1549380232-Automorphic_Forms_and_Applications__Sarnak_

LECTURE 4. THE SUBCONVEXITY PROBLEM 263

(say) for La small power of Q, on average over F. The multiplicative properties of

>..1(n) and >.. 9 (n) convert this product into a linear form in the Fourier coefficients

off of length QL (and into an expression which makes good sense even if f is not

an eigenform):

(4.29)

'""""' , '""""' a

2

Ltx bn >.. (n)

9 (C) = L., cabeμ(a)xx (a)x(b)>.. 9 (b) L., V( Q) )~,:./aenp 1 (aen).

abe~L n;;,1 a bn

Next, we form the mean square of L 1 x 9 (C) over the chosen family; our objective

now is the bound·

(4.30) ui{

1

)' 1Ltxg(CJl

2

~ L H(t1)ILujxg(C)l

2

+ ... «c,g,k (Q1Qg)^0 L ictl^2 ,

' [q,q J j;;, 1 l~L

for, by choosing the amplifier (4.6), we would then obtain (using (2.16))

Liv(! x g, Q) «c,g,t1 (QJQg)E:( ~~~h112.

We apply (2.28) to the middle term of ( 4.30), converting it into the sum of a
diagonal contribution (coming from the diagonal symbol Orn=n in (2.28)), which is
bounded by

plus a sum of Kloosterman sums of the form

(4.31)

called the off diagonal term. (Here, to simplify presentation we have considered only
the part of ( 4.29) where a = b = 1 and e = l.) From the properties of rapid decay
of I, one sees that the crucial range for the c-sum is when c ~ ~Q, so we
assume from now on that c is within this range; to simplify further, we assume that
( c, £ 2 ) = 1. At this point, one could bound all the terms trivially using Weil's bound
for Kloosterman sums, however, we see quickly that such bound is too large by a
factor of q^112 at least; thus we have to place ourselves in a more favorable position.
The next move is to open the Kloosterman sum

Sx(m,n;c) = 2= x(x)e(mx + nx),

c

x(c),(x,c)=l

getting the additive character e(ncx). Now Voronoi's formula applied to then-sum

transforms e( e^2 ~x) into x ' (e2)e(-ner) (since q'I [q, q'] ic) giving

= L A. 9 (m)>.. 9 (n)x'(f2)Sxx'(f1m - R2n, O; c).:J(m, n; c).

m,n~l