FLUCTUATIONS IN THE DISTRIBUTION OF HECKE EIGENVALUES ABOUT THE
SATO-TATE MEASURE
arXiv:1705.04115v1 [math.NT] 11 May 2017
NEHA PRABHU AND KANEENIKA SINHA
Abstract. We study fluctuations in the distribution of families of p-th Fourier coefficients af (p) of normalised holomorphic Hecke eigenforms f of weight k with respect to SL2 (Z) as k → ∞ and primes p → ∞.
These families are known to be equidistributed with respect to the Sato-Tate measure. We consider a fixed
interval I ⊂ [−2, 2] and derive the variance of the number of af (p)’s lying in I as p → ∞ and k → ∞
(at a suitably fast rate). The number of af (p)’s lying in I is shown to asymptotically follow a Gaussian
distribution when appropriately normalised. A similar theorem is obtained for primitive Maass cusp forms.
1. Introduction
The statistical distribution of eigenvalues of the Hecke operators acting on spaces of modular cusp forms and
Maass forms has been well investigated in recent years ([1], [18], [20]). Among the early developments that
motivated this study was a famous conjecture, stated independently by M. Sato and J. Tate around 1960.
This conjecture predicted a distribution law for the second order terms in the expression for the number of
points in a non-CM elliptic curve modulo a prime p as the primes vary. Serre [19] generalised this conjecture
in 1968 to the context of modular forms. The modular version of the Sato-Tate conjecture can be understood
as follows:
Let k be a positive even integer and N be a positive integer. Let S(N, k) denote the space of modular cusp
forms of weight k with respect to Γ0 (N ). For n ≥ 1, let Tn denote the n-th Hecke operator acting on S(N, k).
We denote the set of all newforms in S(N, k) by FN,k . Any f (z) ∈ FN,k has a Fourier expansion
f (z) =
∞
X
n
k−1
2
af (n)q n ,
n=1
where af (1) = 1 and
Tn (f (z))
= af (n)f (z), n ≥ 1.
k−1
n 2
Let p be a prime number such that gcd (p, N ) = 1. By a theorem of Deligne [7], the eigenvalues af (p) lie in
the interval [−2, 2]. One can study the distribution of the coefficients af (p) in different ways:
(A) (Sato-Tate family) Let N and k be fixed and let f (z) be a non-CM newform in FN,k . We consider
the sequence {af (p)} as p → ∞.
(B) (Vertical Sato-Tate family) For a fixed prime p, we consider the families
{af (p), f ∈ FN,k }, |FN,k | → ∞.
(C) (Average Sato-Tate family) We consider the families
{af (p), p ≤ x, f ∈ FN,k }, |FN,k | → ∞, x → ∞.
Serre’s modular version of the Sato-Tate conjecture predicts a distribution law for the sequence defined in
(A). More explicitly, let I be a subinterval of [−2, 2] and for a positive real number x and f ∈ FN,k , let
NI (f, x) := #{p ≤ x : gcd (p, N ) = 1, af (p) ∈ I}.
Date: May 12, 2017.
1991 Mathematics Subject Classification. Primary 11F11, 11F25, Secondary 11F30.
Key words and phrases. Hecke eigenforms, Sato-Tate distribution, distribution of eigenvalues of Hecke operators, Maass
forms.
The research of the first author is supported by a PhD scholarship from the National Board for Higher Mathematics, India.
1
The Sato-Tate conjecture states that for a fixed non-CM newform f ∈ FN,k , we have
Z
NI (f, x)
= µ∞ (t)dt,
lim
x→∞
π(x)
I
where π(x) denotes the number of primes less than or equal to x and
( q
2
1
if t ∈ [−2, 2]
1 − t4
µ∞ (t) := π
0
otherwise.
The measure µ∞ (t) is referred to as the Sato-Tate or semicircle measure in the literature. This conjecture
has deep and interesting generalisations and has been a central theme in arithmetic geometry over the last
few decades. In 1970, Langlands [10] formulated a general automorphy conjecture which would imply the
Sato-Tate conjecture. This general automorphy conjecture is still open. However, using a very special case
of the Langlands functoriality conjecture, M. R. Murty and V. K. Murty [13] have shown that the general
automorphy conjecture follows.
The Sato-Tate conjecture has now been proved in the highly celebrated work of Barnet-Lamb, Geraghty,
Harris and Taylor [1]. The methods in [1] to address the Sato-Tate conjecture are different from the approach of Langlands: the authors prove that the L-functions Lm (s) associated to symmetric powers of l-adic
representations (l coprime to N ) attached to f are potentially automorphic.
If these L-functions are automorphic, then one can also obtain error terms in the Sato-Tate distribution. In
fact, under the condition that all symmetric power L-functions are automorphic and satisfy the Generalized
Riemann Hypothesis, V. K. Murty [14] showed that for a non-CM newform f of weight 2 and square free
level N, we have
Z
p
NI (f, x) = π(x) µ∞ (t)dt + O x3/4 log N x .
I
Building on Murty’s work, Bucur and Kedlaya [5] have obtained, under some analytic assumptions on motivic
L-functions, an extension of the effective Sato-Tate error term for arbitrary motives. Recently, Rouse and
Thorner [17] have generalised√Murty’s explicit result for all squarefree N and even k ≥ 2, further improving
the error term by a factor of log N x.
In 1987, Sarnak [18] shifted perspectives and considered a vertical variant of the Sato-Tate conjecture in the
case of primitive Maass cusp forms. For a fixed prime p, he obtained a distribution measure for the p-th
coefficients of Maass Hecke eigenforms averaged over Laplacian eigenvalues.
In 1997, Serre [20] considered a similar vertical question for holomorphic Hecke eigenforms. For a fixed prime
p, let |FN,k | → ∞ such that k is a positive even integer and N is coprime to p. Let I be a subinterval of
[−2, 2] and
NI (N, k) := #{f ∈ FN,k : af (p) ∈ I}.
Serre showed that
(1)
NI (N, k)
=
|FN,k |→∞ |FN,k |
lim
Z
µp (t)dt,
I
where
µp (t) =
(
(1−t2 /4)1/2
p+1
π (p1/2 +p−1/2 )2 −t2
0
if t ∈ [−2, 2]
otherwise.
That is,
µp (t) =
(p + 1)
µ∞ (t).
(p1/2 + p−1/2 )2 − t2
The measure µp (t) is referred to as the p-adic Plancherel measure in the literature. This theorem was
independently proved by Conrey, Duke and Farmer [6] for N = 1.
Since averaging over eigenforms provides us with the important tool namely, the Eichler-Selberg trace formula, the quantity NI (N, k) becomes easier to approach. Error terms in Serre’s theorem were obtained by
2
M. R. Murty and K. Sinha [15]. They prove that for a positive integer N, a prime number p coprime to N
and a subinterval I of [−2, 2],
Z
|FN,k | log p
NI (N, k) = |FN,k | µp (t)dt + O
.
log kN
I
In this note, we consider the families described in (C),
{af (p), p ≤ x, f ∈ FN,k }
as |FN,k | → ∞ and x → ∞. In other words, this is the Sato-Tate family (A) averaged over all newforms
in FN,k . In fact, in this direction, the following theorem was proved by Conrey, Duke and Farmer [6]: As
x → ∞ and k = k(x) satisfies logx k → ∞, for any subinterval I of [−2, 2],
X NI (f, x) Z
1
lim
= µ∞ (t)dt.
x→∞ |F1,k |
π(x)
I
f ∈F1,k
Nagoshi [16] obtained the same asymptotic under weaker conditions on the growth of k, namely, k = k(x)
k
satisfies log
log x → ∞ as x → ∞. An effective version of Nagoshi’s theorem was proved by Wang [21]. Under
the above mentioned conditions, he proves that
X NI (f, x) Z
1
log x log x log log x
.
= µ∞ (t)dt + O
+
|F1,k |
π(x)
log k
x
I
f ∈F1,k
We also note that although Conrey, Duke and Farmer [6] and Nagoshi [16] state their “average” Sato-Tate
theorems for N = 1, one can easily generalise their techniques to N > 1 satisfying appropriate growth
conditions. One can show that if k runs over all positive even integers, N runs over positive integers coprime
k)
to p and log(N
log x → ∞ as x → ∞, then
X NI (f, x) Z
1
lim
= µ∞ (t)dt.
x→∞ |FN,k |
π(x)
I
f ∈FN,k
When we vary N, we have to exercise care in choosing N coprime to p. Fixing N = 1 gets rid of this additional
obstacle. Therefore, for simplicity of computation and exposition, we assume that N = 1. Henceforth, we
denote F1,k by Fk and |F1,k | by sk . The “average” Sato-Tate theorem tells us that for a fixed interval I, the
expected value of NI (f, x) as we vary f ∈ Fk ,
1 X
NI (f, x),
sk
E[NI (f, x)] :=
f ∈Fk
is asymptotic to
π(x)
Z
I
µ∞ (t)dt
log k
as x → ∞ with log
x → ∞. We therefore ask what is the variance of this random variable and what can
we say about the fluctuations of NI (f, x) about the expected value. In this direction, we prove that under
appropriate conditions on the growth of k = k(x), NI (f, x) has variance asymptotic to
π(x) µ∞ (I) − (µ∞ (I))2 ,
where
µ∞ (I) :=
Z
µ∞ (t)dt.
I
Finally, when appropriately normalised, NI (f, x) has a Gaussian distribution. More precisely, we prove the
following theorem:
3
Theorem 1.1. Let I = [a, b] be a fixed interval in [−2, 2]. Suppose that k = k(x) satisfies
x → ∞. Then for any bounded, continuous, real-valued function g on R, we have


Z∞
NI (f, x) − π(x)µ∞ (I) 
t2
1 X 
1


g(t)e− 2 dt.
g r
lim
h
i  = √2π
x→∞ sk
2
f ∈Fk
π(x) µ∞ (I) − (µ∞ (I))
−∞
In other words, for any real numbers A < B,




NI (f, x) − π(x)µ∞ (I)
 = √1
r
<
B
A
<
lim Prob Fk 
i
h


x→∞
2π
2
π(x) µ∞ (I) − (µ∞ (I))
Z
B
2
e−t
/2
√log k
x log x
→ ∞ as
dt.
A
1.1. Harmonic averaging. We can also consider a weighted variant of the statistical questions posed in this
article. Instead of uniformly averaging over cusp forms in Fk , we consider the case of harmonic averaging.
That is, for f ∈ Fk , we denote
Γ(k − 1)
ωf :=
,
(4π)k−1 hf, f i
where hf, gi denotes the Petersson inner product of f, g ∈ Sk . We define
X
hk :=
ωf .
f ∈Fk
For a function φ : Sk → C, we denote its harmonic average as follows:
1 X
ωf φ(f ).
hφ(f )ihk :=
hk
f ∈Fk
We can prove the following analogue of Theorem 1.1 with harmonic weights attached to the quantities in
consideration.
Theorem 1.2. Let I = [a, b] be a fixed interval in [−2, 2]. Suppose that k = k(x) satisfies
x → ∞. Then for any bounded, continuous, real-valued function g on R, we have


