⊈ g(P). Let δ = min{δ1 , δ2 }. We will prove that δ satisfies the desired property. Let β ∈ Cx,y ([0, 1], 𝕏T ) with d(α, β) < δ. Then d(α(t), β(t)) < δ for every t ∈ [0, 1]. In particular, if t ∈ (a󸀠i , a󸀠󸀠 (16.2.5) i ) and β(t) ∈ g(P) then α(a i ) ∈ g(P) because d(β(t), α(a i )) < 2δ1 . Moreover, 󸀠 if t ∈ [a󸀠󸀠 i−1 , a i ] and β(t) ∈ g(P) then C i ⊆ g(P),

(16.2.6)

16.2 Group presentations | 93

since d(α(t), β(t)) < δ2 . Furthermore 󸀠 ∘ ∘ if C i is a tile then β([a󸀠󸀠 i−1 , a i ]) ⊆ C i = g i (P) .

(16.2.7)

∘ 󸀠 Indeed, as α([a󸀠󸀠 i−1 , a i ]) ⊆ α((a i−1 , a i )) ⊆ C i , it follows that C i = g i (P) is the only tile 󸀠󸀠 󸀠 󸀠 intersecting α([a i−1 , a i ]) and therefore it also is the only tile intersecting β([a󸀠󸀠 i−1 , a i ]). Then (16.2.7) follows. Let D = (b0 , h1 , b1 , . . . , h m , g m ) be a β-adapted list. We enlarge D by inserting 󸀠 󸀠󸀠 󸀠 󸀠󸀠 󸀠 󸀠󸀠 each a󸀠i and a󸀠󸀠 i . More precisely, if we rename the list (a 1 , a 1 , a 2 , a 2 , . . . , a n−1 , a n−1 ) = (c1 , c2 , . . . , c2(n−1) ), then we insert in D the sequence (h j , c i , h j , c i+1 , . . . , h j , c k ) between b j−1 and h j for every i whenever c i−1 ≤ b j−1 < c i < ⋅ ⋅ ⋅ < c k < b j ≤ c k+1 . So we may assume without loss of generality that there is an ascending sequence 0 = j󸀠0 < j1 < j󸀠1 < j2 < j󸀠2 < ⋅ ⋅ ⋅ < j n−1 < j󸀠n−1 < j n = m such that for every 󸀠 i ∈ {1, . . . , n} a󸀠i = b j i and for every i ∈ {0, 1, . . . , n − 1} we have a󸀠󸀠 i = b j i . We claim that L󸀠 = (a0 , h j1 , a1 , . . . , h j n , a n ) is an α-adapted list. For that observe that 󸀠 [b j i −1 , b j i ] ⊆ [a󸀠󸀠 i−1 , a i ] and β((b j i −1 , b j )) is contained in h j i (P). Therefore C i ⊆ h j i (P), by (16.2.6). It remains to prove that Φ((L󸀠 )i ) = Φ((D)i ) for every i ∈ {1, . . . , n}. We argue by induction. As α(0) ∈ g1 (P)∘ , necessarily C1 = g1 (P). Using (16.2.7) it is easy to prove 󸀠 that β([a󸀠󸀠 0 = 0, b j1 = a 1 ]) ⊆ g 1 (P). Hence h j = g 1 for every j ∈ {1, . . . , j 1 }. Therefore 󸀠 Φ((L )1 ) = 1 = Φ((D)1 ). Assume that i > 1 and Φ((L󸀠 )i−1 ) = Φ((D)i−1 ). Let E be the cell generated by α(a i−1 ), i.e. the unique one whose relative interior contains α(a i−1 ). For every j ∈ {1, . . . , m} let E j be the cell generated by β(b j ). Then E ⊆ C i , α(a i−1 ) ∈ E j for every j i−1 ≤ j ≤ j󸀠i−1 , by (16.2.5) and C i is contained in every cell intersecting some 󸀠 󸀠 β([a󸀠󸀠 i−1 , a i ]), by (16.2.6). Therefore, for every j i−1 ≤ j ≤ j i , C i is contained in every tile containing β(b j ) and hence E ⊆ C i ⊆ E j . We conclude that E ⊆ E j for every j i−1 ≤ j ≤ j i . Using that E j ⊆ h j (P) ∩ h j+1 (P) we have

κ α(a i−1 ) (h j i−1 , h j i ) = κ E (h j i−1 , h j i ) = κ E (h j i−1 , h j i−1 +1 ) ⋅ ⋅ ⋅ κ E (h j i −1 , h j i ) = κ E ji−1 (h j i−1 , h j i−1 +1 ) ⋅ ⋅ ⋅ κ E ji −1 (h j i −1 , h j i ) = κ β(b ji−1 ) (h j i−1 , h j i−1 +1 ) ⋅ ⋅ ⋅ κ β(h ji −1 ) (h j i −1 , h j i ) by Lemma 16.2.6 (3). Then Φ((L󸀠 )i ) = Φ((L󸀠 )i−1 )κ α(a i−1 ) (h j i−1 , h j i ) = Φ((D)i−1 )κ β(b ji−1 ) (h j i−1 , h j i−1 +1 ) ⋅ ⋅ ⋅ κ β(b ji −1 ) (h j i −1 , h j i ) = Φ((D)i ), as desired. Proof of Theorem 16.2.5 (Relations). Let g1 , . . . , g n be a list of pairing transformations such that g1 ⋅ ⋅ ⋅ g n = 1. We have to show that [g1 ] ⋅ ⋅ ⋅ [g n ] = 1. We may assume that g1 ⋅ ⋅ ⋅ g n = 1 is a minimal relation, i.e. g i ⋅ ⋅ ⋅ g j ≠ 1 for every 1 ≤ i < j ≤ n with (i, j) ≠ (1, n). For every i ∈ {1, . . . , n} let S i = g1 ⋅ ⋅ ⋅ g i−1 (P) ∩ g1 ⋅ ⋅ ⋅ g i (P), a side of both g1 ⋅ ⋅ ⋅ g i−1 (P) and g1 ⋅ ⋅ ⋅ g i (P).

94 | 16 Poincaré’s Theorem Fix c0 ∈ P∘ and for every i ∈ {1, . . . , n} let b i ∈ (P∩g i (P))r and c i = g1 g2 ⋅ ⋅ ⋅ g i−1 (b i ). Observe that each c i ∈ S i for i ≥ 1. If the sides S i and S i+1 are contained in different essential hyperplanes of the tile g1 g2 ⋅ ⋅ ⋅ g i (P) then (c i , c i+1 ) ⊆ (g1 g2 ⋅ ⋅ ⋅ g i )(P)∘ . However, they might be in the same essential hyperplane, and therefore we introduce some additional elements of 𝕏. For each i ∈ {1, 2, . . . , n − 1}, choose c󸀠i ∈ g1 g2 ⋅ ⋅ ⋅ g i (P)∘ . Consider the geodesic segments, [c0 , c1 = b1 ], [c1 = b1 , c󸀠1 ], [c󸀠1 , c2 = g1 (b2 )], [c2 , c󸀠2 ], [c󸀠2 , c3 = g1 g2 (b3 )], . . . , [c n−1 = g1 ⋅ ⋅ ⋅ g n−2 (b n−1 ), c󸀠n−1 ] [c󸀠n−1 , c n = g1 g2 ⋅ ⋅ ⋅ g n−1 (b n )], [c n , c n+1 = c0 ]. By construction, for each 1 ≤ i < n, (c0 , c1 ) ⊆ P∘ , (c i , c󸀠i ) ⊆ g1 ⋅ ⋅ ⋅ g i (P)∘ , (c󸀠i , c i+1 ) ⊆ g1 ⋅ ⋅ ⋅ g i (P)∘ , (c n , c0 ) ⊆ P∘ . Furthermore, the closure of each of the listed geodesic segments is contained in 𝕏T . Let α : [0, 1] → 𝕏T be the continuous function whose graph is obtained by concatenating all these segments. Then there exists an ascending sequence a0 < a1 < a󸀠1 < a2 < a󸀠2 < a3 < ⋅ ⋅ ⋅ < a n−1 < a󸀠n−1 < a n < a n+1 = 1 with α(a i ) = c i and α(a󸀠i ) = c󸀠i for each i. Furthermore, α((a i−1 , a󸀠i )) = (c i , c󸀠i ) ⊆ g1 ⋅ ⋅ ⋅ g i (P)∘ and α((a󸀠i , a i+1 )) = (c󸀠i , c i+1 ) ⊆ g1 ⋅ ⋅ ⋅ g i (P)∘ . Therefore, L = (a0 , 1, a1 , g1 , a󸀠1 , g1 , a2 , g1 g2 , a󸀠2 , g1 g2 , a2 , g1 g2 g3 a󸀠3 ,

⋅ ⋅ ⋅, g1 g2 ⋅ ⋅ ⋅ g n−1 , a󸀠n−1 , g1 ⋅ ⋅ ⋅ g n−1 a n , g1 ⋅ ⋅ ⋅ g n , a n+1 ) is an α-adapted list. Thus, Φ(α) = [g1 ] [g1−1 g1 ] [g1−1 (g1 g2 )] [(g1 g2 )−1 (g1 g2 )] [(g1 g2 )−1 (g1 g2 g3 )] ⋅ ⋅ ⋅[(g1 ⋅ ⋅ ⋅ g n−1 )−1 (g1 ⋅ ⋅ ⋅ g n − 1)] [(g1 ⋅ ⋅ ⋅ g n−1 )−1 (g1 ⋅ ⋅ ⋅ g n )] = [g1 ][g2 ] ⋅ ⋅ ⋅ [g n ]. On the other hand, let β ∈ Cc0 ,c0 ([0, 1], 𝕏T ) denote the constant path, i.e. β(t) = c0 for every t ∈ [0, 1]. Then (0, 1, 1) is a β-adapted list and Φ(β) = 1. Lemma 16.2.12 yields that [g1 ] ⋅ ⋅ ⋅ [g n ] = Φ(α) = Φ(β) = 1, as desired.

Problems 16.2.1. Determine a formula for H(s, t), the homotopy H defined in (16.2.3) and (16.2.4) in the proof of Lemma 16.2.11. Deduce that H is continuous. (Note that in the hyperbolic case, the continuity can also be deduced by working in the Kleinian model and converting the problem into a Euclidean problem.)

17 Fundamental polyhedra Throughout this chapter 𝕏 is a Riemann manifold of constant curvature and G is a discontinuous group of isometries of 𝕏. In order to be able to apply Poincaré’s Theorem for obtaining a presentation of G, we need to construct a fundamental polyhedron of G and this is not necessarily an easy task. The first two sections are dedicated to introduce two constructions of a fundamental polyhedron of G, the Dirichlet fundamental polyhedron, which is introduced for 𝕏 and G arbitrary, and the Ford fundamental polyhedron which is only introduced in some special cases with 𝕏 either ℍ2 or ℍ3 . Then we study the basic properties of a polyhedron with finite volume. All fundamental polyhedra of G have the same volume (see for example a proof for 𝕏n = ℍ3 in [54, Proposition 2.2.4]). One says that G has finite covolume if it has a fundamental polyhedron of finite volume. A discontinuous group of isometries of 𝕏n is said to be geometrically finite if it has a geometrically finite fundamental polyhedron. A result due to Garland and Raghunathan (see Theorem 2.2.7 in [54]) says that if G has finite covolume then G is geometrically finite. It follows from Theorem VI.C.7 in [159] that if G is geometrically finite then every fundamental polyhedron is geometrically finite. If G has a compact fundamental polyhedron then one says that G is cocompact. Clearly, if G is cocompact then it has finite covolume and hence it is geometrically finite (Problem 17.3.1). In Section 17.4 we present algorithms to calculate a finite subset of G generating a subgroup of finite index H of G and a fundamental polyhedron of H. For the algorithm to work one needs that G has finite covolume. In case G is cocompact, the algorithm is refined to obtain a fundamental polyhedron and a finite generating set for G. These algorithms were introduced for discrete subgroups of PSL2 (ℂ) by Jespers, Juriaans, Kiefer, de A. e Silva and Souza Filho ([109]). They extend an algorithm earlier obtained in [45] by Corrales, Jespers, Leal and del Río. As in [109], the first algorithm is called the Dirichlet Algorithm for Finite Covolume (DAFC). The second algorithm is called the Dirichlet Algorithm for Cocompact Discrete Groups (DACD). In case n is either 2 or 3 then the algorithm is adapted to obtain generators and fundamental polyhedra of discrete subgroups of PSL2 (ℂ), PSL2 (ℝ), PSB2 (ℍ(ℝ)) or PSB2 (ℂ). In the Section 17.5 we study some useful symmetries on Dirichlet fundamental polyhedra.

17.1 Dirichlet fundamental polyhedra In this section, we give one construction of a fundamental polyhedron, the Dirichlet Fundamental polyhedron.

96 | 17 Fundamental polyhedra For every x ∈ 𝕏, let

G x = {𝛾 ∈ G : 𝛾(x) = x},

the stabilizer of x in G. Since G is discontinuous, G x is finite. If 𝛾 ∈ G \ G x then the bisector Bis(x, 𝛾(x)) is the border of the half-space D𝛾 (x) = {y ∈ 𝕏 | d(x, y) ≤ d(y, 𝛾(x))}.

(17.1.1)

We claim that there exists x ∈ 𝕏 with trivial stabilizer. Indeed, we may assume that 𝕏 = ℝn , ℍn or 𝕊n . Then considering 𝕏 as a subset of ℝn or ℝn+1 , one can express every element of Isom(𝕏) as a rational function. Thus, for every 1 ≠ 𝛾 ∈ Isom(𝕏), the set 𝕏𝛾 = {x ∈ 𝕏 : 𝛾(x) = x} is the intersection with 𝕏 of an algebraic variety not containing𝕏. Hence, 𝕏𝛾 has Lebesgue measure 0. As G is countable, ∪𝛾g∈G\{1} 𝕏𝛾 ≠ 𝕏 and therefore, indeed, there is x ∈ 𝕏 such that G x = {1}. So, let x ∈ 𝕏 with G x = {1}. The set D(x) = ⋂ D𝛾 (x) 1=𝛾 ̸ ∈G

is known as the Dirichlet fundamental polyhedron (and also as the Poincaré fundamental polyhedron) of G with center x. The following result shows that the name is justified and also that this notion can be extended to the case when G x is not trivial. As G x is finite, a fundamental polyhedron for G x usually is, for a given group G, more or less easy to construct. Theorem 17.1.1. Let 𝕏 be a Riemann manifold of constant curvature, let G be a discontinuous group of isometries of 𝕏 and let x ∈ 𝕏. If P is a fundamental polyhedron for G x then DP (x) = P ∩ ⋂ D𝛾 (x) (17.1.2) 𝛾∈G\G x

is a locally finite fundamental polyhedron of G. In particular, if G x is trivial then D(x) = D𝕏 (x) is a locally finite fundamental polyhedron for G. Furthermore, if G is cocompact then DP (x) is compact. Proof. Let D = DP (x). As D𝛾 (x) is a closed half-space for every 𝛾 ∈ G \ G x , we have that D is an intersection of closed half-spaces. Since, by assumption, G is discontinuous, any closed ball B(x, 2r) only contains finitely many elements of the type 𝛾(x) with 𝛾 ∈ G. If B(x, r) ∩ Bis(x, 𝛾(x)) ≠ 0 then 𝛾(x) ∈ B(x, 2r). Therefore, only finitely many bisectors Bis(x, 𝛾(x)) intersect B(x, r). This proves that D is a polyhedron. To prove that ∪𝛾 𝛾(D) = 𝕏, we need to show that every orbit O of the action of G on 𝕏 contains an element in D. To do so, we first notice that since, by assumption, G is discontinuous, there is a u ∈ O such that d(u, x) ≤ d(v, x) for every v ∈ O. Moreover, because P is a fundamental polyhedron of G x , there is a 𝛾 ∈ G x such that 𝛾(u) ∈ P. Clearly, d(𝛾(u), x) = d(u, x) and hence we may assume, without loss of generality, that u ∈ P ∩ O. We claim that u ∈ D. Suppose the contrary. Then, there is 𝛾 ∈ G such

17.1 Dirichlet fundamental polyhedra

|

97

that d(u, x) > d(u, 𝛾(x)) = d(𝛾−1 (u), x), in contradiction with the choice of u. Hence, indeed, u ∈ D. Next, assume 𝛾(u) ∈ D∘ ∩ 𝛾(D∘ ), with 1 ≠ 𝛾 ∈ G. Clearly, 𝛾 ∈ ̸ G x as P is a fundamental polyhedron of G x and D0 ⊆ P0 . As 𝛾(u) ∈ D∘ , we get that d(x, 𝛾(u)) < d(𝛾(x), 𝛾(u)) = d(x, u). Since, u ∈ D∘ it then follows that d(x, u) < d(𝛾−1 (x), u) = d(x, 𝛾(u)), a contradiction. Hence, we have shown that D is a fundamental polyhedron of G. To prove that D is locally finite, let K be a compact subset of 𝕏. We need to show that 𝛾(D) ∩ K ≠ 0 for only finitely many 𝛾 ∈ G. To do so, without loss of generality, we may assume that K is a closed ball with center x. Let δ be the radius of K. Suppose 𝛾(D) ∩ K ≠ 0. If 𝛾 ∈ G \ G x and u ∈ D satisfies 𝛾(u) ∈ K then d(x, u) ≤ d(u, 𝛾−1 (x))) = d(𝛾(u), x) ≤ δ. So u ∈ K ∩ 𝛾−1 (K). Since G is discontinuous on 𝕏, this implies that there are only finitely many such 𝛾 ∈ G \ G x . The finiteness of G x then yields that indeed D is locally finite. Finally, we prove the last part of the statement. So, suppose Z is a compact fundamental polyhedron for G. Without loss of generality, we may assume that x ∈ Z. Let B be a closed ball centered at x and of radius r > 0 such that Z ⊆ B. We claim that D ⊆ B (and thus the compactness of D follows). Indeed, suppose u ∉ B and thus u ∈ ̸ Z. Let 𝛾 ∈ G be such that 𝛾(u) ∈ Z. Then, d(𝛾−1 (x), u) = d(x, 𝛾(u)) ≤ r and, consequently, d(u, x) > d(𝛾−1 (x), u). Hence, 𝛾−1 ∈ ̸ G x and u ∈ ̸ D𝛾−1 (x). Thus u ∈ ̸ D, as desired. We give two examples for which we describe a Dirichlet fundamental polyhedron. The first is elementary and the second is classical. Example 17.1.2. Let L be a lattice in ℝn with basis v1 , . . . , v n and consider the subgroup G of Isom(ℝn ) formed by the translations T u of the form x 󳨃→ x + u, with u ∈ L. If x = x1 v1 + ⋅ ⋅ ⋅ + x n v n , with x1 , . . . , x n ∈ ℝ, then G x = 1 and D(x) = {y1 v1 + ⋅ ⋅ ⋅ + y n v n : 2|y i − x i | ≤ 1 for all i}.

Observe that this is the fundamental polyhedron of (5.1.3). Applying Poincaré’s Theorem (Theorem 16.2.5) to this fundamental polyhedron we obtain the following (obvious) presentation L = ⟨T v1 , . . . , T v n | T v i T v j = T v j T v i , for all i, j⟩. Example 17.1.3 (Dirichlet fundamental polyhedron of G = PSL2 (ℤ)). Consider G = PSL2 (ℤ) as a subgroup of Isom+ (ℍ2 ) via the isomorphism PSL2 (ℝ) ≅ Isom+ (ℍ2 ) defined in (15.3.4). A straightforward calculation shows that G i (with i = √−1) is a cyclic group of order 2 generated by the element t ∈ PSL2 (ℤ) represented by T = ( 01 −1 0 ). The 2 isometric circle of t is the unit circle S(0, 1) and the action of t on ℍ interchanges the inner and outer part of S(0, 1). Thus P = ℍ2 \ B(0, 1), the complement in ℍ2 of the Euclidean unit ball, is a fundamental polyhedron of G i . Let D = {x + yi ∈ ℍ2 : x2 + y2 ≥ 1, 2|x| ≤ 1}

(Figure 17.1).

98 | 17 Fundamental polyhedra

D

Q=

−1+√−3 2

−1

P=

− 12

1 2

1+√−3 2

1

Fig. 17.1: Dirichlet Fundamental Polyhedron of PSL2 (ℤ)

We claim that D = DP (i). Let s be the element of PSL2 (ℤ) represented by the elementary matrix S = e12 (1). The action of s on ℍ2 is the translation x 󳨃→ x + 1 and hence D = P ∩ D s (i) ∩ D s−1 (i) ⊇ DP (i). To prove that the equality holds, it is enough to show that D is contained in D g (i) for every g ∈ G \ G i . Indeed, let g ∈ G \ G i be represented by ( ac db ). If c = 0 then a = d = ±1 and b ≠ 0. Therefore g(x) = x + ab and Dg (i) ∩ Dg−1 (i) = {x + yi : 2|x| ≤ |b|} ⊇ D. Assume that c ≠ 0. Then g(i) = ac+bd+i and c2 +d2 hence, by (15.2.1), D g (i) is formed by the elements x + yi ∈ ℍ2 satisfying the following inequality (c2 + d2 ) [(x −

2 ac + bd 2 1 + − (y ) ) ] ≥ x2 + (y − 1)2 . c2 + d2 c2 + d2

In other words D g (i) = {x + yi ∈ ℍ2 : (c2 + d2 − 1)(x2 + y2 ) +

(ac + bd)2 + 1 ≥ 1 + 2x(ac + bd)}. c2 + d2

If c2 + d2 = 1 then d = 0 and b = −c = ±1. Hence D g (i) = {x + yi : a2 ≥ 2acx} ⊇ D, because 2|x| ≤ 1 for every x + yi ∈ D. Assume otherwise that c2 + d2 ≠ 1 and let B = ℍ2 \ D g (i). Then B is an open ball with center in ℝ. Then, to prove that D ⊆ D g (i) (equivalently, D ∩ B = 0), it is enough to show that the two vertices P = 1+√2 −3 and

Q = −1+2√−3 of D belong to D g (i). To prove the latter, we first notice that as c and d are non-zero integers, we have c2 + d2 ≥ 2 and therefore (c2 + d2 − 1 − |ac + bd|)2 ≥ 0 ≥ |ac + bd|(2 − c2 − d2 ). Hence (c2 + d2 − 1)2 + (ac + bd)2 ≥ |ac + bd|(c2 + d2 ) or equivalently c2 + d2 +

(ac + bd)2 + 1 ≥ 2 + |ac + bd|. c2 + d2

It follows that indeed P, Q ∈ D g (i) and therefore D ⊆ D g (i), as desired.

17.2 Ford fundamental polyhedron |

99

Applying Poincaré’s Theorem to D we obtain three sides, namely S t = S t−1 = D ∩ t(D), S s = D ∩ s−1 (D) and S s−1 = D ∩ s(D) and two vertices (edges), namely P and Q. Therefore we obtain three pairing transformations: t, s and s−1 , with the reflection relations t2 = ss−1 = 1 and a unique cyclic relation (ts)3 = 1. Thus PSL2 (ℤ) = ⟨t⟩2 ∗ ⟨u⟩3 , with u = ts. This reproves Theorem 1.4.2. Kulkarni in [139] introduced a system, called Farey symbols, to obtain fundamental polyhedra of the natural images in PSL2 (ℤ) of the congruence subgroups SL2 (ℤ, n). The Farey symbols serve as endpoints of the boundary sides. This work is completed (by eliminating the trial and error method involved in [139]) in [42, 143] by Chan, Lang, Lim and Tan. In [51], Dooms, Jespers and Konovalov give a comprehensive and short proof for this construction, it also has been implemented into a software package [51, 52].

17.2 Ford fundamental polyhedron Another construction of a fundamental polyhedron is that of the Ford fundamental polyhedron. It was introduced by Ford (see [64]). Contrary to the construction of the Dirichlet fundamental polyhedron, this construction is based on Euclidean geometry. The role played by the center for the Dirichlet fundamental polyhedron is replaced by that of ∞ for the Ford fundamental polyhedron. Usually, the Ford fundamental polyhedron only is defined for discrete subgroups G of PSL2 (ℂ) such that G∞ , the stabilizer of ∞, is trivial. However we need to consider also cases where G∞ is not trivial. In order to define the Ford fundamental polyhedron we recall some notation. Let 𝛾 ∈ PSL2 (ℂ) and assume that 𝛾(∞) ≠ ∞. This means that if 𝛾 is represented by the matrix ( ac db ) ∈ SL2 (ℂ) then c ≠ 0. Recall that I𝛾 denotes the isometric sphere of 𝛾. We now consider the intersection of this sphere with ℍ3 which we still denote I𝛾 . So I𝛾 = {x ∈ ℍ3 : ‖cx + d‖ = 1}. The isometric spheres play, for the Ford fundamental polyhedron, the role that the bisectors play for the Dirichlet fundamental polyhedron. In this case the half-space used is I𝛾≥ = {x ∈ ℍ3 : ‖cx + d‖ ≥ 1}. In case 𝛾 ∈ SL2 (ℝ) then the isometric sphere should be replaced by the isometric circle and hence, in this case one uses I𝛾≥ = {x ∈ ℍ2 : ‖cx + d‖ ≥ 1}. Theorem 17.2.1. Let G be a discrete subgroup of PSL2 (ℂ) and let X = {(c, d) ∈ ℂ2 : (

a c

b ) ∈ SL2 (ℂ) represents an element in G for some a, b ∈ ℂ}. d

100 | 17 Fundamental polyhedra If X is a discrete subset of ℂ2 and if F∞ is a locally finite fundamental polyhedron for the action of G∞ on ℍ3 then F = F∞ ∩ ⋂ I𝛾≥ 𝛾∈G\G∞

is a locally finite fundamental polyhedron for the action of G on ℍ3 . It is called the Ford fundamental polyhedron of G. Similarly, for a discrete subgroup G of PSL2 (ℝ), let X = {(c, d) ∈ ℝ2 : (

a c

b ) ∈ SL2 (ℝ) represents an element in G for some a, b ∈ ℝ}. d

If X is discrete in ℝ2 and F∞ is locally finite fundamental polyhedron for the action of G∞ on ℍ2 , then F = F∞ ∩ ⋂ I𝛾≥ 𝛾∈G\G∞

is a locally finite fundamental polyhedron for the action of G on ℍ2 . Proof. We only prove the first part of the theorem, the second is proved similarly. Clearly F is the intersection of half-spaces. To prove that F is a polyhedron we have to show that the set consisting of the isometric spheres of elements in G \ G∞ is locally finite. Each of these isometric spheres is given by I c,d = {x ∈ ℍ3 : ‖cx + d‖ = 1}, with (c, d) ∈ X and c ≠ 0. If K is a compact subset of ℍ3 then the set {(c, d) ∈ X : ‖cx + d‖ = 1 for some x ∈ K} is bounded. Because X is discrete, by assumption, it follows that this set is finite. Thus {(c, d) ∈ X : c ≠ 0 and K ∩ I c,d ≠ 0} is finite, as desired. Hence F indeed is a polyhedron. To prove that F is a locally finite fundamental polyhedron, we now first show that every G-orbit O has an element in F. Let x = z + rj ∈ O, with z ∈ ℂ and r ∈ ℝ+ . Then ‖cx + d‖2 = |cz + d|2 + |c|2 r2 . Since, by assumption, X is a discrete subset of ℂ2 , we have that for every ϵ > 0 there are only finitely many (c, d) ∈ X such that ‖cx + d‖ ≤ ϵ. r Hence, {‖cx + d‖2 : (c, d) ∈ X} has a minimum and therefore { ‖cx+d‖ 2 : (c, d) ∈ X} has a maximum. So, by equation (15.3.3), the set consisting of the third coordinates of the elements of O has a maximum. Assume that this maximum is reached at x = z + rj ∈ O. As, by assumption, F∞ is a fundamental polyhedron of G∞ there exists 𝛾 ∈ G∞ such that 𝛾(x) ∈ F∞ . We know that 𝛾 is represented by some matrix of the type ( 0a db ). As X is a discrete subset of ℂ, the set {d n : n ∈ ℤ} is discrete in ℂ. So, one also has that |d| = 1. Hence x = z + rj and 𝛾(x) = z󸀠 + rj for some z󸀠 ∈ ℂ. Thus, replacing x by r 𝛾(x), one may furthermore assume that x ∈ F∞ and ‖cx+d‖ 2 ≤ r for every (c, d) ∈ X. In particular, ‖cx + d‖ ≥ 1 for every (c, d) ∈ X. Hence x ∈ I𝛾≥ for every 𝛾 ∈ G \ G∞ . Thus x ∈ F, as desired. Next, we prove that if 1 ≠ 𝛾 ∈ G and x ∈ ℍ then {x, 𝛾(x)} is not included in the interior of F. We prove this by contradiction. So, suppose x and 𝛾(x) belong to the interior of F∞ and the interior of I𝛾≥ for every δ ∈ G \ G∞ . The former implies that 𝛾 ∈ ̸ G∞ , because F∞ is a fundamental polyhedron of G∞ by assumption. Let 𝛾 be

17.2 Ford fundamental polyhedron |

101

d −b ). Write, x = z + rj represented by ( ac db ) ∈ SL2 (ℂ). Then 𝛾−1 is represented by ( −c a r 󸀠 󸀠 󸀠 󸀠 and 𝛾(x) = z + r j with z, z ∈ ℂ and r, r > 0. By equation (15.3.3), r󸀠 = ‖cx+d‖ 2 and

r=

r󸀠 . ‖−c𝛾(x)+a‖2

Because x ∈ I𝛾≥ and 𝛾(x) ∈ I𝛾≥−1 we get that r r󸀠 0 such that if z + rj ∈ K, with z ∈ ℂ and r > 0, then ϵ < r and ‖z + rj‖ < δ. So, suppose 𝛾 ∈ G \ G∞ is such that F ∩ 𝛾(K) ≠ 0. Let x = z + rj ∈ K, with z ∈ ℂ and r ∈ ℝ+ , be such that 𝛾(x) ∈ F and let 𝛾 be represented by a matrix ( ac db ) ∈ SL2 (ℂ) (with c ≠ 0). Since 𝛾(x) = (ax + b)(cx + d)−1 we get that (−c𝛾(x) + a)(cx + d) = 1. Hence, 𝛾(x) ∈ I𝛾≥ implies ‖cx + d‖ = ‖c𝛾(x) − a‖−1 ≤ 1 and thus r ≤ ‖z + dc−1 + rj‖ = ‖x + dc−1 ‖ ≤ |c|−1 . So, |c| ≤ r−1 < ϵ−1 and |d| ≤ ‖cx + d‖ + |c| ‖x‖ < 1 + ϵ−1 δ. Hence, the elements (c, d) ∈ X with c ≠ 0 and F ∩ 𝛾(K) ≠ 0 belong to a bounded set. As, by assumption, X is discrete, we get that such (c, d) belong to a finite set. To prove the local finiteness of F for G, it thus is sufficient to show that, for a given 𝛾 = ( ac db ) ∈ G with F ∩ 𝛾(K) ≠ 0, there are only finitely (a󸀠 , b󸀠 ) ∈ ℂ2 such that

−1 ∈ PSL2 (ℂ) and 𝛾1 (K)∩F ≠ 0. Clearly 𝛾𝛾−1 1 ∈ G ∞ and 𝛾(K)∩(𝛾𝛾1 )(F∞ ) ≠ 0. By assumption F∞ is a locally finite fundamental polyhedron for G∞ . Since, 𝛾(K) is compact, we get that the elements 𝛾𝛾−1 1 belong to a finite subset of G ∞ . Hence, the elements 𝛾1 belong to a finite set.

