LOCAL FIELDS AND TREES
15
Lemma.
Let~
be a complete subtree of X with
diam(~) ~
2. Let I) be a complete
subtree not
containing~-
Then there exists a complete subtree
3
properly contained
in~
such that K(3)
C
K(IJ)K(~)
Proof. . We show first that we may assume that
~
n
I) contains an edge. Suppose
~
n
I) does not contain an edge. Then
~
n
I) is empty or contains only one vertex
In this case there exists a vertex
x
E
~
of minimal distance from
1).
This vertex is
unique because if d(x, y)
=
d(x', y') is the minimal distance, with x, x'
E
~'
y, y'
E
IJ,
and x
=1-
x', then [x,
y]
and [x',
y']
intersect
~
and I) only at the end points. Thus
[x, y]
intersects
[x, x']
only at x and
[x', y']
intersects
[y', y]
only at y'. This means
that
[x, x']
U
[x', y']
U
[y', y]
U
[y, x]
is a circuit in
X:
a contradiction. Let now m
be the minimal distance and x the unique vertex of minimal distance. The subtree
1)
1
=
{y
E
X : d(y, IJ) ::; m
+
1} is a complete subtree and
1)1
n
~
contains an edge.
Observe that
1)1
does not contain
~
because if all vertices of
~
had distance at most
m
+
1 from I), then all vertices
of~
would have distance at most one from the vertex
closest to
1),
which would imply that
~
has diameter less than two or that
~
is not
complete. Clearly K(IJ')
C
K(IJ) therefore substituting
1)
1
in the place of I) we may
suppose that
~
n
I) contains an edge. Let 3
=
~
n
IJ. The hypothesis that 3 contains
an edge implies that 3 is a complete subtree. We prove now that K(3)
c
K(I))K(~).
Let x
1
, ... , X
8
be the vertices of 3 which have degree one. We use here the fact that
3 is complete. Let Xi be the (infinite) subtree of X containing Xi and no other vertex
of
Thus Xi is the union of all infinite chains starting at Xi and having no other
vertex in common with Let Ki be the subgroup of Boo which fixes Xi and every
vertex which is not in Xi· Clearly Ki
c
K(3). Furthermore if k E Ki and k' E Kj
with
i =/:- j,
then kk'
=
k'k. In addition every element k
E
K(3) may be written as
the product k
=
k1
...
ks with ki
E
Ki. In other words K(3) is the direct product
K(3)
=
K
1
...
K
8

We show now that for every
i
=
1, ... s, either Ki
C
K(~)
or else
Ki
C
K (I)). Indeed Ki
C
K
(~)
if and only if Xi does not intersect
~
and Ki
C
K (I))
if and only if Xi does not intersect IJ.
It
follows that if Ki is not a subgroup of
either
K(~)
or K(IJ), then Xi must contain vertices of
both~
and IJ. In this case let
x
E
~nXi
andy
E
~nXi.
Then the chain [x,xi] is contained in
~nXi
and the chain
[y, xi]
is contained in
l)nXi.
Let
x'
andy' be respectively the vertex of
[x.xi]
next to
Xi and the vertex of
[y, xi]
next to Xi· Then d(x', Xi)
=
d(y', xi)
=
1 and
since~
and
I) are complete this implies that x', y'
E
~
n
I)= This contradicts the assumption
that Xi has degree one in We have thus proved that either Ki
C
K(~)
or else
Ki
C
K(IJ). Since the groups Ki commute we may write each element k
E
K(3) as
a product k
=
k'k" with k'
E K(~)
and k"
E
K(IJ). Thus K(3)
C K(~)K(IJ).
We are now ready for the proof of the main theorem of this section.
Theorem.
Let
1r
be an irreducible representation of Boo. Let
~
be a minimal tree
associated with
1r.
Suppose that
1r
is not spherical and that consequently
diam(~)
2
Let~ E
H1r. be a vector such that
P1r(~)~
=
~
and
11~11
= 1.
Let
u(g)
=
(n(g)~,
0-
Then
supp u
C
{g E Boo :
g~
C
~}
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