Mathematical Sciences Research Institute Publications

Mathematical Sciences Research Institute Publications

19 Editors

S.S. Chern I. Kaplansky C.C. Moore I.M. Singer

Mathematical Sciences Research Institute Publications

Volume 1 Volume 2 Volume 3 Volume 4 Volume 5 Volume 6 Volume 7 Volume 8 Volume 9 Volume 10 Volume 11 Volume 12 Volume 13 Volume 14 Volume 15 Volume 16 Volume 17 Volume 18 Volume 19 Volume 20

Freed and Uhlenbeck: Instantons and Four-Manifolds Second Edition Chern (ed.): Seminar on Nonlinear Partial Differential Equations Lepowsky, Mandelstam, and Singer (eds.): Vertex Operators in Mathematics and Physics Kac (ed.): Infinite Dimensional Groups with Applications Blackadar: K-Theory for Operator Algebras Moore (ed.): Group Representations, Ergodic Theory, Operator Algebras, and Mathematical Physics Chorin and Majda (eds.): Wave Motion: Theory, Modelling, and Computation Gersten (ed.): Essays in Group Theory Moore and Schochet: Global Analysis on Foliated Spaces Drasin, Earle, Gehring, Kra, and Marden (eds.): Holomorphic Functions and Moduli I Drasin, Earle, Gehring, Kra, and Marden (eds.): Holomorphic Functions and Moduli II Ni, Peletier, and Serrin (eds.): Nonlinear Diffusion Equations and their Equilibrium States I Ni, Peletier, and Serrin (eds.): Nonlinear Diffusion Equations and their Equilibrium States II Goodman, de la Harpe, and Jones: Coxeter Graphs and Towers of Algebras Hochster, Huneke, and Sally (eds.): Commutative Algebra Ihara, Ribet, and Serre (eds.): Galois Groups over Q Concus, Finn, and Hoffman (eds.): Geometric Analysis and Computer Graphics Bryant, Chern, Gardner, Goldschmidt, and Griffiths: Exterior Differential Systems Alperin (ed.): Arboreal Group Theory Dazord and Weinstein (eds.): Symplectic Geometry, Groupoids, and Integrable Systems

Roger C. Alperin Editor

Arboreal Group Theory Proceedings of a Workshop Held September 13-16, 1988

With 62 Illustrations

Springer-Verlag New York Berlin Heidelberg London Paris Tokyo Hong Kong Barcelona

Roger C. Alperin Department of Mathematics and Computer Science San Jose State University San Jose, California 95192 USA

Mathematical Sciences Research Institute 1000 Centennial Drive Berkeley, CA 94720 USA

The Mathematical Sciences Research Institute wishes to acknowledge support by the National Science Foundation. Mathematical Subject Classifications: 20E06, 20E99, 57M05, 57M99

Library of Congress Cataloging-in-Publication Data Alperin, Roger C. Arboreal group theory I by Roger C. Alperin. p. cm. - (Mathematical Sciences Research Institute publications: 19) "During the week of September 13, 1988 the Mathematical Sciences Research Institute hosted a four day Workshop on Arboreal Group Theory."-CIP pref. Includes bibliographical references.

ISBN-I3:978-I-4612-7811-5

1. Group theory-Congresses. 2. Trees (Graph theory)-Congresses. 3. Geometry, Hyperbolic-Congresses. I. Workshop on Arboreal Group Theory (1988 : Mathematical Sciences Research Institute) II. Title. III. Series. QA171.A546 1991 91-232 512' .2-dc20 Printed on acid-free paper. © 1991 Springer-Verlag New York, Inc. Softcover reprint of the hardcover 1st edition 1991

All rights reserved. This work may not be translated or copied in whole or in part without the written permission of the publisher (Springer-Verlag New York, Inc., 175 Fifth Avenue, New York, NY 10010, USA), except for brief excerpts in connection with reviews or scholarly analysis. Use in connection with any form of information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed is forbidden. The use of general descriptive names, trade names, trademarks, etc., in this publication, even if the former are not especially identified, is not to be taken as a sign that such names, as understood by the Trade Marks and Merchandise Marks Act, may accordingly be used freely by anyone. Camera-ready copy prepared by the Mathematical Sciences Research Institute using AW-TEX·

9 8 7 6 5 4 3 2 1 ISBN-13:978-1-4612-7811-5 e-ISBN-13:978-1-4612-3142-4 DOl: 10.1007/978-1-4612-3142-4

Preface During the week of September 13, 1988 the Mathematical Sciences Research Institute hosted a four day Workshop on Arboreal Group Theory. The organizing committee consisted of R. Alperin (chairman), H. Bass, and P. Shalen. This volume is the offspring of that meeting. The program of the conference centered around the topic of the theory of groups acting on trees and the various applications to hyperbolic geometry; topics included the theory of length functions, structure of groups acting freely on trees, spaces of hyperbolic structures and their compactifications, and moduli spaces for tree actions. Below is the summary of speakers and titles at the Workshop: • H. Bass: "Group actions on non-archimedean trees" • M. Bestvina: "Train tracks for free group automorphisms" • I. Chiswell: "Length functions and pregroups" • M. Culler: "The space of free actions on simplicial trees" • M. Feighn: "Bounding the complexity of simplicial group actions on trees" • M. Handel: "Train tracks" • R. Jiang: "Free actions on JR-trees" • R. Kulkarni: "Lattices on trees" • A. Lubotsky: "Trees and discrete subgroups of Lie groups over local fields" • J.P. Otal: "Length spectrum of negatively curved surfaces" • W. Parry: Axioms for translation length functions" • F. Paulin: "The Gromov topology on group actions on JR.-trees" • D. Promislow: "Nonexistence of invariant measures on trees" • F. Rimlinger: "JR-tree dynamics" • P. Sarnak: "Ramanujan graphs" • P. Shalen: "Dendrology of groups in low Q-ranks: I, II" • R. Skora: "Geometric actions of surface groups on A-trees" • M. Steiner: "A natural contraction for spaces of actions of finitely generated free gropus on JR-trees" • K. Vogtmann: "Local structure of the space of free simplicial actions" Many of the topics discussed in these lectures have been included in this volume; additional contributions to the volume have also been made by J. Alonso, S. Basarab, F. Bonahon, M. Cohen, H. Gillet, M. Lustig and J. Stallings. Special thanks are due to the referees for giving their time and assistance in the preparation of this volume. It is also a pleasure to thank the staff at MSRI for providing a congenial atmosphere for the Workshop and also all the necessary technical aid in preparing the manuscript for publication. Thanks also to the organizers of the program on Combinatorial Group Theory and Geometry, K. Brown, S. Gersten and J. Stallings, for their encouragement and support of the Workshop and to my co-organizers H. Bass and P. Shalen for their invaluable assistance in making the Workshop a success.

Roger C. Alperin

California lR-tree, Mt. Diablo State Park . Photo by R.C. Alperin.

Dedicated to the memory of Roger Lyndon {1917-1988}.

Arboreal Group Theory Table of Contents

Preface.............................................................

v

J.M. ALONSO

Growth Functions of Amalgams

1

S.A. BASARAB

On a Problem Raised by Alperin and Bass

H.

BASS

Group Actions on Non-Archimedean Trees

35

69

M. BESTVINA AND M. FEIGHN

A Counterexample to Generalized Accessibility

133

F. BONAHON

Geodesic Currents On Negatively Curved Groups I.M. CHISWELL

Pregroups and Lyndon Length Functions

143

169

M.M. COHEN, M. LUSTIG AND M. STEINER

JR- Tree Actions are Not Determined by the Translation Lengths of Finitely Many Elements ................................... 183

M. CULLER AND K. VOGTMANN

The Boundary of Outer Space in Rank Two

H.

GILLET AND

R.

JIANG

W.

P.

SHALEN

Cohomological Dimension of Groups Acting on JR- Trees Branch Points and Free Actions on JR.-Trees

PARRY

Axioms for Translation Length Functions

F. PAULIN

189 231 251 295

Outer Automorphisms of Hyperbolic Groups and Small Actions on JR- Trees. . . . . . . ... . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 331

x

F.

Table of Contents RIMLINGER

The Structure of Promislow's Continuous Free Product

J.R. STALLINGS

Foldings of G- 'frees

345 355

Growth functions of amalgams JUAN

M.

ALONSO

§O. Introduction. Let G be a finitely generated group with a finite generating set S. Then S determines a length functionf on G (see §l below) which is used to define the growth series of G:

G(z)

=L

zl(g)

gEG

Recently there has been a lot of interest in the study of these series. See for instance [Bel) [Be2) [C) [F-Pl) [F-P2) [G) [Sm) [W). The purpose of this paper is to give a condition (admissibility, see §2) insuring that, if G = G t *A G 2 , then

(1)

1

1

1

1

G(z)

Gt(z)

G 2 (z)

A(z)

--=--+-----

for appropriately chosen generating sets (see Theorem 2 below). By way of motivation, recall that under rather mild assumptions [Br, p.250) there is the similar, well-known formula:

relating the Euler characteristics of these groups. certain groups, one has the remarkable formula:

(2)

Recall also that, for

1

X(H) = H(l)

Serre [Sl) proved (2) for Coxeter groups, and Cannon [C) for closed surface groups. In both cases H(z) is the series of a rational function, and it is this function one evaluates at z = 1 in (2). These facts suggested the plausibility of (1). It should be pointed out, however, that both the growth function and the truth of (2) depend heavily on the choice of generating set. Parry [P) constructed the first example by showing that the free group of rank

2

J .M. Alonso

two has a generating set for which (2) is false. See also [F-PI] where an example is given of a generating set for the closed orient able surface group of genus 4 that violates (2). This difficulty is reflected, when dealing with (1), in the need for the strong admissibility condition. The method of proof is algebraic. Given G and S as above, we use the corresponding length R to define gr RG, a graded version of the group ring RG of G. We then prove (Theorem 1) that if G = G 1 *A G 2 (with appropriately chosen generating sets) and the inclusions of A in G 1 and G 2 are admissible, then gr RG 1 ligr Agr RG 2 ~ gr RG. Formula (1) follows from this by applying a result of Lemaire [L] that computes the Poincare series of an amalgamated product of graded algebras. To use Lemaire's result one must check that his "condition (H)" is satisfied. It was this condition that suggested admissibility. The paper is organized as follows. §I is introductory. Pairs are introduced as a convenient way to formalize the material. Admissibility is defined in §2. In §3 we consider amalgams and prove Theorems 1 and 2. §4 deals with extending these results to more general trees. In §5 we characterize those extensions whose inclusion is admissible. The machinery developed is used in §6 to compute growth functions of several groups, including the closed orient able surface group of genus n + 1 (n :::=: 1). And §7 contains some questions. Finally, it is a pleasure to thank Lennart Borjeson for his assistance with AMS- 'lEX and Graciela Arabaolasa for her fine pictures.

§1. Pairs. In this paper, a pair (G, S) will consist of a group G and a finite subset S which generates G as a semigroup and does not contain 1. Given (G, S) one can define a (word) length function Ron G by letting R(x) (x E G) be the smallest integer q 2 0 such that x is a product of q elements of S. A reduced decomposition of x is any sequence (81, ... , 8 q ) of elements of S such that x = 81 .. . 8 q and q = R( x). This length function induces a filtration {I} = FOG ~ FIG ~ ... of G by defining x E FiG if and only if R(x) :::; i. Note that G = Ui?:O FiG. The set FnG\Fn- 1 G = {x E GIR(x) = n} will be denoted ~(G,s)(n) or simply ~G(n) when S is understood. For any commutative ring R with 1 E R, the R-algebra RG thus inherits a structure of filtered R-algebra. Passing to the associated graded object


3

gr R(G, S), we obtain a connected graded R-algebra, where (gr R(G, S))i = Fi RG / F i - 1 RG, the free R-module generated by ~(G,S)( i).The growth series of a pair (G, S) is defined as the formal series

(G, S)(z) =

L

#

(~(G,s)(n)) zn

n2:0

where # denotes cardinality. When S is understood we write simply G(z). Note that (G, S)(z) equals the Poincare series of gr R(G, S). Given a connected R-algebra B = Bo EEl Bl EEl ••• , put Ibl = i if b E B i , and define the set: M «G, S), B) = {J: G

-t

Blf(l) = 1; If(x)1 = f(x); f(x)f(y) = f(xy) if f(xy) = f(x) +f(y),= 0 otherwise}

The canonical maps 7ri:FiRG

- t FiRG/Fi-1RG give an injection G - t gr R(G, S) defined by setting I-£(x) = 7rl(x)(X). Then 1-£ E M«G, S), gr R(G, S)), since

1-£:

1-£

( ) ( ) x 1-£ y

7rl(xy)(XY), ( ) = 7rl(x)+£(y) xy = { 0,

if f(xy) = f(x) if f(xy) < f(x)

+ fey) + fey).

The algebra gr R(G, S) is characterized by the following universal property: PROPOSITION 1. Let (G, S) be a pair and B a connected graded R- algebra. Tben tbe function

e: HOm(gr_R_alg)(gr R(G, S), B) --+ M «G, S), B) defined by e(1) =

f 1-£ is

a bijection.

PROOF: To check that e(f) E M«G,S),B) it suffices to verify the third condition, the first two being trivially true. For x, y E G, e(f)(x )e(f)(y) = f (l-£(x)I-£(Y)) which equals f (I-£(xy)) = e(f)(xy) if f(xy) = f(x) + fey) and is zero otherwise.

e.

We construct now an inverse v to Let y = 2:xEFi\Fi-l m(x)l-£(x), where Fi\F i - 1 means FiG\Fi-1G and m(x) E R, be a homogeneous element of degree i of grR(G,S). For 9 E M«G,S),B), define v(g)(y) = 2:xEFi\Fi-l

m(x)g(x). If y' =

element of degree j, then

2:zEFi\Fi-1

m'(z)l-£(z) is a homogeneous

4

J .M. Alonso

yy'

m(x)m'(z)J-l(XY)

= xEF'\F'-l zEFi\Fi- 1 l(xz)=l(x)H(z)

Consequently v(g)(yy') = (v(g)(y)) (v(g)(y')) so that v(g) is a graded RI . algebra homomorphism. One checks easily that v = 0

e-

§2. Admissible inclusions. A map of pairs a: (A, SA) -+ (G, S) is a homomorphism a: A -+ G such that a(SA) ~ S. There is a category of pairs whose objects are pairs and whose morphisms are maps of pairs. a will be called an inclusion if a: A -+ G is a monomorphism. In this section we deal only with inclusions. Given a, notice that the corresponding length functions .e A and .ea satisfy .e a (a(a)) :::; .eA(a) for all a E A (the inequality is often strict). Consider the following properties:

(I).e a (a(a)) = .eA(a) for all a E A (H') there is a set T (containing 1) of right coset representatives of G mod a( A) such that

.e a (a(a)t) = .eA(a) + .ea(t) for all a

E A,

t

E T

Note that (H') implies (I). To simplify the notation, we shall often identify A with im a :::; G and, in the presence of (I), .ea and .e A will be simply denoted.e. If the inclusion a: (A, SA) -+ (G, S) satisfies (I), then gr a: gr R( A, SA) -+ gr R( G, S) is a monomorphism of graded R-algebras. Note that gr R(G, S) is a (left) gr R(A, SA)-module via gra. DEFINITION: An inclusion is called

admissible if (H') holds.

EXAMPLES: (1) Suppose that (W, S) is a Coxeter system [B] and let Wx denote the subgroup of W generated by a subset X of S. Then the inclusion (Wx,X) -+ (W,S) is admissible by [B,exc. 26]. (2) An inclusion (GI,S!) -+ (GI x G 2 ,SI II S2), see Corollary 3 below. (3) An inclusion

(GI,S!)

-+

(G I *G 2 ,SIIIS2 ).


5

1. Let G be a finite group and a: A -+ G a monomorphism. Then there are subsets SA C A and S C G such that a: (A, SA) -+ (G, S) is

LEMMA

admissible. PROOF: Choose a set T of right coset representatives of G mod a(A) with 1 E T. Put SA = A \{1} and S = a(SA) U (T\ {1}) and let fA and f denote the corresponding length functions. We have to show that f(a(a)t) = fA(a) + f(t) for all a E A, t E T. This is trivial if a = 1. For t = 1 both sides equal 1 (if a # 1) or 0 (if a = 1). Finally for a # 1 # t both sides are equal to 2. This completes the proof. PROPOSITION 2. The following conditions on an inclusion a: (A, SA) -+ (G, S) are equivalent:

(i) a is admissible; (ii) G has a subset T such that ~a(n) =

(3)

U

c,o (~A(i)

X

Tj)

for all n ;::: 0,

i+j=n i,j~O

where Tj

=

~a(j)nT

and c,o: AxT -+ Gis the map c,o(a, t)

= a(a)t.

'*

PROOF: (i) (ii) Let T be as in (H'), so that c,o is a bijection. Writing x E Gas c,o(a, t) we have, by admissibility, that fa(x) = fA(a) +fa(t). This shows that ~a = Ui+j=n c,o (~A( i) X Tj) and the union is disjoint because c,o is bijective. (ii) (i) Let T be as in (ii) above. We show that T is a set of right coset representatives of G mod a(A); equivalently, that c,o is a bijection. The map is surjective because G = Un~O ~a(n). To see it is injective, suppose that c,o( a, t) = c,o( a', t') for some a, a' E A and t, t' E T. Then t = a( a")t' for a" = a-la' E A, i.e. t E c,o (~A(O) X Ti(t»)nc,o (~A(f(a")) X Ti(t'»). Whence a" = 1 and t = t'. Next, notice that ~a(O) = {1} = c,o ({1} X To) implies that 1 E T. Finally, the length equation in (H') is an immediate consequence of (3). 0

'*

COROLLARY 1. Suppose that a: (A, SA) -+ (G, S) is an admissible inclusion. Then (1) gr R(G, S) ~ gr R(A, SA) ®R T as graded R -modules, where T = To E8 TI E8 ... , Ti = free R-module generated by Ti = ~a(i) n T.

6

J .M. Alonso (2) G(z) = A(z)T(z), where T(z) = L:n2: 0 #(Tn)zn.

REMARK: The Corollary is a "growth functional" version of the following well-known fact. IT A :::; G are groups for which the Euler characteristic X is defined, and if A has finite index [G: A] in G, then X(A) = [G: A]X(G). Indeed, note that T(l) = #(T) = [G: A] and that if T(l) is finite then (G, S)(l) and (A, SA)(l) are both finite or both infinite. IT they are finite, (A,S~)(l) = T(l)(GJ)(l)· It follows that if (A, SA) and [G: A] is finite, then

--+

(G,S) is admissible

1 1 X(A) = (A, SA)(l) if and only if X(G) = (G, S)(l)

PROPOSITION 3. If the inclusion a: (A, SA) gr R(G, S) is a free gr R(A, SA)-module.

--+

(G, S) is admissible then

REMARK: In Lemaire's terminology [L,p.104] the Proposition asserts that gr a satisfies "condition (H)" when a is admissible. PROOF: There is a gr RA-module homomorphism .,,: (gr RA)(T) = EB(gr RA)(t)

--+

gr RG

T

defined by .,,(p.A(a» = (gr(a)(a»p.(t), where p.A:A --+ (grRA)(t). There is also a map g:grRG - t EIh(grRA)(t) defined by g(p.(x) = p.A(a) E (gr RA)(t) , where x E G equals cp(a, t). It is easily verified that g and." are inverse to each other. Thus." is an isomorphism of gr RA-modules. D


7

We also need the following LEMMA 2. Suppose that O:j:(Gi,Si) --+ (Gi+l,Si+l) (i = 1,2) are admissible inclusions. Then /3 = 0:20:1: (G l , SJ) --+ (G 3, S3) is an admissible inclusion.

PROOF: By hypothesis, we have sets Ti+l ~ Gi+l (i = 1,2) and bijections c,oi+l:Gi x Ti+l --+ Gi+l given by c,oiH(X,t) = O:i(X)t. Define T ~ G 3 by T = c,03(T2 X T 3 ), and c,o: G l X T --+ G 3 by c,o(x, t) = f3( x )t, for (x, t) E G l X T. The commutative diagram:

Gl

X

T2

X

T3

'f'2 xid ----+

idXl"31

Gl

X

G2

X

T3

11"3

T

----+

G3

'f'

shows that c,o is a bijection. Obviously 1 E T and for x E G l and t E T, the admissibility of 0:1 and Q2 readily implies that .e 3(c,o(x,t)) = .e l (x) +.e 3(i). This completes the proof. 0

§3. Amalgams. Consider the diagram of inclusions (G l , SI) ~ (A, SA) ~ (G 2 , S2). The push-out of this diagram in the category of pairs consists of a pair

(G,S), to be denoted (G l ,Sd*(A,SA) (G 2,S2) and maps the commutative diagram:

/31,/32

that fit in

(4) (G,S) Equivalently, G

Q2(a)!a E SA}. PROPOSITION 4. Let (G,S) = (G l ,SJ) *(A,SA) (G2,S2) where 0:1, 0:2 are admissible, and suppose that x E G has the normal form x = ati, ... tin (i.e. a E A, iij E Tij \ {I}, ij E {1,2}, and il =I i2 =I ... =I in). Then

8

J.M. Alonso n

Ra(x) = RA(a)

+L

j=l

Rij (tij)

REMARK: We have identified a with f3;a;(a) (i = 1,2), and tij with f3ij (tij ). Let (Sl, ... , sp) be a reduced decomposition of x E G. Thus x = Sl ... SP' P = Ra(x) and Sk E S = Sl U S2. Set parenthesis on (S1, ... , sp), beginning on the left, so that each segment consists entirely of elements of one of Sl, S2, and is maximal with respect to this property. We obtain (Sl, ... , Sp) = (a1, ... , aqJ, where 1 ~ q1 ~ p, aj = (Smj_1+ 1, ... ,Smj)' j = 1, .. ·,q1, mo = 0, m q1 =p. Moreover, aj ~ Sij (meaning Sk E Sij for mj-1 + 1 ~ k ~ mj) and i1 =f i2 =f ... =f i q1 . The product Tj of the elements of aj is in G ij . If q1 = 1, or if q1 > 1 and all Tj E Gi j : = Gij \A, stop here. Suppose that q1 > 1 and Tj E A (for some 1 ~ j ~ qI) . Use then the fact that Rij(Tj) = RA(Tj) to replace aj by a new sequence aj = (ami -1 +1, ... , ami) ~ A. Do the same for every j with Tj E A, to obtain a new sequence, denoted (s~, ... , s~), of the same length and with the same product x. PROOF:

Repeat the above procedure to get, say, (s~, ... , s~) = (a~, ... , a~2)' where aj ~ Sij' i1 =f i2 =f ... =f i q2 • Note that q2 < q1 since we assumed that at least one product Tj belonged to A. The process thus terminates, and we get, after a finite number of steps, a sequence (Sl, ... , sp) = (o-l, ... ,o-r) where either r = 1 and 0- 1 ~ Si (i = lor 2), or r > 1 and

o-j ~ Sii' i1 =f i2 =f ... =f ir, p = #(0-1) + ... + #(o-r), and with Tj E To simplify the notation set Xj: = Tj, so that x = Xl ... Xr .

Gi· J

If r = 1 then x E G i so that it can be expressed uniquely as x = at for t E Ti, and since ai is admissible, we have p = Ra(x) = Ri(at) = RA(a) + Ri(t). If r > 1 we reduce x = Xl ... Xr , Xj E Gi.J to normal form (with respect to T1 and T2):

x ~ Xl ... Xr = Xl· .. Xr-1(ar-1tr)

= Xl

... (X r-1 ar-dtr = Xl ... (ar-2tr-I)tr

t r E T i r \{l} t r -1 E Ti r _ 1 \{1}


9

By admissibility, fir(Xr) = iA(ar -1) +fir(tr) and iij(Xjaj) = fA(aj-1) + fij (tj) for 1 ~ j ~ r - 1. Then fa( x)

= fa( aot1 ... t r ) ~ fa( ao) + fa( t1) + ... + fa(tr)

+ fdh) + ... + iir(tr) = fA(ao) +f;Jx1aI) -fA(ao) + ... +iir(X r ) -fA(ar-l) ~ fil (Xl) + ... + iir (Xr) = ia( X). ~ fA(aO)

D

This completes the proof. With the notation of diagram (4), we now have: LEMMA 3. If the inclusions ai: (A, SA) then so are the inclusions

-+

(Gi, Sd (i

PROOF: By admissibility there are bijections r.pi: A and fi(ati) = fA(a) +fi(ti). Define the set T'

=

{x E Glx

= atil

X

=

1,2) are admissible,

Ti

-+

... tin (n ~ 0) is in normal form, a

G i , where 1 E Ti

=

1 &

i1

= 2}

and let r.p: G 1 X T' -+ G be given by multiplication. Then r.p is a bijection and 1 E T'. If y = at' E G 1 (a E A,t' E T 1) and t = ti2 ... ti n E T', then yt = at'ti 2 ... tin is in normal form. By Proposition 4, ia(yt) = iA( a) +f1 (t') + L:~=2 iir (tir) = i1 (y) +ia( t), showing that /31 is admissible. The proof for /32 is similar. D Suppose that in diagram (4) the ai are admissible inclusions. Then there are injections

and their push-out in the category of connected graded R-algebras will be denoted gr R(G 1 , SI) ilgrR(A,SA) gr R(G 2, S2). We then have: THEOREM 1. Let (G, S) = (G 1 , Sl) *(A,SA) (G 2, S2) where ai: (A, SA) (Gi' Si), (i = 1,2) are admissible. Then the inclusions gr /3i: gr R(G i , Si) gr R( G, S) (i = 1, 2) induce an isomorphism

-+ -+

10

J.M. Alonso

of connected graded R-algebras. PROOF: We prove that gr R(G, S) satisfies the universal property defining the push-out A. To simplify the notation we write maps as ( :..-) instead of gr( -). Diagram (4) yields: gr R(A, SA)

1. Then X is obtained from the subtree X' = X - v by adjunction of some terminal vertex v and geometric edge {e,e-}. We assume that v = tee), w = o(e). By (10),

(13) The inclusions in (11) are admissible by Lemma 4. Then Theorem 1 yields gr R«g,Sh) ~ gr R«g,Sh,)

ligrR(G,S)e

gr R(G, S)v

By the induction hypothesis, grR«g,Sh,) ~ (grR(g,S)h, and then by (12), grR«g,Sh) ~ (grR(g,S)h, as was to be proved. 0 The desired generalization of (1) is given by THEOREM 4. Suppose «g,S),X) be an admissible tree of pairs, and let (g,Sh = lim«g,S),X). If X is finite then ---+

(14)

1

(g, Sh(z)

=2: 1 -2: 1 vEVX (G, S)v(z) eEEX (G, S)e(z)


15

By induction on n = #(VX). If n = 1 there is nothing to prove. Suppose then that n > 1. Let X, X' be as in the proof of Theorem 3. We then have (11) with admissible inclusions, by Lemma 4, so that, by PROOF:

Theorem 2,

on the other hand, by induction

and (14) follows immediately.

o

§5. Extensions. DEFINITION: A map of pairs 7f: (G, S) --+ (Q, SQ) is called an admissible surjection if 7f: G --+ Q is surjective, 7f(S) = SQ U {I}, and there is a section a: Q --+ G of 7f such that

for all k E ker

7f

and x E Q.

REMARKS: (a) We always assume that 17(1) = 1. (b) For any surjection of pairs (not necessarily admissible) eQ(7f(g)) :::; ea(g) for all g E G. In particular, eQ(x) :::; ea(a(x)) for every section a of 7f. (c) When the section a in the definition can be chosen to be a homomorphism, then it is an admissible injection.

DEFINITION:

A short exact sequence of pairs

(15) consists of a short exact sequence of groups (16) A ~ G ..::-.. Q with S =

S1 II S2, where S1 = a(SA) and 7f(S2) = SQ.

16

J.M. Alonso

(15) satisfying, in addition, the condition that SAl = SA and S,/ = SQ. Then a is an admissible injection if and only if 7r is an admissible surjection. LEMMA 5. Suppose given a short exact sequence of pairs

PROOF: Suppose that a is admissible. Let T C G be such that A X T ~ G and fG(a(a)t) = fA(a) + fG(t). Since 7rIT is a bijection, the function 0' = (7rIT)-l: Q -+ T eGis a section of 7r. To show that

(17)

iG(kO'(x)) = fG(k)

+ fQ(x)

write an arbitrary 9 E G uniquely in the form 9 = a(a)t. Then a(a) = k E ker 7r and t = O'(x) for x = 7r(t). By admissibility of a, fG(g) = fG(k) + fG(O'(x)). Hence (17) holds if and only if fQ(x) = fG(O'(x)) for all x E Q. As remarked in (b) above it suffices to show that fG(O'(x)) $ fQ(x). Suppose that fQ(x) = q and let x = 7r(sI) ... 7r(Sq) be a reduced decomposition of x. Then Sl ... Sq = a(a')O'(x) for some a' E A, and q = fQ(x) = fa(Sl ... Sq) = iA(a') + fG(O'(x)) ~ fG(O'(x)). Thus 7r is admissible. To prove the converse, suppose that 7r has a section 0' for which (17) holds. Then O'(SQ) = S2. Write k = a(a). Then, since fQ(x) = fa(O'(x)), (17) gives fG(a(a)O'(x)) = iG(a(a)) + iG(O'(x)). So in order to prove that a is admissible we need only show that fG(a(a)) = fA(a) for all a E A. Moreover, it suffices to show that iA(a) $ fG(a(a)), since the reverse inequality is always true. The proof is by induction on the length of a(a). The inequality is trivial if fG( a( a)) $ 1. Suppose that q = fG( a( a)) ~ 2 and let a( a) = Sl ... Sq be a reduced decomposition. Consider first the case where there is an i, 1 $ i < q such that Sl ... Si E A or Si+1 ... Sq E A. Note that in this case both Sl ... Si and Si+1 ... Sq must belong to A, since their product does. Let a1, a2 E A satisfy a = a1a2, a(aI) = Sl ... 8i, and a( a2) = 8i+1 .. . 8q. Since a1 =f a f a2, we can use the induction hypothesis to obtain:

as was to be proved. Consider now the case where for all i, 1 $ i < q, 81 . . . Si ct. A and 8;+1 .. . 8q ct. A. In particular, we can take i = 1 and write 81 = 0'(X1) and 82···8q = k20'(X2) for appropriate k2 E A, Xl E SQ and 1 f X2 E Q. By


17

(17), q = .eQ(Xl) + .ea(k2) + .eQ(X2). Similarly, we write u(xI)k2 = k3U(X3) and use (17) to obtain .ea(u(xI)k2) = .e a (k3) + .eQ(X3). We then have a(a) = k3u(X3)U(X2) = k3k4 where k4 = U(X3)U(X4) E A. Moreover, q :::; .e a (k3)

+ .e a (k 4) :::; .ea (k3) +.e a (U(X3)) + .ea (U(X2)) = .ea (u(xI}k2) + .ea(U(X2)) :::; .eQ(xI} + .ea(k2) + .eQ(X2) = q

Suppose first that 0 < .ea(k4 ) < q. By induction (with the obvious abuse of notation), q ~ .e A(k 3) +.eA(k4) ~ .eA(k3k4) = .eA(a), as was to be proved. We now show that the two remaining cases cannot actually occur. Indeed, suppose that .ea (k 4 ) = q. Since X3X2 = 1, setting X3 = x, we have kilu(x) = u(x-l)-l and we can compute:

.ea(kilu(x)) = .ea(kil) + .eQ(x) = .ea(u(x-l)-l) = .ea(u(x- l ))

by admissibility by symmetry S = S-l

= .eQ(x-l) = .eQ(x)

That is q = 0, a contradiction. Finally, if .e a (k 4) = 0 then also .eQ(X3) + .eQ(X2) = 0 contradicting the fact that X2 is different from one. This completes the proof. REMARK: Note that the symmetry of the generating sets was needed only to prove that 11" admissible implies a admissible. DEFINITION: A short exact sequence (15) is called admissible if a is admissible. COROLLARY 2. Let (15) be an admissible short exact sequence of pairs. Then

(1) grR(G,S) ~grR(A,SA)®RgrR(Q,SQ) as graded R -modules. (2) G(z) = A(z)Q(z). PROOF: Follows easily from Proposition 2 and the first part of Lemma 5.

18

J.M. Alonso

Next, we find necessary and sufficient conditions for a short exact sequence to be admissible. Recall (see e.g. [B]) that an extension of groups (16) is described as follows. Choose a section a: Q --+ G of 7r satisfying a(l) = 1. Conjugation by a(x) yields an automorphism If I satisfying 7.ii): 7.vi) h(f,s,g) 2:: o for each triple (f,s,g) E E(x,y)xGyxE(y,z),x,y,z E X; 7.vi)' IE(f,s,g)l:::; If I + Igi for each triple (f,s,g) as above.


43

PROOF: 7.vi) ~ 7.vi)': Let A = (j, s,g), B = (/, p(A),c(A)), C = (/,1,1). By 6.v), 6.iii), 6.vi) and 6.vii) we get c(B) = c( c( C), p( 6)-1 s, g) = c(ly, 1, g) = g, and hence 0 :s; 8(B) = If I + Igl -lc(A)1 by 7.ii) and 7.vi). 7.vi)' ~ 7.vi): Applying 7.ii) and 7.vi)' to the triple B above we get 8(.4) ~ O. 0 The following statement is an extension of Lemmas 1.2 and 1.3. PROPOSITION 2.3. Let X be a structure satisfying the conditions 1 )--6) and 7.i), 7.ii). For g E E(y, z), let [g] = {(j,p) : fEE, t(j) = y, pEG J\G y and 8(j,p,g) = O}, and O:g : [g] ~ [O,lgl] be the map given by O:g(j,p) = If I· For (j,s,g) E E(x,y) x G y x E(y,z), let Y(j,s,g) = ((h,t) E [f] :

8(h,tp(j,s,g), c(j,s,g» = 8(c(h,t,j), p(/,BjlrI,h)-ls,g) = O}. [g], O:g and Y(j, s, g) are well-defined according to the remark above. The following statements are equivalent: X is a A-graph of groups. X has the next properties: O:g is bijective for each gEE; Y(j,s,g) is non-empty for each suitable triple (j,s,g). X satisfies the triangle inequality 7.vi) and the following conditions: O:g is an isometry for each gEE, i.e., O:g is bijective and d( (j, s), (h, t» := 1c(j,srl,Ii)1 = Ilfl-lhll for (j,s), (h,t) E [g]; 7.viii)' Y(j, s, g) is a singleton for each suitable triple (j, s, g), depending only on f,g and the double coset GfsG g.

a) b) 7.vii) 7.viii) c) 7. vii)'

PROOF: a) ~ c): Assume X is a A-graph of groups. First let us check 7.vi). Let A = (j, s, g) E E( x, y) x G y x E( y, z), P E p( A) and C = (c( A), 1, c( A) ). We get c(A) = c(A) and c(C) = Ix by 6.i) and 6.iii), and hence 8(C) = 0 by 7.i) and 7.ii). Let F = (g,AA(p),c(A» and D = (j,sq,c(F»), with q E p(F). Applying 6.v) to the quintuple (j, s, g, AA(p), c(A), we get c(D) = dC) = lx, and hence c(F) = /, by 6.viii). It follows 8(D) = 0, by 7.i) and 7.ii). As 8(A) ~ min(8(C), 8(D), by 7.iv), it follows 8(A) ~ 0, as required. Next let us check 7.vii)'. Let g E E(y, z). As O:g is onto by 7.v), it remains to show that d:= Ic(j,st-l,Ii)1 = Ilfl-lhll iff 8(j,s,g) = 8(h,t,g) = O. Indeed, if so, then the injectivity of O:g follows by 7.i) and 6.viii), ([g], d) becomes a A-metric space and O:g is an isometry. Let A = (f, s, g), B = (h,t,g), C = (c(A),p-lq,c(B)) with p E peA), q E pCB), F = (g,q,c(B», D = (j, st-I, c(F» and H = (ly, rlB];\ Ii). By 6.v), 6.iii) and 6.vi), we get dF) = c(c(g, 1, g), t- 18];1, Ii) = c(H) = Ii and p(F) = p(H) = t-lG",

44

S.A. Basarab

and hence D = (f,srl,h). On the other hand, e(C) = e(D) by 6.v), and e(B) = e(B) by 6.i). Consequently, o(C) = le(A)1 +d-le(B)1 = d-Ifl + Ihl since o(A) = o(B) = 0 by assumption, and oeD) = If I + d - Ihl. As 0= o(A) ~ min(o(C),o(D)) by 7.iv), it follows d = Ilfl-lhll, as required. Finally let us check 7.viii)'. Let A = (f,s,g) E E(x,y) x G y x E(y,z). By 7.iii), 7.vi) and 7.vi)', o(f,s,g) = 2')' with')' E A such that 0 ::; ')' ::; min(lfl,le(A)I). As af is onto there exists a pair (h, t) such that t(h) = x, t E G x , o(h,t,f) = 0 and Ihl = ')'. Let us show that (h,Gkt) E Y(f,s,g). Let B = (h,t,f), C = (h,tp,e(A)), D = (e(B),q-ls,g) with p E peA), q E pCB). As 0 = o(B) ~ min(o(C),o(D)) by 7.iv), it remains to show that O(C) = oeD). We get o(C) = Ihl + le(C)1 - le(A)1 = ')' + le(C)1 (2')' - If I + Igl) = Ie(C)1 + If 1- Igl- ')' and oeD) = le(B)1 + le(D)I-lgl = (Ifl-Ihl)+ le(D)I-lgl = le(D)1 + Ifl-Igl-')', and hence O(C) = oeD) since e(C) = e(D) by 6.v). Thus (h,Gkt) E Y(f,s,g). The unicity of (h,Gkt) is immediate thanks to the injectivity of af. Thus Y(f,s,g) is a singleton depending obviously only on f, g and the double coset G /sG y • b) {:} c): We have only to show that b) c) 7.vi) is immediate. Indeed, let (h,t) E Y(f,s,g). Then 21hl = o(f,s,g) and hence o(f,s,g) ~ 0 as Ihl ~ O. Using the same argument and the injectivity of af we get also 7.viii)'. So it remains to show that a g is an isometry for each gEE. In this order we verify the following useful property:

'*

= S(h,t,!) = 0 then o(h,ts,g) = O. Let A = (f,s,g), B = (h,t,!), C = (e(B),q-letp,e(A)), F = (e(B),q-le;l,f), D = (e(F),ts,g), with p E peA), q E pCB). By 6.v), 7.ix). If o(f,s,g)

6.iii) and 6.ii), we get e(F) = e(h, t, It(f») = h and pep) = t- l G k ' By 6.v) applied to the quintuple (e(B),q-le,\f,s,g), it follows e(C) = e(D) = e(h, ts,g). As o(A) = S(B) = 0 by assumption, it follows by 7.vi) and 7.vi)' that 0::; oeD) = Ihl + le(C)I-lgl::; (lfl-le(B)I)+ (le(B)1 + le(A)I) -Igl = o(A) = 0, and hence SeD) = 0, as required.

'*

Now we are ready to finish the proof of the implication b) c). Let (f,s,g), (h,t,g) be such that o(f,s,g) = o(h,t,g) = 0 and If I ::; Ihl. We have to show that o(f, st- l , h) = O. Since ak is onto there exists a pair (u,p) such that o(u,p, h) = 0 and lui = If I· By 7.ix) it follows o(u,pt,g) = O. As a g is injective, we get u = f and pts- l E G j , and hence o(f,srI,h) = o(u,p,h) = o. c) a): 7.iii) and 7.v) are immediate, so it remains to check 7.iv). Let (f,s,g,t,h) be a suitable quintuple, A = (f,s,g), B = (g,t,h), C =

'*


45

(c:(A), AA(p)-lt, h), D = (f,sq,c:(B)) with p E peA), q E pCB), and assume that 8(C) ~ 8(D). Let (l,v) be such that (l,GJV) E Y(D). As 8(D) = 21ll, 8(l,v,f) = 0, c:(l,vp,c:(A)) = c:(c:(l,v,f),p(f,frtv-1,1)-ls,g) and 8( c:(l, v, f), p(f, 9j 1v- 1,1)-1 S, g) ~ 0 it suffices to show that 8(l, vp, c:(A)) = o to conclude that 8(A) ~ 8(D). Since c:(C) = c:(D) and pp(C) = p(D) by 6.v), and 8( l, vp(D), c:(D)) = 0 by assumption, it follows 8(l, vpp( C), c:( C)) =

o by

6.vii). Let (k,w) be such that (k,Gkw) E Y(C). Then III :S: Ikl, as 8(C) ~ 8(D), by assumption, and 8(k,w,c:(A)) = 8(k,wp(C),c:(C)) = O. Since Qg(e) is an isometry, we get 8(l, vpw- 1 , k) = 0, and hence 8(l, vp, c:(A)) = 0 by 7.ix). The case 8(C) :S: 8(D) is quite similar. 0 COROLLARY

2.4. Let X be a A-graph of groups. Then X has the following

property: 7.x) If (f,s,g) E E(x,y) x G y x E(y,z), X,y,z E X, and 8(f,s,g) then sGgs- 1 C G J. PROOF: As 8(f, GrGg,g) = 8(f,s,g) GJs = GJsGg, i.e., sGgs- 1 C G J .

=

0 and

Q

g

=

0

is injective, it follows 0

DEFINITION 2.5: A subpretree T of a A-graph of groups X consists of a non-empty subset T of X and a family (lXY)(X,Y)ET2, where lxy E E(x,y),

such that lxx = lx, Ixy = lyx and c:(A) = lxz, peA) = G xz := G 1xz and AA = Wxz := Wlxz for A = (lxy, 1, lyz), x, y, z E T. DEFINITION 2.6: The A-graph of groups X is strongly connected if there

exists a subpretree T of X such that T = X.

If X is strongly connected then X is connected, i.e., E( x, y) is non-empty for arbitrary X,y E X. The converse is also true if G x is trivial for each x EX, i.e., X is a A-graph as defined in Section 1. In the following we extend to an arbitrary ordered abelian group A the construction given in [SeJ assigning a graph of groups to a group action on a connected graph. Given a connected A-graph X and a group G acting from the left on X, let Y = G\X, E = G\arrow X, E(Ga, Gb) = {Gj: o(f) EGa, t(f) E Gb}, Gj = Gj-I, IGa = Gl a . Choose a section j : Y -+ X of the projection map X -+ Y : a 1--+ Ga and a family (lXy)(x,y)EY2 of arrows lxy : jx -+ jy in X in such a way that lxylyz = lxz for x, y, z E Y; in particular lxx = ljx and l;;J = lyx. This choice is possible since X is connected. Set lxy = Glxy E E(x, y) for x, y E Y. Thus lxx = Ix and lxy = lyx. Extend j to a section j : E -+ arrow X of the projection map arrow X -+ E such

46

S.A. Basarab

that oUf) = jo(f) for each fEE and j1xy = lxy for x,y E Y. Choose a map a : E ---t G in such a way that U f)-I = a(f)j J for each fEE; in particular, t(j f) = a(f)jt(f). We may assume that a(J) = a(f)-1 if

f =J J and a(lxy) = 1 for x,y E Y. Set G f := G jf = {s E G: sjf = jf} for fEE, and G x := G1x for x E Y. Thus G f is a subgroup of Ga(f) and G f = a( f) -1 G f a(f). Define the isomorphism W f : G f ---t G f by wf(s) = a(f)-1 sa(f), and let e f = a(f)a(J) E G f. For (f,s,g) E E(x,y)xGyxE(y,z),lets(f,s,g) E E(x,z)betheG-orbit of the composite arrow (j f)( a(f)sjg) E X(jx, a(f)sa(g )jz). Consequently, there exists p E G x such that (jf)(a(f)sjg) = pjs(f,s,g). Obviously, the element p is uniquely determined modulo Ge(f,s,g), so let p(f,s,g) C G x be the coset pG e(f,8,g)' Let A = (f,s,g),.4 = (g,e;1s-1etJ) and p E peA).

We get (jg)(a(g)e;1 s -1e·t j J) = (jg)(a(g)-1 s -1jJ)

=

a(g )-1 s-1 a(f)-1 [(j f)( O'(f)sjg )]-1 = a(g )-1 s-1a(f)-1pa(s(A) )jc( A) and hence seA) = seA), a(g)-1s-1a(f)-1pa(s(A» E G z and peA) = a(g)-1s-1a(f)-1pa(s(A»G e(A)' Define the bijective map 'xA : peA) ---t peA) by 'xA(P) = a(g)-1s-1a(f)-1pa(s(A». Finally, let us note that

the length function I I: arrow X ---t A induces a map I I: E ---t A given by If I = Ij fl· A routine verification shows that the structure Y = Y( G, X; j, a) = (Y, E,w, s, p,,x, I I) constructed above is a strongly con-

e,

nected A-graph of groups and (Y, (lxy )(x,y)EY2) is a subpretree of Y.

§3. The fundamental group of a strongly connected A-graph of groups. In this section we extend the construction of the fundamental group of a graph of groups given in [Se] to the general case of arbitrary ordered abelian groups. Suppose X

is a strongly connected A-graph of groups and let

T = (X, (lXy)(x,y)EX2) be a subpretree of X. Denote by F the free product of the groups G x for x E X and the free group generated by E = Ex. Let 7rt(X, T) be the quotient of F by the normal subgroup RT generated by the elements lxy for x,y E X and the elements PS(A)'xA(p)-1 g-1 s -1 f- 1 for A = (f,s,g) E E(x,y) x G y x E(y,z), X,y,z EX, P E peA). We shall see (Proposition 3.5) that the group 7r1(X, T) does not depend, up to an isomorphism, on the choice of the subpretree T, and so we may speak on the fundamental group 7r1(X) of X. Choose PA E peA) for each suitable triple A

=

(f,s,g), and let

,x~

=


47

AA(PA)-l E peA). Denote by R'r the normal subgroup of F generated by the elements lxy for x, y E X, and the elements PAc:(A)A~g-ls-l j-1 for each suitable triple A = (f, s, g). LEMMA 3.1.

RT = R'r.

PROOF: We have to show that the generators of RT belongs to R'r. First note that j/Bt E R'r and jWf(a)j- 1a- 1 E R'r for a E Gf, by 6.iii) applied to the triples (f,1,/) and (f,wf(a),f). Now let A = (f, s, g) and p E peA). As AA is compatible with We:(A) we get pc:(A)AA(p)-l == pc:(A)We:(A)(p-1pA)A~

==

PAc:(A)A~

== jsg mod R'r, as required.

D

Let us give an alternative description of the fundamental group of X. Fix a vertex x E X and let S = Sx = G x x E(x, x) x G x . Define an equivalence relation == on S as follows: (s,j,t) == (p,g,q) iff j = g, s-lp E G f , tq-1 E G f and Wf(S-lp) = tq-1. If a = (s,j,t), (3 = (p,g,q) E S, let us put a

0

(3

=

(sPA,c:(A),A~4q), where

A

= (f,tp,g).

