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.
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