Quantitative version of the joint distribution of
eigenvalues of the Hecke operators
Yuk-Kam LAU1 , Yingnan WANG1,∗
Department of Mathematics, The University of Hong Kong, Pokfulam Road, Hong Kong
Abstract
Recently, Murty and Sinha proved an effective/quantitative version of Serre’s
equidistribution theorem for eigenvalues of Hecke operators on the space of
primitive holomorphic cusp forms. In the context of primitive Maass forms,
Sarnak figured out an analogous joint distribution. In this paper, we prove a
quantitative version of Sarnak’s theorem that gives explicitly estimate on the
rate of convergence. The same result also holds for the case of holomorphic
cusp forms.
Keywords: Quantitative version, joint distribution, primitive Maass form,
Hecke eigenvalues
1. Introduction
Let Sk be the set of all primitive holomorphic cusp forms of even integral
weight k for the full modular group Γ = SL(2, Z). Then for each f ∈ Sk ,
its eigenvalue λf (n) under the nth normalized Hecke operator Tn satisfies
the inquality (Deligne’s bound) |λf (p)| 6 2 for all primes p. Indeed λf (p),
p ≤ X, distribute nicely: as X → ∞, they are equidistributed in [−2, 2] with
respect to the Sato-Tate measure
1√
4 − x2 dx,
2π
a proof of which (the Sato-Tate conjecture) is recently announced in BarnetLamb, Gerahty, Harris and Taylor [1].
∗
1
Corresponding author
Email addresses: [email protected] (Y.K. Lau), [email protected] (Y. Wang)
Published in Journal of Number Theory 131 (2011)
Updated on January 27, 2012
As well for a fixed prime p, the values of λf (p), f ∈ Sk , follows some
distribution law as k → ∞. Conrey, Duke and Farmer [3] and Serre [14]
found that they are equidistributed with respect to the measure
√
4 − x2
p+1
dµp :=
dx.
(1.1)
2π (p1/2 + p−1/2 )2 − x2
In [13], Sarnak figured out the same distribution law, as T → ∞, for the
Hecke eigenvalues λj (p) of primitive Maass forms for fixed p and 1 ≤ j ≤ r(T )
where r(T ) denotes the number of Laplacian Q
eigenvalues up to T 2 . Besides
Sarnak derived the joint probability measure p∈S dµp for (λj (p))p∈S where
S is a finite set of distinct primes.
Recently, Murty and Sinha [12] investigated the effective/quantitative
version, which gives explicitly estimate on the rate of convergence.
Theorem. (Murty & Sinha) Let p be a prime. For an interval [α, β] ⊂
[−2, 2],
Z β
log p
1
µp + O
# {f ∈ Sk : λf (p) ∈ [α, β]} =
,
sk
log k
α
where sk is the cardinality of Sk and the implied constant is effectively computable.
Remark 1. In fact, Murty and Sinha considered the case of congruence
log p
,
subgroups Γ0 (N ) (N ≥ 1) and proved a result with the error term O log
kN
where p and N are coprime. The key ingredients of Murty and Sinha’s
method are Deligne’s inequality, a variant Erdös-Turán inequality, BeurlingSelberg polynomials and Eichler-Selberg trace formula.
In this paper we shall consider the joint distribution and study the case
of (primitive) Maass forms, which carries the difficulties arising from the
possible exceptional eigenvalues. To start with, we brief the setting of Maass
forms.
Let H be the open upper plane in C, and consider the eigenfunctions of the
Laplacian ∆ = −y 2 (∂/∂x)2 +(∂/∂y)2 in L2 (Γ\H), called Maass cusp forms,
where
Γ = SL(2, Z). The L2 norm is induced by the inner product hf, gi =
R
−2
y f (z)g(z) dxdy. The Maass cusp forms span a subspace C(Γ\H) in
Γ\H
2
L (Γ\H). The Hecke operators Tn , n = 1, 2, · · · , together with the Laplacian
2
∆ form a commutative family H of Hermitian operators on L2 (Γ \ H). Let
{uj : j > 0} be a complete orthonormal basis for C(Γ\H) consisting of the
common eigenfunctions of H, where u0 is a constant function. Then
∆uj = (1/4 + t2j )uj ,
Tn uj = λj (n)uj
where 0 < t1 6 t2 6 · · · , and λj (n) ∈ R are parameters that determine uj ,
X
√
uj (z) = yρj (1)
λj (n)Kitj (2π|n|y)e(nx)
(z = x + iy ∈ H)
n6=0
where ρj (1) 6= 0 and Kν is the K-Bessel function of order ν. (By a primitive
Maass form we mean ρj (1)−1 uj (z).) Moreover, we have
1. (Weyl’s law)
1
vol(Γ \ H)T 2 + O(T log T ).
4π
Note that vol(Γ \ H) = π/3. (Γ = SL(2, Z).)
2. (Bounds towards Ramanujan’s conjecture) For all primes p,
r(T ) := #{j : 0 < tj ≤ T } =
|λj (p)| ≤ pθ + p−θ
where θ = 7/64. (See Kim and Sarnak [10].) The conjecture asserts
θ = 0.
Our objective is to prove the following.
Theorem 1. There exists a small constant δ > 0 such that for all sufficiently
large T ,
1
# {0 < tj 6 T : (λj (p1 ), . . . , λj (pN )) ∈ I}
r(T )
Z Y
N
N log(p1 p2 · · · pN )
=
dµpn + O
log T
I n=1
holds uniformly for integer N ≥ 1 and distinct primes p1 , p2 , . . . , pN satisfying
N log(p1 p2 · · · pN ) 6 δ log T,
and uniformly for
I=
N
Y
[an , bn ] ⊂ [−2, 2]N .
n=1
Here dµp is defined as in (1.1).