LEMMA 3.2. Let a, (3, a', {3' E S and assume a

a

0

(3

== a' and (3 == (3'. Then

== a' 0 (3'.

PROOF: Let a = (s,j,t), a' = (s',j,t') == a and (3 = (p,g,q). Let us show that a 0 (3 == a' 0 (3. Let A = (f,tp,g), A' = (f,t'p,g). As A' = (f,wf(s,-l s )tp,g), it follows by 6.iv) that c:(A) = c:(A'), p;)s,-l sPA E

Ge:(A) and We:(A) (PA,l S'-l SPA) = A~' AA 1, and hence

we get a

0

(3

==

a

0

(3' if (3

== (3'.

000 (3

==

a' 0 (3. Similarly,

D

Let 7r1(X,X) be the quotient set SI ==. By Lemma 3.2, 7r1(X,X) is equipped with a binary composition law induced by the map S2 ---. S : ( a, ,8) I--t a 0 (3, which does not depend on the choice of the representatives

PA of the cosets p(A.). LEMMA 3.3. 7r1 (X, x) is a group with respect to the composition law above.

PROOF: Let a = (s,j,t), (3 = (p,g,q)" = (r,l,u) E S. We have to show that (a 0 (3) 0 , == 000 ((3 0 , ) . Let A = (f,tp,g), B = (g,qr,l), C = (c:(A), A~ qr, I), D = (f, tPPB, c:(B»). By definition we get (a 0 (3) 0 , = (SPAPC,C:(C), A~U) and ao({3o,) = (sPD,c:(D), A~A~U). The required congruence is immediate by 6.v). By 6.ii), it follows that the class of (1, lx, 1) is the neutral element of 7r1 (X, x), while (r1 ,/, Bf 1s -1) is a representative of the inverse of the class of (s, j, t), by 6.iii). D The next lemma is immediate.

48

S.A. Basarab

LEMMA 3.4. The maps Gx -- S : s 1-+ (s, lx, 1) and E(x, x) -- S : f 1-+ (1, f, 1) identify G x with a subgroup Of7rl(X, x) and E(x, x) with a subset of

7rl(X, x). Let F x be the free product ofG x and the free group generated by E(x,x), and Rx be its normal subgroup generated by Ix and the elements PAc(A)AAg-1s- I f- I for A = (f,s,g) E E(x,x) x G x x E(x,x). The Rx does not depend on the representatives PA of the cosets peA) E Gx/Ge(A) and Fx/Rx, 7rI(X, X) are canonically isomorphic. Let As = (Ixy,s,I yx ), Af = (lxy,I,j), Bf = (c(Af),AA"Izx) for s E G y, f E E(y,z), y,z EX. Note that PA.,AA.,PA"PB"A B, E Gx and c( As), c( B f) E E( x, x), so we may identify them with elements of the group 7rl(X, x). Define a map." : (UYEXGy) U E -- 7rl(X, x) by .,,(s) = PA.c(As)AA., .,,(f) = PA,PB,c(Bf)A B, for s E UYEXGy, fEE. Note that the map." depends on the choice of the pretree T, but not on the choice of the representatives PA., PA" PB,· PROPOSITION

3.5. The

map

."

above

induces

an

isomorphism

"'T,x : 7rI(X, T) -- 7rI(X, x). PROOF:

a) First we have to show that." induces a homomorphism F --

7rl(X,X), i.e., .,,(st) = ",(s)",(t) for s,t E Gy, y E X. Since, by Lemma 3.4, the identity fqg = pc(A)AA(p)-I holds in 7rI(X,X) for A = (f,q,g), f,g E E(x,x), q E G x , p E peA), we get .,,(s).,,(t) = PA.Pcc(C)ACAA., where C = (c( As), AA. PA., c( At». Thus we have to show that c( C) = c(Ast),PA:!.PA,PC E Ge(C) and W~(C)(PA:!.PA,PC) = AA,.A'A:~N(/. Let H = (Iyx,PA"c(At), 1= (Iyx, 1, Ixy), J = (c(I) , t, Iyx) = (Iy, t, Iyx). By 6.vi), we get c(J) = lyx, p(J) = tG yx and AAt) = 1. Since p(I) = G yx and AI = Wyx by assumption, it follows by 6.v) that c(H) = c(J) = Iyx and AH(t) = AA.. Applying the rule 6.v) to the triples C and Ast = (lxy, st, c(H» we get c(C) = c(Ast), PA,p(C) = p(Ast) and NA:~. = AA,. (PA, (PA:! PA •• » = AH(t)-I AC(PA:~ PA,.) = A'A:; A'C/Wt:(C)(Pc/ PA:! PA •• ), as required. b) Next we have to show that the normal subgroup RT of F is contained in the kernel of." : F __ 7rl(X,X), i.e., .,,(I yz ) = 1 for y,z E X, and .,,(f).,,(s).,,(g) = ."(po)",(c(O»,,,(A~) for 0 = (f,s,g) E E(y,z) x G z x E(z,u). The first condition is immediate, while the second one will follow after a long chain of computations. By Lemma 3.4, we get .,,(f).,,(s).,,(g) = PA,PB,PCPDc(D)AVA Bg , where C = (c(Bf), AB,PA.,c(As» and D = (c(C), A~AA,PAgPBg,c(Bg».


49

Consider the triples

F D1

= (E(As), >'~.PA9PB9' E(Bg)),

= (E(B!), >.~,PA.PF' E(F)),

H = (lzx,PA.PF,E(F)), D2 = (E(A!), >.~,PH' E(H)), 1 = (f,PH,E(H)),

D3 = (lxy, PI, E(l)), J

h

= (f, 1, 1 zx ), = (E(J),>'~PA.PF,E(F)),

K = (E(J),>'~PA.,E(As)),

h = (E(K), >'K>'~.PA9PB9' E(Bg)), L

= (E(K), >'K>'~.PA9,E(Ag)),

h

= (E(L), >.~>.~

M

= (E(K), >'K>'~., 1xz ), = (E(M), >'~,g),

L1

9

,lux),

N = (E(As),>'~.,>'~., 1xz ), M1 = (E(J), >'~PA.PN' E(N)), P = (lzx, PA,PN, E(N)), M2 = (f,PP,E(P)), L2 = (f,s,g), 14 = (E(L 2 ), >'~2' lux) and

D4

= (lxy,PL 2PI

4

,E(14))

We get, step by step:

= E(D3),PV!PA,PB,PCPD E Ge(D) and its image through We(D) is >''va>'~>'~>'~>''v1, by 6.v) applied to D, D 1, D 2, D3; II) E(1) = E(13), PI 1PJPKPLPIa E Ge(I) and its image through we(I) is >'~>'~>'~>'~9>.'Ia1, by 6.v) applied to I,11,I2,I3; III) E(L) = E(Ld,PL 1PMPL 1 E Ge(L) and its image through we(L) is I) E(D)

>'~>'~9>.'L:, by 6.v); IV) E(M) = E(M2 ), PM~PJPKPM E Ge(M) and its image through We(M) is >'~2>'P>'~>.'M1, by 6.v) applied to M,M1,M2 ;

50

S.A. Basarab V) c:(N) = 1xz ,PA.PN E G xz and Wxz(PA,PN) = sA'//, by 6.v), 6.ii) and 6.iii); VI) P = (lzx,PA,PN, 1xz) = (lzx,wzx(sA'iV1), 1 xz ), c:(P) = 1z and A'p = -1 \1-1 b 6···) Pp s'" N' Y .lll;

VII) M2 = (f, 1, 1z), c:(M2) = f, P(M2) = G f and A:U2 = wf(PM,)-1 PP' by 6.ii); thus PJPKPM E G f and wf(PJpj.'PM) = sYivl by IV), VI); VIII) L1 = (f, A:U, g), c:(Ld = c:(L2)' PL~ PJPKPMPLI E Ge(L 2) and its image through we(L21 is A~2.x'L:, by 6.iv); consequently PL~ PJPKPL E G e(L 2) and its image through we(L 2) is A~2 NA~ .x'L1, by III); IX) 13 = (c:(L2)' A~A~g' lux), c:(I3) = c:(I4), pJ} PL>JPKPLPIs E Ge(I4) and its image through We(I4) is A~4 A'is1, by 6.iv); thus c:(I) = c:(I4),

PI 1PL 2PI4 E Ge(I) and we(I)(Pr1 PL 2PI.) = A~A~A~A'sg.x'l,,1, by II); X) D3 = (lxy, PJ,c:(I4)), c:(D3) = c:(D4)' PD!PD. E G e(D4) and We(D 4)(PD!PDJ = A'v3A~A~A~A'sg.x'l,,1.x'D!, by 6.iv); consequently, c:(D) = c:(D4), PD!PA,PB,PCPD E Ge(D) and its image through we(Dl

.

\I

\I

\,-1\,-1 b

I)

"'D4"'!4'" Bg'" D' Y . It follows 'f/(f)'f/(s)'f/(g) = PD4c:(D4)A'v4A~4. One may show similarly that 'f/(p L,)'f/( c( L 2) )'f/(.x~2) = PD4 c:( D4 )A'v4A~4' getting the required equality. Thus we obtain a homomorphism 'f/: 7r1(X, T) --t 7r1(X,X). c) In order to show that the 'f/ above is an isomorphism, let us define the map ~ : G x U E(x,x) --t 7r1(X, T) by ~(s) = sRT , ~(f) = fR T for s E G x , f E E(x,x). By Lemma 3.4, the map ~ induces a homomorphism ~

: 7r1(X, x)

IS

--t

7r1(X, T) and it is obvious that 17 and

~

are inverse to each

0

~~.

Proposition 3.5 shows that the fundamental group 7r1 (X) c:::::: 7r1 (X, T) c:::::: 7r1 (X, x) does not depend, up to an isomorphism on the choice of the pretree T or the vertex x. It follows also that for each x EX, the canonic homomorphism G x --t 7r1(X, T) and the canonic map E(x, x) --t 7r1(X, T) are injective. Moreover, we get: COROLLARY 3.6. Let x, y E X. Then i) the canonic map i : E(x,y) --t 7r1(X, T) : f I--> fRT is injective; ii) the map E(x, y) --t G x \7r1(X, T)jG y : f I--> Gxi(f)G y is bijective; iii) thecanonicmapGxxE(x,y)xG y --t 7r1(X,T): (s,f,t) I--> sftRT is onto, and sft == pgq mod RT iff f = g, s-lp E G f and wf(s-lp) =

tq-1. PROOF: As i) and ii) are immediate consequences of iii), it remains to


51

prove iii). First let us prove the surjectivity of the map above. Let a E 7r := 7r1(X, T). We have to show that there exist f E E(x, y), s E G x and t E G y such that the identity a = sft holds in 7L By Proposition 3.5, there exist h E E( x, x) and p, q E G x such that the identity a = phq is true in 7r. We get the sequence of identities in 7r : a = phq = phq1xy = sft, where

s = PPc, f = s(C), t = AC' C = (h,q, 1xy). Next let f,g E E(x,y), s E G x and t E G y • We have to show that the identity sft = g holds in 7r iff f = g, s E G f and wf(s) = t- I . We get the sequence of identities in 7r : sft = sftl yx = sPAs(A)A~, where A = (1, t, 1yx). Similarly we get g = PBs(B)A~, with B = (g, 1, 1yx). By Proposition 3.5 and the definition of 7r1 (X, x) it follows that sft = g is true in 7r iff the following condition is satisfied:

If f

= g, s

E Gf and wf(x)

= rl

then (*) follows by 6.iv). Conversely, assuming (*), let C = (s(A),A~,lxy) and D = (s(B),A~,lxy). By 6.iv) it follows s(C) = seD), PI} SPAPC E Ge(C) and We(C) (Pr} pi/ SPAPC) = I ADA'C . On the other hand, by 6.v) and 6.ii), we get

6/

and Wf(PAPC) = tA'CI, where F = (lyx, 1, 1xy). Similarly, we get seD) = g,PBPD E G g and Wg(PBPD) = )..I r}. Consequently, f = g, s E Gf and W f( x) = r l , as required. D REMARK: If G x is trivial for each x E X, i.e., X is a connected A-graph as defined in Section 1, then 7r1(X) ~ Aut(x) for x EX.

§4. From strongly connected A-graphs of groups to group actions on A-trees. In this section we extend to arbitrary ordered abelian groups the construction of the universal covering relative to a graph of groups given in

[Se]. Let us let X be a strongly connected A-graph of groups and let T = (T, (lxy )x,yEX2) be a subpretree of X. \iVe construct a A-tree X = XT on which the fundamental group 7r1(X, T) acts from the left, in such a way that the pair (7r1 (X, T), XT ) does not depend, up to an isomorphism, on the choice of T.

52

S.A. Basarab

Let X be the disjoint union UxEX7r/Gx, where 7r = 7rl(X, T) and G x is identified with a subgroup of 7r by Proposition 3.5. By Corollary 3.6, we may also identify the sets E( x, y) for x, y E X with subsets of 7r. The group 7r acts canonically from the left on the set X. Given a = aG x , b = bG y E X, the double coset Gxa-1bG y does not depend on the choice of the representatives a and b of the cosets a, b. By Corollary 3.6, there is a unique fa,b E E(x,y) such that Gxa-1bG y = Gxfa,bGy. Note that fa,b = lxy if a-1b E GxG y (in particular, fa,a = Ix) and fb,a = fa,b. Define a map d: X 2 _ A by dCa, b) = Ifa,bl for a, bE X. PROPOSITION 4.1. acts.

X

=

(X, d) is a A-tree on which the group 7r =

7rl(X, T)

°

PROOF: Obviously, dCa, b) 2:: for a, b E X, with equality iff a = b. As fb,a = fa,b, we get d(a,b) = d(b,a). For a,b E X, let [a,b] = [b,a] = {e EX: d(a,e) + d(e, b) = dCa, b)} and ia,b : [a, b] - [0, dCa, b)] be the map e t-+ d(a, c). For a, b, e E X, let Yea, b, c) = [a, b] n [b,e] n [e, a]. To conclude that X = (X, d) is a A-tree, it suffices to show that the map ia,b is bijective and the set Y( a, b, c) is non-empty for arbitrary a, b, e EX. First let us show that ia,b is bijective. Let f = fa,b and consider the map of: [J] - [0, Ifll introduced in Section 2, which is bijective by Proposition 2.3. We define a map f3 : [a, b] -+ [f] in such a way that of 0 f3 = ia,b. Let a = aGx , b = bG y and e = cG z E [a, b], 9 = fe,a, h = fe,b. By assumption If I = Igl + Ihl and there exist p, t E G x , q E G y and s E G z such that the following identities hold in 7r : a-1b = pfq and c-1a = sgt. Consequently, c-1b = sgtpfq = sPAc:(A),A~q, where A = (g,tp,j), and hence c:(A) = h and (g,Ggtp) E [fl. As Of is injective and 191 = d(a,e), it follows that f3( c) := (g, Gytp) is uniquely determined by a, b, e, and of 0 f3 = ia,b. It remains to check that f3 is bijective. Let e' = c'G u E [a, b] be such that f3(e) = f3(e'). Let g' = fe',a,c'-la = s'g't' with s' E G u , t' E G x . By assumption, 9 = g', z = u and tt,-l EGg. It follows c-1c' = sgtt'-lg-ls,-l = swg1(tt'-1)s'-1 E G z , i.e., e = e', and hence

f3 is injective. Let now 9 E E(z,x), v E G x be such that (g,Gyv) E [fl.

Let e = cG z , where c = apv-1g- 1; remind that a-1b = pfq with p E G x , q E G y. Then c-1a = gvp-l and c1b = gvfq = PBc:(B),A'Eq, where B = (g,v,j). Thus fc,a = g, fe,b = c:(B), e E [a,b] and f3(e) = (g,Ggv), and hence f3 is onto. Let now a = aG x , b = bG y, e = cG z E X. We have to show that Yea, b, e) is non-empty. Let f = fa,b, 9 = fb,e, a-1b = pfq, b-1c = sgt


53

with p E G x , q, s E G y, t E G z . Then a-Ie = pfqsgt = pPAE(A)AAt, where A = (f, qs, g), and hence fa,c = E(A). By Proposition 3.2, there exist u EX

and (h,v) E E(u,x) x G x such that Ihl + IE(B)I = If I, Ihl + IE(C)I = IE(A)I and IE(B)I + IE(C)I = Igl, where B = (h, v, f) and C = (h, VPA, E(A». Let e = eG u , where e = apv-Ih- I . It follows e-Ia = hvp-l, e-lb = hvfq =

= hVPAc(A)AAt = PGE(C)A:::'AAt, and hence d(e, a) = d(e,b) = ie(B)1 and d(e,c) = ie(C)I. Consequently, e E Y(a,b,c), as

PBE(B)A~q, e-le

Ihl,

required.

It remains to observe that d(sa,sb) the group

7r

acts on the A-tree

X.

= dCa, b)

for a, bE

X,

s E

7r,

i.e.,

0

PROPOSITION 4.2. The pair (7rI (X, T), XT) does not depend, up to an isomorphism, on the choice of T. PROOF: Let T' = (X, (l~y)(x,Y)Ex2) be another subpretree of X and Xo be a vertex of X. Consider the isomorphisms TIT ,Xo : 7rl (X, T) ~ 7rl (X, xo) and TIT' ,Xo : 7rl (X, T') ~ 7rl (X, xo) given by Proposition 3.5, and let Fl : 7rl (X, T) ~ 7rl (X, T') be the composite isomorphism Tli;,xo 0 TlT,xo' Identifying G x , x E X, with a common subgroup and E(x,y),x, y EX, with a common subset of 7rl (X, T) and 7rl (X, T'), we get Fl (s) = 1xoxs1xxo and FI(f) = 1xoxflyxo for s E C x , f E E(x,y). Define the map F2 : X T ~ XT , by F2(aG x ) = Fl(a)lxoxGx for a E 7rI(X, T); the independence on the representative a is immediate. The map F2 is bijective and F 2- l (bG x ) = FI-l(b)l~oxGx for b E 7rl(X, T'). Clearly, the pair (Fl' F 2 ) is compatible with the group actions. It remains to show that d'(F2(a),F2(b» = dCa, b) for a = aG x , b = bGy EXT. Let f = fa,b and a-lb = sft with s E G x , t E G y. It follows (Fl(a)lxox)-I(Fl(b)lxOY) =

lxxoFl(a-lb)lxoy

= lxxoFl(s)F}(f)F}(t)lxoy = sft,

and hence d'(F2(a),F2(b»

= If I = dCa, b),

i.e., f~2(a),F2(b) as required.

= f, 0

§5. Structure of a group acting on a A-tree. The aim of this section is to extend to arbitrary ordered abelian groups the structure theorem for a group acting on a A-tree [Se] Ch. I, Theorem 13. Consider two pairs (G, X), (G', X'), where X, X' are connected A-graphs and G,G' are groups acting from the left respectively on X,X'. By a morphism from (G, X) to (G', X') we understand a pair F = (F}, F 2 ) where F} : G ~ G' is a homomorphism and F2 : X ~ X' is a morphism of Agraphs which are compatible with the actions of the groups G and G', i.e.,

54

S.A. Basarab

Fz(sh) = FI(S)Fz(h) for s E G, hE arrowX. We say that the morphism F is an epi if F I , Fz are both onto. DEFINITION 5.1: The epi F = (FI,Fz ) : (G,X) -+ (G',X') is a cover (we may also say that (G, X) is a cover of (G', X')) if the following conditions are satisfied:

i) if hI, hz are arrows of X and Fz(hd = Fz(hz) then h2 = sh l for some s E G; ii) FI is locally isomorphism, i.e., for each vertex x of X the homomorphism G x -+ G'p.2(X)' between the stabilizers of x, Fz(x) in G, G', induced by F I , is an isomorphism; iii) Fz preserves the length, i.e., IFz (h) I = Ih I for each arrow h of X. 5.2. IfF = (FI,Fz ): (G,X) ing conditions are satisfied: LEMMA

-+

(G,X') is a cover then the follow-

a) The maps G\X -+ G'\X' and G\ arrow X -+ G'\ arrow X' induced by F2 are bijective. b) For each arrow h of X the homomorphism Gh -+ G'p.2(h) between the stabilizers of the arrows hand Fz (h) in G, G', induced by F I , is an isomorphism. c) Fz is locally bijective, i.e., for each x EX, the map Fz induces a bijection between X(x,-) = UYEXX(X,y) and X'(Fz(x),-) Uy/EXIX'(Fz(x), y'). d) If X' is an A-tree then X is a A-tree and F is an isomorphism. a) is immediate by 5.1.i) since FI and F2 are onto. b) Let h E X( x, y), x, Y EX. The injectivity of the homomorphism G h -+ G'p.2(h) is immediate by 5.l.ii). To check its surjectivity, let t E G'p.2(h) C G'p.2(X)' By 5.l.ii), there is s E G x such that FI(S) = t. Consequently, o(sh) = o(h) = x and Fz(h-I(sh)) = 1 F2 (x)' As F2 preserves the length by 5.1.iii), we get sh = h, i.e., s E G h , as required. c) First let us show that the map Fz,x : X(x,-) -+ X'(F2(x),-) is injective. Let hl,h z E X(x,-) be such that Fz(hd = Fz(h z ). By 5.l.i), hz = shl for some s E G x . It follows FI(S) E G'p.2(h')' and hence s E Gh l by b), i.e., hI = h 2 , as required. Next let us show that the map Fz,x is onto. Let h' E X'(Fz(x), -). As Fz : arrow X -+ arrow X' is onto, there is an arrow h of X such that F2(h) = h'. Let y := o(h). As Fz(y) = o(h') = Fz(x), it follows by 5.l.i) that x = sy for some s E G. Thus PROOF:

FI(S) E G'p.2(X)' and hence, by 5.1.ii), FI(S) = FI(t) for some t E G x·


55

Consequently, rlsh E X(x,-) and F 2(t- l sh) = F I (t-IS)F2(h) = h', as required. d) As F is epi, which preserves the length, it remains to show that FI and F2 are injective. First let us show that F2 : X ---+ X' is injective. Let X,Y E X be such that F2(X) = F2(Y)' Since X is connected, there exists an arrow h E X(x, y). As X' is assumed to be a A-tree, we get F2(h) = I F2 (x) and hence h = Ix by c). Thus x = y, as required. Next let us show that FI is injective. Let s E ker FI and x E X. As F2(sx) = F2(X) it follows sx = x by the first part of the proof. Since s E G x and FI(s) = 1, we get s = 1 by 5.l.ii). Finally let us show that F2 : arrow X ---+ arrow X' is injective. Let hI, h2 be arrows of X such that F 2(h l ) = F2(h2)' In particular, F2(O(hd) = F2(O(h2)' and hence o(hd = O(h2) thanks to the injectivity of F2 : X ---+ X'. According to c), we get hI = h2' as required. 0 The next lemma is immediate. LEMMA 5.3. Let F: (G,X) ---+ (G',X') and F' : (G,X/) ---+ (Gil, X") be epis and assume that F' is a cover. Then F' 0 F is a cover iff F is a cover. DEFINITION 5.4: The cover F : (G, X) ---+ (G, X) is universal if for each cover F' : (G',X/) ---+ (G,X) there is an epi (not necessarily unique) F : (G, X) ---+ (G', X') such that F' 0 F = F. The next lemma plays a key role in the following. LEMMA 5.5. Let F : (G, X) ---+ (G, X) be a cover and assume that A-tree. Then the cover F is universal.

X

is a

PROOF: Consider a cover F' : (G,X/) ---+ (G,X). To construct an epi (in fact a cover by Lemma 5.3) F: (G,X) ---+ (G',X') such that F' o f = F, we proceed step by step as follows. 1) Fix a point Xo of the A-tree X. As the map F~ : X' ---+ X is onto we may choose a vertex x~ of X' such that F~(x~) = F2(XO). Define a map F2 : X ---+ X' with F 2(xo) = x~, as follows. Given x E X, it follows by Lemma 5.2.c) that there exists a unique arrow Zx of the A-graph X' such that o(lx) = x~ and F~(lx) is the image F2(xo,x) of the unique arrow in the A-tree X with origin in Xo and terminus in x. Let us put F2(X) = t(lx). 2) Extend F2 : X ---+ X' to a map F2 : X 2 = arrrow X ---+ arrow X' by F 2 (x,y) = t;IZy. Obviously, o(F2 (x,y» = F2 x), t(F2 (x,y») = F 2 (y), F2 (x,x) = I F2 (x) and F 2(x,y)F2(y,z) = F2(x,z) for x,y,z E X. Thus

56

F2

S.A. Basarab

X

X' is a morphism of A-graphs and F~ 0 F2 = F2. In fact, according to Lemma 5.2.c), F2 is the unique morphism of A-graphs subject to F~ 0 F2 = F2 and F2(xo) = x~. 3) Given sEe, let us show that there exists a unique t E G satisfying: F{(t) = FI(S) and F 2(sx) = tF2(X) for each x E X. Fix for a moment some point x EX. Since F{ : G' -+ G is onto, there is pEG' such tht F{(p) = FI(s). Consequently, F~(pF2(X» = F2(SX) = F~(F2(SX)). As F' is a cover, there is q E G' such that qpF2 (x) = F2 (sx), and hence -+

F{(q) E G p2 (sx)" By 5.l.ii), there exists r E G~2(SX) with F{(r) = F{(q). Let us put t = r-Iqp. Obviously, F{(t) = FHs) and F2(SX) = tF2(X). Now let y E X, may be y = x, and t' E G' be such that F{(t') = FI(S) and F 2( sy) = t' F2(y). We have to show that t = t'. Consider the arrows

and

Since the arrows above have the common origin F2 (x) and F~ (F2 (x, y» = F2(x, y) = F~(el F2(sx, sy», it follows by Lemma 5.2.c) that F2(x, y) = t- I F2(sx, sy) and hence elt' E G~2(Y). As F{(eIt') = 1, we get t = t' by

e

5.l.ii). Thus there exists a unique map FI : -+ G' subject to F{ oFI = FI and F2(S(X, y» = FI (s )F2(X, y) for each sEe and each arrow (x, y) in the A-tree X. The unicity of the map FI satisfying the conditions above implies also that FI is homomorphism. Thus the pair F = (FI' F 2 ) is the unique morphism from (e, X) to (G', X') with the properties F' 0 F = F and F2(xo) = x~. To finish the proof we have to show that the morphism F is an epi. First let us check that the homomorphism FI : -+ G' is onto. Fix some x E X and let t E G'. As X' is connected, there is an arrow h E X'(F2(x), tF2(X». Since F is a cover, there exists, by Lemma 5.2.c), a unique point y E X such that F2 (x, y) = F~(h). As F~(h) E X(F2(X),F{(t). F2(X», it follows by 5.l.i), that y = sx for some sEe. Applying Lemma 5.2.c) to the arrows h and F2(x,sx) in X' having the common origin F2 (x) and the same image through F~, we get h = F2(x,sx), and hence tF2(X) = t(h) = t(F2(x,sx» = FI(S)F2(X). Thus t-IFI(s) E G~2(X)' and hence F{(t)-IFI(S) E G p2 (x). By 5.1.ii), there is

e

p E ex such that FI(P) = F{(t)-I FI(S). Since

e l FI(Sp-l)

E G~2(X) and


57

FHt-IH(sp-I» = 1, we get t = FI(sp-I), by 5.l.ii), proving that FI is onto. It remains to show that F2 : X 2 -+ arrow X' is onto. Let h be an arrow of X'. As F2 : X 2 -+ arrow X is onto there is an arrow (x, y) of X such that F2 (x, y) = FHh). Since the arrows hand F 2 (x, y) in X' have the same image through F~, it follows by 5.1.i) that h = tF2 (x, y) for some t E G ' . As Fl is onto, there is s E G such that FI (s) = t. Consequently, h = FI(s), F 2(x,y) = F 2(sx,sy), as required. D We are now prepared to prove the main result of

t~e

paper.

THEOREM 5.6. Given a group G which acts on the connected A-graph X, there exists a universal cover F : (G, X) -+ (G, X). X is a A-tree and F is unique up to an isomorphism (not necessarily unique). PROOF: Existence: To the pair (G, X) we may assign as in Section 2 a strongly connected A-graph of groups Y = Y(G,X;j,a) and a maximal subpretree T = T(G,X;j,a) = (Y,(l xy )(x,Y)EY2) depending on the choice of the maps j : E = G\ arrow X -+ arrow X and a : E -+ G. Further, to the pair (Y, T) one may assign as in Section 4 a A-tree Y = YT together with an action of the fundamental group 7r = 7r1 (Y , T) on Y. By construction, the maps G x -+ G : s r--t S E G jx C G and a : E -+ G induce a homomorphism FI : 7r -+ G. Note that G x = G jx is identified with a common subgroup of 7r and G and PI is the identity on G x for each x E Y. Moreover, PI is onto. Indeed, let a E G and x E Y. Since X is connected, the set X(jx,ajx) is non-empty, and hence there exist f E E(x,x) and s E G x such that t(sjf) = ajx. As t(jf) = a(f)jx, we get a-Isdf) E G x , i.e., a = sa(f)t for some t E G x' Therefore a = FI (s ft mod R T ), as required. On the other hand, we may define a morphism of A-graphs F2 : Y -+ X as follows. Let us put F2(a) = FI(a)jx for a = aG x E Y; obviously, the definition does not depend on the representative a of the coset a E 7r IGx' For an arrow of the A-tree Y, i.e., an ordered pair (a, b) with a = aG x , b = bGy E Y, we have to define an arrow F 2(a, b) E X(F2(a),F2(b». By Corollary 3.6, there exist fa,b E E(x, y), uniquely determined by the cosets a, b, and some s E G x , t E G y such that the identity a-Ib = Sfa,bt holds in 7r. Let us put F 2(a, b) = F 1 (a)sjfa,b' By Corollary 3.6, the definition does not depend on the choice of a,b,s,t, and F2(a, b) E X(F2(a),F2(b». Obviously, F 2(a,a) = 1 P2 (a) and IF2 (a,b)1 = Ifa,bl = d(a, b), so it remains to show that F2 (a, b )F2 (b, c) = F2 (a, c) for a, b, c E Y to conclude that F2 : Y -+ X is a morphism of A-graphs preserving the length.

58

S.A. Basarab

Let a = aG x , b = bGy, c = eG z , f := fa,b, g := fb,c, a-lb = sft, b-le = pgq with s E G x , t,p E G y , q E G z . We get a-Ie = sftpgq = sPAs(A)A~q, where A = (J,tp,g), PA E peA) and A~ = AA(PA)-l. Thus fa,c = seA) and F2 (a,c) = Fl(a)sPAjs(.4) = Fl(a)s[(jJ)(a(J)tpjg] = (F2 ( a, b ))(Fl ( asftp)jg) = (FHa, b ))(Fl (b )pjg) = F 2 ( a, b )F2 (b, c), as required. Note also that the map F2 : y2 -+ arrow X is onto. Indeed, let h be an arrow of X. Then there exist f E E(x, y), x, Y E Y, and v E G such that h = vjf. Since the homomorphism Fl : Jr -+ G is onto, there is a E Jr such that Fl(a) = v. Let a = aG x , b = bG y E Y with b = af. As a-lb = f it follows F 2 (a, b) = FlCa)jf = vjf = h, as required. Obviously, the pair (Fl' F 2 ) is compatible with the actions of Jr and G. Thus F = (FI, F2 ) : (Jr, Y) -+ (G, X) is an epi which preserves the length. To conclude that F is an universal cover it suffices, by Lemma 5.5 to show that F satisfies the conditions 5.l.i), ii). First let us check 5.l.i). Let a = aG x , b = bG y, a ' = a'Gx" b ' = b'Gyl E Y, f:= fa,b, l' := fal,bl , a-lb = sft, a'-lb' = alI't' with s E G x , t E G y, s' E GXl, t ' E Gyl, and assume that F 2(a, b) = F2(a / , b/), i.e., Fl(a)sjf = Fl(a/)s'j1'. It follows f = 1', x = x', y = y' and p := s'-l Fl(al-la)s E G j. Consequently, Fl(a'-la) = s'ps-l E G x , so we may see it as an element of Jr and consider the element e := aIFl(a,-la)a-l of Jr. We get the identities in 7l' : e = a' s'ps-la- 1 = b't'-l f-1pftb- 1 = blt,-IWj(p)tb- l . It follows ea = eaG x = aIFl(al-1a)G x = a'Gx = a ' and eb = ebG y = blt'-1Wj(p)tG y = b'G y = b/. Thus e(a, b) 5.l.i) is verified.

The condition 5.l.ii) is immediate since given a = aG x E of a and F2(a) = Fl(a)jx in 7l' and G are respectively , , 1 ' G F2 (a) = FI(a)GxFI(a)- , and FI is the identity on G x .

=

(a', b/) and

i< the stablizers 7l'a

=

aGxa- l

,

o

Unicity: Immediate by Lemmata 5.2.d) and 5.3.

COROLLARY 5.7 (Structure theorem for groups acting on A-trees). With the notations above the following statements are equivalent:

i) X is a A-tree. ii) FI : 7l' -+ G is an isomorphism. iii) F2 : Y -+ X is an isomorphism of A-graphs. i) =} ii) is immediate by Lemma 5.2.d) since is trivial since Y is a A-tree. PROOF:

F is a cover.

iii)

=}

i)


59

'*

ii) iii): As F2 is onto and preserves the length, we have only to show that F2 is injective. Let hI, hz be arrows of Y such that F2(hd = F2(h2). Since F is a cover, it follows by 5.l.i) that h z = ahl for some a E 7r. Thus FI (a) E G F2 (hll' and hence a E isomorphism. Consequently, hI

7rh,

= ahl = hz,

since by assumption FI is an i.e., F2 is injective.

0

The most interesting assertion of the corollary above is the implication i) ii) which gives a description of a group acting on a A-tree in terms of generators and relations as in Section 3. We end this section with the following version of Theorem 5.6 which is an immediate consequence of Theorem 5.6 and of the proof of Lemma 5.5.

'*

COROLLARY 5.S. Given a group action on a rooted connected A-graph, i.e., a triple (G, X, xo) with Xo E X, there exists a group action on a rooted A-tree (6, X, xo) and a cover F = (FI, F2 ) = (6, X) -> (G, X) preserving the root, i.e., F2 (xo) = Xo, such that the following universality property holds: Given a group action on a rooted connected A-graph (G' , X', x~) and a cover F' = (F{,FD : (G',X/) -> (G,X) with F~(x~) = Xo, there exists a unique cover F = (Fl' F 2 ) : (6, X) -> (G ' , X') with Fz(xo) = x~ such that F' o f = F.

§6. Examples. To illustrate the general theory above we give in this section two examples. The first one is quite simple, while the second one concerning the SL z of a valued field is more elaborated and susceptible of further developments.

6.1. The isometry group of an ordered abelian group. Given an ordered abelian group A, we may assign to A two significant structures of connected A-graphs. The first one is the A-tree X considered in [M-S], [A-B], whose universe X is the underlying set of the group A and whose metric is given by d(a, (3) = 10' - (31 for 0',(3 E A, with 10'1 = max(a, -a). The second one is the A-graph X having a unique vertex, with arrow X = A. The groupoid structure on X is identified with the additive group structure of A and the length function A -> A is nothing else than the module function a f-+ 10'1. Denote by 6 the isometry group of the A-tree X consisting of the translations ta : x f-+ x + a and the reflections r a : x f-+ Q - x, for a E A. On

60

S.A. Basarab

the other hand, denote by G the cyclic group of order two with generator roo The group G acts on the A-graph X according to the rule 7"oX = -x for x E A. Let 1'1 : ---> G be the homomorphism given by F1(ta) = 1, 1'1 (7" a) = 7"0 for 0: E A. Denote by 1'2 : X ---> X the morphism of A-graphs which sends the points of X in the unique vertex of X and each arrow of X, i.e., an ordered pair (0:,{3) E A2, in the arrow {3 - 0: of X. It is easy to see that the pair I' = (1'1,1'2): (a,X) -+ (G,X) is a cover, and hence a universal cover, by Lemma 5.5. Thus we may obtain a description of the gorup in terms of generators and relations applying to the pair (G, X) the procedure described in the existence part of Theorem 5.6. First we have to assign a strongly connected A-graph of groups Y = Y(G,X;j,a) to the pair (G,X). Obviously, Y has a unique vertex and the space of G-orbits E = G\A may be identified with the subset of nonnegative elements of A. Then the maps 0: f--+ a and 0: f--+ 10:1 are both the identity of E. Let j : E ---> A be the inclusion and a : E -+ G be defined I ifo:=O by a( 0:) = { . . Thus Oa = 1 for each 0: E E, the stabilizer G a 7"0 If 0: f. 0 is trivial if 0: i= 0, respectively G if 0: = 0, and Wa = Ie". The map c : 10: - (31 if s = 1 Ex G x E -+ E is given by c( 0:, s, (3) = Io:+a( 0: )s{31 = { 0: + {3 if s = 7"0 For A = (0:, S, (3) E E x G x E, we get:

a

a


{I}

or

peA) =

8 = 1 and (a > = ro and a oj o.

if either 8

(3 or 0

{ro}

if either a = 1 and 0 < a < (3

G

if eitheT

or

or

8

8

= TO 8

=

= TO

and 0

=a
O.

64

S.A. Basarab

by W(x,a,y) G(x,a,y) -+ G(x,a,y) the isomorphism u(x,a,y)-l su(x,a,y). Now let us define the map c : E(x,y) x G y x E(y,z) -+ E(x,z) for x,y,z E Y, by c«x,a,y),s,(y,{3,z)) = (X",Z), where I = -v(e;I U(X, a,y)su(y, {3,z)ez)' For A = «x,a,y),s,(y,{3,z)) E E(x,y) x Gy x E(y,z), let A = «y,{3,z),f}(y,{~,%)S-If}(x,a,y),(x,a,y)), peA) = G x n u(x, a, y)su(y, {3, z)G zu(c(A))-1 E Gx/G€(A), AA : peA) -+ peA) be the bijection given by AA(P) = u(y, {3, Z )-1 s-lu(x, a, y)-lpu(c(A)). Thus we Denote

s

1-+

constructed a strongly connected A-graph of groups Y and a maximal subpretree (lxy)(x,y)EY2. To get the desired presentation of G it suffices to compute 7rl (Y, xo) for some vertex Xo of Y = A/2A. Take for simplicity Xo = 0 mod 2A. Then E( xo, xo) is identified with A ~ 0,

u(a) = {I Sa

~fa=O,

If a> 0

Ga =

{(a c

morphism Wa of Ga is the identity if a

s=

(~ ~)

db) EGo

:v(b)~2a},

= 0 and waC s) =

the auto-

(-b~';/ -~;c ) if

EGa, a> O. Further, we get c(A) = max(a+{3-v(b), la-{3I)

for A = (a, s, {3) E A;?:o x Go x A;:::o, s putations it follows that G the elements Sa =

(_~;1

~

=

(~ ~).

Doing the explicit com-

7rl(Y'O) is generated by Go = SL 2 (0) and

to)

for 0 < a E A, subject to the relations:

S! =-1

(1)

(2)

(3) for 0 ::; v(b) < 2min(a, {3), I = a + {3 - v(b). In particular, if A = Z, i.e., the valuation v is discrete with v( 7r) = 1 for some 7r E K, the group G = SL 2 (K) is generated by Go :;= SL 2 (0) and the element S

(1')

=

(_O~ ~)

subject to the relations:


=

Let G 1

{ (~ !)

{(:e

7r~lb): (~

!) EGO}

and B

=

Go n G 1

~s

if s

=

and G 1

(:e

-+

F :s

7r~lb),

=

2 I}. Let F be the free product of the group Go

E Go : vee)

and the infinite cyclic group with generator S. Consider the maps Go s

65

f-+

v(b)

s if s E B, s

=

f-+

-+

F :

(~I 7rd~-l) S (~~~ ~b)

O. One checks easily that the maps above

induce an isomorphism from the sum Go *B G 1 of the groups Go and G 1 amalgamated along their intersection B onto the quotient of F by its normal subgroup generated by the relations (11), (2/), (3 / ). Thus we recover Theorem 7 from [Se], Ch. II. REMARK: A quite similar result may be obtained by applying the procedure above to the tree considered by Morgan in [MoJ. Let (K, v) be a valued field with valuation ring 0 and value group A and assume that the residue field is formally real. Consider the non-degenerate symmetric bilinear form xy = XOYl +XIYO+ L~=2 XiYi on the vector space K n+1 , n 21. Let

G = SO(n,I)K be the subgroup of SLn+1(K) consisting of the matrices which preserve the bilinear form above. To an O-lattice L in K n + 1 one assigns its dual lattice L* = {x E Kn+l : x·L CO}. Lis unimodular if L = L*. It is shown in [MoJ that the set X of unimodular lattices in K n +1 has a canonic structure of A-tree on which the group G acts transitively. Denote by Lo the standard lattice on+l and Go = G n SLn+1(O) be its stabilizer in G. The distance function d: X 2 -+ A is given by d(sLo, tLo) = -V(S-lt) for s, t E G, where v(s) = mini,j V(Sij) for s = (sij)OSi,j~n' For n = 1, the A-tree X = (X,d) is identified with A,d(a, (3) = la - (31, and G ~ K X acts on A through translations: sa = v( s ) + a for s E K x , a E A. Thus we may assume in the following that n 2 2. For a E A, a > 0, choose to! E KX such that veta,)

=

a, and denote by

66

S.A. Basarab

So: the matrix in G 1

1 (0

1 n -2

1

t;;l 0 0

1 n-2 0 to: 0 0 0 1n - 2 0 0

1

~ ).

-1

Put So = 1n+1. The map j : A::::o --. X 2 : 0: f-+ (Lo, So:Lo) identifies A::::o with the space of G-orbits G\X2. Note that (j0:)-1 = (So:Lo,Lo) = So:jo: for each 0: E A:::: 0 . Thus the A-graph of groups Y assigned to the pair (G, X) has a unique vertex and the set E = Ey of edges is identified with A::::o. It follows that a = 0:, Go: := Gjo: = Go n So:GoSo: = {s E Go : v(sod ?: 20:}, 80: = S; = 1n+1 and the automorphism Wo: of Go: is given by wo:( s) = So:sSo: for s E Go:. Now we have to describe explicitly the map c: : E x Go x E --. E. CLAIM:

s

=

c:(o:, s, (3)

=

max(lo: - (31,0:

+ (3

- v(sod) for 0:, (3 E E

=

A::::o,

(Sij)O:;i,j:;n EGo.

As the case 0: = 0 or (3 = 0 is trivial we may assume 0:, (3 > O. We have to verify that v(So:sS~) = mine -10: - (31, v(sod - 0: - (3). Taking into account the form of the matrix SasS ~ it suffices to check that v( soo) = v( Sl1) =

0, v(sod = 2min2:;.:;nv(s.d and min2:;i:;n v(so.) ?: min2:;.:;n v(s.d if v(sod > O. Assuming v(sod > 0 it follows 2min2:;i:;n v(s.d = V(SOlS11) since 2s 01 Sl1 + L:7=2 s;l = 0 and the residue field of v is formally real. On the other hand, the identifies SOOSl1 + S10S01 + L:7=z SiOSil = 1 and SOlSIj + Sl1S0j + L:7=2SiISij = 0 for 2::; j ::; n imply v(soo) = V(S11) = 0, v(sod = 2minz:;i:;n v(sid and min2:;i:;n v(so.) ?: min2:;.:;n v(s.d· CLAIM:

The group G is described in terms of generators and relations as

follows: Generators: Go and the matrices Sa for 0:

> O.

Relations:

(1) S; = 1n+1 for 0' > O. (2) Sas = wa(s)Sa for 0: > 0, S EGa. (3) SasS~ = s'S,,{s" for 0:, (3 > 0, s E Go with Soo = Sl1 = 1, Sij = bij for i :I: 0, j :I: 1 (thus Sd = -SOi for 2::; i ::; nand 2801 + L:~z s;l = 0), v(sod < 2min(0:, (3), 'Y = o:+(3-v(sod, . h8'00 = Sl1 , = l ' = Vij i: £or 1.. -r~ 0 ,J. -r~ l - 1 s , E G 0 WIt ,Sij ,8'01 = t Za S01' .I _ t . - 1lor !"? < .< 1 -.I t ,-1' "£ 2 < sd - a8dSOl ~ _ 1. _ n- ,.'i n1 - - o:Snl.'iO I ,SO; _ 1.. < _ n, 1 or

-.'i'


67

and SIt E Go with s~~ = t-;;lti/t-yS018iO, s~i = tatpC;lso/81i for 0 S; i S; n, S~l = t-;;ltpt-y, s~i = _t-;;lt-ySil for 2 S; i S; n - 1, s~n = t-;;lt-ySn1' S~/1 = -tpsilS011 for 2 S; i S; n, s~j = 8ij + Si1Sj1S0/ for 2 S; i S; n, 2 S; j S; n -1, s~/n = -8 in - Si1 Sn1 so/ for 2 S; i S; n. The claim above is a consequence of the general theory and of the following fact: Let S = (Sij)O~i,j~n E Go. We distinguish two cases. ,n 2 C ASE 1: v ( ) > O. A S 2s 00 s lO + "LJi=2 SOO SiO = 0, n

SOOSl1

+ S10 S01 + L

i=2

SiO S i1

=

1

and the residue field of v is formally real it follows min2~i~n v( SiO) > 0 and v( sod = v( S10) = O. Let p E na>OG a and q E Go be the matrices given by POi = S0180;, Pi1 = sl/8i1 for 0 S; i S; n, P10 = -tso/ L~=2(1 + Si1?, PiO = 1 + Si1 for 2 S; i S; n, Pi1 = -so/(1 + sid for 2 S; i S; n, Pij = 8ij for 2 S; i,j S; n; qoo = ql1 = 1, % = 8ij for i i= 0, j i= 1, qOi = -n~l for 1 S; i S; nand qi1 = n~l for 2 S; i S; n. One gets (pq)-ls E na>OGa . 2: v(soo) =). Then S = pqr, where P E na>OG a , q E Go and r E na>OGa are given by PiO = SiO for 0 S; i S; n, Pl1 = SOOl, P1i = -sOOl SiO for 2 S; i S; n, Pij = 8ij for i i= 1, j i= 0; qoo = qll = 1, Q01 = SOol SOl, qOi = -Qi1 = SoOl SOlSiO - Si1 for 2 S; i S; n, qij = 8ij for i i= 0, j i= 1; rij = 8ij for i = 0,1 or j = 0,1 and rij = Sij - SOOl SiOSOj for 2 S; i,j S; n. CASE

In particular, if the valuation v is discrete, the group G is generated by Go and the matrix 5 = 51 subject to the relations 52 = 1n+1 and 5s = wl(s)5 for S E G 1 = {p E Go : v(pod 2:: 2}. Thus G is the group derived from (G 1, Go,wd by a HNN-like construction. It follows that Gis the semi-direct product of the cyclic group of order two generated by 5 and the normal subgroup U generated by Go, consisting of the matrices S for which v( s) == 0 mod 2. Moreover U is the amalgamated sum of Go and its conjugate 5G o5 along the intersection G 1.

