uniform tree lattices

JOURNAL OF THE AMERICAN MATHEMATICAL SOCIETY Volume 3, Number 4, October 1990

UNIFORM TREE LATI1CES

HYMAN BASS AND RA VI KULKARNI

TABLE OF CONTENTS

o. Introduction

1. Graphs of groups of finite index; unimodularity Appendix A: Nonabelian cohomology of a graph

2. Finite groupings of edge indexed graphs; volumes

3. Automorphism groups of locally finite trees; unimodularity

4. Existence, conjugacy, and commensurability of uniform lattices

5. Volumes, Euler characteristics, and ranks

6. Finiteness properties Appendix B: Commensurators

7. Nonfiniteness phenomena

References O. INTRODUCTION

Let X be a locally finite tree. Then G = Aut(X) is a locally compact group; the stabilizers G x are open and profinite. A subgroup r:5 G is discrete if all rx are finite. We then call r a lattice if

Vol(I\ \X) = L Tfl xEf\X x

is finite, and a uniform lattice if the graph I\X is finite. These are the objects of our study.

It is fruitful to think of (G, X, r) as a combinatorial analogue of (PSL2(R), upper half-plane, Fuchsian group). An even more direct analogy

Received by the editors February 28, 1990. 1980 Mathematics Subject Classification (1985 Revision). Primary 05C25, 20F32, 22E40. The authors were partially supported by NSF grants DMS 8802181 (first author) and DMS 8902

214 (second author).

© 1990 American Mathematical Society

843

License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

844 HYMAN BASS AND RA VI KULKARNI

is with PSL2(K) , K a local nonarchimedean field, and its Bruhat-Tits tree. Our results show that, while these analogies are far reaching, they also break down in dramatic ways. For example, the Commensurability Theorem below implies that, if one uniform lattice is "arithmetic," then all are!

The first question is: When does X admit a uniform lattice? The obvi-ous necessary homogeneity condition, that G\X be finite, turns out to be not sufficient.

Existence Theorem. X admits a uniform lattice iff G\X is finite and G is unimodular (in the sense of Haar measure).

In this case we call X a uniform tree. More generally, H :5 G contains a uniform lattice iff H\X is finite and H (closure of H) is unimodular. This gives Lubotzky's results [Lub] on the existence of uniform lattices in rank one simple Lie groups over local fields.

A uniform lattice on X contains a finite index subgroup acting freely on X . Hence, uniform trees are just the universal covers of finite connected graphs.

Let X be a uniform tree. Put Latu(X) = the set of uniform lattices on X.

Commensurability Theorem. If r 0' r 1 E Latu (X), then grog -1 and r 1 are commensurable for some g E G.

This is equivalent to the special case when r 0 and r 1 are free lattices, in which case the result can be reformulated, using covering theory, as

Leighton's Common Covering Theorem [L]. If two finite connected graphs have a common covering, then they have a common finite covering.

Indeed, it was the desire to give a covering space theoretic proof of Leighton's theorem that originally inspired this paper. The second named author produced a strategy for such a proof in [K]. However, the theory of "branched coverings of graphs of groups" needed to execute that argument did not exist at the time. That theory has since been constructed in [B], and the present paper is an exploitation of that theory.

The Commensurability Theorem implies that the commensurator of r E Latu(X) ,

CG(r) = {g E Glgrg -1 and r are commensurable},

depends, up to conjugacy, only on X. We sometimes denote it C(X) , and call it the commensurability group of X. It is countable, and C(X)\X = G\X.

Conjectures. I. C(X) is dense in G. II. Let r :5 G and let Y be a r-invariant subtree of X on which r acts

discretely with finite quotient r\y. Then there exist r E Latu(X) and an injective homomorphism h: r -+ r+ such that h(g)IY = glY for all g E r.

We can show that II (for finite r) implies I, that II holds when r is free, and that I and II hold when X is a homogeneous tree or if G\X is a tree.


UNIFORM TREE LATTICES 845

Finiteness properties of lattices. Let r E Latu(X). Put m(r) = IcmlEl , where F varies over finite subgroups F of r. There are only finitely many conjugacy classes of such F, and r contains a free subgroup of index m(r). Moreover NG(r)/r is finite.

For mo > 0,

umO(r) = {r' ~ GI r ~ r' and [r' : r] ~ mol

is finite. For vo' mo > 0,

Lat~o.mO(X) = {r E Latu(X)1 Vol(r\ \X) ~ Vo and m(r) ~ mol

forms finitely many G-conjugacy classes. Put

L(r, G) = {h E Hom(r, G)I h is injective and h(r) E Latu(X)}.

If X is not "virtually linear" (i.e., if r is not virtually infinite cyclic), then

G\L(r, G) / Aut(r) is finite, where G acts by conjugation and Aut(r) by composition.

Nonfiniteness phenomena. Let Xn denote the n-homogeneous tree; each vertex has index n. For n ~ 3, there exist infinite ascending chains of r E Latu(Xn)' In particular, Vol(r\ \Xn ) -+ 0 as r increases.

For n ~ 5 , given any integer v > 0, v even if n = 5 , there exist infinitely many conjugacy classes of r E Latu(Xn) with Vol(r\ \Xn) = v . Methods. Our method is based on interpreting tree actions as graphs of groups (cf. [S, (1.5)] or [B]). A graph of groups A = (A, oN) consists of a connected graph A, vertex groups ~ (a E A), edge groups ~ = ~ (e E E(A)) , and monomorphisms a£: ~ -+ ~ e' We put i(e) = [~e : ae~] and call o 0 (A, i) = J(A) the corresponding edge-indexed graph. If ao E A, then the fundamental group r = 1l'1(A, ao) acts without inversion (of edges) on the covering tree X = (A~o), with quotient p: X -+ r\X = A so that, for e E E(X) , ap(e): .w;,(e) -+ .w;,(ooe) is isomorphic to the inclusion re ~ r ooe . Conversely, every action without inversion of a group r on a tree X arises this way from a quotient graph of groups r\ \X = (r\X , .N) .

Thus, constructing a tree action (r, X) is equivalent to constructing a graph of groups A = (A,.N). We do this in two steps. First construct J(A) = (A, i). In fact X = (A~o) depends only on (A, i) , so we also write X = (A ,7: ao) . Given (A, i), a graph of groups A with J(A) = (A, i) is called a "grouping" of (A, i); a "finite grouping" if all ~ (a E A) are finite, and an "effective grouping" if r = 1l'1 (A, ao) acts effectively (Le., faithfully) on X. Thus, to construct a uniform latti~n a tree X we first seek a finite edge-indexed graph (A, i) so that X = (A, i, ao) (for a base point ao E A), and we then seek a finite effective grouping A of (A, i). Then r = 1l'1 (A, ao) is the desired uniform lattice, and Vol(r\ \X) = Vol(A) = L:aEA l/I~I. This problem is an interesting mix of combinatorics and finite group theory.


846 HYMAN BASS AND RAVI KULKARNI

Let (A, i) be a finite edge-indexed graph, ao EA. In §§ 1 and 2 we show that (A, i) admits a finite (effective) grouping iff (A, i) is "unimodular" in the sense that, for any closed edge-path y = (e l , ••• ,en) at ao E A, n7=1 i(ei) = n7=1 i(eJ. In any event, the formula dey) = n7=1 i(ei)/i(ei) defines a homo-morphism

d: 7r 1(A, ao) _ QX, which is trivial iff (A, i) is unimodular.

Suppose that (A, i) = I(H\ \X), where H is a group of automorphisms without inversion of a locally finite tree X . There is a natural projection p: H = 7r1 (H\ \X, ao) -+ 7r1 (A, ao), whence the composition

H x d =dop:H-Q .

Let H denote the closure of H in G = Aut(X) . In §3 we show that d H = dHIH and that d H is the modular homomorphism, in the sense of Haar measure. Thus, when H\X is finite, we have the following equivalent conditions:

(a) H is unimodular. (b) I(H\ \X) is unimodular. (c) I(H\ \X) admits a finite effective grouping. (c) There is arE Latu(X) with r\x = H\X.

Under conditions (a) and (b), and hence all of the above conditions when H\X is finite, we say simply that" H is unimodular."

The next task is to try to embed a r as in (d) into H, via a conjugation in G. For this we need an appropriate notion of "covering morphism" <1>: r\ \X -+

H\ \X of graphs of groups, that will give rise to the desired embedding. It is here that we invoke the theory of such covering morphisms developed in [B]. Using that theory, it can be shown that, for a free subgroup r of finite index in r, there is a covering morphism r\ \X - H\ \X , whence a conjugation of r into H. This is contained in the following Conjugacy Theorem.

Let X be a tree, G = Aut(X) , H S G without inversions, and put GH = {g E Gig stabilizes all H-orbits on X}.

Conjugacy Theorem [B, (5.2)]. If r S GH acts freely on X, then grg- I S H for some g E G H .

From this we deduce Uniform Lattice Theorem. Assume that X is locally finite, H\X is finite, and H is unimodular.

(a) There is a lattice S GH such that \X = H\X . (b) If r S G H acts freely on X, then r can be G H-conjugted into Hand

into <1>.

Using conjugation into H gives the Existence Theorem. Using conjugation into CI> gives the Commensurability Theorem. These and many related results are obtained in §4.


In §5 we discuss volumes, Euler characteristics, and other numerical invari-ants attached to uniform trees and their lattices. In §6 we establish the finiteness properties announced above. Following this is an appendix on commensura-bility groups of uniform tree lattices, and on their action on the ends of the corresponding trees.

The nonfiniteness phenomena are illustrated in §7 by a great variety of exam-ples. The game here is to produce finite unimodular edge-indexed graphs (A, i) with a prescribed covering tree X , and then either to find arbitrarily large finite effective groupings of (A, i) or else to prove that there are only finitely many such. For example, consider

(A, i) = 0 rno rn l 0 •

Then X is the bihomogeneous bipartite tree with indices mo and m l • A finite effective grouping of (A, i) is an "amalgam", Go ~ H ~ G1 ' where the Gj are finite groups, [Gj : H] = mj (i = 0, 1), and if N ~ H is normal in each Gj ,

then N = 1 . Assume that mj ~ 2 (i = 0, 1). If mo or m 1 is composite, then there exist infinite ascending chains of finite effective groupings of (A, i) (cf. (7.13)), and hence infinite ascending chains of edge-transitive lattices on X. On the other hand, if mo and m 1 are prime, then there are conjecturally only finitely many effective (mo' m1)-amalgams, i.e., only finitely many conjugacy classes of edge-transitive lattices on X. This follows for mo' m l ~ 3 from a well-known work of Goldschmidt [Go]. It has been proved more generally by Fan [F], whenever H is a p-group for some prime p t- mo or m l •

1. GRAPHS OF GROUPS OF FINITE INDEX; UNIMODULARITY

Our background reference here is [B]. (1.1). Let A = (A,.9I') be a graph of groups. For e E E(A), 80e = a, we put

(1) ~/e = ~jae(~)

and

(2)

We shall assume that" A has/mite index," i.e., that all ofthese indices i(e) are finite. We then call

(3) /(A) = (A, i)

the edge-indexed graph of A, and further define

(4) q(e) = i(e)ji(e) E Q:o for e E E(A). Note that

(5) ) -I q(e = q(e) .

Thus q defines a homomorphism

(6) q: ll(A) -> Q:o'



where, viewing A as a graph oftrivial groups, 1l'(A) is defined as in [B, I, (1.5)] by the presentation

(7) 1l'(A) = (E(A)le = e -I for all e E E(A)}.

If y = (e l , ••• , en) is an edge path in A we put Iyl = el ... en E 1l'(A) and q(y) = q(lyl) = q(el )· .. q(en)· Putting y = (en"'" el ) we have q(y) = q(y)-I.

Choose a base point ao E A and consider

1l'1 (A, ao) = {Iyl E 1l'(A)ly is a closed path at ao}.

By restriction, q defines a homomorphism

.1a (q): 1l'I(A, ao) -+Q;o' o

Composing with the natural surjection 1l'1(A, ao) .!.. 1l'I(A, ao) (cf. [B, (1.5), Example 3]) we obtain

A x .1a =.1a(q)op:1l'I(A,ao)-+Q>o·

o 0

( 1.2) Proposition. The following conditions are equivalent.

(a) .1~ = 1. o (b) For any closed edge path y = (e l ' ••• , en) in A,

q(y) (= q(el ) .. ·q(en» = 1.

(c) For any edge path y = (el , ... , en) in A, q(y) depends only on the end points of y .

(d) There is a function N: A -+ Q;o such that I .

q(e) = N(8oe)N(8Ie)- for all e E E(A).

Under these conditions, the function N in (d) is unique up to a constant factor.

Let y = (e l ' ••• , en) be an edge path in A with vertex sequence bo, bl ' ... ,

bn . Assuming (d) we have q(y) = N(bo)N(bn)-I, whence (c). Moreover, (c) :::> (b) because a closed path at a has the same end points as the empty edge path at a. Obviously, (b):::> (a).

Finally, to prove (a):::> (d), define Na (a) = q(y)-I , where y is any edge o

path from ao to a. Given e E E(A) let Yi be an edge path from ao to 8ie (i = 0,1). Then y = (yo' e, YI) is a closed path at ao so, by (a),

-I 1 = q(y) = q(Yo)q(e)q(yl ) = Nao (8oe) q(e)NOo (8 I e),

so q(e) = Na (8oe)Na (8I e)-1 , whence (d). o 0

Suppose also that q(e) = N'(8oe)N'(8I e)-1 for all e, for some N': A -+ Q;o' Put M(a) = Na (a)-I N'(a). The hypothesis

o


implies that M(8oe) = M(81e) for all e E E(A). Since A is connected, M is constant.

A more general (nonabelian) version of (1.2) is presented in Appendix A, Proposition (A.2).

(1.3) Definitions. Under the conditions of (1.2) we call A and (A, i) unimodu-lar. The functions N: A -+ Q;o such that q(e) = N(8oe)jN(81e) as in (1.2)(d) are called vertex orderings of (A, i). We call N integral if, for all e E E(A) , N(8oe)ji(e) E Z and (hence) N(a) E Z for a EA. We further define

VoIN(A, i) = L IjN(a) (:5 00) aEA

and say that (A, i) has finite volume, denoted Vol(A, i) < 00, if VoIN(A, i) < 00 for some N. If N' is another vertex ordering, then N' = wN for some w E Q;o' and so VolN(A, i) = w VoIN,(A, i). Hence, finiteness of Vol(A, i) does not depend on the choice of N.

Let (A, i) be unimodular with vertex ordering N. For e E E(A) put r(e) = N(8oe)ji(e). We say that (A, i) has bounded denominators if the ra-tional numbers r(e) (e E E(A)) have bounded denominators. As above, this condition does not depend on the choice of N.

( 1.4) Proposition. The following conditions on (A, i) are equivalent. (a) (A, i) admits an integral vertex ordering N: A -+ Z>o. (b) (A, i) is unimodular and has bounded denominators.

Under these conditions there is a unique integral vertex ordering N min such that every other one is of the form N = m . N min for some -integer m > 0 .

Proof. If N is an integral vertex ordering of (A, i), then (A, i) is unimodu-lar, by (1.2), and the rational numbers r(e) = N(8oe)ji(e) are integers, hence have bounded denominators. Conversely, assuming (b), (A, i) admits a vertex ordering N so that the numbers r(e) = N(8oe)ji(e) have bounded denomi-nators. Adjusting by a scalar factor we may assume that N(ao) = 1 for some ao EA. This done, let D be the least common denominator of the numbers r( e) , and put N min = D . N. Then, by construction, N min is an integral vertex ordering. If N' is any integral vertex ordering we can write N' = q . N for some q E Q. Then N' (ao) = qN(ao) = q E Z. Moreover, for all e E E(A), N' (8oe)ji(e) = qN(8oe)ji(e) = qr(e) E Z. It follows that q is a common de-nominator for the r(e)'s; hence q = mD and N' = mNmin for some integer m > o. This proves the proposition.