3
Remark 2. (a) Theorem 1 is a quantitative version of Theorem 1.2 in [13].
The condition N log(p1 · · · pN ) ≤ δ log T is included to give a nontrivial statement, for otherwise, Theorem 1 holds true trivially. Hence one may omit
this condition in the statement. (b) In the course of proof, we showed in
Lemma 4.3 below that the number of ”exceptional” tj ≤ T (in the sense that
|λj (pn )| > 2 for some 1 ≤ n ≤ N ) is
2
log(p1 p2 · · · pN )
r(T )
.
log T
This is analogous to Theorem 1.1 in [13] for N = 1 where Sarnak showed
that the exceptional tj ≤ T for which |λj (p)| ≥ α is
T 2−
log(α/2)
log p
with α > 2.
The same result as in Theorem 1 holds for holomorphic primitive forms,
more precisely, we have
Theorem 2. For all sufficiently large even k, we have
1
# {f ∈ Sk : (λf (p1 ), . . . , λf (pN )) ∈ I}
sk
Z Y
N
N log(p1 p2 · · · pN )
dµpn + O
=
log k
I n=1
Q
N
whenver N log(p1 p2 · · · pN ) 6 δ log k and I = N
n=1 [an , bn ] ⊂ [−2, 2] . Here
sk = |Sk | k, δ > 0 is some suitably small absolute constant and p1 , · · · , pN
denote distinct primes.
We shall omit the proof of Theorem 2 as it is very similar to and even
easier than the proof of Theorem 1. Finally we remark that the method of
our proofs work for primitive forms of higher level, and we shall return to
this case in a suitable occasion.
2. Preliminaries
In this section we cite some results in [2] and [19], with modification to
fit our situation.
4
Let ϕu,v : R/Z → R be the normalized characteristic functions defined
as,

1 if u < x − n < v for some n ∈ Z,



1
ϕu,v (x) =
if u − x ∈ Z or if v − x ∈ Z,


2
0 otherwise,
where u < v < u + 1. For our purpose we take 0 ≤ u < v ≤ 1/2, and define
ϕ
eu,v (x) = ϕu,v (x) + ϕ−v,−u (x) ∈ [0, 1]
for any x ∈ R, since the two intervals (u, v) and (−v, −u) do not overlap in
R/Z. Note that ϕ1−v,1−u (x) = ϕ−v,−u (x) = ϕu,v (−x).
As in [2, §2], we may express ϕu,v as
ϕu,v (x) = (v − u) + ψ(u − x) + ψ(x − v)
with ψ(x) = x − [x] − 1/2 for x ∈
/ Z and ψ(x) = 0 for x ∈ Z.
Moreover, we set
2
X 1
sin π(M + 1)x
|`|
e(`x) =
kM (x) =
1−
M +1
M +1
sin πx
(2.1)
|`|≤M
and
jM (x) =
X
b
J
|`|≤M
`
e(`x)
M +1
(2.2)
b = 1 and J(t)
b = πt(1 − |t|) cot πt + |t| for 0 < |t| < 1 and M is a
where J(0)
positive integer specified up to our disposal. Then we have (from [2, (2.4)]),
|ψ(x) − ψ ∗ jM (x)| ≤ (2M + 2)−1 kM (x)
where
1
Z
ψ ∗ jM (x) =
ψ(y)jM (x − y) dy
0
=
X
(−2πi`)−1 Jb
1≤|`|≤M
5
` e(`x)
M +1
(2.3)
by (2.2). Hence, as in [2, (2.6)], we have
|ϕ
eu,v (x) − α
eu,v (x)| ≤ βeu,v (x)
(2.4)
with
α
eu,v (x) = ϕu,v ∗ jM (x) + ϕ−v,−u ∗ jM (x)
= ψ ∗ jM (u − x) + ψ ∗ jM (x − v)
+ ψ ∗ jM (−v − x) + ψ ∗ jM (x + u)
+ 2(v − u)
(2.5)
and
βeu,v (x) = (2M + 2)−1 kM (x − u) + kM (x − v) + kM (x + u) + kM (x + v) .
Now, we may write α
eu,v and βeu,v in cosine series by direct calculation:
X
α
eu,v (x) = α
bu,v (0) +
α
bu,v (`) cos(2π`x),
1≤|`|≤M
βeu,v (x) = (2M + 2)−1
X
(2.6)
βbu,v (`) cos(2π`x)
|`|≤M
where α
bu,v (0) = 2(v − u), βbu,v (0) = 4 and for ` 6= 0,
`
−1 b
α
bu,v (`) = (πi`) J
(e(−`u) − e(−`v)),
M +1
|`|
βbu,v (`) = 2 1 −
(e(−`u) + e(−`v)).
M +1
b
In view of the definition of J(t),
we have for 1 ≤ |`| ≤ M ,
2 π` |`|
2
|b
αu,v (`)| ≤
cot
1−
+
M +1
M +1
M +1
π(M + 1)
|`|
2 1
≤
1−
+
M +1
πM +1
|`|
≤ 2 1−
=: 2b
kM (`)
M +1
6
(2.7)
since sin θ ≥ 2θ0 /π when θ ∈ [θ0 , π − θ0 ] where 0 < θ0 < π/2. Clearly,
|βbu,v (`)| ≤ 4b
kM (`).