𝛾1 = ( ac

󸀠

b󸀠 ) d

Example 17.2.2 (Ford fundamental polyhedron of G = PSL2 (ℤ)). Let G = PSL2 (ℤ). Clearly G satisfies the assumptions of the second part of Theorem 17.2.1. A straightforward calculation shows that F∞ = {x + yi ∈ ℂ : 2|x| ≤ 1} is a locally finite fundamental polyhedron of G∞ and the Ford Fundamental polyhedron of G provided by Theorem 17.2.1 is F = F∞ ∩ I t≥ (Figure 17.1), 0 1 ). As I ≥ is the external part of the unit circle, F coincides where t is represented by ( −1 t 0 with the Dirichlet fundamental polyhedron D obtained in Example 17.1.3.

If u = z + rj ∈ ℍ3 , with z ∈ ℂ and r ∈ ℝ+ then we call r the height of u and denote this by h(u). By (15.3.3), if 𝛾 ∈ PSL2 (ℂ) is represented by ( ac db ) ∈ SL2 (ℂ) then h(u) ‖cz + d‖2 and therefore, if G is a subgroup of PSL2 (ℂ) then h(𝛾(u)) =

⋂ I𝛾≥ = {u ∈ ℍ3 : h(u) ≥ h(𝛾(u)) for every 𝛾 ∈ G \ G∞ }.

𝛾∈G\G∞

(17.2.1)

(17.2.2)

102 | 17 Fundamental polyhedra Lemma 17.2.3. Let G be a discrete subgroup of PSL2 (ℂ) containing a translation 𝛾t : x 󳨃→ x + t for t ∈ U(ℂ). Then ⋂ I𝛾≥ = {u ∈ ℍ3 : h(u) ≥ h(𝛾(u)) for every 𝛾 ∈ G}.

𝛾∈G\G∞

Moreover this set in invariant under the action of G∞ . Proof. Let F0 = ⋂𝛾∈G\G∞ I𝛾≥ and F1 = {u ∈ ℍ3 : h(u) ≥ h(𝛾(u)) for every 𝛾 ∈ G}. Conversely, let u ∈ F1 and let 𝛾 ∈ G. If 𝛾 ∈ ̸ G∞ then h(u) ≥ h(𝛾(u)) by (17.2.1). Assume otherwise that 𝛾 ∈ G∞ . Then 𝛾 is represented by a matrix of the form A = ( 0a ab−1 ) and 𝛾 and 𝛾t have exactly one common fixed point. By Lemma 15.5.4, 𝛾 is neither hyperbolic nor loxodromic and hence tr(A) = a + a−1 ∈ [−2, 2]. This implies that either |a| = 1. This is clear if a ∈ ̸ ℝ and easy to prove for a ∈ ℝ (using elementary analysis of the real function x 󳨃→ x + x−1 ). Applying (17.2.1) once more we deduce that h(𝛾(u)) = h(u).

Problems 17.2.1. Let G be a discontinuous group of isometries of a Riemann manifold 𝕏 of constant curvature. Suppose 𝛾 ∈ G \ G x , for some x ∈ 𝕏. Prove that 𝛾(D𝛾−1 (x)) = (𝕏 \ D𝛾 (x)) ∪ Bis(x, 𝛾(x)) and 𝛾(Bis(x, 𝛾−1 (x))) = Bis(x, 𝛾(x)). 17.2.2. Let G be a discontinuous group of isometries of a metric space 𝕏 and let 𝛾 ∈ Isom(𝕏). Prove that if P is a locally finite fundamental polyhedron for G then 𝛾(P) is a locally finite fundamental polyhedron for the group 𝛾G𝛾−1 .

17.3 Polyhedra of finite volume Let 𝕏n be a Riemann variety of constant curvature. If X is a subset of 𝕏n then vol(X) denotes the volume of X in the metric of 𝕏n , provided that it exists. For example, if 𝕏n = ℍn then dx1 ⋅ ⋅ ⋅ dx n vol(X) = ∫ . (17.3.1) x nn P

has a well defined volume and vol(P) ≠ 0 if and only if P is Every polyhedron P of thick in 𝕏n . In this section we study the polyhedra of finite volume. If 𝕏n is spherical then it is compact and therefore every polyhedron is compact. If 𝕏n is Euclidean and P is a thick polyhedron of 𝕏n then P is compact if and only if it has finite volume. Thus the only interesting case occurs when 𝕏n is hyperbolic. Therefore in the remainder of the ̂n . section we assume that 𝕏n is either ℍn or 𝔹n and we consider these as subsets of ℝ 𝕏n

Definition 17.3.1. Let P be a polyhedron in 𝕏n . An ideal point of P is a point of P ∩ ∂𝕏n .

17.3 Polyhedra of finite volume

|

103

Lemma 17.3.2. If P is a thick geometrically finite polyhedron of 𝕏n then the set of ideal points of P is either finite or contains an open subset of ∂𝕏n . Proof (Figure 17.2). Let I be the set of ideal points of P and suppose that I is not finite. As ∂𝕏n is compact it has an accumulation point x. Because of Theorem 15.2.3 and since ̂ n−1 ) acts ̂n into itself that maps ℍn to 𝔹n and as GM(ℝ η0 is a homeomorphism of ℝ ̂ n−1 transitively on ℝ (Problem 15.1.4), it easily follows that we may suppose, without loss of generality, that 𝕏n = ℍn and x = ∞. Let H1 , . . . , H m be the hyperspheres ̂n such that H1 ∩ ℍn , . . . , H n ∩ ℍn are the essential hyperplanes of P. Order the of ℝ H i ’s so that ∞ ∈ H i if and only if i ≤ k. Then P = ∩ni=1 H i+ ∩ ℍn , where each H i+ is ̂n with boundary H i . Moreover, if i > k then H + one of the two closed half-spaces of ℝ i is the external part of an Euclidean ball and if i ≤ k then H i ∩ ∂ℍ is an Euclidean ̂ n−1 . Then, there is an Euclidean open ball B containing each H with hyperplane of ℝ i ̂n \ B, Q = ∩k H + ∩ ℍn , P1 = B+ ∩ Q and let i > k (see Figure 15.3.10). Let B+ = ℝ i=1 i I1 and J be the sets of ideal points of P1 and Q, respectively. Thus P1 ⊆ P ⊆ Q and ̂ n−1 = {∞} ∪ R, where R is a geometrically finite I1 = J \ B ⊆ I ⊆ J = ∩ki=1 H i+ ∩ ℝ n−1 polyhedron of ℝ . As P is thick in ℍn so is Q. Hence J is thick in ℝn−1 . Moreover, ∞ is an accumulation point of J. Thus R is not bounded. Hence the interior of R \ B contains an interior point and hence I1 contains an open subset of ℝn−1 . We conclude ̂ n−1 . that I contains an open subset of ∂𝕏n = ℝ

H1 P \ P1

P1

∞

H3 I1

I \ I1

Q\P

H2

J\I B

Fig. 17.2

̂n properly contained Definition 17.3.3. A horoball of 𝕏n is an open half-space of ℝ in 𝕏n whose boundary is tangent to ∂𝕏n . A horosphere of 𝕏n is the boundary of a horoball. The point of intersection with ∂𝕏n of a horosphere S is called the base of S and of the horoball with boundary S. Clearly the Euclidean spheres contained in 𝕏n and tangent to ∂𝕏n are horospheres of 𝕏n . These are all the horospheres of 𝔹n and the horospheres of ℍn with base in ℂ.

104 | 17 Fundamental polyhedra

Fig. 17.3: Horospheres of 𝔹2 and ℍn .

The horoballs of ℍn with base in ∞ are given by equations of the form x n > R for R a positive real number (Figure 17.3). The horoball B of ℍn given by x n > R has infinite volume because +∞

∫ B

dx1 . . . dx n dx n = ∫ dx1 . . . dx x n−1 ⋅ ∫ n = +∞. n x1 xn ℝn−1

R

̂ ̂ n−1 ) acts transitively on ∂𝕏n and GM(ℝ n−1 ) ≅ Isom(ℍn−1 ), all the horoballs As GM(ℝ of ℍn have infinite volume. Observe that the isometry η0 maps horoballs of ℍn to horoballs of 𝔹n . So, we deduce that the horoballs of 𝔹n have also infinite volume. The next lemma shows how “to cut” a horoball with a polyhedron to obtain a set of finite volume. Lemma 17.3.4. Let n ≥ 2, let H be a horoball of 𝕏n with base x, let P be a geometrically finite polyhedron in 𝕏n and let I be the set of ideal points of P. Suppose that x has a ̂n such that I ∩ U = {x}. Then H ∩ P has finite volume. neighborhood U in ℝ Proof. Without loss of generality, we may assume that H, P ⊆ ℍn and x = ∞. Hence, + , where each H + is a H = {x ∈ ℍn : x n > r} for some r > 0. Write P = H1+ ∩ ⋅ ⋅ ⋅ ∩ H m i half-space with boundary H i and assume that ∞ ∈ H i if and only if i ≤ k. Hence H i+ is the complement of the open ball with boundary H i . Let Q = H1+ ∩ ⋅ ⋅ ⋅ ∩ H kk and let B be an open ball containing I k+1 ∪ ⋅ ⋅ ⋅ ∪ H m and P1 = Q \ B. The assumption implies that

P

H

P H K 𝔹2

ℍ2

Fig. 17.4: Polyhedron (blue) and horoball (red) with intersection of finite volume.

17.3 Polyhedra of finite volume

|

105

I \ {∞} is contained in an open ball of ℝn . Without loss of generality, we may assume that I \ {∞} ⊆ B. Then, Q = K × (0, +∞) for some polyhedron K of ℝn−1 and the set of ideal points of Q is K ∪ {∞}. Moreover, the set of ideal points of P1 is {∞} ∪ (K \ B). By assumption, B contains all the finite ideal points of P. As P1 ⊆ P, we have K ⊆ B. Thus K is compact. Then H ∩ P ⊆ K × (r, +∞) and the volume of this set in ℍn is ∞ vol(K) × ∫r dx x n < ∞. Proposition 17.3.5. Let P be a polyhedron in 𝕏n . Then P has finite volume if and only if it is geometrically finite and has finitely many ideal points. Proof. We may assume that 𝕏 = 𝔹n . Let I be the set of ideal points of P. Suppose that vol(P) < ∞. Then P is geometrically finite, by [54, Theorem 2.2.7]. Suppose that I is not finite. By Lemma 17.3.2, there is an open ball B(x, r) in ℝn , with x ∈ ∂𝕏n such that B(x, r) ∩ ∂𝕏n ⊆ I. This implies that the horoball H with base in x and radius 2r is completely contained in P. As H has infinite volume, so does P. Conversely, assume that P is geometrically finite and let I = {x1 , . . . , x n }. For every i = 1, . . . , n, let H i be a horoball with base x i . Then P \ ∪ni=1 H i is compact. Moreover, for every i = 1, . . . , n there is a neighborhood U of ℝn with U i ∩ I = {x i }. Therefore vol(P) ≤ vol(P \ ∪ni=1 H i ) + ∑ni=1 vol(P ∩ H i ) < ∞, by Lemma 17.3.4 (Figure 17.5).

Fig. 17.5

Lemma 17.3.6. Let G and G0 be two discrete subgroups of PSL2 (ℂ). Assume G0 has finite covolume and has a locally finite fundamental polyhedron, say P0 . If P is a fundamental polyhedron of G and G ⊆ G0 then vol(P) = [G0 : G] vol(P0 ). In particular, G is of finite index in G0 if and only if G has finite covolume. Proof. Write G0 = ⋃t∈T Gt, where T is a transversal T of G in G0 . Let P = ∪t∈T t(P∘0 ). Then, P is open in ℍ3 and the members in {𝛾(P) : 𝛾 ∈ G} are mutually disjoint. The set ∪t∈T tP0 is closed in ℍ3 , since P0 is locally finite. Therefore P = ∪t∈T tP0 . Clearly, ℍ3 = ∪𝛾∈G 𝛾(P). Furthermore, as P0 is locally finite, ∂P ⊆ ∪t∈T ∂(tP0 ) = ∪t∈T t∂(P0 ). Because G0 is countable and vol(∂P0 ) = 0, it follows that vol(∂P) = 0. Hence, vol(P) = vol(∪t∈T t(P0 )) = ∑t∈T vol(P0 ) = vol(P0 ) [G0 : G]. Since the finiteness of the volume is independent of the fundamental polyhedron, the result follows.

106 | 17 Fundamental polyhedra

Problems 17.3.1. Prove that a cocompact discontinuous group of isometries of ℍ3 is geometrically finite. 17.3.2. Prove that η0 gives a bijection from the horospheres of ℍ3 with vertex P and the horospheres of 𝔹3 with vertex η0 (P). In particular it maps the horospheres of ℍ3 which are not Euclidean spheres to horospheres of 𝔹3 with vertex j.

17.4 An algorithm to compute a fundamental polyhedron In this section we introduce two algorithms related with the computation of a Dirichlet fundamental polyhedron of a discontinuous subgroup G of isometries of a hyperbolic space 𝕏n . In both cases the input of the algorithm is formed by G and a fundamental polyhedron P of G C , where C is the center of 𝕏n . The output of the first algorithm is a finite set generating a subgroup of finite index in G, provided that G has finite covolume. The output of the second algorithm is a finite set of generators of G and the Dirichlet fundamental polyhedron DP (C), provided that G is cocompact. In the remainder of the section 𝕏n is either ℍn or 𝔹n , and we consider 𝕏n as a ̂n . Moreover, C denotes the center of 𝕏n (i.e. C = j if 𝕏n = ℍn and C = 0 if subset of ℝ n n 𝕏 = 𝔹 ). Observe that if 𝛾 ∈ Isom(𝕏n ) with 𝛾(C) ≠ C then D𝛾 (C) = Bis≥𝕏n (𝛾) ∩ 𝕏n . Therefore, if G is a discontinuous subgroup of Isom(𝕏n ) and P is a fundamental polyhedron of G C then DP (C) = P ∩ ⋂ Bis≥𝕏n (𝛾). 𝛾∈G\G C

We present the first algorithm. Algorithm 17.4.1 (Dirichlet Algorithm for Finite Covolume (DAFC)). Input: A discontinuous subgroup G of Isom(𝕏n ) and a fundamental polyhedron P of G C , where C is the center of 𝕏n . (1) Choose an unbounded strictly increasing sequence of real numbers 0 = r0 < r1 < r2 < . . . ̂n \ P and for N ≥ 1 calculate (2) Set G(0) = G C and B(0) = ℝ H(N) = {𝛾 ∈ G : r n−1 < d(C, 𝛾(C)) ≤ r N }, G(N) = G(N−1) ∪ H(N) , B(N) = B(N−1) ∪ ⋃ Bis 0. (5) Suppose ∂𝕏n ⊆ B(N) . Therefore D(N) has finite volume by (3). Let H be the subgroup of G generated by G(N) and let DH be the Dirichlet fundamental polyhedron of H with center C constructed from P. As D(N) contains DH , we have that vol(DH ) ≤ vol(DH ) < ∞, by Lemma 17.3.6 vol(D(N) ) < ∞ and hence [G : H] = vol(D P (C)) In case G is cocompact we can refine DAFC to obtain a finite set of generators of G and a Dirichlet fundamental polyhedron of G. Algorithm 17.4.3 (Dirichlet Algorithm for Cocompact Discrete Groups (DACD)). Input: A discontinuous subgroup G of Isom(𝕏n ) and a fundamental polyhedron P of G C , where C is the center of 𝕏n . (1) Choose an unbounded strictly increasing sequence of real numbers 0 = r0 < r1 < r2 < . . .

108 | 17 Fundamental polyhedra ̂n \ P and for N ≥ 1 calculate (2) Set G(0) = G C , D(0) = P and B(0) = ℝ H(N) = {𝛾 ∈ G : r N−1 < d(C, 𝛾(C)) ≤ r N }, G(N) = G(N1 ) ∪ H(N) , D(N) = D(N−1) ∩ ⋂ Bis≥𝕏n (𝛾), 𝛾∈H(N)

B(N) = B(N−1) ∪ ⋃ Bis 0, (x − 12 )2 + y2 ≥ 1} is a fundamental Dirichlet polyhedron of PSL2 (ℤ, 2). (See Figure 17.6.)

γ1−2

γ2

γ2−2

γ12 Fig. 17.6: Dirichlet Fundamental Polyhedron of PSL2 (ℤ, 2).

Proof. Let G = PSL2 (ℤ, 2). By the previous paragraph, the half-spaces Bis 0 such that ρ󸀠 < r ≤ ρ for every z + rj ∈ P, with z ∈ ℂ and r > 0. Let r󸀠 be a real number greater than ρ and ρ1󸀠 ϵ . Since P is a fundamental polyhedron of PSL2 (R), there is z + rj ∈ P and 𝛾 ∈ PSL2 (R) such that 𝛾(z + rj) = r󸀠 j. Let 𝛾 be represented by a matrix ( ac db ) ∈ SL2 (R). By Theorem 15.3.2, r󸀠 = |cz+d|2r+r2 |c|2 . If c = 0 then d ∈ U(R) and 1 1 hence |d| = 1. Thus r󸀠 = r ≤ ρ. Otherwise, r󸀠 ≤ r|c| 2 ≤ ρ 󸀠 ϵ 2 . In both cases we have a contradiction with the choice of r󸀠 . So, indeed, P is not compact. Theorem 10.1.2 in [54] shows that if O is an order in either ℚ or an imaginary quadratic extension of ℚ then PSL2 (O) has finite covolume. Combining this with Proposition 18.1.2 we have the following result. Proposition 18.1.3. Let A be an algebra of Kleinian type, O an order in A and let σ be an embedding of the center F of A in ℂ such that σ(O1 ) is discrete. Then σ(O1 ) has finite covolume and hence it is geometrically finite.

Problems 18.1.1. Let 𝕏 be either an Euclidean, hyperbolic or spherical space. Let G be a discontinuous group of isometries of 𝕏. For x ∈ 𝕏 and 𝛾 ∈ G \ G x , prove that the bisector Bis(x, 𝛾(x)) is the hyperplane in 𝕏 that is perpendicular to the line joining x and 𝛾(x).

18.2 Bianchi groups In this section we calculate a presentation and a fundamental polyhedron of PSL1 (O), for some orders O in split quaternion algebras of Kleinian type. By Proposition 18.1.1

120 | 18 Unit groups of orders in quaternion algebras such algebras are of the form M2 (F), with F either ℚ or an imaginary quadratic extension of ℚ. We restrict our attention to the case where O = M2 (R), with R the ring of integers of F. If F = ℚ then the group under consideration is PSL2 (ℤ) which already has been described in Example 17.1.3. Thus we only consider the second case. In the remainder of the section Δ is a positive square-free integer, F = ℚ(√−Δ) and O is the ring of integers of F. By Example 4.1.9, O = ℤ[ω Δ ]

with

√−Δ, ω Δ = { 1+√−Δ 2

if Δ ≢ 3 mod 4; otherwise.

,

We will study the groups Γ = PSL2 (O). These groups are called Bianchi groups because they were first studied by Bianchi in [27]. For more information we refer to [62]. Clearly, Γ is a discrete subgroup of PSL2 (ℂ). The goal is to use Poincaré’s Theorem (Theorem 16.2.5) in order to calculate a presentation for Γ, for some values of Δ. For this purpose, since Γ satisfies the hypothesis of Theorem 17.2.1, we may use a Ford fundamental polyhedron, . Elements of SL2 (ℂ) will be denoted with capital letters and their natural images in PSL2 (ℂ) will be donated with the corresponding lower letters. The following elements of SL2 (O) will play an important role in our presentations: 0 1

T=(

−1 ), 0

S=(

1 0

1 ), 1

1 UΔ = ( 0

ωΔ ). 1

(18.2.1)

Observe that t2 = 1 and also that the isometric sphere of t is the unit sphere. Thus t interchanges the inner and outer parts of the unit sphere S(0, 1). Moreover, s and u acts as translations on ℍ3 . Let Γ∞ = {𝛾 ∈ Γ : 𝛾(∞) = ∞}, the stabilizer of ∞. The elements of Γ∞ are represented by the matrices of the form ( 0α αβ−1 ), with α ∈ U(O) and β ∈ O. A straightforward calculation shows that Γ∞ = ⟨s, u, r⟩ , where ξ R=( 0

0 ξ −1

),

ζ4 = i, { { { with ξ = {ζ3 = −1+2√−3 , { { {1,

if Δ = 1; if Δ = 3;

(18.2.2)

otherwise.

It is easy to see that {z + rj ∈ ℍ3 : − 12 ≤ Re(z), Im(z) ≤ 12 } is a fundamental polyhedron of ⟨s, u⟩. Moreover, if Δ = 1 then r is a rotation of degree π around z = 0, and if

18.2 Bianchi groups

Δ = 3 then r is a rotation of degree D∞

π 3

| 121

around z = 0. It then easily follows that

{x + yi + rj ∈ ℍ3 : − 12 ≤ x ≤ 12 , 0 ≤ y ≤ 12 }, { { { = { {x + yi + rj ∈ ℍ3 : 0 ≤ x ≤ 12 , 0 ≤ x + √3 y ≤ 1}, { { 1 1 3 { {x + yω Δ + rj ∈ ℍ : − 2 ≤ x, y ≤ 2 },

if Δ = 1; if Δ = 3; otherwise.

is a fundamental polyhedron of Γ∞ (Figure 18.1). 1+√−2 2

1+√−7 4

√−3 3

1+i 2

1 2

1 2

Δ=1

−

+

√−3 6

√−3 6

Δ=3

Δ=2

Δ=7

Fig. 18.1: Orthogonal projection of D∞ for Δ = 1, 2, 3 and 7.

Then, by Theorem 17.2.1, F = D∞ ∩ ( ⋂

≥ Iw )

w∈Γ\Γ∞

is a fundamental polyhedron of Γ. We claim that if Δ = 1, 2 or 3 then F = D∞ ∩ I t≥

(Figure 18.2).

We prove this for Δ = 1 and we leave the cases Δ = 2 and Δ = 3 as an exercise for the reader. Assume that Δ = 1 and let F = D∞ ∩I t≥ . Observe that F has four vertices and

(a) Δ = 1

(b) Δ = 2

Fig. 18.2: Ford Fundamental polyhedra for PSL2 (O) for Δ = 1, 2 and 3.

(c) Δ = 3.

122 | 18 Unit groups of orders in quaternion algebras

its only ideal point is infinity. By Theorem 18.1.2, Γ is not cocompact and hence we can use the ideas of Remark 17.4.8. In other words, we only need to prove that the vertices ≥ are contained in I w for finitely many w’s. In fact in this case we can prove easily that F ≥ . If x + yi + rj ∈ F then r2 ≥ 1 − x2 − y2 ≥ 12 . This implies that if is contained in every I w ≥ F is not contained in I w with w represented by ( ac bd ) then |c| = 1. Then the isometric sphere of w has radius 1 and it is centered at −d. If −d = x0 + y0 i ≠ 0, with x0 , y0 ∈ ℤ, ≥ . and x + yi + rj ∈ F then 1 − r2 ≤ x2 + y2 ≤ (x − x0 )2 + (y − y0 )2 and therefore, F ⊆ I w This proves the claim. Now we compute a presentation of Γ for Δ = 1, 2 and 3. Example 18.2.1. PSL2 (ℤ[i]) = ⟨r, v, s, t | r2 = v2 = t2 = (sr)2 = (sv)2 = (tr)2 = (tv)3 = (ts)3 = 1⟩. Proof. The ring of integers of ℚ(i) is O = ℤ[i] and hence Γ = PSL2 (ℤ[i]). Moreover, F has five essential hyperplanes given by the following equations (see Figure 18.2 (a)): y = 0,

y=

1 , 2

1 x=− , 2

x=

1 , 2

x2 + y2 + z2 = 1.

The intersection of F with the essential hyperplanes gives the following five sides of F: S r = F ∩ r(F),

S v = F ∩ v(F),

S s−1 = F ∩ s(F),

S s = F ∩ s−1 (F),

S t = F ∩ t(F),

where i 0

R=(

0 ) −i

and

i V=( 0

1 ). −i

Observe that each S r , S v and S t is paired with itself while S s and S s−1 are paired with each other. So we have three reflection relations: r2 = v2 = t2 = 1. Moreover, F has eight edges E r,s−1 = S r ∩ S s−1 , E r,t = S r ∩ S t ,

E r,s = S r ∩ S s , E v,t = S v ∩ S t ,

E v,s−1 = S v ∩ S s−1 , E s,t = S s ∩ S t ,

E v,s = S v ∩ S s ,

E s−1 ,t = S s−1 ∩ S t .

The edge-side cycles are represented, up to a cyclic permutation or a reversing of the order, by (E r,s−1 , S r , E r,s , S s ),

(E v,s−1 , S v , E v,s , S s ),

(E v,t , S v , E v,t , S t )

and

(E r,t , S r , E r,t , S t ),

(E s,t , S s , E s−1 ,t , S t ).

So, we obtain the cycle relations: (sr)2 = (sv)2 = (tr)2 = (tv)3 = (ts)3 = 1. The result then follows from Poincaré’s Theorem (Theorem 16.2.5). Example 18.2.2. PSL2 (ℤ[√−2]) = ⟨t, s, u | t2 = (ts)3 = [s, u] = [t, u]2 = 1⟩.

18.2 Bianchi groups | 123

Proof. The ring of integers of ℚ(√−2) is O = ℤ[√−2] and the sides of F are S s = F ∩ s−1 (F),

S s−1 = F ∩ s(F),

S u = F ∩ u−1 (F)

and S u−1 = F ∩ u(F),

S t = F ∩ t(F).

(see Figure 18.2 (b)). In this case the only side paired with itself is S t and the edges are E s,u = S s ∩ S u ,

E s,u−1 = S s ∩ S u−1 ,

E t,s = S t ∩ S s ,

E s−1 ,u = S s−1 ∩ S u ,

E t,s−1 = S t ∩ S s−1 ,

E t,u = S t ∩ S u ,

E s−1 ,u−1 = S s−1 ∩ S u−1 , E t,u−1 = S t ∩ S u−1 .

Again the result follows from Poincaré’s Theorem. Example 18.2.3. PSL2 (ℤ[ζ3 ]) = ⟨r, v, w | r3 = v3 = w2 = (vr)3 = (wr)2 = (wv)3 = 1⟩. Proof. In this case, the orthogonal projection of F on ℂ is a rhombus with vertices 0, 12 + √6−3 , 12 − √6−3 and √33 (see Figure 18.2 (c) and Figure 18.3). The last three points are vertices of a regular hexagon centered in 0 and inscribe in a circumference of radius √33 .

v r −1 w

v −1 r

Fig. 18.3: Orthogonal projection of Ford Fundamental Polyhedron of PSL2 (ℤ[ζ3 ])

The sides of F are S r = F ∩ r−1 (F),

S r−1 = F ∩ r(F), −1

S v = F ∩ v (F ) with R=(

ζ3−1 0

0 ), ζ3

and

S w = F ∩ w(F),

S v−1 = F ∩ v(F),

V = TR = (

ζ3−1 0

ζ3 ) ζ3

124 | 18 Unit groups of orders in quaternion algebras

and W = SR = (

0 −ζ3−1

ζ3 ). 0

Again there is only one side paired with itself, namely S w . The edges are E r,r−1 = S r ∩ S r−1 ,

E r,v−1 = S r ∩ S v−1

E r−1 ,v = S r−1 ∩ S v ,

E v,v−1 = S v ∩ S v−1 ,

E w,r = S w ∩ S r ,

E w,r−1 = S w ∩ S r−1 ,

E w,v = S w ∩ S v ,

E w,v−1 = S w ∩ S v−1

and we have the following representatives for the edge-side loops: (E r,r−1 , S r ),

(E r,v−1 , S r , E r−1 ,v , S v , E r,v−1 ),

(E w,r , S r , E w,r−1 , S w , E w,r ),

(E v,v−1 , S v ),

(E w,v , S v , E w,v−1 , S w ).

Applying Poincaré’s Theorem we obtain the result. We explain how to apply the DAFC for an arbitrary Bianchi group. Recall the general notation Δ, ω Δ , F, O and Γ introduced at the beginning of the section. Let {Δ, α Δ = NrF/ℚ (ω Δ ) = { 1+Δ , { 4

if Δ ≢ 3

mod 4;

if Δ ≡ 3

mod 4

and {0, β Δ = TrF/ℚ (ω Δ ) = { 1, {

if Δ ≢ 3

mod 4;

if Δ ≡ 3

mod 4.

Suppose that 𝛾 ∈ Γ is represented by the matrix A = ( ac db ) ∈ SL2 (O) and set: a = a1 + a2 ω Δ ,

b = b1 + b2 ω Δ ,

c = c1 + c2 ω Δ ,

d = d1 + d2 ω Δ .

with each a i , b i , c i , d i ∈ ℤ and xi = (

ai ci

bi ) di

for i = 1, 2. Given matrices M = (m ij ), N = (n ij ) ∈ M2 (ℤ), we define ⟨M|N⟩ = ∑ m ij n ij

and

where n22 N󸀠 = ( −n12

J(M, N) = ⟨M|N 󸀠 ⟩ , −n21 ). n11

Then det(A) = det(x1 ) − α Δ det(x2 ) + (J(x1 , x2 ) + β Δ det(x2 ))ω Δ .