(1.5) Examples. I. Suppose that

(A.i)=a 8e. This is unimodular iff n = m, in which case q(e) = 1, and Nmin(a) = n.


2. Suppose

(A,i) =

is unimodular. The minimal integral vertex ordering is

54Q-----------~36

180 CF------------'O 48 3. For (A, i) unimodular the bounded denominator condition clearly holds

if the graph A is finite, but not in general. For example, if e l e2 e.

A=(~···~···),

then for the indexing i(en ) = 1, i(en ) = 2 for all n, (A, i) has bounded denominators, whereas, for the indexing j(en ) = 2, j(en) = 1 for all n, (A, j) does not. Note that all edge-indexings are unimodular since A is a tree.

ApPENDIX A: NONABELIAN COHOMOLOGY OF A GRAPH

The material here is not needed elsewhere in the paper. (A. 1). Let Q be a (multiplicative) group. For any set A let QA denote the group of set functions A -+ Q .

Suppose that A is a graph, with edge set E(A) and end point maps OJ: E(A) -+ A (i = 0, 1). Then we have a map

~: QA -+ (QE(A))-

defined, for N E QA , by (~N)(e) = N(ooe)N(ole)-1 ,

and

Recall that n(A) = (E(A)le = e -I for e E E(A)} ,

so that we can identify (QE(a))- = Hom(n(A), Q).

Choose ao EA. Then n l (A, ao) is the group of all elements e l ... en E n(A) such that (el , ... ,en) is a closed edge path in A at ao. For g E (QE(A))- we put



Finally, let y: Q -+ QA be the diagonal embedding sending SEQ to the constant function y(s): a 1-4 s (a E A). These maps assemble into the sequence of pointed sets

(1)

where the quotient on the right is by the conjugation action of Q and X is induced by ..1llo ' We shall show that the sequence (1) is "exact" in a strong sense.

(A.2) Proposition. Suppose that A is a connected graph. (a) y is an injective group homomorphism. (b) The fibers of t5 are the y( Q)-right cosets; t5 N' = t5 N iff for some SEQ,

N'(a) = N(a)s for all a EA. (c) The group QA acts on the set (QE(A))- by

(N· q)(e) = N(aoe)q(e)N(ale)-1

for N E QA, q E (QE(A))- , and e E E(A); t5 is the corresponding orbit map at the constant function ql: ql(e) = 1 for all e E E(A). We have

(2) ..1a (N· q) = ad(N(ao)) o..1a (q) . o 0

Moreover, the stabilizer (QA)q of q E (QE(A))- admits the isomorphism (QA)q -+

ZQ(Im(..1llo (q))) sending N to N(ao)' (d) The fibers X are the QA -orbits.

Proof. (a) Obvious. (b) Evidently, t5(Ny(s)) = t5(N) for N E QA and SEQ. Suppose that

N' E QA and t5N' = t5N. Let s = N'(ao)-I N(ao)' Replacing N' by N'y(s) , we may assume that N'(ao) = N(ao)' If e E E(A) and N(aoe) = N(aoe) , then, since t5N' = t5N, we have N'(ale) = N(ale). Thus, N' = N since A is connected.

(c) Let a E A and let 0 = (el ' ... ,en) be an edge path in A from ao to a, with vertex sequence ao' aI' ... , an = a. Let q E (QE(A))- = Hom(1l' (A) , Q) and N E QA. Then N· q: 1l'(A) -+ Q sends 101 = el .. ·en E 1l'(A) to

-I -I (N· q)(lol) = (N(ao)q(el)N(a l ) )(N(al )q(e2)N(a2) ) -I ... (N(an_l)q(en)N(an) )

= N(ao)(q(el )q(e2)··· q(en))N(an)-1

= N(ao)q(loI)N(a)-1 ,

whence formula (2), when a = ao' We further see that

'ff {N(a) = q(lol)-1 N(ao)q(lol) N·q=q 1 whenever 0 is an edge path from ao to a.



Taking a = ao' this condition implies that N(ao) centralizes q(711 (A, ao)) = Im(~a (q)). Further, N(ao) (and q) determine N, whence a monomorphism

o N ..... N(ao) from (QA)q to ZQ(lm(~ao (q))). To show surjectivity, let SEQ centralize q(711 (A, ao»' If a is an edge path from ao to a in A, then Ns(a) := q(lal)-lsq(lal) is well defined, since two such paths differ by compo-sition with a closed path at ao' By construction, Ns E (QA)q, and Ns(ao) = s.

(d) Formula (2) shows that ~ is constant on QA -orbits. Suppose that q, q' E (QE(A»- and

(3) ~a (q') = ad(h) 0 ~a (q) o 0

for some h E Q. For a E A define

(4)

where a is an edge path from ao to a. Any other such path is equivalent to one of the form a' = (P, a), where P is a closed path at ao' and we have q'(IPI) = hq(IPl)h-1 by (3). Hence

q' (la'l) -I hq(la'l) = q' (lal)-I q' (IPI) -I hq(lpl)q(lal)

= q'(lal)-lhq(lal),

so N is well defined. We conclude the proof by showing that N· q = q' . Let e E E(A) , 00e = a, ole = b. Let a be an edge path from ao to a, and P an edge path from ao to b. Then (a, e, P) is a closed path at ao' From (3),

q'(la, e, PI) = q'(lal)q'(e)q'(IPI)-1

coincides with hq(la, e, Pl)h- 1 = hq(lal)q(e)q(IPI)-lh-1

= q'(laI)N(a)q(e)N(bflq'(IPI)-1 ,

by (4). Hence, q'(e) = N(a)q(e)N(b)-1 = (N· q)(e) , as was to be shown.

2. FINITE GROUPINGS OF EDGE-INDEXED GRAPHS; VOLUMES

(2.1) Consider an edge-indexed graph (A, i). Thus, A is a connected graph and i assigns to each edge e E E(A) an integer i(e) > O.

For example, a graph of groups A = (A,.>1') of finite index gives rise to the edge-indexed graph /(A) = (A, iA), where iA(e) = [~e : ae(~)]' which is

o finite by hypothesis. If iA = i, then we call A a grouping of (A, i); a finite grouping if the groups ~ are all finite, and an effective grouping if A is an effective graph of groups (cf. [B, I, (1.24)]), i.e., if, for a E A, 7l 1(A, a) acts effectively on (A";'a). In general, if A' = (A, sf') is the effective quotient of A, in the sense of [B, I, (1.25)], then ;A' = ;A .



When A is a graph of finite groups we define its volume to be

(1) Vol(A) = L 1/1~1 aEA

(which is possibly infinite). We present criteria for the existence of finite groupings of (A, i) .

(2.2) (A, i) admits the following canonical infinite cyclic grouping Z(A, i) = (A, J2f), where ~ = ~ = Z for all a E A, e E E(A), and where Q.e : ~ --+

~ e is multiplication by i(e). In case this is not effective the effective quotient o

Z(A, i)' = (A, J2f') is an effective finite cyclic grouping of (A, i) ..

(2.3) Proposition. 1. If A = (A, J2f) is a finite grouping of (A, i), then N(a) = I~I (a E A) defines an integral vertex ordering of (A, i), in the sense of( 1.2), i.e.,

for all e E E(A) . 2. Let N: A --+ Z be an integral vertex ordering of (A, i). For e E E(A) put

N(e) = N(ooe)/i(e) (E Z). For an integer m > 0 let f.lm denote the complex mth roots ofunity. Then (A, i) admits the finite (cyclic) grouping A = (A, J2f), where ~ = f.lN(a) , ~ = f.lN(e) , and Q e : ~ --+ ~oe is the inclusion for all a E A, e E E(A) .

Proof. (1) Let ooe = a, ale = b. Then

N(a) I~I I~I 1.w;.1 iCe) N(b) = I~I = I~I . I~I = i(e)·

Assertion (2) is clear.

(2.4) Corollary. The following conditions on (A, i) are equivalent. (a) (A, i) admits an integral vertex ordering. (b) (A, i) is unimodular and has bounded denominators. (c) (A, i) admits a finite grouping.

Under these conditions, (A, i) admits an effective finite cyclic grouping A 0 = (A, J2fo) such that, for any finite grouping A = (A, J2f) of (A, i), there is an integer m = m(A/Ao) > 0 such that

o I~ I = m I~ I for all a EA.

Consequently, 1 0 Vol(A) = - Vol(A ). m

Proof. We have (a) ¢} (b) from (1.4) and (b) ¢} (c) from (2.3). Let Nmin be the minimal integral vertex ordering (cf. (1.4)), and let AO be the cyclic grouping of (2.3)(2) with I~ol = Nmin(a). If A = (A, J2f) is another finite grouping, then N(a) = I~I is an integral vertex ordering ((2.3)(1)) so N = m . Nmin for some integer m > 0, by (1.4). It remains to see that A 0 is effective.



Let A' denote the effective quotient of A 0 [B, (1.25») and put N' (a) = I~' I . Then N' = m . Nmin for some integer m > O. But N' (a) = I~'I divides Nmin(a) = I~ol since ~' is a quotient group of ~o. Hence, m = 1 so A' = A 0 • The corollary now follows.

(2.5) Corollary. If the graph A is finite. then (A, i) admits a finite grouping iff (A, i) is unimodular.

In fact the bounded denominator condition is automatic when A is finite.

(2.6) Remark. If A is a finite grouping of (A, i), then it follows from (2.4) that the finiteness of Vol(A) depends only on (A, i), not on A. Moreover, there is a maximum possible value of Vol(A), given by

Volmax(A, i) = L N .1 (a) . aEA mm

We shall later investigate the existence of positive lower bounds for Vol(A) (see §7).

The next result is recalled from [B, I, (3.10)); see also [S, (II, 2.6)).

(2.7) Proposition. Let A = (A,.9I) be a graph of finite groups, ao E A, and r = 7r I (A, ao)' Then r has a free subgroup of finite index iff the orders I~ I (a E A) are bounded. If the latter are bounded put m = m(A) = lcmaEAI~I. Then there is a free subgroup r> of index m in r. Any other free subgroup has index a multiple of m .

(2.8) Corollary. Let r be a group acting without inversion on a tree X with finite stabilizers r x' Then r is virtually free iff the orders Ir x I (x E r\X) are bounded. In this case put m = IcmxEI\xlrxl. Then r contains a free subgroup of index m. and any other free subgroup has index a mUltiple of m . Proof. Apply (2.7) to A = r\ \X , and use [B, I, (3.7)).

3. AUTOMORPHISM GROUPS OF LOCALLY FINITE TREES; UNIMODULARITY

(3.1) We fix the following notation in this section: X = a tree. G = Aut(X). d = the (edge path) distance function on X.

We recall from [B, (6.3)) that there is an exact sequence

o I (1) I-+G -G-+Z/2Z,

where /(s) = [d(sx, x) mod 2) for all x EX, and d acts without inversion on X. (3.2) For x E X we have

Bx(r) = {y E XI d(x, y) :::; r},



the ball of radius r (~O) centered at x. The stabilizer Gx leaves Bx(r) invariant and it is easily seen that

(2)

(3.3) Assume henceforth that X is locally finite, i.e., that

St(x) = {e E E(X)180e = x}

is finite for all x EX. It follows then that Bx(r) is finite for all r ~ 0, and so, the inverse limit (2) shows that

(3) G x is a profinite group.

We topologize G so that two automorphisms are close if they agree on a large finite subtree of X. Then the G x's are compact open subgroups, and G is locally compact. Note that GO in (1) is a closed subgroup. In fact, I: G -+ Z is continuous. (3.4) Let H $ G be a closed, hence locally compact, subgroup of G. (Being closed is equivalent to Hx being closed in Gx for all x.) Let

(4)

be the modular homomorphism of H (cf. [D, (XIV, 3)]). Recall that this is defined as follows. Let )J. be a left-invariant Haar measure on H. For h E Hand S a measurable subset of H, )J.(hS) = )J.(S). Moreover, )J.h(S) := )J.(Sh) defines another left-invariant Haar measure, so )J.h = llH (h)·)J. for some llH (h) > 0; this defines llH. If H O $ H is an open subgroup, then llH1Ho =

HO II .

Since compact open subgroups of H are commensurable, we can scale.)J. so as to take rational values on all of them. (3.5) Let H $ G be a subgroup that acts without inversions on X. Then (cf. [B, (3.2)]) we can form a quotient graph of groups

(5) H\\X = (H\X, K)

and the edge-indexed graph

(6) /(H\ \X) = (H\X, iH),

where iH (e) = [~e : O:e~] is finite because X is locally finite. As in (1.1), ° we obtain, relative to a base point ao E H\X , a homomorphism

H\\X \\ x (7) II : H = 1t\(H X, ao) -+ Q>o'

where we use [B, I, (3.7)] to identify H with 1t\(H\\X,ao). Since Q;o is abelian, llH\\X does not depend on ao (cf. [B, (1.22)]).

Let H denote the closure of H in G. Then it is easily seen that H also acts without inversions on X , that H\X = H\X , and that we can construct H\ \X



so that H\ \X is a full graph of subgroups of H\ \X [B, I, (3.9)]. Moreover, /(H\\X) = /(H\ \X) and .1~\\XIH = .1:\\x. The homomorphism .1:\\X is

-u 0 0 trivial on the (open) subgroup generated by all vertex stabilizers so it is contin-uous, hence determined by .1:\ \X on the dense subgroup H.

o

(3.6) Proposition. Let H:::; G be a closed subgroup acting without inversion on X. Then the modular homomorphism A H: H -+ R;o agrees with .1:\ \X : H =

o 111 (H\ \X, ao) -+ Q;o. Proof. Say H\ \X = (H\X, K) is formed relative to ~ Abtrees X ~ S ~ T and (hX)XES' as in [B, (3.2)]. Let e E E(H\X) , aoe = a, ale = b. Then

coincides with

T S h S where a E T covers a, e E £(S) covers e, and 0 = haoes , so ho(aoe ) = aT; similarly for bT , hi. The groups HaT, Hes, HbT are compact open and hoHesh;;1 :::; HaT with index i(e) , where i = iH . Hence, if J.l is a left-invariant Haar measure on H, we have

J.l(~) = J.l(HaT) = i(e)· J.l(hoHesh;;l)

= i(e).1H(ho)-IJ.l(Hes) = i(e).1H(ho)-IJ.l(Jt;).

Similarly,

J.l(~) = i(e).1H (hl)-I . J.l(,;e;)

(8) _ i(e).1H (ho). ~) _ .1H h j e • ~ - i(e).1H (hi) J.l( a - ( (e) q( » J.l( a),

where he = hoh~1 E Hand q(e) = i(e)ji(e). Let I' = (e l ' ... ,en) be an edge path with vertex sequence ao' a l ' ... , an.

Then II'I = e l ... en E 1l(H\ \X), and q(l') = q(ll'i) = q(el )q(e2) •.. q(en). We also put hl' = he he ... he = ha• h;1 E H. It follows from (8) above that

I 2 n -u n

(9)

Taking for I' a closed path atao we conclude that

( 10) H A (hl') = q(y)

for all closed paths y at ao . In this case q(y) = .1H\ \X (11'i) , by definition ( 1.1). On the other hand, the identification of H with 111 (H\ \X ,ao) is ob-tained (cf. [B, (3.6)]) by restriction to 111 (H\\X, ao) of the homomorphism


"': 1C(H\ \X) ~ H such that ",(s) = s for s E ~ = HaT and ",(e) = he = (ha s)(ha s)-I for e E E(H\X). It follows that h)' = ",(II'I) for I' a closed

oe Ie path at ao' Thus it follows from (10) that

(11) /l.H\\X (h) = /l.H (h)

whenever h = 11'1, I' a closed edge path at ao' On the other hand, both /l.H\\X and /l. H are trivial on all vertex stabilizers in H . Thus ( 11) holds for all h E H , whence the proposition.

From (3.6) and its proof we obtain

(3.7) Corollary. Let H ::::; G, with closure H. Let ~ ::::; H have finite index and act without inversion on X (e.g., H O = H n GO; cf (3.1)(1)).

The following conditions are equivalent. (a) H is unimodular, i.e., /l.H = 1 . (b) H\ \X is unimodular, i.e., /l.~\\X = 1.

