Lectures on the Hardy-Littlewood Singular Series

Takashi Ono

September 3, 2008

0 Introduction

The story begins almost ninety years ago with Hardy and Little-wood, who examined the integer solutions for the equation

xδ1 + xδ2 + ...+ xδn = ν, ν ∈ Z

for an integer δ ≥ 2. Now, instead of finding solutions for x1, x2, ...xnin Zn, it is an equally valid question to ask whether there are solu-tions in Onk for a global field k, or an A-field in the sense of Weil’sBasic Number Theory (1967). Here, by Ok we mean the ring ofintegers of k.

To put this another way, if f is a polynomial function such thatf : kn → k where ν ∈ k, we wish to study the set f−1(ν) (i.e. thefiber of f which maps down to ν). Since we are interested in thenumber of solutions for the Hardy-Littlewood equation, we will de-fine

Nt(ν) = #(f−1(ν)⋂Kt) = #x ∈ Kt : f(x) = ν

such that⋃tKt = knA,

where the Kt is a family of compact sets in knA. The parame-ter t itself will be later explained, but for now it suffices to say thatwe will be interested in the growth of Nt as t likewise grows. Morespecifically, we will consider

1

limt→∞Nt(ν)t∗

,

where the denominator denotes that t is raised to some power. Themeaning of ”t→∞” here is ill-defined but will be properly definedin 2.4.

This type of quantity is known as the singular series, denotedS(ν). Hardy and Littlewood made extensive use of the singularseries and a method which they called the circle method in provingtheir results about the Hardy-Littlewood equation.

Upcoming sections will use these ideas to develop an adelizedversion of the Hardy-Littlewood Theorem.

0.1 Nt(ν)

Now, we will define our terms more formally. Let k be an A-field,f : kn → km a polynomial map, where ν ∈ Omk . We will define

f−1(ν) = γ ∈ kn : f(γ) = ν

to be the fiber of f which maps down to ν. Then

k ⊂ kA := ring of adeles of K ⊃ Kt.

Note that k is discrete in kA, kA is locally compact, and Kt is com-pact. For a given set Kt, we have

Nt(ν) = #(Kt

⋂f−1(ν)).

0.2 Integral Representation of Nt(ν)

In this section, we wish to find an integral which will accuratelydescribe Nt(ν). Let ψt be the characteristic function of Kt. It isobvious, then, that

Nt(ν) =∑

γ∈kn,f(γ)=ν ψt(γ).

We also define the inner product in the usual way:

2

<,>: knA × knA → kA,< x, y >=

∑xiyi.

We let χ be a basic character of k+A which is trivial on k but non-

trivial on k+A . This character is unique up to k×, i.e. if χ′ is another

such character then

χ′(x) = χ(αx), α ∈ k×.

We can use this to define a function Γfψt which maps kmA to C:

(Γfψt)(ξ) =∑

γ∈km ψt(γ)χ(< f(γ), ξ >).

This is a function on kmA /km.

For ν ∈ km, the ν-th Fourier coefficient of Γfψt is∫kmA /k

m Γfψt(ξ)χ(< ξ, ν >)dξ

=∫kmA /k

m Γfψt(ξ)χ(< ξ,−ν >)dξ

=∑

γ ψt(γ)∫kmA /k

m χ(< f(γ)− ν, ξ >)dξ.

Note that < f(γ)− ν, ξ > is a character on ξ. In fact,

< f(γ)− ν, ξ >=

1 if f(γ) = ν,

0 otherwise.

So∑γ ψt(γ)

∫kmA /k

m χ(< f(γ)− ν, ξ >)dξ

=∑

γ∈kn,f(γ)=ν ψt(γ)

= Nt(ν).

Thus, we have reduced the initial question to one about (Γfψt)(ξ).

0.3 Singular Series

Instead of Γ, we will move to G:

(Gfψt)(ξ) =∫knAψt(x)χ(< f(x), ξ >)dx.

3

We assume Gfψt ∈ L1(kmA ). We let

St(ν) = Gfψt(−ν) =∫kmAGfψt(ξ)χ(< ξ, ν >)dξ.

In upcoming sections, we will compare St(ν) and Nt(ν).

0.4 Pair of Representations of an Algebraic Group

We start with a group G and a pair of representations ρ1 and ρ2

Gρ2

##GGGGGGGGρ1 // GLn

GLm

which are commutative and equivariant. In fact, for any s ∈ G,we get the commutative diagram

knf−−−→ kmyρ1(s)

yρ2(s)

knf−−−→ km

,

where the f is the one described in our generalization of the Hardy-Littlewood Theorem such that

f(ρ1(s)x) = ρ2(s)f(x) ∀ x ∈ kn, ∀ s ∈ G.

Next, we also need to define our Ktt∈GA , where each Kt ⊂ knAis a compact set. So, for a place v which determines a valuation | · |v,

Bv = x ∈ kv : |xi|v ≤ 1, 1 ≤ i ≤ n.

If v = ∞ then Bv is an n-dimensional box. If v 6= ∞ then Bv =Onv ⊂ knv , where Onv is the standard lattice points. We relate theseB’s to the language of adeles by taking

K1 = ΠvBv

and taking t = 1 (the 1 adele) to be the origin. This is a standard

4

compact set by Tychonof’s Theorem. Now, we can define the Kt’s as

Kt = ρ1(t)Kt ∀ t ∈ GA.

Recalling our definition of ψt as a characteristic function on Kt,we note that

ψt(x) = ψ1(ρ1(t)−1x).

If we define the adelic valuation

|x|A = Πv|xv|v

then, recalling from the previous section our definition for Gfψt andSt(ν), we can rewrite the definition of Gfψt as

Gfψt(ξ) = |detρ1(t)|Agfψ1((ρ2(t))T ξ),

where |detρ1(t)|A ∈ R×. Similarly, for St(ν),

St(ν) = |detρ1(t)|A|detρ2(t)|A

S1(ρ2(t)−1ν).

Now, we can more formally define ”t→∞” to mean |detρ1(t)|A →∞.

Armed with these definitions, we hope, as Hardy and Littlewooddid previously, that

Nt(ν)|detρ1(t)|A|detρ2(t)|A

= S1(ρ2(t)−1ν) +O(|detρ1(t)|−εA ).

Hardy and Littlewood then went on to find more explicit resultsfor ε.

0.5 Hardy-Littlewood Theorem

Having defined the previous notation, we can describe in this no-tation Hardy and Littlewood’s work from their 1920 paper, A NewSolution To Waring’s Problem

f : Qn → Q,

5

f : x 7→ xδ1 + xδ2 + ...+ xδn for 1 ≤ δ ∈ Z,

and

k = Q,G = Q× = GL1,ρ1(t) = tIn ∈ GLn,ρ2(t) = tδ ∈ GL1.

Plugging these in as our definitions gives us the following infor-mation:

det(ρ1(t)) = tn, det(p2(t)) = tδ,f(ρ1(t)x) = ρ2(t)f(x).

Moreover, in defining our Kt’s, we find in this case that

K1 =

v =∞ B∞ = x ∈ Rn : |xi| ≤ 1, 1 ≤ i ≤ n,v = p Bp = Zn

p ,

K = B∞ × ΠpZnp ,

where, as always, Zp is defined as

Zp = x ∈ Qp : |x|p ≤ 1.

We can even describe t in terms of an element in the group GA =Q×A = A× of ideles in the ring of adeles A by letting t = (τ, 1, 1, 1, ...) ∈GA, where τ ∈ Q∞ = R, τ > 0. Then

Kt = ρ1(t)K1 = B∞(τ)× ΠpZnp .

Additionally, for ν ∈ Z, our choice of t means that

Nt(ν) = #γ ∈ Qn : f(γ) = ν, γ ∈ Kt= #γ ∈ Zn : |γi| ≤ τ, 1 ≤ i ≤ n such that f(γ) = ν,

and St(ν) can be written as

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St(ν) = |t|n−δA J(τ−δν)S(ν),

where |t|n−δA = τn−δ, J(τ−δν) is a Gamma factor, and S(ν) is thesingular series.

Hardy and Littlewood’s theorem can then be written as

Theorem (Hardy-Littlewood): ∃ ε such that

Nt(ν)τn−δ

= J(τ−δν)S(ν) +O(τ−ε).

We give an outline of the proof:

“Proof”: Since m = 1 (i.e. ν ∈ Q1), we found in 2.2 that

Nt(ν) =∫

QA/Q(Γfψt)(ξ)χ(ξν)dξ.

We can express Γfψt(ξ) as

Γfψt(ξ) =∑

γ∈Qn ψt(γ)χ(f(γ)ξ)

=∑

γ∈Zn,|γi|≤τ χ((γδ1 + γδ2 + ...+ γδn))

= (∑

γ∈Z,|γ|≤τ χ(γδξ))n.

So

Nt(ν) =∫

QA/Q[∑

γ∈Z,|γ|≤τ χ(γδξ)]nχ(ξν)dξ.

Let us now define our character χ on QA. Let x = (xv) ∈ QA.For v = p, xp ∈ Qp implies that ∃ l > 0 such that plxp ≡ z (mod pl)for some z ∈ Z. So let

χv(xv) =

e−2πix∞ if v =∞,e

2πi

plz

if v = p.

Then a basic character χ : QA/Q→ T is given by

χ(x) = Πvχv(xv).

In fact, χ∞ yields the exact sequence

7

0→ Z→ R χ∞→ T→ 0.

Analogously, χp yields the sequence

0→ Zp → Qpχp→ T,

where Zp is the kernel of χp.Next, since Q is discrete in QA, we can discuss the fundamental

domain F of QA/Q

F = [0, 1)× ΠpZp.

Of course, the volume of this fundamental domain is 1.Now, we can rewrite Nt as

Nt(ν) =∫F(∑

γ∈Z,|γ|≤τ χ(γδξ))nχ(ξν)dξ.

We note that for any non-archimedean place, ξ ∈ F is integral,and hence γδξ is also integral. So

χ(γδξ) = Πvχv(γδξ) = χ∞(γδξ∞)Πpχp(γ

δξp) = χ∞(γδξ∞),

and similarly

χ(ξν) = χ∞(ξ∞ν).

Then our expression for Nt becomes

Nt(ν) =∫ 1

0(∑

γ∈Z,|γ|≤τ χ∞(γδξ∞))nχ∞(ξ∞ν)dξ∞.

Put α = ξ∞ and

Tτ (α) =∑

γ∈Z,|γ|≤τ χ∞(γδα).

Then

Nt(ν) =∫ 1

0(Tτ (α))nχ∞(αν)dα.

The finite sum Tτ is counted using the circle method. Let x ∈

8

[0, 1)⋂

Q, x = am

, (a,m) = 1. Moreover, let ε and τ > 0. For ourpurposes, ε will be small, while τ will be larger. Define

S(ε, τ) = x ∈ [0, 1)⋃

Q : m ≤ τ ε,Mx = α ∈ [0, 1) : |α− x| < 1

τδ−ε.

Then for x, x′ ∈ S(ε, τ), x 6= x′ implies that Mx

⋃Mx′ = Ø.

Finally, we split the integral into two parts and evaluate eachone. Let

M =⋃x∈S(ε,τ) Mx,

M∗ = [0, 1)−M.

M and M∗ are referred to as the major and minor arcs, respec-tively. We split the interval [0, 1) into the two arcs:∫ 1

0=∫M+

∫M∗ .

For the minor arc, if n ≥ 2δ + 1 then∫M∗ |Tτ (α)|ndα = O(τn−δ−ε

′),

where ε′ > 0 is a function of our chosen ε. For the major arc,the integral was computed explicitly to be∫

M |Tτ (α)|ndα = Nt(ν)tn−δ

= J(τ−δν)S(ν) +O(τ−ε).

1 Finite Fields1

1.1 Gauss Sums

Let Fq be the extension of Fp such that q = pf (i.e. [Fq : Fp] = f)and let T to be the trace in Fq/Fp. Then for x ∈ Fq, we can definethe function

1For details, see A. Weil, The Number of Solutions of Equations in Finite Fields, Bull.A.M.S. 55 (1949)

9

ε(x) = e2πipT (x).

Clearly, ε ∈ F+q . So if ϕ ∈ C(Fq) is a complex valued function

then

ϕ(ξ) =∑

x∈Fq ϕ(x)ε(xξ)

is the Fourier transform of ϕ. Of course, the field Fq has two op-erations, so we must also consider multiplicative characters. For

χ ∈ F×q , we extend it to get χ ∈ C(Fq) by

χ(0) =

1 if χ = χ0 (i.e. χ is trivial),

0 otherwise.

If χ 6= χ0, we set

χ(ξ) =∑

x∈F×q χ(x)ε(xξ) = Gξ(χ).

This Gξ(χ) is called a Gauss sum. The Gauss sum has three basicproperties:

(1) If ξ 6= 0 then χ(ξ) = χ(ξ)−1χ(1).(2)∑

ξ 6=0 χ(ξ) = 0.

(3) If ξ 6= 0 then |χ(ξ)| = √q.

The first two properties are to be expected, but the third is rathersurprising.

1.2 N (q)(ν)

Let f be a function

f : Fnq → Fq,

where ν ∈ Fq. Define

N (q)(ν) = #f−1(ν).

Put

10

Γf (ξ) =∑

x∈Fnq ε(f(x)ξ).

So Γf (ξ) ∈ C(Fq). Then, as before, we can relate N to the ν-thFourier coefficient of Γf :∑

ξ∈Fq Γf (ξ)ε(ξν) =∑

x

∑ξ ε((f(x)−ν)ξ) =

∑f(x)=ν q = qN (q)(ν).

This means that

N (q)(ν) = qn−1 + 1q

∑ξ∈F×q ε(ξν)

∑x∈Fnq ε(f(x)ξ).

This process can be repeated for the more general set of functionsf : Fnq → Fmq by simply replacing the products with inner products(e.g. f(x)ξ would be replaced by < f(x), ξ >).

Let us take the particular case of

f(x) = a1xδ11 + ...+ anx

δnn .

Note that

ε(f(x)ξ) = Πni=1ε(aix

δii ξ).

So

N (q)(ν) = qn−1 + 1q

∑ξ∈F×q ε(ξν)Πn

i=1(∑

xi∈Fq ε(aixδii ξ)).

As in 1.1, we define Gξ(χ) as

Gξ(χ) = χ(ξ) =∑

F×q χ(x)ε(xξ).

This is a generalization of the classical Gauss sum, which wouldbe the case where δ’s are all 2.

Next, if d = (δ, q − 1) then χα ∈ F×p can be defined as

χα(y) = e2πidL(y)α for y ∈ F×q .

Here, L(y) is defined as follows. Choose a generator ω ∈ F×q . Then

y = ωl(y) for some l(y) in Z/(q − 1)Z. Let j be the reduction map

11

from Z/(q − 1)Z to Z/dZ. This gives us the maps

F×ql

→ Z/(q − 1)Z j→ Z/dZ,

where l is an isomorphism and j is surjective. Then L = j l.Using our generalized version of the Gauss sum, we wish to show

the following:

Claim:∑

x∈Fq ε(cxδ) =

∑α∈Z/dZ,α 6=0Gc(χα).

Let us begin with the left side, which we can rewrite as∑x∈Fq ε(cx

δ) =∑

y∈Fq µ(y)ε(cy) = 1 +∑

y∈F×q µ(y)ε(cy),

where

µ(y) = #x ∈ Fq : y = xδ,

meaning that µ(0) = 1. Now, by basic modular arithmetic, weknow that if ax ≡ b (mod m) has a solution then the number ofsolutions is equal to d = (m, a). So

µ(y) =

d if d|l(y)⇔ L(y) = 0,

0 if d - l(y)⇔ L(y) 6= 0.

So

1 +∑

y∈F×q µ(y)ε(cy) = 1 + d∑

L(y)=0 ε(cy).

Additionally, let ed(∗) = e2πid

(∗). So

∑α∈Z/dZ χα(y) =

∑α ed(L(y)α) =

0 if L(y) 6= 0,

d if L(y) = 0.

So

1 + d∑

L(y)=0 ε(cy) = 1 +∑

y∈F×q∑

α∈Z/dZ χα(y)ε(cy)

= 1 +∑

α∈Z/dZ∑

y∈F×q χα(y)ε(cy)

= 1 +∑

y∈F×q ,α=0 ε(cy) +∑

α∈Z/dZ,α 6=0

∑y∈F×q χα(y)ε(cy)

12

= 1 + 0 +∑

α∈Z/dZ,α 6=0

∑y∈F×q χα(y)ε(cy)

=∑

α∈Z/dZ χα(c)

=∑

α∈Z/dZGc(χα).

This proves the claim.Now, we can use this to give a bound for |N (q) − qn−1|. As an

aside, the n = 2 case is essentially the Riemann Hypothesis for func-tion fields, as proven by Deligne in 1974. For a δi, let di = (δi, q−1).By the claim, it follows that∑

xi∈Fq ε(aixiξ) =∑

αi∈Z/diZ,α 6=0 χαi(aiξ).

We can plug this into our expression for N (q) to find

N (q)(ν) = qn−1+1q

∑ξ∈F×q ε(ξν)Πn

i=1(∑

xi∈Fq∑

αi∈Z/diZ,α 6=0 χαi(aiξ)).

Note that |ε(ξν)| = 1 and |χαi(aiξ))| =√q by property (3) of Gauss

sums. So, the above implies that

|N (q)(ν)− qn−1| ≤ 1q(q − 1)Πn

i=1((di − 1)√q)

= qn2−1(q − 1)Πn

i=1(di − 1) < Mqn2

with M = Πni=1(di − 1) independent of q.

2 Weyl Sums

2.1 Weyl Sums over Finite Abelian Groups

Let A be a finite abelian group of order N , and let G be a locally

compact abelian group. Let χ ∈ G = Hom(G,T), the dual of G,and let ϕ : A→ G be any map. We define the Weyl sum

W (ϕ) =∑

x∈A χ(ϕ(x)).

Additionally, let x, h, h1, h2 ∈ A. Then we define

(∆hϕ)(x) = ϕ(x+ h)− ϕ(x),(∆h1h2ϕ) = (∆h2(∆h1ϕ)).

13

More generally, we will write

∆h1h2..hlϕ = ∆hl(∆h1...hl−1ϕ).

We claim that this ∆ action commutes

Lemma 2.1.1: ∆h1h2 = ∆h2h1 .

Proof : Let us denote ψ = ∆h1ϕ. Then

∆h1h2(x) = ∆h2(∆h1ϕ)(x) = ∆h2(ψ(x)) = ψ(x+ h2)− ψ(x)= (∆h1ϕ)(x+ h2)− (∆h1ϕ)(x)= ϕ(x+ h2 + h1)− ϕ(x+ h2)− ϕ(x+ h1) + ϕ(x).

