On Convergence of Singular Series for a Pair of Quadratic
Forms
by
Thomas Wright
A dissertation submitted to The Johns Hopkins University in conformity with the
requirements for the degree of Doctor of Philosophy
Baltimore, Maryland
January, 2009
c
Thomas
Wright 2009
All rights reserved
Abstract
Examining the system of Diophantine equations



f1 (x) = x21 + ...x2n = ν1 ,


f2 (x) = λ1 x21 + ...λn x2n = ν2 ,
with λi 6= λj and νi , λi ∈ Z, we show that the singular series S(ν) converges if n ≥ 6.
Readers: Takashi Ono (Advisor).
ii
Acknowledgments
I would first like to express my deep gratitude to my advisor, Takashi Ono, for all of
his help and guidance.
Additionally, I would like to thank Qiao Zhang for acting as an unofficial second
advisor throughout many of my years here.
Thanks also to Mike Limarzi, who had the Herculean task of proofreading this
entire thesis, and to Ramin Takloo-Bighash for his helpful comments and suggestions.
Finally, thanks to all the graduate students, faculty, and staff. This thesis would
never have happened without the support of the entire community.
iii
Contents
Abstract
ii
Acknowledgments
iii
1 Introduction
1
1.1
The Classical Waring’s Problem . . . . . . . . . . . . . . . . . . . . .
1
1.2
New Formulations for N (t) and S(t) . . . . . . . . . . . . . . . . . .
2
1.3
The Intersection of Two Quadratics . . . . . . . . . . . . . . . . . . .
4
2 v = ∞: f = (f1 , f2 ) over R
7
2.1
Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
7
2.2
Evaluation of S(ν) . . . . . . . . . . . . . . . . . . . . . . . . . . . .
10
2.3
Taylor Series . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
14
2.4
Bessel Functions
. . . . . . . . . . . . . . . . . . . . . . . . . . . . .
19
2.5
Fourier Transforms . . . . . . . . . . . . . . . . . . . . . . . . . . . .
22
3 v 6= ∞: f = (f1 , f2 ) over Qp
24
3.1
Gauss Sums . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
24
3.2
A p-adic Application of the Gauss Sum . . . . . . . . . . . . . . . . .
31
3.3
Convergence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
36
3.4
Evaluation of Sp (ν) for Almost All p . . . . . . . . . . . . . . . . . .
39
iv
3.5
3.4.1
The Integral Outside (p−1 Zp )2 . . . . . . . . . . . . . . . . . .
40
3.4.2
The Integral Over (p−1 Zp )2
. . . . . . . . . . . . . . . . . . .
44
Convergence of the Singular Series . . . . . . . . . . . . . . . . . . . .
51
Vita
55
v
Chapter 1
Introduction
1.1
The Classical Waring’s Problem
The story begins almost ninety years ago with Hardy and Littlewood, who examined
the integer solutions for the equation
xδ1 + xδ2 + ... + xδn = ν,
ν∈Z
for an integer δ ≥ 2. To put this another way, if f is a polynomial function such that
f : Qn → Q where ν ∈ Q, we wish to study the set f −1 (ν) (i.e. the fiber of f which
maps down to ν). Classically, to describe this set, one would first define the ring of
(rational) adeles as follows. Let Qp be the completion of Q with respect to the p-adic
norm, and let Q∞ = R (i.e. the completion of Q with respect to the usual absolute
value). Then for any finite set of primes S (possibly including ∞), we can define
QA (S) =
Y
Qp ×
p∈S
Y
Zp .
p6∈S
The union of these QA (S) over all possible S is called the ring of adeles, denoted QA .
Using this, we can define
Nt (ν) = #(f −1 (ν)
\
Kt ) = #{x ∈ Kt : f (x) = ν} =
X
x∈Kt :f (x)=ν
1
1
such that
[
Kt = QnA ,
t
where the {Kt } is a family of compact sets in QnA which can be parameterized by t
in such a way that limt→∞ Kt = QnA . One would then consider
Nt (ν)
,
t→∞
t∗
lim
where t∗ , loosely speaking, is t to some non-negative power. This type of quantity
is known as a singular series, denoted S(ν). Hardy and Littlewood made extensive
use of the singular series; using a method which they called the circle method, they
showed that the above quantity grew sufficiently quickly as t → ∞ to ensure that
N (ν) itself was guaranteed to be non-zero for sufficiently large ν.
In this paper, while preserving the spirit of the Hardy-Littlewood computation, we
take a slightly different tack in devising our version of the singular series. This allows
us to apply the ideas of Hardy and Littlewood to larger classes of functions, and it
allows us to ask a broader range of questions about these functions. We describe these
methods and the function of particular interest to this paper in the next section.
1.2
New Formulations for N (t) and S(t)
We state here the new ideas for N and S in more generality than is necessary for the
specific example discussed in this paper. Let k be an A-field, let kA be the ring of
adeles over k, and let S 1 denote the group of complex numbers of absolute value 1.
We fix a basic character χ of k, i.e. let χ : kA → S 1 be a homomorphism such that
χ(k) = 1 but χ is non-trivial. Let f : k n → k m be a polynomial map over k. We use
2
the same notation f for maps obtained from f over kv , kA in the natural way.
Next, we consider the function
ϕ : kAn → C
and use it to define
X
Nϕ (ν) =
ϕ(γ).
(∗)
γ∈kn ,f (γ)=ν
This sum can be thought of as Nf,ϕ (ν). We assume that ϕ is a function in what is
known as the Schwartz space, denoted S(kAn ). The assumption of ϕ ∈ S(kAn ) implies
the following properties:
X
|ϕ(γ)| < +∞, i.e. ϕ|kn ∈ L1 (k n ).
(A.1)
γ∈kn
ϕ ∈ L1 (kAn ).
(A.1)A
Let
(Γϕ)(ξ) =
X
ϕ(γ)χ(< f (γ), ξ >).
(∗∗)
γ∈kn
This is clearly integrable, since |χ(< f (γ), ξ >)| = 1. Moreover, because f (γ) ∈
k m , it is clear that Γϕ is a function modulo k m , i.e.
(Γϕ)(ξ + a) = (Γϕ)(ξ), ∀ a ∈ k m
Also , for ν ∈ k m , the ”ν-th Fourier coefficient of (Γϕ)(ξ)” is
Z
(Γϕ)(ξ)χ̄(< ξ, ν >)dξ
m /k m
kA
=
X
γ∈kn
Z
χ(< f (γ) − ν, ξ >)dξ
ϕ(γ)
m /k m
kA
Let
ψ(ξ) = χ(< f (γ) − ν, ξ >).
3
Note that ψ ∈ Hom(kAm /k m , S1 ). Hence we have



Z
1 if ψ = 1,
ψ(ξ)dξ =

m /k m
kA

0 otherwise.
So this integral is 1 if and only if f (γ) = ν. Thus,
Z
X
χ(< f (γ) − ν, ξ >)dξ
ϕ(γ)
m /k m
kA
γ∈kn
X
=
ϕ(γ)
γ∈kn ,f (γ)=ν
= N (ν).
This means that (∗) is the ”ν-th Fourier coefficient of (∗∗)”.
Now, we can move from the sum over k n to the integral over kAn , because of adelized
axiom (A.1)A . For ξ ∈ kAm , we let
Z
Gϕ(ξ) =
ϕ(x)χ(< f (x), ξ >)dx.
n
kA
By (A.1)A , this integral converges. We let
Z
S(ν) =
Gϕ(ξ)χ̄(< ξ, ν >)dξ.
m
kA
c
Clearly, S(ν) = Gϕ(ν),
the Fourier transform of Gϕ. In order for this to be useful,
we require an implicit condition:
Gϕ ∈ L1 (kAm ).
(A.2)
This allows the integral S(ν) to converge.
1.3
The Intersection of Two Quadratics
This new formulation of the singular series allows us to ask the same Waring-type
questions about other polynomials or m-tuples of polynomials as well. In this section,
4
we will consider the specific case of
K = Q,
f : Qn → Q2 ,
x = (x1 , x2 , ..., xn ),
f (x) = (f1 (x), f2 (x)),
where n is even and
f1 (x) = x21 + ...x2n ,
f2 (x) = λ1 x21 + ...λn x2n ,
We will assume λi 6= λj ∀ i 6= j. For future ease, we note that f1 (x) = |x|2 .
Of course, we are interested in whether a pair of natural numbers can be represented by such a pair of functions; i.e., for ν = (ν1 , ν2 ) ∈ N2 , we examine when
f1 (x) = ν1 ,
f2 (x) = ν2 .
We require a Schwartz function ϕ which allows the axioms in the previous section to
be satisfied. We will take
ϕ(x) =
Y
ϕv (x) = ϕ∞ (x)
v
Y
ϕp (x),
p
where
ϕp (x) = the characteristic function on Zp ,
2
ϕ∞ (x) = e−π|x| .
5
Since
T
p
Zp = Z, we note that
Q
p
ϕp (x) is the characteristic function on Z. So
X
Nϕ (ν) =
ϕ(γ)
γ∈kn ,f (γ)=ν
X
=
e−πf1 (γ)
γ∈Zn ,f (γ)=ν
X
= e−πν1
1.
γ∈Zn ,f (γ)=ν
So eπν1 N (ν) gives the exact number of solutions to the equation f (x) = ν.
Now, we approximate N (ν) with our singular series S(ν). First, for xp ∈ Qp , we
let {xp } denote the fractional part of xp ; i.e.
{xp } ≡ xp (mod Zp ).
We let χ be a basic character of Q; in particular, we will let
χ(x) =
Y
χv (xv )
v
be the product over the various valuations of Q, where



