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# An additive problem with the Fourier coefficients of holomorphic non-cusp forms

Authors:
An additive problem with the Fourier coeﬃcients
of holomorphic non-cusp forms
Irina Rezvyakova
Steklov Mathematical Institute, Moscow
1–5 December 2014
S=X
anbm=l,
nN
a(n)a(m),
where a,b,lare positive integers, which can slightly grow with N.
A trivial estimate
SεN1+ε,
if the coeﬃcients a(n) satisfy the Ramanujan-Petersson conjecture.
Applications
In all problems, where we have to evaluate
2T
Z
T
L1
2+it
2
dt,
or 2T
Z
T
L1
2+itη21
2+it
2
dt,
where
L(s) =
+
X
n=1
a(n)
nsfor Re s>1
and η(s) is a short Dirichlet polynomial.
Applications
The last mentioned integral arises when we want
1. To prove an analogue of Selberg theorem, that L-function has
a positive proportion of its non-trivial zeros on the critical
line.
2. To break a convexity bound for L-function on the critical line.
3. To prove that log L(1/2 + it ) has “normal distribution”.
Normal distribution of L-functions on the critical line
Let κa,bdenote the characteristic function of the interval (a,b),
then
2T
Z
T
κa,blog |L(1/2 + it)|
cLπlog log Tdt =T
b
Z
a
eπu2du +o(T),
or even
2T
Z
T
κa,b log |L1(1/2 + it)| − log |L2(1/2 + it)|
p(c1+c2)πlog log T!dt
=T
b
Z
a
eπu2du +o(T).
Selberg density theorem
To prove such value distribution results, we need a special density
theorem for a given L-function.
-
6
01
1
2σ
T
Let N(σ, T) be the number of zeros of
L(s) in the region Re sσ, where
1/2σ1. Then there is a constant
a>0 such that the following estimate
holds:
N(σ, T)T1a(σ1
2)log T.
The proof of Selberg density theorem
Let
Jσ(T) =
2T
R
T|L(σ+it)η2(σ+it)1|2dt.
The Selberg density theorem follows by standart arguments from
the following estimate
Jσ(T)T1a(σ1/2).
Choosing appropriate η(s), we ﬁnd
J1/2(T)T,J3/2(T)T1a.
Now Gabriel’s convexity theorem yields the desired estimate.
Selberg theorem on the zeros of L(s) on the critical line
The Riemann hypothesis: all non-trivial zeros of ζ(s) lie on the
critical line Re s= 1/2.
A. Selberg (1942): a positive proportion of non-trivial zeros of ζ(s)
lie on the critical line Re s=1
2.
Detecting zeros on the critical line
Let F(t) be a real-valued function,
J(t) =
t+H
R
t|F(u)|du,I(t) =
t+H
R
t
F(u)du
. If J(t)>I(t) then F
has a sign change in (t,t+H), and thus a zero of odd order on
this interval.
If E={t(T,2T) : J(t)>I(t)}, then it is easy to show that F
has µ(E)H1zeros on (T,2T).
For a given L-functions we want to establish that
µ(E)T,while H(log T)1.
Detecting zeros on the critical line
We suppose, that L(s) =
+
P
n=1
a(n)
nssatisﬁes the following functional
equation
Λ(s) := HL(s)L(s) = Λ(1 s),
where HL(s) = θAsQk
j=1 Γ(λjs+µj) and |θ|= 1, A>0, λj>0,
Re µj0 and choose
F(t)=Λ1
2+it|η1
2+it|2,
Detecting zeros on the critical line
η(s) = X
ν<X
α(ν)
νs+ 2 X
XνX
α(ν)
νs
log X
log X
and (L(s))1
2=
+
P
n=1
α(n)
ns.
Detecting zeros on the critical line
Deﬁne
M(t) =
t+H
Z
t
L(1/2 + iu)η2(1/2 + iu)du H.
Then
J(t)c
t+H
Z
t|L(1/2 + iu)η2(1/2 + iu)|du c(H− |M(t)|).
Detecting zeros on the critical line
If we show that
|M(t)|<H/3,|I(t)|<cH/3
then |I(t)|<J(t) and, thus, there is a zero of L(1/2 + iu) on
(t,t+H).