68

S.A. Basarab REFERENCES

[A-B] R. Alperin and H. Bass, Length functions of group actions on A-trees, Proceedings of the 1984 Alta Conference on Combinatorial Group Theory and Topology, Ann. of Math. Studies. [M-S] J. W. Morgan and P. B. Shalen, Valuations, trees, and degenerations of hyperbolic structures, I, Annals of Math. 120 (1984), 401-476. [Se] J. P. Serre, "Trees," Springer-Verlag, Berlin-Heidelberg-New York, 1980. [Mo] J. Morgan, Group actions on trees and the compactification of the space of classes of 80(n, I)-representations, Topology, 25:1 (1986), 1-33.

Department of Mathematics, The National Institute for Scientific and Technical Creation, Bd. Pacii 220, 79622 Bucharest, Romania

Group Actions on Non-Archillledean Trees HYMAN BASS

§O. Introduction. Let A be a (totally) ordered abelian group. Let X be a A-tree, as defined in [MSI] (see also [AB] for various equivalent definitions). Let r be a group acting (without inversion) on X. GENERAL PROBLEM: What group theoretic information about drawn from its action on the A-tree X?

r

can be

The motivation for this problem comes from the special case A = I, when r acting simplicially. In that case one knows

X is a simplicial tree with from Trees [Se] that

the fundamental group of a graph of groups (0, Y) relative to a maximal subtree T of Y. Here Y = x/r and if y is a vertex or edge of Y then O(y) is the stabilizer r ii of a suitable lift fj of y to X. The above presentation often yields important combinatorial information about r. The work of Morgan-Shalen [MSI] shows that interesting A-tree actions arise naturally even when A is no longer cyclic, e.g. with A = IR. The group theoretic information carried by such actions is now beginning to be explored (cf. [GS], [M], [CM], [RI], [AM], which deal with IR-trees). The first natural case to consider is that of free actions. To facilitate discussion let us call a group r A-free if r can act freely (without inversion) on a A-tree. This terminology is motivated by the fact that I-free groups are free groups [Se]. In general A-free groups are stable under passage to subgroups (obvious) and free products (Corollary (4.4) below). Moreover A acts freely on X = A by translation, so subgroups of A and their free products are A-free. Lyndon [L] once conjectured (in a different context) that alllR-free groups are free products of subgroups of R Counterexamples have been produced by Alperin-Moss [AM], by Promislow [Prj (cf. [RI]) , and by Morgan-Shalen [MSI], who showed that most surface groups (= fundamental groups of surfaces) are IR-free. It has been tentatively conjectured that these examples explain all finitely presented IR-free groups.

70

H. Bass

Special cases of this have been confirmed in Gillet-Shalen [GS], R. Jiang [J], and J. Morgan [M]. This paper investigates A-tree actions when A is non-archimedean, i.e. not embeddable in R. In that case A admits a nontrivial "convex" subgroup Ao (0 ~ a ~ ao, ao E Ao =} a E Ao), and then A * = AI Ao inherits a natural order. Our main tool is the base change functor, X -+ X* = "A * ®A X", from A-trees to A*-trees. There is a canonical projection X -+ X* whose fibers are Ao-trees, the "balls of radius Ao" in X. If a group r acts on X then it acts also on X*. We consider the following "extension problem". Given a A*-tree X* with r -action, what additional data are needed to construct a A-tree X with r-action so that A* ®A X is r-equivariantly isomorphic to X*? We provide a general solution of this problem when A* = 1. In that case the action on X* is equivalent to a presentation r = 71"1 (9, Y*, T*) where y* = r\x* and T* is a maximal subtree. Thus, if A = 1 xAo (lexicographic) then we can characterize A-free groups r as those admitting a presentation r = 71"1 (9, Y*, T*) where the vertex groups are Ao-free, the edge groups are either trivial or maximal abelian in the adjacent vertex groups, and some further "compatibility conditions" are satisfied. This characterization is effective enough to decide A-freeness for many interesting examples. For example let r i be a free group and Si an element of r i which is not a proper power (i = 0,1). Then the amalgam r = ro *(so=st} r 1 is (1 x 1)free. Such amalgams are more general than surface groups. If ro = r 1 then the HNN-extension r = (ro, t I ts ot- 1 = Sl) is (1 x I)-free iff Sl is not conjugate in ro to SOl, and there is a free action of ro on a I-tree X so that So and Sl have the same hyperbolic length. A variety of more elaborate examples is discussed in detail in Section 4. Section 2 solves the extension problem above for a base change X -+ X*, and Section 3 solves it equivariantly in the presence of a group action. The main results are Theorems (3.5) and (3.8). In passing from one fiber of X over X* to another fiber, one exits the first fiber through one of its "ends", and enters the new fiber through one of its ends. The analysis of this phenomenon requires a fairly detailed general discussion of ends of A-trees; this is the main purpose of Section 1 and Appendix E.


71

Contents of Group Actions on Non-Archimdean Trees

O. Introduction

69

1. A-trees and their ends

71

2. Base change for A-trees; ends between fibers

83

3. Base change for group actions on A-trees

91

4. Free actions on A-free groups

102

5. Appendix on ends

121

72

H. Bass

§1. A-trees and their ends. Much of this section is a review of material in [AB], Ch. 1. Let A be a nonzero (totally) ordered additive abelian group. For a < b in A we put lal = max(a, -a) and [a, b] = [b, a] = {x E A I a :s: x :s: b}. A A-metric space is a set X with "A-metric" d : X X X --+ A satisfying d(x,y) = d(y,x) ~ 0, d(x,y) = iff x = y, and d(x,z) :s: d(x,y) + d(y,z). An example is X = A with d(x,y) = Ix - YI. A metric map between A-metric spaces is one preserving distance; it is necessarily injective. A geodesic from x to y in X is a metric map 8 : [O,d(x,y)] --+ X such that 8(0) = x and 8(d(x,y)) = y. A nonempty A-metric space X is called a A-tree if: (i) For x, y E X there is a unique geodesic from x to y; we denote its image [x,y]. (ii) For X,y,z E X, [x,y] n [x,z] = [x,u] for some u E X. (iii) For x, y, z EX, if[x, y] n [y, z] = {y} then [x, y] U [y, z] = [x, z]. For example A itself is a A-tree. A linear A-tree is any A-tree isomorphic to a subtree of A. The isometries of A are two types: translations tcC x) = x + c; and "reflections" r c( x) = -x + c. Here c is any element of A. If c t/: 2A then r c is called an inversion because its fixed point (c/2) is missing. The translations form a normal subgroup (~A) of Aut(A) of index 2. ([AB], (2.5).)

°

(1.1) LEMMA (Linear Rigidity). Let L be a linear tree and A a subset of

L. (a) Any two metric maps A --+ A differ only by composition with an isometry of A. (b) Any metric map 8A : A --+ A extends to a metric map 8 : L --+ A. If Card( A) ~ 2 then 8 is unique. (Cf. [AB), (2.5).) Let X be a A-tree and x EX. Then X - {x} is the disjoint union of its maximal subtrees. The cardinal number of these we call the index of x in X, denoted index) or indx(x). (Other terminologies used are "valence", "degree", ... .) If ind( x) = 0 then X = {x}. We call x an end point if ind( x) :s: 1, an edge point if ind( x) = 2, and a branch point if ind( x) ~ 3. (1.2) REMARK: Let x be an end point of X. If y, z E X are distinct from x then [x, y] n [x, z] = [x, u] with u # x. For otherwise x is interior to [y, z] = [y, x]U[x, z] and so y and z belong to distinct components of X -{x},


73

contradicting ind( x) :5 1. By induction, if Yl, ... ,Yn are all distinct from x then ndy;, x) = [u, x) with u =f:. x. Let X be a A-tree. For x E X consider linear subtrees L of X having x as an end point. Maximal such are called X-rays from x. It might happen that an X-ray L from x is a closed interval [x, e). In this case e is clearly an end point of X, called the terminal end point of L. Otherwise we call L an open X-ray. Let L, L' be X-rays (from possibly different points). We call L and L' equivalent if either L and L' are closed with the same tenninal end point, or else L n L' is an open X-ray. This is an equivalence relation, whose classes are called the ends of X. If e is the end of an X-ray L we say that "L ends at e." (Cf. [AB), (2.23).) Let e be an end of X. For each x E X we define [x, e) as follows. If e = ee is the "closed end" corresponding to an end point e of X then [x, ee) = [x, e); for x =f:. e this is the X-ray from x ending at ee. If e is an open end then [x,e) is the unique X-ray from x ending at e (which exists). (Cf. [AB), (2.24).) If [x,y) n [y,e) = {y} then clearly [x,y) U [y,e) = [x,e:). (1.3) EXAMPLE: A has two ends, 00 = OOA and -00 = -OOA, such that [a, 00) = {x E A I a:5 x} and [a,-oo) = {x E A I x:5 a}. An end e of a A-tree X is said to be of full A-type if for some (and hence every) x EX, [x,e) ~ [O,OOA). (A discussion of end types in general can be found in Appendix E3 below.) If LeX is a linear subtree from x and L ~ [0, ooA) then L must be an X-ray. For otherwise L C [x, y) for some y, and d(x, y) > a for all a E A, which is impossible. (1.4) LEMMA. Let X be a A-tree and e: an end of X. (a) There is a map 8 : X -+ A such that, for all x E X, 8 I [x,e) is a metric map with least value 8(x). (Such a map is called an "end map toward e".) (b) Let 8' be another end map toward e:. Then 8' = 8+c (8'(x) = 8(x )+c) for some c E A. PROOF: (Cf. [AB], (2.33).) If X = {e} then any 8(e) will do. Otherwise fix some x E X so that Card([x, e» :::: 2 and any metric map 8x : [x, e:) -+ A with least, value 8(x). For y E X let 8y : [y,e:) -+ A be the unique metric map agreeing with 8x on Uy,x = [y,e:)n[x,e:). The existence and uniqueness of8 y follows from linear rigidity (1.1) since, ify ~ [x,e:), CardUy,x:::: 2 (cf. (1.2». Similarly, if also z ~ [x,e:) then Card([z,e) n [x,e:) n [y,e» :::: 2, so 8y and 8z agree on [y,e) n [z,e:). Thus the 8y assemble to give the required

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H. Bass

X --+ A. Since /j is determined by /jx, and /jx is determined up to an additive constant, assertion (b) follows also. /j :

(1.5) LEMMA. Let X be a A-tree, E, E' distinct ends, and /j, /j' : X --+ A end maps toward E, E' respectively. (a) There is an x E X such that [x, E) n [x, E') = {x}. Then [x, E) U [x, E') is a linear tree, consisting of all x as above; we denote it (E', E). (b) /j + /j' I (E',E) is a constant map (E',E) --+ A. ( c) HE, E' ,E" are distinct ends then (E', E) n (c;", E) n (E', E") is a single point, denoted Y(E',E,C;"). PROOF:

For (a) cf. [AB], (2.24)(b).

Since (E',E) = Ux[x,c:) = Ux[X,E'), where x ranges over (E',E), and the unions are nested, it follows that both /j and /j' define metric maps (E', E) --+ A. Since they are oppositely oriented it follows from linear rigidity (1.1) that /j' = -/j + c on (c;', c;) for some constant c E A. To prove (c) choose

n (c;" , E) (E,E') n (E",E') (E,E") n (E',E")

x' E (E', E)

x' E x" E

and put y = Y(x,x', x"), so that

{y} = [x',x] n [x",x] n [x', x"] C (E', E)

n (E", E) n (c;', c;")

c;'

~' E"

(cf. [AB], (2.9) and (2.11». Then

(E', E) n (E", E) n (E', E") = ([y,E') U [y,E» n ([y,E") U [y,E» n ([y,E') U [y,E"»


75

Since any two of [y,c), [y,c'), [y,c") have only y in common, the above intersection is just {y}. We now recall ([AB], §6) the classification of tree automorphisms (self isometries) of X and the hyperbolic (or translation) length function

I· I : Aut(X)

-+

A.

Let 8 E Aut(X). If 82 has a fixed point but 8 has none then we call 8 an inversion, and put 181 = o. Otherwise we define

181 = min{d(x,8x) I x EX}, which exists. In all cases we put

Xs = {x E X

I d(X,8X)

= 181}.

If 181 = 0 then either 8 is an inversion, and Xs = 0, or else Xs is the tree of fixed points of 8; in the latter case we call 8 elliptic. If 181 > 0 we call 8 hyperbolic. In this case Xs is a linear tree on which 8 induces a translation of amplitude 181; we call X. the 8-axis. If Xs -=f. 0 then Xs meets every (8 }-invariant subtree. The group Aut(X) acts on the set Ends(X) of ends of X. Moreover for 8 are precisely the open ends of Xs ([AB] , (6.17», and if 8 fixes an end then 8 is not an inversion ([AB], (6.18». Moreover every end c of X fixed by 8 is an end of Xs; for if c = Ce is a closed end then 8e = e means that 8e = e, so Ce is a closed end of Xs. It may happen however that 8 is elliptic and Xs has closed ends that are not ends of X; for example X = A, 8X = -x, Xs = {O}. If 8 is hyperbolic then the ends of X fixed by 8 are the two ends of the linear tree X •. We define c s (= c s (X» to be the end of X s such that 8 translates toward c s. For t E Aut(X), tXs = Xtst-l. If 8 is hyperbolic then tcs = Ctst-l. 8

E Aut(X), the open ends of X fixed by

(1.6) PROPOSITION. Let X be a A-tree, c an end of X, and b: X

-+

A an

= Aut(X) and G e = {t E G I tc = c}. (a) There i8 a homomorphism Te : G e -+ A, depending only on c, such that, for t E G e , ITe(t)1 = It I and b(tx) = b(x) + Te(t) for all x E X. Moreover G e contains no inversions. end map toward c. Let G

76

H. Bass (b) Let s E G, Se

length

lsi.

i=

nxs

c. Then (s-le,e)

v

is a closed segment [v,sv] of

sv

The function fj 0 s : X -+ A is an end map toward S-le; hence fj 0 s + fj on

(s-le, c) takes a constant value, c(fj,s) = fj(sv)

+ fj(v).

For t E G e we have c(fj,st) = c(fj,s) - Te(t)

and c(fj,ts)

= c(fj,s) + Te(t).

(c) Suppose that s, u E G and e,S-le, Ue are distinct. Put y = Y(S-le, e, ue) (see (1.5)(c»). Then

c( fj, su) = c( fj, s) - c( fj, u) PROOF:

Let s

E

+ 28( u- l y).

G. Then for x EX,

[X,S-lc)

Ii

s

-+ [sx,c) -+

A

is a metric map with least value fj( sx). Hence fj 0 s is an end map toward s-lc. If t E G e then fj 0 t is, like fj, an end map toward c, so fj 0 t = fj + Te(t) for some Te(t) EA. If also t' E Ge then for x E X we have fj(x)

+ Te(tt')

= fj(tt'x) = fj(t'x) = fj(x)

Hence Te : G e

-t

A is a homomorphism.

+ Te(t)

+ Te{t') + Te(t).


77

Let t E G e . Then c: is an end of X t (cf. Remarks above) so t is not

an inversion, and X t = UxEx, [x, c:). If It I = 0 then for x E X t we have 8(x) = 8(tx) = 8(X)+Te(t), so Te(t) = 0 = Itl. If It I > 0 then the linear tree X t is the nested union of the [x, c:) (x EXt) so it follows that 8 : X t ---+ A is a metric map. For x E X t we then have ITe(t)1 = 18(tx) - 8(x)1 = d(tx, x) = Itl. Finally, since 8 is unique up to an additive constant it is clear that Te depends only on c:, not 8; this proves (a). (b) If Y is a "closed" subtree of X not containing c: then there is a unique point v = v(Y,c:) such that [v,c:) n Y = {v} (cf. [AB), (2.24)(e». If s E G then, by transport of structure, sv(Y, c:) = v( sY, sc:). Suppose that sc: i- c:. Put v = v(Xs,s-lc:); then sv = v(Xs,C:). If lsi> 0 then it is easy to see that

v (S-lc:, c:)

sv

n Xs = [v, sv]. If lsi = 0 choose x E [v, s-lc:), x

~

[v, c:). Then (cf.

[AB), (6.1)(b» [x,sx] c [v,s-lc:) U [v,c:) contains a unique (mid) point in X s; this point must be v, so

c:

v

it follows that (s-lc:, c:) = [v, S-lc:) U [v, c:), and the latter meets Xs only at v. We have observed above that 80S is an end map toward s-lc:, hence, by (1.5)(b), 80S + 8 on (s-lc:,c:) takes a constant value c(8,s) = 8(sv) + 8(v). Let t E G e and x E (s-lc:, c:) = (s-lr1c:, c:). Then

+ Te(t) = -8(x) + c(8,s) + Te(t) = -8(x) + c(8,ts)

8(tsx) = 8(sx)

so

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H. Bass

c(c5,ts) = c(c5,s)

+ Te(t).

+ c(c5,s) -c5(y) - Te(t) + c(c5,s) -c5(y) + c(c5,st)

c5(sty) = -c5(ty) = =

so

c(c5,st) = c(c5,s) - Te(t). This proves ( b). (c) Assume that s, u E G and s-l e, e, Ue are distinct. Put y = Y(s-l e, e, Ue) (cf. (1.5)(c)) and z = u- 1y E (U- 1S- 1e,e) n (U- 1e, e). Then

c5(suz) = -c5(z)

+ c(li,su)

II

c5( sy) = -c5(y) + c( c5, s) = -c5( uz) + c( c5, s) = -( -c5(z) + c(c5, u)) + c(c5,s) whence

c(c5,su) = c(c5, s) - c(c5, u)

+ 2c5(z).

This proves (c), and concludes the proof of (1.6). We next investigate normalizers of end stabilizers.

(1. 7) PROPOSITION. Let X be a A-tree and e an end of X. Let r be a group acting on X, re = {s E rise = e}, and Te : re --+ A the homomorphism given by (1.6)(a). (a) If s E r is hyperbolic on X (i.e. lsi> 0) and if sn E re for some n f= 0, then sEre· Let N = Nr(r e), the normalizer of re, and let E denote the set of ends of X fixed by r e. Then E is N -invariant, whence an exact sequence 1 --+ re --+ N --+ Aut(E). Hence if E = {e} we have N = reo (b) If Card(E) 2: 3 then Te(re) = 0, and E is the set of ends of X belonging to the fixed tree X r ., on which N Ir e acts faithfully.


79

(c) Suppose that E = {c, c'}, c .; E'. Then f e; acts on the linear tree (E', E) by translation; t E fe; translates toward E by Te;(t), and Te;,(t) = -Te;(t). Suppose that there is an sEN, s rt. f e;. Then s exchanges c and E', hence induces a reflection on (E', E), and Te;(sts- I ) = -Te;(t) = Te;(t- I ) for t E fe;. We have s2 E fe, Te(s2) = 0, and N = fe;· (s). Suppose furtber tbat f e; acts freely on X. Then Te; : f e; --+ A is injective, so fe; is abelian. Furtber s2 = 1, sts- I = C I for t E fe and N = fe > 0, hence Te is injective and fe is abelian. Since Te;(sts- I ) = -Te(t) = Te;(C I ) we have sts- I = t-I. Since Te(s2) = 0 we have s2 = 1. This proves (c). Finally, if f acts freely on X then Te : f e --+ A is injective, and s E f e whenever sn E fe (n .; 0), by (a). Suppose that fe .; {I}. Each t .; 1 in fe;, being hyperbolic, fixes at most two ends. Hence either E = {E} and N = fe, or E = {c,E'}, [.; E'. In the latter case, if N .; fe, then, as we saw above, N contains an s .; 1 of order 2. Since f acts freely on X such an 5 must be an inversion. Thus N = f e if f e =f. {I} and f acts freely without inversion on X.

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H. Bass

(1.8) PROPOSITION. Let r act freely without inversion on a A-tree X. Let s, t E r - {I}, and let e be an end of X •. Then e is also an end of X, and the following conditions are equivalent (a) s and t commute

(b) tX.

= X.

(c) te = e

(d)X.=Xt (e) s and tst- l commute. PROOF: Since X. is a closed subtree ([AB], Th. (6.6)(b)) and open ends of closed subtrees are also open ends of X ([AB], Prop. (2.24)(d») it follows that e is an end of X. For the same reasons, an inclusion X. c X t would imply that Xs = Xt. Now from the relation

we see that (a) ::::} (b). IftXs = Xs then t must act as a translation on Xs (the action is free and without inversion). It follows then that X. eXt, hence Xs = X t by the remarks above, and further tE = E. Thus (b) implies (c) and (d). Moreover (c) ::::} (a) since r", is abelian «1.7)(d), and (d) ::::} (a) since sts-Ie l is, assuming (d), the identity on X. = X t and the action is free. Thus (a), (b), (c), (d) are equivalent, and evidently (a) ::::} (e). Assuming (e), and applying (a) ::::} (d) to sand tst- l we find that Xs = Xtst-l = tX., whence (e) ::::} (b). This completes the proof.

(1.9) COROLLARY. Let r act freely without inversion on a A-tree X. Let H =1= {I} be an abelian su bgroup of r. (a) H fixes exactly two ends, E', E, of X, and Xs = (E', c) for all s =1= 1 in H. Moreover e and e' are in distinct r -orbits. (b) The abelian group r", (= r e;') is the normalizer (and centralizer) of H in r. More precisely ift E rand tHt- 1 n re; =1= {I} then t Ere:. If two elements of r e; are conjugate in r then they are equal. (c) If s E rand sn E r", for some n =1= 0 then s E r",. PROOF: Choose s =1= 1 in H, and let e',e be the ends of Xs. Let Z = the centralizer of s in r. By (1.8), X t = (e', c) for all t =1= 1 in Z. Since re; is abelian we have H c Z = r", and r '" is the centralizer of H. Moreover e' , e are exactly the ends fixed by H. Suppose that tst- l Ere;. Then t and s commute by (1.8), (c) ::::} (a), so t E r",.


81

(1.10) COROLLARY. Let r act freely without inversion on a A-tree X. Let H #- {I} be a subgroup ofr. The following conditions are equivalent. ( a) H = r e for some end E: of X. (b) H is maximal abelian in r. ( c) H is the centralizer of some s #- 1 in r. (d) H is abelian and self normalizing in r. This follows from (1.9) and (1.7)(d).

(1.11) COROLLARY. Let a countable group r act freely without inversion on a A-tree X. Let H be a subgroup of r. The following conditions are equivalent. (a) H = re for some end c ofr. (b) Either H is maximal abelian in r or H = {I} and r is non-abelian. The only assertion that is not immediate from what precedes is that, if r is non-abelian, then X has an end c such that r e = {1}. Since r acts freely without inversion the action cannot be dihedral. According to Corollary (1.15) below, Ends(X) is uncountable. On the other hand, by (1.10), every maximal abelian subgroup of r is the centralizer of some s #- 1 in r. Hence there are only count ably many maximal abelian subgroups. It follows therefore that:

(1.12)

re = {I} for uncountably many c E Ends(X).

(1.13) REMARK: The count ability assumption in (1.11) is likely unnecessary. (1.14) THEOREM. Let r act minimally on a A-tree X with non-abelian length function (d [AB), (7.3)-(7.6)).

(a) X = Ulsl>oXs. (b) All ends of X are open. (c) Let Ao denote the convex subgroup of A generated by WI = {lsi I s E r}. Then X is a Ao-tree and [O,CXlAo) = UsEr[O, Isll. (d) Assume further that the action is nondihedral (cf. [AB), (7.3) and (7.15»), i.e. that X has> 2 ends. Let x E X and c E Ends(X). Then there is an end c' #- c of X such that x t/:. (c',c). Hence Y(c',x,c) is a branch point of X on [x,c). PROOF: (a) By minimality, it suffices to show that Ulsl>oXs is connected, hence a r-invariant subtree. If Xs n X t = 0 then (see [AB], (8.1)) Istl > 0 and X st meets both Xs and Xt, whence the desired connectivity.

82

H. Bass

(b) This follows from (a) since, for lsi> 0, no point of Xs can be an end point.

(c) For s,t E r we have max(lsl, Itl, Istl, IscII) ? lsi

+ Itl·

°

This is clear if lsi = or It I = 0. Otherwise (cf. [Pa], III), either Istl IsCII > Is I + It I of max(lstl, 1st- I I) = lsi + Itl· Let L = usEdO, lsi]. It follows from (*) that L is an additive monoid, hence L = [0, OOA o ) where Ao is the convex subgroup generated by Iri. Since AI Ao is torsion free we have 2A n Ao = 2Ao. It follows then from [AB], (7.13)(c) that X is a Ao-tree.

(d) Suppose, on the contrary, that x E (e:', e:) for all ends e:' I- e:. Then [x,e:) contains no branch points of X, except perhaps x; since e: is open (by (b» we can advance x toward e: along [x,e:) and so assume that x also is not a branch point. We shall obtain a contradiction by showing that y = X - r· [x, e:) is a r -invariant subtree, so contradicting minimality. Since r-invariance is clear we need only show that Y I- 0 and Y is connected. CLAIM 1: If s E rand X. meets [x,e:) then [x,e:) C Xs. Suppose, on the contrary, that Xs meets [x,e:) and [x,e:) ct. Xs. Then X8 n [x,e:) must be a closed interval (cf. [AB], (2.10)(3) and (6.6)(b»; say Xs n [x,e:) = [u, v] with u E [x, v]. If v I- x then v is an end point of X s (since v is not a branch point) hence fixed by s. But then v is a branch point of [x, v] U [v, e:) U [v, ss); contradiction. This shows that v = x, i.e. X8 n [x,e:) = {x}. Since x is not a branch point x must be an end point of X s , hence sx = x. Then [x,e:) n [x,se:) = {;r} so the linear tree (se:, e:) = [x, e:) U [x, se:) contains no branch points, hence equals X. But then the action must be dihedral, contrary to assumption. This proves Claim 1. CLAIM 2: r· [x,e:)

I- x.

Suppose that r· lx,s) = X. Let s E r. Then Xs meets u[x,e:) = [ux,us) for some u E r, and then X s :J [ux, ue:) by Claim 1. Hence sue: = ue:, so u -1 SU Ere. Thus r is the union of the conjugates of r e. Since the action is nondihedral, i.e. X has> 2 ends, X has a branch point b. From (c) above we see that there is an s E r such that Is I > d( b, x); from the observation above we may assume that se: = e: and, replacing s by s-I, if necessary, that Te(S) > 0, i.e. s translates lx,s) c Xs towarde:. Let [b,b']


83

be the bridge from b to Xs.

r b'

x

sb'

Then clearly b', like b is a branch point of X. Since Is I > d( b, x) ~ d( b', x) and s translates toward E we have sb' E [X,E). But then sb' is a branch point in [x, E). This contradiction gives Claim 2. CLAIM

0.

3: Y

=

X - f . [x, E) is connected.

If not then, by f-invariance, there exist u, v E Y such that [u, v] n [x, E) Say [u, v] n [x, E) = [uo, va] with va E [x, va].

.x

Va

va

i=

E

Since [X,E) contains no branch points we must have va = va = x. But then x itself is a branch point, contrary to assumption. This proves Claim 3. and so completes the proof of (d), and so also of (1.14). (1.15) COROLLARY. Let f act on X with nonabelian and nondihedral length function. Then Ends( X) is uncountable. PROOF: Let Xmin denote the minimal f-invariant subtree (cf. [AB], (7.13)). From Proposition (E1.2) in the Appendix we see that Card(Ends(X)) ~

Card(Ends(Xmin )). Hence we can replace X by Xmin and assume that the action is minimal. Suppose that EI, E2, E3, ... is any sequence of ends of X. Let Xo E X. We shall inductively construct a sequence Xl, 'Y2, x3, ... in X so that for all n

~

1, Xi E [xo, Xn]

ct [xo, Ei)

for i

=

1, ... ,n.

Suppose that Xo, .. . ,Xn-I have been constructed satisfying (*m) for m < n. Embed [XO,Xn-I] in an X-ray [xo,o) from Xo; by (1.14)(b), 0 is an open

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H. Bass

end. If

Xn-l

1-

[XO,En)

we take

Xn

=

Xn-l,

and

(*n)

is clear. Suppose that

Xn-l E [XO, En).

If b 0:;

En

put Y =

Y(En,Xn-l,b)

Xo

Xn-l

y

and choose Xn E [y, b), Xn 0:; y. Again (*n) is clear. Suppose that b = En. By (1.14)(d) there is a branch point y 0:; Xn-l in [Xn-l,E n ), and then y = Y(En,Xn-l,b) for some end b 0:; En. Then [xo, Xn-l] C [xo, y] C [xo, b) so we can choose Xn as in the previous case. Now put L = U n 2::1 [xo, xn]. This is a linear subtree from Xo, hence L C [xo, b) for some end b. Since L rt [XO, En) for all n we have b 0:; En for all n. This proves that Ends(X) cannot be countable, whence (1.15).

§2. Base change for A-trees; ends between fibers. Consider an exact sequence 0-+

p * --to Ao-+A--+A

of ordered abelian groups, where Ao is a nonzero proper convex subgroup of A (a,b E Ao :=;' [a,b] C Ao), p is the natural projection to A* = A/Ao, and A* is given the quotient order: a ~ b in A :=;. p( a) ~ p( b) in A * . Let X be a A-tree. Then we have the quotient A * -tree (cf. [AB], §4)

X* = A* ®A X and the natural projection

p=px:X-+X*,

p(x)=x*,

satisfying

dx·(x*, y*) = p(dx(x, y)).


85

Here px is surjective and its fibers are the "balls of radius Ao:"

p( x) = p( y) d( x, y) E Ao. For x E X, x*

= p(x),

put

X( x*) = p-l (x*) = p-l(p( x))

= {y

E X

I d(x,y)

E Ao},

the fiber through x; X(x*) is a Ao-tree. (2.1) LEMMA. Let X,y E X, x*

# yO.

(a) [x, y] n X(x*) = [x, Ex',y' ) where Ex' ,y' is an open end of full Ao-type in the Ao-tree X(x*), depending only on x*, y*, not x, y. (b) Ifx* f. z* and x* E [y*,z*] then Ex',y' # Ex',z'. If a = d(x,y) then b < a for all b E Ao. Hence L ~ [0, a] n Ao = [O,OOAo). It follows that L = [X,E) where E is an end of full Ao-type in X(x*). Clearly if u ELand v E [x, y] n X(y*) then [u, v] n X(x*) = [u, E). Let x' E X(x*) and y' E X(y*), and let [u,v] = [x,y] n [x',y'] with U E [x,v]. PROOF:

Let L

= [x,y] n X(x*).

Then u = Y(x,y,x') E [x,x'] n [x,y] E X(x*) n [x,y] = L, and similarly v E X(y*). Thus [x', y']nX(x*) = [x', v]nX(x*) = [x', u]U([u, v]nX(x*)) = [x',u] U [U,E) = [X',E). This shows that E depends only on x* and yO, not x and y, so we can denote E by Ex',y'. To prove (b), suppose that x*, yO, and z* = p(z) are distinct, and x* E [y*,z*]. Put v = Y(x,y,z).

86

H. Bass

Then p( v) = Y(x*, y*, z*) = x*, since x* E [y*, z*], so v E X(x*). Thus [x, v] = [x, y] n [x, z] nX(x*) = [x, cx' ,yO ) n [x, Cx' ,z')' It follows, as claimed, that Cx' ,yO # Cx' ,z" We now consider the following "extension problem". Suppose that we are given a A*-tree X*, and, for each x* E X*, a Ao-tree X(x*). Put X = llx' X(x*) and let p: X ----> X* map X(x*) to x*. What is needed to make X into a A-tree inducing the given Ao-tree structure on each X(x*) and so that p is identified with the base change projection X ----> A* ®A X? We shall describe an answer to this when A * '===' I. In this case we can identify A = I x Ao with lexicographic order, and p : A ----> A* = I the first projection. Let X be a I-tree. We define its edges E(X)

=

{(x,y) E X x X

If e = (x,y) E E(X) we put ooe = x, Ole we put

I d(x,y) =

y, and

Ex (= Ex(X)) = {e E E(X)

= 1}.

e=

I ooe =

(y,x). For x E X

x}.

Note that if e, f E Ex and e # f then x E [Ol e, ad]. (This observation is relevant to (2.2)(iii) below, in view of (2.1)(b) above.) (2.2) THEOREM. Let the following be given. (i) A I-tree X*. (ii) For each x* E X*, a Ao -tree X(x*). (iii) For each edge e of X*, an end Ce of X(ooe) of full Ao-type, such that if e # f (and ooe = oof) then Ce # cf· Put X = llx' X(x*). (a) There exists on X a A-metric d (A = I x Ao) satisfying: 1. (X, d) is a A-tree and d induces the given Ao-metrics on each X(x*). 2. If e is an edge of X* and if Xi E X (oie) (i = 0,1) then [xo, xd n X(ooe) = [XO,ce). (b) Choose, for each edge e of X*, an end map De : X(ooe) ----> Ao toward Ce. There is a unique d as in (a) such that, for all e, xo, Xl as in (a)(2), we have d(xo, xd = (1, -De(XO) - De(XI))

(E A = I x Ao).

Group Actions on Non-Archimedean 'frees

87

Moreover every d as in (a) arises tbis way. (c) Let (b~) be a second family of end maps as in (b), and let d' be tbe corresponding A-metric on X. Write b~ = be +c e witb C e E AD (cf (1.4)(b)). Tben d' = d iff C e + C e = 0 for all edges e of X*.

Assertion (c) follows from the formula and uniqueness in (b). Thus it suffices to prove (b), which clearly contains (a). Let be : X (00 e) --t AD be a family of end maps as in (b). Define p: X p(x)

=

--t

X*,

x* for x E X(x*).

Let d* denote the I-metric in X*, and dx. the AD-metric in X(x*) for x* E X*. For x, y E X we now define

d( x, y) E A = I x Ao as follows. {

(2.3)0

If p(x) d(x,y) (2.3)

n

=

= =

If p( x) d(x, y)

x*

oF

y*

= x * = p( y) then = (0, dx. (x, y)). = p(y) then

(n,-~(x,y)),

where

d*(x*,y*), and, if(el .... ,e n )

is the sequence of edges from x* to y* along [x*, y*] in X*, then

~(x, y) = be, (x)

+ L~:/(De; + De ;+,) + Den(Y)·

Here the terms "(De; +De;+,)" signify the following. Put xi = OOei = oOei+l' Since ei oF ei+l hypothesis (iii) implies that the ends Ee; and Ee;+1 of X(xi) are distinct. Hence, by (1.5)(b), De; + De;+1 takes a constant value, which we denote (De; + De;+1)' along (Ee;, Ee;+1)' To show that d is a A-metric first note (clearly) that d( x, y) ~ 0, and d(x, y) > 0 if x oF y. The symmetry, d(x, y) = dey, x), can be easily deduced from (2.3)0 and (2.3). We next show the triangle inequality (~)

d(x, y) :::; d(x, z)

+ d(z, y).

88

H. Bass

Put x* = p(x), y* = p(y), z* = p(z). If x* = y* = z* then (~) follows from the triangle inequality in X(x*). Suppose that x* = y* I=- z*. Then the left side of (~) is in Ao (= 0 x Ao c A), whereas both terms on the right exceed all elements of Ao. Hence we are reduced to the case x* I=- y*. Let

(el,' .. ,en) be the edge path from x* to y* as in (2.3).

We have d*(x*, y*) :::; d*(x*, z*) + d*(z*, y*). If this inequality is strict then, in view of the lexicographic order in A, (~) also holds strictly. Thus we may assume that d*(x*, y*) = d*(x*, z*) + d*(z*, y*), which means that z* E [x*, y*]. CASE 1: z*

=

x*

Say [x,ce,) n [z,c e,) = [v,c e,). We have d(x,z) + d(z,y) - d(x,y) = (O,dx·(x,z)+8e,(x)-8e,(z». vVe further have the formulas dx'(x, z) = dx'(x, v) + dx'(v,z), 8e,(x) = 8e,(v) - dx'(x,v), and 8e ,(z) = 8e,(v) - dx'(v,z), whence

dx'(x,z) CASE

2: z*

=

+ 8e,(x) -

8€,(z)

=

2dx'(v,z)::::: O.

y*.

This follows from Case 1 by symmetry. CASE

3: x* I=- z* I=- y*.

Say z* = 8 0 e r +l for some r (1

X(x*)

0 then on X(x*)tUX(y*)t the common t-translation is by distance Te(t) toward c: in X(x*)t, and by distance Te(t) toward t in X(y*); since these directions on X t are opposite, we must have Te(t) + Te(t) = O. The last assertion of (3.1) is obvious.

(3.2) LEMMA. Put (c:,t)

= n[x,y], (e,e)

X(x*) where (x, y) ranges over X(x*) x X(y*).

X(y*)

(a) If (c:, t) # 0 then (c:, t) is a linear tree on which H acts by translation; t E H translates by distan,ce Te(t) "toward t" in (c:, t). (b) We have (c:,t) = 0 iffd*(x*,y*) is the least element> 0 in A*. In this case we have

94

H. Bass

If t E H then it is clear that (E, E) eXt, and (a) follows easily from this. PROOF:

Clearly (E,t:) = 0 {::} [x*,y*]x. = {x*,y*} {::} d(x*,y*) is the least element > 0 in A *. Suppose that this is the case. To show that H = (f x' )0 we need only show that if sx* = x* and sy* i- y* then SE i- E. Since [x*,y*] = {x*,y*} and [x*,sy*] = {x*,sy*} have only x* in common we have x* E [y*,sy*]. It follows then from (2.1)(b) that E(= Ex"Y') iex'" ,ay·

==

Csx· ,sy·

==

Scx· ,y.

==

Se,

as claimed.

(3.3) PROPOSITION. (a) Let s E f. Then s is an inversion on X iff either s is an inversion on X* or s has a fixed point x* in X* and s is an inversion on X(x*). Hence f acts without inversions on X ifff acts without inversions on X* and, for all x* E X*, f x' acts without inversions on X(x*).

(b) Let x* E X* and x E X(x*). Then fx = (fx')x (clearly). Hence f acts freely on X iff for all x* E X*, f x' acts freely on X(x*). PROOF: (a) Write Sx for the action of s on X, etc. Suppose that Sx IS an inversion, i.e. s3c has a fixed point but Sx doesn't. If sx' is not an inversion, then SX' fixes some x* E X*. Now s3c has a fixed point in X, hence also in the invariant subtree X(x*). It follows that SX(x') is an inversion. Conversely, suppose that SX' is an inversion. Then clearly Sx has no fixed points, so to show that s X is an inversion we must show that s~ has a fixed point. From the classification of automorphisms ([AB], §6) it suffices for this to show that Is Ix = O. Since Isix. = 0 in A * we have Islx E Ao. Suppose lsi X > O. Then the sx-axis Xs is a linear tree on which Sx is a translation of amplitude Islx E Ao. But then sx' fixes all points of (X s)* in X*, contrary to the assumption that s X' has no fixed points. This shows that if s X' is an inversion, so also is s x. Finally suppose that s E f x' is an inversion on X (x*). Then s x has no fixed points in X, since it has none in the invariant subtree X(x*), so Sx is an inversion. This proves ( a).

(b) is obvious. (3.4) REMARK: Let T* C X* contain one point from each f-orbit. Then we clearly have: (a) f acts without inversion on X iff f acts without inversion on X* and, for all x* E T*, f x' acts without inversion on X (x*).

(b) f acts freely on X iff, for all x* E T*, f

x'

acts freely on X(x*).


95

To analyze matters further we shall now specialize to the case A = Z: A = Z x Ao, with lexicographic order. Then X* is a Z-tree, to which we can apply the structure theory in Trees ([Se], (5.4», with the following result.

(3.5) THEOREM. Let r be a group acting on a A-tree X (A = Z x Ao as above) and let X* = Z ®A X. Assume that r acts without inversion on X*, and put y* = r\x*. Then there is a graph of groups (9, Y*) with the following properties. (i) r e:! 7rl (9, Y*, T*) where T* is a maximal subtree of Y*. (ii) For each x* E Y* til ere is a Ao-tree X(x*) on which g(x*) acts. (iii) For each edge e of y* there is an end C e of X ( 00 e) of full Ao - type. These satisfy the following conditions, where a e : g( e) --t g( ooe) denotes the injection given by (9, Y*). (a) For all edges e of Y*, aeg( e) = g( ooe )ee' and (b) Ife =f f are edges ofY* and ooe distinct g( x* )-orbits.

= oaf = x* then

(3.6) REMARK AND PROOF: Any edge

0--,»--0

Ce

and cf lie in

e

:r* x*

= y*)

y*

m Y* (possibly

el

can be lifted to an edge

o---'!>_--o in X* (where xi =f y;).

xi yi From the construction in ([Se l, (5.4» we then have, up to isomorphism,

X(x*) = X(xi), X(y*) = X(yn, C e = c.z;~,y~, C e = Cy~,x~, (cf. (2.1», g(x*) = rx~, g(y*) = ry~, gee) = rei' and a e is the inclusion rei C rx~. The assertions of ( a) thus follow from (3.1) and (3.2). Assertion (b) follows from (2.1) (b) and the fact that Ex' = {e of y* I 00 e = x *} is the quorient of Ex'I by the action of r x'I = g(x*).

(3.7) COROLLARY. Suppose, in (3.5), that r acts freely without inversion onX. (a) For all :r* E Y*, g(x*) acts freely on the Ao-tree X(x*). (b) For all edges e of Y*, Tee 0 a e : g( e) --t Ao is injective, so g( e) is abelian. If s E g( x*) and sn E aeg( e) for some n =f 0 then s E aeg( e). If gee) =f {1} then aeg(e) is its own normalizer in g(x*). This follows from (3.4), (3.6) and (1.9).

96

H. Bass

We now come to our fundamental result, which is a converse to (3.5). This solves the equivariant analogue of the "extension problem" solved by Theorem (2.2).

Let the following data be given. (i) A graph of groups (g, Y*). Put r = 7I"1(g, Y*) (relative to some maximal subtree ofY*). Put r z* = g(x*), identified with a subgroup ofr, for x* E Y*, and re = gee) for e an edge ofY*. Let G: e : re -+ raoe be the given injection. (ii) For each x* E Y*, a Ao-tree X(x*) on which r z. acts. (iii) For each edge e ofY*, an end Ce = c;· of X(ooe) of full Ao-type. Assume that the following conditions are satisfied. (a) lfe is an edge ofY* then G:ere = (raoe)ee and (3.8)

THEOREM.

(b) He =1= f are edges ofY* with One = oof = x* distinct orbits mod r z •. Let X* = X(g, Y*) be the Z-tree on which r acts in [Sel, (5.3). Then there is a A-tree X on which r r-equivariantly identify Z ®A X with X* and so that above arise from X as in (3.5).

then

Ce

and cf lie in

with y* = r\X*, as acts so that we can (g, Y*) and the data

The detailed meaning of the last assertion will be spelled out in the proof, which we now begin. The proof will, in fact, essentially describe all possible A-trees X with r -action as in the theorem. We first recall the constructions from [Sel, (5.3), in a form convenient for our discussion. Let 71" : X* -+ y* = r\x* be the quotient map. Then there are subtrees T* C S* C X* such that 71" :

71" :

T*

E(S*)

-+

-+

Y* is bijective (on vertices), E(Y*) is a bijection of edge sets,

and each edge of S* has at least one end point in T*. We shall adopt the following notational conventions: Identify T* = y* (via 71") as sets (of vertices) Identify E(S*) = E(Y*) (via 71") as sets.


97

For e an edge of X* we shall write of' (e) for its end points in X*. In case e E E(S*) = E(Y*) we write ar(e) E Y* (= T*) for its end points when e is viewed as an edge of Y*. Thus or' (e) is the unique vertex of T* in the same f -orbit as of' (e). Thinking of T* as a maximal subtree of Y*, we have f = 71"1(9, Y*, T*) and X* = X(9, Y*, T*); For each x* E X* there is an element gx' E f so that g;}(x*) E T*; if x* E T* we choose gx. = 1. For e E E(Y*) put ge = gx' where x* = of' (e). Thus

For s E f, of' (se)

= saf' (e),

and so for e E E(Y*), s E f.

(1)

We shall identify fe = gee) with the stabilizer in f of e, viewing e as an edge of S* c X*. With this identification, we can describe

as

Now we can translate condition (a) as follows: If e E ~(Y*), and x*

(2)

{

= ar(e),

then

ge f ege = (f x' )er' and, for t E fe TeY' e

Let x* E Y*. Then f quotient is denoted

x'

(g;1tge)

+ Te Y ' (g;-1tge) = o. •

acts on f x X(x*) by t· (s,x) = (sr 1,tx). The

f xr x * X(x*) = fx.\(f x X(x*)),

98

H. Bass

and we write s· x for the image of (s, x). Clearly f acts on f xr". X(x*) on the left. Put

X(sx*)

= s· X(x*) s

and make this a Ao-tree so that X(x*) = 1 . X(x*) -+ X(sx*) is a Aoisometry. This depends only on sx*, not s, since any other choice is of the form st with t E f x', and, by hypothesis, t is a Ao-isometry of X(x*). We have f xr. X(x*) = X(sx*). sEr /r x'

II

Now define the set

II

x=

f xr x ' X(x*),

x*EY·

with the evident action of f. By construction

x=

II

X(x*)

x·EX·

where each X(x*) is a Ao-tree, and, for s E f,

s : X(x*)

----+

s . X(x*)

= X(sx*)

is a Ao-isometry. Every edge of X* is of the form se with s E f, e E E(Y*). Putting x* = 8r'(e) we have, by (1), 8:'(se) = sgex*, and we define (3)

ct

Here is the end of X(x*) given in (3.8)(iii), so Cse is an end of full Ao-type in X(8t" (se». If t E fe then, by (2) above, g;ltge E (f x' L:r" so -1 Y* Y* .. stgec_yo e = sge(ge tge)c e = sgece . Hence our defimtIOn of Cse depends only on se, and not on the choice of s. Moreover it is clear that u(cse) = Cuse for U E f. From (2) above we further have that, for t' E f€l Tgr* (g;lt'ge)+

Tgr' (g"i 1 t' ge) =

o.

We claim that: (4)

for t E f se.

Here of course f se = sf es-1 is the stabilizer of se. Since Cse = sge(ct) we have Tg,,(t) = Tg;*«sge)-lt(sge» and similarly for To,.. Putting t' = s-lts E fe we see that (4) follows from the equation just above it.


