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London Mathematical Society Lecture Note Series, 162
Harmonic Analysis and Representation Theory for Groups Acting on Homogeneous Trees Alessandro Figa-Talamanca Department of Mathematics, University of Rome "La Sapienza" and Claudio Nebbia Department of Mathematics, University of Rome "La Sapienza"
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C O N T E N T S
Preface
vii
Chapter I
1) Graphs and trees
1
2) The free group as a tree
5
3) Automorphisms of a tree
6
4) The group of automorphisms Aut(X)
10
5) Compact maximal subgroups
12
6) Discrete subgroups
14
7) Cayley graphs which are trees
16
8) Amenable subgroups
18
9) Orbits of amenable subgroups
24
10) Groups with transitive action on the boundary 11) Notes and remarks
26 31
Chapter II
1) Eigenfunctions of the Laplace operator
34
2) Spherical functions
41
3) Intertwining operators
44
4) The Gelfand pair (G,K)
46
5) Spherical representations
50
Vi
6) The resolvent of the Laplace operator and the spherical Plancherel formula
56
7) The restriction problem
63
8) Construction and boundedness of P
66
9) Approximating the projection P0
68
10) The constant 1 is a cyclic vector
74
11) Notes and remarks
80
Chapter III
1) A classification of unitary representations
84
2) Special representations
87
3) Cuspidal representations and the Plancherel
formula of Aut(3E) 4) Notes and remarks
98 114
Appendix 1) p-adic fields
119
2) A locally compact field of characteristic p
120
3) Locally compact totally disconnected fields
122
4) Two-dimensional lattices
125
5) The tree of PGL(2,6)
127
References
138
Symbols
144
Index
147
P R E F A C E
Over the past few years,
we ran a Seminar in Harmonic
Analysis at the Mathematics Department of the University of Rome "La Sapienza". In this seminar many of the talks given by staff
members
and
visitors
were
concerned,
directly
or
indirectly, with infinite trees or tree-like graphs, and their
automorphism groups. Seminar notes were occasionally taken by
one or both of us,
and sometimes written up informally for
distribution to newcomers to the seminar. felt
that
it
would be convenient
After a while,
we
to give a more coherent
organization to these notes. Once this decision was taken it became apparent that, at the cost of some omission, the general
aim of describing the group of automorphisms of a homogeneous tree and its irreducible unitary representations would provide a convenient focus which would include much of the material we had in mind. We felt that this approach would shed light on the
connection between harmonic analysis on trees and harmonic analysis
on
hyperbolic
spaces,
by emphasizing
the
strict
analogy between the group of automorphisms of the tree and real rank 1 semisimple Lie groups. This choice left out a lot of
valuable material specifically concerning free groups and free
products of finite groups.
We felt however that the notes
[F-T P2] and the memoir [F-T S2] could provide an introduction to these topics.
We also decided not to treat the case of a
semihomogeneous tree. Semihomogeneous trees are natural objects
because they are exactly the Bruhat-Tits buildings of rank 1 [BT].
It follows that every rank
1
reductive algebraic group
over a local field is a closed subgroup of the group of auto-
morphisms of a semihomogeneous tree.
We felt however that,
while no major conceptual step is needed to extend the theory of representations of the group of automorphisms of a homoge-
neous tree
to
the case of a semihomogeneous
tree,
from a
practical point of view the notation would have become more burdensome. The connection with matrix groups over local fields is explained in these notes by giving, in the Appendix, Serre's construction of the tree of
where 3 is a local field.
Chapter I contains a description of the geometry of a ho-
mogeneous tree X and its boundary, the group of automorphisms Aut(X) and some of its notable subgroups. Chapter II contains
the boundary theory for eigenfunctions of the Laplace
(or
Hecke) operator on the tree and a complete description of spherical functions and spherical representations which applies
to every closed subgroup of Aut(l) with transitive action on the vertices and the boundary of the tree. It also contains the
proof of an important result due to T. Steger which asserts that
every
spherical
representation
(with
one
possible
exception) of Aut M restricts irreducibly to any cocompact discrete
subgroup.
square-integrable
Chapter III
representations
contains of
a
Aut(X)
description following
of the
beautiful geometric classification due to G.I. 01'shianskii. At
the end of the chapter we give the complete Plancherel formula
for Aut(X). The Appendix,
as already mentioned,
contains a complete
and elementary account of the construction of the
tree of
PGL(2,a) and a discussion of the action of this group on its tree.
ix
These
notes
owe
a
great
deal
to
the
many
friends,
colleagues and students who participated in our seminar.
We
would like to thank especially M.G. Cowling, F.I. Mautner and R. Szwarc for the many critical observations and comments which stimulated our work and often found their way into these notes.
Very special thanks are due to Tim Steger who contributed
in many ways with help and advice in the preparation of these
notes. He also gave his permission to include in these notes his yet unpublished restriction theorem, which was presented at our seminar in 1987.
1
CHAPTER
I
1. Graphs and trees. A
tree
is
a
graph
connected
without
circuits. This definition requires a word of explanation of the
terms graph, connected, and circuit. A graph is a pair (X,l) consisting of a set of vertices 3f and a family 9 of two-element
subsets of X, called edges. When two vertices x, y belong to
the same edge (i.e., {x,y}eO they are said to be adjacent; we also say that x and y are nearest neighbors.
A path in the graph (X,e) is a finite sequence xo,...,x
A graph is called connected if,
such that {xi,x1+l}ET. two vertices x,yc:X,
there exists a path xo,...,xn,
given
with xo x
and x n=y. A chain
is
a path
xo,...,xn,
such
that
xi*xi+z'
for
i=0,...,n-2. A chain xo,...,xn, with x=xo is called a circuit.
In particular if
(3f,1I)
is a tree and x,yel,
there exists a
unique chain x0,...,xn, joining x to y. We denote this chain by [x,y].
We are interested in locally finite trees. These are trees such that every vertex belongs to a finite number of edges. The
number of edges to which a vertex x of a locally finite tree belongs is called the degree of x. If the degree is independent
of the choice of x,
then the tree is called homogeneous.
In
these notes we will treat mainly locally finite homogeneous trees.
The common degree of all vertices of a homogeneous tree is
called the degree of the tree and is generally denoted by q+1.
The reason for this notation is that, as will be shown in the Appendix, the number q may be identified, in many cases, with the order of a certain finite field. Furthermore many of the formulae
appearing
in
the
sequel,
and
especially
in
the
explicit computation of spherical functions (Chapter II, below)
Ch. I
2
involve powers of q, rather than q+i.
There are also nonhomogeneous trees which are of interest. An important example is that of a semihomogeneous tree. Suppose that
1
and q are positive integers.
vertex has degree
1+1
or q+1,
vertices have different degrees,
A tree such that every
and such that
two adjacent
is called semihomogeneous of
degree (l,q).
The set of vertices of a homogeneous or semihomogeneous tree is always infinite. A tree may be represented graphically as shown in Figs 1 and 2.
m q+1 = 3
q+1 = 4 Fig.1 Homogeneous trees
FIg.2. Semihomogeneous tree: q = 2,
1
=3
The set of vertices of a tree is naturally a metric space.
Graphs and trees
3
The distance d(x,y) between any two distinct vertices x and y is defined as the number of edges in the chain [x,y] joining x and y, in other words the length of [x,y].
The metric space structure of suffices
to define
the
set
of vertices
the tree uniquely because
I
two vertices
belong to the same edge if and only if their distance is 1. We
will often think of a tree as a set of vertices with a metric which makes it into a tree.
An infinite chain is an infinite sequence xo,xI'x2,..., of
vertices such that,
for every i,
xixxi+2 and {xi'xi+l}
is an
edge.
We define an equivalence relation on the set of infinite
chains, by declaring two chains xo, xl, 'x 2.... and equivalent
if
(as sets of vertices)
YO'yl'y2....
they have an infinite
intersection. This means that there is an integer nell such that x =y k
k+n
tree
for every k sufficiently large.
The boundary 53 of a
is the set of equivalence classes of infinite chains.
Observe that an infinite chain identifies uniquely a point of the boundary, which may be thought of as a point at infinity.
Sometimes the points of the boundary are called ends of the tree.
An alternative way to define the boundary is by fixing a
vertex x0 and considering all infinite chains which start at x0. A boundary point is associated with a unique infinite chain starting at x0.
A doubly infinite chain is a sequence of vertices indexed
by the integers,
...x_2,x_1'x0,xI'x2,...,
with the properties
that xi*x1+2' and {xi'xi+i} is an edge for every integer ieZ. A doubly infinite chain is also called an infinite geodesic.
It
identifies two boundary points. Conversely, given two distinct boundary points wl,w2ei2, them.
there is a unique geodesic joining
We denote this geodesic by
notation [x,w)
(w1,w2).
We also use the
for an infinite chain starting at x in the
direction of w (that is belonging to the equivalence class w).
Ch.I
4
Since
also
we
infinite
chains
want
consider
to
and
geodesics,
the
direction
(w,x]
of
considered
be
will
chains,
different from [x,w) even though they have the same vertices,
and similarly
formally different
is
[x,y]
from
[y,x]
and
(Wi,W2) is different from (w2,wi). All these concepts are more or less geometrically evident and may be illustrated with a picture
Fig.3 The boundary points WI and W2
Identify a geodesic (Wi,W2)
The space 3full can be given a topology in which 3EuQ is compact, the points of X are open and X is dense in 1uQ. To define
this
topology
it
suffices
to
define
a
basis
of
neighborhoods for each boundary point (because each vertex is open).
Let
we92,
and let x be a vertex.
Let X=[x,w) be the
infinite chain from x to w. For each ye[x,w) the neighborhood
(x,y) of w is defined to consist of all vertices and all end
points of the infinite chains which include y but no other vertex of [x,y] (Fig.4).
Fig.4
The free group as a tree
5
It is not difficult to show that 1:vS2 is compact and 3E is The relative topology on
dense in Q.
S2
(under which
S2
is
compact) is best described by the open sets S2(x,y) consisting
of all boundary points associated to
infinite chains which
start at x and pass through y (in this order). In this way, for
each vertex x,
S2(x,x)=S2 and,
for every positive integer n,
f2=U{S2(x,y): d(x,y)=n}. Thus the family {i2(x,y): d(x,y)=n} is a partition of S2 into (q+1)gn-1 open and compact sets, where q+1 is the degree of the tree. Using these partitions we can define
a measure v
x
on the algebra of sets generated by the sets
by letting v (id(x,y))=1/(q+1)gn_1, if d(x,y)=n. The positive measure v x may be extended to a Borel probability !2(x,y),
measure on Q. 2. The free group as a tree.
We preview in this section an
example to which we will return in Section 7.
Let F2 be the
free group with two generators a and b. An element of F2 is a reduced word in the letters a,a 1,b,b 1.
We denote by e the
empty word, which is the identity of F2.
There is a natural
correspondence between F2 and the vertices of a tree of degree 4, which is obtained by defining two words x and y to be in the
same edge if y-1 x is one of the generators or their inverses. This means that x and y can be obtained one from the other by right multiplication with an element of {a,a 1,b,b 1}. The tree which is so defined is described by Fig.5
Ch. I
6
Other
which
groups
homogeneous
trees
are
may
similarly
be
described
in
associated
Section
6,
as
with
certain
discrete subgroups of the group of automorphisms of a tree.
It is interesting that the boundary of the tree of Fig.5 can be identified with the set of infinite reduced words in the
letters
{a, b, a 1, b 1} .
That
xiox1-1 and xiE{a, b, a
11.
b
is the words w=xIx2x3... , with We may also observe that left
multiplication by elements of F2 on itself gives rise to an isometry of the tree. Left multiplication by a finite word is also
defined on
words
infinite
the
and gives
rise
to
a
homeomorphism of n (see Section 7, below).
An automorphism of a tree
3. Automorphisms of a tree. bijective
map
of
the
of
set
vertices
onto
itself
is a which
preserves the edges. An automorphism is also an isometry of the
metric space
3f
endowed with the natural metric.
every isometry of E is also an automorphism.
Conversely,
We shall give
presently a description of the automorphisms of a tree. We first need a preliminary result.
(3.1)
Let g be an automorphism of the tree and x a
LEMMA.
vertex. Let x=x0'x1,...,xR g(x) be the chain joining x to g(x), and suppose that n>O. If g(x 1)*xn-1 ,
then there exists a doubly
infinite chain,
{...x
-n
,x -n+1
,...,x
,x ,x ,...,x n=g(x0), xn+1,...1v -1
0
1
for every jEZ.
such that g(xj)=xj n,
PROOF.
If g(x1)*xn-1 we can extend the chain (x0,g(x0)]
defining xn+1
g(x1).
Since g is
bijective,
g(x0)=xn
by
implies
that g(x2)*xn. Therefore we can define g(x2)=xn+2' By induction w e define g(xk)=xj+k and similarly xj-g g 1(xj-k).
We obtain in
this way a doubly infinite chain on which g acts according to the formula g(xj)=xn+j for every jE3. 1
Automorphisms of a tree
7
(3.2) THEOREM. Let g be an automorphism of a tree; then one and only one of the following occurs. (1) g stabilizes a vertex.
(2) g stabilizes an edge exchanging the vertices of the same edge.
(3) There exist a doubly infinite chain T={x } and an integer j n
such that g(x )=x n
n+j
for every nEZ.
PROOF. Let g be any automorphism and let j=min{d(x,g(x)): xEy}.
Let xEZ be such that d(x,g(x))=j.
then g(x)=x and g
If j=0,
satisfies condition (1). If j=1, then {x,g(x)} is an edge. In this case g satisfies (2) if g2(x)=x, or (3) if g2(x)*x (this from
follows
(3.1)).
then
x1,...,xj=g(x)},
Finally
if
g(x1)*x_1,
[x,g(x)]={xo x,
and
j>1
because
d(xl,xj_1)=(j-2)<j.
Therefore g satisfies condition (3). The fact that only one of the three conditions holds follows readily. 1
On the basis of (3.2) it is natural to give the following definition.
(3.3)
DEFINITION.
rotation
if
An automorphism of stabilizes
it
a
a
vertex;
an
tree
is
called
inversion
if
a it
satisfies condition (2) of (3.2); and a translation of step j along y if it satisfies condition (3) of (3.2).
In this last
case j=min{d(x,g(x)): xEX} and T={xEX :d(x,g(x))=j}.
If the tree is homogeneous there are always rotations,
inversions and translations of any step on any geodesic. For a locally finite nonhomogeneous tree the situation may be quite different. First of all the existence of a translation implies
the existence
of
an
infinite
geodesic.
The only possible
automorphisms of a finite tree are rotations and inversions. Furthermore, for any automorphism g, the degree of x and g(x)
is the same. This implies, for instance, that in a semihomo-
geneous tree g(a)*b if {a, b} is an edge. In other words every
Ch.I
8
automorphism of a semihomogeneous tree is either a rotation or a translation of even step.
if g and g' are automorphisms of a tree,
We remark that,
then g' and gg'g 1 are both rotations, both inversions or both translations.
It is also obvious that an m-power of a step-j
translation on
a'
is a step-Imjl translation on x, for every
me7L.
We conclude with a technical result which will be useful
later. (3.4)
PROPOSITION.
(1)
The composition of two inversions on
distinct edges is a translation of even step. (2)
The composition of an inversion about
an edge and a
rotation which does not fix both vertices of the edge is a translation of odd step on a geodesic containing that edge.
(3) Every automorphism of a homogeneous tree of degree q+1>2 is a product of translations.
PROOF. Let {a,b} and {c,d} be distinct edges, that is, having at most one vertex in common (Fig.6).
b=c
Fig.6
Let h be an inversion on {a,b} and g be an inversion on {c,d}. If
g
-1
b=c,
then
gh(a)=g(b)=d
and
gh(b)=g(a)*b
(because
a*d=
(b)). This means that gh(b) is a vertex of distance 1 from d
which is not geodesic
b.
By
(3.1)
gh is a step-2 translation on a
' containing {a,b,d}. Suppose that the two edges {a,b}
and {c,d} have no vertex in common. By naming the four vertices
Automorphisms of a tree
9
appropriately we may suppose that d(b,c)=n is the distance of the two edges. This means that the chain [a,d] contains [b,c]. Then g(b)i[b,d] (3.1)
and g(a)O[b,g(b)]
gh
that
(Fig.6).
This
implies by
a translation on a geodesic containing
is
[a,g(a)]. Finally we observe that d(d,g(b))= d(g(c),g(b))=n and d(a,gh(a))=d(a,g(b))=d(a,b)+d(b,c)+d(c,d)+d(d,g(b))=
therefore
2(n+l), which means that gh is of even step. To prove (2) let g
be an inversion on the edge {a,b} and k a rotation. If k(a)=a and k(b)*b,
then kg(b)=k(a)=a, while kg(a)=k(b)*b. By (3.1),
applied to [a,b], we conclude that kg is a translation of step 1.
If k(a)#a,
and k(b)#b,
distance from {a,b} d(x,a)+l.
let
x be
the point
Suppose that d(x,b)=
such that k(x)=x.
Then d(k(a),b)=d(k(a),a)+1=2d(x,a)+l.
k(a) and kg(a)=k(b)V[b,k(a)].
of minimal
Now k(g(b))=
Therefore by (3.1)
applied to
[b,k(a)] we conclude that kg is a translation of step 2d(x,a)+i (Fig.7).
b
k(a)
Fig.7
This proves (2).
Finally assume that the tree is homogeneous
with degree q+1>2. Let g be an inversion on the edge {a,b} and
let 7 be a doubly infinite geodesic containing a but not b. (such a a' exists because q+1>2). Let 'c be a step-1 translation on 7. Then z(b)*a and tg(a)=x(b) (Fig.8). b
2 (a)
'e-1(a)
a Fig.8
t'(b)
r(a)
2(a)
Y-i
Ch.I
10
We apply now (3.1) to [b,z(a)] to conclude that ig is a step-1 translation. This shows that g=i fig is the composition of two
translations. Let now k be a nontrivial rotation. Since k is
nontrivial and fixes a point starting
at
x0
starting at
x0.
xo,
it
a different
into
maps a finite chain V
finite
chain
Let a be the first point of
a*k(a)=b and let x be the last point of
also b°
such that
such that k(x)=x;
Let g be an inversion on {x,a}.
then d(x,a)=d(x,b)=1.
gk 1(b)=x and gk 1(x)=a,
that
is i=gk
1
Then
is a step-1 transla-
tion. Therefore k=r 1g is the product of a translation and an
Since every inversion is the product of transla-
inversion.
tions, (3) follows. '
We assume from now on
4. The group of automorphisms Aut(X). that
is a homogeneous tree of degree q+1,
(X,C)
and we let
Aut(T) denote the group of automorphisms of (X,1l), which is the same as the group of isometries of X as a metric space.
It is not difficult to define on Aut(X) a locally compact
topology under which the group operations are continuous. To define a basis of neighborhoods of gEAut(X), let F be a finite
subset of X, and let UF(g)={hEAut(X): g(x)=h(x), for all xEF}. It
under the topology generated by the sets
is clear that,
the group operations of Aut(o) are continuous.
UF(g),
also
not
compact.
difficult Indeed,
definition K
x
is
to
for open.
show that
But K
presently shown. Every gEK
9S3 ={w : n
x
x
topology
is
locally
K ={gEAut(Z): g(x)=x}. x
is also compact
is
By
as will be
acts as a permutation on the set
d(x,w)=n}, the set of vertices of distance n from x.
This set 5n has rn thought
let
xr=X,
the
It
of
as
(q+l)qn-1
elements.
a subgroup of the
Therefore Kx may be
infinite product
of the
permutation groups S(r ). We will show that in this product K x n is closed. An element gE fl S(r n) is in the complement of K if, x
for some n, and some w1, w2E 3n, g(w1)=w2, while some element in the chain between x and w1 is not mapped by g into the element
The group of automorphisms Aut(X)
11
of the chain between x and w which has the same distance from 2
x. This condition defines an open subset of the product n S(r n)
and therefore the complement of K
x
is open. This shows that K
the group K
is compact. As x describes X,
x
of compact open subgroups of Aut(f) if g(x)=y,
other:
then K =gK g 1. y
group K
x
x
describes a family
conjugate to each
all
We also observe that
the
x
is totally disconnected.
We will now prove that, if d(x,y)=n, then K n K has index x
r =(q+1)qn-1 in K and in K . n
x
each
for
weV
n
y
choose
K =U{g (K n K ): weV }, x w x y n
Indeed,
g E K w
and
x
the
let
such cosets
distinct. The claim follows because iB
n
i4
n
that
y
be as above, and g (w)=y. w
g (K n K ) w x y
are
Then all
has (q+1)qn-1 elements.
Before analyzing in more detail the group Aut(X) for a general homogeneous tree,
we give here a description of the
simplest nontrivial case, that for which q+1=2 (the reader may verify that a homogeneous tree of degree 1 consists of just one edge).
If q+1=2, then 3f consists of a doubly infinite chain,
that is a geodesic {...x_2,x_1,x0,x1,...}=X, with d(xi,xi+1)=1.
We may therefore identify X with the
integers through the
correspondence n4x . It is clear that translations on 3f form a n
group generated by a step-1 translation and isomorphic to Z. For every x EX, n
there is only one nontrivial rotation which
fixes xn and maps xn+3 to xn-J , for jEZ. If k0 is the rotation about x0 and x is a step-1 translation, ink0i n is a rotation which fixes xn. All rotations of Aut(X) are of this form. Similarly there is only one inversion g0 on the edge {x0,x1}
and all the inversions of Aut(f) are of the form Tng0t n,
nEZ.
It is easy to see that Aut(Z) can be written as the semidirect
product of the group generated by a step-1 translation i, and the two-element group generated by k0. The translations of even
step form a subgroup of Aut(l), still isomorphic to Z.
This
subgroup has two orbits. Every edge intersects both orbits, and
therefore the group generated by the translations of even step
and one inversion acts transitively on X. It is easy to see
Ch. I
12
(see (6.4) below, for a more general construction) that this group is generated by two inversions a and b on edges having one vertex in common. Observe that a2=b2=e (the identity) and that the group generated by a and b is isomorphic to the free
A step-1 translation
product 712*712 of two copies of 7L2 7L/27L.
may be obtained as the product of an inversion followed by a rotation. This means that Aut(l) is also the semidirect product
of the group generated by two inversions on adjacent edges and
the two-element group generated by the rotation about their common
vertex.
other
In
words
Aut(X)
is
isomorphic
to
(12*Z2)>i Z2. 5. Compact maximal subgroups.
vertex K , x
K
Besides
stabilizer
the
of
a
we can also consider the stabilizer of an edge This group is also compact
{a,b}={gEAut(.X): g({a,b})={a,b}}.
where g is an inversion
and indeed K(a,b)=(Kan Kb it, g(Kan Kb)
The subgroups K
with g(a)=b and g(b)=a.
are also
all
{a, b)
conjugate to each other, and open. We will show next that the groups Kx and K{a,b}are the only maximal compact subgroups of Aut(f). We first need an elementary observation.
Let G be a subgroup of Aut(X);
(5.1) PROPOSITION.
closure of G is compact if and only if,
then the
for every xel,
the
orbit G(x)={g(x): geG} is finite.
If G is compact, then G(x)
PROOF.
metric space
3f,
and therefore
is compact in the discrete
G(x)
is
finite.
Conversely,
suppose that, for some xcg, G(x)={g1(x),...,g(x)} is finite. n Gc U1{g: g(x)=gj()}.
Then =gjKK
,
But
{g: g(x)=gJ(x)}={gjg: g(x)=x}
which is compact.'
(5.2) THEOREM. Every compact subgroup of Aut(g) is contained in
a group of the type Kx , for some xE3f, or a group of the type K(a,b)'
for
some
edge
subgroups of Aut(X).
{a,b}Eff.
These
groups
are
maximal
Compact maximal subgroups
13
PROOF. Let G be a compact subgroup of Aut(o). Since for each xEZ the orbit G(x) is finite, so is the subtree of 3E generated
This subtree is mapped by G into itself.
by G(x).
Therefore
If 9 contains
there exists a minimal G-invariant subtree 55X.
more than two vertices, then it contains vertices of degree 1
and vertices of degree greater than
1.
Since the degree
is
invariant under G the subtree of 5 obtained by omitting the vertices of degree
1
is still
invariant,
contradicting the
minimality of D. Therefore D consists of one vertex or of one edge, which implies that G is contained in some K K
remains
It
{a,b}
compact
subgroups.
{a, b}
SK . x
and K
K
x
Observe that
and no edge
vertex xeE, K
to show that
x
or in some
are
maximal
that
for no
(a, b)
(3.4)
implies
can we have K SK
{a,b}ELF,
x
or (a,b)
This shows, by the first part of the theorem, that,
if G is compact and K SG, x
implies x=y and G=K
K{a,b}9GSK(c,d) ,
for
which
K SGSK , x
y
if G is compact and K(a,b) SG,
and,
some
for some y,
then,
then
which implies {a, b}={c, d}
{c, d}ELF,
and G=K
(a,b)
section.
(5.3) LEMMA. Let G be a subgroup of Aut(X) and x a vertex. If
the orbit G(x) contains three distinct vertices on the same geodesic, then G contains a translation.
PROOF. If x,y,tEG(x) are in the same chain, we may assume that
the chain [x,yl joining x to y contains t,
that
[x,y] =
is
[x,tlu[t,yl, and [x,tln[t,y]={t}. Let g,hEG be such that g(t)=x
and
h(t)=y.
Suppose
It, Y1,Y2,...,ym y}.
By
(3.1),
if
translation. Similarly, if h(y )*y 1
and
the
g(x1)=xn-1
lemma
and
is
and
[t, x]={t,x1,x2,...,xn x}
proved.
h(y1)=yM-1.
g(x1)*xn-1, ,
g
is
a
then h is a translation
m-1
Therefore But
then
[t, y]=
x1*y1
we
can
implies
suppose
that
hg-1 (xn-1) =
Ch. I
14
Therefore
h(x1)*h(y1)=ym-1.
(3.1)
implies
that
hg _1
is
a
translation because.hg1(x)=y. 1
6. Discrete subgroups. We will now discuss a class of discrete
subgroups of
which have trivial
Aut(3E),
intersections with
every KX. DEFINITION.
(6.1)
We say that a subgroup r of Aut(l)
faithfully and transitively on X,
or that
acts
is a faithful
it
transitive subgroup, if, for every x,yEX, there exists gEr such that g(x)=y, and if r'nK
X
If r
is a faithful transitive group of automorphisms,
given
then,
is the identity for each xEX.
any
vertex
oEX,
the
map
bijective.
is
g->go
Furthermore r is discrete in Aut(X). There is an easy characterization of the faithful transitive subgroups of Aut(X) for which we need first the following technical result.
(6.2) LEMMA. Suppose that there exist a vertex oEX and a subset A of q+1 elements of Aut(E), such that
(a) A=A1 (b) A(o)={a(o): aEA}={y: d(o,y)=1}.
Then (1)
for every finite sequence a1,...,a
of elements of A,
d(o,a1...a(o)):sn, (2)
for
every
sequence
xEX,
with d(o,x)=n,
a ,...,a
of
elements
there
exists
of
A,
a
such
finite that
1
x=a1 a2 ... an (o) and alai+l*1, for i=1,...,n-1.
PROOF.
To prove (1) we use induction on n. The assertion is
true for n=1. If a1,...,an+1EA, then
d(o,a1...an+1(o))= d(a11(o),az...an+1(o))< d(o,a2...an+1(o))+1 = n+1,
because d(a11(o),o)=1. To prove (2) we observe again that the assertion is true for n=1. Supposing that it is true for n>-1,
Discrete subgroups
15
let xe3E, and d(x, o)=n+1. Let fo=x o, x1, ...
be the chain
xn+1=x}
[o,x]. There exists only one element aEA such that a(o)=x1. But d(a 1(x),o)=d(x,x1)=n,
and therefore a-1 (x)=a
with
a 2... a (o), I
alai+l*1,
for i=1,2,...,n-1.
It follows that x=aala2...an(o).
It remains to prove that aa1*1. But aal=1 implies d(o,x):5n-1, by part (1). Since d(o,x)=n, we must have aa1*1. 1
(6.3) THEOREM. Let A be a subset of Aut(3f) which satisfies the
hypothesis of (6.2).
Then A generates a faithful transitive
subgroup r of Aut(X). Furthermore r is isomorphic to the free product of t copies of Z and s copies of the two-element group 7L 2,
Finally every
for some taO and s>_0 satisfying 2t+s=q+1.
is generated by a
faithful transitive subgroup r of Aut(3f) subset A satisfying the hypothesis of (6.2).
PROOF. Let o be a fixed vertex of E, then the map a-a(o) is bijective from A onto
then either
If aeA,
Si={x: d(x,o)=1}.
a2(o)=o, and therefore a is an inversion which leaves the edge {o,a(o)} fixed, or a(o)*o, and by (3.1) a cannot be other than a step-1 translation.
and
since
inversions
A=A 1, of
it
If a is an inversion then a(o)=a 1(0),
be
must
period
2
al, ... as be the
Let
a=a 1. contained
in
A.
Then
A
={al,...,as,Alt,81,..,at}, where q+l=s+2t, and al,...,at are step-1 translations.
Let G=71*... *7L*712*... *7L2 be the free
product of t copies of Z and s copies of Z2.
Then r is the
image of G under the canonical homomorphism which assigns the
generators of order two of G to the elements al, .... , as, the generators of infinite order of G to al...... at.
and
Observe
that the set G of elements of G which have length no greater n
than n has the same number of elemets as X ={xEX: d(o,x):sn}. n
Let r be the image of G under the homomorphism above, that is n
let r
n
n
be the set of elements of r which can be written as
words of length no greater than n in the elements of A. the map g4g(o) maps r
surjectively onto Xn
n at least the same cardinality as 3f
or G . n
n
.
Therefore r
Then n
has
This shows that the
Ch. I
16
canonical homomorphism is injective and that r is isomorphic to G.
It also shows that the map g--)g(o) is injective and therefore
r is a faithful subgroup.
Conversely,
let r be a faithful
transitive subgroup of Aut(3E), and let o be a fixed vertex of A={ger: d(o,g(o))=1}.
Since d(g(o),o)= d(g 1(0),0)
we
have that A=A 1. Since g-*g(o)
is a bijection of r onto X,
it
X.
Let
follows
A
that
is
in
one-to-one
correspondence
with
1531={x: d(x,o)=1}. Therefore A satisfies the hypothesis of (6.3)
and generates a faithful transitive subgroup of r. Clearly this subgroup must be all of r.,
We have shown in particular that, given or=X, any faithful
transitive subgroup r is generated by the set {ger: d(g(o),o)=
1}. Observe that Aut(X) can be written as the product of the
of the vertex o,
and any faithful transitive
Indeed if gEAut(X),
there exists g'er such that
stabilizer K
0
subgroup r.
g'(o)=g(o), which implies g-1 g'E K
0
and g=kg', for some keK ; 0
clearly K nr={e}. 0
A consequence of the decomposition Aut(X)=K r
is
that
0
Aut(l) is unimodular and its Haar measure is the product of the
Haar measure on K (which may be normalized so as to have total 0
mass 1) and the Haar measure on r (which may be normalized so that points have mass 1). A similar decomposition is true for
every closed subgroup of Aut(X)
which contains a subset A
satisfying the hypothesis of (6.3).
7. Cayley graphs which are trees.
In
Section
described the
to
a free group with two
tree associated
2,
Fig.5,
we
generators. Indeed to every homogeneous tree we can associate a
finitely generated group, in such a way that the elements of the group are identified with the vertices of the tree and the
tree is the Cayley graph of the group with respect to a given
set of generators. We recall that, if r is a group and E is a set of generators of r, the Cayley graph of r is a graph with
Cayley graphs which are trees
vertices r and edges xs=y}.
