REDUCTIVE GROUPS 25

a point ofS not lying on the geodesic through s and

sf.

Then d(t9 m) \d(t9 s) +

ld(t, s').

Let a = am, then as =

sf.

We have

2d(t, m) = d(t9 a(t)) g d(t9 s) + d(a(t)9 s) = d(t9 s) + d(i, s%

If the extreme terms are equal we have, by 5.25(ii) that s lies on [t, a(t)]. Then so

does s' = as, and s, s\ t lie on a geodesic, which is contrary to the assumption. The

inequality follows.

If C is a subset of X we denote by 1(C) the subgroup of the group of isometries of

S whose elements stabilize C.

5.28.

LEMMA.

IfC is compact the group 1(C) has a fixed point in S.

Let r = mfxGX$upy(=cd(x, y). Then F = {xeS\ supyeC d(x9 y) = r} is the

intersection ofthe decreasing family of sets Fn = {x e S | sup^c d(x9 y) g r + l/n}

(n = 1,2, •••)• Since these are nonempty and compact (by 5.25 (iii)) it follows that

C is nonempty.

Suppose a, beF, a ¥" b and let m be the midpoint of [ab]. We then have by

5.27, for each yeC, d(m, y) \d(a9 y) + \d(b9 y) = r, which is impossible. Hence

F consists of only one point. It is clearlyfixedby 1(C).

5.29.

THEOREM.

Let M be a compact subgroup ofG. Then M fixes a point ofS.

This follows by applying 5.28 to an orbit of M in S.

5.30.

COROLLARY.

M is conjugate to a subgroup ofK.

5.31.

COROLLARY.

All maximal compact subgroups ofG are conjugate.

5.32.

REMARKS.

(1) In our discussion of the symmetric space S we wanted to

stress, more than is usually done, the analogy with the Bruhat-Tits building & of

a/?-adic reductive group. We mention a few features of this analogy.

(a) It is clear from our discussion that S can be obtained, like ^ , by gluing

together apartments (see [21,2.1]).

(b) We have introduced metric and geodesies in S in the same way as is done in

the case of ^ (see [9,2.5] and [21, 2.3]). In the case of & an important role is played

in the discussion of the metric, by the retractions onto an apartment [loc. eit, 2.2].

Such retractions can also be introduced in S (an example is the map p used in the

appendix).

(c) Thefixedpoint Theorem 5.29 has a counterpart for ^ % 3.2.4J.

(d) I owe the proof of 5.29, using the strong convexity property 5.27, to J. J.

Duistermaat. 5.29 can also be proved via the argument used in [8] (see also

[21, 2.3]) to prove its counterpart for @. This requires the inequality (m is the mid-

point of [x,y])

d(xf z)2 + d(y9 zf £ 2d(m9 z)2 + $d(x, yf9

which can also be established in our situation (e.g. by using that the exponential

map 3 — S increases distances).