Lecture Notes in Analytic Number Theory

Home , Dirichlet character, Dirichlet convolution, Gauss sum

Lectures by Dr. Sheng-Chi Liu

Throughout these notes, signiﬁes end proof, and N signiﬁes end of example.

Table of Contents

Table of Contents i

Lecture 1 Background and Introduction 1 1.1 Basic questions ...... 1 1.2 Landau’s problems ...... 1 1.3 Notation ...... 2 1.4 Counting prime numbers ...... 3

Lecture 2 Diﬀerent proofs of the inﬁnitude of primes 3 2.1 Prime counting ...... 3

Lecture 3 Chebyshev’s idea continued 6 3.1 Primes in factorials ...... 6

Lecture 4 Lower bound of Chebyshev’s theorem 9 4.1 Chebyshev ﬁnalised ...... 9 4.2 Averages of arithmetic functions ...... 11

Lecture 5 Dirichlet convolution 12 5.1 Approximation of average functions by integrals ...... 12 5.2 Dirichlet convolution ...... 14

Lecture 6 M¨obiusfunction 15 6.1 More convoluting ...... 15

Lecture 7 M¨obiusinversion formula 16 7.1 M¨obiusinversion ...... 16 7.2 Multiplicative functions ...... 18

Lecture 8 Dirichlet series 19 8.1 Multiplicativity of M¨obiusfunction ...... 19 8.2 Dirichlet series ...... 20 Notes by Jakob Streipel. Last updated July 29, 2019.

i TABLE OF CONTENTS ii

Lecture 9 L-functions of convolutions 22 9.1 L-functions associated with Dirichlet convolutions ...... 22 9.2 Dirichlet series and multiplicative functions ...... 24

Lecture 10 Euler products 25 10.1 L-series continued ...... 25

Lecture 11 Primes in arithmetic progressions 28 11.1 Arithmetic progressions ...... 28

Lecture 12 Characters 29 12.1 Analogue of Merten’s theorem ...... 29 12.2 Characters of ﬁnite abelian groups ...... 29

Lecture 13 Fourier analysis on ﬁnite abelian groups 31 13.1 Orthogonality of characters ...... 31 13.2 Fourier analysis on ﬁnite abelian groups ...... 32 13.3 Characters on cyclic groups ...... 33 13.4 Dirichlet characters ...... 33

Lecture 14 Dirichlet characters 33 14.1 L-functions attached to Dirichlet characters ...... 33

Lecture 15 Mertens’ theorem for arithmetic progressions 36 15.1 Proof ﬁnished ...... 36 15.2 Dirichlet L-functions at 1 ...... 37

Lecture 16 Landau’s lemma 38 16.1 Landau’s lemma ...... 38

Lecture 17 Riemann’s memoir 40 17.1 Riemann’s memoir ...... 40 17.2 Review of Fourier analysis ...... 41

Lecture 18 Poisson summation 42 18.1 Fourier analysis ...... 42

Lecture 19 The functional equation for ζ(s) 45 19.1 Mellin transformation ...... 45 19.2 Gamma function ...... 46 19.3 The functional equation for ζ(s) ...... 47

Lecture 20 The functional equation for L-functions 48 20.1 Theta series ...... 48 20.2 Trivial zeros of ζ(s)...... 49 20.3 Functional equations of L-functions ...... 49 20.4 Gauss sums ...... 50

Lecture 21 The functional equation for L-functions, continued 51 21.1 Gauss sums ...... 51 TABLE OF CONTENTS iii

Lecture 22 Functions of ﬁnite order 54 22.1 Proof concluded ...... 54 22.2 The Hadamard factorisation theorem ...... 55

Lecture 23 Jensen’s formula 56 23.1 Jensen’s formula ...... 56 23.2 Application of Jacobi’s formula ...... 58

Lecture 24 Hadamard factorisation theorem 59 24.1 Hadamard factorisation ...... 59

Lecture 25 The infinite product for ζ(s) 60 25.1 Hadamard factorisation finished ...... 60 25.2 The infinite product for ζ(s) and the explicit formula ...... 61 25.3 Infinite product for L(s, χ)...... 62 25.4 Application to counting zeros of ζ(s) ...... 62

Lecture 26 Counting zeros 63 26.1 Application to counting zeros of ζ(s) ...... 63 26.2 Applications to countinjg zeros of L(s, χ) ...... 64

Lecture 27 Weil’s explicit formula 66 27.1 Application to counting zeros of L(s, χ) ...... 66 27.2 Weil’s explicit formula ...... 67

Lecture 28 Weil’s explicit formula 68 28.1 Proof continued ...... 68

Lecture 29 The theorem of Hadamard and de la Vall´ee-Poussin 70 29.1 Zero free region ...... 70

Lecture 30 Exceptional zeros 73 30.1 Zero free region for L(s, χ)...... 73

Lecture 31 Landau’s theorem 75 31.1 Exceptional zeros ...... 75 31.2 Siegel’s theorem ...... 77

Lecture 32 Siegel’s theorem 78 32.1 Proof ...... 78

Lecture 33 The Prime number theorem in Arithmetic progressions 80 33.1 Prime number theorem ...... 80

Lecture 34 Bombieri-Vinogradov 82 34.1 Prime number theorem in arithmetic progressions ...... 82 34.2 The Bombieri-Vinogradov theorem ...... 83

Lecture 35 Bombieri-Vinogradov, continued 85 35.1 The trivial case ...... 85 35.2 The nontrivial part ...... 86 TABLE OF CONTENTS iv

Lecture 36 The Large Sieve 87 36.1 The Large siece inequality ...... 87

Lecture 37 Vaughan’s Identity 90 37.1 Basic Mean Value Theorem, again ...... 90 BACKGROUND AND INTRODUCTION 1

Lecture 1 Background and Introduction

Number theory is the study of numbers, a natural starting point of which is the study of the integers Z. The integers are equipped with addition and multiplication—the opposite of addition, i.e. subtraction, doesn’t move us out of the integers, but the opposite of multiplication, being division, does. Hence another type of numbers of great interest is the rational numbers Q. On the other hand, to understand the integers one can, seemingly, settle for understanding the prime numbers ℘, since with those we can build the integers. We want to study these prime numbers which, despite being the simplest building blocks, turns out to be the hardest part of these sorts of questions.

