UNIFORM TREE LATI1CES

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

UNIFORM TREE LATI1CES HYMAN BASS AND RAVI 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 Gx 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

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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 obvious 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 offinite connected graphs. Let X be a uniform tree. Put Latu(X) = the set of uniform lattices on X. Commensurability Theorem. If r 0' r commensurable for some g E G.

1 E

Latu (X), then grog -1 and r

1

are

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£: ~ -+ ~oe' We put i(e) = [~e : ae~] and call 0 (A, i) = J(A) the corresponding edge-indexed graph. If a o 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 a o 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.

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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, i(ei ) = i(eJ. In any event, the formula dey) = i(ei)/i(ei ) defines a homomorphism d: 7r 1(A, ao) _ QX,

n7=1

n7=1

n7=1

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 d

H

x

=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" : 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 for some g E GH .

r

S GH acts freely on X, then grg- I S 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 GH acts freely on X, then r can be GH-conjugted into Hand into . 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.

UNIFORM TREE LATTICES

847

In §5 we discuss volumes, Euler characteristics, and other numerical invariants attached to uniform trees and their lattices. In §6 we establish the finiteness properties announced above. Following this is an appendix on commensurability 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 examples. 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), 80 e (1)

= a, we put

~/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 /(A)

(3)

= (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)

q(e)

= q(e) -I .

Thus q defines a homomorphism (6)

q: ll(A)

->

Q:o'

848

HYMAN BASS AND RAVI KULKARNI

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)}.

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

If y

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

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

A

.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 '

q(y) (= q(e l )

..

••• ,

en) in A,

»= 1.

·q(en

(c) For any edge path y = (e l , ... , en) in A, q(y) depends only on the end points of y . (d) There is a function N: A -+ Q;o such that

q(e)

= N(8oe)N(8I e)-

I

.

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 , b l ' ... , 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 8 i e (i = 0,1). Then y = (yo' e, YI) is a closed path at ao so, by (a), 1 = q(y)

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

so q(e) = Na (8oe)Na (8 I 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

849

UNIFORM TREE LATTICES

implies that M(8oe) = M(8 1e) 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) unimodular. 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 rational 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 unimodular, 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 denominators. Adjusting by a scalar factor we may assume that N(a o) = 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' (a o) = 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 denominator 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

This is unimodular iff n

8e.

= m, in which case

q(e)

= 1, and

Nmin(a)

= n.

HYMAN BASS AND RAVI KULKARNI

850

2. Suppose

(A,i)

=

is unimodular. The minimal integral vertex ordering is 54Q-----------~36

180 C F - - - - - - - - - - - - ' 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 el 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)

so that we can identify

= (E(A)le = e -I

for e E E(A)} ,

(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 (e l , ... ,en) is a closed edge path in A at ao . For g E (QE(A))- we put

UNIFORM TREE LATTICES

851

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

ifffor some SEQ,

(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

..1a (N· q) o

(2)

= ad(N(ao)) o..1a (q) . 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(a o)' 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 = (e l ' ... ,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 = e l .. ·en E 1l'(A) to (N· q)(lol)

= (N(ao)q(el)N(a l )- I )(N(a l )q(e2 )N(a2 ) -I ) ... (N(an_l)q(en)N(a n)

-I

)

= N(a o)(q(e l )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.

852

HYMAN BASS AND RAVI KULKARNI

Taking a Im(~a

N(ao) centralizes q(711 (A, ao)) = (and q) determine N, whence a monomorphism

= ao' this condition implies that

(q)). Further, N(ao)

o

N ..... N(ao) from (QA)q to ZQ(lm(~ao (q))). To show surjectivity, let SEQ centralize q(71 1(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 ~a (q') = ad(h) 0 ~a (q)

(3)

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 a o ' 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 = 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),

e E E(A) , 00e

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, i A), where iA(e) = [~e : ae(~)]' which is o

finite by hypothesis. If i A = 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 .

UNIFORM TREE LATTICES

853

When A is a graph of finite groups we define its volume to be (1)

L

Vol(A) =

aEA

1/1~1

(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.l m 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

N(a) N(b)

= b.