Z∞
t2
1
ωf NI (f, x) − π(x)µ∞ (I) 
1 X 


= √
g r
g(t)e− 2 dt.
lim
i
h

x→∞ hk
2π
2
f ∈Fk
π(x) µ∞ (I) − (µ∞ (I))
−∞
√log k
x log x
→ ∞ as
1.2. Maass cusp forms. The case of primitive Maass cusp forms admits a similar analysis to what we
present in this article for holomorphic modular cusp forms. We therefore make some observations in this
case.
Let C(Γ\H) denote the space of Maass cusp forms with respect to Γ = SL2 (Z). Let {fj : j ≥ 0} denote an
orthonormal basis for C(Γ\H), which consists of the simultaneous eigenforms of the non-Euclidean Laplacian
operator ∆ and Hecke operators Tn , n ≥ 1. Here, let f0 denote the constant function. For an eigenform fj ,
we have
1
2
+ tj fj , Tn fj = aj (n)fj .
∆fj =
4
For z = x + iy ∈ H, each fj has the Fourier expansion
fj (z) =
∞
X
√
y̺j (1)
aj (n)Kitj (2π|n|y)e(nx),
n=1
where aj (n) ∈ R, ̺j (1) 6= 0 and Kν is the K-Bessel function of order ν. We order the fj ’s so that 0 < t1 ≤
t2 ≤ t3 ≤ . . . . It is well known, by a result of Weyl, that
1 2
T + O(T log T ).
(2)
r(T ) := #{j : 0 < tj ≤ T } =
12
4
The Ramanujan-Petersson conjecture, which is still open, is the assertion that for all primes p,
|aj (p)| ≤ 2.
For an interval I = [a, b] ⊂ R and for 1 ≤ j ≤ r(T ), let us define
NI (j, x) = #{p ≤ x : aj (p) ∈ I}.
We have the following analogue of Theorem 1.1 for Maass cusp forms.
T
→ ∞ as x → ∞. Let I = [a, b] be a fixed interval in
Theorem 1.3. Suppose that T = T (x) satisfies √log
x log x
R. Then for any bounded, continuous, real-valued function g on R, we have


Z∞
r(T )
NI (j, x) − π(x)µ∞ (I) 
t2
1
1 X 


g r
g(t)e− 2 dt.
lim
i  = √2π
h
x→∞ r(T )
2
j=1
π(x) µ∞ (I) − (µ∞ (I))
−∞
In other words, for any real numbers A < B,

lim Prob
x→∞



1
NI (j, x) − π(x)µ∞ (I)


1≤j≤r(T ) A < r
i < B  = √2π
h
π(x) µ∞ (I) − (µ∞ (I))2
Z
B
2
e−t
/2
dt.
A
1.3. Remarks on proofs. The main technique used in the proof of Theorem 1.1 is the approximation of
NI (f, x) by certain trigonometric polynomials called the Beurling-Selberg polynomials. We then estimate
the exponential sums associated to Hecke eigenvalues that arise in these polynomials via the Eichler-Selberg
trace formula. These polynomials were used by M.R. Murty and Sinha [15] and by Wang [21] to obtain error
terms in families (B) and (C) respectively. In this article, we use this technique in a more refined way: we
compute moments of functions arising from the Beurling-Selberg polynomials which give approximations to
higher moments of NI (f, x)− π(x)µ∞ (I). The moments of these modified approximating functions are shown
to match those of the Gaussian distribution after suitable normalisation. This refined technique owes its
origin to the work of Faifman and Rudnick [8], who used it to prove a central limit theorem for the number
of zeros of the zeta functions of a family of hyperelliptic curves defined over a fixed finite field as the genus of
the curves varies. The ideas of Faifman and Rudnick have since been fruitfully adapted by various authors
(for example, [3], [4], [22]) to study similar statistics for different families of smooth projective curves over
finite fields.
k
Nagoshi [16] proved another remarkable theorem. He showed that if k = k(x) satisfes log
log x → ∞ as x → ∞,
then for any bounded continuous real function h on R,
!
P
Z∞
t2
1
1 X
p≤x af (p)
p
h(t)e− 2 dt.
= √
lim
h
x→∞ |Fk |
2π
π(x)
f ∈F
−∞
k
P
In this article, we consider the statistics of NI (f, x) as opposed to p≤x af (p) as f is picked up at random
from Fk .
The proofs of Theorems 1.2 and 1.3 are very similar to that of Theorem 1.1. The key difference is in the
trace formulas used to estimate the exponential sums arising from the Beurling-Selberg approximation for
these families. Hence, we shall omit the proofs. For Theorem 1.2, one uses a trace formula of Petersson (see
[9, Section 2]). For the case of Maass forms, one uses an unweighted version of the Kuznetsov trace formula.
This has been derived by Lau and Wang ([11, Lemma 3.3]). We also make a note that we do not assume the
Ramanujan-Petersson conjecture in Theorem 1.3. Therefore, in treating the case of Maass forms, we have
to take adequate care of the contribution of the “exceptional” eigenvalues aj (p), that is, those eigenvalues
which could possibly lie outside the interval [−2, 2]. This is done with the help of a result of Sarnak ([18,
Theorem 1]) which estimates the density of such exceptional eigenvalues.
5
1.4. Outline. In Section 2, we set up some notation and review some important properties of Hecke eigenvalues that will be needed in the proof of Theorem 1.1. In Section 3, we describe the Beurling-Selberg
polynomials and prove some results about the asymptotics of their Fourier coefficients. In Section 4, we use
the Beurling-Selberg polynomials to derive the expected value of NI (f, x) for f ∈ Fk and obtain error terms
in the theorem of Nagoshi. In Section 5, we derive the second central moment of NI (f, x) and in Section
6, the corresponding asymptotics. In Section 7, we describe the strategy for the proof of Theorem 1.1. We
show that in order to prove Theorem 1.1, it is sufficient to derive the higher odd and even moments of our
modified approximating functions for NI (f, x) − π(x)µ∞ (I). In Section 8, we derive these higher moments
and deduce Theorem 1.1.
2. Preliminaries
In this section, we state fundamental results about modular forms and eigenvalues of Hecke operators that
will be needed in the proof of Theorem 1.1. We start by recalling the following classical lemma which
describes multiplicative relations between af (p)’s.
Lemma 2.1. Let f ∈ Fk . For primes p1 , p2 and non-negative integers i, j,
(
af (pi pj2 )
if p1 =
6 p2
j
i
af (p1 )af (p2 ) = Pmin1 (i,j)
i+j−2l
a
(p
)
if
p
f 1
1 = p2 .
l=0
The recursive relations between af (pm )’s for m ≥ 0 can be elegantly encoded by the following lemma [20,
Lemma 1].
Lemma 2.2. For a prime p and f ∈ Fk , let θf (p) be the unique angle in [0, π] such that af (p) = 2 cos θf (p) .
For m ≥ 0,
af (pm ) = Xm (af (p)),
where the m-th Chebyshev polynomial is defined as follows:
Xm (x) =
We observe that for m ≥ 2,
sin(m + 1)θ
, x = 2 cos θ.
sin θ
2 cos mθ = Xm (2 cos θ) − Xm−2 (2 cos θ).
Thus, we have the following corollary to the above lemma.
Corollary 2.3. With the same notation as in Lemma 2.2, for m ∈ Z, m 6= 0,
(
af (p)
if |m| = 1
2 cos(m θf (p)) =
|m|
|m|−2
af (p ) − af (p
) if |m| ≥ 2.
Proposition 2.4. Let k be a positive even integer and n be a positive integer. We have
(
k−1 √1
X
+ O (nc ) if n is a square
12
n
af (n) =
O (nc )
otherwise.
f ∈Fk
Here, c is an absolute constant and 0 < c < 1. The implied constant in the error term is also absolute.
Proof. This proposition follows from the Eichler-Selberg trace formula for Hecke operators Tn , n ≥ 1 acting
on Sk . The Eichler-Selberg trace formula (see [15, Sections 7, 8] and [20, Section 4]) states that for every
integer n ≥ 1,
4
X
X
Bi (n),
af (n) =
i=1
f ∈Fk
where Bi (n)’s are as follows:
B1 (n) =
(
k−1
12
0
√1
n
6
if n is a square
otherwise.
B2 (n) = −
1
1
2 n(k−1)/2
X
t∈Z, t2 <4n
̺k−1 − ̺k−1
H(4n − t2 ).
̺−̺
Here, ̺ and ̺ denote the zeroes of the polynomial x2 − tx + n and for a positive integer l, H(l) denotes the
Hurwitz class number.
(b)
X
1
dk−1 .
B3 (n) = − (k−1)/2
n
d|n
0≤d≤
√
n
The notation (b) on top of the summation denotes that if there is a contribution from d =
multiplied with 1/2. Finally,
(
P
1
if k = 2,
(k−1)/2
d|n d
n
B4 (n) =
0
otherwise.
√
To estimate B2 (n), we observe that |̺| = n. Thus,
k−1
̺
− ̺k−1 2n(k−1)/2
≤ √4n − t2 .
̺−̺
√
n, it should be
Following a classical estimate of Hurwitz, we have
H(4n − t2 ) ≪
the implied constant being absolute. Thus,
p
4n − t2 log2 (n),
|B2 (n)| ≪
√
n log2 n.
One can immediately observe that
|B3 (n)| ≪
and
|B4 (n)| ≪
X
1
d|n
√
d≤ n
√ X
1.
n
d|n
Combining the above estimates, we prove Proposition 2.4. In particular, n = 1 in the above trace formula gives us
(3)
sk =
k−1
+ O(1).
12
We also record the following important estimate:
X1
= O(log log x).
(4)
p
p≤x
In particular, using Proposition 2.4, we have the following lemma:
Lemma 2.5. Suppose k = k(x) runs over positive even integers such that
any positive integer m and and positive real number a, we have
X X
1
(5)
lim
af (pm ) = 0.
x→∞ (π(x))a sk
log k
log x
→ ∞ as x → ∞. Then, for
p≤x f ∈Fk
Furthermore, for non-negative integers m1 , m2 , . . . mr not all zero,
X
X
1
mr
1 m2
af (pm
(6)
lim
1 p2 . . . pr ) = 0,
a
x→∞ (π(x)) sk
p1 , p2 ,...pr ≤x f ∈Fk
where p1 , p2 , . . . pr are distinct primes not exceeding x.
7
Proof. From Proposition 2.4, equations (3) and (4), one deduces, for m ≥ 1, the following:
P
mc p
1