e−2πix∞ if v = ∞,
χv (xv ) =


e2πi{xp } if v =
6 ∞, v = p.
In this case, given our definitions above, we have
Z
Gϕ(ξ) =
ϕ(x)χ(< f (x), ξ >)dx,
Qn
A
which means that
Z
Gϕ(ξ)χ̄(< ξ, ν >)dξ.
S(ν) =
Q2A
In the upcoming parts of this section, we will consider the localizations of this function
for the various places of Q. Such a localization is
Z
Sp (ν) =
Gϕ(ξ)χ̄(< ξ, ν >)dξ.
Q2v
We will consider separately the places where v is infinite and v = p is finite.
6
Chapter 2
v = ∞: f = (f1, f2) over R
2.1
Convergence
First, we consider the case where v = ∞, which means that Qv = R. Before evaluating
Sp (ν), we must show that |Gf ϕ(ξ)| ∈ L1 (R2 ), i.e.
Z
|Gf ϕ∞ (ξ)|dξ < ∞.
R2
This will allow us to use Fubini’s Theorem.
Throughout this section, we will use Gf ϕ and Gf ϕ∞ interchangeably when the
context is clear. By our definitions,
Z
2
Gf ϕ(ξ) =
e−π|x| χ(f1 (x)ξ1 + f2 (x)ξ2 )dx
n
Z R P
Pn
Pn
n
2
2
2
=
e−π i=1 xi e−2πi( i=1 xi ξ1 + i=1 λi xi ξ2 ) dx
n
ZR
Pn
2
=
e−π i=1 xi (1+2i(ξ1 +λi ξ2 )) dx
=
Rn
n
YZ
i=1
2
e−πxi (1+2i(ξ1 +λi ξ2 )) dx.
R
So
Z
Z
|Gf ϕ(ξ)|dξ =
R2
n Z
Y
e−πx2i (1+2i(ξ1 +λi ξ2 )) dxdξ.
2
R i=1
R
7
By a computation of T. Ono, we know that
Z
in
2
e−πxi (1+2i(ξ1 +λi ξ2 )) dx = e− 2
1
tan−1 (2(ξ1 +λi ξ2 ))
1
(1 + 4(ξ1 + λi ξ2 )2 ) 4
R
.
So the above integral is
Z
n
Y
1
1
(1 + 4(ξ1 + λi ξ2 )2 ) 4
n
Z Y
2
1
dξ
R2 i=1
=
R2 i=1
1
1
4
1
(1 + 4(ξ1 + λi ξ2 )2 ) (1 + 4(ξ1 + λi+ n2 ξ2 )2 ) 4
dξ.
Now, let us review Hölder’s inequality. As always, we let || · ||p denote the Lp -norm.
The basic form of Hölder’s is the following statement:
Hölder’s Inequality. Let f ∈ Lp and g ∈ Lq , where
1
p
+
1
q
= 1. Then
||f g||1 ≤ ||f ||p ||g||q .
Now, instead of using Hölder’s for two equations f and g, we require a generalization to r equations f1 , ..., fr :
Generalized Hölder’s Inequality. Let f1 , ...fr be such that fi ∈ Lpi ∀ i, where
Pr 1
i=1 pi = 1. Then
r
r
Y
Y
||
fi ||1 ≤
||fi ||pi .
i=1
i=1
The proof is a simple exercise in induction.
Now, we can apply this in our case by letting r = n2 , and
fi =
1
1
(1 + 4(ξ1 + λi ξ2
1
)2 ) 4
1
(1 + 4(ξ1 + λi+ n2 ξ2 )2 ) 4
Since the fi are all positive, it is clear that
n
||
2
Y
n
Z
fi ||1 =
2
Y
R2 i=1
i=1
8
fi dξ.
.
So
Z
|Gf ϕ(ξ)| =
n
n
2
Y
2
Y
fi dξ ≤
R2 i=1
n
Z
2
Y
2
n
||fi || n2 =
( (fi ) 2 dξ) n ,
i=1
i=1
R2
or, equivalently,
n
Z
2
Y
|Gf ϕ(ξ)| ≤
( (
R2
i=1
1
1
1
4
(1 + 4(ξ1 + λi ξ2 )2 ) (1 + 4(ξ1 + λi+ n2 ξ2 )2 )
n
1
4
2
) 2 dξ) n
n
Z
2
Y
( (
=
i=1
R2
n
n
2
1
1
)8(
) 8 dξ) n .
2
2
n
(1 + 4(ξ1 + λi ξ2 ) )
(1 + 4(ξ1 + λi+ 2 ξ2 ) )
Let us change variables to let
z = ξ1 + λi+ n2 ξ2 ,
w = ξ1 + λi ξ2 .
Let Ji be the Jacobian of the transformation above. It is easy to see that
Ji =
1
.
λi − λi+ n2
By our assumption that the λ’s are unequal, this change of variables is non-degenerate.
Moreover, let
n
J=
2
Y
Ji .
i=1
Then the above integral is
n
Z
2
Y
(
i=1
Ji (
R2
n
n
2
1
1
)8(
) 8 dzdw) n
2
2
(1 + 4(w) )
(1 + 4(z) )
n
Z
2
Y
=J
( (
2
n
i=1
R2
n
2
n
n
2
1
1
8(
8 dzdw) n
)
)
(1 + 4(w)2 )
1 + 4(z)2
YZ =J
(
2
n
i=1
R
n
2
YZ
=J
( (
2
n
i=1
R
1
1 + 4(w)2
n8
Z
dw
n
4
1
) 8 dw) n .
2
1 + 4(w)
9
(
R
n
2
1
) 8 dz) n
2
1 + 4(z)
But we know the integral
Z
(
R
n
1
8 dw
)
(1 + 4(w)2 )
converges if n ≥ 6 and diverges if n ≤ 4. Thus, if n ≥ 6 and n is even then
Gϕ(ξ) ∈ L1 (R2 ).
As a corollary, if n is odd and n ≥ 6 then Gϕ ∈ L1 (R2 ). This follows from the
fact that, if n odd,
Z Y
n
R2 i=1
n−1
Y
Z
1
1 dξ ≤
(1 + 4(ξ1 + λi ξ2 )2 ) 4
1
1
R2 i=1
(1 + 4(ξ1 + λi ξ2 )2 ) 4
dξ,
since
1
1
(1 + 4(ξ1 + λn ξ2 )2 ) 4
≤ 1 ∀ ξ.
Hence if the integral converges for n − 1 then it converges for n, and we know that
n − 1 ≥ 6.
2.2
Evaluation of S(ν)
Naturally, throughout this section we will assume n ≥ 6. The goal of this section is
to prove that, in the case described previously, we have the following:
Theorem 2. Let
n−2
1
K = e−π(ν1 ) (ν1 ) 2 .
2
Additionally, let J0 denote the 0-th Bessel function of second type, and let rect denote
the usual rectangle function; both of these terms will be defined more explicitly later in
the chapter. Moreover, let ui vary over the unit ball, and let dV be the usual measure
of this ball in hyperspherical coordinates. Then
ν
S(ν)
=π
K
2·
Z
S n−2
|π(ν1 )
Pn−1
i=1
rect( 12
(λn − ν2 )
1
Pn−1
2)
i=1 (λi −λn )ui
+
r
(λi −
λn )u2i |
10
1 − (1 +
dV.
ν
2(λn − ν2 )
Pn−1
i=1
1
(λi −λn )u2i
)2
Proof. To begin the proof of this theorem, we write the definition of S(ν):
Z
S(ν) =
Gf ϕ(ξ)χ̄(< ξ, ν >)dξ
R2
Z Z
=
ϕ(x)χ(< f (x), ξ >)χ̄(< ξ, ν >)dxdξ
R2 Rn
Z Z
Pn
Pn
2
2
2
e−π|x| e−2πi(( i=1 xi )ξ1 +( i=1 λi xi )ξ2 ) e2πi<ξ,ν> dxdξ
=
2
n
ZR R
Z
Pn
Pn
2
2
−π|x|2
=
e
e−2πi(( i=1 xi )ξ1 +( i=1 λi xi )ξ2 ) e2πi(ξ1 ν1 +ξ2 ν2 ) dξdx.
Rn
R2
Now, the first step will be to get rid of one of the integrals, specifically the one with
respect to ξ1 . To facilitate this process, let us change variables so as to eliminate ξ1
by using
z = ξ1 ν1 + ξ2 ν2 ,
i.e.
z − ξ2 ν2
,
ν1
1
dξ1 = dz.
ν1
ξ1 =
Then
Z Z
Z
P
Pn
ξ2 ν2
z
2
2
1
−2πi(( n
−π|x|2
i=1 xi )( ν1 − ν1 )+( i=1 λi xi )ξ2 ) 2πi(z)
e
S(ν) =
e
e
dzdξ2 dx
ν1 Rn
R R
Z
Z Z
P
Pn
ν2
2 z
2
1
−2πi(( n
−π|x|2
i=1 xi ) ν1 +( i=1 (λi − ν1 )xi )ξ2 ) 2πiz
e
e
e dzdξ2 dx
=
ν1 Rn
R R
Z
Z
Z
P
ν2
2
1
−2πi( n
−2πiz(|x|2 ν1 −1)
−π|x|2
i=1 (λi − ν1 )xi )ξ2
1
=
dzdx.
e
e
dξ2 e
ν1 Rn
R
R
2
Now, we will use the fact that the Fourier transform of e−πX is itself. Define I
by
1
I =
ν1
Z
e
Rn
−π|x|2
Z
P
ν2
2
−2πi( n
i=1 (λi − ν )xi )ξ2
e
1
R
Z
dξ2
e
−2πiz(|x|2 ν1 −1) −π(z)2
1
R
Then S(ν) = lim→0 I . Isolating the last part of I , we let
Z
|x|2
−2πiz( ν −1) −π(z)2
1
J =
e
dz.
e
R
11
e
dzdx.
If we let X = z, Y =
|x|2
−1
ν1
, then
1
J =
Z
2
e−2πi(XY ) e−πX dX.
R
2
But this integral is just the Fourier transform of e−πX evaluated at Y . As noted
2
above, the Fourier transform of e−πX is itself. So
1
2
J = e−πY
|x|2
1 −π( ν1 −1 )2
= e
.
Plugging this back into the definition for I gives
Z
Z
|x|2
P
ν2
ν −1 2
2
1
−2πi( n
−π|x|2
)
−π( 1
i=1 (λi − ν1 )xi )ξ2
I =
e
e
dξ2 e
dx.
ν1 Rn
R
Next, let us change to hyperspherical coordinates, i.e. let
x1 = t cos φ1 ,
x2 = t sin φ1 cos φ2 ,
···
xn−1 = t sin φ1 sin φ2 ... sin φn−2 cos φn−1 ,
xn = t sin φ1 sin φ2 ... sin φn−2 sin φn−1 .
So |x|2 = t2 . For ease of notation, let ωi = xti . Then
Z
Z ∞Z
t2
Pn
ν
ν −1
2
2
1
−πt2 −2πit ( i=1 (λi − ν21 )wi )ξ2 −π( 1 )2 n−1
I =
e
e
e
t
ν1 S n−1 0
R
· sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dξ2 dtdφ.
Let
y=
t2 − ν1
.
ν1
Then
yν1 + ν1 = t2 ,
ν1 dy = 2tdt.
12
So
Z Z
Z
∞
e−π(yν1 +ν1 ) e
I =
R
−2πi(yν1 +ν1 )(
ν2
2
i=1 (λi − ν1 )wi )ξ2
Pn
e−πy
2
− 1
S n−1
· (yν1 + ν1 )
n−2
2
sinn−2 φ1 sinn−3 φ2 ... sin φn−2
dy
dφdξ2 .
2
If we take the limit as → 0,
1
lim I =
→0
2
Z Z
R
Z
S n−1
∞
P
ν2
2
2
−2πi(ν1 )( n
i=1 (λi − ν1 )wi )ξ2 −πy
e−π(ν1 ) e
e
(ν1 )
n−2
2
−∞
· sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dydφdξ2
Z Z
P
ν2
2
n−2
1
−2πi(ν1 )( n
i=1 (λi − ν1 )wi )ξ2
=
e−π(ν1 ) e
(ν1 ) 2
2 R S n−1
· sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dφdξ2 ,
where the last line comes from the well-known fact that
Z
∞
2
e−πy dy = 1,
−∞
since this is the integral of the Gaussian distribution over all R. So
n−2
1
S(ν) = e−π(ν1 ) (ν1 ) 2
2
Z Z
e
R
−2πi(ν1 )(
ν2
2
i=1 (λi − ν1 )wi )ξ2
Pn
sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dφdξ2 .
S n−1
For future ease, we write out the bounds for integration over S n−1 explicitly. If we
let
n−2
1
K = e−π(ν1 ) (ν1 ) 2
2
then
Z Z 2π Z π Z π
P
ν2
2
S(ν)
−2πi(ν1 )( n
i=1 (λi − ν1 )wi )ξ2
=
...
e
sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dφdξ2
K
Z ZR π Z0 π 0 Z π 0
P
ν2
2
−2πi(ν1 )( n
i=1 (λi − ν1 )wi )ξ2
sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dφdξ2
=
...
e
R 0
0
Z Z 2π
Z π0 Z π
P
ν2
2
−2πi(ν1 )( n
i=1 (λi − ν1 )wi )ξ2
+
...
e
sinn−2 φ1 sinn−3 φ2 ... sin φn−2 dφdξ2 .
R
π
0
0
13
2.3
Taylor Series
The goal of this section is to rewrite the expression for S(ν) with the Taylor Series
expansion for e; this allows us to take the integral more effectively. First, though,
we wish to remove the last variable wn ; to do this, we change from hyperspherical
coordinates to rectangular coordinates and then back to hyperspherical.
To this end, let wi be as above. So
w1 = cos φ1 ,
w2 = sin φ1 cos φ2 ,
···
wn−1 = sin φ1 sin φ2 ... sin φn−2 cos φn−1 ,
wn2
=1−
n−1
X
wi2 .
i=1
∂φi
|. For 1 ≤ i ≤ n − 1,
To change variables, we must compute the Jacobian J| ∂w
j
∂φi
1
=−
.
∂wi
sin φ1 sin φ2 ... sin φi
Note that
∂φi
∂wj
= 0 for i > j, so the Jacobian matrix is upper triangular. Thus, if we