=

I~I

I~I

Then I~I

=

I~I

1.w;.1 . I~I =

iCe) 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~

Consequently,

I

for all a EA.

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

Proof. We have (a) ¢} (b) from (1.4) and (b) ¢} (c) from (2.3). Let N min 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 A0 is effective.

854

HYMAN BASS AND RAVI KULKARNI

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' = A0 • 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

aEA

N .1 (a) . 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)).

= (A,.9I) be a graph offinite groups, a o E A, and Then r has a free subgroup offinite 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 of index m in r. Any other free subgroup has index a multiple of m . (2.7) Proposition. Let A

r = 7r I (A, ao)'

r>

(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 Irx 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 (1)

o I I-+G -G-+Z/2Z,

where /(s) = [d(sx, x) mod 2) for all x EX, and on X. (3.2) For x E X we have

Bx(r)

d

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

acts without inversion

UNIFORM TREE LATTICES

855

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)180 e = x}

is finite for all x EX. It follows then that Bx(r) is finite for all r so, the inverse limit (2) shows that (3)

~

0, and

Gx 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 Gx'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 = II

HO

.

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 i H (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 (7)

IIH\\X : H

x = 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

HYMAN BASS AND RAVI KULKARNI

856

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 continuous, 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 AH: H -+ R;o agrees with .1:\\X : H = 111 (H\ \X,

o

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 where a T E T covers a, eS E £(S) covers e, and h0 = haoes , so ho(aoeS )

=

aT; similarly for b T , hi. The groups HaT, Hes, HbT are compact open and hoHesh;;1 :::; HaT with index i(e) , where i = i H . 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).1 H (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(e l )q(e2 ) •.. q(en). We also put hl' = he I he2 ... hen = ha-u• h;1 E H. It follows from (8) above that 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 obtained (cf. [B, (3.6)]) by restriction to 111 (H\\X, ao) of the homomorphism

857

UNIFORM TREE LATTICES

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

xEf\X

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

~) < x

00,

858

HYMAN BASS AND RAVI KULKARNI

(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

discrete subgroup.

(a) If the natural is finite, then is finite. (b) Suppose that H-Iattice, i.e.,

r

:5 H is a

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

r is an X-lattice, i.e., Vol(r\\X) < 00. Then r is an 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 representatives of r\HjU: H = IlsEsrsu. Then r\H

= lip(sU). SES

Now p(sU) rs\(sUs-

I) ,

= p(sUs- 1). sand sUs- 1 ~ p(sUs- 1) where rs = rnsus- I . Therefore, J.l(p(sUs

and so J.l(p(sU))

-I

))

1

= Irsl . J.lH(SUS

= ~J.lH(U),

-I

)

1

is-equivalent to sUs- 1 ~

= IrslJ.lH(U)J1.

H-I (s)

Thus,

J.l(r\H)

= f1.H(U), L iF!' 1

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) .

Now for h E H we have

L in:5 J.lH(Hx) L in < y yEI"\X y

yEI"\(H"x)

00.

UNIFORM TREE LATTICES

and so I1H (h)

=1

859

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 Glattice, 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 ° /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>: :r1I1ef>x l

by (ii). It follows that [ef> : fl] proof of (4.7).

= mM.

= [ef>: rO]/mlef>~I,

This proves (v), and completes the

(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 commensurable for some g E GH .

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

UNIFORM TREE LATTICES

861

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

r;,

°

(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 equivalent. (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) 2. Then X = Xn ' the n-homogeneous tree, and G acts with inversion on X. Consider the barycentric subdivision X' = X n , 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 "deformed" 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}.

UNIFORM TREE LATTICES

For H

=G

we also write

871

= Latu(G).

Latu(X)

(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 E

LatU 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(v 3 (A)+2v 4 (A)+3v s(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).

879

UNIFORM TREE LATTICES

(S.20) Example. For

follows:

r

=2

CD

we must have IAI

e

~

2. The possibilities are as

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) 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.

=0 .

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 X is not finite.

r

E Latu(X) , with centralizer Z = ZG(r). Assume that

(a) rand G have the same minimal subtree: Xr = XG (cf [B, II, (7.5}]). (b) Either Z is finite and acts trivially on X G , 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 without inversion on X. By (4.7), G\X = \X for some 0 E Latu(X). Then by [B, II, (7.S)}, XG = X~o. Let r E Latu(X). By (4.8), g0 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 = X G , whence (a).