Furthermore, it is established in [2] some nice inequalities under the condition (v − u)(M + 1) ∈ Z; from now on, we impose the stronger condition of
both v(M + 1), u(M + 1) ∈ Z. By [2, (2.10)], when (v − u)(M + 1) ∈ N, we
have 0 ≤ ϕu,v ∗ jM (x) ≤ 1 for all x. The same inequality holds for ϕv,1−v ∗ jM ,
ϕ1−v,1−u ∗ jM and ϕ1−u,u+1 ∗ jM , in light of the new condition. Following the
argument in [2, (2.14)], we have
Z u+1
ϕu,v ∗ jM + ϕv,1−v ∗ jM + ϕ−v,−u ∗ jM + ϕ1−u,u+1 ∗ jM =
jM (x) dx
u
= 1.
Consequently we have the analogous inequality
0≤α
eu,v (x) = ϕu,v ∗ jM (x) + ϕ−v,−u ∗ jM (x) ≤ 1
(2.8)
and plainly, 0 ≤ βeu,v (x) ≤ 2.
Next we turn to a multidimensional version of (2.4): let
Φu,v (x) =
N
Y
ϕ
eun ,vn (xn ),
(2.9)
n=1
where u = (u1 , u2 , · · · , uN ) and v = (v1 , v2 , · · · , vN ) with 0 ≤ un < vn ≤ 1/2
and un (M + 1), vn (M + 1) ∈ Z for all n = 1, 2, . . . , N . For simplicity we
en (x), and similarly for α
eun ,vn and βeun ,vn . Instead
abbreviate ϕ
eun ,vn (x) with ϕ
of using Theorem 7 in [2], we apply the following auxiliary.
Lemma 2.1. We have
N
Y
α
en −
n=1
N
X
βen ≤ Φu,v =
n=1
N
Y
n=1
ϕ
en ≤
N
Y
α
en +
n=1
N
X
βen .
n=1
This can be easily proved by induction with (2.4) and (2.8), for instance,
ϕ
eN +1
N
Y
n=1
α
en − ϕ
eN +1
N
X
n=1
βen ≥
N
+1
Y
α
en − βeN +1
n=1
Hence we have the following result for later use.
7
N
Y
n=1
α
en −
N
X
n=1
βen .
Proposition 1. Let Φu,v : (R/Z)N → R be defined as in (2.9) where 0 ≤
un < vn ≤ 1/2 and un (M + 1), vn (M + 1) ∈ Z. Then for x = (x1 , · · · , xN ),
e
|Φu,v (x) − α(x)|
≤ B(x)
where
e
α(x)
=
N
Y
n=1
B(x) =
N
X
X
α
en (xn ) =
βen (xn ) =
n=1
b
α(`)cos(2π`
x),
`∈([−M,M ]∩Z)N
N
X
X
1
βbn (m) cos(2πmxn ).
2(M + 1) n=1
|m|≤M
Here
b
α(`)
=
N
Y
α
bn (`n ),
cos(2π` x) =
n=1
N
Y
cos(2π`n xn )
(2.10)
n=1
and ` x := (`1 x1 , · · · , `N xN ) denotes the Hadamard product; the values of
α
bn (`) and βbn (m) are defined in (2.7), moreover,
|b
αn (`)| ≤ 2b
kM (`),
|βbn (m)| ≤ 4b
kM (m)
(2.11)
where b
kM (`) = (1 − |`|/(M + 1)).
3. An application of the trace Formula
The following lemma is an analogous result of Theorem 6 in [11] with a
similar proof.
Lemma 3.1 (Kuznetsov). Let m, n be positive integers. Then we have for
arbitrarily small > 0,
X
tj 6T
αj λj (m)λj (n) =
T2
δm,n + O T 1+ (mn)7/64+ + (mn)1/4+ ,
2
π
where αj = |ρj (1)|2 / cosh πtj and δm,n is the Kronecker symbol.
8
Proof. Let H(r, t) = cosh πr/ cosh π(r − t) cosh π(r + t). By a special case of
Kuznetsov trace formula ([11], formula (4.50) or [5], formula (4.8)), we have
∞
X
1
αj λj (m)λj (n)H(tj , t) +
π
j=1
Z∞
σ2ir (n)σ−2ir (m)
−∞
∞
X S(n, m; c)
2t
tδn,m
+
Φ
= 2
π sinh πt π sinh(2πt) c=1
c
m ir
n
H(r, t)
dr
|ζ(1 + 2ir)|2
√
4π mn
,t ,
c
(3.1)
where σν (n) denotes the sum of the νth powers of the positive divisors of
n, ζ(·) is the Riemann zeta-function, S(n, m; c) is the classical Kloosterman
sum and for x > 0,
Z∞
du
Φ(x, t) = x (J2it (u) + J−2it (u))
u
x
with Jν being the J-Bessel function of order ν. We multiply sinh πt on both
sides of (3.1) and integrate over t from 0 to T + 2 log T ; noting that
Z
X
hX (r) =
H(r, t) sinh(πt)dt
cosh πX
1
1
coth πr arctan
− arctan
.
=
π
sinh πr
sinh πr
0
Quite plainly we have for all r > 1,
hX (r) =
1
arctan eπ(X−r) + O(e−πr ),
π
and for 1 6 r 6 X − log X,
hX (r) =
1
+ O(X −3 + e−πr )
2
while for r > X + log X,
hX (r) e−π(r−X) .