99

We next claim that:

(5)

Ifte and uf (t,u E f,e,f E E(Y*)) are

edges of X* and Cte

P u t s = t - 1 u. Then Ce showing that:

=

t- 1Cte Ce

=

= cuf

then te

t- 1cuf

= csf *

e

=

= uf.

d t0 csf, so we are red uce

= sf·

First we have geor(e) = ot*(e) = o{"(sf) = s9forU). Since distinct vertices of y* lie in distinct f -orbits we must have or (e) = or U); call this vertex x*. Then we have gex* = sg fX*' so v = g-;1 sg f E f x", From the . Y" Y" Y· y., relatIOns Ce = ge(c e ) = csf = sge(c f ) we see that v(c f ) = Ce . Smce or (e) = x* = or (f) it follows from hypothesis (3.8)(b) that if e # f, ct and e j* cannot be in the same f x. -orbit. Thus we conclude that e = f. Put C = cr· = cj*; we have v E (f x.)", with v = g-;1 sge. By (2) above

It follows that s E fe, and so sf = se = e, as claimed. Now for each e E E(Y*) choose an end map toward ct, Y·

be

y.

: X(oo (e))

--+

Ao·

Fix coset representatives S f-+ S for f/fe, using the same sets of representatives for e and c. For each edge se of X* (s E f, e E E(Y*)) define

(6)

Thus b se is an end map toward Cse. Now we have all the data and conditions of Theorem (2.2)(b), which thus furnishes us with a A-metric d on X. We claim that f acts on X by d-isometries. Once shown, then (X, d) clearly furnishes the A-tree with f-action promised by Theorem (3.8). Let u E f. For each e E E(X*) define c5~ = b ue 0 u. This is an end map X(o{· (e)) --+ Ao toward Ce, and hence b~ = c5 e + C e for some constant

100

H. Bass

E Ao. According to (2.4), u : X -+ X is a A-isometry (w.r.t.d) iff = 0 for all e E E(X*). To show this, write e = sj, s E r, j E E(Y*)j we may assume that s = s is its own representative mod r f· Then oe = osf = or 0 (sgf)-l, and similarly oue = ous! = or 0 (usgf)-l. Let x E X(of* (e». Then

Ce

Ce

+ Cl

where us = ust, t E rf where to = g"ttgf E (ra';-U»ei" and Xo = (sgf)-lx E X(06"U)) by (1.6)(a) by (6). Thus Ce

= rei" (gfltgf),

where t = us-lus E r f.

Since we used the same coset representatives modr f = r f for j and follows that we similarly have

J it

Now the desired equation C e + C e = 0 follows from (2) above. This completes the proof of Theorem (3.8). (3.9) REMARKS: 1. Concerning the uniqueness of (X, d) in (3.8) we observe first that X as a set with r-action, and the map X -+ X* with given Aotree fibers, are essentially unique. The possible variation lies in the choice of d. In our construction d was determined by the choice of end maps by* : X (06" (e» -+ Ao towardet for each e E E(Y*). It is the (complete) freedom to choose these that explains all possible variation in d. 2. Let r = 1l"l(g, Y*) as in Theorem (3.8). The data required for a A-tree X with r-actions, as in (3.8) are: (ii) For all x* E Y*, a Ao-tree X(x*) with r x*-actionj and (iii) For all e E E(Y*) an end ee of full Ao-type in X(ooe). We slightly rephrase the conditions these must satisfy. For x* E y* put E x * = {e E E(Y*) I ooe = x*}. Then we must have the following conditions, for all x* E Y*.

ot


101

(ax' ) If e E Ex' then ae(r e) = (r x' )ee' (a~.) If e, e E Ex' then Tee 0 a e + Te. 0 a e = 0 on r e. (b x') If e, f E Ex' and e f= f then Ce and Cf lie in distinct r x·-orbits. (a' ) For all e E E(Y*), Tee 0 a e + Teo 0 a e = 0 on reo Of course (a ' ) subsumes (a~.). The point of this formulation is simply to isolate those conditions (practically all of them) which are local at a given vertex x* E Y*.

(3.10) ILLUSTRATION: We illustrate here what Theorem (3.8) is telling us for amalgams and for HNN -extensions. As above, A = 1 x Ao (lexicographic).

(1) Let

r

~ r,

'H

r,

~" (~o x~

with a fundamental domain

»

0

~'). Then r act., on a l·t,ee X'

H xi' 0

so that r i

= r xi (i = 0, 1) and

e

H = reo To lift this to an action on a A-tree X as in (3.8), we require exactly the following. (i) For i = 0,1, a Ao-tree Xi on which r i ads. (ii) For i = 0,1, an end Ci of full Ao-type in Xi so that H = (ri)e" and so that

(2) Let r

= rO*H = HNN(H

00

~ ro)

= 7rl(rO

H)

(We can identify H with aoCH) to make ao an inclusion.) Then r acts on a I-tree X* with quotient

O.

For a suitable edge

xi'

x~ 0

,..

e

0

of X* we

have ro = r x~, H = r e, gx~ = xi'. To lift this to an action on a A-tree X as in (3.8) we require exactly the following. (i) A Ao-tree Xo on which ro acts. (ii) Ends Eo, El of full Ao-type in X o, such that

(a) H = (rO)eD and al(H) = (ro)e" and

(b) Eo and Cl lie in distinct r o-orbits.

102

H. Bass

§4. Free actions on A-trees. Let A be an ordered abelian group. is A-free if r acts freely without inversion on tree-free if it is A-free for some A.

(4.1) DEFINITION: A group some A-tree X. We call

r

r

(4.2) EXAMPLES: 1. A acts freely on X = A by translation, so A is A-free. Hence all torsion free abelian groups are tree-free. 2. Subgroups of A-free groups are evidently A-free. 3. If A' is a subgroup of A then A' -free groups are A-free. This follows by base change. 4. A group is I-free iff it is free (cf. [Se], Ch. I, Th. 4). (This motivates the terminology.) Hence free groups are A-free for all A (# 0). 5. Corollary (4.4) below shows that free products of A-free groups are A-free. 6. If a A-free group r has two generators then r is either abelian or free. This is due to Harrison [H] for A C fil, and to Steiner [St] in general. Hence a A-free group is either abelian or else contains a nonabelian free group. (Cf. Cor. (4.5) below.) Let a group r act on a A-tree X and let Xo EX. Then we have the "Lyndon length function"

Lxo(s)

= d(xo, sXo) = lsi + 2d(xo,Xs).

(Cf. [AB], §5.) It satisfies Lxo(s-l) = Lxo(s) and

By induction it follows that

These properties will be used in the proof of the next Theorem. mula (4.3.1) below is a generalization of Parry's Lemma 1.2 in [Pal.

(4.3) THEOREM. Let r = ro * rl. (1) Let X be a A-tree on which

For-

r acts and, for i = 0,1, let Xi be a ri-invariant closed subtree such that, for all s # 1 in r i , Xs C Xi. Assume that Xo n Xl = 0, let [xo, xd denote the bridge from Xo to Xl, and put

Group Actions on Non-Archimedean Trees ~

=

d(xo, Xl)' Then for Sl, ... , Sr E

ro -

{I} and t l

, ... ,

tr E

rl

-

103 {I} we

have r

( 4.3.1)

r

=L

ISltl ... srtrl

ISitil

i=l

= L(Lxo(s;) + L xt (t;) + 2~)(> 0). i=l

Moreover r acts faithfully on X. If r i acts freely (without inversions) on Xi (i = 0,1) then r acts freely (without inversions) on X. (2) For i = 0,1, let Xi be a A-tree on which ri acts, and let Xi E Xi. Let ~ E A, ~ > O. Then there is a A-tree X on which r acts, containing Xi as a r i invariant subtree, so that Xo n Xl = 0, [xo, Xl] is the bridge from Xo to Xl, and d(xo, xd = ~. Moreover X = r· (Xo u [xo, xtJ u Xt), and for all S E ri - {I}, Xs C Xi. These properties determine X up to a unique r-equivariant isometry preserving the Xi and Xi. PROOF OF (1): Let

S

Then Xs C Xo and X

E

t

ro -

{I} and t E

rl

{I}.

-

E Xl are disjoint, so, by [AB], Prop. (8.1),

Now d(Xs, Xt) = d(Xs, xo) + d(xo, xd + d(Xl' Xt), Lxo(s) = d(xo, sxo) lsi +2d(xo,Xs ), and similarly Lst(t) = It I +2d(xl,Xt}. Thus

=

This is the case r = 1 of (4.3.1), which we now prove by induction on r. For r > 1 put u = Slt l ... Sr-ltr-l and v = srtr. By induction, lui> 0 and Ivl > O. Hence (cf. [Pa]' III), either

luvl = luv-ll > lui

or

max(luvl, luv-ll)

Hence, to show that luvl to show that

+ Ivl lui + Ivl·

=

= lui + lvi, whence (4.3.1) luv-ll < lui

+ Ivl·

by induction, it suffices

104

H. Bass

Writing

If SI

rv

::I Sr lui

for conjugacy in r, we have

and t r- I

::I tr

+ Ivl- luv-II

then, applying induction, we obtain

= (Lxo(sJ)

+ Lxo(Sr) - Lxo(s;:-I SI»

+ (Lx1(tr-J) + LX1(tr) -

LxJtr_It;:-I»

+ 2~.

As noted above (cf. [AB], (5.3», the two expressions in parentheses are ~ 0, and ~ > 0, whence (*). Suppose that SI = Sr. Then uv- I rv sztz ... Sr_I(tr_It;:-ItI) so if tr_It;:-ItI ::I 1 then lui + Ivl-Iuv-II = Lxo(sJ} + Lxo(Sr) + 2~ + (LXl (tI) + Lx1(tr-J) + Lx1(t r ) - LX1(tr_It;:-Itd), and again this is > 0 since the expression in parentheses is ~ O. If tr_It;:-ItI = 1 then we notice that uv- I rv (s;:-2 I sz)tz ... sr-ztr-z, and we argue similarly. The situation is analogous when t r - I = t r . Every element u of r 0 * r 1 is conjugate to an element of r our 1 or to an element of the form sItI ... srtr as in (4.3.1). In the latter case lui> 0 by (4.3.1), so u acts without fixed points on X. If u E r 0 - {1} then Xu C Xo so u acts nontrivially on X and without fixed points if lui> 0; similarly if u E r l - {I}. Hence r = ro * r l acts faithfully on X, and freely (without inversions) if each r i acts thusly on X;. SKETCH OF PROOF OF (2): Let (O,~) = {a E A that X is forced up to isomorphism as a r -set:

I0 < a
E H) is r-invariant. If so, Sl E H and Xso n X Si = 0 then it follows from (4.3)(1), and applied to r i = (Si) and Xi = Xsp that SO,Sl is a free basis for (SO,Sl), whence the Corollary. Thus we need only derive a contradiction from the assumption that Xs n X t i= 0 for all s, t E H. If Y = nsEHX s i= 0 then Y is a r-invariant linear subtree, so the length function is abelian or dihedral, contrary to assumption. Hence nsEHX s = 0. It then follows from [AB], (2.28) that one of the following occurs. (i) There is a unique end c: common to all Xs (s E H). In this case r must fix c: so, by (1.6), lsi = 17,,(s)1 for s E r, and the length function is abelian, contrary to assumption. (ii) There is a r-invariant cut X = Xo 11 Xl and ends C:i of Xi such that Xs n Xi contains C:i (i = 0,1) for all s E H. Since s E H must preserve {Xo,Xd and s translates along Xs it follows that s must preserve each Xi and hence each C:i. The stabilizer r' of X o in r must have index S 2, and Her' = r~o; hence lsi = 17eo (s)1 for s E r' so the length function is abelian on r', and abelian or dihedral on r, contrary to assumption. This proves (4.5). PROOF:

O}. The family Xs (s

REMARKS: Lyndon [L] conjectured, in a different guise, that every R-free group is a subgroup of a free product of copies of R. Counterexamples to this have been given by Alperin-Moss [AM], Morgan-Shalen [MSI], and Promislow [Prj (see llimlinger [RI]). In fact Morgan-Shalen [MS2] show that if 5 is a compact surface then 71"1 (5) is R-free ... with three exceptions 71"1 (5;) (i = 1,2,3), where 5 i is the connected sum of i real projective


107

planes. They conjecture that every finitely presented R-free group is a free product of free abelian groups and nonexceptional surface groups. Results supporting this conjecture have been obtained by Morgan [M), Jiang [J], and Gillet-Shalen [GS]. We here discuss A-free groups for A nonarchimedean. To bring matters into focus we first record the following observation.

(4.6) PROPOSITION. Let r act freely without inversion on a A-tree X. Let Ao denote the subgroup of A generated by all lsi (s E r). Then r is Ao-free. PROOF: Replacing A by !A and X by !A0 A X (cf. [AB], §4) we can assume that Ao C 2A. Put Al = !Ao C A. By [AB), Th. (7.13)(c), X = A 0Al Xl for some AI-tree Xl, on which r acts freely without inversion. Hence r is AI-free. Since Al = !Ao ~ Ao, r is Ao-free. Suppose that A is nonarchimedean, so that there is an exact sequence 0-+

p * Ao-+A-+A

where Ao i- 0 is a convex subgroup, and A * X we have the base change projection

-+0

i- 0 also.

If r acts on a A-tree

p:X-+X*=A*0AX,

If r acts freely without inversion on X then, for x* E X*, r x* acts freely without inversion on the Ao-tree X(:r*) = p-I(X*). Moreover r is somehow built out of the r x* 's with a structure encoded in the action of r on X*. The only case where this structure is well understood is the case A * = Z, which we treated in §3. So we now take

A= lxAo

(lexicographic order).

We first record the results of Theorems (3.5) and (3.8). For a I-tree X* and x* E X* we write E(X*) for the set of edges of X* and Ex' (= Ex'(X*) = {e E E(X*) I ooe = x*}. (4.7) THEOREM. For A = Z x Ao, the following conditions on a group are equivalent. (a) r is A-free. (b) There exist the following.

r

108

H. Bass

(i) A graph of groups (9, Y*) such that r = 71"1(9, Y*) (relative to some maximal tree in Y*). (We write r x' = g( x*) c r and r e = g( e) for x* E Y*, e E E(Y*).) For all x* E Y*: (ii) A Ao-tree X(x*) on which r x' acts freely without inversion. (iii) For all e E Ex., an end Ce offull Ao-type in X(x*) such that (a) aere = (rx·)e. «(3) Tee 0 a e = - Te • 0 a e : re --+ Ao (,) If e 1= f in Ex. then Ce and Cf lie in distinct r x. -orbits. We now try to make these conditions more purely group theoretic, i.e. without explicit reference to the trees X(x*) and ends Ce. The following terminology will be useful for this purpose. (4.8) DEFINITION: If a group r is tree-free, i.e. A' -free for some ordered abelian group A', then a subgroup H of r will be called an end subgroup if either H is maximal abelian in r or else H = {I} and r is not abelian. In the next theorem we use the same notational conventions as in Theorem (4.6). (4.9) THEOREM. With A = Z x Ao and r a group, consider the following conditions. (a) r is A-free. (b) There is a graph of groups (9, Y*) such that (i) r = 71"1(9, Y*); and, for all x* E y* (ii) r x ' is Ao-free; (iii)(a) For e E Ex., aere is an end subgroup ofr x • (cf. Definition (4.8»); and (iii)(T) Ifel,e2,e3 E Ex· are distinct and reI 1= {I} then aeIre" a e2 r e2' and aear ea are not all conjugate in r x., Then (a) =} (b), and we have (b) =} (a) in the following cases: y* is a tree, Ao C Q, and either Ao = Q or Y* is finite. To show (a) =} (b) consider the data (9, Y*) X (x*), ... of (4. 7)( b). Then (b) (i) and (ii) of (4.9) follow from (b) (i) and (ii) of (4.7). Moreover (iii) (a) of (4.9) follows from (iii)(a) of (4.7) plus Corollary (1.11). To show (iii)(T) of (4.9) put Hi = ae;r e; (i = 1,2,3) and assume that HI 1= {I}. Assuming that H 1 ,H2 ,H3 are conjugate in r x • we shall obtain a contradiction. Say HI = SiHiSi1 for some Si E r x' (i = 2,3). Let Ci = Ce;. Then from (iii)( a) of (4.7), HI is the stabilizer of Cl,S2C2, and S3C3' But since H 1= {I}, H


109

stabilizes only two ends in X(x*) «l.9)(a». Hence two of C1,C2,C3 lie in the same r x. -orbit, and this violates (iii)CI) of (4.7). To prove that (b) ~ (a) we shall establish the conditions of Theorem (4.7)(b) (i), (ii), and (iii)(a), CI). First let r = 7r1(g, yo.) as in our hypothesis (b)(i). We may dismiss the trivial cases when r = {1} or when y* is a single point. Let x· E yo.. By hypothesis (b)(ii) r x. is Ao-free; moreover Ex. -# 0 since y* #- {x*}. We claim that: (ii) There is a Ao-tree X(x*) such that: (a) r x. acts freely without inversion on X(x*);

un

(1) Card({c E Ends(X(xo.)) (Ex. = Ex.(Y*»; and

I (r x.)"

= {1}})

> Card(r x Ex·)

CI) Every end of X(x*) is offull Ao-type. In fact, an X(x*) satisfying (ii)(a) is given to us by hypothesis (b)(ii). To achieve (ii)(,8), take a nonabelian free group r 1 with Card(rd > Card(r x Ex.). By Proposition (4.3) and Corollary (4.4) X(x*) can be rx.-equivariantly embedded in a Ao-tree X on which r x. * r 1 acts freely without inversion. Then the set of ends of hyperbolic axes of elements -# 1 in r 1 already has cardinal:::: Card(r 1) > Card(r x Ex. ), and no such end can be fixed by an s -# 1 in r x. (since s cannot commute with a nontrivial element E r1). Replacing X(x*) by X we now have (a) and (;3) of (ii) above. Now we replace X(x*) by its Ao-fulfillment (Appendix E, (El.7) and (El.S» and we further obtain (ii)CI). Next we claim: (iii) There is a map

Ex.

-+

Ends(X(x*»

such that (a') Putting He = aer e for e E Ex., we have He = (r x· )"e· CI') If e, f E Ex·, e -# f, then Ce and C f lie in distinct r x. -orbits. To prove this first put E~.

= {e E Ex. Ire -# {1}}.

r;' = {c E Ends(X(x*)) I (r x.)" -# {1}}. r;

= Ends(X(x*»

-

r;'.

110

H. Bass

It clearly suffices to separately define E~. --+ ~' satisfying (a') and (,'), and (Ex. - E~.) --+ ~ satisfying (-y'). For the latter it clearly suffices that Card(E x • - E~.) :::::: Card(rx• \~), and this is evidently a consequence of (ii)(;1) above, which says that Card(~) > Card(r x. X Ex.).

To define E~. --+ ~' note first that, for e E E~. , He = a erefixes exactly two ends of X(x*) (cf. (1.9)(a». Choose one of them arbitrarily, call it Ce, and denote the other end c;. Since, by hypothesis (iii)(a) He f:. {I} is an end subgroup of r x• we have, by (1.10), He = (rx.)ee' whence (iii)(a'). If (iii)(,') is also satisfied we are done. Otherwise there exist e f:. f in E~. such that Ce = sCf for some S E r x., Then sHfs-1 = He. We can eliminate this violation of (iii)(,') by replacing cf by cf (keeping Ce fixed), and noting that, by (1.9)(a), Ce and c; lie in distinct rx.-orbits. Now such adjustments can be compatibly made unless we encounter three distinct edges el, e2, e3 in E~. with cel' ce2' cea in the same r x. -orbit. But this would imply that H el , H e2 , Hea are conjugate in r x., in violation of hypothesis (iii)('Y) of (4.9)(b). Thus we have now established (iii)(a') and (") above. Of the conditions required by (4.7)( b) it remains now only to achieve (iii)(;1): (;1') For all e E E(Y*),

where we write Te for Tee 0 a e . Achieving (;1') will require some possible modification of our trees X(x*) and ends Ceo Consider the following condition: 1;1'1

la e(s)IX(8o e) = la e (s)IX(8o e) for all e E E(Y*), sEre.

CLAIM 1: If 1;1'1 is satisfied then there is a new end assignment e 1--+ E~ satisfying (iii)( a') and (") as well as (iii)(;1'): T~ = -T~ : r e --+ Ao, where T! = Te~ 0 a e. (We shall later show that 1;1'1 can be achieved by suitable rescalings of the trees X(x*).)

a»,

Recall that, for s Ere, ITe(s)1 = ITe.( ae(s»1 = lae(s )IX(8o e) (cf. (1.6)( and similarly for Te(S). Assuming 1;1'1 it follows that ITe(s)1 = ITe(s)1 for all sEre' It then follows from [AB], (1.4) that Te = ±Te . Thus 1;1'1 implies (±;1')

Te

=

±Te

for all e.


111

For x* E y* recall that Ex. = {e E E(Y*) I ooe = x*}. Pick a base point x~ E Y*. The new end assignment e I-t c:~ will consist of maps Ex. -+ Ends(X(x*» which we shall define by induction on n = d(x*, xo) and so that the desired condition ((3')

T~ = -T~

holds whenever both ooe and ooe have distance < n from x~, and so that (iii)(a') and b') are preserved. For n = 0 we have x* = x~ and we take c:~ = C: e for e E Ex~. Suppose n > 0 and the new end assignment is defined with the above properties on all E y• with d(y*,x~) < n. Say d(x*,x~) = n. Since y* is a tree there is a unique edge f E Ex. such that y* = oo! has distance n - 1 from x~; hence c:, and have been defined, and by (±P') we must have T/ = ±Tf. If T/ = -Tf then we take c:~ = C: e for all e E Ex., and all conditions are met. Suppose now that Tj f:. -Tf, hence T/ = Tf and r f f:. {I}. Then H f = afrf C r x • fixes exactly two ends of X(x*), C:f and another, denoted c:i. Since Te, = -Tel on Hf, we can set c:i = c:i and so achieve the desired

T/

T/ -Te"

condition, = for f. Now let e E Ex., e f:. f. If He is not conjugate to H f then C: e is not in the r x. -orbits of c: f or c: and we put c:~ = c: e · By hypothesis (4.9)(iii)(-y) there is at most one e E Ex. such that He is conjugate to H f . If e is such an edge, say He = sHfs- 1 (s E r x . ) then we must have €e = SC:f (by (iii)(,'», and we set c:~ = sc:i. It is easily seen now that this definition of c:~ (e E Ex.) preserves (iii)(a') and b'). This completes the inductive proof of Claim l.

i,

To complete the proof of Theorem (4.9) we now have to arrange condition IP' I, on which Claim 1 is based. Let e : y* -+ Q be a map such that, for all x* E Y*, e(x*) > 0 and e(x*)Ao C Ao. Put X'(x*) = Ao ®Ao X(x*), where the base change is c(x·)

relative to the homothety Ao - - Ao. Then for s E

(e)

r x.

we have

Isix/(x.) = e(x*)lslx(x·).

According to Proposition (E2.1) of Appendix E, all ends of X'(x*) remain (like those of X(x*» offull Ao-type, and there is a r x. -equivariant bijection c: I-t c:' from Ends(X(x*» to Ends(X'(x*». Now replacing X(x*) by X'(x*) and C: e by c:~ for e E Ex., we preserve conditions (iii)( a') and b'). Therefore to complete the proof of Theorem (4.9) it suffices to show that we can choose the e(x*) so that

112

H. Bass

1,B/le

lae(s)IXI(8oe) = lae(s)IXI(8oe)

In view of (e) above we can rewrite

1,B/le

V sEre'

1,B/le

in the fonn

e(ooe)lae(s)IX(8oe) = e(ooe)la e (s)IX(8oe)

V sEre'

Define q : E(Y*) -+ Q* as follows. If r e = {1} put q( e) = q( e) = 1. Suppose that re 1= {1}. Then Te and Te are two injective homomorphisms r e -+ Ao. We now invoke our hypothesis that

Ao

c

Q.

From this it follows that Te = ±q(e)Te for a unique q(e) > 0 in Q. This defines q(e), and clearly q(e) = q(e)-l. Let sEre, s 1= 1. Then

lae(s)IX(8oe) = ITe(s)1 = q(e)ITe(s)1 = q(e)la e(s)IX(8oe)' rewrite

1,B/le

I,B" Ie

e( aoe )q( e) = e( 01 e)

Thus we can

in the fonn:

1= {1}. If re = {1} then 1,B/le is automatic for any choice of c(aoe), e(aoe), since both sides equal O. Now assume that y* is a tree. Then,

if re

according to Proposition (4.10) and Corollary (4.11) below, we can find e : y* -+ Q* so that e(x*) > 0 and 1,B"le is satisfied for all e E E(Y*). It remains only to see that we can further arrange that

(A)

e(x*)Ao C Ao

for all x* E Y*. If Ao = Q then (A) is automatic. On the other hand, if y* is finite then we can multiply all of the (finitely many) e(x*) by a common denominator d, and so arrange that all c(x*) are integers. Then again (A) is automatic. This completes the proof of Theorem (4.9).

(4.10) PROPOSITION. Let Y be a connected graph, let G be a group, and let q: E(Y) -+ G be a map such that q(e) = q(e)-l for e E E(Y). The following conditions are equivalent. (a) There is a map e : Y -+ G such that eCole) = e(ooe)q(e) for all e E E(Y). (b) For all closed edge paths (el. e2,"" en) in Y,


113

He' is another map as in (a) then there is a d E G such that c' (x) = de( x) for all x E Y. PROOF: If a = (el, ... , en) is an edge path in Y put q(a) = q(et} ... q(e n ) and a = (en, ... ,et}; thus q(a) = q(a)-l. We also put ooa = oOel and

ala = Olen. Assuming (a), we have

e(8:ta) = e(ooen)q(e n ) = e(olen-dq(e n ) = c(80en-dq(e n -l)q(en )

Thus,

(a)

=}

e(ola) = e(80a)q(a) for all paths a.

If a is a closed path (80a = ala) we conclude that q(a) = 1, whence (a) =} (b). Conversely, assuming (b), we can define e as follows. Pick a base point Xo. For x E Y put e(x) = q(a) where a is any path from Xo to x. If j3 is another such path then (a, jj) is a closed path at Xo so, by (b), 1

= q(a,jj) = q(a)q(jj) = q(a)q(j3)-1

whence q( a) = q(j3), so c is well defined. If, is a path from x to y then (a,,) is a path from Xo to y so

e(y) = q(a)q(!) = c(x)q(!) I.e.

e( 01/) = e( 80, )q(!). When, is a single edge this gives (a). If c' is another such map and a is a path from Xo to x, then e'(x) = e'(xo)q(a) = c'(xo)c(x), whence the uniqueness assertion.

(4.11) COROLLARY. IfY is a tree then a map c as in (4.10)(a) exists for any given q. (4.12) COROLLARY. Suppose that M is a submonoid ofG and q: E(Y)-+ G satisfies (b): q( a) = 1 for all closed paths a in Y. The following conditions are then equivalent.

114

H. Bass

(a) There is a map c : Y

-+

M such that

for all e E E(Y). ( b' ) There is a d E G such that dq( a) E M for all (loop-free) paths a in

Y. Let Xo E Y and define co(x) = q(a) where a is a path from Xo to x. Any other solution of the equations must be of the form c( x) = dco (x) = dq(a) for some dE G. To make c(x) take values in AI it suffices therefore PROOF:

that dq( a) E M for all paths a from Xo. Since q( I) = 1 if I is a loop (= closed path) q( a) is unchanged if we excise any loops from a. Thus (b' ) {:} (a). We shall now give some applications of Theorem (4.9) as well as an extension of it to certain cases where y* is no longer assumed to be a tree. First a word of caution. Suppose that f = 71"1(9, Y*, T*) where (Q, Y*) is a graph of groups and T* a maximal subtree of Y*. Our results assert that f is A-free (A = Z x Ao) iff f admits some such presentation where (9, Y*) satisfies the conditions of (4.7). However this does not exclude the possibility that f = 71"1 (Q, Y*, T*) as above, (Q, Y*) does not satisfy the conditions of (4.7), and yet f is A-free. We shall investigate the case where all edge groups of (9, Y*) are assumed to be abelian, and ask under what conditions f is or isn't A-free for some A. (4.13) PROPOSITION. Let f = 71"1(Q,Y*,T*) where (Q,Y*) is a graph of groups, T* is a maximal subtree, and all edge groups f e are abelian. For e E E(Y*) put He = aef, C faoe.

Assume that f is A-free for some

ordered abelian group A. (1) For all x* E Y*, f x ' is A-free. (2) For all e E E(Y*) either He (in faoe) or He (in faoe) is an end subgroup, i.e. either trivial or maximal abelian, and isomorphic to a subgroup of A.

(3) Suppose that e is a loop, say ooe = ooe = x*. If for some u E f x', uHeu- l n He 1= {1}, then ua e (8)u- l = a e(8) for all 8 E fe' In particular, if So, 81 E f e and ae(so) and ae(sd are conjugate in f x' then 80 = 81· Assertion (1) is immediate. For (2) and (3) there is no loss in replacing y* by the subgraph consisting of e, e and their end points. MorePROOF:

over (2) and (3) are trivial if fe

=

{1}, so assume that fe

1= {l}.

Group Actions on Non-Archimedean Trees CASE 1: y*

=

0

e

?

o

1. Then f 0

= fo *H fl

with H abelian,

115

i= {I}.

If H is not maximal abelian in either fo or fl then the centralizer Zr(H) is not abelian, so (Rplies that f can't be A-free.

CASE 2: y* = 0

U f

e. Then f is an HNN extension,

= (fo,t I tsc 1 = s

for all s E H),

where H = O:efe is an abelian subgroup i= {I} in f o, and s f----+ S is the isomorphism 0:.00:;1 : H -+ H = o:.f e C fo. In f we have the subgroup (fo, tfot-l) ~ fo *F! tfot-l. If f is A-free so also is this subgroup, and so, by Case 1, either H is maximal abelian in f o, or H (= tHt- l ) is maximal abelian in tf oC l , i.e. H is maximal abelian in fo, whence (2). To prove (3), suppose that U E fo and uHu- l nH i= {I}. Putting v = t- l u we then have H n vHv- l i= {1}. But then it follows from (1.9) that v centralizes H, so U8U- l = t8t- l = s for all 8 E H. If 80,81 E H - {I}, U E fo, and U80U-l = SI, then uHu- l n H i= {I}, so usou-1 = SO, by what we just proved, so 80 = 81. (4.14) EXAMPLES: Let f o, fl be free groups, Si E f

i , 8i

i= 1, and

Consider the conditions (for i = 0,1): (ai) Si is not a proper power in rio 1. Assuming (ao) and (ad, f is (I x I)-free. This is a special case of Theorem (4.9), (b) ~ (a), with Ao = I and y* = 0 - - - 0 . 2. If f is A-free for some A then either (ao) or (al) holds. This follows from (4.13). 3. Say fi has basis Xi,Yi (i = 0,1) and we have So = Xo and 81 = x~. Then f = (xO,Yo,Xl,Yl I Xo = x~) = (YO,Xl,Yl 1-), a (I)-free group, even though (ad fails. The next result deals with an HNN-extension f = 7rl(fo (fo,t I tst- l = s for all 8 E H). Here H = O:efe and isomorphism 0:.00:;1 : H -+ H = o:.f e •

Q 8

(4.15) PROPOSITION. With f as above, assume that: (a) fo acts freely without inversion on a Ao-tree Xo so that for all S E H; ( b) H and H are maximal abelian in f 0; and

S

~e) = IS

the

Islxo = Islxo

116

H. Bass

(c) For some s i= 1 in H, s is not conjugate in ro to s-l. Then r is A-free for A = Z x Ao (lexicographic). PROOF: According to (1.9), H fixes exactly two ends, c, c-, of X o, and = (ro)" = (ro),,-. Similarly H = (ro)e = (roh-, where e, e- are the two ends fixed by H. Put Ce = c. For s E H we have ITe(s)1 = Islxo = Islx o = ITe(s)l, and so, by [AB], (1.4), there is a W = ±1 so that Tl(S) = WT,,(S) for all s E H. Choose Ce = e if W = -1, and Ce = e if W = 1. Then we have T".(S) = -Te(S) for all s E H. Finally we claim that Ce and Ce lie in distinct ro-orbits. For suppose, on the contrary, that Ce = UCe for some u E roo Then we have H = (ro),,_ = u(r o)".u- 1 = uHu- 1 and, forsEH,

(b) implies that H

Hence usu- 1 = s-l for all s E H, contradicting (c). Now we have established the conditions of (4.7)(b), from which it now follows that r is A-free.

(4.16) COROLLARY. Let ro be a Ao-free group, H a maximal abelian subgroup, and r

= (ro, tits =

st for all s E H),

the corresponding "benign" HNN -extension. Then r is A-free for A = Z x Ao (lexicographic). PROOF: This is the special case of (4.15) when s = s for all s E H. Conditions (a) and (b) of (4.15) follow directly from our hypotheses, and condition (c) follows from (1.9).

(4.17) EXAMPLES: Let ro be a free group and SO,Sl elements i= 1 in roo Consider the HNN-extension r = (r o, t I tsor1 = Sl), and consider the following conditions. (ai) Si is not a proper power in roo (b) ro acts freely on some Ao-tree Xo (for some Ao) so that Isolxo =

IS 11xo •

(c) Sl is not conjugate to sol in roo 1. If r is A-free for some A then (b), (c) and either (ao) or (ad hold, by (4.13). 2. If (ao), (a1), (b), and (c) hold then r is (Z x Ao)-free, by (4.15). 3. Suppose that ro has basis x, y, So = x, and Sl = y2. Then r = (x, y, t I txt- 1 = y2) = (y, t I -) is (Z)-free, yet condition (a1) fails.


117

4. If So = Sl and (ao) holds then (ad, (b), (c) are automatic, so f is (Z x Z)-free. (Cf. (4.16).)

(4.18) EXAMPLE (W. Parry): Condition (b) in (4.17) raises the following question. If s, t are elements -# 1 in a A-free group f when can we find a free action without inversions of f on a A-tree so that lsi = It I (where I I = I Ix)7 Walter Parry (private communication) pointed out that this is not always possible. More precisely: CLAIM: If S,u E f and v

= s-lu-lsu -# Isv-ll

=

lsi

1 then

+ Ivl >

lsi-

In fact it follows from Parry's Axiom III [Pa) (cf. [AB], (8.1) and (8.3) for general A), that either

or max(lsvl, Isv-ll) Since sv

= u-lsu

we have Isvl

= lsi
_mlnr,r,r c: ) ,

which follows from RTE2. Now according to [AB], §3 there is a rooted A-tree T = T(U, /\) with base point 0 and a map a : U -> T such that u /\ v = a(u) 1\0 a(v) for u,v E U. For u = (c:,r) E U put c:(r) = a(u) E T. Then we have c:( r) = c:' (r') iff r = r' :::; c: 1\ c:'. Moreover r f---+ c:( r) defines an isometry [O,OOA) -> Twithimage [O.C:A) in T. Fore E Ends(X), x E [xo,c:), and d(xo,x) = r, put 'P(x) = Err). Suppose also that x E [xo,s'). Then r :::; E 1\ s' so s(r) = E'(r). Thus we have a well defined map T, and

bo(B, C) =} bo(B, C) = boCA, C), where boCA, B) = 1/2( d(O, A)+ d(O,B) - d(A,B». A long diagram of actions D consists of a surjective homomorphism of groups 1> : H -; G and a commutative diagram of surjective maps of Rtrees:

5=

lo

1

To

5=

iI1

T1

+-

+-

5=

121

T2

5 = ...

h1

+-

T3

+- ...

91 92 93 94 where f n is 1>-equivariant and 9n is both G-equivariant and distance nonincreasing. For sand s' in 5, define dv(s,s') = sup{dnUn(s),fn(s'» In E N}. It is easy to check that the function d v defines a pseudometric on 5 satisfying the tree condition of Alp erin-Bass. Denote by (T( D), d v ) the metric space obtained from (5, d v ) by identifying points of distance zero. Thus we have the following proposition.

The metric space (T(D), d v ) is an R-tree with a G-action and fits into a long diagram: PROPOSITION.

T(D)

1

To

= T(D)

1

+-

T1

gl

= T(D)

1

T2

+-

= T(D) = ...

1

+-

g2

g3

T3

+- ...

g4

We will now construct a long diagram £ of actions. Let Ok be a sequence of positive numbers with sum one. Let 1> : P * Q -; G(P, Q) be the natural map and let the R-tree Tn be indicated by the Bass-Serre diagram: G(Pn.Qnl

P n - 1 EBQn-2

o

0

Pn-lEBQn-l

P n - 2 EBQn-3

P n -24Qn-2

0


139

where the kth edge has length Ci'k (the kth edge is the one with label PkffiQk). Let gn : Tn -+ T n - 1 be the map obtained by collapsing all lifts of edge n - 1 in Tn. Finally, let in be the composition of maps of A-trees indicated by the following diagram and denote by S the P * Q-tree which is the domain of in. P {e} Q o------------------rle-n~gt~h-,l----------------~O

1

Subdivide

~

{e}

{e}

Q

{e}

~r-Ie~n~gtUh~a~k+7-a-k+-l~+~... ~o~--~le~n~g~th~a-o+7Q~1~+~.-..~+-a-k-_-'----'O

1

Collapse Edge

P

Q

{e}

O----l~e~ng~t~h~a-o-+~a~,~+-..-.+~a-k---,----O

1

Folds

P*Qn

o

Pn-l*Qn-2

Pn-2*Qn-3

0

0

Pn-1*Qn-l

P n -2*Qn-2

1

G('Pn.Qn)

P n -l(J)Qn-2

P n -2iBQn-3

P,$Qo

o

0

0

o

Pn-1EfJQn-l

P n -2(J)Qn-2

Po$Q Po$Qo

0

THEOREM 2. The A-tree T(£) has a G(P, Q)-actioll. Allllolldegenerate arc stabilizers are small. The tree T( E) has illfillitely mall,r G(P, Q )-equivalellce classes of vertices. Further, all directioll-to-vertex mOllomorpllisms are proper. PROOF: The first statement follows from the last proposition. The second follows because any nondegenerate arc of T( E) has nondegenerate image in Tn for large enough n and all edge stabilizers of Tn are abelian. The third follows because the subtree of Tn indicated by: P n - 1 (IJQn-2

o

P n -2(IJQn-3 0

embeds isometrically in T m for m 2: n and so embeds in T( £) as well. Let V be the set of vertices of S with stabilizer conjugate to P. Let 4> : S -+ T( E) denote the quotient map. The final statement follows from the facts that a point x E T(E)\¢(V), has neighborhoods in T(E) which embed in Tn for

140

M. Bestvina and M. Feighn

large n and if x E (V) then a conjugate of t E P (notice that P embeds in G(P, stabilizes x but doesn't stabilize any nondegenerate arc containing x (again choose n large enough so that the arc is nondegenerate in Tn and notice that t doesn't stabilize the image of the arc in Tn). 0

Q»

Let SA(G) denote the space of actions of G on R-trees with small arc stabilizers topologized by length functions. By [eM], SA( G) is compact. An action of G on an R-tree T is said to be simplicial if the set of vertices of T is discrete in T. Let G('R, S) be as in section 2. The next corollary uses an idea of Michael Steiner [Ste]. (to Theorems 1 and 2). The subspace of simplicial actions in SA( G('R, S» is infinite dimensional. Furthermore, the Hilbert cube embeds COROLLARY

in SA(G('R,S». By choosing arbitrary positive numbers for the lengths of the nth graph of groups diagram for G('R,S) in Theorem 1, we embed the open (n - 1 )-simplex into the space of actions. The second statement follows similarly. D PROOF:

QUESTION:

Is the previous corollary true for any finitely presented group?

Is there an example of a finitely-presented group G and a real G-tree with small edge stabilizers and infinitely many G-equivalence classes of vertices with proper direction-to-vertex stabilizers? QUESTION:


141

REFERENCES

[AB] R. Alperin and H. Bass, Length functions of group actions on A-trees, in "Combinatorial Group Theory and Topology", ed. S. M. Gersten and J. R. Stallings, Ann. Math. Stud. 111 (1987), 265-378. [BF] M. Bestvina and M. Feighn, Bounding the complexity of simplicial group actions on trees, preprint. [eM] M. Culler and J.W. Morgan, Group actions on R-trees, Proc. London Math. Soc. (3) 55 (1987), 571-604. [D] M.J. Dunwoody, The accessibility of finitely presented groups, Inv. Math. 81 (1985),449-457. [SW] G.P. Scott and C.T.C. Wall, Topological methods in group theory, in "Homological Group Theory" , ed. C.T.C. Wall, London Math. Soc. Lecture Notes 36 (1979), 137-203. [Se] J.P. Serre, "Trees," Springer-Verlag, 1980. [Sta] J .R. Stallings, Topology of finite graphs, Inv. Math. 71 (1983), 551-565. [Ste] M. Steiner, Gluing data and group actions on A-trees, Thesis, Columbia University (1988). [W] C. T .C. Wall, Pairs of relative cohomological dimension one, J. Pure AppJ. Algebra 1 (1971), 141-154.

M. Bestvina, Department of Mathematics, University of California, Los Angeles, Los Angeles, CA 90024 and Institute for Advanced Study M. Feighn, Department of Mathematics, Rutgers University, Newark, NJ 07102 M. Bestvina's research supported in part by NSF and an lAS fellowship. M. Feighn's research supported in part by a Rutgers University fellowship. Research at MSRI supported in part by NSF Grant DMS-8505550.

Geodesic currents on negatively curved groups FRANCIS BONAHON

University of Southern California Abstract. A negatively curved or hyperbolic group, as introduced by M. Gromov, is a finitely generated group whose Cayley graphs asymptotically behave at infinity like a tree. Considering the action of a negatively curved group on one of its Cayley graphs, we study asymptotic directions in the set of conjugacy classes of this group. We then discuss some applications to group actions on R- trees and manifolds of negative curvature.

§o. Introduction. In topology, one often deals with the set C of conjugacy classes of a finitely generated group r, usually occurring as the set of free homotopy classes in some space. There are many situations where the topologist would like to use limiting processes, and to define asymptotic directions in this space C. For instance, if we want to study a foliation of a manifold M, a coarse invariant of a leaf L is the image of the map 'iT] (L) --> 'iT] 0'.1). Since the leaf may very well be simply connected, this does not give much information on how the leaf wraps around the manifold M. One way to improve this data is to consider the set Co of those conjugacy classes of r = 'iT] (M) which can be represented by closed curves made up of one arc in L and of one arc of length less than c. If we let c tend to 0, Co in general goes to infinity in C, and the corresponding asymptotic directions give much more information on the homotopic behavior of the leaf L. A similar example occurs when the group r acts on an R- tree T. Then, the asymptotic directions corresponding to elements of C whose translation distance tends to 0 give a lot of information on this action. For instance, when T is the R- tree dual to a complete measured geodesic lamination 0' on a surface S and when r = 'iT1 (S), we will see in §4 that these asymptotic directions are associated to all measured geodesic laminations which have the same support as 0'; thus, we can recover the support of 0' from these asymptotic directions. To define asymptotic directions, it is necessary to have a notion of straight line in C. An obvious candidate is offered by the sequences n ~ "In associating to n E N the conjugacy class "In E C of the n- th power of

144

F. Bonahon

any element of r representing l' E C, at least when l' has infinite order. However, such power sequences are relatively scarce. We can extend them at least formally by introducing the "tensor product" C Q9 R+ , defined as the quotient of C x R+ = C x [O,oo[ by the equivalence relation which identifies (1'n, A) to (1', An) for every l' E C, n EN, A E R+. For obvious reasons, let 1'A E C Q9 R+ denote the equivalence class of (1', A). Observe that 1'A = 1 in C Q9 R+ if and only if A = 0 or l' E C is represented by some torsion element of r. Our goal is to construct a completion C (r) of this space C Q9 R+, so that the multiplicative action of R+ on C Q9 R+ extends to C (r) (thus defining rays through the identity 1) and so that the "projective space" peer) = (C(r) - {l})/R+ is compact. Then, given a sequence Crn)nEN going to infinity in C C C Q9 R+ C C (r), the limit points of the images of the 1'n in PC (r) are asymptotic directions to the 1'n in C (r), and describe how this sequence goes to infinity in C. Such a completion C (r) of C Q9 R+ can be constructed when the group is negatively curved, or hyperbolic, in the sense of Gromov [Gro]. For the sake of this introduction, we will say that r is negatively curved if it admits a co compact properly discontinuous action on a space X whose behavior at infinity is similar to that of an infinite tree. This includes in particular the case where X is a simplicial tree, in which case r is the fundamental group of a graph of finite groups (see [Serl)o Others examples of negatively curved groups are fundamental groups of compact negatively curved Riemannian manifolds with convex boundary, or small cancellation groups [LySe]. When r is a negatively curved group, it is possible to embed C Q9 R+ in a certain space C (r) of measures, which has the properties required and where C Q9 R+ is dense. We can illustrate what these asymptotic directions are in the case of a free group r. In this case, every conjugacy class l' E C is represented by a unique reduced cyclic word in the generators of r , and therefore defines a periodic reduced bi-infinite word in these generators. Then, the two sequences Crn)nEN and Cr~)nEN in C are asymptotic if, for every finite word, this word appears with approximately the same frequency in the bi-infinite words respectively associated to 1'n and 1'~, for n sufficiently large. The completion C (r) was first introduced in [Bonl][Bon2] in the case where r was the fundamental group of a compact surface of negative Euler

Geodesic Currents on Negatively Curved Groups

145

characteristic, and was inspired by Thurston's theory of measured geodesic laminations [Thu]. There, the elements of C (r) occurred as geometric currents, in the sense of [SuI], carried by the geodesic flow of the surface; for this reason, they were called geodesic currents. The main purpose of the present paper is to emphasize the group theoretic, as opposed to geometric, nature of these geodesic currents. In the last section §4, we also discuss some applications of these geodesic currents to the study of actions of negatively curved groups on negatively curved spaces. The article is mostly expository. The only non-trivial result which is proved here is the property that C 0 R+ is dense in the space C (r) (Theorem 7). In the proof of this result, we follow the rough lines of the argument of [Bon3, pp. 134-137], which was itself closely related to [Sigl][Sig2] (although we were not aware of this at the time). The passage from negatively curved manifolds to negatively curved groups however involves some additional technical difficulties, indicated in the remark after Lemma 1. We rely heavily on some technical results (as well as on the general philosophy) of [Gro], but we have tried to stay at an elementary level so that only a superficial reading of (or faith in) [Gro] is required. We would like to thank Thomas Delzant and the referee for useful comments on the manuscript.