18.2 Bianchi groups |

125

Therefore det(x1 ) − α Δ det(x2 ) = 1 and J(x1 , x2 ) + β Δ det(x2 ) = 0. Moreover ‖𝛾‖2 = ‖x1 ‖2 + α Δ ‖x2 ‖2 + ⟨x1 |x2 ⟩ =

1 (‖2x1 + x2 ‖2 + (4α Δ − 1)‖x2 ‖2 ). 4

By (15.3.10), this is an integer greater than or equal to 2. Hence, we can layer the elements of Γ by non-negative integers n so that if 𝛾 is parametrized with x1 , x2 ∈ M2 (ℤ), as above and ‖𝛾‖2 = 2 + n then the following equations are satisfied: 4‖𝛾‖2 = ‖2x1 + x2 ‖2 + (4α Δ − 1)‖x2 ‖2 = 8 + 4n,

(18.2.3)

det(x1 ) − α Δ det(x2 ) = 1,

(18.2.4)

J(x1 , x2 ) + β Δ det(x2 ) = 0,

(18.2.5)

Therefore, when applying the DAFC to Γ, it is convenient to take the sequence (r n = 2 + n). The following lemma describes Γ j . Recall that t, s and u are the elements of Γ represented by the matrices T, S and U Δ given in (18.2.1) and r is the element represented by the matrix R of (18.2.2). Clearly ⟨t, r⟩ ≅ C2 × C2 and ⟨s, u⟩ is free of rank 2. Lemma 18.2.4. {⟨t, r⟩ ; if Δ = 1 or Δ = 3; Γj = { otherwise. ⟨t⟩ ; { Proof. Let 𝛾 ∈ Γ j be represented by x1 and x2 . By Proposition 15.3.4, x1 and x2 satisfy (18.2.3), (18.2.4) and (18.2.5) for n = 0. If x2 = 0 then two entries of x1 are 0 and the other two are ±1. Then 𝛾 is either 1 or t. If Δ ∈ ̸ {1, 2, 3, 7} then there are no more solutions. Suppose that x2 ≠ 0. If Δ is either 2 or 7 then α Δ = 2 and hence (18.2.3) takes the form ‖2x1 + x2 ‖2 + 7‖x2 ‖2 = 8. As x2 ≠ 0, x2 has a unique non-zero entry and 2x1 + x2 = ±x2 . Then x1 has a unique non-zero entry and therefore det(x1 ) = det(x2 ) = 0, in contradiction with (18.2.4). This proves the lemma for Δ ∈ ̸ {1, 3}. If Δ is either 1 or 3 then α Δ = 1 and hence (18.2.3) for n = 0, takes the form ‖2x1 + x2 ‖2 + 3‖x2 ‖2 = 8. Thus either ‖x2 ‖2 = 1 and ‖2x1 + x2 ‖2 = 5 or ‖x2 ‖2 = ‖2x1 + x2 ‖2 = 2. However, in the former case both x1 and x2 have a unique non-zero entry. Then det(x1 ) = det(x2 ) = 0, contradicting (18.2.4). Thus both x2 and 2x1 + x2 have two zero coordinates and two coordinates equal to ±1. Moreover, the zero entries of x2 and 2x1 + x2 have to be located in the same coordinate. Therefore ‖x1 ‖2 ≤ 2 and in the coordinates where the entry of x2 is 0 the entries of x1 are also 0. If ‖x1 ‖2 = 2 then |det(x1 )| = |det(x2 )| in contradiction with (18.2.4). If ‖x1 ‖2 = 1 then J(x1 , x2 ) = ±1 in contradiction with (18.2.5). Thus x1 = 0 and hence det(x2 ) = −1. Thus x2 = ±(1, 0, 0, −1) or x2 = ±(0, 1, 1, 0). Therefore 𝛾 = r or tr. This shows that Γ j = ⟨t, r⟩.

126 | 18 Unit groups of orders in quaternion algebras

Using Lemma 18.2.4 and Proposition 17.5.1 one can easily prove that the following is a fundamental polyhedron for Γ j : θi 3 2 2 {{ρe + rj ∈ ℍ : ρ + r ≥ 1, 0 ≤ θ ≤ π}, { { P = {{ρe θi + rj ∈ ℍ3 : ρ2 + r2 ≥ 1, 0 ≤ θ ≤ 2π 3 }, { { 3 {ℍ \ B(0, 1),

if Δ = 1; if Δ = 3; otherwise.

We are in a position of performing the DAFC with input Γ and P. One can use (18.2.3), (18.2.4), and (18.2.5) to calculate the elements 𝛾 of Γ with a given norm. For example, Table 18.1 describes the elements 𝛾 ∈ Γ for Δ = 19 with ̂ In the first step ‖𝛾‖2 ≤ 48. The DAFC calculates the sets B(N) until its closure covers ℂ.

B(0)

B(0, 1) ∪ {ρe θi + rj ∈ ℍ3 : π < θ < 2π}, { { { = {B(0, 1) ∪ {ρe θi + rj ∈ ℍ3 : 2π 3 < θ < 2π}, { { {B(0, 1),

if Δ = 1; if Δ = 3; otherwise.

If N ≥ 1 then B(N) is the union of B(N−1) with the union of the half-spaces Bis≤ℍ3 (𝛾) with ‖𝛾‖2 ≤ r n . Thus B(0) covers one half of the unit ball if Δ = 1, one third of the unit ball if Δ = 3, and the unit ball otherwise. The half-spaces Bis≤ℍ3 (𝛾) are calculated with ̂ is a half-space of ℂ. ̂ the formulas of Lemma 15.4.4. Its intersection B𝛾 with ℂ For example, assume that Δ = 19 and let ω = ω19 . Then B(0) = B(0, 1) which covers the unit ball B(0, 1) of ℂ. We use Table 18.1 and Lemma 15.4.4 to calculate the sets B(N) . If ‖𝛾‖2 = 3 then B𝛾 is either B s = {(x, y) : x ≥ 12 }, B s−1 = {(x, y) : x ≤ − 12 }, B ts = B(−1, 1) or B ts−1 = B(1, 1). The first two cover everything except the strip − 12 ≤ x ≤ 12 and the last two do not contribute because the part of this strip in B(1, 1) ∪ B(−1, 1) is contained in the unit ball, which was already covered by B(0) . The new bisectors obtained in the next three steps do not contribute, i.e. B(4) = B(1) . However in the fifth step we encounter the bisectors B u , B u−1 , B su−1 and B s−1 u , which are the four halfspaces = {(x, y) : ±(x± √19y) < 5}. We observe in this example some of the symmetries of Proposition 17.5.1. Indeed, Γ is normalized by ρ4 and B ρ4 𝛾ρ−1 is the result of rotating 4 B𝛾 by π radians about 0. Moreover, Γ is closed under the map 𝛾 → 𝛾 and B(𝛾) is the reflection in the line y = 0. Thus B(N) is invariant under reflections in x = 0 and y = 0. This covers a big part of the strip, and it only remains to cover X = H \ B(0, 1), where H is the hexagon {(x, y) : − 12 < x < 12 , −5 < x + √19y < 5, −5 < x − √19y < 5} (see Figure 18.4). A lengthly calculation shows that B𝛾 ∩ X = 0 for every 𝛾 with ‖𝛾‖2 < 18. However, X is covered by B v = B( ω2 , 12 ) and the result of applying the symmetries to this ball (see Figure 18.4). Therefore the DAFC stops here producing the list of elements of Table 18.1. These elements generate a subgroup of finite index in Γ. However, we have observed that to cover ℂ, we only need the B𝛾 ’s with 𝛾 running through the following list: s, s−1 , u, u−1 , su−1 , s−1 u and v, v−1 , tvt, tv−1 t. As Γ j = ⟨t⟩ we deduce that {t, s, u, v} generates a subgroup of finite index in Γ.

18.2 Bianchi groups

| 127

In general, the output of DAFC is a finite set Y Δ of Γ generating a subgroup of finite index Γ. The next Example shows the output obtained in [109] for Δ = 19, 23 and 27. Figure 18.5 represents a Dirichlet fundamental polyhedron of ⟨Y Δ ⟩ in these cases. Tab. 18.1 n

‖γ‖2

γ

0

2

0 1, t = ( −1

1

3

s±1 = (

4

6

(

5

7

7

9

9

11

10

12

11

13

1 0

1 0

1 ) 0 ±1 0 ),( 1 −1

±2 0 ),( 1 −1

1 ±1 ),( ±1 −1

1 ±2 ),( ±2 −1

1 1 ),( 0 ±1

1 1 ),( 0 ±2

0 ) 1

0 ) 1

±ω 0 1 ±ω 1 1 0 ),( ),( ),( ) 1 −1 ±ω −1 0 ±ω 1 1 ±(1 − ω) 0 1 ±(1 − ω) (su−1 )±1 = ( ),( ),( 0 1 −1 ±(1 − ω) −1 1 −2 1 −1 1 −1 1 1 ( ),( ),( ),( ), 1 −1 −1 2 2 −1 −2 −1 1 1 1 2 2 −1 2 1 ( ),( ),( ),( ) 1 2 −1 −1 −1 1 1 1 1 0

u±1 = (

1 1 ),( 0 ±(1 − ω)

±(1 − 2ω) 1 0 1 )( ) −1 0 −1 ±(1 − 2ω) 1 ±(1 − 2ω) 1 0 ( )( ) 0 1 ±(1 − 2ω) 1 1 0 ±(1 + ω) 1 ( )( ) ±(1 + ω) 1 −1 0 1 ±(1 + ω) 0 1 ( )( ) 0 1 −1 ±(1 + ω) (

0 −1

(

1 1 ),( ±3 0

±3 1 ),( 1 ±3

0 ±3 ),( 1 −1

1 ) 0

ω −1 1−ω 1 ω 1 1 − ω −1 )( )( )( ) ω 1 1−ω 1 −ω 1 ω−1 1 1−ω ω ω 1−ω 1 − ω −ω ω ω−1 ( )( )( )( ) −1 1 −1 1 1 1 1 1 1 −1 1 1 1 −1 1 1 ( )( )( )( ) ω 1−ω −ω 1 − ω 1−ω ω −1 + ω ω 1 −1 + ω 1 ω 1 −ω 1 1−ω ( )( )( )( ) 1 ω −1 1 − ω 1 1−ω −1 ω (

±(3 − ω) 1 0 )( −1 0 −1 0 1 1 ( )( −1 ±(2 + ω) 0 (

1 1 ±(3 − ω) 1 )( )( ±(3 − ω) 0 1 ±(3 − ω) ±(2 + ω) 1 0 ±(2 + ω) )( )( 1 ±(2 + ω) 1 1

0 ) 1 −1 ) 0

0 ) 1

128 | 18 Unit groups of orders in quaternion algebras Tab. 18.1 (continued) n

‖γ‖2

13

15

16

18

γ 3 −1 1 ( 1 1 ( 3 1 ( 2 (

−2 3 −1 3 1 3 2 )( )( )( ) 1 −2 1 2 1 1 1 −3 1 3 2 −3 2 3 )( )( )( ) −2 −1 −2 1 −1 −1 −1 −1 1 1 2 −1 2 1 )( )( )( ) −2 −3 −2 3 −1 −3 −1 1 1 2 1 −2 1 −1 )( )( )( ) 3 1 3 −1 3 −2 3

ω 2 1−ω 2 ω −2 1 − ω −2 )( )( )( ) 2 1−ω 2 ω −2 1 − ω −2 ω 2 ω 2 −ω 2 1−ω 2 −1 + ω ( )( )( )( ) −1 + ω −2 1 − ω −2 −ω −2 ω −2 0 1 1 ±4 1 0 ±4 −1 ( )( )( )( ) −1 ±4 0 1 ±4 1 1 0 v=(

s− u u v

v−

s−

s

t

tv − t

tvt u− su Fig. 18.4

18.2 Bianchi groups

Example 18.2.5. Let Δ ∈ {19, 23, 27}, let w Δ = (18.2.1). Let 1−w { { { {T, S, U Δ , ( { { { 2 { { { { { { −2 + w Y Δ = {{T, S, U Δ , ( { −1 − w { { { { { { 2 { { { {{T, S, U Δ , ( 1 − w {

2 w

1+√−Δ 2

and let T, S and U Δ be as in

if Δ = 19;

)}, 3

),(

−3 −w −3

| 129

−3 + w

2+w

−2 − w

−3 + w

)},

if Δ = 23; if Δ = 27;

)},

Then ⟨Y Δ ⟩ generates a subgroup of finite index of SL2 (ℤ[ω Δ ]). Observe that ∞ is the only ideal point of the polyhedra in Figure 18.5. By Proposition 18.1.2, Γ is not cocompact and therefore, in these cases, one can use Remark 17.4.7 to obtain a finite generating set and a fundamental polyhedron of Γ. We are going to indicate how to do this for Δ = 19. We start calculating the vertices of the polyhedron Q = D(16) which are 2 ), (± 12 , ± 2√719 , √ 19

(a) Δ = 19

7 (± 12 , ± 2√919 , √ 38 ),

7 (0, ± √519 , √ 38 ),

3 (0, ± √419 , √ 19 ).

(b) Δ = 23

Fig. 18.5: Fundamental polyhedron of Y Δ for Δ = 19, 23 and 27. (c) Δ = 27

130 | 18 Unit groups of orders in quaternion algebras

The maximum of the distances of these points to j is smaller than 2.15. We have 2 cosh(4.3) < 74, and we have taken the sequence r n = 2 + n. Therefore, by Remark 17.4.7, D(72) is a fundamental polyhedron of Γ and G(72) generates Γ. It remains to solve the equations (18.2.3), (18.2.4) and (18.2.5) for every n ≥ 72 and calculate the bisectors Bisℍ3 (𝛾) with the formulas in Proposition 15.4.4 (Problem 18.2.2). We finish this section explaining how one can organize the calculations of the DAFC for a congruence subgroup of SL2 (O). Let m ≥ 2 and let G = PSL2 (O, m), the image of SL2 (O, mℤ(Δ)) in PSL2 (O). By Lemma 18.2.4, G j = 1. We have to adapt slightly the considerations for PSL2 (ℤ, m) given at the end of Section 17.4 Every element of SL2 (O, m) is of the form A = I + mB,

with B = (

a c

b ) ∈ M2 (O). d

and 1 = det(A) = 1 + m(tr(B) + m det(B)) and therefore SL2 (O, m) = {I + mB : B ∈ M2 (ℤ), tr(B) + m det(B) = 0}. If a ∈ O then a + a = TrK/ℚ (a) and ‖a‖2 = NrK/ℚ (a) and these are integers. Therefore n = ‖B‖2 − TrK/ℚ (det(B)) = ‖B‖2 +

TrK/ℚ (tr(B)) m

is an integer and 2 ≤ ‖A‖2 = 2 + m2 n. Thus, n ≥ 0 and hence, when applying the DAFC, we may consider the sequence r n = 2 + m2 n, with n ≥ 0 and to run the algorithm for n we have to solve the equations tr(B) + m det(B) = 0, TrK/ℚ (tr(B)) = n. m The elements 𝛾 of PSL2 (O, m) for which the bisector is a vertical plane are those represented by the elementary matrices of the form e12 (mb) with b ∈ O \ {0}. These half-spaces cover the complement in ℂ of an Euclidean polyhedron and, up to finitely many points, one has to cover this polyhedron with the remaining half-spaces. The following theorem collects the output of applying this process to PSL2 (ℤ[i], 2) and PSL2 (ℤ[√−2], 2). We use the notation X ∗ introduced at the end of Section 17.4. ‖B‖2 +

Example 18.2.6 ([109]). Let X1 = {(

1 0

−2 1 ), ( 1 0

−1 + 2i 2

(

−2i −1 ), ( 1 −2i

2 −1 − 2i ), ( −1 − 2i −2i

0 −1 ), ( −1 2 2i )} . −1 + 2i

0 ), −1

18.2 Bianchi groups

1.0

0.5

–1.0

–0.5

0.5

1.0

–0.5

–1.0 Fig. 18.6: Dirichlet fundamental polyhedron of Γ1 .

1.5 1.0 0.5

–1.5

–1.0

–0.5

0.5

1.0

1.5

–0.5 –1.0 –1.5 Fig. 18.7: Dirichlet fundamental polyhedron of Γ2 .

and 1 X2 = {( 0 (

−2 1 ), ( 1 0

−3 −2√−2

1 − 2√−2 2√−2 ), ( −2√−2 −3

−1 + 2√−2 4

(

−2√−2 −1 ), ( 1 2

0 −1 ), ( −1 −2√−2

0 ), −1

2√−2 ), −1 − √−2

2 −1 + 2√−2 ), ( −1 − √−2 2

4 ), −1 − √−2

| 131

132 | 18 Unit groups of orders in quaternion algebras −3 + 2√−2 4

−3 − 2√−2 4 ), ( −3 − 2√−2 −4√−2

−3 − 2√−2 −2√−2

5 − 2√−2 4√−2 ), ( −3 + 2√−2 −4√−2

(

(

2√−2 ), −3 + 2√−2 4√−2 )} 5 + 2√−2

(1) The group Γ1 = ⟨X1∗ ⟩ is of finite index in SL2 (ℤ[i]). The fundamental Dirichlet polyhedron of the image of Γ1 in PSL2 (ℂ) is (Figure 18.6) {z + rj = x + yi + rj : |x|, |y| ≤ 1, |z − a|2 + r2 ≥ 14 , for all a ∈ {± 12 , ± 2i , ±1 ± 2i , ± 12 ± i}}. (2) The group Γ2 = ⟨X2∗ ⟩ is of finite index in SL2 (ℤ[√−2]). The fundamental Dirichlet polyhedron of the image of Γ2 in PSL2 (ℂ) is (Figure 18.7) {z + rj = x + yi + rj : |x| ≤ 1, |y| ≤ √2, ‖z − a‖2 + r2 ≥ R2 , for all (R, a) as in the table below }. R

a

1 2

± 12 + a2 √2

√2 4 1 4 √2 8

a = a1 +

+

a2 ∈ {0, ±1} a1 ∈ {0, ±1}, a2 ∈ {±1, ±3}

√2 2

a1 ∈ {±1, ±3}

a2 √82

a2 ∈ {±3, ±5}

a1 14 ± ± 12

a2 √42

Problems 18.2.1. Let Γ = PSL2 (ℤ[ζ3 ]) and let Γ j and Γ∞ be the stabilizers in Γ of j and ∞ respectively. Prove that (1) Γ j ∩ Γ∞ = ⟨r⟩ and D j,∞ = {x + yi + rj : x ≥ 0, x + √3y ≥ 0} is a fundamental polyhedron of Γ j ∩ Γ∞ . (2) D j = D j,∞ ∩ Σ r is a Ford fundamental polyhedron of Γ j . (3) The Ford fundamental polyhedron F = D∞ ∩ I t≥ of Γ given in this section is also the Dirichlet fundamental polyhedron centered at j obtained from D j . 18.2.2. Complete the DACD for PSL2 (ℤ[√−19]), as indicated in this section, and give a fundamental polyhedron and generators for this group. 18.2.3. Calculate presentations of the groups ⟨Y Δ ⟩ of Example 18.2.5 and of the groups Γ1 and Γ2 of Theorem 18.2.6. 18.2.4. For i = 1, 2, let Q1 and Q2 be the fundamental polyhedra Q i of the group ⟨X ∗ ⟩i of Example 18.2.6. Calculate the set P i of ideal points of Q i . Prove that Q1 is the convex hull of P1 . Prove that Q1 is also the Ford fundamental polyhedron of PSL2 (ℤ[i], 2). Deduce that PSL2 (ℤ[i], 2) = ⟨X ∗ ⟩1 .

18.3 Calculating fundamental polyhedra for orders in division algebras of Kleinian type | 133

18.3 Calculating fundamental polyhedra for orders in division algebras of Kleinian type In this section we present three examples of application of the DACD (Algorithm 17.4.3) to calculate a presentation of PSL1 (O) or SL1 (O) for some order O in a division algebra D of Kleinian type . In the first example the center of D is ℚ. Example 18.3.1. PSL1 ( −1,3 ℤ ) = ⟨i, 2 + j, 2 + k, 2 + 3i + 2j, 2 + 3i − 2j, 2 + 3i + 2k⟩. −1,3 Proof. Let D = ( −1,3 ℚ ) and let O = ( ℤ ). Because of Example 2.1.7 (3) one easily verifies that D is a division algebra. We consider D as a subalgebra of M2 (ℚ(√3)) via the identification: √3 0 0 1 i=( ). ) and j = ( √ 0 − 3 −1 0

Let G = PSL1 (O). Every element of 𝛾 ∈ O is represented by an element of the form a = x0 + x1 i + x2 j + x3 k, with x0 , x1 , x2 , x3 ∈ ℤ. Then, RNrD/ℚ (a)2 = x20 + x21 − 3(x22 + x23 ) and ‖𝛾‖2 = 2(x20 + x21 + 3(x22 + 3x23 )). We associate to 𝛾 an integer n such that ‖𝛾‖2 = 2 + 12n, x22 + x23 = n, x20 + x21 = 1 + 3n. In Table 18.2 we describe the elements 𝛾 ∈ G with ‖𝛾‖2 ≤ 200: Tab. 18.2 n

‖γ‖2

0 1 4 5 8 13 16

r0 r1 r2 r3 r4 r5 r6

=2 = 14 = 50 = 62 = 98 = 158 = 194

{|x0 |, |x1 |}

{|x2 |, |x3 |}

{0, 1} {0, 2} {2, 3} {0, 4} {5, 0}, {3, 4} {2, 6} {7, 0}

{0, 0} {0, 1} {0, 2} {1, 2} {2, 2} {3, 2} {4, 0}

134 | 18 Unit groups of orders in quaternion algebras Observe that there are no elements in G with n = 2 or 3 because 3 and 7 are not sums of two squares. Thus G(0) = G i = ⟨i⟩ and it is easy to see that P = B(0, 1) ∩ ℍ2 is a fundamental polyhedron for G i . Now we apply DACD (Algorithm 17.4.3) with input G and P. As ‖𝛾‖2 = 2 + 12n we can select r n = 2 + 12n. However, we have observed that some of these values are not realized as ‖𝛾‖2 . So we select the first four r n ’s as in Table 18.2. One starts the algorithm ̂2 \ P. with D(0) = P and B(0) = ℝ Then we calculate the bisectors Bisℍ2 (𝛾) with ‖𝛾‖2 = r1 = 14. Using Table 18.2 we obtain that these 𝛾’s are the ones represented by the elements ±2 ± j,

±2 ± k,

±2i ± j,

±2i ± k.

If two of these elements differ up to a change of sign or up to multiplication by i on the right then they have the same bisector. Thus it is enough to consider the elements r = 2 − j,

r−1 = 2 + j,

s = 2 − k,

s−1 = 2 + k.

This gives four elements which are represented by the four matrices (

2 ± √3 0

0 ), 2 ∓ √3

(

2 ±√3

±√3 ). 2

Making use of the formulas in Lemma 15.4.4 we obtain the following four bisectors S(0, 2 ± √3), Thus

√3 2 S (± √3, ). 3 3

√3 √3 2 2 B(1) = B(0, 2 − √3) ∪ B ( √3, ) ∪ B (− √3, ) \ B(0, 1) 3 3 3 3

and hence ∂ℍ2 \ B(1) = [−

√3 √3 , √3 − 2] ∪ [ , 2 − √3] . 3 3

(See Figure 18.8. Observe that the bisector S(0, 2 + √3) does not contribute in the calculation of B(1) .) As ∂ℍ2 ⊈ B(1) , the algorithm continues. We now repeat the same procedure to calculate the bisectors Bisℍ2 (𝛾) for ‖𝛾‖2 = r2 = 50 and it is enough to consider the following eight elements 2 ± 3i ± 2j

and

2 ± 3i ± 2k.

Table 18.3 displays the center P and radius R of the bisectors obtained so far. We have that D(2) is the difference between P = ℍ3 ∩ B(0, 1) and the union of the other seven open balls displayed in Figure 18.8.

18.3 Calculating fundamental polyhedra for orders in division algebras of Kleinian type |

135

Tab. 18.3 γ r −1 r

2+j 2−j

s−1

2+k

s t −1 u−1 u t v −1 v w w −1

P

≈P

0 0

0 0

3 − 2√ 3

−1.1547

2√3 3

2−k 2 + 3i + 2j 2 − 3i + 2j 2 + 3i − 2j 2 − 3i − 2j 2 + 3i + 2k 2 − 3i − 2k 2 + 3i − 2k 2 − 3i + 2k

1.1547

− 34 (1 + √3) 3 √ 4 (1 + 3) 3 √ ( 3 − 1) 4 − 34 (√3 − 1) − 23 (2√3 − 3) 2 √ 3 (2 3 − 3) 2 √ 3 (2 3 + 1) 2 √ − 3 (2 3 + 1)

−2.04904 2.04904 0.549038 −0.549038 −0.309401 0.309401 4.3094 −4.3094

R2

≈R

7 + 4√3 7 − 4√ 3

3.73134 0.267953

1 3 1 3

0.577358

1 √ 8 (2 + 3) 1 √ 8 (2 + 3) 1 √3) (2 − 8 1 √3) (2 − 8 1 √ 3 (7 − 4 3) 1 √ 3 (7 − 4 3) 1 √3) (7 + 4 3 1 √3) (7 + 4 3

0.683022 0.683022 0.183015 0.183015 0.154703 0.154703 2.15472 2.15472

s−1

0.577358

s v −1 t

r v u

Fig. 18.8: Fundamental polyhedron of PSL1 ( −1,3 ℤ ).

Observe that B(2) contains ∂ℍ2 . Therefore the second step of the DACD has been completed. Next we have to calculate the set V of vertices of D(2) . By a straightforward calculation we obtain 1 1 V = { (±√3, 1), (±(8√3 − 6), 4 − √3) , 2 13 1 1 (±(10√3 − 12), 5 − 2√3) , (±(2√3 − 3), 2 − √3)} 13 2 Calculating the distances d(i, v) for all the elements v ∈ V we obtain that the max1 imum value is obtained for v = 13 (±(10√3 − 12), 5 − 2√3) and if r = d(i, v) then cosh(r) = 5. Then cosh(2r) = 4 cosh2 r − 2 = 196 and according to the DACD we also need to calculate G(m) and D(m) for m = 3, 4, 5, 6 (see Table 18.2). Now a computer aided calculation shows that all the vertices belong to Bis≥ℍ2 (𝛾) for every 𝛾 ∈ G(6) . This is visualized in Figure 18.9 which represents some of the bisectors occurring in this calculation. Some of the bisectors have not been represented to avoid a cumbersome figure.

136 | 18 Unit groups of orders in quaternion algebras

Fig. 18.9: Fundamental polyhedron of PSL1 ( −1,3 ℤ ).

This finishes the algorithm concluding that D = D(2) is a fundamental polyhedron of PSL1 ( −1,3 ℤ ). By Remark 17.4.6 we have PSL1 (

−1, 3 ) = ⟨i, r, s, t, u, v⟩ . ℤ

This also shows that −1, 3 SL1 ( ) = ⟨i, 2 + j, 2 + k, 2 + 3i + 2j, 2 + 3i − 2j, 2 + 3i + 2k⟩ . ℤ For the second example D is a Hamiltonian quaternion algebra over a quadratic imaginary extension of ℚ. We start with considering a generic example of this form. Let d be a square free positive integer, let R be the ring of integers of K = ℚ(√−d) and let D = ℍ(ℚ√−d)). We assume that D is a division algebra. Because of Theorem 4.5.11, this is equivalent with d ≡ −1 mod 8. From Example 4.1.9, we know that √ R = ℤ[ω] with ω = 1+ 2 −d . Let N = RNrD/K , G = SL1 (ℍ(R)) = {x ∈ ℍ(R) : N(x) = 1}, and Γ the image of G in PSL1 (ℍ(R)). We write every element u ∈ ℍ(K) as u = u0 + u1 i + u2 j + u3 k, with u t = x t + y t w and all x t , y t ∈ ℚ. Put x = (x0 , x1 , x2 , x3 )

and

y = (y0 , y1 , y2 , y3 ).

So u is determined by the vector (x, y) and the elements of ℍ(R) are those with x, y ∈ ℤ4 . Moreover ω2 = ω − 1+d 4 . It follows that u2i = x2i − y2

1+d + (y2 + 2xy)ω. 4

Hence N(u) = u20 + u21 + u22 + u23 = ‖x‖2 − ‖y‖2 and therefore

1+d + (‖y‖2 + 2 ⟨x|y⟩)ω 4

x, y ∈ ℤ4 , { { 2 u ∈ G if and only if {‖x‖ = 1 + ‖y‖2 1+d 4 and { 2 ‖y‖ + 2 = 0. ⟨x|y⟩ {

(18.3.1)

18.3 Calculating fundamental polyhedra for orders in division algebras of Kleinian type | 137

In order to see G as a discrete group of SL2 (ℂ) we consider the algebra embedding ℍ(K) 󳨀→ M2 (ℂ) u0 + u1 i u = u0 + u1 i + u2 j + u3 k 󳨃→ 𝛾u = ( −u2 + u3 i

(18.3.2) u2 + u3 i ). u0 − u1 i

A straightforward calculation yields ‖𝛾u ‖2 = |u0 + u1 i|2 + |u0 − u1 i|2 + |u2 + u3 i|2 + |u2 − u3 i|2 = 2‖x‖2 + 2 ⟨x|y⟩ +

1+d 2 ‖y‖ 2

If u ∈ G then, by (18.3.1), this takes the form ‖𝛾u ‖2 = 2 + d‖y‖2 = 2(1 − ⟨x|y⟩).