1.1 Basic questions It is true, and has been known for a very long time, that there are inﬁnitely many primes (more on this soon). With that in mind, there are a few basic questions we will ask—and answer— in this course. 1. How many primes are there up to some number x? In other words, how does π(x) = #{ p ∈ ℘ | p ≤ x } behave as a function of x? To answer this question precisely is very hard; there is essentially no option apart from simply counting one by one. On the other hand, answering this question approximately, in a very precise sense, is entirely doable! In particular, the asymptotic behavious of this quantity is known as the Prime number theorem, which we will endeavour to prove in this course. 2. How many primes are there in a given arithmetic progression? That is to say, given a, n > 0 with (a, n) = 1, how does

π(x; a, n) = #{ ℘ 3 p ≤ x | p ≡ a (mod n) }

behave? This is a much tougher question. That π(x; a, n) is inﬁnite is Dirichlet’s theorem. This, as well as understanding how this grows asymp- totically in x, is the second goal of this course.

1.2 Landau’s problems At the 1912 International Congress of Mathematicians, Edmund Landau listed four basic problems about prime numbers. All four of these remain unsolved to this day. Listing them demonstrates how easy some of the fundamental problems in number theory are to understand, and yet how inconceivably hard they are to solve.

1. Goldbach’s conjecture: Can every even integer greater than 2 be written as the sum of two primes?

Date: January 7th, 2019. BACKGROUND AND INTRODUCTION 2

2. Twin prime conjecture: Are thjere inﬁnitely many primes p such that p+2 is prime?

3. Legendre’s conjecture: Does there always exist at least one prime between consecutive perfect squares? 4. Are there inﬁnitely many primes p such that p − 1 is a perfect square? In other words, are there inﬁnitely many primes of the form n2 + 1?

We’ll give some brief information about the current state and progress on these. The biggest step toward the Goldbach conjecture is due to Chen, who proves that for n sufficiently large, we can write 2n = p + q where q is either prime or the product of two primes. There is also a weaker version of the Goldbach conjecture, naturally known as the Weak Goldbach conjecture. This states that every odd number greater than 5 can be written as the sum of three primes. This was proved by Vinogradov and Helfgott. The Twin prime conjecture is on similar grounds as Goldbach’s conjecture: the best known is by Chen, proving that there are infinitely many primes p such that p + 2 is either a prime or a product of two primes. A related sort of problem, in a different direction, is asking about gaps between primes. In particular, are there infinitely many prime pairs with a bounded gap M, i.e. is

${ (p, q) ∈ ℘ × ℘ | 0 < |p − q| ≤ M } = ∞?

There has recently been tremendous progress in this problem. Some founda- tional results are due to Goldston–Pintz–Y1ld1r1m. Recently it was proven for M = 70 million by Yitang Zhang, and shortly thereafter a very similar result was proven by Maynard using a completely different method. The gap has since been reduced to M = 246 by the Polymath Project, led by Tao. Under the assumption of the Elliot-Halberstam conjecture, M has been reduced to 6. The closest known result to Legendre’s conjecture is due to Ingham, proving that there is a prime between n3 and (n + 1)3 for every sufficiently large n. Finally, the best result on the road to the fourth of Landau’s problems is due to Iwaniec, proving that there are infinitely many n ∈ Z such that n2 + 1 is a prime or a product of two primes.

1.3 Notation We will use following notation when discussion asymptotic behaviour over and over. Let f(x) and g(x) be two functions on R, with g(x) 6= 0 for suﬃciently large x. We write f ∼ g to mean that

f(x) lim = 1, x→∞ g(x) i.e. the values of the two functions are the same as x grows. We write f(x) = O(g(x)), or f(x) g(x), to mean that |f(x)| ≤ Cg(x) for some constant C and all x ∈ R, i.e. f is bounded by g (with some constant). DIFFERENT PROOFS OF THE INFINITUDE OF PRIMES 3

Finally we write f(x) = o(g(x)) meaning that for any ε > 0 there exists a constant N > 0 such that |f(x)| ≤ ε|g(x)| for all x ≥ N, or in other words g(x) is much bigger than f(x). We might also put it as

|f(x)| lim = 0. x→∞ |g(x)|

1.4 Counting prime numbers As claimed earlier, it is a fact that Theorem 1.4.1. There are inﬁnitely many prime numbers.

There are are many known proofs of this. The oldest among them is due to Euclid: Proof. Suppose there aren’t inﬁnitely many primes, meaning that there are ﬁnitely many of them, say p1, p2, . . . , pn. Now consider the number m = p1p2 ··· pn + 1. This number is not divisible by pi for any i, since its remainder is always 1. Hence m = p1p2 ··· pn + 1 does not have a prime factor pi, which is a contradiction. This is a very elegant proof. Indeed, it is too elegant, because it doesn’t tell us anything more!

Lecture 2 Diﬀerent proofs of the inﬁnitude of primes

2.1 Prime counting Define the prime counting function X π(x) = #{ p ∈ ℘ | p ≤ x } = 1, p≤x where when indexing a sum over p we mean that it sums only over prime numbers. The above theorem therefore says that π(x) → ∞ as x → ∞. A natural question is then at what rate this goes to infinity. Gauss (about 1792, around age 15) and Legendre (1798) gave a conjectural asymptotic formula for π(x). It was proved by Hadamard and de la Vallée- Poissin independently in 1896, known as

Theorem 2.1.1 (Prime number theorem). As x → ∞, we have x π(x) ∼ . log(x) Date: January 9th, 2019. DIFFERENT PROOFS OF THE INFINITUDE OF PRIMES 4

Remark 2.1.2. In the 1830’s, Dirichlet reformulated the (then) Prime number conjecture as Z x dt π(x) ∼ Li(x) := . 2 log(t) Note that using integration by parts,

Z x x Z x Z x dt t t x 1 Li(x) = = − − 2 dt = + 2 dt + O(1) 2 log t log(t) 2 2 t log(t) log(x) 2 log(t) x x Z x 1 x x x = + 2 + 2 dt + O(1) = + 2 + O 3 log(x) log(x) 2 log(t) log(x) log(x) log(x) and so forth, whence Li(x) ∼ x/ log(x). Remark 2.1.3. The proof of the Prime number theorem actually gives