HYMAN BASS AND RAVI KULKARNI

880

(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, X r . 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.

n

Proof. Form A = \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 finite and that each rx be finitely generated.

(6.3) Corollary. Let r, r' E Latu(X) and NG(r, r') = {g Then r'\NG(r, r') is finite.

E

nX

be

Glgrg- 1 :5 r'}.

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 N G ( (Y), r'), we

UNIFORM TREE LATTICES

can, without loss, assume that f r' , then, by (5.6),

= (y).

If hE NG(f,

881

r') and fh = hfh- I :5

[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: U m (r) - t S defined by per') = gfg -I = Ker(r' - t Aut(r' jf» ,

n

gEl"

and per') 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 G H as in (4.7) so that tp\X = H\X. Let f E Lat~O,mO(H). By (2.8) of index m(r) (:5 m o)' After GH-conjugation f contains a free subgroup (cf. (4.2)) we may assume that ro:s tpo . Then

r>

[tpo : ro]Vol(tp\\X)

= Vol(f\ \X) = m(f)Vol(f\\X) :S movo

UNIFORM TREE LATTICES

883

[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 For each given rD we have r E UmO(rD) , and the many possibilities for latter is finite by (6.5). SO

r>.

(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 x o} lies in finitely many GH-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' ... , h n 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 GH-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\Lat u(r, X) JAut(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)

so

= K(X)Vol(~ \X) = K(X) ·IAhl

884

HYMAN BASS AND RAVI KULKARNI

is independent of h. If a E Aut(r) and g E G, then Ah = A hoo and Aghg-, ~ A h . 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 ~ A h • 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;:-a o) 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 vertices 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: COMMENSURATORS

(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 ) '" (a 2 , 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,

UNIFORM TREE LATTICES

885

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(SL 2 (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 CG(r) = {g E Glgrg -I and r are commensurable}. Then ¢G: CG(r) -+ C(r) gl-+[ad(g),rng

is a homomorphism, and Ker( ¢G) =

-I

rg]

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)

886

HYMAN BASS AND RAVI KULKARNI

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 Gt 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

(loc. cit.). If y E Gt and x

E Xl' '

for y E Gt

then [x, e) c Xl' .

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

E Latu(X).

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.

UNIFORM TREE LATTICES

887

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: (2)

If 1', 1"

E

r

are hyperbolic, then

= 1::"1 1iff I' In = I' mfior some n, m > 0 . HE(r) = the set of r-hyperbolic ends of B/

We shall write 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' 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'd m and O:(I')O:(d)m are primitive in r, and so I'd m = y'l'd m = y'O:(l'd m) = 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 , .

=..;yn

The following lemma was suggested to us by Geoffrey Mess.

888

HYMAN BASS AND RAVI KULKARNI

(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, yt5 m 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(t5 m ) = mL(t5) for m ~ O. Choose mo > 0 so that moL(t5) > L(y). Then for m ~ mo we have

L(yt5 m )

= L(yt5 mO ) + (m -

mo)L(/ ) .

Suppose that (m - mo)L(t5) > L(yt5 mO ). Let e = Vyt5 m and write yt5 m = en . We claim that, for m sufficiently large, n = 1. Suppose, on the contrary, that n ~ 2. Since the product yt5 m = (yt5 mO ) • 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), m and c5 1 =F 1. From yc5 = en = (t5 l c5 n we obtain

t

sm-,-I

yu

~

'Uo

= en-I •

Write 151 = a· 15: • a -I (reduced) with c5: cyclically reduced. Then e = 15115' = a· (c5: • a -I • t5 r - 1 .150 , a· 15:) • a -I = a·.1· a -I (reduced), with .1 cyclically reduced. Thus yt5 m - r- 1 • 150 = a . .1n-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 yt5 m = (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

, d(e, e )

1

= IEne 'I + 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

UNIFORM TREE LATTICES

889

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 xo does not depend on the choice of xo. If g E Aut(X) , then g: Ends(X) -+ Ends(X) is isometric from dx to d gX ,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 effective finite graph offinite groups.