(18.3.3)

Thus, ‖y‖2 is even and hence a good choice for the sequence (r n ) is r n = 2(1 + nd) such that ‖y‖2 = 2n, ‖x‖2 = 1 +

1+d n, 4

⟨x|y⟩ = n, ‖𝛾‖2 = 2(1 + nd). We now suppose d = 7 and specialize the previous calculations to this case (the cases d = 15 and d = 23 have been studied in [109]). Table 18.4 can be used to calculate the elements 𝛾 ∈ Γ with ‖𝛾‖2 ≤ 30. In particular, G j = {±1, ±i, ±j, ±k} = Q8 . Observe that 𝛾i and 𝛾j are the transformations ρ4 and t of Proposition 17.5.1. Using this proposition it easily follows that P = {x + yi + rj ∈ ℍ3 ∩ B(0, 1) : y > 0} is a fundamental polyhedron of Γ j . Tab. 18.4 n

‖γ‖2

{u0 , u1 , u2 , u3 }

0 1 2

r0 = 2 r1 = 16 r 2 = 30

{1, 0, 0, 0} {±(1 + ω), ±1, ±1, ±ω} {±(1 + 2w), ±2, 0, 0}

138 | 18 Unit groups of orders in quaternion algebras

Therefore we are in position to apply the DACD (Algorithm 17.4.3) with input Γ and P. Observe that Γ is invariant under the transformations 𝛾 󳨃→ 𝛾 and 𝛾 󳨃→ 𝛾󸀠 of Proposition 17.5.1. Hence ⋂𝛾∈Γ j Bis≥ℍ3 (𝛾) is invariant under reflections in the planes x = 0, y = 0 and the unit sphere. After performing the algorithm for n ≤ 2, i.e. calculating the bisectors of the elements represented in Table 18.4, we obtain the polyhedron D(2) . It is the represented in Figure 18.10, where the big ball is part of the unit ball and the smaller balls represent the bisectors Bisℍ3 (𝛾) with 𝛾 running thought the elements in (18.3.4). These bisectors can be calculated using Lemma 15.4.4. Hence D(2) is formed by the elements in P outside the union of the small balls in Figure 18.10. This completes the first part of the DACD, since D(2) is compact. Completing the calculations (i.e. calculating the maximum distance r from i to the vertices and calculating the bisectors of the elements 𝛾 ∈ Γ with ‖𝛾‖2 ≤ 2 cosh(2r)) shows that each vertex belongs to Bis≥ℍ3 (𝛾) for every 𝛾. Thus D(2) is a fundamental polyhedron of Γ, and hence the elements in Theorem 18.3.2 generate Γ. Applying Poincaré’s Theorem, one obtains a presentation of Γ, from which one can obtain the presentation of G given in Theorem 18.3.2. In the concrete realization of the presentation of SL1 (ℍ(R)), P, Q and J represent i, j and −1, respectively. The interested reader can consult [45] for further details.

Ch Wl Dl

Vl Sl

Wr

Vr C

Sr

Dr

Fig. 18.10: Fundamental polyhedron for −7 PSL1 (ℍ(ℤ + ℤ[ 1+√ 2 ])) and its orthogonal projection. The symbols are chosen to remind an orchestra: Conductor, Soloists, Violins, Drums, Wings, Chorus.

18.3 Calculating fundamental polyhedra for orders in division algebras of Kleinian type |

139

Example 18.3.2. Let ω = 1+√2 −7 and let R = ℤ[ 1+√2 −7 ]. Then SL1 (ℍ(R)) is generated by the following elements: C l = (2 + i − w + iw)j,

C r = (1 − 2i + w + iw)j,

S l = iw + (−2 + i − iw)j,

S r = −i + iw + (2 + iw)j,

V l = −i + iw + (2i − w)j,

V r = iw + (1 + 2i − w)j,

W l = −2i + (1 − w + iw)j,

W r = −2i + (−i + w + iw)j,

Ch l = −i − iw + (−2i + iw)j,

Ch r = 2i − iw − (i + iw)j,

D l = 2i − iw + (1 + w)j,

(18.3.4)

D r = i + iw + (−2 + w)j.

The group SL1 (ℍ(R)) has a presentation given by the generators J, P, R, C l , C r , S l , S r , V l , V r , W l , W r , Ch l , Ch r , D l , D r and the relations P2 = R2 = C2l = C2r = S2l = S2r = V l2 = V r2 = W l2 = W r2 = Ch2l = Ch2r = D2l = D2r = (S l PS r P)2 = (W l RW r R)2 = (V l V r )2 = (PR)2 = J, S l C l V l Ch l W l D l = S r C r V r Ch r W r D r = PD r RD l = V l W l S l = S r W r V r = C r PC l = Ch r RCh l = J 2 = 1. In the third example D is a non-Hamiltonian quaternion algebra over a quadratic imaginary extension of ℚ. More precisely, we study the group G = PSL1 (

2, 5 ) ℤ[ζ4 ]

along the lines used in [109]. This group also has been investigated in [54, Section X, page 474]. Put ξ = ζ4 , R = ℤ[ξ], F = ℚ(ξ) and D = ( 2,5 F ). (Caution: we do not denote ζ 4 as i because we reserve this symbol for one of the basis elements. So i2 = 2, j2 = 5 and 2 2 2 ξ 2 = −1.) Observe that D = ( 2,5 F ) is a division ring because the equation 2x + 5y = z does not have non-trivial solutions in F. Indeed, assume that (x, y, z) is a non-zero solution. Since √2, √5, √10 ∈ ̸ F, we have xyz ≠ 0. Therefore we may assume that x, y, z ∈ ℤ[ξ]. Then 2x2 ≡ z2 mod 5ℤ[ξ]. Using the isomorphisms ℤ[ξ]/5ℤ[ξ] ≅ (ℤ[ξ]/(1 + 2ξ)ℤ[ξ]) × (ℤ[ξ]/(1 − 2ξ)ℤ[ξ]) ≅ ℤ/5ℤ × ℤ/5ℤ, we deduce that 2x2 = z2 mod 5, which yields a contradiction because 2 is a not a square modulo 5. Consider the embedding (

a, b ) 󳨀→ M2 (ℂ) F

u = u0 + u1 i + u2 j + u3 k 󳨃→ 𝛾u = (

u 0 + u 1 √2 u2 √5 − u3 √10

u2 √5 + u3 √10 ). u0 − u1 √2

140 | 18 Unit groups of orders in quaternion algebras 2,5 As in Example 18.3.2, each u ∈ ( ℤ[ξ] ) is of the form u = u0 + u1 i + u2 j + u3 k with each u t = x t + y t ξ for

x = (x0 , x1 , x2 , x3 ),

y = (y0 , y1 , y2 , y3 ) ∈ ℤ4 .

Consider the bilinear form B((x0 , x1 , x2 , x3 ), (y0 , y1 , y2 , y3 )) = x0 y0 − 2x1 y1 − 5x2 y2 + 10x3 y3 and the corresponding quadratic form q(x) = B(x, x). Then q = RNrD/F and for u as above we have q(u) = q(x) − q(y) + 2B(x, y)ξ. Each element 𝛾 of G is hence represented by x, y ∈ ℤ4 satisfying q(x) = 1 + q(y)

and

B(x, y) = 0.

(18.3.5)

Moreover, we set r1 (a, b) = a2 + 10b2 ,

r2 (a, b) = 2a2 + 5b2 ,

and r(a, b, c, d) = r1 (a, d) + r2 (b, c). Then ‖𝛾‖2 = 2(r(x) + r(y)). As r(x) + r(y) ≡ q(x) − q(y) = 1 mod 2, we have that n = integer and we have

‖𝛾‖2 −2 4

is a non-negative

r1 (x0 , x3 ) + r1 (y0 , y3 ) − r2 (x1 , x2 ) − r2 (y1 , y2 ) = 1, r1 (x0 , x3 ) + r1 (y0 , y3 ) + r2 (x1 , x2 ) + r2 (y1 , y2 ) = 1 + 2n Thus we can layer the elements of G by non-negative integers n such that if 𝛾 is represented by the tuples x and y then x20 + y20 + 10(x23 + y23 ) = 1 + n, 2(x21 + y21 ) + 5(x22 + y22 ) = n, x0 y0 − 2x1 y1 − 5x2 y2 + 10x3 y3 = 0, ‖𝛾‖2 = 2 + 4n. This can be used to calculate the elements of G of small norm (see Table 18.5). It follows that G j = 1. Therefore, we can apply DACD with P = ℍ3 . Observe that the element ρ8 represented by diag(ζ8 , ζ8 ) belongs to D and G is normalized by ρ8 . Moreover, G is invariant under the transformations 𝛾 󳨃→ 𝛾 and 𝛾 󳨃→ 𝛾󸀠 . By Proposition 17.5.1, D(N) is invariant under a rotation of 2π radians and under reflections in

18.3 Calculating fundamental polyhedra for orders in division algebras of Kleinian type |

Tab. 18.5 n

‖γ‖2

{|x0 |, |y0 |}

{|x1 |, |y1 |}

{|x2 |, |y1 |}

{|x3 |, |y3 |}

0 4 8

r0 = 2 r1 = 18 r2 = 34

{1, 0} {1, 2} {3, 0}

{0, 0} {1, 1} {2, 0}

{0, 0} {0, 0} {0, 0}

{0, 0} {0, 0} {0, 0}

3

2

1

–3

–2

–1

1

2

3

–1

–2

–3

2,5 Fig. 18.11: Fundamental polyhedron of the subgroup of finite index in PSL1 ( ℤ[i] ) from Theorem 18.3.3.

141

142 | 18 Unit groups of orders in quaternion algebras the planes x = 0 and y = 0. After lengthy calculations one obtains a compact polyhedron D(N) after intersecting all the half-spaces Bis≥ℍ3 (𝛾) with ‖𝛾‖2 ≤ 406. The resulting polyhedron and the bases of the contributing bisectors is displayed in Figure 18.11. As a consequence, one obtains the following example which provides generators for a subgroup of finite index of G. The example and the fundamental polyhedron obtained 2,5 reflect the fact that PSL1 ( ℚ(ξ) ) contains ρ8 and t and it is closed under the map 𝛾 󳨃→ 𝛾 (see Proposition 17.5.1). 2,5 Example 18.3.3 ([109]). Let Y be the subset of ( ℤ[ζ ) consisting of the elements 4]

ζ4 − ζ4 i,

2ζ4 − ζ4 j,

3ζ4 + k,

3 + ζ4 i − ζ4 k,

3 − 3i + k,

2 − ζ4 i − j,

2 + ζ4 i − ζ4 j − ζ4 k,

2 − 3i + ζ4 j + k,

2 + 4ζ4 i − ζ4 j − 2ζ4 k,

2 − 2i + ζ4 j, 4ζ4 − ζ4 i − ζ4 j + k,

3 − ζ4 i − 2j − k,

6 − 3j − k,

6 − ζ4 j − 2ζ4 k,

2 − 4ζ4 i − 3j − k,

(1 − 4ζ4 ) + (4 + 2ζ4 )i + (2 + 2ζ4 )j + 2k, (1 + 4ζ4 ) + (4 − 2ζ4 )i + (−2 − 2ζ4 )j − 2k,

3 − 7i + 4ζ4 j + k,

(8 + 3ζ4 ) + (−4 + 2ζ4 )i + (−2 − 2ζ4 )j + (1 − 2ζ4 )k. Then ⟨t, ρ8 , g, g | g ∈ Y⟩ 2,5 is a subgroup of finite index in SL1 ( ℤ[ζ ). 4]

Problems 18.3.1. Calculate a presentation of PSL2 ( −1,3 ℤ ) using the fundamental polyhedron obtained in Example 18.3.1

18.4 Generators of unit groups of group rings with exceptional components For many finite groups G it was shown in Chapter 11 that the Bass units together with the bicyclic units generate a subgroup of finite index in U(ℤG). The restrictions on the group G are on the existence of nonabelian fixed point free epimorphic images and the existence of exceptional simple components ℚGe (with e a primitive central idempotent). More precisely, it is shown in the proof of Theorem 11.3.2 that for each simple component ℚGe (with e a primitive central idempotent) of ℚG, the group generated by the Bass units and the bicyclic units contains a subgroup of finite index in each 1 − e + U((ℤG)e) (note that ℤGe is an order in ℚGe), provided the group Ge is not-fixed point free when Ge is nonabelian and ℚGe is not exceptional. Actually,

18.4 Generators of unit groups of group rings with exceptional components |

143

the group generated by the bicyclic units contains a subgroup of finite index in the reduced norm one subgroup of the order ℤGe. If Ge is nonabelian and fixed point free and ℚGe is not exceptional then one has to add a collection of generalized bicyclic units (see Corollary 11.2.6). In case G is nilpotent the extra needed generalized bicyclic have been precisely described in Theorem 11.5.4. Because of the results in the previous sections one can, on a case by case basis, also deal with exceptional simple components provided they are of Kleinian type. We explain this in case the finite group G is nilpotent (and thus G is strongly monomial by Theorem 3.5.10). Recall that in this case, the group generated by the Bass units contains a subgroup of finite index in Z(U(ℤG)). Actually from Corollary 13.1.4 we know that the central subgroup ⟨c(b) : b a Bass unit in ℤG⟩ is of finite index in Z(U(ℤG)). Therefore, using the ideas from Chapter 11, to obtain a subgroup of finite index in ℤG we need units in U(ℤG) which generate subgroups of finite index in each 1 + SL1 (Oe ), where e runs through the primitive central idempotents of ℚG. If neither Ge is nonabelian fixed point free nor ℚGe is exceptional then the bicyclic units will deliver a subgroup of finite index in 1 + SL1 (Oe ). In Theorem 11.5.4 we have seen how to deal with the case that Ge is nonabelian fixed point free and ℚGe is not a division ring. A characterization of the exceptional simple components of ℚG via epimorphic images of G is given in Corollary 12.1.2 (components that are matrices over fields), Corollary 12.2.2 (components that are matrices over a rational quaternion division algebra; M2 (ℍ(ℚ))) and Corollary 13.5.4 (components that are division algebras of one of the following forms: ℍ(ℚ(ζ2m + ζ2−1 m )) with m ≥ 3 and ℍ(ℚ(ζ n )) with n an odd integer greater than 1 and such that 2 has odd order modulo n). By the mentioned results, if e is a primitive central idempotent of ℚG then ℚGe is an exceptional component of Kleinian type if and only if Ge is isomorphic to one of the groups D8 , D−16 , D+16 , D, D8 × C3 , Q8 × C3 or E, yielding respectively the simple components M2 (ℚ), M2 (ℚ(√−2)), M2 (ℚ(i)), M2 (ℚ(i)), M2 (ℚ(√−3)), M2 (ℚ(√−3)) and M2 (ℚ(i)). Furthermore, the rational group algebra of each of these groups has a unique primitive central idempotent e such that the map g 󳨃→ ge is injective (Problem 18.4.1). Therefore, the map e 󳨃→ N e = ker(g 󳨃→ ge) gives a one-to-one correspondence between the primitive central idempotents e of ℚG such that ℚGe is exceptional of Kleinian type and the normal subgroups N of G such that G/N is isomorphic to one of the groups D8 , D−16 , D+16 , D, D8 × C3 , Q8 × C2 or E. In order to deal with these exceptional components ℚGe we will use the geometric methods explained in the previous sections. Hence, one needs a description of the rational representation of G determined by this component. This has been done in Corollary 13.5.6 by describing a set of matrix units, say E11 , E12 , E21 , E22 in case ℚGe = M2 (K). It is this set of matrix units that will be used to describe a congruence subgroup Γ of level m in an order of type M2 (O) in ℚGe, with m a positive integer such that 1 − e + Γ ⊆ U(ℤG). The existence of this m is a consequence

144 | 18 Unit groups of orders in quaternion algebras

of Lemma 4.6.9 (4). (By the description of the matrix units given in Corollary 13.5.6, it follows that if O is chosen so that M2 (O) ⊆ ℤGe then m = |G| satisfies the desired condition.) The concrete choice of matrix units provides an isomorphism ℚGe ≅ M2 (ℂ) which maps Γ to a discrete subgroup of SL2 (ℂ). The DAFC algorithm allows one to calculate a finite set of generators for such discrete group and hence for Γ. Generators of these groups are the units that are used in part (4) of the following result. The units listed in part (3) are to deal with orders determined by fixed point free groups (see Theorem 11.5.4). So we obtain the following result due to Jespers, Juriaans, Kiefer, de A. e Silva and Souza Filho [109]. Theorem 18.4.1. Let G be a finite nilpotent group. Assume that the rational group algebra ℚG does not have simple components of the type M2 (ℍ(ℚ)) or ℍ(ℚ(ζ p )), with p an odd integer such that the multiplicative order of 2 modulo p is odd (equivalently G does not have epimorphic images of the type Q8 × C p for such p nor any of the groups H with ID(H) ∈ {[32, 8], [32, 44], [32, 50], [64, 37], [64, 137], [129, 937]} (see Corollary 12.2.2). Then the group generated by the following units is of finite index in U(ℤG): (1) c(b), with b a Bass unit in ℤG (see Corollary 13.1.4). (2) the bicylic units in ℤG. (3) the generalized bicyclic units of types b(x, f) and b(f, x) with f running through the set of idempotents F of the form f p,N , where N is a subgroup of G with G/N ≅ Q8 ×C q with q an odd integer such that the order of 2 modulo q is odd and p a prime divisor of q (see (11.5.1)). (4) generators for the groups 1 − e N + Γ N , where N runs through the normal subgroups of G such that N is isomorphic to one of the following groups: D8 , D−16 , D+16 , D, D8 × C3 , Q8 × C2 or E; e N is the unique primitive central idempotent of ℚG such that N is the kernel of the map g 󳨃→ ge N , and Γ N is the congruence subgroup of level m for an order in ℚGe N and a positive integer m such that 1 − e N + Γ N ⊆ U(ℤG). The groups listed in Theorem 18.4.1 (4) are discrete groups of finite covolume (coarea) but they are not cocompact. A set of generators can be calculated using the DAFC. However, for most of them it gives a too long list of generators (see Example 17.4.10). We illustrate Theorem 18.4.1 calculating a presentation of the unit group U(ℤD), where D = ⟨x, y, z | x2 = y2 = z4 = 1, yx = z2 xy, (x, z) = (y, z) = 1⟩, a group of order 16 that is faithfully embedded in the exceptional simple component M2 (ℚ(i)) (see Corollary 12.1.2). This result is due to Pita, del Río and Ruiz [177]. Recall from Proposition 12.5.5 that U(ℤD) = V ⋊ (±D)

with V = U(ℤD) ∩ (1 + ker(ω) (1 − z2 ))

18.4 Generators of unit groups of group rings with exceptional components |

145

with ω the augmentation map of ℤD. Furthermore, the normal complement V is a torsion-free group. Because D󸀠 = ⟨z2 ⟩ and D/D󸀠 is an elementary abelian 2-group, we obtain that ℚDẑ2 ≅ ℚ8 . Moreover, (H, K) = (⟨x, z⟩ , ⟨x⟩) is a strong Shoda pair of G. Applying Theorem 3.5.5 (3), we deduce that ℚGe(G, H, K) ≅ M2 (ℚ(i)). Note that 2 ND (⟨x⟩) = H and hence, this isomorphism associates z 1−z 2 with the scalar matrix i. A dimension count yields that ℚD = ℚ8 × M2 (ℚ(i)) and the primitive central idempotent of the unique non-commutative simple compo2 nent is e(G, H, K) = 1 − ẑ2 = 1−z 2 . Corollary 13.5.6 yields a set of matrix units of M2 (ℚ(i)) that allows to give an explicit isomorphism between ℚGe and M2 (ℚ(i)) and hence yields a concrete description of the normal complement as a group of matrices to which the geometric methods can be applied. Making use of this information we will prove the following result. Proposition 18.4.2 ([177]). Let D = ⟨x, y, z | x 2 = y2 = z4 = 1, yx = z2 xy, (x, z) = (y, z) = 1⟩.

Let V = U(ℤD) ∩ (1 + ker(ω) (1 − z2 )), let W be the subgroup of SL2 (ℤ[i]) formed by the matrices of the form 1 + 2( ac 2b d ) with a, b, c, d ∈ ℤ[i] and let Γ be the image in 2

2

1−z PSL2 (ℂ). Then, there is an isomorphism χ : ℤ ⟨z⟩ 1−z 2 ≅ ℤ[i], with χ(z 2 ) = i and an isomorphism V ≅ Γ induced from the map φ : V → W defined by

φ(1 + (α0 + α1 x + α2 y + α3 xy)(1 − z2 )) = 1 + 2(

α0 − α2 + i(α1 + α3 ) α2 − iα1

2i(α1 + α3 ) ), α0 + α2 − i(α1 + α3 )

2

where each α i ∈ ℤ ⟨z⟩ and α i = χ(α i 1−z 2 ). 2

Proof. Let e = 1−z 2 . We know that there is an isomorphism ℚD e ≅ M 2 (ℚ(i)), which restricts to isomorphism χ : ℚ ⟨c⟩ e ≅ ℚ(i) mapping ce to i. The following elements form a set of matrix units of ℚDe: 1 − ixy e 2 y + ix = e 2

y − ix e 2 1 + ixy = e 2

e11 =

e12 =

e21

e22

Furthermore, xe = i(e12 − e21 ),

ye = e12 + e21

and

xye = i(e11 − e22 ).

Hence, if u = 1 + α(1 − z2 ) with α ∈ ℤD, then u = 1 + 2(α0 + α1 x + α2 y + α3 xy)e = 1 + 2(α0 (e11 + e22 ) + α1 (ie12 − ie21 ) + α2 (e12 + e21 ) + α3 (ie11 − ie22 )) = 1 + 2[(α0 + iα3 )e11 + (iα1 + α2 )e12 + (−iα1 + α2 )e21 + (α0 − iα3 )e22 ],

146 | 18 Unit groups of orders in quaternion algebras where each α i ∈ ℤ ⟨z⟩. Let U2 = U(ℤD) ∩ (1 + ℤD(1 − z2 )). It easily is verified that U2 = ⟨z2 ⟩ × V. The following mapping ψ : U(ℤD) ∩ (1 + ℤD(1 − z2 )) → M2 (ℤ[i]) defined by u = 1 + α(1 − z2 ) 󳨃→ 1 + 2 (

α0 + iα3 iα1 + α2

iα1 + α2 ) α0 − iα3

is injective and preserves multiplication. For u ∈ U2 , define 1 0

φ(u) = (

−1 1 ) ψ(u) ( 1 0

1 ). 1

It follows that φ(u) = 1 + 2 (

α0 + iα1 − α2 + iα3 −iα1 + α2

2(iα1 + iα3 ) ). α0 − iα1 + α2 − iα3

Since U2 is a group, φ(U2 ) ⊆ GL(2, ℤ[i]), i.e. det(φ(u)) ∈ U(ℤ[i]) = ⟨i⟩, for u ∈ U2 . Therefore, φ(U2 ) ⊆ {A = 1 + 2 (

a c

2b ) : a, b, c, d ∈ ℤ[i], det(A) ∈ ⟨i⟩}. d

Moreover, if a, b, c, d ∈ ℤ[i], one has that A = 1 + 2( ac 2b d ) ∈ φ(U2 ) if and only if det(A)) ∈ U(ℤ[i]) and there is a ℤ[i]-solution of the following linear system 1 0 ( 0 1

i i −i −i

−1 0 1 1

i x0 a i x1 b )( ) = ( ). 0 x2 c −i d x3

The latter system has solutions in ℤ[i] if and only if so does 1 0 ( 0 0

i i 0 0

−1 0 1 2

a i x0 x1 b i ). )( ) = ( b+c x2 i x3 d − a + 2b 0

Hence, the system has a solution if and only if a ≡ d mod 2. As det(A) = 1 + 2(a + d) + 4ad − 8bc, one easily verifies that det(A) ∈ U(ℤ[i]) and a ≡ d mod 2 if and only if det(A) = 1. So we have proved that φ(U2 ) is formed by the elements of SL2 (ℤ[i]) of 2 2 −1 0 the form 1 + 2( ac 2b d ). Since φ(c ) = ( 0 −1 ) and because U2 = V × ⟨1, c ⟩, the result follows.

18.4 Generators of unit groups of group rings with exceptional components |

147

By Proposition 18.4.2, to obtain a presentation of U(ℤD) it is enough to find a presentation Γ. We start calculating a Ford fundamental polyhedron of this group. We have Γ∞ = ⟨(

1 0

4 1 ),( 1 0

4i )⟩ 1

and a fundamental polyhedron of Γ∞ is F∞ = {x + yi + zj : |x|, |y| ≤ 2}.

Let X be the set formed by the matrices of the form 1 + 2( ac 2b d ) in W with c = ±1 or 3 ±i. The isometric spheres (in ℍ ) of these matrices are the spheres with − 1+2d 2c and radius 12 . The set of its centers is precisely { 12 , 2i } + ℤ[i]. Figure 18.12 represents the intersection of these spheres with ∂ℍ3 . Let F = ⋂ I𝛾≥ 𝛾∈X

= {z + rj ∈ ℍ3 : |z − P|2 + r2 ≥

1 i 1 , for every P ∈ { , } + ℤ[i]} . 4 2 2

It easily is verified that F is the convex hull of the lattice L of ℂ generated by 1+i 2 and 1−i 2 . Each of the points of this lattice belong to four of the isometric spheres (see Figure 18.12). We claim that the Ford fundamental polyhedron of Γ is F = F∞ ∩ F. To prove this it is enough to show that L does not intersect any of the open balls of ℂ with boundary an isometric circle I𝛾 of some 𝛾 ∈ Γ \ Γ∞ . Indeed, the latter holds because if 1 + 2( ac 2b d ) ∈ W and q ∈ L then 2cq + 1 + 2d ∈ 1 + (1 + i)ℤ[i] ⊆ ℤ[i] \ {0} and hence |2cq + 1 + 2d|2 ≥ 1. As a consequence of Proposition 18.4.2 and after applying Poincaré’s Theorem to the fundamental polyhedron of Γ we obtain the following description, due to Pita, del Río and Ruiz [177]. Example 18.4.3. Let D = ⟨x, y, z | x2 = y2 = z4 = 1, yx = z2 xy⟩. Then U(ℤD) = V ⋊ (±D)

and V is a torsion-free group isomorphic to the group with a presentation given by the generators A0 , A1 , B0 , B1 , C0 , C1 , W0 , W1 , W2 , W3 , X0 , X1 , X2 , X3 , Y0 , Y1 , Y2 , Y3 , Z0 , Z1 , Z2 , Z3

148 | 18 Unit groups of orders in quaternion algebras –2 0 2

–2 0 2

Fig. 18.12: Fundamental polyhedron of Γ .

and subject to the relations (A0 , A1 ) = (B0 , B1 ) C−1 0 Z2 C0 Z0 C−1 1 Z3 C1 Z1

= =

(W0 , X0 ) = (W1 , X1 ) = (W2 , X2 ) = (W3 , X3 ) =

A−1 0 Z2 A0 Z0 A−1 1 Z3 A1 Z1 B1 W1−1 Z0 X0−1 B0 W2−1 Z1 X1−1 B1 W3−1 Z2 X2−1 B0 W0−1 Z3 X3−1

= Y3 Y2 Y1 Y0 = = = = = =

=

A−1 0 Y2 A0 Y0 A−1 1 Y3 A1 Y1

=

X0−1 C1 W1−1 Y0 X1−1 C0 W2−1 Y1

=

X2−1 C1 W3−1 Y2 X3−1 C0 W0−1 Y3

=

= = = 1.

18.4 Generators of unit groups of group rings with exceptional components |

149

The elements in the presentation of Example 18.4.3 are constructed as follows. (Recall that ρ n is the element of PSL2 (ℂ) represented by diag(ζ n , ζ n ).) For 0 ≤ i ≤ 3, put i M i = ρ−i 8 Mρ 8 , where M is represented by a matrix in the following list 1 0

A=(

3 −2

C=(

−4 ) 1 −4 ) 3

−1 − 2i −2

X=( Z=(

1 B=( −2

1 + 4i 2i

0 ) 1

3 − 2i W=( −2i 3 + 4i 2i

4i ) 3 + 2i

Y=(

−4 ) −1 + 2i −12i ) 3 − 4i

−8i ) 1 − 4i

Example 18.4.3 in this section commenced with a reduction to a torsion-free normal complement that is contained in GL2 (ℂ). In [111], Jespers and Leal have done this for an infinite class of non-commutative finite groups, i.e. the computation of the unit group is reduced to subgroups of GL2 (ℂ). As an application, one obtains a description of U(ℤG) in terms of matrix groups for all groups G of order 16 (see also [122]). One can now apply the methods explained in this chapter to obtain a presentation for these unit groups. In [176], Pita and del Río have done this for the group U(ℤD−16 ). For some of this work we refer the reader to the exercises.

Problems 18.4.1. For each of the following groups G prove that the given e is the unique primitive central idempotent e of ℚG for which the map g 󳨃→ ge is injective: (1) G ≅ D8 , D−16 , D+16 , D8 × C3 , Q8 × C3 , and e = ε(A, 1), for A a cyclic subgroup of G of index 2. (2) D = (⟨a⟩2 × ⟨z⟩4 ) ⋊ ⟨b⟩2 , with z a = z b = z, a b = z2 a, and e = e(G, ⟨z, a⟩ , ⟨a⟩). (3) E = (⟨x⟩4 × ⟨y⟩4 ) ⋊ ⟨b⟩2 , with x b = y and e = e(G, ⟨x, y⟩ , ⟨x⟩). 18.4.2. Prove that the group Γ of Proposition 18.4.2 has index 4 in PSL2 (ℤ[i], 2). 18.4.3. Let G be a finite group such that G/Z(G) = C2 × C2 . Prove that if Z(G) is cyclic, say generated by s, then there exists x, y ∈ G such that either G = ⟨x, y, s | x2 = y2 = 1⟩ or G = ⟨x, y, s | x2 = y2 = s⟩. n

18.4.4. Let G = ⟨x, y, s | x2 = y2 = 1, s2 = 1, s central, xy = s2 2n−1

n−1

yx⟩, n > 1. Let

e= . Prove the following properties. n−1 (1) V = {u = 1 + α(1 − s2 ) | α ∈ ω(ℤG), u ∈ U(ℤG)} is a torsion free normal complement of ±G in U(ℤG). (2) e is a primitive central idempotent. 1−s 2

150 | 18 Unit groups of orders in quaternion algebras (3) ℚGe ≅ M2 (ℚ(ζ2n )). (4) Let W denote the subgroup of GL2 (ℤ[ζ2n ])/{I, −I} determined by the matrices A = 1 + 2(

a c

2b ) d

with a, b, c, d ∈ ℤ[ζ2n ] and such that Nrℚ(ζ2n )/ℚ (det(A)) = 1 and 2 | (a + d). Then, the following mapping is a group isomorphism φ : V → W, with φ(1 + (α0 + α1 x + α2 y + α3 xy)(1 − e)) = 1 + 2( each α i ∈ ℤ ⟨s⟩ and α i = α i e.