change variables from φ to w, we see that the change of measure is
n−1
Y ∂φi
∂φi
J|
|=|
|
∂wj
∂wi
i=1
=
1
n−1
n−2
sin
φ1 sin
φ2 ... sin φn−1
1
1
=
.
n−2
n−3
|wn | sin
φ1 sin
φ2 ... sin φn−2
14
Then
S(ν)
=
K
1
Z Z
Z
1−w12
....
−1+w12
Pn−1 2
1− i=1 wi
−1
R
Z
...
P
2
−1+ n−1
i=1 wi
1
Z Z
Z
Pn−1 2
ν2
ν2
2
1 −2πi(ν1 )(Pn−1
i=1 (λi − ν1 )wi +(λn − ν1 )(1− i=1 wi ))ξ2
e
dwn−1 ...dw1 dξ2
|wn |
1−w12
+
...
−1+w12
Pn−1 2
1− i=1 wi
−1
R
Z
...
P
2
−1+ n−1
i=1 wi
Pn−1 2
ν2
ν2
2
1 −2πi(ν1 )(Pn−1
i=1 (λi − ν1 )wi +(λn − ν1 )(1− i=1 wi ))ξ2
e
dwn−1 ...dw1 dξ2 ,
|wn |
where we must split the integrals separately into 0 ≤ φn−1 < π and π ≤ φn−1 < 2π so
that the change of variables is an injection and hence is well-defined. So the above is
Z Z 1 Z 1−w12
...
2
R
−1
−1+w12
Z 1−Pn−1
2
i=1 wi
...
P
2
−1+ n−1
i=1 wi
Z Z
1
Z
Pn−1 2
ν2
ν2
2
1 −2πi(ν1 )(Pn−1
i=1 (λi − ν1 )wi +(λn − ν1 )(1− i=1 wi ))ξ2
e
dwn−1 ...dw1 dξ2
|wn |
1−w12
...
=2
R
−1+w12
Pn−1 2
1− i=1 wi
−1
Z
...
P
2
−1+ n−1
i=1 wi
Z Z
1
Z
P
Pn−1 2
ν2
ν2
2
1
−2πi(ν1 )( n−1
i=1 (λi − ν1 )wi +(λn − ν1 )(1− i=1 wi ))ξ2
q
e
dwn−1 ...dw1 dξ2 ...
Pn−1 2
1 − i=1 wi
1−w12
=2
R
−1
1
Z
...
−1
−1+w12
P
ν2
2
1
−2πi(ν1 )( n−1
i=1 (λi −λn )wi +(λn − ν1 ))ξ2
q
e
dwn−1 ...dw1 dξ2 .
Pn−1 2
1 − i=1 wi
Now, we will put this back into hyperspherical coordinates, except in one fewer variable. So
w1 = t cos φ1 ,
w2 = t sin φ1 cos φ2 ,
···
wn−2 = t sin φ1 sin φ2 ... sin φn−3 cos φn−2 ,
wn−1 = t sin φ1 sin φ2 ... sin φn−3 sin φn−2 .
15
For notational convenience, we write
v
u n−1
uX
w2 ,
t=t
i
i=1
ui =
wi
,
t
dV = sinn−3 φ1 sinn−4 φ2 ... sin φn−3 du.
Then we have
Z Z
2
R
Z
1
Pn−1
ν
1
−2πi(ν1 )(t2 i=1
(λi −λn )u2i +(λn − ν2 ))ξ2
1
√
e
dtdV dξ2
1 − t2
S n−2 0
Z
Z
Z 1
Pn−1
ν
1
2
2
−2πiν1 (λn − ν2 )ξ2
1
√
=2
e
e−2πi(ν1 )(t i=1 (λi −λn )ui )ξ2 dtdξ2 dV.
1 − t2
S n−2 R
0
Examining only the integral over dt, we write the exponential in terms of its Taylor
series expansion:
1
Z
0
Pn−1
∞
X
(λi − λn )u2i ξ2 ))k
(−2πi(ν1 )(t2 i=1
1
√
dt,
k!
1 − t2 k=0
where we have used here the fact that
x
e =
∞
X
xk
k=0
k!
.
Now, the sum is absolutely convergent, and the integral is over a finite interval. Thus,
we can rearrange summation and integration and find the integral of each summand.
So the above is equal to
∞ Z
X
k=0
Pn−1
(−2πi(ν1 )(t2 i=1
(λi − λn )u2i ξ2 ))k
1
√
dt
k!
1 − t2
0
Pn−1
Z
∞
X
(−2πi(ν1 )( i=1
(λi − λn )u2i ξ2 ))k 1 t2k
√
=
dt.
2
k!
1
−
t
0
k=0
1
For this integral, we will integrate by parts repeatedly and then use a trigonometric
16
substitution. Performing the integration by parts first, we let
u = t2k−1 ,
du = (2k − 1)t2k−2 dt,
√
v = 1 − t2 ,
dv = √
t
dt.
1 − t2
So
Z
0
1
t2k
dt
1 − t2
Z 1
√
√
2k−1 1
2
= ( 1 − t )t
(2k − 1)t2k−2 1 − t2 dt
]0 −
0
Z 1
√
=−
(2k − 1)t2k−2 1 − t2 dt.
√
0
Again using integration by parts, this time with
1 √
v = ( 1 − t2 )3
3
√
dv = t 1 − t2 dt
u = −(2k − 1)t2k−3
du = −(2k − 1)(2k − 3)t2k−4 .
we see that the above integral is
Z
√
(2k − 1) √
1 1
3 2k−3 1
2
( 1−t ) t
]0 +
(2k − 1)(2k − 3)t2k−4 ( 1 − t2 )3 dt
−
3
3 0
Z
(2k − 1)(2k − 3) 1 2k−4 √
=
t
( 1 − t2 )3 dt.
3
0
Continuing this process iteratively, we finally find that
Z 1
t2k
√
dt
1 − t2
0
Z 1 √
(2k − 1)(2k − 3)...(1)
k
=
(−1)
( 1 − t2 )2k−1 dt
(1)(3)(5)...(2k − 1)
0
Z 1 √
= (−1)k
( 1 − t2 )2k−1 dt.
0
17
Now, to evaluate this, we require a trigonometric substitution:
t = sin u,
dt = cos udu.
Then
Z
0
1
√
( 1 − t2 )2k−1 dt
Z π
2
(cos u)2k−1 cos udu
=
0
Z π
2
=
(cos u)2k du
0
π
1
2k − 1
=
(cos u)2k−1 sin u]02 +
2k
2k
Z π
2k − 1 2
=
(cos u)2k−2 du.
2k
0
π
2
Z
(cos u)2k−2 du
0
where the integration here is again by parts. Following this process iteratively, we get
Z
1
√
( 1 − t2 )2k−1 dt
0
(2k − 1)(2k − 3)...(1)
=
2k(2k − 2)...(2)
(2k)! π
= k
( ).
(2 (k!))2 2
Z
π
2
du
0
So our sum is
P
∞
X
(−2πi(ν1 )( n−1 (λi − λn )u2 ξ2 ))k
i
i=1
k=0
k!
(2k)! π
( )
(2k (k!))2 2
Pn−1
∞
(λi − λn )u2i ξ2 ) k (2k)!
π X −2πi(ν1 )( i=1
=( )
(
)
2 k=0
4
(k!)3
Pn−1
∞
(λi − λn )u2i ξ2 ) k (2k)!
π X −πi(ν1 )( i=1
=( )
(
)
.
2 k=0
2
(k!)3
18
2.4
Bessel Functions
In this chapter, we review the applicable basic properties of Bessel functions.∗ Consider the differential equation
x2 y 00 + xy 0 + x2 y = 0.
This is known as the 0-th Bessel differential equation. It is an easy exercise to show
that one solution to this differential equation, which we will define to be J0 (x), is
J0 (x) =
∞
X
(−1)k x2k
k=0
(2k)!2
.
Another solution, denoted Y0 (x), is
∞
k
X
(−1)k+1 x2 k X 1
Y0 (x) = J0 (x) log x +
(
).
2
(2k)!
j
j=1
k=1
J0 (x) is known as the 0-th Bessel function, and Y0 is known as the 0-th modified BesselNeumann function. Importantly, any other solution to this differential equation is a
linear combination of J0 and Y0 , so if y(x) is a solution then
y(x) = a1 J0 (x) + a2 Y0 (x)
for constants a1 , a2 . Note that
lim Y0 (x) = −∞,
x→0
lim J0 (x) = 1.
x→0
So if limx→0 y(x) = 1 then a2 must be zero and a1 must be 1. Hence if limx→0 y(x) = 1
then y(x) = J0 (x).
Recall from last section that we were left with the sum
Pn−1
∞
(λi − λn )u2i ξ2 ) k (2k)!
π X −πi(ν1 )( i=1
( )
(
)
.
2 k=0
2
(k!)3
∗
For a more expansive overview, one can examine [1]
19
We show that this is related to the J0 described above; the following lemma will help
illustrate the relationship:
Lemma 2.4.1. Let J0 (x) denote the 0-th Bessel function. Then
∞
X
(2k)!
(ix)k 3 = J0 (2x)e2ix .
k!
k=0
Proof. First, note the equivalence between this statement and
e−ix
∞
X
ix (2k)!
( )k 3 = J0 (x).
2
k!
k=0
Let
y=e
−ix
∞
X
ix (2k)!
( )k 3 .
2
k!
k=0
The lemma will be proven in two steps; we show that y solves the Bessel Differential
Equation above, and then we prove that limx→0 y = 1. From above, this implies the
lemma.
In order to show that y solves the necessary differential equation, we must first
compute the various derivatives for y:
0
y = −ie
00
y = −e
−ix
−ix
∞
∞
X
ix k (2k)! 1 −ix X ix k−1 (2k)!
( )
+ ie
k( )
2
k!3
2
2
k!3
k=0
k=1
∞
∞
∞
X
X
ix k (2k)!
ix k−1 (2k)! 1 −ix X
ix
(2k)!
−ix
( )
+e
k( )
− e
k(k − 1)( )k−2 3 .
3
3
2
k!
2
k!
4
2
k!
k=1
k=2
k=0
From here, we must plug y and its derivatives into the Bessel Differential equation
20
and show that it is satisfied; this is merely an exercise in basic algebra:
x2 y 00 + xy 0 + x2 y
ix
ix 2 00 2 ix 0
) y + ( )y − 4( )2 y
2
i 2
2
∞
∞
X
X
ix 2
(2k)!
(2k)!
ix
ix
= −4( ) (−e−ix
( )k 3 + e−ix
k( )k−1 3
2
2
k!
2
k!
k=0
k=1
= −4(
∞
(2k)!
1 −ix X
ix
− e
k(k − 1)( )k−2 3 )
4
2
k!
k=2
∞
∞
X
ix k (2k)! 1 −ix X ix k−1 (2k)!
2 ix
−ix
( )
k( )
+ ie
)
+ ( )(−ie
i 2
2
k!3
2
2
k!3
k=0
k=1
∞
ix 2 −ix X ix k (2k)!
− 4( ) e
( )
2
2
k!3
k=0
= −4e−ix
∞
X
k(
k=1
− 2e
−ix
− 4e
−ix
− 2e−ix
∞
∞
X
X
ix k+1 (2k)!
ix (2k)!
−ix
( )
+e
k( )k 3
3
2
k!
2
k!
k=0
k=1
∞
X
ix (2k)!
ix k+1 (2k)!
−ix
+e
k(k − 1)( )k 3
k( )
3
2
k!
2
k!
k=1
k=0
∞
X
∞
∞
X
X
ix
(2k)!
ix (2k)!
( )k+1 3 + e−ix
k( )k 3
2
k!
2
k!
k=0
k=1
∞
X
ix (2k)!
ix k+1 (2k)!
−ix
+e
k(k − 1 + 1)( )k 3
(2k + 1)( )
3
2
k!
2
k!
k=1
k=0
−ix
∞
X
−ix
∞
X
−ix
∞
X
= −2e
= −2e
= −2e
∞
X
ix (2k)!
ix k+1 (2k)!
−ix
)
+
e
k(k − 1)( )k 3
3
2
k!
2
k!
k=2
∞
X
ix k+1 (2k)!
ix
(2(k + 1))!
−ix
(2k + 1)( )
+e
(k + 1)2 ( )k+1
3
2
k!
2
(k + 1)!3
k=0
k=0
(2k + 1)(
k=0
∞
X
−ix
+e
ix k+1 (2k)!
)
2
k!3
(k + 1)2 (2k + 2)(2k + 1)(
k=0
= −2e−ix
∞
X
(2k + 1)(
k=0
∞
X
−ix
+e
k=0
ix k+1 (2k)!
)
2
k!3
2(2k + 1)(
ix k+1 (2k)!
)
2
(k!)3
= 0.
21
ix k+1
(2k)!
)
2
(k + 1)3 (k!)3
So y satisfies the differential equation. Additionally, it is easy to see that
lim e
−ix
x→0
∞
X
ix (2k)!
( )k 3 = 1.
2
k!
k=0
Thus, y = J0 (x).
As a result of this lemma, we have
P
∞
2
π X −πi(ν1 )( n−1
i=1 (λi − λn )ui ξ2 ) k (2k)!
( )
)
(
2 k=0
2
(k!)3
n−1
X
Pn−1
π
2
= ( )J0 (−π(ν1 )( (λi − λn )u2i ξ2 ))e(−πi(ν1 )( i=1 (λi −λn )ui ξ2 )) .
2
i=1
2.5
Fourier Transforms
Using the findings from the previous chapter, we first rewrite S(ν):
π
S(ν)
=2
K
2
Z
Z
S n−2
n−1
X
J0 (−π(ν1 )( (λi − λn )u2i ξ2 ))
R
i=1
Pn−1
ν
2
(−πi(ν1 )( i=1 (λi −λn )ui ξ2 )) −2πiν1 (λn − ν12 )ξ2
·e
Z
e
Z
=π
S n−2
R
dξ2 dV
n−1
P
X
ν2
2
(−πi(ν1 )( n−1
i=1 (λi −λn )ui )+2(λn − ν1 ))ξ2
J0 (−π(ν1 )( (λi − λn )u2i ξ2 ))e
dξ2 dV.
i=1
Note that the inner integral is the Fourier transform of
n−1
X
J0 (−π(ν1 )( (λi − λn )u2i ))
i=1
evaluated at
P
2
ν1 (( n−1
i=1 (λi − λn )ui ) + 2(λn −
ν2
))
ν1
2
.
This type of quantity has been computed classically; here, we state the result of this
computation without proof† :
2 · rect(πx)
J0 (t)e−2πixt dt = √
1 − 4π 2 x2
R
Z
†
For a reference about this result, see [2]
22
where rect is the rectangular function given




