LOCAL FIELDS AND TREES 9

It is clear that R is an equivalence relation. The equivalence classes of R are called

the horocycles of X with respect to the boundary point w.

Now if

X

is the tree associated with

F

and we let

w

= oo, (the boundary point

identified by an ascending chain of balls) then the horocycles of X with respect to

oo are exactly the sets of balls of a given radius. Accordingly the horocycles with

respect to oo may be numbered in such a way that the balls of radius qn belong to

the horocycle

Hn.

In other words we stipulate that

p-no

E

Hn

(Fig.l).

THE GROUP OF ISOMETRIES AND ITS ACTION ON THE TREE

Let g

E

Aut(X) be an automorphism of the tree X associated to the local field

F.

Suppose that g has the property that goo= oo. In other words suppose that g

maps every chain of the type [x, oo) into a chain of the same type. Then g defines

a transformation of

0 \ {

oo} = F onto F.

In accordance with the terminology used in

[2]

we shall denote by

Goo

the group

{g

E

Aut(X): goo= oo}. We shall presently describe the action of G

00 •

We start

with the subgroup

Boo

which leaves invariant every horocycle with respect to oo.

The action of

Boo

on

F

is easy to describe. Every element of

Boo

maps balls of a

given radius onto balls of the same radius. In other words an element of

Boo

defines

an isometry of F. Conversely every isometry of F determines a map on the set of

closed balls, hence on X, which leaves every horocycle invariant. In other words

the group

Boo

may be identified directly with the group I som(F) of all isometries

of

F.

We give now an alternative description of B

00

•

Theorem. Boo

is the group of all rotations which

fix

oo.

Proof. Let g E Aut( X) and suppose that goo= oo, and for some x EX, gx = x. We

shall prove that for every y

E

X,

gy belongs to the same oo-horocycle as y. Observe

that if gx = x, then g fixes every element of

[x,

oo). Let

[z,

oo) =

[y,

oo)

n

[gy, oo)

and

[t,

oo) =

[x,

oo)

n [z,

oo ). Then gt = t and d(y, z) = d(y,

t)

-d(t, z) = d(gy, gt)-

d(t, z) = d(gy, t)- d(t, z) = d(gy, z). Thus every rotation of

Goo

preserves the oo-

horocycles. Conversely if g

E Boo,

then g cannot be an inversion, because no

inversion can fix an element of the boundary, and

g

cannot be a translation on an

infinite geodesic

(w,

oo) because every such geodesic intersects each horocycle in

exactly one point. We conclude that

Boo

consists of all rotations which leave oo

fixed.

Corollary.

The group

Boo

is normal in

G00

•

Furthermore

Goo

is generated by

Boo

and a step one translation along the geodesic

(0,

oo)

=

{p-kO}.

Therefore

Goo/ Boo

is isomorphic to the group of integers Z.

Proof. Let g

E

B

00

,

and h

E

G

00

•

We show that hgh-

1

is a rotation. Let x be a

vertex fixed by g. Then hgh-

1

fixes oo and hx therefore hgh-

1

is an element of

B

00

.

To complete the proof it suffices to show that

Goo

is generated by

Boo

and a

single translation. Let (0, oo) be the infinite geodesic consisting of the sequence of

balls

p-ko.

Let

g

be an element of

Goo

which is not in

B

00

.

Then

g

is a translation

along a geodesic (a, oo) with a

E

F . Let [x, oo) = (a, oo)

n

(0, oo). Let

hE

Boo

be a rotation which fixes every point of [x, oo) and moves (a, oo) into (0, oo). Then

hgh-

1

is a translation along the geodesic (0, oo). As such it is an appropriate power

It is clear that R is an equivalence relation. The equivalence classes of R are called

the horocycles of X with respect to the boundary point w.

Now if

X

is the tree associated with

F

and we let

w

= oo, (the boundary point

identified by an ascending chain of balls) then the horocycles of X with respect to

oo are exactly the sets of balls of a given radius. Accordingly the horocycles with

respect to oo may be numbered in such a way that the balls of radius qn belong to

the horocycle

Hn.

In other words we stipulate that

p-no

E

Hn

(Fig.l).

THE GROUP OF ISOMETRIES AND ITS ACTION ON THE TREE

Let g

E

Aut(X) be an automorphism of the tree X associated to the local field

F.

Suppose that g has the property that goo= oo. In other words suppose that g

maps every chain of the type [x, oo) into a chain of the same type. Then g defines

a transformation of

0 \ {

oo} = F onto F.

In accordance with the terminology used in

[2]

we shall denote by

Goo

the group

{g

E

Aut(X): goo= oo}. We shall presently describe the action of G

00 •

We start

with the subgroup

Boo

which leaves invariant every horocycle with respect to oo.

The action of

Boo

on

F

is easy to describe. Every element of

Boo

maps balls of a

given radius onto balls of the same radius. In other words an element of

Boo

defines

an isometry of F. Conversely every isometry of F determines a map on the set of

closed balls, hence on X, which leaves every horocycle invariant. In other words

the group

Boo

may be identified directly with the group I som(F) of all isometries

of

F.

We give now an alternative description of B

00

•

Theorem. Boo

is the group of all rotations which

fix

oo.

Proof. Let g E Aut( X) and suppose that goo= oo, and for some x EX, gx = x. We

shall prove that for every y

E

X,

gy belongs to the same oo-horocycle as y. Observe

that if gx = x, then g fixes every element of

[x,

oo). Let

[z,

oo) =

[y,

oo)

n

[gy, oo)

and

[t,

oo) =

[x,

oo)

n [z,

oo ). Then gt = t and d(y, z) = d(y,

t)

-d(t, z) = d(gy, gt)-

d(t, z) = d(gy, t)- d(t, z) = d(gy, z). Thus every rotation of

Goo

preserves the oo-

horocycles. Conversely if g

E Boo,

then g cannot be an inversion, because no

inversion can fix an element of the boundary, and

g

cannot be a translation on an

infinite geodesic

(w,

oo) because every such geodesic intersects each horocycle in

exactly one point. We conclude that

Boo

consists of all rotations which leave oo

fixed.

Corollary.

The group

Boo

is normal in

G00

•

Furthermore

Goo

is generated by

Boo

and a step one translation along the geodesic

(0,

oo)

=

{p-kO}.

Therefore

Goo/ Boo

is isomorphic to the group of integers Z.

Proof. Let g

E

B

00

,

and h

E

G

00

•

We show that hgh-

1

is a rotation. Let x be a

vertex fixed by g. Then hgh-

1

fixes oo and hx therefore hgh-

1

is an element of

B

00

.

To complete the proof it suffices to show that

Goo

is generated by

Boo

and a

single translation. Let (0, oo) be the infinite geodesic consisting of the sequence of

balls

p-ko.

Let

g

be an element of

Goo

which is not in

B

00

.

Then

g

is a translation

along a geodesic (a, oo) with a

E

F . Let [x, oo) = (a, oo)

n

(0, oo). Let

hE

Boo

be a rotation which fixes every point of [x, oo) and moves (a, oo) into (0, oo). Then

hgh-

1

is a translation along the geodesic (0, oo). As such it is an appropriate power