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).
Previous Page Next Page