X
X
if m is even
+
O
m
1
k
p2
p≤x
af (pm ) =
cm
P
p
O
sk
if m ≥ 1, m is odd.
p≤x f ∈Fk
p≤x k

cm

(7)
O(log log x) + O x sπ(x)
if m = 2



cm
k
x π(x)
= O(1) + O
if m > 2, m is even

sk
cm


x π(x)

if m ≥ 1, m is odd.
O
sk
Since
log k
log x
→ ∞ as x → ∞,
xr
=0
x→∞ k
lim
for any real power r > 0. Moreover,
log log x = o(π(x))a ,
for any a > 0. This proves equation (5). Equation (6) follows by a similar argument. Remark 2.6. We note that the proof outlined above gives us a stronger statement, which is of independent
interest. Let us assume the same growth conditions on k as stated above. Equation (7) tells us that for any
a > 1, with ,
X X
1
af (pm ) = 0.
lim
x→∞ (log log x)a sk
p≤x f ∈Fk
Furthermore, for non-negative integers m1 , m2 , . . . mr not all zero,
X
X
1
mr
1 m2
af (pm
lim
1 p2 . . . pr ) = 0,
a
x→∞ (log log x) sk
p1 , p2 ...pr ≤x f ∈Fk
where p1 , p2 , . . . pr are distinct primes not exceeding x.
3. Beurling-Selberg polynomials
The Beurling-Selberg polynomials are trigonometric polynomials which provide a good approximation to the
characteristic functions of intervals in R. The strength of these polynomials is that they reduce the estimation
of counting functions to evaluating finite exponential sums. We briefly review important properties of these
polynomials in this section and refer the reader to a detailed exposition by Montgomery (see [12, Chapter
1]).
−
Let I = [α, β] ⊆ [− 12 , 12 ] and M ≥ 1 be an integer. One can construct trigonometric polynomials SM
(x) and
+
SM (x) of degree less than or equal to M, respectively called the minorant and majorant Beurling-Selberg
polynomials for the interval I such that
−
+
(a) For all x ∈ R, SM
(x) ≤ χI (x) ≤ SM
(x)
(b)
Z 1/2
1
±
SM
(x)dx = β − α ±
,
M
+1
−1/2
(c) For 0 < |m| ≤ M,
1
.
M +1
Henceforth, we will use the following notation: for an interval I = [a, b] ⊆ [−2, 2], we choose a subinterval
1
I1 = [α, β] ⊆ 0,
2
(8)
±
Ŝ M (m) − χ̂I (m) ≤
such that
θ ∈ I1 ⇐⇒ 2 cos(2πθ) ∈ I.
8
±
For M ≥ 1, let SM
(x) denote the majorant and minorant Beurling-Selberg polynomials for the interval I1 .
We denote, for 0 ≤ |m| ≤ M,
±
±
±
ŜM
(m) = ŜM
(m) + ŜM
(−m).
By equation (8), we have, for 1 ≤ |m| ≤ M,
±
ŜM
(m) =
e(−mα) − e(−mβ)
+O
2πim
1
M
and
±
ŜM
(−m)
e(mβ) − e(mα)
+O
=
2πim
Thus,
±
ŜM
(m) =
(9)
1
M
sin(2πmβ) − sin(2πmα)
+O
mπ
.
1
M
.
Proposition 3.1. For [α, β] ⊆ [0, 12 ] and an integer M ≥ 1, we have
(10)
2
M
X
m=1
and, for M ≥ 3,
2
(11)
M−2
X
m=1
±
ŜM
(m)2 = 2(β − α) − 4(β − α)2 + O
±
±
ŜM
(m) SˆM
(m + 2) = −
+
log M
M
1
(sin(2πβ) − sin(2πα))2
π2
1
(1 − 4(β − α))(sin(4πβ) − sin(4πα)) + O
2π
log M
M
.
Proof. We start by making the following observation:
M X
sin(2πmβ) − sin(2πmα) = O(log M ).
mπ
m=1
We have
2
M
X
m=1
±
ŜM
(m)2
2
M X
sin(2πmβ) − sin(2πmα)
1
+O
mπ
M
m=1
M 2 X sin2 (2πmβ) sin2 (2πmα)
sin(2πmβ) sin(2πmα)
log M
+
O
= 2
+
−
2
π m=1
m2
m2
m2
M
=2
2
= 2
π
2
= 2
π
M
M
M
M
X
sin2 (2πmβ) X sin2 (2πmα) X cos(2πm(β − α)) X cos(2πm(β + α))
+
−
+
m2
m2
m2
m2
m=1
m=1
m=1
m=1
∞
∞
∞
∞
X
sin2 (2πmβ) X sin2 (2πmα) X cos(2πm(β − α)) X cos(2πm(β + α))
+
−
+
m2
m2
m2
m2
m=1
m=1
m=1
m=1
We now use the trigonometric sum (see [2, p. 360]),
∞
X
sin2 (mθ)
1
= θ(π − θ),
2
m
2
m=1
9
0 ≤ θ ≤ π.
!
!
+O
log M
M
+O
log M
M
.
We also have (see [2, p. 370]),
∞
X
cos(2πmθ)
1
= θ2 − θ + ,
2
(mπ)
6
m=1
Therefore, we have
2
0 < θ < 1.
M
X
1
±
ŜM
(m)2 = 2β(1 − 2β) + 2α(1 − 2α) − 2 (β − α)2 − (β − α) +
6
m=1
log M
1
+O
+ 2 (β + α)2 − (β + α) +
6
M
log
M
= 2(β − α) − 4(β − α)2 + O
.
M
This proves equation (10).
In order to prove equation (11), we observe,
2
M−2
X
m=1
=2
±
±
ŜM
(m) ŜM
(m + 2)
M−2
X
m=1
=
sin(2πmβ) − sin(2πmα)
mπ
sin(2π(m + 2)β) − sin(2π(m + 2)α)
(m + 2)π
+O
log M
M
M−2
M−2
2 X sin(2πmβ) sin(2π(m + 2)β)
2 X sin(2πmβ) sin(2π(m + 2)α)
−
π 2 m=1
m(m + 2)
π 2 m=1
m(m + 2)
M−2
M−2
2 X sin(2πmα) sin(2π(m + 2)α)
log M
2 X sin(2πmα) sin(2π(m + 2)β)
.
+ 2
+O
− 2
π m=1
m(m + 2)
π m=1
m(m + 2)
M
By applying standard trigonometric identities that allow us to write a product of sines as a difference of
cosines and simplifying, we arrive at the following equation:
2
M−2
M−2
M−2
1 X cos(4π(m + 1)β)
1 X cos(4πα)
1 X cos(4πβ)
±
±
− 2
+ 2
ŜM
(m) SˆM
(m + 2) = 2
π m=1 m(m + 2) π m=1
m(m + 2)
π m=1 m(m + 2)
m=1
M−2
X
M−2
M−2
1 X cos(4π(m + 1)α) 2 cos(2π(α + β)) X cos(2π(m + 1)(β − α))
−
π 2 m=1
m(m + 2)
π2
m(m + 2)
m=1
M−2
2 cos(2π(β − α)) X cos(2π(m + 1)(β + α))
log M
+
O
+
π2
m(m + 2)
M
m=1
−
We use the following trigonometric sum (see [2, p. 368]):
∞
X
cos((m + 1)θ)
1 cos(θ) π − θ
= +
−
sin(θ),
m(m
+
2)
2
4
2
m=1
Using the above equation and the following identity:
∞
X
1
m=1
m(m + 2)
we deduce the following:
P cos(4πβ)
3
1 M−2
1
.
= 2 cos(4πβ) + O
(A) 2
π m=1 m(m + 2)
4π
M
10
=
3
4
0 < θ < 2π.
(B) −
(C)
1 cos(4πβ) (π − 4πβ)
1
.
+
−
sin(4πβ) + O
2
4
2
M
P cos(4πα)
3
1 M−2
= 2 cos(4πα) + O
2
π m=1 m(m + 2)
4π
(D) −
(E)
P cos(4π(m + 1)β)
1
1 M−2
=− 2
2
π m=1
m(m + 2)
π
P cos(4π(m + 1)α)
1 M−2
1
=− 2
2
π m=1
m(m + 2)
π
1
M
.
1 cos(4πα) (π − 4πα)
1
.
+
−
sin(4πα) + O
2
4
2
M
M−2
2 cos(2π(α + β)) X cos(2π(m + 1)(β − α))
π2
m(m + 2)
m=1
2 cos(2π(α + β)) 1 cos(2π(β − α)) (π − 2π(β − α))
1
=−
.
+
−
sin(2π(β − α)) + O
π2
2
4
2
M
−
(F)
M−2
2 cos(2π(β − α)) X cos(2π(m + 1)(β + α))
π2
m(m + 2)
m=1
1
2 cos(2π(β − α)) 1 cos(2π(α + β)) (π − 2π(α + β))
.
+
−
sin(2π(α + β)) + O
=
2
π
2
4
2
M
From equations (A)-(D) above, we obtain
M−2
M−2
M−2
M−2
1 X cos(4π(m + 1)β)
1 X cos(4πα)
1 X cos(4π(m + 1)α)
1 X cos(4πβ)
−
+
−
π 2 m=1 m(m + 2) π 2 m=1
m(m + 2)
π 2 m=1 m(m + 2) π 2 m=1
m(m + 2)
1
cos(4πβ) − 1 cos(4πα) − 1
1
.
+
+
((1 − 4β) sin(4πβ) + (1 − 4α) sin(4πα)) + O
=
2π 2
2π 2
2π
M
From equations (E) and (F) above, we get
−
M−2
M−2
2 cos(2π(α + β)) X cos(2π(m + 1)(β − α) 2 cos(2π(β − α)) X cos(2π(m + 1)(β + α)
+
π2
m(m + 2)
π2
m(m + 2)
m=1
m=1
1
1
(cos(2π(β − α)) − cos(2π(α + β))) + (1 − 2(β − α)) cos(2π(α + β)) sin(2π(β − α))
π2
π
1
1
− (1 − 2(α + β)) cos(2π(β − α)) sin(2π(β + α)) + O
π
M
(1 − 2β)
2α
1
2
.
sin(4πα) +
sin(4πβ) + O
= 2 (sin(2πβ) sin(2πα)) −
π
π
π
M
=
Therefore, putting all of it together we have
2
M−2
X
m=1
=
±
±
ŜM
(m) ŜM
(m + 2)
cos(4πβ) − 1 cos(4πα) − 1
1
+
+
((1 − 4β) sin(4πβ) + (1 − 4α) sin(4πα))
2π 2
2π 2
2π
(1 − 2β)
2
2α
log M
.
+ 2 (sin(2πβ) sin(2πα)) −
sin(4πα) +
sin(4πβ) + O
π
π
π
M
Simplifying, we get equation (11). This proves the proposition.
11
For M ≥ 3 and 1 ≤ m ≤ M, let
±
ÛM
(m)
(
:=
±
±
ŜM
(m) − ŜM
(m + 2),
±
ŜM (m),
if 1 ≤ m ≤ M − 2
if m = M − 1, M.
We record the following bound, which is not optimal, but good enough for our purposes.
Lemma 3.2. Let I = [α, β] be a fixed interval and mr = (m1 , . . . , mr ) be an r-tuple of positive integers such
±
±
±
(m1 ) · · · ÛM
(mr ).
that 1 ≤ mi ≤ M. Let ÛM
(mr ) = ÛM
(3)
X
mr
Here,
(3)
P
±
| ÛM
(mr )| = Or (log M )r .
denotes that the sum is taken over r-tuples of positive integers lying between 1 and M.
mr
Proof. From equation (8), we observe that for any 1 ≤ m ≤ M,
±
| ÛM
(m)| ≤
2
2
+
.
π|m| M + 1
Thus, for a fixed r-tuple (mr ),
±
| ÛM
(mr )| ≪
r Y
1
1
+
mj
M +1
j=1
r
≪
Hence,
(3)
X
mr
X
1
1
+
k
(M + 1)
(M + 1)r−k
±
| ÛM
(mr )| ≪
k=1
(3)
X
mr
(3)

r
X
j1 ,j2 ,...,jk
r
X
1
1

+
k
(M + 1)
(M + 1)r−k
k=1
1
.
mj1 mj2 · · · mjk
=1
r
X
j1 ,j2 ,...,jk
(3)
r
r
X
X
X
1
1
1
+
k
r−k
m
(M
+
1)
(M
+
1)
m
j1 j2 · · · mjk
mr j1 ,j2 ...,jk =1
mr
k=1
r X
r
1
≪
M r−k (log M )k ≪r (log M )r .
k (M + 1)r−k
≪
X
k=0
4. First moment
For an interval I = [a, b] ⊆ [−2, 2], we define
NI (f, x) := # {p ≤ x : af (p) ∈ I} .
Denoting af (p) = 2 cos θf (p), with θf (p) ∈ [0, π], we consider the families
θf (p)
θf (p)
,−
, f ∈ Fk .
2π
2π
As before, we choose a subinterval
1
I1 = [α, β] ⊆ 0,
2
so that