Note that this would still be the case if h1 and h2 were switched. Sothe lemma holds.

Proposition 2.1.1: |W (ϕ)|2 =∑

h∈AW (∆hϕ).

Proof : Since W (ϕ) ∈ C,

|W (ϕ)|2 = W (ϕ)W (ϕ)

=∑

y χ(ϕ(y))∑

x χ(ϕ(x))

=∑

x,y χ(ϕ(y)− ϕ(x)).

Let y = x+ h. Then∑x,y χ(ϕ(y)− ϕ(x)) =

∑x,h χ(ϕ(x+ h)− ϕ(x))

=∑

h

∑x χ(∆hϕ(x))

=∑

hW (∆hϕ).

Next, we know that for x, y ∈ Cn,

| < x, y > | ≤ |x||y|

by the Cauchy-Schwartz inequality, where x = (x1, ..., xn), y =(y1, ..., yn), < x, y >=

∑i xiyi is the usual Hermitian, and |x| =√

|xi|2 is the usual norm. Then

14

|∑

i xiyi|2 ≤ (∑

i |xi|2)(∑

i |yi|2).

If x = (1, ..., 1) ∈ CN then the above yields

(1) |∑

i yi|2 ≤ N∑

i |yi|2.

We can apply this to our ∆ by the following. Let

yhl−1=∑

h1,h2,...,hl−2W (∆h1h2,...hl−2hl−1

ϕ).

Then

|∑

h1,...,hl−1W (∆h1...hl−1

ϕ)|2 = |∑

hl−1

∑h1,h2,...,hl−2

W (∆h1...,hl−2hl−1ϕ)|2

≤ N∑

hl−1|∑

h1,h2,...,hl−2W (∆h1...,hl−2hl−1

ϕ)|2

by (1).

Theorem 2.1.1: If l ≥ 1 then

|W (ϕ)|2l ≤ N2l−l−1∑

h1...hlW (∆h1...hlϕ).

Proof : If l = 1 then this is Proposition 2.1.1 (with equality). Theproof is then an induction on l, and is left as an exercise to thereader.

Now, we amend the previous setup such that R is a locally com-

pact ring, χ ∈ R+, A ⊂ R+ is a finite subgroup of order N , andϕ : A→ R is any mapping from A to R. In particular, we will takeϕ(x) = xδξ, ξ ∈ R, δ > 0. So

W (ϕ) =∑

x∈A χ(xδξ).

Note that if δ = 1 then

W (ϕ) =∑

x∈A χ(xξ) =

N if χ(xξ) = 1 ∀x ∈ A,0 otherwise.

So we will assume that δ > 1. This means that

15

∆hϕ(x) = ϕ(x+ h)− ϕ(x) = (x+ h)δξ − xδξ= δhxδ−1ξ+(lower order terms in x).

So

∆h1...hδ−1ϕ(x) = δ!ξh1...hδ−1x+ c

for some c ∈ R. If we put l = δ − 1 then, by Theorem 2.1.1,

• |W (ϕ)|2δ−1 ≤ N2δ−1−δ∑h1...hδ−1

|∑

x∈A χ(δ!ξh1...hδ−1x)|,

since |χ(c)| = 1.

2.2 Riemann-Roch Theorem

Let k be a function field defined over Fq, and let kA be its ring ofadeles. We let χ : kA → T be a basic character of k.

Now, we get the following diagram.kA

γ

000000000000000 0

k k#

0 kA,

where

k# = χ ∈ kA : χ(k) = 1

is the annihilator of k. We know that an isomorphism γ existsbecause of the self-duality of kA. More specifically, γ must be of theform

γ(x)(a) = χ(ax)

for a ∈ k×A . For another character χ′ ∈ kA,

16

χ′(x) = χ(ax)

for some a ∈ k×. The character χ can be expressed as

χ(x) = Πvχv(xv),

where x = (xv) ∈ kA. We define ordv χv ∈ Z by

P−ord χvv = x ∈ kv : χv(xy) = 1 ∀ y ∈ Ov,

where Pv = (πv) is the prime ideal of Ov. Let

π(χ) = (π−ord χvv ) ∈ k×A .

Additionally, for t ∈ k×A , define

t∗ = π(χ)t−1.

It is obvious that t∗∗ = t. Next, let

P (t) = Πv(tvOv) ⊂ kA,L(t) = k

⋂P (t) ⊂ kA.

Note that P (t) is compact, while k is discrete in kA. So L(t) isa finite abelian group. Note that P (t) is defined equivalently by

P (t) = x ∈ kA : |x|v ≤ |t|v,∀ v.

P (t) contains Fq, which can be defined by

Fq = x ∈ k : |x|v ≤ 1.

Finally, since L(t) can be viewed as a vector space over Fq, let

l(t) = dimFqL(t).

Hence |L(t)| = ql(t).Now, we come to an important theorem in algebraic geometry

and complex analysis.

17

Riemann-Roch Theorem: Let t ∈ k×A and let g be the genus ofk. Then

|t|A = ql(t)−l(t∗)+(g−1).

Proof : To start, we consider the following diagrams:

kA

γ

""EEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEEE 0

k + P (t)

IIIIIIIII

xxxxxxxxxx(k + P (t))#

MMMMMMMMMM

ssssssssss

k

FFFFFFFFFF P (t)

uuuuuuuuuk#

KKKKKKKKKK P (t)#

qqqqqqqqqq

L(t) k# + P (t)#

0 kA.

We wish to prove four properties about the elements of the abovediagrams:

(1) (k + P (t))# = k#⋂P (t)#.

(2) (P (t))# = γ(P (t∗)).(3) (k + P (t))# = γ(L(t∗)).(4) (L(t))# = γ(k + P (t∗)) = k# + P (t)#.

The proofs are as follows:

(2) We know that γ : kA → kA is an isomorphism. So

γ(x) ∈ P (t)# ⇔ γ(x)(P (t)) = 1⇔ χ(xP (t)) = 1⇔ χ(x · ty) = 1 ∀ y ∈ P (1) = ΠvOv⇔ ∀ v, χv(xvtvyv) = 1 ∀ yv ∈ Ov⇔ ∀ v, xvtv ∈ P−ord χvv

18

⇔ ∀ v, xv ∈ π−ord χvv t−1v Ov

⇔ x ∈ P (t∗).

(3) This is proven by the following string of equalities

(k + P (t))# = k#⋂P (t)#

= γ(k)⋂γ(P (t∗))

= γ(k⋂P (t∗))

= γ(L(t∗)).

(4) First, we note that

γ(k + P (t∗)) = γ(k) + γ(P (t∗)) = k# + P (t)#.

This shows the latter of the two equalities. Now, kA/k is compact,which means that kA/(k + P (t)) is compact (and, in fact, finite).Moreover,

kA/(k + P (t)) ≈ (kA/(k + P (t))) = (k + P (t))# = γ(L(t∗)),

where we define the notation A to be the same as A. Now weprove the former of the two equalities by showing inclusion in eachdirection:

(⊃) L(t)# = (k⋂P (t))# ⊃ k# + P (t)# = γ(k + P (t∗)).

(⊂) Consider the exact sequence

(#) kA/γ(k + P (t∗))→ kA/L(t)# → 0.

Note that

L(t) ≈ L(t) ≈ kA/L(t)#,

where the former is because L is a finite group, and the latter isby Pontrjagin duality. So

[kA/L(t)#] = ql(t).

19

On the other hand,

[kA : γ(k + P (t∗))] = [kA : k + P (t∗)] = ql(t∗∗) = ql(t).

Since l(t∗∗) = l(t), (#) is an isomorphism. So L(t)# = γ(k+P (t∗)).

Now, let µ be the Haar measure of the additive group kA such thatµ(πvOv) = µ(P (1)) = 1. So

µ(kA/k) = ql(t∗)µ(k + P (t)/k)

= ql(t∗)µ(P (t)/L(t))

= ql(t∗)−l(t)µ(P (t)),

kA

ql(t∗)

k + P (t)

JJJJJJJJJ

wwwwwwwwww

k

GGGGGGGGGG P (t)

ttttttttt

L(t)

ql(t)

0.

But µ(P (t)) = |t|A, and, since L(1) = Fq,

µ(kA/k) = qgµ(k + P (1)/k) = qgµ(P (1)/L(1)) = qg−1.

We combine these two expressions for µ(kA/k) to get

|t|A = ql(t)−l(t∗)+(g−1).

Q.E.D.

Corollary 2.2.1: |π(χ)|A = q2g−2, l(π(χ)) = g.

Proof : From the Riemann-Roch Theorem,

20

|t∗|A = ql(t∗)−l(t)+(g−1).

So

|tt∗|A = q2(g−1).

The corollary then follows from the definition

t∗ = π(χ)t−1.

Now, put t = 1. We know l(1) = 1 since L(1) = Fq. Then wehave the diagram

kA

Fgq

k + P (1)

PPPPPPPPPPPP

vvvvvvvvvv

k

HHHHHHHHHH P (1) = ΠvOv

nnnnnnnnnnnn

L(1) = Fq

0.

Then

1 = q1−l(1∗)+g−1.

Since 1∗ = π(χ), this means that

q0 = q−l(π(χ))+g,

i.e. g = l(π(χ)).

Corollary 2.2.2: If |t|A < 1 then l(t) = 0.

21

Proof : Recall that l(t) = dimFqL(t) by definition. If L(t) 3 x 6= 0then |x|v ≤ |tv|v ∀ v, and hence, by the product formula in k,

1 = Πv|x|v ≤ Πv|tv|v < 1,

a contradiction.

Corollary 2.2.3: Let |t|A > g2g−2. Then l(t∗) = 0. Moreover,|t|A = ql(t)+(g−1) and kA = k + P (t).

Proof : First, if |t|A > g2g−2 then

|t∗|A = |π(χ)t−1|A < 1.

So l(t∗) = 0 by Corollary 2.2.2. By the Riemann-Roch Theorem,this means that |t|A = ql(t)+(g−1). Additionally, the space kA/k+P (t)has dimension l(t∗), so kA = k + P (t).

Finally, we will consider the set of valuations v. First, we notethat in the completion of k by v, Fq extends to Fqv = Ov/Pv, wherePv is the maximal prime ideal. We can define the degree of v by

qdeg v = qv.

Moreover, let us define a divisor A as

A =∑

v a(v)v,

where a(v) ∈ Z and all but finitely many a(v) = 0. Similarly,we define

deg A =∑

v a(v)deg v.

The group of divisors will be denoted D(k), and the group of degreezero divisors will be denoted D0(k). So we have the following exactsequence

0→ D0(k)→ D(k)deg→ Z→ 0.

22

To t = (tv) ∈ k×A , we associate a divisor

At =∑

v ordv(t−1v )v.

This yields another exact sequence

0→ ΠvO×v → k×A → D(k)→ 0.

Additionally, it can be shown that |t|A = qdegAt (an exercise tothe reader).

Let

t = π(χ) = (π−ord χvv ).

So

ord χv = ord(t−1v ).

and hence

At = Aπ(x) =∑

v(ord χv)v.

In general, Aπ(x) will be called the canonical divisor and denotedC.

Proposition 2.2.1: At∗ = C− At.

Proof : Follows from t∗ = π(χ)t−1.

Proposition 2.2.2 (The Classical Form of the Riemann-RochTheorem): For a divisor A =

∑v a(v)v, let

Λ(A) = k⋂

ΠvP−a(v)v ⊂ k

and

λ(A) = dimFqΛ(A).

Then

23

degA = λ(A)− λ(C− A) + (g − 1).

Proof : Note first that

Λ(At) = L(t),λ(At) = l(t).

Then by the Riemann-Roch Theorem

qAt = |t|A = ql(t)−l(t∗)+(g−1) = qλ(At)−λ(At∗ )+(g−1).

But any divisor can be written as At for some t. So the Propo-sition follows from Proposition 2.2.1.

2.3 Weyl Sums Over L(t)

Let k be a function field and χ be a basic additive (nontrivial) char-acter on kA which is trivial on k. Again, we will define f : kA → k by

f(x) = a1xδ11 + ...+ anx

δnn .

Let δ = l.c.m.(δi). We let

ρ1(t) =

tδδ1 .... 0...

. . ....

0 ... tδδn

, ρ2(t) = tδ, t ∈ k×A .

In this case,

K1 = ΠvOnv ,

Kt = ρ1(t)K1 = Πni=1P (t

δδi ) ∈ knA,

Kt

⋂kn = Πn

i=1L(tδδi ) ⊂ kn.

In this case, using the notation in 2.1

ΓfΨt(ξ) =∑

α∈Kt⋂kn χ(f(α)ξ)

24

= Πni=1(∑

αi∈L(tδ/δi ) χ(aiαδii ξ)), ξ ∈ kA.

We define

W(a,δ)t (ξ) =

∑α∈L(t) χ(aαδξ).

This is a Weyl sum with R = kA and A = L(t). Recalling theintegral representation in 2.2 for Nt(ν) for ν ∈ k, we can write

Nt(ν) =∫kA/k

(Πni=1W

(ai,δi)

tδ/δi(ξ))χ(ξν)dξ.

Now, if |t|A > q2g−2, then |t|A = ql(t)+(g−1), which means thatl(t) > (g − 1). This gives us the following diagrams

kA = k + P (t)

NNNNNNNNNNN

rrrrrrrrrrrrkA

ql(t)

k

LLLLLLLLLLLL P (t)

ppppppppppppk + P (t∗)

LLLLLLLLLL

vvvvvvvvvv

L(t)

ql(t)

k

HHHHHHHHHH P (t∗)

rrrrrrrrrr

0 L(t∗) = 0.

Note that k + P (t∗)/k ∼= P (t∗)/L(t∗) since L(t∗) = 0. So∫kA/k

= q−l(t)∫P (t)

.

Plugging this into our definition for Nt(ν) gives

Nt(ν) = q−l(t)∫P (t)

(Πni=1W

(ai,δi)

tδ/δi(ξ))χ(ξν)dξ.

By the inequality (•) at the end of 2.1,

|W (a,r)(ξ)t |2r−1 = (ql(t))2r−1−r∑

h1,..,hr∈L(t) |∑

χ∈L(t) χ(r!aξh1...hr−1)|=∑

x Ψh1...hr−1(ξ)(x)

=

ql(t) if Ψh1...hr−1(ξ) = 1,

0 otherwise.

25

From now on, we assume that p = char(k) > r. So r! 6= 0 ink. Note that

Ψh1...hr−1(ξ) = 1 ⇔ γ(r!aξh1...hr−1)(x) = 1 ∀ x ∈ L(t)⇔ γ(r!aξh1...hr−1)(x) ∈ L(t)# = γ(k + P (t∗)⇔ r!aξh1...hr−1 ∈ k + P (t∗)⇔ aξh1...hr−1 ∈ k + P (t∗).

By the Riemann-Roch Theorem, ql(t) = |t|Aq1−g. So

∑x Ψh1...hr−1(ξ)(x) =

|t|Aq1−g if Ψh1...hr−1(ξ) = 1,

0 otherwise.

So

♥ |W (a,r)t(ξ)|2r−1 ≤ (ql(t))2r−1−r+1#(h1, ..., hr−1) ∈ L(t)r−1 :ξah1...hr−1 ∈ k + P (t∗).

Note that

ξah1...hr−1 ∈ k + P (t∗) ⇔ ξh1...hr−1 ∈ k + a−1P (t∗)

and

k + a−1P (t∗) = k + P (a−1t∗) = k + P ((at)∗).

This additional a can be manipulated via the following:

L(at) = aL(t),l(at) = l(t),l((at)∗) = l(a−1t∗) = l(t∗) = 0.

Now, if we define

N = #(h1, ..., hr−1) ∈ L(t)r−1 : ξh1...hr−1 ∈ k + P ((at)∗),N0 = #(h1, ..., hr−1) ∈ L(t)r−1 : ∃ i such that hi = 0

and note that L(t)× = L(t)⋂k×, then ♥ can be rewritten as

26

|W (a,r)(ξ)t |2r−1 ≤ (ql(t))2r−1−r+1N

and

N0 = #(Lr−1)−#(L(t)×)r−1

= (ql(t))r−1 − (ql(t) − 1)r−1

= ((ql(t)) − (ql(t) − 1))((ql(t))r−2 − (ql(t))r−3(ql(t) − 1) + ... +(ql(t) − 1)r−2)

= (1)((ql(t))r−2 − (ql(t))r−3(ql(t) − 1) + ...+ (ql(t) − 1)r−2)≤ (r − 1)ql(t)(r−2).

So

N0 ≤ (r − 1)ql(t)(r−2).

Next, let us define

N1 = #(h1, ..., hr−1) ∈ (L(t)×)r−1 : ξh1...hr−1 ∈ k + P ((at)∗).

Then

N ≤ (r − 1)ql(t)(r−2) +N1.

Putting all this together with ♥ gives

♣ |W (a,r)(ξ)t |2r−1 ≤ (ql(t))2r−1−r+1N ≤ (r−1)(ql(t))2r−1+N1(ql(t))2r−1−r+1.

Now we define Φ : L(t)r−1 → L(tr−1) by

Φ(h1, ..., hr−1) = h1...hr−1.

Then we can express N1 in terms of Φ:

N1 =∑

x∈L(tr−1)×, ξx∈k+P ((at)∗) #Φ−1(x).

Given this expression for N1, we have two goals

Problem I: For x ∈ L(tr−1)×, estimate #Φ−1(x).

27

Problem II: Estimate #x ∈ L(tr−1)× : ξk ∈ k + P ((at)∗).

To begin with, we state a lemma which is an analogue to Diophan-tine approximation in kA:

Lemma 2.3.1: Let ξ ∈ kA, t ∈ k×A , a ∈ k×, |t|A > q2g, and r ≥ 3.Then ∃ x, y ∈ k, x 6= 0 such that x ∈ L(tr−1) and ξx−y ∈ P ((at)∗).

Proof : We start again with a diagram

kA

ql(at)=ql(t)

k + P ((at)∗)

OOOOOOOOOOO

ttttttttttt

k

KKKKKKKKKKKKK P ((at)∗)

nnnnnnnnnnnnn

0.

Now, we know that #L(tr−1) = ql(tr−1). Moreover,

|t|A = ql(t)+(g−1) > q2g

⇒ l(t) > g + 1 ≥ 1⇒ l(t) ≥ 2.

Additionally,

|tr−1|A = ql(tr−1)+(g−1) = ql(t)(r−1)+(g−1)(r−1),

which means that

l(tr−1) = l(t)(r − 1) + (g − 1)(r − 2) > l(t).