Assume these conditions and let f.l be a Haar measure on H, rational on compact open subgroups. For a E A put N(a) = f.l(jf~), where x is any vertex of X over a. Then N is well defined and is a vertex ordering of I(H\ \X) (cf (1.3)).

(3.8) Definition. Under the conditions of (3.7) we say that H is unimodular.

4. EXISTENCE, CONJUGACY, AND COMMENSURABILITY OF UNIFORM LATTICES

(4.1) We fix the following notation. X = a locally finite tree. G = Aut(X). H ::::; G is a subgroup acting without inversions on X. H\ \X = (A, JI!') , a quotient graph of groups and p: X ~ A = H\X is the quotient morphism. GH = {g E Glgx E Hx, ge E He Vx EX, e E E(X)}.

Thus, GH is the largest subgroup H' of G such that H'\X = H\X (= A). We recall [B, I, (5.2)]

(4.2) Theorem ("Conjugacy Theorem"). Ifr::::; GH acts/reelyon X, then there is agE GH such that grg- I ::::; H.

(4.3) Definitions. Let r::::; G. We call r discrete if r is a discrete subgroup of G, i.e., if r x is finite for all x EX. In this case r\ \X is a graph of finite groups. We call r an (X-)lattice if

Vol(r\ \X) (= L ~) < 00, xEf\X x

and a uniform (X-)lattice if r\X is finite.

( 4.4) Recall that if L is a locally compact group, then a discrete subgroup r:5 L is called an (L-)lattice if I\L carries a finite positive L-invariant measure, and a uniform (L-)lattice if I\L is compact. If L admits a lattice, then L must be unimodular, i.e., J1.L = 1 (cf. [Rag, Chapter I, Remark 1.9]).

(4.5) Proposition. Suppose that H is a closed subgroup of G and r :5 H is a discrete subgroup.

(a) If the natural morphism p: r\X ~ H\X has a finite fiber, e.g., if r\X is finite, then I\H is compact. Hence, if H is also discrete, then r\H is finite.

(b) Suppose that r is an X-lattice, i.e., Vol(r\\X) < 00. Then r is an H-Iattice, i.e., Vol(r\H) < 00, and so H is unimodular.

Proof. (a) Choose x E X so that H·x consists of finitely many r-orbits. Then I\H j Hx is finite, so H = r . S . Hx for a finite set S. Since S· Hx is then compact, so also is r\H.

(b) Let J.lH be a left-invariant Haar measure on H, and let p: H ~ r\H, p(h) = [h], be the natural projection. There is a unique measure J.l on r\H such that if A cHis measurable and piA is injective, then J.l(P(A)) = J.lH(A). For h E H we then have J.l(P(A)h) = J.l(P(Ah)) = J.lH(Ah) = J1.H (h)J.lH(A). Thus, J.l(Bh) = J1.H(h)J.l(B) for measurable Be I\H.

Let U be a compact open subgroup of H, and let S be a set of representa-tives of r\HjU: H = IlsEsrsu. Then

r\H = lip(sU). SES

Now p(sU) = p(sUs- 1). sand sUs- 1 ~ p(sUs- 1) is-equivalent to sUs- 1 ~ rs\(sUs- I ) , where rs = rnsus- I . Therefore,

-I 1 -I 1 H-I J.l(p(sUs )) = Irsl . J.lH(SUS ) = IrslJ.lH(U)J1. (s)

and so J.l(p(sU)) = ~J.lH(U), Thus,

1 J.l(r\H) = f1.H(U), L iF!'

SES s

Now let x E X and take U = Hx' Then SUS-I = Hsx so rs = rsx' We can identify H j Hx with H· x , and then the elements sx (s E S) represent the r -orbits in H" x . Thus,

1 I J.l(r\H) = J.lH(Hx) . L in:5 J.lH(Hx) L in < 00.

yEI"\(H"x) y yEI"\X y

Now for h E H we have

and so I1H (h) = 1 for all h, i.e., H is unimodular, as was to be shown.

We shall see examples below where G itself is discrete, hence a uniform G-lattice, yet not an X-lattice (Example (4.12)4), and also one where G is a free uniform X-lattice (Example (4.12)2).

(4.6) Theorem. Under conditions (4.1), the following conditions are equivalent. (a) There is a discrete subgroup <1»::; GH such that <1»\X = H\X (= A). (b) (U) H is unimodular (cj. (3.7»; and (BD) I(H\ \X) has bounded

denominators (cj. (1.3)). (c) I(H\ \X) admits an integral vertex ordering (cj. (1.3». (d) I(H\\X) admits a finite grouping.

Under these conditions, <1» is a lattice (resp. a uniform lattice) iff I(H\ \X) hasfinite volume (cj. (1.3)) (resp. H\X is/mite). Proof. If <1» satisfies (a) then 1(<1»\ \X) = I(H\ \X) so <1»\ \X is a finite group-ing of I(H\ \X), whence (d). In view of (3.7), the equivalence of (b), (c), and (d) follows from (2.4). Finally, the implication (b) => (a) is contained in the next result.

Note that when H is unimodular and H\X is finite, I(H\ \X) automatically has bounded denominators.

(4.7) Theorem. Assume that H is unimodular and that I(H\ \X) has bounded denominators, e.g., that H\X is finite. Then there is a subgroup <1»0 ::; G H with the following properties:

(i) <1»°\X = H\X and <1»~ is finite cyclic for all x EX. (ii) If <1» ::; GH is discrete and <1»\X = H\X, then there is an integer m

such that l<1»xl = ml<1»~1 for all x EX. (iii) If r ::; GH acts freely on X, then r is GH-conjugate to a subgroup of

H and to a subgroup of <1»0 . Suppose that the orders 1<1»~1 (x E X) are bounded, e.g., that H\X is finite.

° Put M = lcmXEXI<1»xl. (iv) There is a free subgroup rD ::; <1»0 of index M and a go E GH such that

gorD gol ::; H. (v) With <1» and m as in (ii), there is agE GH such that grDg- 1 ::; <1»,

and [<1»: grDg-l] = mM.

Proof. Since I(H\\X) is unimodular with bounded denominators, it follows from (1.4) that there is an integral vertex ordering JI3 = Nmin of I(H\\X) such that any other integral vertex ordering is an integral multiple of NO. Let AO=(A,Jaf°) be the finite cyclic grouping of I(H\\X) with l~ol=No(a) for all a E A, as in (2.3)(2), or (2.4). Choose ao E A and put <1»0 = 111 (Ao , ao), which acts on (AO:ao) with quotient A. Since I(Ao) = I(H\\X) , there exist

860 HY:V1AN BASS AND RA VI KULKARNI

bijections

(1) ° ~/e -?~/e ~

for each e E E(A) (ooe = a). Since (Ao, ao) depends, up to isomorphism over ~, only on the pointed sets ~~e (cf. [B, (1.18)]), and similarly X = (H\ \X, ao) depends only o~e ~/e' we can use the bijections (1) to ~duce

an isomorphism of trees (Ao, ao) -? X over A. Thus, identifying (Ao, ao) = X , we obtain ef>0 ~ G Hand ef>°\X = A. If ef> ~ G H is discrete and ef>\X = A , then ef>\ \X = (A,..w') is a finite grouping of I(H\ \X) so N(a) = I~I defines an integral vertex ordering of I(H\ \X); hence N = m~ for some integer m > O. If x EX, then Ief>xl = N(P(x)) = mN°(p(x)) = ml<l>~1 , whence (ii).

Now suppose that M = lcmxExlef>~1 = lcmaEANo(a) is finite. Then by (2.7), ef>0 contains a free subgroup fl of index M, and any other free subgroup has index divisible by M. Clearly, fl acts freely on X. If r ~ GH = G¢>o acts freely on X, then, by the Conjugacy Theorem (4.2), r is conjugate in GH to a subgroup of H and to a subgroup of ef>0. In particular, gofl g; I ~ H for some go E GH ·

Finally let ef> be as in (ii): ef> ~ GH is discrete, ef>\X = H\X, and Ief>xl = mlef>~1 for x EX. By the Conjugacy Theorem (4.2) applied to ef> and fl we can replace ef> by a G H-conjugate and assume that :r<> ~ <l>. Let x EX. Then Ifl\ef>° . xl = Ifl\ef>° /ef>~1 = [ef>0 : fl]/Ief>~1 = M/Ief>~1 since, fl being free, the finite group ef>~ acts freely on fl\ ef>0. Since ef>\X = H\X = ef>\X we have ef>. x = ef>0 . x. Thus, by the same reasoning, we have.

M/Ief>~1 = Ir\ef>° ·xl = Ifl\ef>.xl = [ef>: :r<>1I1ef>xl = [ef>: rO]/mlef>~I,

by (ii). It follows that [ef> : fl] = mM. This proves (v), and completes the proof of (4.7).

(4.8) Corollary. Assume that H is unimodular and H\X is finite. (a) H contains a free uniform lattice fl. (b) Let r ~ H be discrete. Assume that r has" bounded torsion," i.e., that

the finite subgroups of r have bounded orders (e.g., that r is free, or that r is finitely generated (cf [B, II, (8.3)])). Then there is a free subgroup r' offinite index in r and agE GH such that gr' g-I ~ fl.

(c) If r o, r l ~ H are uniform lattices. then grog-I and r l are commen-surable for some g E G H .

Proof. Let ef>0, fl be as in (4.7), whence (a). Assertion (b) is trivial if r is finite. Otherwise, the existence of a free r 1 of finite index in r comes from [B, II, (8.3)]. After GH-conjugation, using (4.7)(iii), we can assume that r l ~ ef>0 . Then r' = r l n fl responds to (b). To prove (c), use (b) to get free ~ in rj

of finite index (i = 0, 1) and use (4. 7)(iii) again to conjugate ~,r; into <1>0 • Being uniform lattices, they have finite index in <1>0. Hence, ~ and r;, so also r ° and r 1 ' are commensurable.

(4.9) Corollary. If H\X is finite and H contains a lattice, then H contains a uniform lattice.

It suffices to show that H is unimodular. But H (the closure of H in G) contains a lattice, so H is unimodular by (4.5)(b). Hence H is unimodular by (3.8).

(4.10) Corollary (Finite covering criterion). The following conditions are equiv-alent.

(a) X covers a finite graph. (a') G contains a free uniform X -lattice. (b) (F) G\X isfinite; and (U) G is unimodular.

Proof. If X covers a finite graph A, then r = n l (A) is a free X-lattice in G. Thus (a) <=? (a'). The group GO (cf. (3.1) or [B, II, (6.3)]) of index :5 2 in G acts without inversions on X, and satisfies (F) and (U) iff G does. By (4.6), (a') for GO <=? (b) for GO, whence the corollary.

(4.11) Definition. A locally finite tree X satisfying the conditions of (4.10) will be called a uniform tree.

(4.12) Examples. 1. We shall exhibit a locally finite tree X with G = Aut(X) such that G\X is finite yet X is not uniform.

Consider the edge-indexed graph

(A. i) =

3 cr-:---------:-O C

1

Note that it is not unimodular: t· t . t = ~ =I 1 . Let A = Z(A, i) = (A , .SIf) be the canonical infinite cyclic grouping of (A, i) (see (2.2)). Thus every vertex and edge group is Z, and ~ --+ ~ e is multiplication by i(e). Let

o

~ ( 2 -I 3) X=(A,a), r=n l (A,a)2: s,tits t =s , G = Aut(X).

The vertices in the three r -orbits have indices 2 (type a), 3 (type b), and 4 (type c), so the quotient G\X of A = r\X must coincide with A. It follows that G acts without inversion and (A, i) = I(G\ \X); hence G is not unimodular; hence X is not uniform, as claimed.


2. Consider X = (A ~o) and r = n I (A, ao)' where (cf. [B, Appendix, Example (A.14)1]),

(A) =

a7 a8 a9

Vertex aj i(a) {i(a)la E St(a)} a l 3 4,5,3 a2 3 3,5,6 a8 3 5,4,6

ao 4 3,5,5,5 as 4 5,5,4,6 a6 4 5,4,3,6

a3 5 4,3,3,6,5 a4 5 4,5,6,4,5 a7 5 4,5,4,4,3

a9 6 3,4,4,5,5,3

Since the covering X -+ A induces isomorphisms on stars of vertices, it follows from the table that G = Aut(X) cannot collapse two r-orbits on X. Since A is a combinatorial graph (no two edges have the same end points) it follows that G\X = A, hence G = Gr. But since r acts freely, X -+ A is a covering and Gr = r (cf. [B, (5.5)]). Thus,

X is a uniform tree such that G = Aut( X) is a free (uniform) lattice.

Smaller examples, due to Tits, are given in [B, Appendix, Examples (A.14)2, 3].

3. Let



where P = Zp' the p-adic integers, and for x E P, 0e(x) = x and 0e(x) = px. Thus,

I(A)= ~O

is not unimodular, but its universal cover is the uniform (p + I)-homogeneous tree Xp+ 1 • Put

H = 1C 1 (A, 0) == (P, elexe -I = px for all x E P) == Qp )q (e),

where Qp denotes the p-adic numbers. Then H\Xp+1 = Ois finite, XP+I is uniform, and H is a closed cocompact subgroup of G = Aut(Xp+I ) ' yet H is not unimodular.

4. Let

A=/SA ... A ... where, for n ~ 4, An denotes the graph

A = n

n vertices

Put r = 1C I (A, *) and X = (A:'"'*). Then, by combinatorial arguments as in Example 2, one can show that G = Aut(X) coincides with r. Thus, X admits no lattices, yet G, being discrete, is a uniform lattice in itself. Moreover, G acts minimally on X [B, (7.4)].

(4.13) Corollary. Let X be a uniform tree. Let r:5 G = Aut(X). (a) The following conditions are equivalent.

(i) A subgroup offinite index in r is contained in a uniform lattice on X. (ii) r is discrete with bounded torsion (cf (4.8)(b)).

(iii) A subgroup of finite index in r acts freely on X.


(b) If r acts freely without inversion on X, then r is contained in a uniform lattice on x. Proof. Taking a barycentric subdivi5ion, if necessary, we may assure that G acts without inversion. With H = G, let <1>0 be as in (4.7); <1>0 is a uniform lattice. With r as in (b), by (4.7)(iii) some conjugate of <1>0 contains r, whence (b). For assertion (a), (iii) ~ (i) follows from (b), (i) ~ (iii) follows from (4.8)(a) and (c), (iii) ~ (ii) is trivial, and (ii) ~ (iii) follows from (4.8)(b).

(4.14) Corollary (Lubotzky [Lub]). Let H be a rank one simple algebraic group over a locally compact nonarchimedean infinite field F. Then H contains un-countably many conjugacy classes of cocompact lattices.

Proof. Let X be the (locally finite) Bruhat-Tits tree of H (cf. [BTl). Then H is a closed subgroup of G = Aut(X), H is unimodular, and H\X is finite. Hence, by (b) ~ (a) of (4.6), H contains uniform X-lattices, which are uniform H-lattices by (4.5). It remains to show that these lattices form uncountably many H -conjugacy classes.