{{x,y}:x,yer,
17
there exists sEE,
with
We observe that the Cayley graph as defined above is
always connected and that left multiplication by an element of
r is an automorphism of the Cayley graph of r (it maps edges into edges). Furthermore r acts faithfully and transitively on
its graph. Therefore (6.3) tells us that the graph of r is a
locally finite homogeneous tree if and only if r is the free product of t copies of 71 and s copies of 72, Figs 9-12
illustrate
the
graphs
of
71,
with 2s+t=q+1. 7*712,
g2*Z2'
Z2*g2*g2. The graph of Z*Z was given in Section 2, Fig.5.
-2
-3
a
a
Fig.9 r = 71,
bab
ba
Fig.10
_1
e
a
a
a
e
a
ab
aba
ba
\
/
\aba'i /aba
_1
a b
Fig. ii
3
r = 712*712, a and b are generators of order 2
bat
-2
a
a is a generator of infinite order
b
arl b a-'\ a l b a/
a
2
ab
e
a
2
a
r = 71*712, a is of infinite order and b is of order 2
and
Ch.I
18
r = Z2*
Fig.12
Z2, a, b and c have order 2
if the graph of r is a tree X, then the
We observe that,
length of an element of r as a reduced word in the generators
and their inverses is nothing but the distance in the tree of the vertices corresponding to that element and to the identity.
8. Amenable subgroups.
order
In
other
describe
to
notable
subgroups of AutM, we need to consider the action of Aut(X)
on the boundary 0 of
We recall
X.
that
equivalence classes of infinite chains,
when their
being equivalent
S2
is
the set
of
two infinite chains
intersection
infinite
is
(and
therefore cofinite in each of the two chains). An automorphism maps
gEAut(Z)
preserving
an
the
naturally on n.
infinite
equivalence
chain
into
classes.
We observe that,
an
infinite
Therefore
if xE.T,
chain,
acts
Aut(3E)
the group K
x
acts
transitively on Q. Each element wE2 has a representative of the type
[x,w),
and 0 may be
infinite chains
starting
identified with the set at
Given two
x.
distinct
of all chains
{x, ti, t2, ... } and {x, si, s...I we can define kEKx such that k(s )=t n
,
n
by specifying the action of k as a permutation of
iiin ={y: d(x,y)=n}. Let wE.Q and define Gw {gEAut(Z): g(w)=w}. It is clear that
Gw is a closed subgroup of Aut(o).
We define now Bw {gEGw:
there exists xeX, such that g(x)=x}.
In other words B
w
consists of all the elements of G which
are rotations about some vertex.
w
It
is not difficult to see
Amenable subgroups
that Bw is a group.
19
the infinite chains
Indeed, given x,yE ',
[x,w) and [y,w) have infinite intersection. elements of B
If g and g'
are
and g(x)=x, g'(y)=y, for every tE[x,w)n[y,w), we
Ln
The group Bw is the union of compact
have g'g 1E GwnKts Be,,.
subgroups BW = U KxnG0 , XEX
and indeed,
infinite chain belonging to
if {x0,x1,x2,...}
the equivalence class
is any w,
then
00
Bw
n
0Kx nGw. By definition KxnG0 is open in Gw and in Bw , for
each x.
n
Therefore Bw is the union of countably many compact
open subgroups,
which implies that Bw is open in Gw and is
amenable. We will show now that Bw is normal in Gw.
Let gEGw
and suppose that g is not a rotation. Since an inversion cannot fix any point of S2, g cannot be an inversion. Therefore we may
assume that g
is a translation along a geodesic y=(w,w')=
{ ... , x_1, xo, x1.... } . gkg EGwrK9(x
).
and,
But Bw U GwAKx , n
if kEGwnKx , then
n
n
Therefore Bw is normal in Gw. Let now T=(w,w'
n
be a fixed geodesic and let Ml
Then M
K
is a compact
xEy
group contained in B
Observe that if t is a fixed step-1
translation on y, moving x in the direction of w, and g any step-n translation on d, then xng 1EM*Y . Thus every translation on y may be written as the product of an element of My and an element
of
the group generated by t
(which of course
is
isomorphic to Z). If gEGw, and g4Bw, then g is a translation on
a geodesic
where
w"
may be different from W.
Replacing g with g-1 if necessary, we may assume that g moves y' towards
w.
,-n
9(x)=x.
Let
xeXny'
and suppose
Therefore t ngeBw.
that
d(x,g(x))=n.
Then
We have proved therefore that
every element of Gw may be written as the product of an element
of Bw and an element of the group generated by a step-1 translation on the geodesic y=(w',w).
The
latter group
is
abelian, and isomorphic to Z. In conclusion we have shown that Gw is the semidirect product of a copy of Z and the group Bw. In particular Gw is amenable.
20
Ch. I We saw that any two translations along the same geodesic z
differ by an element
of
MId .
The quotient
Bw/M7 does not
correspond to any subgroup of Bw, but, as will be seen later,
may be naturally identified with the orbits of B
which are
called the horocycles, relative to w.
We should also observe that the set of all automorphisms which are translations along the geodesic T=(w',w), or are the identity on this geodesic, is a group containing M7 as a normal subgroup.
Thus
Mly
is the kernel of a homomorphism of this
group onto Z.
Given the geodesic G
{gEAut(X): g(7)=z}.
=(w',w) If
K
is
we may consider the group either
inversion interchanging w and w', then G
a
rotation
or
an
(GwnGw,)u K(GwnGw,).
It follows that the closed amenable subgroup GwnGw, has index 2 and therefore GId is amenable. We will now prove that the
in G
only
amenable
closed
subgroups
of
Aut(X)
are
the
closed
subgroups of the groups Kx, K{a,b}' Gw, and G1,. (8.1) THEOREM. Let G be a subgroup of Aut(E), and suppose that
G contains no translations. Then GSK
x
for some vertex x,
or
GSK(a,b) for some edge {a, b}, or else GSBw for some WES2. PROOF.
By (5.3) we know that no orbit of G contains three
vertices on the same geodesic. If G has compact closure, it is
contained in a maximal compact subgroup which,
by (5.2),
is
either a Kx or a K{a,b>. It suffices to show, by (5.1) that,
if
G(x)
is infinite for some vertex x,
then GSGw for some wEfl.
Indeed if G has no translations and GSGw , of
rotations,
boundary.
since
no
then G consists only
inversion can fix a point
of the
This implies that GSBw. Suppose now that one,
and
therefore every, orbit of G is infinite. Then every orbit has a limit point in the compact space JfuQ. We want to show that, for
some vertex x, G(x) has only one limit point. Since the closure
of G(x) in lug is G-invariant, this will show that there is a point wEf2,
which is fixed by G,
that is that GSGw. We shall
Amenable subgroups
reason by contradiction observing that,
21
if w and w'
are two
distinct limit points of G(x), then G(x)n(w,w')=o. Indeed let t
belong to the intersection of
z be
and G(x), and let y,
vertices of (w,w') at distance 1 from t and in the direction of
w and w',
respectively. Then C(t,z) and C(t,y) are disjoint
neighborhoods of w and w' respectively. Therefore there exist vertices y'EC(t,y)nG(x) and z'EC(t,z)nG(x). The chain [y',z'l includes the point t and therefore,
by (5.3),
G contains a
translation. Let now G(x ) be the denumerable set of infinite n
orbits
into
which
X
partitioned.
is
We
are
assuming
the
contrary of what we want to show, that each of the orbits has
more than one limit point in 0.
Let
i21
be the set of limit
points of G(x1), and let V1 be the subtree consisting of all
the chains (w, w' ) with w, w' ESa1. Observe that 51 is G-invariant, and
Therefore
)1nG(x1)=m.
sequence of subtrees qk'
can
we
construct
and a sequence of vertices yk,
where
and y k xn , k
if
Xn EDk-1 and finally, k G(yk),
k-i
first
index such that
and Qk5;Q k-i
closed nested sets
Clearly V knG(x)=D,
Furthermore
large.
such
the set of limit points of
is
with w,w'EQ .
(w,w')
sufficiently
c
S2k
the
is
we let Vk to be the tree consisting of all
geodesics
k
nk
a
It suffices indeed to let
that gJknG(yk)=E, ykEDk-1, and n 9k m. x1=y1,
inductively
G(yk)SDk-1,
infinite
if k
which
is
implies
This means that the intersection of the S2k
is nonempty.
This intersection is an
invariant subset of S2 which cannot consist of more than one point because the intersection of the Vk is empty. Thus there
exists a point of
S2
which
is
invariant under G and G is
contained in Gw., (8.2) LEMMA. Let
'
vertex in common. Then
and X' be two geodesics having at most one Let
-t
and T'
be translations on 7 and T'
the subgroup generated by z and
't'
is
PROOF. Let xEa' be the point of minimal distance from X'.
By
respectively.
discrete and isomorphic to a free group.
22
Ch.I
hypothesis x is uniquely defined, 7n"'={x}.
Let
C'(x)
the
be
because either XnX'=m,
union of
chains
infinite
all
or
starting at x and containing no other point of y. Let C(x) be the complement
of C'(x)
together with the
vertex
x.
Then
C'(x)nC(x)={x}, and y'gC'(x). Furthermore any chain connecting a point of C' (x) with a point of C(x) goes through x.
Observe
that, if "" is a geodesic contained in C'(x) (resp. C(x)), then t(ar")
does
(resp. t'('")) not
intersect
is contained in C(x) ''
(resp. y').
(resp. C'(x))
and
now w be a reduced
Let
nonempty word in t and t' which contains the letter t' at least
once. We observe that w(v) is disjoint from T.
Indeed w is the
product of powers of t and t'; a power of t maps
' onto itself;
therefore we may assume that w has a t' as its last letter. A nonzero power of t' maps a' onto a geodesic contained in C' (x)
which does not intersect T. Therefore the first application of
a power of t' and all successive applications of powers of t and t' map z into geodesics which do not meet T. We have proved
that w(T)n'=r, if w contains a power of T'. On the other hand if w is a nonzero power of t, then w(x) is not the identity on T. We have thus proved that any reduced word in t and t' is not
the identity as an element of Aut(X), and therefore that t and t' generate a free group. In addition the intersection of this group with any stabilizer of a point of
'
is the identity, and
therefore this free group is discrete. '
We shall
characterize now the maximal
closed amenable
subgroups of Aut (X) . (8.3) THEOREM. Let G be a closed subgroup of Aut(3E). Then G is
amenable if and only if one of the following occurs: (i) G is compact; (ii) GSGWfor some X52; (iii) G9G7 for some geodesic T.
PROOF. We have observed already that the groups GW and G7 are amenable.
Compact groups are of course amenable.
Conversely,
suppose that G is amenable. If G has no translation, then by
Amenable subgroups
23
(8.1) it is compact, or it is contained in Bw.
We may assume
therefore that G has a nontrivial translation t on a geodesic T. For the rest of the proof the translation t and the geodesic ' will be fixed. We observe first that, if gEG, then a'ng(a') is infinite.
Indeed,
geodesic g(T).
is a nontrivial
gtg 1
But G
is
translation on the
and therefore
amenable,
it
cannot
contain a discrete free group with two generators. This means, by (8.2), that g(')nd*0. Furthermore g(,X)na' cannot be finite,
because otherwise,
for some
which implies g(y)nThg(')=0, (8.2),
integer
which
h*O,
th(g('X)n'X)ng(T)=0,
again
is
impossible,
by
because thgtg 1i h is a translation belonging to G on
the geodesic thg(,). We have proved that g(T)nT is infinite for
every geG. On the other hand if h is another element of G, we also have that h(')ng(') is infinite because hth
1,
and gtg
1
are translations on h(') and g(a'), respectively. We shall prove
now that either G9G7 or G stabilizes one of the end points of '.
and E_{w,w'}.
Let
If g(g)=T then g(E)=E, but if
g(T)*7 then g(7)nq is an infinite chain and g(E)nE consists of only one point.
then G9G7. Assume now
If for every g g(')=-d,
that there exists an element g of G such that g(')*T, and let w be the point in g(E)nE. We shall prove that for every hEG, such
that h(')*T, we have h(E)nE={w}. In other words we shall prove that h(')ng(')n-y is an infinite chain. This is clear because,
if h(')ng(')ny is finite, then either h(')ng(') is finite, or h(T)ng(7)ny h(-X)ntg(')
consists is
empty.
of
only
Both these
one
point,
things are
and
therefore
impossible and
therefore n g(E)={w} is a G-invariant subset of 0, which means geG
that GSG. ,
We can deduce from the results of this section that a nonamenable closed subgroup of Aut(o) necessarily contains a
free discrete group in two generators.
Indeed if G is not
24
Ch.I
amenable,
must contain a translation z on a geodesic 7
it
(8.3). But the proof of (8.3) shows that if g(')n'*e, for every gEG,
then G9G1X or GSGw for some
w.
if G is not
Therefore
amenable, for some gEG, g(T)n'=a, which implies, by (8.3), that t and gig-1 generate a free discrete subgroup.
We should also remark that Gw acts transitively on X and on SZ\{w}.
It is clear that the stabilizer of a vertex or of an
edge or of a geodesic does not act transitively on X. In other
words we may say that a closed subgroup of Aut(X), transitively on X,
acting
is amenable if and only if it stabilizes a
point of Q. 9. Orbits of amenable subgroups.
In
describe the orbits on the set
this
section
we
shall
of vertices X and on the
boundary S2 of the subgroups of Aut(X) which were described
in
the previous sections.
If xEZ the orbits of K
are exactly the sets Z ={y: X
d(x,y)=n}.
a
Furthermore,
closed
subgroup
K
of
X
n acts
transitively on SZ if and only if it acts transitively on 5n for every n. This follows immediately from the fact that S2 can be described as the set of infinite chains starting at x. If {a,b}
is an edge,
then K
nK {a,b}
a
acts transitively on the chains
which start at a and pass through b.
inversion exchanging a and
b,
it
Since K{a, b} contains an follows
that
K
(a,b)
acts
transitively on Q. The orbits of K{a,b} on X are {a,b}, the sets {y: d(a,y)=1, y*b}u{y: d(y,b)=1, yxa}, and in general, for n>1, {y: d(y,a)=n, bit[a,y]}u{y: d(y,b)=n, ai6[b,y1}.
Let i' be a doubly infinite geodesic. The group G, has ' as
an orbit and the other orbits are the sets {y: d(y,T)=n}. On
the boundary S2, G7 has two orbits:
and SZ\{w, w' } , where w
and w' are the two ends of T. For w a point of the boundary, we have that Bw , fortiori Gw ,
and a
act transitively on Q\{0. In addition Gw acts
transitively on X. This is proved as follows. Given x,yEX, the
Orbits of amenable subgroups
25
chains [x,w) and [y,w) intersect in the chain [z,w). Suppose d(y,z)_d(x,z), then there is a rotation k, which fixes z and w and maps x into a point of [y,z]. Thus k(x)e[y,wl. Let 7 be any
geodesic containing [y,w), and let T be a translation along 7 and such that i(k(x))=y. Then TeGw and therefore rkeGw.
The group Bw does not act transitively on X.
Indeed,
if x
and y are different elements lying on the same geodesic having
w as one of the ends, no element of Bw can map x into y.
To
describe the orbits of Bw we introduce an equivalence relation on X as follows. Let x,yEX and suppose that [x,w)n[y,w)=[z,w).
Then we say that x is equivalent to y if d(z,x)=d(z,y). The equivalence classes of this relation are called horocycles of w.
The
horocycles
exactly
are
that
=(w',w)={.... s_1,s0's1,...},
the is
orbits assume
of
Let
Bw.
Then
lim sn w.
H ={gs gEB } is a horocycle and U H =X. Each horocycle H n n n n w divides X into two subsets, intersecting in the horocycle H : :
n
the subtree
"inside"
the horocycle,
U H
k
,
and the subset
k2tn
"outside" the horocycle,
U Hk.
Observe that every geodesic
kin
having w as one of the end points intersects each horocycle exactly
once.
Fig.13
gives
a
horocycles, for the case q=2.
geometric
picture
of
the
H _y
Fig.13
This picture makes it clear why horocycles are sometimes
said to represent the generations with respect to a common mythical ancestor w.
26
Ch. I
10. Groups with transitive action on the boundary. We have seen
that compact maximal subgroups of Aut(X) act transitively on the boundary. The following proposition deals with noncompact groups which act transitively on Q.
(10.1) PROPOSITION. Let G be a closed subgroup of Aut(X) and suppose that G is not compact. Then G acts transitively on 0 if
and only if there exists xEX such that GnK
acts transitively
x
on Q. PROOF.
Observe that,
if G contains no translation,
then by
(8.1) it either is compact or fixes a point of the boundary. Therefore, under the hypothesis that G is noncompact and acts transitively on 0, there exist a geodesic I and an element TEG which is a translation along a'.
Choose xEy. The subgroup K nG X
is open in G and has countable index. This means that, if wEi2, then,
by the transitive action of
G,
D-G(w)=y hi(GnKx)(w),
where hi is a complete set of coset representatives. Since 0 is
a complete metrizable space and hi(GnK)(w)
is compact,
it
follows that hi(Gr& )(w) has an interior point. Therefore the
orbit (GnK )(w) has an interior point, and hence it
is open.
X
But if the orbits of GnK
in 0 are open, they must be finitely x
many. Let w' and w" be the two ends of y. Then the orbits of w'
and w"
under GnK
are
This
open.
x
means
that
there
exist
x',x"Ea', with x'E[x,w') and x"E[x,w") such that {w: x'E[x,w)}S (K nG)(w') and {w: x"E[x,w)}S(K nG)(w") (Fig.14). x
x
Fig. 14
Groups with transitive action on the boundary
27
But G contains a translation t on the geodesic (w',w"). Suppose
that t moves x in the direction of w".
Then there exists an
index N such that, for n>_N, x"E[x,tnx]. Therefore, if w*w', we have
[x,tnw)=[x,tnxlu[wnx,tnw)
Let kEK nG
and tnWE(K nG)(w").
x
x
be such that ktn(w)=w". Then [x,ktnw)=[x,ktnx)u[ktnx,w"). This implies
that
ktnx=tnx.
Thus
and i nktnEK nG.
t nkxnX=x
and therefore wE(K nG)(w").
t nkcn(w)=w",
x
S2\{w'}c(K nG)(w"). x
But
x
We have shown that
Similarly f2\{w"}c(K rG)(w'). x
if qa2,
But,
then S2\{w' } and S2\{&)"} intersect, which implies that Kx nG acts
transitively on Q. Obviously, if KxnG acts transitively on 92, for some x, then so does G. '
The next result shows that a group which acts transitively
on S2 and is not compact has at most two orbits on X.
These
orbits are nothing but the equivalence classes of the relation
"d(x,y) is an even number". The fact that this is an equivalence relation follows from the observation that,
if x,y,z,EX
and t is the point of [x,y] of minimal distance from z, then d(x,y)=d(x,z)+d(z,y)-2d(z,t). If we fix or=X, these equivalence
classes may be written as X+={x:d(o,x) is even}
and
3f ={x:
d(x,o) is odd}. With this notation we prove the following.
(10.2) PROPOSITION. Let G be a closed noncompact subgroup of Aut(T) which acts transitively on Q. Then either G is transitive on the vertices, or G has exactly the orbits 3f PROOF.
By
(10.1)
there
exists
x
such
and X+.
GnK
that
x
acts
transitively on Q. Therefore GnK acts transitively on the sets x
Six={y: d(x,y)=n}, for every n. n
Let yEG(x), and let n=d(x,y).
Then SixcG(x). Therefore G(x) is the union G(x)= U Sixn n
,
where E
nEE is a set of nonnegative integers. Let k be the smallest integer
other than zero occurring in E, Furthermore K
and let
Then yEG(x).
is a conjugate subgroup of K nG. Therefore K nG x
y
acts transitively on 0 and on the sets Siy={z: d(z,y)=n}. This n means that every element having distance k from y belongs to
28 G(x).
Ch. I It follows that E must contain all multiples of k. We
show now that k is at most 2.
If k=1 then G(x)=3f and G acts
transitively on X. Suppose now that k>1. Let ye8fk, and let x'
be the first element after x in the chain [x,y]. Let z be a vertex, not in the chain [x,y], having distance 2 from x and distance 1 from x' (Fig.15).
Fig.15
Then d(z,y)=k
and
therefore
zelS3'SG(x).
This
implies
that
5X9G(x) and therefore k=2. Since U Ekj5 G(x), we conclude that j_o
G(x)={veX: d(x,v) is even}. By the proof of (10.1), that
there
exists
transitively on Q. d(t,v) is
even}.
with d(x,t)=1
tEl
such
that
it follows GrKt
acts
A similar argument proves that G(t)={vEX: Since
k>1
and
d(x,t)=1
it
follows
that
G(x)nG(t)=0, and the proposition follows. '
A notable subgroup of Aut(X) which acts transitively on 0, Aut(X)+,
but not on 1,
is the group
generated by all rotations.
The next result concerns this group.
(10.3) PROPOSITION. The orbits of the group Aut(l)+ generated by all rotations are exactly the equivalence classes
X+
and 3f
of the equivalence relation "d(x,y) is even". Furthermore, if x is any vertex, Aut(X)+=K r+, where r is any faithful transitive
subgroup and
r+
X
is the subgroup of r which leaves invariant
X+
Groups with transitive action on the boundary
29
and X-. Finally Aut(3f)+ is the only noncompact proper subgroup of Aut(X) which contains K
x
may be defined with reference
PROOF. Observe that the 3f+ and 3f
to any vertex o. Clearly they are invariant under the group of rotations about o. Since o is arbitray they are invariant under any rotation. By (5.2), Aut(3f)+ is noncompact, and therefore by
(10.2) its orbits are exactly 3f+ and X-. Let Aut(o)=Kx r, where r
is
a faithful
subgroup
transitive
(6.3).
Let
r+={ger:
d(x,g(x)) is even}. Let ger+, then there exists hEAut(3f)+ such that
g(x)=h(x).
Thus
g-1 hEK
and
,
x
gEAut(3f)+.
Therefore
r+SAut(3f)+. Since r acts transitively on 3f, so does r+ on 3f+. This means that a similar argument shows that Aut(3f)+=Kx r+. Let now G be a noncompact proper subgroup of Aut(X), containing K . X
Then, by (10.2), G(x)=3f, or G(x)=,3:+. there exists hEG,
In the first case, if gEr,
such that g(x)=h(x),
which implies g-1 hEK
x
and gEG. Thus G contains every faithful transitive subgroup and
it must be all of Aut (If) . In the second case G contains r+ and therefore G=Aut(3f)+ because Aut(X)+ has index 2 in Aut(30 (in fact Aut(30+={gEAut(X): d(x,g(x)) is even}). '
We remark that the proposition above implies that Aut(3f)+
is generated by Kx and a nontrivial rotation not belonging to K . In particular, if {a,b} is an edge, Aut(X)+ is generated by x the subgroups Ka and Kb. It can actually be shown that Aut (3f )+ is, in a natural way, the amalgamated product of Ka and Kb over their
intersection
K nK a
.
b
The
relationship
between
groups
acting on a tree and amalgamated products is studied in [Se] where the reader will find the proof of the statement above. A group G is said to act doubly transitively on 3f if, for
every two
pairs
(x,y),
(z,t)E3fx3E
with d(z,t)=d(x,y),
there
exists an element g of the group such that g(x)=z and g(y)=t. A
closed subgroup G of Aut(X) is doubly transitive on X, if and only if G is transitive on transitively on the
sets
3f
and,
Ex={y: n
for every x,
d(x,y)=n}.
GnK
x
acts
Equivalently,
a
30
Ch. I
closed subgroup G acts doubly transitively on X if and only if it acts transitively on X and Q.
There is an analogous notion
of doubly transitive action on Q. A closed subgroup G of Aut(X) doubly
acts
transitively
on
and
if
S2
only
if
G
acts
transitively on 9 and GnGW ={gEG: g(w)=w} acts transitively on M{w},
for
every
WED.
We
observe
that
Aut(X)+
is
doubly
transitive on 0 because, for each w, it contains the group B w which acts transitively on Q\{w} and because it contains a full
group of rotations about a vertex, which is transitive on Q. Nevertheless Aut(X)+ is not transitive on X.
We conclude this chapter with a necessary condition for a group of automorphisms to contain faithful transitive subgroups
of every isomorphism type. We observe first that a group G is
transitive on T, the set of edges,
if it is transitive on Z
and, for every x, KXnG is transitive on lt31x={y: d(x,y)=1}. With
this in mind we can prove the following proposition.
(10.4) PROPOSITION. Let G be a closed subgroup of Aut(X) and suppose that (a) G acts transitively on the set of edges T, (b) G contains an inversion of order 2.
Then, for all integers t, s such that 2t+s=q+l, G contains a faithful transitive subgroup isomorphic to the free product of t copies of Z and s copies of Z2. PROOF.
Since G
is transitive on the
edges,
condition
(b)
implies that there are inversions of order 2 on every edge. Therefore, by (6.3),
it
is sufficient to show that,
for any
three vertices x,x',x", with x'*x" and d(x,x')=d(x,x")=1, there
exists gEG such that g(x')=x and g(x)=x",
in other words G
contains a step-1 translation on a geodesic containing [x',x"]. By
condition
(a)
there
exists
OEG
such
that
O(x)=x
and
O(x')=x", and there exists ¢EG such that O(x')=x and O(x)=x'. Thus g=00-1EG, g(x')=x and g(x)=x". ,
Notes and remarks
31
Homogeneous trees and their automorphism
11.Notes and Remarks.
groups come up naturally in many areas of mathematics.
The
interest in trees over the past 15 years was kindled by the lecture notes by J.P.Serre [Se] which were made available as mimeographed
notes
well
before
their
actual
publication.
P.Cartier initiated the study of spherical functions on trees [C3] (see Chapter II, below).
Many of our definitions and simple geometrical ideas on trees
are
taken
from
the
work
of
P.Cartier
[C1,C2].
The
contributions of J.Tits to the study of groups of automorphisms
of trees are very important.
In his paper [Tit] he shows that
the group Aut(X)+ is simple. The simple classification of the automorphisms of a tree given in Section 3 and Lemma (5.3) are taken from [Tit].
Theorem (6.3)
is taken from [Ch] which in
turn is based on [F-T P3]. An earlier result of the same type is contained in [BP].
The characterization of amenable groups
given by Theorem (8.3) is taken from [N4], where, as in these notes, it is deduced from Theorem (8.1) which is due to J.Tits [Tit].
The fact that a solvable subgroup of Aut(X) satisfies
one of the conditions (1), (ii) and (iii) of Theorem (8.3) also
follows from a result of [Ti2] concerning more general trees (O2-trees). A characterization of amenable subgroups of PSL(2,R) similar to that of Theorem (8.3) is given in [NS].
Horocycles on trees were introduced by P.Cartier [C2]. The results
of
condition
Section (b)
of
10
are
taken from
Proposition
(10.4)
[N2]
is
and
not
[Ni].
The
necessarily
satisfied by every group acting transitively on 3f and Q.
The
existence of counterexamples was kindly communicated to us by T. Steger.
We
describe
here
a
counterexample
under
the
hypothesis q>2. A similar more complicated example for the case q=2 was also found by T. Steger. We consider the collection E of all subsets o<-3f consisting
of exactly q vertices all adjacent to the same fixed vertex. For each o-1 we choose once and for all a labelling, that is a
32
Ch. I {1,2.......
coordinate mapping Jo
Every automorphism of
3f maps E into E and the map Jg(,.)ogo5Q, defines a permutation on
q objects, i.e. an element of the symmetric group S(q). We say
that gEAut(X) is J-even if the permutation JgIQ)ogo5a is even
In order to define an J'-odd automorphism we
for every o-EE.
consider an edge {a,b}
and two elements
EE consisting
b
a
respectively of all vertices adjacent to a except b, and all
vertices adjacent to b except a. For g to be J-odd we require that
the
permutations
Jg1
)ogoJoa
1
and
9'g(
have
b) 0g09'
b
different parity for every edge {a,b}El.
G-H+v
We now let
H
where H+ and H are defined as follows. The set H consists of
all J-odd elements gEAut(fl such that d(g(x),x)
is odd for
every xEX (the latter condition implies that g is either an inversion or an odd-step translation). The set H+ (which is a subgroup) consists of all J-even elements g such that d(g(x),x)
is even for every xE3f (this means that g is a rotation or an even-step translation). It is easy to verify that G is a closed
subgroup without inversions of order 2.
To show that G is
transitive on Q it suffices to prove that, given x,yEX at equal
distance from a vertex oEX, there exists an element of H+ which
maps x into y leaving o fixed. This element may be constructed
step by step so that the condition of belonging to respected;
it
H+
is
is in this step-by-step construction that the
condition q+1>3 is used. A similar construction allows us to find a translation of odd step in H and this is sufficient (by Proposition (10.2)) to prove that G acts transitively on X. The counterexamples and Theorem (6.3) imply that if q+1 is odd
then
there
transitively
on
exist X
transitive subgroups.
and
closed 52
subgroups
which
do
not
of AutW contain
acting faithful
If q+1 is even, condition (a) of (10.4)
is enough for the existence of a faithful transitive subgroup
isomorphic to a free group with (q+1)/2 generators. other
hand,
F. Bouaziz-Kellil
[BK2, Ex. f, p.19]
On the
gives
for
Notes and remarks
33
every q an example of a discrete subgroup acting transitively on 3E which fails to contain a faithful transitive subgroup.
34
CHAPTER II
1. Eigenfunctions of the Laplace operator.
Let f be a function
defined on 1, the set of vertices of a homogeneous tree.
We
let Lf(x) denote the average of the values of f on the nearest neighbors of x. In other words, f(y)
Lf(x) = (q+1)-1 (x,Y)ELF
The operator L which is defined on the vector space of all complex-valued functions on 3E is called the Laplace operator on X.
(Sometimes,
perhaps more appropriately,
the name Laplace
operator is given to the operator (Lf-f).) In this section we will characterize the eigenfunctions of
the Laplace operator, that is,
the functions f which satisfy
Lf=µf, for some eigenvalue peC.
First of all we define certain elementary eigenfunctions associated to points of the boundary.
Recall that i2(x,y)={w:
[x,y]
if
[x,y]
is any (oriented) chain the sets
is a subchain of [x,w)} are open in 0, and
indeed form a basis. If xeX, the sets 92(x,t) with d(x,t)=n form
a partition of 12 into (q+1)qn-1 disjoint open and closed sets.
Therefore there exists a unique Borel probability measure v
x
on
Q such that vx(Q(x,t)) =
The measures v
x
q q+1
-d(x,t)
q
are all absolutely continuous with respect
to each other. This is most easily seen by computing the Radon-
Nikodym derivative dv /dv (w), Y
x
which will be shown to be an
everywhere positive function of w assuming only finitely many values.
Let weQ and x,yEX. Let {x,s1,s2,...}=[x,w) be the infinite
Eigenfunctions of the Laplace operator
chain from x to w.
The sets f2(x,s) are a basic system of n
))=(q+l)-1q n+1
neighborhoods of w, and v (12(x,s
n
x
Consider
the
sequence
horocycles
of
indexed so that xEH0 (Fig. 1) , and sEH suppose
yEHk,
kEZ.
35
Choose
n>O
and
of
w,
{H : n
nE7L},
for n>O. Let ye.T and n>k.
Then
vy(S2(x,sn))_
P (S2(x,s ))qk. Letting n go to infinity, we obtain dv /dv (w)= n
x
y
x
q k
Fig.1
To give a more concise description of the Radon-Nikodym derivative, we introduce the function S(x,y,w) which is defined
as the distance of the w-horocycle H(x) containing x and the w-horocycle H(y) containing y, taken with the positive sign if H(y)
is closer to w than H(x),
and with the negative sign
otherwise. With this notation dv /dv Y
Observe that,
(w)=gSCx,y,w)
x
for fixed x and y,
the function S(x,y,w)
takes values in Z, and -d(x,y):58(x,y,w):5d(x,y). This means that dv /du (w) can assume only finitely many values. We shall use y
x
the notation P(x,y,w) =
dv dvy(w). x
Let now o be a fixed vertex, w an element of 9, and zEC. Define fz(x)=PZ(o,x,w). We shall prove that
z
1-z
Lf (x) = µ(z) f (x), z z
where µ(z)=qq +q1 If t is a nearest neighbor of x, then S(o,t,w)=S(o,x,w)±1.