α0 − α2 + iα1 + iα3 α2 − iα1

2(iα1 + iα3 ) ), α0 + α2 − iα1 − iα3

19 Virtually free-by-free groups Recall that a group is said to satisfy virtually a group theoretical condition if it has a subgroup of finite index satisfying the given condition. For example, G is virtually abelian if and only if G has an abelian subgroup of finite index. The main result of this chapter deals with the following general question: classify the finite groups G for which U(ℤG) virtually satisfies some group theoretical property. For example, by Theorem 1.5.6 and Theorem 14.2.4, if G is a finite group then U(ℤG) is virtually abelian if and only if G is either abelian or a Hamiltonian 2-group. Jespers classified the finite groups G for which U(ℤG) is virtually free nonabelian [102]. Jespers, Leal and del Río classified those for which U(ℤG) is virtually a direct product of free groups [106, 115, 146]. In this chapter we classify the finite groups G for which U(ℤG) is virtually a direct product of free-by-free groups. This is a result of Jespers, Pita, del Río, Ruiz and Zalessskii [125] (earlier results were obtained in [177]). As an application, we also obtain the results mentioned above. These structural results cover all the finite groups G for which the structure of U(ℤG) is known up to finite index. It also fits in with the following philosophy of Kleinert on what a unit theorem should be. Let A be a finite dimensional rational simple algebra. A generic unit group of A is a subgroup of finite index in the group of reduced norm 1 elements of an order in A. Then, according to Kleinert [137], “a unit theorem for A consists of the definition, in purely group theoretical terms, of a class of groups C(A) such that almost all generic unit groups of A are members of C(A)”. This approach has an obvious generalization to finite dimensional semisimple rational algebras, such as the rational group algebra ℚG of a finite group G and its orders, for example ℤG.

19.1 Free-by-free groups Recall that the free product of two groups G1 and G2 is denoted G1 ∗ G2 . In particular, if G1 and G2 are infinite cyclic groups then G1 ∗ G2 is a free group of rank 2. If H is a group and f1 : H → G1 and f2 : H → G2 are group homomorphisms then we denote by G1 ∗H G2 the amalgamated free product, i.e., the group G1 ∗H G2 = (G1 ∗ G2 )/N where N is the normal subgroup of G1 ∗ G2 generated by all the elements f1 (h)f2 (h)−1 with h ∈ H. Let G be a group and let ϕ : H → K be an isomorphism between two subgroups of G. The HNN extension G∗ϕ (relative to ϕ) is the group given by the following presentation G∗ϕ = ⟨G, u | uhu−1 = ϕ(h), h ∈ H⟩. More precisely, a presentation of G is given by taking as generators one symbol [g] for each element g of G and one new symbol u, and as relations [g1 g2 ] = [g1 ][g2 ], for each g1 , g2 ∈ G and u[h] = [ϕ(h)]u for each h ∈ H.

152 | 19 Virtually free-by-free groups

Both groups, amalgamated free products and HNN extensions, can be seen as groups acting on trees (see [48, 209]). The following two propositions are particular cases of a more general result on subgroups acting on trees (see for example [48, Proposition 7.9]). Proposition 19.1.1. Let G = G1 ∗A G2 be an amalgamated free product and let N be a normal subgroup of G. If N ∩ G1 = N ∩ N2 = 1 then N is free. Proposition 19.1.2. Let N be a normal subgroup of an HNN extension H ∗ϕ . If N∩H = {1} then N is a free group. Recall that a group G is said to be free-by-free if G contains a normal subgroup N so that both N and G/N are free groups. As every short exact sequence of groups 1 → N → G → F → 1 with F free is split, a free-by-free group is a semidirect product N ⋊ F where both N and F are free groups. We consider the trivial group as a free group and therefore the class of free-by-free groups contains the class of free groups. From the Nielsen-Schreier Theorem [196, 6.1.1] one knows that subgroups of free groups are again free. It follows that the class of free-by-free groups is closed under taking subgroups. Indeed, let H be a subgroup of a free-by-free group G. Suppose N is a normal subgroup of G such that both N and G/N are free groups. Then, H ∩ N ≤ N and H/H ∩ N ≅ NH/N ≤ G/N. Consequently, by the Nielsen-Schreier Theorem, both H ∩ N and NH/N are free groups. As H is a normal subgroup of N we conclude that H is free-by-free. In contrast with the Nielsen-Schreier Theorem, subgroups of direct product of free groups are of a quite different nature (see for example [19, 161]). Nevertheless, one has the following property. Proposition 19.1.3. If C is a class of groups which is closed under taking subgroups of finite index then the class consisting of the finite direct products of groups in C is closed under taking subgroups of finite index. Proof. Let G = ∏ni=1 G i with each G i in C and let H be a subgroup of finite index in G. Then [G i : G i ∩ H] < ∞ and hence G i ∩ H belongs to C. Moreover [H : ∏ni=1 G i ∩ H] ≤ [G : ∏ni=1 G i ∩ H] = ∏ni=1 [G i : G i ∩ H] < ∞. Corollary 19.1.4. The class of groups consisting of the finite direct products of free groups (respectively, finite direct products of free-by-free groups) is closed under taking subgroups of finite index. In particular, if G1 and G2 are commensurable subgroups of a group and G1 is virtually a direct product of free groups (respectively, free-by-free groups) then so is G2 . We say that a group G is nearly indecomposable if it is not virtually the direct product of two subgroups of infinite order. The following lemma will be needed to prove the main result of this chapter.

19.1 Free-by-free groups |

153

Lemma 19.1.5. If G is a free-by-free group which is nearly indecomposable and not virtually abelian then Z(G) = {1}. Proof. Suppose G = N ⋊ F, with N and F free groups. We first prove that Z(G) ⊆ N. Suppose the contrary. Then Z(F) is not trivial and thus F is a cyclic group. Because ⟨Z(G), N⟩ properly contains N we thus obtain that ⟨Z(G), N⟩ has finite index in G. By assumption, G is not virtually abelian and thus it follows that N is not an abelian group. Hence ⟨Z(G), N⟩ = Z(G)×N, with both Z(G) and N of infinite order. This contradicts the hypothesis. So, indeed, Z(G) ⊆ N. If Z(G) ≠ {1} then N is cyclic and Z(G) × F has finite index in G, again a contradiction. In order to give examples of free-by-free groups the following presentation of the alternating group A4 of degree 4 will be useful. A4 = ⟨a1 , a2 , b | a21 = a22 = (a1 , a2 ) = b3 = 1, ba1 = a2 b, ba2 = a1 a2 b⟩. Put c = b2 a1 . Then c3 = 1 and one obtains the following alternative presentation A4 = ⟨b, c | b3 = c3 = (bc)2 = 1⟩.

(19.1.1)

Example 19.1.6. If d = 1, 2, 3 and R is the ring of integers of ℚ(√−d) then both SL2 (R) and PSL2 (R) are virtually free-by-free. Proof. Observe that PSL2 (R) = SL2 (R)/(SL2 (R) ∩ U(R)) and SL2 (R) ∩ U(R) is finite. Hence, SL2 (R) is virtually free-by-free if and only if so is PSL2 (R). We consider the three values of d separately. For d = 1 and 2 we use arguments from [229]. In Example 18.2.1 we obtained the following presentation PSL2 (ℤ[i]) = ⟨r, v, s, t | r2 = v2 = t2 = (sr)2 = (sv)2 = (tr)2 = (tv)3 = (ts)3 = 1⟩. Put A = tv, B = tr, C = s−1 t and D = s. It is then easily verified that one obtains the alternative presentation PSL2 (ℤ[i]) = ⟨A, B, C, D | (BD)2 = (AD)2 = (BC)2 = (AC)2 = 1, B2 = A3 = C3 = D2 = 1⟩. Hence, PSL2 (ℤ[i]) = G1 ∗M G2 , where M = ⟨C, D : D2 = C3 = 1⟩ = C2 ∗ C3 , G1 = ⟨A, C, D | A3 = C3 = D2 = (AD)2 = (AC)2 = 1⟩ = ⟨A, C | A3 = C3 = (AC)2 = 1⟩ ∗⟨A⟩3 ⟨A, D | A3 = D2 = (AD)2 = 1⟩ = A 4 ∗C2 D 6 ,

154 | 19 Virtually free-by-free groups G2 = ⟨B, C, D | B2 = C3 = (BD)2 = D2 = (BC)2 = 1⟩ = ⟨B, C | B2 = C3 = (BC)2 = 1⟩ ∗⟨B⟩2 ⟨B, D | B2 = D2 = (BD)2 = 1⟩ = D6 ∗C2 (C2 × C2 ). From Theorem 5.5.2 we know that PSL2 (ℤ[i]) is virtually torsion-free and hence so are G1 and G2 . Proposition 19.1.1 then implies that every torsion-free normal subgroup of G i is free. Thus G1 and G2 are virtually free. From the presentations of G1 and G2 we deduce that G2 = M ⋊ ⟨B⟩ and that G1 has an automorphism ϕ of order 2 given by ϕ(A) = A−1 ,

ϕ(C) = C−1 ,

ϕ(D) = D.

Moreover ϕ(x) = x B for every x ∈ ⟨C, D⟩ and therefore there is a homomorphism G2 → G1 ⋊ ⟨ϕ⟩ mapping B to ϕ and restricting to the identity on M. This isomorphism extends to a surjective homomorphism τ : PSL2 (ℤ[i]) = G1 ∗M G2 → G1 ⋊ ⟨ϕ⟩ . Let K = ker(τ) and let N1 be a free subgroup of finite index of G1 . Put N = τ−1 (N1 ). Then, N is a subgroup of finite index of PSL2 (ℤ[i]) and N/K ≅ N1 . Since K ∩ G1 = K ∩ G2 = {1}, Proposition 19.1.1 implies that K is free. Therefore, N is a free-by-free subgroup of finite index of PSL2 (ℤ[i]), as desired. For d = 2 we make use of the presentation obtained in Example 18.2.2: PSL2 (ℤ[√−2]) = ⟨t, s, u | t2 = (ts)3 = usu−1 s−1 = (utu−1 t)2 = 1⟩. Put v = utu−1 . One obtains an alternative presentation PSL2 (ℤ[√−2]) = ⟨t, s, u, v | t2 = v2 = (ts)3 = (vs)3 = (vt)2 = 1, usu−1 = s, utu−1 = v⟩. Then, from the presentation of A4 given in (19.1.1), we obtain that PSL2 (ℤ[√−2]) = G∗ϕ , with G = ⟨t, s, v | t2 = v2 = (vt)2 = (ts)3 = (vs)3 = 1⟩ = (C2 × C2 ) ∗C2 A4 and ϕ : ⟨t, s⟩ → ⟨s, v⟩ is given by ϕ(s) = s and ϕ(t) = v. From Proposition 19.1.2 we know that every torsion-free normal subgroup of G is free. As SL2 (ℤ[√−2]) is virtually torsion-free, we hence get that G is virtually free. Let N be a free normal subgroup of finite index in G. So, ⟨N, u⟩ is free-by-cyclic and has finite index in PSL2 (ℤ[√−2]). Therefore PSL2 (ℤ[√−2]) is virtually free-by-free, as claimed. For d = 3 we use a different approach. Let G be the subgroup of SL2 (ℤ[ζ3 ]) generated by the matrices A=(

1 0

1 ) 1

and

B=(

1 ζ3

0 ). 1

19.1 Free-by-free groups |

Let Y = BA = (

1 ζ3

1 ) −ζ32

and

X = (A−1 , Y −1 ) = (

0 ζ32

155

−ζ3 ) 1 − ζ3

and let N = ⟨X, Y⟩. Then AYA−1 = XY

and

AXA−1 = XYX

Thus N is a normal subgroup of G and G/N is cyclic. Also let C1 = (A, B) = (

0 −ζ32

ζ3 ), 1 − ζ3

and A1 = C−1 1 C2 = (

−1 0

C2 = (A−1 , B−1 ) = (

0 ζ32

−ζ3 ), 1 − ζ3

2√−3 ). 1

2(ζ32 − ζ3 ) 1 ) = −( −1 0

Clearly 1 ⟨A, A1 ⟩ = ( 0

ℤ + ℤ2√−3 ). 1

−1 −1 Let U = XA−1 1 , V = XY = AYA , Z = AY and thus B = A V. Then

1 − ζ3 −ζ32

X=(

0 ζ32

−ζ3 ), 1 − ζ3

X −1 = (

Y=(

1 ζ3

1 ), −ζ32

Y −1 = (

Z=(

U −1 = (

−ζ32 ζ3

1 ), 1

V −1 = (

−ζ32 ζ3

1 − ζ32 ), −ζ32

Z −1 = (

B=(

1 ζ3

0 ), 1

−ζ3 ), 0

1 − ζ3 ζ32

ζ3 ), 1 − ζ3

V=(

−1 ), 1

−ζ32 −ζ3

0 −ζ32

U=(

ζ3 ), 0

1 −ζ3

−1 ), −ζ32

−ζ32 −ζ3

ζ32 − 1 ), 1 + ζ3

1 −ζ3

0 ) 1

B−1 = (

We write the elements of the Poincaré model of the 3-dimensional hyperbolic space in the real coordinates (x, y, r) with r > 0. We first consider the Dirichlet fundamental polyhedron F∞ with center j of G∞ , the stabilizer of ∞. The natural image of a matrix M ∈ SL2 (ℂ) in PSL2 (ℂ) we denote by the corresponding lower letter m. The bisectors determined by a, a−1 , a1 and a−1 1 are the planes with equations x=

1 , 2

1 x=− , 2

y = 2√3,

and

y = −2√3

respectively. Let 1 1 ] × [−√3, √3] × ℝ+ 2 2

E∞ = [− ,

156 | 19 Virtually free-by-free groups

and ≥ E = E∞ ∩ (∩U∈L I U ),

where L = {x, x−1 , y, y−1 , u, u−1 , v, v−1 }. The orthogonal projection of E on the plane is represented in Figure 19.1. The outside rectangle is the projection of E∞ and the segments inside the rectangle are the projections of the intersections of two isometric spheres, except that the four lines touching one of the black dots are included in only one isometric sphere. The black dots represent the centers of the isometric spheres. The segments represent the edges, except that there are four vertical segments which are represented by ×. The polygons limited by the segments represent the projection of the sides, except that there are four vertical sides which projects on the four edges of the rectangle. The letter inside one of the sides or aside one of the four vertical sides label the pairing transformations. The pairing between two paired sides is made in such a way that the colors and full line edges or dashed line edges should much. The color and continuous-dashed of the edges represent the edge loops. Observe that the isometric spheres of the matrices in L have radius 1. The center of their isometric circles is displayed in Table 19.1. Any letter x displayed in Figure 19.1 correspond to the element x ∈ PSL2 (ℂ) such that S x = E ∩ x−1 (E). Tab. 19.1: Center of the isometric circles of the elements of L. g Center

x ζ32 − ζ3

x −1 0

y ζ3

y −1 ζ32

u ζ3 − ζ32

u−1 0

v −ζ32

v −1 −ζ3

Let F denote the Ford fundamental polyhedron of G using F∞ . So ≥ F = E∞ ∩ (∩U∈G\G∞ I U ).

As ⟨A, A1 ⟩ ⊆ G∞ , we have F∞ ⊆ E∞ and hence F ⊆ E. It is easy to see that F can be covered with 12 isometric copies of the fundamental polyhedron of PSL2 (ℤ[ζ3 ]) obtained above. This implies that [PSL2 (ℤ[ζ3 ]) : G] ≤ 12. On the other hand, the natural projection of PSL2 (ℤ[ζ3 ]) in SL2 (ℤ[ζ3 ]/4ℤ[ζ3 ]) is surjective while the projection of G has index 12 in SL2 (ℤ[ζ3 ]/4ℤ[ζ3 ]) (this has been verified using GAP). Consequently, E = F and [PSL2 (ℤ[ζ3 ]) : G] = 12. Moreover −I ∈ ̸ G because the projection of −I does not belong to the projection of G in SL2 (ℤ[ζ3 ]/4ℤ[ζ3 ]) (again this has been verified making use of GAP). Therefore G ≅ G. For every pairing transformation α we represent the corresponding side by S α = F ∩ Fα−1 . In Figure 19.1, we simply write α. The edge contained in two sides S α and S β is represented by E α,β . We now apply Poincaré’s Theorem to the fundamental polyhedron represented in Figure 19.1 and obtain a presentation of G. For this, one simply reads the edge relations from the full colored line edges and the dashed colored line edges. For example, the three full red lines form the edge loop E a−1 −1 , S a 1 , E a −1 ,u , S u , E u,x , S x −1 1 ,x 1

19.1 Free-by-free groups |

157

a1 ×

× u−1 z−1 v −1 y −1 b−1 u

a−1

a x b v y z x −1 ×

a−1 1

× Fig. 19.1: Orthogonal projection of E on ℂ.

which induces the relation x = a1 u. Additionally, also the four vertical lines and the four vertical sides form the edge loop , S a , E a,a−1 , S a1 , E a,a1 , S a−1 , E a−1 ,a1 , S a−1 E a−1 ,a−1 1 1 1 and this gives rise to the relation aa1 = a1 a. Continuing in this way we obtain the presentation. G = ⟨a, x, y, b, a1 , u, v, z | y = ba, v = xy, z = ay, xz = au, x = ua1 , z = va, v = ab, v = yu, b x = a−1 , b u = a−1 , aa1 = a1 a⟩. We claim that the above presentation is equivalent to the following presentation: H = ⟨a, x, y | y a = xy, x a = xyx⟩ . Indeed, in G, we have aya−1 = ab = v = xy and a−1 = x−1 bx = x−1 ya−1 x. Hence x = ay−1 xa−1 = y−1 x−1 axa−1 . This proves that a, x and y satisfy the relations of H and therefore there is a homomorphism f : H → G mapping the generators a, x and y of H to the generators of G with the same name. The homomorphism f is surjective because b = ya−1 , v = xy, z = ay, u = x a y and a1 = y−1 (a, x). To prove that f is injective it is enough to check that the 11 relations of the presentation of G are satisfied by the elements a, x, y, b = ya−1 , v = xy, z = ay, u = x a y and a1 = y−1 (a, x)

158 | 19 Virtually free-by-free groups

of H. This is clear for the first five relations. The sixth and seventh relation of G takes the form ay = xya and xy = aya−1 . The eight relation takes the form xy = ya−1 xay or equivalently axa−1 = aya−1 x = xyx. Using this, we can check the 9-th relation: b x = x−1 ya−1 x = a−1 ax−1 ya−1 x = a−1 x−1 y−1 x−1 xyx = a−1 . The 10-th relation follows from a(ab u )a−1 = a(ay−1 a−1 x−1 aya−1 a−1 xay)a−1 = (ay−1 x−1 ya−1 )x(aya−1 ) = y−1 x−1 x−1 y−1 x−1 xyxxy = 1. Finally the last relation is a consequence of a1 = u−1 x and the 9-th and 10-th relations. Summarized, we have proved that G is a subgroup of index 12 in SL2 (ℤ[ζ3 ]) given by the following presentation G = ⟨x, y, a | x a = xyx, y a = xy⟩. Clearly this also is the presentation of a semidirect product ⟨x, y⟩ ⋊ ⟨a⟩ where ⟨x, y⟩ is a free group of rank 2 and ⟨a⟩ is an infinity cyclic group. Therefore G is free-by-infinite cyclic and, in particular, it is a free-by-free group.

19.2 Orders and free-by-free unit groups In this section we prove that the unit group of an order in a finite dimensional rational semisimple algebra A is virtually a direct product of free-by-free groups if and only if for each simple component S of A the unit group of an order O in S is virtually freeby-free. In order to show this, we first need to prove two results on such an order O. In the first it is shown that S is generated, as an algebra, by any subgroup G of finite index in U(O). In the second it is proved that often any subgroup of finite index in O1 is not the direct product of two subgroups of infinite order. Theorem 19.2.1 (Kleinert [136]). Let A be a finite dimensional rational simple algebra, let O be an order in A and let G be a subgroup of finite index in U(O). Assume that A is neither a quadratic imaginary extension of a totally real number field nor a totally definite quaternion algebra. Then A = ℚ[G], i.e. G generates A as ℚ-algebra. Proof. Write A = M n (D) with D a division algebra and let E = ℚ(G). Assume first that n ≥ 2. Because D is a division ring and the equality (9.2.4) it is easy to see that SLn (D) is generated by the elementary matrices and that every element of M n (D) is a product of elementary matrices and diagonal matrices of the form diag(a, 1, . . . , 1, 0, . . . , 0), with a ∈ D. Thus, it is enough to show that E contains all the matrices of these two types. This is easy for an elementary matrix e = e i,j (r) because e m ∈ G for some m > 0 1 and e = 1 + m (e m − 1). Thus SLn (D) ⊆ E. Let now A = diag(a, 1, . . . , 1, 0, . . . , 0). Put 0 ( B=( (

(−1)n−1

1 0

1 .. .

..

. 0

) ) 1 0)

19.2 Orders and free-by-free unit groups |

159

and C = diag(a, 0, 0, . . . , 0). Then, B ∈ SLn (D) and if A has a zero diagonal entry then A − B ∈ SLn (D). Therefore, if A has a zero diagonal entry then A ∈ E. Applying this to C, we deduce that C, A − C ∈ E, whether or not A has a non-zero entry. Thus A = C + (A − C) ∈ E. Assume now that n = 1 and suppose that E ≠ D, i.e. suppose E is a proper subdivision algebra of D. We have to prove that D is either a quadratic imaginary extension of a totally real number field or a totally definite quaternion algebra. Let O1 = O ∩ E. Then O1 is finitely generated as ℤ-module because so is O. On the other hand as O is a full lattice in D, every element x of E is a rational linear combination of elements of O and hence mx ⊆ O1 for some positive integer m. Thus E = ℚ(O1 ). This shows that O1 is an order in E. Moreover, [U(O) : U(O1 )] ≤ [U(O) : G] < ∞. If D is commutative then, by Corollary 5.2.7, E is totally real and D is an imaginary quadratic extension of E, as desired. Suppose that D is non-commutative. Let Dℝ = ℝ ⊗ℚ D and Eℝ = ℝ ⊗ℚ E. Let d be the degree of D. Then, by of Proposition 2.1.11 and Problem 2.2.1, Dℝ ≅ ∏ki=1 M n i (D i ), where for each i either n i = d and D i = ℝ or ℂ, or n i = 2d and D i = ℍ(ℝ). We have to show that D is a totally definite quaternion algebra. This is equivalent with showing that n l = 1 for every 1 ≤ l ≤ k. We claim that if n l ≠ 1 then Eℝ contains every simple component of Dℝ . As M n l (D l ) is generated over ℝ by the matrices of the form Eij (x) and having in mind the equality Eii (x) = Eij (x)Eji (1), we may assume, without loss of generality, that w ∈ ̸ Eℝ for some w = Eij (x) ∈ M n l (D i ) with i ≠ j. Then e = 1 + w is an elementary matrix and hence the sequence (e m ) is contained in the group B formed by the elements of Dℝ of reduced norm 1. Fix a basis V of Eℝ over ℝ and extend it to a basis W of Dℝ containing V ∪ {w}. Let δ be the Euclidean metric in Dℝ having W as an orthonormal basis. By Hey’s Theorem (Theorem 5.2.1), B/O1 is compact. Hence, because [O1 : O1 ∩G] < ∞, also B/O1 ∩G is compact. Therefore, there is a subsequence (e m h ) that is convergent modulo G. In other words, there is b ∈ B and a sequence (a h ) of elements of G such that (δ(e m h , b + a h )) converges to 0. Hence, there is a positive integer h0 such that δ((m u − m v )w, a u − a v ) = δ(e m u − e m v , a u − a v ) < 12 for all integers u, v ≥ h0 . However, if u ≠ v then m u ≠ m v and a u − a v ∈ Eℝ . Furthermore, by the choice of the metric, δ((m u − m v )w, a u − a v ) ≥ 1. This yields a contradiction and hence it finishes the proof of the claim. As by assumption, E ≠ D, Eℝ ≠ Dℝ and therefore M n l (D l ) ⊈ Eℝ for some l. Then, by the claim, n l = 1 and D l = ℍ(ℝ). Thus D is a quaternion algebra and the center F of D is not totally complex. As G ∩ F has finite index in the group of units of the ring of integers of F and because the result holds in the commutative case, we deduce ℚ(G ∩ F) = F. Thus F ⊆ E ⊂ D. So, E is a field, because D is a quaternion algebra over F. Thus Eℝ does not contain M2 (ℝ). We deduce that n i = 1 for every i = 1, . . . , k, as desired.

160 | 19 Virtually free-by-free groups Given a finite dimensional semisimple rational algebra A such that ℝ ⊗ℚ A = ∏ki=1 M k i (𝕂) with 𝕂 ∈ {ℝ, ℂ, ℍ(ℝ)} we set k

rr(A) = ∑ (k i − 1). i=1

Let O be an order in A and H a subgroup of A1 commensurable with O1 . Assume N is a normal subgroup of H. It follows from the result of Margulis (Theorem 12.4.3) that if rr(A) ≥ 2 then either N ⊆ Z(H) or [H : N] < ∞. Proposition 19.2.2 (Kleinert-del Río [138]). If O is an order in finite dimensional simple rational algebra A then O1 is nearly indecomposable. Proof. Let H1 × H2 be a subgroup of finite index of O1 and assume that both H1 and H2 have infinitely many elements. Then the normalizer N i of H i has finite index in O1 and H i has not finite index in N i . If rr(A) ≥ 2 then, by the result of Margulis (Theorem 12.4.3), each H i is abelian and hence O1 is virtually abelian. Hence, U(O) is virtually abelian, by Proposition 5.5.1. Thus O1 is finite by Theorem 5.5.6, a contradiction. Therefore, rr(A) is either 0 or 1. If rr(A) = 0 then A is either a field or a totally definite quaternion. In both cases O1 is finite, because of Proposition 5.5.6, and the result follows. So assume that rr(A) = 1. Let K be the center of A. Then K(H1 ) and K(H2 ) are distinct mutually commuting algebras of A that each properly contain K. This is not possible in a quaternion algebra and hence the degree k of A is at least 3. Then ℝ⊗ℚ A ≅ M k (ℝ)r1 × M k (ℍ(ℝ))r2 × M k (ℂ)s , for some integers r1 , r2 and s such that (r1 + s)(k − 2

1) + r2 ( 2k − 1) = rr(A) = 1. Therefore, k = 4, r1 = s = 0 and r2 = 1. Thus K = ℚ. By Proposition 19.2.1, A = ℚ(H1 , H2 ) = A1 A2 , with A i = ℚ(H i ) for i = 1, 2. So, the natural homomorphism A1 ⊗ℚ A2 → A is surjective. Thus the center of A i is contained in the center of A and hence each A i is a central simple rational algebra. By the Double Centralizer Theorem (Theorem 2.1.10) we have dimℚ (A) ≤ dimℚ (A1 ) dimℚ (A2 ) ≤ dimℚ (A1 ) dimℚ (CenA (A1 )) = dimℚ (A). Consequently, A ≅ A1 ⊗ℚ A2 and A1 and A2 are quaternion algebras over ℚ. Moreover, M2 (ℍ(ℝ)) ≅ ℝ ⊗ℚ (A1 ⊗ℚ A2 ) ≅ (ℝ ⊗ℚ A1 ) ⊗ℝ (ℝ ⊗ℚ A2 ). This implies that ℝ ⊗ℚ A1 and ℝ ⊗ℚ A2 are non-isomorphic quaternion algebras over ℝ and therefore one of them is isomorphic to M2 (ℝ) and the other is isomorphic to ℍ(ℝ). Hence one may assume that A1 is a totally definite quaternion algebra over ℚ. Moreover, O1 = O ∩ A1 is an order in a subalgebra B of A1 and H1 ⊆ U(O1 ). Then B is either ℚ, A1 or a totally imaginary quadratic extension of ℚ. As H1 has infinite order, this is in contradiction with Corollary 5.5.8. We are now in a position to prove the main result of this section.

19.2 Orders and free-by-free unit groups |

161

Theorem 19.2.3. Let A = ∏ki=1 A i , with each A i a finite dimensional simple rational algebra. Let O be an order in A and, for each i, let Oi be an order in A i . Then U(O) is virtually a direct product of free-by-free groups (respectively, of free groups) if and only if each O1i is virtually free-by-free (respectively, virtually free). Proof. By Corollary 5.5.3, U(O) has a subgroup of finite index of the form U(ℤ(O))×V1 × ⋅ ⋅ ⋅ × V k , with V i ≤ O1i . By Dirichlet’s Unit Theorem, U(ℤ(O)) is virtually free abelian of finite rank. If each Oii is virtually free-by-free then so is each V i and hence U(O) is virtually a direct product of finitely many free-by-free groups. Conversely, assume that the direct product T = ∏x∈X T x is a subgroup of finite index in U(O), where every T x is a non trivial free-by-free group. By Corollary 19.1.4 we may assume, without loss of generality, that T is normal in U(O). Recall from Theorem 1.5.6 that U(O) is a finitely generated group. Hence, X is finite and each T x is finitely generated. Moreover, again without loss of generality, we may assume that every T x is nearly indecomposable and that either T x is infinite cyclic or it is not virtually abelian. For each x ∈ X, let π x : T → T x denote the natural projection. For every i = 1, . . . , k, let Z i = U(Z(Oi )) ∩ T, S i = O1i ∩ T and put n

Ŝ i = (∏ Z i ) × (∏ S j ) i=1

and

S = S i × Ŝ i .

j=i̸

By Proposition 5.5.1, Z i ∩ S i is finite and hence it is trivial because T is torsion-free. So, by Corollary 5.5.3, S is a subgroup of finite index in T. Therefore, π x (S) is a subgroup of finite index of T x and hence it is nearly indecomposable and free-by-free. We claim that if π x (S i ) is not abelian, with x ∈ X and 1 ≤ i ≤ k, then π x (Ŝ i ) = 1. Indeed, if π x (S i ) is not abelian then, by the choice of the T x ’s, we have that T x is nearly indecomposable and not virtually abelian. Hence, π x (S) is free-by-free, nearly indecomposable and not virtually abelian. Then, by Lemma 19.1.5, π x (S i ) ∩ π x (Ŝ i ) ⊆ Z(π x (S)) = 1. Therefore, π x (S) = π x (S i ) × π x (Ŝ i ). Consequently, one of the two factors is trivial. So π x (Ŝ i ) = 1, as desired. This proves the claim. We have to show that each O1i is virtually free-by-free or, equivalently, that so is S i . If S i is virtually abelian then U(Oi ) is virtually abelian and thus O1i is finite, by Proposition 5.5.6; consequently, S i = 1 and thus it is free-by-free. So, in the remainder of the proof we assume that S i is not virtually abelian and thus there is an element x ∈ X such that π x (S i ) is nonabelian. Let Y = {x ∈ X : T x is nonabelian}, Y i = {x ∈ X : π x (S i ) is nonabelian} and Y i󸀠 = Y \ Y i . For every subset X1 of X, let π X1 : T → ∏x∈X1 T x be the natural projection map. If y ∈ Y i󸀠 then there is j ≠ i such that π y (S j ) is nonabelian. So, by the claim, Ŝ j ⊆ ker(π y ) and thus we have shown that S i ⊆ ker(π Y i󸀠 ). Therefore S i ∩ker(π Y i ) ⊆ ker(π Y ) ⊆ Z(T). Let s ∈ S i ∩ker(π Y i ) and u ∈ U(O). As T U(O) and Z(T) is characteristic in T, we have (s, u) ∈ Z(T)∩ O1i . Moreover, since [U(O) : T] < ∞, u m ∈ T for some m > 0. Then (s m , u) = (s, u)m = (s, u m ) = 1. This proves that

162 | 19 Virtually free-by-free groups s m ∈ O1i ∩ Z(U(O)). As S i is torsion-free and O1i ∩ Z(U(O)) is finite we deduce that s = 1. This shows that the restriction of π Y i to S i is injective. Therefore S i is isomorphic to a subgroup of finite index of a direct product of free-by-free groups. Hence S i is also a direct product of free-by-free groups, by Corollary 19.1.4. By Proposition 19.2.2, S i is nearly indecomposable. Since S i also is torsion-free, it follows that S i is free-by-free, as desired.