Now, denote the elements of L(tr−1) by xi, indexed for 1 ≤ i ≤ql(t

r−1) and consider ξx1, ξx2, ..., ξxql(tr−1) . We know that

ql(tr−1) > ql(t) = [kA : k + P ((at)∗)].

28

So by pigeonhole principle, ∃ i 6= j such that

ξxi − ξxj ∈ k + P ((at)∗)

or, equivalently,

ξ(xi − xj) ∈ k + P ((at)∗).

Since xi 6= xj, we can take x = xi − xj. So ∃ y ∈ k such that

ξx− y ∈ P ((at)∗).

Q.E.D.

Next, let

x = Φ(h1, h2, ..., hr−1) = h1h2...hr−1.

We fix a valuation v. Then

|h1|v...|hr−1|v = |x|v,|hi|v ≤ |t|v, 1 ≤ i ≤ r − 1

since hi ∈ L(t). Then

|x|v ≤ |h1|v|t|r−2v ,

which means that

|x|v|t−1|r−2v ≤ |h1|v ≤ |t|v.

By the definition of | · |v, we know that |h1|v = q−ordvh1v , qv = qdeg(v).

So there exists an e such that

(∗) |x|v|t−1|r−2v ≤ q−ev ≤ |t|v.

We define

29

Tv(x) = #q−ev such that (∗) holds.

We let α : k× → ΠvR×>0 be defined by

α(h) = (|h|v).

Clearly, ker α = F×q . Then

#Ψ−1(x) ≤ (q − 1)r−1(ΠvTv(x))r−1.

So we must study Tv more carefully. Let us write (∗) in termsof qv.

q−ordvxv q(r−2)ordvtv ≤ q−ev ≤ q−ordvtv ,

which means that

ordvx− (r − 2)ordvt ≥ e ≥ ordvt.

The number of possible e’s that satisfy this are

#e’s= ordvx− (r − 1)ordvt+ 1 ≥ 1.

If we let

λv = ordvx− (r − 1)ordvt

then

#e’s= λv + 1 = Tv(x).

Now, we can compare ΠvTv(x) to |t|εA for ε > 0. Recall that, forx ∈ k×, |x|A = 1. So

|tr−1|A = |tr−1|A|x−1|A= Πvq

ordvx−(r−1)ordvt = Πvqλv .

Then

30

ΠvTv(x)

|t|ε(r−1)A

= Πvλv+1

qλvεv.

We determine what this means for two cases:

(i) qεv > 2,(ii) qεv ≤ 2.

Note that the number of v which satisfy case (ii) is finite.Case (i): First,

qλvεv ≥ 2λv ≥ λv + 1

which means that

λv+1

qελvv≤ 1.

So

Πvλv+1

qελvv≤ Πv

λv+12ελv≤ C(ε)

for some constant C(ε) which does not depend upon λ, since, forany λ, ∃ a constant B(ε) such that

λ+12ελ≤ B(ε).

So

ΠvTv(x) ≤ C(ε)|t|ε(r−1)A .

Using this in our bound for #Φ−1(x) gives

#Φ−1(x) ≤ (q − 1)r−1C(ε)r−1|t|ε(r−1)2

A .

Let us write ε for ε(r − 1)2. Then

#Φ−1(x) ≤ (q − 1)r−1C(ε)r−1|t|εA ∀ ε > 0.

Then

31

N1 = #(h1, ..., hr−1) ∈ (L(t)×)r−1 : ξh1...hr−1 ∈ k + P ((at)∗)=∑

x∈L(tr−1)×,ξx∈k+P ((at)∗) #Φ−1(x)≤ C(ε)|t|εA#x ∈ L(tr−1)× : ξx ∈ k + P ((at)∗).

Finally, define

∆(a,r)t (ξ) = #x ∈ L(tr−1) : ξx ∈ k + P ((at)∗).

Using ♣ and the fact that ql(t) = |t|Aq1−g, we have the followingresult for Problem 1:

Problem 1: Let char(k) > r ≥ 3, a ∈ k×, ξ ∈ P (t), and assumethat |t|A > q2g. Then

|W (a,r)t (ξ)|2r−1 ≤ |t|2r−1−r

A (c1|t|r−1A + c2|t|1+ε

A ∆(a,r)t (ξ)),

where c1 and c2 are independent of t.

Having dealt with Problem 1, we turn our attention to Problem 2:

Problem 2: Let ξ ∈ P (t), a ∈ k×, r ≥ 3, |t|A > 22g, and let

∆(a,r)t (ξ) ⊂ k ⊂ kA as above. Define ϕξ : L(tr−1) → kA/(k +

P ((at)∗)) by

ϕξ(x) = ξx (mod k + P ((at)∗)).

Note that ϕ = ϕξ is an additive homomorphism. Then

∆(a,r)t (ξ) = #Ker ϕ− 0.

Recall that under the above hypotheses, we have l(tr−1) > l(t)(Lemma 2.3.1). Note that

L(tr−1)/Ker ϕ → kA/(k + P ((at)∗)).

Since the cardinality of the former group is ql(tr−1)/#Ker ϕ and the

cardinality of the latter is ql(t), the first divides the second and hence

ql(tr−1)−l(t)|#Ker ϕ.

32

Let A = L(tr−1) and B = kA/(k + P ((at)∗)). So #A = ql(tr−1)

and #B = ql(t). This yields the exact sequence

0→ Ker ϕ→ A→ B → Cok ϕ→ 0,

which implies that

#Ker ϕ ·#B = #A ·#Cok ϕ.

So

#Ker ϕ = #Cok ϕ · (ql(tr−1)−l(t)).

Now, recall from page 28 that

l(tr−1) = l(t)(r − 1) + (g − 1)(r − 2).

So

ql(tr−1)−l(t) = q(l(t)+(g−1))(r−2) = |t|r−2

A ,

where the last step is by Riemann-Roch. Then

#Ker ϕξ = |t|r−2A ·#Cok ϕξ.

Let 0 < η < 1 such that #Cok ϕξ ≤ |t|1−ηA . So

#Ker ϕξ ≤ |t|r−1A ,

which means that

∆(a,r)t (ξ) = #Ker ϕξ − 1 ≤ |t|r−1−η

A .

This gives us an upper bound for Problem 1. Let us substitutethis upper bound into the equation from Problem 1:

|W (a,r)t (ξ)|2r−1 ≤ |t|2r−1−r

A (c1|t|r−1A + c2|t|1+ε

A ∆(a,r)t (ξ))

≤ c3|t|2r−1−(η−ε)

A ,

33

where 0 < η − ε < η < 1. If we redesignate ε as ε2r−1 , we can

then rewrite the above as

|W (a,r)t (ξ)| ≤ c4|t|

1− η

2r−1 +ε

A .

As before, we let

f(x) = a1xδ11 + ...+ anx

δnn .

We relate this to the findings above by letting r = δi, t = tδδi , and

|tδδi |A > 22g. Then we define ϕ

(i)ξ : L(t

δ− δδi ) → kA/(k + P ((ait

δδi )∗))

in the obvious way. Moreover, we choose 0 < ηi < 1 in such a way

that #Cok ϕ(i)ξ ≤ |t|

1−ηiA . So

|W (a,δi)

tδδi

(ξ)| ≤ c(i)4 |t

δδi |

1− ηi

2δi−1 +ε

A .

Then

Πni=1|W

(a,δi)

tδδi

(ξ)| ≤ c5Πni=1|t

δδi |

1− ηi

2δi−1 +ε

A .

Examining the exponent in the above expression,∑ni=1

δδi

(1− ηi2δi−1 + ε) =

∑ni=1

δδi− δ

∑ni=1( ηi

δi2δi−1 ) + ε∑n

i=1δδi

Let us define ε0 = ε∑n

i=1δδi

. Then the above is rewritten as∑ni=1

δδi− δ

∑ni=1( ηi

δi2δi−1 ) + ε0

=∑n

i=1δδi− δ + δ − δ

∑ni=1( ηi

δi2δi−1 ) + ε0= δ((

∑ni=1

1δi

)− 1)− (δ(∑n

i=1ηi

δi2δi−1 − 1)− ε0).

Define the latter expression by ε′, i.e. let

ε′ = δ(∑n

i=1ηi

δi2δi−1 − 1)− ε0).

Then

34

Πni=1|W

(a,δi)

tδδi

(ξ)| ≤ c5|t|δ((

∑ni=1

1δi

)−1)−ε′

A .

We will choose ηi in a slightly different manner. Let 0 < ηi < 1be such that∑n

i=1ηi

δi2δi−1 > 1.

Then we define

Et(ηi) = ξ mod k ∈ kA/k : #Cok ϕ(i)ξ ≤ |t|

1−ηiA .

So ∫Etηi |Π

ni=1W

(ai,δi)

tδδi

(ξ)|dξ ≤ c|t|δ((

∑ni=1

1δi

)−1)−ε′

A .

Using this, we have the following proposition

Proposition 2.3.1:∫Etηi |Π

ni=1W

(ai,δi)

tδδi

(ξ)|dξ ≤ c |det ρ1(t)|A|det ρ2(t)|A

|t|−ε′A ,

where

ρ1(t) =

tδδ1 0 ... 0

0 tδδ2 ... 0

0 0 ... tδδn

,

ρ2(t) = tδ,

and hence

det ρ1(t) = tδ

∑ 1δi ,

det ρ2(t) = tδ.

2.4 k = Fq(T )

We now examine the special case where k = Fq(t) is a rational func-tion field. We specify that

f(x) = xδ1 + ...+ xδn,

35

where δ ≥ 2. So

G = k×,

ρ1(t) =

tδδ1 0 ... 0

0 tδδ2 ... 0

0 0 ... tδδn

,

ρ2(t) = tδ,ν ∈ k,t = (τ, 1, 1, ...) ∈ k×A , τ ∈ k×∞,

As before, we let

Nt(ν) =∫kA/k

(∑

α∈L(t) χ(αδξ))nχ(ξν)dξ.

We wish to consider the various valuations on k. Note first that

O = Fq[T ] ⊂ k.

Now, the completion of k at the valuation v associated to a primepolynomial π ∈ O is

kπ = x =∑

ν≥ν0 cνπν , cν ∈ O, deg cν < deg π,

where cν0 6= 0. So we can say that

|x|π = q−ν0π , qπ = qdeg π.

For the place v =∞,

k∞ = x =∑

ν≥ν0 cνT−ν , cν ∈ Fq ⊃ O = Fq[T ],

where the inclusion is analogous to the case where R ⊃ Z. So

|x|∞ = q−ν0 ,

and

|x|∞ = qdeg x.

36

when x ∈ O. For any valuation v, we get the inclusion

kv ⊃ Ov ⊃ Pv = x ∈ Ov : |x|v < 1,

where

Ov = x ∈ kv : |x|v ≤ 1.

Note that for v =∞,

O∞ = x ∈ k∞ :∑

ν≥0 cνT−ν,

P∞ = x ∈ k∞ :∑

ν≥1 cνT−ν.

We also have the relation that⋂v(k ∩ Ov) = Fq,⋂v 6=∞(k ∩ Ov) = O.

Now, in the case of k = Fq(t), we have the diagram

kA

QQQQQQQQQQQQQQ

k

>>>>>>>> k∞ × ΠπOπ = kA(∞)

nnnnnnnnnnnnnn

O,where the top right and the bottom left are isomorphic as com-pact abelian groups, and the notation kA(S) means that for a finiteset of primes S,

kA(S) = Πv∈SkvΠv 6∈SOv.

For x ∈ k∞ with the expansion

x =∑

ν≥ν0 cνt−ν ,

37

we put

c1 = Res x,

the residue of x. Note that an element x ∈ kA(∞) is of the formx = (xv), where x∞ ∈ k∞ and xπ ∈ Oπ.

Now, we will discuss characters on kA, beginning with the char-acter χ1:

χ1(x) = ep(trFq/Fp(Res x∞)),

where

ep(z) = e−2πizp .

If x ∈ O then Res x∞ = 0 and hence χ1(x) = 1. Hence, χ1 be-comes a character of kA(∞)/O ∼= kA/k.

Denote by χ the basic character of kA thus obtained. The char-acter can also be restricted to v; in our case of x ∈ k(∞), χv = χ|kvis defined by

χ∞(x∞) = ep(trFq/Fp(Res x∞)),χπ(Oπ) = 1,

where the latter equality comes from the fact that χπ(O) = 1 andχπ is continuous.

Next, let

k(π) = aπn, n ≥ 0, a ∈ O.

Then we have the following diagram

38

kπ

AAAAAAAA

k(π)

CCCCCCCC Oπ

||||||||

O

0.

Oπ is the integral part of kπ and k(π) is the fractional part. Weextend the character χπ to k(π) via the following claim:

Claim: Let ξ = aπn∈ k(π). Then

χπ(ξ) = ep(tr Res (ξ)).

Proof : Note that ξ ∈ Oπ′ ∀ π′ 6= π. Since ξ ∈ k, χ(ξ) = 1, andχ(x) = Πvχv(x). So

1 = χ(ξ) = χ∞(ξ)Ππ′ 6=π(χπ′(ξ))χπ(ξ)= ep(trFq/Fp(Res ξ))(Ππ′ 6=π1)χπ(ξ)

and the claim follows.

Given a character χ, we can now determine ord χv, where

P−ord χvv = x ∈ kv : χv(xy) = 1 ∀ y ∈ Ov.

We prove the following

Lemma 2.4.1: ord χπ = 0 and ord χ∞ = −2.

Proof : We show first that ord χπ = 0. Note that χπ(Oπ) = 1. So

Oπ ⊂ P−ord χππ .

For x 6∈ Oπ, ∃ l > 0 such that πlx ∈ O×π . Choose a constantc ∈ Fq such that tr c 6= 0 and let δ =deg π. Then we define

y = cTlδ−1

πlx.

39

Note that

χπ(xy) = ep(tr c) 6= 1.

Thus, x 6∈ Oπ ⇒ x 6∈ P−ord χππ .Next, we show that χ∞ = −2, or, equivalently, that

P2∞ = x ∈ k∞ : χ∞(xy) = 1 ∀ y ∈ O∞.

We show containment in both directions:

(⊂) We know that x ∈ P2∞ implies

x = c2T 2 + c3

T 3 + ....

If y ∈ O∞ then

y = b0 + b1T

+ b2T 2 + ....

Multiplying these expressions together, it is clear that Res xy = 0and hence

χ∞(xy) = 1.

(⊃) Assume that χ∞(xy) = 1 ∀ y ∈ O∞. Let

x = c−ν0Tν0 + ...+ c−1T + c0 + c1

1T

+ ....

We claim that

c−ν0 = ... = c−1 = c0 = c1 = 0.

Then x ∈ P2∞ as required. In fact, let

y = λT ν0+1 ∈ O∞,

where λ ∈ Fq. Then

40

xy =λc−ν0T

+ λc−ν0+1 + ...

By assumption on x,

1 = χ∞(xy) = ep(tr (λc−ν0)).

So

tr (λc−ν0)) = 0 ∀ λ ∈ Fq,

and hence c−ν0 = 0. Similarly, we get

c−ν0+1 = ... = c−1 = c0 = c1 = 0.

Q.E.D.

Recall that we defined

π(χ) = (π−ord χvv ) ∈ kA,

where

πv =

π if v = π,

T−1 if v =∞.

Then

q2g−2 = |π(χ)|A = Πv|πv|−ord χvv = |T−1|−ord χ∞∞= |T |−ord χ∞∞ = (qdeg T )−2 = q−2.

Thus, g = 0, which corresponds with the standard definition ofgenus of a rational function field.

Now, let

t0 = (T−1, 1, 1, ....) ∈ k×A .

Then

|t0|A = |T−1|∞ = q−1,

41

P (t0) = P∞ × ΠπOπ ⊂ kA,L(t0) = k

⋂P (t0) = 0.

Note that

t∗0 = π(χ)t−10 = (T−2, 1, 1, ...)(T, 1, 1, ...) = (T−1, 1, 1, ...) = t0,

l(t∗0) = l(t0) = 0.

Because this last line is true, our diagram from p.20 is shrunk asshown:

kA

LLLLLLLLLLL

uuuuuuuuuuu

k

HHHHHHHHHH P (t0)

rrrrrrrrrr

L(t0) = 0.

Assume that∫kA/k

dx = 1.

For each v, we denote dxv the measure given by∫Ov dxv = 1.

For a fixed character χ, we denote by dχvxv the Haar measure on kvwhich is consistent with the self-duality kv ∼= kv. Then we have

dχvxv = q−ord χv

2 dxv,dx = Πvd

χvxv = qΠvdxv,

where the last equality comes from Lemma 2.4.1. (For more de-tails, see Weil, Basic Number Theory p.108).

We can use these for the case of

t = (T d, 1, 1, ....),

where d ≥ 1. Accordingly, we have

42

|t|A = qd > q2g = 1,P (t) = x ∈ kA : |x|v ≤ |t|v ∀ v

= P−d∞ × ΠπOπ,

L(t) = k ∩ P (t) = x ∈ O = Fq[T ], |x|∞ ≤ qd= x ∈ O = Fq[T ], deg x ≤ d.

Note that the last expression indicates that L(t) is simply the poly-nomials of degree less than d. So we can adjust the notation toreflect this fact. Define

M(d) = x ∈ O = Fq[T ], deg x ≤ d = L(t).

Then

l(t) = dimFqM(d) = d+ 1.

If we let

ξ ∈ P (t0) = P∞ × ΠπOπ

then

Wt(ξ) =∑

x∈L(t) χ(xδξ).

Of course, L(t) = M(d) ⊂ O. Note that for any non-infinite place,

πδξπ ∈ Oπ, χπ(xδξπ) = 1.

So

Wt(ξ) =∑

x∈M(d) χ∞(xδξ∞).

Let us denote

Wd(ξ∞) =∑

x∈M(d) χ∞(xδξ∞).

Recalling the definition for Nt(ν),

43

Nt(ν) =∫kA/k

(∑

α∈L(t) χ(αδξ))nχ(ξν)dξ

= q∫kA/k

(Wd(ξ∞))nχ∞(ξ∞ν)dξ∞,

where∫O∞ dξ∞ = 1 and

∫P∞ qdξ∞ = 1.

From the previous section, we saw that

|Wt(ξ)|2δ−1 ≤ c1|t|2

δ−1−1A + c2|t|2

δ−1−δ+1+εA ∆t(ξ),

where

∆t = #Ker ϕξ − 1

and ϕξ is described in the previous section. We let

Ed(η) = ξ ∈ P (t0) : #Cok ϕξ ≤ |t|1−ηA .