9
With the estimate (3.10) below and λj (n) n7/64+ε , we deduce that the
first summation on the left-hand side of (3.1) gives
∞
X
αj λj (m)λj (n)hT +2 log T (tj )
j=1
r(T )
1X
αj λj (m)λj (n) + O (mn)7/64+ T 1+ .
=
2 j=1
(3.2)
Since by [17, §14.2],
Z X
dr
ζ(2)
=
X + o(X) and |ζ(1 + 2ir)| log(1 + 2|r|)−1 ,
2
|ζ(1 + 2ir)|
ζ(4)
0
the second summation leads to
Z∞
m ir h
2
T +2 log T (r)
σ2ir (n)σ−2ir (m)
dr
π
n
|ζ(1 + 2ir)|2
−∞
T +2
Z log T
τ (m)τ (n)
dr
T 1+ (mn) ,
|ζ(1 + 2ir)|2
(3.3)
0
To handle the sum involving Kloosterman sums, we make use of the
formula (5.15) in [11]: uniformly in T > 0,
T +2 log T
√
Z
if a > 1,
 a,
t
Φ (a, t) dt (3.4)
1
 a 1 + log
cosh πt
,
if 0 < a < 1,
a
0
whose proof is included at the end for the convenience of readers.
With Weil’s bound S(m, n, c) τ (c)(m, n, c)1/2 c1/2 , we see that
∞
X
S(n, m; c)
c=1
(mn)1/4
T +2
Z log T
c
t
Φ
cosh πt
√
4π mn
, t dt
c
0
X
√
c64π mn
|S(n, m; c)|
+ (mn)1/2
c3/2
(mn)1/4+ .
10
X
√
c>4π mn
|S(n, m; c)| log c
c2
This proves Lemma 3.1 with (3.1), (3.2) and (3.3).
Finally, we give the proof of (3.4) in [11]. By [6], §7.12, formula (14), we
have the integral representation
Z ∞
4
sin(u cosh ξ) cos(2tξ)dξ.
J2it (u) + J−2it (u) = cosh(πt)
π
0
Dividing the interval [0, ∞) into two and integrating by parts, we get
Z
sin(2tT )
u T
sin(u cosh T ) −
cos(u cosh ξ) sin(2tξ) sinh ξdξ + RT (u, t)
2t
2t 0
(3.5)
where
Z
RT (u, t) =
∞
sin(u cosh ξ) cos(2tξ)dξ.
T
We treat it by an inequality of Landau ([17], Lemma 4.4),
Z ξ 2
−1/2
iF
(ξ)
00
e
dξ 6 8 min |F (ξ)|
.
ξ1
ξ1 6ξ6ξ2
Since (cosh ξ)00 = cosh ξ, it follows that for any u > 0,
RT (u, t) u−1/2 e−T /2
uniformly in t. Therefore, we have from (3.5), for any positive b > a > 0 and
X > 0,
Z b
Z X
du
t
(J2it (u) + J−2it (u)) dt
cosh πt a
u
0
Z b
sin(u cosh T )
cos(2T X) − 1
du
=
πT
u
a
Z
1 T sin(b cosh ξ) − sin(a cosh ξ)
sinh ξ
+
(cos(2Xξ) − 1)
dξ
π 0
cosh ξ
ξ
+ O a−1/2 Xe−T /2 .
Taking T → +∞ and then b → +∞, we have for positive a and X,
Z X
Z
Φ(a, t)
1 ∞ tanh ξ
t
dt =
(1 − cos(2Xξ)) sin(a cosh ξ)dξ. (3.6)
cosh πt a
π 0
ξ
0
11
Now we apply a generalized version of Landau’s inequality, namely, the second derivative test. To this end, we divide the interval of integration into
the subintervals [k, k + 1], k = 0, 1, · · · . Over each subinterval, the function
ξ −1 tanh ξ has a bounded total variation (k + 1)−1 . The second derivative test (see [8], Lemma 5.1.3) gives an estimate O(a−1/2 k −1 e−k/2 ) since
(a cosh ξ ± 2Xξ)00 aek . Consequently the right-hand side of (3.6) is
X
∞
1 −k/2
1
+
min 1, √ e
min(a−1/2 , 1 + | log a|),
min 1, √
a
k
a
k=1
and hence we obtain (3.4).
Next we want to remove the weight αj = |ρj (1)|2 / cosh(πtj ) in Lemma 3.1.
The key is some underlying relation between αj and λj (n)’s. The eigenvalue
λj (n) carries arithmetic structure, for instance,
X
λj (m)λj (n) =
λj (mn/d2 )
(3.7)
d|(m,n)
for any m, n ≥ 1. Like many arithmetic functions, its associated L-function
is endowed with rich analytic properties. Furthermore, its Rankin-Selberg
L-function L(s, uj ×uj ) factorizes into L(s, uj ×uj ) = ζ(s)L(s, sym2 uj ) where
L(s, sym2 uj ) = ζ(2s)
∞
X
λj (n2 )n−s
(3.8)
n=1
is the symmetric square L-function. An evaluation of the residue of the pole
of L(s, uj × uj ) at s = 1 gives
L(1, sym2 uj ) = 2αj−1 .
(3.9)
This relation is ingeniously exploited in [7] to derive
ε
t−ε
j αj tj .
(3.10)
The Dirichlet series
is absolutely convergent in <e s > 1, because in
P in (3.8)
2
[9] it is proved
λj (n) tj N ; thus
n6N
X
n≤N
|λj (n2 )| XX
n≤N d|n
λj
X λj (d)2
n 2
N
tj N 1+ .
d
d
d≤N
12
(3.11)
In addition, Shimura [15] proved that L(s, sym2 uj ) is entire and satisfies the
functional equation
3
s
s
s
Λ(s) := π − 2 s Γ( )Γ( + itj )Γ( − itj )L(s, sym2 uj ) = Λ(1 − s).