36
Ch. I I
As t varies among the q+1 nearest neighbohrs of x, the sign + occurs once and the sign - occurs q times. Therefore (q+1)-1
Z f(t)= (q+1)-'(f (x)gZ + f() qi-z)= A(z)f Z. z
{x,t}OF
This shows that f
is an eigenfunction. We will show that every
z
eigenfunction with eigenvalue µ(z) can be obtained from the elementary
eigenfunctions
f z
by
an
appropriate
limiting
process.
Let X(SZ) be the linear space of continuous
(1.1) DEFINITION.
functions on 0 which take only a finite number of values, i.e.
which are linear combinations of characteristic functions of the sets S2(o,x).
The elements of X(12) are called cylindrical
functions. Let X'(12) be the space of finitely additive complex-
valued set functions defined on the algebra of subsets of 12 generated by the open and compact sets Q(o,x). The elements of X'(Q) are called finitely additive measures.
It is not difficult to show that X' W) is the dual space of X(c).
Indeed the sets of constancy of an element of X(S2)
belong to the algebra of subsets of Q which is the domain of the elements of X'.
Observe in particular that P(o,x,w)EX, if oE3f is fixed. It
makes sense therefore, for zEC, to define, for mEX', PZ(o,x,w) dm(w).
P m(x) = z
112
? m(x) is a µ(z)-eigenfunction of L. This fact is easily proved z
by observing that, if x varies in a finite set, PZm(x) = Eim(Ai) PZ(o,x,wi),
where the A1SQ are sets on which the functions PZ(o,x,w) are
constant and w EA . The assertion then follows from the fact i
that,
for
fixed
i
w,
PZ(o,x,w)
as
a
function
of
x
is
a
Eigenfunctions of the Laplace operator
g(z)-eigenfunction.
The
following
theorem
says
37
that
every
µ-eigenfunction may be expressed as P m(x), for some mEX' and z
some zEC, such that µ(z)=µ. It is important to observe that the map z-
(z) is surjective.
(1.2) THEOREM.
Let f be a function defined on 1 and satisfying
Lf=µf, for some peC. Let oeX, and let z be a complex number µ=µ(z)=(qz+q 1-z
such that
)/(q+l), and z*kin/ln q, for keZ. Then
there exists mEX' such that f(x) = P m(x) = IS2PZ(o,x,w) dm(w). z PROOF. For simpler notation, we shall write P(x,w) in the place
of P(o,x,w). Let S be a finite subtree of 1, containing o.
We
say that a vertex of 3 is an interior point if each of its q+l
A vertex of S which is not an
nearest neighbors lies in S.
interior point is said to belong to the boundary 8S of St, or to
be a boundary point. We call a complex-valued function on S a µ-eigenfunction, if Lf(x)=pf(x), for every interior point xeS. Observe that,
if S has no interior point, then every function
on S is a µ-eigenfunction. Now every boundary point of & is the
last element of the chain [o,w)nr for some wE12. every we
Conversely,
identifies uniquely the last element yw of the chain
[o,w)nS. The restriction to S of the function 6(o,x,w) may be It follows that the restriction to
written as d(o,yw)-d(x, yw).
of the function P(x,w) may be written as P(x,w)I
Therefore
the
functions
= gd(o,yw)-d(x,yw z
P (x,y)=q
z(d(o,y)-d(x,y))
with
yEa ,
are exactly the restrictions to S of the functions PZ(x,w). These functions are µ-eigenfunctions on 3, before
the
statement
of
the
because,
theorem,
PZ(x,w)
as shown is
a
µ-eigenfunction on 1, for every we.2. We shall prove that the set B={PZ(x,y): yEBj}
µ-eigenfunctions on
is a basis for the vector space of the S.
We prove first
that
it
is
linearly
independent. We use induction on the number of vertices in S.
38
Ch. I I
Assume that B is linearly independent, and let S' be the tree obtained from & by adding one point. If the new point is added in such a way that some boundary point of S becomes an interior
then the number of elements of 8g'
point of T',
remains the
if y is the element of 8S which becomes an
same. Furthermore,
interior point and y'
is the added boundary point then y'
the last element of the chain [o,w)nq'
last element of the chain [o,w)ri.
is
if and only if y is the
Therefore, for every xES,
the element P(x,y) of B has the same value as the element P(x,y') of B'. This means that the induction hypothesis implies
that the restrictions to S of the elements of B' are linearly
In this case the elements of B'
independent.
are a fortiori
If each point of 8S is also a point of 8S' ,
independent of mot' .
let 8S'={t,yI'...,yn}, where 8S={yl,...yn}, and t to y1.
is adjacent
If, for xES', n
oPz(x,t) + E ciPz(x,yi) = 0, i=1
then the same holds for xES, where Pz(x,y1)=Pz(x,t). It follows that, for xESt, n
(co+cl)Pz(x,yl)+ E ciPz(x,yi) = 0. i=2
co+c=0,
This implies,
by the induction hypothesis,
c3 ..... =cR 0.
Therefore co z(x,t)+cIPz(x,y1)=0,
and c2
for xEs'.
In
particular, co z(t,t) + c1Pz(t,y
The chain [o,yl] include t.
chain
c0+cIq
2z=0,
.
and therefore does not
But yI and t are nearest neighbors, therefore the
[o,t]
Pz(t,y) =
is contained in 3,
= 0
contains
gzd(o,yl)-zd(t,yl)
y1.
=
Pz(t,t)=gzd(o,t)
Therefore
gz(d(o,t)-2)
It
follows
and
that
which together with c0+c1=O is possible only with
co cl=O, unless q
2z=q2z=1.
The case q2z=1, that is z=ika/ln q,
Eigenfunctions of the Laplace operator
has been excluded by hypothesis.
with kEZ,
39
We conclude that
co c1=0, and the set B is linearly independent. We will prove
now that the space of µ-eigenfunctions on S has dimension n, where n is the number of elements of 8S={yl...... yy}. Again we use induction on the size of S. If S'
is obtained by adding one
then t must be adjacent to some point
point,
say t,
ylEB3.
It follows that either yl remains a boundary point in
S',
or
it
to
3,
becomes an interior point.
In the first case a
µ-eigenfunction on S may be extended to a ii-eigenf unction on 8S',
by
assigning
an
arbitrary
value
at
t,
because
the
condition of being a µ-eigenfunction applies only to interior
points and t
is not adjacent to an interior point.
In the
second case the number of boundary points remains the same and
a g-eigenfunction f on S extends uniquely to the point t.
In
fact the unique extension is f(x)
f(t) = (q+1)N.f(yl) -
(x,Yl)ELF
This shows that the dimension of the space of g-eigenfunct ions
on S is the same as the number of points of 8S.
Thus,
if S is
any subtree containing o, the subset B is a basis for the space
={xEX: d(o,x):5n}. Then of g-eigenf unct ions of L on X. Let 3E n
n is a subtree with boundary 8X ={xEX: d(o,x)=n}. By what we have n
just proved fly can be expressed uniquely as
f(x) = E
my PZ(x,y)
d(y,o)=n for some complex numbers m . y
The
numbers
m y
determined and we may define m(S2(o,y))=m .
are
uniquely
This definition
y
determines a finitely additive measure, because
my = E mt where the sum is taken over all t such that d(o,t)>d(o,y), and {y,t}Eg. By definition, for each x
40
Ch. I I
f(x) =
f PZ(x,w)dm(w) J2 The uniqueness of the numbers {m } implies that the measure m .
v
is unique.'
The
relationship between the
eigenfunction
and
f
finitely additive measure m for a given eigenvalue µ(z) given by the following proposition.
the is
We write for simplicity
P(o,x,w)=P(x,w), and c(o,x)=S2(x). Let f be a µ-eigenfucntion with µ=µ(z)=
(1.3) PROPOSITION.
(qZ+ql-Z)(q+1)-1,
and
let
mr=X'(Q)
be
finitely additive
the
measure such that
f(x) = f PZ(x,w) dm(w. Then m(S2)=f(o).
if x' is the vertex at distance 1
Moreover,
from x in the chain [o,x], then m(S2(x)) = (qZ-q Z)-iq Zd(o>x')(f(x) - gZf(x' )). PROOF. The fact that m(f2)=f(o) follows directly from the fact that P(o,x)=1. Observe that PZ(x,w)=q ZPZ(x',w) if woS2(x), and PZ(x,w)=gzPZ(x',w) if wecZ(x). Therefore
f(x)= I PZ(x,w)dm=
2
qZ
PZ(x,w)dm +
S2J\(x)
PZ(x',w)dm +
=
L"WX)
J(x )
PZ(x,w)dm
c2
W
qZ
PZ(',w)dm. S2(x)
On the other hand PZ(x',w)
is
equal
to
gZd(X',o)
for every
weS2(x). Therefore, PZ(W '
f
JS2\52(x)
=
dm = m(52(W )
PZ(x',w)dm=
)qZd x 'o), while
PZ(',w)dm 52
f(x')-m((2(x))gZd(x',0)
PZ(W ',w)dm
52(x)
Spherical functions
41
Therefore, z(f(x')=(qz-q z)gzdcx',0)) + q zd(x',o)m(O
f(x)-q
mf(x)=(qz-q z)q zd(x',o)(f(x)-q zf(x'
and
x)),
)).,
As before let oe3E be a fixed vertex.
2. Spherical functions.
We shall call a function defined on X radial (with respect to
the vertex o) vertex
o.
if
That
it depends only on the distance from the f
is,
radial
is
if
d(o,x)=d(o,y)
implies
f(x)=f(y). We study in this section the radial eigenfunctions of L.
As in the proof of (1.2), we write again for simplicity of
notation P(o,x,w)=P(x,w). Then by (1.2) a µ-eigenfuntion of L may be written as
f(x) =
f2 Pz(x,w) dm
,
J
where m is a finitely additive measure and µ=µ(z).
We should
therefore identify the finitely additive measures which yield radial eigenfunctions.
(2.1) THEOREM. A µ-eigenfunction of L is radial if and only if it is a constant multiple of the function
0(x) z =
Pz(x,w) dv (w), in is the positive measure on Q defined by
where µ=µ(z) and v=v
I
0
q v(f2)=1, v(S2(o,x))=q d(o,x)q+1 PROOF. We observe first of all that the measure v defined above is
(up to multiplication by a constant)
additive
measure
which
invariant
is
the only finitely
under
K=K .
That
is
0
v(kE)=v(E), if E is a basic open set and kEK. Next we observe that, if m is a finitely additive measure, the measure defined on the basic open sets by mK(E) = [Km ( kE) dk _
42
Ch. II
where dk is the normalized Haar measure on K, is K-invariant. Therefore mK=m(S2)v. Recalling now the definition of P(x,w), and
given in Section
8(o,x,w),
we observe
1,
that
8(o,kx,kw)=
8(o,x,w), provided that ko=o; hence 8(o,kx,w)=8(o,x,k 1w),
if
keK, which implies P(kx,w)=P(x,k 1w). Thus Pz(x, klw) dv =
0z(kx) = to
because
v
On
K-invariant.
is
Pz(x,w) dv =
(x),
trip.
the
other
hand
f
if
is
a
µ-eigenfunction which is K-invariant, then by (1.2) f(x) =
Pz(x,w) dm(w)
,
J
and
JIKPZ(x,
f(x)=
klw)dm(w)dk= fir (x,w)dmK(w)
JKf(kx)dk=
J
= m(S2)4z (x).' (2.2)
DEFINITION.
Let
be
0
a
eigenfunction
radial
of
L
is
a
satisfying 0(o)=1. Then 0 is called a spherical function.
With
this
terminology we
have now
spherical function with eigenvalue µ,
such that
µ(z)=µ.
We
that,
then ¢_IOz ,
now compute
will
if
the
¢
for some z
values
explicitly. To simplify notation we write
lxl=d(o,x).
of ¢z
A radial
function is then a function of lxl alone. Since lxl takes only integral values,
function
on
we may think of a spherical function as a nonnegative
the
Oz(n)=Oz(x), when lxl=n. elements
having distance
integers.
We
therefore
write
Observe that the value of ¢Z on the
one from o
exactly
is
µ(z).
By
definition qz(o)=1. Therefore the following lemma allows us to compute all the values of
(2.3) LEMMA.
If 0
z
by induction.
is a spherical function, then, for n>--1,
z
z
(n+1) =
q+1
q
0 (1)¢ (n) z z
q1
z
(n-1).
Spherical functions
43
If lxl=n, the equality Loz()=µ()oz(x) implies
PROOF.
q+1 0z(n-1) + qq+1 ¢z(n+1) =
Let zEC, and let h (x)=q Zl"l, then
(2.4) PROPOSITION.
z
(i) if
q2z-1*1,
then, for every xe1,
() = c(z) zh() + c(1-z) h()1_z,
z
where
q2z-1=1,
(ii) if
(q1-z-qz-1)(q z-qz-1)-1'
1
c(z) =
q+1
then, for every xEX,
Oz(x) = (1+ q+1lxl) hz().
PROOF. Consider the system (in the unknowns c and c') c+c'=1, cq z+c'qz-1=µ(z). Let
If
q2z-1*1,
and c'(z)
c(z)
then the system is nonsingular.
be the solutions of the system.
Since
µ(z)=µ(1-z), we also have c'(z)=c(l-z). The expression of the solution c(z) is given in the statement. The function O(x) = c(z) hz(x) + c(1-z) h1-z(x) is
and satisfies
radial,
L0(o)=µ(z)=µ(z)0(o).
O(x)=¢z(x),
for every
it
x,
q(x)=µ(z)
0(o)=1,
particular
In
order
suffices
to
for
lxl=1.
show
to
show that
In
that
Lo(x)=
µ(z)0(x) for x*o. Observe that if x*o, and y varies among the nearest neighbors of x, h
and once
the
z
assumes q times the value qzh
value gzh(). z
z
1-z z ++q
Therefore
Lhz (x)=qq
µ(z)hz(x). Since 0 is a linear combination of hz and h1-z
(x),
hz (x) =
,
and
it follows that LO(x)=µ(z)O(x). We conclude that
µ(z)=µ(1-z),
Assume now that
q2z-1=1,
and let k (x)=lxlh W. Suppose z
z
that x*o; then Lkz(x)
_
(q+1)-1(q1 z( lxl+1)+gz(lxl-1)hz(x)
= µ(z)kz+ (1+q+ilxl)hz().
Then ¢
is
a
linear
combination of
h
z
and
k . z
Therefore
44
Ch. I I
L¢(x)=p(z)4,(x) for x*o.
is the value of 0 on the
But L4,(o)
elements of distance 1 from o, and therefore L4,(o)=q Z(1+q+i)= µ(z)=p(z)¢(o).
This
shows
that
L4,=µ(z)4,,
and,
since
¢
is
radial, ¢=¢Z., 3. Intertwining operators. Let ZEC be such that q2i*± 1. Let o
be a fixed vertex and,
write P(o,x,w)=P(x,w) and
as before,
c(o,x)=S2(x). Recall that v is the unique probability measure on S2,
which is
invariant under Ko K.
Let X(0) be the linear
subspace of X(c) generated by {xit(x): d(x,o)=n}. transform
The Poisson
is defined on the space X'(c) of finitely additive z
measures as 7, m(x) =
z
Observe that both
and
z
µ(z)-eigenfunctions
of
Pz(x,w) dm(w).
I
J s2
map the space X' onto the space
1-z
where
µ(z)=(qz+ql-z)(q+1)-1.
This
follows from (1.2) and our hypothesis on zEC.
For
(3.1) DEFINITION.
EX , define I gEX n
z
T (I 9dv)(x)= z z
n
satisfying the conditions above, and
to be such that _
I
Pz(x,w)I 9(w)dv= z
ci
I
Q
= P1-z(edv) W.
To show that the definition makes sense we must prove that ICEX(S2). But if CEXn , then C=E f(y)Pz(y,w), with f supported in 3F ={x: d(o,x):sn}. Therefore n
1 z (x,w)9(w)dv= Ef(y)J J12P
=
p1-z(x,w)Pz(y,w)dv s2
[Pz(X) [f(y)P1_z(y,w)Jdv.
We have used the fact that the measure P1-z(x,w)Pz(y,w)dv has
the same distribution as the corresponding measure where the roles of x and y are exchanged. Indeed the first measure may be
Intertwining operators
written as (dv /dv )Zdv z
which changing the role of x and y
Thus we can write I
becomes (du /dv )Zdv . x y y
In other words I
,
45
(w)=Zf(y)Pi-Z(y,
z
may also be defined as the linear extension
of the map PZ(x,w) identity operator.
P1-Z(x,w).
The operators
Observe that and
I
Z
I
1-z
I
is
I
the
1-z are called the Z
intertwining operators relative to the eigenvalue µ(z).
We shall now find a common set of eigenvectors for the intertwining operators. Let x*o and let x' be the vertex of the
chain [o,x) which has distance 1 from x. Define %=l and, for x*o, define gX (q/q+1)(q
-
The functions
(3.2) LEMMA.
gd(O,x')x,(x'))
d(0,x)x2(x)
.
xEX, span
and, moreover,
9((S2)
Izoo' and for x*o I
(q1-z-qz-1)(gz_q z)-lq (2z-1)d(x',o)gx
g
Z X
In particular, if z=1/2+it,
extends to a unitary operator on
I
z
L2(52,v), and I1/2 is the identity operator.
PROOF.
is
It
easy to see by induction that
the
set
ex
d(o,x):sn, spans ?C n (12). It is also obvious that I z O_%O. Let d(x,o)=n+1, with n>-0. Observe that PZ(x,w)= r)1(w)+cixi2(x) where
711EXn and ci*O. We can write therefore ax r)2(w)+c2PZ(x,w), with 71 E?Cn .
2
Therefore
x
(w)=Ef(y)PZ(y,w), with
may be written as
x
f supported on {y: d(y,o)sn}v{x}. Since Izgx(w)=E'(y)P1-Z(y,w), it follows that Izgx r)3 (w)
then
L(y)9xdv=0.
Therefore
where
713EKn.
Tz('xdv)
and
But
if d(y,o)=n
?1-Z('xdv)
identically zero on 3f n={y: d(o,y)5n}. So, for any yE3f Pz('Q3dv)(y)=
=
fQ(y)713(w)dv=0 for
and therefore r)3 0.
compute
xdv)(y)
P1-z(gxdv)-c3Rz(gxdv)(y)=0.
This implies that
to
are
n,
c3.
every YEln.
We have proved that Izgx c3'x.
Let
But
71 3E?Cn
It remains
0(y)=P1-z(gXdv)(y)=Pz(Iz'xdv)(y).
Using
46
Ch. I I
(3.1) and the fact that 0(y)=0 for d(y,o)sn, we compute
f I
dv=(qz-q z)-iq zd(o> x') (0(x)-q zo (x')
dv=(ql-z-qz-1)-iq(1-z)d(o,x')O(x). J
S2 x
It
follows
the
proof
of
the
-iq(2z-i)d(x',o),
(q1-z-qz-1)(qz-q
c3
that
assertion
first
coefficients multiplying a
x
The
complete.
is
assertion now follows when we observe that
4.
and
z)
are of modulus
last
if z=1/2+it the 1
and are
1
if
Let G be a closed subgroup of
The Gelfand pair (G,K).
Aut(X) and suppose that G acts transitively on X. If oEX is a fixed vertex, then the orbit Go is all of X. Therefore 3f may be
identified through the map g-go with the quotient G/K, where K=GrKo {gEG: go=o}. This means that every function on X may be lifted to a function on G, by defining f(g)=f(go). The function f
has
the
property
that
for
f(gk)=f(g),
every
kEK,
and
conversely a K-right-invariant function on G may be identified with a function on X.
If f is a measurable function on G then
fK(g) = is
a
K-right-invariant
JKdk
function
on
G,
and
map
the
f4fK
preserves continuity, maps compactly supported functions onto compactly
supported
functions,
and
is
a
norm-decreasing
projection of LF(G) onto its subspace consisting of K-rightinvariant
elements,
K-bi-invariant,
if
for
1-sp:5e.
A function on G
f(kgk')=f(g),
for
every
called
is
k,k'EK.
A
K-bi-invariant function may be identified with a function on X which
is
invariant
transitively on
under
the
action
of
K.
If
K
acts
the boundary of X, then K-bi-invariant functions on G are constant on the sets S3 ={x: d(x,o)=n} and n S2,
therefore correspond to radial functions on X.
The Gelfand pair (G, K)
47
We shall assume from now on that the action of K on 0 is Under
transitive.
these
conditions
have
we
the
following
result.
(4.1) LEMMA. The subspace of L1(G) consisting of K-bi-invariant functions
is
commutative
a
subalgebra
of
under
L1(G)
convolution.
PROOF. Let u and v be K-bi-invariant elements of L1(G) and let k,k'EK then u*v(k'gk)=J u(k'gh)v(h 1k 1)dh=u*v(g). G
This shows that the space of K-bi-invariant elements is an algebra.
To prove that this algebra is commutative,
Indeed d(go,o)=d(o,g 1o), therefore there exists
that g-1 EKgK. kEK,
such that
therefore
we show
This means
go=kg lo.
g 1=k'gk-1.
follows
It
that
that,
g if
kg 1=k'EK,
and
u
are
-1 1
and
v
K-bi-invariant,
u*v(g)={ u(gh)v(h l)dh=J u(gh)v(h)dh G G
=1 u(h)v(g lh)dh=J v(g 1h)u(hl)dh=v*u(g 1)=v*u(g).I JG J
G
Observe that, if f is a locally integrable function on G, we may define Kf K
(g) = f
IKf (kgk') dkdk' ,
J
which is a K-bi-invariant function. f->KfK
We observe that the map
defines a norm-decreasing projection on each of
the
spaces LP(G), for 1:5p:5w.
A pair (G,K), where G is a locally compact group and K a compact subgroup, is called a Gelfand pair if the convolution
algebra L1(K\G/K) commutative.
of integrable K-bi-invariant functions
We have thus proved that,
is
under the hypothesis
48
Ch. II
that K acts transitively on 12, (G,K) is a Gelfand pair.
It is
not difficult to prove that the transitive action of K on i2 is
also a necessary condition for (G,K) to be a Gelfand pair. We will now characterize the multiplicative linear functionals on the algebra L1(K\G/K).
Observe that,
if t is a continuous linear functional on
L1 (K\G/K),
then 1D
is the restriction of a continuous linear
functional
on L1(G).
Therefore there exists an essentially
bounded measurable function 0 such that, for every ueLI(K\G/K),
Cu)=
I J
u(g)¢(g)dg.
G
But
t(u)=JGJJKu(k'gk)o(g)dgdk'dk=fGu(g)K0K(g)dg. J
Therefore we may choose the function ¢ to be K-bi-invariant. In
case ¢
this
is
necessarily continuous,
because
it
is
constant on the disjoint open sets KgK.
LEMMA.
(4.2)
Let ¢ be a bounded,
continuous K-bi-invariant
function; then the functional 'D (u)G u(g)0(g)dg
is multiplicative on L1(K\G/K) if and only if, for every g and
g'eG 0(g)¢(g')=J 0(gkg')dk.
(1)
K Suppose that the functional equation (1) holds and let
PROOF.
u,ve L1(K\G/K). Then
(u)4(v)= fGJfGu(g)v(g)0(g)0(h)dhdg J =
JIGJGu(g)v(h)O(gkh)dhdgdk=
fGJfGu(g)v(h)o(gh)dhdg
J
1)dh= Vv*u). I J (We have used the fact that a K-bi-invariant function has the =
I
fG u(g)v(g lh)¢(h)dgdh=
J
same value at g and
g-1.
) Conversely, if 7(v*u)=4(uWv), then
The Gelfand pair (G,K)
I
49
fG v(h)u(gh 1)O(g)dgdh=JGJGv(h)u(g)O(h)¢(g)dgdh,
J GJ which implies, for every kEK, fGJ fGv(h)u(gh'k
1)0(g)dgdh= fGJ Gv(h)u(g)o(gkh)dhdg J
J
= fGJfGv(h)u(g)0(h)O(g)dgdh. J
Integrating over K, one obtains
v(h)u(g)o(h)¢(g)dgdh. JGJGJKO(gkh)v(h)u(g)dkdgdh=JGJG The function (gkh)dk is defined on the Cartesian product of K f0 the double-coset spaces K\G/K X K\G/K, because it is separately
K-bi-invariant as a function of g and as a function of h. Since
v and u are arbitrary K-bi-invariant integrable functions we conclude that
0(gkh)dk=¢(g)f(h). '
I
JK
(4.3)
LEMMA.
Let
0 be
a
bounded
K-bi-invariant
function
satisfying the functional equation 0(gkh)dk=0(g)0(h).
K Let 0 be the radial function on X defined by s(x)=¢(g) with J
go=x. Then 0 is spherical in the sense of (2.2).
PROOF. Since ¢ is K-bi-invariant, 0 is radial. We only have to
show that Lo(x)=po(x) for some µEC, and all xEX. Let µ be the value of 0 on the nearest neighbors of o.
Let go=x, and let
y--ho be a nearest neighbor of x. Observe that gKg 1 stabilizer of x=go,
is the
and that as k runs over gKg1, kho runs
over the nearest neighbors of x. Therefore,
0(kh)dk=f O(gkglh)dk=o(g)o(g 1h)=µO(g),
Lo(x)=J
gKg1 because
JK
d(g1ho,o)=d(ho,go)=d(x,y)=1.
follows that L0(x)=p0(x). ,
Since
0(g)=oi(x),
it
50
Ch. II
With
a
slight
abuse
of
language
we
shall
call
a
K-bi-invariant function satisfying the functional equation (1)
a spherical
function,
by
since,
preceding
the
result,
it
identifies uniquely a spherical function on X. Indeed it is not
difficult to show that a spherical function on X gives rise to
a continuous K-bi-invariant
function on
G,
satisfying
the
condition (1). Unbounded K-bi-invariant functions on G may be associated commutative
multiplicative
to
algebra
of
linear
on
functionals
compactly
K-bi-invariant
the
supported
functions. However, from now on we shall only encounter bounded spherical functions.
5. Spherical representations. In this section we shall identify the
irreducible unitary representations of G which have a
matrix coefficient which is a spherical function.
Recall that a unitary representation n of the group G is a
homomorphism of G into the group of unitary operators on a Hilbert
space
such
3fn
that,
for
all
g,nE3e
the
function
u(g)=(n(g)g,n) is continuous on G.
A unitary representation is said to be irreducible if closed subspace of 3fn which is
there exists no nontrivial,
preserved by the action of all unitary operators n(g),geG. other words it is irreducible if, for every nonzero geX
,
In
the
71
closed span of
{n(g)g: gEG}
is
all
of
3{n.
We
shall
characterize certain irreducible representation of G.
now
Recall
that our hypothesis on G is that its action on 3E and on SZ is transitive.
As before we let K denote the compact subgroup
which fixes a vertex o of X.
DEFINITION.
(5.1)
representation of G;
Let
then
it
it
be
an
irreducible
unitary
is said to be spherical (with
respect to K) if there exists a nonzero K-invariant vector, that
is
if there exists a nonzero vector ene3fn such that
Spherical representations
51
n(k)Cn n for every keK. We first show that the irreducibility of it implies that the space of K-invariant vectors is at most one-dimensional.
(5.2) LEMMA. Let it be an irreducible unitary representation of
n(k)g=e for all keK}. Then dim It K :51.
G and let
Observe that if fE L1(K\G/K) then n(f)CE3{n, because
PROOF.
n(k)n(f)eJG =J f(g)n(k)u(g)gdg=J f(k 1g)n(g)edg=n(f)e. G
This means that n defines a representation of the commutative involution algebra L1(K\G/K) on the space than
one-dimensional,
reducible.
then
3QKR.
If 3en
representation
this
is more must
be
In other words there exists a nontrivial proper
subspace W53fKK which is invariant under n(f) for fEL1(K\G/K). K
Let PK be the projection of 3en onto 3fn
,
then, for CE3fn
PKg=f n(k)gdk. JJK
Let 0*iE3fit
,
and suppose that ip W.
We shall prove that,
for
every fELI(G), n(f)j.LW, which is a contradiction, because it is irreducible.
Indeed
let
eeW.
Since
both
g
and
i
are
K-invariant, (9,n(f)7?)=
I
f(g)(n(k)9,n(g)n(k')71)dg
JG
=
=
JJJf()(,it(kk' ))dgdkdk' JKfK(g)(it(g))dg= (9,n(KfK)7)= 0,
because the orthogonal complement of W in X,KK is invariant under
n(KfK).'
We can now show that if gn is a K-invariant vector of the spherical representation it, and I19,rt11=1, then (n(g) n,en) is a spherical function.
52
Ch. I I Let n be a spherical representation; then there
(5.3) THEOREM.
is one and only one positive-definite spherical function 0
which is a matrix coefficient of it, that is (n(g)grt,9n)=qb(g), where gn is a K-invariant vector of norm 1. Conversely, if ¢ is
function then 0 is a matrix
a positive-definite spherical coefficient
of
spherical
a
representation.
particular
In
different spherical functions are coefficients of inequivalent representations. PROOF.
u
If
a
is
representation
spherical
and
4(g)=
then
(r(g)9n,
¢(kgk')=(u(g)n(k' )gn, n(k 1)gn)=O(g) ¢
Therefore
K-bi-invariant,
is
obviously
and
bounded.
FurthermoreJK II9,RII=1, which implies 0(e)=1. Let g,hEG; then (
n(gkh)gnen)dk=fK(n(k)n(h)gn,n(g
)dk
PKi(h)g a(h)gn
Since 3en is one-dimensional and Thus
fK y(gkh)dk= fK (n(gkh)f;, 9X)dk=a(h) (n(g)9,1, 9R)=a(h)0(g) J one obtains a(h)=(x(h)t;,11,971 )=¢(h). Thus Letting g=1G, J
function
satisfies
0
spherical
the
K-bi-invariance
a
be
0
of
matrix
spherical
implies
(5.2)
by
Thus
1=0(1G)=IInhI2 di(g)=O(g).
can
(4.2 (1)),
other
No
function.
(n(g)ri,rl)=0(g)
condition
that
is
a
coefficient the
K-invariant,
and
with
r1=cf;,t
is
because
function, 11
and
the
IcI=1
and
Conversely let ¢ be a positive-definite spherical
function on G.
Then 0 defines,
by the Gelfand-Naimark-Segal
construction [D1,Prop.2.4.4], a unitary representation on the Hilbert
space
translates
of
3f
generated with
j (g, 71)cidi0(gigjl),
by
inner
linear
product
when g+,0(gig)
combination
of
defined
follows:
and
as
71=Edj0(gjg).
left
For
Spherical representations
=EcI0(gig),
define
lglg).
53
Then
it
is unitary if
the inner product is defined as above. With these definitions, 0
a cyclic
is
vector
of
the
representation and
¢--g
is
n
K-invariant. It remains to prove that the representation it is in other words that every nonzero vector
irreducible,
cyclic
vector.
then
g=Ecio(gig),
If
is a
PKg $ ciK¢(gIkg)dk= J
(Ecl0(gi))n. Since linear combinations of left translates of 0
are dense, we conclude that, for every PKC is a constant multiple of fin. On the other hand, if f;*O and PKn(g)F;=O for then n(g)etgn for every
every gEG,
because gn is cyclic.
g,
which implies
f;=O,
Therefore, for some gEG, PKn(g)g*O.
In
order to show that f; is cyclic it is enough to prove that n(g)g
is cyclic for some g.
We may assume therefore that PKg*O.
We
conclude that PKr=) n(k)edk=c(e)
K
n
is a nonzero multiple of n and since PK is a limit of convex combinations of vectors u(k)e, with kEKSG, we conclude that belongs to the closed linear space generated by n(G)e. Since
n n
is cyclic so is a and we have proved that it is irreducible.'
It remains to characterize the spherical functions which are positive-definite.