π(x) = Li(x) + O(x exp(−Cplog(x))) for some constant C > 0. Additionally, assuming the Riemann hypothesis we get substantial power savings, namely π(x) = Li(x) + O(x1/2 log(x)2). Let us now prove the infinitude of the primes again, but this time with a very different method, courtesy of Euler. Define the (Riemann) zeta function

∞ X 1 ζ(s) = . ns n=1 As a historical anectode, this is named after Riemann despite Euler considering it before Riemann was around because Euler did it only for s ∈ R, s > 1, whereas Riemann considered it for all of C, continued it analytically and found a functional equation for it. Consider similarly for each prime p the sum

1 1 X 1 ζ (s) = 1 + + + ... = = (1 − p−s)−1 p ps p2s pαs α≥0 since it’s a geometric series, again assuming s > 1. Now by the unique factorisation of integers into primes,

Y Y −s −1 ζ(s) = ζp(s) = (1 − p ) , p p called the Euler product of ζ(s). Now by considering the sum for ζ(s) above as a Riemann sum for the integral of 1/x2, we have ∞ X 1 Z ∞ 1 1 ζ(s) = ≥ dx = , ns xs s − 1 n=1 1 whereby as s → 1+, ζ(s) → ∞. DIFFERENT PROOFS OF THE INFINITUDE OF PRIMES 5

Now if we suppose there are only finitely many primes, then Y Y lim ζ(s) = lim (1 − p−s)−1 = (1 − p−1)−1, s→1+ s→1+ p p is finite, which is a contradiction. Hence there are infinitely many primes. Now suppose we want to play this same game, but instead of counting the primes all with weight 1, count them with weight 1/p. X 1 Theorem 2.1.4 (Euler). The sum diverges. p p Proof. For s > 1, X X log ζ(s) = (1 − p−s)−1 = − log(1 − p−s). p p

Recalling the Taylor expansion of

x2 x3 log(1 − x) = −x + + + ... 2 3 (which is easy to see as the integral of 1/(1 − x) = 1 + x + x2 + ...), we have

X 1 1 log ζ(s) = + O . ps p2s p

On the other hand 1 log ζ(s) ≥ log = − log(s − 1), s − 1 whence X (p−s + O(p−2s)) ≥ − log(s − 1) p which goes to ∞ as s → 1+, so

X 1 + O(1) = ∞. p p

A more interesting question, which we will answer later, is what the asymptotic behaviour of X 1 . p p≤x Another approach is due to Chebyshev (1850),

Theorem 2.1.5. There exist constants 0 < c1 < c2 such that for x ≥ 2 x x c ≤ π(x) ≤ c . 1 log(x) 2 log(x)

To prove this we need a bit of groundwork. CHEBYSHEV’S IDEA CONTINUED 6

Definition 2.1.6 (p-adic valuation). Let p be a prime. For n ∈ Z \{ 0 } define α the p-adic valuation νp(n) to be the largest integer α ≥ 0 such that p | n (i.e. pα+1 - n). Moreover we define νp(0) = +∞. Hence by factoring into prime factors, for any n we have Y n = pνp(n). p

Note also that νp(mn) = νp(m) + νp(n) for all m, n ∈ Z. Chebyshev’s idea is based on the property that n! is divisible by all the primes less than or equal to n, and not by any other primes. We then have Y n! = pνp(n!), p≤n which when taking logarithms becomes X log(n!) = νp(n!) log p, p≤n but on the other hand n X log(n!) = log k = n log n − n + O(log n). k=1

Lecture 3 Chebyshev’s idea continued

3.1 Primes in factorials

Continuing the calculations above, we need to get a sense of how νp(n!) behaves. To this end, note that by deﬁnition

α X νp(k) = max{ α | p | k } = 1, α≥1 pα|k which means that n n X X X νp(n!) = νp(k) = 1. k=1 k=1 α≥1 pα|k By switching the order of summation we get X X X n ν (n!) = 1 = . p pα α≥1 1≤k≤n α≥1 pα|k By bxc we mean the largest integer less than or equal to x, or alternatively bxc = x − {x} where {x} is the fractional part of x. Crucial for our purposes is that bxc ≤ x, so X n n n n ν (n!) ≤ = = + O p pα p(1 − p−1) p p2 α≥1

Date: January 11th, 2019. CHEBYSHEV’S IDEA CONTINUED 7 since 1/(1 − p−1) = 1 + 1/p + O(1/p2). Hence   X X n n X log p X log p ν (n!) log p ≤ log p + O = n + O n . p p p2 p  p2  p≤n n≤p p≤n p≤n Note that the sum in the error term is bounded by the same sum over all k ≤ n instead of p ≤ n, which is the Riemann sum of a convergent integral, so the error term is O(n). Combining our estimates we therefore have X log p n log n − n + O(log n) ≤ n + O(n), p p≤n and dividing by n gives us X log p ≥ log n − 1 + O(1), p p≤n the right-hand side of which goes to ∞ as n → ∞. Hence there are inﬁnitely many primes. We’re now ready to start proving Chebyshev’s theorem: Proof. We start with the upper bound. First let us split π(x) into two sums: X X X π(x) = 1 = 1 + 1. p≤x p≤x1/2 x1/2

The ﬁrst of these sums is trivially bounded by x1/2. For the second one, note that X 1 X 1 ≤ log p log(x1/2) x1/2

2n n 2n X X log = log((2n)!) − 2 log(n!) = log k − 2 log k n k=1 k=1 = (2n) log(2n) − 2n + O(log(2n)) − 2(n log n − n + O(log n)) = (log 2)2n + O(log(2n)). CHEBYSHEV’S IDEA CONTINUED 8

On the other hand, by the Binomial theorem,

2n X 2n (1 + 1)2n = > n k=0

2n 2n 2n so that n < 2 , meaning that log n < (log 2)2n. 2n It is also true that, by the same sort of argument as we used above, n is divisible by primes in (n, 2n]. Hence

2n X 2n X log = ν log p ≥ log p = ϑ(2n) − ϑ(n) < (log 2)2n, n p n p≤2n n

0 ≤ ϑ(x) − ϑ(x/2) − (ϑ(2n) − ϑ(n)) ≤ log x, wherein we bound the parenthesis by (log 2)2n < (log 2)x, so

ϑ(x) − ϑ(x/2) ≤ (log 2)x + O(log x).