2
2
2
With these knowledge we may obtain the following lemma.
Lemma 3.2. For arbitrarily small > 0, we have
X λj (n2 )
n6y
n
=
2
12
+ − 43 +
7
224
+
O
t
y
.
j
π 2 αj
Proof. By Perron’s formula (see [18], Theorem II.2.2),
X λj (n2 )
n6y
n
+iH
Z
1
=
2πi
−iH
X
y |λj (n2 )|
L(s + 1, sym2 uj ) y s
).
ds + O(
1+ (1 + |H log n |)
ζ(2s + 2)
s
n
y
n≥1
The error term is
X
y X |λj (n2 )|
|λj (n2 )|
+
y
H
n1+
n1+ (1 + H|n − y|/n)
|n−y|> y2
|n−y|6 y2
7
25
y 32 H −1 + y − 32 (ytj )
by (3.11) and the bound λj (n) n7/64+ . We move the line of integration
to <e s = − 12 + . It follows with (3.9) that
1
2πi
+iH
Z
−iH
1
12
1
= 2 +
π αj 2πi
− 12 +−iH
Z− 2
+
− 12 +−iH
+iH
Z
Z
+
−iH
− 12 ++iH
As L(s, sym2 uj ) tj on the line <s = 1 + (by (3.11)), the standard
argument with Stirling’s formula, the functional equation and PhragmenLindelöf principle ([16], §5.65) yields
3
L(s, sym2 uj ) t1−β+6
|1 + s| 2 (1−β)+6
j
13
in the strip − 6 <e s = β 6 1 + . Consequently,
±iH
Z
Z
− 12 +±iH
and
3
t−β+
H − 2 β+ y β
j
1
1 1
dβ
H −1 + H − 4 y − 2 tj2 (Hytj )
H
− 21 +
− 12 ++iH
Z
y
1
− 21 + 2 +
ZH
tj
0
− 12 +−iH
23
1
3
dt
+ − 21 + 2 +
4
y
t
.
H
j
(1 + t)1/4
−2
Taking H = y 56 tj 7 , the statement is proved.
Below is an unweighted version of Lemma 3.1 - one of our main tools.
11
, η0 =
Lemma 3.3. Let κ0 = 155
arbitrarily small > 0, we have
X
λj (m)λj (n) =
tj 6T
where σ(`) =
P
d|`
43
620
and m, n be any positive integers. For
1 2
σ((m, n))
T δmn= √
+ O T 2−κ0 + (mn)η0 + ,
12
mn
d and δ`= = 1 if ` is a square and δ`= = 0 otherwise.
Proof. We may assume mn T 2 ; otherwise the expression inside the Oterm T 2.06 dominates and it is trivially valid in view of Lemma 3.1 and
(3.10). By Lemma 3.2, we have
X
tj 6T
λj (m)λj (n) =
X
αj λj (m)λj (n)
X π 2 λj (h2 )
12h
h6y
X
ν+ −τ +
+O T y
|λj (m)λj (n)|
tj 6T
tj 6T
where ν = 2/7 and τ = 43/224. The O-term is T 2+ν+ y −τ + by CauchySchwarz’s inequality, Lemma 3.1 and (3.10). Using (3.7) and Lemma 3.1
with the a relaxation of the O-terms to O(T 1+ (mn)1/4+ ) for simplicity, the
14
main term equals
mn X π2 X X
αj λj
λj (h2 )
2
12h
d
t 6T
h6y
d|(m,n)
j
2
1/4+ T X1 X
=
δ mn2 ,h2 + O T 1+ y 2 mn
d
12 h6y h
d|(m,n)
=
1/4+ T2
σ((m, n))
δmn= √
+ O T 1+ y 2 mn
,
12
mn
(3.12)
2/(1+2τ )
√
where we have assumed y > mn. Setting y = T 1+ν (mn)−1/4
so
√
2(1+ν)/(1+τ )
2
1/4
2+ν −τ
that T (y mn)
= T y and y ≥ mn whenever mn ≤ T
2(1+ν)
which is valid as 1+τ = 192
> 2, we complete the proof of this lemma with
84
2τ −ν
τ
κ0 = 1+2τ and η0 = 2+4τ .
A direct application of Lemma 3.3 gives the following.
Lemma 3.4. Let S be a finite set of distinct primes. Then we have for
mp ≥ 1 (p ∈ S),
X Y
ap λj (pmp ) + bp λj (pmp −2 )
1≤j≤r(T ) p∈S
Y
bp T2 Y
ap
#(S) 2−κ
mp η
T
=
δ2|mp m/2 + m/2−1 + O 2
p
max(|ap |, |bp |)
12 p∈S
p
p
p∈S
where δ2|h = 1 if 2|h and 0 otherwise, ap , bp are constants depending only on
43
11
p, #(S) denotes the cardinality of S, 0 < κ < κ0 = 155
and η > η0 = 620
are
mp −2
any absolute constants. When mp = 1, λj (p
) denotes 0.