0




rect(t) = 1
2






1
by
if |t| > 12 ,
if |t| = 12 ,
if |t| < 12 .
So if we let
x=
Pn−1
(ν1 )(( i=1
(λi − λn )u2i ) + (λn −
ν2
))
ν1
2
n−1
X
k = −π(ν1 )( (λi − λn )u2i ),
,
i=1
w = kξ2 ,
dw = kdξ2 ,
then
Z
n−1
Pn−1
X
ν
(−πi(ν1 )( i=1
(λi −λn )u2i )+2(λn − ν2 ))ξ2
1
dξ2
J0 (−π(ν1 )( (λi − λn )u2i ξ2 ))e
R
i=1
Z
J0 (kξ2 )e(−2πix)ξ2 dξ2
R
Z
w
1
J0 (w)e(−2πix) k dw
=
|k| R
2 · rect(π xk )
= p
.
|k| 1 − 4π 2 ( xk )2
=
where we have |k| because if k < 0 then the change of variables causes the bounds
to go from ∞ to −∞, and reorienting the bounds cancels the negative in front of k.
Thus,
S(ν)
=π
K
Z
S n−2
2 · rect(π xk )
p
dV.
|k| 1 − 4π 2 ( xk )2
Plugging in x and k, we get
ν
S(ν)
=π
K
2 · rect( 12 +
Z
S n−2
|π(ν1 )
Pn−1
i=1
(λn − ν2 )
1
Pn−1
2)
i=1 (λi −λn )ui
r
(λi − λn )u2i | 1 − (1 +
which is as stated in Theorem 2.
23
dV.
ν
2(λn − ν2 )
Pn−1
i=1
1
(λi −λn )u2i
)2
Chapter 3
v 6= ∞: f = (f1, f2) over Qp
3.1
Gauss Sums
Throughout the upcoming sections, we will use extensively the classical calculation
of Gauss sums. As such, we first show how such a calculation is performed. Here,
the actual method given is that of Siegel, and the result is a generalization of that of
Gauss.
Let a, b ∈ N, λ ∈ Q be such that
ab + 2aλ ≡ 0 (mod 2).
We will evaluate the sum
b
1 X πi a h2 +πiah
√
e b
b h=1
(cf. [7], p. 50). This is actually more general than is necessary; for our purposes, it
will be sufficient to let a = 2r, b = pl , which gives the sum
l
p
− 2l
p
X
2πi 2
x r
e pl
.
a
2
x=1
Next, we let z ∈ C and define
f (z) =
eπi b (z+λ)
.
e2πiz − 1
24
Note that the numerator is never zero, and the denominator has a pole at every
integer h (and at no other points). We show that the poles are all of order 1:
Resz=h f (z) = lim (z − h)f (z)
z→h
(z − h)
z→h e2πiz − 1
a
1
2
= eπi b (h+λ) lim
z→h 2πie2πiz
a
2 1
= eπi b (h+λ)
2πi
a
2
= eπi b (h+λ) lim
where the penultimate step is by L’Hospital’s rule. Thus, if we have a contour C
which contains the integers 1, 2, ..., b (and no others) then
Z
f (z)dz = 2πi
C
b
X
Resz=h f (z) =
b
X
a
2
eπi b (h+λ) .
h=1
h=1
In particular, we will take C to be the contour in Figure 2, where L is the line which
goes through the real axis at a 45◦ angle and L0 is the line parallel to L which goes
through the real axis at b + 21 . We claim that as the horizontal lines on the top and
bottom of the contour go toward infinity, the integrals over these lines go to zero.
This is because
a
2
2
eπi b ((x+λ) +2iy(x+λ)−y )
f (x + iy) =
.
e2πix−2πy − 1
a
So f (z) → 0 if e−π b (2y(x+λ)) → 0 and e2πix−2πy 6→ 1. For the upper horizontal line,
y → ∞ and x + λ is positive; for the lower horizontal line, y → −∞ and x + λ
a
is negative. Thus, in both cases, e−π b (2y(x+λ)) → 0 and e2πix−2πy 6→ 1, and hence
f (z) → 0.
Now, note that if z is on the line L then z + b is on the line L0 . So
25
Figure 3.1: Contour of integration
26
Z
C
Z
Z
f (z)dz =
f (z)dz − f (z)dz
L0
L
Z
= (f (z + b) − f (z))dz
L
Z πi a (z+λ)2
a
e b
2
[eπi b (2b(z+λ)+b ) − 1]dz.
=
2πiz
−1
L e
Now,
a
(2b(z + λ) + b2 ) = 2aλ + ab + 2az ≡ 2az (mod 2)
b
by our assumption earlier, which means that
a
2
eπi b (2b(z+λ)+b ) = e2πiaz .
Then
Z
a
Z
πi ab (z+λ)2
f (z)dz =
C
e
L
Z
=
πi ab (z+λ)2
e
L
=
=
[
a−1
X
2
eπi b (2b(z+λ)+b ) − 1
]dz
[
e2πiz − 1
e2πihz ]dz
h=0
a−1 Z
X
h=0
a−1
X
a
2 +2πihz
eπi b (z+λ)
dz
L
e
πi ab h2 −2πihλ
Z
a
b
2
eπi b (z+λ+ a h) dz.
L
h=0
Here, the first step comes from the expansion
a−1
xa − 1 X h
=
x .
x−1
h=0
Let us define
b
Z = z + λ + h.
a
Then the line L instead shifts to another parallel line which we can denote L1 , i.e.
a−1
X
e
πi ab h2 −2πihλ
Z
L1
h=0
27
a
2
eπi b Z dZ.
Let L0 be the line parallel to L1 which goes through the origin. Since the integrand
above has no poles, it follows from Cauchy’s Theorem that we can shift the contour
of integration from L1 to L0 . The integral in the above expression is then
Z
a
2
eπi b Z dZ.
L0
We make the change of variables
r
ib
t
Z=−
ra
ib
dZ = −
dt.
a
Note that the integral over L0 , like the integrals over L and L1 , goes downward along
the line. So the change of measure allows us to change the orientation of the integral,
i.e.
Z
e
r
πi ab Z 2
dZ =
L0
ib
a
Z
r
∞
e
−πt2
dZ =
−∞
ib
.
a
Thus, in summary,
Z
f (z)dz =
C
r
=
r
=
ib
a
ib
a
b
X
a
2
eπi b (h+λ)
h=1
a−1
X
−πi ab h2 −2πihλ
e
h=0
a
X
b
2 −2πihλ
e−πi a h
h=1
where the shift in parameter in the last step is allowed because
ab + 2aλ ≡ 0 (mod 2).
We arrive, then, at the theorem
Theorem 3.1.1. Let a, b, and λ be as above. Then
r a
r b
1 X πi a (x+λ)2
i X πi b x2 −2πixλ
e b
=
e a
.
b x=1
a x=1
28
As a corollary, we can prove the special case which will be of use to us throughout
the remainder of the chapter:
Corollary 3.1.2. Let a = 2, λ = 0, b = p be a prime number. Then