= (A~o) iff A

is an

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,

HYMAN BASS AND RAVI KULKARNI

890

recall, ___ i(e)

= I~ e : ae~] 0

for e E E(A). Observe that the tree X

= (A--:ao) =

(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 grouping 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) = X 2m , 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

W, = {x E V,lx(O) = 1 E M}

,-I

(~M

).

Define A, (or AM,,) = (L,.9I('») by .9Ia(,) = v" ~(') = .>4(') = w" a e = 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 a~( W,) = {I} .

n;=o

(7.5). Suppose that r divides r'. Let p: Zjr'Z tion. This defines an embedding . V - M(z/,Z) 'I' . , -

'I'(x)(i)=x(p(i))

----+

----+

ZjrZ be the natural projec-

V _ M(Z/r'Z) ,'-

forxEV"

,

iEZjr'Z.

UNIFORM TREE LATTICES

Clearly

I/f(~):$ ~, .

I/f(ar(x))(i)

891

Moreover,

= 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 : Ar ~ A r,· Since the ",-induced maps on cosets Y,.I ~ ~ y", I~' and Y,.I ar ~ ~ v;., Ia,'~, 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 X 2m , and Vol(rr \ \X2m ) = Ilmr . If r divides r' , then rr :$ rr' with index mr' -r. If m, r ~ 2, then, in Latu (X2m ).

r r'

(s = 1 , 2, 3, ... ) is an infinite ascending chain

(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).

.wo =Y,.xP, (r)

v;. x Q , N(r) =~(r) = W e e r ..J(r)

,)4(1

=

..J(r) _

,)4(f

..J(r) -

-,)4(7

-

V

r•

x

P

'

892

HYMAN BASS AND RAVI KULKARNI

All edge monomorphisms are the obvious inclusions except for Cte = Ct r X Idp : u-:. x P -+ ~ x P . The effectiveness of Ar follows easily from that of example (7.4) (cf. [B, I, (1.23)]). Put X = (A:t: 0) and rr = 7r 1 (Ar , 0) E Latu(X). Then rr\X = A and

Vol(r

r

\\X) __1_r + _1_r = _1r -

pm

qm

m

(.!.p + .!.) = p +q q pqm

r '

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

-+

= X q • This proves

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

~

1. Then there exist

is independent of r, and

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

= 1,2,3, ... )

is an

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

~

with indexing i(e)

= i(e) = i(f) = i(7) = 1,

(A, i)

r------1€}

~o)_I----'f.....

A = o>-2_........._ _

i(g)

= 2, and

i(g)

=0 -3------2~O)_----I-OII .. .

= 3.

UNIFORM TREE LATTICES

Then (A, i, 0) = X 3 • Let M and a r as in (7.4). Putting Q

= Z/2Z. For r ~ = Z/3Z, define

893

1 define

V, = M(ZjrZ) ,

0

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

v,.

Wr

v,.

v,.

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

1C I

rr\X3

=

=ar .

(Ar' 0) E Lat)X3 )

~ (Q x Then

u-:.,

u-:.) *~ v"

A and Vol(rr\\X3 )

elexe- I = ar(x) for all x E V,).

1

1

1

= 2r + 2r + 3. 2r- 1 = 3. 2r- 3 •

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

->

Ar' inducing an

(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/(3.2 r - 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 Lat u (X3 )·

(7.11) Lattices generated by finite groups. Consider a graph of groups A = (A, .91), a o 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.

894

HYMAN BASS AND RAVI KULKARNI

e

(7.l2) Amalgams. Let A = 0 0--+---0 1 with indexing i (e) = rn o ' i (e) assume that rn i 2: 2 (i = 0, 1). A

mO ,m 1

= (A i) = '

0

mo

ml

= rn 1 ;

0

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 0' m' I

the bihomogeneous bipartite tree with indices rn o ' 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 can form the wreath products M S ~ P and M s' ~ Q. With these we define the

graph of groups

where the edge monomorphisms are the obvious inclusions. To show effectiveness note that 1 = I(S,So) =

1l: 1(A(S,So)'

0)

= (M

s

~ P) *M S ' (M

s'