§1. Negatively curved spaces. We begin by briefly reviewing some facts about negatively curved spaces, referring the reader to [Gro] for more details. Other useful references include the expositions of Gromov's work in [Bow][CoDP][GhH][MSRI]. Consider a metric space X, with metric d. A geodesic arc is a map a from an interval [a, b] of R to X which is isometric, namely such that d(a(t),a(u)) = It - ul for every t, u E [a,b]. The space X is said to be geodesic if any two points in X can be joined by such a geodesic arc. For instance, a connected complete Riemannian manifold, a connected locally finite graph (where each edge has length 1) or an R-tree are geodesic spaces. Given a number {j :::: 0, Gromov defines a metric space X to be {j- hyperbolic, or {j- negatively curved, if it is complete and geodesic and if the following property holds: For every triangle in X made up of three geodesic arcs, each point of an edge of the triangle is at distance at most

.146

F. Bonahon

8 from the union of the two other edges. The space X is (uniformly) negatively curved, or hyperbolic, or if it is 8- negatively curved for some 8. (This definition slightly differs from Gromov's, because he does not require X to be geodesic. Also, a space which is 8 - hyperbolic in our sense is probably only 88- hyperbolic in Gromov's sensei compare [Gro, §6].) For instance, an R- tree is 0- negatively curved. A simply connected complete Riemannian manifold whose curvature is negative bounded away from 0 is 8- negatively curved for some 8 depending on the bound on the curvature. A 8-negatively curved space X has a well-defined boundary aX, which can be defined as follows. Given a positive function fl on X, define the fl-length of a path Q to be the integral fa 11, and let the /-l- distance between two points x, y be the infimum d" (x, y) of the /-l-lengths of the paths joining x to y. In other words, the metric dl' is obtained by "conformally" multiplying the metric d by the function /-l. In particular, we can consider the function /-lp (x) = e-pd(r,x o ) where p > 0 and Xo E X. Gromov shows that, provided that p is less than a certain constant depending only on 8, the metric dl'p is independent of p and Xo up to Holder equivalence (see [Gro, §7.2]). The completion of X for the metric d"p , where p is chosen less than the above mentioned constant, is therefore a topological space canonically associated to X. The boundary aX is defined to be the space of limit points of this completion X u aX. For instance, when X is a simply connected complete Riemannian manifold without boundary and whose curvature is negative and bounded away from 0, the boundary aX is exactly the sphere at infinity of X in the sense of [EbON]. When X is a locally finite simplicial tree, aX is the space of topological ends of X. When X is an R- tree, aX is the space of geodesic rays Q : [0, oo[ ......., X originating from an arbitrary base point .TO E X (a situation reminiscent of the sphere at infinity of a negatively curved manifold). An important ingredient for the study of negatively curved spaces is the notion of quasi-geodesic arc. Given k ~ 1 and L ~ 0, a path Q : [a, b] ......., X is (k, L) - quasi-geodesic if, for every a', b' E [a, b] for which the length of Q I [a ' ,b ' ] is at most L, this length is actually at most kd (Q ( a' ) , Q (b')) + k . Recall that the length of the arc Q : [a, b] ......., X is defined as the supremum of the sums L~:Ol d(Q (t;) ,Q (t i + J )) for all subdivisions a = to, t J , .•. , t n = b of the interval [a, b].


147

The fundamental property of quasi-geodesic arcs is that they are close to geodesic arcs: LEMMA 1. For every b :2: 0 and every k :2: 1, there are constants L :2: 0 and f{ :2: 0 such that every (k, L) - quasi-geodesic arc 0 and given k :2: 1 and L :2: 0 as in Lemma 1 there is a (large) constant A. :2: 0 with tile following property: Consider two (k, L) - quasi-geodesic rays 0 such that, for every bi-infinite (k, L) - quasi-geodesics

a

:

]-CXl,+CXl[ -+ X and (3: ]-CXl,+CXl[ -+ X with d pp (a (-CXl),(3(-OO)) < T/ and d pp (a ( +CXl) ,(3 (+00)) < T/, then the part of a which is contained in the ball of radius A around Xo stays at distance at most D from (3. Again, we can restrict attention to geodesics a, (3 : ]-CXl, +CXl[ -+ X. If d pp (a ( +CXl) ,(3 (+CXl)) < T/ then, by definition of d pp , a d pp - geodesic arc joining a(t) to (3(t) must stay at d-distance at least E from Xo for t sufficient close to +00, for some constant E which is large if T/ is small enough. Since, by [Gro, Lemma 7.2.L]' every d pp - geodesic arc stays at uniformly bounded distance from a d- geodesic arc, the same property holds for any d- geodesic arc joining a (t) to (3 (t), as well as for any d- geodesic arc joining a ( u) to (3 ( u) for u sufficient close to -CXl. PROOF:

Consider the two geodesic triangles formed by five geodesic arcs joining (3 ( u) to a ( u ), a ( u) to a (t), a ( t) to (3 ( u ), (3 (u) to (3 (t), and (3 (t) to a (t), respectively. Applying to these triangles the fundamental property of geodesic triangles in (the definition) of 8- negatively curved groups, we

Geodesic Currents on Negatively Curved Groups find that €X stays at distance at most 26 from centered at Xo.

149

f3 in the ball of radius E - 26 0

§2. Geodesic currents on negatively curved groups. Consider a finitely generated group r. To each set of generators of r, we associate the Cayley graph X (r) whose vertices are the elements of r and where two such vertices " " E r are joined by an edge if and only if or is one of these generators. Assigning length 1 to all of its edges, we can turn X (r) into a metric space, which is geodesic since the graph X (r) is locally finite. By definition, the group r is negatively curved if the Cayley graph X (r) is negatively curved, for some set of generators. The set of generators we choose for r actually does not matter in this definition. Indeed, let r act (isometrically) on a geodesic metric space X in such a way that the orbit of a point is discrete and that the orbit of some ball covers X. (This holds for instance if the action is properly discontinuous and cocompact; in particular, this is true for the action of r on any Cayley graph X (r) defined by right multiplication). Then, it can been shown that X is negatively curved if and only if r is negatively curved. The idea is to r-equivariantlyembed the Cayley graph X (r) in X, and to observe that the metric of X (r) is Lipschitz equivalent to the one induced by the metric of X; compare [Gro, §8.2]. Also, if r acts on X as above and if we identify r to one of its orbits, the boundary oX is also the set of limit points of the completion of r for the metric d,.p defined in §l. Changing the space X changes the metric d,.p on r only by a Holder equivalence. It follows that the space oX depends only on the group r. It is called the boundary of the group r and denoted by or . The space r u or is compact and has finite topological dimension; see [Gro, §3.1]. Observe that the action of r on X extends to an action on oX = or.

,,'-1 ,',-1

Consider, E C, represented by an infinite order element 9 E r. Then, 9 fixes two distinct points 9- 00 and 9+ 00 of or, namely the limits of 9n in r u or as n tends to -00 or +00, respectively (see [Gro, §8.1], and compare the proof of Lemma 4 below). Let a (9) denote the point (9- 00 ,9+ 00 ) in the space 2 r of pairs of distinct points in or (namely, o2r is the complement of the diagonal in or x or).

a

4. The a (9), where 9 ranges over all elements of r representing the conjugacy class, E r, form a discrete subset A (,) of o2r .

LEMMA

150

F. Bonahon

PROOF: First, we give a more geometric description of A (1'). Consider a free properly discontinuous cocompact action of r on a 8 - negatively curved group X; for instance, we can consider the action of r on one of its Cayley graphs. A closed loop in X Ir determines a conjugacy class l' E C. For l' E C, there exists by compactness of X Ir a closed loop 1(1') in X Ir which represents l' and has minimum length for this property. Observe that the lifts of 1C1') to X are (1, L) - quasi-geodesic for any L which is less than the length of 1(1') .

Let L 2: 0 and J( ~ 0 be associated by Lemma 1 to k = 1 and 8. Gromov proves in [Gro, §8.1] that, if l' E C has infinite order, there is an n such that the length of 1hn) is larger than L. Then, 1hn) lifts to a r - invariant family of (1, L) - quasi-geodesics in X. By Lemma 1, each of these quasi-geodesics stays at bounded distance from a geodesic, and consequently has two limit points in ax defining an element of a2 r. Such limit points associated to all the lifts of 1hn) form a r - invariant subset of 0 2 r which is clearly equal to A ( 1') . Assume that A ( 1') is not discrete. Then, by Lemma 3, there are for every A 2: 0 two lifts gl and g2 of 1(I'n) which stay at distance less than a fixed constant D from each other over a distance of at least A, and for which (gl (-00) ,gl (+00)) #(g2 (-00), g2 (+00). By construction, gl and g2 are respectively invariant under elements 9f and 9!f E r representing I'n E C and acting by translation of the length of 1C1'n) on these lifts. Since (gl (-00), gl (+00» # (g2 (-00), g2 (+00», there is no positive p, q such that I'~p = I';q. We conclude that there is a large number, depending on A, of elements of the orbit of a base point in X which are at bounded distance, independent of A, from the base point. For A large enough, this contradicts the fact that the action of r on X is properly discontinuous.

o

Essentially the same proof gives the following result. LEMMA 5. The two subsets A (1'1) and A (1'2) are equal if and only if there exists two positive integers p and q such that I'f = I'i.

o Now, let 1'>' E C 0 R+ . The proof of Lemma 4 shows that l' E C can be written as l' = 1'[; where n is a positive integer and where 1'1 is primitive, namely does not admit a non-trivial decomposition of this type. Then, by

Geodesic Currents on Negatively Curved Groups Lemmas 4 and 5, we can identify "'{A to the discrete A ("'(0) of 2 r , endowed with the weight An.

a

r-

151

invariant subset

To analyze a discrete subset endowed with a real weight, it is natural to use a probabilist approach and to consider the Dirac measure it defines. Therefore, consider the space C (r) of those positive Radon measures on a2 r which are invariant under r. (Recall that a measure is Radon if, for every compact subset K, it induces a continuous linear form on the space of continuous functions with support contained in K, endowed with the topology of uniform convergence; a typical example is the Dirac measure of weight A 2 0 defined by a discrete subset A, which associates the mass A EaEA ep (a) to any function ep with compact support). The elements of C (f) are, by definition, geodesic currents of the group r. The terminology is motivated by the case where r is the fundamental group of a compact Riemannian manifold of negative curvature, in which case the elements of C (r) can be interpreted as geometric currents, in the sense of [SuI], supported on the geodesic flow of the manifold; see [BonI]. We should mention that the geodesic currents considered here are slightly different from those discussed in [BonI] [Bon2] in the sense that they carry a certain orientation. However, the geodesic currents in the sense of [Bon I] can readily be identified with those elements of C (r) which are invariant under the canonical involution T : C (r) --+ C (r) induced by the homeomorphism of a2 r c ar x ar which exchanges the two factors of ar x af. (Observe that T (-y) = ",{-I for every "'{ E C C C (r». If we associate to "'{A, with "'{ primitive, the Dirac measure of weight A associated to A ("'{), we now have an embedding of the set C 0 R+ into the space C (r) of geodesic currents of r.

The space C (f) is endowed with the weak* uniform structure, defined by the family of semi-norms dcp(a.,(3) = Ia.(ep) - (3 (ep)1 where ep ranges over all continuous functions ep : 82 r --+ R with compact support. In particular, this topologizes the set C 0 R+ . PROPOSITION 6. The uniform space C (r) is complete. Considering the multiplicative action of Rt = ]0, co[ on C (r), the projective space PC (f) = (C (f) - {I}) jRt is compact for the quotient topology. PROOF: These are classical properties of spaces of Radon measures. See for instance [Bou, Chap. III, §I]. 0

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We therefore obtain a completion of C ® R+ by considering its closure in C (r). Actually, this completion turns out to be the whole space C (r), by the following result. THEOREM

7. The space C ® R+ is dense in C (r).

The proof of Theorem 7 will occupy §3.

§3. Proof of Theorem 7. Let r act isometrically, properly discontinuously and freely (say) on a negatively curved space X, in such a way that the quotient X jr is compact. The typical example to keep in mind is that of the action of a negatively curved group on one of its Cayley graphs. We will consider the geodesic flow of the negatively curved space X, defined as follows. As above, let a geodesic be an isometric map 9 : R -> X. Following Gromov [Gro, §8.3]' we endow the space ex of geodesics with the metric

1

+00

d(g,h)=

-00

d(g(t),h(t)) 2- ltl dt.

Observe that a sequence gn E ex converges to 9 for this metric if and only if the restriction gn 1 [-K,+K] uniformly converges to 9 1 [-K,+K] for every K 2:: O. The translations in R define an action of R on ex, called the geodesic flow of X. There is a natural continuous projection p: ex -> X defined by peg) = 9 (0). Observe that, for the action of R on ex, the orbit map t 1-+ tg E e X lifts the geodesic t 1-+ 9 (t) EX. The action of r on X lifts to an action on ex, which is easily seen to be properly discontinuous. In addition, the quotient eXjr is compact. Each geodesic 9 : R -> X has two limit points

g(+oo) = lim get) t-+oo

and

9 ( - 00)

=

lim 9 (t )

t-+-oo

in the boundary {)X, and it follows from Lemma 2 that the map IT : ex -> {)2 X associating (g ( -00) ,g (+00)) E {)2 X to gEe X is continuous. When X is a manifold of negative curvature or a tree, the geodesic flow has the very convenient property that two geodesics g, g' such that IT (g) = IT (g') E {)2 X are in the same R- orbit. This is unfortunately not always true for an arbitrary negatively curved space, for instance for many


153

Cayley graphs. To deal with this problem, Gromov defines in [Gro, §8.3] a construction which identifies two R-orbits of GX when they have the same limit points in Ej2 X, in a way which is natural with respect to the action of r. This provides a quotient space of GX which is foliated by lines, where each IR-orbit in GX projects homeomorphically to one of these leaves, and where two orbits IRg and IRg' in GX have the same projection if and only if 7r (g) = 7r (g'). In addition, the action of r on G X descends to a properly discontinuous action on and the map 1? : ---t Ej2 X induced by 7r is a locally trivial bundle map with fiber homeomorphic to R. See [Gro, §8.3]. Since 1? : ---t fj2 X is a locally trivial bundle, every point 9 E has a neighborhood ii which is a flow box: Namely, ii admits a homeomorphic parametrization ii ~ A X [0,1] where each leaf of meeting B meets it along an arc a x [0,1]. By compactness of /r, there is a family

ax

ax,

ax

ax

ax

ax

{iii; i

E

I}

of flow boxes in

ax

ax such that:

ax.

(a) the union of the Bi is equal to (b) there is an action of the group r on the index set is finite and, for each 'Y E (c) the projections 1? compact.

(.Hi)

C

r,

'Y

(iii)

=

ii-Y(i) .

I

such that

I/r

8 2 X are (Borel) measurable and relatively

ax,

By (b) and by the proper discontinuity of the action of r on each iii meets at most finitely iij. Subdividing the iii and adjusting their lengths if necessary, we can therefore require in addition that: (d) the intersection of two iii ~ Ai X [0,1] and contained in Ai X {O, I} and Aj x {O, I}.

iij

~

Aj

x [0,1]

IS

Let Bi C GX denote the preimage of .Hi under the quotient map q : GX ---t Each Bi ~ Ai X [0,1] is a flow box for the geodesic flow on GX. Also, the Bi satisfy the conditions obtained by removing the hats ~ from (a), (b), (c), (d) above. We are now ready to prove the following result, which constitutes the core of the proof of Theorem 7.

ax.

8. The subset V (C ® R+) consisting of all finite sums of elements of C ® R+ is dense in C (r) . PROPOSITION

Let a E C (r). We want to show that a IS 1D the closure of V (C ® IR+). By definition of the topology of C (r), this amounts to prove PROOF:

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F. Bonahon

that, for every finite family of continuous functions 'PI, .,. ,'Pn: with compact support, there is a arbitrarily close to

i

f3

h

[J2 X

E V (C 0 1R +) such that each

-+

IR

'Pj is

'Pj.

Endow [)X with a metric d lLp as in §1, using a base point endow [)2 X C [)X x [)X with the metric d lLp x d lLp •

Xo

E X and

Let {Bi; i E I} be the family of flow boxes in G X which we just constructed. The proof will make use of suitably chosen constants, namely 'small' constants E: > 0, 7] > 0 and 'large' constants D ~ 0, k ~ 1 and N ~ 0, with N an integer. The precise properties required for these constants will be specified when necessary, but we can already announce that the choice of 7] will depend on E:, that k will depend on D, and that N will depend on both k and 7]. Let L ~ 0 be associated to the constant k by Lemma 1. It follows from Lemma 3 that, for every compact subset K of [)2 X, there is a ball B in GX such that every (k,L) -quasi-geodesic whose end points are in K meets B. Consequently, there is a finite subset 10 of I such that every (k, L) - quasi-geodesic whose end points are in the support of one of the o. Then, because the tree X is uniquely geodesic at infinity, every representative 9 E r of "( E C respects some geodesic of X, called its axis. In addition, it is not difficult to find two representatives 91 and 92 of "( whose axes are disjoint. Then, if "(~ E C denotes the conjugacy class of 9l'9;n E r, the same argument as in the proof of Propo-

sit ion 9 proves that ("tn') n E C 0 R+ converges to " + ,,(-1 E C (r) as n tends to infinity. However, because the axes of 91 and 92 are disjoint, a classical result on translation distances in trees shows that tx ("tn') = 2ntx ("t) + A where A is the distance between the two axes (see [CuMo][Pau] for instance). In particular, if tx extends continuously I

to C (r), then tx ("( + ,-1) tradicting our first estimate.

= limn~oo tx (( "(n')*") = 2tx b) >

0, con0

J .-P. Otal has carefully analyzed this example in detail in rOta]. For instance, he has considered the restriction of the map tx to C s 0 R+ , where C s consists of those elements of C of the form + "(-1) where, can be represented by a simple closed curve on the surface S. The closure of C s 0R+ in C (r) has a natural identification with Thurston's space MC (S) of measured laminations on the surface. Otal has identified the maximal open subset of MC(S) on which the restriction tx : C s 0 R+ -+ R+ has a continuous extension. This open subset is the Masur domain introduced in [Mas], and is some kind of domain of discontinuity for the action on MC (S) of the subgroup of those elements of r which act trivially on X.

t ("(


163

When f be the fundamental group of a compact surface 8 of negative Euler characteristic, there is another case of interest where the function tx : C ® R+ -+ R+ extends continuously to C (f), but where the action is not properly discontinuous. Indeed, in this case, the space C (f) has an interesting additional feature (see [Boo1][Bo02] for details). Namely, there is a continuous bilinear form i : C (r) x C (f) -+ R+ such that, when "I, "I' are two distinct elements of C C C (f), i C'Y, "I') is the minimum of the cardinal of 9 n 9' where '7 and 9' range over all closed curves on 8 representing "I and "I' , respectively. Also, in this case, the space Me (8) of measured geodesic laminations has a natural identification with the space ofthose 0' E C (f) such that i (0',0') = 0 and T (0') = 0', where T : C (r) -+ C (f) is the orientation reversal involution such that T C'Y) = "1-1 for every "I E C C C (r). Then, consider the action of f on the R- tree X dual to a measured geodesic lamination 0' E Me (8) c C (f) (see [MoSh] for definitions, and see [Sko1][Sk02] for a characterization of these actions of f on R-trees).Then, it follows from definitions that txC'Y) = i(O',"I) for every "I E C ® R+. By continuity of the intersection form i, this implies that t x continuously extends to all of C (r) . We can apply these ideas to the problem mentioned in the introduction. Let f still be the fundamental group of a surface 8 of negative Euler characteristic, acting on the dual R- tree of a measured lamination 0' on 8. One way to reconstruct the lamination from this action is the following. Let "In E C - {1} be a sequence such that lim n _ oo t X C'Yn) = O. Since the projective space PC (f) is compact, after subsequencing, we find a sequence "In ~n E C ® R+ such that lim n ..... oo "I;n = "100 i- 1 E C (r) where the An are bounded from below; in other words, the line R+"Ioo is asymptotic to the sequence "In in C (f). We conclude that tx boo) = i (0', "100) = O. It easily follows that no (xoo, Yoo) E {)2 X in the support of "100 E C (f) crosses any (xO', yO') E {)2 X in the support of 0: E Me (r) c C (f), namely x oo , xO/' Yoo, yO' never occur in this order on the circle ()X. Considering all such sequences "In E C, one easily reconstruct the support of 0:. For instance, if 0: fills the surface, in the sense that no "I E C has translation distance t x ("I) = 0, any such "100 has the same support as 0:. Similar techniques were used in [Boo1] to associate certain invariants, which were geodesic laminations on surfaces, to actions of a surface group f on hyperbolic 3- space H3 . An important problem is to understand how much of the above can be

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extended to actions of an arbitrary negatively curved group r on an R- tree X. A particular case of interest is when all the edge stabilizers of this action are 'small', namely contain cyclic subgroups of finite index; see [MoSh] for motivations. Indeed, R. Skora proved in [Sko2] that, among all actions of the fundamental group of a surface S on an R- tree, those which are dual to a measured lamination on S are characterized by the property that their edge stabilizers are small. In view of what we saw in the case of surface groups, it seems quite reasonable to conjecture that, given an action of a negatively curved group r on an R-tree X, the map tx continuously extends to C (r) if and only if the action has small stabilizers. This problem does not seem that easy to tackle, but understanding it may help understanding which negatively curved groups can act with small edge stabilizers on R- trees. Now, let us consider another problem. When a negatively curved group r acts properly discontinuously and co compactly on a negatively curved space, we have seen that the space aX, endowed with the action of r, depends only on the algebraic structure of r. On the other hand, this is not true any more if the action is not co compact. One can therefore try to analyze the action of r on aX in order to get some information on the action of r on X. In view of applications, the example we have in mind here is the case of a non-compact complete Riemannian manifold M of negative curvature bounded away from 0, whose fundamental group r is abstractly isomorphic to the fundamental group of a compact manifold of negative curvature with convex boundary; the second hypothesis guarantees that r is negatively curved and, by analyzing the action of r on the universal covering X of M, we want to detect how much the geometric behavior of M differs from that of a compact negatively curved manifold with convex boundary. Such a problem occurs for instance in the study of the space of Kleinian groups which have a given algebraic type (see [Thu][Mor][Bonl]). So, consider a negatively curved group r acting properly discontinuously on a negatively curved space X. Let ,Ee be a conjugacy class of r. Analyzing what subsists of the proof of Lemma 4 in this case, we find that (exactly) one of the following occurs:

(i) the translation distance t X (,) is equal to o. (ii) the translation distance tx CI) is positive. In this case, the sequences ;:ynx and ::y-nx have limits ::y+oo and ::y-oo as n tends to 00, for every


165

x E X and every representative ::y E f of I. In addition, these limits are independent of x and the union of the points (::y+ 00 , ::Y-OO) E 0 2 X, as ::y ranges over all representatives of I, forms a discrete subset A (,) of ()2 X.

In this way, we can again associate to l E e a f - invariant measure on the space 0 2 X, namely the measure 0 in case (i) and the Dirac measure of weight n defined by A (,) in case (ii), where n is the positive integer such that I = 1[; for some primitive 10. These measures are supported on the limit set Ar C oX of the action of f on X, defined as the set of limit points of an(y) orbit of f in X. It therefore makes sense to consider the space C(X,f) of f-invariant positive Radon measures on 0 2 X C oX x oX whose support is contained in Ar x Ar . The elements of C (X, f) will be called geodesic currents for the action of f on X. By the above considerations, we have associated an element of C (X, f) to each conjugacy class l E e . This map 'linearly' extends to a map p : C ® R+ ~ C (X, f). It is then natural to ask whether this map IS continuous, and extends to a continuous map C (f) ~ C (X, f). We will say that a E C (f) is realized in X/f if the map p : C ® R+ ~ C (X, f) can be continuously extended at a, namely if there exists an element p (a) E C (X, f) such that p( a) = limn~oo p (,~n) in C (X, f) for every sequence I~n E C ® R+ converging to a in C (f). In this case, pea) E C(X,f) is the realization of a E C(f) in X/f. The introduction of this notion is motivated by Thurston's theory of realization of measured laminations through pleated surfaces, which constitutes the backbone of his analysis of limits of quasi-Fuchsian groups in [Thu]. It can be shown that Thurston's definition of realizability is equivalent to ours, although we find our point of view conceptually more satisfactory. For instance, the main result of [Bonl, §5] can be paraphrased by saying that, if f is the fundamental group of a compact surface of negative Euler characteristic and if Hn /f is a negatively curved manifold with the same fundamental group, then every measured lamination a E Me (S) c C (f) on S can be realized in Hn /f. In addition, if n = 3, the measured laminations a E Me (S) such that p (a) = 1 are characterized in [Bonl, §6], in terms of the cusps and of the ending laminations of the ends of H3 /f . One reason why this realizability problem is interesting is the following. If the negatively curved group f acts properly discontinuously on the

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F. Bonahon

negatively curved space X, it is usually conjectured that the limit space Ar c ax is a certain f-equivariant quotient of af. For instance, if X is the hyperbolic 3- space H 3 , this property was proved by W. Floyd [Flo] when the action of f is geometrically finite, possibly with cusps, and by J. Cannon and W.P. Thurston [CaTh] for some very specific geometrically infinite actions. However, little is known in the general case. If there is such a continuous quotient map, then it certainly induces a continuous extension p : C (f) - t C (X,f). Conversely, understanding the space C (X, r) of invariant measures on the dynamical system (Ar x Ar - ~,f) would give some information on the space Ar endowed with the action of f. For these reasons, the problem of realizing geodesic currents seems to be a good approach to the analysis of limits sets. When X is the hyperbolic n- space Hn, using some extension of the arguments of [Bonl, §5], we have been able to construct such a continuous map p : C (r) - t C (X, r) at least when the injectivity radius of the orbifold Hn If is bounded from below. The details will appear elsewhere. A related result appears in rOta], for the situation of Proposition 11 (see our first discussion of Otal's results after the proof of that statement). Indeed, to prove that the map tx : C s ® R+ - t R+ continuously extends at the Masur domain n, Otal shows that the representation map p : C s 0 R+ - t C (X, r) defined as above continuously extends at n. If fa denotes the subgroup of those elements of f which act trivially on X, then C (X, f) = C (X, r If a ) and the continuity if tx on f now follows from Proposition 10. In addition, the corresponding continuous map n - t C (X, r) clearly factors through a map n/fo - t C (X, f), and Otal proves the remarkable result that this map n/fo - t C (X, f) is one-to-one.


167

REFERENCES [Bon 1] F. Bonahon, Bouts des varieUs hyperboliques de dimension 3, Ann. of Math. 124 (1986), 71-158. [Bon2] F. Bonahon, The geometry of Teichmuller space via geodesic currents, Invent. Math. 92 (1988), 139-162. [Bon3] F. Bonahon, "Structures geometriques sur les varietes de dimension 3 et applications," These d'Etat, Universite d'Orsay, 1985. [Bou] N. Bourbaki, "Elements de mathematiques," livre VI (Integration), Hermann, Paris, 1965. [Bow] B. Bowditch, "Notes on Gromov's hyperbolicity criterion for path-metric spaces," manuscript, University of Warwick, 1989. [CaTh] J. Cannon, W.P. Thurston, Group invariant Peano Curves, unpublished article (1982). [CoDP] M. Coornaert, T. Delzant, A. Papadopoulos, "Notes sur les groupes hyperboliques de Gromov," manuscript, Institut de Recherches Mathematiques Avancees, Strasbourg, 1989. [CuMo] M. Culler, J.W. Morgan, Group actions on R- trees, Proc. Lond. Math. Soc. 55 (1987), 571-604. [EbON] P. Eberlein, B. O'Neill, Visibility manifolds, Pac. J. Math. 46 (1973), 45-109. [Flo] W.J. Floyd, Group completions and limit sets of Kleinian groups, Invent. Math. 57 (1980), 205-218. [GhH] E. Ghys, P. de la Harpe et aI., "Sur les groupes hyperboliques d'apres Gromov," (with contributions of W. Ballman, A. Haefliger, E. Salem, R. Strebel, M. Troyanov), manuscript, Ecole Normale Superieure de Lyon and Universite de Geneve, 1989. [Gro] M. Gromov, Hyperbolic groups, in "Essays in group theory," (S.M. Gersten ed.), Springer-Verlag, Berlin Heidelberg New York, 1987, pp. 75-263. [LySe] R.C. Lyndon, P.E. Schupp, "Combinatorial group theory," Springer-Verlag, Berlin Heidelberg New York, 1977. [Mas] H. Masur, Measured foliations and handlebodies, Erg. Theory and Dyn. Syst. 6 (1986), 99-116. [MoSh] J.W. Morgan, P.B. Shalen, Valuations, trees, and degenerations of hyperbolic structures, I, Ann. of Math. 120 (1984),401-476. [Mor] J .W. Morgan, On Thurston's uniformization theorem for the three-dimensional manifolds, in "The Smith Conjecture," (H. Bass, J.W. Morgan ed.), Academic Press, 1984. [MSRI] J. Alonso, T. Brady, D. Cooper, T. Delzant, V. Ferlini, M. Lustig, M. Mihalik, M. Shapiro, H. Short, "Notes on negatively curved groups," manuscript, Mathematical Sciences Research Institute, Berkeley, 1989. [at a] J.-P. Otal, "Courants geodesiques et surfaces," These d'Etat, Universite de Paris-Sud, Orsay, 1989. [Paul F. Paulin, The Gromov topology on R- trees, Topol. Appl. 32 (1989), 197-221. [Ser] J .-P. Serre, "Arbres et amalgames," Asterisque, Societe Mathematique de France, 1977. [Sigl] K. Sigmund, Generic properties of invariant measures for Axiom A- diffeomorphisms, Invent. Math. 11 (1970), 99-109. [Sig2] K. Sigmund, On dynamical systems with the Specification Property, Trans. Amer. Math. Soc. 190 (1974), 285-299. [Skol] R. Skora, Geometric actions of surface groups on A -trees, preprint. [Sko2] R. Skora, Splittings of surfaces, preprint.

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[SuI] D. Sullivan, Cycles for the dynamical study of foliated spaces and complex manifolds, Invent. Math. 36 (1976), 225-255.

[Thul W.P. Thurston, "The topology and geometry of 3-manifolds," Lecture notes, Princeton University, 1976-79.

Department of Mathematics, University of Southern California, Los Angeles, CA 900891113, U.S.A. The author gratefully acknowledges partial support from N.S.F. and from the A.P. Sloan Foundation


LM.

CHISWELL

Dedicated to the memory of Roger Lyndon

The term pregroup was introduced by Stallings in [SI] and [S2], and arose from his work on groups with infinitely many ends. The ideas in the definition go back to Baer [Ba]. A pregroup is a set with a partial multiplication having certain group-like properties, to which one can associate a group (the universal group of the pregroup), and there is a normal form for the elements of the group in terms of the pregroup. This generalises the construction of free groups, free products and HNN-extensions. The definition of pregroup is designed to enable the well-known and elegant argument of van der Waerden to be used to prove the normal form theorem (this argument was originally used in [W] for free products). The structure of arbitrary pregroups is mysterious, and the purpose of this article is to give a brief survey of recent progress in understanding them. This involves constructing actions of the universal group on simplicial trees, making use of the order tree of a pregroup, a (of a kind well-known in set theory) generalised tree with basepoint invented by Stallings and introduced in Rimlinger's book [R2; Sect. 2]. We begin with the precise definition of pregroup. Let P be a set with a partial multiplication, that is, a mapping D -+ P, (x, y) 1-+ xv, where D is a subset of P x P. When it will not cause confusion, we shall say that xy is defined instead of (x, y) ED. We assume that there is a mapping P -+ P, X 1-+ x-I, and a distinguished element e E P satisfying the following three aXIoms:

(PI) for all x E P, xe

= ex = x;

(P2) for all x E P, x-Ix = xx- I = e; (P4) for all elements x, y and z of P, if xy and yz are defined, then x(yz) is defined if and only if (xy)z is defined, in which case x(yz) = (xy)z.

170

I.M. Chiswell

Of course, written in full, (PI) should read "xe and ex are defined and xe = ex = x", and similar comments apply to (P2). The reader may well wonder what happened to Axiom (P3)j this is the following: (P3) if xy is defined, then so is y-lx-l, and (xy)-l = y-lX- l . It was noticed by Squier that this follows from the other axioms by a simple argumentj see [Hoj Prop. 1.2). The usual elementary arguments used with the axioms for a group show that (x-l)-l = x. There is one more axiom for a pregroup which can be written in several equivalent forms, and we give three of them. For x E P, put L(x) = {a E P I ax is defined}, define x :S y to mean L(y) ~ L(x), and x < y to mean L(y) ~ L(x). DEFINITION: (P, D) is a pregroup if PI, P2, P4 and the following equivalent conditions are satisfied: P5(i) if WX, xy and yz are all defined, then either wxy or xyz is definedj P5(ii) if x-la and a-ly are defined but x-ly is not defined, then a < x and a < Yj P5(iii) if x-ly is defined, then either x :S y or y :S x. For the proof of equivalence (in part due to Rimlinger), see Hoare [Hoj Theorem 1.5). We give two simple examples of pregroups which are useful to bear in mind. Let F be a free group with basis X, and put P = Xu X-l U {I}. Define D to be the set of pairs (x,y) in P X P such that xy E P, where xy means the product in F, and multiplication is obtained by restriction from F. Of course e = 1, the identity element of F, and it is obvious how X-l is defined. The second example is an amalgam of groups P = A Uc B, that is the union of two groups A and B intersecting in a common subgroup C. We define (x, y) E D if and only if either x and y both belong to A or x and y both belong to Bj the product xy is then the product in A or B, as appropriate. These two examples can be generalised. Let I be an integer-valued Lyndon length function on a group G (this means [ is required to satisfy Lyndon's axioms AI', A2 and A4j see [Ly)). Put P = {x E G I [(x) :S I} and D = {(x, y) E P X Pixy E P}, where xy means the product in G, and the product in P is the restriction to D of the product in G. For a proof that P is a pregroup, see Lemma 7 in [C2). This generalises the previous examples, first taking [ to be the usual length with respect to the basis X of the free group F, then taking [ to be the length on the free product with


171

amalgamation A *0 B defined in Sect. 1 of [C3] (it is shown to be a length function, and to be obtained from the action of the group on a tree in a standard way, in Lemma 3 and Corollary 2 of [C3]). For further examples of pregroups see [82; 3.A.5] and [R2]. Note also that any group is a pregroup. A mapping cp : P ~ Q of pregroups is a morphism if whenever xy is defined in P, cp(x)cp(y) is defined in Q and equal to cp(xy). (It follows easily that cp must preserve inverses and the identity element). This gives a category whose objects are pregroups, having the category of groups and homomorphisms as a full subcategory. There is a forgetful functor from groups to pregroups, and this functor has a left adjoint which we denote by U, and we call U(P) the universal group of the pregroup P. Thus U(P) has the following universal property: let P be a pregroup; there is a morphism of pregroups L : P ~ U(P), such that, if cp : P ~ G is any morphism with G a group, there is a unique group homomorphism 'IjJ : U(P) ~ G such that 'ljJL = cpo An explicit construction of U(P) is given by Stallings [82], and his construction also gives a normal form theorem describing its structure. A word is a finite sequence (Xl, •.. ,xn) with n ~ 1 and Xi E P for 1 ~ i ~ n. The word is reduced if (x;, Xi+l) t/: D for 1 ~ i ~ n -1, and the element of U(P) which it represents is LeXI) ... L(Xn). Any element of U(P) is represented by a reduced word and two reduced words (XI, ... ,X n ) and (YI, ... ,Ym) represent the same element of U(P) if and only if m = n and there exist elements al, ... , an-l in U(P) such that Yi = ai=!lxiai for 1 ~ i ~ n, where ao = an = e. NORMAL FORM THEOREM.

To clarify the last part of the statement, we mean that Xl al is defined and equal to YI, that a1lx2, X2a2 and a11(x2a2) are defined and Y2 = a11(x2a2) and so on. Parentheses can be omitted by Axiom P4. In particular, L is injective, and in view of this we may suppress it in the notation. If necessary we shall call a reduced word P-reduced to avoid confusion. If P is the pregroup X U X-I U {I} where X is a basis for the free group F, then U(P) is just F. This can be seen using the normal form theorem or more simply using the universal mapping properties of U(P) and F. Similarly the universal group of the amalgam A Uo B is the free product A *0 B. In the length function example, the universal group is G, provided G is generated by {x E G I lex) ~ I} (we can always replace G by the subgroup generated by this set). See [C2; Lemma 6] for the proof of this.

172

I.M. Chiswell

We now introduce the order tree of a pregroup P. For x, y in P, define x '" y to mean x :::; y and y :::; x, that is, L(x) = L(y). This is an equivalence relation on P, and the quotient set PI '" is a partially ordered set, defining [x] :::; [y] if and only if x ~ y, where [x] denotes the equivalence class of x. Moreover (see [R2; Lemma 2.2] or [Ho; Prop. 1.6]) : (1) PI'" has a least element (namely [e]); (2) for all O!, (3 and, in PI "', if O! :::; , and (3 :::; " then either (3 :::; O!.

O! :::;

(3 or

We call a partially ordered set satisfying (1) and (2) an order tree. An example of an order tree is given by fixing a point Xo in a A-tree X (A being any ordered abelian group) and defining x :::; y if and only if x lies on the geodesic from Xo to y. Returning to the pregroup P, suppose that, for each x E P, there are, up to equivalence, only finitely many u in P such that u < x. Then the order tree can be made into a simplicial tree with vertex set PI "', joining [x] and [y] by an edge when [x] < [y] and there is no vertex [u] such that [x] < [u] < [y]. This simplicial tree defines a corresponding I-tree, and PI'" is obtained from this I-tree as described above for arbitrary A-trees, with x = [e]. For more details see [Ho; Sect. 2]. We shall describe this situation by saying that the order tree is simplicial. Guided by the example of an amalgam of groups, we can define a notion of length of elements of U(P), where P is an arbitrary pregroup. First, let

that is (using P3), a E A if and only if xa and ax are defined for all x E P. Then define a mapping p on reduced words by: _ {O

p ( x!, .. . ,x n ) -

n

if n = 1 and . otherwIse.

Xl

EA

By the normal form theorem p induces a mapping U(P) ~ I, which we continue to denote by p. (This differs from Rimlinger's definition of length [R2; 1.5]). It is easy to see that p is a I-semigauge, that is it satisfies: (1) pee) = 0 (2) p(x- l ) = p(x) for all x E U(P) (3) p(xy) :::; p(x) + p(y) for all x and y in U(P). These are Lyndon's axioms AI', A2 and A3 (see [Ly]). The question arises whether or not p is a Lyndon length function on U(P), that is, does his


173

axiom A4 hold? The answer is no in general. It was shown by the author [C4] and by Nesayef [N] that p is a Lyndon length function if and only if P satisfies an extra condition which we call P6, and there are examples of pregroups not satisfying P6. We shall exhibit some of them later. There are several formulations of Axiom P6 (see [C4j Sect. 1]), and it is convenient here to use one given in [C4j Prop. 2]. (P6) for all X,y and a in P, if x-Ia and a-Iy are defined but x-Iy is not defined, then a '" e. It turns out that P6 can be characterised in terms of the order tree of P. This has been noted by Kushner and Lipschutz [KLj p. 171].

PROPOSITION 1. A pregroup P satisfies P6 if and only if the order tree is simplicial and all vertices other than [e] are adjacent to [e]. PROOF: Assume P satisfies P6 and suppose x < y (where x and yare in P). Then y-l E L(y) by P2, so y-l E L(x), i.e. y-Ix is defined, so x-Iy is defined (using the missing axiom P3). Also, there is an element z E L(x)\L(y), so zx is defined but zy is not. By P6, x '" e, and it follows that the order tree is as described in the proposition. Conversely, assume the order tree is as described, and suppose x-Ia and a-Iy are defined but x-Iy is not. Then X-I E L(a)\L(y), so not (y ::; a). By version (iii) of P5, a < y. Hence a '" e and so P6 holds. 0 Thus it seems at first sight that the assumptions needed to carry out Nielsen cancellation arguments (Lyndon's axioms) are stronger than those needed to prove a normal form theorem (the axioms for a pregroup). However, this viewpoint is too naive, and using a clever and intricate cancellation argument, Hoare [Ho] has shown that it is possible to define an integer-valued Lyndon length function on U(P) assuming only that the order tree is simplicial. In this case, the tree structure defines an integervalued metric d on the set of vertices, giving a l-tree. The Hoare length is defined by: if g E U(P) is represented by the reduced word (Xl, ... , x n ), then n+l

leg) = Ld(xi~I,Xi) i=l

where d(x, y) means d([x], [y]) and Xo = Xn+l = e. If we have a pregroup satisfying P6, we have two Lyndon length functions, 1 and p, on U(P) and the question arises of the relation between them. This has a simple answer.

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LM. Chiswell

PROPOSITION 2. If the pregroup P satisfies P6, then 1 = 2p. PROOF: We divide the proof into three steps. (1) First observe that, if [x] = [e] (i.e. x '" e), then L(x) = L(e) = P, i.e. xy is defined for all yEP, but also yx is defined for all yEP by the last part of Prop. 2 in [C4] (the assertion that P6 implies P7). It follows by P3 that x-Iz is defined for all z E P, i.e. [x-I] = [e]. Hence [x] = [e] if and only if [X-I] = [e] since (x- l )-1 = x. Thus, [x] = [e] if and only if x E A, where A is the set used in the definition of p. (2) If x and yare in P, then xy is defined if and only if either [x] = [e], or [y] = [e], or [x-I] = [y]. For by Step 1, if [x] = [e], or [y] = [e] then xy is defined, while if [x-I] = [y], i.e. x-I", y, then x E L(x- l ) = L(y) and again xy is defined. Conversely, assume xy is defined and [x] i=[e] i=- [y]. Then by version (iii) of P5, either [X-I] ::; [y] or [y] ::; [X-I], and by Step 1, [x-I] i=- [e]. By Proposition 1, [x-I] = [y]. (3) Now take g = Xl ... xn in U(P), where (Xl, ... , x n) is a P-reduced word. By definition, n

leg) = dee, Xl) + d(x;;l, e) +

L

d(xi~l' Xi).

i=2

If n > 1, then by Step 2 [Xi] i=- [e] and [Xi~l] i=- [Xi] since the word is reduced, so [Xi~l] and [x;] are distinct vertices of the order tree adjacent to [e] by Proposition 1. Thus leg) = 1 + 1 + 2(n - 1) = 2n = 2p(g). If n = 1, then

leg) = d(e, xI)

+ d(xll, e)

and using Step 1 and Proposition 1, if [Xl] i=- [e] then leg) 2p(g), while if [Xl] = [e] then leg) = 0 = 2p(g).

= 2= 0

We now consider two specific examples of pregroups with simplicial order tree, and for each we shall compute the Hoare length function 1. The first example is the "up-down pregroup" of Stallings [S4]. Part of the analysis needed has already been given in [S4; Prop. 3.1], but we give a self-contained argument. Let F be a non-abelian free group with basis X, and let 5 be the submonoid of F generated by X (i.e. all positive words on X). Put P = 5.5- 1 and D = {(x, y) E P x Pixy E P}, where xy is the


175

product in F, with multiplication obtained by restriction from F. Stallings shows that P is a pregroup with universal group F. In using terms such as cancellation and initial segment, we are thinking of elements of F as reduced words relative to the basis X. Also, in writing elements of P as uv- l , with U and v in S, we tacitly assume that there is no cancellation (of letters in X±l) in forming this product in F. Now let ulv 1 l and u2vil be elements of P, where Ul, Vb u2 and V2 are in S. Then the product (ulvll)(u2vil) is defined in P if and only if, in forming the product v11u2 in F, either vII is completely cancelled by U2, or U2 is completely cancelled by vII. Consequently, if Ul is an initial segment of U2 then ulv 1 1 :$ u2vil. Moreover, if UI is a proper initial segment of U2 then Ul vII < u2vil. For in this case if U2 = UIXW where x E X and W E S, let y be an element of X other than x, and put Z = UlY. Thenz E L(ulvll)\L(U2Vil). It follows from this that conversely if Ul vII :$ u2vil then Ul is an initial segment of U2. Hence the order tree is simplicial, but not of finite diameter. Let A denote the usual Lyndon length function on F, that is, A( u) is the length of U when viewed as a reduced word relative to X. If d is the distance function in the order tree of P, then d( e, uv- I ) = A( u), where e is the identity element of F (the empty word), and u, V are in S. Suppose that the element g E F is represented by the P-reduced word W = (Ul vII, ... , unv;l). Then -I )-1 , Ui V-I) = d( Vi-lUi_I' -I UiVi-I) d« Ui_1 Vi_I i

where Ci is the length of the common part of the paths joining [el to [uivill and to [Vi-l uilll in the order tree, which is the length of the largest common initial segment of Ui and Vi-I relative to the basis X. Thus

This is illustrated by the diagram:

Co

[el----
0 there is a unique minimal action x,y,z of F(a, b) on an R-tree with Ilbll = x, IIcll = z and distance between the translation axes of band c equal to y. This action is free and properly discontinuous. It corresponds to the following point in Culler-Vogtmann space

length = z

c

length

y

length = x.

b

It is easy to see that if W

=

=

bn1 c"'lb n2 c"'2

••. bnqcl!q,

with

C;

= ±l and

ni

=/::.

0,

then

Now, let (x, Yb Zl) and (x, Y2, Z2) be two distinct triples of positive real numbers. If 2Yl + Zl = 2Y2 + Z2, it follows that the translation length functions of X,Yl,Zl and X,Y2,Z2 agree on the Wi but not on the word Wi = C.

o

(1) Precisely the same arguments as in the first proof give the Theorem with F(a, b) replaced by any non-trivial free product G = A * B, when there is at least one element of infinite order in A or B. (2) In [C-Mj, §O, Culler and Morgan asked when the spaces P LF( G) and S LF( G) of projective translation length functions are finite dimensional. The first demonstration that SLF(F2 ) is infinite dimensional appears in [Stj. It is pointed out in [C-Lj that, by iterating the amalgamated product construction of the first proof above, the spaces SLF( G) are infinite dimensional for all groups G = A * B as in (1) above. (See also [B-Fll, [B-F2j.) To be specific, if b is an element of infinite order in the group B and p is a positive integer, the graph of groups decomposition of A * B given by REMARKS:

A

•

{I}

•

•

B

•

186

M.M. Cohen, M. Lustig and M. Steiner

leads to a p-simplex of simplicial actions in S LF( G), as one allows the lengths of the p + 1 edges of groups to vary. (3) It is shown in [C-L] that the R-tree actions constructed using proper powers as in the first proof and in (2) above are not limits of free R-tree actions. [Here "(free) R-tree actions" is short for "projective classes of translation length functions of (free) R-tree actions".] It would be of great interest to know under which conditions the closure in S LF( G) of the space of free R-tree actions of G is finite dimensional. Notice that the second proof of the Theorem demonstrates that: The projective space of a space Z of translation length functions may be finite dimensional even though, given any finite set of n words, there exists a pair of elements of Z with distinct projective classes which agree on these n words. For the Culler-Vogtmann space [C- V] is finite dimensional, and no finite set of words can distinguish all pairs of elements in this projective space of length functions. (4) The distinct free simplicial actions which we give in the second proof cannot be normalized so that they come from markings of graphs which have the sum of the lengths of the edges equal to one. This is achieved in a sharpening of our results for the free group of rank n (n ~ 3) by Smillie and Vogtmann [8-V].

