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An additive problem with the Fourier coeﬃcients
of holomorphic noncusp forms
Irina Rezvyakova
Steklov Mathematical Institute, Moscow
1–5 December 2014
An additivetype problem
An additive problem
S=X
an−bm=l,
n≤N
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 RamanujanPetersson 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 Lfunction has
a positive proportion of its nontrivial zeros on the critical
line.
2. To break a convexity bound for Lfunction on the critical line.
3. To prove that log L(1/2 + it ) has “normal distribution”.
Normal distribution of Lfunctions 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
e−πu2du +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
e−πu2du +o(T).
Selberg density theorem
To prove such value distribution results, we need a special density
theorem for a given Lfunction.

6
01
1
2σ
T
Let N(σ, T) be the number of zeros of
L(s) in the region Re s≥σ, where
1/2≤σ≤1. Then there is a constant
a>0 such that the following estimate
holds:
N(σ, T)T1−a(σ−1
2)log T.
The proof of Selberg density theorem
Let
Jσ(T) =
2T
R
TL(σ+it)η2(σ+it)−12dt.
The Selberg density theorem follows by standart arguments from
the following estimate
Jσ(T)T1−a(σ−1/2).
Choosing appropriate η(s), we ﬁnd
J1/2(T)T,J3/2(T)T1−a.
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 nontrivial zeros of ζ(s) lie on the
critical line Re s= 1/2.
A. Selberg (1942): a positive proportion of nontrivial zeros of ζ(s)
lie on the critical line Re s=1
2.
Detecting zeros on the critical line
Let F(t) be a realvalued function,
J(t) =
t+H
R
tF(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)H−1zeros on (T,2T).
For a given Lfunctions 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 µj≥0 and choose
F(t)=Λ1
2+itη1
2+it2,
Detecting zeros on the critical line
η(s) = X
ν<√X
α(ν)
νs+ 2 X
√X≤ν≤X
α(ν)
νs
log X/ν
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
tL(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 Lfunctions 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
0M(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
+(t→t+H)+O(X−a/4)
with abeing an arbitrary positive number less than 1.
Proof of the meanestimate 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
2−σ)dσ·
3/2
Z
1/2
Xa
2(σ−1
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 Lfunctions
If we have a linear combination of two (or more) Lfunctions
c1L1(s) + c2L2(s),
with a functional equation but without an Euler product, then
it has many nontrivial zeros outside the critical line.
Selberg theorem for linear combination of Lfunctions
But if we can prove the meanvalue 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 Lfunctions has a
positive proportion of nontrivial zeros on the critical line.
Additive divisor problem
X
m−n=l,n≤N
τ(m)τ(n) = NPl(log N) + R.
History:
A. Ingham (1927): ﬁrst main term.
T. Estermann (1931): RN11/12+ε.
D.R. HeathBrown (1979): RN5/6+ε.
J.M. Deshouillers, H. Iwaniec (1982): RN2/3+ε.
The case ambn = l:
W. Duke, J.B. Friedlander, H. Iwaniec (1994): Ra,b,lN8/9+ε.
Additive problem for modular forms
X
m−n=l,n≤N
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 ambn = l:
J.L. Hafner (198387): both cases (implicit result).
V. Blomer, G. Harcos, P. Sarnak, P. Michel. E. Kowalski,
J. Vanderkam (20012007): congruence subgroups (holomorphic or
Maass).
Circle method
Theorem
Let δ(k) = (1,if k= 0,
0,else.
Then
δ(k) = X∗
X
q≤Q<a∗≤q+Q
1
qa∗
Z
0
2 cos(2πk(a/q−α))dα,
where aa∗= 1( mod q), Q≥1.
Proof
Let f:R→Cbe 1periodic. 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
q−1
q(q+q0),a
q+1
q(q+q00), where
a0
q0<a
q<a00
q00.
We ﬁnd
µ(f) = X∗
X
0<a<q≤Q
1
q(q+q00)
Z
−1
q(q+q0)
fa
q+xdx.
Proof
µ(f) = X∗
X
q≤Q<a∗≤q+Q
0
Z
−1
qa∗
fa
q+xdx +
1
qa∗
Z
0
f−a
q+xdx
.
If f(−x) = f(x), then
µ(f) = 2 Re X∗
X
q≤Q<a∗≤q+Q
1
qa∗
Z
0
fa
q−xdx.