√1
if p ≡ 1 (mod 4),

p


p

1 X 2πix2 p1
= √i if p ≡ 3 (mod 4),
e
p

p x=1





0
if p = 2.
Proof. We know that
√
iπ
1+i
i=e4 = √ .
2
Then
p
p
1 (1 + i)
1 X 2πix2 p1
e
= √ √ (1 + e−πi 2 ).
√
p x=1
2
2
p
Naturally, e−πi 2 depends on p. In particular,




1 − i if p ≡ 1 (mod 4),




p
1 + e−πi 2 = 1 + i if p ≡ 3 (mod 4),






0
if p = 2.
from which the corollary follows.
Now, we can derive a useful theorem about Legendre characters. For an odd prime
c+ = Hom(Fp , S 1 ) such that
p, let us choose a ζ ∈ F
p
c+ =< ζ > .
F
p
We define
R(ζ) =
X
ζx
2
x∈Fp
=1+
X
ζx
2
x∈F×
p
=1+
X
2
ζx +
x<0
29
X
x>0
2
ζx ,
where an element in F×
p can be described as being > 0 if it is between 0 and
p−1
2
or
< 0 otherwise. Then
R(ν) = 1 + 2
X
ζx
2
x>0
X
=1+2
ζy.
2
y∈(F×
p )
If we define
0 =T (ζ) =
X
=1+
X
ζy
y∈Fp
ζy
y∈F×
p
X
=1+
X
ζy +
2
y∈(F×
p )
ζy
× 2
y∈F×
p ,y6∈(Fp )
then we can express R(ζ) as
R(ζ) = R(ζ) − T (ζ)
X
X
=
ζy −
2
y∈(F×
p )
=
ζy
× 2
y∈F×
p ,y6∈(Fp )
X y
( )ζ y ,
p
y∈F
p
where ( yp ) is the Legendre symbol with ( yp ) = 0 if p|y. Thus, we arrive at the following:
Theorem 3.1.3.
p−1
X y
X
2
y
( )ζ =
ζx .
p
x=0
y∈F
p
Now, if p - r then we can replace ζ by ζ r . Then, by the theorem,
p−1
X
x=0
2
ζ rx =
X x
( )ζ rx
p
x∈F
p
r X rx rx
=( )
( )ζ
p x∈F p
p
p−1
r X x2
=( )
ζ .
p x=0
30
Throughout the chapter, this fact will be applied in the following form. Let
ζ x = χ(x).
Then
p−1
r X
χ(x2 ).
χ(rx ) = ( )
p x=0
x=0
p−1
X
3.2
2
A p-adic Application of the Gauss Sum
Now that we have our Gauss sum computation, we can relate it to the singular series.
Throughout this section, for a ∈ Zp − pZp , ( ap ) will be used to denote the Legendre
symbol (similarly, ( pa2 ) will denote the Jacobi symbol). Of course, we know that if we
define ã to be a (mod p) then
ã
a
( ) = ( ).
p
p
This remark augurs an important motif of this section: in many instances, the only
term in the p-adic expansion of a which matters is the first one. One such example of
this is given in the following lemma, which will be used repeatedly to relate classical
Gauss sums to the singular series computations:
Lemma 3.2.1. Let p 6= 2, and assume a ∈ Zp − pZp . Moreover, let



 √1 if p ≡ 1 (mod 4),
p
G=


 √i if p ≡ 3 (mod 4).
p
be the Gauss sum from Corollary 3.1.1. Then



Z
p− 2e
if e is even,
a 2
χ( e x )dx =

p
Zp

a
p− e−1
2 ( )G
if e is odd.
p
31
Proof. For ease of notation, we will define
Z
a
A(e, a) =
χ( e x2 )dx.
p
Zp
We proceed by induction. If e = 1 then we have
Z
XZ
a 2
a
χ( x )dx =
χ( x2 )dx.
p
p
Zp
b+pZp
b∈F
p
Now, we change variables, letting x = b+py, where y ranges over Zp . Then dx = p1 dy.
So the above is
Z
a
a
1X
χ( b2 + (2bpy + p2 y 2 ))dy
p b∈F Zp p
p
p
Z
1X
a
=
χ( b2 )χ(a(2by + py 2 ))dy
p b∈F Zp p
p
Z
X
a 2
1
χ( b )
dy
=
p b∈F
p
Zp
p
a
=( )G,
p
where the penultimate step comes from the fact that χ is trivial on Zp , and the last
step is merely the evaluation of a Gauss sum.
For e = 2, we will change variables, letting x = b + py, where y ranges over Zp
and b ranges over Fp . So
Z
XZ
a 2
a
χ( 2 x )dx =
χ( 2 x2 )dx
p
p
Zp
b∈Fp b+pZp
X1Z
a
=
χ( 2 (b + py)2 )dy
p Zp p
b∈Fp
X1Z
a
=
χ( 2 (b2 + 2bpy + p2 y 2 ))dy
p Zp p
b∈Fp
Z
X1 a
a
2
=
χ( 2 (b ))
χ( 2 (2bpy + p2 y 2 ))dy
p p
p
Zp
b∈Fp
Z
X1 a
a
2
=
χ( 2 (b ))
χ( 2by)dy.
p p
p
Zp
b∈F
p
32
If b 6= 0 then the integral is zero (since the integral of a non-trivial character over Zp
is zero); otherwise, it’s 1. So the above is p1 .
Now, if e > 2, we can again let x = b + py. So
Z
XZ
a 2
a
χ( e x )dx =
χ( e x2 )dx
p
p
Zp
b∈Fp b+pZp
X1Z
a
χ( e (b + py)2 )dy
=
p Zp p
b∈Fp
X1Z
a
=
χ( e (b2 + 2bpy + p2 y 2 ))dy
p Zp p
b∈Fp
Z
X1 a
a
2
χ( e−1 (2by + py 2 ))dy.
χ( e (b ))
=
p p
p
Zp
b∈F
p
If b 6= 0, we can let u = 2by + py 2 . Then
du = |2b + 2py|p dy = dy
since 2b + 2py is a p-adic unit. But
a
χ( e (b2 ))
p
Z
χ(
Zp
a
pe−1
u)du = 0.
So the integral is zero unless b = 0, in which case the integral is 1. Then
Z
a
1
χ( e x2 )dx =
p
p
Zp
Z
a
1
χ( e p2 y 2 )dy =
p
p
Zp
Z
χ(
Zp
a
pe−2
y 2 )dy.
Thus, proceeding by induction, we achieve the desired result.
Remark . We highlight here the fact that, for e = 1, we have
Z
XZ
a
a 2
a
χ( x )dx =
χ( b2 )dx = ( )G.
p
p
p
Zp
b+pZp
b∈F
p
This will be used repeatedly throughout the latter part of this chapter.
Next, we derive the analogous result for p = 2:
33
Lemma 3.2.2. Assume a ∈ Z2 − 2Z2 . Then