Observe that if ¢ is spherical then 0(g_1 )=O(g). Therefore
if 0 is positive-definite it is also real-valued. Let p be the
on the elements such that d(go,o)=1.
value of
Then if we
as a function defined on the vertices of the tree 3E,
think of
we have that L0(x)=p0(x) (2.3). Furthermore lµlsl. This means that positive-definite spherical functions are µ-eigenfunctions of L, for real p, satisfying -Isgs1. Observe that Aut(3f)+nG is a
subgroup
invariant
of
index
character of
2.
G.
Therefore
which
is
(-1)d(go,o)
is
a
K-bi-
of course a positive-
definite spherical function. Likewise the function identically 1 is a positive-definite spherical function.
54
Ch. I I We shall prove that a spherical function associated with a
real eigenvalue -1
Observe that
µ=µ(z)=(gz+g1-z)(q+1)-1
satisfies -1
if
and only if z=1/2+it, with tdR, or Im(z)=O and OsRe(z)sl. Let K(S2)
be the space of cylindrical functions.
For gEK(S2),
and
zEC, define n (g)g(w)=Pz(g,w)g(g_1 W).
Observe that the identity P(gh,w) = P(g,w)P(h,g 1w),
(1)
which follows directly from the definition of P(g,w) implies that u
Radon-Nikodym derivative,
z
z
(g)(a
as a
z
In addition, if z=1/2+it, o12dv=J
P(gw)(g w)12d.
Therefore, for Re(z)=1/2, n is a unitary representation of G t z in the space L2(SZ,v). Let 1 be the function identically equal
to 1 on S3. Then P
Oz(g) cz in
Therefore
is
a
coefficient
diagonal
a
of
unitary
z
representation and so it is a positive-definite function. This means
that
eigenvalues definite.
the
functions
spherical
µ=µ(1/2+it)=(q+1)-1q1/229Ze(gIt)
Observe that
as
t
assumes
[-1,11.
values
the
all
are
all
to
the
positive-
ranges over the reals Re(gIt)
assumes all values in the interval µ(1/2+it)
associated
This means that
in
the
interval
1-2g1/2(q+1)-1,2g1/2(q+l)-1l.
Next we consider the spherical functions associated to the real
eigenvalues
-q1/2(q+1)-1.
µ
Observe
satisfying that
in
q1/2 (q+1)-1<µ<1,
this
case
or
-1
µ=±(qs+q1-s)(q+1)-1,
with O<s<1, and s#1/2. We may assume that µ>O, because, if Lo L4s,s
=µ(s)o
s'=s+in/ln q,
-µ(s)s,=µ(s' )s,
and
(g)=(-l) d(9o,o)0
,
then
Spherical representations
55
Recall that the identity (1), the definition of P(g,w) as a
Radon-Nikodym 1
P- (g_
1
the
and
fact
P(g,gw)=
that
imply (g lh)=
=
derivative
P$ (g-1h, w)du=
I
P$
I
(g-1, W) pB
f Ps(g l,g lw)P$(h,w)P(g,w)dv=
Jn
(h, gw) dv
l-$(g,w)P$(h,w)dv. LP
In other words, in terms of the intertwining operators I ¢s (g 1h)=J ISPs(g,w)Ps(h,w)dv.
n
Therefore, if e_E ciPs(gi,w), then cicj¢s(gilgi) =
(I.C)Zdv. J
It follows that in order to show that 4s is positive-definite it suffices to show that, for every CE3((n),
1( Ise) W Jn
By (3.2) Is has a set of eigenvectors x which spans X(n).
If
O<s<1, the corresponding eigenvalues are the positive numbers (ql-s-qs-i)(qs-q s)-lq (2s-1)d(x',O)
Therefore (I g,g)-0, which proves that the spherical functions s
are positive-definite.
As we saw before the spherical functions Os define
irreducible
unitary
K-invariant vector.
representations
,
of
with O<s<1,
G
with
a
We shall prove that the representation
associated to Os is equivalent to a representation defined on a
which is the completion, under an appropriate
Hilbert space 3f s
norm, of the space of cylindrical functions 3((n).
We observe first that the sesquilinear form defines an inner product on K(Q). We let 3f
n)s
be the completion s
of 3((n) with respect to this inner product. The space 3((n) is in one to one correspondence through the Poisson transform
1-s
56
Ch. II
with the space of linear combinations of left translations of s
and the relationship = (9,9)$
E cicj0 (giigj) =
J(2(Is9)()9()dv
shows that the norm of X
for 9=E c1Ps(gi,w),
corresponds on
K(92) to the norm defined by 9s on the linear span of its left translates. Left translation on the latter space (which defines
the unitary representation associated to 0 ) s
corresponds on
J((SZ) to the representation n (g)g(w)=Ps(g,w)g(g iw) s As a last observation we notice that if gE?(((2), and Ak
the orthogonal
projection of g on the space generated by
{X: d(x,o)=k},
then
orthogonal
are
and
is
g=EkAk
.
The
eigenfunctions
Ak are
functions of
with
I
all
respect
to
s
the eigenvalue c(k,s) =
(qi-s_qs-1)(gs_q s)-iq (k-1)(2s-1)
for k2:1, and c(O,s)=l. Therefore 11g1l
(Isg,g) = Ekc(k,s) IIAkgII22 L ((2)
n
We should observe that
E Akg is nothing but the orthogonal k=0
projection of g into the subspace spanned by the characteristic functions of the sets (2(x) with Ixl=n.
(5.4)
DEFINITION.
{R1i2+it: tER}
is
The
series
called
the
of
unitary
unitary
representations
principal
series
of
representations of G. The series of representations
{n: s s*1/2 Osssl} u {n
s+irt/ln: a
s#1/2 Osssl}
is called the unitary complementary series.
6.
The resolvent of the Laplace operator and the spherical
Plancherel formula.
In this section we compute the resolvent
of the Laplace operator L, acting on the Hilbert space P2(X).
The resolvent of the Laplace operator
57
Define, for fE22(X) and n>O,
Lnf(W ) = (q+l)-1q n+1 Observe that L=L1.
f(Y) d x,y)=n
For completeness, we may define Lo I,
the
identity operator.
We shall prove that if Rez>1/2, and µ(z)=(qz+q1-z)(q+l)-1, then the series W HZ L0+(q+1)q 1L
qn(1-z)
Ln
n=1
converges absolutely in the operator norm, and that (L-µ(z))-1 = (q+1)(q z-qz)-1H
.
z
Let o be a fixed vertex and define 6 ={x: d(o,x)=m}. m
We
the characteristic function of the set 0
denote by x
m
m
Let f be a function supported on 6k; then for any
(6.1) LEMMA.
positive integers m and n, IIXm(Lnf)112<
(q1/(q+1))1"2 11f112.
Moreover, if x (L f)*0, then n+k-m is even and In-kl:sm:5n+k. m
PROOF.
n
If, for some xEIf3m
,
we have L f(x)*0, then there exists n
tESk such that d(t,x)=n. Let y be such that [o,tln[o,x]=[o,y]. Then n+k-d(o,x)=d(t,x)+d(t,o)-d(o,x)=2d(t,y). Moreover m=d(o,x)
and the triangle inequality implies Id(t,x)-d(t,o)l:5 d(o,x) s d(t,x)+d(t,o). This proves the second part of the lemma. For the first part we first assume that m=n+k.
Let d(o,x)=m. The
assumption implies that there is only one element yxEok such
that d(y x,x)=n. Hence (L nf)(x)=q-n+1 (q+1)-1f(yx ). On the other hand, for each yeSk there are qn elements x such that d(o,x)=m, and yE[o,x]. Therefore,
llxm(Lnf)112
ILnf(x)12
=
=
(qn+1/(q+1) )2qn
(q/(q+l) )2Q n11f 112`
Suppose now that d(y,o)xk-j,
If(y) l2 d o,y)=k
d o,x)=m
(qn+1/(q+l)) llf 112
m=n+k-2j,
with j>O.
We
define F(y)=O
if
58
Ch. I I
If (t) 121/2, if d(y,o)=k-j.
F(y)=41 v>t)=1
Then IIFII2 IIfII2,
and F is supported in itik-j. Therefore by the
first part of the proof
llxm(Ln_jF) II2`(q/(q+1) )1"q `v4
IIFII2.
Let
d(o,x)=m and let yx be the unique vertex satisfying yxe[o,x] yxe[x,tI for some tesk. Then d(yx,o)=k-j and and L f(x)=q n
n+1
/(q+1)
f(t).
'
the
by
Thus,
Cauchy-Schwarz
d Yx ,t)=j
inequality,
IL f(x)Isq
n+i(q+1)-1gji2
I
n
= q n+1
f (t)
12)12
(q+1)-igji2F(yx).
Since there is only one element yx in the support of F and at distance n-j from x,
Ln-jF(x)=
Lnf(x) I:q J 2ILn-jF(x)
< q n+l(q+1)
1122
I.
(q/(q+1))q n+JF(yx).
that
conclude
We
Therefore
llxm(Lnf)(x) ll2
The proof is completed by observing that
IIFII=IifII.1
Let fe2(x); then
(6.2) LEMMA.
IILnf II2` qi-n(q+1) (n+1) IIf II2. PROOF.
Let fk xkf, then f=E fk and
IIxm(Lnf) II2` E IIxm(Lnfk) II2.
Ilfk1I2 = EIIfkI12.
We have
xm(Lnfk)*0 only if im-nlsksm+n,
But
k
in other words only if k=min(n,m)-2j,
with j=O,...,m+n.
follows that mi n(n, m)
Ilxm(Lnf) II2: J
L
Ilxm(Lnfm+n-2j) 112
min(n,m) q(1-n)i2(q+1)-1(n+1)1/20 I`l
j =o
(I21/2 IIfm+n-2 j
2J)
n
Finally,
IILnf II2=
Emlixm(Lnf) 112:
qn-1(q+1)-1(n+1)2llfII2
qi-n(q+1)-1(n+1)E j=o
1
o0
E k=n-j
if II2
It
The resolvent of the Laplace operator
59
The previous lemmas imply that the operator
qn(1-z)Ln HZ Lo+(q+1)q 12; 00 n=1
is well defined by the absolute convergence of the series, and zd(x,y)
This implies that,
Note that H S (y)=q
is bounded.
it
z x
for y*x,
(q+1)-1(q z(d(x,y)-1)+qq z(d(x,y)+1))
L(H (3 )(y)= Zx
= µ(z) (H a)(y). Zx a In other words the operator (L-p(z))H applied to Sx x z multiple of 8 indeed ((L-p(z))H 6 ,8 )= q Z-µ(z) _ and z x x (q
x
Z-qz)/(q+l).
It follows that Z-qZ)-1
(L-µ(z))-1 = (q+l)(q
HZ.
We summarize this result in a proposition.
If Rez>1/2,
PROPOSITION.
(6.3)
and µ(z)=(qZ+ql-Z)/(q+1),
the
operator L-µ(z) is invertible in e2(A) and (L-µ)-1= (q+1)(q Z-qZ)-1HZ.
spectrum
The
of
in
L
E2(3f)
the
is
interval
[ -q/ q+1). q/ q+1 ). PROOF.
The first part of the proposition is proved in the
foregoing remarks. Since every p which does not belong to the interval above satisfies µ=µ(z), that
spectrum
the
[- q(, q(q+1 1. interval
which
of
L
is
with Rez>1/2, contained
we also have the
in
interval
Suppose now that µ is a number in this
does
not
belong
to
the
spectrum.
Then
µ=µ(1/2+it) for some real t. For e>0,
(L-µ(1/2+c+it))-1(Vy) (q+1)(q1/2+C+it-q 1/2-C-it)-1 (1/2+C+it)d(x,y)
_ As
c-O
q
the
q(1/2+e+it)d(x,y)
the
norm
in
constant
which
multiplies
is bounded away from zero. I
of
the
same
function
But,
the
function
for fixed x,
becomes
unbounded,
contradicting the fact that the resolvent is continuous in its domain of definition. ,
60
Ch. I I is easy to see, using the expression for
It
(µ(z)-L)-1(6 0 ) (x)=(q+1)
rµ(x)=(µ-L)-1(So)(x).
Let
Plancherel
formula of
formula [DS,p.920]:
H
the
prove
We
that
spherical
the
the Carlemann
recall
We
tree.
now
,
z
(qz-q z)-1q d(x,o)z
if T is a bounded self-adjoint operator on
a Hilbert space 3f and [a,b] is the spectrum of T then for every ,l1ER the following equality holds: S lim+(1/2ni)
+((Rµ+ia-Rµ-ie)g,n)dp
where Rµ (µ-T) -1.
In particular if T=L, R= 2(fl, =So and ii=f is a function on Z with finite support, then we obtain -S
q(
f(o)=- lim+ lim(1/2ni)f S-)0
_ E f(x)
J
c--)O+
Cr
q (+S q(
n
EGO+
-S
(1/2ni)J
IS-)O+
+ie-rµ ie'f)dp
(r +iE-r
)(x)dµ
q(q+1)+S
We observe that if c-3O+ and µ=µ(1/2+it) then µ+ic=µ(z with Rez >1/2 and z -1/2+it for c-0+. Therefore E
C
rµ+iE(x)=(q+l)(q1/2+it-q1/2-it)-1q d(x,o)(1/2+it)
u rn
On the other hand µ-is=µ(wE) with Rewe>1/2 and Therefore
l
+r iE
(q+1)(qi/2-it-qi/2+it)-lgd(x,o)(1/2-it)
For µ=µ(1/2+it) O
q (-S (1/2ni)f
f(o)=E f(x)
5im
+ic-r
- q(+(3
ic)(x)dµ
The resolvent of the Laplace operator
/Inq-S
f(x)Ilim x
116-)0
61
(1/2ni)J((q+l)(q1/2+1t_g1/2-it)-1 gd(x,o)(1/2+it)
a
-(q+1)(q1/2-it_q 1/2+it)-1g d(x,o)(1/2-it) )u (1/2+it)dt
/1nq (A(t)gd(x,o)(1/2+it)+B(t)
_E f(x) (lnq/2n) f J
x
where
-d(x,o)(1/2-it) )dt
0
A(t)=
(q1/2+it_g1/2-it)/(q1/2+it_g1/2-it)
B(t)=
(g1/2-it_g1/2+it)/(g1/2-it_ -1/2+it)
and
We have that
A(t)q d(x,o)(1/2+it)+ B(t)q d(x,o)(1/2-it) _ (q/(q+1)) [c(1/2_it)1g d(x,o)(1/2+it)+ c
(1/2+it)-1gd(x,o) (1/2-1t)
(q/(q+1))Ic(1/2+it)I-2o 1/2+it
W.
We recall that c(1/2+it)=c(1/2-it) (see Proposition (2.4)).
It
follows that
/1nq f(o)= E f(x)(glnq)(2n(q+1))-1 x
0
/lnq (0
= (glnq)(2n(q+i))-1
01/2+it(x)Ic(1/2+it)I-2dt
1/2+it'
f)Ic(1/2+it)I-2dt.
0
Therefore we have proved the following theorem.
(6.4) THEOREM. Let J be the interval [O,n/lnq] and dm(t) be the measure on J defined as follows: dm(t)=(glnq)(2n(q+1))-1Ic(1/2+it)I-2dt.
Then for every f on 3f with finite support
f(o)=1 (f,0 J
J
1/2+1t
)dm(t).
62
Ch. I I Let r be a faithful transitive subgroup of Aut(3f); then r
Theorem (6.4) implies that, for f=g *g, 2
(g**g, 1/2+it )dm(t)=
IIgI12= 2
J
Ji
dm(t)
11i1/2+it (g)1 f L2(SZ)
for every function g on r with finite support.
Also, Theorem
(6.4) implies that
01/2+it(x)dm(t)=SX(o).
J
J
We deduce from Theorem
(6.4)
the spherical Plancherel
formula in the sense of Gelfand pairs [F,Th.IV,2] for any closed subgroup of Aut(l) acting transitively on I and Sl (see Section 4).
(6.5) COROLLARY. Let G be a closed subgroup of Aut(X) acting
transitively on 3f and 52. Fix the Haar measure on G so that K has measure equal to 1. Let J and dm(t) be as in Theorem (6.4).
Then for every K-left-invariant function f on G with compact support we have 2
IIf112 2
dm(t)=
11'1/2+it(f)1112
J
1.
PROOF. By [F,Th.IV,2] it
IIn1/2+it(f)II2Sdm(t).
(S2)
is enough to prove the equality for
every K-bi-invariant function f with compact support. Theorem (6.4)
it
1/2+it
implies that ]l,2+i()dm(t)=xK(). We have
MN 1/2+it (k) (f) =n
it
1/2+1t
1/2+it
for
(f)P 0
every where
kEK.
P
is
This the
it
1/2+it
implies
(f)=
that
one-dimensional
0
orthogonal projection onto the space of constant functions on
fl.
Therefore 11W 1/2+1t()II HS=II r1/2+it(f)111
One obtains
The restriction problem
63
2
f**f(g)n1/2+it(g)1 dgdw
I
G
f**f(g)01/2+it(g) dg.
= I
* Since f *f is K-bi-invariant, it follows that (f) 111 2 IIn
JJ
L2(S?)
J
I
1/2+it
(g) dm(t)dg
JGf
f**f(g)XK(g)dg= f**f(1G)= IIf*II2°
=
IIf II2
1
G
7. The restriction problem. that,
*f(g)fJ
dm(t)=
1/2+it
if
G
a
is
transitively on
3f
closed
In the previous sections we saw subgroup
of
Aut(X)
which
acts
and its boundary, then (G,K) is a Gelfand
pair, where K={geG: go=o}. This implies the irreducibility of spherical representations. Recall that, if Rez=1/2 or Smz=O and
O
n (g)(w)=PZ(g,w)9(g iw),
where
g
is
an
element
of
a
suitable Hilbert space completion of X(0). For the definition of n
there is no need to assume that G acts transitively on
the
boundary.
z
We
may
for
instance
define
spherical
representation for a finitely generated free group acting on
its Cayley graph. This was done in [F-T P2], where it is also proved that spherical representations of such a free group are irreducible. A much stronger result is in fact true.
(7.1) THEOREM.
Let r be a discrete subgroup of Aut(X) with the
property that Aut(X)/F is compact. =(qZ+qi Z)(q+1)-1.
Let
it
z
representation. Then
be
the
Let -1
with µ(z)
corresponding
spherical
64
Ch. I I
(1) if µ(z)*O then n restricted to r is irreducible. z
(2) if
µ(z)=0
and
r Aut(X)+
then
restricted
nz
r
to
is
irreducible.
(3) if µ(z)=0 and FcAut(l)+ then n restricted to r is the sum z
of exactly two irreducible subrepresentations.
The complete proof of this theorem is contained in [St2].
In these notes we shall give a proof of this result under a
number of simplifying assumptions which allow us to avoid certain technical details. This simpified version will contain
however the main ideas of the proof. A discussion of how the
proof may be completed to yield the full strength of the theorem is contained,
in the section of notes and
in part,
remarks to this chapter.
Our purpose here
is to prove the
following version of the theorem.
THEOREM.
(7.2)
Let
r be a discrete subgroup of Aut(X).
Suppose that (gEr: gx=x}={e}, for every x. Suppose further that
Aut(x)/r is compact, or, what is the same, has finitely many orbits.
Let
g21t*t1,
that r acting on X
and let n=n
Then
.
1/2+1 t
the restriction of n to r is irreducible.
Comparison between the hypotheses of (7.1) and (7.2) shows that (7.2) consists of part (1) of (7.1) for representations of
the principal series, under the assumption that no element of r except the identity leaves a vertex fixed.
W, with eEL2(f,v),
We write
where
v is the isotropic measure defined with reference to a fixed vertex
oEX.
this
In
identically 1 on 12,
context
the
function
1,
which
is
is the K-invariant vector.
Let itr be the restriction to r of it. We must prove that every vector eeL2(i2) is a cyclic vector for fir. The proof is carried out in three steps.
The restriction problem
65
Construct bounded operators Pe in the C -algebra of nr
STEP A.
with
IIPEIr=0(1).
Show
STEP B.
converges
to
for
that,
each
(P0n(g1)1,ir(g2)1),
(P£n(gl)1,n(g2)1)
g1,g2Er,
where P
is
the
0
orthogonal
projection onto the space of constant multiples of 1. STEP C.
Show that 1 is a cyclic vector for nr.
We prove now that these three steps are sufficient for the
proof of the theorem. vector
there
exists
if C is a nonzero
Indeed, by STEP C,
such that
a'Er,
Therefore
(n(')e,1)*0.
P0n(-y)g*0. But, by STEP A and STEP B, PO may be approximated in
the weak operator topology by elements in the span of n(r). This means that P0n(')9, which is a nonzero multiple of 1,
is
in the weakly closed linear span of n(F)g. But STEP C says that 1
is a cyclic vector and therefore every vector
is
in the
closed linear span of
We conclude this section with a remark on the spherical representation of the principal series which Given two unitary representations it
is
used later.
and n2 of a locally
compact group G, we say that the first is weakly contained in
the other if, for every fELI(G), then n1
is weakly contained in n2 whenever the positivemay be approximated uniformly on
definite function compact
We know from
if nl is a cyclic representation with cyclic vector
[D1] that, g,
llnl(f) II<_IIn2(f) II.
sets
by
coefficients
positive-definite
of
the
representations n2.
(7.3)
series
Every representation
LEMMA.
of
Aut(f)
weakly
is
ir
1/2+it
contained
of in
the principal the
regular
representation. PROOF.
By the foregoing remarks it suffices to show that the
spherical function
1/2+it
may be approximated on compact sets
by positive-definite coefficients of the regular representa-
tion. By (2.4)
2+E
1/2+1t
EL,
for every a>O. On the other hand,
66 if
Ch. I I
1/2<s<1,
¢ E
lim 4s(g)=1,
L1/(1-s)(Aut(.T)).
on
on
compact
Therefore
s
uniformly
uniformly
compact
sets,
lim
s41
and
¢ ¢ s
1/2+it
condition implies [D1,Th.13.8.6] that
and
sets,
s 1/2+1t
1/2+1t
The
latter
EL2.
is a positive-
1/2+it
definite coefficient of the regular representation. I
8. Construction and boundedness of P£ (STEP A). Throughout this section, for notational convenience, let G=Aut(X). Let c>O, and z=z(E)=1/2+c+it. Let
IG/rl be the volume of G/r,
the number of orbits of r
the same,
qzd(go,o)
in
what is
or,
Define h (g)=
1.
z
and
PE IG/rI nr(9Ze(D(£)hz))=IG/rI V 9te[D(e)hz(y)] a(z) a'E
where Ghz(g)P1/2+it(g,w)
D(e)-1=t
Observe that,
since h
z
dgdv = (n(hz)1,1).
is K-bi-invariant,
)XE
the integral above
may be written as a sum: E'
hz(x) IQ P1/2+1t(x,w) dv = El
XEX
hz ()0(W
But
O(x)=c(1/2+it)q
(1/2+1t)d(0,x)+c(1/2-it)q (1/2-1t)a(o,x)
Therefore, 1=(q+1)q 1 [c(1/2+it)E0q-n(E+2it)
D(s)
l
+c(1/2-it) E q -En
n=0
n=0
l
J J
(q+1)q 1rc(1/2+it) (1-q (E+21t))-1+ c(1/2-it)(1-q E)-11.
This implies that a/D(c) is bounded, as well as bounded away
from zero as c-O.
Notice that
co.
belongs to the C -algebra of nr ,
Our next IIPEII=O(1).
task
Therefore nr(hz
yE
*
is
to show,
and so does PE.
according to STEP A,
that
Because of what we know of D(E) it suffices to show
Construction and boundedness of P
that
I1nr(hz)II=0(E).
Let pr be the right regular representation
We observe that nr
of r.
from
follows
is weakly contained in pr.
Therefore
(7.3).
I1pr(hZ)II=0(E).
67
C
suffices
it
This
show
to
that
The function hz(') acting by right convolution
on E2(r) may be thought of as an operator Hr defined on the z
subspace
onto e2(ro).
matrix
Let
coefficients
restriction
to
matrix
coefficients
Let Q be the orthogonal projection of
(HZSTo,6 ,o)=hz(, e2(3E)
having
and
e2(ro)se2(E)
t2(ro)
H
be
the operator on
(HzSx , Sy )=q of
za(x,y)
Then
Therefore,
QH Q-1.
22(I) having
is
the
order
to
Hr z
in
estimate the norm of Hr it suffices to estimate the norm of H . z z Recall that, according to (6.3), (L-µ(z))-1= (q+l)(qz-qz)-1H
z
Our hypotheses imply that the coefficient multiplying H
z
in the
equation above is bounded away from zero as c-O. Therefore in
order to estimate the norm of nr(hz) it suffices to estimate the norm of the resovent (L-µ(z))-1 of the Laplace operator on the tree 3f,
where µ(z)=(q+1)(qz+ql-z), z=1/2+c+it, and c is a
small positive number.
We shall use now a classical estimate
for the resolvent of a self-adjoint operator.
(8.1) LEMMA.
Let L be a self-adjoint operator on a Hilbert
space, and µ a complex number. Let d(µ) be the distance of µ from the convex hull of L. Then
II (L-µ)-1II
(8.2) COROLLARY. For z=1/2+£+it, and under the hypothesis of (7.2),
PROOF.
remarks
I1nr(h)II=O(1/c), and IIPEII=0(1), as c-O.
The first estimate follows from the lemma above and the preceding
Sm(µ(z))=E.
it,
as
soon
as
we
observe
that
d(µ)=
The estimate for PE follows from the fact that Pe=IG/rlnr(Pe(D(e)hz),
and D(c)=e. I
68
Ch. I I
9. Approximating the projection P0 (STEP B).
Having completed
STEP A of the proof in the previous section, our next task is to show that, if g1,g2Er, Eio (P£n(g1)1,n(g2))= (Pon(g1)1,x(g2)). Recall that PE=71(Re(Dhz )). Therefore
(PEn(g1n(g2)1)= IG/rl
ReDhz(T)((zg1)1,n(g2)1) aE
G/r(
ReDhz(a') ¢(g21ag1)= JG/rJ a'E
ReDhz(g2T)O(Tg1). 3"E
We also observe that P
0
may be written as
Po n(ReDhz)=J5eDhz(g) n(g) dg. G
Indeed n(ReDh) is a self-adjoint bounded operator which maps z
every vector into a constant multiple of 1 (this is because h z is
K-bi-invariant).
But
D=D(E)
Therefore
n(ReDh )
was
chosen
so
that
must be the orthogonal z projection onto the space of multiples of 1. We write then (n(ReDh )1,1)=l. z
(P0N(g1)1,n(g2)1) _ (n(9teDht )ir(g1)i,n(g2)M) =
JRe(Dh)(g)(g1gg1)
We observe that
l
dg=
D(e)=0.
It
GRe(Dhz)(g2g)o(gg1) dg.
suffices therefore to prove
that G/rI
p"ehz(g2-Y)O(argl)-]
To prove (1)
9Zehz(g2g)O(gg1)dg=0(1).
(1)
it suffices to prove the assertion when Reh is z
replaced by h
.
We are thus
led to consider the function
z
h(g1g)0(gg2) as co 0. Since all the functions appearing in (1) are uniformly bounded for c-> 0, it suffices to prove (1) when g
and X are restricted outside a given compact subset of G. Furthermore the expression of ¢ as a linear combination of
h
1/2+it
and h 1/2-it makes (1) true as soon as we establish that
the quantities
Approximating the projection P
69
0
IG/rlaeIhz (g
1
-d) h
('g 2 )-IJ Ghz (g1 g)h1/2-i t (gg2)dg
(2)
1/2+it (2'g 2 )-JG hz(g1 g)h 1/2+i t (gg2)dg
(3)
1/2-i t
and
IG/ri aE
hz (g1 2-' )h
are uniformly bounded.
find
shall
We
a convenient
first
expression for
the
functions hz (g g)h1/2+1 t (gg2 ) and hz (g g)h1/2-it (gg2 ) outside a 1
1
compact subset of G. We will find finitely many cones C(o,xj), j=1,...,r, and finitely many vertices y1,...,ys, such that,
vi(x)=h1+£(x)xC(o,xthen hz(g1g)h1/2-it(gg2)
after
is,
if
cor-
rection on a compact set, a linear combination of the functions
A perfectly analogous statement
vi(gyj).
hz(g1g)h1/2+1t(gg2)
the functions
if
functions u1=h1+£+21txC We may write
vi
will
true for
be
are replaced by the
(o, xi ) '
hz(g1g)h1/2-it(gg2) (1+£)d(gO,O)q z(d(gIgO,O)-d(gO,O))
q (1/2-it)(d(gg2 ,O)-d(gO,O))
q
The function d(gIgo,o)-d(go,o) as
soon
as
d(x,o)ad(g1o,o).
if
d(g1o,o)=d(g-l o' o):5
then the chain connecting g11o to go
and goEC(o,x),
d(x,o),
is constant for
Indeed,
goes through x. Therefore d(glgo,o)-d(go,o) = d(go,gl1o)-d(go,o) = d(go,x)+ d(x,g11o) - d(go,x)-d(o,x) = d(x,g_1o)- d(o,x),
which
is
d(x,o)>d(g2 o, o).
shows
reasoning
that
soon as
Let N be the greater of the two
integers
distance N from o; the
hz(g1g)h1/2+1t(gg2)
Similar
constant when g-1 OET(O,X),
is
and d(g2,o)
Then outside
g.
as
d(gg2 o,o)-d(go,o)
d(g1o,o)
of
independent
and
x1,...,x
be
the
points
at
Eij={g: goEC(o,xi)}n{g: g 1oEC(o,xj)}.
let
compact
and
let
set
{g: d(go,o):5N}
hz(g1g)h1/2-1t(gg2)
are
the
functions
respectively
70
Ch. I I combinations
linear
of
the
functions
and
h1+E(g)xE(g) ij
h
1+E+21t
(g)x (g). We shall consider one of the functions above, E
tj
relative to two vertices x and x then
such
x'E[o,x
that
2
at distance N from o. We fix
d(xxi)=l,
for
i=1,2.
Let
A=
{g: gx2El(o,x1)} and B={g: gx2EUo,xl}. Then (g)
h1+E(g)xE 12
2(1+e))-1h
=(l+q
1+E
(x
2
) (h
1+E
(gx )x 2
A similar formula holds for h
- q (1+c)h 1+E
A
1+E+2it
xE
,
(gx2'))x
B
).
if 1+c on the right
12
hand side is replaced by 1+e+2it. But xA(g)=xl(o,x )(gx2) and xe(g)=xLr(o'x1)(gx2). Therefore h1+cxE12 as a function of g may
be written as a linear combination of functions of the type v1(gyj)
where
v1(x)=h1+E(x)xg(0,x(x).
We
conclude
that
hz(g1g)h1/2-1t(gg2), after correction on a compact set, may be expressed as a linear combination of the functions v1(gyi). The corresponding claim about hz(g1g)h1i2+1t(gg2) is also proved.
In conclusion, in order to prove that the expressions (2) and (3) are uniformly bounded as c-0, it suffices to show that,
if x1 and x2 are arbitrary vertices v(x)=h1+E(xNr(o,x1)(x) and u(x)=h1+E+21t(x)XEy(0,x)(x),
IG/rI
v(rx 2J)-fv(gxz ) dg=0(1),
(4)
u(2'x)-J u(gx) dg=0(1). z 2) G
(5)
zE IG/rI
Observe that
v(gx) dg = E v(x)m({g:gx =x}) z 2 xEx where m is the Haar measure of G. Notice that, f
JG
g2 =x2,
{g: gx2 x}={g:
x })=l.
It
g11gg2 =o}=g1Kg21.
follows that
Therefore m({g: gx2
JGv(gx2)dg = Iv(x). xE
if g1o=x and
We must prove
Approximating the projection P
71 0
therefore that v(x) = 0(1)
v(7x2) - El
lG/rl
XE
7C
and u(a'x2) - E u(x) = 0(1)
iG/rl
7C
XEX
Let v and u be the functions defined above.
(9.1) LEMMA.
