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 G 00 • 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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