Let r ~ H be a free uniform lattice. Let

 = ('I', (y»: r\\X = B = (B, 9) -. H\\X = A = (A,~) be a corresponding covering morphism (as in [B, §4l). Since r acts freely on X, B is a graph of trivial groups. Let bo E B, ao = rp(bo). Then we can identify both (B--:ho) and (A-:ao) with X so that <l>b : 7l\ (B, bo)

o (= 7l1(B, bo» -. 7l\(A, ao) is identified with the inclusion r ~ H (loc. cit.). Recall that (y) consists of elements Yb E 7l\ (A, rp(b» (b E B) and Ye = YbJe (e E E(B), ooe = b) with Je E .w;,(b). If bE B, rp(b) = a = oof, f E E(A), then the fact that is a covering is expressed by the condition that the maps

<l>b/!: rp~; (f) -. ~/a~!

sending e to [Je ]! are all bijective (cf. [B, I, (2.6)]). Recall that qJ~;(f) = {e E E(B)looe = band rp(e) = f}. Choose (arbitrarily) an element Be E a~! ~ ~ whenever e E E(B), 00e = b, rp(e) = f, and a = qJ(b). Put ~,~ , ~, d' Th ih' ( ('» B A· h ue = ueee , Ye = Ybue' an Yb = Yb· en 'V = '1', Y : -. IS anot er morphism, and ~/f = <l>b/f for all (b, f) as above, so <1>' is again a covering and it defines a new embedding ~ : r -. H.

o Let T be a maximal subtree of B. For bE B let gb = e1e2 •• ·en E 7l(B),

where (e\, e2 , ••• ,en) is the reduced edge path in T from bo to b. For e E E(B) put ge = gaeeg;;~ E 7l1(B, bo) = r. Let S consist of one edge of

o I

E(B) - E(T) from each pair e, e. Then r is free with basis (ge)eES. For e E E(B) put J(e) = Je'I' (e)J'; I E 7l(A). Then J(e) = J(e)-I , so

this defines a homomorphism J: 7l(B) -. 7l(A). Moreover, <l>b : 7l\ (B , bo) -. o

7l 1(A, Ao) is given by <l>b (g) = Yb J(g)y;1 (cf. [B, (2.9)]). o 0 0

Choose (ee)eES with eeEO'fAf, f=q>(e). Define Be= 1 for eEE(B)-S. Then define <1>' = (q>, (y'» as above: y~ = Yb' Y; = Ybo;, 0; = 0eee' For the basis elements ge = ga eeg;;; (e E S) of f we have

o I

I -I -I -I <l>b (ge) = Yb o(ga e)oeeeq>(e)0-e o(ga e) Yb ' o 0 0 I 0

which differs from <l>b (ge) only by insertion of the factor Be' Thus we have a o

family of distinct lattice embeddings and f -+ H parametrized by the families (ee)eES E TIeES Q ~, where f = q>(e). Since SIIf is an open subgroup of H, this F -manifold has F -dimension d r , where d is the F -dimension of Hand r = lSI is the rank of f. The conjugation action of H yields a quotient of F -dimension at least d r- I > O. Thus, we have uncountably many conjugacy classes of lattice embeddings of f into H. Two of these with the same image differ by composition with an automorphism of f. Since Aut(f) is countable, the last assertion of the corollary now follows.

Remarks. 1. Let f be free on (ge)eES as in the last part of the proof. Then we can identify Hom(f, H) with H S , and the last part of the proof shows that

{h E Hom(f, H)lh is injective and h(r) is an X-lattice}

is open in Hom(f, H) . 2. The last assertion of (4.14) should be contrasted with Corollary (6.8)

below.

(4.15) Corollary ("Commensurability Theorem"). If fo' fl are uniform X-lattices in G = Aut(X), then gfog- I and fl are commensurable for some gEG. .

Let H be a subgroup of finite index in G acting without inversion on X (e.g., H = GO, as in [B, II, (6.3)]). The corollary follows from (4.8)(b) applied to the groups fj n H.

(4.16) Corollary (Leighton's "Finite Common Covering Theorem" [Ln· If Ao and Al are finite graphs with a common covering, then they have a common finite covering.

Let X be the common universal cover of Ao' Al and write Aj = fj\X with fj = 1l I (A). After conjugation (4.15) we may assume that fo and fl are commensurable. Then (f 0 n f I) \X is the desired common finite cover.

(4.17) Corollary. If f 0 ' flare uniform X -lattices, then their commensurators CG(fo) and CG(f I ) are conjugate in G.

Recall that CG(r) = {g E Glgfg- I and f are commensurable}. Clearly we have CG(r) = CG(r') if f and r' are commensurable, and CG(gfg- I ) = gCG(f)g-i. Hence the corollary follows from (4.15).


(4.18) Definition. Let X be a uniform tree and G = Aut(X). Put

Latu(X) = the set of uniform X-lattices.

Put C(X) = CG(r) for r E Latu(X). This is a subgroup of G defined up to conjugacy, which we call "the" commensurability group of X.

(4.19) Remarks. 1. C(X) is countable (cf. Appendix B, (B.6)). 2. C(X) contains a conjugate of every uniform lattice on X (4.15). 3. C(X) = G iff G is discrete, because a profinite group is countable iff it

is discrete. This occurs, for example, if X is a linear tree. 4. C(X)\X = G\X. In view of Remark 2 it suffices to observe that cl>\X =

G\X for some cl>0 E Latu(X). If G acts without inversion on X this follows from (4. 7)(i). Otherwise, pass to the barycentric subdivision X' of X, where G acts without inversion. By [B, II, (6.5)], G = Aut(X') unless X is a linear tree, in which case C(X) = G (Remark 3).

(4.20) Conjectures. Let X be a uniform tree, G = Aut(X) . I. C(X) is dense in G. II. Let r ~ G and let Y be a subtree of X on which r acts discretely

with finite quotient r\y. Then there exist r E Latu(X) and an injective homomorphism h: r -+ r such that h(g)IY = glY for all g E r.

We shall show below that Conjecture II (even for finite r) implies Conjec-ture I (see (4.23)). We shall verify Conjecture II (and therefore also I) for X homogeneous or bihomogeneous bipartite (Corollary (4.25)).

(4.21) Proposition. Let X be a tree, x EX, r> 0, and let Bx(r) = {y E Xld(x, y) ~ r} and Sx(r) = {y E Xld(x, y) = r} be the ball and sphere of radius r about x, respectively. Let G = Aut(X) and

Gx(r) = Ker(Gx ~. Aut(Bx(r))).

There is a subgroup F ~ Gx such that Gx = Gx(r) )4 F. If X is locally finite, then F is finite. Proof. We can represent X as Bx(r) to which we attach a rooted tree (~, s) at each s E Sx(r) , and then Gx(r) = llSESx(r) Aut(Ys ' s). Each g E Gx defines isomorphisms gs: (~, s) -+ (YgS ' gs) for s E Sx(r). Let S c Sx(r) be a set ofrepresentatives of the Gx-orbits on Sx(r) , and put ~ = Gx·s. Choose (ar-bitrary) isomorphisms to identify (Yt ' t) with (Ys ' s) for all t E Ts' Then for g E G x we can identify gt: (Yt ' t) -+ (Ygt ' gt) with an element of Aut(~, s) . Define hE Gx(r) by ht = gt- 1 : (¥;, t) -+ (¥;, t) for all t E Sx(r).

Then gh E F, where

F = {g E Gxlgt = Id(y,S) for all t E Ts ' all s E S}.

Clearly F is a subgroup of Gx ' F n Gx(r) = {I}, and, as shown above, Gx = GJr) . F, whence the first part of the proposition. Since F injects


into Aut(Bx(r)) and Bx(r) is finite if X is locally finite, F is also finite in this case. (4.22) Corollary. Let X be a locally finite tree, G = Aut(X) , C ~ G. Assume that:

(i) C\X = G\X; and (ii) Let x EX, r > 0, and let F ~ Gx be a finite subgroup such that

FIBx(r) = GxIBx(r). Then there is a monomorphism h: F --+ C such that h(F)IBx(r) = FIBx(r) .

Then C is dense in G. Proof. Let x EX. By (i) we have C· x = G· x and so G = C· G x' It suffices therefore to show that Cx is dense in Gx ' The groups Gx(r) (r> 0) form a base for neighborhoods of the identity in G x ' so it suffices to show that Gx = Cx . Gx(r) for each r> O. Write Gx = Gx(r) ><l F as in (4.21). Apply assumption (ii) to F and Bx(r) to obtain h: F --+ C so that h(F) and F coincide on Bx(r). Then clearly h(F) ~ Cx and Gx = Gx(r).h(F) ~ GJr)·Cx ' thus completing the proof.

Let X be a locally finite tree, G = Aut(X). Recall from (4.20): Conjecture II. Let r ~ G and let Y be a subtree of X on which r acts discretely with finite quotient r\y. Then there exist r E Latu(X) and an injective homomorphism h: r --+ r such that h (g) I Y = g I Y for all g E G .

(4.23) Corollary. If X satisfies Conjecture II for finite groups r ~ G, then C(X) is dense in G. Proof. If suffices to verify (i) and (ii) of (4.22) for C(X). Condition (i) is supplied by Remark 4 of (4.19). For (ii), consider x EX, r > 0, and F ~ Gx a finite subgroup such that FIBx(r) = GxIBx(r). From Conjecture II for r = F and Y = Bx(r) we get a monomorphism ho: F --+ r+ E Latu(X) so that h(f)IBx(r) = fIBx(r) for all f E F. By (4.17), gor gol ~ C(X) for some go E G. Let hI = ad(go) 0 ho: F --+ C(X). Then go: Bx(r) --+ Bgox(r) is equivariant for hI: F --+ hI (F). By condition (i), cgox = x for some c E C(X). Put h = ad(c) 0 hI: F --+ C(X). Then cgo: Bx(r) --+ Bx(r) is equivariant for h: F --+ h(F). Since FIBx(r) = GxIBx(r) , it follows that the same is true of h(F), whence condition (ii) of (4.22). This completes the proof. (4.24) Theorem. Let X be a uniform tree, G = Aut(X). Let r ~ G and let Y c X be a subtree on which r acts discretely with finite quotient r\ Y . Assume that G acts without inversion on X and that G\X is a tree. Then there exist r E Latu(X) and a monomorphism h: r --+ r such that h(g)IY = glY for all gEr. Proof. Since r is discrete, Ker(r --+ rl Y) is finite. Hence, if we enlarge Y to admit all points at distance ~ r from Y, we can, for r sufficiently large, make the action of r on Y effective (= faithful). Assume now that r acts effectively on Y.

From the inclusion (f, Y) - (G, X) we obtain an immersion of quotient graphs of groups

Cf> = (¢, (1')): B = (B, g) - A = (A,.w'),

where B = f\ \Y and A = G\\X (cf. [B, I, §4)). Consider the underlying edge-indexed graphs /(B) = (B, j) and /(A) = (A, i). Put

U={(b,/)lbEB, IEE(A), ¢(b)=aof}

and, for (b, I) E U ,

S(b, J) = {e E E(B)laoe = b, ¢(e) = f} (= ¢~~ (f)).

The fact that is an immersion implies that

(2) db(f) := i(f) - I: j(e)? 0 eES(b ,f)

for all (b, I) E U (cf. [B, I, (2.8)]). By assumption, A = G\X is a tree, so each IE E(A) is a separating edge.

We shall schematically write

(A, i) = i(f) f

where I(A, i) denotes the portion (A, i) joined to I at all (= a07) , while similarly 7(A, i) denotes the portion joined to I at aol. We now enlarge (B, j) to an edge-indexed graph (B+, j) as follows. For each (b, I) E U with db(/) > 0 we attach the edge-indexed graph

to (B, j) by identifying aofb E B(b,f) with bE B. Henceforth, we shall view B(b,f) , via this identification, with a subgraph of

(B+ , j) = (B, j) u U B(b,f)' dh(/»O


Next we extend ¢: (B, j) -> (A, i) to ¢+: (B+ ,j) -> (A, i) as follows: On B(b,!) , ¢+(~) = f and ¢+ induces the inclusion morphism on f(A, i) C B(b.!) into (A, i).

Now let b E B+, f E E(A) , ¢+(b) = 0of, and put S+(b, f) = {e E E(B+)looe = band ¢+(e) = f}. We claim that

(3) i(f) = L j(e) . eES+(b ,f)

For b E B this has been expressly arranged: Either S+ (b, f) = S(b, f) and (3) holds, or else S+(b, f) = S(b, f) u {~}, and j(~) = db(f) was chosen to validate (3). If, on the other hand, biB, then bE f(A, i) C B(bo,f) for some (bo' f) E U . In this case, ¢ + is defined so as to give a local isomorphism of edge-indexed stars, (StB+ (b), j) -> (StA(¢+(b)) , i), so (3) is evident.

Equation (3) says that ¢+: (B+ , j) -> (A, i) is a covering of edge-indexed graphs, and so they have isomorphic covering trees (cf. [B, I, (2.8)]). We propose now to construct an effective finite grouping B+ = (B+ ,g+) of (B+ ,j) such that B is a subgraph of subgroups and, for e E E(B), ooe = b, the map gb/a.eBe -> fBb+ /a.;ge+ is bijective, ~hat B c B+ is an immersion. This done, for bo E B , we can identify (B+, bo) with X (as observed above) and then r = 1t) (B, bo) is embedded in the uniform lattice r = 1t) (B+ ,bo) on X.

First observe that (B+, j) is unimodular. This follows since (B, j) is uni-modular and (B+ ,j) is obtained by attaching edge-indexed trees to (B, j) at a single vertex, so that 1t) (B+) = 1t) (B). Now it follows from (2.4) that (B+, j) admits an integral vertex ordering N: B+ -> N :

N(ooe) _ N(ooe) z j(e) - j(e) E

for all e E E(B+). Moreover, N is unique up to a constant factor. Since b 1-+ IBbl is a vertex ordering of (B, j) (cf. (2.3)), the uniqueness, applied to (B, j) , implies that

N(b) = mlBbl for all b E B

for some rational constant m. Define M: B+ -> Q by

+ { m if b+ E B, M(b ) = + . +

N(b )/lfBbl If b E B(b,f) , (b, f) E U.

Note that, on B(b, f)' M is a constant multiple of N, hence a vertex ordering of (B(b,f) ' j). Multiplying N by a constant we can make m and all of these vertex orderings integral, i.e.,

M(~e) + M(e) = j(e) E Z for all e E E(B ) - E(E).



For each integer n ~ 1 let J.ln denote the group of complex nth roots of unity. If n' divides n, then J.ln' ::::: J.ln • With this notation we define the grouping B = (B+ ,SI+) of (B+ ,j) as follows. For b E B, Slb+ = Sib X J.lm •

If (b, f) E U and b+ E B(b,f) , then Sib+" = Sib X J.lMW)' If e E E(B) , then Sle+ = Sle x J.lm and 0:;: Sle x J.lm ~ Sla e x J.lm is O:e X Idp. . Suppose that

o m

e E E(B(b,f)) ((b, f) E U). Then Sle+ = Sib x J.lM(e) and 0:;: Sib x J.lM(e) ~ Sib X J.lM(8oe) is Id!#b x inclusion.

Now, as promised, B+ = (B+ ,SI+) is a finite grouping of (B+ ,j) contain-ing B = (B, SI) as a subgraph of subgroups so that Be B+ is an immersion. However, we have not taken care to make B+ effective. For this we just replace B+ by its effective quotient (cf. [B, I, (1.25)]). This does not compromise any of the other conditions above, in particular the immersion B ~ B+ , because, by construction, B was already effective.

(4.25) Corollary. Let X = Xm,n' the bihomogeneous bipartite tree with indices m, n ~ 2. Then X satisfies Conjectures I and II of (4.20). Proof. By (4.23), Conjecture II implies Conjecture I, so we need only verify the former.

Case 1. n =I m. Then clearly G acts without inversion on X and G\X = 0--0 is a tree, so the corollary follows from (4.24).

Case 2. m = n = 2. Then X is a linear tree and G itself is a lattice, so we can take r = G and h the inclusion in Conjecture II.

Case 3. m = n > 2. Then X = Xn ' the n-homogeneous tree, and G acts with inversion on X. Consider the barycentric subdivision X' = Xn , 2' By [B, II, (6.5)], G = Aut(X'). Thus, we can apply Case 1 to conclude the proof.

(4.26) Remark. Let X be a uniform tree, G = Aut(X) , and y E G. In order for y to belong to some uniform X -lattice, (y) must be discrete, i.e., either

(i) y is hyperbolic (l(y) > 0) , or (ii) y has finite order.

If y is hyperbolic, then y belongs to a uniform lattice, by (4.12)(b). On the other hand, if y has finite order, then, in general, y need not belong to a uniform lattice, nor even to the commensurability group of a uniform lattice. Examples demonstrating this have been produced by Ying-Sheng Liu (Columbia dissertation, in progress). Nonetheless, according to (4.25), y can be "de-formed" outside a large finite subtree to an element belonging to a uniform lattice.