µ=µ(1+2it)=(q+l)-1(q1+21t+q Zit)
Let
Let L be the Laplace operator
on X. Then lv(x) - Lv(x)l = 0(1),
(6)
lµu(x) - Lu(x)l = 0(1).
(7)
XE XEX
PROOF. =0.
and d(x,xI)>1, then v(x)-Lv(x)=u(x)-Lu(x)
If
Since there are only a finite number of vertices
at
distance not greater than 1, it suffices to show that
lv(x)-Lv(x)l= 0(1), and Y
Y
But,
lpu(x)-Lu(x)l = 0(1).
d x,x )>1
d x,x )>1 xEl(Olxi)
xEff(Olxi)
if d(x,x1)>1 and
v(x)=g
Lv(x)=Lh1+e (x)=(q+1)-1(qi+E+q
£)hi+e(x).
(1+s)d(o,x)=h1+E
(x) and
Therefore
v(x)-Lv(x)=
hi+£(x)(1-(q+1)-1(qi+e+q
E) and lv(x)-Lv(x)l= d(x,xi)>1 xET(O,x ao
l1-(q+1)-l
gnq(1+E)n
(qi+e+q £)Ih1+C(xi)
Z n=2
(q+1)-1(1-q £)-1g2e(q(1-q£)+(l-q E)) hi+E(xi). As c-30,
the expression above converges to (q-1)(q+1)-l hI(xi).
This proves the estimate (6).
uECo,xI)
and
d(x,xI)>1,
As for (7),
observe that, for
Lu(x)=(q+1)-1 (q1+e+21t+gE-21t)u(x)
=A(1+2it+c)u(x). Therefore, lµu(x)-Lu(x)J=lµ(1+2it+E) - µ(1+2it)llu(x)l (q+1)-1
=
(q1+21t(q£-1)
+ q 21t(q -1))lu(x)l s Me.
72
Ch. II
lu(x)I=h1+E(x1)q ZE/(l-q £),
But
and
the
estimate
(7)
d x ,x)>1
follows. 1
Given a summable function on X it is possible to define a
function F on the set of r-orbits on X,
as F(rx)= 'f(Tx). aE
Since the function v is summable on 3f we may define in this fashion on the set of orbits r\3E the function V(rx) = )v('x), a'E
and a similar function U defined on
r\3E
associated to
Therefore (4) and (5), keeping in mind that
IG/ri
u.
is exactly
the number of orbits Ir\3EI, may be expressed as
v(rx2) - 1/Ir\xI E v(rx) = 0(1), rxEr\X u(rx2) - 1/Ir\xl E u(rx) = 0(1). rxEr\x In other words we must show that, as c-O,
(8)
(9)
the values of
the functions U and V at a given point of r\X are not too distant from their average values on the entire set of orbits. To prove (8) and (9) we shall introduce a Laplace operator for functions defined on r\.. We define F(ry).
LF(rx) = (q+1)-1 d'x,y)=1
Observe that L may be thought of as the ordinary Laplace operator defined on r-invariant functions on X. A basis for the finite-dimensional
vector space of complex-valued functions
defined on r\X is given by the functions 6rx which are 1 on the
orbit rx and zero on every other orbit. With respect to this basis the matrix coefficients of L are given by 8rx(rz)=(q+1)-11{': d(2'x,y)=1}I.
(L8rx,8ry)=(q+1)-1 d y,z)=1
But d(,Xx,y)=d(x,' 1y),
and therefore L has a symmetric matrix
with nonnegative elements.
We shall now use a classical result
Approximating the projection P
73
0
on nonnegative matrices [S], for which we need a definition.
(9.2) DEFINITION. A nonnegative self-adjoint matrix A is called
irreducible if for every pair of indices (i,j) there exists k such that (Ak)ij>0.
We observe that L
is
irreducible with respect
to
its
canonical basis. Indeed, if rx and ry are given, there exists k such
that
S )>O
(Lk6 x
(it
y
suffices
to
take
k=d(x,y)).
It
follows then that (Lkarx,ary)>0.
(9.3) THEOREM.
Let A be a nonnegative irreducible self-adjoint
matrix; then there exists a unique eigenvalue r such that (a) r is a real positive number,
(b) an eigenvector of r has strictly positive entries, (c) r--JAI for every eigenvalue t,
(d) the eigenspace associated to r is one-dimensional,
(e) if A is any eigenvalue and IAI=r,
then A is a simple root
of the characteristic equation of A. PROOF.
[S, Theorem 1.5 and Theorem 1.7].
We now apply the previous theorem to the matrix L. Observe that,
if 1
is the function identically 1 on r\x,
because (LBrx, ary)=(q+1)
I {2"eI':
d(,xx, y)=1} I ,
then L1=1,
and E (Larx, ary)
ry =(q+1)/(q+1)=1. On the other hand the norm of L is 1. Therefore 1 is the maximum eigenvalue of L. Since the eigenspace of 1 has dimension 1, the operator (I-L)-1 is bounded on the subspace of t2(T\X)
which is orthogonal
to
1,
that
is on the space of
functions which have mean value zero. Denote this space by 5R. We can then write the function V of orthogonal
functions
V1
and
V2,
(8)
where
as the sum of two V1
is
the
constant
74
Ch. I I Ir\3f-1E
function (I-L)
and V2 has mean value
V(rx),
rx invertible
is
on
because
and
9R,
zero.
of
(9.1(6)),
Therefore
IIV2II`CII(I-L)V2II=CII(1-L)VII=0(1).
Since
IIV-V1II=11V2II=0(1).
The
space t2(r\x) is finite-dimensional, and therefore this implies
that, for every x2, V(rx2)-V1=0(1),
In order
to
show that µ=µ(1+21t)=(ql+2it+q -21t
in other words (8) holds.
true,
is
(9)
we
observe
that,
if
then under the hypothesis of
)/(q+l),
the theorem µ is a strictly complex number. Therefore (µ-L) is invertible. Therefore (9.1(7)) implies that
IIUII`CII(µ-L)UII=0(1).
This implies that both terms on the left-hand side of (9) are bounded
and
therefore
the
estimate
have
We
holds.
thus
concluded the proof of STEP B.
In order to
10. The constant 1 is a cyclic vector (STEP C).
complete the proof of (7.2) we must now show that 1 is a cyclic
vector in L2(f,v) for the representation nr. In other words we must prove the following.
(10.1) PROPOSITION.
Let g be a nonzero element of L2(f2,v),
then there exists at least one element a'Er, such that
(e,X(2')1)=J
pl/2-it(2,w)
(w) du * 0.
Recall that the Poisson transform relative to the complex
number z=1/2-it is defined on the space of finitely additive P1i2-it(x,
measures as P (m) (x)= f Z
w)dm.
Given an element g in
S2
L2(S2,v), we may apply T
Z
to the measure cdv, and define, for
xEX, T9(x) = PZ(gdv)(x) =
pli2-it(x,w)g(w) I
dv.
_c2
In order to prove (10.1) we must show that,
if Tg(ro)=0 for
every a'Er, then g=0.
Recall also the definition of the intertwining operator
The constant 1 is a cyclic vector
I
introduced
,
z
in
notation we set
Section
3,
for
simplicity
For
zEC.
and recall
I1/2+it It
75
that
It
of
is a unitary
operator on L2(sa,v), with the property that
dv =1,2p
la p
1/2+1t(x,w)It (w) dv.
We now apply (1.3) to obtain a convenient expression for P:;.
be the
Let x*O, and, as in the statement of (1.3),let x'
vertex at
distance
1
from x in the chain
then an
[o,x];
application of (1.3) yields
Tg(x) - q (1/2+it) T (x') _ (q1/2+it-q 1/2-it)q(1/2+it)d(x',O)
jO(x) (w)d v.
Likewise, .`Pe(x) - q(1/2-it) Pe(x') = (q1/2-it-q 1/2+it)q(1/2-it)d(x',0)
I g(w)dv.
SaJ(x)t Multiplying the first equation by q(1/2-it),
(q
and the second by
and subtracting, one obtains, ? (x)=
1/2-it -q -1/2+it
) (q-1/2-it-q-1/2+it)
-
q1/2+1t,
(q+l)-'[q(1/2+it)d(O,x)v(Q(X))-1 JQ(x)
Mdv
q(1/2-it)d(O,x)v(i2(x))-1ISa(x) I g(w)dv].
t
In other words,
7)9(x) = C(t)(q (1/2+it)d(o,x)v(Sa(x))-1J
t;(w)dvfa(x)
q(v2-it)d(o,x)v(sa(x))-1J sa(x)
I(w)dvl t
J
.
(1)
76
Ch. I I Let g be a square-integrable function on 0;
(10.2) LEMMA.
then, for almost every wEQ, g(w)dv = g(w),
lim v(f2(x))-lI
JO(x)
x-)w
as x approaches w keeping within a bounded distance from [o,w) (nontangential convergence).
PROOF. Let [o,w) be the chain {o,xl,x2,...}. Then the VitaliLebesgue theorem [R] directly implies that, for almost every w,
lim v(S2(xn ))-ll
C(w) dv = C(w)
JS2(x ) n
Let
d(x,[o,w))=d(x,x ) =m<M, n
integer;
for
then,
where
M
wE12(x)912(x ),
and,
n
a
is
fixed
positive
the
notation
with
introduced at the end of Section 5,
Iv(s2(x))-1 J S2(x)
g(w) du - v(S2(x))-1 f g(w) dv I n Jf(xn) W)I < (m+1) sup
< M Vj sup II An+k II 4 0. O:Sk:Sm
This concludes the proof of the lemma. I
It follows now that (1) implies the estimate gd(o,x)i2P9(x)=(q itd(o,x)g(w)_ q+itd(o,x)It9(w))+o(1)
(2)
for almost every wES2, when x approaches w nontangentially.
Choose now w ES2, 0
estimate (2),
such that
0
)*0.
Keeping in mind the
in order to prove (10.1) we must show that, as
d('o,o) becomes large,
subject,
for some M,
to
the condition
d('o, [o,w0)):sM, the quantity -itd(O,a'O)
q
t(w)-q
+itd(O,70) (3)
does not become arbitrarily small. We may assume that Indeed, II9(wo)I-IIt9(w0)11,
qit=ei2n9
if not,
(3)
is bounded away from zero by
and cannot become arbitrarily small.
Let
Then the quantity (3) may be written as a constant
The constant 1 is a cyclic vector
77
multiple of sin(2nOd('o,o)+90), where 9O=(91-02)/2, and 01 and
92 are respectively the arguments of t(wo)
and
In
other words we are reduced to showing that there exist c>O, and
a positive
integer
such
M,
for
that,
infinitely
many
''
satisfying d(yo,[o,w ))<-M, 0
Isin(2n0d(zo,o)+90)I> c.
(4)
We will consider now two different cases.
the number 0 is irrational.
Case (1):
first choose a number M'
In this case we
large enough that
infinitely many
points of ro satisfy d('o,[o,w))sM'. This is possible because t\X is finite, and a fundamental region of r (a set containing exactly one element of each orbit of F) is likewise finite. We
let S=Fn{x: d(x,[o,w))sM'}. Next, again using the fact that r has finitely many orbits, we choose N so that each of the q+1 disjoint
cones
neighbors of
where
T'(o,xj),
o,
xl,...,xq+1
contains an element
are
the
nearest
of Fo at distance no
Finally we let M=M'+N. By construction
greater than N from o.
every ball of radius N centered at an element To of S contains elements of ro inside each cone T'('o,'xi). This means that for
each ToeS there is a vertex "'o, such that the chain [o,ao] is
properly contained in
and d('o, "' o):5N. Now, for aER,
[ o, 2'' of ,
define f (a)=sin(a+2im9). The fact that 0 is irrational implies n
that fn (0)#0.
Choose
c>O,
whenever
Let 8 be such that,
such that
Isinal<e.
8>c,
and
for n=1,...,N Ifn(0)I>8>0. Ifn(a)I>8>c,
for n=l,...,N
Let Xo be an element of S such that
fails and let a=2n0d('o,o)+90. Then Isinal
(4)
and therefore,
for every n=1,...,N Isin(a+2nn9)I>c. But by (10.3) there exists g'o,
0<nsN.
such that d("'o,[o,w))<-M, It
follows that
Isin(2n0d(-d'o,o)+90)Iac.
and d("'o,o)=d(yo,o)+n,
27E0d(y'o,o)+O
a+2Tm9,
with
and therefore
We have thus proved that for every
element 'oeS which fails to satisfy (4) there exists at least
one other element
x,orro,
still within distance M from the
78
Ch. I I chain
infinite
(o,w),
and
at
distance
N
from
which
To,
satisfies (4). This concludes the proof for the case in which 0 is irrational.
Case (2): that
r
and
hypothesis
s
the number 0 is rational, 0=r/s. We may assume are
g21t*±1
integers
of
(7.2)
with no
common
implies
that
The
divisors.
s*2.
A
careful
scrutiny of the proof for the case in which 0 is rational shows
that the same arguments work if we can prove that each of the q+1 cones T(o,xi) contains vertices zio, such that d(o,Tio) is
not an integral multiple of s.
We shall presently prove that
this is indeed the case. We shall treat a slightly more general case.
(10.3) DEFINITION. for every xE3f,
A subgroup of Aut(X) is called cofinal if,
the set T(o,x)Mo is nonempty.
Observe that the hypothesis that r is cocompact implies that it is cofinal, because a cone has infinite diameter.
(10.4) LEMMA.
Let rcAut(X) be a cofinal sugroup. Suppose that
there exist an integer sa3,
and an element x
El,
such that
0
d(To,o) is an integral multiple of s for every Toeg(o,x ). Then 0
there exists w ESZ, such that rw ={w 0 0 0 PROOF.
Let 9I={ '-10:
(EF, a'oEC (o, x ) } . 0
We shall prove first of
all that this set has exactly one limit point w ED. It suffices 0
to show that given 71,a'2Ell, such that d(Tio,o)_d(x0 o)+N+1, for j=1,2,
becomes
[o,j11o1n[o,,d21o1
is a chain of at least N edges. As N
large this identifies an infinite chain [o,w
).
To
0
prove the assertion suppose that [o,T-1o]n [o,-dzlo] consists of N
0
edges with N0
such that the chain [o,x] has exactly N
edges in common with
0
[o,gllo] and N0+1 edges in common with [o,T-1o]. Let that yoOy(o,x) (Fig. 2).
7 be such
The constant 1 is a cyclic vector
79
Y o)
_I Y
(0)
2
0
Fig. 2 Observe
(Fig.
2)
that
d(z To,o)=d(zo,T11o)=d(o,To)+d(o,j110)
-2No, by construction of x and X. d(To,o)-2N
0
(mod s). A similar argument leads to the conclusion
d(y2yo,o)=d(To,o)-2(No+1)
that
This means that d(rl'do,o)=
(mod s).
conclude
We
that
d(TITo,o)*d(T2yo,o) (mod s), and therefore 71To and T2To cannot both belong to Ii(o,x ). We will reach a contradiction and prove
the assertion if we show, on the contrary, that 71To and T2T0 are
in
T'(o,xo).
Apply now Tj
to
the
chains
and
[o,T11o]
[yj-logo] (Fig. 2). We obtain the chains [o,Tlo] and [o,TiTo] (Fig. 3).
The total number of edges of [o,ijo] is at least d(x0,o)+N+1,
while only No edges lie in the portion of the chain [o,Tlo] which is not in common with [o,ija'o]
It follows that
(Fig. 3).
must be in the portion of the chain [o,yo] which is in
x 0
common
with
[o,Tjyo].
In
other
contradiction proves our assertion.
words,
We
This
7 1ToEC(o,x ). 0
let w
be the limit
0
point of V. If y'er, then "'w0=lim {X'T-lo: a'EF, yoeco,x0 )}_ lim{j1o: yy'oeT(o,x )}. 0
But yy'oel(o,x ), 0
when 'o0r(o,x ), 0
soon as d('o,o)ad(o,x )+d(y'o,o). It follows that v'w =w . ' 0
0
0
as
80
Ch. II
We go back now to the hypothesis of (7.2) for the discrete group r. Then r is certainly cofinal. We prove that it does not fix any boundary point. With the notation of Chapter I, we must
prove that r is not a subgroup of GW for any wEe. But since l contains no rotation TnBw contains only the identity.
If MGw
then r leaves invariant a geodesic and is isomorphic to Z. This
cannot happen because r is cocompact (and hence cofinal).
We
conclude now by (10.4) that, for every vertex x, there exists
such that zoEC(o,x) and d(yo,o)*0 (mod s).
yEr,
This proves
(10.1) and concludes STEP C, and the proof of (7.2).
Theorem (1.2)
11. Notes and remarks.
is essentially the same
as Theorem A of [NZ1] dealing with eigenfunctions of the simple random walk on a free group. The simpler proof given here which is
based only on the geometry of the tree
is
taken from
[F-T S2] where a different and more general result is proven for a nearest-neighbor anisotropic random walk on a homogeneous tree.
The results of [K1,2] on affine buildings of arbitrary
rank are also more general
but
in a different
direction.
Observe that (1.2) is the analog of a result of S.Helgason for
symmetric spaces
EHe].
Proposition (1.3)
is a special case
of Lemma 2.1 of [F-T S2]. Related to this proposition are the
more general results of
[KP],
[KPT] and [PW].
The theory of
spherical functions and spherical representations as described
in Section 2 and 4 is classical. We have especially benefited
from the expositions given in H and [F]. Spherical functions and spherical representations for a group acting transitively on a tree and its boundary were first considered by [C21, but spherical functions for PGL(2) over a p-adic field were studied by G. van Dijk [Di]. Observe that spherical representations and spherical
functions
in
this
context
are
nothing
but
the
Notes and remarks
restrictions
of
spherical
81
representations
functions of the full group Aut(o).
and
spherical
Indeed much of the theory
may be developed intrinsically for the free group or any other
group which acts simply transitively on the vertices of the tree [F-T P1,2]. An interesting characterization of spherical
representations of free groups
in terms of a condition of
symmetry of a coefficient viewed as a function on X is given in [KS2].
The explicit formulas contained in (2.3) and (2.4) are
taken
from
[F-T P1,2].
operators
Intertwining
used
were
explicitly by [MZ1] in order to define the representations of the
complementary
series.
However
convenient
the
diago-
nalization of the intertwining operators given in Lemma (3.2)
is taken from [F-T S2] where this result is proved in a more general context. The inequalites of Lemma (6.1) and Lemma (6.2)
are particular instances of more general convolution theorems due to U.Haagerup [H]. The formula of Proposition (6.3) for the
explicit computation of the resolvent was given in [F-T P1,2]. The spherical Plancherel formula was first computed by Cartier [C3]. A different computation in the context of free groups was
given in [F-T P1,2] (see also [CdM]). Our immediate source for (6.4)
is
the simple computation given
in
[FP].
The direct
ancestor of Theorem (7.1) is the theorem on the irreducibility
of spherical representations of free groups
[F-T P1,2].
should mention however the earlier result of T. Pytlik
One 101
yielding irreducibility for almost every representation of the
principal series of a free group.
Related results are the
theorem on the irreducibility of the representations associated
to anisotropic nearest-neighbor random walks [F-T S2], and the
theorem of finite reducibility for representations associated
to an arbitrary group-invariant symmetric finitely supported
random walk
[Sti].
An exposition of
the
latter result
is
contained in [A]. Theorem (7.1) has an analog for lattices of SL(2,U2)
[St3]. An important rigidity theorem which makes use of
82
Ch. I I
(7.1) is the following.
Let r1=i1(r)
be two embeddings of a free
and r2 12(r)
group r as cocompact subgroups of Aut(X). Let 71 1and it
2
restrictions
to
representation
rl
of
and
of
respectively,
r2,
Suppose
Aut(3f).
that
nl
be the
a
spherical
is
unitarily
equivalent to n2. Then for some gEAut(X) grIg 1=r2.
This
from
follows
the
techniques
of
[BS1,2]
(which
contains the analogous result for embeddings of a free group in
SL(2, I2)) and Theorem 3.7 of [CM]. Observe that this implies that spherical representations of a free group, as defined in [F-T P1,2], depend on the choice of the generators. In other words if two sets of generators are
not mapped one into the other by an inner automorphism of the they
group
free
give
rise
to
inequivalent
spherical
representations.
We now describe briefly the tools required for the proof of (7.1) in addition to those which were used for the proof of (7.2).
hypothesis
The
that
{2'er: xx=x }={e}
inessential
is
because of a result of H. Bass and R. Kulkarni [BKu] which asserts
that
contains
every discrete
cocompact
a subgroup of finite
subgroup
index satisfying
of
Aut(l)
the
above
hypothesis. Observe however that the proof does not work if we only assume that Aut(X)/r has finite volume. Discrete groups r
such that Aut(x)/r has finite volume but is not compact do exist
[Lu].
To extend the proof to
representations of the
complementary series we only need an estimate, (8.1)
for
1/2:sRez'1.
(a (L)-µ)-1,
when
p
is
a
analogous to
complex
number
This estimate holds once the spectrum of n (L)
and is
determined to be contained in an interval with end points ±(qs+ql-s)/(q+1).
For STEP B the hypothesis that
IrxI=1
is
used, but not in an essential way, in some of the formulas. The
hypothesis that
g2it*±1
is also used at the end of the section
but it may be done away with by inverting I+L on the complement
Notes and remarks
of the eigenspace of -1.
83
The proof of irreducibility breaks
down in general at STEP C without the hypothesis that g21t*±1. A separate argument is needed to treat this case so as to yield (2) and (3) of (7.1).
following
The
general
result
on
restrictions
of
irreducible representations was recently proved by M.G. Cowling
and T. Steger [CS]. THEOREM.
Let G be a separable unimodular locally compact group
and t be a lattice of G such that the quotient space G/l has finite volume. Let p be the restriction of the quasi-regular representation of G on LZ(G/I') to the orthogonal complement of the
space
of
functions
constant
on
G/t'.
irreducible unitary representation of G.
Let
n
be
an
it ® p does not
If
contain it then the restriction of it to t is irreducible.
This theorem is applied in [CS] to prove the analog of Theorem 7.1 for lattices of semisimple Lie groups. Using the
same result and the theory of representations of Aut(l) as described in Chapter III,
it is also possible to establish the
conclusion of Theorem (7.1) for a lattice of Aut(E)
(without
the hypothesis of cocompactness). The direct proof of Theorem (7.1)
given here
has
the
advantage of
not
requiring any
knowledge of the representation theory of Aut(l). This suggests that
its
arguments
may
also
be
applied
to
nonspherical
irreducible unitary representations of free groups and their restrictions to subgroups of finite index.
84
CHAPTER III
1. A classification of unitary representations. In this section we classify the irreducible unitary representations of a closed
group G acting transitively on a tree and its boundary. The representations
divided
are
into
three
disjoint
classes:
spherical representations, special representations and cuspidal representations.
While
all
definitions
make
sense
in
this
general context the class of cuspidal representations becomes significant only when G contains sufficiently many rotations. Accordingly,
studied
in Section 3,
when cuspidal representations are
we restrict
in detail,
our attention to
the
case
Assume now that G is a closed subgroup of Aut(o)
G=Aut(3E).
acting transitively on 3f and Q. This means by (I,10.1) that the
compact group K C={gEG: g(x)=x} x
also
notice
(G, GrK
space
by
acts transitively on Q.
[F,Prop.I,1]
G
is
unimodular
We
because
is a Gelfand pair (III, 4.1).
)
x
Let
that
it
H,X.
be a unitary representation of G on the Hilbert only
We
representations
consider
it
which
are
continuous in the weak operator topology, which means that, for
every g,n e 3{rz
,
the function (x(g)g,n) is continuous on G.
If K is a compact subgroup of G and dk its normalised Haar
measure, we define, for g E Its
,
dk
P,N(K)9 =
.
JK
It is easy to verify that P,it(K) is an orthogonal projection and
that, for each kEK, n(k)PN(K)=PIT(K). Conversely, if it(k)g= for
every kEK, then Pn(K)g=g. In other words the range of P,t(K) is
the space of all x(K)-invariant vectors.
(We shall usually
refer to this space as the space of all K-invariant vectors, when the representation it is fixed.)
A classification of unitary representations
Let 3 be a finite subtree of X.
85
Define K(S)={geG: gx=x,
for all xE } and let P,TE(s) be the projection on the space of K(s)-invariant
Since K(s)
vectors.
is
open
its Haar
in G,
measure is a multiple of the Haar measure m of G restricted to K(s). Therefore,
Pn(s) = 1/m(K(S))
n(g) dg.
A finite subtree S is called complete if it consists of a single vertex, or if every vertex has degree either q+l or 1. We denote by
(the interior of S) the subtree of 3 consisting
of vertices of homogeneity q+l. The boundary of S is the set 8S=S\S0. Clearly every finite subtree is contained in a finite
complete subtree. More precisely, for every m>O, TcV (S)={xEX: M
d(x,S)<m} and Vm(s) is a finite complete subtree. As S varies among all finite complete subtrees, K(s) describes a basis of neighboroods of the identity of G.
It follows that,
if
it
is a unitary representation and
O*gE3fn then, for some finite complete subtree S, dg * 0,
JK(3 ) which implies Pn(S)g*O. The remarks above justify the following definition.
(1.1)
DEFINITION.
representation of G.
Let
it
be
an
irreducible
unitary
Let en be the smallest positive integer
for which there exists a finite complete subtree S having tW vertices and such that Pn(s)*O. Then it is called spherical if eR 1,
is called special
if
and is called cuspidal
71=2,
if
t >2. it
We observe that by (11,5) every spherical representation in the
sense of (11,5.1).
the definition above We
shall
is
therefore
spherical not
as
discuss
defined
in
spherical
86
Ch. III
representations in this chapter.
If Pn(S)*O and card(s)=tn then we say that & is a minimal tree associated to n.
If & is a minimal tree then gS is also a
minimal tree for every gEG because P71
(gs)=n(g)Pn(S)n(g)-1
and
card(s)=card(gS).
is any finite complete subtree,
If S
let
3fn(S)
be the
subspace of Rn consisting of all K(s)-invariant vectors.
In
other words 3f71 (S) is the range of P71 (s).
We
consider
now
the
subspace
This
V71 U3en(3).
is
a
S
nontrivial linear space. Observe that Vn is invariant under the
action
of
n(G)
K(gS)=gK(S)g 1.
because
n(g)3fn(S)=3fn(gS)
and
of
course
It follows that Va is dense in 3fn.
Note that the space Vn is dense for an arbitrary unitary representation n, Indeed,
without the hypothesis of
irreducibility.
if M is a nontrivial invariant closed subspace of 3fn
and C*O E3fn then, for some subtree S, PnWC*O. Thus MnVn*0. Since the orthogonal complement of Vn is invariant, this shows that V
is dense.
be a sequence of finite complete subtrees such that
Let jT
S nCsn+1
for every n and
particular, some n0;
dim(n)<+oo,
Then
X=USn. n
then V71
is closed and
this means that 7IK(S
)
If,
in
3f71 =3enn )
for
UnH ,rt(Sn)=Vn.
is trivial.
0
We have thus
n 0
proved the following proposition.
(1.2) PROPOSITION. Let it be a unitary continuous representation
of G. If dim(n)<+w then there exists a complete finite subtree S such that nIK(S) is trivial.
Let it be a unitary representation and let S be a finite
complete subtree such that the number of vertices of S
is
exactly n. This means that if q is a proper complete subtree
Special representations
then Pn(n)C=0 for every Celt (s).
of S,
87
follows that,
It
if
the coefficient function u(g)=(u(g)g,i,)
g,714Vn and Pn(S)g=g,
has the following properties: (1) u(gk) =
u(g), for keK(S);
(2) there exists S' such that u(kg) = (n(g)e,n(k 1)q) = u(g) for all
keK(S')
(mot'
is
any
complete
subtree
for
which
P71 (S'
(3) if s is a complete subtree properly contained in & then 1/m(K(s))
u(gk) dk = (P71 (s)C,n(g-1 )n)=O for all geG. K(s)
The remarks above justify the following definition.
(1.3) DEFINITION. Let S be a complete finite subtree of X. We let
`.Y(S)
be
the
linear
space
of
continuous
functions
u
satisfying the following properties: (1) u is K(s)-right-invariant;
(2) u is K(S')-left-invariant for some finite complete subtree which depends on u; (3) for every proper complete subtree s9S, u(gk) dk = 0 for all geG.
It is clear that .(3) is G-left-invariant. Therefore,
if
'(S)*0 then it makes sense to consider the restriction AS of the left regular representation to .(3): A3(h)u(g)=u(h 1g).
It
is easy to see that p(g)`.'(S)=.`'(gS) for every geG.
2. Special representations.
In this section we assume again
that G acts transitively on T and its boundary (,
and we
classify, up to equivalence, the special representations of G.
We shall in fact prove that there exist only two inequivalent special
representations,
which
are
the
restrictions
to
G
88
Ch. III
of the special representations of Aut(3f).
Let it be a special representation. This means according to
(1.1) that the minimal tree & associated to it is an edge. Let
=e={a,b}. Define as before K(e)={geG: ga=a, gb=b}, K({a})=K = a
{geG: ga=a}, K({b})=Kti {geG: gb=b},
n(k) dk * 0
Pn(e)
,
K(e)
and Pn({a}) =
n(k) dk = J Ka
n(k) dk = Pn({b}) = 0.
I
Kb
We observe now that the hypotheses on G imply that its action is doubly transitive. This means that we may replace the
edge e with any other edge e' ga=a'
and
then
gb=b',
because,
if gEG is such that
P71 (e')=P71 (ge)=n(g)Pn(e)n(g 1),
and
similarly Pn({b'}) = n(g)Pn({b})a(g 1).
Pn({a'}) = n(g)Pit ({a})n(g 1),
K(ge)=gK(e)g-1,
This assertion is proved using the fact that
and changing projections.
the
variables
in
the
integrals
We now fix an edge e ={a,b},
defining
the
and consider the
0
space .'(e
).
0
We observe first of all that °(e )
is G-left-invariant.
0
Therefore the condition u(kg) dk = 0
JK. for elements of
'(e) implies that, for every hEG, 0
I.
u(h 1kg) dk = 0. a
Therefore u(h 1khg) dk = 0,
JK for every h and every g.
a
In other words, for every hr=G,
Special representations
89
u(kg) dk = 0.
JK
ha
Furthermore, the condition that there exists a finite complete subtree 3 such that u is K(S)-left-invariant may be replaced by the condition (2)
there exists n such that,
if kx=x for all vertices at
distance n from a, then u(kg)=u(g).
In other words the complete subtree 3 may be replaced by the complete subtree 3E n={x: d(x,a):Sn}. Since the action of G is doubly transitive on X, G acts transitively on the set of oriented edges [x,y] where {x,y}El.
Since [x,y]*[y,x] we may identify the set of oriented edges with the union l'uf
of two copies of T.
This identification
may be made in such a way that whenever [x,y]Ee+ (respectively
[x,y]Eff) all the other q oriented edges [x,t] also belong to l+ (respectively C). This is possible as shown in Fig.1.
Fig.1
The stabilizer of [a,b]
in ff+uff-
is the compact group K(e
).
0
Accordingly the map g - [ga,gb] defines a bijection from the
quotient space G/K(e)
to the set of oriented edges f+u1 .
0
Therefore every K(eo)-right -invariant
function on G may be
_.
identified with a function on
ff+
In particular .ff(e) may be 0
90
Ch. I I I
identified with a space of functions 5 defined on
LF+ .
A
function u on ff+uT- may be identified of course with a pair of
functions (u+,u) defined on !, whose components correspond to the restrictions of u to l+ and f
respectively. The space $0
,
may be characterized as follows.
(2. 1) LEMMA. Let u be a function on
T+ ,
then uE$ if and only
if the following conditions are satisfied.
(1) For every xE3f the sum of the values of U+, and separately of u , over the edges having a preassigned distance from x, is zero. (2) There exists a positive integer n such that if d(a,x)=n then the restrictions of the functions u+ and u
to the edges
contained in the cones Va,x) depend only on the distance of the edge from a.
