1549380232-Automorphic_Forms_and_Applications__Sarnak_

LECTURE 3. KNOWN BOUNDS FOR THE CUSPIDAL SPECTRUM 71

First of all, an easy argument [19a,§ 5.1] shows that we may increase Sin order

to prove our theorems. In particular we may assume S 3 oo; we will also assume

that G'(As) is non-compact for each Q-simple Q-factor G' of G.

If K c G(As) is compact-open, where As = IJ 'QP (restricted product), we

pf/:.S

have the associated S-congruence group r = G(Q)nK seen as a discrete subgroup of

G(As). By the arguments sketched at the end of Lecture 1, it will suffice to consider

the action of Gs on L^2 (f\G(As)) for all congruence groups, and the similar actions

of Hs. Note that

G(Q)\G(A)/ K ~ r\G(As)

by strong approximation. This is not true for H, and not necessary.

We begin with Theorem 3.9. By the previous remarks, it suffices to consider

Do c Hs (Greek groups will be congruence subgroups) and to show that the sup-

port of ind~~ (L^2 (D. \Hs)) is automorphic. By induction through stages this is

L^2 (Do\ Gs). We use Theorem 3.4; by Lemma 3.5 we may consider a coefficient c

associated to f E Cc(D. \Gs).

We must show that

c(g) = (gf, f) = r f(x)f(xg)dx
lt:..\es
is uniform limit of coefficients of L^2 (f\Gs) for varying r.
Let Wn be an increasing and exhaustive family of compact a in Gs, and let
WJ C D.\Gs be the support off. We can assume WJ C WJWn· We will construct
subgroups r as follows. We have Do = H(Q) n KH where KH C H(As). Then
KH is contained in a maximal compact subgroup Kg of G(As), and there exists
a decreasing sequence (Kg)°'~ 1 of compact open subgroup of Kg with intersection
KH. Set r °'=Kg n G(Q), thus r °' c Gs and Doc r <>·
Consider the projection 1T°' : Do \G----> r °' \G, and, for n:::: 1, WjWn c Do \G. It is
an easy exercise in the adelic topology to show that (for n fixed) ?T°' is a bijection
(thus a homeomorphism) from WJWn to its image if a>> 0.
Choose such an a= a(n), define f °'(?T°'x) = f(x) for x E WJWn, and extend f°'
by zero tor°' \G. Then for g E Wn:

(gf°',f°')r,,\e = 1 f°'(xg)f°'(x)dx = 1 f°'(1T°'(xg))f(x)dx

~ 0 (w1) Wf

because f°' is supported on 1T°'(w1). But xg E WJWn, so f°'(1T°'(xg)) = f (xg) and

(gf°',f°')r 0 \e = 1 f(xg)f(x)dx = c(g).

Wf

This proves the theorem. The reader is invited to visualize the proof, at least for

S = { oo} and for the quotients of the symmetric spaces. Thus XH is the symmetric

space of H(IR.), Xe that of G(IR.); Do \Xe is a "tube variety", fibered over its totally

geodesic subvariety Do \XH by the fibers of orthogonal projections. For a----> oo the

r °'\Xe coincide with Do \XH along this subvariety, but approximate Do \X along

the fibers. For this see Bergeron [5, Ch. I].

We now prove Theorem 3.10. Assume r E G(Q). Recall that A ·-

L2(G(Q)\G(A)/ K) = L^2 (f\G(As)). Then r defines on A :