1549380232-Automorphic_Forms_and_Applications__Sarnak_

54 L. CLOZEL, SPECTRAL THEORY OF AUTOMORPHIC FORMS

Lemma. - Assume G is a finite group and r an irreducible representation of G;

let 8 be its character. Then

1

8(j8(g)I 2: 1 + c) :::; (l + c) 2

Of course 6 is the density of the indicated set, i.e., its cardinality divided by
IGI. The proof is left to the reader. We refer to Ramakrishnan's paper for more
delicate results relative to the distribution of the ap, following from his method.
In Ramakrishnan's approach, the knowledge of a symmetric power of 7f allows
one to improve the lower bound on ~(Sf). If all symmetric powers were known, we
would get 6(SJ) = 1 (Dirichlet density), which is weaker than the Ramanujan con-
jecture^3. In the meantime, however, results such as Theorem 1.2 and Theorem 1.3
are complementary. (See Shahidi's lectures for more information on Ramakrish-
nan's results).

1.8. The ad e lic metho d

Assume G 1 = SL(2)/Q, and let r c SL(2, Z) be a congruence subgroup. Write

A for the adeles of Q, At for the finite adeles, Of for the integers in At· Then the

closure of r is a compact-open subgroup Krl in G 1 (0 f ). The strong approximation

theorem [4 7] implies

(1.14) r\ SL(2, JR) ~ SL(2, Q)\ SL(2, A)/ Kf,.

Assumer= f 0 (N) and write

Kr = {g E GL(2, 0 f) : g = G : ) [N]}.

Let G = GL(2)/Q. Then we still have

(1.15) JR~G(Q)\G(A)/ Kr = r\ SL(2, JR)

(see Gelbart [23, Ch. 3]). The (classical or Maass) automorphic forms can be seen

as functions on r\ SL(2, JR), so also on G(Q)\G(A).

For a general reductive group G /Q, and K C G(AJ) as before, (1. 1 5) is replaced

by

(1.16) G(Q)\G(A)/K = IIri\G(JR)

(finite union) where the r i are congruence arithmetic groups.
Thus automorphic forms on f\G(JR) - for congruence subgroups of G(Q) -
can be seen as defined on G(Q)\G(A).
Contrary t o intuition, this adeli c theory is easie r. Assume Z is the center of
G - a Q-group. We will fix a unitary character w of Z(A)/Z(Q) and consider the
space

A= L^2 (G(Q)\G(A),w)

of functions f such that f(zg) = w(z)f(g) (z E Z(A)). This carries a unitary
representation of G(A) on the r ight. Using (1.16), one finds that:

(^3) which would be known, of course, by Langlands's approach.