With this in mind, let us write ϑ(x) as a telescoping sum: X x x ϑ(x) = ϑ − ϑ 2k 2k+1 log x 0≤k≤b log 2 c X x ≤ (log 2) + O(log(x/2k)) 2k log x 0≤k≤b log 2 c = 2x log 2 + O((log x)2), since the last sum is geometric. Returning to π(x), with this in hand we get

X X 1 X π(x) = 1 + 1 ≤ x1/2 + log p log(x1/2) p≤x1/2 x1/2

Hence we can take c2 large enough that π(x) ≤ c2x/ log x. It also establishes that ϑ(x) ≤ cx for some c. It is important and interesting to note that this does not tighten to c2 = 1, meaning that Chebyshev’s argument is not suﬃcient to prove the Prime number theorem. LOWER BOUND OF CHEBYSHEV’S THEOREM 9

Lecture 4 Lower bound of Chebyshev’s theorem

4.1 Chebyshev ﬁnalised To get a lower bound for Chebyshev’s theorem we need a more precise estimate 2n for νp( n ). Recall how

2n n 2n X X ν = ν ((2n)!) − 2ν (n!) = ν (k) − 2 ν (k) p n p p p p k=1 k=1 X2n X n X 2n n = − 2 = − 2 pα pα pα pα α≥1 α≥1 α≥1 X n = β . pα log 2n 1≤α≤ log p

Here we mean ( 1, if {x} ∈ [1/2, 1) β(x) = b2xc − 2bxc = 0, if {x} ∈ [0, 1/2) and the range of summation can be cut oﬀ at log(2n)/ log(p) since overwise the integer parts above are 0 anyway. The trivial upper bound here is log(2n)/ log p, naturally, since β(x) ≤ 1. Now recall

2n X 2n X log 2n log = ν log p ≤ log p = (log 2n)π(2n). n p n log p p≤2n p≤2n

On the other hand,

2n n 2n X X log = log((2n)!) − 2 log(n!) = log k − 2 log k n k=1 k=1 = 2n log(2n) − 2n + O(log 2n) − 2(n log n − n + O(log n)), hence 2n π(2n) ≥ (log 2) + O(1), log n and so for x ≥ 2,

x log x π(x) ≥ (log 2) 1 + O . log x x

Ergo there exists some suﬃciently large c1 so that π(x) ≥ c1x/ log x.

Date: January 14th, 2019. LOWER BOUND OF CHEBYSHEV’S THEOREM 10

We can also extract from this a lower bound for ϑ(x): X X X ϑ(x) = log p = log p + log p p≤x p≤x1/2 x1/2

Hence there exists some suﬃciently large c0 such that ϑ(x) ≥ c0x. X log p Theorem 4.1.1 (Mertens). = log x + o(1). p p≤x This result is, maybe surprisingly much easier than the Prime number theorem, even though it would seem that counting the primes with the strange weight log p/p would be harder than weight 1. Proof. We have

n X log n! = log k = n log n − n + O(log n), k=1 but on the other hand X X n log n! = log p . pα p≤n α≥1 pα≤n

The main term here comes from α = 1, since for α ≥ 2 we have

X X n X X n X log p log p ≤ log p ≤ n = O(n) pα pα p2 p≤n α≥2 p≤n α≥2 p pα≤n pα≤n where for the last inequality we notice that the inner sum is n/(p2(1 − 1/p)) since it’s geometric. x Moreover = x + {x}, so

X n X n log p + O(n) = log p + O(1) + O(n) p p p≤n p≤n X log p X = n + O log p + O(n). p p≤n p≤n

But the error term is ϑ(n), which is order n, so this is

X log p n + O(N), p p≤n and dividing through by n gives us what we want. LOWER BOUND OF CHEBYSHEV’S THEOREM 11

4.2 Averages of arithmetic functions Deﬁnition 4.2.1. An arithmetic function is a real or complex valued function on N, i.e. f : N → R or C. Example 4.2.2. The constant function 1(n) = 1 for all n ∈ N is an arithmetic function. The delta function (at 1), ( 1, if n = 1 δ(n) = 0, if n 6= 1 is another one we’ll use. Of particular interest is the characteristic function of the primes, ( 1, if n ∈ ℘ 1℘(n) = 0, if n 6∈ ℘.

A ﬁnal one we’ll have occasion to play with is the von Mangoldt function ( log p, if n = pα, α ≥ 1 Λ(n) = 0, if n 6= pα. N

Deﬁnition 4.2.3. Let f be an arithmetic function. The average function Mf (X) of f is the function deﬁned on R≥0 by X Mf (X) = f(n). 1≤n≤X

Note that this is not quite the usual arithmetic average; for that we’d divide by X. However if we know the behaviour of one, we automatically get the behaviour of the other, so there is no need to complicate things further by that extra division. The main question we ask ourselves about such things is what, given an arithmetic function f, the behaviour of Mf (X) is as X → ∞. Example 4.2.4. Some of the functions we’re used to are averages of arithmetic functions. For instance, π(x) = M1℘ (x). Similarly ϑ(x) = Mf (x) where f(n) = log(n)1℘(n). Another function we will have reason to care about is X ψ(x) = MΛ(x) = log p. N pα≤x

All of these three are closely related in the problem of counting primes. Indeed we would, and should, suspect that ϑ and ψ are close to one another, since one is counting primes with weight log p and the other is counting prime powers with the same weight. Indeed

Proposition 4.2.5. We have ψ(x) = ϑ(x) + O(x1/2 log x). Hence by Chebyshev’s theorem, ψ(x) ϑ(x) x, by which we mean that they are the same order. DIRICHLET CONVOLUTION 12

Proof. We write X ψ(x) = ϑ(x) + log p pα≤x α≥2 So   X X  X  log p = log p  1   pα≤x p≤x1/2 pα≤x α≥2 α≥2 where the inside is bounded by log x/ log p, so

X log x ≤ log p = O(log(x)x1/2), log p p≤x1/2 where we’ve used the weaker bound; Chebyshev could give us something tighter.