Problems 19.2.1. Let A be a finite dimensional simple algebra over ℚ. Prove (1) rr(A) = 0 if and only if A is either a field of a totally definite quaternion algebra. (2) rr(A) = 1 if and only it either A is a quaternion algebra which is not ramified at exactly one infinite place, or A = M 2 (D) with D a totally definite quaternion algebra over ℚ or A is a division algebra with center ℚ and degree 4 and ℝ ⊗ℚ A ≅ M2 (ℍ(ℝ)).

19.3 Virtual cohomological dimension and finite groups of Kleinian type In this section we first revise some of the classical background on virtual cohomological dimensions. For details and proofs we refer to [37]. First we recall that a virtually free-by-free group has virtual cohomological dimension at most two and second we determine and prove when a finite dimensional simple rational algebra has an order O such that the virtual cohomological dimension of O1 is at most two. In the second part of the section we relate the virtual cohomological dimension of some groups with the finite groups G such that ℚG is of Kleinian type. Such groups will be called groups of Kleinian type. Let G be a group. As in Section 2.5, we consider ℤ as a left ℤG-module via the trivial action of G, that is gn = n for g ∈ G and n ∈ ℤ. The projective dimension of this module is called the cohomological dimension of G and it is denoted cd(G). In other words cd(G) = inf{n ≥ 0 : there is a projective ℤG − resolution 0 → P n → P n−1 → . . . P1 → P0 → ℤ → 0} = inf{n ≥ 0 : ker(f n ) is projective for every projective ℤG − resolution f n−1

fn

f n−1

f1

f0

f−1

⋅ ⋅ ⋅ → P n+1 → P n → ⋅ ⋅ ⋅ → P1 → P0 → ℤ → 0} = inf{n ≥ 0 : Hn (G, −) = 0} = max{n ≥ 0 : Hn (G, M) ≠ 0 for some M ∈ ℤG-Mod}.

19.3 Virtual cohomological dimension and finite groups of Kleinian type |

163

The cohomological dimension of G can be calculated using any projective resolution of ℤ which finishes with the augmentation map ω : ℤG → ℤ. Examples 19.3.1. (1) cd(G) = 0 if and only if ℤG ℤ is projective. Equivalently, the augmentation map ω : ℤG → ℤ splits. So cd(G) = 0 if and only if G = 1. (2) If G = ⟨g⟩n , the cyclic group of order n, then we have a projective resolution g−1

̃ G

̃ G

g−1

g−1

ω

⋅ ⋅ ⋅ → ℤG → ℤG → ℤG → ℤG → ℤG → ℤ → 0 where, for an element α ∈ ℤG, we also represent by α the map that is left multiplication by α. Since neither g − 1 : ℤG → ℤG(g − 1) nor ̃ G : ℤG → ℤG ̃ G split, we deduce that cd(G) = ∞. (3) If G is a free group on the set S then we have the following exact sequence δ

ω

0 → ⊕s∈S ℤGe s → ℤG → ℤ → 0 where ⊕s∈S ℤGe s is the free left ℤG-module with basis {e s : s ∈ S} and δ(e s ) = s−1. Therefore every non-trivial free group has cohomological dimension 1. In fact, a deep theorem of Swan (and Stallings in the finitely generated case) shows that the converse of this statement holds, i.e. every torsion-free group of cohomological dimension 1 is free [213, 214]. Proposition 19.3.2. Let H be a subgroup of G. (1) cd(H) ≤ cd(G). (2) If both cd(G) and [G : H] are finite then cd(H) = cd(G). (3) If H is normal in G then cd(G) ≤ cd(H) + cd(G/H). As a consequence of Proposition 19.3.2 (3) we obtain at once the following property. Corollary 19.3.3. If G is a free-by-free group which is not free then cd(G) = 2. Observe that the two hypotheses of Proposition 19.3.2 (2) are needed. For example, if G is finite and non-trivial and F is a non-trivial free group then cd(G × F) ≠ 1 = cd(F) < ∞, by Examples 19.3.1. The following remarkable theorem shows that for torsion-free groups the first hypothesis is not needed. Theorem 19.3.4 (Serre [205]). If G is a torsion-free group and H is a subgroup of finite index of G then cd(G) = cd(H). The virtual cohomological dimension of a virtually torsion-free group G is vcd(G) = cd(H), with H a torsion-free subgroup of finite index of G. Because of Theorem 19.3.4, this definition does not depend on the choice of H. Clearly, the earlier stated results can be summarized as follows.

164 | 19 Virtually free-by-free groups

Corollary 19.3.5. Let G be a non-trivial group. Then G is virtually free if and only if vcd(G) = 1. If G is virtually free-by-free then vcd(G) ≤ 2. The virtual cohomological dimension of the group O1 consisting of the elements of reduced norm 1 in an order O of a simple algebra is given by the following theorem. Theorem 19.3.6 ([135, Theorem 4]). Let A be a finite dimensional simple rational algebra with index d and degree nd. Assume that ℝ ⊗ℚ A ≅ M nd (ℝ)r1 × M nd (ℍ(ℝ))r2 × 2 M nd (ℂ)s . If O is an order in A then vcd(O1 ) = r1

(nd − 2)(nd + 1) (nd + 2)(nd − 1) + r2 + s(n2 d2 − 1) − n + 1. 2 2

Corollary 19.3.7. Let A be a finite dimensional simple rational algebra. The following statements hold for an order O in A. (1) vcd(O1 ) = 0 if and only if A is a field or a totally definite quaternion algebra. (2) vcd(O1 ) = 1 if and only if A = M2 (ℚ). (3) vcd(O1 ) = 2 if and only if A = M2 (K) with K an imaginary quadratic extension of the rationals or A is a quaternion division algebra over a totally real number field which is not ramified at exactly one infinite place. Proof. Let K denote the center of A, d = Ind(A) and n = Deg(A)/Ind(A), the reduced degree of A. By Proposition 2.1.11 and Problem 2.2.1, ℝ⊗ℚ A ≅ M nd (ℍ(ℝ))r1 ×M nd (ℝ)r2 × 2 M nd (ℂ)s and K has r = r1 + r2 real embeddings and s pairs of complex embeddings. (Notice that if d is odd then r1 = 0.) By Proposition 5.5.6, vcd(O1 ) = 0 (i.e. O1 is finite) if and only if A is either a field or a totally definite quaternion algebra. If A = M2 (ℚ) then O1 is virtually free by Corollary 1.4.4 (2). If A = M2 (K), with K an imaginary quadratic extension of ℚ, then n = 2, d = 1, r1 = r2 = 0 and s = 1. Theorem 19.3.6 thus yields that vcd(O1 ) = 2. If A is a quaternion division algebra over a totally real field and r1 = 1 then n = 1, d = 2 and s = 0. So, again by Theorem 19.3.6, we obtain that vcd(O1 ) = 2. Conversely, assume that A is not a field and vcd(O1 ) ≤ 2. By Theorem 19.3.6 one has (nd − 2)(nd + 1) (19.3.1) + s(n2 d2 − 1) ≤ n + 1. r2 nd + r 2 Since A is not a field, nd ≥ 2 and therefore the three summands on the left hand side of (19.3.1) are non-negative, which implies that each summand is at most n + 1. Hence, s(nd − 1)(nd + 1) = s(n2 d2 − 1) ≤ n + 1 ≤ nd + 1 and thus it follows that either s = 0 or d = s = 1 and n = 2. In the latter case r1 = 0 and since s(n2 d2 −1) = n +1, one has that r2 nd = 0 and therefore r2 = 0. Thus A = M2 (K) where K is an imaginary quadratic extension of ℚ. Assume now that s = 0, that is, K is totally real. From r2 nd ≤ n + 1 we deduce that either (a) r2 = 0, (b) r2 = d = 1 or (c) n = r2 = 1 and d = 2. We deal with each case separately.

19.3 Virtual cohomological dimension and finite groups of Kleinian type |

165

(a) If r2 = 0 then r = r1 ≠ 0, and hence d is even. Furthermore (nd − 2)(nd + 1) (nd − 2)(nd + 1) ≤r ≤ n+1 2 2 and therefore (nd−2)(nd+1) ≤ 2n+2. Thus nd(nd−1) ≤ 2n+4 and so n(d(nd−1)−2) ≤ 4. If n ≥ 2 then n(d(nd − 1) − 2) ≥ 2(2 ⋅ 3 − 2) = 8. Thus n = 1, that is A is a division algebra. Further d(d − 1) ≤ 6 and thus d = 2, because d is even. We conclude that A is a totally definite quaternion algebra. (b) Assume that r 2 = d = 1. Then r1 = 0, that is, K = ℚ and r2 nd = n, so that (n−2)(n+1) ≤ 1 and one deduces that n = 2. Thus A = M2 (ℚ). 2 (c) Finally if n = r2 = 1 and d = 2 then A is a quaternion algebra over a totally real number field and which is not ramified at exactly one infinite place, as desired. Let G be a finite group. We say that G is of Kleinian type if the rational group algebra ℚG is of Kleinian type, that is, for every primitive central idempotent e ∈ ℚG the simple algebra ℚGe is either a number field, or a quaternion algebra over a number field F such that the group σ(ℤGe)1 is a discrete subgroup of SL2 (ℂ) for some embedding σ of F in ℂ (see Section 18.1). Because of Proposition 18.1.1, there are six types of simple algebras of Kleinian type: (1) number fields; (2) totally definite quaternion algebras; (3) M2 (ℚ); (4) M2 (K), where K is an imaginary quadratic extension of the rationals; (5) quaternion division algebras over totally real number fields which are not ramified at exactly one infinite place; and (6) quaternion division algebras with exactly one pair of complex (non-real) embeddings which are ramified at all the real places. Because of Corollary 19.3.7 the first five types correspond to the simple finite dimensional rational algebras A such that vcd(O1 ) ≤ 2 for some (any) order O in A. Because of Theorem 19.3.6, in the sixth case vcd(O1 ) = 3. Therefore we have the following property. Corollary 19.3.8. Let A be a finite dimensional simple rational algebra and O an order in A. If vcd(O1 ) ≤ 2 then A is of Kleinian type. Suppose that F is a finite Galois extension of ℚ and let A be a Schur algebra over F. Then F is totally real or totally complex. Therefore, if A is of type (6), then A is a division quaternion algebra over an imaginary quadratic extension of ℚ. Moreover, by the Benard-Schacher Theorem (Theorem 4.8.9), if A is of type (5) then F = ℚ, i.e. A is a division quaternion algebra over ℚ such that ℝ ⊗ℚ A ≅ M2 (ℚ). So, we have at once the following properties. Proposition 19.3.9. Let F be an abelian number field and let A be a non-commutative Schur algebra over F. Then A is of Kleinian type if and only if A is a quaternion algebra such that one of the following conditions hold: A is totally definite, F = ℚ, or F is an imaginary quadratic extension of ℚ.

166 | 19 Virtually free-by-free groups Recall that if G is a finite group then the center of every simple component of ℚG is a Galois extension of ℚ (see Theorem 3.3.1 and Problem 3.3.1). Hence, because of the previous result and Corollary 12.1.2, we obtain the following consequence. Corollary 19.3.10. Let G be a finite group. Then, G is of Kleinian type if and only if every non-commutative simple component A of ℚG is a quaternion algebra satisfying one of the following conditions: A is totally definite, Z(A) = ℚ, or Z(A) is an imaginary quadratic extension of ℚ. Moreover, if G is of Kleinian type and A is a simple component of ℚG which is not a division algebra then the center of A is ℚ(√−d) with d ∈ {0, 1, 2, 3}. We finish this section with a lemma on the class of finite groups of Kleinian type. For a finite group G we denote by C(G) a set of representatives of the isomorphism classes of the non-commutative simple Wedderburn components of ℚG. Lemma 19.3.11. (1) The class of algebras of Kleinian type is closed under epimorphic images and semisimple subalgebras. (2) The class of finite groups of Kleinian type is closed under subgroups and epimorphic images. (3) The class of finite groups G with C(G) ⊆ {Q, M2 (ℚ), M2 (ℚ(√−1), M2 (ℚ(√−2), M2 (ℚ(√−3) :

Q a totally definite quaternion algebra} is closed under subgroups and epimorphic images. Proof. (1) Obviously the class of algebras of Kleinian type is closed under epimorphic images. Let A be an algebra of Kleinian type and B a semisimple subalgebra of A. If B1 is a simple quotient of B then B1 is isomorphic to a subalgebra of a simple quotient of A. In order to prove that B is an algebra of Kleinian type we thus may assume that A and B are simple and B is not a field. Since A is of Kleinian type and B is nonabelian, it is clear that A and B are quaternion algebras and that there is an order O in A and an embedding σ : A → M2 (ℂ) such that σ(O1 ) is a discrete subgroup of SL2 (ℂ). Then S = O ∩ B is an order in B and σ(S1 ) ⊆ σ(O1 ) is a discrete subgroup of SL2 (ℂ). (3) follows by similar arguments and (2) is an immediate consequence of (1).

19.4 Examples of groups of Kleinian type In this section we give an infinite class of finite groups of Kleinian type. We begin with some technical lemmas. Of course, every finite abelian group is of Kleinian type. Some of the nonabelian groups we already have encountered are of Kleinian type. This follows at once from

19.4 Examples of groups of Kleinian type |

167

their Wedderburn description that has been obtained in previous chapters (see also Corollary 12.1.2 for the notation used). ℚD2n ≅ ℚ(D2n /D󸀠2n ) ⊕ ⊕d|n,2 2. Again as in the proof of part (1), replacing y i by y1 y i if needed, one may assume that each y i has order 8 and applying Lemma 19.5.2 (4), one deduces that y4i = (y i , x) for every i. It follows that G is an epimorphic image of V2n . (3) Assume G3 = 1 and G󸀠 ⊈ Z(G). We prove that G is an epimorphic image of either T × A, T1n , T2n or T3n for A an elementary abelian 2-group. By Lemma 19.5.2 (1), the exponent of G divides 8. By parts (1) and (6) of Lemma 19.5.1, G has an abelian subgroup of index 2, say Y = ⟨y1 , . . . , y n ⟩, and G󸀠 has exponent 4 (and its elements of order 2 are central ). Then G = ⟨Y, x⟩ for some x ∈ G and G󸀠 = ⟨t1 , . . . , t n ⟩, where t i = (x, y i ) for i = 1, . . . , n. Without loss of generality, we may assume that t1 is of order 4 (and thus t1 is not central). Since G󸀠 ⊆ Y, (t i , y j ) = 1 for all 1 ≤ i, j ≤ n. If t j is not central then, by Lemma 19.5.3, (x, t j ) = t2j . If t j is central, we also get that (x, t j ) = 1 = t2j . So, in all cases we have (x, t j ) = t2j . We now show that we may assume that ⟨t1 ⟩ ∩ ⟨t i ⟩ = 1 for every i ≥ 2. Because the order of t i divides 4, this is clear if t2i ∈ ̸ ⟨t1 ⟩. If t i ∈ ⟨t1 ⟩, say t i = t1a then we replace −a −a y i −a y i by y󸀠i = y−a 1 y i and we notice that (x, y i ) = (x, y 1 y i ) = (x, y i )(x, y 1 ) = t i t 1 = 1. 2 2 2 2 2 In the remaining case t i ∈ ̸ ⟨t1 ⟩ and t i ∈ ⟨t1 ⟩. Then either t i = 1 or t i = t1 . If t2i = 1 then the claim is clear. If t2i = t21 then replacing y i by y󸀠i = y1 y i we obtain that (x, y󸀠i ) = y (x, y1 y i ) = (x, y i )(x, y1 )y i = t i t1i = t i t1 ∈ ̸ ⟨t1 ⟩ and (x, y󸀠i )2 = 1, which finishes the proof of the claim. So, from now on we assume that for i ≥ 2, ⟨t1 ⟩ ∩ ⟨t i ⟩ = 1. Since the order of t i divides 4 and the order of t1 is 4, this implies that ⟨t i ⟩ ∩ ⟨t1 t i ⟩ = 1 for i ≥ 2. Put F i = ⟨x, y1 , y i ⟩, for 1 ≤ i ≤ n. We prove three claims.

19.5 Nilpotent groups of Kleinian type | 183

Claim 1: If y41 = 1 then y4j = 1 for every j (with 1 ≤ j ≤ n). Indeed, suppose y41 = 1. If t j = 1 then y j is central in G and thus, by Lemma 19.5.1 (5), we get that y4j = 1. So assume that t j ≠ 1. Apply Lemma 19.5.2 (4) to the group F j . Since ⟨t1 ⟩ ∩ ⟨t j ⟩ = 1, one has (F j )y j = ⟨t j ⟩ ≠ F 󸀠j ≠ ⟨y1 y j ⟩ = (F j )y1 y j and hence y4j = (y1 y j )4 ∈ ⟨t j ⟩ ∩ ⟨t1 t j ⟩ = 1, as desired This proves Claim 1. −2 Claim 2: If x4 = 1 and t1 = y−2 1 (and thus y 1 has order 8) then t j = y j for every j with 1 ≤ j ≤ n. 2 2 Indeed, suppose x4 = 1 and t1 = y−2 1 . Let Z = ⟨x , t 1 ⟩. Then Z is a subgroup of i 2 Z(F1 ). Moreover y−2 1 = t 1 ∈ ̸ Z(F 1 ) and (xy 1 , t 1 ) = t 1 ≠ 1 for every i. This shows that 2 Z = Z(F1 ). Hence Z(F1 ) = 1. Let j with 2 ≤ j ≤ n. Recall from Lemma 19.5.1 (6) that ⟨t j ⟩ is a normal subgroup of the group F j and this is of Kleinian type. Since t21 ∈ ̸ ⟨t j ⟩ (because ⟨t1 ⟩ ∩ ⟨t j ⟩ = 1 and t1 has order 4) we may apply Lemma 19.5.5 to the elements x1 = x, x2 = y1 and x3 = y j 2

of the nonabelian group F j / ⟨t j ⟩ of Kleinian type and deduce that y j 2 ∈ Z = 1. Hence y2j ∈ ⟨t j ⟩

(19.5.4)

We now proceed by considering the possible orders of t j . If t j = 1 then (19.5.4) implies that y−2 j = 1 = t j , as desired. If t j has order 4 then, again because ⟨t 1 ⟩ ∩ ⟨t j ⟩ = 1, the second part of Lemma 19.5.6 is applicable to the group F j /⟨t−1 1 t j ⟩, for x 1 = x, 2 x2 = y1 and x3 = y j . It follows that t21 y21 y2j ∈ Z = 1. Hence t21 y21 y2j = t1 y2j ∈ ⟨t−1 1 t j ⟩. −1 −1 −1 2 Combining this with (19.5.4), we obtain y j ∈ ⟨t j ⟩ ∩ t1 ⟨t1 t j ⟩ = {t j }, as desired. If t j has order two then again (19.5.4) implies that either y2j = t j or y2j = 1. The former is as desired. In the second case we may apply Lemma 19.5.5 to the nonabelian group F j /⟨t21 t j ⟩ of Kleinian type (note that t1 y j is central in this group). It follows that (t1 y j )2 ∈ Z 2 = 1 and thus t21 = t21 y2j ∈ ⟨t21 t j ⟩. Hence t21 = t21 t j , a contradiction. This finishes the proof of Claim 2. Claim 3: If G󸀠 is not cyclic then y4i ∈ ⟨t i ⟩ for every i with 1 ≤ i ≤ n. If, furthermore, 2 t i = y4i ≠ 1 for some i ≥ 1, then x4 = 1. Assume that G󸀠 is not cyclic. Then G y i = ⟨t i ⟩ ≠ G󸀠 , for each i ≥ 1. Hence, by Lemma 19.5.2 (4), y4i ∈ ⟨t i ⟩, as desired. If, furthermore, x4 ≠ 1 and t2i = y4i ≠ 1 for some i ≥ 1 then x and y i satisfies the hypotheses of case (2) in Lemma 19.5.3, and neither (2a), nor (2b) nor (2c) holds. Hence the claim follows. It is now sufficient to deal with the following three cases. Case 1. Suppose y41 = 1. Because of Claim 1 we obtain that y4j = 1 for every j. Hence we conclude that G is an epimorphic image of T1n . Case 2. Suppose x4 = 1 and t1 = y−2 1 . Because of Claim 2 we conclude that G is an epimorphic image of T2n . Case 3. Suppose that neither Case 1 nor Case 2 hold. Claim 4. One may assume that, for every i ≥ 1, if t2i ≠ 1 then one has y4i ≠ 1, 2 ±2 or x 2 ∈ {t 2 , y 4 }. t i ≠ y−2 i and either t i y i = x i i

184 | 19 Virtually free-by-free groups In order to prove this, suppose that t2i ≠ 1. Then x and y i satisfy condition (2) of Lemma 19.5.3 and hence one of the three cases (a), (b) or (c) of this statement holds. If y4i = 1 then interchanging the roles of y1 and y i one may assume that Case 1 holds. So one may assume that y4i ≠ 1 and hence (a) does not hold. Suppose now that t i = y−2 i . If x4 ≠ 1 then (c) does not hold and from (b) we get that 1 = t i y2i = x±2 ≠ 1, a contradiction. Thus in this case x4 = 1 and interchanging again the roles of y1 and y i one may assume that Case 2 holds. So one may assume that y4i ≠ 1 and t i ≠ y−2 i . Then neither (a) nor (c) holds. Thus (b) holds and this finishes the proof of Claim 4. We now finish the proof in case G󸀠 is cyclic. Then ⟨t i ⟩ ⊆ ⟨t1 ⟩. Since we already know that ⟨t1 ⟩ ∩ ⟨t i ⟩ = 1, for i ≥ 2, this implies that t i = 1. Hence y i is central for i ≥ 2. By Lemma 19.5.5, we get that y2i = z2i for some z i ∈ Z(F1 ). Then, replacing y i by y i z i , we may assume that y2i = 1, for i ≥ 2. Thus G is an epimorphic image of F1 × C2n−1 . We are going to show that F1 is an epimorphic image of T and therefore G is an epimorphic image of T × C2n−1 . First assume that x4 ≠ 1 and so, by Claim 4, t1 y2i = x±2 . If t1 y2i = x−2 then (xy1 )2 = 2 2 2 t−1 1 y 1 x = t 1 and t 1 = (y 1 , xy 1 ). Then, replacing x by xy 1 , one sees that F 1 is an epimorphic image of T . A similar computation shows that if t1 y21 = x2 then replacing x 4 by xy−1 1 one deduces that F 1 is an epimorphic image of T . Second assume that x = 1. 4 2 2 2 2 2 2 2 By Claim 4, either t1 y1 = x , x = y1 or x = t1 . If x = t1 then H is clearly an epimorphic image of T . If x2 = y41 then x 󳨃→ xy21 and y 󳨃→ y1 induces an epimorphism T → H and if t1 y21 = x2 one gets an epimorphism T → H given by x 󳨃→ xy1 and y 󳨃→ y1 . This finishes the proof if G󸀠 is cyclic. Finally, assume that G󸀠 is not cyclic. By Claim 3, y4i ∈ ⟨t i ⟩ ∩ Z(G) for every i. In particular, if t2i ≠ 1 (for example for i = 1), then y4i = t2i (because by assumption y4i ≠ 1 by Claim 4). Because of Claim 3, we then get that x4 = 1. Moreover Z(⟨x, y i ⟩) = ⟨x2 , t2i , t i y2i ⟩ (see Lemma 19.5.3) and so Z(⟨x, y i ⟩)2 = 1 for every i such that t2i ≠ 1. We claim that y2i ∈ ⟨t i ⟩ and t2i = 1 for every i ≥ 2. Clearly the image y i of y i in H = F i / ⟨t i ⟩ is central. Because ⟨t i ⟩ ∩ ⟨t1 ⟩ = 1 and y41 = t21 , Lemma 19.5.5 is applicable to the group H. Indeed, H = ⟨x1 = x, x2 = y1 , x3 = y i ⟩, x3 ∈ Z(H), (x1 , t) = t2 ≠ 1 and (x2 , t) = 1, where t = (x1 , x2 ) = t1 , because t21 ∈ ̸ ⟨t i ⟩. Therefore we get that x23 ∈ Z(⟨x1 , x2 ⟩)2 = 1, or equivalently y2i ∈ ⟨t i ⟩. This proves the first part of the claim. Assume now that t2i ≠ 1. Then, by the previous paragraph, y4i = t2i and hence y2i = t i , (because the option y2i = t−1 i is excluded by Claim 4). The last part of Claim 4 now implies x2 = t2i . Interchanging the role of y1 and y i in the above reasoning, we get that t21 = x2 . Hence t21 = t2i ≠ 1, contradicting with ⟨t1 ⟩ ∩ ⟨t i ⟩ = 1. This proves that t2i = 1 and finishes the proof of the claim. Let i ≥ 2. The natural image of t1 y i is central in the nonabelian group F i /⟨t21 t i ⟩ which is of Kleinian type. Hence, applying Lemma 19.5.5, to this group, we obtain that t21 y2i ∈ Z(⟨x, y1 ⟩)2 = 1. Consequently t21 y2i ∈ ⟨t21 t i ⟩. Thus y2i ∈ {1, t i } ∩ {t21 , t i } = {t i }, i.e. y2i = t i . Moreover, by Claim 4, either x2 = t21 or t1 y21 = x2 . In the first case, G is a quotient of T3n . In the second case, setting x󸀠 = y1 x and y󸀠1 = y1 , one has t󸀠1 =

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2 (x󸀠 , y󸀠1 ) = t1 and t󸀠i = (x󸀠 , y i ) = t i = y2i , for every i ≥ 2, and x󸀠 = y21 x2 t−1 1 = t 1 . This implies that again G is a quotient of T3n . 2

19.6 Groups of Kleinian type that are not nilpotent Throughout this section G is a finite group that is not nilpotent. We will characterize when such a group is of Kleinian type. To do so, we will prove several technical lemmas under the assumption that G is of Kleinian type and for this it is convenient to first introduce some notation. Because of Lemma 19.5.1 (4), we know that G is a semidirect product G3 ⋊ G2 of an elementary abelian 3-group G3 and a 2-group G2 . Moreover, since by assumption G is not nilpotent, Lemma 19.5.1 (1) implies that G has an abelian subgroup G3 × N2 such that N2 has index 2 in G2 . Thus G2 = N2 ∪ N2 x, for every x ∈ G2 \ N2 . Let K = G3 ∩ Z(G). Then G3 = K × M, for some subgroup M of G3 . Note that M is not trivial because, by assumption, G is not nilpotent. For every m ∈ M, let k m = mm x

and

̃ = k m m. m

Since x2 centralizes G3 and G3 is abelian, k m ∈ K and thus ̃ −1 . ̃ x = k m m x = k2m m−1 = m m Furthermore, k m1 m2 = m1 m2 (m1 m2 )x = k m1 k m2 and hence ̃ ̃1 m ̃2 . m 1 m 2 = k m1 m2 m 1 m 2 = k m1 m 1 k m2 m 2 = m So,

̃ ̃ | m ∈ M} M = {m

is an elementary abelian 3-group and M. G3 = K × ̃ Consequently, replacing M by ̃ M, we may assume that ̃ M = M, and ax = {

a a−1

if a ∈ K if a ∈ M.

186 | 19 Virtually free-by-free groups

So, G = K × (M ⋊ G2 ) = K × (M ⋊ ⟨N2 , x⟩), where K and M are elementary abelian 3-groups, G2 = ⟨N2 , x⟩ = N2 ∪ N2 x is a 2-group, ⟨N2 , M⟩ = N2 × M is abelian and x acts on M by inversion. Notice that this group is completely determined by N2 , G2 and the ranks k and m of K and M respectively. To emphasize this information we use the following notation G = G k,m,N2 ,G2 = K × (M ⋊ G2 ) = K × ((M × N2 ) : ⟨u⟩2 ),

(19.6.1)

with k and m non-negative integers, m ≥ 1, and K = C3k ,

M = C3m ,

u2 ∈ N2

and

w u = w−1 for w ∈ M.

So, N2 = CenM⋊G2 (M). By Lemma 19.3.11, the nilpotent group K × G2 also is of Kleinian type. Hence, it is one of the groups described in Theorem 19.5.8. In particular, if K ≠ 1 and G2 is nonabelian, then K × G2 satisfies condition (1) and so the exponent of G2 is 4, G󸀠2 ⊆ Z(G2 ) and the exponent of Z(G2 ) is 2. In the following four lemmas we prove more restrictions on k, m, N2 and G2 . Lemma 19.6.1. If G = G k,m,N2 ,G2 is of Kleinian type then the exponent of G2 divides 8. Furthermore, if k ≠ 0 then the exponent of G2 divides 4 and the exponent of G2 ∩ Z(G) is 1 or 2. Proof. First we prove that the exponent of G2 divides 8. If G2 is nonabelian then this follows from Lemma 19.5.2 (1). If G2 is abelian and g ∈ G2 then g2 ∈ N2 and therefore it is central. Lemma 19.5.1 (5) implies that the order of g2 divides 4 and thus the order of g divides 8. Assume now that k ≠ 0, or equivalently K ≠ 1. If G2 is nonabelian then G2 has exponent 4 and Z(G2 ) has exponent 2 and thus G2 ∩ Z(G) has exponent 1 or 2, as desired. On the other hand, if G2 is abelian then N2 = G2 ∩ Z(G). Since K has a central element of order 3, the exponent of Z(G) is either 3 or 6 and therefore the exponent of G2 ∩ Z(G) is either 1 or 2. Furthermore, g2 ∈ G2 ∩ Z(G) for every g ∈ G2 and so g 4 = 1. Notice that if M1 is a maximal subgroup of M then G/(K ×M1 ) ≅ G0,1,N2 ,G2 . So, without loss of generality, in order to obtain restrictions on G2 and N2 one may assume that k = 0 and m = 1. This will be used in the proof of the next three lemmas.