Besides, for M ≥ 1, we have
X
Y
λj (pM )2 r(T )
1≤j≤r(T ) p∈S
Y σ(pM )
p∈S
pM
+ T 2−κ
Y 2M η
p
.
p∈S
4. Further Preparation
The Hecke eigenvalue λj (p) can be expressed in terms of the Satake parameters of an automorphic representation, consequently λj (p) = αj,p + βj,p
with αj,p , βj,p ∈ C and αj,p βj,p = 1. It was known quite long time ago that
15
|αj,p | ≤ p1/2 ; the Ramanujan conjecture says |αj,p | = 1 and the state of arts
is |αj,p | ≤ pθ with θ = 7/64. Hence, we may write αj,p = eiθj (p) so that
λj (p) = 2 cos θj (p)
where θj (p) ∈ [0, π] ∪ i(0, θ log p] ∪ π + i(0, θ log p]. (Recall that λj (p) ∈ R.)
Besides we have
λj (pn ) =
sin(n + 1)θj (p)
=: Xn (2 cos θj (p)),
sin θj (p)
(4.1)
i.e. Xn is the nth Chebychev polynomial.
The value of θj (p) is uniquely determined. Consider
λj (p) = 2 cos θj (p) ∈ (a, b) ⊂ [−2, 2].
Then λj (p) ∈ (a, b) is equivalent to θj (p)/(2π) ∈ (u, v) ⊂ [0, 1/2], or equivalently, ϕ
eu,v (θj (p)/(2π)) = 1. Therefore,
X
ϕ
eu,v (θj (p)/(2π))
#{1 ≤ j ≤ r(T ) : λj (p) ∈ [a, b]} ∼
1≤j≤r(T )
θj (p)∈[0,π]
or more precisely,
X
1≤j≤r(T )
λj (p)∈(a,b)
1≤
X
ϕ
eu,v
1≤j≤r(T )
θj (p)∈[0,π]
1
θj (p) ≤
2π
X
1.
(4.2)
1≤j≤r(T )
λj (p)∈[a,b]
Let p1 , · · · , pN be distinct primes, and θj (p) = (θj (p1 ), · · · , θj (pN )) where
θj (pn ) is defined as above. We shall consider
Φu,v
1
θj (p)
2π
defined in Section 2. In light of Proposition 1, we are led to prove the lemmata
below.
Lemma 4.1. For ` = (`1 , · · · , `N ) where |`n | ≤ M , we have
X
cos(` θj (p)) − r(T )cp (`) T 2−κ (p1 · · · pN )M η
1≤j≤r(T )
16
(4.3)
where cp (`) =
QN
n=1 cpn (`n )
with cp (0) = 1, cp (`) = 0 for odd ` and
1
cp (`) := p−|`|/2 (1 − p)
2
for all even ` 6= 0. Here η is chosen as in Lemma 3.4 and the implied constant
in (4.3) depends only on η.
Proof. Using 2 cos(`θ) = X|`| (2 cos θ) − X|`|−2 (2 cos θ) for |`| ≥ 2, we have
 1
 2 λ(p|`| ) − λ(p|`|−2 ) , |`| ≥ 2,
1
cos(`ϑ(p)) =
λ(p),
|`| = 1,
 2
1,
` = 0.
Therefore, if we denote Q(X, Y ) = 12 (X − Y ), then
Y
Y
Q(λj (pn ), 0) ×
Q(λj (p|`n n | ), λj (pn|`n |−2 ))
cos(` ϑj (p)) =
n: |`n |=1
n: |`n |≥2
Applying Lemma 3.4, the product of the main term is
Y
δ2|`n
|`n |≥2
1
1 1
1 −
|`
|/2
|`
n
n
2p
2 p |/2−1
Q
which is equal to `n cpn (`n ) = cp (`). The O-term in Lemma 3.4 is plainly
bounded by the right-hand sided of (4.3) for |`n | ≤ M and |ap |, |bp | ≤ 1/2
and #(S) = N . Our result follows.
b
Lemma 4.2. Let α(`)
and cp (`) be respectively defined as in Proposition 1
and Lemma 4.1. We have
X
b
α(`)c
p (`) =
N Z
Y
n=1
`∈([−M,M ]∩Z)N
vn
2Fpn (y) dy + O(
un
where Fp (y) is defined as in (4.4) below.
Proof. The left-hand side equals
N X
Y
α
bn (`)cpn (`),
n=1 |`|≤M
17
N
)
M
where by (2.7), each local factor is given by
X
` Σp := 2(v − u) +
(πi`)−1 Jb
(e(−`u) − e(−`v))cp (`)
M +1
1≤|`|≤M
= ϕ
eu,v ∗ JM (0)
if we take
JM (x) =
X
` cp (`)e(`x).
M +1
Jb
|`|≤M
R 1/2
b
(Compare with (2.5).) As by (2.2), J(`/(M
+ 1)) = −1/2 jM (t)e(−`t) dt and
R 1/2
j (t)e(−`t) dt = 0 for |`| > M , we see that JM (x) = jM ∗ Fp (x) where
−1/2 M
Fp (y) :=
∞
X
cp (`)e(`y) = 1 + (1 − p)
p−`/2 cos(2π`y)
`=1
2|`
`=−∞
=
∞
X
2(p + 1) sin2 (2πy)
(p1/2 + p−1/2 )2 − 4 cos2 (2πy)
(4.4)
R1
following the calculation in [12, p.698]. Note that Fp (y) ≥ 0 and 0 Fp (y) dy =
1, yielding a probability density function on the space R/Z. Thus Σp =
ϕ
eu,v ∗ jM ∗ Fp (0) = α
eu,v ∗ Fp (0); by (2.8), we infer 0 ≤ Σp ≤ 1 and by (2.4),
|Σp − ϕ
eu,v ∗ Fp (0)| ≤ βeu,v ∗ Fp (0).