Lecture 5 Dirichlet convolution

5.1 Approximation of average functions by integrals

In the special case where f : R → R, the average Mf (X) is just a restriction to N, so we want to compare X Z x f(n) ∼ f(t) dt. n≤x 1

Proposition 5.1.1 (Monotone comparison). Let f : R≥0 → R be continuous. Suppose f is monotone. Then we have

Z X Mf (X) = f(t) dt + O(|f(1)| + |f(X)|). 1 Proof. Suppose f is monotonically increasing. For n ≥ 2, we have

Z n Z n+1 f(t) dt ≤ f(n) ≤ f(t) dt. n−1 n Summing this over 2 ≤ n ≤ bXc, we are done. If f is monotonically decreasing we do precisely the same, except the in- equalities above are reversed.

Example 5.1.2. Take f(x) = log x, and this immediately gives us

X Z X log n = log t dt + O(|log 1| + |log X|) = X log X − X + O(log X) n≤x 1 with a little bit of integration by parts thrown in. N Date: January 16th, 2019. DIRICHLET CONVOLUTION 13

Theorem 5.1.3 (Integration by parts). Let g be an arithmetic function and let f be a continuously differentiable function on [1,X]. Then Z X X 0 Mfg(X) = f(n)g(n) = f(X)Mg(X) − Mg(t)f (t) dt. n≤x 1 Proof. By using integration by parts with Riemann-Stieltjes integrals, Z X X Z X X 0 f(n)g(n) = f(t) dMg(t) = f(t)Mg(t) − f (t)Mg(t) dt − − − n≤x 1 1 1 Z X 0 = f(X)Mg(X) − Mg(t)f (t) dt. 1− x Proposition 5.1.4. (i) ϑ(x) = (log x)π(x) + O . log x x (ii) π(x) ∼ if and only if ϑ(x) ∼ x if and only if ψ(x) ∼ x. log x Proof. (i) Writing X X ϑ(x) = log p = log n1℘(n) p≤x n≤x and then using the above, we get Z x ϑ(x) = (log x)π(x) − π(t)(t) dt, 1 and by Chebyshev’s theorem Z x π(t) Z x 1 t x dt ≤ C dt = O . 1 t 1 t log t log x (ii) The first equivalence follows directly from the above, and the second equivalence we have proved before, namely Proposition 4.2.5. Remark 5.1.5. Both Hadamard’s and de la Vallée-Poussin’s theorems of the Prime number theorem actually proved ψ(x) ∼ x. Corollary 5.1.6 (Euler-Maclaurin formula). Let f be a C1 function on (0, ∞). Let B1(x) = x − bxc − 1/2 = {x} − 1/2. For x > 1, we have Z X Z X X 0 Mf (X) = f(n) = f(t) dt+ B1(t)f (t) dt−B1(1)f(1)−B1(X)f(X). n≤X 1 1 In particular

Z X Z X 0 Mf (X) − f(t) dt ≤ |f (t)| dt + |f(1)| + |f(X)|. 1 1

Proof. Use summation by parts with g(n) = 1 and Mg(t) = btc. Remark 5.1.7. This formula does not require f to be monotone.

Remark 5.1.8. The function B1(x) is called the ﬁrst Bernoulli function and B1 := B1(0) is the ﬁrst Bernoulli number. DIRICHLET CONVOLUTION 14

5.2 Dirichlet convolution Deﬁnition 5.2.1 (Dirichlet convolution). Let f and g be arithmetic functions. Deﬁne the Dirichlet convolution f ? g by X n X f ? g(n) = f(d)g = f(a)g(b). d d|n ab=n

Proposition 5.2.2. Dirichlet convlution (i) is commutative, f ? g = g ? f; (ii) is associative, (f ? g) ? h = f ? (g ? h); (iii) has an identity, namely the function δ(n) which is 1 if n = 1 and 0 otherwise; (iv) sometimes has inverses. Speciﬁcally f is invertible under ?, i.e. there exists some g such that f ? g = δ, if and only if f(1) 6= 0, in which case its ?-inverse g is determined by the recurrence relation g(1) = 1/f(1),

1 X n g(n) = − f(d)g . f(1) d d|n d>1

Proof. (i) This follows directly by deﬁnition. If we switch f and g, the order of summation is reversed, but the summands are the same.

(ii) We compute X X X X (f ?g)?h(n) = (f ?g)(a)h(b) = f(c)g(d)h(b) = f(c)g(d)h(b). ab=n ab=n cd=a bcd=n Doing precisely the same thing to f ? (g ? h)(n) we get the same sum, whence the two are equal for all n and so equal as functions.

(iii) Note that f ?δ(n) is the sum of f(d)δ(n/d), but δ(n/d) = 1 onlt if n/d = 1, i.e. d = n, so f ? δ(n) = f(n) for all n.

(iv) For the forward direction, suppose f is ?-invertible. Then f ? g = δ for some g, and evaluating this at n = 1 we get f ? g(1) = δ(1) = 1. Moreover

X 1 f ? g(1) = f(d)g = f(1)g(1), d d|1 so we have f(1)g(1) = 1, which has a solution g(1) if and only if f(1) 6= 0. For the converse direction, suppose f(1) 6= 0. Let us solve the equation f ? g = δ for g. First f(1)g(1) = 1, by the above, so g(1) = 1/f(1). For n > 1, X n X n 0 = δ(n) = f ? g(n) = f(d)g = f(1)g(n) + f(d)g . d d d|n d|n d>1 Solving this for g(n) gives us the recurrence we want. MOBIUS¨ FUNCTION 15

Remark 5.2.3. Since n/d < n for d > 1, g(n) is completely determined by f(1), f(2), . . . , f(n) and g(1), . . . , g(bn/2c). Example 5.2.4. We deﬁne X d(n) := 1 ? 1(n) = 1, d|n which counts the number of (positive) divisors of n. Similarly X d3(n) := 1 ? 1 ? 1(n) = 1 abc=n is the number of representations of n as a product of 3 positive integers. In general, X dk(n) := 1 ? 1 ? ··· ? 1(n) = 1 | {z } k times a1a2···ak=n is the number of representations of n as a product of k positive integers. N