Detecting zeros on the critical line
For certain L-functions one can show (and, moreover,
simultaneously) that
2T
Z
T
|I(t)|2dt =OTH
log T,
2T
Z
T
|L(1/2+it)η2(1/2+it )|2dt =O(T).
From the last estimate we deduce later that
T
Z
0|M(t)|2dt =OTH
log T.
Thus,
|I(t)| ≤ cH
3and |M(t)| ≤ H
3
outside a set of measure OT
Hlog T. Choosing H=A
log Twe get
that µET.
Detecting zeros on the critical line
We now prove the estimate
2T
Z
T
|M(t,H)|2dt =OTH
log T.
Using the Cauchy inequality we get
|M(t,H)|=
t+H
Z
t
(L(1/2 + iu)η2(1/2 + iu)1)du
3/2
Z
1/2
(L(σ+it)η2(σ+it)1)dσ
+|(tt+H)|+O(Xa/4)
with abeing an arbitrary positive number less than 1.
Proof of the mean-estimate for M(t)
Also,
3/2
Z
1/2
(L(σ+it)η2(σ+it)1)dσ
2
3/2
Z
1/2
Xa
2(1
2σ)dσ·
3/2
Z
1/2
Xa
2(σ1
2)L(σ+it)η2(σ+it)1
2dσ.
Setting, for example, a=1
2, we obtain from the estimate on Jσ(T)
2T
Z
T
|M(t,H)|2dt =OT/log2X=O(TH/log T).
Selberg theorem for linear combination of L-functions
If we have a linear combination of two (or more) L-functions
c1L1(s) + c2L2(s),
with a functional equation but without an Euler product, then
it has many non-trivial zeros outside the critical line.
Selberg theorem for linear combination of L-functions
But if we can prove the mean-value estimate
2T
Z
T
|Lj(1/2 + it)η2
j(1/2 + it)|2dt =O(T),
and Lj(s) are independent, then we can prove the value
distribution result for
log |L1(1/2 + it)| − log |L2(1/2 + it)|
which implies that our linear combination of L-functions has a
positive proportion of non-trivial zeros on the critical line.
X
mn=l,nN
τ(m)τ(n) = NPl(log N) + R.
History:
A. Ingham (1927): ﬁrst main term.
T. Estermann (1931): RN11/12+ε.
D.R. Heath-Brown (1979): RN5/6+ε.
J.-M. Deshouillers, H. Iwaniec (1982): RN2/3+ε.
The case am-bn = l:
W. Duke, J.B. Friedlander, H. Iwaniec (1994): Ra,b,lN8/9+ε.
X
mn=l,nN
a(m)a(n)R.
History:
A. Good (1982): RN2/3+ε(holomorphic modular forms).
T. Meurmann (1987): RN2/3+ε(Maass forms).
H. Iwaniec, J.B. Conrey (2001): RN8/9+εcusp forms for
congruence groups.
The case am-bn = l:
J.L. Hafner (1983-87): both cases (implicit result).
V. Blomer, G. Harcos, P. Sarnak, P. Michel. E. Kowalski,
J. Vanderkam (2001-2007): congruence subgroups (holomorphic or
Maass).
Circle method
Theorem
Let δ(k) = (1,if k= 0,
0,else.
Then
δ(k) = X
X
qQ<aq+Q
1
qa
Z
0
2 cos(2πk(a/qα))dα,
where aa= 1( mod q), Q1.
Proof
Let f:RCbe 1-periodic. We shall evaluate
µ(f) =
1
R
0
f(x)dx.
Split the unit interval 1
Q+1 ,Q
Q+1 iinto Farey segments
Ma
q=a
q1
q(q+q0),a
q+1
q(q+q00), where
a0
q0<a
q<a00
q00.
We ﬁnd
µ(f) = X
X
0<a<qQ
1
q(q+q00)
Z
1
q(q+q0)
fa
q+xdx.
Proof
µ(f) = X
X
qQ<aq+Q
0
Z
1
qa
fa
q+xdx +
1
qa
Z
0
fa
q+xdx
.
If f(x) = f(x), then
µ(f) = 2 Re X
X
qQ<aq+Q
1
qa
Z
0
fa
qxdx.