1

m
j
1 mj2 · · · mjk
=1
θf (p)
∈ I1 ⇐⇒ 2 cos θf (p) ∈ I.
2π
12
We denote I2 = (α, β]. Thus,
NI (f, x) =
X
χI1
p≤x
θf (p)
2π
θf (p)
+ χI2 −
,
2π
since
θf (p)
= 0.
−
2π
Following the notation and properties of the Beurling-Selberg polynomials from the previous section, we
have
X
X
θf (p)
θf (p)
θf (p)
θf (p)
−
−
+
+
(12)
SM
+ SM −
≤ NI (f, x) ≤
+ SM −
.
SM
2π
2π
2π
2π
χI2
p≤x
p≤x
We now denote
Xf (x) = NI (f, x) − π(x)
Z
µ∞ (t)dt
I
and explore the moments
lim E[Xf (x)r ] = lim
x→∞
x→∞
1 X
(Xf (x))r .
sk
f ∈Fk
Our strategy is to use equation (12) to approximate Xf (x) by certain trigonometric polynomials and evaluate
the moments of these polynomials.
We observe
X
θf (p)
θf (p)
+
+
NI (f, x) ≤
SM
+ SM −
2π
2π
p≤x
X X
m θf (p)
m θf (p)
+
+e −
ŜM (m) e
=
2π
2π
p≤x |m|≤M
i
X X h
+
ŜM
(m) (2 cos(m θf (p)))
=
(13)
p≤x |m|≤M
=2
X
+
ŜM
(0) +
M
X
+
+
(ŜM
(m) + ŜM
(−m))
m=1
p≤x
2 cos(m θf (p))
p≤x
2
M +1
+
2
NI (f, x) ≥ π(x) 2(β − α) −
M +1
+
= π(x) 2(β − α) +
X
M
X
+
ŜM
(m)
M
X
−
ŜM
(m)
m=1
X
2 cos(m θf (p)).
X
2 cos(m θf (p)).
p≤x
By a similar argument, we derive
(14)
m=1
p≤x
Let us denote
S ± (M, f )(x) :=
2
X
m=1
±
ŜM
(m)
X
af (pm ) +
M
X
m=3
p≤x
±
ŜM
(m)
X
p≤x
af (pm ) − af (pm−2 ) .
By combining equations (13), (14) and Corollary 2.3, we get
h
i
π(x)
+
(15)
NI (f, x) − π(x) (2β − 2α) − ŜM
(2) ≤ S + (M, f )(x) + 2
M +1
and
h
i
π(x)
−
(16)
−2
+ S − (M, f )(x) ≤ NI (f, x) − π(x) (2β − 2α) − ŜM
(2) .
M +1
13
We are now ready to calculate the first moment of NI (f, x). Henceforth, for any function φ : Sk → C, we
denote the average
1 X
hφ(f )i :=
φ(f ).
sk
f ∈Fk
r
In order to derive the moments h(Xf (x)) i, we explore the moments of S ± (M, f )(x). In this direction, we
state the following proposition, which shows that the Sato-Tate conjecture is true on average as x → ∞.
Proposition 4.1. Let k = k(x) be a positive even integer. Then, for any interval I = [a, b] ⊆ [−2, 2],
Z b
π(x) log x
hNI (f, x)i = π(x)
µ∞ (t)dt + O
+ log log x .
log k
a
k
Thus, if k = k(x) runs over positive even integers such that log
log x → ∞ as x → ∞, then
Z b
1
lim
hNI (f, x)i =
µ∞ (t)dt.
x→∞ π(x)
a
Remark 4.2. Proposition 4.1 is essentially due to Y. Wang [21, Theorem 1.1]. He proves an analogous
result for primitive Maass forms and indicates that a similar technique works for the average Sato-Tate
family. We provide a brief proof of this proposition as a first step in evaluating moments of the polynomials
S ± (M, f )(x).
Proof. We have, by equation (9),
±
ŜM
(2)
sin 4πβ − sin 4πα
+O
=
2π
1
M
.
Combining the above with equations (15) and (16), we can find constants C and D such that
sin 4πβ − sin 4πα
π(x)
−
≤ NI (f, x) − π(x) (2β − 2α) −
S (M, f )(x) + C
M +1
2π
(17)
π(x)
≤ S + (M, f )(x) + D
.
M +1
We observe, for [α, β] ∈ [0, 1/2] as chosen before,
(2β − 2α) −
sin 4πβ − sin 4πα
=2
2π
Z
=4
=
Z
β
α
Z β
a
(1 − cos 4πθ)dθ
sin2 2πθdθ
α
b
µ∞ (t)dt.
Thus, for every positive integer M,
Z b
π(x)
π(x)
µ∞ (t)dt ≤ S + (M, f )(x) + D
≤ NI (f, x) − π(x)
.
(18)
S − (M, f )(x) + C
M +1
M +1
a
By equation (7),

(
M
X p1 ,
X
±
±

S (M, f )(x) = O
| ŜM (m)|
1
m=1
m even
±
Since | ŜM
(m)| ≪
1
m,
p≤x
p
m −1
2
,
we get, for every positive integer M,
M
X
m=1
xcm π(x)
±
| ŜM
(m)|
k
=O

if m = 2
+O
if m ≥ 4
M
π(x) X xcm
k m=1 m
14
!
=O
M
X
m=1
xcm π(x)
±
| ŜM
(m)|
π(x)xcM
k
k
.
!
.
Thus,

cm
x
π(x)
±

+
| ŜM
(m)|
k
m=1
p≤x
cM
x π(x)
= O log log x +
.
k

X
±
S (M, f )(x) = O 
∞
1 X 1
+
p m=2 pm
M
X
!
That is, for every positive integer M, by equation (18), we have
*
NI (f, x) − π(x)
Z
b
µ∞ (t)dt
a
+
xcM π(x)
π(x)
= O log log x +
.
+
k
M +1
We now choose
d log k
M=
c log x
for some 0 < d < 1. This proves the proposition. 5. Second moment
In this section, we will compute
1 ±
(S (M, f )(x))2 ) .
x→∞ π(x)
lim
Proposition 5.1. Let [α, β] be a fixed interval in [0, 1/2]. Then, for every M ≥ 3,
±
(S (M, f )(x))2 )
(19)
= π(x)
M−2
X
m=1
±
ŜM
(m)
−
±
ŜM
(m
2
±
±
+ 2) + ŜM
(M − 1)2 + ŜM
(M )2
!
π(x)2 x2Mc
.
+ O (log log x)2 +
k
Thus, if k = k(x) runs over positive even integers such that
log k
log x
→ ∞ as x → ∞, then, for every M ≥ 3,
1 ±
(S (M, f )(x))2
x→∞ π(x)
lim
=2
M
X
m=1
±
ŜM
(m)2 − 2
M−2
X
m=1
±
±
±
±
ŜM
(m) ŜM
(m + 2) − ŜM
(1)2 − ŜM
(2)2 .
Proof. Recall, for M ≥ 3 and 1 ≤ m ≤ M,
±
ÛM
(m)
:=
(
±
±
ŜM
(m) − ŜM
(m + 2),
±
ŜM (m),
15
if 1 ≤ m ≤ M − 2
if m = M − 1, M.
We start by observing that

S ± (M, f )(x))2 = 
2
X
m=1

(20)
=

=
=
±
ŜM
(m)
X
±
±
(ŜM
(m) − ŜM
(m + 2))
M
X
±
ÛM
(m)
m=1
m=1
M
X
m1 ,m2 =1
X
p≤x
af (pm ) +
m=3
p≤x
M−2
X
2
M
X
X
p≤x
±
ŜM
(m)
X
p≤x
±
af (pm ) + ŜM
(M − 1)
X
p≤x
2
af (pm ) − af (pm−2 ) 
±
af (pM−1 ) + ŜM
(M )
X
p≤x
2
af (pM )
af (pm )
±
±
ÛM
(m1 ) ÛM
(m2 )
X
m2
1
af (pm
1 )af (p2 ).
p1 ,p2 ≤x
Applying Lemma 2.1 and Proposition 2.4, we deduce that if p1 6= p2 then
M
X
m1 ,m2 =1
±
±
ÛM
(m1 ) ÛM
(m2 )
M
X
=
m1 ,m2 =1
1 X
sk
X
m2
1
af (pm
1 )af (p2 )
f ∈Fk p1 6=p2 ≤x
±
±
ÛM
(m1 ) ÛM
(m2 )
X
p1 6=p2 ≤x
δ(m1 )δ(m2 )
+O
(p1 )m1 /2 (p2 )m2 /2
1 cm2
pcm
p2
1
k
±
where δ(n) = 1 if n is even and is zero otherwise. Using the estimate | ÛM
(m)| ≪
,
1
m,
the above is equal to
!
[M/2]
M
1 cm2
X δ(m1 )δ(m2 )
X
X
pcm
p2
±
±
±
±
1
| ÛM (m1 )|| ÛM (m2 )|
ÛM (2m1 ) ÛM (2m2 )
+O
m1 /2 m2 /2
k
p2
m1 ,m2 =1
m1 ,m2 =1
p1 6=p2 ≤x p1



[M/2]
M
1 cm2
X
X
X
X
1
pcm
p2 
1


≪
+
O
m1 m2
m1 m2 p 1 p 2
km1 m2
m1 ,m2 =1 p1 6=p2 ≤x
p1 6=p2 ≤x m1 ,m2 =1
π(x)2 x2Mc
.
= O(log log x)2 + O
k
Here, as in the first moments case, we have used the elementary estimates