0



Z

a
√ ( a )iπ − e
χ( e x2 )dx =
2e 4 2 2

2
Z2




√

 2e{ a8 }iπ 2− 2e
if e = 1
if ≡ 0 (mod 2),
if ≡ 1 (mod 2), e > 1.
where { a8 } denotes as before the fractional part of a8 , and ( a4 ) is the Jacobi symbol.
Proof. For e = 1, this is Corollary 3.1.1. For e = 2,
Z
3 Z
X
a 2
a
χ( x )dx =
χ( x2 )dx.
4
4
2
Z2
b=0 b+2 Z2
Setting x = b + 4y gives dx = 41 dy, and hence the above is
3
Z
=
1X a 2
χ( b )
4 b=0 4
1X
4 b=0
a
χ( (b + 4y)2 )dy
4
Z2
3
1
a
= (2 + 2χ( ))
4
4
1
a
= (1 + ( )i)
2√
4
2 ( a )iπ
e4 .
=
2
For e = 3,
7
Z
X
a
χ( x2 )dx =
8
Z2
b=0
Z
a
χ( x2 )dx.
8
b+23 Z2
Setting x = b + 8y gives dx = 81 dy, and hence the above is
7
1X
8 b=0
Z
a
χ( (b + 8y)2 )dy
8
Z2
7
1X a 2
=
χ( b )
8 b=0 8
1
a
= (4χ( ))
8
8
1 { a }iπ
= e 8 .
2
34
where the penultimate step is achieved by explicitly computing the sum.
Finally, if e ≥ 4,
Z
Z
Z
a 2
a 2
a
χ( e x )dx =
χ( e x )dx +
χ( e x2 )dx
2
2
2
Z2
2Z2
1+2Z2
Z
Z
X
a
a
=
χ( e x2 )dx +
χ( e x2 )dx.
2
2
2
2Z2
b=1,3 b+2 Z2
We wish to show that the sum over b is zero; this will allow us to proceed by induction.
Examining only the latter integral, we set x = b + 4y, which gives dx = 41 dy, and
hence
Z
a
(b + 4y)2 )dy
e
2
Z
a
=
χ( e (b2 + 23 by + 24 y 2 ))dy
2
Z2
Z
a
a 2
χ( e (23 by + 24 y 2 ))dy.
= χ( e b )
2
2
Z2
χ(
Z2
Set u = by + 2y 2 , which makes du = dy (since b is odd). Then
Z
a 3
(2 by + 24 y 2 ))dy
2e
Z
a
=
χ( e−3 (u))du = 0
2
Z2
χ(
Z2
since this is the integral of a non-trivial character over Z2 . Thus,
Z
a
χ( e x2 )dx =
2
Z2
Z
χ(
2Z2
a 2
x )dx.
2e
If we change variables, letting u = 2x and du = 21 dx, then
Z
a
1
χ( e x2 )dx =
2
2
Z2
The lemma thus follows from induction.
35
Z
χ(
Z2
a
2e−2
u2 )du.
3.3
Convergence
We now consider the case where v = p, which means that we examine solutions to
f = ν over Qp . We must first show that, as in the real case, |Gf ϕ| ∈ L1 (Q2p ) if n ≥ 6.
We know that
Z
dξ
||Gf ϕ||1 =
χ(<
f
(x),
ξ
>)dx
n
Q2p
Zp
Z Z
n
n
X
X
2
2
dξ
χ(
x
ξ
+
λ
x
ξ
)dx
=
1
i
2
i
i
Z
Q2p
Zn
p
Z Z
=
2
=
χ(
Zn
p
Qp
Z
i=1
n
X
i=1
(ξ1 +
i=1
n Z
Y
2
χ((ξ1 +
Zp
Qp i=1
λi ξ2 )x2i )dxdξ
λi ξ2 )x2i )dxi dξ.
Now, we will use the same trick with Hölder’s inequality that was used in the real
case. Let us group the product into pairs, with
Z
χ((ξ1 + λi ξ2 )x2i )χ((ξ1 + λi+ n2 ξ2 )x2i+ n )dxi dxi+ n2 .
fi =
2
Z2p
From Hölder’s inequality, we know that
||
r
Y
fi ||1 ≤
r
Y
i=1
||fi ||pi .
i=1
We apply this to mean that
Z
n Z
2
Y
2
χ((ξ1 +
Z2p
Qp i=1
λi ξ2 )x2i )χ((ξ1
+λ
i+ n
2
ξ2 )x2i+ n )dxi dxi+ n2 dξ
2
n
2 Z
Y
≤(
i=1
Z
2
Qp
Z2p
χ((ξ1 +
λi ξ2 )x2i )χ((ξ1
+
Now, we change variables, letting
zi = ξ1 + λi ξ2 ,
wi = ξ1 + λi+ n2 ξ2 .
36
n
2
λi+ n2 ξ2 )x2i+ n )dxi dxi+ n2 2
2
dξ) n .
Since the λ’s are unequal, this change of variables is non-degenerate (i.e. the Jacobian
is not zero). If we let Ji be the Jacobian of the transformation from (ξ1 , ξ2 ) to (zi , wi )
and let
n
J=
2
Y
Ji ,
i=1
the right side of the above inequality is then
n
2 Z
Y
(J
Qp
i=1
n
2
Z
2
Z2p
χ(zi x2i )χ(wi x2i+ n )dxi dxi+ n2 2
2
dzi dwi ) n .
Since no terms contain both zi and wi (or xi and xi+ n2 ), we can split the integrals,
giving
n
(J
2 Z
Y
i=1
Qp
n
2
Z
Zp
χ(zi x2i )dxi n
2
YZ
=(J
(
i=1
Qp
Z
Zp
Z
dzi
Qp
n
2
χ(zi x2i )dxi Z
Zp
n
2
χ(wi x2i+ n )dxi+ n2 2
2
dwi ) n
2
dzi )2 ) n .
(∗)
Now, let us evaluate
Z
Qp
Z
Zp
n
2
χ(zi x2i )dxi dzi .
First, we note that we can break Qp into annuli as follows:
Z
n
n
Z Z
∞ Z
X
2
2
2
2
χ(z
x
χ(z
x
)dx
)dx
dz
+
(
i i
i i
i
i dzi )
i
−e
−e+1
Zp
Zp
Zp
p Zp −p
Zp
e=1
Z
n
Z
∞
X
2
2
dzi .
=1+
χ(z
x
)dx
i
i
i
e=1
p−e Zp −p−e+1 Zp
Zp
We evaluate the above separately for p 6= 2 and p = 2. The two cases evaluate to
similar (but not identical) expressions.
Case 1. Assume p 6= 2. From Lemma 3.2.1, we know that, for zi ∈ p−e Zp − p−e+1 Zp ,

Z

 −e
if e is even,
|p 2 |
2
=
χ(z
x
)dx
i
i
i
 e−1
Zp

|p− 2 ( a )G| if e is odd,
p
e
= p− 2 .
37
1
since |G| = p− 2 and |( ap )| = 1. So
1+
∞ Z
X
e=1
p−e Zp −p−e+1 Zp
=1+
=1+
Z
Zp
n
2
χ(zi x2i )dxi ∞ Z
X
e
dzi
n
|p− 2 | 2 dzi
p−e Zp −p−e+1 Zp
e=1
∞
X
pe
e=1
p − 1 − ne
p 4
p
∞
p − 1 X − (n−4)e
p 4 .
=1+
p e=1
Case 2 : Assume p = 2. Here, we use Lemma 3.2.2 to see that