Taking f(x) = e2πikx , we obtain the theorem.
Let a(n) correspond to coeﬃcients of Hecke L-function with
complex (or real) ideal class group character of Q(D). In case
of a real character we have the identity
a(n) = X
d|n
χv(d)χw(n/d),vw =D.
Suppose (m1,m2) = 1. Then
X
m2n2m1n1=l,
n1<N
a(n1)a(n2) = c(m1,m2,l,v,w)·N+R,
where Rm1,m2,lN7/8+ε,
c(m1,m2,l,v,w) = 2π2h2(D)
m2ζ(2) ×
×(δ((m1m2,v)1)Av+δ((m1m2,w)1)Aw),
Av= (v,l)µv
(v,l)
X
d|l,(d,D)=1
1
d
Y
p|m1m2,(p,D)=1 1 + 1
p×
×Y
p|D
(D,m1m2)11
p21Y
p|(v,l)11
pα1
pα+1 ,
l=Y
p
pα(p).
For simplicity, we consider the case m1=m2= 1. It is possible to
divide this sum into ln Nsmoothed sums of the form
X
n2n1=l,
1n1<N
a(n2)a(n1) =
K
X
k=KX
n2n1=l
a(n2)a(n1)gk(n1)
where 0 gk(x)1, vol gkMk,g(j)
kjMj
k,
K
X
k=K
gk(n) = 1,1nN1
0,nNand n0.
If vol gkN7/8+ε, we estimate trivially using Deligne’s result.
The worst situation left is for a smooth function gsatisfying
0g(x)1,
supp g(N/2,N),vol gMN7/8+ε,g(j)jMj.
We shall proceed as follows:
X
n2n1=l
a(n2)a(n1)g(n1) = X
n2n1=l
a(n2)a(n1)g(n1)g(n2) + O(lNε)
and the error term ﬁts our needs if lN7/8.
Letting
S(α) = X
n
a(n)g(n) exp(2πinα),
we have
S:= X
n2n1=l
a(n2)a(n1)g(n1)g(n2)
= 2 Re X
X
qQ<aq+Q
1
qa
Z
0|S(a/qα)|2e2πil(a/qα)dα.
Voronoi summation formula
Let (a,q) = 1, aa1( mod q), k(t) is a smooth function with a
compact support on (0,+). Then
+
X
n=1
τv,w(n)k(n)e2πia
qn=2πi
qrh(D)θv(a)
2Zk(x)dx
+
+
X
n=1
τv,w(n)e2πi(ar)
qn˜
k(n)!,
where ˜
k(n) =
+
R
0
k(t)J04πnt
qdt,r=D/(D,q), v=w(q,v)
(q,w),
vw=D,|θv(a)| 6= 0 if (q,D) = vor (q,D) = w.
We come to the following sum
σ=X
qQ
1/(qQ)
Z
0
e2πilαVq(α)dα,
where
Vq(α) = X
Q<amin(q+Q,1/(qα))
e2πila/q|S(a/qα)|2.
We apply Voronoi formula with
k(t) = g(t)e2πinα.
Main term
Vq(α) = c
q2X
a( mod q)
e2πila/q|ˆg(α)|2+. . . .
S= 2 Re X
qQ
c
q2rq(l)
1/(qQ)
Z
0
e2πilα|ˆg(α)|2+. . .
=X
qQ
c
q2rq(l)
+
Z
0
g(α)g(α+l)dα+. . . .
Error term
RNεX
qQ
q
q2 +
X
n=1 Zg(n)e2πinαJ04πnα
qdα!2
.
J0(y) = W(y)e2πiy +W(y)e2πiy ,
yνW(ν)(y)νy1/2.
Taking QN, the integrals won’t have stationary points. Thus,
Z···  M3/2N1/2q2N
nM2ν
2+1
4
.
If q2N
nM2Nε, then the contribution from those nis negligible.
Error term
For q2N
nM2Nεwe apply the estimate for the integral with
ν= 0 and obtain that
RNεX
qQ
1
q3/2(qQ)1
X
nq2N1+ε
M2q2N
nM21
4
M3/2N1/2
2
N
MQ3/2.
Since we should take QN, and MN7/8, we obtain
RN7/8+ε.
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