X [M/2]
X 1
X1
 = O(log log x)
= O
m
mp
p
m=1
p≤x
p≤x
and


M
X pM
XX
π(x)xM
pm


.
=O
=O
mk
k
k
m=1
p≤x
p≤x
If p1 = p2 = p, then,
M
X
m1 ,m2 =1
=
±
±
ÛM
(m1 ) ÛM
(m2 )
M
X
m1 ,m2 =1
=
X
p≤x


1 X X
af (pm1 )af (pm2 )
sk
±
±
ÛM
(m1 ) ÛM
(m2 )
M
X
m1 ,m2 =1
f ∈Fk p≤x
min{m1 ,m2 }
X
1 X X
af (pm1 +m2 −2i )
sk
i=0
f ∈Fk p≤x
±
±
ÛM
(m1 ) ÛM
(m2 )
min{m1 ,m2 }
X
i=0
16
δ(m1 + m2 )
+O
p(m1 +m2 −2i)/2

p(m1 +m2 −2i)c 
.
k
In the expression
M
X
X
p≤x m1 ,m2 =1
±
±
ÛM
(m1 ) ÛM
(m2 )
min{m1 ,m2 }
X
i=0
δ(m1 + m2 )
,
p(m1 +m2 −2i)/2
the innermost sum contributes 1 exactly when m1 = m2 = i.
Otherwise, if m1 6= m2 , then we have 1 ≤ i ≤ min{m1 , m2 } implying that m1 + m2 − 2i > 0. Thus,
±
±
ÛM
(m1 ) ÛM
(m2 )
min{m1 ,m2 }
X
i=0
Therefore we may write,
X
M
X
p≤x m1 ,m2 =1
= π(x)
M
X
M
X
m1 =1
Moreover,
M
X
m1 ,m2 =1
min{m1 ,m2 }
±
±
ÛM
(m1 ) ÛM
(m2 )
m1 =1
= π(x)
1
δ(m1 + m2 )
≪
(m
+m
−2i)/2
1
2
m1 m2
p
X
i=0

X
±
ÛM
(m1 )2 + O 
p≤x
min{m1 ,m2 }
X
i=1
δ(m1 + m2 )
.
p(m1 +m2 −2i)/2
δ(m1 + m2 )
p(m1 +m2 −2i)/2
M
X
m1 ,m2 =1
m1 6=m2
min{m1 ,m2 }
1
m1 m2
X
i=0

δ(m1 + m2 ) 

p(m1 +m2 −2i)/2
±
ÛM
(m1 )2 + O(log log x).
±
±
ÛM
(m1 ) ÛM
(m2 )
min{m1 ,m2 } (m +m −2i)c X
1
2
p
k
i=0
=O
p2Mc
k
.
Combining the above information with equation (20), we deduce
1 ±
(S (M, f )(x))2 )
π(x)
M−2
2
X
±
±
±
±
(M − 1)2 + ŜM
(M )2
ŜM
(m) − ŜM
(m + 2) + ŜM
=
m=1
+O
(log log x)2
π(x)x2Mc
+
π(x)
k
.
This proves equation (19). Suppose k = k(x) runs over positive even integers such that
Then, given any A > 0, we have, for sufficiently large values of x,
log k
> A, that is, k > xA .
log x
log k
log x
→ ∞ as x → ∞.
Thus, given M ≥ 3, we may take, for example, A = 2(M c + 1) and deduce,
x2Mc π(x)
= 0.
x→∞
k
lim
Thus,
(21)
X
M−2
1 ±
±
±
±
±
(S (M, f )(x))2 ) =
(SˆM
(m) − ŜM
(m + 2))2 + ŜM
(M − 1)2 + SˆM
(M )2
x→∞ π(x)
m=1
lim
=
M−2
X
m=1
=2
±
±
±
±
±
±
(SˆM
(m)2 + ŜM
(m + 2)2 − 2 ŜM
(m) ŜM
(m + 2)) + ŜM
(M − 1)2 + ŜM
(M )2
M
X
m=1
±
ŜM
(m)2 − 2
M−2
X
m=1
17
±
±
±
±
ŜM
(m) SˆM
(m + 2) − ŜM
(1)2 − ŜM
(2)2 ,
proving the proposition. 6. Asymptotic formula for the variance
We compute an asymptotic formula for
±
VM
(22)
M
X
:= 2
m=1
±
ŜM
(m)2
−2
M−2
X
m=1
±
±
±
±
ŜM
(m) ŜM
(m + 2) − ŜM
(1)2 − ŜM
(2)2 .
Proposition 6.1. Let I = [a, b] ⊂ [−2, 2] be a fixed interval and M ≥ 3 be an integer. Then, we have
log M
±
VM
= µ∞ (I) − µ∞ (I)2 + O
.
M
Proof. We have, by equation (9),
±
±
− ŜM
(1)2 − ŜM
(2)2 = −
(23)
1
1
(sin(2πβ) − sin(2πα))2 − 2 (sin(4πβ) − sin(4πα))2 + O
π2
4π
1
M
.
Combining equations (10) and (11), we have, after a suitable rearrangement,
2
(24)
M
X
m=1
±
ŜM
(m)2 − 2
M−2
X
m=1
1
(sin(2πβ) − sin(2πα))2
π2
1
log M
−
.
(1 − 4(β − α))(sin(4πβ) − sin(4πα)) + O
2π
M
±
±
ŜM
(m) ŜM
(m + 2) = 2(β − α) − 4(β − α)2 +
Finally, putting equations (23) and (24) together, we arrive at the following:
2(β − α)
(sin(4πβ) − sin(4πα)
− 4(β − α)2 +
(sin(4πβ) − sin(4πα))
2π
π
log M
1
.
− 2 (sin(4πβ) − sin(4πα))2 + O
4π
M
±
VM
= 2(β − α) −
Since
2(β − α) −
(sin(4πβ) − sin(4πα)
= µ∞ (I),
2π
we have,
±
VM
2
= µ∞ (I) − µ∞ (I) + O
log M
M
.
The following proposition immediately follows from Propositions 5.1 and 6.1.
Proposition 6.2. Let I = [a, b] ⊂ [−2, 2] be a fixed interval. Suppose k = k(x) runs over positive even
log k
integers such that log
x → ∞ as x → ∞. Then, for M ≥ 3,
1 ±
log M
2
2
.
lim
(S (M, f )(x)) = µ∞ (I) − µ∞ (I) + O
x→∞ π(x)
M
Proof. For any M ≥ 3, we have, by Proposition 6.1 and equation (19),
±
log M
2
2
(S (M, f )(x)) = π(x) µ∞ (I) − µ∞ (I) + O
M
(25)
2 2Mc
π(x) x
.
+ O (log log x)2 +
k
The proposition follows. 18
Remark 6.3. By almost exactly the same process, one can show that
+
log M
−
2
S (M, f )(x)S (M, f )(x) = π(x) µ∞ (I) − µ∞ (I) + O
M
(26)
2 2Mc
π(x) x
+ O (log log x)2 +
.
k
7. Strategy for proof of main theorem
The proof of Theorem 1.1 depends on the following fundamental steps.
(1) We first show that for a suitable choice of M = M (x),
S ± (M, f )(x)
p
π(x)(µ∞ (I) − µ∞ (I)2 ))
converges in mean square to
NI (f, x) − π(x)µ∞ (I)
p
π(x)(µ∞ (I) − µ∞ (I)2 ))
as x → ∞. This convergence holds as we vary the families Fk under certain growth conditions on k.
(2) For the above choice of M = M (x), we then derive, for every n ≥ 1, the limit of the moments
!n +
*
S ± (M, f )(x)
p
π(x)(µ∞ (I) − µ∞ (I)2 ))
as x → ∞. In the next section, we show that these converge to the Gaussian moments under the
growth conditions on weight k imposed in the previous step.
(3) Since convergence in mean square implies convergence in distribution and the Gaussian distribution
is characterized by its moments, one deduces Theorem 1.1.
Towards the first step, we prove the following proposition:
p
Proposition 7.1. Let I = [a, b] ⊂ [−2, 2] be a fixed interval. Let M = ⌊ π(x) log log x⌋. Suppose k = k(x)
k
→ ∞ as x → ∞. Then,
runs over positive even integers such that √log
x log x
+
*
N (f, x) − π(x)µ (I) − S ± (M, f )(x) 2
I
∞
p
lim = 0.
x→∞ π(x)(µ∞ (I) − µ∞ (I)2 )
Proof. From equation (17), we deduce the following two equations:
π(x)
π(x)
−
+
−
(27) C
≤ NI (f, x) − π(x)µ∞ (I) − S (M, f )(x) ≤ S (M, f )(x) − S (M, f )(x) + D
M +1
M +1
and
−
+
(28) S (M, f )(x) − S (M, f )(x) + C
π(x)
M +1
+
≤ NI (f, x) − π(x)µ∞ (I) − S (M, f )(x) ≤ D
Thus, for M ≥ 1 and a suitable positive constant E,
π(x)
M +1
.
h(NI (f, x) − π(x)µ∞ (I) − S ± (M, f )(x))2 i
(
2 *
2 +)
Eπ(x)
Eπ(x)
+
−
≤ max
,
S (M, f )(x) − S (M, f )(x) +
M +1
M +1
2
Eπ(x)
Eπ(x)
hS + (M, f )(x) − S − (M, f )(x)i
+ max 0, h(S + (M, f )(x) − S − (M, f )(x))2 i + 2
≤
M +1
M +1
19
We observe,
π(x)xMc
hS + (M, f )(x) − S − (M, f )(x)i = O log log x +
.
k
Moreover, combining equations (25) and (26), we know that for any M ≥ 3,
h(S + (M, f )(x) − S − (M, f )(x))2 i
= hS + (M, f )(x)2 + S − (M, f )(x)2 − 2S + (M, f )(x)S − (M, f )(x)i
log M
π(x)2 x2Mc
= O π(x)
.
+ (log log x)2 +
M
k
From the above, we deduce
h(NI (f, x) − π(x)µ∞ (I) − S ± (M, f )(x))2 i
log M
π(x)2 x2Mc
π(x)
π(x)2
2
+
π(x)
+
(log
log
x)
+
+
≪
(M + 1)2
M
k
(M + 1)
π(x)xMc
log log x +
k
Thus,
(29)
+
*
N (f, x) − π(x)µ (I) − S ± (M, f )(x) 2
I
∞
p
π(x)(µ∞ (I) − µ∞ (I)2 )
π(x)
log M
(log log x)2
π(x)x2Mc
1
≪
+
+
+
+
(M + 1)2
M
π(x)
k
(M + 1)
π(x)xMc
log log x +
k
We now choose
Thus,
M =⌊
lim
x→∞
p
π(x) log log x⌋.
π(x)
= 0.
(M + 1)2
Suppose k = k(x) runs over positive even integers such that
√log k
x log x
→ ∞ as x → ∞.
Let us fix 0 < d < 1. The above growth condition on k tells us that for sufficiently large values of x,
p
2c π(x) log log x log x + 1 < d log k.
Thus,
and
π(x)x2cM ≪ k d
π(x)x2cM
= 0.
x→∞
k
lim
This proves the proposition. We note that so far, we were able to derive the asymptotics of the first and second moments for any fixed
log k
M ≥ 3 under the condition that log
x → ∞. However, for (2) outlined above, we require asymptotics of the
n-th moments for n ≥ 1 when M grows fast enough as a function of x.
More precisely, aspseen in equation (29), mean square convergence in Step (1) requires M to grow to an
order larger than π(x). We now need to check if the higher moments of
*
!n +
S ± (M, f )(x)
p
π(x)(µ∞ (I) − µ∞ (I)2 ))
converge to Gaussian moments for such a choice of M.
20
8. Higher moments
Henceforth, we set
±
TM
(x) :=
and evaluate the moments
S ± (M, f )(x)
p
π(x)
n
1 X
±
TM
(x)
sk
f ∈Fk
for positive integers n ≥ 3.
By definition, we have