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Lemma 19.6.2. Assume that G = G k,m,N2 ,G2 is of Kleinian type. Let a ∈ N2 . If L is a normal subgroup of G2 contained in N2 such that G2 /L ≅ D8 then a2 ∈ L. In particular, if G2 = D8 and a is an element of order 4 in G2 then a ∈ ̸ N2 . Proof. One may assume that k = 0 and m = 1. Then G/L ≅ C3 ⋊ D8 . If a2 ∈ ̸ L then G/L = ⟨c⟩3 ⋊ (⟨a⟩4 ⋊ ⟨b⟩2 ) = ⟨ca⟩12 ⋊ ⟨b⟩2 ≅ D24 . However, this yields a contradiction because of Lemma 19.5.1 (3). Lemma 19.6.3. Assume that G = G k,m,N2 ,G2 is of Kleinian type. Let L be a normal subgroup of G2 contained in N2 . Then G2 /L is not isomorphic to any of the groups Q16 , D−16 , D+16 , D, E. If, moreover, G2 /L is a nonabelian group of order 16 then G2 /L has exponent 4 and the exponent of Z(G2 /L) is 2. Proof. Again, we may assume that k = 0 and m = 1 and thus H = G/L = G0,1,Q,P = ⟨w⟩3 ⋊ P, where P = G2 /L and Q = N2 /L. First assume that P = Q16 = ⟨a, b | a8 = b2 a4 = 1, ba = a−1 b⟩ or P = D−16 = ⟨a, b | a8 = b2 = 1, ba = a3 b⟩. Then P/⟨a4 ⟩ ≅ D8 and a2 ∈ ̸ ⟨a4 ⟩. By Lemma 19.6.2, a ∈ ̸ Q. However (⟨w, a2 ⟩, 1) is a strong Shoda pair of H and [H : ⟨w, a2 ⟩] = 4, contradicting Lemma 19.5.1 (2a). Second assume that P = D+16 = ⟨a, b | a8 = b2 = 1, ba = a5 b⟩. If b ∈ Q then a ∈ ̸ Q and (A = ⟨w, a 2 , b⟩ , B = ⟨b⟩) is a strong Shoda pair of H with [A : B] = 12 and B is not normal in G, contradicting Lemma 19.5.1 (2b). On the other hand, if b ∈ ̸ Q then, interchanging generators if needed, we may assume that a ∈ Q and hence (A = ⟨w, a⟩ , 1) is a strong Shoda pair of H. Let e = e(H, A, 1), a primitive central idempotent of ℚH. Then b2 e = e but be ≠ ±e (because be cannot be central in ℚHe). Hence ℚHe is split. Since |A| = 24 we obtain a contradiction with of Lemma 19.5.1 (2c). Third, assume that P = D = ⟨a, b, c | ca = ac, cb = bc, a2 = b2 = c4 = 1, ba = c2 ab⟩. Since ab is of order 4 and ⟨a, b⟩ = D8 , Lemma 19.6.2 implies that ab ∈ ̸ Q. It thus follows that either a ∈ ̸ Q or b ∈ ̸ Q. By symmetry, we may assume that a ∈ ̸ Q and b ∈ Q. If c ∈ Q then (A = ⟨w, c, b⟩ , B = ⟨b⟩) is a strong Shoda pair of H, [A : B] = 12 and ⟨b⟩ is not normal in H, contradicting Lemma 19.5.1 (2b)). If c ∈ ̸ Q then (A = ⟨w, c2 , b⟩ , B = ⟨b⟩) is a strong Shoda pair of H such that [H : A] = 4, again in contradiction with Lemma 19.5.1 (2a). Fourth, assume that P = E = ⟨a, b, c | ca = ac, cb = bc, a4 = b2 = c4 = 1, ba = ca3 b⟩. Then a2 c ∈ P󸀠 ⊆ Q and a2 ∈ Q. Thus c ∈ Q. Moreover P/ ⟨c⟩ ≅ D8 . By Lemma 19.6.2, a ∈ ̸ Q. So (A = ⟨M, a2 , c⟩ , B = ⟨a2 ⟩) is a strong Shoda pair of H with [H : A] = 4. This again yields a contradiction with Lemma 19.5.1 (2a)). Finally, to prove the second statement of the lemma assume that P is nonabelian and of order 16. By Lemma 19.5.1 (3) we know that P is not isomorphic to D16 . By the first part of this lemma, we also know that P is not isomorphic to any of the groups: Q16 , D−16 , D+16 , D−16 , D. Hence, from well-known description of the nonabelian groups

188 | 19 Virtually free-by-free groups of order 16 we obtain that P is isomorphic to one of the groups: Q8 × C2 , D8 × C2 , W21 or ⟨a, b | a4 = b4 = (ab)2 = (a2 , b) = 1⟩. Hence the result follows. Lemma 19.6.4. Let G = G k,m,N2 ,G2 be a finite group of Kleinian type. If G2 is nonabelian then its exponent is 4, G󸀠2 ⊆ Z(G2 ) and Z(G2 ) has exponent 2. In particular, Q16 , D+16 , D−16 , D and E are not epimorphic images of G2 . Proof. Again we may assume that G = G0,1,N2 ,G2 = ⟨w⟩3 ⋊ G2 . We first prove two claims. Claim 1. Let x, y ∈ G2 with t = (x, y) ≠ 1. If y has order 8 then t has order 4. In order to prove this we may assume that G2 = ⟨x, y⟩ and argue by contradiction. So, suppose that t does not have order 4. By Lemma 19.5.1 (6) and Lemma 19.5.2 (2), t ∈ Z(G) and t has order 2. Let V = (⟨s⟩2 ×⟨y21 ⟩4 ×⟨y22 ⟩4 ) : (⟨y1 ⟩2 × ⟨y2 ⟩2 ), with s = (y1 , y2 ) and Z(V) = ⟨s, y21 , y22 ⟩ (this is the same group V as defined in Section 19.4 but with generators renamed to avoid confusions with the elements t, x and y of G). Then, there is an epimorphism V → G2 mapping y1 to x and y2 to y. Since V/⟨y22 , sy41 ⟩ has order 16 and exponent 8, G2 /⟨y2 , tx4 ⟩ has order at most 16. However, if |G2 /⟨y2 , tx4 ⟩| = 16 then G2 /⟨y2 , tx4 ⟩ ≅ V/⟨y22 , sy41 ⟩ and hence G2 /⟨y2 , tx4 ⟩ has exponent 8, contradicting Lemma 19.6.3. This implies that G2 /⟨y2 ⟩ has order at most 16 and G2 /⟨tx4 ⟩ has order at most 32. Since the latter is nonabelian of exponent 8, it has order 32, by Lemma 19.6.3. This implies that y2 ∈ ̸ ⟨x, t⟩. Indeed, for otherwise |G2 | ≤ 32 and hence |G2 | = |G2 /⟨tx4 ⟩| = 32. So t = x4 and therefore y2 ∈ ⟨x, t⟩ = ⟨x⟩. Thus |G2 | = 16, a contradiction. We thus obtain that G2 /⟨y2 , tx4 ⟩ has order 8 because we have seen that this group has order at most 8 and G2 /⟨tx4 ⟩ has order 32. Moreover, since |G2 /⟨y2 ⟩| ≤ 16, Lemma 19.6.3 yields that the group G2 /⟨y2 ⟩ is either abelian or has exponent 4 and thus either t ∈ ⟨y2 ⟩ or x4 ∈ ⟨y2 ⟩. Since y2 ∈ ̸ ⟨t, x⟩, either t = y4 or x4 = y4 . So in both cases we get x2 y2 ∈ ̸ ⟨tx4 ⟩ and x4 y4 ∈ ⟨tx4 ⟩. This implies that G2 / ⟨tx4 , x2 y2 ⟩ has order 16 and exponent 8, because x4 ∈ ̸ ⟨tx4 ⟩ ∪ ⟨tx4 ⟩ x2 y2 = ⟨tx4 , x2 y2 ⟩. Lemma 19.6.3 therefore yields that G2 /⟨tx4 , x2 y2 ⟩ is abelian, that is, t ∈ ⟨tx4 , x2 y2 ⟩. Since t ∈ ̸ ⟨tx4 ⟩ we conclude that y2 ∈ ⟨x, t⟩ a contradiction. This proves the claim. Claim 2. If x ∈ G2 has order 8 then (x, (x, G2 )) = 1. It is sufficient to show that if y ∈ G2 and t = (y, x) ≠ 1 then (x, t) = 1. Assume the contrary, then by Lemma 19.5.3, (x, t) = t2 ≠ 1. Hence Claim 1 implies that both t and t2 have order 4, a contradiction. This proves Claim 2. We now first prove by contradiction that G2 has exponent 4. So assume x ∈ G2 has order 8. Because of Lemma 19.5.1 (5), we know that x ∈ ̸ Z(G2 ). Let y ∈ G2 be such that t = (x, y) ≠ 1. As before, we may assume that G2 = ⟨x, y⟩. Because of Claim 1, t has order 4 and by the second claim (x, t) = 1. By Lemma 19.5.1 (6), ⟨t2 ⟩ is a normal subgroup of G contained in N2 . Then, applying Claim 1 to G/⟨t2 ⟩ = G0,1,N2 /⟨t2 ⟩,G2 /⟨t2 ⟩ , we get that x4 ∈ ⟨t2 ⟩. This implies that t2 = x4 . Since t is not central in ⟨x, y⟩ (as t has order 4), we get that (y, t) ≠ 1 and (xy, t) = (xy, (xy, x)) ≠ 1. Because of Claim 2 we obtain that y4 = (yx)4 = 1. Moreover, by Lemma 19.5.2 (5a)), (x2 , y) = t2 and, by part (5).(b) of the same lemma, (y2 , x) = 1. This implies that y2 , tx2 ∈ Z(G2 ). Since

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189

t ∈ ̸ Z(G2 ), we thus have that G2 /⟨y2 , tx2 ⟩ is a nonabelian quotient of D16 . Since D16 is not of Kleinian type, G2 /⟨y2 , tx2 ⟩ has order 8 and, from t2 = x4 and y4 = 1, we have that G2 has order at most 32. By Lemma 19.6.3, it follows that G2 has order exactly 32. Therefore ⟨y2 , tx2 ⟩ has order 4 and thus ⟨y2 ⟩ ∩ ⟨tx2 ⟩ = 1. Hence, both G2 /⟨y2 ⟩ and G2 /⟨tx2 ⟩ are nonabelian groups of order 16. Therefore, by Lemma 19.6.3, both groups have exponent 4. Thus x4 ∈ ⟨y2 ⟩ ∩ ⟨tx2 ⟩ = 1, a contradiction. This finishes the proof of the fact that the exponent of G is 4. We now prove that G󸀠2 ⊆ Z(G2 ). We argue by contradiction. So, because of Lemma 19.5.1 (6), there exist x, y ∈ G2 such that t = (y, x) has order 4. Without loss of generality, one may assume that x and y satisfy condition (1) of Lemma 19.5.3 (see Remark 19.5.4), that is (x, t) = (y, t) = t2 and x2 , y2 ∈ Z(G). Then 1 = (yx)4 = (yx2 yt)2 = t2 , a contradiction. It remains to show that Z(G2 ) has exponent 2. Again we argue by contradiction. So, assume that there exists z ∈ Z(G2 ) of order 4. Since G2 is not abelian, there exist x, y ∈ G2 with (x, y) = t ≠ 1. As before, one may assume that G2 = ⟨x, y, z⟩. Since t has order 2 and z has order 4, H = ⟨t, z2 , x2 , y2 ⟩ is an elementary abelian 2-subgroup of Z(G) and so there is a subgroup L of index 2 in H which contains tz2 but does not contain t. We will use the bar notation for the natural images of the elements of G2 in G2 /L. If x2 ∈ L we set x1 = x, and otherwise we put x1 = tx. Similarly, define y1 = y if y2 ∈ L, and y1 = ty otherwise. Then G2 /L = ⟨x1 , y1 , z⟩ is a nonabelian epimorphic image of D with a central element z of order 4. This yields a contradiction with Lemma 19.6.3, because L ⊆ N2 . This finishes the proof. Theorem 19.6.5. Let G be a finite group that is not nilpotent. The group G is of Kleinian type if and only if it is an epimorphic image of A × H, where A is an abelian group and H = M ⋊ P, with M a non-trivial elementary 3-group such that [P : CenP (M)] = 2 and m p = m−1 , for every p ∈ P \ CenP (M), and one of the following conditions hold: (1) exp(A) = 4 and P = C8 , (2) exp(A) = 6, P = W1n = (∏ni=1 ⟨t i ⟩2 × ∏ni=1 ⟨y i ⟩2 )⋊⟨x⟩4 and CenP (M) = ⟨y1 ,. . ., y n , t1 , . . . , t n , x2 ⟩, (3) exp(A) = 2. P = W21 = ⟨y1 ⟩4 ⋊ ⟨x⟩4 and CenP (M) = ⟨y21 , x⟩. Proof. That the conditions are sufficient has been shown in Lemma 19.4.6. For the converse, assume G is of Kleinian type. So, write G = G k,m,N2 ,G2 as in (19.6.1). Of course, G2 may be abelian or nonabelian. Assume first that G2 is abelian. Then Z(G) = K×N2 . Let u be an element of minimal order in G2 \ N2 . Then G2 = L × ⟨u⟩ and N2 = L × ⟨u2 ⟩. Because of Lemma 19.6.1, the exponent of G2 divides 8 and, by Lemma 19.5.1 (5), the exponent of Z(G) divides 4

190 | 19 Virtually free-by-free groups or 6. We separately deal with the cases K = 1 and K ≠ 1. Assume that K = 1. Then G = L × (M ⋊ ⟨u⟩) and the exponent of L divides 4. Therefore G is an epimorphic image of A × H, where A is abelian of exponent 4 and H satisfies condition (1) of the statement. Assume now that K ≠ 1, then the exponent of Z(G) divides 6 and thus the order n of u divides 4. Thus G is an epimorphic image of A × H1 , with A abelian of exponent 6 and H1 = M ⋊ C n = G0,m,⟨u2 ⟩,⟨u⟩n . Then H1 is an epimorphic image of H = M ⋊ W11 = G0,m,⟨y1 ,t,u2 ⟩,W11 . We conclude that G is an epimorphic image of A × H, where A and H satisfy condition of (2) of the statement. In the remainder of the proof we assume that G2 is not abelian. Notice that Z(G2 ) ⊆ N2 because N2 is abelian and [G2 : N2 ] = 2. By Lemma 19.6.4, G2 has exponent 4, G󸀠2 ⊆ Z(G2 ) and G󸀠2 has exponent 2. Furthermore, if T is a proper subgroup of G󸀠2 then G/T ≅ G k,m,N2 /T,G2 /T and hence, again by Lemma 19.6.4, the exponent of Z(G2 /T) is 2. It thus follows from Lemma 19.5.7 that G2 is an epimorphic image of either C2n × W, W1n or W2n for some n. Assume that G2 is an epimorphic image of C2n × W, but not of Win for i = 1, 2 and some n. This implies that G2 = C2r × W for some r. Because Z(G2 ) ⊆ N2 , one has that G = A × G0,m,Q,W for A an abelian group of exponent dividing 6 and Q and abelian subgroup of index 2 in W. Let L1 = ⟨x2 , y2 ⟩, L2 = ⟨x2 , (xy)2 ⟩ and L3 = ⟨y2 , (xy)2 ⟩. Then L i ⊆ Q and W/L i ≅ D8 . Further, the image of xy (respectively y, x) in W/L1 (respectively W/L2 , W/L3 ) has order 4. Thus, xy, x, y ∈ ̸ Q, by Lemma 19.6.2, contradicting the fact the [W : Q] = 2. So, in the remainder of the proof, we also assume that G2 is an epimorphic image of W1n or W2n . For simplicity, the symbols used for the generators in the definition of the groups W1n and W2n (before Lemma 19.4.3), also will be used for their images in G2 . Hence, we write G2 = ⟨x, y1 , ⋅ ⋅ ⋅ , y n , t1 = (y1 , x), . . . , t n = (y n , x)⟩ with the respective relations. Then G󸀠2 = ⟨t1 , . . . , t n ⟩ is an elementary abelian 2group. Assume that |G󸀠2 | = 2r . Then, reordering the y i ’s, one may assume that α α G󸀠2 = ⟨t1 , . . . , t r ⟩. Let r < i ≤ n and let t i = t11 ⋅ ⋅ ⋅ t r r , with α i = 0 or 1. Then α1 αr 󸀠 y i = y i y1 ⋅ ⋅ ⋅ y r ∈ Z(G2 ) ⊆ N2 , for i > r. Thus, replacing y i by y󸀠i for i > r one has that G2 = B × P, where B = ⟨y r+1 , . . . , y n ⟩, an elementary abelian 2-group, and P an epimorphic image of W1r or W2r such that P󸀠 = ⟨t1 ⟩2 × ⋅ ⋅ ⋅ × ⟨t r ⟩2 . Then the map α α α α f : P󸀠 → ⟨y1 , . . . , y r ⟩ given by f(t11 ⋅ ⋅ ⋅ t r r ) = y11 ⋅ ⋅ ⋅ y r r , with α i = 0 or 1, is well 2 defined. Moreover, (x, f(s)) = s and therefore (xf(s)) = sx2 f(s)2 , for every s ∈ P󸀠 . Assume that P is an epimorphic image of W1r . Let A1 = K × B, an abelian group of exponent dividing 6. Then G = A1 × H1 , where H1 = G0,m,Q,P and Q is an abelian subgroup of index 2 in P. We will show that G is an epimorphic image of A × H, with A and H satisfying the conditions of part (2). For this, it is enough to show that one may assume that y1 , . . . , y r , t1 , . . . , t r , x2 ∈ Q. Obviously t1 , . . . , t r , x2 ∈ Q. Assume that y i0 ∈ ̸ Q, with 1 ≤ i0 ≤ r. We separately deal with the cases x2 ∈ ⟨t i0 ⟩ and x2 ∈ ̸ ⟨t i0 ⟩. If x2 ∈ ̸ ⟨t i0 ⟩ then both K1 = ⟨x, y i0 ⟩/⟨x2 ⟩ and K2 = ⟨x, y i0 ⟩/⟨x2 t i0 ⟩ are isomorphic to

19.6 Groups of Kleinian type that are not nilpotent |

191

D8 . Moreover x has order 4 in K2 and xy i0 has order 4 in K1 . Hence x, xy i0 ∈ ̸ Q, by Lemma 19.6.2, and y i0 ∈ ̸ Q, yielding a contradiction. Suppose now that x2 ∈ ⟨t i0 ⟩. Then, replacing x by xt i0 , if needed, one may assume that x2 = 1. Then, for every i = 1, . . . , r, we have ⟨x, y i ⟩ ≅ D8 and xy i has order 4. Therefore xy i ∈ ̸ Q, by Lemma 19.6.2. Since y i0 ∈ ̸ Q, one gets that x ∈ Q. For every 1 ≠ y ∈ ⟨y1 , . . . , y r ⟩, the group ⟨x, y⟩ is not abelian. Then y ∈ ̸ Q. That is ⟨y1 , . . . , y r ⟩ ∩ Q = 1 and hence r = 1. Thus P = ⟨x, y1 ⟩ ≅ D8 , with x2 = 1 = y2i and Q = ⟨x, t⟩. Interchanging the roles of x and y1 one may assume that y1 ∈ Q as desired. Finally, assume that P is an epimorphic image of W2r , with |P󸀠 | = 2r . Hence, 2 f(s) = s for every s ∈ P󸀠 . We also assume that P is not an epimorphic image of W1h for any h ≥ 1. We claim that if r = 1 then the exponent of A1 divides 2. Notice that P is nonabelian, W21 has order 16 and D8 is an epimorphic image of W11 . Then P is isomorphic to either W21 or Q8 . This implies that K × H2 is an epimorphic image of G, where H2 = G0,1,⟨a⟩,Q8 and a is an element of order 4 in Q8 . Then H2 has a cyclic subgroup K2 of index 2 and so (K2 , 1) is a strong Shoda pair of H2 . Thus e = e(H2 , K2 , 1) is a primitive central idempotent of ℚH2 and, applying Proposition 3.5.5, one has ℚH2 e ≅ ℍ(ℚ(√3)). Therefore, if K ≠ 1 then ℚG has a simple component isomorphic to ℚ(ζ3 ) ⊗ℚ ℍ(ℚ(√3)) ≅ M2 (ζ3 , √3), contradicting Lemma 19.5.1 (5). This proves the claim. Now we separately deal with the cases x2 ∈ ̸ P󸀠 \ {1} and x2 ∈ P󸀠 \ {1}. In both cases we will show that r = 1 and hence, by the above, K = 1 and G = B × H1 , where B is an elementary abelian 2-group and H1 = G0,m,Q,P with Q is an abelian subgroup of index 2 in P. Then, in order to show that G is an epimorphic image of A × H with A and H satisfying condition (3), it is enough to prove that one may assume that x ∈ Q. Suppose x2 ∈ ̸ P󸀠 \ {1}. Then ⟨x, f(s)⟩ /⟨x2 ⟩ is isomorphic to D8 , for every s ∈ P󸀠 \ {1}. By Lemma 19.6.2, f(s) ∈ ̸ Q because the natural image of f(s) in ⟨x, f(s)⟩ / ⟨x2 ⟩ has order 4. This implies that r = 1, because otherwise y1 , y2 , y1 y2 ∈ ̸ Q, contradicting the fact that [P : Q] = 2. Then, replacing x by xy1 if needed, one may assume that x ∈ Q. Assume now that x2 ∈ P󸀠 \ {1}. We claim that one may assume that x2 = t1 . If α α x2 ∈ ̸ ⟨t2 , . . . , t r ⟩ this is obtained by replacing y1 by f(x2 ). Otherwise x2 = t22 ⋅ ⋅ ⋅ t r r , for some α1 , . . . , α r ∈ {0, 1} with α i = 1 for some i. Then, replacing {y1 , . . . , y r } by {f(x2 ), y1 , . . . , y i−1 , y i+1 , . . . , y r }, one obtains the desired conclusion. So we assume that x2 = t1 . Then f(s) has order 4 in ⟨x, f(s)⟩ / ⟨x2 ⟩ ≅ D8 for every s ∈ P󸀠 \ ⟨t1 ⟩ and therefore (P󸀠 \ ⟨t1 ⟩) ∩ Q = 1. This implies that r ≤ 2. If r = 2 then y1 y2 , y2 ∈ ̸ Q and therefore y1 ∈ Q. Replacing x by xy2 if needed, one may assume that x ∈ Q. So Q = ⟨x, y1 , y22 ⟩. Let 1 ≠ m ∈ M. Then (U = ⟨m, x, y22 ⟩ , ⟨y22 ⟩) is a strong Shoda pair of H = ⟨m, P⟩ and [H : U] = 4, contradicting Lemma 19.5.1 (2). Thus r = 1 and x2 = t1 = y21 . Therefore P ≅ Q8 and either x or y1 does not belong to Q. By symmetry, one may assume that y1 ∈ ̸ Q and, replacing x by xy1 if needed, one may assume that x ∈ Q. This finishes the proof.

192 | 19 Virtually free-by-free groups

19.7 Structure theorem In the previous sections we have classified all the finite groups of Kleinian type and we have proven several equivalent conditions. We summarize these results in the following theorem. Theorem 19.7.1 (Jespers, Pita, del Río, Ruiz and Zalessskii [125]). The following properties are equivalent for a finite group G. (A) U(ℤG) is virtually a direct product of free-by-free groups. (B) For every simple component A of ℚG and some (every) order O in A, O1 is virtually free-by-free. (C) For every simple component A of ℚG and some (every) order O in A, O1 has virtual cohomological dimension at most 2. (D) G is of Kleinian type. (E) Every simple quotient of ℚG is either a field, a totally definite quaternion algebra or M2 (K), where K is either ℚ, ℚ(i), ℚ(√−2) or ℚ(√−3). (F) G is either abelian or an epimorphic image of A × H, where A is abelian and one of the following conditions holds: (1) A has exponent 6 and H is one of the groups W, W1n or W2n . (2) A has exponent 4 and H is one of the groups V, V1n , V2n , U1 , U2 . (3) A has exponent 2 and H is one of the groups T , T1n , T2n or T3n . (4) H = M ⋊ P = (M × Q) : ⟨u⟩2 , where M is an elementary abelian 3-group, P = Q : ⟨u⟩2 , m u = m−1 for every m ∈ M, and one of the following conditions holds: – A has exponent 4 and P = C8 . – A has exponent 6, P = W1n and Q = ⟨y1 , . . . , y n , t1 , . . . , t n , x2 ⟩. – A has exponent 2, P = W21 and Q = ⟨y21 , x⟩. The non-nilpotent groups of Kleinian type are those listed in part (4) with M non-trivial. Proof. Theorem 19.2.3 shows that (A) and (B) are equivalent. That (B) implies (C) is shown in Corollary 19.3.5. Corollary 19.3.8 yields that (C) implies (D). Let d ∈ {1, 2, 3}. From Example 19.1.6 on the description of PSL2 (ℤ[√−d]) it follows that O1 is virtually free-by-free for O an order in ℚ(√−d). From Corollary 19.3.5 and Corollary 19.3.7 we also know that SL2 (ℤ) is virtually free. Because of Proposition 5.5.1 and Proposition 5.5.6 we thus obtain that (E) implies (B). In Theorem 19.5.8 and Theorem 19.6.5 we have shown that (D) implies (F). That (F) implies (E) follows from Lemma 19.4.3, Lemma 19.4.4, Lemma 19.4.5 and Lemma 19.4.6. Hence, the result follows. Example 19.1.6, Corollary 19.3.5 and Corollary 19.3.7 yield that the groups SL2 (√−d), with d ∈ {1, 2, 3} are free-by-free but not free. This combined with Theorem 19.2.3 gives the following consequence. Corollary 19.7.2 (Jespers, del Río [106]). Let G be a finite group. The following conditions are equivalent:

19.7 Structure theorem

| 193

(1) U(ℤG) is virtually a (finite) direct product of free products of (finitely many finitely generated) abelian groups. (2) U(ℤG) is virtually a (finite) direct product of (finitely generated) free groups. (3) Every nonabelian simple quotient of ℚG is isomorphic to either M2 (ℚ), ( −1,−3 ℚ ) or ℍ(K) with K = ℚ, ℚ(√2) or ℚ(√3). A precise description of the finite groups G such that U(ℤG) is virtually a direct product of free products of free abelian groups also easily can be deduced from Theorem 19.7.1. Corollary 19.7.3 (Jespers, Leal, del Río [106, 115, 146]). The following properties are equivalent for a finite groups G. (1) U(ℤG) is virtually a (finite) direct product of free products of (finitely many finitely generated) abelian groups. (2) G is either abelian or isomorphic to H × C2k , where H is one of the following groups: (a) ⟨x, y | x4 = y4 = (x2 , y) = (x, y2 ) = (x, (x, y)) = (y, (x, y)) = 1⟩, (b) ⟨x, y1 , . . . , y n | x4 = y2i = (y i , y j ) = (x2 , y i ) = ((x, y i ), y j ) = ((x, y i ), x) = 1⟩, (c) ⟨x, y1 , . . . , y n | x4 = y4i = y2i (x, y i ) = (y i , y j ) = (x2 , y i ) = (y2i , x) = 1⟩, (d) ⟨x, y1 , . . . , y n | x2 = y2i = (y i , y j ) = ((x, y i ), y j ) = (x, y i )2 = 1⟩, (e) ⟨x, y1 , . . . , y n | x2 = y4i = y2i (x, y i ) = (y i , y j ) = ((x, y i ), x) = 1⟩, (f) ⟨x, y1 , . . . , y n | x4 = y4i = x2 y21 = y2i (x, y i ) = (y i , y j ) = (y2i , x) = 1⟩, (g) ⟨x, y1 , . . . , y n | x4 = x2 y4i = y2i (x, y i ) = (y i , y j ) = 1⟩, (h) Z ⋊ ⟨x⟩ where Z is an elementary abelian 3-group, x has order 2 or 4 and z x = x−1 zx = z−1 for every z ∈ Z, (i) Z ⋊ H where Z is an elementary abelian 3-group, H = ⟨x, y⟩ ≅ Q8 and z x = x−1 zx = z−1 = y−1 zy = z y for every z ∈ Z. One now also easily deduces the following consequence. Corollary 19.7.4 (Jespers [102]). The following are equivalent for a finite group G: (1) U(ℤG) is virtually a free product of abelian groups. (2) U(ℤG) is either virtually abelian or virtually nonabelian free. (3) ℚG is a direct product of fields, division rings of the form ( −1,−3 ℚ ) or ℍ(K) with K = ℚ, ℚ(√2) or ℚ(√3) and at most one copy of M2 (ℚ). (4) One of the following conditions hold: (a) G = Q8 × C2n (in this case U(ℤG) is finite), (b) G is abelian or (c) G is one of the following groups: D6 , D8 , Q12 or W21 (in this case U(ℤG) is virtually nonabelian free).

Problems 19.7.1. Prove Corollary 19.7.3 and Corollary 19.7.4

194 | 19 Virtually free-by-free groups

This finishes our story on Group Ring Groups. But . . . darn . . . we forgot the murder . . .

