14

ALESSANDRO FIGA-TALAMANCA

Definition.

A finite subtree

~

of X is called a complete tree if it reduces to a single

vertex or else the degree of each vertex is either one or q

+

1.

Clearly every finite subtree is contained in a finite complete subtree. Given a

complete tree

~

we consider the group

K(~)

=

{g

E

Boo:

gx

=

x for all x

E ~}.

This is a compact open subgroup of

Boo.

Let now

1r

be an irreducible representation of

B

00

defined on the Hilbert space

H1r.

Let

~

be a finite complete subtree. Define a projection on the space

Htr

as

follows:

As~

varies among finite complete subtrees,

K(~)

describes a basis of neighborhoods

of the identity. Therefore if 0

=/=-

~

E

H.1r

then, for some finite complete subtree

~'

which implies

Ptr(~)~

=/=-

0. This justifies the following definition.

Definition.

Let

1r

be an irreducible unitary representation of

Boo.

Denote by

f1r

the smallest positive integer for which there exists a finite complete subtree having

etr

vertices and such that

Ptr(~)

=/=-

0.

Proposition.

If

etr

=

1 then

1r

is {unitarily equivalent to)a spherical representa-

tion. If

etr 1

then

etr 2.

Proof.

f1r

=

1 means that there exists a vertex x, such that t.he projection

P1r( {

x})

is not zero. But any vector in the range of this projection is invariant under the

action of tr(k) for every k

E

Kx, the compact group which stabilizes

x.

Recall

that x corresponds to a closed ball of radius qn. Therefore there exists an element

h

E

Boo

such that hx is the ball of diameter qn which contains 0, in other words the

ball

p-no.

This means that the group Khx = hKxh-

1

maps the ball

p-no

in itself

and therefore contains the group

Ko

which fixes 0

E

F.

Define

tr1(g)

=

tr(h-

1

gh).

Then tr' is unitarily equivalent to

1r.

Let

~

be a Kx-invariant element under the

action of

1r.

Then~

is hKxh- 1-invariant, and hence K

0

-invariant under the action

of

1r1

.

This means that

tr'

is a spherical representation, which proves the first part

of the assertion. Suppose now

etr

1. We shall prove that

etr

=/=-

2. Indeed if

etr

=

2

there exists an edge e

= {

x,

y}

of the tree such that the projection

Ptr(

e) is not zero.

The vertices x and y of e belong to different horocycles. We may suppose, without

loss of generality, that the infinite chain [ x, oo) contains y. This means that every

element of

Boo

which fixes x fixes also y. In other words K(e)

=

K({x}). This

means that

etr

=

1 contrary to the hypothesis that

etr

=

2

The following lemma provides the key geometrical argument for the theorem

asserting that every nonspherical irreducible representation of

Boo

has a coefficient

of compact support.