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
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