n
M−2
X
X
X
X
1 
n
±
±
±
±
±
TM
(x) =
(ŜM
(m) − ŜM
(m + 2))
af (pm ) + ŜM
(M − 1)
af (pM−1 ) + SˆM
(M )
af (pM ) .
n
π(x) 2 m=1
p≤x
p≤x
p≤x
For a prime p, let
±
YM
(p)
M
X
=
m=1
±
ÛM
(m)af (pm ),
where, as before, we denote, for M ≥ 3 and 1 ≤ m ≤ M,
(
±
±
ŜM
(m) − ŜM
(m + 2),
±
ÛM (m) :=
±
ŜM (m),
if 1 ≤ m ≤ M − 2
if m = M − 1, M.
Therefore,
n
±
TM
(x) =
1
n
π(x) 2


X
p≤x
n
±
YM
(p) .
Using the multinomial formula, we may write the above equation as follows.
(30)
where,
n
±
TM
(x) =
(a) The sum
1
π(x)
n
2
(1)
P
n
X
(1)
X
u=1 (r1 ,r2 ,...,ru )
n!
1
r1 !r2 ! . . . ru ! u!
(2)
X
±
±
±
(pu )ru ,
(p2 )r2 . . . YM
YM
(p1 )r1 YM
(p1 ,p2 ,...,pu )
is taken over tuples of positive integers r1 , r2 , . . . , ru so that
(r1 ,r2 ,...,ru )
r1 + r2 + · · · + ru = n, that is, a partition of n into u positive parts.
(b) The sum
(2)
P
is over u-tuples of distinct primes not exceeding x.
(p1 ,p2 ,...,pu )
We first focus on the inner sum in equation (30),
(2)
X
(31)
±
±
±
(pu )ru
(p2 )r2 . . . YM
YM
(p1 )r1 YM
(p1 ,p2 ,...,pu )
for a fixed partition (r1 , r2 , . . . , ru ) of n.
By repeated use of Lemma 2.1, we may write, for each 1 ≤ i ≤ u,


(3)
X
X


±
±
YM
(pi )ri =
ÛM
(mi ) Dri ,mi (0) +
Dri ,mi (t)af (pti )
mi
(32)
=
±
CM
(i)
t∈I(mi )
t≥1
+
(3)
X
mi
±
ÛM
(mi )
21
X
t∈I(m i )
t≥1
Dri ,mi (t)af (pti ),
where
1. mi denotes an ri -tuple (mj1 , . . . , mjri ).
2.
(3)
P
mi
denotes that the sum is taken over ri -tuples mi where 1 ≤ mjl ≤ M for each 1 ≤ l ≤ ri .
3. The term Û ± (mi ) denotes the product Û ± (mj1 ) . . . Û ± (mjri ).
4. For each ri -tuple mi , I(mi ) denotes the set of non negative integers t that occur in the power of pi on
using Lemma 2.1 and for each t ∈ I(mi ), Dri ,mi (t) denotes the coefficient of af (pti ) so obtained. Note that
I(mi ) is a finite set for each mi . We observe that Dri ,mi (t) is independent of the prime pi .
±
5. CM
(i) is the sum of the coefficients of af (1) = 1, coming from the the expansion using Lemma 2.1. That
is,
(3)
X
±
±
ÛM
(mi )Dri ,mi (0).
CM (i) =
mi
±
CM
(i)
Observe that
is independent of the prime pi and is in fact a polynomial expression in Ŝ ± (m), 1 ≤
m ≤ M.
We now prove the following proposition:
Proposition 8.1. Let 1 ≤ i ≤ u and mi be an ri -tuple as specified above. Then, for t ∈ I(mi ),


0,
if ri = 1, t = 0





1,
if
ri = 1, t ≥ 1

Dri ,mi (t) = O(1),
if ri = 2, t ≥ 0


ri −2

O
M
,
if
ri ≥ 3, t ≥ 1



O M ri −3 , if r ≥ 3, t = 0.
i
Proof. While focusing on an ri -tuple mi , we may also denote Dri ,mi (t) as Dri (t) for brevity.
The cases ri = 1, 2 are clear. In fact, for ri = 2, we have
YM (p)2 =
±
±
ÛM
(m1 ) ÛM
(m2 )
M
X
±
±
ÛM
(m1 ) ÛM
(m2 )
m1 , m2 =1
=
m1 , m2 =1
t
min{m1 , m2 }
M
X
X
af (pm1 +m2 −2i )
X
af (pt ),
i=0
t∈I(m1 ,m2 )
so that the coefficient of af (p ) = 1 if t ∈ I(m1 , m2 ) and zero otherwise. In particular, if t = 0,
(
1 if m1 = m2
(33)
D2,(m1 ,m2 ) (0) =
0 otherwise.
Observe that if m1 + m2 is even, then
I(m1 , m2 ) ⊆ {0, 2, . . . m1 + m2 }.
On the other hand, if m1 + m2 is odd, then
In either case,
I(m1 , m2 ) ⊆ {1, 3, . . . m1 + m2 }.
|I(m1 , m2 )| ≤
m1 + m2
2
22
+ 1 ≤ M + 1.
We now address the case ri = 3. Let l ∈ I(m1 , m2 , m3 ). The product
af (pm1 )af (pm2 )af (pm3 )
equals
af (p
m3
)
min{m1 , m2 }
X
af (pm1 +m2 −2i ).
i=0
We observe that in the above product, af (pl ) can occur at most in all possible expansions
af (pm3 )af (pj ), j ∈ I(m1 , m2 ).
Since D2 (t) = 1 for all t ∈ I(m1 , m2 ) and |I(m1 , m2 )| ≤ M + 1, we deduce
D3 (l) ≤ M + 1.
ri −2
This proves D3 (ri ) = O(M
) for ri = 3.
We now proceed by induction. Assume that for some k ≥ 3, Dk (l) = O(M k−2 ). We observe that for each
k-tuple mi ,
m1 + m2 + · · · + mk
+1
|I(mi )| ≤
2
(34)
kM
+ 1 = Ok (M ).
≤
2
Now, in the expansion
(af (pm1 )af (pm2 ) · · · af (pmk ))af (pmk+1 )
X
Dk (t)af (pt ),
= af (pmk+1 )
t∈I(m1 ,m2 ,...,mk )
l
any af (p ) can occur at most in all possible expansions
By induction hypothesis,
af (pmk+1 )af (pj ), j ∈ I(m1 , m2 , . . . , mk ).
Dk (j) = Ok (M k−2 ), j ∈ I(m1 , m2 , . . . , mk ).
Thus, by equation (34), we have
Dk+1 (l) ≤ |I(mi )||Dk (l)| = Ok (M k−1 ).
(35)
Thus, by induction, we have proved that if ri ≥ 3, t ≥ 0,
Dri (t) = O M ri −2 .
Note that the implied constant depends on ri . We now use these estimates to get a better estimate for Dri (0)
for ri ≥ 3. We prove
Dri (0) = O M ri −3 , ri ≥ 3.
Equation (33) tells us that for ri = 2, Dri (0) ≤ 1.
For ri = 3, looking again at the expansion
X
D2 (j)af (pj )
af (pm1 )af (pm2 )af (pm3 ) = af (pm3 )
j∈I(m1 ,m2 )
=
X
j∈I(m1 ,m2 )
min{j,m3 }
X
D2 (j)af (pm3 +j−2i ),
i=0
we observe that m3 + j − 2i = 0 if and only if i = j = m3 . Thus,
D3 (0) ≤ D2 (m3 ) = O(1).
In general, for ri ≥ 3,
af (pm1 ) · · · af (pmri −1 )af (pmri ) = af (pmri )
23
X
j∈I(m1 ,...,mri −1 )
Dri −1 (j)af (pj )
min{j,mri }
X
=
X
i=0
j∈I(m1 ,...,mri −1 )
Dri −1 (j)af (pmri +j−2i ).
As before, mri + j − 2i = 0 if and only if i = j = mri . Therefore,
Dri (0) ≤ Dri −1 (mri ) = O(M ri −1−2 ) = O(M ri −3 ).
Here, the implied constant depends on ri . This proves the proposition. We record the following lemma.
±
±
±
Lemma 8.2. For ri = 2, CM
(i) = VM
, where VM
is as defined in equation (22).
Proof. Observe that for ri = 2, from equation (33), it follows that
±
CM
(i) =
M
X
m=1
±
ÛM
(m)2 ,
±
which is exactly the right hand side of equation (21), denoted by VM
. The claim follows. ±
Taking the product of YM
(pi )ri over i = 1, . . . , u, we may write (31) as
(2)
X
(36)
±
±
(pu )ru =
YM
(p1 )r1 . . . YM
(p1 ,p2 ,...,pu )
(2)
X
X
(p1 ,p2 ,...,pu ) (m1 ,...,mu )
where
1.
(4)
P
(t1 ,...,tu )
±
ÛM
(m1 , . . . , mu )
(4)
X
Dr,m (t)af (pt11 . . . ptuu ).
(t1 ,...,tu )
denotes that the sum is taken over u-tuples t = (t1 , . . . , tu ), where each ti ≥ 0, unless oth-
erwise specified and ti ∈ I(mi ).
2. We abbreviate the notation by setting
±
±
±
(m1 ) . . . ÛM
(mu )
ÛM
(m1 , . . . , mu ) := ÛM
and for a given tuple m = (m1 , . . . , mu ),
Dr,m (t) := Dr1 ,m1 (t1 )Dr2 ,m2 (t2 ) . . . Dru ,mu (tu ).
We now prove the following proposition:
Proposition 8.3. Suppose k = k(x) runs over positive even integers such that
p
M = ⌊ π(x) log log x⌋. For each partition (r1 , r2 , . . . , ru ) of n,
√log k
x log x
→ ∞ as x → ∞. Let
(2)
X
1 X
±
±
±
(pu )ru
(p2 )r2 . . . YM
YM
(p1 )r1 YM
lim
n
x→∞ π(x) 2 sk
f ∈Fk (p1 ,p2 ,...,pu )
(
(µ∞ (I) − µ∞ (I)2 )n/2 if (r1 , r2 , . . . , ru ) = (2, . . . , 2)
=
0
otherwise.
1
Proof. From equation (36), we have, for each partition (r1 , . . . , ru ) of n,
1 X
1
π(x)n/2 sk f ∈F
k
1 X
1
π(x)n/2 sk
(2)
X
(p1 ,p2 ,...,pu )
±
±
(pi )ru =
YM
(p1 )r1 · · · YM
(2)
X
X
f ∈Fk (p1 ,p2 ,...,pu ) (m1 ,...,mu )
±
ÛM
(m1 , . . . , mu )
24
(4)
X
(t1 ,...,tu )
Dr,m (t)af (pt11 · · · ptuu ).
For each tuple (m1 , . . . , mu ), on applying Proposition 2.4, we have