PROOF. The first property is the exact translation of property
For the second
(1) for K(e )-right-invariant functions on G. 0
property recall the definition of the cones i'(a,x) (I,1).
then clear that edges contained distance
from
x
are
in
the
having the same
in C(a,x)
same
It is
orbit
of
subgroup
the
K(X ) ={gEG: gx=x, if d(a,x)=n}. Hence the second property is a n
translation of (2) for the corresponding K(e )-right-invariant 0
functions.,
The characterization above implies that each element of $0
is uniquely determined by the values it assumes on a finite number of edges.
To be precise assume that [a,b]Eff+ and suppose that ud, and that u is K(X n)-left-invariant, where 3E ={y: d(y,a):Sn}. Let n d(x,a)=n and let d(t ,x)=1. x
t
x
be the vertex such that d(a,t )=n-1 and
Suppose that u+([t ,xl)=a.
x
Then u+ has the value
x -a/q on the positively oriented edges having distance n from a,
and containing x; in general u+ has the value (-1)k+1 a/gk+1 on
Special representations
91
the positively oriented edges contained in C(a,x) and having
Thus the values of u+ outside X
distance n+k from a.
n
are
determined by the values on the positively oriented edges having distance n from a.
In particular the subspace of S(e
consisting of all
)
0
K(X )-left-invariant functions is finite-dimensional. With this n
in mind we can prove the following theorem.
ud
If
THEOREM.
(2.2)
E
then
(lu+(n)I2+lu (v)I2)<00.
In
or=LY
addition, if u is not identically zero, then
E (lu+(n) I+lu (n) I )_ Co. 1 EC
Therefore
space
the
We0
contained
is
in
L2(G)
and
Y(e )nL1(G) ={0}. 0
PROOF.
Assume that u is K(X )-left-invariant. Then n
00
Ill+(n) l2 Iu+(n) =n=oZ vsl(a,x) T Z vSl(a,x)
d(t),a)=n
00
-
q
qk+1
lu([t ,x])l2
_
l2
lu+([tX,xl)l2 q-1
q
n=o
Therefore,
q lu+(n) I2+
Elu+(n) I2 = ro-Xn
q-
1
(t),x)=n
lu+([t,x]) 12<00.
The same calculation is true for u . We have thus proved the first part of the theorem. Let now u+*0, and suppose that u+ is K(X )-left-invariant.
is
easy
to
first
part
of
It
n
notation
as
u+([t ,xl)=0, x
in
the
for all
u+(y)=0, for all yE3f
.
n
vertices
see the
x such that
proof)
the that
d(x,a)=n,
same if
then
We may assume therefore that for some
vertex x, with d(x,a)=n, u+([t ,x1)=a:0. Then X
00
lu+(ro)I = Z
(a, x)
(using
k=O
(lal/gk+i) qk+1
92
Ch. I I I
The L2-norm of an element of .(e
is a multiple of the sum
)
0
lU-(.0) 12 1
E1JGey where u+ and u
1/2
lu+(n)l 2 +
r
Ld
roG ff
are the restrictions to
ff+
and l ,
respective-
of the function of S corresponding to that element.
ly,
similar statement
true for the L1-norm and the
is
A
theorem
follows.,
The subspace of f(e
(2.3) PROPOSITION
)
consisting of all
0
K(e )-left-invariant functions is a two-dimensional subspace. 0
For every K(e )-left-invariant fE.(e 0
0
the following equality
)
holds:
l1fll2 = (m(K )/(q-1)) (if+(e)o l2+ if (e)0 l2). 2 a M
Let
PROOF.
be
subspace
the
of
g'(e
)
consisting
of
0
If fEM then f is constant on
K(e )-left-invariant functions. 0
the edges at distance n from e 0. We shall presently show that there exist exactly two linearly independent elements of Y(e ) 0
which
are
K(e )-invariant.
Indeed
if
uE°(e )
0
invariant,
with
0
and
a0={a, b}
and
[a, b]eLV+,
is
if
K(e )0
u+(e
0
u([a,b])=a, u (e )=u([b,a])=13, then u+ takes the value -a/q on 0
the q positively oriented edges starting with b, and on the q positively oriented edges ending with a. The other values of u+
may be inductively determined. Similarly the values of u
may
be determined from the value of u
the
on
e .
Let
0
v
be
1
function which is zero on f and determined as above on is+, starting with the value v1([a,bl)=1, and let v2 be zero on 9+ and
similarly
determined
on
v2([a,bl)=1. Then u=av1+13v2.
ff
,
starting
with
the
value
This shows that v1 and v2 form a
basis for the space of K(e )-invariant functions of °(e ). 0
0
Therefore the map f-(f+(e ),f (e )) is a linear isomorphism of 0
0
M onto C2. We have that
11f 2112 =m(K(eo))(E 1007
( if+O l2+lf O l2));
Special representations
as
proof
the
in
of
can
we
(2.2),
93
easily,
prove,
that
If+(n)I2=If+(e
)I2((q+1)/(q-1))
v5 and I2=If (e)I2((q+1)/(q-1)).
I
E vcl
G acts transitively on
S2
has index q+1 in K a; and K(e) o
therefore m(Ka)=(q+1)m(K(eo)) and the proposition follows.'
Theorem (2.2) and Proposition (2.3) imply that '(e) is a nontrivial left-invariant subspace of L2(G). Let At(e) be the 0
closure of .(e0) in L2(G). Let Ae be the subrepresentation of 0
the left regular representation of G on L2(G) which corresponds to the invariant subspace At(e0 ). Since p(g).`'(e0)=`,'(ge0 ), we
are unitarily
have that p(g)At(e0)=At(ge0); therefore Ae and Age 0
0
equivalent. The space At(e ) can be regarded as the subspace of 0
Z2(9+t
)
consisting of functions with the property To prove this,
Lemma (2.1).
it
(1)
of
is enough to see that every
function fe@2(L+vC ) with the property (1) of Lemma (2. 1) limit in 22 of functions of ,°(e 0 ). The proof
is a is
straightforward: if u is the function in °(e ) such that u is 0 n n hx=x for every x such constant on the orbit of K(X )=(heG :
n
that d(a, x)=n} and un =f on In, then it is easy to see, as in the proof of Theorem (2.3), that IIf-uII
<
If+(v) I2+If (v) I2)
m(K(e))(1+(q-1)-112)2(E
n-1 because
E U¢X
(Iu n+(D)I2+Iun
(n)I2
n
_ (q-1)-1 E vc3f /3E n
(If+(v)I2+If (v)I2). n-1
94
Ch. I I I
Since
e>O
n such
exists
there
that
As in Theorem (2.2), we have that M(e0)nLI(G)={0}.
IIf-unII2<E.
It
every
fEE2(ff+uf ),for
is also clear that every K(e )-left-invariant function of 0
is in S°(e ); therefore Proposition (2.3) is true for
At(e) 0
At(eo). This implies that the range of the projection PA(eo) on is two-dimensional.
that is the subspace PA(eo)(A((eo)),
At(eo),
Since K(e ) 0
has index 2 in the stabilizer k(e )={gEG:
g0 Me )=K(e )g 0
ge =e
0
0
for every inversion g0 on e0.
0
for every fEAK(eo), PA(eo)f(1G)=f(1G)
0
0
In particular,
(1G is the identity of G)
and PA(eo)f(go)=f(go). This means that the function PA(eo)f is the K(e )-left-invariant function of °(e ) which corresponds to 0
0
is the pair (f+(e ),f (e )) for every fEAt(e0 ), and so feAt(e) 0 0
orthogonal f -(e
0
0
in L2(G)
if and only if f+(eo)=
to PA(eo)(A((eo))
)=0, that is f is identically zero on K(e ).
Every nontrivial closed left-invariant subspace
LEMMA.
(2.4)
of We ) contains a nontrivial We )-left-invariant function of 0
0
We 0 ). PROOF.
Let
H
be
a
nontrivial
invariant under left translation. suppose
that
translates.
u(1G)*0,
Since
H
replacing
is
Let uEH, u
of
subspace
closed
left-invariant,
0
we may also
u*O;
one
by
At(e ),
of
its
left
PA(ea)HcH
and
PA(eo)u(l )=u(1G)*0 because u is Me0) -right- invariant, and so u is constant on K(e ). This proves the lemma.' 0
Lemma (2.4) implies that e0 is a minimal tree for Ae and 0
for
every
subrepresentation
of
In
Ae
every
particular
0
irreducible subrepresentation of
Ae
is
special.
Since
the
0
space
of
K(e )-left-invariant 0
functions
of
At (e
0
)
is
two-dimensional, Lemma (2.4) and Proposition (2.3) imply that
Special representations
95
either is irreducible or is the sum of two irreducible
A 0
subrepresentations.
Every unitary special irreducible representation
(2.5) LEMMA.
of G is unitarily equivalent to a subrepresentation of Ae . 0
Let it be a unitary special irreducible representation
PROOF.
of G and gER
n a nontrivial K(eo )-invariant vector. (n(g)9,9) is
a
nontrivial
function
K(e )-bi-invariant 0
This implies that
particular (n(g)t;,e)EL2(G).
of it
°(e ); 0
is a square-
Since G is unimodular,
integrable representation [D1,14.1.3].
every coefficient of it is in L2(G); therefore for every REHH. By [B1,14.3.3] (n(. do is the formal dimension of
in
0
where
IIZ III
the operator U: 3fn-4 At(e0), defined by U(71)(.)=( fin) 11911 1(n,n(. )t;), is a Hence
it.
unitary operator intertwining the representations it and A. The lemma follows., (2.6) THEOREM.
is the orthogonal sum of
The representation Ae 0
and a2. The
two inequivalent irreducible subrepresentations representations
2 are L2 but not L1-representations,
and
and their formal dimensions are do.=do.=(q-1)/(2m(Ka)). 2
1
Let f0 be the unique K(e0 )-left-invariant function of `.°(e) such that f+(e )=f0 (e0 )=1 and let h0 be the K(e0 )-left0 0 0 )=1 and h 0(e 0)=-1. invariant function of Y(e 0) such that h+(e 0 0
PROOF.
The vectors (1,1) and (1,-1) are orthogonal in C2;
therefore,
by Proposition (2.3), (f0 h0)=0 in L2(G) and {f 0 h0 } is a basis of the space of K(e0 )-left-invariant functions of At(e0 ). Let g0 be an inversion on the edge e0 ; since f+=f , A(g0 )f0=f 0 and 0 0 Mg )h =-h 0
0
0
because
h+=-h 0
0
orientation of every edge. A((eo)
generated by the set
and
an
inversion
changes
the
Let M1 be the closed subspace of {A(s)f0: 5EG)
and let M2 be the
96
Ch. I I I
closed subspace of Al(e) generated by the set {a(s)h : sEG}. 0
0
The subspaces Ml and M2 are closed and nontrivial. Let o2 be the subrepresentations of Ae
and
corresponding to M1 and M2,
0
respectively.
invariant
now F(t)=(A(t)f h
Let
function
.(e 0)
of
a Me0 )-leftF+ (e0 )=F (e0 (f 0,h )0 =0; F
);
0
0
and
is
F is identically zero.
therefore, by Proposition (2.3),
This
implies that (v,w)=0 for every vEMI and for every wEM2. This means that the closed nontrivial invariant subspace M1®M2 is
the orthogonal sum of M1 and M2; function of
every K(e)-left-invariant
is contained in M1eM2 because this space
At(eo)
contains the basis
{f
therefore Proposition (2.3)
h }; 0
and
0
Lemma (2.4) imply that the orthogonal complement of M1®M2 must
be zero.
This means that M1eM2 At(eo) and Ae =o'1eo2 is the 0
orthogonal sum of Lemma (2.4),
and o2. Moreover, by Proposition (2.3) and
there are at most two irreducible
as observed,
and o2 are irreducible
and so
subrepresentat ions of Ae ,
o'1
0
special
representations.
show now
We
that
and
(r
are
o2
1
Suppose that,
inequivalent.
unitary
Ae (g)U=UAe (g) 0
U
operator for
mapping
every
there exists a
on the contrary,
gEG.
onto
M1
This
M2
and
such
that
implies
that
PA(ea)U=
0
U takes K(ea)-left-invariant functions of M1
into Me )-left-invariant
functions
of
M2.
But Me )-left-
invariant functions form one-dimensional subspaces of M1 and M2,
respectively.
Therefore Uf=ch
for some constant
c.
In
particular Vg )Uf =-Uf because Mg0)h 0=-h0 (recall that g0 is 0 0 0 an
inversion
A(g0)Uf0=-Uf0
,
on
edge
the
e).
Uf =UA(g )f =
Finally,
0
0
0
0
that is Uf =0. This contradicts the injectivity 0
to M1 and M2
of U and we conclude that the restrictions of le 0
are inequivalent. We recall that d71 the formal dimension of an where Ilt;11=1 L2-representation is equal to II(n(. )e,e) II22 F(t)= where In [DI, 14.3.3]. particular do, =IIFII22 1
Special representations
The function F is Me )-left-invariant and
(A(t)f,f )0I1f II2. 0 0 02
0
moreover
F4=-M(e)
0
0
for every tEG. Hence d =II similarly
proved. do, =do, =(q-1)/2m(Ka).
means
this
F`(e )=F (e )=1
0
1
97
011
F(t)=f (t)
that
0
2
2
The fact that
.
dQ, =11ho11 22
2
Proposition
that is
implies
(2.3)
The fact that
(the proof for
o'1
is
(r 2
2
similar) is not an L1-representation is a consequence of the fact that ht(e)nL1(G)=0.
Indeed,
(A(.)g,g)EL1(G) for geM1
let
and t;*O; as we may replace t; by one of its left translates we may assume that g(1G)*0 (G is unimodular and hence the function
PA(eo)t;(1G)=t;(1G)*0 K(eo)-left-invariant,
every
for
(A(.)A(t)t;,A(t)g)EL1(G)
and
since
PA(eo)Z;EM1;
An
PA(eo)C=e(lG)fo.
have
We
teG).
that
PA(eo)t;
easy
is
computation
shows that JG(A(g)PA(e)g,PA(e)C)dg =
we have that
Since f aO, 0
(A(.)f
JG
(A(g)9,9) dg.
This is a
f )E L1(G)r4i(e ). 0
0
0
contradiction.,
Theorem (2.6) and Lemma (2.5) imply that, for any closed subgroup of Aut(X) acting transitively on X and 0, there exist only
inequivalent
two
unitary
irreducible
special
representations.
A similar statement
is
true
a closed noncompact
for
subgroup G of Aut(fl acting transitively on 0 but not on X, that is a subgroup G which has exactly the two orbits f+ and 3f
(see Proposition (10.2) of Chapter I); for instance Aut(X)+ or In this case G/K(e ) can be
PSL(2,3) where 3 is a local field.
0
regarded as the set of nonoriented edges. This implies that the subspace functions
of is
ht(e )
0
consisting
one-dimensional
of
and
K(e )-left-invariant
all A
0
is
e
irreducible.
0
Proposition (2.3) and Lemma (2.4) are true with the same proof.
98
Ch. I I I
Hence,
for such a G,
irreducible
there exists only one unitary special
representation.
L1-representation
and
This
but
L2-
dimension
formal
its
an
is
not
equal
is
to
(q-1)/m(K ). a
3.
formula of
Plancherel
In this section we suppose that G=Aut(X). The results
Aut(.T). we
representations and the
Cuspidal
present
are
here
transitively on
3E
of
thought
always
not
and on as
fl. a
true
for
acting
groups
In particular (3.2) fails for group
of
automorphisms
of
its
It will be clear, however, from the proof of
canonical tree.
(3.1), that all the results of this section remain true for a closed
subgroup
of
containing
Aut(X)
many
sufficiently
in particular they remain true for Aut(Z)+,
rotations;
the
group generated by all rotations.
(3.1) LEMMA. Let 3 be a complete subtree of 3f with diam(S)22. Let ) be a complete subtree not containing S. Then there exists
a proper complete subtree acS such that K(g)9K(n)K(S).
PROOF. We first show that we may assume that Snr) contains an edge.
Indeed,
if Snh is empty or consists of only one vertex,
then there exists a unique vertex xE
q.
of minimal distance from
For if m is the minimal distance of S and r and m=d(x,y)
=d(x', y '
)
intersect
with x, x' ES, y, y' etg and x*x', then [ x, y] and [ x' , y' ] S or g only at the end points.
But
[x,x'193 and
is a
[y,y' ]sr). This means that [x,x' ]u [x' ,y' ]u [y' ,y]v [y,x] circuit: a contradiction. Let g'={xE
:
d(x,q):5m+1}. Then t)'
is
a complete subtree and t)' nS contains an edge. Observe that n' does not contain S.
Indeed,
if
all
the vertices of
3 had
distance at most m+1 from p, then all vertices of S would have
distance at most 1 from the vertex closest to imply that complete.
S
has diameter
Clearly
less than 2
K(q')9K(g).
Therefore
or
r),
that
which would is
not
substituting,
if
S
Cuspidal representations
needed, a=i)nl
for r) we may suppose that i)rq contains an edge. Let
r)'
then
itself.
99
We
a complete subtree of
is
a
prove
that
K(a)cK(r))K(S).
vertices of a which have degree
1.
T
Let
different
from
x1,...,x
be
Let t'1=Lr(x,xi),
with xEa,
be the cones of vertex x1 which intersect a only in {x1}. K l
Clearly KicK(a),
and in particular every element of
kikj=kjki
for
i*,j.
is the direct product of the groups K1,...,K
prove now that,
KISK(r)).
Ki
In addition every element kEK(a) may be written
a product k=kl...kwith kiEKi and
Therefore K(a) We
Let
be the group of rotations which fix every element outside
stabilizes x1. as
S
the
Indeed
for every
K1cK(S)
if
and
either
1=1,..,s,
only
if
T'I\{x1}
KIcK(S)
does
or not
intersect S. Similarly KI9K(r)) if and only if TI\{x1} does not intersect r).
Therefore, if K1 is contained neither in K(s) nor
in K(r)), then T i\{x1} must contain vertices of both jT and r). Let xESnM i\{x1}) and yEI)n(T'i\{xi}). Then [x,xI]csnE and vertices x' ESr 1 [ y, xi ] cr)ncr 1. Therefore there exist and y' Er)l1Lri, such that d(x', xi)=d(y' , xi)=1. Since S, r) and Sr) are complete subtrees, this implies that x',y'ESnr), a contradiction
since we assumed that the degree of x1 proved that,
for each i,
is
1.
KICK(S) or K1cK(r)).
We have thus If kEK(a)
then
k=kI...ks with kIEKI. Since the k1 commute, and each belongs to either K(s) or K(r)),
we may write k=k'k",
with k'EK(S),
and
k"EK(r)). In other words K(a)cK(S)K(r))., (3.2) PROPOSITION. Let & be a finite complete subtree of I and
suppose that the maximum distance of two vertices of S is at least 2,
in symbols diam(S)a2. Let ncz be a complete subtree.
If u is a K(r))-left-invariant element of .(S), then supp(u)c{gEAut(X): gScr)}.
PROOF. We must prove that, if uE°(S) and u is K(r))-invariant, then u(g)=O, for all automorphisms g such that gS¢i). According
to (3.1) this is true for g=e. Indeed if 0 does not contain S
100
Ch. I I I
there exists a complete subtree 3 properly contained in S such that
Since
K(3)9K(S)K(n).
u
K(s)-right-invariant
is
and
K(g)-left-invariant u must be constant on K(3). But, since 3 is properly contained in S, the average of u on K(3) must be zero, which implies that u is zero on K(3). In particular u(e)=0. Let
now
gEAut(X)
such
be
that
g
does
not
contain
then
gS;
p(g)uEY(gS) and p(g)u is K(h)-left-invariant. We also have that
g3 is a complete subtree of diameter at least 2.
This means
that p(g)u(e)=u(g)=0.' (3.3)
COROLLARY.
If
diameter at least 2,
then every element of Y(S) has compact
particular
In
support.
a finite complete subtreee with
is
T
every
irreducible
cuspidal
representation of Aut(l) is an L1-representation. PROOF.
If
uE°(S)
then
there
exists
such
t
u
that
is
K())-left-invariant. This means by (3.2) that the support of u
is contained in the compact set
If n
gSS)}.
is
cuspidal and g is a nontrivial K(s)-invariant vector, where S is a minimal tree for n, then
is in SO(S) and u is an
L1-representation. , (3.4)
COROLLARY.
sentation
of
Let
Aut(l)
it
be a
with
irreducible repre-
cuspidal
minimal
tree
Let
S.
[s]={gs:
geAut(X)}; then S' is a minimal tree of a iff S'E[s]. PROOF.
As observed for special representations,
if S
is a
minimal tree of u then also gS is a minimal tree for every gEAut(X)
because card (gS)=card (s)
Conversely
if
S'
is
a
minimal
and n(g)P71 (S)7E(g)-1=P71 (gs).
tree
of
n
then
card(s)=
card(S')=2n; let g be a nontrivial K(s)-invariant vector and -1 be a nontrivial K(s')-invariant vector.
Then
K(s')-left-invariant nontrivial function of .(S). (3.2) implies that the support of set {geAut(l): nontrivial.
is a Proposition
is contained in the
gScq'} which is nonempty because (n(.)g,R)
is
Therefore there exists gEAut(X) such that gScq';
Cuspidal representations
101
this means that gS=S' because card(gt)=card(s)=card(S').'
Assume now, as before, that S is a finite complete subtree with diam(s)a2. Let K(it)={geAut(S):gS=mot}. Then K(s) is a normal
subgroup of k(S) and the finite group K(S)/K(,t) is isomorphic
to the group of automorphisms of the finite tree
is
It
S.
which is
interesting to observe that every function of °(St)
K(s)-left-invariant is supported on K(,T) and may be identified
with a function on K(S)/K(S).
Similarly an element of $(S)
which is K(p)-left- invariant may be identified with a function
on the cosets gK(S) which is zero on the cosets which do not
Since only finitely many
intersect the compact set {g:gSQ)}.
cosets intersect a compact set we conclude that the space of K(g)-left-invariant elements of '(S) is finite-dimensional. Before going on with the study of cuspidal representations
we want to show that for every complete finite tree & there exists at least one cuspidal representation having S as its minimal tree. In other words we shall prove that .(S)*{0}. This
will be accomplished in several steps.
We first need a few
observations. Let
be a complete proper maximal subtree of
S'
follows that S\S'ce S=V1({xo})
for some
If diam(S)=2,
and card(S\S')=q. xoEX,
It
S.
that
is
then the maximal proper complete
subtrees of S are the q+l edges of S.
If diam(St)>2 then the
maximal proper complete subtrees of S correspond, bijectively,
to the vertices of 8i ; in fact if vEB is a complete maximal subtree,
then S\{al,a2,...,aQ} are the q
where al,a2,...,a
vertices of 8S at distance 1 from v.
is maximal proper
If S'
and ge K(s) then also gS' is a maximal proper complete subtree,
in fact gf = that
the
.
inner
This proves that K(,t°)
automorphisms
K(S ),K(S2),...,K(Sj)
of
k(S)
is normal in k(S) and permute are
where
groups
the
the
complete
1
maximal
proper
subtrees
of
S.
Also
it
is
clear
that
102
Ch. III
K(1)cK(11)ck(1) for every i=1,2,...,j.
(3.5) DEFINITION.
A unitary representation of k(S) is called
standard if it has no nonzero K(11)-invariant vectors for every i=1,2,...,j. Let (K(S))" be the set of all unitary irreducible 0
standard representations of k(S) which are trivial on K(S).
The finite group K(it)/K(1)
is isomorphic to Aut(1),
the
group of all isometries of the finite tree S, and K(S i)/K(1) is isomorphic
Therefore
to
the group
{heAut(1): h(x)=x for every x511}.
(IC(s))" can be regarded as the finite set of all
irreducible standard unitary representations
of
finite
the
We prove now that (K(1))"*m for every complete
group Aut(1).
0
finite subtree S of X with diam(1)-2.
Let G be a finite group and HI,H2,...,Hi be j
(3.6) LEMMA.
subgroups of
G.
Then there exists a
irreducible
(unitary)
representation n of G, such that, for every i=1,2,...,j, it has
no nontrivial Hi-invariant vectors if and only if there exists a function f on G not identically zero such that E f(ght)=0 for hr=Hi
every g,teG and i=1,2,...,j. Let
PROOF.
M
that E f(ght)=0 heH
be
for
the
space
of
functions
every g,teG and
f
G
on
i=1,2,...,j.
M
such is
a
t
bi-invariant finite-dimensional space of functions on G. Hence
M is a finite sum of irreducible invariant subspaces.
It
is
clear that if f is a H -left-invariant function of M then f=0.
This proves that if M:*0 then there exist irreducible unitary representations of G without nonzero H1-invariant vectors for every i=1,2,...,j. Conversely,
if it
is such a representation,
then it is equivalent to a subrepresentation of AG ;
let N be the
left-invariant space of functions on G which corresponds to it.
Cuspidal representations
PA(H1)=
A(h)
IH1I-1E
the
is
103
projection
on
function.
Since
space
the
hEH consisting
of
Hi-left-invariant
nonzero Hi-invariant vectors
follows
it
has
i
no
PA(H1)f=0
that
for
every fEN. Since N is left-invariant, PA(H1)A(g)f=0 for every gEG, fEN and i=1,2.... j. Thus NcM and so M*0. This proves the
lemma.' (3.7) LEMMA.
Let G be a finite group and H=HixH2x...xHjcG the
If the inner
direct product of j nontrivial subgroups of G. the subgroups
automorphisms of G permute
then
H1, H2,...,Hj
there exists an irreducible (unitary) representation
of G
it
such that, for every i=1,2,...,j, it has no nonzero Hi-invariant vectors.
PROOF. First, we prove the lemma in the special case H=G; by Lemma (3.6) it is enough to prove that there exists a function
E f(ht)=0 for every teG
f*0 on H=Hix H2x....x Hj such that
hEH i
and i=1,2,...,j because, for every i Hi is a normal subgroup of
H and so gHit=gtH1. Let E1={x1' yi} where x1, y1EH1 and xi*yi for every i. Let E=E1xE2x... xEj and for vEE let N(v) be the number of
xi
occurring
in
the
coordinates
of
define
We
vEE.
for vEE and f(v)=0 elsewhere. If g=(a1 a 2 ...,a j ) then the coset H g={ (a , a2 ,. .. , a i-I, h, a i+i, ... , a ) : hEH 1. If f(v)=(-1)Nv
j
1
i
there exists k,
identically zero on H1g.
is
i
1:sk<j k*i, such that ak*Ek then H1gnE=8 and f
HgnE={(a ,..,ai-1 ,xI 1 1
akEEk for every k*i,
If
a 1+1 ,..,a j ),(a ,..,a i-1 ,y 1
1
a 1+i''
then
.,a j)}.
Since
N((ai, .. , ai-i' xi ai+1'
.
, a aj ))=N((ai, .. , ai-1, yi, ai+i' ..,a ))+1, J
it follows that E f(ht)=0 for every t and i=1,2.... j. hEH
This
i
proves the lemma in the special case G=H. (3.6)
it
is
enough
to
find
a
function
Finally,
FOO
on
by Lemma G
such
104
Ch. III
that E F(ht)=0 for every tEG and i=1,2,...,j. This is because hEH i
the assumptions imply that every set gHit is a left coset of Hk
for some k (depending on g and
1).
defined
H=H1xH2x...xHj,
above
for
group
the
Let f be the function then
F=fxH
satisfies the requirements.'
(3.8) LEMMA. Let S(q+1) be the group of all permutations of the set (1,2,...,q+1)
(q+1>-3). Let S(q) be the stability subgroup
of a point of (1,2,...,q+1}.
Then there exists a (unitary)
irreducible representation of S(q+1) which has no nontrivial S(q)-invariant vectors.
PROOF. This is a consequence of a standard result in the theory
if a
of finite groups [I,Corollary 5.17] which asserts that
finite group G with IGI>2 acts doubly transitively on a set X
and H is the stability subgroup of a point of X then there exists
an
contained
representation
irreducible in
the
quasi-regular
of
G
which
representation on
Is
G/H.
not
We
include a direct proof for S(q+1) for completeness. First, we
if G is a finite nonabelian group and H is a
observe that
subgroup of G such that every irreducible representation of G has a nontrivial H-invariant vector then IGI/IHI>IGI
(where G
This is a consequence of the fact
is the dual object of G).
that the irreducible representations of G having a nontrivial H-invariant
vector
irreducible
subrepresentations
sentation the
AG/H,
up
are,
equivalence,
to
of
the
exactly
quasi-regular
the
repre-
that is the representation which corresponds to
left-invariant
space
consisting
of
H-right-invariant
functions on G (see the proofs of Lemmas (2.4) and (2.5), or the Frobenius reciprocity theorem
AG/H a ^nGa vector.
If
where no,-1 n a 1
for
[I, Lemma 5.2]).
Therefore
iff o has a nontrivial H-invariant every
o
then
IGI/IHI=
E oeG
Cuspidal representations
E
Since G is nonabelian there exists a o' such
IGI.
and so
that
Therefore
to
105
prove
This
IGI/IHI>IGI.
lemma
the
proves
suffices
it
to
the
claim.
show
that
IS(q+1)I/IS(q)I=(q+1)!/q!=q+1:sIS(q+1)I. This is obvious because
G corresponds, bijectively, to the set of conjugacy classes of G;
if ujES(q+1),
j=0,1,...,q,
is
a permutation which fixes
exactly j points of {1,2,...,q+1} and CJ is the conjugacy class of uJ ,
then CInCJ=m for i*j and so IS(q+1)I>q+l, and the lemma
follows.'
For
THEOREM.
(3.9)
every
complete
finite
subtree
S
with
diam(S)-2 the space (K(S))-#m. In particular Y(S)*0. 0
PROOF.
If diam(s)=2 then S=V1({x0}) for some xoEH;
K(3)/K(S)=
Aut(S)=S(q+1). The maximal proper complete subtrees of & are
the q+1 edges of S. Aut(S) acts transitively on the edges and are conjugate to each other.
so the groups K(SI)/K(S)=S(q)
Since v is H-invariant iff u(g)v is gHg 1-invariant, by Lemma (3.8)
Let now T be a complete finite subtree with
(K(S))"*m. 0
diam(j)>2. Let
be the subtree of S consisting of vertices of
homogeneity q+l.
As in the proof of Lemma (3.1),
K(30)/K(S)
decomposes into the direct product of its nontrivial subgroups
are the maximal proper complete
K(Si)/K(S) where S1'
subtrees of S. Since the inner automorphisms of Aut(S) permute the
subgroups
by
K(Si)/K(S),
Lemma
o'E(K(S))o and eE3f01 a*0; then (o(.)
,
(3.7) )
(K(S))o*o.
Let
is a K(S) -bi -invariant
nontrivial function on k(S) such that the right-averaging over K(sI)
is zero for every i.
Because K(S)cK(SI)cK(S)
that the function defined by f(x)=0 elsewhere is in .DV(S).
it follows
for xEK(S)
and
Thus .`'(S)*0 and the theorem is
proved., Let
At(S)
be
the subspace of L2(Aut(E))
consisting of
106
Ch. I I I
K(it)-right-invariant functions such that the right-averaging
over KW )
is zero for every complete proper subtree S'
of
jT
(or, equivalently, for every maximal proper complete subtree mot'
of
S).
We
have
that
and
Y(S)cAt(S)
A(S)
is
a
closed
left-invariant nontrivial subspace of L2(Aut(3E)). Let AS be the
subrepresentation obtained by restricting to
At(s)
the
left
regular representation of Aut(X). Obviously PA(S')At(S)cY(S) for
every 3'. (3.10) PROPOSITION. Let xS,S, be the characteristic function of
the set {gEAut(X): gS cS'}.
If
and S'
jT
are complete finite
subtrees of 3f with diam(s)-2 then
(1) for every fEAt(S), PA(S' (S')f=fx
S'S'
C (Aut(X)) is the space c continuous functions with compact support on Aut(T);
where
(ii) At(S)nC (Aut(X))=S(S) c
of
(iii) `.°(S) is a dense subspace of Ai(S). PROOF.