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Index of Notation Numbers that are not bolded represent pages from volume 1. F(χ), field of character values 86 ( a,b F ), quaternion F -algebra 35 (g, h) = g−1 h−1 gh 285 (X, Y) = ⟨(x, y) : x ∈ X, y ∈ Y⟩ 285 (E/F, f), classical crossed product 66 (E/F, σ, a) or (E/F, a), cyclic algebra 70 |g|, |X|, |z|, order of a group element g, cardinality of a set X, modulus of a complex number z 2 ‖ ‖, Euclidean norm 170 ‖A‖, norm of a matrix A 58 (M), isomorphism class of a module 282 [M], the element of K 0 (R) determined by a module M 282 M[m] = {f(m) : f ∈ HomR (M, R)} 292 [a]n , class of a in ℤ/nℤ 29 1 [ϕ, ψ] = |G| ∑g∈G ϕ(g)ψ(g−1 ), ϕ, ψ, class functions of the finite group G 76 ⌈a, b⌉I1 ,I2 = ( ac bd )E2 (I1 ≀ I2 ) 319 R ∗ G, a crossed product of a ring R over a group G 64 G1 ∗ ⋅ ⋅ ⋅ ∗ G n , free product of groups G1 , . . . , G n 15 ∗i∈I G i , free product of groups G i 15 G1 ∗H G2 , amalgamated product of G1 and G2 with amalgamated subgroup H 151 f ∗ = HomR (f, U) 57 M∗ = HomR (M, U) 57 R ∝ I = {(a, b) ∈ R × R : a − b ∈ I} 286 N ⋊ H, semidirect product 26 A ⊗F B, the tensor product of two F -algebras A and B 36 ⊗F E, the Brauer homomorphism Br(F) → Br(E) 55 X −1 M, module of fractions 123 O1 = SL1 (O), Group of elements with reduced norm one over the center 198 h g = g−1 hg and H g = {h g : h ∈ H}. 82 x T , transpose of x ∈ R n 295 x g the (right) action of a group element g on x 180 R G , the fixed subring of a ring under the action of a group G 65 ϕ G , character of G induced by ϕ 83

g GF , the F -conjugacy class of g in G 231 g G , conjugacy class of g in G. 86 χ H , restriction of character χ to a subgroup H 82 ρ H , restriction of a representation ρ to a subgroup H 82 R 0 S, contracted semigroup ring of S with coefficients in R 267 ρ B , representation of a module with respect to the basis B 47 M P = X −1 M for X = R \ P, with P prime 123 f P = X −1 f for X = R \ P, with P prime 123 g p , the p-th part of a periodic group element g 115 ∼F , F -conjugate in a group 231 X̃ = ∑x∈X x 22 ̃ 22 ̃ = ⟨g⟩ g 1 ̃ X̂ = |X| X 22 ̂ 22 ̂ = ⟨g⟩ g ̂ or D ̂ v , completion of D with respect D to a the valuation v 146 1G , trivial character 75 A F (χ), simple component of FG with χ(e) ≠ 0, (χ absolutely irreducible character of G) 85 A n , alternating group in n symbols 233 AnnR (x) annihilator of x in R 138 ̃ = 1 + (1 − h)g h, ̃ bicyclic unit 4 b(g, h) ̃ ̃ b(h, g) = 1 + hg(1 − h), bicyclic unit 4 𝔹n , ball model of the hyperbolic n-space 48 Bass(G), group generated by the Bass units in ℤG 6 Bass(ℤ[ξ]G), group generated by the Bass units in ℤ[ξ]G 341 Bassk (G), group generated by Bass units of the form u k,m (g), with g ∈ G 250 Bassm (G), group generated by the Bass units of the form u k,m (g), with g ∈ G 264 Bic(G), group generated by the bicyclic units in ℤG 4 Bis(x, y), bisector of x and y 52 Bis𝕏n (γ), hypersphere of 𝕏n with Bis𝕏n (γ) ∩ 𝕏n = Bis(C, γ(C)), for C the centre of 𝕏n 61

206 | Index of Notation

B n (G, U), group of n-coboundaries 60 Br(F), Brauer group of F 55 Br(E/F), relative Brauer group of E/F 56 c(b), central unit based on a Bass unit 397 cd(G), cohomological dimension of G 162 C n = ⟨g⟩n , cyclic group of order n ∈ ℕ ∪ {∞} generated by g 2 C(G), non commutative simple components of ℚG 166 ̂ γ = {z ∈ ℂ ̂ : γ(z) = z} 66 ℂ CenR (X), centralizer of X in a ring or group R 38 Cen∗G (a), extended centralizer of a in G 30 CharM/F (a), characteristic polynomial of a over F with respect to a module M 47 Cl(R), class group of a Dedekind domain or a number field R 220 CorFK , corestriction map between Brauer groups 70 D2n , dihedral group of order 2n 5 D±4n , semidihedral groups of order 4n 102 Deg(A), degree of the central simple algebra A 43 D γ (x) = {y ∈ 𝕏 | d(x, y) ≤ d(y, γ(x))}, with γ ∈ PSL2 (ℂ) 96 D(x), Dirichlet fundamental polyhedron with center x 96 DP (x), Dirichlet fundamental polyhedron with center x 96 D Q , decomposition subgroup of the maximal ideal Q 144 D = (⟨a⟩2 × ⟨z⟩4 ) ⋊ ⟨b⟩2 , z a = z b = z, a b = z2 a 360 Δ(O/R), discriminant of an R-order O 157 Δ F (x1 , . . . , x n ), discriminant of x1 , . . . , x n over F 157 e(G, H, K), sum of G-conjugates of ε(H, K) (K  H ≤ G) 93 e(χ), primitive central idempotent e of ℂG with χ(e) ≠ 0, (χ ∈ Irr(G) 78 e F (χ), primitive central idempotent e of FG with χ(e) ≠ 0, (χ absolutely irreducible character) 85 e(E/P), ramification index, for Galois extension 145 e(Q/R), ramification index 142 e P , exponential P-adic valuation 133

e P (I), multiplicity of the maximal ideal P in the factorization of I 131 e ij (r) = I n + E ij (r), (1 ≤ i ≠ j ≤ n 3 E ij (r), matrix with r at (i, j) entry and 0 elsewhere 3 En (I), group generated by elementary matrices e ij (r), 1 ≤ i ≠ j ≤ n, r ∈ I 285 En (R, I), normal closure of En (I) in En (R) 285 E2 (I1 ≀ I2 ) 312 E(I) = ⋃n≥1 En (I) 286 E(R, I) = ⋃n≥1 En (R, I) 286 End(M), endomorphism ring of M 32 Exp(G) = exponent of a group G 26 E = (⟨x⟩4 × ⟨y⟩4 ) ⋊ ⟨b⟩2 , x b = y 360, 361 ̂ − D), ̂ if N ≠ G, ̂ ε(G, N) = ∏D/N∈M(G/N) (N G, otherwise 89 f(E/P), residue degree, for Galois extension 145 f(Q/R), residue degree 142 F n , free group of rank n 15 Φ n = Minℚ (ζ n ), n-th cyclotomic polynomial 205 Fr(F), group of fractional ideals 136 φ, Euler function 21 g(E/P) = F

[E:F] e(F/P)f(F/P)

145

GBic (OG), group generated by generalized bicyclic units of OG defined by the set of idempotents F 342 Gen(G), group of generic units of ℤG 262 GLn (R), general linear group of R 3 GL(R) = ⋃n≥1 GLn (R) 286 GL(n, q) = GLn (𝔽q ) 360 GLn (R, I) = {A ∈ GLn (R) : A − 1 ∈ M n (I)} 285 GL(R, I) = ⋃n≥1 GLn (R, I) 286 ̂n ), group of Möbius transformations 46 GM(ℝ ℍ3γ = {w ∈ ℍ3 : γ(w) = w} 66 ℍ1 = ( −1,−1 ℚ ) 366

ℍ3 = ( −1,−3 ℚ ) 366

ℍ5 = ( −2,−5 ℚ ) 366 ℍn , upper half-space model of the hyperbolic n-space 47 Hn (M), n-th homology group of a positive complex M 57 Hn (G, U), n-th cohomology group of G with coefficients in U 58 Hoe(G), the group generated by the Hoechsmann units of ℤG 262

Index of Notation

I A , I γ isometric circle or isometric sphere of A ∈ GL2 (ℂ) and γ ∈ PSL2 (ℂ). 53 ID(G), GAP identification of the finite group G 120 Ind(A), index of the central simple algebra A 43 Inf, inflation map 69 Irr(G), irreducible complex characters of G 78 Irrp (G), absolutely irreducible characters in characteristic p 78 Isom(𝕏), group of isometries of a metric space 𝕏 44 Isom+ (𝕏), group of orientation preserving isometries of 𝕏 44 ι, map on ℂ with ι(z) = z for z ≠ 0 and ι(0) = 1 253 J(R), Jacobson radical of R 33 K0 (R), Grothendieck group of R 282 K1 (R, I), Whitehead group of R 288 K1 (R, I), Whitehead group of R relative to the ideal I 288 𝕂n , Klein model of the hyperbolic n-space 48 L(m) = ∑d|m μ ( m ) [d] ∈ ℤ0 Mn , d logarithmic map 267 L∗ (m), image of L(m) in ℤ0 M∗n 267 Lin(G, K), linear characters of G with kernel K 92 L(z, χ), Dirichlet function. 217 m F (χ), the Schur index of χ over F 85 M(G), set of minimal non-trivial normal subgroups of G 89 M n (R), n × n matrix ring over a ring R 3 ̂n ), group of orientation preserving Möbius M(ℝ ̂n ) 46 transformations of GM(ℝ Max(R), set of maximal ideals of R 123 MinF (a), minimal polynomial of a over F 39 MinM/F (a), minimal polynomial of a over F with respect to a module M 47 Mn , multiplicative monoid of ℤ/nℤ 267 μ, Möbius function 98 N G (H), normalizer of H in G 102 nr, componentwise reduced norm over the center (for separable algebras) 304 NrM/F (a), norm of a over F with respect to a module M 47 ξ k −1 η k (ξ) = ξ−1 , cyclotomic unit 5

| 207

o n (a), multiplicative order of a modulo n. 29 ord(M), order of the finitely generated torsion module M 140 ℙ, set of positive prime integers 222 Pr(R), group of principal fractional ideals of a domain R 220 PSB2 (ℂ) = SB2 (ℂ)/ ⟨−I⟩ 58 PSB2 (ℍ(ℝ)) = SB2 (ℍ(ℝ))/ ⟨−I⟩ 57 Q4n , quaternion group of order 4n 5 ℚp , field of p-adic numbers 150 r F (A), number of simple components of F ⊗ℚ A 226 rkR (M), rank of a finitely generated module 137 ℝn , n-dimensional Euclidean space 44 ̂n , one point compactification of ℝn 44 ℝ ResFK , restriction map between Brauer groups 69 ResGH : H n (G, U) → H n (H, U), restriction map 58 Resm : Gal(ℚ(ζ n )/ℚ) → Gal(ℚ(ζ m )/ℚ), restriction map, for m ∤ n 210 str(R), stable range of R 292 S n , symmetric group on n symbols 3 S(a, r) = {x ∈ ℝn :| x − a |= r} 45 SB2 (ℂ) 58 SB2 (ℍ(ℝ)) 57 𝕊n , n-dimensional unit sphere 44 SLn (R) = {x ∈ GLn (R) : det(x) = 1}, (R commutative ring) 3 SLn (O) = {x ∈ GLn (O) : nr(x) = 1}, (O subring of separable algebra) 304 SL(n, q) = SLn (𝔽q ) 360 SLn (O, I) = SLn (O) ∩ GLn (O, I) 304 SL2 (A, I1 ≀ I2 ) 312 Spec(R), set of prime ideals of R 123 SU2 (ℂ), group of unitary complex 2 × 2 matrices 58 trE/F (χ), trace of the E-character χ, finite Galois extension) 86 trE/F (a), trace over F of a group algebra element a ∈ EG, with E/F a finite Galois extension 86 T Q , inertia group of Q 145 TrM/F (a), trace of a over F with respect to a module M 47

208 | Index of Notation

T HG , transfer map 59 τ ϕ = ∑ρ∈G c ϕ c (ρ)ρ(ζ c ), Gauss sum of a linear character ϕ with conductor c 211 m

̃ u k,m (g) = (1 + g + g2 + ⋅ ⋅ ⋅ + g k−1 )m + 1−k |g| g , Bass unit 6 U(T), group of units of T 1 U1 (ℤG), group of units of ℤG of augmentation 1 228 UF (n) = {r ∈ ℤ/nℤ : σ(ζ n ) = ζ nr , for some σ ∈ Gal(F(ζ n )/F)} 230 v P , (non-exponential) P-adic valuation 148 vol(X), volume of X 170 vol(V/L), the volume of a lattice L in V 172

W(I1 , I2 ) = {(a, b) ∈ R2 : a − 1 ∈ I1 I2 , b ∈ I1 , aR + bR = R} 319 ω : RG → R, augmentation map 6 ω N,R = ω N : RG → R(G/N), relative augmentation map. 22

χ ϕ , Dirichlet character of the linear character ϕ of Gal(ℚ(ζ n )/ℚ) 210

Z(R), center of a group or ring R 35 Z n (G, U), group of n-cocycles 60 ̂ p , ring of p-adic integers 150 ℤ ζ K , Riemann ζ function. 217

Index Numbers that are not bolded represent pages from volume 1. absolutely irreducible – representation 78 – character 78 action – regular 351 adapted – α 86 admissible lattice 188 Albert, A. A. 166 Aleev, R.Zh. 237 algebra – central 35 – central simple 35 – cyclic 70 – cyclotomic 114 – exceptional 342 – group 1 – Hamiltonian quaternion 17 – of Kleinian type 117 – quaternion 35 – Schur 166 – semisimple 35 – separable 38 – simple 28 – symbol 42 – truncated group 348 algebraically independent 159 Allen, P. J. 26 alternating unit 8 alternating-like unit 276 amalgamated free product 151 antipode 51 Archimedean valuation 147 Artin’s Induction Theorem 99 Artin, E. 92, 232 attractor 16 augmentation – ideal 22 – map 6 – modulo a normal subgroup 22 – relative 22 Ayoub, Ch. 228 Ayoub, R. G. 228

Bächle, A. 366 Bak, A. 342 Banieqbal, B. 359 base of a horosphere 103 basis of a lattice 171 Bass Independence Theorem 253 Bass unit 6 – generalized 400 Bass Unit Theorem 338 Bass, H. 6, 226, 248, 252–254, 295, 337, 340, 342, 343 Benard, M 166 Benard-Schacher Theorem 166 Berman, S. D. 21, 73, 112, 113, 227, 230, 231 Bianchi groups 120 Bianchi, L. 120 bicyclic unit 4 – generalized 9 bisector 52 Borel, A. 179 Brauer – equivalent 44 – group 55 – relative 55 Brauer Splitting Theorem 114 Brauer, R. 73, 113, 115, 166 Brauer-Witt Theorem 115 Broche, O. 104, 255, 258, 261

Caicedo, M. 366 cell of a tessellation 73 center of a ring or a group 35 central – algebra 35 – simple algebra 35 Chan, S. P. 99 character 75 – absolutely irreducible 78 – irreducible 75 – table 78 characteristic polynomial 47 – reduced 50 chordal metric 44

210 | Index

class – function 75 – group 220 – sum 74 classical – crossed product 66 – involution 1 Cliff, G. H. 228, 265 coboundary 60 cocompact discontinuous group 95 cocycle 60 – condition 60 codimension of a polyhedron 71 cohomological dimension 162 – virtual 163 cohomology group 58 commensurable subgroups 155 complement 26 – Frobenius 349 – normal 26 complete valuation 146 completion 146 complex – embedding 41 – place 149 – prime 149 compositum – in a field 38 – over a field 41 conductor 210 cone 52 congruence subgroup 17, 304 conjugacy class – F - 231 conjugate – F - 231 constituent – of a character 79 – of a representation 79 constructible unit 8 contracted semigroup ring 267 convex 52 core 111 corestriction 59 Corrales, C. 95 crossed product 64 – classical 66 cycle of edges 82 cyclic algebra 70

cyclotomic – algebra 114 – class 91 – field 205 – polynomial 205 – shifted 258 – unit 6 DACD 107 DAFC 106 decomposition – T - 18 – subgroup 144 Dedekind domain 130 define (polynomial) – a unit on an order 255 – generic units 255 degree – of a central simple algebra 43 – of a character 79 – of a representation 46, 74 – of a unit 276 diagonal change of basis 65 dihedral group 5 dimension – Krull 293 – of a polyhedron 71 Dirichlet – Algorithm – for Cocompact Discrete Groups 95 – for Finite Covolume 95 – character 210 – fundamental polyhedron 96 Dirichlet’s Unit Theorem 176 Dirichlet, J. P. G. L. 170, 176 discontinuous group 65 – cocompact 95 – geometrically finite 95 – with finite covolume 95 discrete – group 65 – subset 10 – valuation 149 – ring 130 discriminant – of a number field 157 – of a sequence 157 – of an order 157 division ring 2

Index |

Dooms, A. 26, 99 Double Centralizer Theorem 38 double coset 83 dual basis 54 edge – loop 79 – of a tesellation 73 Eisele, F. 359, 360, 366 elementary – abelian group 26 – matrix 3 elliptic Möbius transformation 66 embedding – complex 41 – real 41 equivalent – Brauer 44 – extensions 63 – representations – of a group 74 – of an algebra 46 – Shoda pairs 394 – valuations 146 essential – half-space 71 – hyperplane 71 – ideal 304 Euclidean – norm (order) 373 – space 44 – sphere 45 exceptional algebra 342 exponential P-adic valuation 133 extended centralizer 30 extension of groups 63 factorization 131 Farey Symbols 99 Ferraz, R. A. 227, 231, 420, 423, 1, 10, 12, 13 field – local 161 – number 9 – of characters 86 – totally complex 177 – totally real 177 finite – covolume – discontinuous group with 95

211

– place 148 – prime 148 Fitzgerald, R. W. 373 fixed subring 65 Ford fundamental polyhedron 100 Ford, L. 99 fractional ideal 135 Franz Independence Lemma 225 Franz, W. 225 free – companion 21 – monoid 15 – point 4 – product 15 Frobenius – automorphism 164 – complement 349 – group 349 – kernel 349 Frobenius, G. 349 Fuchsian group 66 full – R-lattice 152 – lattice 171 fundamental – domain 38 – polyhedron 80 – Dirichlet 96 – Ford 100 – of a lattice 172 – units 179 Gauss sum 211 Gauss’ Lemma 127 general linear group 3 generalized – Bass unit 400 – bicyclic units 342 generic units – group of 262 – polynomial defining 255 geodesic 48 – segment 51 geometrically finite – discontinuous group 95 – polyhedron 71 Giambruno, A. 356 Gonçalves, J. Z. 1, 2, 10, 14, 18, 20–22, 28, 31, 35, 37

212 | Index

Grothendieck group 282 group – abelian-by-supersolvable 107 – algebra 1 – Bianchi 120 – Brauer 55 – center of a 35 – cohomology 58 – dihedral 5 – discontinuous 65 – cocompact 95 – geometrically finite 95 – with finite covolume 95 – discrete 65 – elementary abelian 26 – fixed point free 345 – Frobenius 349 – Fuchsian 66 – general linear 3 – Grothendieck 282 – Hamiltonian 26 – homology 57 – inertia 145 – Kleinian 66 – metabelian 110 – monomial 84 – nearly indecomposable 152 – of generic units 262 – of Kleinian type 165 – p󸀠 112 – periodic 2 – quasi-elementary 112 – quaternion 5 – ring 1 – semidihedral 102 – special linear 3 – strongly monomial 104 – supersolvable 107 – topological 65 – torsion 2 – torsion-free 2 – unitary 58 – Whitehead 288 – relative 288 – Z- 354 group algebra 1 – truncated 348 group ring 1 – skew 65

– twisted 65 Guralnick, R. M. 37 half-space 45 – closed 69 – essential 71 – open 184, 69 Hamiltonian – group 26 – quaternion algebra 17 Harish-Chandra, 179 Hartley, B. 1, 13, 20 Hasse Norm Theorem 150 Hasse, H. 150, 166 height 101 Hensel’s Lemma 161 Hensel, K. 161 Herman, A. 168 Hey’s Theorem 174 Hey, K. 170, 173, 174 Higman, G. 20, 21, 26, 204, 227, 228 HNN extension 151 Hobby, C. 26 Hoechsmann unit 8 Hoechsmann, K. 8, 264, 265 homology group 57 homotopic 195 homotopy 195 horoball 103 horosphere 103 Hurwitz quaternion 128 Hurwitz, A. 179 hyperbolic – Möbius transformation 66 – space 47 hyperplane 69 – essential 71 hypersphere 45 ideal – essential 304 – maximal 123 – order 140 – point 102 – prime 123 index 43 – local 165 – Schur 85

Index |

induced – character 83 – representation 83 inertia group 145 infinite – place 148 – prime 148 inflation 69 injective module 33 integral – closure – of a domain 126 – over a ring 126 – over a ring 125 integrally closed – domain 126 – in a ring 125 invariant 165 – local 165 invariants – list of local 166 involution – classical 1 irreducible – F -character 75 – representation 75 isometric – circle 19 – sphere 53, 62 isometry – orientation preserving 44 – orientation reversing 44 Jacobson radical 33 Janssens, G. 92, 100 Jespers, E. 26, 92, 226, 252, 337, 343, 345, 358, 366, 379, 383, 385, 388, 390, 392, 394, 397, 400, 402, 408, 416, 420, 2, 69, 95, 99, 144, 149, 151, 192, 193 Juriaans, S. O. 95, 144 Kargapolov, A. V. 237 kernel – Frobenius 349 – of character 79 Kiefer, A. 359, 360, 366, 69, 95, 144 Klein, F. 1 Kleinert, E. 173, 174, 179, 201, 359, 375, 379, 158, 160

213

Kleinert, K. 379 Kleinian – group 66 – type – algebra of 117 – group of 165 Konovalov, A. 99 Krull dimension 293 Kulkarni, R. S. 99 Kuroš’ Theorem 16 Kuroš, A. 16 Lam, T.-Y. 339 Lang, M. L. 99 lattice 152, 171 – full 171 – full R- 152 Leal, G. 337, 343, 345, 358, 366, 383, 95, 149, 151, 193 Liehl, B. 312, 342, 343 Lim, C. H. 99 linear – character 82 – representation 82 list of local invariants 166 local – field 161 – index 165 – invariant 165 – ring 124 local index – at a prime of ℚ (for Schur algebras) 167 locally finite 70 logarithm map 176 loop 81 loxodromic Möbius transformation 67 Mackey, G. 83, 84 Mandel, A. 1, 10 Marciniak, Z. S. 255, 258, 263, 265, 270, 272, 273, 276, 280, 1, 2, 8 Margulis, G. A. 376, 378 Maschke 74 Maschke’s Theorem 74 matrix units 343 maximal – ideal 123 – order 153 measurable 170

214 | Index

Milnor, J. 226, 248, 252, 254, 337, 340 minimal polynomial 39, 47 Minkowski’s Theorem 173 Minkowski, H. 173 Möbius – function 98 – Inversion Formula 256 – transformation 19, 46 module – injective 33 – projective 33 – semisimple 32 – simple 32 – torsion 137 – torsion-free 137 monomial – character 84 – group 84 – representation 84 multiplicative order 29 multiplicatively closed 123 nearly indecomposable 152 Noether, E. 166 Noether-Skolem Theorem 37 non-Archimedean valuation 147 norm 47 – Euclidean order 373 – in complex vector space 14 – of a matrix 58 – of an element of PSL2 (ℂ) 58 – of an ideal 141 – reduced 51 normal – basis 417 – complement 26 – element (of a field extension) 417 – form 15 normalized unit 6 number field 9 Olivieri, A. 73, 92, 93, 102 Olteanu, G. 73, 92, 115, 379, 392, 394, 397, 400, 402, 408, 416, 420 O’Meara, O. T. 179 opposite ring 33 order 152 – maximal 153

orientation – preserving isometry 44 – reversing isometry 44 Ostrowski’s Theorem 148 Ostrowski, A. 148 P-adic – exponential valuation 133 – valuation 148 p-adic – integers 150 – numbers 150 parabolic Möbius transformation 66 Parmenter, M. M. 385, 390, 400 partial tessellation 73 partition of a positive integer 233 Passman, D. S. 1, 2, 14, 18, 20, 21, 28 Pell equation 14 periodic – element of a group 2 – group 2 Perlis, S. 90, 91 permutation matrix 3 p 󸀠 -group 112 Pickel, P. F. 1, 13, 20 Pita, A. 117, 144, 147, 149, 151, 192 place 148 – complex 149 – finite 148 – infinite 148 – real 148 Poincaré – extension 49 – upper half-space 47 Poincaré’s Theorem 82 Poincaré, H. 194, 80 Polcino Milies, C. 383 polyhedron 71 – fundamental 80 – locally finite 80 – geometrically finite 71 positive – complex 57 – definite – bilinear form 181 – quadratic form 181 prime – above 143, 149 – complex 149

Index |

– finite 148 – ideal 123 – infinite 148 – of a number field 148 – real 148 primitive – central idempotents 34 – element of a finite field 31 – root modulo an integer 29 projective – module 33 – subset 15 prolongation of a valuation 146 p-th part 115 quaternion – algebra 35 – totally definite 200 – conjugation 35 – group 5 – Hurwitz 128 ramification index – over a Dedekind domain 142 – over a local field 163 ramified – at a prime 165 – in a field extension 143 – over a local field 163 – with respect to a Dedekind domain 143 ramifies – at a prime 165 – at a prime of ℚ (for Schur algebras) 167 – in a field extension 143 rank – of a lattice 171 – of a module 137 real – embedding 41 – place 148 – prime 148 reduced – characteristic polynomial 50 – degree 43 – norm 51 – trace 50 reflection in a hypersphere 45 regular – action 351

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– representation – of a group 75 – of an algebra 47 regulator – of a list of units 179 – of a number field 179 Rehmann, U. 342 relative – augmentation map 22 – interior 71 representation – F - 74 – absolutely irreducible 78 – induced 83 – of algebra 46 – with respect to a basis 47 – of group 74 – regular – of a group 75 – of an algebra 47 residue – degree – of a discrete valuation 163 – over a Dedekind domain 142 – division ring 149 – field 149 restriction 58 ring – center of a 35 – contracted semigroup 267 – discrete valuation 130 – division 2 – local 124 – of integers 127 – opposite 33 – semigroup 267 – semilocal 292 – semisimple 33 – simple 28 – skew group 65 – torsion-free 4 – twisted group 65 – valuation 149 Río, Á. del 73, 92, 93, 102, 104, 226, 252, 255, 258, 261, 366, 379, 392, 394, 397, 400, 402, 408, 416, 420, 2, 21, 22, 31, 35, 37, 69, 95, 117, 144, 147, 149, 151, 160, 192, 193 Ritter, J. 4, 8, 227, 232, 233, 265, 337, 346, 347, 383, 389, 390

216 | Index

Roquette, P. 414 Ruiz, M. 2, 117, 144, 147, 151, 192 Salwa, A. 2, 8 Sanov’s Theorem 4 Sanov, I. N. 1, 4 Schacher, M. M. 166 Schur – algebra 166 – index 85 Schur’s Lemma 32 Schur, I. 32 Sehgal, S. K. 4, 227, 228, 232, 233, 255, 258, 263, 265, 270, 272, 273, 276, 337, 346, 347, 356, 383, 389, 390, 1, 2, 8, 13 semibasis 185 semidihedral group 102 semigroup ring 267 semilocal ring 292 semisimple – algebra 35 – module 32 – ring 33 separable algebra 38 Serre, J.-P. 163 shifted cyclotomic polynomial 258 Shirvani, M. 1, 10 Shoda pair 93 Shoda pairs – equivalent 394 Shoda, S. 84 side of a tesellation 73 Siegel, C. L. 170, 179 Silva, A. de A. e 95, 144 Simón, J. J. 73, 92, 93, 102, 420, 423 simple – algebra 28 – components 34 – module 32 – ring 28 simply connected 195 skew group ring 65 Sokolov, V. V. 237 Souza Filho, A. C. 95, 144 special linear group 3, 200 splitting field – of a finite group 76 – of an algebra 42 stabilizer 96

stable range 292 Stallings, J. R. 163 standard resolution 59 Steinitz’s Theorem 138 Steinitz, W. 138 strong Shoda pair 104 strongly monomial – character 104 – group 104 subgroup – congruence 17, 304 – decomposition 144 subgroups – commensurable 155 submodule – torsion 137 subspace – generated by a subset in a Riemann manifold of constant curvature 52 – of a Riemann manifold of constant curvature 51 supersolvable group 107 support 1 Swan, R. G. 194, 163 symbol algebra 42 symmetric 229 system of fundamental – (central) units 237 Tan, S. P. 99 tensor product 36 tessellation 73 – partial 73 thick 71 Thompson, J. G. 351 tiles 73 Tits Alternative 203 Tits, J. 1, 14, 21 topological group 65 torsion – element of a group 2 – group 2 – module 137 – submodule 137 torsion-free – group 2 – module 137 – ring 4

Index |

totally – complex field 177 – definite quaternion algebra 200 – real field 177 trace 47 – function 7 – in group rings 8 – reduced 50 transfer map 58 trivial units 2 truncated group algebra 348 twisted group ring 65 uniformizer 149 unimodular 292 unipotent matrix 285 unit 1 – alternating 8 – alternating-like 276 – Bass 6 – bicyclic 4 – generalized 342, 9 – constructible 8 – cyclotomic 6 – Hoechsmann 8 – normalized 6 unit ball model for hyperbolic geometry 48 Unit Conjecture 2 unit sphere 44 unitary – group 58 – matrix 58 Universal Property – of Group Rings 1 – of Tensor Products 36 unramified – in a field extension 143

– over a local field 163 – with respect to a Dedekind domain 143 valuation 146 – Archimedean 147 – complete 146 – discrete 149 – non-Archimedean 147 – prolongation of 146 – ring 149 – trivial 147 valuations – equivalent 146 Van Gelder, I. 226, 252, 359, 360, 366, 379, 392, 394, 397, 402, 416, 420 Vaseršte˘ın, L. N. 312, 342, 343, 376, 377 Venkataramana, T. N. 342, 343 virtual cohomological dimension 163 v-metric 146 volume 170 v-topology 146 Walker, L. 90, 91 Wedderburn – components 34 – decomposition 34 Wedderburn-Artin Theorem 33 Weiss, A. 228, 265, 382 Weyl, H. 179 Whitehead group 288 – relative 288 Whitehead’s Lemma 287 Whitehead, A. N. 287 Witt, E. 73, 112, 113, 115, 230, 231 Z-group 354 Zalessskii, P. 151, 192 Zassenhaus, H. 174

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