The Translation Lengths of Finitely Many Elements

187

REFERENCES

[B-Fl] Bestvina, M. and Feighn, M., Bounding the complexity of simplicial group actions on trees, preprint.

[B-F2] Bestvina, M. and Feighn, M., A counterexample to generalized accessibility, this volume. [C-L] Cohen, M. M. and Lustig, M., Very small group actions on IR-trees and Dehn twist automorphisms, preprint. [C-M] Culler, M. and Morgan, J., Group actions on IR-trees, Proc. London Math. Soc. 55 (1987), 571-604. [C- V] Culler, M. and Vogtmann, K., Moduli of graphs and automorphisms of free groups, Invent. Math. 84 (1986), 91-119. [He] Helling, H., Diskrete Untergruppen von SL2(~)' Inventiones Math. 17 (1972), 217-229. [S-W] Scott, G P. and Wall, C.T.C., Topological methods in group theory, in "Homological Group Theory", ed. C.T.C. Wall, London Math. Soc. LN 36 (1979), 137-203. [St] Steiner, M., Glueing data and group actions on A-trees, to appear. [S- V] Smillie, J. and Vogtmann, K., Length functions and outer space, to appear.

M.M. COHEN: Department of Mathematics White Hall, Cornell University Ithaca, NY 14853

M. LUSTIG: Institut fur Mathematik Ruhr- U niversitiit Bochum Postfach 10 21 48, D-4630 Bochum 1, West Germany M. STEINER: Department of Mathematics University of Texas at Austin Austin, TX 78712 M.M. Cohen's research partially supported by NSF Grant DMS-8803561. Research at MSRI supported in part by NSF Grant DMS-8505550.

The Boundary of Outer Space in Rank Two MARC CULLER AND KAREN VOGTMANN

§1. Introduction In [4] a space Xn was introduced on which the group Out(Fn) of outer automorphisms of a free group of rank n acts virtually freely. Since then, this space has come to be known as "outer space." Outer space can be defined as a space of free actions of Fn on simplicial IR-trees; we require that all actions be minimal, and we identify two actions if they differ only by scaling the metric on the IR-tree. To describe the topology on outer space, we associate to each action a: Fn X T -+ T a length function I· I",: Fn -+ IR defined by Igl", = inf d(x,gx) xET

where d is the distance in the tree T. We have Igl", = Ih-1ghl", and 1·1", == 0 if and only if some point of T is fixed by all of Fn. Thus an action with no fixed point determines a point in IRc - {O}, where C is the set of conjugacy classes in Fn. Since actions differing by a scalar multiple define the same point of outer space, we have a map from Xn to the infinite dimensional projective space pC = IRc - {O} /IR*. It can be shown that this map is injective (see [3] or [1]). We topologize Xn as a subspace of pC. Part of the motivation for the definition of outer space was the idea of developing an analogy between the action of Out(Fn) on outer space and the action of the mapping class group of a surface on the Teichmiiller space of that surface. In particular, the action of the mapping class group on Teichmiiller space was exploited by Thurston in his classification of automorphisms of surfaces. Thurston gives an embedding of Teichmiiller space into an infinite dimensional projective space and shows that its closure in this projective space is a finite-dimensional ball; he then uses the fact that the ball has the fixed point property to analyze the action of a single automorphism on the closure of Teichmiiller space. We would like to know how much of Thurston's theory can be adapted to automorphisms of free groups. It is shown in [4] that outer space is contractible of dimension 3n - 4, and in [3] that its closure X n in pC is

190

M. Culler and K. Vogtmann

compact. M. Steiner and R. Skora have recently announced proofs that X n is contractible. It is not known whether the closure of outer space is finite dimensional or whether it is an ANR; if it is an ANR, this together with contractibility of X n would imply that X n has the fixed point property. In this paper we restrict ourselves to the case n = 2. We give an explicit description of the closure X = X 2 of outer space in rank 2. In particular, we show that X is contractible, and give an imbedding of X as a twodimensional subset of 1R3 which makes it clear that X is an ANR. The paper is organized as follows. In section 2 we recall some basic definitions and properties of actions on IR-trees. In sections 3-5 we determine what lies on the "boundary" aX = X -X. As a starting point, we have by [3] that points in aX correspond to non-trivial actions of F2 on IR-trees with cyclic arc stabilizers. In addition, we note in section 2 that the stabilizer of an arc in any limit of free actions is either trivial or is a maximal cyclic subgroup of the stabilizer of each of its endpoints. We remark that Cohen and Lustig [2] have given criteria for deciding when an action of Fn on a simplicial tree is a limit of free actions. For n = 2 these criteria consist of the above conditions on arc stabilizers. (For n > 2 an extra condition must be imposed.) In section 6 we describe the imbedding of X into 1R3. One consequence of our analysis in sections 3-5 is that all actions in X are geometric, in the sense that they are isomorphic to actions of the fundamental group of a punctured torus or twice-punctured disk on trees which are dual to projective measured laminations on those surfaces. Our embedding demonstrates how the projective lamination spaces for various surfaces fit together to form outer space. A description of these projective lamination spaces is given in Hatcher [6]. We would like to thank Peter Shalen for helpful conversations and for inventing the name "outer space," Richard Skora for explaining the proof of his realization theorem, and Curt McMullen for supplying the computer programs used to generate the portrait of outer space.


191

§2. Group actions on trees. In this section we establish some notation and conventions that we will use throughout the paper. For the general theory of group actions on IR-trees, we refer to [3] or [1]. By a tree we will always mean an IR-tree, with metric d. A germ at a point p of a tree is an equivalence class of rays from p, where two rays are equivalent if they agree in some neighborhood of p. An IR-tree is simplicial if it is homeomorphic to a simplicial tree, i.e. to a connected, I-connected, I-dimensional simplicial complex. Our actions on trees will always be left actions. Two actions al: G x Tl TI and a2: G x T2 - T2 are isomorphic if there is an equivariant isometry from TI to T 2 . An action a: G x T - T is simplicial if T is a simplicial tree. The action is minimal if T has no proper invariant sub-tree. The translation length of an element 9 EGis given by Igl" = inf d(p,gp). pET

We will often omit the subscript a, when no confusion will result. H Igl" = 0, then 9 fixes a subtree Fix(g) of Tj in this case 9 is called elliptic. H 9 has no fixed point, 9 is called hyperbolic and has a translation axis Axis(g). By the characteristic set of 9 we will mean Fix(g) if 9 is elliptic and Axis(g) if 9 is hyperbolic. Definition. An action of a group on an IR-tree will be said to have maximal cyclic arc stabilizers if any non-trivial stabilizer of an arc is a maximal cyclic subgroup of the stabilizers of each endpoint of the arc. Lemma 2.1. Let a: F2 X T - T be an action which is a limit of free simplicial actions. Then a has maximal cyclic arc stabilizers. Proof. Let e be an arc of T with endpoints v and w. As noted in the introduction, [3] show that any limit of free actions has cyclic arc stabilizers, so the stabilizer of e is cyclic. H it is not a maximal cyclic subgroup of the stabilizer of v, there is an element 9 in the stabilizer of v such that ge i:- e but gke = e for some k > O. Let {a n :F2 x Tn - Tn} be a sequence of free simplicial actions such that the associated (non-projective) length functions I . I"n converge to the length function I . I,,· By [11], the topology on X 2 as a subset of pC is the same as the Gromov topology. In particular, we can find points Wn E Tn such that d( Wn, gw n ) _

192


d(w,gw) and d(Wn,gk Wn ) trees Tn, we have

-+

d(w,gkw). Since 9 has an axis in each of the

This gives a contradiction, since the left-hand side converges to d( w, gkw) = 0, while the right-hand side converges to d(w,gw) > o. 0


193

§3. Simplicial actions in the boundary. In this section we describe all actions in X with the property that some primitive element of F2 has a fixed point. When we complete our analysis of all actions in X, we will see that these are exactly the simplicial actions in aX =X -X. Definition. Let G be a group acting on an R-tree T. A subtree H of T is a fundamental domain for the action if GH = T and if for any g E G, gH n H is either empty, a single point, or equal to H n Fix(g). The following lemma gives a criterion for an action of F2 on a tree to be simplicial, and produces a fundamental domain for the action. The lemma is easily generalized to free groups of higher rank; we give the statement for rank 2 to avoid complicated notation. Lemma 3.1. Let {a, b} be a basis for F2, let F2 x T _ T be a minimal action, and let H be a non-empty closed subtree ofT. Suppose that T - H is the disjoint union of open sets Sa, Sa-1, Sb, and Sb-1 with finite boundary. Assume, for x,y E {a,a-l,b,b- 1}, that 1. xHnH -:f:. 0 2. xH C Fix(x) U Sx; 3. x(Sx U Sy U Sy-l) c Sx, for x =1= y, y-l. Then H is a fundamental domain for the action of F2 on T, and T is a simplicial tree. Proof. We may assume that Fix(a) n Fix(b) = 0, since otherwise, by minimality, T is a point and the lemma is trivial. In order to show that F2H = T, it suffices to show that F2H is connected; F2H is then an invariant subtree of T, which must be all of T since T is minimal. Let Tn denote the union of all translates gH, where g E F2 has word length less than or equal to n in the generators a and b. We show that Tn is connected for all n, by induction on n. If n = 0, this is just the statement that H is a subtree. If n = 1, this follows from hypothesis 1 of the lemma. If n > 1, T n +1 = aTn U a-1Tn U bTn U b-1Tn . Each of these four sets is connected by induction, and each contains H so their union is connected. To show that gH n H is of the right form for any g E F 2 , we use the following fact.

194


Claim. Let 9 E F2, let w be the unique reduced word in a and b representing g, and let x be the :first letter of w. Then (i) gH C Sx U (Fix(x) n H), and (ii) if gH is not contained in Sx, then 9 = xk for some k. Proof. The proof is by induction on the length n of w. If n = 1, statement (i) follows from hypothesis 2 and statement (ii) is trivial. If n > 1, we have w = xv, where v is a word with first letter y ¥- x-I. Let p be a point of H. By induction, we know that vp E Sy U (Fix(y) n H). If vp E Sy, then wp = xvp E Sx by hypothesis 3. If vp is not in Sy, then vp E Fix(y)nH, and we have two cases to consider. If y ¥- x, then vp is not in Fix( x), since Fix( x) and Fix(y) are disjoint. Thus by hypothesis 2, wp = xvp E Sx . If y = x, then wp = xvp = vp E Fix( x); since vp is not in Sx, statement (ii) implies inductively that v, and hence w, is a power of x. 0 We now show that gH n H is either empty, a single point of H, or IS contained in Fix(g). If 9 is not a power of x, then gH C Sx by part (ii) of the claim, so gH n H C asx . Since gH n H is connected, and asx is discrete, gH n H is either empty or a single point in this case. If 9 = xk for some k > 0, the claim shows that gHnH C (Fix(x )nH)USx . We now claim that Fix(x) n H = Fix(x k ) n H, from which it follows that gH n H C Fix(g). The inclusion Fix(x) n H C Fix(x k ) n H is trivial. To show the opposite inclusion, let p be a point of H with xkp = p. By hypothesis 2, xp E Fix(x) U Sx. If xp E Fix(x), we are done. If xp E Sx, then by hypothesis 3, xip is in Sx for all i. Thus p = xkp E asx . It follows that xp and x 2p are also in the asx ; if they were in the interior of Sx, then xkp would be as well. The arc [p, xp] is in H since p and xp are; therefore x[p, xp] = [xp, x 2p] is in SxnH = asx . But asx is a discrete set, so [xp, x 2p] is a single point, i.e. xp = x 2p = p. To see that T is a simplicial tree, consider the minimal subtree K of H which contains the boundary of H. This subtree K is a simplicial tree since the boundary of H is finite. We claim that F2K is a simplicial tree; by minimality this will imply that F2K = T, and hence that T is simplicial. We have shown above that H n gH is either empty, a single point, or equal to Fix( x) n H for x = a or x = b; the last case occurs only if 9 = xk. Thus, if K n gK C H n gH is non-empty, then it is either a single point or one of the two trees Fix(x) n K. This implies that F2K is simplicial; it


195

remains only to show that F2K is connected. For this it suffices to show that K n xK is not empty for x E {a, a-I, b, b- I }. If Fix( x) n H is empty, then x maps a boundary point of H to a boundary point of H, so K n xK is non-empty. Thus we need only show that if Fix( x) n H is non-empty then so is Fix(x) n K. In fact, if x fixes a point of H then it fixes a boundary point. To see this consider a point p of ()Sx and let [p, qj be the bridge from p to Fix( x) n H. The interior points of [p, qj are not fixed by x, so the image of the interior of [p, qj under x must be contained in Sx. Thus q E Sx, proving that a point of ()H is fixed by x. 0 Proposition 3.2. Let F2 x T -+ T be a minimal action with maximal cyclic arc stabilizers. If some primitive element a E F2 has a fixed point, then T is simplicial. Proof. Since a is primitive, there is an element b E F2 such that a and b generate F 2 • We divide the proof into cases, applying Lemma 3.1 in each case to find a fundamental domain H for the action. Case 1. b has a fixed point in T. Since the action is non-trivial, we have Fix(a)nFix(b) = 0. Let H = [p,qj be the bridge from Fix(a) to Fix(b). Let TJp and TJq be the germs at p and q determined by H. We take Sa (resp. Sa-I) to be the union of all open rays in T emanating from p with germ anTJp for some n > 0 (resp. n < 0). Similarly, let S6 (resp. 8 6-,) be the union of all open rays in T emanating from q with germ bnTJq for some n > 0 (resp. n < 0). (See Figure 1)

Figure 1

196


Note that aH n H = {p}, since otherwise a would fix a sub arc of H, and H would not be the bridge from Fix(a) to Fix(b). Similarly, bHnH = {q}. Also, an H n H = {p} and bn H n H = {q} for all n 1= 0, since otherwise the generator of an arc stabilizer would be included into a vertex stabilizer as a proper power, contradicting our hypothesis that the action have maximal cyclic arc stabilizers. This shows that the sets Sa, Sa-l, Sb, Sb-l, and Hare disjoint. The other hypotheses of Lemma 3.1 are easily verified, showing that H is a fundamental domain for the action. Case 2. b is hyperbolic and Fix(a) n Axis(b) = 0. Let (p, qj be the bridge from Fix(a) to Axis(b), let [r, brj be a segment of Axis(b) containing q in its interior, and let H = (p,qj U [r,brj. Let'r/p be the germ at p determined by [p, q], let 'r/r be the germ at r determined by b- 1[r, br],and let 'r/br be the germ at br determined by b[r, brj. We take Sa (resp. Sa-l) to be the union of all open rays in T emanating from p with germ an'r/p for some n > 0 (resp. n < 0). We take Sb (resp. Sb-l) to be the union of all open rays in T emanating from br (resp. r) with germ 'r/br (resp. 'r/r). (See Figure 2)

p

TIp

Axis(b)

q

br

Figure 2 Case 3. b is hyperbolic and Fix(a) n Axis(b) is a point. Let {p} be the intersection of Fix(a) and Axis(b), and let [r,brj be a segment ofAxis(b) with p as its midpoint. If an[p,brj = [p,rj for some n, then ban stabilizes br, and we are in Case 1 with b replaced by ban. If an[p, brj is never equal to [p, r], set H = [r, brj. Let 'r/p,1 be the germ at p determined by [p, brj, and 'r/p,2 be the germ at p determined by [p, rj. Let 'r/r be the germ at r determined by b- 1[br,pj, and 'r/br be the germ at br determined by b[r,pj. We take Sa (resp. Sa-l) to be the union of all open rays in T emanating from p with germ a n'r/p,1 or a n'r/p,2 for some n > 0


197

(resp. n < 0). We take Sb (resp. Sb-1) to be the union of all open rays in T emanating from br (resp. r) with germ 'fIbr (resp. 'fir). (See Figure 3)

1lr

Axis(b) br

Figure 3 Case 4. b is hyperbolic and Fix(a) n Axis(b) is an interval of length less than Ibl. Let [p,q] = Fix(a) n Axis(b), and let H = [r,br] be a segment ofAxis(b) containing [p, q] in its interior. We may assume q E [p, br]. Let 'fIp be the germ at p determined by [p, rJ, and 'fIq be the germ at q determined by [q,br]. Let 'fir be the germ at r determined by b-1[br,q], and 'fIbr be the germ at br determined by b[r,p). We take Sa (resp. Sa-1) to be the union of all open rays in T emanating from p with germ an'flp or from q with germ an'flq for some n > 0 (resp. n < 0). We take Sb (resp. Sb-1) to be the union of all open rays in T emanating from br (resp. r) with germ 'fIbr (resp. 'fir). (See Figure 4)

1lr

•r

"';»p

qf'

1l br

•

br

•

Axis(b)

Figure 4 Case 5. b is hyperbolic and Fix(a) n Axis(b) is an interval of length equal to Ibl. Let H = [p, bp) = Fix(a) n Axis(b). Let 'fIp be the germ at p determined by b- 1 H, and 'fIbp the germ at bp determined by bH. We take Sa (resp. Sa-1) to be the union of all open rays in T emanating from p with germ an'f/p or from bp with germ an'flbp for some n > 0 (resp. n < 0). We take Sb

198


Axis(b)

Figure 5 (resp. 5 b -1) to be the union of all open rays in T emanating from p (resp. bp) with germ 'f/p (resp. T'/bp). (See Figure 5) Case 6. b is hyperbolic and Fix( a) n Axis( b) is an interval of length greater than

Ibl.

Let [p,q) = Axis(b)nFix(a). Then a and b-lab both fix an initial segment of [p, q). But a and b-lab do not commute, so the action has non-cyclic arc stabilizers, contradicting the fact that the action is a limit of free simplicial actions. 0 Definition. Let F2 x T -+ T be a simplicial action. The quotient diagram for the action is the quotient T / F2 with vertices and edges labelled by the isomorphism type of their stabilizers. Two quotient diagrams are isomorphic if there is an isometry between the graphs so that corresponding edges and vertices have the same labels. Given the fundamental domain for a simplicial action, one can easily construct the quotient diagram. Figure 6 shows the quotient diagrams, without specifying the lengths of the edges, in each case of the previous lemma. We will need to be able to determine when two actions which satisfy the hypotheses of the previous proposition are actually isomorphic. Recall that two actions Ul: G x Tl -+ Tl and U2: G X T2 -+ T2 on simplicial R-trees are isomorphic if there is an equivariant isometry

~(a,

b).

Proof. We consider separately the cases when the axes for a and b are disjoint and when they have non-empty intersection.


205

Case 1. AXis(a) n Axis(b) = 0. Let [p, qj be the bridge from Axis( a) to Axis( b), and let H = [p, apj U [p,qj U [q,bqj. Let'T/p be the germ at p in the direction a-l[p,apj, let 'T/ap be the germ at ap in the direction a[p, apj, let 'T/q be the germ at q in the direction b- l [q, bq], and let 'T/bq be the germ at bq in the direction b[q, bqj. We take Sa (resp. Sa-1) to be the union of all open rays in T emanating from ap (resp. p) with germ 'T/ap (resp. 'T/p). We take Sb (resp. Sb-1) to be the union of all open rays in T emanating from bq (resp. q) with germ 'T/bq (resp. 'T/q). We now apply Lemma 3.1 to see that H is a fundamental domain for the action of Fz on T. The action in this case is free and simplicial, with quotient a "barbell" as shown in Figure 11.

Figure 11 Case 2. Axis(a) n Axis(b) -I- 0. Let p be the initial point of the overlap and set H = [p, apj U [p, bpj. Let 'T/p,l be the germ at p in the direction a-l [p, ap], let 'T/ap be the germ at ap in the direction a[p, apj, let 'T/p,z be the germ at p in the direction b- l [p, bpJ, and let 'T/bp be the germ at bp in the direction b[p, bpj. We take Sa (resp. Sa -1) to be the union of all open rays in T emanating from ap (resp. p) with germ 'T/ap (resp. 'T/p,I). We take Sb (resp. Sb-1) to be the union of all open rays in T emanating from bp (resp. p) with germ 'T/bp (resp. 'T/p,z). Lemma 3.1 again applies to show that H is a fundamental domain for the action of Fz on Tj the action is free and simplicial, with quotient a "theta graph" as shown in Figure 12. If ~( a, b) = 0, the theta graph degenerates to a "rose".

o ReInark. If the division process does not terminate, the lengths of the basis elements approach zero, while the difference ~(a, b) -Ial-Ibl remains constant.

206


Figure 12 For the rest of the section we fix a basis {a, b} of F2 • Let a be any action of F2 on an IR-tree T.

Definition. If a and b are hyperbolic and Axis( a) n Axis( b) is a nontrivial arc, we define the slope of a to be ±Iai/lbl, where the sign is positive if and only if the translation directions of a and b agree on the overlap. If the overlap is empty or a point, or if a or b is not hyperbolic, then the slope is zero if

lal < Ibl, infinite if lal > Ibl

and undefined if

lal

=

Ibl·

Note that the division process always terminates if the slope is rational.

Proposition 4.3. Let F2 X Tl -+ Tl and F2 x T2 -+ T2 be actions on IR-trees. Assume that for each action there is a primitive element which has a fixed point and that the quotient diagrams are isomorphic and homeomorphic to a circle (Cases 3, 4, and 5 of Proposition 3.2). The actions are isomozphic if and only if they have the same slope. Proof. By Proposition 3.3, actions of this type are determined up to isomorphism by the conjugacy class of the primitive element of length zero and the quotient diagram. Thus we must show that the slope determines which primitive element has a fixed point. This is clear if the slope is 0 or 00. Otherwise a and b are hyperbolic in each action, so the division process can be started. The slope is rational because both lal and Ibl are integer multiples of the total length of the quotient circle. Thus the division process must terminate. Since the actions are not free, it terminates by producing a primitive element with a fixed point. We claim that for both actions, the division process produce the same primitive element. This follows from the observation that the division process, and the associated sequence of basis changes (Nielsen transformations) are completely determined by the slope. The sign of the slope determines


207

whether b should be replaced by b- I at the beginning of the process. The generators a and b must be interchanged if and only if the absolute value of the slope is less than 1. From that point onward the process is determined by the ratio lai/ibl.

0

Next we use the division process to classify actions with overlap which is

+ Ibl. If ~( a, b) < lal + Ibl,

either shorter or longer than lal Proposition 4.4.

the action is simplicial.

Proof. By Proposition 4.2, it suffices to show that the division process

always terminates if ~(a, b) < lal + Ibl. Assume that the division process does not terminate. Then we have a sequence {ai, b;} of bases for F2 such that the lengths of ai and bi approach zero. But ~(ai, bi) - lad - Ibd = ~(a, b) - lal - Ibl is constant and less than zero for all i. This is a contrao diction, since ~(ai' bi) is positive for all i.

Proposition 4.5. If ~(a, b) > lal + Ibl, there is an arc stabilizer containing a free subgroup of rank two in F 2 • Proof. Since lal+ Ibl < ~(a, b), we can apply the division process to obtain a sequence {ai, bd of bases; since ~(ai, b;)-Iai I-Ibd remains (constant and) positive, the only way the process can terminate is if bn is elliptic for some n. In this case, the basis {an-I,bn-d has lan-II = Ibn-II < ~~(an_l,bn-d. Now note that b~':'l an-I and the commutator [a~':'I' b~':'ll each fix an initial segment of the overlap Axis(an_d n Axis(bn-d. Since these two elements generate a free group of rank two, we have a non-cyclic arc stabilizer. If lai/lbi is not rational, the division process does not terminate. In this case the lengths of ai and bi approach zero, while ~i = ~(ai, bi) approaches

a positive constant. Thus, for n sufficiently large, we have Ian I+2lb n I < ~n. Since lanl + Ibn I < ~n' the commutator [an, bnl fixes an initial segment of the overlap Axis( an)nAxis(bn), and since Ian 1+21bn I < ~n' the commutator [an, b;] fixes a (smaller) initial segment of the same overlap. Since these two commutators generate a free group of rank two, we again have a non-cyclic arc stabilizer. 0

208


§5. Critical overlap: Ll(a, b) = lal

+ Ibl

Throughout this section we let F2 x T ---+ T be a minimal action and we assume that F2 is generated by hyperbolic elements a and b, where the axes of a and b meet in a segment of length .Ll = lal + Ibl. Note that laba -1 b -11 = 0, so this action is not in X. As in section 4, we define the slope of the action to be the extended real number m = ±Iai/lbl,where the sign is positive if and only if the translation directions of a and b agree on the overlap of their axes. For convenience we will assume that for the action on T the translation directions agree; there is no loss of generality since this condition holds after replacing b by b- 1 • We begin by giving a simple construction, for each m, of a tree GTm with a natural F 2 -action. We then show that T is equivariantly isometric to GTm. The tree GTm is the dual tree to the measured foliation of the punctured torus by lines of slope m. Specifically, it is constructed as the space of leaves of a PL measured foliation of a simplicial complex E, which is equivariantly homeomorphic to the universal cover of the punctured torus with a countable number of points added at infinity. The general theory of such measured foliations and their relation to actions on IR-trees is developed in [5]. While our construction is self-contained, we use results from [5] to show that the space which we construct is an IR-tree. IT m is rational then the division process described in the last section, when applied to {a, b}, terminates at a basis containing one elliptic generator which fixes a fundamental domain for the action of the other generator on its axis. Such actions are described in Case 5 of Proposition 3.2: the action is simplicial, the quotient graph is a circle with one vertex, the stabilizer of the edge is infinite cyclic, and the vertex stabilizer is free of rank 2. The tree GTm, for m rational, is the dual tree to a foliation of the punctured torus by parallel simple closed curves, which is a simplicial tree. One can check that Tis equivariantly isometric to GTm . However, the details of the argument in the rational case differ from those in the irrational case; since the rational case can be handled by the methods of section 3, we assume for the rest of this section that m is irrational. Let S be the unit square [0,1] x [0,1] foliated by line segments of slope m. To construct the simplicial complex E, we take a copy wS of S for each reduced word w in the generators a and b of F2 , and glue them along their edges according to:


209

There is a natural left action of F2 on~. The foliations on the squares wS patch together to give a PL foliation Fm of~. There is a transverse measure on Fm which is inherited from the transverse measure on the square, in which the length of a monotone arc transverse to the leaves is the length of its projection onto the direction of slope -l. Deleting the corner points of S and identifying opposite sides gives a punctured torus. The complement in ~ of the set of corner points of the squares wS can be identified in a natural way with the universal cover of this punctured torus. We also have a natural identification of the universal abelian cover of this torus with 1R2 - Z2, where b acts by shifting downward and a acts by shifting to the right. The covering projection from the universal cover to the universal abelian cover extends to a map 71": ~ -+ 1R2 which sends S to [0,1] X [0,1]. A leaf of Fm meets a square wS in a line segment. This gives each leaf a natural simplicial structure in which these line segments are the edges. Following Gillet and Shalen, a leaf of Fm is called classical if does not contain a corner of any translate of S, and singular if it does contain a corner. A classical leaf is homeomorphic to the real line, while a singular leaf is homeomorphic to a simplicial tree, with a vertex of infinite valence coming from each corner which it contains. Note that the map 71": ~ -+ 1R2 takes leaves of Fm to lines of slope m. It follows that a singular leaf has only one vertex which is not bivalent, since a line of irrational slope passes through at most one point of the integer lattice in 1R2. In the language of [5], ~ is a simply-connected uniform IR-foliated surface with points at infinity. It is proved in [5, Theorem 5.20 and Proposition 5.25] that the leaf space of such a foliation is an IR-tree. We define GTm to be the leaf space of F m. Thus the underlying topological space of GTm is the quotient ~/ "', where p '" q if and only if p and q lie on the same leaf of the foliation. The metric on GTm is induced by the transverse measure on Fm; the distance between two leaves is the measure of an arc joining them which meets each leaf of :Fm in a connected set. We will call a point of GTm a vertex point if it corresponds to a singular leaf of the foliation, and an edge point if it corresponds to a classical leaf. This agrees with the usual notion of a vertex point as a point from which more than two germs of arcs emanate. An edge point has exactly two germs of arcs emanating from it. This implies that if an edge point is

210


contained in the interior of two arcs then it is contained in the interior of their intersection. The next step is to define an equivariant map

2(al

+ ... + an), l\a,b

is contractible.

First observe that if Ek,s is a lattice cube, then 7I"o:(Ek,S) is contained in [a, b) if and only if k . a E [a, b) and (b - k . a) ~ EiES ai. Let m = (a + b)/2, and let H be the hyperplane in l\a,b which is the inverse image of m under the map 71"0:' Since 7I"a(Ek) is a closed interval of length E~=l ai, which by the hypothesis of the theorem is less than or equal to (b - m), it follows that any cube having a nonempty intersection with H lies in Ka,b. Therefore, since nn is the union of the E k , He Ka,b. Since H is contractible, it will suffice to show that H is a deformation retract of Ka,b' Set La,b = Ka,b n 71";;1 ([a, m». By symmetry it will suffice to show that H is a deformation retract of La,b' Now consider the flow on PROOF:

240

H. Gillet and P. Shalen

R n with gradient vector aj we shall see that this flow provides the necessary

defonnation. We claim that if x E La,b then the line segment

e=

{x + talt

~

(m - x . a)/(a .

an

is contained in La,b' To see this, it suffices to show that any y E e lies in Ka,b. Observe that if k E zn, then x E Ek if and only if, for all i = 1, ... ,n, ki ~ Xi ~ k i + 1. In particular, if [x) is the vector in zn with i-th coordinate the greatest integer [Xi) less than or equal to Xi, for i = 1, ... ,n, then x E E[x]' Since x E La,b, it follows that 1I'0/([x)) E [a, m). For y E e, since the 0:; are all positive, [Yi) ~ [Xi) for all i = 1, ... ,n. Hence a ~ 1I'0/([x)) ~ 1I'0/([y)) ~ m. Thus 11'0/ maps [y) into [a,m), i.e., [y) E La,b' It follows by the hypothesis on b - a, that if k E zn is mapped into [a, m) by 11'0/, i.e., if k E La,b, then the entire lattice cube Ek is contained in Ka,b. Therefore E[y], which contains y, is contained in Ka,b. 0 §3. Construction of the complex. Let T be a A-tree. Given a fixed basis A of A, in 3.1 we define a family of cubes which are fitted together in 3.2 to form a cubical C-W complex K(T,A). In 3.3 we define a continuous map 6n+1 : K(T,A(n!») -+ K(T, A«n H )!»), where A(m) is the basis obtained from A by dividing the elements of A by m. In 3.4 we construct the complex K of theorem 1 as the infinite mapping telescope of the sequence of maps {6 n }, and in 3.5 we show that K is contractible. 3.1 Bricks. Given an A-segment There is a map

0'

in T with vertex set S, we write

Ecr

for the cube

I:(cr).

1I'IT : Ecr -+

f

I--t

RT

L

af(a) E RO' C RT

O/EA(cr)

Recall here that Ecr = {flf : A(O') -+ Is} and that f(a) is a function S -+ [0,1). Hence f determines a function 9 = L:O/EA(cr) af( a) : S -+ [0,10'11 such that g(x) + g(y) = 10'1, i.e., an element of !s(IO'D. By Lemma 2.5.1, 9 corresponds canonically to a point 1I'cr(f) in RO'. Clearly 1I'cr is continuous.

Cohomological Dimension of Groups Acting on JR.-Trees THEOREM

241

3.1.1. Suppose that a is an A-segment.

i) There is an isomorphism Fu between the partially ordered set of sub-A-segments of a and the partially ordered set of cells in the CW complex Eu. ii) If 7 is a sub-A-segment of a, there is a canonical map of cubical cell complexes fU,T : ET -+ Eu which is an isomorphism onto Fu(7).

iii) If A iv) If 7

of basic segments, then fU,TfT,>. ~ a, we have an equality of maps from ET to RT: ~

7

~ a is a chain

7r a fu,r

v) If A and

7

=

=

fu,>..

7fT

are sub-A-segments of a, then either Fu(7) n Fu(A)

empty, or there is a sub-A-segment A 1\

Furthermore

7

7 ~

a such that

1\ A is the maximal sub-A-segment

a sub-A-segment of both A and

1S

of a which is also

7.

PROOF: Let us write {x,y} for the endpoints of a. To prove i) observe that by 2.5 and 2.7.1 both sets are bijective to the set offunctions A( a) -+ {x,y,I{x,y}}' and that under these bijections both sets induce the same partial order on this set of functions. To prove ii, let us write {x', y'} for the endpoints of 7, with x' the endpoint closest to x. Then ET = I1EA(T) I{XI,y/} and Fu(7) is canonically homeomorphic to I1EA(T) I{x,y}' We may therefore define the map fU,T to be the product of A(7) copies of the unique isometry I{XI,y/} -+ I{x,y} which identifies x with x' and y with y'. Since the natural isometry I{XIl,yll} ~ I{x,y} is the composition of the natural isometries I{XIl,yll} ~ I{XI,y/} ~ I{x,y}, assertion iii) follows from the definition of the maps fu,>., fT,>., fU,T' For the proof of iv), we start by observing that by lemma 2.5.1, for any a we may regard 7ru as a map Eu -+ I{x,y}(lai). If we interpret f E EfT as a map f: A(a) -+ I{x,y}, then 7r fT (J) = LEA(fT) oJ(O'). Now if 7 -( a and g E En then

f.,,(g)(a)(x)

~ { ~(a)(x')

if [7)u(O')

=

x

if a E A(7) if [7)u(O')

=Y

242

H. Gillet and P. Shalen

Hence 7r

u(fu,r(g))(X) =

7r

r(g)(x)

L

+

a

[r] .. (a)=x

=

7r

r (g)(x)

+ d(x, x')

But this is precisely the image of 9 E RT under the inclusion RT C RI7. (Note that a point in Ro- is determined uniquely by its distance from either of the endpoints x or y.) Finally we turn to the proof of part v). Recall that

Fu(T) =

II

[TJu(a).

aEA(u)

Hence

Fu(T) n Fu(A)

=

II

[TJu(a) n [AJu(a)

aEA(u)

This is empty unless [TJu(a) n [AJu(a) is non-empty for all a. Define '!/J(a) = [TJu(a) n [AJu(a) C I{x,y}, for a E A(I7). Then Fu(T) n Fu(A) = TIaEA(u) '!/J(a) is a cell in E u , and hence equals Fu(T /\ A) where T /\ A is the unique sub-A-segment of 17 with [T /\ I7Ju = 'I{-'. It follows from part i) that T /\ A is the meet of T and A in the partially ordered set of sub basic segments of 17. 0 REMARK: In the theorem above we do not assume that the segment 0is non-degenerate; though the cube Eu will be zero dimensional if 17 is degenerate, the results still hold.

3.2 Mortar. Let Y = lluEB Eu. This is a CW complex. We define a relation on Y as follows: x '" y if x E E u , y E E r , and there exists Z E E)., with A ::S T and A ::S 17, such that fu,).(z) = x and fr,).(z) = y. LEMMA 3.2.1. '" is an equivalence relation. PROOF: The relation is clearly reflexive and symmetric. Now suppose that

x E Eu, y E E r , and Z E E,.., with x '" y and y '" z. Then there exists pEE). with A ::S 17 and A ::S T such that fu,).(p) = x and fr,).(P) = y. Similarily there exists q E E
n, we have the following definitions. DEFINITION 2.3: For any

u, v

E

X(m - 1),

U

and v are called (m - 1,p)-

equivalent if (1)

U

and v are m-equivalent.

(2) Suppose that

U

=

Uo

~

(m - 1 )-equivalent, then

•••

~

Ui-1, Ui

Ut

=

v,

E U}=1 C'/.

if

Ui-1

and

Ui

are not

and v are (m - 1,p)-equivalent, C E 7l"o(X(m) - {u}), C' E 7l"o(X(m) - {v}), CnX(m -1) =10, and C' nX(m1) =I 0; then C and C' are called (m - 1,p)-equivalent if DEFINITION 2.4: Suppose that

U

(1) C and C' are m-equivalent. (2) Suppose that C = Co ~ ... ~ C t = C'; if C;-1 and C; are not (m - 1 )-equivalent, then Ui-1, Ui E U}=1 Cj'. (Recall definition 2.2, Cj E 7l"o(X(mj) - {Uj}),j = 1, ... ,k). 2.5. (1) The (m-l, l)-relation is the same as the (m-1)-relation. (2) The (m - 1, k )-relation is the same as the m-relation.

LEMMA

PROOF:

CLAIM 1: IT U and v are (m - 1, 1)-equivalent, then U and v are (m - 1)-

equivalent. By definition 2.3, U and v are m-equivalent. Suppose that U = Uo ~ ... ~ Ut = v; if claim 1 is not true, suppose that Ui-1 and Ui are not (m - 1)-equivalent. Then by (2) of definition 2.3, U;-1, Ui E q'. Thus Ui = Ui-1Q'; E q'ncf'Q'; =10. But this is impossible, since Cf' E 7l"o(X(m1) - X(m - n)) and Q'; E Am - {I} imply that Cf' n q'Q'; = 0. This proves claim 1. CLAIM 2: IT

C and C' are (m-1, 1)-equivalent, then C and C' are (m-1)-

equivalent. The proof is similar to that of claim 1 and is left to the reader. Note that (m - I)-equivalence implies (m - 1, 1)-equivalence and (m 1, k )-equivalence implies m-equivalence.

266

R. Jiang

3: If u,v E X(m -1) and are (m - 1, k)-equivalent. CLAIM

U

and v are m-equivalent, then

U

and v

Suppose that U = Uo ~ ... ~ Ut = v. If Ui-1 and Ui are (m - 1)equivalent for all i, then U and v are (m - 1)-equivalent, and they are also (m -1, k)-equivalent. We prove that Ui-l, Ui E Uj=l C'/ whenever Ui-1 and Ui are not (m - 1)-equivalent. By definition 2.3, this implies that U and v are (m - 1, k)-equivalent. First we prove that Qi E Am. Recall definition 2.1 Ui-1Qi = Ui, Qi E Ami and mi :S m. If mi :S m - 1, then by definition 2.1, Ui-1 and Ui are (m -1)-equivalent, which contradicts the assumption above. Thus mi = m and Qi E Am. Secondly, we prove that Ui-1,Ui E X(m - 1) - X(mn). Since Qi E Am and Ui-1Qi = Ui, Ui-1 E X(m - n) if and only if Ui E X(m - n) = X(m - n)Am. If Ui-1,Ui E X(m - n), then they are (m - n)-equivalent, which gives a contradiction as before. So Ui-1, Ui E X(m - 1) - X(m - n). Finally we prove that Ui-1,UiE Uj=lCj'. Since Ui-1,Ui E X(m - 1) - X(m - n), there are C(Ui-I),C(Ui) E 7ro(X(m1) - X(m - n)) such that Ui-1 E C(Ui-1),Ui E C(Ui). Since Ui-1Qi = Ui E C( Ui-1 )Qi n C( Ui) =I- 0, C( Ui-1) and C( Ui) are Am-equivalent. Thus C(Ui-1),C(Ui) E {q', ... ,Cn. Therefore Ui-1,Ui E Uj=lCj'. So U and v are (m - 1, k)-equivalent. 4: If C E 7ro(X(m) - {u}), C' E 7ro(X(m) - {v}), C n X(m -1) =I- 1) =I- 0, C and C' are m-equivalent, then C and C' are (m - 1, k )-equivalent. CLAIM

0, C' n X(m

Suppose C = Co ~ ... ~ C t = C'. We shall prove that Ui-1, Ui E Ci-1 and C i are not (m -I)-equivalent. This will imply that C and C' are (m - 1, k)-equivalent. (Here we use the notation of definition 2.4.) Suppose that Ci-1 and Ci are not (m - 1)-equivalent for some i. Arguing as before, Qi E Am.

Uj=l Cj' whenever

Consider first the case that Ui-1, Ui E X(m - n). Since Ci-1 Q i n Ci =I-

0, we have Ci-1 n X(m - n) = 0 if and only if Ci n X(m - n) = 0. If Ci-1 n X(m - n) =I- 0 and Ci n X(m - n) =I- 0 then C i - 1Q i n Ci =I- 0 implies that Ci-1 and Ci are (m - n )-equivalent, contrary to the fact that Ci-1 and Ci are not (m -1)-equivalent. Hence Ci-1 nX(m -n) = 0 and Ci nX(mn) = 0. Recall definition 2.2, Ci-1 n X(mi-I) =I- 0 and Ci-1 n X(mi) =I- 0. By claim 1 of lemma 2.4, we may assume that Qi-1,Qi+1 ct. Am. Since Qi-1 E A mi_ u mi-1 =I- m. Thus mi-1 :S m - 1 and Ci-1 n X(m - 1) =I- 0.

Branch Points and Free Actions on JR- 'frees

267

By the same reasoning, Ci nX(m -1) =I- 0. Since Ci-1ai n Ci =I- 0, we may assume that there are C',C" E 7ro(X(m - 1) - X(m - n)) such that C' and C" are Am-equivalent and C' n C;-l =I- 0, C" n Ci =I- 0. So Ui-1, Ui E

(C' U C") C Uj=l OJ'.

Consider next the case that Ui-1, Ui ~ X(m- n). Then Ui-1, Ui E X(m1) - X(m - n). We may suppose that U;-l E C(Ui-1),Ui E C(Ui) for some C(Ui-I), C(Ui) E 7ro(X(m - 1) - X(m - n)), then ui-1ai = Ui E C(ui-l)ai n C(Ui) =I- 0. Therefore C(Ui-l) and C(Ui) are Am-equivalent, and Ui-l, Ui E (C(ui-d U C(Ui)) c Uj=lOj'. This proves claim 4, and lemma 2.2 follows immediately. Let [U](m-l,p) be the (m -l,p)-class containing u. Write ind[u](m_l,p) is equal to the cardinality of

{C E 7ro(X(m)-{v})lv E [u](m_1,p),CnX(m-1) =I- 0}/(m-1,p)-relation. Let

f(m - 1,p) = 2:)ind[ul 1 for all t. {Ui .. ui,+d is a bridge between [Vt] and [Vt+1], ai,+l E Am. Since ii+1 it > 1 is minimal for all t, we may assume that ai" lXi,+2 rJ. Am and lXi,+2lXi,+3 ... ai'+l is a reduced word of G for all t. Then this contradicts a2 ... lXin+1 = 1. This proves claim 1. We now proceed to give a proof of the inequality of the lemma. CASE

1: [u], [v] are not adjacent, 6 = O.

[u] E V, [v] E U - V, [u] and [v] are not adjacent. Recall the choice of [u] and [v], we know that [v] is not adjacent to any [u'] E V. Furthermore, [v] n [U']em-1,p) =

0 for all

u' E V, otherwise we can find [u'] E V and

[v] E U - V such that [u'] and [v] are adjacent, which contradicts the choice of [u] and [v]. So [U'](m-1,p),V

=

[U'](m_l,p)nV

= [U']em-1,p)n(Vu{[v]}) =

[U'hm-1,p),VU{[v]}

for all u' E V. Therefore ind[u']em-1,p), v = ind[u'](m-1,p),vU{[v]}' On the other hand, [Vhm-1,p),VU{[v]} = [v] = [V](m-1,p-1)' Since (m I,p - I)-equivalence implies (m - I,p)-equivalence, ind[v]e m - 1,p),Vu{[v]) ::; ind[v](m_1,p_1)' Then f(V U ([v]}) =

(ind[u](m-1,p),VU{[v]) - 2) [U]( ... -l,1'), Vu{ [v]}

EeVu{[v]} )/(m-1 ,p)-relation (ind[u'hm_1,p),V - 2)

= [U']( ... -l,1'), v

::;f(V) CASE

+ (ind[v]e m- 1,p),VU{[v]) -

2)

EV/(m-1,p)-relation

+ (ind[v](m_1,p_1) -

2: b([u], [v])

=

2).

2 - t, 6 = 0, [u] and [v] are adjacent.

Let {u',v'} be a bridge between [u] and [v] and u' E definition of b,

C;'.

Recall the

Branch Points and Free Actions on JR.-Trees

273

Suppose that

then there exist (3j E Am, j = 1, ... , t such that u' (3j E U~:tc~'. Furthermore, there exist Cj E 7ro(X(m -1) - {u'(3j}) such that Cj(3j n Cj =1= 0, j = 1,'" , t. By claim 2 of lemma 2.2, Cj and Cj are not m-equivalent for i =1= j. Since Cj(3j n Cj =1= 0 and u' E C;, we have that Cj and Cj are (m - 1,p)-equivalent but not (m - I,p - I)-equivalent. Therefore C: and Cj are not m-equivalent for i =1= j, neither are Ci and Cj. So Card( {C1 ,'"

,

Ct, C~,,·· , CD/(m -I,p -1) - relation)

Card( {C 1 ,'"

,

= 2t,

Ct, c~,··· , CD/(m -I,p) - relation) = t.

Without loss of generality, we may assume that u' E [u], v' E [v]. For any [w] E V, if [Whm-1,p) n [v] = 0, then [Whm-1,p),V = [W](m-1,p),VU{[v]}' Therefore ind[whm-1,p),V = ind[w](m-1,p),vu{[v]}' If [W](m-1,p) n [v] =1= 0, then ind[whm-1,p),vu{[v]} ::; ind[whm-1,p),v + ind[v] - t. The reason for the inequality is the following. The contribution of [v] to f(V U [v]) is less than or equal to ind[v] - t, since at least t representatives (for example

CL .. · ,CD

of

{C E 7ro(X(m) - {w} )Iw E [v], C nX(m -1) =1= 0}/(m -I,p -1) - relation are (m - I,p)-equivalent to t representatives (for example C 1 , ... , Ct) of

{C E 7ro(X(m) - {w'}1 w'

E [W](m-1,p)

n V, C n X(m -1)

=1= 0}/(m - I,p) - relation.

Thus

f(V U [v]) ::; f(V)

+ ind[v] -

2 + 2 - t = f(V)

+ (ind[v]

- 2)

+ b([u], [v]),

and the result follows. CASE

3: {)

=

1.

Then C; and C;'_l are Am-equivalent and wp and v are (m - I,p - 1)equivalent. Note that and 1 are not (m - I,p - I)-equivalent, but they are (m - I,p)-equivalent.

C;'

C:_

R. Jiang

274

If C~ and C~_l are not (m-I,p-I)-equivalent to any Ci, 0:, i = 1,··· , t, then

Card( {Cl

,··· ,

Card( {Cl

Ct, Cf,··· Ci, c;', C~_d/(m -I,p -1) - relation) = 2t + 2,

, ... ,

Ct, Cf, ... , Ci, c~, C~_d/(m - I,p) - relation) ::; t

+ 1.