if e = 1
0

Z


2
= |√2e( a4 )iπ 2− 2e | if ≡ 0 (mod 2),
χ(z
x
)dx
i
i
i

Z2




√

| 2e{ a8 }iπ 2− 2e | if ≡ 1 (mod 2), e > 1,


0

if e = 1,
=


2− e−1
2
if e > 1.
So
1+
∞ Z
X
e=1
2−e Z2 −2−e+1 Z2
=1+
=1+
Z
n
2
2
dzi
χ(z
x
)dx
i
i
i
Z2
∞ Z
X
e=2
∞
X
(2−
e−1
2
2−e Z2 −2−e+1 Z2
n(e−1)
1
2e ( )2− 4 .
2
e=2
If we re-index, the above is
1+
∞
X
ne
1
2e+1 ( )2− 4
2
e=1
=1+
∞
X
e=1
38
2−
(n−4)e
4
.
n
) 2 dzi
For the second case,
p−1
p
= 12 . So the first and second cases differ only by the fact
that, in the second case, the infinite sum has been multiplied by 2.
Thus, returning to (∗), the
n
Z
2
Y
(J
(
i=1
above tells us that
Z
n
2
2
dzi )2 ) n2
χ(z
x
)dx
i
i
i
Qp
Zp
n
2
∞
Y
p − 1 X − (n−4)e 2 2
≤(J
(1 + 2(
)
p 4 ) )n
p
e=1
i=1
∞
p − 1 X − (n−4)e n 2
)
p 4 ) )n .
p
e=1
P∞ − (n−4)e
Thus, the integral is finite if the sum e=1 p 4 is convergent. For this to happen,
=(J(1 + 2(
the exponent must be negative, which occurs when
−
(n − 4)
< 0,
4
or n > 4. Since n is even, this means that n ≥ 6.
3.4
Evaluation of Sp(ν) for Almost All p
In this section, we will evaluate the expression Sp (ν) for all but finitely many p. This
will allow us to test S(ν) for convergence; since we are only eliminating a finite set
of p’s, the product of the Sp (ν) over the remaining p will converge if and only the
product over all p converges.
In particular, we will assume that p does not divide any of the λ’s or ν’s and that
the λ’s are all unequal. Moreover, we assume that p does not divide λi − λj ∀ i 6= j
and p - ν1 λi + ν2 ∀ i. Additionally, we assume either that
ν2
6≡ λi (mod p) ∀ i
ν1
or that
ν2
= λi
ν1
39
for exactly one i. Note that the assumption of p not dividing λi − λj ∀ i, j means
that p 6= 2.
Now, recall that
Z
Z
Sp (ν) =
χ̄(< ξ, ν >)
Q2p
χ(< f (x), ξ >)dxdξ
Zn
p
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
=
Qp
Zn
p
Qp
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ1 dξ2 .
i=1
We can break up the integral over Q2p into a sum over its various annuli:
Z
Z
=
Qp
Qp
+
∞
X
e2 =1
∞ X
∞ Z
X
e1 =1 e2 =1
Z
Z
+
p−e1 Zp −p−e1 +1 Zp
p−e2 Zp −p−e2 +1 Zp
∞ Z
X
e1 =1
p−e1 Zp −p−e1 +1 Zp
Z
Zp
Z
.
Zp
p−e1 Zp −p−e1 +1 Zp
The key to the evaluation of Sp (ν) will be to show the integral over any annulus is
zero if either ξ1 or ξ2 is outside of p−1 Zp . After this, the evaluation of the integral
over Q2p is merely the evaluation of the integral over (p−1 Zp )2 .
3.4.1
The Integral Outside (p−1 Zp )2
Here, we prove the earlier claim that if the ξ 6∈ (p−1 Zp )2 then the integral as ξ ranges
over an annulus is zero. This will be achieved using two lemmas:
Lemma 3.4.1. If max{e1 , e2 } > 1 and e1 6= e2 then
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
p−e1 Zp −p−e1 +1 Zp
p−e2 Zp −p−e2 +1 Zp
Zn
p
=0.
40
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
Proof. Let us assume first that e1 > e2 . Note that
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
p−e1 Zp −p−e1 +1 Zp
p−e2 Zp −p−e2 +1 Zp
·
+p−e1 +1 Zp
b
p e2
+p−e2 +1 Zp
(ξ1 + λi ξ2 )x2i )dxdξ2 dξ1 .
χ(
Zn
p
χ̄(ξ1 ν1 + ξ2 ν2 )
a
pe1
a∈Z/pZ× b∈Z/pZ×
Z
n
X
i=1
Z
X Z
X
=
Zn
p
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
If we let
a
,
pe1
b
ξ20 = ξ2 − e2 ,
p
ξ10 = ξ1 −
a measure-invariant change of variables, then the above can be rewritten as
Z
X
X Z
a
b
χ̄(( e1 + ξ10 )ν1 + ( e2 + ξ20 )ν2 )
−e1 +1 Z
p
p
p−e2 +1 Zp
p
a∈Z/pZ× b∈Z/pZ× p
n Z
Y
a
b
· (
χ((( e1 + ξ10 ) + λi ( e2 + ξ20 ))x2i )dxi )dξ20 dξ10
p
p
Zp
i=1
X
=
X
a∈Z/pZ× b∈Z/pZ
Z
n
Y
a
b
χ̄( e1 ν1 + e2 ν2 )
A(a, e1 )
p
p
×
i=1
Z
·
p−e1 +1 Zp
=(A(a, e1 ))n
p−e2 +1 Zp
X
χ̄(ξ10 ν1 + ξ20 ν2 )dξ20 dξ10
X
χ̄(
a∈Z/pZ× b∈Z/pZ×
Z
Z
χ̄(ξ10 ν1 + ξ20 ν2 )dξ20 dξ10
·
p−e1 +1 Zp
n
=(A(a, e1 ))
p−e2 +1 Z
X
b
a
n
ν
1 + e ν2 )Ae1
e
p1
p2
p
X
b
a
χ̄( e1 ν1 + e2 ν2 )
p
p
×
a∈Z/pZ× b∈Z/pZ
But since e1 > 1, we know that
Z
p−e1 +1 Z
Z
p−e1 +1 Zp
χ̄(ξ10 ν1 )dξ10
Z
p−e2 +1 Zp
χ̄(ξ10 ν1 )dξ10 = 0.
p
For e2 > e1 , the proof is exactly the same, except the (A(a, e1 ))n is replaced with
Qn
i=1 A(bλi , e1 ).
41
χ̄(ξ20 ν2 )dξ20 .
Lemma 3.4.2. Let e1 > 1 and e1 = e2 = e. Then
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
p−e1 Zp −p−e1 +1 Zp
p−e2 Zp −p−e2 +1 Zp
Zn
p
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
=0.
Proof. As before, we rewrite p−e Zp as a sum:
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
p−e Zp −p−e+1 Zp
p−e Zp −p−e+1 Zp
X
X
a∈Z/pe Z×
b∈Z/pe Z×
=
Z
·
Zn
p
Zn
p
Z
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
a
+p−e+1 Zp
pe
b
+p−e+1 Zp
pe
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1 .
i=1
We change variables as in the previous lemma, making the above expression into
X
X
a∈Z/pe Z× b∈Z/pe Z×
n Z
Y
Z
Z
χ̄((
p−e+1 Zp
p−e+1 Zp
a
b
+ ξ10 )ν1 + ( e + ξ20 )ν2 )
e
p
p
a
b
+ ξ10 ) + λi ( e + ξ20 ))x2i )dxi )dξ20 dξ10
e
p
p
Zp
i=1
Z
Z
X
X
b
a
χ̄( e ν1 + e ν2 )
=
p
p
−e+1 Z
p−e+1 Zp
p
a∈Z/pe Z× b∈Z/pe Z× p
n Z
Y
a
b
0
0
· (ξ1 ν1 + ξ2 ν2 ) (
χ((( e + λi e ) + (ξ10 + λi ξ20 ))x2i )dxi )dξ20 dξ10 .
p
p
Zp
i=1
·
(
χ(((
a
pe
Now, for a given a and b, if
+ λj pbe 6∈ p−e Zp − p−e+1 Zp then
a + λj b ≡ 0 (mod p).
But then
a + λi b 6≡ 0 (mod p) ∀ i 6= j
else λj ≡ λi , contradicting our assumption that p - λj − λi . So, for all but at most
one λ (which we denote λj ),
a
b
+ λi e ∈ p−e Zp − p−e+1 Zp .
e
p
p
42
This means that, for these i 6= j,
Z
χ((
Zp
a
b 2
+
λ
)x ) = A(a + λi b, e).
i
pe
pe i
So let λj be as above. If no such λ exists, choose any of the λ’s to be λj . Then our
integral expression is
X
X
Y
A(a + λi b, e)
a∈Z/pe Z× b∈Z/pe Z× 1≤i≤n,i6=j
Z
Z
·
χ̄(
p−e+1 Zp
Z
·(
χ(((
Zp
p−e+1 Zp
b
a
ν
+
ν2 )χ̄(ξ10 ν1 + ξ20 ν2 )
1
e
e
p
p
a
b
+ λj e ) + (ξ10 + λi ξ20 ))x2j )dxj )dξ20 dξ10 .
e
p
p
If we let
z = ξ10 ν1 + ξ20 ν2 ,
w = ξ10 + λj ξ20
then the change of measure is
|ν1 λj + ν2 |−1
p .
By assumption, |ν1 λj + ν2 |−1
p is non-zero. Thus, the integral is
|ν1 λj +
Y
ν2 |−1
p
Z
χ̄(
A(a + λi b, e)
p−e+1 Zp
1≤i≤n,i6=j
Z
·(
Zp
χ((
Z
b
a
+ λj e + w)x2j )dxj )dwdz.
e
p
p
Since e > 1, we know that
Z
χ̄(z) = 0.
p−e+1 Zp
Thus the lemma holds.
43
p−e+1 Zp
a
b
ν
+
ν2 )χ̄(z)
1
pe
pe
3.4.2
The Integral Over (p−1 Zp )2
Now that we have determined that the Sp (ν) is entirely determined by the annuli
where ei ≤ 1, we will calculate the integral over each of the remaining annuli individually.
Lemma 3.4.3. Let e1 = e2 = 1. Assume that n is even and
ν2
6≡ λi ∀ i.
ν1
(Here, all congruences are modulo p.) Additionally, define
−1
H = {u ∈ F×
∀ i, u 6≡ −ν1 ν2−1 },
p : u 6≡ −λi
and let G be the classical Gauss sum as defined before. Then
Z
Z
Z
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
χ̄(ξ1 ν1 + ξ2 ν2 )
p−1 Zp −Zp
p−1 Zp −Zp
Zn
p
i=1
n
n
X
Y
XY
1 + λi − λ−1
−ν1 + λ−1
1 + λi u n
j
j ν2
n
(
)G − G p
)(
)
=−
(
p
p
p
j=1 1≤i≤n,i6=j
u∈H i=1
n
Y
1 − λi ν1 ν2−1 n
+ (p − 1) (
)G .
p
i=1
Proof. First, we rewrite the above integral as
Z
n Z
X XZ
Y
a
b
a + λi b 2
χ̄( ν1 + ν2 )(
χ(
xi )dxi )dξ1 dξ2
a
p
p
p
+Zp pb +Zp
×
×
i=1 Zp
p
a∈Fp b∈Fp
=
n Z
Y
a
b
a + λi b 2
χ̄( ν1 + ν2 )(
xi )dxi ).
χ(
p
p
p
Z
p
×
i=1
X X
a∈F×
p b∈Fp
×
Now, since a, b ∈ F×
p , we can write b = au for some u ∈ Fp . So the above can instead
be written as
n Z
Y
a
a(1 + λi u) 2
χ̄( (ν1 + uν2 ))(
χ(
xi )dxi )
p
p
×
i=1 Zp
X X
a∈F×
p u∈Fp
=
n Z
Y
a
a(1 + λi u) 2
χ̄( (ν1 + uν2 ))(
χ(
xi )dxi ).
p
p
Z
p
×
i=1
X X
u∈F×
p a∈Fp
44
Now, since the λ’s are incongruent modulo p, we can break the u’s into three sets.
These sets are
−1
H = {u ∈ F×
∀ i, u 6≡ −ν1 ν2−1 },
p : u 6≡ −λi
−1
for some i, u 6≡ −ν1 ν2−1 },
B = {u ∈ F×
p : u ≡ −λi
D = {u ≡ −ν1 ν2−1 }.
We deal first with u ∈ H. Then 1 + λi u 6≡ 0. Here,
Z
a(1 + λi u)
a(1 + λi u) 2
xi )dxi = (
)G
p
p
a (1 + λi u)
= ( )(
)G.
p
p
χ(
Zp
So
n Z
Y
a
a(1 + λi u) 2
χ̄( (ν1 + uν2 ))(
χ(
xi )dxi )
p
p
Z
p
×
i=1
XX
u∈H a∈Fp
n
=G
n
Y
a
a 1 + λi u
χ̄( (ν1 + uν2 )) ( )(
)
p
p
p
×
i=1
XX
u∈H a∈Fp
n
Y
XX a
a
1 + λi u
n
=G
).
( ) χ̄( (ν1 + uν2 )) (
p
p
p
×
i=1
u∈H
n
a∈Fp
If n is even, the above is
n
G
n
Y
1 + λi u
a
χ̄( (ν1 + uν2 )) (
)
p
p
×
i=1
XX
u∈H a∈Fp
n
XY
1 + λi u X a
=G
(
)
χ̄( (ν1 + uν2 ))
p
p
×
i=1
u∈H
n
a∈Fp
=Gn
X
n
Y
u∈H i=1
(
1 + λi u
)(−1).
p
since the sum of the a’s is the sum of all of the p-th roots of unity except 1. This
gives us the first summand from the right side of the equality in the statement of the
45
lemma.
Next, we deal with the case of u ∈ B. In this case, ∃ i such that 1 + uλj ≡ 0. So,
for this j,
Z
χ(
Zp
a(1 + λj u) 2
xj )dxi = 1.
p
Then, ∀ i 6= j,
Z
χ(
Zp
a (1 + λi u)
a(1 + λi u) 2
xi )dxi = ( )(
)G
p
p
p
as before. So
n Z
Y
a
a(1 + λi u) 2
χ̄( (ν1 + uν2 ))(
χ(
xi )dxi )
p
p
Z
p
×
i=1
XX
u∈B a∈Fp
= Gn−1
X
Y
(
u∈B 1≤i≤n,i6=j
= Gn−1
X
Y
a∈Fp
(
u∈B 1≤i≤n,i6=j
a
1 + λi u X a
)
χ̄( (ν1 + uν2 ))( )n−1
p
p
p
×
1 + λi u X a
a
χ̄( (ν1 + uν2 ))( ).
)
p
p
p
×
a∈Fp
Now, from Theorem 3.1.2, we know that if p - r then
p−1
X x
r X rx
r X
r
( )χ(rx) = ( )
( )χ(rx) = ( )
χ(x2 ) = ( )Gp.
p
p x∈F p
p x=0
p
x∈F
p
p
In our case,
a
a
−(ν1 + uν2 )
χ̄( (ν1 + uν2 ))( ) = (
)Gp.
p
p
p
×
X
a∈Fp
So the above is
Gn p
X
Y
u∈B 1≤i≤n,i6=j
(
1 + λi u −(ν1 + uν2 )
)(
).
p
p
Since u ∈ B means u ≡ −λ−1
j for some j, the above is
n
X
1 + λi − λ−1
−ν1 + λ−1
j
j ν2
(
G p
)(
).
p
p
j=1 1≤i≤n,i6=j
n
Y
This is the second summand in the lemma.
Finally, we deal with the case where u ≡ −ν1 ν2−1 . In this case, we know that
46
1 + uλi 6≡ 0 ∀ i. So
Z
χ(
Zp
a(1 + λi u)
a(1 + λi u) 2
xi )dxi = (
)G ∀ i.
p
p
and hence
n Z
Y
a
a(1 + λi u) 2
χ̄( (ν1 + uν2 ))(
xi )dxi )
χ(
p
p
×
i=1 Zp
XX
u∈D a∈Fp
n
XY
1 + λi u X a
a
(
=G
)
χ̄( (ν1 + uν2 ))( )n
p
p
p
×
u∈D i=1
n
a∈Fp
= Gn
n
Y
(
1−
λi ν1 ν2−1
p
i=1
= (p − 1)Gn
)
X
χ̄(0)
a∈F×
p
n
Y
1 − λi ν1 ν2−1
).
(
p
i=1
This is the third summand in the lemma.
Lemma 3.4.3.1. Let e1 = e2 = 1. We make the same assumptions and definitions
as in Lemma 3.4.3, except that
ν2
= λk
ν1
for exactly one k. Then
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
p−1 Zp −Zp
p−1 Zp −Zp
=−
X
Zn
p
n
Y
u∈H i=1
+ (p − 1)
(
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
X
Y
1 + λi − λ−1
−ν1 + λ−1
1 + λi u n
j
j ν2
)G − Gn p
(
)(
)
p
p
p
1≤j≤n,j6=k 1≤i≤n,i6=j
Y
1≤i≤n,i6=k
(
1 − λi ν1 ν2−1 n−1
)G .
p
Proof. : The proof for the first term (i.e. for u ∈ H) is exactly the same. For the
−1
second term (where u ∈ B), we note that B does not include −λ−1
k since −λk =
ν1 ν2−1 . For the term where u ∈ D, we note that
Z
χ(
Zp
a(1 + λi u) 2
xk )dxk ) = 1.
p
47
So
n Z
Y
a
a(1 + λi u) 2
χ̄( (ν1 + uν2 ))(
xi )dxi )
χ(
p
p
×
i=1 Zp
XX
u∈D a∈Fp
X
= Gn−1
Y
(
u∈D 1≤i≤n,i6=k
n
Y
= Gn−1
(
1−
a
1 + λi u X a
)
χ̄( (ν1 + uν2 ))( )n
p
p
p
×
a∈Fp
λi ν1 ν2−1
p
i=1
X
(
1≤i≤n,i6=k
1 − λi ν1 ν2−1
).
p
Lemma 3.4.4. Let e1 = 1, e2 = 0. Then
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
p−1 Zp −Zp
Zn
p
Zp
=
χ̄(0)
a∈F×
p
Y
= (p − 1)Gn
)