1
n
π(x) 2
f ∈Fk (p1 ,p2 ,...,pu ) (t1 ,...,tu )
(4)
X
(2)
X
1
n
π(x) 2
(4)
X
(2)
X
X
1
sk

Dr,m (t)af (pt11 · · · ptuu ) =
δ(t1 , . . . , tu )
Dr,m (t)
+O
1
(pt11 · · · ptuu ) 2
(p1 ,p2 ,...,pu ) (t1 ,...,tu )
(pt11 · · · ptuu )c
k
!
,
where δ(t1 , . . . , tu ) = 1 if 2|ti for every ti > 0 and δ(t1 , . . . , tu ) = 0 otherwise. Observe that for each
1 ≤ i ≤ u, ti is even if and only if the sum of the components of the corresponding mi is even.
The sum
1
1 X
n
2
sk
f ∈Fk π(x)
(37)
=
(p1 ,p2 ,...,pu )
1
n
2
π(x)

±
±
(pi )ru
YM
(p1 )r1 · · · YM
(⋆)
X
(2)
+ O
where
(⋆)
P
(2)
X
X
(p1 ,p2 ,...,pu ) (m1 ,...,mu )
(2)
X
1
π(x)
n
2
±
ÛM
(m1 , . . . , mu )
X
(p1 ,p2 ,...,pu ) (m1 ,...,mu )
(4)
X
Dr,m (t)
(t1 ,...,tu )
±
| ÛM
(m1 , . . . , mu )|
(4)
X
1
(pt11
1
· · · ptuu ) 2
Dr,m (t)
(pt11

· · · ptuu )c 
k
(t1 ,...,tu )
,
denotes that the sum is over those tuples such that δ(t1 , . . . , tu ) = 1.
(m1 ,...,mu )
Now we consider the first term on the right hand side of (37) , which is
1
π(x)
=
1
n
π(x) 2
n
2
(p1 ,p2 ,...,pu ) (m1 ,...,mu )
(2)
X
(p1 ,p2 ,...,pu )
We write each
(⋆)
X
(2)
X


(⋆)
X
m1
±
ÛM
(m1 )


as
(⋆)
X
mi
(⋆)
X
mi
±
ÛM
(m1 , . . . , mu )
(4)
X
Dr1 ,m1 (t1 )
t /2
p11
t1 ≥0
±
ÛM
(mi )
±
ÛM
(mi )Dri ,mi (0) +
(4)
X
(t1 ,...,tu )


···
(⋆)
X
(4)
X
Dri ,mi (ti )
t /2
ti ≥0
(⋆)
X
mi
pi i
±
ÛM
(mi )
mu
(⋆)
X
mi
±
ÛM
(mu )
(4)
X
Dri ,mi (ti )
t /2
ti ≥2
pi i
1
. . . ptuu ) 2
(4)
X
Dru ,mu (tu )
t /2
tu ≥0

±
ÛM
(mi )Dri ,mi (0),
25
1
(pt11

Therefore, denoting
±
CM
(i) =
Dr,m (t)
.
puu

.
we have, for a partition (r1 , r2 , . . . ru ) of n,




(⋆)
(4)
(4)
(2)
(⋆)
X
X
X
X
X
(t
)
(t
)
D
D
1
r1 ,m1 1
ru ,mu u
±
±


···
ÛM
(m1 )
ÛM
(mu )
n
t1 /2
t /2
2
π(x) (p1 ,p2 ,...,pu ) m1
p1
puu
mu
t1 ≥0
tu ≥0
!
(2)
u
Y
X
1
±
CM (i)
=
(38)
n
π(x) 2 (p1 ,p2 ,...,pu ) i=1

εi
(4)
(⋆)
(2)
u
X
X
X
X Y
D
(t
)
1
1−εi
ri ,mi i
±
±

 .
CM
(i)
+
ÛM
(mi )
n
t /2
π(x) 2 (p1 ,p2 ,...,pu ) (ε1 ,...,εu ) i=1
pi i
mi
ti ≥2
Here, in the second term on the right hand side, (ε1 , ε2 , . . . , εu ) runs over all u-tuples such that for each
i = 1, . . . , u, the corresponding εi ∈ {0, 1} and at least one εi is non-zero. The tuple (0, . . . , 0) is accounted for
±
±
by the first term. We also follow the convention that if CM
(i) = 0, then εi is fixed to be 1 and CM
(i)1−εi = 1.
Let
D̃mi (ri ) := max{Dri ,mi (ti ) : ti ∈ Imi }.
Then, we have
(⋆)
X
mi
From this, we derive,
(2)
X
1
π(x)
n
2
(p1 ,p2 ,...,pu )


±
ÛM
(mi )
(⋆)
X
m1
(39)
+ O
n
π(x) 2
≪
(4)
X
Dr1 ,m1 (t1 )
t /2
t1 ≥0
1
n
π(x) 2
(2)
X
1
t /2
pi i
ti ≥2
±
ÛM
(m1 )
=

(4)
X
Dri ,mi (ti )
X
p11
(2)
X
mi
±
ÛM
(mi )
D̃mi (ri )
.
pi


(⋆)
(4)
X
X
Dru ,mu (tu )
±

···
ÛM
(mu )
t /2
puu
mu
tu ≥0

u
Y
!
±
CM
(i)
i=1
(p1 ,p2 ,...,pu )
(p1 ,p2 ,...,pu ) (ε1 ,...,εu )
(⋆)
X

εi 
(⋆)
u
Y
± 1−εi X
(r
)
D̃
i
m
±
i
C (i)

 ,
| ÛM
(mi )|
M
p
i
m
i=1
i
Consider the error term in the above equation. We prove that this term vanishes as x → ∞ for our choice
of M by showing that for each tuple (ε1 , . . . , εu ),

εi
(⋆)
(2)
u
Y
X
± 1−εi X
D̃mi (ri )
±
C (i)

 = 0.
| ÛM
(mi )|
lim
M
x→∞
pi
m
i=1
(p1 ,p2 ,...,pu )
i
First, for each tuple (ε1 , . . . , εu ) observe that we may write
εi

(⋆)
(2)
u
X
Y
± 1−εi X
D̃mi (ri )
±
C (i)


| ÛM
(mi )|
M
pi
m
i=1
(p1 ,p2 ,...,pu )
as
(2)
X
(p1 ,p2 ,...,pu )
For ε = (ε1 , . . . , εu ), we define
i




(⋆)
u
u
X
Y
Y
D̃mi (ri )

±
C ± (i)

.
| ÛM
(mi )|


M
pi
i=1
i=1
m
εi =0
εi =1
i
α(ε) := α(ε1 , . . . , εu ) := #{1 ≤ i ≤ u : εi = 0}.
26
±
We observe that if ri = 1, then CM
(i) = 0. In general, for ri ≥ 2, we have
(⋆)
X
±
|CM
(i)| ≪
mi
±
| ÛM
(mi )||Dri ,mi (0)|.
If ri = 2, then for each mi ,
Dri ,mi (0) = O(1).
Thus, by Lemma 3.2,
±
|CM
(i)| ≪
(⋆)
X
mi
±
| ÛM
(mi )| ≪ (log M )ri .
On the other hand, if ri ≥ 3, then, by Proposition 8.1, for each mi ,
Dri ,mi (0) = O(M ri −3 ).
Once again, by Lemma 3.2,
±
|CM
(i)| ≪
(40)
≪
(⋆)
X
mi
(
±
| ÛM
(mi )||Dri ,mi (0)|
(log M )ri
M ri −3 (log M )ri
if ri = 1, 2
if ri ≥ 3.
Similarly, by another application of Proposition 8.1 and Lemma 3.2 , we have
( (log M)
(⋆)
X
if ri = 1
D̃mi (ri )
±
(41)
| ÛM (mi )|
≪ M rpii−2 (log M)ri
p
if ri ≥ 2.
i
m
pi
i
The partition (r1 , r2 , . . . , ru ) can be of two types as described below.
Case 1: The partition (r1 , . . . , ru ) satisfies the condition ri > 1 for i = 1, . . . , u. Observe that this means
u ≤ n2 .
In this case, by equations (40) and (41), for each tuple (ε1 , . . . , εu ), we have




(⋆)
(2)
u
u
X
Y
Y
X
(r
)
D̃
i
m

±
i
C ± (i)


| ÛM
(mi )|


M
p
i
i=1
i=1
m
(p1 ,p2 ,...,pu )
≪
(2)
X
εi =0
i
εi =1


u
u
ri
Y
Y
(log M ) 


M ri −2
M ri −2 (log M )ri  


pi
i=1
i=1

(p1 ,p2 ,...,pu )
εi =1
εi =0
(42)
≪M
n−2α(ε)−2(u−α(ε))
n
(log M )
(2)
X
(p1 ,p2 ,...,pu )
1
u
Q
pi
i=1
εi =1
≪ M n−2α(ε)−2(u−α(ε)) (log M )n π(x)α(ε) (log log x)u−α(ε)
≪ M n−2u π(x)α(ε) (log log x)u−α(ε) (log M )n .
p
We now choose M = ⌊ π(x) log log x⌋. The above error term is
n
≪ π(x) 2 −u π(x)u−1 (log log x)u (log x)n ,
27
since α(ε) ≤ u − 1. Thus, for each tuple (ε1 , . . . , εu ),




(⋆)
(2)
u
u
X
Y
X
±  Y
(r
)
D̃
1
i
m

±
i
C (i)


lim
| ÛM
(mi )|

M
x→∞ π(x)n/2
p
i
i=1
i=1
m
(p1 ,p2 ,...,pu )
≪ lim
x→∞
εi =0
εi =1
i
n
1
2 −1 (log log x)u (log x)n = 0.
n π(x)
π(x) 2
Case 2: The partition (r1 , . . . , ru ) has at least one component ri equal to 1. Let l be the number of 1’s
in the partition. Without loss of generality, we may assume that the last l parts are equal to one while
±
r1 , . . . , ru−l are at least 2. By our convention, since CM
(i) = 0 if ri = 1, we have εi = 1 for u − l + 1 ≤ i ≤ u.
Also, if ri = 1, D̃mi (ri ) = 1. For ε = (ε1 , . . . , εu ) = (ε1 , . . . , εu−l , 1, . . . , 1), let
αl (ε) = #{1 ≤ i ≤ u − l : εi = 0}.
Therefore, if the partition in consideration has l components equal to 1, we have