This is essentially our definition of Et(η) from before. We will take

1 > η > δ2δ−1

n. Then∫

Ed(η)Wd(ξ)

nχ(νξ)dξ = O(|t|n−δ−ε′A ),

where ε′ = ε′(ε) > 0. This integral is analogous to the minor arcestimation in the initial work of Hardy and Littlewood.

Since we have t = (T d, 1, 1, ...)

t∗ = π(χ)t−1 = (T−d−2, 1, 1, 1, ...),

which gives us the diagram

kA

IIIIIIIII

rrrrrrrrrrr

k + P (t∗)

KKKKKKKKKK

vvvvvvvvvvP (t0),

vvvvvvvvv

k

IIIIIIIIIII P (t∗)

rrrrrrrrrrr

0

44

where

P (t0) = P∞ × ΠπOπ,P (t∗) = Pd+2

∞ × ΠπOπ.

Then

P (t0)/P (t∗) = P∞/Pd+2∞ ,

with

|P∞/Pd+2∞ | = qd+1.

Note that

kA/(k + P (t∗)) ∼= P (t0)/P (t∗).

We can describe α ∈ kA in terms of an integral part [α] ∈ k and afractional part α ∈ P (t). Moreover, if α∞ ∈ P∞ is defined inthe obvious manner then we can project from α ∈ kA to P∞/Pd+2

∞ by

α 7→ α∞ (mod Pd+2∞ ).

With this mapping, we can create a commutative diagram. Letξ ∈ P (t0). Then

L(tδ−1) = M(d(δ − 1))ϕξ //

++VVVVVVVVVVVVVVVVVVVVkA/(k + P (t∗))∫

P∞/Pd+2

∞

.

It is a trivial exercise to show that

ξx∞ = ξ∞x,

which shows that the diagram commutes.Next, let

Ed(η)∞ = α ∈ P∞ : #Cok ϕα ≤ qd(1−η).

45

Then

Ed(η) = Ed(η)∞ × ΠπOπ.

So in our integration over Ed(η), we can ignore every place exceptthe infinite one. Then

(♠)∫Ed(η)∞

Wd(ξ)nχ(νξ)dξ = O(|t|n−δ−ε′A ).

Now, for x ∈ k∞, we note that

k∞ = O + P∞

and hence we can again break x into an integral part [x] ∈ O and afractional part x.

For α ∈ P∞, we can define ϕα : M(d(δ − 1))→ P∞/Pd+2∞ by

ϕα(x) = αx (mod Pd+2∞ ).

This is the simply the previous definition for ϕξ combined with theisomorphism from kA/(k + P (t∗)) to P∞/Pd+2

∞ . Clearly,

x ∈ ker ϕα ⇔ |αx|∞ ≤ 1qd+2 .

This gives us the following approximation theorem:

Lemma 2.4.2: ∀ λ > 0 and ∀ α ∈ k∞, ∃ (x, y) ∈ O2 where x 6= 0such that |x|∞ ≤ qλ (i.e. deg x ≤ λ) and |αx− y|∞ < 1

qλ.

Proof : Let N = [λ]. So N ≤ λ ≤ N + 1. Then we defineψ : M(N)→ P∞/PN+1

∞ by

ψ(x) = x∞ (mod PN+1∞ ).

Clearly,

M(N)/Ker ψ → P∞/PN+1∞ .

46

We know that #M(N) = qN+1 and #(P∞/PN+1∞ ) = qN . So Ker

ψ 6= 0. Let x ∈Ker ψ such that x 6= 0. Then

αx− [αx] = αx ∈ PN+1∞ .

Let y = [αx] ∈ O. Then

|x|∞ ≤ qN ≤ qλ

by assumption and

|αx− y|∞ = |αx− [αx]|∞ ≤ 1qN+1 <

1qλ

.

Corollary 2.4.1: Let α ∈ P∞ and λ > 0. Then ∃ (m, a) ∈ O2 withm 6= 0, (m, a) = 1, and deg a <deg m such that |m|∞ ≤ qλ and|mα− a| < 1

qλ.

Proof : Take x, y as in Lemma 2.4.2. Then we can find a,m suchthat (a,m) = 1 by letting a

mbe the reduced fraction for y

x.

Now, we can finally ascertain bounds for the kernel and cokernel ofϕα. Recall that

#Ker ϕα = |t|δ−2A ·#Cok ϕα = qd(δ−2)#Cok ϕα.

Let µ = deg m ≤ δ. Then

Proposition 2.4.1: #Ker ϕα ≤

qµ−(d+1) if µ− 1 ≥ d(δ − 1),

qdδ−2d if d(δ − 1) > µ− 1 > d,

qdδ−d−µ+1 if µ− 1 ≤ d.

Proposition 2.4.2: #Cok ϕα ≤

qµ−1−d(δ−1) if µ− 1 ≥ d(δ − 1).

1 if d(δ − 1) > µ− 1 > d.

qd−(µ−1) if µ− 1 ≤ d.

Proof : We only prove the first case of Proposition 2.4.1; the otherproofs are similar. So let µ−1 ≥ d(δ−1). SoM(d(δ−1)) < M(µ−1).

47

Then

#Ker ϕα ≤ #x ∈M(µ− 1) : αx ∈ Pd+2∞ .

Using Corollary 2.4.1, we note that x ∈M(µ− 1) means that

αx ≡ amx (mod Pd+2

∞ ).

So the above bound can be rewritten as

#Ker ϕα ≤ #x ∈M(µ− 1) : amx ∈ P d+2

∞ .

The above is a set of representatives for O/mO (i.e. ax ≡ u (modm)). So

#Ker ϕα ≤ #u ∈M(µ− 1) : um ∈ P d+2

∞ .

But

um ∈ P d+2

∞ ⇒ | um|∞ ≤ 1

qd+2

⇒ qdeg u

qµ≤ 1

qd+2

⇒ deg u ≤ µ− (d+ 2)⇒ # Ker ϕα ≤ qµ−(d+1).

Q.E.D.To clarify the separation into major and minor arcs, we define

0 < θ ≤ 12,

d ≥ 2θ.

Later, it will suffice to consider θ = 12. Define Ed(η) as before,

and let a,m be such that am∈ k

⋂P∞, where

a,m ∈ O = Fq[t],deg a <deg m,(a,m) = 1,µ =deg m ≤ dθ.

48

With these restrictions, we can define

M am

= M am

(d, θ) = α ∈ P∞ : |mα− a|∞ < 1qd(δ−θ)

.

We require the following proposition:

Proposition 2.4.3: If am6= a′

m′then M a

m

⋂M a′

m′= ∅.

Proof : Let α ∈M am

⋂M a′

m′. Since | · |∞ is non-archimedean,

| am− a′

m′|∞ ≤ max(| a

m− α|∞, |α− a′

m′|∞) < 1

qd(δ−θ).

Note that

| am− a′

m′|∞ = |am′−a′m

mm′|∞

= qdeg (am′−a′m)

qdeg m+deg m′

≥ 1q2dθ

.

because deg m ≤ dθ. So

d(δ − θ) < 2dθ⇒ δ − θ < 2θ⇒ δ < 3θ ≤ 3

2.

But δ ≥ 2, which is a contradiction.

Let

M(d, θ) =⋃

am

M am

(d, θ).

This is the major arc on P∞. The minor arc is then

M∗(d, θ) = P∞ −M(d, θ).

Proposition 2.4.4: If n > δ2δ

θthen M∗(d, θ) ⊂ Ed(

θ2)∞.

As an immediate corollary to the Proposition, we get

49

Corollary 2.4.2: Let θ = 12, n > δ2δ+1. Then, for ν ∈ O,∫

M∗(d,θ)Wd(α)nχ∞(να)dα = O(|t|n−δ−εA ).

This corollary follows from Proposition 2.4.4 and the equation de-noted (♠).

Proof of Proposition 2.4.4: Let α ∈M∗(d, θ). Putλ = d(δ− θ) > 0. Then ∃ a

m∈ k

⋂P∞ such that µ =deg m < λ and

|mα− a|∞ < 1qd(δ−θ)

.

Then µ > dθ (since µ ≤ dθ implies α ∈M am

). So dθ < µ ≤ d(δ−θ).Now, recall that we had three cases for the cokernel of ϕα:

#Cok ϕα ≤

qµ−1−d(δ−1) if µ− 1 ≥ d(δ − 1),

1 if d(δ − 1) > µ− 1 > d,

qd−(µ−1) µ− 1 ≤ d.

Recall that n > δ2δ

θ, and let η = θ

2. Then 1 > η > δ2δ−1

n. In

this case, we have

Ed(θ2)∞ = α ∈ P∞ : #Cok ϕα ≤ qd(1−η).

We will prove the first case; the others are similar and are left as anexercise to the reader.

Case 1 : #Cok ϕα ≤ qµ−1−d(δ−1).

But

µ− 1− d(δ − 1) = µ− 1− dδ + d≤ d(δ − θ)− 1− dδ + d= d(1− θ − 1

d)

≤ d(1− θ)≤ d(1− θ

2)

= d(1− η).

Now, to garner a result which parallels that of Hardy and Little-

50

wood, we define analogues for some of the concepts used in theoriginal proof. Let

S am

=∑

x∈M(µ−1) χ∞( amxδ).

S am

serves as an analogue to the Gauss sum. Additionally, let

β = α− am

,

and let

I(β) =∫|ξ|∞≤qd χ∞(βξδ)dξ.

I(β) will help us to define our Gamma factor. First, however, weshow how it relates to Weyl sums:

Proposition 2.4.5: Let S am

and I(β) be as above. Then

Wd(α) =∑

x∈M(d) χ∞(αxδ) = 1qµ−1S a

mI(β).

Proof : First, we know that

µ ≤ dθ < d2< d.

So we can divide polynomials in M(d) by m to get a quotient yand remainder z, i.e.

M(d) = m(M(d− µ)) +M(µ− 1)

such that, for x ∈M(d), ∃ y ∈M(d− µ), z ∈M(µ− 1) where

x = my + z.

We use the above, along with the definition of β, to find that

αxδ = (β + am

)(my + z)δ

≡ β(my + z)δ + amzδ (mod O).

So

51

Wd(α) =∑

x∈M(d) χ∞(αxδ)

=∑

y∈M(d−µ)

∑z∈M(µ−1) χ∞(β(my + z)δ)χ∞( a

mzδ)

=∑

z∈M(µ−1) χ∞( amzδ)∑

y∈M(d−µ) χ∞(β(my + z)δ).

Let us fix z. Then we define

J(β) =∫|η|∞≤qd−µ χ∞(β(mη + z)δ)dη.

Note that

η ∈ k∞ : |η|∞ ≤ qd−µ = M(d− µ) + P∞,

where the sum on the right is actually a direct sum. This meansthat, for η in the set above, ∃! y ∈ M(d − µ), u ∈ P∞ such thatη = y + u. Then

J(β) =∑

y∈M(d−µ)

∫y+P∞ χ∞(β(my +mu+ z)δ)du.

We claim that

χ∞(β(my + z)δ−ν(mu)ν) = 1

for ν ≥ 1. In view of ord(χ∞) = −2, this is equivalent to show-ing that

|β(my + z)δ−ν(mu)ν |∞ ≤ 1q2

,

or, equivalently,

|β(my + z)δ−ν(mu)ν |∞ < 1q.

Since | · |∞ is non-archimedean,

|my + z|∞ ≤ maxqµ+d−µ, qµ−1 = qd.

Moreover, u ∈ P∞ implies that |mu|∞ ≤ qµ−1. So

|β(my + z)δ−ν(mu)ν |∞ ≤ ( 1qd(δ−θ)

)(qd(δ−ν))(qν(µ−1))

52

= q−dδ+dθ+d(δ−ν)+ν(µ−1)

= qdθ−dν+ν(µ−1)

= qdθ−ν(d−(µ−1))

≤ qdθ−(d−(µ−1))

≤ qd2−d+µ−1

< q−1,

which proves the claim. This claim tells us that

J(β) =∑

y∈M(d−µ) χ∞(β(my + z)δ)∫y+P∞ du

= q−1∑

y∈M(d−µ) χ∞(β(my + z)δ).

Then

Wd(α) =∑

z∈M(µ−1) χ∞( amzδ)∑

y∈M(d−µ) χ∞(β(my + z)δ)

= q∑

z∈M(µ−1) χ∞( amzδ)J(β).

Now, in the original expression for J(β), we can change variablesfrom η to ξ by letting

ξ = mη + z.

Then

|η|∞ ≤ qd−η ⇔ |ξ|∞ ≤ qd.

Note that, for the measures defined by dξ and dη, we have therelationship

dξ = |m|∞dη = qµdη.

So

J(β) =∫|η|≤qd−µ χ∞(β(mη + z)δ)dη

= |m|−1∞∫|ξ|≤qd χ∞(β(ξ)δ)dξ

= |m|−1∞ I(β).

Q.E.D.

53

Finally, we can state our analogue to the Hardy-Littlewood resultfor Nt(ν). Define

Sd,θ =∑

m∈M([dθ])

∑a∈M(µ−1),(a,m)=1(|m|−1

∞ S am

)nχ∞(ν am

)

and

Γd,θ(νT−dδ) =

∫|γ|∞<qdθ(

∫|ζ|∞≤1

χ∞(γζδ)dζ)nχ∞(νT−dδγ)dγ

Applying proposition 2.4.5 to the expression for Nt(ν) on page 44:

Nt(ν) = q∑

m∈M([dθ])

∑a∈M(µ−1),(a,m)=1

∫M a

m (d,θ)Wd(α)nχ∞(να)dα

+O(qd(n−δ−ε′))= qqn(1−µ)

∑m∈M([dθ])

∑a∈M(µ−1),(a,m)=1

∫M a

m(d,θ)

SnamI(β)nχ∞(να)dα

+O(qd(n−δ−ε′)).

If we let

βT dδ = γξT−d = ζ

then the associated changes in measure will be

qdδdβ = dγq−ddξ = dζ.

Let us define (∗) to be

(∗) =∫|β|∞< 1

qd(δ−θ)(I(β))nχ∞(νβ)dβ

= q−dδ∫|γ|∞<qdθ(I(T−dδγ))nχ∞(νT−dδγ)dγ

= qd(n−δ) ∫|γ|∞<qdθ(

∫|ζ|∞≤1

χ∞(γζδ)dζ)nχ∞(νT−dδγ)dγ.

From this equality, we see that

Nt(ν) = qn+1qd(n−δ)Sd,θ(ν)Γd,θ(νT−dδ) +O(qd(n−δ−ε′)).

54

If we let

Γd,θ =∫|γ|∞<qdθ(

∫|ζ|∞≤1

χ∞(γζδ)dζ)ndγ

then, using the above expression for Nt(ν), we obtain the follow-ing theorem:

Theorem 2.4.1: Let 0 < θ ≤ 12, ch(k) > δ, d > deg ν

δ−1, d ≥ 2

θ, and

n > δ2δ

θ. Then

Nt(ν)

|t|n−δA= qn+1Sd,θ(ν)Γd,θ +O(q−dε(θ)).

We wish to define the limits

S(ν) = limd→∞Sd,θ(ν),Γ = limd→∞ Γd,θ

where the latter will be postponed until the discussion of Kelvin’sprinciple in the next section. We are aided by the fact that, for ν,we can find d such that νT−dδγ ∈ P2

∞ ∀ γ which satisfy |γ|∞ < qdθ;this means that χ∞(νT−dδγ) = 1. So we can define Γ as∫

k∞(∫O∞ χ∞(γξδ)dξ)ndγ.

We require a proposition:

Proposition 2.4.6: ∃ a c independent of m such that

|S am| ≤ c|m|

1− 1

2δ−1 +ε∞ ∀ ε > 0.

Proof : First, we will let

t = (T µ−1, 1, 1, 1, ...),ξ = ( a

m, 1, 1, 1, ...).

Note that, by our definitions,

S am

= Wµ−1( am

).

55

So

|S( am

)| = |Wµ−1( am

)| = |Wt(ξ)| ≤ c|t|1− η

2δ−1 +ε

A

for some 0 < η < 1, and η > ε is such that

#Cok ϕξ ≤ |t|1−ηA ,

or, equivalently,

#Cok ϕ am≤ (qµ−1)1−η.

We recall from Proposition 2.4.2 that

#Cok ϕα ≤

qµ−1−d(δ−1) if µ− 1 ≥ d(δ − 1),

1 if d(δ − 1) > µ− 1 > d,

qd−(µ−1) µ− 1 ≤ d.

Since d = µ− 1, we use the third case. So

#Cok ϕξ ≤ qd−(µ−1) = 1,

which means that 1− η = ε′ > 0 can be arbitrarily small. Then

|S am| = |Wt(ξ)| ≤ c|t|

1− 1−ε′

2δ−1 +ε

A .

Let

ε′′ = ε+ ε′

2δ−1 .

Then

|S am| = |Wt(ξ)| ≤ c|t|

1− 1

2δ−1 +ε′′

A

= c(qµ−1)1− 1

2δ−1 +ε′′

< cq−1+ 1

2δ−1 |m|1− 1

2δ−1 +ε′′

∞

= c′|m|1− 1

2δ−1 +ε′′

∞

56

where c′ = cq−1+ 1

2δ−1 .

Q.E.D.

Let

S(ν) =∑

m∈Fq [t], monic∑

a∈M(µ−1),(a,m)=1(|m|−1∞ S a

m)nχ∞(ν a

m).

We wish to show two things: that S(ν) converges, and that it islimd→∞Sd,θ(ν).

Claim: S(ν) converges.

Proof : From Proposition 2.4.6,

(|m|−1∞ |S a

m|)n ≤ c|m|−

n

2δ−1 +ε

for some ε > 0. So∑a∈M(µ−1),(a,m)=1(|m|−1

∞ S am

)n ≤ c1|m|1− n

2δ−1 +ε

A

since the number of possible a’s is bounded by qµ−1 < |m|∞. Werecall that

n > δ2δ

θ> δ2δ ≥ 2δ+1.

So

− n2δ−1 < −2− 1

2δ−1 .

Then∑a∈M(µ−1),(a,m)=1(|m|−1

∞ S am

)n ≤ c1|m|−1− 1

2δ−1 +ε∞

= c1

|m|1+ε′∞,

and∑m

1

|m|1+ε′∞< +∞,

57

thus proving the claim.

Claim: limd→∞Sd,θ(ν) = S(ν).

Proof : By definition,

|S(ν)−Sd,θ(ν)| ≤∑

deg m>dθ

∑a∈M(µ−1),(a,m)=1(|m|−1

∞ |S am|)n

≤ c∑

deg m>dθqµ

qµ(1+ε′)

= c∑

deg m>dθ1

qµ(ε′)

≤ c2q−ε′′d.