By Proposition (3.2),
supp PA(3')fc{gEAut(3f): gScS'}.
Therefore it suffices to prove that PA(S')f(g)=f(g) if gTcs'. We have that PA(s')f(g)= (m(K(S')))-1J f(tg)dt _(S (m(K(,T')))-1I
If
gscs'
then
)
(m(K(S')))-1I
f(gt)dt. JK(g 1S')
f(gt)dt = g-1K(s')
Scg 133'
K(s)-right-invariant,
'
f
and is
K(g 1S')cK(S). constant
on
Since
f
is and
gK(g 1S')
m(K(g 1S' ))=m(K(S' )). It follows that PA(3')f(g)=f(g). To prove (ii) it is enough to show that every function fEAt(S)nCc(Aut(X))
is K(S')-left-invariant for some complete finite subtree
St'.
Since K(s) is compact open there exist g1,g2,...,gnEAut(3f) such n
that supp fc U giK(s); i=1
this means that the set U he supp f
hS is
Cuspidal representations
107
n
finite because it is contained in U giS. In particular there 1=1
exists a complete finite subtree S'
U
such that
hScS'.
he supp f
This implies that supp fc{gEAut(X): gTcS'}.
By
(i)
f=fxS,S,=
PA(S')f; this proves (ii).
Let
a fixed vertex
x
be
d(y,xo)-n}.
Let
o
E ={gEAut(l): n
of
1
g'TcSn }
and Sn=Vn ({xo})={yE.l: and for fEAt(1), let
fn fxE. By (i) fn PA(Sn)fE`.°(s). For every n, En is compact open n and
m
EncEn+1,
U En Aut(X).
This
implies
fn- f
that
in
n=1
L2(Aut(X)) and the proposition follows.'
Proposition
implies
(3.10)
that
a function fEM(j)
is
In particular the functions of
K(S)-bi-invariant iff f=fx K(it)
At(s) with compact support in gK(S) are exactly the functions
{fx
_ : gK()
fEht(S)}=A(g)({fx
:
fEft(S)}) because At(1)
is left-
K(S)
invariant. This proves that every function in
'(5)
is a finite
sum of left translates of K(S) -bi -invariant functions of .°(5) n
because supp fc U gik(S) for some g1192' ... ' gn. 1=1
The following
results
two
are
proved exactly as
corresponding results were proved in Section 2
the
for special
representations.
(3.11) LEMMA. Every closed left-invariant nontrivial subspace of M(5) contains a nontrivial K(S)-bi-invariant function.
PROOF. (See Lemma (2.4).) 1
(3.12) LEMMA. Every cuspidal irreducible unitary representation
of Aut(E) with minimal subrepresentation of AS
tree S is unitarily equivalent .
to a
108
Ch. III
PROOF. (See Lemma (2.5).) 1
As in Section 2, Lemmas (3.11) and (3.12) and the fact that the subspace of .(S) consisting of K(S)-bi-invariant functions
is finite-dimensional imply that S is a minimal tree for each
subrepresentation of
AS
that
and
irreducible subrepresentation.
AS
is
Moreover,
a
finite
sum
of
up to equivalence,
the cuspidal irreducible representations of Aut(3f) with minimal
tree & are the inequivalent irreducible subrepresentation of AS.
If
then AS and AS, have no common components (see
Corollary (3.4)), while AS is unitarily equivalent to A9s for every gEAut(X). We prove now that the map
o' -*
I nd
o-
K(S)TAut(X)
is a bijection from (K(S))" onto the classes of inequivalent cuspidal
irreducible representations of Aut(X)
with minimal
tree S. We recall, briefly, a few basic results concerning induced
representations of groups in the special case of unimodular separable locally compact groups and compact open subgroups (for proofs and more details about induced representations we refer
the
separable
reader
to
[H]
locally compact
subgroup of G.
and
[G]).
Let
G be a unimodular
group and K be
a compact
open
Let o be a unitary representation of K; leto
be the space of functions f: G -4 X0' such that (i) f(xh)=
1)f(x) for every xEG and hEK
(ii) J Ilf(g) II2dg < +-. G
We have that J Ilf(g) ll2dg= m(K) E G
11f (g) Ill and
xEG/K
space with the following scalar product
517
is a Hilbert
Cuspidal representations
(f,g)=
(f(x),g(x))dx
f
109
for f,gEb°'.
JG
Observe that b°, is invariant under left translations. The unitary left regular representation of G on 5°' is called the
induced representation of G by o and is denoted by Hence
02 then also
is unitarily equivalent to
If
lt).
is unitarily equivalent to
Ind(o-1)®Ind(o2). Let ge3fo; for xEK and fe(W)=0
It
IIf1I=IIeII m
for
xVK.
It
and
Ind((r 2)
we define f(x)=o-(x follows
that
f9E)°
and
is easy to see that
for xEK and
This
for xtK.
proves that every coefficient of o- is a coefficient of Ind((-) with support in K. In particular this proves that if are
inequivalent
inequivalent.
If
then
also is
Ind(o-)
and irreducible
and o2 are
Ind(o2)
then
is
it
an
L1-representation and it has a dense subset of coefficients with
compact
support;
formal
its
In fact if Eelfo,, d
inda'=(J
dimension
equal
is
to
IIEII=1 then
I (0.(g) g, g) I2dg)-1= [m(K) J I (0_(g) g, g) I2(dg/m(K))
]-1
I
K
= d /m(K) = dim((Y)/m(K).
o
0 Let 5K be the subspace of 5°1 consisting of function with
support
in K.
It
is clear that bR is a closed nontrivial
subspace of
the map ->f
The map
is the projection of b onto bK. This implies
that every function of
ba,
is an isomorphism of 3f onto bK.
with compact support
is a linear 01
combination of left translates of functions in bR
(see the
remarks after Proposition (3.10) ). In particular the subspace §o For completeness of generated by U A(g)og is dense in gEG
we include here the proof of the following proposition.
110
Ch. III
(3.13) PROPOSITION. Let o- be a unitary irreducible representation of K; then closed
is irreducible if (and only if) every invariant
nontrivial
subspace
bo
of
contains
a
nontrivial function of bR PROOF.
It is enough to prove that if M is a closed nontrivial
invariant subspace of b°, such that b M4x0 then M=b°y. Since M is
closed and
suffices to prove that b0 CM
it
because, as observed, the subspace generated by U gEG
is dense in 5°. Let N={f(1G): fEMnb°}. We recall that if febK then f=0 iff f(l )=0; so MnbT#0 implies that N*0. Moreover, for
every kEK we have that o-(k)f(1G)=f(k this
means
and
that
nontrivial o'-invariant subspace of 3fT.
it follows that N=3e
and N
is
a
Since (Y is irreducible
Let gEb0; then there exists fEMtt0 such
.
that f(1G)=g(iG); therefore f(k)=g(k) for every kEK and so f=g.
This means that bTcM and the proposition follows.,
(3.14) THEOREM. Let 3 be a finite complete subtree of 3: with diam(s)a2.
Let
o=
then
0 0
equivalent
to
is unitarily
A
`
O'E(-K (3))"
For
Ind(o-)= e
every
0 E(K(ST))
is irreducible and the map o----Ind(or) is a bijection from (K(S)Y0 onto the inequivalent irreducible oe(K(S))0",
subrepresentat ions of AS
,
that
is the classes of distinct
irreducible cuspidal representations of Aut(X) with minimal tree
PROOF.
S.
The
formal
dimension
of
is
equal
to
Let M be the subspace of .(j) consisting of K(St)-bi-
invariant functions. Then M is a nontrivial k(S) -bi -invariant
subspace of functions with support in k(S) (by (3.2)). Let oS
be the subrepresentation of A
relating to M. i(s)
Then M is
Cuspidal representations
finite-dimensional and so
111
is a finite sum of irreducible
representations. Finally, oS is standard and it is trivial on so
K(S),
every
irreducible
Conversely,
(IC(s))
if
trivial on K(s),
(S
is
in
is a standard subrepresentation of
o-
0
A
subrepresentation of
then the left-invariant subspace of
K(it)
L2(K(S)) which corresponds to o is contained in M and so o' is contained in oS; in particular
0
Since = ®
A K(it)
(dim(n))n
'tE(Kq))^
it follows that o = e
We prove now that A
)o-.
is
o
o unitarily equivalent to
be the closed
Let
Ind(o-).
xK(S)
o subspace of
consisting of functions with support in xk(S).
o We
put
Uf=A(x)[f(x)]
for
S
fE
(we
recall
that
xK(S)
f:Aut()-M).
In
particular,
Uf
is
a
function
in
°(S)
which is K(xS)-left- invariant. This operator is well defined,
in fact A(x) [f(x)]=A(xk) [f(xk)] o bijective operator from 5
if kd(S).
U
is a linear
onto the subspace of K(xS)-left-
invariant functions of .(S) which is equal to A(x)M. In fact if
BEM the function f(xk)=A(k 1)g for kEK(S) and f=0 elsewhere is
o a function in 5
such that If=A(x)g, which proves that U is O,
surjective. The functions in 5 &
depend only on the value in
xK(St) x,
so Uf=Ug implies that f(x)=g(x)
and U is
injective.
We
choose the Haar measure of Aut(l) in such a way that K(it) has
112
Ch. III
measure equal to 1.
every
function
In this way n is isometric. We recall that
of
°(S)
a
is
combination
linear
of
is dense in M(&). On
K(xS)-left-invariant functions and .9(S)
o S
the other hand the space of functions of 5
with compact
o support is dense in 5 S and every such function is a finite sum
o
o
o1
of functions in
_
.
n
then
If xK(,T)*yk(s)
={O} yK(S)
xK(S)
xK(ST)
Q and so
it
S
extends to a unitary operator from 5
intertwining
Ind(o3)
and
Under
AS.
onto Af(S)
action of
the
It,
the
condition of Proposition (3.13) is exactly Lemma (3.11), and hence Ind((Y)
The rest of
is irreducible for every oE(IC(S))".
the statement follows from the previous remarks. ,
Putting together the results of Chapter II and Section 2 for G=Aut(X), we have a complete classification of all unitary irreducible representations of Aut(X).
We observe now that Aut(X) is a type
group. A unitary
I
representation it is called admissible if the spaces Rit(.) are
for every
We prove
all
unitary
irreducible representations of Aut(X) are admissible.
Indeed,
finite-dimensional
if
S
is
a
minimal
tree
S.
of
K(s)-invariant vector, then the
it
and
C
that
is
map 'q-(n (.)
a
, ii)
nontrivial
Is a linear
injective map from RX in the space of K(s)-right-invariant functions, as is easily seen because
it
is irreducible and g
is a cyclic vector. In particular if it is a special or cuspidal
representation then, for every S',
If n(S')
is isomorphic to a
subspace of the space of K(S')-left-invariant functions of `,°(S)
which is finite-dimensional
(remarks after Lemma
(2.1)
and
Corollary (3.4)). The proof of the fact that every spherical irreducible representation of Aut(f) is admissible is similar.
d(x, o)sn} and, with the terminology and Indeed let S9X ={xe3f: n
Cuspidal representations
notation of (11,5), let V={gEJ((S2):
Then V has dimension
(q+1)qn-1.
113
it (k)g=g for every keK(X )}. z n Therefore it is admissible. z
Since every spherical representation is equivalent to
for
z
some zeC, such that -lsµ(z)51,we conclude that every spherical representation is admissible.
These facts imply that n(L1(Aut(X))) contains a nontrivial
indeed if S is a minimal tree of
compact operator,
for f=(m(K(S))-1%K(T)
OOPH(S)=n(f)
By
we
remark
that,
in
the
then
(D1,13.9.4] Aut(X)
postliminal and so Aut(X) is a type I group [DI,
Finally
71
is
18.8.2].
terminology
of
[DI,
13.9.4], Aut(3f) is liminal, that is n(f) is a compact operator for every fEL1(Aut(X)). This is a consequence of the facts that the linear space F generated by the characteristic functions of
the sets gK(S'), for every geAut(T) and every complete subtree
3t',
is dense in L1(Aut(X)), and it(f) is a compact operator for
every feF because it is admissible. In particular the closure of x(L1(Aut(l))) in the norm operator topology is . P(R1) the space
of all compact operators on ItIT. Every spherical
irreducible or
representation
square-integrable;
of
Aut(X)
therefore
the
is
either
Plancherel
spherical formula (II,(6.5)) and the expressions for the formal
dimensions of special and cuspidal irreducible representations of
Aut(g)
yield
the
[D1; 18.8.2, 18.8.5]. Indeed,
Plancherel
let
formula
[[]={gS: geAut(X)};
of
let
Aut(l) 71+,
'a
be the special representations of Aut(l). We choose the Haar measure of Aut(X) in such a way that the stability subgroup of a vertex has measure equal to 1. Therefore for every continuous function f on Aut(X) with compact support we have
114
Ch. III
IlfII2-
nii2+it(f)11122 J
dm(t)
L(12)
J
+[(q-1)/2][tr(n+(f)n+(f)*) + tr(n (f)n (f)*] E
+
(m(K(s)))-1[ E_
di am(3t)a2
4. Notes and Remarks.
The content of the present Chapter is
basically due to G.I. 01'shianskii and is taken from [02]; in particular Proposition (3.2), which is the crucial step for the
classification of all cuspidal irreducible representations of Aut(X),
is due to G.I. Ol'shianskii
classification
of
algebraic,
who gives
admissible
and
in
[02]
the
algebraically
irreducible representations of Aut(3f) in a vector space V (a priori with no topology). Our exposition however is more direct
insofar as it does not make use of the notion of admissible representation.
We
have
chosen
also
representations for a group more general
to
treat
special
than Aut(Z).
Our
treatment applies for instance to PGL(2,a). We now give a brief description of 01'shianskii's original
approach and the connection between the present chapter and
[02]. A representation n in V is called algebraic if each vector
v is K(s)-invariant for some S depending on v.
An admissible
representation it is such that the subspace of K(s)-invariant
vectors is finite-dimensional for every fixed complete finite
Notes and remarks
subtree S.
115
Finally, n is said to be algebraically irreducible
if there are no nontrivial invariant subspaces of V.
We have
observed that if a is a unitary irreducible representation of Aut(X)
in a Hilbert space 3{n then Vn is a dense nontrivial
invariant subspace of Hn (see
(1. 1) )
and it
follows
that
the remark after Definition
is admissible. On the other hand by every
algebraic,
it
[02]
algebraically
admissible,
irreducible representation of Aut(3f) is unitary. Let now n be a
unitary
representation
irreducible
in
let
3fn;
the
be
no
restriction of it to the dense invariant subspace Vn.
We have
that no is an algebraic, admissible, algebraically irreducible representation
of
Aut(l)
(in
fact
if
M
a
is
nontrivial
is
invariant subspace of Vn then M is dense in R because
it
irreducible; therefore Pit (S)M is dense in HH(S)=Pn(S)3fn;
this
71
means that Pn(S)M=3fn(S) for every S because Hn(s)
is finite-
dimensional. Since for every & and vEVn, Pn(S)v is a finite linear
combination of
n(k)v with kEK(S),
it
follows
that
P71 (S)McM for every S and so VX =M). Also if n1 and n2 are two unitary
irreducible
representations
then
ni
is
unitarily
equivalent to n2 if and only if ni is algebraically equivalent
to n2, that is iff there exists a linear isomorphism T of Vn 1
onto Vn such that Tn1(g)=n2(g)T for every gEAut(3f). In fact if 2
such a T exists, then T(3fn (S))=Hn (S) and TPn (s)=Pn (S)T for 1 2 1 2 every finite complete subtree S (we recall that if v is
K(s')-invariant for some S' then, without loss of generality, we can suppose that ScS' and so K(S') has finite index in K(it),
therefore P71 (S)v is a finite linear combination of n(k)v with kEK(S)
and TPn (s)v=Pn (S)Tv). 2 1
This
proves
vn -30 and Tvn -w then w=0 (in fact Pn2 so Pn (S)w=O for every mot). 2
that
vnEVn
if
1
=TP
n
n1
--)00
n
and
By [Y, Prop.2, p.77] there exists a
116
Ch. III
closed operator T defined on a domain D which includes Vit
and 1
which agrees with T on Vi .
In addition Ta1(g)=n2(g)T on D. The
operator T may be uniquely written as the product T=US where U
is a partial isometry and S=I T
T.
The irreducibility of nl
and n2 implies that U is a unitary operator intertwining x1 and
n2 (see also [JL, Lemma 2.6, p.30]). It follows that the classes (up to algebraic equivalence) of
algebraic,
sentations
admissible,
of
a vector
in
Aut(lf)
algebraically
irreducible
space
repre-
exactly
are
the
representations 71 where n varies among all classes (up to of unitary
unitary equivalence)
(topologically)
irreducible
representations of Aut(fl in a Hilbert space. For instance, the
invariant irreducible subspaces of ht(S)
are the closures in
L2(Aut(X)) of the algebraically irreducible invariant subspaces
of `.'(J). A similar result is true for special representations of a closed subgroup acting transitively on X and on Q.
We recall that G.I.O1'shanskii in [01] proves that Aut(X) is a type I group.
Observe
that
O1'shianskii's
to
the
classification
other
notable
effectively representations
of
method
apply
does
not
of
irreducible
subgroups
of
Aut(3E).
As
noticed at the beginning of Section 3, Proposition (3.2) does not apply to PGL(2,6) because it does not contain sufficiently many
rotations.
Thus
representation may
be
while
the
formally
notion
considered
of
"cuspidal"
for
any closed
subgroup of Aut(X), when the group considered does not contain
sufficiently many rotations, representations"
fail
to
it
have
may happen that coefficients
with
"cuspidal
compact
support. This is exactly what happens for PGL(2,6) since there
exist nonspherical irreducible representations which are not square-integrable
(the
nonspherical
principal
series).
The
Notes and remarks
117
classification of irreducible representations of
be achieved using the machinery of [GGP]
or
may
[Si].
Of course
spherical and special unitary representations of PGL(2,3) are exactly the restrictions of the corresponding representations of
Aut(fl M.
principal
But
representations
the
series
cannot
be
obtained
of as
the
nonspherical
restrictions
of
irreducible representations of Aut(f).
On the other hand Ol'shianskii's ideas may be applied to classify the irreducible representations of the group Bw which
fixes a point w of the boundary and stabilizes the horocycles
with an appropriate choice of w
[N4]. Observe that
is the group of matrices
with a,bE3 and
a-il
la'=1
(see
LO
Appendix Proposition (5.6), below). We
remark
that
cuspidal
and
special
representations,
unlike spherical representations, do not restrict irreducibly
to any discrete subgroup of Aut(X); the reason being that no irreducible representations of an infinite discrete group may be contained in the regular representation [C F-T]. A.M.Mantero
and A.Zappa [MZ2] also study the same restriction to a free group
proving
a
very
weak
form
of
The
irreducibility.
decomposition of the restriction of special representations to a simply transitive subgroup of a tree was analyzed in full by
G.Kuhn and T.Steger [KS1]. We summarize the results on irreducibility of restrictions of representations of Aut(3f) as follows.
(1) Spherical representations restrict irreducibly to any closed unimodular subgroup acting transitively on X [B-K1] and
to lattices (except for the exceptions indicated in Theorem (11,7.1) [CS] [St2].
(2) Special
representations
restrict
irreducibly
to
a
closed subgroup acting transitively on X and 11 (see Section 2).
(3) Cuspidal representations do not restrict irreducibly,
118
Ch. I I I
in general, to a group acting transitively on X and Q. We know
for instance that some cuspidal representations must reduce Indeed every coefficient of the
when restricted to PGL(2,3).
regular representation of PGL(2,3) coefficient [Hz2].
of
the
regular
is the restriction of a
representation
of
Aut(X)
[Hz1]
On the other hand one can consider a subrepresentation
of the regular representation of PGL(2,6) which is a direct integral series.
of
representations
This representation
representations
can
therefore
of
nonspherical
the
orthogonal
is
and
to
the
only
be
obtained
principal
the spherical
representations
special
by
to
restricting
and
cuspidal
representations.
Unitary representations of the group of automorphisms of a (or semi homogeneous)
homogeneous
tree which is not
locally
finite have also been studied by 01'shianskii [03]. This group is not locally compact but turns out to be of type I.
The full Plancherel formula for Aut(3E) is given here for the first time as a consequence of Olshianskii's work and the spherical SL(2,a)
Plancherel was
given
formula.
by
The
P.J. Sally
Plancherel
and
formula
J.A. Shalika
for [SS].
C.L. Gulizia [Gu] made use of the full Plancherel formula to prove the Kunze-Stein phenomenon for SL(2,a) where 3 is a local
field such that the finite field O/P has characteristic p#2 (see Appendix below). Notice however that only the spherical Plancherel formula is needed for the Kunze-Stein phenomenon as
shown in [N3] where the Kunze-Stein phenomenon is established for any closed noncompact subgroup acting transitively on Q and therefore
also
for
PSL(2,a)
without
restrictions
on
the
characteristic of D/P.
A curious fact
is
that
the Kunze-Stein phenomenon
is
equivalent to transitive action on 0 when q+1=3 [N4]. We thank J. McMullen who made available to us his notes
[MIA] on 01' shianski i' s work.
119
APPENDIX
1. p-adic fields. We will define in this section a class of locally compact fields which can obtained by completing the field of rational numbers with respect to a metric different
Let 0 be the field of rational
from the ordinary metric.
the set of nonzero elements. Let p be a fixed
numbers and 0 prime.
If
then
xEQi
x=phm/n,
where
and
n>O
h,m,nEZ,
(m,n)=(n,p)=(m,p)=1.
In other words m and n have no common divisors and neither of them
is
divisible by
p.
Such a triple h,m,n
and therefore the map
identified by x,
is
x---) Ixl =p
-h
uniquely
is well
P
defined for xsO . We complete the definition by letting 101 =0. P
The number Ixl
p
is sometimes called the p-adic norm or modulus
of the rational number x.
It is not difficult to show that the
norm satisfies the following properties: IxIPlylp
(1)
;
Ixylp (ii)
Ix + yIPS max(IXIp,IyI ).
Property (ii)
is called the ultrametric inequality
It
.
obviously implies that (iii)
Ix + YIPS Ixlp+ IYlp.
It follows that the norm induces a metric d(x,y)=Ix-yl
P,
with respect to which the field operations in 0 are continuous.
As a consequence the completion of 0 with respect to this metric
{xE)p:
is
a
field
which
is
denoted
by
Op.
The
sets
Ixl P
Notice that the integers ZcQp have norm less than or equal to 1,
and that the integers not divisible by p have norm 1. The
closure of the integers in Op is denoted by Zp. We will show
Appendix
120
now that Zp is compact and that Zp={x:
To prove the
s1}.
lxl P
is sufficient to show that Z is totally
first assertion it
bounded in Op. Indeed, let c>O and p n
let
lx-klP
If hEZ, then, for some k=0,1,2,...,p-1, h-k=0 mod pn. Therefore p-1
heS(c,k). This shows that Zc U S(c,k) and that Z is totally k=0
To show that
bounded.
Zp={x:
lxl
si},
lxl Psl
let
p
x=prm/n with (n,m)=(n,p)=(m,p)=1, n>0 and r>-0.
xEO;
then
We prove that
given c>0 there exists an integer at distance less than c from x.
and observe
p k
Let
invertible modulo pk. such that hn=1 mod pk.
lxl Ps1}.
Zp={x:
that
(n,p)=1
implies
that
n
is
Therefore there exists an integer hEZ
Observe
spr-k
lx-phml
Thus
that
This means that
p
{x: lxl spn}={x: lxl
because the norm only attains the values pn, nEZ. Therefore the
base for the neighborhood system of zero consists of compact
In conclusion Op
open subsets.
totally disconnected
a
is
locally compact field of characteristic zero.
Let Sp be the
2. A locally compact field of characteristic p.
with a EZ(p),
ring of formal power series E an Xn,
n
the prime
n2ts
field
of
order
representing
p.
the
We
class
denote
by
a EZ(p). n
a
the
n With
Osa
integer
this
map E a Xn- E anpn is well defined with domain Sp and range n
nzs
nl-"s
Indeed the series
contained in Op.
anpn converges in Op. n?s
Furthermore every element of Op can be expressed as a series
For the proof of this fact we may assume that
x= E a
pn. n
n;-Is
lxlp 1.
{ xEZp:
choose
Suppose
Then 1 x - k 1 psp
for -1
uniquely by
kE{O,1,2,...,p-1}
the
spheres
S(p ,k)=
} are disjoint and cover Zp. Therefore we can a 0e(O,1,2,...,p-1},
induction hypothesis
such
that
that
lx-ii 0lpsp 1.
numbers a0a1...,aE n
{0,1,2,...,p-1} have been found to satisfy
lx- E ajpjlp
A locally compact field of characteristic p
121
then
n
x- E ajpj= pn+1y, with IyIPS 1. j=o
But y belongs to one and only one of the spheres S(k,p-1 ), in other words there exists an+1e{O,...,p-1} such that Iy-an+1l P
ix - E a pi I p=
ypn+1_
I
J =o
an+1 pn pI< pn-2
E an pn is surjective.
In conclusion the map E an nzs
We have
nzs
also proved that it is injective because the construction above a e{O,1,2,...,p-1} and each a n n identifies only one element of the residue classes modulo p. A
a
gives
unique
power
formal
sequence
series
can
given
be
norm
the
I E anX°IP n=s
i E anpnI=P8.
The map defined above becomes in this fashion
n2ts
an isometry and Sp is a complete metric space with respect to
the distance map E na Xn nzs
Ix-yip.
It is important to keep in mind that the
E a pn is not a homomorphism. The operations of nzs
n
sum and product in Sp are those of the ring of power series with coefficients in Z(p). In particular Sp has characteristic p.
On
the
series E a
npn
contrary
on
operations
the
convergent
the
elements a
are performed considering the
n
nzs
as
in Sp satisfies
ordinary integers. Nevertheless the norm I.I P
the
ultrametric
inequality
< max(IxIP,IyI
Ix+yl P
P
immediately clear from the definitions. property
),
as
is
multiplicative
The
IxylP=IxIP IyI P is also evident. This means that Sp is
a locally compact ring, which is complete in the metric defined
by I.I
We will presently show that Sp is a field.
P.
Observe
first that elements of the type a e, with neZ, are invertible, the
inverse being simply a
1X
n
n,
n
where
a-1 a =1
n
n
in Z(p).
It
suffices therefore to prove that every element of the type a xn
x=1+
n
na:1
is
invertible.
But,
for
such
an
element
x,
Appendix
122
w I1-xI < p 1, and therefore the series E (1-x) k converges in Sp, p k=0 n
I(1-x)n+il < p n
Ix E (1-x)k_lI =
and
p
p
k=0 OD
This shows that E (1-x)kESp is the inverse of x.
We have
k=0
now proved that Sp is a totally disconnected locally compact field of characteristic p.
3. Locally compact totally disconnected fields.
The fields Op
and Sp are not the only examples of locally compact totally disconnected
fields.
However,
nondiscrete
all
totally
disconnected fields are finite algebraic extensions of Op or Sp.
A precise statement of this result will be given later
without proofs
.
We shall now describe a locally compact field
in general. For complete statements and proofs we refer to [W].
Observe first that every field is locally compact in the discrete topology. We exclude this trivial case and from now on we shall assume that a is a nondiscrete locally compact field. s
We
let
denote the multiplicative group of nonzero
elements; let m be a translation-invariant measure on a,
that
is a nontrivial Borel measure m such that m(E+x)=m(E) for every
xEa. If xES therefore,
then m (E)=m(xE) is also translation-invariant and x
by the uniqueness of the Haar measure on locally
compact groups [WI], mx(E)=Ixlm(E), where 1. I
is a positive real
number. The definition of 1.1 shows that
(1)
Ixyl = IxIlyl
(2)
1.1 is a continuous function on 3
>
We extend the definition of
1. 1
.
to all of a by letting
101=0. It is not difficult to show that (3)
{xE5:
lxlsl} is compact,
and that indeed the sets {xEa:
IxI`r} are a base for the neigh-
borhood system of zero. As a consequence of the compactness of {xE
:
lxlsl} and the continuity of 1.1, one obtains
Locally compact totally disconnected fields
123
lx+yl s A max(lxl,Iyl),
with
A=sup{lx+11:
conclusion that
6=02
or
26=C.
leads
A>1
In the first case
in the second case
absolute value of R,
x=a+ib, xeC and a,
condition
The
lxl:51}.
jxj
to
the
is
the
where
Ixl= a2+ b2,
bEIR.
We shall assume from now on that A=1, which implies the ultrametric inequality : Ix+y1 < max( xl,Iyl) < jxj + lyl.
(4)
Observe that
and therefore I-xl=lxl. This
I-1I2=I(-1)21=1,
implies that
Ix1=I(x+y)-yl:5max(Ix+yl,Iyl) As a consequence
1x+yl = lxl
(5)
if
lyl <
1x1-
Another consequence of the ultrametric inequality is that
is a discrete subgroup of R+.
the set B={lxl: xE3*}
Indeed
suppose that 1 is not an isolated point in B; then there exist xnE
such that
Ixn1*1
We may
for every n and lim Ixn1=1.
assume (taking if needed xn1) that Ixn1>1 and that Ixn1:52.
Since {x:
IxI:52}
is compact,
there exists aE
lim xn a. Clearly lim Ixnl=lal, that is
is a neighborhood of a. x E{a+x:
lxlsl},
la1=1.
such that
But {a+x:
IxI:51}
Therefore for n sufficiently large
and therefore by the ultrametric inequality
lxnl:sl, contrary to the assumption that lxn1>1. It follows that
the sets Ur={x:lxlsr} are open and compact.
Therefore 3
is
totally disconnected. A discrete subgroup of 0t+ is generated by
a positive number q. In other words, for each xE
,
there is
nEZ such that lxl=qn and
{x: (3.1) DEFINITION. P={xE3:
Let
Ixl:sgn}= {x: lxl
1xI:51},
B1={xES:
lxl=1}
and
lx1<1}.
We observe that 0 is compact and open, that it is closed
under addition and multiplication and that
it
contains all
Appendix
124
IxnI=Ixln
elements such that
compact maximal subring of and
Ixl:sl
If xE0 is invertible in 0 then
, .
and therefore xE01.
1I=Ixl-1--1,
Ix
is bounded. It follows that 0 is a
invertible in D.
Ix-1 1=1 and therefore x is
IxI=1 then
R+
(qn: nEZ} be the multiplicative subgroup of
if
Let
onto which I.I
Let pE0 be such that
and suppose p>1.
maps 3`,
Conversely,
lpl=q
1.
Then
P=p0, and ) is a unique maximal ideal of the ring D. It follows
that 0/9=f is finite and, if we let m denote the Haar measure,
m(p)=m(p0)= Iplm(0)=g1m(0). Therefore q is a positive integer
which is exactly the index of P in D. Since 0/P is a finite field, q=ph for some prime p and positive integer h. An element pE0 such that Ipl=q 1 is called a prime element. Let
fixed elements
be
1,2 ,...,
q
different classes of 0/u. show that every xE n
x= E anp
where
n>s
I
Ix- E a
as+k+1 e{
9 1' 2'
-s-k-1 -s-k-'
Indeed
pn I s+n
s+k
n
such
a
s
and
that
implies
that
q
a
then
In other
}.
have been
s+1,...,as+k
ix- E
therefore
p
there
s-k-1 (x-s+k Ea n=s
s +k+1
mod P which
IxI=q $;
s-k
E as+np )E0 n=s
.. , g} q
let
where a s E{
Suppose that
n=s
p
We will
mod p.
q
mod P
s+k
Then
In other words 91
anE{1,
Ix-psasI:5q s-1
found such that
belonging to
0
can be expressed uniquely as a power series
sxI=1 and ps=as
words
of
a
s+npnI
exists
pn) =a s+k+1
s+n
This proves
n=s
that x= E a npn. n?s
We now state the structure theorem for locally compact totally disconnected field [W].
(3.2)
THEOREM.