5. VOLUMES, EULER CHARACTERISTICS, AND RANKS

Let X be a uniform tree (cf. (4.11)), G = Aut(X) , and, for H::::: G, put

Latu(H) = {r::::: Hlr is a uniform lattice on X}.


For H = G we also write Latu(X) = Latu(G).

(5.1) Definition. A multiplicative invariant on Latu(X) is a function rp: Latu(X) - R which is invariant, i.e., rp(grg- I ) = rp(r) for all r E Latu(X) , g E G, and multiplicative, i.e.,

rp(r') = [r: r']rp(r) whenever r, r' E Latu(X) and r' :5 r. (5.2) Proposition. Let rp: Latu(X) - R be a multiplicative invariant. Let r o' r l E LatU<X).

(a) rp(r l ) = qrp(ro) for some rational number q > O. (b) Suppose that r 0 and r I are commensurable.

Let r' :5 ron r I have finite index. Then

fro : r'] rp(r l ) = [rl : r']rp(ro)'

(c) Suppose that rp =I 0, i.e.Jor some (hence every) r E Latu(X), rp(r) =I O. If rp': Latu(X) - R is any multiplicative invariant, then rp' = crp for some constant c E R .

Assertion (b) follows since rp(r') = Wi : r']rp(ri) (i = 0, 1). Assertion (a) follows from (b) and the Commensurability Theorem (4.15). The latter plus (b) further show that rp is uniquely determined by rp(r) for anyone r E Latu(X) , and this gives assertion (c): c = rp'(r)jrp(r) for any r. (5.3) Proposition. Let J1. be a G-invariant discrete measure on X, i.e., J1.: X -Rand J1.(gx) = J1.(x) for all x EX, g E G. For r E Latu(X) define

J1.(r) (or J1.(r\ \X)) = L fir· xEI"\X x

Then J1.: Latu(X) - R is a multiplicative invariant. Proof. Let r E Latu(X), g E G. By transport of structure

J1.(grg- l ) = L J1.(gX~1 = J1.(r) XEr\X Igrxg I

since J1. is G-invariant. Suppose that r' :5 r has finite index. Consider first the case when r' is torsion free. Then

J1.(r') = L J1.(x) (all r'x = 1) xEr'\X

= L L J1.(x) = L J1.(y)Ir'\r. yl (J1. is r-invariant) yEI"\X xEr'\r·y yEI"\X

= L J1.(Y)Ir'\rj r y l= L J1.(Y~r'\rj = [r:r']J1.(f), yEI"\X yEI\X I y I

the last equations following since, with r' torsion free, the finite groups ry act freely on r'\r.

Now if r' is not torsion free we can choose a torsion free r l :$ r' of finite index «4.13)(a)). By the previous case we have

[r' : rdtt(r') = [r : rdtt(r).

Dividing by [r' : r I] then gives the desired result.

(5.4) Remark. The G-invariant measures on X can be ;dentified with R(G\X) ,

whereas, if rp is a nonzero multiplicative invariant on Latu(X) , the set of multiplicative invariants is the one-dimensional space R·rp (by (5.2)(c)). Thus, if IG\XI =I- 1, i.e., if X is not a homogeneous tree, then the association to an invariant measure tt of the corresponding multiplicative invariant on Latu(X) is not an injective map.

(5.5) Example. Let ttl denote the unit measure, ttl (x) = for all, x EX. Then we have

1 ttl (r) = L Tfl = Vol(r\ \X) > O.

xEI\X x

(5.6) Corollary. The map r ~ Vol(r\ \X) is a multiplicative invariant on Latu(X). If rp: Latu(X) -+ R is any multiplicative invariant, then, for some (unique) constant crp E R,

rp(r) = crp . Vol(r\ \X) for all r E LatU<X) .

(5.7) Example. Recall that for x EX,

i(x) = ISt(x)1 = Card{e E E(X)looe = x}.

Define the Euler-Poincare measure ttEP on X by ttEP(X) = 1 - i(x)/2. Thus, for r E Latu(X) ,

(5.8) Proposition. (a) For r E Latu(X),

ttEP(r) = X(r) ,

the rational Euler characteristic of r (cf [S2]). (b) There is a constant K(X) E Q such that

X(r) = K(X) . Vol(r\ \X) for all r E Latu(X) . Proof. (a) Since both terms define multiplicative invariants of r (see [S2], for X) it suffices to show that they agree for a single r, by (5.2). Thus we can

UNIFORM TREE LATTICES

assume that r is free, in which case, putting A = r\X, we have

1 1 X(r) = X(A) = IAI- 2IE(A)1 = L 1 - 2 L 1

aEA eEE(A)

= L (1 - ~ L 1) = L ( 1 - i~)) = ,uEP(r) . aEA 0oe=a aEA

Assertion (b) follows from (5.2)(c) and (5.5).

(5.9) Corollary. If X covers a finite graph A, then

1~~1)1 = 2(1 - K(X)) ,

which is independent of A. Proof. Write A = r\X with r a free lattice. Then

IAI - iIE(A)1 = X(A) = X(r) = K(X) . Vol(~ \X) = K(X) ·IAI ,

whence the formula.

873

(5.10) Definitions. Let A be a finite connected graph, ao E A, r = 1C) (A, ao)' and X = (A-:ao)' The rank r of the free group r will be called the rank of A ; it is also the first Betti number of A. Thus

rank(A) = 1 - X(A) = 1 - X(r).

(5.11) Examples. O. rank(A) = 0 iff A is a tree (i.e., X = A, or r = 1). 1. rank( A) = 1 iff r is infinite cyclic. In this case A is "virtually cyclic," in

the sense that A is obtained from a cyclic graph (Z/ nZ, with edges (i, i ± 1) or a loop if n = 1) by attaching a finite tree at each vertex. Then X is "virtually linear", in the sense that X is obtained from a linear tree (Z, with edges (i , i ± 1)) by attaching finite trees to each vertex so that these finite trees are periodically isomorphic (with period n).

2. If rank(A) > 1 we call A a hyperbolic finite graph, and X a hyperbolic uniform tree.

(5.12). Since 1 - r = X(r) = K(X)Vol(~ \X) = K(X)IAI ,

we conclude that, in the respective cases 0, 1,2 above, we have K(X) = 1/IXI > 0, K(X) = 0, and K(X) < 0, respectively. Moreover the number of (open) ends of X is 0, 2, and 00, respectively.

Thus we have the following trichotomy for uniform trees X.

virtually Type of X finite linear hyperbolic Types of finite graphs covered finite tree virtually cyclic hyperbolic by X (rank 0) (rank 1) (rank ~ 2) K(X) K(X) = fxr > 0 K(X) = 0 K(X) < 0 Number of open ends of X 0 2 00

Types of uniform virtually virtually free lattices on X finite infinite cyclic nonabelian

(5.13) Examples. 1. Let X = Xp,q' the bihomogeneous bipartite tree with indices p and q. The free product r = (Z/pZ) * (ZqZ) acts on X with

r\\X = (Z/pZ) 1 (Z/qZ).

Thus, 0_----00

x(r) = (~+ ~ - 1) = K(X)· Vol(r\\X) = K(X) (~+ ~) so

K(X) = 1 _ (.!. + .!.) -1 = P + q - pq . p q p+q

Thus,

(1) K(X ) = p + q - pq . p,q p + q

When p = q, Xp,q is just the p-homogeneous tree Xp' and we obtain

(2) K(Xp ) = 1 - p/2.

2. The following example shows that uniform lattices on virtually linear trees need not be commensurable. Consider the virutally cyclic graph

with universal cover X: (

Then G = Aut(X) is isomorphic to the semidirect product z

(*) (Z/2Z) )(j Doo'

The dihedral group Doo = (t, ala2 = 1, ata- I = C l ) acts on the horizontal axis, identified with Z, by t( n) = n + 1 and a (n) = - n. The action defining the semidirect product (*) is defined, for g E D 00 and e E (Z/2Z)z, by g(e)(n) = e(g-I(n)). If, for hE Z, we write eh(n) = e(n-h), then th(e) = eh . Putting ge = (e, t) E (Z/2Z)z )(j Doo ' we then have

N N-I NN ge =(e+t(e)+···+t (e),t )=(e+e l +···+eN_l't) for N> O. Choose e so that e+e l + .. ·+eN_ 1 -::j:. 0 for all N> 0, e.g., e(O) = 1 and e(n) = 0 for n -::j:. O. Then g: -::j:. g: for all N> O. Thus (go) and (ge) are uniform X-lattices (with quotient A) which are not commensurable. (5.14) Definition. Let X be a uniform tree. Define

r(X) = the minimum rank of a free lattice on X

and

v(X) = minimumlAI, where A varies over finite graphs covered by X.

(5.15) Examples. O. If X is finite, then r = 1 is the unique free lattice on X, A = X is the unique graph covered by X, and so r(X) = 0, v(X) = IXI.

1. r(X) = 1 iff X is virtually linear. If X is actually linear, then v(X) = 1 ; take

A=O Otherwise v(X) > 1 .

2. Suppose that r(X) > 1, i.e., that X is hyperbolic, and hence K(X) < O. Let r be a free lattice on X of rank r and A = r\x. Then 1 - r = X(r) = K(X)IAI, so we see that rand IAI are simultaneously minimal: r = r(X) iff IAI = v(X).

3. v(X) = 1 iff X covers a bouquet Am of m circles, for some m ~ O. In this case X = X2m ' the 2m-homogeneous tree, so we see that

v(X2m ) = 1 and r(X2m ) = m.

4. A connected graph with two vertices must be of the form

Amp" "8.-------- p -------<8


i.e., bouquets of m and n circles, respectively, joined by p (geometric) edges; here m, n ~ 0 and p > O. Let X = X m, p , n denote the universal cover of Am,p,n'

Case (i). m = n. Put N = 2m + p = 2n + p. Then X = X N. For N even (i.e., p even) we saw (Example 3) above that veX) = 1 and reX) = N/2. If N (and p) are odd, then we must have veX) = 2 and so reX) = rank(Am,p,n) = m + p + n - 1 = N - 1 . Thus:

For N odd, v(XN ) = 2 and r(XN ) = N - 1.

Case (ii). m f. n. Then it is easily seen that Xm,p,n and X m' ,p' ,n' are iso-morphic iff p = p' and {m, n} = {m', n'}. We have v(Xm,p,n) = 2 and r(Xm,p,n) = m + p + n - 1 .

5. Let X = Xm,n' the bihomogeneous bipartite tree with indices m f. n. The group r = (Z/mZ) * (Z/nZ) is a uniform lattice on X with edge-indexed quotient graph (A, i) = o!?!........!!.o. Clearly also A = G\X, where G = Aut(X) , so G acts without inversion on X and G = Gr. It follows therefore from the Conjugacy Theorem (4.2) that every free lattice on X is conjugate to a subgroup of r.

Let d = gcd(m, n), m = mod, n = nod, and M = lcm(m, n) = monod. According to (2.8), r contains a free subgroup ~ of index M, and every other free subgroup has index divisible by M. If r = rank(~) , then

1 - r = x(~) = M . X(r) = M· (! + ~ - 1) .

It follows that r is minimal, i.e., r = r(X) , so

reX) = 1 -M (! + ~ - 1) = 1 -mo - no + monod .

Recall from (5.13) that

K(X) = m + n - mn = mo + no - monod m+n mo+no

and so, since 1 - reX) = K(X)V(X) ,

veX) = K(X)-I(1 - reX»~

( mo + no - monod)-I = (mo + no - monod) = mo + no'

mo + no, In summary: For m f. n ,

r(Xm,n) = 1 - mo - no + monod

and v(Xm,n) = mo + no'

where d = gcd( m, n), m = mod, n = nod.



(In case m = n, so X m • n = X m ' this still gives the correct answers for m odd (Example 4(i», but not for m even.)

A minimal graph A covered by Xm n is obtained as follows: A = Mil N, where M and N have mo and no' elements, respectively. Connect each element of M to each element of N by d (geometric) edges. Example: m = 6, n=4,so d=2:

(5.16). Let X be a uniform tree and cl> E Latu(X). Put m(cl» = IcmlFI, where F varies over all finite subgroups of cl>. Assume that cl> acts without inversion on X. (This can be achieved either by subdividing X or by passing to a subgroup of index :::; 2 in cl>.) Then (cf. [B, II, (7.3)]) every finite subgroup of cl> fixes a vertex of X; hence

m(cl» = IcmxE~\XIcl>xl. According to (2.8), cl> contains a free subgroup r of index m(cl». If r = rank(r) ,then 1 - r = m(cl»x(cl» , so r = r(cl» is independent of the choice of r, and

1 - r(cl» = m(cl»x(cl» . If r' :::; cl> is free of rank r' and index m', then m' = nm for some integer n > 0 (loc. cit.). Hence l-r' = X(r') = nmx(cl» = n(1-r) , so r' = l+n(r-l). Thus either r' = r = 0 or 1 or else r> 1 (i.e., X is hyperbolic, cf. (5.11)(2» and r' ;::: r. It follows that

r(cl» = the least rank of a free subgroup of finite index in cl> and

m(cl» = the least index of a free subgroup of cl>. If cl>\X = G\X, where G = Aut(X) , then we have G = G~, so it follows

from the Conjugacy Theorem (4.2) that every free lattice on X is conjugate to a subgroup of cl>. Therefore:

If cl>\X = G\X , then r(X) = r(cl» and v(X) = m(cl»Vol(cl>\ \X) .

(5.17) Example. If r is a maximal free lattice on X, then it is not necessarily the case that rank(r) = r(X). For example, let cl> = Go * G) ,where IGil = mi and mo oF m) . Then cl> is a uniform lattice on X = Xm m with edge-indexed

a· I

quotient 0 rna ml 0, and clearly cl>\X = G\X, where G = Aut(X). Hence every free lattice on X is conjugate to a subgroup of cl> (Conjugacy Theorem (4.2», r(X) = r(cl» , and

v(X) = m(cl»Vol(cl>\\X) = (mdm )) (~o + ~J = mo; m) ,


where d = gcd(mo' m l). Let

«1>' = Ker(< = Go * GI --+ Go x GI ),

a free subgroup of index mom l . Let r be a maximal free subgroup of «I> containing «1>' , hence a maximal free lattice on X. Suppose that rank(r) = r(X) , and hence I«I>/rj = m(<<I» = moml/d . The transitive action of «I> on «I>/r factors through Go x G1 ' and each Gj acts freely. Put T = GI \ (</r) . Then T is a transitive Go-set (since the actions of Go and G1 on «I>/r commute) and ITI = ~ I«I>/rl = moml/m1d = mo/d. Hence the isotropy groups of Go

I

on T have order d = gcd(mo' m l ).

Now to prevent this, choose a group Go of order mo with a divisor d of mo such that Go has no subgroup of order d. (For example, Go = V)<I (t), where V is elementary of order 4, t3 = 1, and t acts as an automorphism of order 3 on V; then IGol = 12 and Go has no subgroup of order d = 6.) Choose any integer m~ relatively prime to mo/d, and any group GI of order m l = dm~. Then «I> = Go * GI gives the desired example: A maximal free subgroup r of «I> containing «1>' is a maximal free lattice on X, yet rank(r) must be > r(X).

(5.18). Let A be a finite graph. Recall that, for a E A,

i(a) = He E E(A)180e = a}l.

Put vn(A) = I{a E Ali(a) = n}l. Then

IAI = 2:: vn(A) and IE(A)I = 2:: nVn(A). n~O n~O

Hence, 1 1

X(A) = IAI- "2IE(A)1 = "22::(2 - n)vn(A). n~O

Suppose A is connected of rank r(A) = rank(1t1 (A». Since X(A) = 1 - r(A) we have

1 r(A) = 1 + "2 2::(n - 2)vn(A).

n~O

If vo(A) > 0, then A, being connected, is just a single vertex. Thus, if A is connected and E (A) =f. 0 we have

(3) r(A) = 1 + t( -VI (A) + v3(A) + 2v4 (A) + 3vs(A) + ... ).

(5.19) Proposition. Among connected fznite graphs A with no vertices of index :::; 2, there are only finitely many of a given rank r.