Thus, S(ν) is as required.

2.5 Kelvin’s Principle in P-adic fields

For this section, we let K be a P-adic field, O be the unique max-imal compact subring of K, and P the unique maximal ideal in O,where O/P = Fq. Also, define

dx = the basic Haar measure on k, where∫O dx = 1,

P = (π),|π|K = q−1,χ = a basic character of K+,S(Kn) = ϕ : Kn → C : ϕ is locally constant and compactly

supported.

S(Kn) is known as the Schwartz space, and ϕ ∈ S(Kn) is trueif and only if ∃ lattices L,M ⊂ Kn such that L ⊃ M , supp ϕ ⊂ L,and ϕ is constant on each coset in L/M .

For ϕ ∈ S(Kn), define Gfϕ by

Gfϕ(ξ) =∫Kn ϕ(x)χ(< f(x), ξ >)dχx,

where

dχx = q−n2ord χdx,

dx = dx1dx2...dxn.

58

For simplicity, we write dx for dχx most of the time. For a finitefield, this is a kind of Gauss sum, so we can say that this is theGauss sum generalized to the adeles. We will let f : Kn → Km bea polynomial map composed of a series of fi : Kn → K such that

f(x) = (f1(x), f2(x), ..., fn(x)).

We will let dfx denote the Jacobian matrix for f at x. Additionally,we let

Cf = x ∈ Kn : rank dfx < m,

the critical set for f . A natural question arises as to when Gfϕ ∈S(Km). To answer this question, we use the following version ofKelvin’s Principle:

Theorem 2.5.1 (Kelvin’s Principle): If Cf⋂supp ϕ = ∅ then

Gfϕ ∈ S(Km).

Proof : First, if n < m then Cf = Kn. So supp ϕ = ∅, which meansthat ϕ = 0, and hence Gfϕ = 0. Hence, it suffices to consider thecase n ≥ m.

Now, we can assume that ϕ is a characteristic function of a com-pact open set C such that

(i) Cf⋂C = ∅.

(ii) Let Mλ be an m ×m minor of dfx determined by λ, whereDλ is its determinant. Then |Dλ(x)|K = b 6= 0 on C for some λ.

(iii) Let h : Kn → Kn be defined by

h(x) = (f1(x), ...fm(x), xm+1, ..., xn).

Then h induces an analytic homeomorphism between C and N =a+ παOn for some a ∈ Kn.

Let

59

(παOn)′ = ξ ∈ Kn : χ(< t, ξ >) = 1 ∀ t ∈ παOn.

Then it is enough to show that ξ 6∈ (παOm)′ implies Gfϕ(ξ) = 0. Solet ξ 6∈ (παOm)′, and let

t = (t1, ..., tm),u = (um+1, ..., un).

So

Gfϕ(ξ) =∫cχ(< f(x), ξ >)dx

=∫Nχ(< t, ξ >)|Dλ(x)|−1

K dtdu= b−1

∫a′′+παOn−m

∫a′+παOm χ(< t, ξ >)dt,

where b is from (ii) and a = (a′, a′′) ∈ Kn. Then∫a′+παOm χ(< t, ξ >)dt = χ < a′, ξ >

∫παOm χ(< t, ξ >)dt = 0

since ξ 6∈ (παOm)′. Q.E.D.

We can use this in the proof of the following theorem

Theorem 2.5.2: Let f : Kn → Km be be a homogeneous polyno-mial map of degree δ such that Cf = 0, and let |ξ| = max1≤i≤m|ξi|K .Then

Gfϕ(ξ) = O(|ξ|−nδ ),

for all ϕ ∈ S(Kn).

Remark: This theorem can be applied when m = 1, f(x) =xδ1 + ... + xδn, and p - δ, δ ≥ 2, where p is the characteristic ofK. In this case,

Cf = x : ∂f∂xi

(x) = 0 ∀ i = x : δxδ−1i = 0 ∀ i = 0.

If we let ϕ0 be the characteristic function of On ⊂ Kn then

< f(x), ξ >= (xδ1 + ...+ xδn)ξ,

60

which means that

Gfϕ0(ξ) =∫On χ(ξ(xδ1 + ...+ xδn))dx

= (∫O χ(ξxδ)dx)n

= O(|ξ|−nδ ).

Proof of Theorem 2.5.2: Let

ψ(x) = ϕ(x)− ϕ(0)ϕ0(x),

where ϕ0 is the characteristic function of On. Then

ψ(0) = ϕ(0)− ϕ(0)ϕ0(0) = 0.

Now, since ψ ∈ S(Km) by Theorem 2.5.1, it is enough to showthat

Gfϕ0(ξ) =∫On χ(< f(x), ξ >)dx = O(|ξ|−nδ ).

Let ψ0 be the characteristic function of On−πOn. Then ψ0(0) = 0.So Gfψ0 ∈ S(Km) again by Theorem 2.5.1. Hence it has a compactsupport. So

0 = Gfψ0(ξ) =∫On−πOn χ(< f(x), ξ >)dx

for |ξ| = ql ≥ ql0 for some l0. So∫On =

∫πOn . Inductively, we

can make this smaller and smaller if l is sufficiently greater than l0.To be more precise, choose N so that

l −Nδ ≥ l0 ≥ l − (N + 1)δ,

and let

x = πy,

which means that

61

dx = q−ndy.

Then∫On χ(< f(x), ξ >)dx =

∫πOn χ(< f(x), ξ >)dx

= q−n∫On χ(< f(πx), ξ >)dx

= q−n∫On χ(< f(x), πδξ >)dx

= q−nGfϕ0(πδξ)

= q−2nGfϕ0(π2δξ)= ....= q−NnGfϕ0(πNδξ),

if l ≥ l0 +Nδ. Since we have

|Gfϕ0(η)| ≤∫On ϕ0dx ≤ 1,

it follows that

|Gfϕ0(ξ)| ≤ q−Nn.

On the other hand,

l0 > l − (N + 1)δ

which means that

−Nn < − lδn+ l0

δn+ n.

So

|Gfϕ0(ξ)| ≤ q−Nn ≤ ql0δn+n(ql)−

nδ

= c|ξ|−nδ

if |ξ| ≥ ql0 .

Q.E.D.

We leave the following as an exercise:

62

Exercise: Let m = 1. Then

|Gfϕ(ξ)| ≤ c 1

|ξ|nδ

for |ξ| > qν0 for some ν0 ≥ 0. Hence, if we define

Jν =∫|ξ|>qν

dξ

|ξ|nδ

then

Jν = c1(q1−nδ )ν

2.6 k = Fq[T ] (continued)

Let us return to the function field k = Fq(T ), and let us put K = k∞as in section 2.5. Then we have the following:

Theorem 2.6.1: Let 0 < θ ≤ 12, ch(k) > δ, d > deg ν

δ−1, d ≥ 2

θ, and

n > δ2δ

θ. Then

Nt(ν)

|t|n−δA= qn+1S(ν)Γ +O(q−dε

′).

For this theorem, we have finally eliminated the dependence of Sand Γ on d and θ.

Proof : We have already shown that S is the limit of Sd,θ as d→∞.Now, we must do the same for Γ. We can see that

Γ =∫k∞

∫On∞

χ∞(γ(ζδ1 + ...+ ζδn))dζdγ

=∫k∞

(∫O∞ χ∞(γ(ζδ))dζ)ndγ

=∫k∞Gfϕ0(γ)dγ.

where ϕ0 is the characteristic function on On∞. Similarly, by ourdefinition of Γd,θ,

Γd,θ =∫|γ|∞<qdθ(

∫O∞ χ∞(γζδ)dζ)ndγ

=∫|γ|∞<qdθ Gfϕ0(γ)dγ,

From this and from the exercise above, it is obvious that limd→∞ Γd,θ =

63

Γ.

Q.E.D.

3 Singular Series

3.1 N(ν) and S(ν)

Let k be an A-field, and let kA be the ring of adeles over k. We fix abasic character χ of k. Let f : kn → km be a polynomial map overk. We use the same notation f for maps obtained from f over kv,kA in the natural way.

Next, let

ϕ : knA → C

be a function in S(knA), the Schwartz space (this space will be dis-cussed in more detail later). Then we define

(∗) N(ν) =∑

γ∈kn,f(γ)=ν ϕ(γ).

This sum can be thought of as Nf,ϕ(ν); this sum converges as aresult of the fact that ϕ is a Schwartz function.

Now, we will limit our study by introducing the following axioms:

(A.1)∑

γ∈kn |ϕ(γ)| < +∞, i.e. ϕ|kn ∈ L1(kn).

(A.1)A ϕ ∈ L1(knA).

Again, we let

(∗∗) (Γϕ)(ξ) =∑

γ∈kn ϕ(γ)χ(< f(γ), ξ >).

This is clearly integrable, since |χ(< f(γ), ξ >)| = 1. Moreover,because f(γ) ∈ km, it is clear that Γϕ is a function modulo km, i.e.

(Γϕ)(ξ + a) = (Γϕ)(ξ), ∀ a ∈ km

Recall also that, for ν ∈ km, the ”ν-th Fourier coefficient of (Γϕ)(ξ)”

64

is ∫kmA /k

m(Γϕ)(ξ)χ(< ξ, ν >)dξ

=∑

γ∈kn ϕ(γ)∫kmA /k

m χ(< f(γ)− ν, ξ >)dξ

Let

ψ(ξ) = χ(< f(γ)− ν, ξ >).

Note that ψ ∈ kmA /km. Hence we have

∫kmA /k

m ψ(ξ)dξ =

1 if ψ = 1,

0 otherwise.

So this integral is 1 if and only if f(γ) = ν. Thus,∑γ∈kn ϕ(γ)

∫kmA /k

m χ(< f(γ)− ν, ξ >)dξ

=∑

γ∈kn,f(γ)=ν ϕ(γ)

= N(ν).

This means that (∗) is the ”ν-th Fourier coefficient of (∗∗)”.Now, if we move from the sum over kn to the integral over knA,

we require the adelized axiom (A.1)A. For ξ ∈ kmA , we let

Gϕ(ξ) =∫knAϕ(x)χ(< f(x), ξ >)dx.

By (A.1)A, this integral converges. We let

S(ν) =∫kmAGϕ(ξ)χ(< ξ, ν >)dξ.

Clearly, S(ν) = Gϕ(ν). In order for this to be useful, we requireanother axiom:

(A.2) Gϕ ∈ L1(kmA ).

This allows the integral S(ν) to converge.

65

3.2 Nt(ν) and St(ν)

Let G be an algebraic group over k; i.e, let G be a group which is analgebraic variety over k such that there exists a multiplication mapG×G→ G and an inverse map G→ G which are both morphismsover k. We take a universal field Ω that contains k, and we let

GLn = GLn(Ω) = s = (sij), sij ∈ Ω : det(s) 6= 0.

Define ρ to be a representation over k of G:

ρ : G→ GLnt 7→ ρ(t).

Hence ρ(st) = ρ(s)ρ(t). We let Gk be the k-rational points of G. Asusual, one can define

ρk : Gk → GLn(k)ρA : GA → GLn(A).

When the context is clear, we may use ρ interchangeably with ρk orρA. If we define a representation of G in GLm by

ω : G→ GLm,

then it is clear that G acts both on Ωn and Ωm, since, for x ∈ Ωn

and y ∈ Ωm, we can define the action on each by

tx = ρ(t)xty = ω(t)y.

As before, f : kn → km is a polynomial map. We will assumethat

f(tx) = tf(x)

or, equivalently,

f(ρ(t)x) = ω(t)f(x).

66

Next, for a function ϕ : knA → C, we define

ϕt(x) = ϕ(t−1x).

Note that

ϕst(x) = (ϕt)s(x) = ϕt(s−1x) = ϕ(t−1s−1x) = ϕ((st)−1x).

Hence the left action of G on ϕ is well-defined.We state three propositions and leave their proofs as exercises:

Proposition 3.2.1: If a ∈ Gk and t ∈ GA then Nat = Nt(a−1ν)

and Sat(ν) = St(a−1ν).

For the second proposition, let

∆ρ(t) = |det ρ(t)|A∆ω(t) = |det ω(t)|A

Proposition 3.2.2: St(ν) = ∆ρ(t)∆ω(t)−1S(ω(t)−1ν).

Proposition 3.2.3: Let GA(ϕ) = u ∈ GA : ϕu = ϕ. Then, fort ∈ GA and u ∈ GA(ϕ)

Ntu(ν) = Nt(ν)Stu(ν) = St(ν).

Using the definition from Proposition 3.2.2, we can define our thirdand final axiom:

(A.3) ∆ρ(t) : t ∈ GA ⊂ R×>0 is unbounded.

Now, we concatenate these five pieces of information (the group andthe various functions and representations) into a single symbol A:

A = (G, ρ, ω, f, ϕ).

Definition: We say that A is admissible if (A.1), (A.1)A, (A.2),

67

and (A.3) hold and

f(ρ(t)x) = ω(t)f(x).

Additionally, we say that A enjoys an asymptotic formula if ∃ r > 0such that

Nt(ν)∆ρ(t)∆ω(t)−1 = S(ω(t)−1ν) +O(∆ρ(t)

−r).

3.3 Poisson Summation Formula

For ϕ ∈ S = S(knA), its Fourier transform is

ϕ(ξ) =∫knAϕ(x)χ(< x, ξ >)dx ∈ S

and then we get

ϕ(x) =∫knAϕ(ξ)χ(< x, ξ >)dξ ∈ S,

i.e. ϕ(x) = ˆϕ(−x), the inversion formula.

Here, we state two versions of the Poisson summation formula:

Theorem 3.3.0: Let ϕ ∈ S. Then∑γ∈kn ϕ(γ) =

∑γ∈kn ϕ(γ).

Note that

ϕ(0) =∫knAϕ(x)dx.

A generalization of this is the following:

Theorem 3.3.1: Let ϕ ∈ S. Then∑γ∈kn ϕ(x+ γ) =

∑γ∈kn ϕ(γ)χ(< x, γ >).

Remark: Using this, we can actually compute the difference be-tween St(ν) and Nt(ν):

68

St(ν)−Nt(ν) =∫kmA /k

m Ψt(ξ)χ(< ξ, ν >)dξ

where

Ψt(ξ) =∑

γ∈km Gϕt(ξ + γ)−∑

γ∈km ϕt(γ)χ(< f(γ), ξ >).

Proof of Theorem 3.3.1: First, let us write

α(x)(ξ) = χ(< x, ξ >)

and

Φ(x) =∑

γ∈kn ϕ(x+ γ).

Φ is a function on knA/kn, since adding an element in kn to x leaves

Φ unchanged. We can consider the ”γ-th” Fourier coefficient Φ(γ)to be

Φ(γ) =∫knA/k

n Φ(x)α(x)(ξ)dx

=∫knA/k

n(∑

δ∈kn ϕ(x+ δ))α(x)(ξ)dx

=∫knA/k

n(∑

δ∈kn ϕ(x+ δ))α(x+ δ)(ξ)dx

=∫knAϕ(x)α(x)(ξ)dx

= ϕ(γ)

where the third line comes from the fact that, for δ, γ ∈ kn,

α(x+ δ)(ξ) = α(x)(ξ)α(a)(ξ) = α(x)(ξ),

and the fourth line is Fubini’s theorem applied to the exact sequence

0→ kn → knA → knA/k → 0.

Note that ϕ(γ) is then the ”γ-th Fourier coefficient”. Using thisto construct the Fourier series for Φ,

Φ(x) =∑

γ∈kn ϕ(γ)χ(< x, γ >).

69

But

Φ(x) =∑

γ∈kn ϕ(γ + α)

from which the theorem follows.

Corollary 3.3.1:∑

γ∈kn ϕ(x+ γ) =∑

γ∈kn ϕ(γ)χ(< x, γ >).

Proof : Follows from taking the Fourier transform of both sidesin Theorem 3.3.1.

3.4 f = x21 + ...+ x2

n over R

We elucidate the process of how one calculates S(ν) with an exam-ple. Let us take R as our field k and let

f = x21 + ...+ x2

n

be our polynomial over R. Moreover, let

χ(x) = e−2πix = e(x).

Additionally, the Schwartz space S(R) is defined by

S(R) = ϕ : R→ C : ϕ is infinitely differentiable,supx∈R|xβ( d

dx)αϕ| <∞ ∀ α, β ∈ Z+.

In the simplest version, we see what happens when n = 1; hence,f(x) = x2. In this example, we will let

ϕ(x) = e−πx2.

It is clear that ϕ ∈ S(R). One knows that

ϕ(x) = ϕ(x) = e−πx2.

Now, we write explicitly Gϕ:

70

Gϕ(ξ) =∫

R ϕ(x)χ(f(x)ξ)dx

=∫∞−∞ e

−πx2(1+2ξi)dx

= 2∫∞

0e−πx

2(1+2ξi)dx.

Let x′ = x2. Then

Gϕ(ξ) =∫∞

0e−πx

′(1+2ξi)(x′)12−1dx′.

Note the similarity of the above to the Gamma function Γ(s). Infact, let us express the integral above as∫∞

0e−(p+qi)xxs−1dx.

Then

p = π,q = 2πξ,s = 1

2.

To compute this integral, then, we integrate over the complex plane.For ease of notation, let

r =√p2 + qi

and let

g(z) = e−rzzs−1.

We will define our contour of integration using

tanϕ = qp

where 0 ≤ ϕ ≤ π2

(i.e. ϕ is the principal branch of tan−1( qp)).

Then, integrating over the contour in Figure 1, if we let C denotethe path which goes through ABB′A′A then, by Cauchy’s IntegralFormula,

0 =∫Cg(z)dz = e−iϕπΓ(s)

rs.

71

Figure 1: Contour of integration

72

where the latter step follows from much non-trivial computation.Thus,

Gϕ(ξ) = e−πx2(1+2ξi)dx =

e−iϕ2 Γ( 1

2)

√π(1+4ξ2)

14

= e−iϕ2

(1+4ξ2)14

.

Recall that assumption (A.2) required that Gϕ ∈ L1(kmA ). Clearly,in order for this to be true, we must have that Gϕ ∈ L1(R) (wherem = 1). But, for sufficiently large ξ,

|Gϕ(ξ)| = 1

(1+4ξ2)14> 1

1+ξ

which means that∫∞c|Gϕ(ξ)|dξ >

∫∞cdξ 1

1+ξ= log |1 + ξ|]∞c .

So the integral diverges.However, let us define

f(x) = x21 + ...+ x2

n.