εi
(⋆)
(2)
u
X
X Y
± 1−εi X
D̃mi (ri )
±
C (i)


| ÛM
(mi )|
M
pi
m
i=1
(p1 ,p2 ,...,pu ) (ε1 ,...,εu )
(2)
X
=
X
(p1 ,p2 ,...,pu ) (ε1 ,...,εu−l )
i

εi 
(⋆)
u−l
X
Y (r
)
D̃
i
m
1−ε
i
±
±
i

CM
 
| ÛM
(mi )|
(i)
p
i
m
i=1

i
u
Y
i=u−l+1


(⋆)
X
1
±

| ÛM
(mi )|  .
pi
m
i
Again, using equations (40) and (41) as well as Lemma 3.2, for each tuple (ε1 , . . . , εu−l , 1, . . . , 1) we have






(⋆)
(⋆)
(2)
u
u−l
u−l
Y
X
X
Y
Y
X
D̃mi (ri )
D̃mi (1)

±
±
C ± (i)




| ÛM
(mi )|
| ÛM
(mi )|


M
p
pi
i
i=1
i=1
m
m
(p1 ,p2 ,...,pu )
≪
(2)
X
εi =0
(p1 ,p2 ,...,pu )

εi =1
i=u−l+1
i

u−l
i

u−l
  Y ri −2 (log M )ri  (log M )l
 Y ri −2
M
M
(log M )ri  


pi
pu−l+1 · · · pu
i=1
i=1
εi =1
εi =0
(43)
≪ M n−l−2αl (ε)−2(u−l−αl (ε)) (log M )n
(2)
X
(p1 ,p2 ,...,pu )
1
u
Q
pi
i=1
εi =1
≪ M n−2αl (ε)−2(u−l−αl (ε)) (log M )n π(x)α(ε) (log log x)u−αl (ε)
≪ M n−l−2(u−l) π(x)αl (ε) (log log x)u−αl (ε) (log M )n .
Substituting our chosen value for M and using the bound αl (ε) ≤ u − l, the above error term is
n
Therefore,
1
lim
n
x→∞ π(x) 2
l
≪ π(x) 2 − 2 (log log x)u (log x)n .
(2)
X
(p1 ,p2 ,...,pu )



u−l
Y
i=1
εi =0



(⋆)
Y X
±  u−l
(r
)
D̃
mi i
±
C (i)


| ÛM
(mi )|
M
pi
i=1
m
εi =1
≪ lim
x→∞
noting that l ≥ 1.
i
u
Y
i=u−l+1
1
n
1
2 − 2 (log log x)u (log x)n = 0
n π(x)
π(x) 2
28


(⋆)
X
mi

(1)
D̃
mi
±

| ÛM
(mi )|
pi
From the analysis in Cases 1 and 2, we deduce that for any partition (r1 , r2 , . . . ru ) of n, the error term in
equation (39) vanishes in the limit. That is,
(44)
lim
x→∞
1
n
π(x) 2
(2)
X
u
Y
X
(p1 ,p2 ,...,pu ) (ε1 ,...,εu ) i=1

εi
(⋆)
(4)
X
X
(t
)
D
1−ε
i
r
,m
i
i
±
±
i

 = 0,
CM
(i)
ÛM
(mi )
ti /2
p
mi
ti ≥0
i
where we are summing over all tuples (ε1 , ε2 , . . . εu ) with at least one εi is non-zero.
From equations (38) and (44), we deduce that for a partition (r1 , r2 , . . . ru ) of n,


 
(4)
(4)
(⋆)
(2)
(⋆)
X
X
X
X
X
D
D
(t
)
(t
)
1
r1 ,m1 1
ru ,mu u
±
±


···
ÛM
(m1 )
ÛM
(mu )
lim
n
t1 /2
t /2
x→∞ π(x) 2
p
puu
m1
mu
t1 ≥0
tu ≥0
1
(p1 ,p2 ,...,pu )
(45)
!
(2)
u
Y
X
1
±
= lim
CM (i) .
n
x→∞ π(x) 2
(p1 ,p2 ,...,pu ) i=1
We now study the term
(2)
X
1
n
π(x) 2
(p1 ,p2 ,...,pu )
as x → ∞.
We know, from Lemma 8.2, that for ri = 2,
±
±
CM
(i) = VM
=2
M
X
m=1
±
ŜM
(m)2 − 2
M−2
X
m=1
u
Y
!
±
CM
(i)
i=1
±
±
±
±
ŜM
(m) ŜM
(m + 2) − ŜM
(1)2 − ŜM
(2)2 .
Once again, the partitions (r1 , r2 , . . . ru ) are of three different types as described in the three cases below.
Case 1: If (r1 , r2 , . . . ru ) = (2, 2, . . . 2), then u = n/2 and
!
!
(2)
u
u
Y
Y
X
1
π(x)(π(x) − 1)(π(x) − 2) . . . (π(x) − n/2 + 1)
±
±
CM (i)
CM (i) =
n
n
2
π(x) (p1 ,p2 ,...,pu ) i=1
π(x) 2
i=1
± n/2
= (VM
)
π(x)(π(x) − 1)(π(x) − 2) . . . (π(x) − n/2 + 1)
n
π(x) 2
,
log M
.
M
±
Case 2: If ri = 1 for some ri in the given partition, then the corresponding CM
(i) is 0. Thus,
!
(2)
u
Y
X
1
±
CM
(i) = 0.
n
π(x) 2 (p1 ,p2 ,...,pu ) i=1
±
where VM
= µ∞ (I) − µ∞ (I)2 + O
Case 3: Each ri ≥ 2 with at least one ri ≥ 3. Without loss of generality, for some 1 ≤ l ≤ u, suppose we
have r1 , r2 , . . . rl ≥ 3 and rl+1 = · · · = ru = 2.
Thus, (r1 + · · · + rl ) + 2(u − l) = n. By equation (40),
u
Y
Pl
±
CM (i) ≪ M i=1 (ri −3) (log M )n
i=1
≪ M n−2(u−l)−3l (log M )n = M n−l−2u (log M )n .
p
Choosing M = ⌊ π(x) log log x⌋,
(2)
u
Y
X
1
±
C
(i)
M
π(x)n/2
(p1 ,p2 ,...,pu ) i=1
29
n
l
1
(π(x)) 2 − 2 −u+u (log x)n .
n/2
π(x)
≪
Since l ≥ 1,
(2)
X
1
lim
n
x→∞
π(x) 2
(p1 ,p2 ,...,pu )
u
Y
!
±
CM
(i)
i=1
= 0.
p
From the above three cases, we deduce that for M = ⌊ π(x) log log x⌋,
!
(2)
u
Y
X
1
±
CM (i)
lim
n
x→∞ π(x) 2
i=1
(p1 ,p2 ,...,pu )
(46)
=
(
(µ∞ (I) − µ∞ (I)2 )n/2
0
if (r1 , r2 , . . . , ru ) = (2, . . . , 2)
otherwise.
This concludes the analysis of the main term in equation (37). We now look at the error term of the same
equation, which is


(4)
(2)
t1
tu c
X
X
X
1
(p
·
·
·
p
)
u
±
.
O
Dr,m (t) 1
| ÛM
(m1 , . . . , mu )|
n
k
π(x) 2 (p1 ,p2 ,...,pu ) (m ,...,m )
(t1 ,...,tu )
1
u
We observe that for each i,
X
ti ≥0
ti ∈I(mi )
iM
.
(pci )ti ≪ pcr
i
Thus, by Proposition 8.1 and Lemma 3.2, the above error term from equation (37) becomes


(2)
cr1 M cr2 M
cru M
X
p1
p2
. . . pu
1
u
n
n−2u

= O
n π(x) (log M ) M
2
k
π(x)
(p1 ,p2 ,...,pu )
!
cnM
1
x
u
n
n−2u
=O
π(x)u
.
n π(x) (log M ) M
k
π(x) 2
p
For M = ⌊ π(x) log log x⌋, this is
√
!
cn π(x) log log x
x
1
u
n
n/2−u
(log log x)n π(x)u
.
(47)
O
n π(x) (log x) (π(x))
k
π(x) 2
Let
log k
√
→ ∞ as x → ∞.
x log x
Then, for 0 < d < 1,
(cn
p
π(x) log log x + u) log x ≤ d log k
for sufficiently large values of x.
p
In particular, given n ≥ 1 and M = ⌊ π(x) log log x⌋,
lim
x→∞
1
π(x)
n
2
π(x)u (log M )n M n−2u π(x)u
This proves Proposition 8.3.
30
xcnM
= 0.
k
Using Proposition 8.3 in equation (30), we deduce, under the same assumptions on M and k as above,
n
1 X
±
lim
TM
(x)
x→∞ sk
f ∈Fk
(48)
(1)
X
(2)
X
n!
1
±
±
±
(pu )ru
(p2 )r2 . . . YM
YM
(p1 )r1 YM
= lim
n
x→∞ π(x) 2
r
!r
!
.
.
.
r
!
u!
1
2
u
u=1 (r1 ,r2 ,...,ru )
(p1 ,p2 ,...,pu )
(
(1)
n
X
X
1 (µ∞ (I) − µ∞ (I)2 )u if (r1 , r2 , . . . , ru ) = (2, 2, 2, . . . , 2)
n!
=
r !r ! . . . ru ! u! 0
otherwise
u=1 (r ,r ,...,r ) 1 2
1
1
=





2
n
X
u
n!
2 n/2
n (µ∞ (I) − µ∞ (I) )
( n2 )!2 2



0
if n is even
otherwise.
Thus, by equation (48), we have proved the following theorem:
p
Theorem 8.4. Let I = [a, b] be a fixed interval in [−2, 2]. Let M = ⌊ π(x) log log x⌋ and suppose k = k(x)
k
→ ∞ as x → ∞. Then, for a positive integer n ≥ 1,
runs over positive even integers such that √log
x log x
!n + 
*
if n is odd
0
S± (M, f )(x)
n!
p
=
lim
x→∞
if n is even.
 n n
π(x)(µ∞ (I) − µ∞ (I)2 )
( 2 )!2 2
Thus, by Proposition 7.1 and Theorem 8.4, the proof of Theorem 1.1 follows since convergence in mean
square implies convergence in distribution and the Gaussian distribution is characterized by its moments.
Acknowledgements
We are very grateful to Zeév Rudnick for valuable inputs and guidance during the preparation of this article.
We would like to thank Amir Akbary, A. Raghuram, Baskar Balasubramanyam, Abhishek Banerjee and M.
Ram Murty for helpful discussions and comments on a previous version of this article.
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Neha Prabhu, IISER Pune, Dr Homi Bhabha Road, Pashan, Pune - 411008, Maharashtra, India
E-mail address: [email protected]
Kaneenika Sinha, IISER Pune, Dr Homi Bhabha Road, Pashan, Pune - 411008, Maharashtra, India
E-mail address: [email protected]
32