Therefore the contribution of [v] to f(V U {[v]}) is less than or equal to ind[v] - ((2t

+ 2) -

(t

+ 1)) =

ind[v] - (t

+ 1).

If C;' is (m -I,p -I)-equivalent to some Ci, then C~_l cannot be (mI,p - I)-equivalent to any Cj or OJ, i "# j. Otherwise Ci and Cj will be (m - I,p)-equivalent, which contradicts claim 2 of lemma 2.2. If C~_l is (m -I,p-I)-equivalent to Ci, then Ci and 0: are (m-I,p-I)-equivalent, which also gives a contradiction. Thus Card( {Cl , ... , Ct, CL ... , ci, c;', C;'_l }/(m -I,p -1) - relation) = 2t + 1. Cl

, ... ,

C t represent t distinct (m - I,p)-equivalence classes, so

Card( {Cl

, ... ,

Ct, CL ... , ci, c;', C~_d/(m -I,p) - relation) = t.

Then [v] brings in at most ind[v] - ((2t + 1) - t) (m - 1, p )-equivalence classes of directions, thus

=

ind[v] - t - 1 distinct

+ ind[v] - t -

1

= f(V)

+ (ind[v]

+ (2 - t) -

= f(V)

+ (ind[v] -

f(V U {[v]}) ::; f(V)

- 2) 2)

1

+ b([u], [v]) -

8

This proves lemma 2.8. LEMMA

2.9.

f(m - I,p) ::; f(m -I,p -1)

+

b([u], [v]) - 1. ([u],[v])EE(U)

PROOF:

Beginning with V =

f(U)::;

L [u]EU

0 , we

(ind[u]- 2)

get

+

L ([u],[v])EE(U)

b([u], [v]) - 1

Branch Points and Free Actions on R-Trees

275

by applying lemma 2.8 repeatedly. CLAIM 1:

L

(ind[u)- 2) = f(m - l,p - 1).

[ujEU Part (2) of definition 2.7 points out that ind[u) = 2 for any [u) E V(r(p ))V(U). Consequently

L

(ind[u)- 2)

[ujEU

= (ind[u)- 2) = f(m - l,p - 1)

[ujEX(m-1)/(m-1,p-1)-relation CLAIM 2: If [v) f/:. U, [u) and [v) are adjacent, then the 2, then [U)(m-1,p) ind[uhm-1,p) = ind[uhm-1,p),u. This will yield

feU)

L

=

nU

(ind[u)(m_1,p),U -

=I- 0 and

2)

[uj (m-l,p),U EU/( m-1 ,p) -relation

L

[uj (m-l,p) EX(m-1)/( m-1,p )-relation = f(m -l,p)

(ind[uhm-1,p) - 2)

276

R.Jiang

Suppose ind[u)(m_l,p) > 2 and 3v E [Uhm-l,p) such that ind[vhm_l,p_l) > 2, then by (2) of definition 2.7, [Vhm-l,p-l) E U , thus [U)(m-l,p) n U =1= 0. Suppose ind[uhm_l,p) > 2 and ind[vhm-l,p-l) = 2 for all v E [U)(m-l,p). By claim 2, this can only happen when 3[V](m-l,p-l) E [U](m-l,p) n U. So [Uhm-l,p) n U =1= 0. Claim 2 says that if [Vhm-l,p-l) ~ U, then [V)(m-l,p-l) does not bring in any new (m -l,p)-class representatives to ind[u](m_l,p). Thus ind[u](m_l,p) is equal to the cardinality of

{C E 7ro(X(m) - {v})1

vE

[U](m-l,p),

C n X(m -1) =1= 0}/(m - 1,p) - relation)

which is equal to the cardinality of

{C E 7ro(X(m) - {v})1

v E

[U](m-l,p)

n U, C n X(m -1)

=1=

0}/(m - 1,p) - relation)

which is equal to ind[uhm-l,p),U which is a consequence of claim 2. This proves claim 3. Combining

f(U):::;

L

(ind[u)- 2)

+

[u]EU

L

b([u], [v)) - 1,

([u],[v])EE(U)

claim 1 and claim 3 gives lemma 2.9. To complete the proof of f(m -l,p) :::; f(m -l,p -1), we need to show that

L

b([u], [v)) :::; l.

([u],[v])EE(U)

LEMMA

2.10.

L([u],[v])EE(U)

b([u], [v)) = LU'EC;,(2 - f3(u'».

PROOF: We use the notations of definition 2.6. First we show that 'l/J is injective. If 'l/J([Ui],[Vi)) = u',i = 1,2, we need to prove that ([uI],[vd) and ([U2), [V2)) are the same as non-ordered pairs. Let {u',vD be a bridge between [u;] and [Vi], u' E u' E lUi], vl E [Vi], i = 1,2. So lUll = [U2]. It is enough to prove that [vI) = [V2]. Since {u', vn is a bridge and

c;,

Branch Points and Free Actions on R-Trees

277

p - 1 c,,· C" p' we h ave t h at u' OIj = Vj, f or A OIj E d m an ' Vj E u s=i s, J = 1 , 2 . 1 p ' -1 , -1 A " U C" Th ' ' are Then V1011 012 = V2, 011 012 E m, V1,V2 E s=i s· US V1 an dV2 (m - I,p - I)-equivalent; therefore [V1] = [V2], and 'ljJ is injective. Next we prove that 'ljJ is surjective. For any u' E C; n (uj:;Cj'OIj), there exists a E Am such that U'OI E uj:;Cj'. Then u' and U'OI are not (m-I,p1 )-equivalent. Therefore [u'] and [u' 01] are adjacent and 'ljJ([u'], [u' 01]) = u'. Hence 'ljJ is a bijection between E(r(p)) and C;' n (uj:;Cj'OIj). By the

u, E

definition of U, bee) = 0 for all e E E(r(p)) - E(U). Thus

L

b(e)=

L

bee)

L

(2 - IN( e))

eEE(r(p))

eEE(U)

eEE(r(p))

L

u' EC~'n(uf;;;;)

=

L

cr

(2-f3(u')) (1))

(2 - f3( u')).

u'EC;'

This proves lemma 2.10. LEMMA 2.11. L:u'ECII(2 - f3(u')) p

S 1.

PROOF: Since'ljJ is bijective, then lemma 2.7 gives Card{u' E C;'If3(u') -:f. 2} < (Xl. We prove lemma 2.11 by induction on s = Card{u' E C;If3(u') = I}. The case s = 1 is trivial since f3(u') 2: 1. For s > 1, choose u E C;' such

-:f. 2, then v E c; n (Uj:;Cj'OIj), uj:; Cj'. Recall that C;: = C; u {wp }.

that f3(u) = 1. For any v E C;', if f3(v) and there is a E Am such that va E If u E (wp, v), then u

E (w p , v ) C Cp" n (p-1 Uj=i C j" 01).) ,

thus f3( u) 2: 2, a contradiction. So Y( wp, u, v) Thus there is U1 E (wp, u) such that

d(U1,U) = min{d(Y(wp,u,v),u) where d is the distance function on X. 'Tro(C;' - {uI}) such that u E C2 and put

-:f. u for u -:f. v

and f3( v)

f=. 2.

f=. 01f3(v) -:f. 2},

Note that U1 -:f. u. Let C2 E C1 = C;' - C 2 • Replacing by

C;

R. Jiang

278

C1 in the definition of f3, we get f3 C l. f3 C l(V) = f3(v) for all v E C1 - {Ul}. f3c l (U1) = f3( uI) - 1; to see this, note that if Ul E (w p, u ) C Gp" n (PUj=i1C" j O!j ), then f3(Ul) =11. Ul

= Y(wp,u,v) for some v =I u and f3(v)

=12. Since

itfollows that f3(Ul) > 2. Since (ut, ulnC1 = 0, we have ,8cl (Ul) = f3(Ul)-1. Note that f3( v) = 2 for all v E C2 - { u}, since {v E C~If3( v) =l2}-{u} C C1 • Combining the above facts gives

L

(2 - ,8(v)) =

L

(2 - f3(v))

+2 -

f3(U1)

+2 -

f3(u)

vEC1-{Ul}

= L(2-f3cl (v)) VECl

Since Card{u' E C1 1f3cl (u,) = I} then the hypothesis implies that 2.11.

< Card{u' E C;If3(u')

L:vECl (2-,8c l

(v))

~

= I},

1, which gives lemma

Combining lemmas 2.9,2.10 and 2.11 gives f(m-l,p) which completes the proof of theorem 1 for m > n.

~

f(m-l,p-l),

For m ~ n, replace X(m - n) by X m, we can define Am-equivalence relation on 1I"o(X(m-l)-Xm). Lemma 2.4 is still true. Define C~ 1 ~ p ~ k as before. Definition2.3 is replaced by the following. DEFINITION 2.3*.: For any u, v E X(m -1), u and v are called (m -1,p)equivalent if

(1) u and v are m-equivalent. al a. IT Ui-l and (2) Suppose t h at u = Uo --+ ... --+ Ut = v. (m - 1 )-equivalent, then Ui-l, Ui E U}=1 C'/.

Ui

are not

Branch Points and Free Actions on JR- Trees

279

The rest of the proof needs only a little change. This completes the proof of theorem 1.

§3. Groups acting freely on R-trees. In this section we prove theorem 2. Lemma 3.1 is needed for the proof of theorem 2, although in a slightly more general form. Recall that X a , is the translation axis of at. LEMMA 3.1. Let G be a group acting on an R-tree X. Let al, ... , ak be hyperbolic elements ofG (lx(ai) > 0) and let nl, ... ,nk be non-zero integers. For 1 SiS j S k put gij = a~i ... a'ji , and g = glk. Let Y denote the subtree of X spanned by X a1 , ... , X ak . Assume that these axes are not all the same, and let t (1 S t S k) be such that X a1 = ... = X a, =I X a.+1. Assume that BP(Y)gij n BP(Y) = 0 for t

< i S j S k. Then g =11. In fact we may choose

Vl

E X a• such that

Vlglt f/:. Xa'+l and then Vlg ~ Y.

PROOF: Choose Vl E X a• such that vlglt f/:. X a.+1. Let gj = glj' Let vp be the end point of [Vlgp-l, Xa p ] on Xa p ' We prove the following by induction onp. (a)p vlgp f/:.Yfort+1:::;p. (a)~ vlgp f/:. X ap+1 for t :::; p. (b)p vp E BP(Y) U BP(Y)gi,p-l for t

+ 1 :::; i,p.

(a)~ follows from the choice of Vl. Since Vlgt E X a• and Vlgt f/:. Xa'+l' we see that Vt+l is the end point of the bridge [Xa., X a.+1] if X a• n X a.+ 1 is empty, otherwise Vt+l is an end point of X a• n X a.+1. Thus Vt+l E BP(Y), which yields (b)t+l'

CLAIM 1: (a)~ and (b)p+l imply (a)p+l for t S p S k-1. By (a)~, vlgp f/:. X ap+u thus vlgp+l = vlgpa;.+1 1 f/:. X ap +1' Vp+l is the end point of [Vlgp , X ap+1] on X ap+l l in particular, Vp+l E X ap+1 and vp+la;.+1' E X ap+1' Therefore vlgpH =I vp+la;.+1'. By the definition of Vp+l, [Vlgp,Xap+1] = [Vlgp,Vp+l], hence np+l] = [vlgp, Vp+l ]ap+ np+l = [Vlgp+l, vp+lap+l l [Vlgp,Xap+1]a;.+1' = [Vlgp+l,Xap+1]'

280

R. Jiang

Then the assumption BP(Y)gij n BP(Y) a contradiction. CLAIM

2: (b)p implies (b)p+I for t

= 0 implies vp+la;.ti' fj. BP(Y),

+ 1 ::; p

::; k - 1.

Recall that [VIgp, Xap+J = [VIgp, Vp+l] and [VIgp, Xa p] = [VIgp, vpa;P]. By (b)p, vp E BP(Y) U BP(Y)gi,p-l, so vpa;P E BP(Y)a;P U BP(Y)gip. Then the assumption BP(Y)gij n BP(Y) = 0 implies vpa;P fj. BP(Y). If Xa p n X ap +1 = 0, then one of Vp+1 and vpa;P is in BP(Y). So Vp+1 E BP(Y). Suppose Xa p n X ap +1 i- 0. If vpa;P E [VIgp, Vp+l], then

[VIgp,Vp+l]

= [VIgp,Xap+ = [VIgp,vpa;p] 1]

U [vpa;p,Xap+1 ] =

[VIgp, vpa;p] U [vpa;p, Vp+I]. If vpa;P fj. Xa p n Xap+u then vpa;P i- Vp+l, thus Vp+1 is an end point of Xa p n X ap+1 • Hence Vp+1 E BP(Y). If vpa;P E Xa p n X ap+1 ' then vpa;P = Vp+l. By (b)p, vp E BP(Y) U BP(Y)gi,p-l, thus

If vpa;P fj. [VIgp, Vp+I] , then vpa;P is an end point of Xa p n X ap+1 ' thus vpa;P E BP(Y), a contradiction. This proves (b)p+l. Note that (a)p implies (a)~ for all p. Hence (ah follows from claims 1 and 2. Combining (a)k with VI E Y proves lemma 3.1. 3.2. Let G act on an IR-tree X. Suppose the action is freely branched relative to A. Then G = *iEI < Ai > is a free product of free abelian groups, where the Ai'S are equivalence classes of A, and < Ai > is a free abelian subgroup of G with common translation axis (i = 1, ... , n). LEMMA

Since BP(XA) i- 0, BP(XA) n Xa i- 0 for all a E A. If Xa = Xb, then vaba-Ib- I = v for all v E Xa, in particular for all v in BP(XA) n Xa. Since the action is freely branched, aba-Ib- I = 1. If Xa i- Xb, let v be the PROOF:


281

end point of [Xa, Xb] on Xa if XanXb is empty, and otherwise let v be an end point of XanXb. Then v is not fixed by one of aba-Ib- l , ab-Ia-Ib, a-Ibab- l and a-Ib-Iab, which yields ab =I- ba. So ab = ba if and only if Xa = Xb for all a, bE A. Recall that a is equivalent to b iff Xa = X b, thus < Ai > is an abelian subgroup of G. Consider the exact sequence

1 ---+ N where I-" is the natural a~l ... aZ·, ai E A} for N = {I}. If not, choose lemma 3.1 to 9 to prove

----+ *iEJ

< Ai >~ G

---+

1,

map. Let I-"(g) = g, and let Igl = min{klg = all 9 in *iEI < Ai >. It remains to show that 9 =I- 1 in N with Igl = k minimal. We shall apply that 9 =I- 1 in G, which is a contradiction.

Suppose that 9 = a~l ... a~·, ai E A, i = 1, ... ,k. First, we need to show that there exists t such that X a1 = ... = X a, =I- Xa'+l' It suffices to show that X ai =I- Xaj for some i and j; suppose not, then X a1 = ... = X a., so aI, ... ,ak are in the same equivalence class of A, say Ai, hence 9 is in < Ai >. This is impossible since 9 =I- 1, while clearly Nn < Ai >= {I}. Note that lx(ai) > 0, ni =I- 0, for i = 1, ... , k. To apply lemma 3.1, we only need to show that BP(XA) n BP(XA)gij is empty for all t + 1 :::; i :::; j :::; k, since the subtree spanned by {X au " " X ak } is contained in XA. If this is false, it may be assumed that va~i .. , a'ji is in BP(X A ) for some v in BP(XA) and 2 :::; i :::; j :::; k. Since the action is freely branched, we have . In . N . Btl th a t a ini ... ajnj -- l 'In G , thus a ini ... ajnj IS u aini ... ajnj 1 < k . B y the minimality of Igl = k, we have a~i ... a'ji = 1 in N, thus

a contradiction. This proves lemma 3.2. LEMMA 3.3. Let G act on an IR-tree X. Suppose the action branched relative to A. Then the G-action is free.

1S

freely

PROOF: If it is false, then it may be assumed that vg = v for some v in X and some 9 in G. Without loss of generality, we may assume that X is the minimal tree XAG; then it may be assumed that v is in XA. Let 9 = al . . . ak in reduced form, ai E< Ani >, i = 1, ... , k, (recall lemma 3.2, G = *iEI < Ai », put gp = al . . . ap,p = 0, ... , k, and let [vgp-I,Xnp ] = [Vgp-I,Vp] where Xn p is the translation axis of < An p >.

282

R. Jiang

CLAIM 1: If vgp-l is in XA but not in X n", then (a)m vgm is not in XA for p :::; m :::; k. (b)m Vm is in BP(XA) U BP(XA)gi,m-l for p :::; i, m.

=I- vp since Vgp-l is not in X n". Combining vgp-l E XA, [vgp-l,X n ,,) = [Vgp_1,Vp) with Vgp-l =I- vp shows that vp is in BP(XA), which yields (b)p. Freely branched action implies that vpG.p is not III

Vgp-l

BP(XA). Since XnpG.p = X np and vgp-l ¢:. X np ' we have that vgp

=

vgp_1G.p ¢:. Xn p and

If vgp E XA, then vpG.p E BP(XA), a contradiction. Thus vgp = vgp_1G.p is not in XA, which is (a)p. By claims 1 and 2 of lemma 3.1, (b)m implies (b)m+l, (a)m and (b)m+l imply (a)m+l, for p :::; m :::; k - 1. Combining these facts yields claim 1. CLAIM 2: vgk = v is in X nk n X nll and vgp is in Xn p n X np +1 for p = 1, ... , k - 1. Note that G.i is in < Ani> with axis X ni , i = 1, ... , k. It suffices to prove that vgp-l is in X np for p = 1, ... , k. If not, then there exists p such that vgp_l is in XA but not in X np ' 1 :::; p :::; k, since v = vgo is in XA. Then (a)k of claim 1 implies that vgk = v is not in XA, which is a contradiction. This proves claim 2. By claim 2, vgp E XA for all p. We may replace v and 9 by vgp and g;l ggp respectively, if necessary, and assume that 9 is cyclically reduced and v is in XA. If vgp is in BP(XA) for some p, then combining vg p (g;lggp) = vgp with the freely branched property of the action shows that 9 = 1, which gives a contradiction. Hence vgp is not in BP(XA) for p = 1, ... , k. Let

Again after replacing v and 9 by vgp and gil ggp respectively, if necessary, we may assume that do = min{dpIO:::; p :::; k -I}, say do = d(v,w) where w is in BP(XA) n X n1 • CLAIM 3: Suppose that u E X n1 and d(u,v) = do, then ug E X nk n X n1 • We prove that ugp E Xn p n X np +1 for 0 :::; p :::; k - 1 by induction on p. Suppose Ugp-l E X np ' If ugp ¢:. Xn p n X np _ l l then combining


283

ugp E X np - X np +i with vgp E Xn p n X np +i (claim 2) shows that there exists u' in BP(XA) n X np +i such that dp :::; d(u',vgp) < d(ugp, vgp) = do, a contradiction. So the induction hypothesis implies that ugk-l E X nk _ i n X nk • The same argument shows that ug E X nk n X ni • By the definition of do, we can assume that do = d(w,v), where w is in BP(XA) n X ni • By claim 3, wg is in X ni n X nk • Note that d(wg,v) = d( wg, vg) = do. IT wg = w, then the freely branched property of the action yields 9 = 1, which is a contradiction. IT wg =1= w, since wg E X ni and d(wg, v) = do, claim 3 shows that wg 2 is in X ni nxnk • Since d(wg 2 , v) = do, this implies that either wg 2 = wg or wg 2 = w, thus the freely branched property of the action shows that 9 = 1 or g2 = 1. Since G is a free product of groups and 9 is cyclically reduced, g2 = 1 implies that k = 1 and 9 = al E < Ani >. Recall that < Ai > is an abelian subgroup of G with common axis Xi, and that Ai consists of hyperbolic elements. So every element of Ai is a translation on Xi, then every element of < Ai > is a translation on Xi. If a E< Ai > is an identity on Xi, then combining BP(XA) n Xi =1= 0 with the freely branched property of the action implies that a = 1. Hence every non-trivial element of < Ai > is a hyperbolic element, in particular a 2 = 1 implies that a = 1 for all a in < Ai > and all i. So g2 = 1 implies 9 = 1, which is a contradiction. This proves lemma

3.3. Since the translation action of < Ai > on its axis gives an injective homomorphism < Ai >---+ R, we have that < Ai >'s are free abelian groups. Hence lemma 3.2 and lemma 3.3 give theorem 2.

§4. Structure of the minimal tree. Suppose that G =< Al > * ... * < An >, the Ai's are equivalence classes of A, each < Ai > is a free abelian subgroup of G with common axis, and X is the minimal tree XAG. Let Am = Ar for m == r mod n, 1 :::; r :::; n. Let

and X(m) = XAG m for m ;::: 1. There is a natural decomposition of the minimal tree, X = U~=IX(m). Before stating our results we now establish some terminology and notation. To make X(m) and X(m-n) stable under

R. Jiang

284

< Am >-action for m

~

1, let X(i - n) = Xi, i = 1, ... , n. Let

Pm = (BP(X(m - 1)) - BP(X(m - n))) n (X(m - n)) for m ~ 1 (when m = 1, replace X(O) by X A ). Lemma 4.1 shows the structure of the minimal tree. (b)m of lemma 4.1 shows that Pm are "connecting points" of X(m-n) and X(m-1)-X(m-n). For each point v in Pm, let C(v, X(m - 1), X(m - n)) be the union of components of X(m -1) - {v} which do not meet X(m - n). Put

C(v,X(m -l),X(m - n)) = C(v,X(m -l),X(m - n)) U {v}.

(b)m of lemma 4.1 shows that X(m - 1) can be obtained by attaching C(v,X(m -l),X(m - n)) to v on X(m - n) for all v in Pm. Since X(m - n) is stable under < Am >, X(m) can be obtained by translating C(v, X(m-1), X(m-n))'s on X(mn) by < Am >-actions, which is (a)m of lemma 4.1. The freely branched property of the action makes sure of the uniqueness of the resulting space X(m). (e)m of lemma 4.1 shows that BP(XA), which is a finite set, is a fundamental set of BP(X). (J)m of lemma 4.1 shows that BP(X(m)) is contained in the subgroup generated by 8BP(XA) and lx(A), which gives theorem 4. 4.1. Let G act minimally on an JR-tree X. Suppose that the action is freely branched relative to A. Then for m ~ 1, (when m = 1, replace X(O) by XA), LEMMA

(a)m

X(m) = X(m - n) U (UvEP"..,aE.

This proves

§5. The proofs of theorem 3 and theorem 4. THE PROOF OF THEOREM 4: It is easy to see that BP(X) is contained in U~=lBP(X(m)), then (f)m of lemma 4.1 yields that 8BP(X) is contained in < 8BP(XA), lx(A) >. For any 9 E G, since BP(XA) =I- 0, there is

Branch Points and Free Actions on JR- Trees v E Xg nBP(X). Thus lx(g) This proves theorem 4.

= d(v,vg)

289

E 6BP(X) and lx(G) C 6BP(X).

LEMMA 5.1. Suppose that T is an !nY-tree spanned by n copies of the real line, n > 1. Then 6BP(T) is contained in a 2n - 3 generated subgroup of

IR. The proof is left to the reader. 5.2. < 6BP(XA), lx(A) > is contained in a 3m - 3 generated subgroup of !nY, where m is the number of elements of A. COROLLARY

COROLLARY 5.3. Let G act minimally on an iii-tree X and suppose the action is freely branched relative to A. Iflx(G) is contained in the rational numbers then the action is simplicial.

Before proving theorem 3, we establish some terminology and notation. Let diam(Xg n Xh), if Xg n X h =I 0 D.x(g,h) = { -2d(Xg,Xh ), otherwise for all g, h in G. Following R. Alperin and H. Bass we say that 9 and h meet incoherently if lx(g) > 0, lx(h) > 0, D.x(g, h) > 0, and 9 and h translate points of Xg n Xh in opposite directions. Otherwise we say that 9 and h meet coherently. Let EX(g, h) = { 1 -1

if 9 and h meet coherently otherwise

If lx(g) > 0, lx(h) > 0, let the end point of [Xg,X h] in X g, if D.x(g, h) < 0; the g-initial point of Xg n Xh, ifO :::; D.x(g, h) < 00; ux(g, h)

=

+00, if Xg = Xh,g and h meet coherently; -00, if Xg = Xh, 9 and h meet incoherently; the end point of Xg n Xh , if Xg n Xh is a half line.

If one ofux(g,h) and ux(g,J) is not ±oo, let D.x(f,g,h) ux(g, J).

=

ux(g,h)-

290

R. Jiang

The proof of (a) of theorem 3. CLAIM 1: There exists a unique

< Al >-equivariant isometry Pl : X(1)

--t

X'(1) extending PO. CLAIM 2: Given < Ap >-equivariant isometrys Pp : X(p) --t X'(P) extending Pp-l, for p = 1, ... , m. then there exists a unique < A m+l >equivariant isometry pm+l : X(m + 1) --t X'(m + 1) extending pm. Since the proofs of claim 1 and claim 2 are the same, we only prove claim 2. Recall (a)m+l of lemma 4.1, and let pm+l be given by

upm+1 =

{

pm ua-lpma

U

ifuisinX(m+1-n) if u is in C(v,X(m),X(m + 1- n»a

Pm extends < Ap >-equivariant isometry Pp for all p ::; m - 1, so vgpvPog for v E XA and 9 E Gp,p ::; m. Since

PTn = [BP(XA)(Gm - G m +l -

n )]

n X(m + 1- n)

P:" = [BP(X~)(GTn - G m + l -

n )]

n X'(m + 1 -

and

n)

then (e)Tn of lemma 4.1 implies that PmPTn = P:". Consequently the following diagram commutes:

C(v,X(m),X(m

Ipm

+ 1- n»

C(vPTn,X'(m),X'(m + 1- n»

~

C(v,X(m),X(m + 1- n»a

Ipm+l

~

C(vPTn,X'(m),X'(m + 1- n»a

Then PTn+l is an < Am+l >-equivariant isometry extending PTn. To see that PTn+l is unique, let P' : X(m + 1) --t X(m + 1) be an < A m+l >equivariant isometry such that P' restricts to X(m) being an identity. Then up' = ua-lap' = ua-lp'a = ua-la = u, (ua- l E X(m», for all u in C( v, X(m), X(m + 1 - n»a, which shows that P' is an identity and that PTn+l is unique. Let up. = UPTn for u in X(m), then P : X --t X' is a unique G-equivariant isometry extending Po, which proves (a) of theorem 3.

The proof of (b) of theorem 3. To see the "only if" side, note that Po induces < a >-equivariant isometry on Xa for all a in A, which implies that lx(a) = lx,(a) and XiPO = X: for

Branch Points and Free Actions on lR-1fees

291

all i and a. Thus [Xi,Xi]fLO = [XLXj] if Xi n Xi is empty, if not empty, (Xi n Xi)fLO = Xi n Xj, which gives ~x(a, b) = ~x,(a, b) for all a, bin A. Note that c:x(a, b) = c:x,(a, b) and ux(a, b) = ux,(a, b) follows from the fact that flo preserves the orientation of X a , hence ~x(a, b, c) = ~x,(a, b, c) for all a, b, c in A. To see the "if" side, suppose that {Aili E I} = {AI, ... , Am}. Let Xi be the axis of < Ai >-action on X, and let Y(i) = span{XI' ... ,Xi}, i = 1, ... , m. Let Xi and Y'(i) be similarly defined. We prove that there exists an isometry fL(P) : yep) ~ Y'(p), which induces < a >-equivariant isometry on Xi for a E Ai, i = 1, ... , p, and extends fL( i) for i < p. Furthermore ux(a,b)fL(p) = ux,(a, b) for a,b E Uf=IA i and lux(a,b)1 < 00. We prove this by induction on p. The case p = 1 is trivial. Suppose there is fL(P - 1) satisfying the above conditions. Without loss of generality, we can assume that Ap = {a}. First suppose yep - 1)

n Xp i= 0.

To get fL(P), it suffices to prove that

(Y(p - 1) n Xp)fL(P - 1) = Y'(p - 1) n X;

CASE

1: 0::;

~x(a,

ai) for some ai E Ai, i ::; p - 1.

We first prove that ux(a,ai)fL(p -1) = ux,(a,ai). If p = 2, then Y(p1) is a line, so it may be assumed that ux(a, ai)fL(p - 1) = ux,(a, ai). If p > 2, then there exists ai E uf':} Ai such that ~x(a,ai,aj) i= 00. Thus ux(ai,a)fL(p - 1) = ux,(ai,a) follows from ux(ai,aj)fL(p - 1) = ux,(ai, ai)· Combining ux(a, ai)fL(p-1) = ux,(a, ai), €x(a, ai) = €x'(a, ai) with ~x(a, ai) = ~x'(a, ai) we have (Xp n Xi)fL(P - 1) = X; n Xi. If both end points of Xp n Xi are not in BP(Y(p - 1)) then Xp n Xi = Xp n yep - 1) and X; n XI = X; n Y'(p - 1), and we are done. If that is not the case, without loss of generality, suppose ux(a,ai) E BP(Y(p-1)). If yep - 1) n Xp = Xi n X p, then

If yep - 1) n Xp i= Xi n X p, we may assume that ux(a, ai) is an end point of XjnXj for some aj E Aj, ~x(ai,aj) > 0, and €x(a,ai) = 1. Then (Xi nXi) n(Xi nXm) is a non-degenerate interval if and only if €x(a, aj) =

292

R. Jiang

€x(aj, ai). Hence (Xp n Xj)f-t(p - 1) = X; n Xj. Since BP(Y(p - 1)) is a finite set, we can prove inductively that (Y(p - 1) n Xp)f-t(p - 1) = Y'(p1) n X;. Combining ~x(aj, a, ai) = 6.x,(aj, a, ai) with ux(a;, a)f-t(p -1) = ux,(ai,a) yields ux(a,aj)f-t(p - 1) = ux,(a,aj) for aj E Ai, and the result follows.

Ur,:;

CASE

2: ~x(a,ai) < 0 for all ai E U~':iAj.

The proof is similar. Secondly suppose Y(p-1)nXp =

d(Y(p -l),Xp)

0. To get f-t(p), it suffices to prove that

= d(Y'(p -l),X;)

and Uf-t' = u', where [u,Xp] = [Y(p -l),Xp] and [u',X;] = [Y'(p -l),X;]. The proof is analogous to that of case 2. This completes the proof of theorem 3.


293

REFERENCES

[AB] R. Alperin and H. Bass, Length functions of group actions on A-trees, Combinatorial group theory and topology, Ann. of Math. Studies III (1987), Princeton Univ. Press. [AM] R. Alperin and K. Moss, Complete trees for groups with a real-valued length function, J. London Math. Soc. (2) 31 (1985). [B] H. Bass, Group actions on non-archimedean trees, Proceedings of the Workshop on Arboreal Group Theory, MSRI, Sept., 1988, to appear. [BF] M. Bestvina and M. Feighn, Bounding the complexity of simplicial group actions on trees, preprint. [CM] M. Culler and J.W. Morgan, Group actions on IR-trees, Proc. London Math. Soc. (3) 55 (1987), 451-463. [CV] M. Culler and K. Vogtmann, Moduli of graphs and automorphisms of free groups, Invent. Math. 84 (1986), 91-119. [GS] H. Gillet and P.B. Shalen, Dendrology of groups in low Q-ranks, preprint, J. Diff. Geom. (1988) (to appear). [J] R. Jiang, On free actions on IR-trees, Dissertation, Columbia University. [L] R.C. Lyndon, Length functions in groups, Math Scand. 12 (1963), 209-234. [Ml] J.W. Morgan, Ergodic theory and free group actions on IR-trees, preprint. [M2] J.W. Morgan, UCLA lecture notes, CBMS Series in Mathematics (to appear). [M3] J.W. Morgan, Group actions on trees and the compactifications of the space of classes of SO(n, I)-representations, Topology 25 (1986), 1-34. [MSl] J.W. Morgan and P.B. Shalen, Valuations, trees, and degenerations of hyperbolic structure,!, Ann. of Math. 120 (1984),401-476. [MS2] J.W. Morgan and P.B. Shalen, Degenerations of hyperbolic structures III, Ann. of Math. 127 (1988),457-519. [P] W. Parry, Pseudo-length functions on groups, preprint. [SB] J.P. Serre and H. Bass, Arbres, Amalgames, SL2, Asterisque 46 (1977).

Department of Mathematics, Columbia University

Axioms for Translation Length Functions WALTER PARRY

§o. Introduction. This paper arose from the paper [3] of Culler and Morgan concerning group actions on R-trees. An R-tree is a nonempty metric space T such that every two points in T are joined by a unique arc (image of an injective continuous function from a closed interval in Ii to T) and that arc is the image of an isometry from a closed interval in R to T. This generalizes the usual notion of simplicial tree. Roughly speaking, the difference is that in a simplicial tree, the branching occurs at a discrete set of points, namely, the vertices, whereas in an R-tree, there is no restriction on where branching may occur. The notion of IR-tree arose indirectly in the work of Lyndon [5] and Chiswell [2] concerning Lyndon length functions. The first definition was given by Tits in [7]. Let G be a group which acts on an R-tree T by isometries. The translation length function for this action of G on T is the function II II: G ---+ R defined by

IIgil = Inf{d(p,gp) : pET}, where g is an element of G and d is the metric on T. The following two assertions are in 1.3 of [3]. First, IIgil = 0 if and only if g has a fixed point. Second, given g in G with IIgil > 0, the axis of g is by definition the set {p E T : d(p,gp) = IIgll}; it is a subtree of T isometric to R and the action of 9 on it is translation by IIgli. At this point the following problem arises. Which functions" II: G ---+ R are translation length functions? Namely, what conditions on a function from G to the nonnegative real numbers imply that it is the translation length function for an action of G on some R-tree? This appears as the first question at the end of the introduction of [3]. Culler and Morgan propose the following conditions in 1.11.

(0.1)

296

W. Parry 1.

II.

III.

11111 = o. Ilg- 1 11 = IIgll for every g in G., Ilghg- 1 11 = Ilhll for every g, h in G.

IV. Given g, h in G, either or V. Given g, h in G with

Ilgll > 0 and Ilhll > 0, either or

It is easy to see that the first two are redundant. Indeed, if 11111 > 0, then apply condition V with g = h = 1. A contradiction ensues. To see that

Ilg- 1 11 = Ilgll,

apply condition IV twice; once using the pair of elements (1, g) and once using (1, g-I). This leads to the following equivalent list of conditions. These will be called axioms.

(0.2) 1.

Ilghg- 1 11 = Ilhll for every g, h in G.

II. Given g, h in G, either

Ilghll = Ilgh -111

or

III. Given g, h in G with

Ilgll > 0 and Ilhll > 0, either or

All of the above generalizes to A-trees. Here, A is an arbitrary ordered Abelian group. Roughly speaking, a A-tree T is a tree modeled on A rather than IR. Thus T is equipped with a A-metric, namely, a function d from TxT to the nonnegative elements of A which satisfies the usual metric properties. In addition to being a A-metric space, T satisfies certain properties which imitate those of IR-trees. The notion of A-tree was introduced by Morgan and Shalen in [6) for their work on compactifying character varieties of fundamental groups of manifolds. Both A-trees and, especially, actions of groups on A-trees are studied extensively by Alperin and Bass in [1). In particular, the notion of an isometry of a A-tree is meaningful. Thus it is possible for a group G to act on a A-tree by isometries. Suppose given


297

such an action. Let 9 be an element of G. In this generality, it is possible for 9 to be an inversion, namely, 9 has no fixed points in T but g2 does. If 9 is not an inversion, then Min {d(p, gp) : pET} exists. Thus it is possible to define a translation length function II II: G -+ A so that

IIgil = {

0 . Mm { d(p, gp) : pET}

if 9 is an inversion otherwise.

Generalizing the case in which A = IR, Theorem 11.2.3 of [6] and, see also, Theorem 6.6 of [1] give the following two assertions. First, IIgil = 0 if and only if g is an inversion or 9 has a fixed point. Second, given 9 in G with IIgil > 0, the axis of 9 is by definition the set {p E T: d(p,gp) = IIgll}; it is a linear subtree of T (isometric to a subtree of A) and the action of 9 on it is translation by IIg II. The problem of finding axioms for translation length functions is also meaningful iIt this generality. Alperin and Bass devote §9 of [1] to this problem. Along the way they establish the axioms in (0.2) for general translation length functions: Axiom I is (a) of Corollary 6.13, Axiom II follows easily from (a)( ii) of Theorem 8.4 and Axiom III follows easily from (a)( ii) and (d)( i) of Theorem 8.4. (Theorem 8.4 assumes that G contains no inversions, which is not assumed here. The quoted results are nonetheless valid because they hold for A.) Furthermore, a careful examination of the first case in the proof of (a) of Proposition 8.1 shows that given g, h in G with IIgil > 0 and IIhil > 0, then

t

Max{O, IIghil - IIgil - IIhll} E 2A. (Beware that Proposition 8.1, like Theorem 8.4, assumes that G contains no inversions.) This is automatic for A = IR but not in general. Hence the list of axioms in (0.2) must be amended to the following.

(0.3) O. Given g, h in G with IIgil

> 0 and IIhil > 0, then

Max{O, IIghil - IIgil - IIhll} E 2A. I. IIghg- 1 II = Ilhil for every g, h in G. II. Given g, h in G, either

or

298

W. Parry

III. Given g, h in G with Ilgll > 0 and Ilhll > 0, either Ilghll = Ilgh-111 > Ilgll

+ Ilhll

or

Following Culler and Morgan, define a pseudo-length function to be a function II II: G --T A from a group G to the nonnegative elements of an ordered Abelian group A which satisfies the axioms in (0.3). The object of this paper is to prove that every pseudo-length function is a translation length function. The main theorem can be formally stated as follows. MAIN THEOREM. Let G be a group, and let II II be a pseudo-length function from G to an ordered Abelian group A. Then there exists a A-tree T and an action of G on T by isometries such that II II is the translation length function for this action. Most of the remainder of the introduction is a rough outline of the proof of the main theorem. Suppose given a group G, an ordered Abelian group A and a pseudo-length function II II: G -+ A. These will be fixed for the rest of the paper. Much of the terminology of group actions on A-trees carries over immediately to the present situation. An element x in G is said to be hyperbolic if and only if IIxll > o. The pseudo-length function is said to be Abelian if and only if IlxY11 ~ Ilxll + lIyll for all x, y in G. Otherwise it is non-Abelian. It is dihedral if and only if it is non-Abelian but Ilxyll ~ Ilxll + Ilyll for all hyperbolic elements x, yin G. If it is neither Abelian nor dihedral, then it is of general type. First suppose that II II is either Abelian or dihedral. Alperin and Bass prove an analog of the main theorem in this case in §9 of [1]. However, to apply their result for the present purpose it is necessary to verify that II II satisfies all of their axioms. It seems preferable to follow the approach of Culler and Morgan in [3]. They prove the main theorem in this case for A = IR in §6. Their arguments generalize to arbitrary ordered Abelian groups with little change. Rather than essentially repeat these arguments here, it will instead be indicated in the next paragraph where modifications are needed. Lemmas 6.1, 6.2, 6.4, 6.5, 6.6 are valid in general as stated and proved. Corollary 6.7 uses the Archimedean property of IR and is invalid, so disregard it. Essentially as before, Lemma 6.6 easily implies that it is possible

Axioms for Translation Length Functions to define Tu : Su

---+

A so that for every s in Su and integer m

~

299

0,

and Having done this, the proof of Lemma 6.8 must be slightly modified. The main part of this argument deals with elements a, b in Suo First suppose that Tu(a) + Tu(b) ~ O. Modify the identities in the argument of Lemma 6.8 by simply inserting the following absolute value signs.

+ Tu(a)1 m)lIuli + Tu(b)1

lIaumll = Imllull Ilu-mbukll = I(k -

It follows essentially as before that ab E S .. and Tu (a b) = T.. (a) + T.. (b). It is easy to see that if c lies in Su, then so does c- 1 and Tu(C- 1) = -Tu(C). Combining the last two statements concludes the proof of Lemma 6.8. The rest of §6 is valid in general. This proves the main theorem if II II is either Abelian or dihedral. For the remainder of the paper assume that II II is of general type. As in [3], a good pair of elements of G consists of hyperbolic elements x, y for which

0< Ilxll

+ lIyll-lIxy-111

< 2 Min{lIxll, Ilyll},

equivalently, In the geometric situation, this means that the axes of x and y meet, the length of this intersection is positive and x and y translate in the same direction by lengths greater than the length of the intersection. This suggests the following diagram. y

x

-----'I'------r_

x

y

Proposition 6.3 of [3], the proof of which holds in general, shows that since II II is of general type, there exists a good pair of elements of G. Hyperbolic elements x, yare said to meet if and only if IIxYIl S IIxll + lIyll. In the geometric situation, the axis of x meets the axis of y if and only if IIxyll S IIxll + lIyll. For example, the elements of a good pair meet. Define

300

W. Parry

a relation on the set of good pairs as follows. Good pairs {x, y} and {z, w} are related if and only if each of the elements x, y, xy-l meets each of the elements z, w, zw- 1 • Theorem 2.1 proves that this is an equivalence relation. The resulting equivalence classes are called branch points. The branch point represented by the good pair {x, y} is denoted by [x, y]. In the geometric situation, given a good pair {x, y}, the axes of x, y, xy-l meet in a configuration as in the following diagram. xy -1

y

p

_--.L..-___

X _ _ _- ' -_ _.....

X

xy -1

y

The intersection of the axes of x, y, xy-l is the branch point p. If x, y, xy-l all meet z,w,zw-t, then the point of intersection of the axes of z,w,zw-1 is also p. The set of all branch points is denoted by X. The following prepares for the definition of a A-metric on X. Define the distance between hyperbolic elements x, y by (0.4)

d(x,y) = -Max{O,lixyli-lixll-llyli}.

This is precisely where Axiom 0 is needed. It ensures that the values of d lie in A. In the geometric situation, d(x, y) is the distance from the axis of x to the axis of y. If {x, y} is a good pair and z is a hyperbolic element, define the distance from {x, y} to z by

(0.5)

d( {x, y}, z) = d(z, {x, y}) = Max{d(x, z), dey, z), d(xy-l, z)},

and if {x, y} and {z, w} are good pairs, define

(0.6)

d( {x, y}, {z, w}) = Max{d( {x, y}, z), d( {x, y}, w), d( {x, y}, zw)}.

Corollary 2.19 shows that the last two distance functions are actually defined on branch points. In particular, this gives a A-distance function on X. The letter d will denote all of these distance functions. The arguments make the meaning clear. Corollary 2.40 establishes the triangle inequality for the A-distance function on X, and so it is a A-metric.


301

The above gives the set of branch points X the structure of a A-metric space. Axiom I shows directly from the definitions that for every x in G the map

{y,z}

f-t

{xyx-1,XZX- 1}

on good pairs {y, z} induces an isometry

of X. This gives an action of G on X by isometries. Theorem 2.42 applies Theorem 3.17 of [1], which is essentially the generalization by Alperin and Bass of Chiswell's main theorem in [2], to prove that there exists a A-tree T on which G acts by isometries and a G-equivariant isometry of X to a subspace of T. The paper concludes by verifying that the translation length function associated to this action of G on T coincides with II II. This completes the proof of the main theorem. Section 1 is a collection of basic lemmas. Little or no motivation or explanation for these is (or can be!) given. However, it might be noted that they are ordered roughly by complexity. Lemma 1.1 deals with one element, the intermediate lemmas deal with two and three elements and, finally, Lemma 1.16 deals with four elements. The fact that the maximum is four seems to reflect the fact that four points are used in the axioms for Lyndon length functions--the base point and three others. Similarly, consider the four-point condition of Imrich [4]. Section 2 consists of the main body of the proof of the main theorem for pseudo-length functions of general type. Recall from I and II of (0.1) that 11111 = 0 and IIx- 1 II = IIxil for every x in G. Also, III of (0.1) easily implies that IIxyil = IIyxil for every x, yin G. More generally, II II is invariant under cyclic permutation of the factors of a product. These and similar trivialities will be used freely, usually without express mention of them.

§1. Basic lemmas. LEMMA

1.1. Let x be a hyperbolic element. Then for every integer n,

According to I and II of (0.1), it suffices to prove this for n > O. Induct on n. There is nothing to prove for n = 1, so suppose that n > 1 and PROOF:

302

W. Parry

that the lemma is true for smaller values of n. The induction hypothesis and Axiom III applied to x n - 1 and x easily conclude the proof. 0 LEMMA

1.2. Suppose given hyperbolic elements x, y which do not meet.

Then

IIxmlynl ... xmkynk II

= kllxYl1

k

k

;=1

;=1

+ IIxll ~)Imil- 1) + lIyl12)lnd -

for every positive integer k and nonzero integers

1)

m1, ... , mk, n1, ... ,nk.

PROOF: Induct on k. First suppose that k = 1. Since by Axiom II Iixyll = Ilxy- 1 11 = IIx- 1 yll = Ilx- 1y- 111, it may be assumed that m1 > 0, n1 > o. Now induct on m1 + n1. If m1 + n1 = 2, then there is nothing to prove. So assume that m1 + n1 > 2 and that the assertion is correct for smaller values of m1 + n1. Without loss of generality, n1 ~ 2. Applying Axiom III to x m, y n , -1 and y shows that

or By induction and Lemma 1.1,

and so as desired. Now suppose that k > 1 and that the assertion is correct for smaller values of k. Applying Axiom III to xmlynl ... xmk-lynk-l and xmkynk yields

or

Max{llxm1yn1... xmkynk II, Ilxml-mkynl ... xmk-lynk-l-nk II} = Ilxmlynl ... xmk-lynk-lll + IIxmkynk II}.

Axioms for 'Translation Length Functions

303

By induction,

$ (k - l)(l/xyl/

k

k

;=1

;=1

-i/xl/-llyi/) + I/xll L Imil + lIyl/ L Inil

and so the lemma is proven.

D

REMARK 1.3: Using Lemmas 1.1 and 1.2 it is possible to give a simple proof of the main theorem in a very special case. Suppose that G is generated by hyperbolic elements x and y which do not meet and for simplicity suppose that A = R. First of all, Lemmas 1.1 and 1.2 show that IIwll > 0 for every nontrivial word w in x and y, and so G is a free group. Let F be a subset of R2 as indicated below.

Let the top line segment have length I/xll, the bottom line segment have length lIyll and the other have length ~(lIxyll-lIxll-lIyll). For every element in G take a copy of F, and identify these copies of F in a straightforward way to construct an R-tree on which G acts so that x, y, xy have translation lengths IIxl/, I/yl/, IIxyl/. Since this translation length function satisfies Lemmas 1.1 and 1.2, it must agree with 1/ 1/ on every element of G. This proves the main theorem in this case. LEMMA 1.4. Suppose given hyperbolic elements x and y. If xn meets y for some nonzero integer n, then x meets y. PROOF: IT not, then Lemmas 1.1 and 1.2 show that

which is a contradiction.