We can use the steps above to determine how large n must be inorder for Gϕ(ξ) ∈ L1(R). First, we must define the Schwartz spacefor Rn:

f ∈ S(Rn) = f ∈ C∞(R) : supx∈Rn|xβDα(f(x))| < ∞ ∀α ∈ Zn

+, β ∈ Z+

where

α = (α1, ..., αn)

and

Dαf = ∂α1+...+αn

∂xα11 ...∂xαnn

.

As before, we let

ϕ(x) = e−π|x|2.

73

Note that, in this case,

e−π|x|2

= Πnj=1e

−πx2j = e−πf(x).

Clearly, ϕ(x) ∈ S(Rn).

So

Gϕ(ξ) =∫

Rn ϕ(x)χ(f(x)ξ)dx

=∫

Rn e−πf(x)e(f(x)ξ)dx

= (∫

R e−πx2−2πix2ξdx)n

= e−ni2

tan−1(2ξ) 1

(1+4ξ2)n4

.

where the last step comes from our earlier computation that∫R e−πx2−2πix2ξdx

= e−iϕ2

(1+4ξ2)14

= e−i tan−1(2ξ)

(1+4ξ2)14

.

It is easy to see from this computation that

Gϕ(ξ) ∈ L1(R) ⇔ n ≥ 3.

Thus, for the remainder of the computation, we will assume thatn ≥ 3. This will allow us to make frequent use Fubini’s theorem.Recall that

S(ν) =∫

R Gϕ(ξ)χ(ξν)dξ.

Plugging in the definitions chosen for our example,

S(ν) =∫

R

∫Rn e

−π|x|2−2πi|x|2ξe2πiξνdxdξ

=∫

Rn e−π|x|2 ∫

R e2πiξν−2πi|x|2ξdξdx.

Let us define

Jε =∫

R e2πiξν−2πi|x|2ξ−π(εξν)2dξdx,

74

Iε =∫

Rn e−π|x|2 · Jεdx.

Then

S(ν) = limε→0 Iε.

We examine Jε. Set

u = εξνdu = ενdξ.

Then

Jε = 1εν

∫R e

πu2+ 2πiuεν

(|x|2−ν)du.

Now, by our choice of ϕ, we recall that ϕ = ϕ, i.e.∫R e−πX2−2πiXY = e−πY

2.

In our case, we let

X = uY = 1

εν(|x|2 − ν).

Then

Jε = 1ενe−πY

2= 1

ενe−π (|x|2−ν)2

(εν)2 .

So

Iε = 1εν

∫Rn e

−π|x|2e−π (|x|2−ν)2

(εν)2 dx.

This can be more easily evaluated by changing to hypersphericalcoordinates. Set

x1 = t cos(φ1),x2 = t sin(φ1) cos(φ2),x3 = t sin(φ1) sin(φ2) cos(φ3),

75

· · ·xn−1 = t sin(φ1) · · · sin(φn−2) cos(φn−1),xn = t sin(φ1) · · · sin(φn−2) sin(φn−1),

where the φ’s range over the (n− 1)-dimensional unit sphere, and tranges from 0 to ∞. Then

dx = tn−1 sinn−2(φ1) sinn−3(φ2) · · · sin(φn−2), dt, dφ1, dφ2 · · · dφn−1

= tn−1dtdω,

where∫Sn−1 dω = π

n2

Γ(n2

+1),

the volume of the n− 1-dimensional sphere. So

Iε = 1εν

∫Sn−1

∫∞0e−πt

2e−π (t2−ν)2

(εν)2 tn−1dtdw.

We change variables once more, letting

y = t2−νεν

,ενdy = 2tdt.

SoIε =

∫Sn−1 dw

∫∞− 1εe−π(yε+ν)e−πy

2 tn−2

2dy

= 12

∫Sn−1 dw

∫∞− 1εe−π(yε+ν)−πy2(yε+ ν)

n−22 dy

= πn2

2Γ(n2

+1)

∫∞1εe−π(yε+ν)−πy2(yε+ ν)

n−22 dy.

If we take the limit as ε→ 0, we get

S(ν) = limε→0 Iε = πn2

2Γ(n2

+1)

∫∞−∞ e

−π(ν)−πy2(ν)n−2

2 dy

= πn2

2Γ(n2

+1)e−πν(ν)

n−22

∫∞−∞ e

−πy2dy.

But∫∞−∞ e

−πy2dy = 1

76

is the Gauss distribution integrated over all R. Thus,

S(ν) = πn2

2Γ(n2

+1)e−πνν

n−22 .

3.5 f(x) = x21 + ...+ x2

n over Qp

Next, we study the same computation over the p-adic fields. In thiscase, we will define ϕ(x) to be the characteristic function of Zn

p . Asbefore, this allows ϕ to be in S(Qn

p ), the Schwartz space over Qnp .

Additionally, a character of Qp is defined as follows:

χp(x) = e2πix

where x is the fractional part of the p-adic expansion of x ∈ Q+p .

Notationally, this means that ∃ l, r such that l ≥ 0, 0 ≤ r ≤ pl − 1,and

x = rpl

.

Throughout this section, we will use χp and χ interchangeably ifthe context is clear.

This definition of character allows us to define our various terms.We begin with Gϕ;

Gϕ(ξ) =∫

Qnpϕ(x)χ(f(x)ξ)dx

=∫

Znpϕ(x)χ(f(x)ξ)dx, ξ ∈ Qp.

This means that the p-adic singular series, denoted Sp(ν), is de-fined as

Sp(ν) =∫

Qp Gϕ(ξ)χ(νξ)dξ.

Now, we first determine for which n we have Gϕ(ξ) ∈ L1(Qp). Todo this, let us define

Il = 1pl

Zp,

Jl = plZp.

There are three important properties about these sets. First, we

77

note that the inclusion of the sets is given by

... ⊂ Jl ⊂ ...J1 ⊂ J0 = Zp = I0 ⊂ I1 ⊂ ... ⊂ Qp.

Second,

Il/Zp∼= Zp/Jl = Zp/p

lZp = Z/plZ,

Finally, we can write Il as

Il =⋃pl−1r=0 (Zp + r

pl)

where the above union is disjoint. From the first and third proper-ties, we can write:∫

Qp Gϕ(ξ)dξ = liml→∞∫IlGϕ(ξ)dξ

= liml→∞∑pl−1

r=0

∫Zp+ r

plGϕ(ξ)dξ

= liml→∞∑pl−1

r=0

∫Zp+ r

plGϕ( r

pl)dξ.

From this, it follows that∫Qp |Gϕ(ξ)|dξ = liml→∞

∑pl−1r=0

∫Zp+ r

pl|Gϕ( r

pl)|dξ.

Therefore, it remains to evaluate Gϕ( rpl

). From the second property

above, we can write the representatives of the coset group Znp/J

nl as

a = (a1, ..., an), 0 ≤ ai ≤ pl − 1,

which means that

Znp =

⋃a(J

ml + a).

Then

Gϕ( rpl

) =∫

Znpχ(f(x) r

pl)dx

=∑

a∈Znp/Jnl

∫a+Jnl

χ(f(x) rpl

)dx.

Now, since elements of Jnl can be written as ply for some y ∈ Znp , it

78

follows that, for x ∈ a+ Jnl ,

f(x) = f(a+ ply) = f(a) + plz ≡ f(a) (mod Jl).

for some z ∈ Znp . So∑

a∈Znp/Jnl

∫a+Jnl

χ(f(x) rpl

)dx =∑

a∈Znp/Jnl

∫a+Jnl

χ(f(a) rpl

)χ(rz)dx

=∑

a∈Znp/Jnlχ(f(a) r

pl)∫Jnldx

= p−nl∑

a∈Znp/Jnlχ(f(a) r

pl)

= p−nl∑

a∈Zn/Jnle

2πif(a) rpl .

Now, let us denote

cl =∫IpGϕ(ξ)dξ =

∑pl−1r=0 Gϕ(ξ)

= p−nl∑pl−1

r=0

∑a∈Znp/Jnl

χ(2πif(a) rpl

)

= p−nl∑

a∈Znp/Jnl

∑pl−1r=0 χ(2πif(a) r

pl).

If we define

ψa(r) = χ(f(a) rpl

)

then

∑pl−1r=0 ψa(r) =

pl if ψa = 1,

0 otherwise.

So

p−nl∑

a∈Znp/Jnl

∑pl−1r=0 χ(2πif(a) r

pl)

= p−nl∑

a∈Znp/Jnl ,f(a)≡0 (mod p) pl

= p−ln+l∑

a∈Znp/Jnl ,f(a)≡0 (mod pl) 1.

Let us put

Apl(f) = #a ∈ (Zp/plZp)

n : f(a) ≡ 0 (mod pl).

Then

79

cl =Apl

(f)

p(n−1)l .

Now, we note that

Gϕ( rpl

) = p−ln∑

a e2πi

plf(a)r

= (p−l∑

a e2πi

pla2r

)n.

Let us define

(3.5.1) G( rpl

) = p−l∑

a e2πi

pla2r

= p−l∑pl−1

a=0 e2πi

pla2r

.

Note that G is the classical Gauss sum. So

Gϕ( rpl

) = G( rpl

)n.

Now, we evaluate G( rpl

):

|G( rpl

)|2 = G( rpl

)G( rpl

)

= p−2l∑pl−1

a=0

∑pl−1b=0 e

2πia2−b2

plr

= p−2l∑pl−1

a=0

∑pl−1b=0 e

2πi(a−b)(a+b)

plr.

If we substitute c = a− b then the above is equal to

p−2l∑pl−1

c=0

∑pl−1b=0 e

2πic(c+2b)

plr

= p−2l∑pl−1

c=0 e2πi c

2

plr∑pl−1

b=0 e2πi

c(2b)

plr.

Let

ψ(b) = e2πi

c(2b)

plr.

Since ψ(b) is a character mod pl, it follows that

∑pl−1b=0 ψ(b) =

pl if ψ(b) = 1 ∀ b,0 otherwise.

Now,

80

ψ(b) = 1 ∀ b ⇔ 2crb ≡ 0 (mod pl) ∀ b⇔ 2cr ≡ 0 (mod pl).

If p - r and p 6= 2 then this is equivalent to

c ≡ 0 (mod pl),

which means that

|G( rpl

)|2 = p−2lpl = p−l

and hence

(3.5.2) |G( rpl

)| = p−l2 .

If p = 2 and 2 - r then

c ≡ 0 (mod 2l) or c ≡ 0 (mod 2l−1).

So

|G( r2l

)|2 = 2−l∑

c≡0 (mod 2l−1)e2πi c

2

2lr

= 2−l(1 + e2πi 22l−2

2lr)

= 2−l(1 + e2πi(2l−2)r).

Note that

e2πi(2l−2)r =

1 if l ≥ 2,

−1 if l = 1.

So

(3.5.3) |G( r2l

)| =

2−

l−12 if l ≥ 2,

0 if l = 1.

Thus, we arrive at the final expression for |G( rpl

)|. For 1 ≤ r ≤ pl, let

v(r) = maxk ∈ Z : pk|r.

81

Then

|G( rpl

)| =

p−

l−v(r)2 if p 6= 2,

2−l−v(r)−1

2 if l − v(r) ≥ 2, p = 2

0 if l − v(r) = 1, p = 2,

1 if l = v(r), p = 2.

Now, we recall that |Gϕ( rpl

)| = |G( rpl

)|n and

Gϕ ∈ L1(Qp) ⇔ liml→∞∑pl

r=1 |Gϕ( rpl

)| <∞.

Case 1: Assume p 6= 2. Then let

sl =∑pl

r=1 |Gϕ( rpl

)| =∑pl

r=1 p−n(

l−v(r)2

) = p−nl2

∑lµ=0 aµp

nµ2 ,

where

aµ = #r, 1 ≤ r ≤ pl : v(r) = µ.

Clearly,aµ = φ(pl−µ) = pl−µ−1(p− 1),

if µ < l, and a = 1 when µ = l; here, φ is the Euler phi func-tion. So

sl = p−nl2

∑lµ=0 aµp

nµ2 = 1 + p−

nl2

+l−1(p− 1)∑l−1

µ=0 p(n2−1)µ

= 1 + (1− 1p)(1−p−(n2−1)l

pn2−1 ).

From this, we conclude that

liml→∞ sl =

∞ if n = 1, 2,pn2 −1

pn2 −p

if n ≥ 3.

Case 2: Let p = 2. The work, which is analogous to the previouscase, is left as an exercise. Instead, we merely state the results

82

liml→∞ sl =

∞ if n = 1, 2,

2n2 −1

2n2−1−1

if n ≥ 3.

Thus, from the work above, we see that∫Qp Gϕ(ξ)dξ = liml→∞ sl,

which implies that L1(Qp) for n ≥ 3.

3.6 Determination of G( rpl

)

The determination of G( rpl

) is a calculation which has a long mathe-

matical history; as was noted in the previous chapter, its beginningscan be traced back to Gauss. However, we will use the Siegel nota-tion, where, for a, b ∈ N, we define

G(a, b) = 1√b

∑bh=1 e

πiabh2+πiah

(cf. p. 50 C.L Siegel, Analytische Zahlentheorie II, Gottingen1963/64). Recall that, in our case, we had defined

G( rpl

) = p−l∑pl

x=1 e2πi

plx2r

.

This is nothing more than the special case that a = 2r, b = pl,i.e.

G(2r, pl) = pl2G( r

pl).

More generally, though, we will use only the assumption that, fora, b ∈ N, λ ∈ Q, we have the relation that

ab+ 2aλ ≡ 0 (mod 2).

Next, we let z ∈ C and define

f(z) = eπiab(z+λ)2

e2πiz−1.

Note that the numerator is never zero, and the denominator hasa pole at every integer h (and at no other points). We show that

83

the poles are all of order 1:

Resz=hf(z) = limz→h(z − h)f(z)

= eπiab

(h+λ)2 limz→h(z−h)e2πiz−1

= eπiab

(h+λ)2 limz→h1

2πie2πiz

= eπiab

(h+λ)2 12πi

where the penultimate step is by L’Hospital’s rule. Thus, if wehave a contour C which contains the integers 1, 2, ..., b (and no oth-ers) then∫

Cf(z)dz = 2πi

∑bh=1Resz=hf(z) = 2πi

∑bh=1Resz=he

πiab

(z+λ)2 .

In particular, we will take C to be the contour in Figure 2, whereL is the line which goes through the real axis at a 45 angle. Notethat as the horizontal lines go toward infinity, the integrals overthese lines go to zero, since, if we let z = x+ iy, the real part of theexpression above is determined by i(iy2) = −y2, which goes to zeroas y →∞. So∫

Cf(z)dz =

∫L(f(z + b)− f(z))dz

=∫Leπi

ab(z+λ)2

e2πiz−1[eπi

ab

(2b(z+λ)+b2) − 1]dz.

Now,

ab(2b(z + λ) + b2) = 2aλ+ ab+ 2az ≡ 2az (mod 2)

by our assumption earlier, which means that

eπiab

(2b(z+λ)+b2) = e2πiaz.

Then∫Cf(z)dz =

∫Leπi

ab

(z+λ)2 [ eπi ab(2b(z+λ)+b2)−1e2πiz−1

]dz

=∫Leπi

ab

(z+λ)2 [∑a−1

h=0 e2πihz]dz

=∑a−1

h=0

∫Leπi

ab

(z+λ)2+2πihzdz

=∑a−1

h=0 eπi bah2−2πihλ

∫Leπi

ab

(z+λ+ bah)2dz.

84

Figure 2: Contour of integration

85

Here, the first step comes from the expansion

xa−1x−1

=∑a−1

h=0 xh.

Let us define

Z = z + λ+ bah.

Then the line L instead shifts to another parallel line which wecan denote L0, i.e.∑a−1

h=0 eπi bah2−2πihλ

∫L0eπi

abZ2dZ.

Let L1 be the line parallel to L0 which goes through the origin.Since the integrand above has no poles, it follows from Cauchy’sTheorem that we can shift the contour of integration from L0 to L1.The integral in the above expression is then∫

L1eπi

abZ2dZ.

We make the change of variables

Z =√

ibat

dZ = −√

ibadt.

Note that the integral over L1, like the integrals over L and L0,goes downward along the line. So the change of measure allows usto change the orientation of the integral, i.e.∫

L1eπi

abZ2dZ =

√iba

∫∞−∞ e

−πt2dZ =√

iba

.

Thus, in summary,∫Cf(z)dz =

∑bh=1 e

πiab

(h+λ)2

=√

iba

∑a−1h=0 e

−πi bah2−2πihλ

=√

iba

∑ah=1 e

−πi bah2−2πihλ

86

where the shift in parameter in the last step is allowed because

ab+ 2aλ ≡ 0 (mod 2).

We arrive, then, at the theorem

Theorem 3.6.1:√

1b

∑bx=1 e

πiab

(x+λ)2 =√

ia

∑ax=1 e

πi bax2−2πixλ.

As a corollary, we can prove the special case of the sum which Gaussconsidered:

Corollary 3.6.1: Let a = 2, λ = 0, b = n ∈ N. Then

1√n

∑nx=1 e

2πix2 1n =

1 + i if n ≡ 0 (mod 4),

1 if n ≡ 1 (mod 4),

0 if n ≡ 2 (mod 4),

i if n ≡ 3 (mod 4).

Proof : We know that

√i = e

iπ4 = 1+i√

2.

Then

1√n

∑nx=1 e

2πix2 1n = 1√

2

(1+i)√2

(1 + e−πin2 ).

Naturally, e−πin2 depends on n. In particular,

1 + e−πin2 =

2 if n ≡ 0 (mod 4),

1− i if n ≡ 1 (mod 4),

0 if n ≡ 2 (mod 4),

1 + i if n ≡ 3 (mod 4).

from which the corollary follows.

Now, we can derive a useful theorem about Legendre characters.

For an odd prime p, let us choose a ζ ∈ F+p such that

87

F+p =< ζ >.

We define

S(ζ) =∑

x∈Fp ζx2

= 1 +∑

x∈F×p ζx2

= 1 +∑

x<0 ζx2

+∑

x>0 ζx2

,

where an element in F×p can be described as being > 0 if it is be-

tween 0 and p−12

or < 0 otherwise. Then the above is

= 1 + 2∑

x>0 ζx2

= 1 + 2∑

y∈(F×p )2 ζy.

If we define

0 = T (ζ) =∑

y∈Fp ζy

= 1 +∑

y∈F×p ζy

= 1 +∑

y∈(F×p )2 ζy +

∑y∈F×p ,y 6∈(F×p )2 ζ

y

then we can express S(ζ) as

S(ζ) = S(ζ)− T (ζ) =∑

y∈(F×p )2 ζy −

∑y∈F×p ,y 6∈(F×p )2 ζ

y

=∑

y∈Fp(yp)ζy,

where (yp) is the Legendre symbol with (y

p) = 0 if p|y. Thus, we

arrive at the following:

Theorem 3.6.2:∑

y∈Fp(yp)ζy =

∑p−1x=0 ζ

x2.