Lecture 6 M¨obiusfunction

6.1 More convoluting Proposition 6.1.1. We have log n = Λ ? 1(n), i.e. X log n = Λ(d). d|n Since Λ(n) is multiplicative, which we will explore more later, this very neatly translates log into a multiplicative function. Proof. Write n as its prime factorisation Y n = pαp , p and take logarithms, giving X X X X X log n = αp log p = log p 1 = log p = Λ(d). p p α pα|n d|n pα|n Definition 6.1.2 (Möbiusfunction). The Möbiusfunction µ is defined as the ?-inverse of 1, i.e. µ(1) = 1 and X n µ(n) = − µ . d d|n d>1 We will show later that this does in fact coincide with the definition of µ commonly found in elementary number theory texts, namely  0 if n is not square free  r µ(n) = (−1) if n = p1p2 ··· pr, pi all distinct primes 1 if n = 1. Date: January 18th, 2019. MOBIUS¨ INVERSION FORMULA 16

Lecture 7 M¨obiusinversion formula

7.1 Möbiusinversion Recall that µ is the ? inverse of 1, i.e. µ ? 1 = δ. Theorem 7.1.1 (Möbiusinversion formula). Let f and g be arithmetic functions. Then the following identities are equivalent X (i) f(n) = g(d) d|n X n (ii) g(n) = µ(d)f . d d|n Proof. Note that the first property says that f = g ? 1, and the second says g = f ? µ. With this in mind the proof is obvious: f = g ? 1 if and only if f ? µ = g ? 1 ? µ if and only if f ? µ = g ? δ = g. Example 7.1.2. In Proposition 6.1.1 we showed that X log n = Λ(d), d|n i.e. log n = Λ ? 1(n). This is equivalent to X n Λ(n) = log ?µ(n) = µ(d) log . d N d|n

We can use this to count primes: X Λ(n) Theorem 7.1.3 (Mertens). (i) = log x + O(1). n n≤x

X log p (ii) = log x + O(1). p p≤x X 1 (iii) = log log x + O(1). p p≤x Proof. We write X Λ(n) X log p X log p = . n p pα n≤x p≤x pα≤x α≥2 Now X log p X X 1 α = log p α , α p p p ≤x p≤x1/2 2≤α≤ log x α≥2 log p

Date: January 23rd, 2019. MOBIUS¨ INVERSION FORMULA 17 where the innermost sum is bounded by 1/(p2(1 − p−1)), where the geometric term is bounded, whence this is

X log p X log p = O(1). pα p2 pα≤x p≤x1/2 α≥2

Hence the first two assertions are equivalent, and we have proved the second in Theorem 4.1.1. We’ll give a direct proof of the first statement as well. Let us compute X log n n≤x in two different ways. First, we know from before that it is x log x + O(x). On the other hand, X X X X X log n = Λ(d) = Λ(d) 1 n≤x n≤x d|n d≤x n≤bx/dc   X x X X Λ(d) = Λ(d) + O Λ(d) · 1 = x + O(x), d   d d≤x d≤x d≤x since 0 ≤ x/d − bx/dc < 1 and the remainder is by Chebyshev. Combining these we see that

X Λ(x) x + O(x) = x log x + O(x), d d≤x which when dividing by x gives us

X Λ(d) = log x + O(1). d d≤x

Finally, for the last statement, let f(t) = 1/ log t and ( log p if n = p prime g(n) = p 0 otherwise.

Then Z x x Z x X 1 X 0 = f(n)g(n) = f(t) dMg(t) = f(t)Mg(t) − f (t)Mg(t) dt p − − − p≤x n≤x 2 2 2 1 Z x 1 = (log x + O(1)) + O(1) + 2 (log t + O(1)) dt log x 2− t(log t) Z x 1 Z x 1 = O(1) + dt + O 2 dt 2− t log t 2− t(log t) = log log x + O(1), since the second integral is bounded. MOBIUS¨ INVERSION FORMULA 18

7.2 Multiplicative functions Deﬁnition 7.2.1. A nonzero arithmetic function f is called multiplicative if, for (m, n) = 1, we have f(mn) = f(m)f(n). Moreover f is called completely multiplicative if f(mn) = f(m)f(n) for all m and n.

Hence if we write n = Q pαp , for a multiplicative function f, Y f(n) = f(pαp ), p so f is completely determined by its value on prime powers. Similarly, for a completely multiplicative function f, Y f(n) = f(p)αp , p meaning that f is completely determined by its value on primes. Note also that f(1) = f(1 · 1) = f(1)f(1), whence f(1) = 1 for all multiplicative functions. Hence they always have ?- inverses. In the sequel we will use the notation pα k n to mean that pα | n but pα+1 - n, i.e. pα divides n precisely. Proposition 7.2.2. Suppose f and g are multiplicative functions. Then f ? g and the ?-inverse of f are both multiplicative. Proof. Let (m, n) = 1. There is a bijection

{ d | d | mn } ←→ { (d1, d2) | d1 | m, d2 | n } by d 7→ (gcd(d, m), gcd(d, n)) and d1d2 ← (d1, d2). Under this bijection, we compute [

X mn X m n (f ? g)(mn) = f(d)g = f(d1d2)g d d1 d2 d|mn d1|m d2|n X m n = f(d1)f(d2)g g d1 d2 d1|m d2|n     X m X n =  f(d1)g   f(d2)g  d1 dd d1|m d2|n = (f ? g)(m)(f ? g)(n).

Now let h be the ?-inverse of f. First h(1) = 1/f(1) = 1, and further X n h(n) = − f(d)h . d d|n d>1 DIRICHLET SERIES 19

Let (m, n) = 1. We want to show that h(mn) = h(m)h(n), which we will prove by induction. Suppose for all (m0, n0) = 1 with m0n0 < mn we have h(m0n0) = h(m0)h(n0). Then

X mn X m n h(mn) = − f(d)h = − f(d1d2)h d d1 d2 d|mn d1|m d>1 d2|n d1d2>1 X m n = f(d1)f(d2)h h d1 d2 d1|m d2|n d1d2>1     X n X m = −  f(d1)h   f(d2)h  + f(1)f(1)h(m)h(n) d1 d2 d1|n d2|m = −(f ? h)(m)(f ? h)(n) + h(m)h(n) = h(m)h(n) since the ﬁrst two factors at the end are δ(m) and δ(n) respectively.