−Gn
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
if n is even,


Gn+1 ( −ν1 )
p
if n is odd.
Proof. In this case, the above integral is rewritten as
Z
Z
Z
n
X
χ̄(ξ1 ν1 )
χ( (ξ1 )x2i )dxdξ2 dξ1
p−1 Zp −Zp
Zn
p
Zp
i=1
Z
n
X
a
a 2
χ̄( ν1 )
χ(
=
· xi )dxdξ2 dξ1
a
p
p
+Z
Z
Z
p
p
p
×
i=1
p
a∈Fp
Z
Z
XZ
a
a
χ( · x2i )dxi )n dξ2 dξ1
=
χ̄( ν1 )(
a
p
p
+Zp Zp
Zp
×
p
XZ
Z
a∈Fp
=
a
a
χ̄( ν1 )(( )G)n .
p
p
×
X
a∈Fp
If n is even, the above equals
Gn
a
χ̄( ν1 )
p
×
X
a∈Fp
= − Gn .
48
If n is odd, the integral expression instead equals
a
a
χ̄( ν1 )( )Gn
p
p
×
X
a∈Fp
a2
a
χ̄( ν1 )( )
p
p
×
X
n
=G
a∈Fp
= Gn+1 (
−ν1
).
p
Lemma 3.4.5. Let e1 = 0, e2 = 1. Then
Z Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
Zp
p−1 Zp −Zp
=
Zn
p


Q

−( n
λi
n
i=1 p )G

Q

( n
i=1
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
if n is even,
λi
)Gn+1 ( −νp 2 )
p
if n is odd.
Proof. Here, we note that the above integral is equal to
Z
n
X
XZ Z
b
b
χ( (λi )x2i )dxdξ2 dξ1
χ̄( ν2 )
p
p
Zn
Zp pb +Zp
×
p
i=1
b∈Fp
=
n Z
Y
b
b
χ̄( ν2 )
χ((λi )x2i )dxi
p
p
×
i=1 Zp
X
b∈Fp
=
n
Y
λi b
b
χ̄( ν2 )( ( )G)
p
p
×
i=1
X
b∈Fp
n
Y
λi X b
b
=G
( )
χ̄( ν2 )(( ))n .
p
p
p
×
i=1
n
b∈Fp
If n is even then
X b
b
b
χ̄( ν2 )(( ))n =
χ̄( ν2 ) = −1.
p
p
p
×
×
X
b∈Fp
b∈Fp
If n is odd then
X b
b
b
b
−ν2
χ̄( ν2 )(( ))n =
χ̄( ν2 )( ) = (
)G.
p
p
p
p
p
×
×
X
b∈Fp
b∈Fp
Thus, the lemma follows.
49
Lemma 3.4.6. Let e1 = e2 = 0. Then
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
Zp
χ(
Zn
p
Zp
n
X
(ξ1 + λi ξ2 )x2i )dxdξ2 dξ1 = 1.
i=1
Proof. If e1 = e2 = 0 then ξ ∈ Z2p . Since νi , xi ∈ Zp and χ(Zp ) = 1, it follows that
Z
Z
Z
χ̄(ξ1 ν1 + ξ2 ν2 )
Zp
Zn
p
Zp
Z
Z
n
X
χ( (ξ1 + λi ξ2 )x2i )dxdξ2 dξ1
i=1
Z
=
dxdξ2 dξ1
Zp
Zp
Zn
p
= 1.
Finally, we arrive at the main theorems of this section. From the above lemmas,
we have proven the following:
Theorem 3.4.7. Let p - λi , ν1 , ν2 , p - λi − λj , p - ν1 λ1 + ν2 . Moreover, as above, let
−1
H = {u ∈ F×
∀ i, u 6≡ −ν1 ν2−1 }.
p : u 6≡ −λi
Assume that n is even and
λi 6≡
ν2
.
ν1
Then
n
n
XY
X
Y
1 + λi − λ−1
−ν1 + λ−1
1 + λi u n
j
j ν2
n
Sp (ν) = −
(
)G − G p
(
)(
)
p
p
p
i=1
j=1
u∈H
1≤i≤n,i6=j
n
n
Y
Y
1 − λi ν1 ν2−1 n
λi
n
n
+ (p − 1) (
)G − G − G ( ( )) + 1.
p
p
i=1
i=1
Proof. If we make the same assumptions as in Theorem 3.4.1 except
λk =
50
ν2
,
ν1
then
Sp (ν) = −
n
XY
X
Y
−ν1 + λ−1
1 + λi − λ−1
1 + λi u n
j
j ν2
)G − Gn p
)(
)
(
(
p
p
p
u∈H i=1
1≤j≤n,j6=k 1≤i≤n,i6=j
+ (p − 1)
Y
1≤i≤n,i6=k
3.5
(
n
Y
λi
1 − λi ν1 ν2−1 n−1
)G
− Gn − Gn ( ( )) + 1.
p
p
i=1
Convergence of the Singular Series
Here, we will use the computations of the previous section to determine whether our
expression for S(ν) converges. Recall that the singular series S(ν) can be found by
S(ν) =
Y
Sp (ν)
p
where Sp are the local singular series as above. We prove the following theorem:
Theorem 3.5.1. If λk 6=
ν2
ν1
and n ≥ 6 then S(ν) converges.
Proof. First, from Theorem 3.4.7, the product
n
n
X
Y
XY
1 + λi − λ−1
−ν1 + λ−1
1 + λi u n
j
j ν2
n
(
)G − G p
)(
))
T (ν) =
(−
(
p
p
p
j=1 1≤i≤n,i6=j
p prime
u∈H i=1
Y
n
n
Y
Y
1 − λi ν1 ν2−1 n
λi
n
n
+ (p − 1) (
)G − G − G ( ( )) + 1
p
p
i=1
i=1
differs from the product
S(ν) =
Y
Sp (ν)
p
by only a finite number of terms. So if we can show that T (ν) converges then S(ν)
51
must also converge. So
n
n
X
Y
Y XY
−ν1 + λ−1
1 + λi − λ−1
1 + λi u n
j ν2
j
n
)G | + |G p|
)(
)|
|T (ν)| ≤
|(
|(
p
p
p
j=1 1≤i≤n,i6=j
p prime u∈H i=1
n
n
Y
Y
1 − λi ν1 ν2−1 n
λi
n
n
+ (p − 1)
|(
)G | + |G | + |G ||( ( ))| + 1
p
p
i=1
i=1
≤
Y X
n
X
n
n
n
n
n
p− 2 + p1− 2 (
1) + (p − 1)p− 2 + p− 2 + p− 2 + 1
p prime u∈H
=
Y
j=1
n
(p − n − 1)p− 2 + np−
n−2
2
n
+ (p + 1)p− 2 + 1
p prime
=
Y
n
−np− 2 + (n + 2)p−
n−2
2
+ 1.
p prime
Q
Now, we use the fact that the product ∞
i=1 (1 + ai ) converges if and only if the sum
P∞
i=1 ai does as well. Then T (ν) converges if
∞
X
n
− ni− 2 + (n + 2)i−
n−2
2
i=1
= −n
∞
X
i
−n
2
+ (n + 2)
i=1
∞
X
i−
n−2
2
i=1
P
−n
2 converges
converges. We know that a sum like i=1 i converges if k > 1. So ∞
i=1 i
P
− n−2
2
if n > 2, and ∞
converges if n > 4. Since we assumed that n is even, n ≥ 6
i=1 i
P∞
k
is required for both summations to converge.
Theorem 3.5.2. If λk =
ν2
ν1
for one k and n ≥ 6 then S(ν) converges.
Proof. Again, we consider T (ν) and note that it differs from S(ν) by only a finite
number of terms. In this case,
n
X Y
X
Y
1 + λi − λ−1
−ν1 + λ−1
1 + λi u
j
j ν2
n
n
|
|(
|T (ν)| ≤
|(
)||G | + |G |p
)(
)|
p
p
p
u∈H i=1
1≤j≤n,j6=k 1≤i≤n,i6=j
n
Y
1 − λi ν1 ν2−1 n−1
λi
n
n
|(
+ (p − 1)
)G | + |G | + |G ( ( ))| + 1
p
p
i=1
1≤i≤n,i6=k
Y
≤
Y
n
n
(p − n)p− 2 + (n − 1)p1− 2 + (p − 1)p−
n−1
2
p prime
=
Y
p−
n−3
2
+ np−
n−2
2
− p−
n−1
2
n
+ (2 − n)p− 2 + 1.
p prime
52
n
n
+ p− 2 + p− 2 + 1
As before, this converges if and only if
X
p−
n−4
2
+ np−
n−2
2
− p−
n−1
2
n
+ (2 − n)p− 2
p prime
=
X
p−
p prime
+
X
n−3
2
+
X
np−
n−2
2
−
p prime
n
(2 − n)p− 2
p prime
converges. But these four sums all converge for n ≥ 6.
53
X
p prime
p−
n−1
2
Bibliography
[1] F. Bowman. Introduction to Bessel Functions. Dover Publications, Inc., New
York, 1958.
[2] R. N. Bracewell. The Fourier Transform and Its Applications. McGraw-Hill
Series in Electrical Engineering. Circuits and Systems. McGraw-Hill Book Co.,
New York, 2000. Third Edition.
[3] G. H. Hardy and J. E. Littlewood. A new solution to waring’s problem. Q. J.
Math., 48:272–293, 1919.
[4] G. Lachaud. Une présentation adélique de la série singuliére et du probléme de
waring. Enseign. Math., 28:139–169, 1982.
[5] T. Ono. Gauss transforms and zeta-functions. Ann. of Math., 91:332–361, 1970.
[6] T. Ono. Lectures on the Hardy-Littlewood singular series. Transcribed by T.
Wright, 2008.
[7] C. L. Siegel.
Analytische Zahlentheorie : Vorlesung[en] gehalten im Som-
mersemester 1963 [und im Wintersemester 1963/64] an der Universitt Gttingen.
Mathematisches Institut der Universität Göttingen, Göttingen, 1964.
54
Vita
Thomas Wright was born in September, 1980 and was raised in North Easton, Massachusetts. In 2003, he received a Bachelor of Arts in both Mathematics and Economics from Bowdoin College. He enrolled at Johns Hopkins University in the fall of
2003. He defended his thesis on February 20, 2009.
55