Taking f(x) = e2πikx , we obtain the theorem.
The additive problem
Let a(n) correspond to coeﬃcients of Hecke Lfunction with
complex (or real) ideal class group character of Q(√−D). In case
of a real character we have the identity
a(n) = X
dn
χv(d)χw(n/d),vw =D.
Suppose (m1,m2) = 1. Then
X
m2n2−m1n1=l,
n1<N
a(n1)a(n2) = c(m1,m2,l,v,w)·N+R,
where Rm1,m2,lN7/8+ε,
The additive problem
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
dl,(d,D)=1
1
d
Y
pm1m2,(p,D)=1 1 + 1
p×
×Y
pD
(D,m1m2)1−1
p2−1Y
p(v,l)1−1
pα−1
pα+1 ,
l=Y
p
pα(p).
The additive problem
For simplicity, we consider the case m1=m2= 1. It is possible to
divide this sum into ln Nsmoothed sums of the form
X
n2−n1=l,
1≤n1<N
a(n2)a(n1) =
K
X
k=−KX
n2−n1=l
a(n2)a(n1)gk(n1)
where 0 ≤gk(x)1, vol gkMk,g(j)
kjM−j
k,
K
X
k=−K
gk(n) = 1,1≤n≤N−1
0,n≥Nand n≤0.
The additive problem
If vol gkN7/8+ε, we estimate trivially using Deligne’s result.
The worst situation left is for a smooth function gsatisfying
0≤g(x)1,
supp g⊆(N/2,N),vol gM≥N7/8+ε,g(j)jM−j.
We shall proceed as follows:
X
n2−n1=l
a(n2)a(n1)g(n1) = X
n2−n1=l
a(n2)a(n1)g(n1)g(n2) + O(lNε)
and the error term ﬁts our needs if l≤N7/8.
The additive problem
Letting
S(α) = X
n
a(n)g(n) exp(2πinα),
we have
S:= X
n2−n1=l
a(n2)a(n1)g(n1)g(n2)
= 2 Re X∗
X
q≤Q<a∗≤q+Q
1
qa∗
Z
0S(a/q−α)2e−2πil(a/q−α)dα.
The additive problem
Voronoi summation formula
Let (a,q) = 1, aa∗≡1( 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
q√rh(−D)θv(a)
2Zk(x)dx
+
+∞
X
n=1
τv∗,w∗(n)e−2πi(ar)∗
qn˜
k(n)!,
where ˜
k(n) =
+∞
R
0
k(t)J04π√nt
qdt,r=D/(D,q), v∗=w(q,v)
(q,w),
v∗w∗=D,θv(a) 6= 0 if (q,D) = vor (q,D) = w.
We come to the following sum
σ=X
q≤Q
1/(qQ)
Z
0
e2πilαVq(α)dα,
where
Vq(α) = X
Q<a∗≤min(q+Q,1/(qα))
e−2πila/qS(a/q−α)2.
We apply Voronoi formula with
k(t) = g(t)e−2πinα.
Main term
Vq(α) = c
q2X
a( mod q)
e−2πila/qˆg(α)2+. . . .
S= 2 Re X
q≤Q
c
q2rq(l)
1/(qQ)
Z
0
e2πilαˆg(α)2+. . .
=X
q≤Q
c
q2rq(l)
+∞
Z
0
g(α)g(α+l)dα+. . . .
Error term
RNεX
q≤Q
√q
q2 +∞
X
n=1 Zg(n)e−2πinαJ04π√nα
qdα!2
.
J0(y) = W(y)e2πiy +W(y)e−2πiy ,
yνW(ν)(y)νy−1/2.
Taking Q√N, the integrals won’t have stationary points. Thus,
Z··· M3/2N−1/2q2N
nM2ν
2+1
4
.
If q2N
nM2≤N−ε, then the contribution from those nis negligible.
Error term
For q2N
nM2≥N−εwe apply the estimate for the integral with
ν= 0 and obtain that
RNεX
q≤Q
1
q3/2(qQ)−1
X
n≤q2N1+ε
M2q2N
nM21
4
M3/2N−1/2
2
N
MQ3/2.
Since we should take Q√N, and MN7/8, we obtain
RN7/8+ε.