Lecture 8 Dirichlet series

8.1 Multiplicativity of M¨obiusfunction Remark 8.1.1. The previous proposition is not true if we replace multiplicativity with complete multiplicativity. In other words, for f and g completely multiplicative, neither f ? g nor the ?-inverse of f need be completely multiplicative (though of course by the proposition they must be multiplicative). Example 8.1.2. The constant function 1 is completely multiplicative. Thus d(n) = 1 ? 1(n) is multiplicative (but in fact not completely multiplicative). Similarly dk(n) is multiplicative. α α 0 α In particular d(p ) = α + 1 since the divisors of p are p , p1, . . . , p . Hence if n = Q pαp , then

Y αp Y d(n) = d(p ) = (αp + 1). N p p

Example 8.1.3. Since µ is the ?-inverse of 1 it is multiplicative. In particular, to compute µ(pα), let us consider µ ? 1(pα) = δ(pα), which is 0 if α ≥ 1 and 1 if α = 0. On the other hand X X µ ? 1(pα) = µ(d) = µ(pβ). d|pα 0≤β≤α

For α = 0, µ(1) = 1. For α = 1, µ(1)+µ(p) = 0, so µ(p) = −1. Furthermore for α = 2, µ(1)+µ(p)+µ(p2) = 0, but the ﬁrst two terms add to 0, so µ(p2) = 0, and in general µ(pα) = 0 for all α ≥ 2.

Date: January 25th, 2019. DIRICHLET SERIES 20

Hence if n = Q pαp ,  0 if α ≥ 2 for some p, i.e. n is not square free  p Y αp k µ(n) = µ(p ) = (−1) if n = p1p2 ··· pk, pi distinct p 1 if n = 1.

8.2 Dirichlet series Let f be an arithmetic function. The Dirichlet series associated with f is

∞ X f(n) L(s, f) = , s ∈ . ns C n=1 Remark 8.2.1. This series might not converse. We need some control of the size of f. Deﬁnition 8.2.2. An arithmetic function f is of polynomial growth if it satisﬁes one of the following equivalent conditions:

(i) There exists a constant A ∈ R, depending on f, such that |f(n)| = O(nA). (ii) There exists σ ∈ R such that the series L(σ, f) is absolutely convergent. In this case we let

σf := inf{ σ ∈ R | L(σ, f) converges absolutely } ∈ R ∪ { −∞ }.

We call σf the abscissa of convergence of L(σ, f). That (i) and (ii) are equivalent is pretty clear; in the forward direction our P nA series looks like ns , so it converges absolutely for Re s > 1 + A, and we can make a similar argument backwards. Hence |f(n)| = O(nσf −1). Proposition 8.2.3. Let f be an arithmetic function with polynomial growth. Let σf be its abscissa of convergence. Then for all σ > σf , the series L(s, f) converges absolutely and uniformly in the halfplane { s ∈ C | Re(s) > σ }. In this domain, X − log(n)f(n) L0(s, f) = = L(s, − log ·f), ns n≥1 which also has abscissa of convergence σf . s σ Proof. Let σ > σf , and consider Re(s) ≥ σ. Then |f(n)/n | ≤ |f(n)|/n . Summing this over n ≥ 1,

X f(n) X |f(n)| ≤ < ∞ ns nσ n≥1 n≥1 since L(σ, f) converges absolutely by deﬁnition since σ > σf . This bound is uniform for any Re(s) ≥ σ. Finally by the absolute convergence we can switch the limit process in the series and the derivative, and hence diﬀerentiate termwise. DIRICHLET SERIES 21

An arithmetic function f determines its Dirichlet series uniquely, and a converse result is true as well.

Lemma 8.2.4. Let f and g be two arithmetic functions of polynomial growth. Suppose L(s, f) = L(s, g) for all s in

{ s ∈ C | Re(s) > max{ σf , σg }}. Then f = g.

Hence the function is uniquely determined by its Dirichlet series. Proof. Without loss of generality we can assume g = 0 (by replacing f by f −g, i.e. move L(s, g) to the other side). So we have L(s, f) = L(s, g) = 0 for all σ −1 s ∈ { s ∈ C | Re(s) > σf } (since σg = −∞). Let h(n) = f(n)/n f . Then

X f(n) X f(n)/nσf −1 L(s, f) = = = L(s − σf + 1, h). ns ns−σf +1 n≥1 n≥1

Hence L(s, h) = L(s + σf − 1, f). 1−1 For Re(s) > 1, Re(s + σf − 1) > σf , whence σh ≤ 1, so h(n) = O(n ) = O(1). Now suppose h(n) 6= 0 for some n, and consequently let n0 be the smallest such n. For Re(s) > 2,   s X h(n) h(n0) n0 X h(n) 0 = s = s 1 + s  . n n0 h(n0) n n≥n0 n≥n0+1

Note that both of the h(n0) terms are nonzero by choice of n0, so

s n0 X h(n) 1 + s = 0. h(n0) n n≥n0+1

But X h(n) X 1 Z ∞ 1 1 dt = . s Re(s) Re(s) Re(s)−1 n n n0+1 t (Re(s) − 1)(n0 + 1) n≥n0+1 n≥n0+1

Hence

Re(s) ! ns X h(n) n 1 1 + 0 = 1 + O 0 = 0 s Re(s)−1 h(n0) n h(n0) (Re(s) − 1)(n0 + 1) n≥n0+1 so 1 1 + O = 0 Re(s) − 1 which is a contradiction as Re(s) → ∞ since the remainder term vanishes. L-FUNCTIONS OF CONVOLUTIONS 22

Lecture 9 L-functions of convolutions

9.1 L-functions associated with Dirichlet convolutions Theorem 9.1.1. Let f and g be arithmetic functions of polynomial growth with abscissae of convergence σf and σg respectively. Then σf?g ≤ max{ σf , σg }. Moreover for Re(s) = σ > max{ σf , σg } we have

L(s, f)L(s, g) = L(s, f ? g).