Let 6 be a nondiscrete totally disconnected
field. If 6 has characteristic p>0, then 6 is isomorphic to the
field of power series E a e with coefficients in a finite n?s
n
Two-dimensional lattices
field of order ph.
125
The modulus of the series E a XI' with a *0 n?s
is
If 6 has characteristic zero
phs.
n
s
is a finite
then 3
algebraic extension of the p-adic field Op. Let q be the order of the finite field 0/P and let VEP be such that n
q=P -p ,
Ipl=q-1.
Then
0P is a finite extension of the field of order p.
Every element of 3 can be expressed as x= E a pn where the n?s
n
coefficients a are in 0 and are uniquely determined modulo
Finally Ixl=q s, if as x0 mod P. 4. Two-dimensional lattices.
Let
3
a
be
locally
compact
totally disconnected field with modulus I.I.
Let 0={xe6:
IxJ`1}
and P={xe
:
IxI<1}.
Let
pr=p be such
that p0=P. We let V denote a vector space of dimension 2 over
. Observe that V is locally compact and totally disconnected in the product topology. A subset of V is called a module over
0 if it is a subgroup of V (with respect to addition) and is closed under multiplication by elements of D.
(4.1) DEFINITION. A compact open subset of V which is a module over 0 is called a lattice of V.
If {el,e2}
is a basis of V,
then the set el0ee20 is a
lattice. A consequence of our next result is that every lattice
is of this form. Observe that every lattice generates V as a vector space.
Therefore
it
contains 2
linearly independent
elements.
(4.2) LEMMA. Let :C be a lattice. Then the quotient Vp2 is a two-dimensional vector space over the finite field f=0/P. If elE2 and e1itV2, then there exists e2E2 such that 2=el0oe20. PROOF.
The quotient 2/p2 is a vector space over the finite
field f=0/P. Indeed,
The dimension of this vector space cannot be 1.
if this were the case then 2 could be written as
Appendix
126
2=eD+p2, for some nonzero eE2. Inductively one could then write
since k is arbitrarily large,
that
2=eD, contradicting the assumption that 2 is open in V.
This
2=eD+pk2,
which implies,
implies that 2/p2 is of dimension 2.
As a consequence there
exists e2E2 such that
is a basis for 2/p2.
{eI+p2,
e2+p2}
Therefore 2=e1D+e2D+p2 and, again by induction, 2=e1D+e2D+pk2, which implies 2=e1Dee2D. 1
The
next
result
the
is
two-dimensional
version of
a
theorem valid for n-dimensional vector spaces over 3, which is
known as the theorem of elementary divisors. The proof in the general case may be obtained using induction on the dimension of the vector space.
(4.3) THEOREM. Let 2 and 2' be two lattices of V. Then there exist vectors eland e2 and integers n and m such that 2= e 0 e e 0, and 2'=e1pO e e2pm0. Moreover the set {n,m} depends 1
2
only on 2 and 2'.
PROOF. Since 2 is an open set containing the origin and pk2' is
a basis of neighborhoods of the origin,
for some integer j,
pi2'S2. Let j be the smallest such integer, and define 2"=pj2'.
Then 2"52, but 2" is not contained in p2. Let e1 be an element of 2" which is not in p2. Then, by (4.2), there exists e2, such that
2=e1Doe2D.
pie2E2".
We
Let
assert
i
be
that
2=e1Dee2D, then v=oce I+(3e2,
fore
Let
fte2E2".
implies
Therefore
s>-i.
2"=e1Doe2pID,
13=Aps,
the
smallest
integer Indeed,
2"=e1Doe2piD.
But ae1E2",
with a,(3ED.
with
IA1=1.
Then
3e2 Aps-ipie2EpID.
and 2=epnDoe2pmD,
We
such
that
vE2"5
if
and therewhich
pse2E2",
conclude
with n---j and m=i-j.
that
Observe
now that -min(n,m) is the smallest integer j such that pj2'S2, and that gIn_ml
is the index of p'12'
in 2.
{n,m} is uniquely determined by 2 and 2'.1
Therefore the set
The tree of
5. The tree of PGL(2,3).
Let
127
V
a fixed
be
two-dimensional
vector space over 3. If 2 and 2' are lattices in If, we say that
2 and 2' are equivalent when there exists aE3
such that a2=2'.
The set of equivalence classes of lattices will be denoted by
X. Given two elements A and A' of X, and two lattices 2EA and 2'EA', by (4.3) there exists a basis {e1,e2}EV and two integers n,
m such that 2=e1Dee2D, and 2'=pneIDopme2D. Clearly 2'S2 if
and only if n,m20. We observe that the integer only on the equivalence classes A and A'.
then {xe1, xe2}
for
y2'.
q s=lyx ll.
in-ml
Indeed,
depends
if x,yE
is a basis for x2 and {ypne1, yp e2} is a basis
But
ps+nxeID
ypneID a yp1°e2D =
where
a ps+01xe20,
We shall say that the number In-ml=d(A,A') is the
distance between A and A'. We have not shown yet that d even defines
a metric.
We
prove
shall
instead
graph
the
that
structure introduced in X by the following definition yields a tree. The fact that d is a metric will then follow immediately.
(5. 1) DEFINITON. Let A, A' e3f; we say that {A, A'j is an edge or that {A, A' }Ef, if d(A,A')=l. Our next task is to show that (X,1) is a tree. We need a result
which
is
essentially
nothing
but
a
convenient
reformulation of (4.3).
(5.2) LEMMA. Let 20 be a lattice in V, and A an element of X. Suppose that A'
is another element of X,
and A'*A;
then the
following conditions are equivalent for a lattice 2eA. (1) 2c2o,
and
for
some
n>O
and
some
basis
{el,e2}
2 = pneDoeD c e1Dee2D = 2o. (2) 292 and 20p2 (3) 2 is the maximal sublattice of 20 which belongs to A'. (4) 20 /SE is generated by one element.
Moreover there exists only one 2eA satisfying these conditions.
Appendix
128
PROOF.
It is clear that
(1) implies (2).
If
(2) holds and
2'=aY, with llsa292 o, then a=fpk, with I13I=1, and 2'=pkg. Since 2¢p2o and :Cfp:C, we must have k=0. Thus (2) implies (3). Suppose
(3) holds and let {e1,e2} be a basis chosen according to (4.3) so that 2 =e Dee 0 2and 2=pkelDephe2 D. Since 2S2 o, we must have h,
k>-0.
If j=min(h,k), then 29pj20 S2 0. This implies j=0 and (1)
follows. Thus (3) implies (1). Now (4) is clearly implied by
(1). On the other hand if (4) holds 2 cannot be contained in is two-dimensional. The existence of a because o/p$ lattice of A satisfying (1) follows directly from (4.3). It
p2 ,
suffices
to
observe where
p°e1Dep e2D,
pn-m
eIDope 2D
satisfies
any
that
20 e1Dee2D, (1).
lattice
and
Uniqueness
of
A has
therefore, is
also
the
form
if
n>-m,
easy for a
lattice of A satisfying (2). Indeed if 2 and :C'=a2 both satisfy (2), it follows that
1a1=1.1
We observe that, if 2' satisfies one of the conditions of the lemma, then 212'
In particular
is isomorphic to D/pnD, where n=d(A,A').
d(A,A')=O
if
and only
if
A=A';
furthermore
d(A,A')=l if and only if there exist 2EA and S'EA', such that 2'92 and 2/2' is isomorphic to the field D/pD of order q.
The graph 3f is a tree . PROOF. We show first that X is connected. Let A, A' el, and let $EA and 2'r=A' be such that !'92 and 2'#2. By (5.2), there (5.3) THEOREM .
exists a basis {el,e2} such that 2=e1Dee2D and V =phelDee20. For i=0, 1, ... , h, let 2i=pielDee2D. If Ai is the class of JCi, we have that d(Ai, Ai+1)=1, for 1=0,1,... , h. Therefore {A=A0 AI' . .
_Ah=A' l
is a path in the graph connecting A and A'. We show
now that X contains no circuits. We will prove the following statement (which directly implies the absence of circuits):
if
Ao, Al, ... , Ah is a sequence such that d(Ai, Ai+1)=1 for every i, and Ai*A1-2 for i=2, ... , h-2, then d(A0, Ah)=h. This statement is obvious if h=0 or 1. We make the induction hypothesis that it
The tree of PGL(2,a)
is
true for h-1.
Choose 20EA0,
element of Ai
contained in 2
and 2i¢p2i-1'
This means
therefore
2j
has
and
qj
be the maximal
Y
let
In other words, 2 22 2...2.2 0
that 2
index
129
i
index q
has
in
every
for
Y0,
d(A0 , Ah-1 )=h-1, by induction hypothesis, and 2
0
This means that if, according to (4.3),
n
2i-1
and
j.
Now
with index
9`.2
h-1
qh-1.
1
in
2h-1=pneIDep e2D,
with 2 eIDee20, it must be the case that n,m>-0, and n+m= h-1 =
It
In-ml.
satisfies
follows that either n or m is conditions
the
of
zero,
and 2h-1
in particular
(5.2),
Since 2h-1 /2h is a field of order q,
2h-1¢p20.
is generated by 2h
2h-1
and a one- dimensional subspace. But also p2h-2 92h-1 and (5.2) implies that
is generated by p2
2h-i
subspace. In other words
h
and p2
h-2
h-2
and a one-dimensional
are inverse images of two
one-dimensional subspaces of the finite plane h2
respect to the field f=0/pD. A =A h
It
h-2
,
contrary to
follows that
But
p:C
our assumption.
2h p20,
h-2
*2
,
h
-1
/p2
,
h-1
with
because otherwise
Therefore 2
=2 +p2 h-1
because otherwise
2h-15p20.
h
h-2'
We have
proved therefore that 2h satisfies the conditions of (5.2) with
reference to 20. This means that the distance of Ah and A0 is the logarithm in base q of the index of 2h in 2. Therefore
d(A0,Ah)=h. ,
The proof of (5.3) shows that d(A, A') coincides with the distance of A and A' as vertices of the tree X. This implies, of course, that d satisfies the triangle inequality. We shall
prove now that
{A: d(A,A0)=1} consists of q+1
elements.
We
known, by (5.2), that if 20EA0 there exists one and only one
2EA such that 292
and 2¢p20.
Therefore the classes A with
d(A,A0)=1 are just as many as the lattices contained in 20 and satisfying the conditions of (5.2), in fact, by ((5.2),2)), as
many as the lattices 2 such that p20 92920 and p20 2x20. Every such
lattice
is
in
one-to-one
correspondence
with
a
one-dimensional subspace of 2 /p20, which is a two-dimensional
Appendix
130
vector space over the finite field f=0/p0 of q elements.
The lines passing through the origin in the finite plane (D/p0)x(O/pD) are as many as the points in the projective line over the field O/pD, that is q+1. This shows that there are q+1
distinct A such that d(A,A0)=1. Let now . 0 DxO be the lattice containing the canonical vectors (1,0) and (0,1), and A0 be the
We shall prove that the infinite
class to which 20 belongs.
chains {A o,A1,...,An.... } are in one-to-one correspondence with
the lines in the plane ax6=V.
(5.4) PROPOSITION. Let R be the set of one-dimensional subspaces of V, and let weS2 be a point of the boundary of X associated to the infinite chain (Ao, A1, ... , An, ... ). Let 2 eAi be
such
that
and
2 92 j
Then
2 Jp2
j-1
j
generates
n 2
a
J=o J
i-i '
one-dimensional subspace n of V and the correspondence W-' is
a bijection between 0 and R . PROOF. We will prove that the inverse correspondence n9(n)=tv
is a bijection between
R and Q.
following sequence of
Let
we consider the
'ER;
lattices Lj=nn20+pj20.
vector v0Enn20 such that nn50 =0v0.
Indeed,
There exists a
it is obvious that
Ov9nn20 for every vEnnlo; observe further that, if v,wEnn20 and
v=aw with aE, then OvcDw if and only if lalsl. Therefore all
the sets Ov with vE'u
form a chain under inclusion.
0
This
means by the compactness of na o that there exists a maximal
set Ovo such that na Ovo. Observe that exists
woE o\p:Co
such
therefore
Lj=Dvo®piDw0;
that it
By (4.2) there
20 Ovo 0wo.
is clear that
that Lj+1 has index q in Lj.
that
means
This Lj+19 Lj,
L0 =2 0
and
In particular Lj+1 is maximal in
Lj and if Aj=[Lj] then d(Aj,Aj+1)=1 for every j. To show that {Ao Al ... , Aj, ... }
is an infinite chain it suffices to prove
that Aj+2#Aj for every j. k,
pkLj+2 Lj
,
that
is
But if Aj+2 Aj then, for some integer pkD o®pk+j+2Owo
O o@pjOw0.
contradiction because pkv0ELj implies that pkED,
This
is
a
that is k_O,
The tree of PGL(2,3)
while v0epkLj+2 implies that p ke0,
and
a
Lj=Lj+2,
a
is
Therefore k=0
that is k:50.
contradiction.
6(n,)-{AA...,Aj,...} j
131
This We
chain.
proves
that
observe
that
oLj=Dvo nn2o, and therefore the linear subspace generated by
is n. We prove now that the map n.-(n,) is injective.
AL j=0 j Indeed
then
if
=nruw 0+PJ20
equivalent
is
to
L'=a'r%2 +pj2O for every j. Since Ljand L' are sublattices of 2
it follows that Lj=Lj for
both of index qj and equivalent, every j. prove
In particular
that
surjective.
is
a9
to={Ao, A... , Aj, ... } .
Let 2
contained
for
that
in
J=
because
2
j-1
Let
every
and so
.
Finally we
n=n.'.
be
tv
in
and
it
the maximal element of
be
2j generates a line 'z d(Aj,Aj-1)=1
and so
n L = n L' j=o j J=O j
j=1,2,...
now
prove
We
.
Aj
Indeed 2 has index q in 2j-1 index
has
21
in
qj
20.
Since
d(A0Aj)=j this implies that 2j is the maximal element of Aj
contained in 2. In particular 21#20. Let vje21\p20. Passing to
a
sub-sequence
may
we
assume
that
converges.
vi
Let
lim vj=vo ; then vo o\p2 and v0 x0. Since vke2j for k>j and 2 is closed it follows that v0e0j for every j and so v0Ejno2j*0.
But jno2j is a module with respect to D; since 2j has index
qj
00
in 20 the index ofjn02j in 20 must be infinite. Therefore by (4.1)
jno2j cannot contain a basis,
and the subspace a it
generates must be one-dimensional. Also we observe that v0E2j
for every j and Dv =nn2 0
that
a9(a)=w;
indeed
0
0
2 0voeiJwo
+9(n)={A0, [L1], [L2], .... . [L j].... } 2 jc2o Dv(DOW
0
and
v0 E2j,
We prove now
because v E(2 n0\p2 .
there
for
where
0
0
some
woE 2 \p2o,
L j=£lvaopjDw0.
exists
a
k
such
and
Since
that
Appendix
132
C=2
i
\:P2
J,
It follows that 2i=Dv0 epkDw0. Because the index of
21 in 20 is equal to qi we have that k=j and 2i=Li for every j. This means that i(n,)=&) and the proposition follows. ,
We consider the following groups of two-by-two matrices with
entries
matrices;
in :
GL(2,3)
the
is
group
of
nonsingular
is the quotient of GL(2,6) modulo its center
which consists of nonzero scalar multiples of the identity; is the group of matrices with determinant
1
modulo
{+I,-I}. The groups GL(2,6), PGL(2,a) and PSL(2,3) are locally
compact groups when endowed with the usual topology. Choosing
the matrix entries
in the ring D we obtain the subgroups
PGL(2,D) and PSL(2,D) which are open and compact,
GL(2,D),
and PSL(2,3), respectively.
in GL(2,3),
We will describe now the action of PGL(2,3) and PSL(2,6)
on the tree X.
We fix a basis {el,ez}cV,
and let 20 be the
lattice generated by this basis, and A0 its equivalence class.
With respect to the basis {e1,e2} each nonsingular matrix A defines a linear transformation on V which maps a lattice into a lattice. Furthermore, if aE
then A(a2)=aA(f). Therefore if
2 and 2' are equivalent so are A(:C) and A(Y'); in other words A
acts on the equivalence classes of lattices and therefore on X.
Suppose now that A, A' E.T and d(A, A')=n. Assume that 2EA and 2'r=A'
is the maximal element of A'
contained in 2.
Then
A(2')cA(2) and A(2') is maximal in its class with respect to this property. Therefore d(A(2'),A(2))=n. Thus A is an isometry of
1.
It
follows that every nonsingular matrix defines an
automorphism of X. Observe however that, if aI, with aEa ,
nonzero multiple of the identity matrix, each element of X into itself. Conversely,
which acts as the identity on X, Therefore
subgroup of Aut(X).
It
is
not
then aI(2)=a2 maps
if A is a matrix
then A=(xI
aE6 }
is
difficult
is a
for some aE6
isomorphic
to see
subgroup is closed and that the natural topology of
that
to
a
this is
The tree of
133
the same as its relative topology in Aut(X). We
will
now describe
boundary of X.
action of PGL(2,j)
the
on
the
By (5.4) the boundary fl of X is in one-to-one 9
correspondence
with
subspaces of V.
The set R has a natural compact topology once
the
set
of
one-dimensional
vector
it is identified with the projective line relative to 6, i.e. the space of equivalence classes of nonzero vectors (a,I3), two
vectors being equivalent when one is a nonzero multiple of the other. The group GL(2,3) acts naturally on the projective line,
because it maps every nonzero vector into a nonzero vector and preserves equivalence classes. Therefore PGL(2,6) is a group of continuous trasformations of 9t, transformations.
indeed the group of projective
Since PGL(2,j) acts continuosly on S2
it
is
natural to ask whether 0 and R have the same topology and whether the action of PGL(2,3)
is the same.
The answer
is
affirmative on both counts. The correspondence n.
)
{ [a o 20+ pn2 o]: n e W}
is continuous, and therefore bicontinuous, and A(4) corresponds to
the
sequence
of
lattices
{[A(')n A(Y0) + pnA(20)]} _
acts transitively on X,
{A([4n20+pn20])}. Observe that
but also on R and therefore on Q. The group PSL(2,D)
of matrices with entries
determinant 1 is exactly the subgroup of
in D and
which maps 20
onto itself. Thus PSL(2,D) fixes A0 and is a closed subgroup of the group KA . We show that PSL(2,D) acts transitively on Q.
In
0
view of the remarks above it suffices to show that PSL(2,D) acts transitively on the projective line R. because every element of
9Z
This is obvious
is determined by a column vector
with one entry in 0 and the other entry equal to the identity of 6.
It is not difficult to show that PGL(2,3) is a proper
subgroup of Aut(3F).
Indeed the identification of S2 with the projective line R and the resulting identification of the action of PGL(2,6) on S2
Appendix
134
with its natural action on
9Z
imply that every element of
PGL(2,6) which fixes three points of R is the identity. On the
other hand we can find proper rotations in Aut(3E) which fix every element of any finite set. This shows that the inclusions
and PSL(2,O) in Aut(X) and KA , respectively, are
of
0
proper.
We
turn now to
consider
the
which
subgroup
containing PSL(2,D) acts transitively on Q. We observe that PSL(2,3) is noncompact and therefore by (1,10.2) it has at most two orbits on X. We shall prove that the orbits are exactly two by showing that no element of PSL(2,a) can map e1D®e20=20 onto
is a matrix then
bbl
a multiple of 21=pe1Dee2D520. If A= rte
A(20)c2I if and only if a,bEp£) and c,dED, while A(21)c 0 if and only if a,cE 10 and b,dED. If AEPSL(2,a) and A(20) is a multiple of 21, we must have, for some nEZ, pnA(.`$0)=`.Q1, which implies
pnA -1(21)92
and
pnA(20)S.C1
pn-la, pn-1bED;
that, if A= L d] , then pen-1 (ad-bc)=p2n-1 ED,
that
pnc, pndED.
This implies
which means that n>0.
On the other
hand p nA 1(2 1)c20 implies that means
this
p1-nd, p1-ncED
p1-2n (ad-bc)=p1-2n
10
then w,
ol
so
naE);
ns0, X+
has two orbits
It follows that
Let now w--
and p nb,
and
ED
that
contradiction. 3E
These conditions imply
0.
a and
considered as an element of
Lorder
PGL(2,a), is of
2: w2=[0
Li
and therefore w(2o)c Conversely
Lv 1
that w(-W 1)5-W 0. of 2 1,
o1
If a, RED,
LLJ= La i E peI
® e2 D,
pelDee20.
0l [pa] =
Thus,
0J=I.
[13]
,
and since
I
1
0I= w we have
if A0 is the class of Y0 and Al the class
we have that w(A0)=A1 and w(A1)=A0. Since d(A0,A1)=1, we
The tree of PGL(2,a)
135
obtain that w is an inversion. We can now prove the following proposition.
(5.5) PROPOSITION Let t and s be nonnegative integers such that q+1=2t+s. Then there exists a discrete subgroup r'cPGL(2,3) such that
r acts faithfully and transitively on X
(i)
,
(ii) r is isomorphic to the free product of s copies of Z2 and t copies of Z
,
(iii) PGL(2,3)=r'.PGL(2,9).
PROOF. Observe that PGL(2,3) acts transitively on X and on 0, and therefore it acts doubly transitively on X and hence it acts transitively on the edges of X. But as we have just proved above PGL(2,6) contains an inversion of order 2. It follows by (I,10.4) that PGL(2,5) contains for every t and s with 2t+s=q+1
a faithful transitive subgroup satisfying (1) and (ii). As to (iii), we observe that
PGL(2,D) =
g(A0)=A0}. 0
If
there exists one and only one g'Er such that
g(A0)=g'(A0).
Thus
g=g'
(glg')-1,
We
will
PGL(2,j)nG
and
g lg'(A0)=A0
Since
g lg'EPGL(2,D).
(iii) follows. ' now
identify
where
wEA and
the
PGL(2,a)nGw
subgroups is
a doubly
and
infinite
geodesic.
(5.6) PROPOSITION. (i) Let well;
basis (el, e2} in V
then there exists a choice of
with respect
PGL(2,3) which fixes W
to which every element of
can be written as A= LO
If,
in addition,
d, , with ad*0.
A is a rotation then we may also choose
JaJ=JdJ=1.
(ii) Let
be a doubly infinite geodesic;
then there
Appendix
136
exists a basis of V with respect to which every element A of =
ra
01 (for Aw=w and Aw'=w') or d
PGL (2 , 6)nG can be written as A- 0 A=[a
(when A interchanges w and w').
0
PROOF. Let A0 be a vertex of X and let w be identified by the
infinite chain maximal
{ Ao, A1,
... , An, ... } .
satisfying
lattice
{e1,e2} be a basis for if.
matrices
as
C S n
Let DEAD and 2 r=An Let
n-1
e e n 2 ,
1 n=1 n
be a and
e #0 1
If we write the elements of PGL(2,3)
with respect
to
the11 basis
condition AEGw implies that
{e1,e2},
then
the
in other words A=[0
d].
weLLmayLLLllI
assume that An is the class
Without loss of generality
it
is easy to see that
is the set of matrices
dJ with IaI=IdI=1
of Dmp'D for every nEz. PGL(2,j)nGwnKA
Therefore
o
J
and IbI51 while the set PGL(2,a)nGwn(KA \KA o
matrices La
d]
with
IaI=IbI=IdI=1.
)
w=r0
Let
0];
p ndl
dlw n =[aa JJ
infi-
the
nite geodesic {...,A_n,...,A 1,A0 ,A1,...,An,...}.
L
w(A)=An+1
Jdoubly
onLLL
for every n, so w is a step-1 translation
wn[
is the set of
-1
We have that
It follows that PGL(2,3)nBwn(KA \KA n
JJ
p
is equal to the set of matrices ro while PGL(2,S)nBwnKA
is
the set
-n b
d, with IaI=IbI=IdI=1
of matrices
n
IaI=IdI=1 and Ib'I
)
n-1
In particular AA =A n
n
a
b'
0
d
with
for every neZ if
r
and only if A=[0
d] with IaI=IdI=1. This proves M. To prove
(ii), we observe that PGL(2,3) acts doubly transitively on 0,
and therefore, as in part (i),
without loss of generality we
can suppose that a'={ .... A_n, _A A 1, Ao, A1, ... , An, ... } . The fact that A=(a
dl
fixes w and w'
(that is the points [] and [] of
The tree of PGL(2,6)
137
the projective line) is equivalent to the fact that c=b=O. part
(i) we have proved that A=(0
dl
In
fixes An for every n
(that is A is a rotation of PGL(2,3)nGly fixing w and w') if and
only if
lal=ldl=1.
This implies that if A=[
dl
and lal#idl
then A is a translation on the doubly infinite geodesic T. The
condition that A interchanges w and w'
is equivalent to the
fact that a=d=0. The matrices of this type are of order 2 and are exactly the rotations and the inversions of PGL(2,3)nG7 which interchange w and W. It is easy to see that the matrix A=rd
bl
(and so
is a rotation if and only if inglbl-inglcl A
is an inversion interchanging
w
and
inglbl-inglcl is odd). This proves the proposition. 1
is even co'
iff
138
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144
SYMBOLS L
p. 34
1
S(x,y,(0)
" 35
1
P(x,y,w)
" 35
µ(z)
" 35
P(o,x,w)
" 36
X(Q)
" 36
X' (S2)
" 36
pz
" 36
P(x,w)
" 37
1
p. 1
LF
"
[x,yl
"
d(x,y)
" 3 3
S2
[x,w)
" 3
(w,w2) T'(x,y)
"4
c(x,Y)
3
v
" 5 5
v
"5
F
x
"5
2
Aut(X)
"
10
K
"
10
as
37
mK
41
oz Ixl
h z
x
U(g)
10
6n
10
S(q)
10
{ 3E
a, b}
n
X (c)
" 44
"
12
x
"
15
fK
Bw Mw "
18
L1(K\G/K)
" 47
19
K\G/K IfK n P
" 49
20
" 25
n I+,
X-
Aut(1)+ 0x n
46
" 47
7
H
45
KfK
7
G
" 43 " 44
" 18
G
" 42
IZ
n
K
" 41
27
51 " 51
K
" 54
Ak
56
1
" 57
" 28
L
" 29
H z
n
" 57
Symbols
n
z
"r
145
p. 54
At(s)
p. 105
" 64
Indor
"
108
"
108
bo.
rz
" 67
H-
°'
" 109
R
L
" 72
P (K)
" 84
K(s)
" 85
Lb (3f n )
113
n
113
119
P
I.
85 Z
,as
" 85
0
tm
85
Vn
" 86
86
31
119
"
119
P
" 85
V (S)
" p
S(c,k) Sp
120 "
z(P)
120
120 122
87 87
A
I
122
I
123 +,
" 123
89
lF
U
"
123
P
"
124
93
f
" 124
1194
V
"
" 90 " 93
.`Y
fY (e) 0
At(e) 0 K(e
r
A
94
co
a-
125 125
0
130 1'0-
95
2
" 100
[] K(s)
"
(K(s))" 0
101
102
GL(2,6) PGL(2,6) PSL(2,6)
"
132
"
133
GL(2,0) PGL(2,D) PSL(2,O)
147
INDEX action
faithful 14
on the boundary 26 doubly transitive 29 adjacent vertex 1 admissible representation 112,114 algebraic representation 114 algebrically irreducible representation 115 amenable group 18,20,22,31 automorphism of a tree 6 Bass H. 82 Bouaziz-Kellil F. 32 boundary of a tree 3 boundary of a finite subtree 85 boundary point 3 Bruhat F. viii building 80 Cayley graph 16 Carleman formula 60 Cartier P. 31,81 chain 1 infinite 3 doubly infinite 3 circuit 1 cocompact subgroup 78,82 cofinal group 78 compact maximal subgroup 12 compact maximal subring 124 complementary series 56,81 complete maximal subtree 101 complete subtree 85,98,100,105 connected graph 1 Cowling M. G. ix,83 cuspidal representation 84,85,98,107,114,116,117 cylindrical function 36,54,55 van Dijk G. 80 degree of a homogeneous tree 1 degree of a vertex 1 discrete subgroup 14,63,82,135
distance 3 doubly infinite chain 3
Index
148
doubly transitive action 29 edge 1 oriented 89 elementary divisors 126 elementary eigenfunction 34 equivalent lattice 127 end 3 faithful action 14 faithful transitive group 14,135 finite volume 82 finitely additive measure 36 formal dimension 95 formula
Carleman 60 Plancherel 113,118 spherical Plancherel 56,60,62,81,113,118 free group 5 function cylindrical 36,54,55 positive-definite spherical 52,53 radial 41,46,49,54 spherical 41,42,50,51,52,80 Gelfand-Naimark-Segal construction 52
Gelfand pair 46,62 generation 25 geodesic 3 infinite 3,135,136 graph 1 Cayley 16 connected 1 group of automorphisms 10 group amenable 6 cofinal 78
faithful transitive 14,135 free 5
liminal 113 permutation 10,104 postliminal 113 type I 112,118 Gulizia C. L. 118 Haar measure 16,42 Haagerup U. 81 Helgason S. 80 homogeneous tree 1 horocycle 20,25,31,35 induced representation 108 infinite chain 3 infinite geodesic 3,135,136
Index
infinite reduced word 6 interior of a subtree 85 interior point 37 intertwining operators 44,45,74,81 inversion 7 irreducible matrix 73 irreducible representation 50,51,64 K-bi-invariant 46 K-invariant vector 50,51,52,84 K-right-invariant 46 Kuhn G. 117 Kulkarni R. 82 Kunze-Stein phenomenon 118 Laplace operator 34 lattice 81,83,117,125,127,129,131 equivalent 127 length 3 liminal group 113 local field 118 locally compact fields 122 locally finite tree 1
Mantero A.
M.
117
Mautner F. I. ix McMullen J. 118 minimal tree 86,100 module 125 modulus 119 mythical ancestor 25 nearest neighbor 1 nonhomogeneous tree 2 nonspherical principal series 116,117 nontangential convergence 76 01'shianskii G. I . viii, 114, 116, 117, 118 orbit 20,24,25,72 oriented edge 89 p-adic fields 119 p-adic norm 119
path 1 permutation group 10,104 Plancherel formula 113,118 Poisson transform 44,74 positive-definite spherical function 52,53 postliminal group 113 prime element 124 principal series 56,65,81 projective line 130,133,137 projective transformation 133 Pytlik T. 81 radial function 41,46,49,54
149
Index
150
Radon-Nikodym derivative 34,35,54
random walk 80,81 representation 50 admissible 112,114 algebraic 114 algebraically irreducible 114 cuspidal 84,85,98,107,114,116,117 induced 108 irreducible 50,51,64 standard 102,111 special 84,85,87,95,97,98,112,116,117 spherical 50,51,52,80,81,82,84,85,112,117 unitary 50,84 resolvent 56,59,67 rotation 7 Sally P.
118
J.
semihomogeneous tree 2 Serre P. J. 31
Shalika J.
A.
118
simple random walk 80 special representation 84,85,87,95,97,98,112,116,117 spherical function 41,42,50,51,52,80 spherical Plancherel formula 56,60,62,81,113,118 spherical representation 50,51,52,80,81,82,84,85,112,117 spectrum 59 stabilizer 12 standard representation 102,111
Steger T. viii, ix,31,83, 117 step of translation 7 subgroup cocompact 78,82 compact maximal 12 discrete 14,63,82,135 subtree complete 85,98,100,105 complete maximal 101
Szwarc R. ix
Tits J. viii, 31 type I group 112,118 theorem of elementary divisors 126 totally bounded 120 totally disconnected 11,120,122,123,124
translation 7 tree
homogeneous 1 locally finite 1 minimal 86,100 nonhomogeneous 2 semihomogeneous 2
Index
ultrametric inequality 119,121,123 unitary complementary series 56,81 unitary principal series 56,65,81 unitary representation 50,84 vertices 1 Vitali-Lebesgue theorem 76 weakly contained 65 Zappa A.
117
151
