1549380232-Automorphic_Forms_and_Applications__Sarnak_

LECTURE 4. THE SUBCONVEXI1Y PROBLEM 253

limitation for many analytic applications. By exploiting the symmetry of the set

of divisors^3 of n, the 6-symbol method is capable of providing an expression for

6 ( n) in terms of additive characters of much smaller moduli (i.e. basically of size :::;;

fa). The basic identity is the following: one starts with w(x) a smooth, compactly

supported, even function satisfying

w(O) = 0, :Lw(r) = l ;

r;;;,l

for n an integer, the symmetry of the set of its divisors implies that

o(n) = :L[w(k) - w(~)].

kin

We may now detect the condition kin by means of additive characters:

~1 n ~ an ~ ~ an

(4.9) 6(n) = L k(w(k)-w(k)) L e(k) = LD.r(n) L e(~)

k;;;,l a(k) r;;;,l a(r)

(a,r)=l

where
~1 n
D.r ( n) = L kr ( w ( kr) - w ( kr)).
k;;;,l
In practice the formula is applied to lnl :::;; U /2 (say), for w(x) supported on
R/2 < lxl < R (say) with derivatives satisfying wUl(x) «j R-J-^1 ; if R is chosen
equal to U^112 , one sees that D.r(n) vanishes unless 1 :::;; r:::;; Max(R,U/R) = U^112.
This yields a representation of o(n) for the integers lnl :::;; U/2 in terms of addi-
tive characters of moduli r :::;; U^112 ; moreover, one has (by the Euler-MacLaurin
formula) good control on the derivatives of D.r:

Lemma 4.4.1. For j ~ 1 and lul :::;; U /2 we have

(') 1

D.,! (x) «j (rR+ lxi)(rR)i+^1.

We are now in a position to describe how Duke/ Friedlander/ Iwaniec used the
identity ( 4. 9) to solve the Shifted Convolution Problem [DFil, DFI3]. Applying
(4.9) to the Dirac symbol 6(f 1 m - f2n - h) in (4.8), one gets rid of the constraint
t 1 m-f 2 n-h = 0, which was our primary objective. However, for technical reasons
it is useful (see (4.12)) to keep partial track of this condition, namely that f 1 m -
t 2 n - h cannot be large. This is done by introducing a localization factor ¢(t 1 m -
f 2 n - h) in :Ew(g, t 1 , f 2 , h), where¢ is a smooth function compactly supported in
[-U/2, U/2], satisfying ¢(il(x) «i u-i and ¢(0) = 1; the last condition implies that

the sum :Ew(g,t 1 ,t 2 , h) remains unchanged. Applying (4.9) one obtains

:Ew(g,f1,f2,h)

= :L

-ah~- f1ma-f2na

L e(-) L -\ 9 (m)>. 9 (n)e( )Er(m, n, h),

r r

l,,:;;n;;R a(r),(a,r)=l m,n

where

(^3) if djn then (n/d)jn