In fact, from (5.18) above we have r = 1 +t(v3(A)+2v4(A)+3vs(A)+· .. ) ~ 1 + tlAI, so IAI :::; 2(r - 1). Further, 1 - r = IAI - tIE(A)1 so IE(A)I = 20AI + (r - 2» :::; 6(r - 1).

(S.20) Example. For r = 2 we must have IAI ~ 2. The possibilities are as follows:

CD e 0-0 (5.21) Corollary. Given an integer r> 0, there are (up to isomorphism) only finitely many uniform trees X of rank reX) = r having no vertices of index ~2.

(5.22) Remark. For an arbitrary finite connected graph A, we can "trim" A as follows. If A has at most one (geometric) edge, leave it be. Otherwise let peA) denote the result of pruning away all terminal vertices (index 1) (along with their incident edges). After finitely many steps, pn(A) = Core(A) either has no more terminal vertices, or else has at most one edge. If all vertices of Core(A) have index 2, then A must be a circuit {ZjnZ, with edges (i, i± I)}, and we then suppress all but one of the vertices to obtain a loop. If Core(A) has branch poiIits (vertices of index ~ 3), then we suppress all vertices of index 2, amalgamating the two incident edges at each such vertex. The resulting graph T(A) now has no vertices of index 1, unless T(A) = 0 --0, and no vertices of index 2 unless T(A) =0 . Moreover, it is evident that T(A) has the same homotopy type as A, so rank(T(A» = rank(A). For fixed r = rank(A), Proposition (5.19) allows only finitely many possibilities for T(A). On the other hand, A can be obtained from T(A) by the following two operations.

1. Subdivide various edges of T(A}, to obtain B. 2. At each vertex of B. graft a finite tree.

In passing to the universal covers, these correspond to similar operations on the covering trees.

6. FINITENESS PROPERTIES

As in §5, X denotes a uniform tree, G = Aut(X). and Latu(X) (= Latu(G» denotes the set of uniform lattices r on X.

(6.1) Proposition. Let r E Latu(X) , with centralizer Z = ZG(r). Assume that X is not finite.

(a) rand G have the same minimal subtree: Xr = XG (cf [B, II, (7.5}]). (b) Either Z is finite and acts trivially on XG, or else X is virtually linear

(cf (5.11)1) and Z is a lattice, commensurable with r. Proof. (a) After subdividing X, if necessary, we may assume that G acts with-out inversion on X. By (4.7), G\X = \X for some <1>0 E Latu(X). Then by [B, II, (7.S)}, XG = X~o. Let r E Latu(X). By (4.8), g<l>0 g - [ and rare commensurable for some g E G. By [B, (7.7)} (which is applicable since X is not finite), Xr = Xg4}Og-1 . But X~Og-1 = gX~o = gXG = XG, whence (a).


(b) We first claim that Z is discrete. Let Y c X be a finite subtree such that r· Y = X. If a E Z fixes all vertices of Y, then the tree Xo: of fixed points of a contains Y and is r-invariant, hence Xo: = X, i.e., a = 1 . Hence Z is discrete, as claimed.

Moreover, for y E Y and y E r we have Zyy = yZyy-l = Zy' so Z has only finitely many vertex stabilizers (Y being finite and X = r· Y)'. It follows therefore (cf. [B, (7.2)]) that if I(Z) = 0, then Z has a fixed points; hence Z is finite.

Suppose finally that Z is infinite, so that I(a) > 0 for some a E Z. Then the a-axis Xo: is r-invariant, so Xo: contains, therefore equals, Xr . Thus, X is virtually linear and r is virtually cyclic. Let Yo E r generate a subgroup (Yo) of finite index; thus (Yo) is a lattice. The first part of the proof above shows that Zo = ZG(Yo) is discrete. Since Yo E Zo' it follows that Zo is a lattice, so [Zo : (Yo)] is finite. Since Z :5 Zo (clearly) and Z is infinite, Z must have finite index in the virtually cyclic group Zo. Thus Z is a lattice, commensurable with r, and the proof is complete.

(6.2) Proposition ("Geometric Rigidity"). Let r E Latu(X) and r' E Latu(X' ) , where X' is another uniform tree. Let

L(r, r') = {h E Hom(r, r')13 an h-equivariant graph morphism X --+ X'}.

Then r'\L(r, r') is finite, where r' acts by conjugation. Proof. Form A = n \X and A' = r'\ \X'. Every h E L(r, r') arises from a morphism A --+ A' via identification of rand r' with fundamental groups of A and A' , respectively [B, I, (4.4)]. According to [B, I, (2.10)], there are only finitely many such homomorphisms modulo conjugatjon by r' .

Remark. The proof does not require r to be discrete, but only that nX be finite and that each rx be finitely generated.

(6.3) Corollary. Let r, r' E Latu(X) and NG(r, r') = {g E Glgrg- 1 :5 r'}. Then r'\NG(r, r') is finite. Proof. Define a: NG(r, r') --+ LG(r. r') by a(g) = ad(g)lr; then a(g): r--+ r' is a homomorphism such that g: X --+ X is a (g)-equivariant. Moreover, a is r' -equivariant, and factors to give an injective r' -equivariant map

NG(r, r')/zG(r) --+ LG(r, r'),

hence an injection

r'\NG(r, r')/ZG(r) --+ r'\LG(r, r').

The latter quotient is finite by (6.2). If ZG(r) is finite, we thus conclude that r'\NG(r, r') is finite.

If Z = ZG(r) is not finite then, by (6.1), X is virtually linear, and r contains a cyclic group (y) of finite index. Since NG(r, r') c NG ( (Y), r'), we


can, without loss, assume that f = (y). If hE NG(f, r') and fh = hfh- I :5 r' , then, by (5.6),

[r' : f h] = Vol(I\ \X)jVol(r'\ \X) = v is independent of h. The finitely generated group r' has only finitely many subgroups of index v. Thus, fixing h as above, we want the finiteness mod r' of the set of g E NG(r, r') such that gfg- I = f h . Any such g = nh, where n E Nh = NG(fh). Since fh is cyclic, Nh contains Zh = ZG(fh) with index :5 2. By (6.1), [Zh : f h] is finite. Hence, from Nh ;::: fh :5 r' , we see that Nh and r' are commensurable, and so r'\r' . Nh ~ (r' n Nh)\Nh is finite, as required.

(6.4) Corollary. If f E Latu(X), then NG(r)jf is/mite.

Here, of course, NG(r) = NG(f, r), because if gfg- I :5 f then Vol(f\ \X) and [f: gfg-I]Vol(I\ \X) both equal Vol(gfg-\ \X), so gfg- I = f.

(6.5) Theorem. Let f E Latu(X) and m > O. Then

Um(r) = {r' :5 GIf:5 r' and [r' : n :5 m}

is finite. Proof. To see this let S denote the set of subgrou:s of f of index dividing m!. Since r is finitely generated, S is finite. We have a map p: Um (r) -t S defined by

per') = n gfg -I = Ker(r' -t Aut(r' jf» , gEl"

and per') <J r' . If per') = ME S, then M:5 r' :5 NG(M) , and NG(M)jM is finite by hypothesis. It follows that the fibers p-I(M) ofp are finite. We saw above that S is finite; hence U m (r) is finite as claimed. (6.6) Corollary. Let r :5 G be discrete and m > O. Let robe a finitely generated infinite subgroup of r. Then

U;(ro) = {r' :5 flro :5 r' and [r' : fo] :5 m} is finite. Proof. The hypotheses on fo imply that [Cfo) =F 0, where I is the hyper-bolic length function (cf. [B, (7.1) and (7.3)]). Let Y = Xr ' the minimal

o ro-invariant subtree [B, (7.5)]. If r' E U;,(ro) ' then Xr' = Y also [B, (7.7)]. Hence, r' :5 r y := {g E fig Y = Y}. Since f y is discrete, the restriction homomorphism r: r y -t Aut( Y) has finite kernel and discrete image r(f y ) • Since ro is finitely generated, it follows from [B, (7.9)(b)] that ro \Y is finite. Hence, rCf 0) :5 r(r y) are uniform Y -lattices, so [rCf y ): r(r 0)] is finite. Since Ker(r) is finite, it follows that [r y : fo] is finite. Finally,

U;(ro) c {r'lro:5 r':5 fy} is also finite.


(6.7) Proposition. Let f E Latu (X). Let f 0 ' f I :5 f be finitely generated subgroups. Then

forms finitely many f-conjugacy classes. Proof. Since f has only finitely many conjugacy classes of finite subgroups (cf. [B, (8.3)]) it suffices to consider those intersections which are infinite. Further, modulo f-conjugation, we can restrict to intersections of the form fp = fo n pflP-1 (P E r).

Case 1. f is free Then the proposition is a result of Imrich [1m] (cf. also Stallings [St, (5.7)(b)]).

Suppose that ~:5 fo has finite index m. Define 1.fI: Ir(ro' f l ) -+ Ir(~' f l )

f-equivariantly by f//(rp) = I"'p := ~ n f p . Then rrp : I"'p] :5 m so fp E

U;'(I"'p) (notation of (6.6)); thus (6.6) implies that 1.fI-I(I"'p) is finite when I"'p is infinite. Thus:

(*) If Ir(r'o, f l ) is finite modf then so also is Ir(ro' f l ).

Now let I'" 4f be a free normal subgroup of finite index and put ~ = I'" nfj (i = 0, 1). Let A c f be a set of representatives of fir'. For a E A, Case 1 implies that I r' (a~a -I , r;) is finite mod I'" -conjugation, say with representatives a~a-I n p~p-I (P E BOo c I"', BOo a finite set). If )I E f, write )I = a -1)1' with a E A and )I' E I'" . Then

~ ~ -I -I ~ -I ,~'-I 1 0 n )11 1)1 = a (a1 oa. n)l 1 1)1. )a

is f-conjugate to a~a -I n pr;p-I for some p E BOo' This shows that I r(~' ~) is finite mod f -conjugation. Now applying (*) above twice we con-clude that I r(f 0' f I) is finite and mod f -conjugation.

(6.8) Theorem. Let H:5 G, Vo > 0, and mo > O. Put

Lat~O,mO(H) = {f E Latu(H)IVol(I\ \X) :5 Vo and m(r) :5 mol,

where (cJ (5.16)) m(r) = lcmlFl, F varying over all finite subgroups of f. Then Lat~o.mO(H) lies infinitely many GH-conjugacy classes. Proof. After subdividing X if necessary we may assure that H acts without inversion on X. We may clearly also assume that Latu(H) ¥- 0. Choose tpo :5 GH as in (4.7) so that tp\X = H\X. Let f E Lat~O,mO(H). By (2.8) f contains a free subgroup r> of index m(r) (:5 m o)' After GH-conjugation (cf. (4.2)) we may assume that ro:s tpo . Then

[tpo : ro]Vol(tp\ \X) = Vol(f\ \X) = m(f)Vol(f\ \X) :S movo



SO [q,O : rD] is bounded by movo Vol(q,o\ \X)-I . The finitely generated group q,o contains only finitely many subgroups of a given index, whence only finitely many possibilities for r>. For each given rD we have r E UmO(rD) , and the latter is finite by (6.5).

(6.9) Corollary. If X is not virtually linear, then, for H:5 G, xo > 0, mo > 0,

{r E Latu(H)lm(r) :5 mo and Ix(r)1 :5 xo} lies in finitely many G H-conjugacy classes. Proof. There is a constant K(X) such that X(r) = K(X)Vol(r\ \X) for all r E Latu(X) (cf. (5.8)). For X not virtually linear, K(X) =F 0, so bounding Ix(r)1 is equivalent to bounding Vol(~ \X), and then (6.9) follows from (6.8).

Remark. When mo = 1, the condition m(r) :5 mo above is equivalent to "r is free." We shall see in §7 that, in (6.8) and (6.9), one cannot drop the bound on mer) without losing finiteness.

(6.10) Notation. Let H:5 G. For any group r put

Latu(r, H) = {h E Hom(r, H)lh is injective and h(r) E Latu(H)}.

There is a natural action (right composition) of Aut(r) on Latu(r, H).

(6.11) Corollary ("Weak Rigidity"). Assume that X is not virtually linear. Then GH\Latu(r, H)JAut(r) is finite, in the following sense. There exist elements hi' ... , hn E Latu(r, H) such that any h E Latu(r' H) is of the form h = ad(g) 0 hi 0 a for some i, a E Aut(r) , and g E GH. Proof. The Aut(r)-orbits on Latu(r, H) are just the sets of homomorphisms h with a common image, h(r). For any h E Latu(r, H), m(h(r)) = m(r) and X(h(r)) = X(r). Thus, by (6.9) the possible images her) are finite modulo G H-conjugacy.

(6.12). Consider next the case H = G. Then from (6.8),

Lat~O,mO(X) = {r E Latu(X)IVol(r\ \X) :5 vo and mer) :5 mol

forms finitely many G-conjugacy classes. Moreover, putting, for a fixed group r,

Latu(r, X) = {h E Hom(r, G)lh is injective and her) E Lat)X)} ,

we see that, when X is not virtually linear,

G\Latu (r, X) J Aut(r)

is finite. Suppose here that r is a free group of rank r. For h E Latu (r, X) put Ah = h(r)\X , a finite graph covered by X. We have

1 - r = X(r) = K(X)Vol(~ \X) = K(X) ·IAhl

so



is independent of h. If a E Aut(r) and g E G, then Ah = Ahoo and Aghg-, ~ Ah . Thus, we have a map (h) --+ (Ah) from G\Latu(r, X)/Aut(r) to the set Graphn(X) of isomorphism classes (A) of graphs A with n vertices covered by X. We claim that this map is bijective. In fact, let (A) E Graphn(X). Then we can write A ~ 1"\X, where 1" ~ 1C 1 (A), and

1 - rank(r') = K(X) ·IAI = K(X) . n, so rank(1") = r. Hence, r' = h(r) for some h E Latu(r, X) and so A ~ Ah • Finally, suppose that h, h' E Latu(r, X) and there is an isomorphism 'II: Ah --+ A~. Choose ao E Ah and put a~ = 'II(ao). Identifying X with (A;:-ao) and (A;':--a~), we obtain an isomorphism g: X --+ X covering'll so that gh(r)g-I =h'(r). It follows that h' =ad(g)ohoa for some aEAut(r).

We can formulate our conclusion as follows. (6.13) Proposition. Let X be a uniform tree which is not virtually linear. Let rr be a free group o/rank r, and put n = (1 - r)/K(X). Let

Latu(rr' X) = {h E Hom(rr' G)lh is injective and h(rr) E Latu(X)} , where G = Aut(X). The map h ....... Ah = h(rr)\X induces a bijection from G\Latu(rr' X)/Aut(rr) to the set o/isomorphism classes 0/ graphs with n ver-tices covered by X. (0/ course these sets are empty if n ¢. Z.)

(6.14) Examples. 1. For the homogeneous tree X k (k ~ 3) we have K(Xk ) = 1 - k/2 (cf. (5.13)) and so n = (1 - r)(2/(2 - k») = 2(r - 1)/(k - 2), and r = 1 + n(k - 2)/2. Thus the n vertex connected k-regular graphs are parametrized by G\Lat(rr,X)/Aut(rr),where rr is free of rank r= l+n(k-l)/2. (When r ¢. Z there are no such graphs.)

2. Let X = Xp,q' the bihomogeneous bipartite tree with indices p, q. We have K(X) = (p + q - pq)/(p + q) (cf. (5.13») and consequently n = (1 - r)(p + q)/(p + q - pq), whence r = 1 + n(pq - p - q)/(P + q). Thus G\Latu(rr' Xp,q)/Aut(rr) parametrizes (p, q)-biregular connected bipartite graphs.

