1549380232-Automorphic_Forms_and_Applications__Sarnak_

LECTURE 3. KNOWN BOUNDS FOR THE CUSPIDAL SPECTRUM 77

3.6. A soft^6 proof of the Gelbart-Jacquet estimates.

To illustrate the strength (and perhaps the limitations) of Thro. 3.10 we use it

to give a very short proof of the estimates of [24] for Hecke eigenvalues.

Write G = GL(2)/Q.

7f = n= 0 @np is a cuspidal representation of G(A). We assume w"' trivial

p

for simplicity. For almost all p we have the Hecke matrix tp = (ti tz).

The normalization is Langlands's, so Hp tempered {==} ltil = l.

Theorem 3.13 (Gelbart-Jacquet).^7 -

p-1/4:::; ltil:::; pl/4

So we improve the "trivial estimate" by half (see § 3.1). Of course if n is
associated to classical modular forms this is known, so the point is to prove this for
Maass forms.
Choose a complex quadratic field E with p split in E and a quaternion algebra
B/Q with
B= ~IHI (Hamilton quaternions)
B ®E ~ M2(E).
IQ!
We now denote by G' the Q-group associated with Bx, and by H = SL(l, B)
its derived group, and by He = H 0 E the group 81(2)/ E, which we view as a
IQ!
Q-group.
By base change (Langlands [42b]), n yields a representation ne of GL(2, Ae),
which is cuspidal unless n is obtained from a Grossencharakter of E - but then the
Ramanujan Conjecture is known.
Note that He(Qp) = 81(2, Qp) x 81(2, Qp)· We choose a subrepresentation of
nelsL(Z,Ae ) having a vector fixed by 81(2, Zp) x 81(2, Zp)· It is still automorphic
and cuspidal for He. We simply denote it by ne. Note that by the formalities of
base change:
(ne)p ~Hp 0 Hp
all representations being seen as representations of 81(2).
Now choose S = {oo,p} and apply Theorem 3.10. We see that
(3.15) (ne,= 0 ne,p)IH(IRxlQiv)

has automorphic support. We have H(IR) ~ SU(2), H(Qp) ~ 81(2, Qp)· However,

Lemma 3.14. -Assume T is an automorphic representation of H(A) (precisely:
Tc L^2 (H(Q)\H(A)). If T= ~ C, T satisfies the Ramanujan conjecture.

Indeed, we can essentially think of T as a representation of Bx (A). By Jacquet-
Langlands it transports to a representation of GL(2, A), say T^1. But then T:X, is
associated to classical cusp forms (it cannot be an Abelian character) and we are
reduced to the "holomorphic" Ramanujan conjecture.

6The term aroused some mirth at the conference. It was suggested to me by Jacquet. I stand by

it, and will explain why at the end of the §.

(^7) Gelbart and Jacquet prove in fact strict inequality.