D

304

W. Parry

LEMMA 1.5. 1) For all elements x, yin G and evezy integer n ~ 1,

IIxnyll

~ Max{(n -

2)llxll + Ilyll, (n - l)ll xll + Ilxyll}·

2) Given hyperbolic elements x and y, if x meets y, then x meets y for every nonzero integer n.

PROOF: First consider 1). This will be proven by induction on n. The case n = 1 is trivial, so assume that n > 1 and that the statement is correct for smaller values of n. Axiom II applied to xn-ly and x implies that either = Ilx n - 2 yll or Ilxnyll ~ Ilxll + Ilx n- 1 yll. The induction hypothesis easily gives 1). Statement 2) follows easily from 1) and Lemma 1.1. This proves Lemma 1.5. 0

Ilxnyll

LEMMA 1.6. Suppose given hyperbolic elements x and y with

Ilyll

IlxY11 = IIxll+

and nonzero integers ml, ... ,mk, nl, ... ,nk for some positive integer

k.

1) Then

IIxm1yn 1... xmkynk II

~

k

k

i-I

i=1

Ilxll L Imil + Ilyll Lin;!.

2) Equality holds if ml, ... , mk, nl, .. . , nk have the same sign. 3) If the signs are not the same and equality holds, then Ilxy-- 1 11

Ilxll + Ilyll·

PROOF: To begin, 2) will be proven by induction on n = L:~=I(lml + In!). The case n = 2 is trivial, so assume that n > 2 and that 2) holds for smaller values of n. If some exponent has absolute value at least 2, then after an inversion, a cyclic permutation and interchange of x and y if necessary, it may be assumed that ml ~ 2. The induction hypothesis and Lemma 1.1 used with Axiom III applied to x and xml-lynl ... xmkyn k prove 2) in this case. If all of the exponents have absolute value 1, then xmlynl ... xmkynk = (xy)k or (X- 1y-l). In this case 2) follows from Lemma 1.1. Now 1) and 3) will be proven simultaneously by induction on k. Statement 2) implies that after a cyclic permutation and interchange of x and y if necessary, it may be assumed that ml and nl have opposite signs.


305

First suppose that k = 1. To prove 1) in this case, apply Axiom III to x m1 and ynl and use 2) applied to xm1y-n 1 and Lemma 1.1. Next consider 3) in the case k = 1. It is easy to see that it suffices to prove that if Ilxy-1 II < IIx II + IIyll, then IIx m y-1 II < mllxll + IIyll for every positive integer m. The desired result follows easily from 1) of Lemma 1.5. Now suppose that k> 1 and that 1) and 3) hold for smaller values of k. Apply Axiom II to xmlynl and xm2yn2 ... xmkynk. Since the induction hypothesis applies to these two elements and x m2 -ml yn2 ... x mk ynk -n 1 , statement 1) follows easily. As for 3), consider the reduced form of the element xm2-mlyn2 ... xmkynk-nl. If its exponents have the same sign, then because m1 and n1 have opposite signs, there must be cancellation among the original exponents. Thus the induction hypotheses for both 1) and 3) imply that IIxm2-mlyn2 ... xmkynk-nl

II < IIxil

k

L

k

Imil

+ IIyll L

Inil

i=l

i=l

unless IIxy-1 II = IIxil + IIyll. The desired result now follows easily. This completes the proof of Lemma 1.6. 0 The next lemma is unfortunately complicated, but almost all later arguments use it in some way. The conclusion which follows from assumptions 2) and 3) is actually equivalent with that which follows from assumption 1). This is because 2) or 3) implies that

2 Max{IIxYII, IIxzll, IIyzll} ~ IIxYil

+ IIxzll + IIyzll·

1.7. Let (x, y, z) be a 3-tuple of hyperbolic elements. Assume that one of the following sets of conditions holds. LEMMA

1) IIxYil ~ IIxil

+ IIyll

IIxzll ~

IIxil

2 Max{IIxYII, IIxzll, IIyzll}

2) IIIxil - IIylll < IIxYil ~ IIxil IIyll + IIzil 3) 0 < IIIxil - IIylll = IIxyil liz II

+

+ IIzil

=I

IIxYil

+ IIxzll + IIyzil

IIYII

IIxzll =

IIxil

IIxz- 1 11

< IIxil + IIzil

+

IIzil IIyz- 1 11

IIyzil =

< IIyll +

306

W. Parry

If 1) holds then,

IIxyzll

+ Ilxzyll

;:::

Max{2I1 xyll,21Ixz ll,21Iyzll, IlxYIl + Ilxzll + Ilyzll}·

If 2) or 3) holds then,

Ilxyzll

+ IlxzYl1

;:::

IlxY11 + Ilxzll + Ilyzll·

In every case, if xyz and xzy are hyperbolic, then the inequality becomes an equality. The three cases will be handled simultaneously. First make two normalizations. The lemma is symmetric in x and y, and so it may be assumed that Ilyzll ~ Ilxzll. Furthermore, if 1) holds, then it may similarly also be assumed that Ilxyll ~ IIxzll. In all three cases xz and yz are hyperbolic. In applying Axiom III to xz PROOF:

and yz, the value Ilxz(yz)-lll = Ilxy-lll appears. If IlxY11 > Ilxll + Ilyll, then 1) holds. Hence the second normalization gives IlxY11 ~ IIxzll, and so

Ilxy-lll = IlxY11 If

IlxY11

~

~

IIxzll < Ilxzll + Ilyzll·

Ilxll + Ilyll, then

Ilxy-lll

~

Ilxll + IIYII < Ilxll + Ilzll + Ilyll + Ilzll

~

Ilxzll + Ilyzll·

Thus Axiom III implies that

Ilxzyzll =

Ilxzll + Ilyzll·

Next, in applying Axiom III to the hyperbolic elements xzyz and xy, the value Ilxzyz(xy)-lll = Ilzyzy-lll appears. However,

where the first inequality is given by Lemma 1.2 and 1) of Lemma 1.6 and the second is given by the first normalization. Thus Axiom III implies that

II(xzy)(zxy)11

= Ilxyll + Ilxzll + Ilyzll·

Finally, in applying Axiom II or Axiom III to xzy and zxy, the value lI(xzy)(zxy)-lll = Ilxzx-1z-11l appears. If Ilxzll > Ilxll + Ilzll, then 1)


307

holds. Thus Lemma 1.2 shows that Ilxzx- 1z-lll = 211xzll, and 211 xz ll =f IIxYl1 + Ilxzll + lIyzll by assumption and normalization. The conclusion of the lemma follows easily in this case. If Ilxzll = Ilxll + IIzlI, then using 1) of Lemma 1.6 gives IIxzx- 1 z-lll :=:; 211xzll

=

211 x ll

+ 211 z l1

+ IIxll + Ilzll + lIyll + IIzll IlxYIl + IIxzll + Ilyzll·

:=:; IlxYIl :=:;

If 1) or 2) holds, then the second inequality is strict. If 3) holds, then 3) of Lemma 1.6 shows that the first inequality is strict. The desired result in this case follows easily. This proves Lemma 1.7. 0 The next lemma clarifies the situation in which IlIxll-llylll is not covered by Lemma 1.7.

> Ilxyll, which

LEMMA 1.8. Given hyperbolic elements x, y such that Ilxll then {x, y-2} is a good pair.

> IIxYIl + Ilyll,

PROOF: Use Axiom II applied to xy and y to conclude that IIxYl1 = Ilxll. This easily implies that the pair {x, y-2} is good. 0

LEMMA 1.9. Suppose given hyperbolic elements x, y, z such that x meets y,z and Ilxyzll

> Ilxll + Ilyll + Ilzll. Then the values Ilxyozell for 8,c = ±1

are all equal. PROOF:

Since Ilxyzll > Ilxll

+ IIYII + IIzll

~ IlxY11

+ IIzlI,

Axiom II applied to xy and z shows that IIxyzll = Ilxyz-111. Minor variations on this argument complete the proof of the lemma. o LEMMA 1.10. Let ex, y, z) be a 3-tuple of hyperbolic elements such that x meets y, z and y does not meet z. Then the following statements hold.

1) If Ilxyzll equal.

=

IIxll

+ Ilyzll,

then the values Ilxyozell for 8,c

2) Min{llxyzll, Ilxzyll} < IIxll + Ilyzll 3) Min{lIxyzll, IIx-1yzll} < Ilxll + lIyzll 4) Max{lIxyzll, IIxzyll} = Ilxll + lIyzll 5) x meets yz

=

±1 are all

308

W. Parry

PROOF: The argument begins by noting the following temporary strengthening of 1). (Strict inequality is impossible by 4).)

(1.11)

If Ilxyzll 2:: IIxll

+ Ilyzll,

then the values IIxyozell for 8, c = ±1

are all equal.

This follows from Lemma 1.9 because IIxyzll2::

Ilxll

+ lIyzll >

Ilxll

+ lIyll + IIzll·

This proves 1). Now consider 2). To begin, (1.11) shows that y and z may be replaced by their inverses. Consequently, it may be assumed that (1.12)

Ilxyll = Ilxll

+ Ilyll

and Ilxzll = Ilxll

+ IIzll·

This gives the first three conditions of assumption 1) of Lemma 1.7. Now consider the last condition. Lemmas 1.4 and 1.6 imply that x may be replaced by any positive power of x. Thus by (1.12) it is possible to replace x by x 2 if necessary to assume that lIyzll

=f

IlxYIl

+ IIxzll·

This and (1.12) give the last condition of assumption 1) of Lemma 1.7, and so assumption 1) of Lemma 1.7 is satisfied. If xyz and xzy are not both hyperbolic, then 2) is clearly true. Otherwise, Lemma 1. 7 implies that Ilxyzll

+ Ilxzyll = Max{21I x yll, 2l1xzll, 211y z ll, IlxYIl + IIxzll + Ilyzll} = Max{2I1y z ll, Ilxll + lIyll + Ilxll + IIzll + Ilyzll} < 2(ll x ll + Ilyzll)·

This proves 2). Now consider 4). Since 2) is proven, y and z may be interchanged if necessary to assume that IIxzyll < Ilxll + lIyzll. Then Axiom III applied to x and zy implies that Ilxy-l z-lll = Ilxll + Ilyzll, and so 1) easily gives IIxyzll = Ilxll + lIyzll. This proves 4). Now consider 3). Since 4) is proven, it may be assumed that IIxyzll = IIxll + Ilyzll. Then IIxyzll

=

IIz-1y-1x-11l = IIx-1z-1y-111 = IIx-1zyll,


309

where the last equality comes from 1). Thus Ilx-Iyzll < Ilxll + Ilyzll by 2), which proves 3). Statement 5) follows directly from 4). This proves Lemma 1.10. D LEMMA

1.13. Let (x,y,z) be a 3-tuple of byperbolic elements sucb tbat x

meets y, z and PROOF:

Ilyzll = Ilyll + Ilzll.

Tben x meets yz.

Suppose that x does not meet yz. Then

(1.14) From this Lemma 1.9 easily implies that (1.15)

the values

Ilxy6z"ll, Ilxz"y 611 for ii,c

= ±1 are all equal.

Lemmas 1.4 and 1.5 show that the lemma is unaffected by replacing x by x 2 • Thus for the later application of Lemma 1.7, it may be assumed that

Ilxll =I- lIyll· By replacing x by x-I if necessary, it may also be assumed that Ilxzll = Ilxll + Ilzll· Now suppose that IlIxll-llylll :::; IIxyll, and consider assumptions 2) and 3) of Lemma 1.7. If either Ilxz-III = Ilxll + Ilzll or Ilyz-III = lIyll + Ilzll, then

for some choice of ii, c = ±1. Thus 2) of Lemma 1.7 is satisfied for this

choice of ii, c. If Ilxz-III < Ilxll + Ilzll and Ilyz-III < lIyll + Ilzll, then either 2) or 3) of Lemma 1.7 is satisfied. So, if Illxll -IIYIII :::; Ilxyll, then Lemma 1.7 and (1.15) imply that 1

Ilxyzll = "2(llx y6 1 + Ilxz"lI + III z"ll) :::; IIxll + lIyll + Ilzll for some choice of ii and c, which contradicts (1.14). Finally, if Illxll-llylll > IlxYII, then Lemma 1.8 implies that either {x, y-2} or {x- 2 , y} is a good pair. Since the lemma is unaffected by replacing x by x 2 , the second possibility reverts to the case of the previous paragraph. If {x, y-2} is a good pair, then the previous paragraph applies to the 3-tuple (x,y2,z). Thus x meets y2Z. Now apply Axiom III to zxy and y, and note that

Ilxyzll > Ilxll + Ilyzll Ilxyzll + Ilyll

=

implies that

Ilxyzll + lIyll > IIxzll.

IIxy2z 11 :::; Ilxll + IIy 2z li

=

Thus

Ilxll + 211yll + Ilzll·

310

W. Parry

o

Again, this contradicts (1.14). This proves Lemma 1.13.

The next lemma can be viewed as a result on the notion of meeting coherently. Namely, hyperbolic elements x and y are said to meet coherently if IIxYIl = IIxll + Ilyli. In the geometric situation, this means that the axes of x, y meet and x, y translate in the same direction relative to this intersection. Given three hyperbolic elements which meet one another such that two of the pairs meet coherently, it is not always true that the third pair meets coherently. Lemma 1.16 provides conditions for which this is true. 1.16. Let x,y,z,w be hyperbolic elements which satisfy the following three conditions. LEMMA

1)x and y meet z and w 3)lI xzw ll

Then

2)z does not meet w

= IIxll + IIzwll

IlxYIl = IIxll + Ilyll if and only if lIyzwll

=

lIyll + Ilzwll·

The forward implication will be proven first. Assume that IIxYIl = IIxll + lIyll and that the assertion is false. By 5) of Lemma 1.10, y meets zw, hence lIyzwll < Ilyll + Ilzwll. Lemma 1.13 implies that xy meets z and w, hence 4) of Lemma 1.10 implies that either PROOF:

Ilxyzwll

=

IIxyll + Ilzwll

or

IIxywzll =

Ilxyll + IIzwll·

The next paragraph will handle the former case, and the paragraph after that will handle the latter. Assume that IIwxyzll

=

IIxyzwll = IIxyll + IIzwll.

IIxyzwll

Then

= Ilxyll + Ilzwll > IIxll + Ilyll + IIzll + IIwll > IIwxll + Ilyzll,

and so Axiom II applied to wx and yz gives wise,

IIwxz-1y-11l =

IIxyzwll. Like-

Ily-1wxz-111 = IIwxz-ly-lll > lIy-1wll + IIxz-11l, and so Axiom II applied to y-1w and xz- 1 gives lIy- 1wzx- 111 = IIxyzwll. Since lIyzwll < lIyll + Ilzwll, Axiom III applied to y and zw gives Ily-l zwll = Ilyll + IIzwll· Now 2) of Lemma 1.10 gives lIy- 1wz ll < Ilyll + IIzwll. Thus Ily-1wzx-11l

=

Ilxyzwll = IIxll + Ilyll + IIzwll > lIy-1wzll + Ilxll,


311

and so Axiom II applied to y-1wz and x gives Ily-1wzxll = IIxyzwll. Next, 2) of Lemma 1.10 and condition 3) yield IIxwzll < IIxll + IIzwll. Hence much as before,

and so Axiom II applied to wzx and y gives IIwzxyll = IIxyzwll, namely, IIxywzil = IIxyzwll = IIxYIl + IIzwll. This contradicts 2) of Lemma 1.10. Now assume that IIxywzll = IIxYIl + IIzwll. The strategy of the argument in the previous paragraph applies in this case as well. Again, four applications of Axiom II give a contradiction to 2) of Lemma 1.10 as follows. Axiom II applied to zx and yw gives IIzxw-1y-11l = IIxywzll. Axiom II applied to y-l z and xw- 1 gives lIy-l zwx- 1 II = IIxywz II. Axiom II applied to zwx- 1 and y gives lIyzwx-11l = IIxywzil. Axiom II applied to yzw and x gives IIyzwxll = IIxywzll, namely, IIxyzwil = IIxywzil. This contradiction completes the proof of the forward implication. For the converse, assume that IIyzwil = IIYII + IIzwil. At first, suppose that x does not meet y. Since the current problem is symmetric in x and y, 4) of Lemma 1.10 applied to (zw, x, y) implies that (1.17)

IIxyzwil

=

IIxYil

+ IIzwil

may be assumed. Hence IIyzwxll

=

IIxyzwil

=

IIxYil

+ IIzwil > IIxil + IIYII + IIzil + IIwil 2::

Thus Axiom II applied to yz and wx gives

and likewise Axiom II applied to zx- 1 and w-1y gives

Hence Now 1) of Lemma 1.10 applied to (wz,X-t,y-l) gives (1.18)

IIxywzll =

IIxYIl + IIzwll·

IIyzil

+ IIwxll·

312

W. Parry

Hence 2) of Lemma 1.10 applied to (xy,z,w) combines with (1.17) and (1.18) to give a contradiction. Thus x meets y.

So, if Ilxyll i- Ilxll + IIYII, then Ilxy- 111 = Ilxll + Ilyli. But then the converse implies that Ily-l zwll = Ilyll + Ilzwll· Since Ilyzwll = IIYII + Ilzwll, 3) of Lemma 1.10 gives a contradiction. This completes the proof of Lemma 1.16. D

§2. Proof of the ITIain theorem in the case of general type. The main theorem will be proven in this section for pseudo-length functions of general type as roughly outlined in the introduction. The first thing to be considered is the relation on good pairs. Recall that good pairs {x, y} and {z, w} are related if and only if each of the elements x, y, xy -1 meets each of the elements z, w, zw- 1 • THEOREM 2.1. This relation is an equivalence relation. PROOF: It is clearly symmetric, and it is clear that every hyperbolic element meets itself. The facts that 1) x meets y 2) x-I, hence x, meets xy-l and 3) y meets xy-l follow easily from the definition of good pair. Thus the relation is reflexive, and so what must be shown is that it is transitive. This is implied by the following lemma, and so the lemma completes the proof of Theorem 2.1. D LEMMA 2.2. Let x, y, z, w be hyperbolic elements. Suppose that {x, y} is a good pair and that z, w meet x, y, xy-l. Then z meets w. PROOF: Assume that z does not meet w. Since x meets z and w, x meets

zw by 5) of Lemma 1.10. Statement 4) of Lemma 1.10 shows that it may be assumed that 1.16 implies that

Ilxzwll = IIxll + Ilzwll.

(2.3) It is easy to see that implies that

(2.4)

Since

Ilxyll = Ilxll + Ilyll,

Lemma

Ilyzwll = Ilyll + Ilzwll· Ilx(xy-l)1I = Ilxll+llxy-111, and so Lemma 1.16 likewise Ilxy-l zwll = Ilxy-111 + Ilzwll.


313

Now (2.3) and (2.4) show by applying Lemma 1.16 with x replaced by xy-l that

o

a contradiction. This proves Lemma 2.2.

The equivalence classes of good pairs established by Theorem 2.1 are called branch points. The branch point represented by a good pair {x, y} is denoted by [x, Yl. 2.5: A hyperbolic element x is said to meet a branch point if x meets y, z, yz-l.

DEFINITION

[y, zl

Lemma 2.2 shows that this condition is independent of the choice of the good pair representing [y, zl, and so this notion is well-defined. In the geometric situation, to say that x meets a branch point simply means that the branch point lies on the axis of x. It will eventually be shown that the set of branch points X is a A-metric space with special properties. Leading to this, the following results develop the properties of X and the A-distance functions defined in (0.4), (0.5) and (0.6). THEOREM 2.6. Let p = [x, yl be a branch point, and let z be a hyperbolic element which does not meet p. Suppose that

(2.7)

d({x,y},z) = d(x,z).

Then all of the elements x±l z±l, z±l x±l meet p. PROOF: The assumptions imply that x does not meet z. Using Lemma 1.2, it is easy to see that it suffices to prove that the given elements meet y and xy-l.

First consider xz and y. In this case, what is needed from (2.7) is that (2.8)

x does not meet z

and

where the inequality follows from the inequality d(x, z) replace assumption (2.7) for the proof of this case. Now suppose that xz does not meet y, namely,

(2.9)

IIxzyll > IIxzll

+ lIyll·

~

dey, z). Let (2.8)

W. Parry

314

Since x does not meet z and y does not meet xz, statement 5) of Lemma 1.10 with x and y interchanged implies that y does not meet z. Combining this with (2.8) and (2.9) gives II xz yli

> IIxzll

+ Ilyll

~ lIy z ll

+ Ilxll >

Ilxll

+ Ilyll + Ilzll

= IlxY11

+ Ilzll·

From this,

can be gotten, briefly, as follows. The second, third, resp., fourth identities are easily derived from Axiom II applied to z and yx, y and zx- 1 , resp., x and yz. Next consider assumption 1) of Lemma 1.7 for the 3-tuple (x, y, z). The first three inequalities clearly hold. Also, IlxY11

= Ilxll + Ilyll
O. If i is big enough, there exist Wi, Xi, Vi, Zj in X j and an 6-approximation Ri between {w, x, y, z} and {Wi, Xi, Vi, z;} such that WRiWj, XRiXi, yRiYi, ZRiZi. Hence

36

I(Xi

I Zi)w;

- (x

I z)wl < 2'

I(Xi

I Yi)Wi

- (x

I Y)wl < 2

36

I(Yi I Zi)Wi - (y I z)wl
0.

°

THEOREM. Let Mo = {f E (M) : If I is continuous and L(f) = = R(f)}. Then (M) is the free product of Mo and the free group on the set of positive real numbers. Choose a germ (\' of a nontrivial function in Mo. The following formula defines a subgroup MOl of Mo.

MOl

=

{f E Mo : f

= 1 or both

are in a and f(t)

i=

°

f and f- 1

for t E (0, b(f))}

Let F be the free group on the set 9 - [a], where 9 is the set of germs in Mo. Then Mo

= F * MOl'

346

F. Rimlinger

§1. Prornislow's continuous free product. The following group is defined by Promislow in [P) section 10. In order to describe this group we first consider the set of functions f : [0, b) -+ R such that

(i) bE R, b ~ 0, (ii) f is continuous at all but a finite number of points in (0, b), (iii) at each point of discontinuity of fin (0, b), the one-sided limits exist, and (iv) the appropriate one-sided limits of f exist at and at b.

°

Henceforth we shall identify two such functions if their domains are the same and they agree at all but a finite number of points. Let l( G) be the set of identifications classes of functions satisfying (i), (ii), (iii), and (iv). We continue to refer to these classes as functions. Given f E l( G) with domain [0, b) for some b ~ 0, we adopt the following conventions: (i) If t E (0, b) is a point at which the one-sided limits of f agree, then we define f( t) to be the common value of these limits. Similarly, f(O) and feb) are determined by the appropriate one-sided limit, provided b i- O. The unique class of leG) with domain {O} is denoted l. Vacuously, the function 1 is well defined except at a finite set of points. (ii) We denote the value of b by 8(f). We set 8(1) = O. (iii) We define L(f) = f(O) and R(f) = f( 8(!)). (iv) Given t E [0,8(!)), we say f(t) i- y if no one-sided limit of f at t equals y. (v) The function f is continuous at t E (0,8(f)) if the one-sided limits agree at t, and discontinuous if they are unequal at t. A function f is continuous if f is continuous at each t E (0, 8(f)). In particular, 1 is continuous. Given functions

f and

g in l( G), we define

f*g=f(t) = get - 8(f))

f * g by concatenation:

forO::=;t::=;8(f) for 8(f) < t ::=; 8(f)

+ 8eg)·

Clearly f * g is again in leG). We define the inverse of f, denoted f- 1 , by f- 1 et) = - f(8(f) -t). Notice that f and f- 1 have the same domain. This inverse has a group theoretic significance and clearly does not coincide with the function theoretic notion of inverse.

The Structure of Promis low's Continuous Free Product

347

1. DEFINITION: A function f E I( G) is reduced if there does not exist a triple of functions U', u, f") such that u -I=- 1 and f = f' * u * U -1 * f". Let G denote the set of reduced functions of I( G).

Notice that identity for *.

*

is an associative binary operation on I( G) and 1 is an

2. DEFINITION: Given f,g E G, there exists a unique element u E G such that (i) f = f' * u for some f' E G, (ii) 9 = u-Ig' for some g' E G, (iii) if v E G satisfies f = f" * v and 9 = then bU) :S b( u). We denote the element u by the notation u

v-I

* g"

for some f",g" E G,

= uU,g).

Suppose that f, 9 E G, say f = f' * u(j, g) and 9 = u(j, 9 )-1 g' for some f', g' E G. Notice that f' * g' is reduced in the sense of definition 1. Thus f' * g' is in G. We define multiplication in G by the law fg = f'g'. This multiplication gives G the structure of a group, with inverses defined as above and identity element 1. 3. DEFINITION: A subgroup H of G is closed under segmentation if for all E H and for all a, b such that 0 :S a :S b :S b(j), we have the restriction of f to [a, b] is again in H.

f

§2. Pregroups and partial groups. Partial groups were defined by Stallings in [SI]. This notion was later generalized by Stallings in [S2]. Our brief discussion of pregroups here is taken from [Rl]. Partial groups provide the appropriate machinery for proving the main theorems of the paper. Let P be a set. Let i : P ~ P be an involution, denoted x ~ X-I. Let 1 E P be a distinguished element. Let D C P X P and let m : D ~ P be a set map, denoted (x,y) ~ xy. Suppose that (PI) for all x E P, (x, 1) and (l,x) are in D, and xl = Ix = x, and (P2) for all x E P, (x,x- I ) and (x-I,x) ED, and xx- I = x-IX = 1. The sets (P, i, 1, D, m), usually referred to as (P, D), form a category C whose morphisms 'P E C((P,D),(Q,E)) are set maps 'P: P ~ Q such that (x,y) E D implies ('P(x),'P(y)) E E and 'P(xy) = 'P(x)'P(y). There is a

348

F. Rimlinger

forgetful functor from groups to C whose left adjoint U(P) is presented by (P : {xy(m(x, y)-l) : (x, y) E D}). Let F(P) be the free groups on the set P. Then the map l: P --+ U(P) is the restriction of the quotient map F(P) --+ U(P), and for every C morphism 'P : P --+ G whose codomain is a group, there is a unique group homomorphism h : U(P) --+ G such that ht = 'P. U(P) is called the universal group of P. In certain cases, the map l : P --+ U (P) may be injective. By adding two more axioms to (PI) and (P2) above, we can guarantee t is injective and say interesting things about the structure of U(P). A few more preliminaries are required. Given an object (P,D) of C, a P-word (Xl, ... , Xn) E pn of length n represent., Xl ... Xn E U(P). This P-word is P-reduced if each adjacent pair (Xi,Xi+l) ~ D . .We also write (Xl, ... ,Xn)D if each pair (xi,xi+d E D. If (W,X,Y)D, both (W,XY)D and (WX,Y)D, and w(xy) = (wx)y then (w, X, y) associates. Given P-words X = (Xl, ... , xn) and A = (ao = l,al, ... ,a n -},a n = 1) such that (ai_'\,xi,ai) associates for i = 1, ... ,n, define X interleaved by A to be the P-word (xla},a1IX2a2,"" -1 ) an_Ix n .

1. DEFINITION (Stallings [S2]): Let (P, D) be an object of C. Suppose that

(P3) for all w,X,y E P, if (W,X,Y)D, then (W,XY)D or (WX,Y)D implies ( w, x, y) associates, and (P4) for all w,X,y,z E P, (w,x,y,Z)D implies (W,XY)D or (Xy,Z)D' Then (P, D) is a pregroup. (Originally there were five axioms altogether, but one of them was found to be redundant.)

2. THEOREM (Stallings [S2]). Let (P, D) be a pregroup. Let X and Y be P-reduced representations of an element ofU(P). Then X and Y have the same length, and there is a P-word A such that Y is X interleaved by A. In particular, t : P --+ U(P) is injective. There is a converse to theorem 2 which asserts that if a group has a notion of length derived from reduced words, then it may be realized as the universal group of a pregroup. We shall use this result in the sequel. 3. THEOREM [Rl]. Let G be a nontrivial group. Suppose PeG, P = p- 1 , and P generates G. Suppose D C P x P, and suppose that (x, y) E D implies xy E P. Clearly (P, D) E C. By abuse of notation, the P-words


349

represent elements ofG. Suppose all the P-reduced words representing the same element of G are of the same P-length. Then (P, D) is a pregroup and the natural map h : U(P) -+ G is an isomorphism. There is a strong connection between pregroups and graphs of groups. See [R2] for details. Hoare [H] has generalized some of the results in [R2]. In this regard, all we need for this paper is the oldest result of Stallings [SI] on this subject concerning the structure of partial groups. 4. DEFINITION: Let (P, D) be a pregroup. Suppose P actually satisfies a stronger version of (P4), namely: (P4') for all W,x,Y E P, if x

#

1 and (W,X,Y)D, then (WX,Y)D.

Then (P, D) is called a partial group.

5. LEMMA. Suppose (P, D) is a pregroup. Then (P, D) is a partial group {::} for all P-words X and Y representing the same element ofU(P) we have X=Y. PROOF: ( 0, there exists Xt E PH such that L(xt) = t and R(xt) = o. Let F be the free group on the set of positive non-zero real numbers. Let Ho = {f E PH : L(J) = 0 = R(J)}. Then H = F * Ho. PROOF: Suppose x, y E PH - 1. Then x-1y E PH {::} IL(x)1 = IL(y)1 by lemma 1. From definition 2.8, x and y are in the same vertex of PH {::} x-1y E P {::} IL(x)1 = IL(y)l. For each t ~ 0, let Vi be the vertex Vi = {x E P : IL(x)1 = t}. The hypothesis on H implies that PH has exactly one component. Theorem 2.13 now yields the result. 0 We can obtain further splitting results by restricting our attention to the subgroup of G generated by the monotonic functions. 9. DEFINITION: We say f EGis monotonic if (i) for all x, y E [O,8(J)] such that x ::; y, we have f(x) ::; fey), or (ii) for all x,y E [O,8(J)] such that x ::; y, we have f(x) ~ fey). In particular, 1 ::; G is monotonic. Let M be the set of monotonic elements of G, and let (M) be the subgroup of G generated by M. Let

Mo

= {f E

(M) : L(J)

= 0 = R(J)

and If I is continuous}.

Evidently (M) satisfies the hypotheses of corollary 8, so (M) = F * Mo, where F is the free group on the positive real numbers. We construct a partial group structure for M o, further splitting (M). DEFINITION: P = {f E Mo; f(t) {(J, g) E P x P : fg E Pl·

10.

#-

0 for t E (0,8(J»)}, and D =

Before examining the proof of lemma 11 below, the reader may wish to contemplate the functions of the form f = o. Notice any such function satisfies the property f(t) = f(8(J) - t) for all t E [O,8(J)] and therefore is not reduced in the sense of definition 1.1 in the event 8(J) > O. Thus the only function of the form f = 0 representing an element of Promislow's group G has domain of the form [0,0]. By the discussion at the beginning of section 1, all functions with domain [0,0] represent the identity element of G. Similarly, no element of Promislow's group has a reduced factorization in which one of the factors is of the form f = O.

11. LEMMA. P is a partial group structure for Mo. PROOF: Let f E Mo. Let Zf = {t E [O,8(J)] : f(t) = O}. Since Mo eM, and M is generated by monotonic functions, the discussion directly above


353

lemma 11 indicates that Z f is finite. Let to, ... , tn be the elements of Z f ordered by increasing size. Notice to = 0 and tn = 6(f). For each i, i = 1, ... , n, let Xi(t) = J(t; - t), for t E [0, t; - t;_I]. Then (Xl> ... , xn) is P-reduced representing J. It is clear that this representation is unique, and hence the result follows from corollary 2.6. 0

In order to apply the structure theorem for partial groups, we need the usual notion of germs of functions. 12. DEFINITION: Let S c G be any set of functions. Then the set of germs of S is SI "', where b '" h if there exists e > 0 such that 6(fd ~ e, 6(f2) ~ e, and the restriction of Jl to [0, $] equals the restriction of h to

[O,e]. 13. THEOREM. Let 9 be the set of germs of P. Let a E 9 be a fixed germ. Let F(g - {a}) be the free group on the set 9 - {a], and let

Me< = {f E P : J = 1 or both J and J- 1 are in a}. Then Mo ~ F(g - {a}) * Me: r --+ ~ be a map of G-graphs. Suppose that r is an edge-free G-tree. Then there exists an edge-free G-tree maps ofG-graphs E: r --+ rand ¢: r --+ ~, such that ¢> = ¢E, and such that E is surjective and ¢ admits no free fold.

r,

SKETCH OF PROOF: Roughly, we have the situation that ¢> factors through a surjective map E and a map ¢ which admits no free fold, in that ¢ does not collapse any pair of edges (,8, ')') which are in distinct orbits and which have a common initial vertex. The idea of the proof of this is that if ¢> admits a free fold, we fold; by Theorem 4, the resulting G-graph is still a G-tree, and by Lemma 3,

360

J.R. Stallings

there is a factoring as in the statement of Theorem 5. We can do this any finite number of times; in fact, we can continue transfinitely if necessary: At limit ordinals, we create the approximation to as a direct limit of edge-free G-trees. Now, a direct limit of trees is a tree, since "tree" is homologically defined and homology commutes with direct limit; and a direct limit of edge-free things has the edge-free property, by Lemma 1. At some transfinite time, this process has to stop and the theorem will be established. 0

r

The reader may notice that the above argument is simply the transferral to the geometric, tree situation, of the topological argument in [St]. What was a "binding tie" in [St] is a ''free fold" here.

6. THEOREM. Suppose that : r --+ ~ is a map of G-graphs, where both ~ and r are trees, and where r is edge-free. Furthermore, suppose that admits no free fold. Then at least one of the two following alternatives holds: (1) riG --+ ~/G is injective. Or, (2) There is an edge e ofr, with initial vertex v, such that E(v) n E((e))

( v) and 4>( w) belong to the same orbit; that is, there exists 9 E G such that 4>(v) = g( w) = 4>(gw). If we replace w by gw, we see the following: (*) There are vertices v, w E V(r), such that v and ware in different orbits, and such that 4>( v) = 4>( w). Since r is connected, v and w are joined in r by a path p of some length n. We can now minimize n among all pairs of vertices satisfying (*). The number n is not zero, since the path p connects vertices in different orbits, and this implies that v -# w. The path 4>(p) is a closed path in the tree ~, of length n -# 0, and therefore it must have a "back-tracking." Thus, one can write

Foldings of G-trees where e l and e 2 are edges of

r

361

with the same initial vertex, such that

¢(e l ) = ¢(e 2 )· If e l and e l were in different orbits of the edges of r, this would be a free fold, which is contrary to hypothesis. Since it is indeed a fold, it is untwisted, and therefore there is 9 E G, such that e l = ge 2 • This element 9 stabilizes the initial vertex of e l and it also stabilizes the image edge ¢( e l ), since g¢(e l ) = ¢(ge l ) = ¢o(e 2 ) = ¢(e l ). Now, if alternative (2) of the Theorem were not the case, then 9 would belong to K(fl). In other words, 9 would stabilize everything in fl. Under this supposition, we can examine the paths PI and gP2 in r; the rightmost endpoint of PI is r( e l ), and the leftmost endpoint of gP2 is gr(e 2 ) = r(ge 2 ) = reel)' Thus we can define:

and we see that p is a path in r between v and gw, which are still in distinct orbits; and ¢( v) = ¢( to) = g¢( to) = ¢(gw), since 9 stabilizes ¢( to) in fl. But this path p would have length n - 2, contradicting the assumption on the minimality of n. 0 The hypotheses in Theorem 6 can be weakened slightly. The exact properties which were ui:>'ed were that r is connected, and that fl is a forest.

§5. Relating graphs to groups. If r is a G-tree, then Serre [Se] has described how to relate the structure of G to stabilizers of representative vertices and edges, in terms of amalgamated free products and HNN extensions. We shall single out three specific cases of this: Amalgamation: riG consists of one edge-pair and two vertices. Then let e be any edge of r, and consider the diagram of inclusions of stabilizers: ~(l(e))

f-

~(e)

-+

~(r(e)).

Then G is the pushout of this diagram, and is conventionally called the free product with amalgamated subgroup, written thus: G = A *B C. In this, A = ~(l(e)), B = ~(e), and C = ~(r(e)). The choice of a different edge than e, simply changes the picture by an inner automorphism of G.

362

J.R. Stallings

HNN: riG consists of one edge-pair and one vertex, thus giving the appearance of a circle. Let e be any edge of r. Define v = tee). Then there exists t E G such that r( e) = tv. Define A = E( v) and B = E( e). Then E (r( e)) = tAt- l and B = A n tArl. The group G can be presented by adding a formal generator T to a presentation of A, and for every b E B a relation b = Ta( b)T- l , where a: B --+ A is the homomorphism given by a(b) = rlbt. This construction is due to Higman, Neumann, and Neumann; called an HNN extension, and conventionally written something like G = A * B a. Edge-free case: Suppose that r is an edge-free G-tree. In the quotient graph riG, select a maximal tree T and lift that tree back to a tree T' contained in r. Furthermore, pick an orientation of the edges of (r IG) \ T, and for each such oriented edge e, pick a preimage edge e' in r, such that tee') is in T', and choose an element ge E G, such that ger(e') is in T'. Then the collection of elements {gel form a basis of free subgroup F of G. Let {vi} be the collection of all vertices of T'; define E; = E(v i ); let Ui Ei denote the free product of these subgroups of G. The structure of G in this case is G = F * Ui E i . Conversely, if we are given, abstractly, a group G which can be written in this form, for F free, we can invent an edgefree G-tree r, in which riG consists of a wedge of circular arcs (perhaps divided into several edges), one for each basis element.of F, together with a wedge of simple arcs (likewise divided into many edges) attached at a basepoint; the other endpoint of such a simple are, opposite the base-point, corresponds to a vertex in r, whose stabilizer is E i ; the preimages in r of all vertices except for those at the ends of these simple arcs have trivial stabilizers.

§6. The Kneser-Grushko-N eumann-Wagner-Stallings-Higgins-etc. Theorem. Consider an edge-free H -tree 6, for which we write, according to the previous section, H = F*Ui E;, for F free. And now, suppose that we have another group G of similar form, G = F' * Uj Ej. Suppose, furthermore, that we have a surjective homomorphism -I(B), and so that

~(¢(L(e»)) =

A or C.

We note that

K(b..), in the context that ~ is considered an F-tree, contains the kernel of 4>: F ~ H. Finally, we note that, since r is an edge-free F-tree, ~(L(e») is a free factor of F. Thus, ~ (L( e)) is the factor Fp such that 4> of it, which is contained in FI n 4>-I(B)

ct

~ (1/' (L( e») ), is contained in A or in C. We have

ker 4>, which translates to say that 4>(Fd n B

i- {I}.

0

9. THEOREM. Let H = A*Ba; thus H is generated by its subgroup A and an additional element t, such that B = An tArl, with relations saying that for all b E B, b = ta(b)r l . Suppose that B i- {I}. Let F be a free group, and let 4>: F ~ H be a surjective homomorphism. Then F has a free factorization F = FI * F 2 , such that one of the following symmetric alternatives is true: (1) FI C A and FI n B (2) FI C tAt-I, and FI

i- {I}. Or, n B i- {I}.

The proof is very similar to that of Theorem 8.

§8. Unsplittable subgroups. A subgroup S of a free group F is called "unsplittable in F," when: For every free factorization F = FI * F2 , if S n FI i- {I}, then S c Fl· 10. PROPOSITION. Suppose that SeT c F are groups, where F is free. Then: (i) If T is unsplittable in F, then S is ullsplittable in F. ( i i) If S is unspJi t table in T, then S is unspli t table in F. PROOF: (i) Suppose F = FI *F2 and SnFI i- {I}. Since TnFI contains S n F I , then Tn FI i- {1}. Since T is unsplittable in F, we have TeFl' and so S C Fl. (ii) Suppose F = Fl * F2 and S n FI i- {I}. By the Kurosh subgroup theorem applied to T as a subgroup of FI * F 2 , we see that T = (T n F I ) * G for some G. Then Sn(TnF1 ) = SnFI i- {I}; and so, since Sis unsplittable in T, we have S c TnFI , and thus S c Fl. 0

11. PROPOSITION. Every cyclic subgroup S of a free group F is unspJittable in F.

366

J.R. Stallings

PROOF: This follows from lO(ii), since S is unsplittable in itself; i.e., a o cyclic group has no non-trivial free factorization. 12. PROPOSITION. Suppose that w E F, a free group of rank at least 3, and that the l-relator group H = (F I w = 1) has no non-trivial free factorization. Then the normal subgroup «w}) , generated by w and its conjugates, is unsplittable in F. (And hence, by lOCi), any subgroup of «w}) is unsplittable in F.) PROOF: Let F = FI * F2 be a non-trivial factorization of F. The Freiheitssatz implies that «w)} n FI = {1}, or else that a cyclically reduced conjugate w' of w written in terms of generators of FI and F2 will not have any of the F 2 -generators in it. In the latter case,

Fj«w}} = Fj«w'}} = (FIj«w'}})

* F2

would be a non-trivial factorization. (If FI were cyclic, generated by w', then the quotient H would be isomorphic to F 2 , which has, since the rank of F is supposed to be at least 3, non-trivial factorizations.) What we have in fact shown is that under the hypotheses of this proposition, every non-trivial free factor of F intersects «w}) trivially, and this is rather more than to say that «w}) is unsplittable. 0 The hypothesis that a I-relator group has no non-trivial free factorization can be verified, for instance, if it can be computed that that group has only one end. Closed 2-manifold groups are examples of this. I wonder if there are any interesting examples derived from 3-dimensional topology.

§9. Theorems of Shenitzer and Swarup. 13. THEOREM [Shenitzerj. Suppose that a free group F is an amalgamated free product F = A *Be, in which the amalgamated subgroup B is cyclic. Then B is a free factor of A or a free factor of C. PROOF: By Theorem 8, taking the identity map from F to itself, we know that one of two symmetric alternatives is true; let us suppose the alternative is that F = FI *F2 , such that FI C A and FI nB -1= {1}. Now we apply the Kurosh subgroup theorem to C as a subgroup of FI * F 2 ; this implies that C has a free factor of the form C n Fl' Since Bee, that free factor of C contains B n F I , which is non-trivial; since B is cyclic, it is unsplittable in

Foldings of G-trees

C (Proposition 11); and therefore Bee free factor C n FI of C is in fact B.

n FI c en

A

=

367

B. Thus, the 0

14. THEOREM [Swarup]. Suppose that a free group F is an HNN extension F = A *B a, in which the amalgamated subgroup B is cyclic. As usual, we express F in terms of A and an extra generator t, such that B = A n tAt-I. Then A has a free product structure A = Al * A 2, in such a way that one of the following symmetric alternatives holds: (1) Be Ap and there exists a E A such that t- l Bt = a-I A 2a. Or, (2) rl Bt c AI' and there exists a E A such that B = a-I A 2a. PROOF: We apply Theorem 9, using the identity map F -+ F = A *B a. Suppose that alternative (1) in the conclusion of Theorem 9 is the case: Thus, F = FI *F2, FI C A, and FI nB 1= {I}. Apply the Kurosh subgroup theorem to the subgroup A C FI * F2 • Then A has a factorization with several factors, two of which are

Al

= A n FI = F I ,

A2

=

An 8FI 8- 1 .

There is a delicate point here: 8 is the representative of the coset At- l III the Schreier system of representatives which are chosen with respect to the free product structure FI * F 2. Thus, 8 = at- I for some a E A. Now, Al contains B n F1 , a non-trivial subgroup of B; and since B is unsplittable in A (by Proposition 11, since B is cyclic), it follows that B is contained in AI; and, of course, Al = Fl. Furthennore, C l Bt

= An c

l

At :::> An

c

l

FI t :::> An c

l

Bt

=c

l

Bt.

Thus, A2

=

An aC I FI ta- l

=

a(A n c l Flt)a- l

=

aC I Bta- l ,

or t- l Bt = a-I A 2a. This means that we have established alternative (1) of the conclusion of the theorem. The other alternative comes out from the case of alternative (2) of Theorem 9. 0 As an example to think about, here is an HNN extension which is free: (a,b,t I taC I

=

baba-Ib- l ).

This gives an example where the mysterious "a" in the conclusion of Theorem 14 has to be taken into account. Of course, we could look at it differently, changing the stable letter of the HNN -extension in this instance from t to a-Ib-It.

368

J.R. Stallings REFERENCES

[K] H. Kneser, Geschlossene Fliichen in dreidimensionalen Mannigfaltigkeiten, Jahresb. der Deut. Math.-Verein. 38 (1929), 248-260. [G]I.A. Grusko, On generators of a free product of groups, Matern. Sbornik N. S. 8 (1940), 169-182. [Le] F.W. Levi, On the number of generators of a free product, and a lemma of Alexander Kurosch, Indian Math.Soc. N. S. 5 (1941), 149-155. [N] B.H. Neumann, On the number of generators of a free product, J. London Math. Soc. 18 (1943), 12-20. [Sh] A. Shenitzer, Decomposition of a group with a single defining relation into a free product, Proc. Amer. Math. Soc. 6 (1955), 273-279. [W] D.H. Wagner, On free products of groups, Trans. Amer. Math. Soc. 84 (1957), 352-378. [Ly] R.C. Lyndon, Grushko's theorem, Proc. Amer. Math. Soc. 16 (1965), 822-826. [St] J.R. Stallings, A topological proof of Grushko 's theorem on free products, Math. Z. 90 (1965), 1-8. [H] P.J. Higgins, Grushko's theorem, J. Algebra 4 (1966), 365-372. [Z] H. Zieschang, tiber die Nielsensche Kurzungsmethode in freien Produkten mit Amalgam, Inv. Math. 10 (1970), 4-37. [C]I.M. Chiswell, The Grushko-Neumann theorem, Proc. London Math. Soc. (3) 33 (1976),385-400. [Sill] N. Smythe, A generalization of Grushko 's theorem to the mapping cylinder group, Notices Amer. Math. Soc. 23 (1976), A-419. [B] H. Bass, Some remarks on group actions on trees, Comm. Aig. 4 (1976), 1091-1126. [Se] J.-P. Serre, "Arbres, Amalgames, SL2," Asterisque, 1977. [Sc] G.P. Scott, Strong annulus and torus theorems and the enclosing property of characteristic submanifolds of 3-manifolds, Quart. J. Math. Oxford (2) 35 (1984), 485-506. [Sw] G.A. Swarup, Decompositions of free groups, J. Pure and Appl. Alg. 40 (1986), 99-102.

Keywords. free group, free product, amalgamation 1980 Mathematics subject classifications: 20E05, 20E06, 05C25,05C05 Mathematics Department, University of California, Berkeley CA 94720 This work is based on research partly supported by the National Science Foundation under Grant No. DMS-8600320.