ApPENDIX B: COMMENSURA TORS

(B.l) Abstract commensurators. We discuss here a notion pointed out to us by J.-P. Serre and by Walter Neumann. Let r be a group. Let FI(r) denote the set of subgroups of finite index in r. Let V = V (r) denote the set of pairs (a, 1"), where r' E FI(r) and a: 1" --+ r is a monomorphism with ar' E FI(r). Define an equivalence relation on V by (aI' r l ) '" (a2 , r 2) if a l jrJ = a21r3 for some r3 E FI(r1 n r 2). Put [a, r'] (or [a]) = the "'-class of (a, 1") and

C(r) = ([a, r']I(a, r) E V} = V/ '" . We can make C(r) into a group, with composition

[a, r,,] 0 [p , r p] = [y, r;,1,



where ry = p-I(prp n r,J and y = 0: 0 P on r y. The inverse is [0:, r"r l = [0: -I , o:r"J. This group C(r) is called the (abstract) commensurator of r.

(B.2) Examples. 1. C(Zn) = GLn(Q) (= Aut(Qn)). 2. For n ~ 3, C(SLn(Z)) = PGLn(Q) )q (0-) , where 0- = transpose inverse.

This uses rigidity properties of SLn(Z) (cf. [BMS)). 3. If r' E FI(r), then evidently C(r') = C(r). Hence C(SL2 (Z)) = C(Fn)

for all n ~ 2 , where Fn denotes the free group on n generators. Problem. What interesting things can be said about this group?

(B.3). There is an evident homomorphism ¢: Aut(r) -+ C(r), and ¢(o:) = 1 iff the fixed points of 0: form a subgroup of finite index. The composite

r ~ Aut(r) !.. C(r)

has kernel Ur'EFI(r) Zr(r'). Example. If r is an infinite group with no proper subgroups of finite index (e.g., an infinite simple group), then ¢ is evidently an isomorphism. (B.4). Suppose that r is a subgroup of a group G. The commensurator of r in G is

Then CG(r) = {g E Glgrg -I and r are commensurable}.

¢G: CG(r) -+ C(r) -I gl-+[ad(g),rng rg]

is a homomorphism, and

Ker( ¢G) = U ZG(1") . r'EFI(r)

(B.5) Proposition. If r is finitely generated, then C(r) is countable, as is FI(r) . Proof. For r' E FI(r) put

Monc(1" , r) = {0:1(0:, 1") E U} c Hom(1" , r). Then

C(r) = lim Monc(1", r), ---+

r'EFI(r) where the maps in the inductive system are restrictions to smaller subgroups. Since r is finitely generated, there are only finitely many r' ::; r of a given finite index; hence FI(r) is countable. If r' E FI(r) , then r' is finitely generated, so Hom(r', r) is countable, and hence so also is Monc (r' , r). Whence the proposition.

(8.6) Corollary. If X is a uniform tree, G = Aut(X) , and r E Latu(X) , then CG(r) is countable. Proof. If X is finite, G is finite. If X is infinite we have

¢G: CG(r) -> C(r)



with C(r) countable, by (B.5). Moreover Ker(¢G) is the union of ZG(f") (r' E PIer» , and PI(r) is countable (B.5), so it suffices to see that each ZG(f') is countable. The latter follows from (6.1).

(B.7) The commensurator of a uniform tree lattice. Let X be a uniform tree «4.11» and G = Aut(X). Assume that X is hyperbolic «5.11», i.e., infinite and not virtually linear. Assume further that X is minimal, i.e., that Gleaves no proper subtree invariant.

Let r E Latu(X). Then it follows from (B.4) and (6.1) that

CGer) - C(r) is injective.

Walter Neumann pointed out to us that this cannot be surjective, because NG(r)/r is finite «6.4» whereas NC(r)(r) = Aut(r) , with r::; Aut(r) ::; c(r) , again using (6.1). Taking r a free lattice, for example, we see that Outer) = Aut(r) /r is infinite.

(B.8) End stabilizers. Let X be a tree and e an (open) end of X. If x E X let [x, e) denote the ray (infinite half-line) from x to e. For Y E X define

(x - Y)e = d(y, z) - d(x, z)

for any Z E [x, e) n [y, e) ; this is independent of z:

x

z

y

In G = Aut ( X) , the stabilizer G t of e admits the homomorphism

,£:Gt-Z, 'e(Y) = (yx - x)t for any x E X

(cf. [B2, (1.6)]). We have

I(y) = l't(y)1 for y E Gt

(loc. cit.). If y E Gt and x E Xl' ' then [x, e) c Xl' .

(B.9) r-hyperbolic ends. Let X be a uniform tree, G = Aut(X) , r E Latu(X). Let e be an end of X and consider the exact sequence

I-f>-r ~Z. I: t

Since I(~) = 0 and r E Latu(X) has bounded torsion, it follows from [B, (7.2)] that

(1) ~ is finite.

We call ear-hyperbolic end of X if 't(rt ) ::/: o.



If I' E r is hyperbolic we put:

B"I = the end of XI' toward which I' translates.

We have By = B iff I' Ere and 'l"e(l') > O. In this case (I') has finite index in r e, by (1). Consequently, if 1" E r is hyperbolic and el" = B"I' then I'ln = I'm for some n, m > O. The converse is trivial. Thus:

If 1', 1" E r are hyperbolic, then (2) iff In mfi 0 B/ = 1::"1 1 I' = I' or some n, m > .

We shall write HE(r) = the set of r-hyperbolic ends of X. This is group theoretically definable as the equivalence classes of elements of infinite order in r for the relation defined by (2).

If rand r' are commensurable, then HE(r) = HE(r'). In fact, if I' E r, then I'n E r' for some n > 0, so HE(f) c HE(r') , whence HE(r) = HE(r') by symmetry.

(B.IO) The action of C(r) on HE(r). As above, let X be a uniform tree, G = Aut(X), r E Latu(X). Let g E C(f) , say g = [0:, r'] with r' E FI(r) . Then 0: defines a map

Bo(y)

-> HE(o:r') 1/

HE(r) It is easily seen, using (B.9), that this is well defined, depends only on g, and defines an action h: C(f) -> Aut(HE(f)), where the latter refers to set automorphisms of HE(r).

If X is finite, C(r) = {I}. If X is virtually linear, then C(r) ~ C(Z) = QX, HE(f) has two elements, and h is surjective with kernel the positive rationals.

(B. I I ) Proposition. If X is hyperbolic, then h: C (r) -> Aut(HE(r)) is injective. Proof. We are at liberty to replace r by a subgroup of finite index, and so assume that r is free. Call I' in r primitive in r if I' generates its own centralizer Zr(I'). For any I' t= 1 in r, there is a unique primitive element .jY (generating Zr(l')) such that I' =..;yn with n an integer> O.

Now let [0:, r'] E Ker(h). Then for I' =F I in r' we have By = BO(i) , hence O:(I')n = I'm for some n, m > 0, hence y'0:(1') = .jY in r. Let 1', d be distinct members of a basis of r' . Such exist because X is hyperbolic. Then, by (B.I2) below, for all sufficiently large m, I'dm and O:(I')O:(d)m are primitive in r, and so I'dm = y'l'dm = y'O:(l'dm) = O:(I')o:(d)m . Using the analogous equation with m + 1 in place of m we conclude that O:(d) = d. Hence 0: fixed each basis element of r' ,so 0: = Idr, .

The following lemma was suggested to us by Geoffrey Mess.



(B.12) Lemma. Let r be a free group. Let y, 15 be a basis for a free subgroup. For all sufficiently large m > 0, yt5m is primitive in r. Proof. Fix a basis B of r. For y E r let L(y) denote the length of the reduced word in B representing y. Call a product y. 15 in r reduced if L(yt5) = L(y) + L(t5) .

Now let y, 15 be a basis for a free subgroup of r. After conjugation we can assume that 15 is cyclically reduced, i.e., L(t5m ) = mL(t5) for m ~ O. Choose mo > 0 so that moL(t5) > L(y). Then for m ~ mo we have

L(yt5m ) = L(yt5mO ) + (m - mo)L(/ ) . Suppose that (m - mo)L(t5) > L(yt5mO ). Let e = Vyt5m and write yt5m = en . We claim that, for m sufficiently large, n = 1. Suppose, on the contrary, that n ~ 2. Since the product yt5m = (yt5mO ) • c5 m - mo is reduced and the second factor is longer than the first, we must have

e = 15115', r ~ 0, 15 = c50 ' 151 (reduced product), and c51 =F 1. From yc5m = en = (t5 l c5nt we obtain

sm-,-I ~ n-I yu 'Uo = e •

Write 151 = a· 15: • a -I (reduced) with c5: cyclically reduced. Then e = 15115' = a· (c5: • a -I • t5r- 1 .150 , a· 15:) • a -I = a·.1· a -I (reduced), with .1 cyclically reduced. Thus yt5m- r- 1 • 150 = a . .1 n-I . a -I . If 150 =F 1 it follows that 150 , a is not reduced. This contradicts the fact that 15 = 150 .151 = t5o.a.c5:.a -I is reduced. Thus 150 = 1, i.e. e = 15'+1, so yt5m = (t5 r+l )n ,whence y E {t5}, contrary to our hypothesis. (B.13) The metric topology on Ends(X). Let X be a tree and let Ends(X) denote the set of (open) ends of X. Fix a base point Xo EX. For e, e' E Ends(X) put

ene' = [xo' e)n [xo' e'), Ie n e'l = length(e n e') (~oo),

and , 1 d(e, e ) = I 'I . Ene + 1

We claim that this defines a metric (of diameter 1) on Ends(X). In fact, if e, e', e" are ends, then two of ene' , ene" , and e' ne" are equal and contained in the third:

e'

xoo-------o-------~-------

en


Hence, the two smaller of Ie n e'l, Ie n e"l ,and Ie' n e"l are equal, so the two larger of d(e, e'), d(e, e"), and dee' ,e") are equal; hence,

d(e, e") S max(d(e, e'), dee' , e")). It is easily seen that the topology defined by d = d does not depend on the xo

choice of xo. If g E Aut(X) , then g: Ends(X) -+ Ends(X) is isometric from d x to dgX ,hence continuous.

o 0

(B.14) Questions. Let X be a unifonn tree, G = Aut(X) , and C(X) = CG(ro) for some ro E Latu(X) .

I. Is the action of qro) on HE(ro) c Ends(X) (cf. (B.IO)) continuous for the induced topology (cf. (B.13))?

II. If r E Latu(X) and CG(r) = CG(ro)' then must r be commensurable with ro?

III. If r S C(X) is finitely generated, how can we (group theoretically in qX)) recognize when r has a fixed point in X? When r E Latu(X)?

IV. Does the group qX) (or G) detennine X up to isomorphism?

7. NONFINITENESS PHENOMENA

We shall here construct examples of unifonn lattices which illustrate, for example, the following phenomena.

(7.1) Theorem. Let Xn denote the n-homogeneous tree. (a) For n ~ 3, Latu(Xn) contains infinite ascending chains r, < r 2 < r3 <

.... In particular, Vol(ri\ \Xn ) -+ 0 as i -+ 00.

(b) Suppose that n ~ 5. Given an integer v > 0, v even if n = 5, there exist infinitely many conjugacy classes of r E Latu(Xn) such that Vol(r\ \Xn) = v.

Assertion (b) responds to a question posed to us by A. Borel.

(7.2) A method for constructing lattices on X. Let A = (A,.st') be a graph of groups, ao E A, X = (A~o)' r = 1f, (A, ao)' and G = Aut(X). Then r acts on X, with quotient X .!!... r\X = A , whence a homomorphism p: r -+ G. The action is discrete iff ~ is finite for all a E A (since r x 2:: ~(x), it is cocompact iff A = r\X is finite, and it is effective, i.e., p: r -+ G is injective, iff A is effective in the sense of [B, I, (1.24)]. Thus:

r = 1f, (A, ao) is a uniform lattice on X = (A~o) iff A is an effective finite graph of finite groups.

Further, from [B, I, (2.4), (2.7), and (3.6)] we have: If H S G acts without inversion on X, then constructing g E G so that g r g -, S H is equivalent to constructing a covering morphism A -+ H\ \X .

(7.3) We shall construct A instead of r. In practice this is done in two steps. I. Construct the underlying edge-indexed finite graph (A, i) = I(A) , where,


recall, i(e) = I~ e : ae~] for e E E(A). Observe that the tree X = (A--:ao) = ___ 0

(A, i, ao) is determined by (A, i) up to isomorphism (cf. [B, I, (1.18)]). For a E A put i(a) = 2:aoe=a i(e). Then X = X n , the n-homogeneous tree, iff i(a) = n for all a EA. Similarly, X = Xm,n' the bihomogeneous bipartite tree with indices m, n iff A is bipartite with i(a) = m, respectively n, in the components of the bipartition.

II. Given the finite edge-indexed graph (A, i), we wish to find a finite group-ing A = (A,.9I) of (A, i) such that A is effective. According to (2.4), a finite grouping exists iff (A, i) is unimodular. In an infinite family of such finite groupings A the sizes of the groups ~ must increase, whereas this largeness of the ~ makes it more difficult to make A effective. Resolving this issue becomes, in each case, an interesting problem in finite group theory (see, e.g., (7.16)).

(7.4) Example. Consider the loop L with indices m, m:

It has vertex 0 and indices i(e) = m = i(e). (Equality of indices is required for unimodularity.) We have (L:;: 0) = X2m , the 2m-homogeneous tree. Let M be a group of order m. For each integer r >0 put v,. = M(Z/rZ) , the group of set functions x: ZjrZ ----+ M. Define a, E Aut(v,.) by a,(x)(i) = xU + 1). Let

,-I W, = {x E V,lx(O) = 1 E M} (~M ).

Define A, (or AM,,) = (L,.9I('») by .9Ia(,) = v" ~(') = .>4(') = w" ae = a,l W,: W, ----+ v" and ae = inclusion. Then clearly A, is a finite grouping of L m , r, = 7l 1(A, , 0) = (v,., elexe- I = a,(x) for all x E w,}, r,\X2m = L, and Vol(r,\ \X2m ) = 1jm'. Finally, we claim that A, is effective. According to [B, I, (1.23)] we must show that if N 5 W, is normal in v,. and a,(N) = N, then N = {I}. This follows from the obvious fact that n;=o a~( W,) = {I} .

(7.5). Suppose that r divides r'. Let p: Zjr'Z ----+ ZjrZ be the natural projec-tion. This defines an embedding

. V - M(z/,Z) V _ M(Z/r'Z) 'I' . ,- ----+ ,'- ,

'I'(x)(i)=x(p(i)) forxEV" iEZjr'Z.


Clearly I/f(~):$ ~, . Moreover,

I/f(ar(x))(i) = ar(x)(p(i» = x(P(i) + 1) = x(P(i + 1» = I/f(x)(i + 1) = ar,(I/f(x))(i) ,

i.e., I/f 0 a r = ar' 0 I/f. Hence I/f defines an inclusion of graphs of groups <1>: Ar ~ Ar,· Since the ",-induced maps on cosets Y,.I ~ ~ y", I~' and Y,.I ar ~ ~ v;., I a,'~, are clearly bijective, it follows from [B, I, (2.6) and (2.7)] that is a covering, and hence defines an inclusion rr :$ rr.' .

This proves

(7.6) Proposition. Let M be a group of order m. There are lattices rr E Latu(X2m ) (r ~ 1) such that for all r, rr \X2m is a loop, (rr) ~ M r for all x E X2m , and Vol(rr \ \X2m ) = Ilmr . If r divides r' , then rr :$ rr' with index mr' -r. If m, r ~ 2, then, r r' (s = 1 , 2, 3, ... ) is an infinite ascending chain in Latu(X2m ).

(7.7) Example. Consider next the graph

A=l~e

with indexing iCe) = m = i(e), i(f) = p, i(7) = q:

Choose groups M, P, Q of orders m, p, q, respectively. For integers r ~ 1 define groupings Ar = (A, N(r» of (A, i) as follows: Let v;. = M(Z/rZ) , ~, and or E Aut(Y,.) be as in (7.4).