~ 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 follows that

-+

Mlx(g(so)) = I}. Since P

ngEP*Q

gM

s'

g-

1

*Q

acts transitively on S, it

r'. Now suppose that N::; M s' ::; 1 ~ Q , hence normal in 1. Projecting to r'

= {I} in

is normalized by M S ~ P and M S ' we see from the calculation above that N = {I} . Thus, have I(S ,so) \X = 0 - - - - 0 and, with 5 = lSI, Vol(1

(S,so)

\ \X)

=

A(S ,so)

is effective. We

_1_ + _1_ _1_ (_1_ + .!.) = _1_ (_1 + _1_) rnsp rns-1q rn s- 1 rnp rn S- 1 rno rn q =

1

•

UNIFORM TREE LATTICES

895

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 + 1j q) for some r > 0 (depending on r), there exist 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) = 2j m S 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) 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.

900

HYMAN BASS AND RAVI KULKARNI

(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. ~

@

M r M r Mrx T

M r 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) so N

= 3 . 2,-I . v, -

= 2' +

3 . 2'

= 3·2,-I'

. 6. Let v be any mteger > 0 ,and put N,

3 2'- 1 .v =.

6.

UNIFORM TREE LATTICES

901

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 (B N , 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) = X 6+t' Putting r t " = n 1 (B,) E Lat u (X6+t ) we have Vol(r t ,,\\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 helpful 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.

HYMAN BASS AND RA VI KULKARNI

902

REFERENCES R. Alperin and H. Bass, Length functions of group actions on A-trees, in Combinatorial Group Theory and Topology, Ann. of Math. Stud., no. Ill, Princeton Univ. Press, Princeton, NJ, 1987, pp. 265-378. H. Bass, Covering theory for graphs of groups, J. Pure Appl. Algebra, submitted. [B] __ , Group actions on non-archimedean trees, Proc. MSRI Conf. on Arboreal Group The[B2] ory,1988. [BMS] H. Bass, J. Milnor, and J.-P. Serre, Solution of the congruence subgroup problem for SLn (n ~ 3) and SP2n (n ~ 2), Inst. Hautes Etudes Sci. Publ. Math. 33 (1967),59-137. A. Borel, Commensurability classes and volumes of hyperbolic 3-manifolds, Ann. Scuola [Bo] Norm. Sup. Pisa Cl. Sci. (4) 8 (1981),1-33. F. Bruhat and J. Tits, Groupes algebriques simples sur un corps local, Proc. Conf. on Local [BT] Fields (T. A. Springer, ed.), Springer-Verlag, Driebergen, 1967, pp. 23-36. J. Dieudonne, Treatise on analysis, vol. II, Academic Press, New York, 1976. [D) P. Fan, Amalgams of prime index, J. Algebra 98 (1986), 375-421. [F] D. Djokovic, A class offmite group amalgams, Proc. Amer. Math. Soc. 80 (1980), 22-26. [Dj] D. M. Goldschmidt, Automorphisms of trivalent graphs, Ann. of Math. (2) 111 (1980), [Go] 377-406. w. Imrich, Subgroup theorems and graphs, Combinatorial Mathematics V, Lecture Notes [1m] in Math., vol. 622, Springer-Verlag, Berlin and New York, 1977. R. Kulkarni, Lattices on trees, automorphisms of graphs, free groups, surfaces, preprint, [K] CUNY, 1988. F. T. Leighton, Finite common coverings of graphs, J. Combin. Theory Ser. B 33 (1982), [L] 231-238. [Lub] A. Lubotzky, Trees and discrete subgroups of Lie groups over local fields, Bull. Amer. Math. Soc. (N.S.) 20 (1988), 27-31. [Rag] M. S. Raghunathan, Discrete subgroups of Lie groups, Ergebnisse der Math., vol. 68, Springer-Verlag, 1972. J.-P. Serre, Trees, Springer-Verlag, New York, 1980. [S] __ , Cohomologie des groupes discrets, Ann. of Math. Stud., no. 70, Princeton Univ. Press, [S2] Princeton, NJ, 1971, pp. 71-169. J. Stallings, Topology offinite graphs, Invent. Math. 71 (1983), 551-565. [St] [AB]

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