Now, if p - r then we can replace ζ by ζr. Then, by the theo-rem,∑p−1

x=0 ζrx2

=∑

x∈Fp(xp)ζrx

= ( rp)∑

x∈Fp(rxp

)ζrx

= ( rp)∑p−1

x=0 ζx2

.

88

If we let

ζ = e2πip

then the above string of equalities tells us that

G( rp) = ( r

p)G(1

p)

where

G(1p) =

√p p ≡ 1 (mod 4),

i√p p ≡ 3 (mod 4).

3.7 Gauss Sums (Classical)

Gauss sums have been studied by many mathematicians throughouthistory. Here, we give a brief history of the notation used by thevarious mathematicians who studied Gauss sums. We begin withDirichlet, who defined the following sum:

φ(h, n) =∑n

x=1 e2πix2 h

n ,

where h ∈ Z and n ∈ N. His student, Paul Bachmann, deviseda notation which could facetiously be referred to as a generalizationof his teacher’s:

ϕ(m,n) =∑n

x=1 eπix2m

n

where m ∈ Z and n ∈ N. Note, then, that

ϕ(2h, n) = φ(h, n).

A variant of this notation was also used by Hardy; he let

Sh,n = φ(h, n).

On the other hand, Gauss (and later Siegel) used the notation

89

G(a, b) = 1√b

∑bx=1 e

πiabx2

.

It is worth noting that Siegel, not Gauss, chose the letter G, asGauss did not name the sums after himself. So

G(2r, n) = 1√nφ(r, n) = 1√

nϕ(2r, n).

In particular,

G(2r, pl) = p−l2φ(r, pl) = p

l2G( r

pl) = p−l

∑pl

x=1 e2πix2r

pl .

where 1 ≤ r ≤ pl.Let us stay with Bachmann’s notation for ϕ. Then we state the

following fifteen properties, most of whose proofs are left as an ex-ercise to the reader.

(1) If m ≡ n ≡ 1 (mod 2) then ϕ(m,n) = 1.

For the remaining properties, we will assume that mn ≡ 0 (mod2).

(2) ϕ(m, 1) = 1 ∀ m ∈ Z.

(3) ϕ(m, 2) = 1 + im ∀ m ∈ Z.

(4) ϕ(m′, n) = ϕ(m,n) if m′ ≡ m (mod 2n).

(5) Let m = 2µ and (k, n) = 1. Then ϕ(k2m,n) = ϕ(m,n).

(6) Let m = 2µ and (n′, n) = 1. Then ϕ(mn′, n)ϕ(mn, n′) =ϕ(m,nn′).

(7) Let h ∈ N. Then ϕ(hm, hn) = hϕ(m,n).

The next several properties are in the specific case where n is apower of a prime, often 2.

(8) Assume µ is odd and α ≥ 2. Then ϕ(2µ, 2α) = 2∑2α−1−1

x=0 eπix22 µ

2α .

90

(9) With notations as above and α ≥ 4, ϕ(2µ, 2α) = 2ϕ(2µ, 2α−2).

(10) If α = 2 then ϕ(2µ, 22) = 2(1 + iµ).

(11) If α = 3 then ϕ(2µ, 23) = 4πi4 µ.

(12) For given n and p, let αp be such that pαp ||n and P = npαp

.

Then ϕ(2µ, n) = Πp|nϕ(2µP, pαp).

(13) If p 6= 2 then

ϕ(2µ, 2α) =

pα2 if α ≡ 0 (mod 2),

pα−1

2 ϕ(2µ, p) if α ≡ 1 (mod 2).

(14) If µ is odd then

ϕ(2µ, 2α) =

2α2 (1 + iµ) if α ≡ 0 (mod 2),

2α+1

2 iµ2 if α ≡ 1 (mod 2).

Now, we apply this to our specific case of

Sp(ν) =∫

Qp Gϕ(ξ)χp(ξν)dξ

= liml→∞∫IlGϕ(ξ)χp(ξν)dξ

= liml→∞∑pl

r=1G( rpl

)ne−2πi r

plν.

We can use Dirichlet notation to write

G( rpl

) = p−lϕ(2r, pl)

We can let r = ptr′, p - r′. Then, if we use (7), letting m = 2r′,n = pl′ where l′ = l − t, and h = pt, we find

ϕ(2r, pl) = ptϕ(2r′, pl′)

and thus

G( rpl

) = G( r′

pl′).

Thus, we can assume that p - r.

91

Now, we apply (13) to Gϕ. In our case, we have µ = r and α = l.For p 6= 2,

G( rpl

) = p−l ·

pl2 if l ≡ 0 (mod 2),

pl−12 ϕ(2r, p) if l ≡ 1 (mod 2).

Of course, we know that

ϕ(2r, p) =

( rp)p

12 if p ≡ 1 (mod 4),

i( rp)p

12 if p ≡ 3 (mod 4).

Combining this information,

G( rpl

) =

p−

l2 if l ≡ 0 (mod 2),

( rp)p−

l2 if l ≡ 1 (mod 2), p ≡ 1 (mod 4),

i( rp)p−

l2 if l ≡ 1 (mod 2), p ≡ 3 (mod 4).

We note that this is easily evaluated in the case of 4|n:

(15) (p 6= 2) If n ≡ 0 (mod 4) then G( rpl

)n = p−ln2 .

Next, we consider when p = 2. So we will apply (14). Since

G( r2l

) = 2−lϕ(2r, 2l)

and

ϕ(2r, 2l) =

2l2 (1 + ir) if l ≡ 0 (mod 2),

2l+12 i

r2 if l ≡ 1 (mod 2),

it follows that

G( r2l

) =

2−

l2 (1 + ir) if l ≡ 0 (mod 2)

2−l−12 i

r2 if l ≡ 1 (mod 2)

=

2−

l−12

1+ir√2

if l ≡ 0 (mod 2),

2−l−12 i

r2 if l ≡ 1 (mod 2).

This gives us a proposition similar to (15) above:

(15) (p = 2) If n ≡ 0 (mod 8) then G( r2l

)n = 2−( l−12

)n.

92

Now, recall that

Sp(ν) = liml→∞∑pl

r=1 G( rpl

)ne−2πi r

plν.

We can expand

r = atpt + ...+ al−1p

l−1 = ptu,

with 0 ≤ ai ≤ p− 1, at 6= 0, and

u = at + ...+ al−1pl−1−t.

and thus p - u. So we can rewrite Sp as

Sp(ν) = liml→∞∑l

t=0

∑0≤u<pl−t,p-uG(p

tupl

)ne−2πi u

pl−tν.

If, as before, we let Sl be the double sum above (where Sp =liml→∞ Sl) then, by (15), we have

Sl =

∑lt=0 p

− l−t2n∑

0≤u<pl−t,p-u e−2πi u

pl−tν

if p 6= 2,∑lt=0 2−

l−t−12

n∑

0≤u<2l−t,2-u e−2πi u

2l−tν if p = 2.

To better understand Sl, we review the theory of Ramanujan sums.Let us define the ring

< = f : N→ C

with operations (+, ∗), where ∗ is Dirichlet convolution, which isdefined by

f ∗ g(n) =∑

d|n f(d)g(nd).

< is then a commutative ring with identity e = 1< defined by

e(n) =

1 if n = 1,

0 if n ≥ 2.

So

93

f ∗ e = e ∗ f = f .

Moreover, we can define

f ′(n) = f(n) log n

which gives us

(f ∗ g)′ = (f ′ ∗ g) + (f ∗ g′).

Note also that this is a unique factorization domain. Additionally,we define in the natural way

<× = f : ∃ g such that f ∗ g = e.

We claim that

<× = f : f(1) 6= 0.

To prove this claim, we note that if f ∈ <×, ∃ g such that

f ∗ g(n) =∑

d|n f(d)g(nd) =

1 if n = 1,

0 if n ≥ 2.

If n = 1, f(1)g(1) = 1. If n > 1,

0 =∑

d|n f(nd)g(d) = f(1)g(n) +

∑d|n,d<n f(n

d)g(d).

So

g(n) = − 1f(1)

∑d|n,d<n f(n

d)g(d).

Thus, f(1) 6= 0 if and only if such a g exists.Next, let us define the functions η, ε ∈ < be such that

η(n) = 1 ∀ n ∈ N.ε(n) = n.

94

Additionally, we define ϕ and µ to be the classical Euler and Mobiusfunctions, respectively:

ϕ(n) = #(Z/nZ)×,

µ(n) =

1 if n = 1,

0 if ∃ p such thatp2|n,(−1)k if n = p1p2...pk is square-free.

With these definitions, we introduce the following properties:

(1)∑

d|n µ(d) = e.

To prove this, note that∑d|n µ(d)

= µ(1) + µ(p1) + ...µ(pk) + µ(p1p2) + ...+ µ(pk−1pk)+...+ µ(p1....pk)

=∑k

i=1(−1)i(ki

)= (1− 1)k

= 0.

(2) µ ∗ η = e.

This is true because µ ∗ η(n) =∑

d|n µ(d).

(3) ϕ ∗ η = ε.

This follows from the fact that∑d|n ϕ(d) = n = ε(n).

Note that (2) and (3) imply that µ−1 = η and

ϕ = ε ∗ η−1 = ε ∗ µ.

(4) ∀ f, g ∈ <, f = g ∗ µ−1 ⇔ g = f ∗ µ.

In addition to the above properties, it is clear that ε, η, ϕ, µ ∈ <×. So

95

< ε, η, ϕ, µ >⊂ <×.

By the relations from (2) and (3), we note that

< ε, η, ϕ, µ >=< ϕ, µ >.

The above information can be related to our singular series usingthe aforementioned Ramanujan sum ck, which is defined as

ck(n) =∑

m (mod k),(m,k)=1 e2πimn/k.

We will prove the following theorem about these sums

Theorem 3.7.1: ck(n) =∑

d|(n,k) dµ(kd).

As an obvious consequence of the theorem, we have the followingcorollaries:

Corollary 3.7.1: ck(1) = µ(k).

Corollary 3.7.2: If k|n then

ck(n) =∑

d|k dµ(kd) = ε ∗ µ(k) = ϕ(k).

We note that ck, k ≥ 1, k square-free generates a subgroup of<×. Additionally, any ck has period k, i.e. ck(k + n) = ck(n) ∀ k.In fact, we have the following theorem about periodic functions:

Theorem 3.7.2: If f ∈ < has period k then ∃ g with period k suchthat

f(m) =∑k−1

n=0 g(n)e(mnk

),

where

g(n) = 1k

∑k−1m=0 f(m)e(mn

k).

This theorem can be easily verified and lends itself to a more im-

96

portant theorem:

Theorem 3.7.3: Let f, g ∈ < and

Sk(n) =∑

d|(n,k) f(d)g(kd),

where Sk is periodic with period k. Then

Sk(n) =∑

m (mod k)ak(m)(mn

k),

with

ak(m) = 1k

∑n (mod k)

Sk(n)e(mnk

)

= 1k

∑kn=1

∑d|n,d|k f(d)g(k

d)e(mn

k).

Note that, for given d, we can write n = cd, where 1 ≤ c ≤ kd.

Additionally, we can make the substitution d→ kd. So∑k

n=1

∑d|n,d|k f(d)g(k

d)e(mn

k) =

∑d|k f(k

d)g(d)

∑dc=1 e( cm

d)

=∑

d|k,d|m f(kd)g(d)d.

where the last line comes from the fact that

∑dc=1 e( cm

d) =

d if d|m,0 otherwise.

Thus, to prove Theorem 3.7.1, we take the special case of f(k) = k,g(k) = µ(k). So

Sk(n) =∑

d|(n,k) f(d)g(kd)

=∑

m (mod k ak(m)e(mnk

).

where

ak(m) = 1k

∑d|(m,k)

kdµ(d)d

=∑

d|(m,k) µ

=

1 if (m, k) = 1,

0 otherwise..

97

Now, recalling that

Sl =

∑lt=0 p

− l−t2n∑

0≤u<pl−t,p-u e−2πi u

pl−tν

if p 6= 2,∑lt=0 2−

l−t−12

n∑

0≤u<2l−t,2-u e−2πi u

2l−tν if p = 2,

we can use the fact that

ck(n) =∑

m (mod k),(m,k)=1e2πimn/k =

∑d|(n,k) dµ(k

d).

to rewrite Sl as

Sl =

1 +

∑lλ=0 p

−λn2

∑d|(ν,pl) dµ(p

λ

d) if p 6= 2,∑l

λ=0 2−(λ−1)n

2

∑d|(ν,2λ) dµ(2λ

d) if p = 2.

Now, we can examine the specific case of ν = 1. In this case, d = 1.So

∑d|(ν,pλ) dµ(2λ

d) =

−1 if λ = 1,

0 if λ ≥ 2,

and hence

Sl =

1− p−n2 if p 6= 2,

2−n2 − 1 if p = 2.

Note that Sl is independent of l. So

Sp(1) = liml→∞ Sl = Sl.

Thus, we have

Πp6=∞Sp(1) = 2n2

ζ(n2

),

S∞(1) = πn2

Γ(n2

+1)· 1

2e−π.

Thus, for 8|n, we have

S(1) = ΠvSv(1) = 2n2−1π

n2 e−π

Γ(n2

+1)ζ(n2

).

Recalling that

98

ζ(2k) = 22k−1Bk(2k)!

π2k

where Bk is the k-th Bernoulli number, the above is then

S(1) = e−πn!Γ(n

2+1)Bn

4

.

In the case where n = 8,

S(1) = e−π8!Γ(5)B2

= e−π 8!·304!

= 50400e−π.

3.8 Estimate of Gauss Sums (Local Fields)

The goal of this section is to show that the definition from this chap-ter for S(t) gives the same estimate for N(t) as does the definitionfrom the previous chapter. As such, we return to the original nota-tion; let K be a local field, where

K ⊃ O ⊃ P = (π) = Oπ

are as before. Additionally, let

κ = O/P ∼= Fq

for

q = pe.

We require a measure dx such that∫O dx = 1.

Let χ be a basic character of K. Recalling the definition of ordχ, given by

P−ord χ = x ∈ K : χ(xy) = 1 ∀ y ∈ O,

we know from before that ord χ = 0.

99

Now, for a ∈ O, define

a ≡ a (mod P).

Then a ∈ κ. Let

χ(a)def= χ( a

π).

We check that χ is well-defined and non-trivial. First, to provethat this is well defined, we show that a = b implies χ(a) = χ(b).But

a = b⇔ a ≡ b (mod P)⇔ a− b = πc for some c ∈ O⇔ a

π− b

π∈ O

⇔ χ( aπ− b

π) = 1

⇔ χ( aπ) = χ(− b

π)−1 = χ( b

π).

where the penultimate step is because ord χ = 0.To show that χ is non-trivial, note that χ(a) = 1 ∀ a means that

χ( aπ) = 1 ∀ a ∈ O. Since ord χ = 0, this implies that 1

π∈ O, which

is a contradiction.Next, define

Gxδ(β) = q−1∑

α∈κ χ(αδβ)

where p - δ. If we let Gxδ denote Gfϕ where f(x) = xδ and ϕ is thecharacteristic function on O then we have the following relationship:

Proposition 3.8.1: Gxδ( ξπ ) = Gxδ(ξ).

Proof : To begin, let us denote A = a ∼= O/P to be a set ofrepresentatives of the cosets of O/P . Then

Gxδ( ξπ ) =∫O χ(xδ ξ

π)dx

=∑

a∈A∫a+P χ(xδ ξ

π)dx

=∑

a∈A χ(aδ ξπ)∫a+P dx

100

= q−1∑

a∈A χ(aδ ξ)

= q−1∑

α∈κ χ(αδ ξ)

= Gxδ(ξ).

Proposition 3.8.2: If β 6= 0 then

|Gxδ(β)| ≤√

2δq−12 =

√2qδ.

Proof : Put

σ =∑

η∈κ |Gxδ(η)|2

=∑

η∈κGxδ(η)Gxδ(η)

= q−2∑

α∈κ∑

γ∈κ∑

η∈κ χ((α2 − γ2)η)

= q−1#α, γ ∈ κ× κ : αδ ≡ γδ≤ q−1(q)δ= δ,

where the antepenultimate line comes from the fact that the sumover η is zero if the character is non-trivial and q otherwise. Soσ ≤ δ.

Next, for λ ∈ κ×, we define the action of λ on κ+ by

λ η = λδη.

We show that this action does not change G:

Lemma 3.8.1: G(λ η) = G(η).

Proof : G(λ η) = q−1∑

α∈κ χ(αδλδη)= q−1

∑α∈κ χ((αλ)δη)

= q−1∑

α∈κ χ(αδη)= G(η)

where the second-to-last step is because λ ∈ κ×. So let us denotean orbit of β under this action by λ β. Then

δ ≥ σ ≥∑

η∈λβ |G(η)|2 = |G(β)|2#λ β.

101

Now,

#λ β = |κ×|#λ∈κ×:λδβ=β = q−1

(δ,q−1).

So

δ ≥ |G(β)|2 · q−1(δ,q−1)

≥ |G(β)|2 · q−1δ

≥ |G(β)|2 · q2δ

since q ≥ 2. So

|G(β)|2 ≤ 2q−1δ2,

and hence

|G(β)| ≤√

2q−12 δ,

which proves the proposition.With this relation between G and Gf , we can use the bound on

the former to give us information about the latter. To do this, westart by letting θ ∈ S(K) be the characteristic function of O − Pand f(x) = xδ. Note that if f ′(x) = δxδ−1 = 0 then x = 0. Since0 6∈ supp θ, we see that Cxδ

⋂supp θ = ∅. Thus,

Gxδθ ∈ S(K).

This allows us to apply our knowledge of Kelvin’s principle. Weknow that ∃ lattices L,M ⊂ K such that L ⊃ M , supp Gxδθ ⊂ L,and Gxδθ is constant on each coset in L/M . We claim that

(∗) L = π−1O = P−1 ⊃M = O ⊃ P .

Now,

Gxδθ(ξ) =∫O−P χ(xδξ)dx.

To show (∗), it suffices to show that Gxδθ(ξ) = 0 for ξ outside L. So

102

Gxδθ(η) = 0 if η 6∈ P−1

⇔ η = πνξ for ν ≥ 2, ξ ∈ O − πO.

Now, let

A = (O −P)/Pν ,B = (O −P)/Pν−1,C = O/P .

So

A = a = b+ πν−1c, b ∈ B, c ∈ C.