(r) .wo =Y,.xP, ..J(r) Q ,)4(1 = v;. x , N(r) =~(r) = W x P

e e r ' ..J(r) _ ..J(r) - V

,)4( f -,)4(7 - r •


All edge monomorphisms are the obvious inclusions except for Cte = Ctr X

Idp : u-:. x P -+ ~ x P . The effectiveness of Ar follows easily from that of exam-ple (7.4) (cf. [B, I, (1.23)]). Put X = (A:t: 0) and rr = 7r1(Ar , 0) E Latu(X). Then rr\X = A and

Vol(r \\X) __ 1_ + _1_ = _1 (.!. + .!.) = p + q r - pmr qmr m r p q pqmr '

If r divides r' , then, just as in (7.5), we obtain a covering morphism Ar -+

Ar' inducing an inclusion rr ~ rr' .

Note finally that if q = 2m + p, then X = Xq • This proves

(7.8) Proposition. Let n = 2m + p with m, p integers ~ 1. Then there exist rr E Latu(Xn) (r ~ 1) such that

is independent of r, and

If r divides r', then rr ~ rr" If m, r ~ 2, then rrs (s = 1,2,3, ... ) is an infinite ascending chain in Lat)Xn).

Taking p = 1 and m ~ 2, (7.6) and (7.8) produce infinite ascending chains in Lat)Xn) for all n ~ 4. For part (a) of Theorem (7.1) it remains to treat X3 ' which we do as follows.

(7.9) Example. Let

A = o>-2_ ......... ~ __ ~o)_I----'f ..... r------1€} with indexing i(e) = i(e) = i(f) = i(7) = 1, i(g) = 2, and i(g) = 3.

(A, i) = 0 . . . -3------2~O)_----I-OII


Then (A, i, 0) = X 3 • Let M = Z/2Z. For r ~ 1 define V, = M(ZjrZ) , u-:., and a r as in (7.4). Putting Q = Z/3Z, define

A,. = (A, "" (r» = 0 Q x w,. Vr 0 ,)I¥ ,......::..--'-------<0>----'----- v,. v,. Wr v,.

where all edge monomorphisms are the obvious inclusions except for a = a . e r Then just as in (7.4) we see that Ar is effective. Put

rr = 1C I (Ar' 0) E Lat)X3)

~ (Q x u-:.) *~ v" elexe- I = ar(x) for all x E V,). Then rr\X3 = A and

1 1 1 Vol(rr\ \X3) = 2r + 2r + 3. 2r- 1 = 3. 2r- 3 •

If r divides r', then, as in (7.5), we produce a covering Ar -> Ar' inducing an inclusion r r $ r r' •

(7.10) Proposition. There exist rr E Lat)X3 ) (r ~ 1) with the same quotient

r;.\X3 = 0----0----0 and

Vol(rr\\X3) = 1/(3.2r - 3).

If r divides r', then rr $ rr" Hence, if r ~ 2, then r r' . (s = 1, 2, 3, ... ) is an infinite ascending chain in Latu(X3)·

(7.11) Lattices generated by finite groups. Consider a graph of groups A = (A, .91), ao E A, X = (A-;Qo) , and r = 1C 1 (A, ao)' There is a natural exact sequence

1 -> N --+ r -> 1C 1 (A, ao) --+ 1,

where N is the subgroup of r generated by all r x (x E X) (cf. [B, I, (1.5), Example 3]). It follows that r is generated by the stabilizers rx iff A = r\X is a tree. When r is discrete, e.g., a lattice, this is equivalent to r being generated by finite groups (cf. [B, II, (7.3)]).

Our construction of ascending chains of lattices made important use of a loop in the graph A; the automorphism a r was decisive in establishing effectiveness. When, on the contrary, A is a tree, such constructions are slightly more subtle.

We begin with an edge.


e (7.l2) Amalgams. Let A = 0 0--+---0 1 with indexing i (e) = rno ' i (e) = rn 1 ;

assume that rn i 2: 2 (i = 0, 1).

A = (A i) = 0 mo m l 0 mO,m1 '

A grouping of Am m is an "amalgam" 0' I

A=(10?le~11)

with [Ii: Ie] = rn i (i = 0, I), where we identify Ie with its image in Ii (i = 0, 1). We have

and X=(A,O)=Xm m'

0' I

the bihomogeneous bipartite tree with indices rno ' mi' A is effective, i.e., 1 acts effectively on X, iff Ie contains no subgroup N i- {I} which is normal in each Ii'

(7.13) Composite amalgams. Suppose that one of rno and rn 1 is composite, say rno = pm with p, rn 2: 2. Following an idea offered by Paul Fan (for which he referred to Djokovic [Dj]), we shall construct large families of lattices associated with Am m . Writing q for rn 1 , choose groups M, P, Q of respective orders

0' I rn, p, q. Consider pairs (S, so) , where S is a finite set with a transitive action of P * Q, and So E SQ is a fixed point of Q; put Sf = S - {so}, Then we

S s' can form the wreath products M ~ P and M ~ Q. With these we define the graph of groups

where the edge monomorphisms are the obvious inclusions. To show effective-ness note that

s s' 1 = I(S,So) = 1l: 1(A(S,So)' 0) = (M ~ P) * MS ' (M ~ Q)

has as quotient the wreath product r' = M S ~ (P * Q) (obtained by making Q centralize the so-factor of M S). If g E P * Q, then in r', gMS' g-l = Mg(S') = {x: S -+ Mlx(g(so)) = I}. Since P * Q acts transitively on S, it

s' 1 s' follows that ngEP*Q g M g - = {I} in r'. Now suppose that N::; M ::; 1 is normalized by M S ~ P and M S ' ~ Q , hence normal in 1. Projecting to r' we see from the calculation above that N = {I} . Thus, A(S ,so) is effective. We have I(S ,so) \X = 0----0 and, with 5 = lSI,

Vol(1 \ \X) = _1_ + _1_ = _1_ (_1_ + .!.) = _1_ (_1 + _1_) (S,so) rnsp rns-1q rns- 1 rnp q rnS- 1 rno rn 1 •


Suppose that T is another transitive (P * Q)-set, that to E T2, and that p: T -+ S is a (P * Q)-equivariant map with p(to) = so. Then we obtain an embedding M S -+ M T , inducing M S' -+ MT' , and which is (P * Q)-equivariant. Hence, we obtain a morphism A(S,So) -+ A(T,to) which is easily seen to be a covering, and so we have an inclusion r(S,So) S; r(T, to) .

It remains only to produce such pairs (S, so) as above. These correspond bijectively to finite index subgroups Hs in G = P * Q such that Q S; Hs. We then have S = GjHs and So = 1 . Hs E S. The situation (T, to) -+ (S, so) above corresponds to an inclusion HT S; Hs. Since the group G is infinite and residually finite, there exist (S, so) with s = lSI arbitrarily large. In fact, the intersection of all such Hs is just Q. This proves

(7.14) Proposition. Let m, p, q be integers 2:: 2 and X = Xmp,q. Among r E Latu(X) with a common edge-indexed quotient I(r\ \X) 0 mp q 0 and Vol(~ \X) = ~ ( 1 j mp + 1 j q) for some r > 0 (depending on r), there ex-ist infinite ascending chains.

We can aIso use the construction of (7.13) to prove

(7.15) Proposition. Let q = m + 1 with m an integer 2:: 2. Among r E Latu(Xq) with a common edge-indexed quotient I(~ \Xq) = ~ and Vol(~ \Xq) = 2jmS for some s > 0, there exist infinite ascending chains. Proof. Choose groups M, Q of respective orders m, q , and put P = Q. As in (7.13), consider pairs (S, so) with S a transitive (P * Q)-set and So E SQ , and put Sf = S - {so}. Then define

where the edge monomorphisms are the obvious inclusions. This is a grouping of

~

with (A, i) = X,

and it follows easily from the discussion in (7.13) that it is effective. Put r(S,So) = n1(A(S,So)' *). If (T, to) -+ (S, so) is a (P * Q)-morphism of such pairs, then we obtain inclusions MS)<l P -+ MT )<Ip , etc., whence a covering mor-phism, hence an inclusion r(S,So) S; r(T,to). We have I(r(S,so)\\Xq) = (A, i) and, putting s = lSI,

1 1 1 Vol(r(S s )\ \Xq) = -s- + -s + ~

'0 mq m m q 1 2q 2 = -s-(1 +q+m) = -s- = -S· mq mq m

Now Proposition (7.15) follows, as did (7.14), from the observations at the end of (7.13).


(7.16) Prime amalgams. Consider the edge-indexed graph

A = o-f!..-...!Lo p,q ,

where we now assume that p and q are primes. The covering tree is Xp,q = ~,q. Finite effective groupings of Ap,q correspond to lattices r E Latu(Xp,q) which are transitive on geometric edges, but not on vertices.

Conjecture. There are (up to isomorphism) only finitely many effective finite groupings of Ap, q .

Equivalently, there are only finitely many conjugacy classes of r E Latu(Xp,q) which are transitive on geometric edges, but not on vertices.

Let A = (0 ro r, 0) be an effective finite grouping of Ap,q: [ro: r, r e] = p, [r 1 : r e] = q. The corresponding lattice is r = r 0 *r r l' The conjecture above has been proved by P. Fan [F] under the additional assumption that re is an I-group for some prime I different from p and q. When p = q = 3, re is automatically a 2-group, and this case was first treated in a well-known paper of Goldschmidt [Go]. The case p = q = 2 is trivial: X2 2 = X2 is the linear tree and the unique effective (2, 2)-amalgam is the infinite dihedral group (Zj2Z) * (Zj2Z). More generally, consider: (7.17). The case Ap ,2 = oE-2o with p > 2. Then Xp ,2 is the barycen-tric subdivision of Xp' Let r E Latu(Xp,2) with r\Xp,2 = 0--0. Then r defines (by suppressing vertices of index 2), a lattice r' E Latu(Xp) that is transitive on both vertices and oriented edges; hence r' contains an inversion. According to [B, II, (6.3)], r' contains an index 2 subgroup ~ without in-versions and even length function. Consider (A, i) = /(~\ \Xp )' The group r' j~ = {I , (J} acts on (A, i) , transitively on vertices and on oriented edges, preserving indices. If follows that A has only one geometric edge, and so

(A, i) = oL..E..o .

Conclusion. Let r E Latu(Xp,2) with /(r\ \Xp,2) = oE-2o. If p is odd, then there is a ~ E Latu(Xp) with /(~\ \Xp) = oL..E..o and ~ is isomorphic to a subgroup of index 2 in r.

In this way, for example, (3.2)-amalgams can be described in terms of (3, 3)-amalgams, covered by Goldschmidt's work.

(7.18) Example. Let p be a prime. The following construction provides lattices re E LatU(Xp+1) such that



with e + 2 vertices, and with volume

where r(e) = (pe - 1)/(p - 1). Thus, ve - 0 as e - 00. Explicitly, re = 7r! (Ae' *), where

and where Q is any group of order p + 1, P = Po is a Sylow p-subgroup of the symmetric group Sp" Pi+! has index p in Pi' and Pe is a point stabilizer in P (acting on {l, 2, ... , pe}). Then IFI = p,(e) , and Pe contains no normal subgroup "I 1 of P, so Ae is effective.

(7.19) Example. The following constructions provide many uniform lattices on the same tree with the same covolume, thus establishing part (b) of Theorem (7.1 ).

Let M be a group of order IMI = m 2:: 2. For r 2:: 1 , write M' for M(z/,Z) and Q, for the automorphism Q,(x)(i) = xU + I) for x E M' . Let

abbreviate the loop of groups

where M~ = {x E M'lx(O) = I}, and for x E M~, Qe(x) = x and Qe(x) = Q,(x). We have seen (Example (7.4)) that this is an effective finite grouping of



Let N be an integer > 0, and consider the following edge-indexed graph (A N' i) , where unlabeled edges are understood to have index 1:

N vertices , m m mm m m m m + I

The notation

on the left signifies 2m copies of

~, all issuing from the same right-hand vertex, and similarly on the right with

Note that (A~ i) = X 2m+1 ' AN has 2m + 1 + 2N + m + 2 = 2N + 3(m + 1) vertices. Now we define a finite grouping Ar = (A, .sar(r») of (AN' i) as follows, using the abbreviation

introduced above.

m+ I



Here the edge monomorphisms outside the loops are either identities or the obvious isomorphism M' ~ M~+I C M,+I . Since

o is effective, so also is A, (cf. [B; I, (1.24)]). Let r, = 71:1 (A,) E Latu(X2m+I ).

Then the covolume is

\ 2m+ 1 2N m+2 v, = Vol(r,\\X2m+ l ) = ,+ r:;:T +--r:+:T m m m

1 2 = m,+1 (2m + m + 2N + m + 2)

2 2 = m'+ 1 (m + m + 1 + N) ,

so

Fix an even integer v = 2vo > 0, and define

For r ~ 2, N, > o. Taking N = N, above we get v, = v for all r;::: 2.

Conclusion. Let m be an integer ;::: 2 ,and v an even integer > o. There exist lattices r, E Latu(X2m+l ) (r ~ 2) such that Vol(r,\\X2m+l ) = v but r, and r,' are nonconjugate for r", r'; in fact, r, \X2m+1 has vm,+1 - 2m2 + m + 1 vertices.



(7.20) Example. To handle Xn for n eVen consider the following edge-indexed graph (AN' i) :

~

Note that (AN' i) = X6 and AN has 2N + 4 vertices. Let M = Z/2Z, T = Z/3Z, and consider the following grouping A, of (AN' i) , where we use the abbreviation @ of (7.19) above.

Mr Mr Mrx T Mr Mrx T

Mr

Mrx T

Put r, = 11:1 (A) E Latu(X6). Then the covolume is 4 2N 6+N

v, = Vol(r,\\X6) = 2' + 3 2' = ,-I' . 3·2 ,-I . 0 d N 3 2'- 1 6 so N = 3 . 2 . v, - 6. Let v be any mteger > ,an put , =. . v - .



Then N, > 0 for r 2: 2 , and even for r = 1 if v > 1. With N = N, we then see that v, = v , independently of r.

Now add new edges to (AN' i) to obtain (BN , j) as follows:

Here, for a chosen integer t 2: 0 , each broken edge denotes t geometric edges, each with index 1 for both orientations. In A" the groups at the ends of each of these new edges are identical, so we can put the same group on the new edge to extend A, to a grouping D, of (B tL:.-j ) with the same vertex groups, and hence the same volume. Note that (B N ' j) = X6+t' Putting rt " = n 1 (B,) E Latu(X6+t ) we have Vol(rt ,,\\X6+t ) = Vol(r,\\X6) = v,. Choosing N = N, as above we make these volumes equal v, independently of r.

Conclusion. For n 2: 6 and v any integer > 0, there exist lattices r, E Latu(Xn) (r 2: 2) such that Vol(r,\ \Xn) = v and r,\Xn has 3·2' . v - 8 vertices. Hence the r, are pairwise nonconjugate.

ACKNOWLEDGMENTS

Alex Lubotzky first proposed many of the questions addressed here, and he was a constant source of stimulation. Robert Edwards first brought Leighton's paper [L] to our attention. J.-P. Serre and Walter Neumann furnished help-ful observations on commensurability groups. For background information and constructions of finite effective groupings we received valuable assistance from Robert Guralnick and Paul Fan, and helpful references from Michael Aschbacher. Geoffrey Mess suggested Lemma (B.12). Armand Borel proposed the question answered by (7.1)(b); his paper [Bo] was very helpful to us. We have had helpful conversations with and encouragement from Roger Alperin, Peter Sarnak, Isaac Effrat, and J. Tits. We are grateful to all of the above. Finally, we thank Ying-Sheng Liu for a careful reading of the manuscript and several important corrections.


902

[AB]

[B] [B2]

[BMS]

[Bo]

[BT]

[D) [F] [Dj] [Go]

[1m]

[K]

[L]

[Lub]

[Rag]

[S] [S2]

[St]

HYMAN BASS AND RA VI KULKARNI

REFERENCES

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DEPARTMENT OF MATHEMATIcs, COLUMBIA UNIVERSITY, NEW YORK, NEW YORK, 10027

DEPARTMENT OF MATHEMATICS, CITY UNIVERSITY OF NEW YORK GRADUATE CENTER, NEW YORK, NEW YORK 10036


uniform tree lattices

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