Observe that
v
Z
ϕ
eu,v ∗ Fp (0) =
2Fp (y) dy ∈ [0, 1]
u
and by the nonnegativity of βeu,v and (2.6),
Z 1
e
|βu,v ∗ Fp (0)| ≤ max |Fp (y)|
βeu,v (y) dy y∈[0,1]
0
1
.
M +1
In summary we have, similar to the proof of Lemma 2.1,
N X
N
N Z vn
Y
Y
Y
N
α
bn (`)cpn (`) =
Σpn =
2Fpn (y) dy + O( ).
M
n=1
n=1
n=1 un
|`|≤M
18
In contrast with (4.2), the complete sum in (4.3) leads to the problem of
controlling the “exceptional” eigenvalues. When θj (p) = iϑj (p) or π + iϑj (p)
for some real ϑj (p), we do not have the inequalities (2.8). However, we have
from (2.6), (2.11) and (2.1),
X
b
|e
αu,v (θj (p)/(2π))| ≤ 2
kM (`) cosh(`ϑj (p))
|`|≤M
≤ 2
X
b
kM (`) cosh(2`ϑj (p))
|`|≤M
as cosh(φ) ≤ cosh(2φ) for real φ, thus by (2.1) the last line gives
|e
αu,v (θj (p)/(2π))| ≤ 2kM
θj (p) π
2
XM (2 cos(θj (p)))2
M +1
2
=
λj (pM )2
M +1
=
(4.5)
by (4.1). Similarly,
|βeu,v (θj (p)/(2π))| ≤
4
λj (pM )2 .
(M + 1)2
(4.6)
Next we prove that for each prime p, almost all eigenvalues λj (p) fall in
the interval [−2, 2]. Let us write
ΘT (p) = {1 ≤ j ≤ r(T ) : θj (p) ∈ [0, π]}.
(4.7)
Lemma 4.3. Let p be a prime. Then for all sufficiently large T ,
2
1
log p
|#ΘT (p) − r(T )| r(T )
log T
where #ΘT (p) denotes the cardinality of ΘT (p) and the implied constant is
absolute.
Proof. By (4.2) and (2.4), we have
X
r(T ) ≥ #ΘT (p) ≥
ϕ
e0,1/2 (θj (p)/(2π))
1≤j≤r(T )
θj (p)∈[0,π]
≥
X
1≤j≤r(T )
θj (p)∈[0,π]
θj (p)
θj (p)
e
α
e0,1/2 (
) − β0,1/2 (
) .
2π
2π
19
(4.8)
We need to include the exceptional θj (p) to complete the sum, but beforehand, we make a crucial observation: in this case of u = 0, v = 1/2, we have
by (2.6),
α
e0,1/2 (x) − βe0,1/2 (x) = 1 −
X
2
b
kM (`) cos(2π`x)
M + 1 |`|≤M
(4.9)
2|`
as in view of (2.7), α
b0,1/2 (−`) = −b
α0,1/2 (`) for ` 6= 0, and
βb0,1/2 (`) = 2(1 + (−1)` )b
kM (`) = 4δ2|` 1 −
|`| .
M +1
Now we exploit the observation. Assume M = 2L + 1 is odd. By (2.1)
and (4.1) again, the right-hand side of (4.9) is equal to
1−
1 Xb
1 Xb
kL (`) cos(4π`x) = 1 −
kL (`)e(`θj (p)/π)
L+1
L+1
|`|≤L
|`|≤L
1
= 1−
XL (2 cos θj (p))2
(L + 1)2
1
λj (pL )2
= 1−
(L + 1)2
when x = θj (p)/(2π), and furthermore, for exceptional λj (p),
1−
1
1 Xb
L 2
λ
(p
)
=
1
−
kL (`) cosh(2`ϑj (p))
j
(L + 1)2
L+1
|`|≤L
1 Xb
kL (`) = 0.
≤ 1−
L+1
|`|≤L
Consequently, the last sum in (4.8) is
X ≥
1−
1≤j≤r(T )
1
λj (pL )2
(L + 1)2
= r(T ) 1 − O(L−2 ))
if we set
κ log T
M = 2L + 1 = 2
+ 1.
8η log p
20
(4.10)
Here we have used
X
λj (pL )2 T 2 (1 + T −κ p2Lη )
(4.11)
1≤j≤r(T )
from Lemma 3.3. This finishes the proof.
5. Proof of Theorem 1
When N log(p1 · · · pN ) log T , we have N Θ :=
∞
\
√
log T as pi ≥ 2. Define
ΘT (pn )
n=1
and Θ0 := {j : 1 ≤ j ≤ r(T )} \ Θ. (Here we suppress symbols for the
dependence on T and p1 , · · · , pN , as no ambiguity will arise.) With the
notation in Sections 2 and 4, we prove the following.
Q
N
satisfy the conditions in ProposiLemma 5.1. Let N
n=1 [un , vn ] ⊂ [0, 1/2]
tion 1. We have
N Z vn
Y
N log(p1 · · · pN )
1 X
1
2Fpn (y) dy + O
θ (p) =
Φu,v
. (5.1)
r(T ) j∈Θ
2π j
log T
n=1 un
Proof. By Lemma 4.3, we have
2
N
r(T ) X
log(p1 · · · pN )
2
|Θ | (log pn ) r(T )
.
