1549380232-Automorphic_Forms_and_Applications__Sarnak_

26 ARMAND BOREL, AUTOMORPHIC FORMS ON REDUCTIVE GROUPS

the union of the translates etXj 6 for 0 ::::; t ::::; l. There exists a constant c > 0,

independent off, such that

(67) l(fj - fj+1)(x)I::::; ca(x)>.-/3j L sup IYifj+i(Y)la(y)->.

i yE6'

In particular, if f E C^00 (fN\G) is of uniform bounded growth, bounded by>., then

(68) l(fj - fj+i)(x)I--< a(x)>.-/3i (x E 6).

Proof. It is quite similar to that of 7.4 in [6]. Since rj \Nj is fibered by rj+i \Nj+i

over the quotient of Nj \Nj+l by the cyclic group generated by the element exp Xj

(which leaves f invariant), we can write

(69) fj(x) = 1

1

dt!J+i(etxi .x)

(70) fj(x) - fj+1(x) = 1

1

dt(fj+1(etXix) - fj+1(x))

hence

(71)

However, the derivative is along e^5 xi viewed as a right-invariant vector field,

whereas we consider usually derivatives with respect to le.ft-invariant fields (see

Section 1.2). Set Ys = e^5 xix; it lies in 6'.

( 72 ) ds d f j+l ( e sX.^3 x ) = dr d f j+l ( e r X.^1 Ys ) I r=O = dr d f j+l ( Ys·Ys -1 e r X.^3 Ys ) I r=O

Since the exponential commutes with G, we have:

we get

(73)

Write x = n.a(x).m.k using the Langlands decomposition. Note that Ys =

e^5 Xix has a Langlands decomposition differing from that of x only by the N-factor.

In particular, a(ys) = a(x).

For any u EN, we have

a(x)-

1

uxj = a(x)-/3iXj +X', (X' E nj+1)

Since fj+i is invariant under NJ+ 1 , we have X' !J+i = 0, hence

a(x)-ln-l Xjfj+l = a(x)-/3i Xjfj+l, y:;

1

Xjfj+1 = a(x)-!3i. (mk)-

1

XjfJ+l·

There exist smooth functions Cji on G such that for any g E G, we have

g Xj = L Cji(g).Yj.

Now

(74)