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

3· Thus Xi is the union of all infinite chains starting at Xi and having no other

vertex in common with 3· 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)= 3· This contradicts the assumption

that Xi has degree one in 3· 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

~}