Then

Gxδθ(π−νξ) =∫O−P χ(xδπ−νξ)dx

=∑

a∈A∫a+Pν χ(xδπ−νξ)dx

= q−ν∑

a∈A χ(xδπ−νξ).

But

aδ = (b+ πν−1c)δ

= bδ + δbδ−1πν−1c+

(δ2

)c2π2(ν−1)bδ−2 + ....

≡ bδ + δbδ−1πν−1c (mod Pν)

since ν ≥ 2 implies that 2(ν − 1) = 2ν − 2 ≥ ν. So

π−νaδξ ≡ π−νbδξ + δbδ−1π−1cξ (mod O),

which means that

χ(π−νaδξ) = χ(π−νbδξ)χ(δbδ−1π−1cξ).

Plugging this in for our expression above Gxδθ(π−νξ) above gives

Gxδθ(π−νξ) =∑

b∈B∑

c∈C χ(π−νbδξ)χ(δbδ−1π−1cξ)=∑

b∈B χ(π−νbδξ)∑

c∈C χ(δbδ−1π−1cξ)

103

=∑

b∈B χ(π−νbδξ)∑

γ∈κ χ(δbδ−1ξγ).

Now, p - δ, b 6∈ P , and ξ ∈ O − πO. So δbδ−1ξ ∈ O − P , whichmeans that∑

γ∈κ χ(δbδ−1ξγ) = 0,

which shows that Gxδθ(π−νξ) = 0 when ν ≥ 2.

We can combine this information about Gxδθ with that of Gxδ togive a bound for |Gxδθ|. To this end, we present the following theo-rem

Theorem 3.8.1: |Gxδθ(ξ)| ≤√

2δ|ξ|−1δ

K ∀ ξ ∈ K.

Proof : Let ϕ be the characteristic function of O. So

Gxδϕ(ξ) =∫O χ(xδξ)dx

=∫O−P χ(xδξ)dx+

∫P χ(xδξ)dx

= Gxδθ(ξ) +∫P χ(xδξ)dx.

Let

x = πy ∈ P

for y ∈ O. Hence

dx = q−1dy.

Plugging into the above gives

Gxδϕ(ξ) = Gxδθ(ξ) + q−1∫O χ(yδπδξ)dy

= Gxδθ(ξ) + q−1Gxδϕ(πδξ).

For notational expediency, let

G = Gxδϕ,G0 = Gxδθ.

104

So

G(ξ) = G0(ξ) + q−1G(πδξ)= G0(ξ) + q−1G0(πδξ) + q−2G0(π2δξ) + ...+ q−(N−1)G0(π(N−1)δξ)+q−NG(πNδξ).

Recall that χ(O) = 1 since ord χ = 0. So if ξ ∈ O then χ(ξ) = 1.Thus, we assume that ξ 6∈ O, and we let N ≥ 1 be the smallest Nsuch that πNδξ ∈ O. So

πNδξ = πν0ε, ε ∈ O×

for some 0 ≤ ν0 < δ. Then

π(N−1)δξ = πν0−δε,

and for r ≤ N − 1,

πrδξ = π−(N−r)δπNδξ= π−(N−r)δπν0ε= π−((N−r)δ−ν0)ε.

Case 1 : 0 ≤ r ≤ N − 2.

Then

(N − r)δ − ν0 > 2δ − δ = δ ≥ 2.

So

πrδξ = π−((N−r)δ−ν0)ε 6∈ P−1

⇒ G0(πrδξ) = 0.

Case 2 : r = N − 1.

Then

πrδξ = π−((N−r)δ−ν0)ε = π−(δ−ν0)ε.

105

Subcase 2.1 : δ − ν0 > 1.

In this case,

G0(π(N−1)δxi) = G0(π−(δ−ν0)ε) = 0.

Subcase 2.2 : δ − ν0 = 1.

Here,

G0(π(N−1)δξ) = G0(π−1ε)=∫O−P χ(xδ ε

π)dx

=∫O χ(xδ ε

π)dx−

∫P χ(xδ ε

π)dx

= G(π−1ε)−∫P χ(xδ ε

π)dx.

Now, if x ∈ P then xδ επ∈ O. So∫

P χ(xδ επ)dx = q−1.

Thus,

G0(π(N−1)δξ) = G(π−1ε)− q−1

= Gxδ(ε)− q−1.

So

G(ξ) =

q−NG(πNδξ) if δ − ν0 ≥ 2,

q−(N−1)(Gxδ(ε)− q−1) + q−NG(πNδξ) if δ − ν0 = 1,

=

q−N if δ − ν0 ≥ 2,

q−(N−1)Gxδ(ε) if δ − ν0 = 1.

since G(πNδξ) = 1. Now, we know that

|Gxδ(ε)| ≤√

2δq−12 .

So

106

|G(ξ)| ≤

q−N if δ − ν0 ≥ 2,

q−(N− 12

)√

2δ if δ − ν0 = 1.

Recall that πNδξ ∈ O. So

q−Nδ|ξ|K ≤ 1,

or

q−N ≤ |ξ|−1δ

K .

which proves the theorem for δ − ν0 ≥ 2. In the case of δ − ν0 = 1,we have

πNδξ = πν0ε.

Thus

q−(Nδ−ν0) ≤ |ξ|−1K

⇔ q−(N− ν0δ

) ≤ |ξ|−1δ

K .

But

N − ν0δ≤ N − δ−1

δ≤ N − 1

2.

So

q−(N− 12

) ≤ q−(N− ν0δ

) ≤ |ξ|−1δ

K .

if ξ 6∈ O. This proves the theorem.

3.9 Back to Fq(t)

Let us return now to the case where k = Fq(t) is a rational functionfield. Let χ be a basic character of k+

A such that

χ(x) = Πvχv(xv),

where x = (xv), xv ∈ kv, xv ∈ Okv ∀′ v, and χv ∈ k+v . As be-

107

fore, the order of χv is given by

P−ord χvv = x ∈ kv : χv(xy) = 1 ∀ y ∈ Ov.

Moreover, we put O = Fq[t], and for v = π 6=∞,

kv = kπ = x =∑

ν≥ν0 cνπν : cν ∈ O, cν0 6= 0 for x 6= 0,

deg cν < deg π, ν0 ∈ Z,|x|π = q−ν0π

where qπ = qdeg π. Additionally, for v =∞,

kv = k∞ = x =∑

ν≥ν0 cνt−ν : cν ∈ Fq, ν0 ∈ Z,

|x|∞ = q−ν0 .

Recall that the character at the infinite place was

χ∞(x∞) = ep(Tr(Res x∞))

where

ep(α) = e2πipα.

For non-infinite places, we had

k(π) = aπn, n ≥ 0, a ∈ O

χπ( aπn

) = ep(Tr(Res ( aπn

)))

with the diagram

kπ

AAAAAAAA

k(π)

CCCCCCCC Oπ

||||||||

O

0

108

Now,

ord χπ = 0ord χ∞ = −2.

Hence, if we use the Haar measure dx with∫kA/k

dx = 1

then

dχπxπ = dxπ,dχ∞x∞ = qdx∞,

and thus

dx = qΠvdxv.

We define f : kn → k such that

f(x) = xδ1 + xδ2 + ...+ xδn.

Using the same polynomial, one can define fv : knv → kv andfA : knA → kA. We will use fA and f interchangeably when thecontext is clear.

Now, in order to compute the singular series, we require a defi-nition for ϕ. Let

ϕ(x) = Πvϕv(xv),ϕv =the characteristic function of Onv .

In order for this ϕ to be admissible, we must ensure that the variousconditions for admissibility are satisfied; specifically, we check that

(A.1)A ϕ ∈ L1(knA),(A.2)A Gfϕ ∈ L1(kA)

hold, where

109

Gfϕ(ξ) =∫knAϕ(x)χ(f(x)ξ)dx.

For the first one,∫knAϕ(x) = Πv

∫knvϕv(xv)d

χvxv

=∫kn∞ϕv(x∞)dx∞Ππ

∫knπϕv(xπ)dχπxπ

=∫kn∞ϕv(x∞)qndx∞Ππ

∫knπϕv(xπ)dχπxπ

= qn.

For the second one, we can write

Gfϕ(ξ) = ΠvGfvϕv(ξv).

Now, it will aid in our computation to note that, for ξv ∈ Ov,

Gfvϕv(ξv) =∫Onπχπ(f(xπ)ξπ)dxπ = 1.

More generally, we can see that

Gfvϕv(ξv) =∫Onvχv(f(xv)ξv)dxv

= (∫Ov χv(x

δvξv)dxv)

n

= (Gxδ(ξv))n.

From Theorem 3.8.1, we know that for all but finitely many v,

|Gxδθ(ξv)| ≤√

2δ|ξv|− 1δ

v ,

or, equivalently

|Gfvϕv(ξv)| ≤ 2n2 δn|ξv|

−nδ

v .

We denote S to be the finite set of primes for which the theoremwas not proven. By definition, putting Gf = Gfϕv for simplicity,∫

kA|Gf (ξ)|dξ = Πv∈S

∫kv|Gf (ξv)|dξvΠv 6∈S

∫kv|Gf (ξv)|dξv.

Since S is a finite set, we see that the integral over kA is conver-gent by showing that the latter product on the right side converges

110

(cf. Theorem 2.5.2 (Kelvin’s Principle)). For v 6∈ S,∫kv|Gf (ξv)|dξv =

∫Ov |Gf (ξv)|dξv +

∫kv−Ov |Gf (ξv)|dξv

= 1 +∫kv−Ov |Gf (ξv)|dξv.

This means that

Πv 6∈S∫kv|Gf (ξv)|dξv = Πv 6∈S(1 +

∫kv−Ov |Gf (ξv)|dξv).

We will use the fact that, when an ≥ 0, Π(1 + an) converges ifand only if

∑an does as well. So convergence of the above is equiv-

alent to convergence of the sum∑v 6∈S∫kv−Ov |Gf (ξv)|dξv.

Now, we note from the theorem that∫kv−Ov |Gf (ξv)|dξv ≤ c

∫kv−Ov |ξv|

−nδ dξv.

We rewrite kv −Ov with

kv−Ov = (P−1v −Ov) + (P−2

v −P−1v ) + ...+ (P−νv −P

−(ν+1)v ) + ....

So ∫kv−Ov |ξv|

−nδ dξv =

∑∞ν=1 q

−nδν

v

∫P−νv −P

−(ν+1)v

dξv

=∑∞

ν=1 q−nδν

v (q−νv − q−(ν+1)v )

= (1− 1qv

)∑∞

ν=1 q(1−n

δ)ν

v

= (1− 1qv

) 1

qnδ−1

v

.

Let us absorb the 1− 1qv

’s into a constant c1. Then∑v 6∈S∫kv−Ov |Gf (ξv)|dξv ≤ c1

∑v 6∈S

1

qnδ−1

v

.

This converges if nδ− 1 > 1, which occurs if n > 2δ. This re-

striction on n is slightly weaker than the one we found in applyingthe classical circle method to rational function fields.

Now, the goal of this chapter is to show that the quantity

111

S(ν) = Gfϕ(ν) =∫kAGfϕ(ξ)χ(ξν)dξ,

derived from the methods of this chapter, is similar to the quan-tity S(ν) (p. 55), the classical singular series. Let

t = (T d, 1, 1, 1, 1, ....) ∈ k×A

where the first coordinate is the ”infinite” place and the others arethe finite ones. Additionally, we let

ρ(t) = tIn,ω(t) = tδ

where In denotes the n× n identity matrix. Then

∆(ρ(t)) = |t|nA,∆(ω(t)) = |t|δA.

With these definitions,

St(ν) =∫kAGϕt(ξ)χ(ξν)dξ

where

ϕt(x) = ϕ(t−1x),

and hence

Gϕt(ξ) =∫knAϕt(x)χ(f(x)ξ)dx.

Earlier, we showed that

St(ν) = ∆ρ(t)∆ω(t)−1S(ω(t)−1ν)= |t|n−δA S(t−δν).

In our case, |t|A = qd. Now,

P (t) = x ∈ kA : |x|v ≤ |t|v ∀ v

112

= P−d∞ × ΠπOπ,

and

L(t) = k⋂P (t) = x ∈ O = Fq[T ] : |x|∞ ≤ qd def= M(d)

where

l(t) = dimFqL(t) = d+ 1.

Let

η = t−δν = (T−δdν, ν, ν, ...).

So

η∞ = T−δdν,ηπ = ν, ∀ π.

We evaluate

S(η) = ΠvS(ηv).

In the case where v =∞,

S(η∞) =∫k∞

(∫On∞

χ∞(f(x∞)ξ∞)dχ∞x∞)χ∞(ξ∞t−dδν)dχ∞ξ∞

= qn+1∫k∞

(∫O∞ χ∞(ζγ)dζ)nχ∞(γt−dδν)dγ

= qn+1Γ0.

which means that

S(η∞) = qn+1Γ0.

Let

Γ =∫k∞

(∫O∞ χ∞(γζδ)dζ)ndγ

=∫k∞

(Gϕ∞(γ))ndγ,

and let

113

Γd,θ =∫|γ|∞<qdθ(

∫O∞ χ∞(γζδ)dζ)ndγ

for 0 < θ ≤ 12. From Theorem 2.6.1, if ch(k) > δ and n > δ2δ

θthen

|Γ− Γd,θ| ≤ c1q−ε′d.

Similarly,

|Γ0 − Γ0,d,θ| ≤ c2q−ε′′d.

So

|Γ− Γ0| ≤ |Γ− Γd,θ|+ |Γd,θ − Γ0,d,θ|+ |Γ0,d,θ − Γ0|.

The first and the third terms on the right side go to zero. Wecheck to see that the second term does as well. Evaluating this term,

Γd,θ − Γ0,d,θ =∫|γ|∞≤qdθ Gfϕ(γ)(1− χ∞(γt−dδν))dγ.

Since d >> 1, we know that γt−dδν → 0. So the integral goesto zero. Thus,

|Γ− Γ0| ≤ cq−εd,

i.e.

S(η∞)− qn+1Γ = O(q−εd).

Next, for the cases where v = π 6=∞, we claim that

(#) ΠπS(ν)π = S(ν).

We show what this claim implies. First, evaluating S(ηπ), we have

S(ηπ) =∫kπGfϕπ(ξπ)χ∞(ξπν)dξπ.

Since we assume that n > 2δ2δ > 2δ, we know not only that

114

Gfϕπ ∈ L1(kπ)

but more generally that

Gfϕ ∈ L1(kA).

Recall that in Section 2.4, we defined

S(ν) =∑

m monic

∑a∈M(µ−1),(a,m)=1(|m|−1

∞ S am

)nχ∞(ν am

)

where µ = deg m and

S am

=∑

x∈M(µ−1) χ∞( amxδ)

= Wµ−1( am

).

Additionally, we defined

Sd,θ(ν) =∑

m monic,µ≤dθ∑

a∈M(µ−1),(a,m)=1(|m|−1∞ S a

m)nχ∞(ν a

m).

We know that

|S(ν)−Sd,θ(ν)| ≤ c3q−ε′′′d.

In the case of the rational function field, the classical Hardy-LittlewoodTheorem states that, given our assumptions from before, ∃ ε suchthat

Nt(ν)

|t|n−δA= qn+1S(ν)Γ +O(q−dε)

for some ε > 0. If (#) is true,

qn+1S(ν)Γ = qn+1ΠπS(ν)πΓ∼ S(η∞)ΠπS(ν)π

if we ignore the error term of size O(q−εd). But

S(η∞)ΠπS(ν)π = S(T−dδν)∞ΠπS(ν)π = S(t−δν).

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So the classical Hardy-Littlewood Theorem can be restated as

Nt(ν)

|t|n−δA= S(t−δν) +O(q−dε).

All that remains, then, is for us to check whether (#) is true.To test this, we have

S(ν)π =∫kπGf (ξπ)χπ(ξπν)dξπ

with

Gf (ξπ) =∫Onπχ(f(xπ)ξπ)dxπ.

Now, since

Oπ ⊂ P−1π ⊂ ... ⊂ P−lπ ⊂ ... ⊂ kπ,

we can rewrite S as

S(ν)π = liml→∞∫P−lπ Gf (ξπ)χπ(ξπν)dξπ.

Note that,

P−lπ /Oπ ≈ Oπ/P lπ = O/P l = M(deg(πl)− 1),

where the last equality follows from

P−lπ =⋃a∈M(deg(πl)−1)(Oπ + a

πl).

So

S(ν)π = liml→∞∑

a∈M(deg(πl)−1)

∫Oπ+ a

πlGf (ξπ)χπ(ξπν)dξπ

= liml→∞∑

a∈M(deg(πl)−1) Gf (aπl

)χπ( aπlν).

We see that

Gf ( aπl ) = (∫Oπ χπ(xδ a

πl)dx)n.

By the equivalence between Oπ/P lπ and M(deg(πl)− 1),

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Oπ =⋃γ∈M(deg(πl)−1)(γ + P lπ).

So we change variables, letting y ∈ Oπ be such that

x = γ + πly,dx = |π|lπdy = q−lπ dy.

This means that∫Oπ χπ(xδ a

πl)dx =

∑γ∈M(deg(πl)−1)

∫γ+Plπ

χπ(xδ aπl

)dx

=∑

γ∈M(deg(πl)−1) |πl|−1π χπ(γδ a

πl).

Thus,∑a∈M(deg(πl)−1) Gf (

aπl

)χπ( aπlν) =

∑a∈M(deg(πl)−1)(

∑γ∈M(deg(πl)−1) |πl|−1

π χπ(γδ aπl

))nχπ( aπlν).

On the flip side, let us examine S. Define

S am

=∑

γ∈M(µ−1) χ∞(γδ am

),

G am

= |m|−1∞ S a

m.

Then

S(ν) =∑

m monic

∑a∈M(µ−1),(a,m)=1 G

namχ∞(ν a

m).

If we let

Sm(ν) =∑

a∈M(µ−1),(a,m)=1Gnamχ∞(ν a

m)

then

S(ν) =∑

m monic Sm(ν).

Note that

Smm′(ν) = Sm(ν)Sm′(ν).

So, for m = Πππlπ ,

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Sm(ν) = ΠπSπlπ (ν).

Therefore

S(ν) = Ππ(1 + Sπ(ν) + ...+ Sπl(ν) + ...).

But

S(ν)π = liml→∞∫P−lπ Gf (ξπ)χπ(ξπν)dξπ

= liml→∞∑

a∈M(deg(πl)−1)(G a

πl)nχπ( a

πlν)

= 1 + Sπ(ν) + ...+ Sπl(ν) + ....

Thus,

S(ν) = ΠπS(ν)π.

Q.E.D.

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