(log T )2 n=1
log T
0
In view of the function B(x) in Proposition 1, we consider |βen (θj (p)/(2π))|
which is, by the line below (2.8) and (4.6), 1 or λj (pM )2 /(M +1)2 according
to θj (p) ∈ R or not. We thus deduce that
N
X
X
1
N
1
M 2
N+
)
N |Θ0 | + r(T ) 2 ,
B
θj (p) λ
(p
j
n
2
2π
(M + 1) n=1
M
j∈Θ0
j∈Θ0
X
by (4.11). Applying Lemma 4.1 to the complete sum in the splitting
X
X
X
=
−
,
j∈Θ
1≤j≤r(T )
21
j∈Θ0
we deduce that
X
j∈Θ
B
N
1
r(T ) X X b
θj (p) =
βn (m)cpn (m)
2π
2M + 2 n=1
|m|≤M
+O T
2−κ
N
X
η
p2M
n
n=1
N |Θ0 | + r(T )
N
+ N |Θ | + r(T ) 2
M
0
N
M
(5.2)
if
M≤
κ
log T
.
4η log(p1 · · · pN )
(5.3)
Next we deal with
X
j∈Θ
e
α
1
θj (p) .
2π
In view of its definition (in Proposition 1), we see by (2.8) and (4.5) that for
j ∈ Θ0 ,
M
Y
1
2
2
e
|α
θj (p) | ≤
λj (pM
1+
n ) ,
2π
M +1
n=1
whence with the second part of Lemma 3.4 and the fact σ(pM )/pM ≤ 2,
1
θj (p)
2π
j∈Θ0
N XY
2
M 2
1+
λj (pn )
≤
M +1
j∈Θ0 n=1
N X
4 h
N
0
2
|Θ | + T
+ 2N T 2−κ (p1 · · · pN )2M η
h
M +1
h=1
X
e
α
|Θ0 | + r(T )
N
M
under the assumption (5.3). Note that the sum over h is (1+ M4+1 )N −1 N/M .
22
Invoking Lemma 4.1, it follows that
X
X
1
e
b
θ
(p)
−
r(T
)
(`)
α
α(`)c
p
j
1≤j≤r(T ) 2π
`∈([−M,M ]∩Z)N
X
Y
2kM (`n )
T 2−κ (p1 · · · pN )2M η
|`1 |,··· ,|`n |≤M n=1
N
(2M + 2) T
2−κ
(p1 · · · pN )2M η .
Together with Lemma 4.2, we infer that
N Z vn
X
Y
1
e
α
2Fpn (y) dy θj (p) − r(T )
2π
n=1 un
j∈Θ
(2M + 2)N T 2−κ (p1 · · · pN )2M η + r(T )
N
+ N |Θ0 |.
M
(5.4)
Combining (5.2) and (5.4) with Proposition 1, we conclude that
N Z vn
X
Y
N log(p1 · · · pN )
1
2F
(y)
dy
Φ
(p)
−
r(T
)
θ
,
r(T )
pn
u,v
j
2π
log
T
u
n
n=1
j∈Θ
with the choice
log T
κ
.
M=
4η log(p1 · · · pN )
(5.5)
Now weQ
are ready to complete the proof. Let T be sufficiently large, and
write I = N
n=1 [an , bn ] with [an , bn ] ⊂ (−2, 2), n = 1, · · · , N . We choose
0
[un , vn ] ⊂ [un , vn0 ] (⊂ [0, 1/2]) such that u(M + 1), v(M + 1) ∈ Z for (u, v) =
(un , vn ) and (u0n , vn0 ), the complement has a small measure
0 0
[un , vn ] \ [un , vn ] 1/M
where M takes the value as in (5.5), and also,
ϕ
eun ,vn (θ/(2π)) ≤ χ[an ,bn ] (2 cos θ) ≤ ϕ
eu0n ,vn0 (θ/(2π))
where χ[a,b] denotes the characteristic function over [a, b]. We denote u =
(u1 , · · · , uN ), v = (v1 , · · · , vN ), and write u0 , v 0 similarly. Applying Lemma 5.1
23
to Φu,v and Φu0 ,v0 , we obtain lower and upper bounds of the form in the rightside of (5.1) for
1
#{1 ≤ j ≤ r(T ) : (λj (p1 ), · · · , λj (pN )) ∈ I}.
r(T )
Write u = (u1 , · · · , uN ) and v = (v1 , · · · , vN ). It remains to show
N Z
Y
n=1
vn
2Fpn (y) =
Z Y
N
un
I n=1
dµp + O(
N
)
M
(5.6)
for (u, v) = (u, v) and (u0 , v 0 ). By a change of variable x = 2 cos 2πy, we see
that (with the subscript n suppressed)
Z v
Z v
4(p + 1) sin2 (2πy)
2Fp (y) dy =
dy
1/2 + p−1/2 )2 − 4 cos2 (2πy)
u (p
u
√
Z
4 − x2
p + 1 2 cos 2πu
dx.
=
2π 2 cos 2πv (p1/2 + p−1/2 )2 − x2
As [2 cos 2πv, 2 cos 2πu] ⊂ [a, b] ⊂ [2 cos 2πv 0 , 2 cos 2πu0 ], we get
Z
v
Z
2Fp (y) dy =
u
b
dµp + O(1/M )
a
and (5.6) follows. Finally we relax the condition [an , bn ] ⊂ (−2, 2) to [−2, 2]
with (5.6). The proof of Theorem 1 is complete.
Acknowledgements
The authors wish to thank the referee and Dr. Jie Wu for their helpful
comments. This work is supported by a grant from the Research Grants
Council of Hong Kong Special Adminstrative Region, China (HKU702308P),
and a GRF incentive award of The University of Hong Kong (10209047).
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