Proof. For Re(s) > max{ σf , σg }, P ∞ | f(a)g(b)| ∞ X f ? g(n) X ab=n X X |f(a)||g(b)| = ≤ ns nRe(s) |ab|Re(s) n≥1 n=1 n=1 ab=n ∞ ∞ ! ∞ ! X X |f(a)||g(b)| X |f(a)| X |g(b)| = = < ∞ |a|Re(s)|b|Re(s) |a|Re(s) |b|Re(s) 1=1 ab=n a=1 b=1 so σf?g ≤ max{ σf , σg }. For Re(s) > max{ σf , σg } we have absolute convergence, so we can rearrange terms in the series as desired, so

∞ ∞ ∞ X f(m) X g(k) X 1 X L(s, f)L(s, g) = = f(m)g(k) ms ks ns m=1 k=1 n=1 mk=n ∞ X f ? g(n) = = L(s, f ? g). ns n=1 Corollary 9.1.2. Suppose f is ?-invertible. Let g be the ?-inverse of f. Assume σf , σg < ∞. Then for Re(s) > max{ σf , σg }, we have L(s, f)L(s, g) = 1. Proof. By the last theorem, since f?g = δ, which is supported only on n = 1. In particular L(s, f) does not vanis on

{ s ∈ C | Re(s) > max{ σf , σg }}. Example 9.1.3. For f = 1, we have

∞ X 1 L(s, 1) = = ζ(s), ns n=1 converging on Re(s) > 1, so σ1 = 1. The M¨obiusfunction µ is the ?-inverse of 1, so ∞ X µ(n) L(s, 1)L(s, µ) = ζ(s) = 1. ns n=1 0 We have |µ(n)| ≤ 1 = O(n ), so σµ ≤ 1.

Date: January 28th, 2019. L-FUNCTIONS OF CONVOLUTIONS 23

In fact it is equal to 1; let s = 1, so

∞ X |µ(n)| X 1 ≥ = ∞ ns p n=1 p so σµ = 1. Hence L(s, 1)L(s, µ) = 1 for Re(s) > 1, and ζ(s) 6= 0 for Re(s) > 1. Also

∞ 1 X µ(n) = L(s, µ) = ζ(s) ns n=1 for Re(s) > 1. N Example 9.1.4. Let d(n) = 1 ? 1(n). Then

L(s, d) = L(s, 1 ? 1) = L(s, 1)L(s, 1) = ζ(s)2 for Re(s) > 1. In general ∞ X dk(n) ζ(s)k = ns n=1 for Re(s) > 1. N Example 9.1.5. Euler’s ϕ-function is deﬁned by

ϕ(n) = #{ 1 ≤ m ≤ n | (m, n) = 1 }.

This is multiplicative (which follows immediately from the Chinese remainder theorem), and ϕ(n) = µ ? Id(n), where Id(n) = n. It’s easy to check this on prime powers. Now ∞ X n L(s, Id) = = ζ(s − 1), ns n=1 so σId = 2, and ζ(s − 1) L(s, ϕ) = L(s, µ)L(s, Id) = ζ(s) for Re(s) > max{ σµ, σId } = max{ 1, 2 } = 2. Hence ∞ ζ(s − 1) X ϕ(n) = ζ(s) ns n=1 for Re(s) > 2. N Example 9.1.6. The von Mangoldt function Λ(n) = log ?µ(n), so

L(s, Λ) = L(s, log)L(s, µ).

Now ∞ X 1 L(s, 1) = ζ(s) = , ns n=1 L-FUNCTIONS OF CONVOLUTIONS 24 and so long as Re(s) > 1, so that we have absolute convergence, we can diﬀer- entiate termwise, whence ∞ X − log n ζ0(s) = ns n=1 meaning that L(s, log) = −ζ0(s). Then

∞ −ζ0(s) X Λ(n) L(s, Λ) = = ζ(s) ns n=1 for Re(s) > 1. Note also that the fraction is a fraction of holomorphic functions, the bottom nonvanishing, so it is holomorphic and has an antiderivative, so

∞ X Λ(n) −(log ζ(s))0 = ns n=1 for Re(s) > 1. N

9.2 Dirichlet series and multiplicative functions Theorem 9.2.1. Let f be a multiplicative function of polynomial growth. Then for Re(s) = σ > σf , we have (i) For all primes p, X f(pα) L (s, f) := p pαs α≥0 converges absolutely and uniformly in Re(s) ≥ σ.

This Lp(s, f) is called the local factor of f at p. (ii) We have Y L(s, f) = Lp(s, f), p which is an inﬁnite product, so technically we mean Y lim Lp(s, f), P →∞ p≤P

which converges uniformly in Re(s) ≥ σ.

(iii) Conversely, let f be an arithmetic funciton f such that σf < ∞ and f(1) = 1. Suppose Y L(s, f) = Lp(s, f) p for Re(s) suﬃciently large. Then f is multiplicative. Proof. (i) This is trivial: the local factors are subseries of an absolutely and uniformly convergent series. EULER PRODUCTS 25

(ii) The intuition is that we can factor n into primes, and collecting them as appropriate, the two sides are equal. More precisely, for the convergence, let P ≥ 2 and write p1 < p2 < . . . < pk ≤ P , the set of primes less than or equal to P . Then α α Y X f(p 1 ) ··· f(p k ) L (s, f) = 1 k p α1 αk s (p1 ··· pk ) p≤P α1,...,αk≥0 α α X f(p 1 ··· p k ) X f(n) = 1 k = α1 αk s s (p1 ··· pk ) n α1,...,αk≥0 n≥1 p|n⇒p≤P

Hence Y X |f(n)| L(s, f) − L (s, f) ≤ → 0 p ns p≤P n>P as P → ∞ since the left-hand side becomes the sum over n with all prime factors exceeding P , which is a smaller set than n > P , and the right-hand side is the tail of a convergent sum.

Lecture 10 Euler products

10.1 L-series continued

Proof continued. (iii) Let f˜ be the multiplicative function deﬁned by Y fˆ(n) = f(pα). pαkn

σ For σ > σf , |f(n)|/n = o(1), so for n large enough, say n > N, we have |f(n)|/nσ < 1. Hence

|f˜(n)| Y |f(pα)| Y |f(pα)| Y |f(pα)| = = = O(1) nσ pασ pασ pασ pαkn pαkn pαkn pα≤N pα>N with the implied constant depending on f, since the ﬁrst product is a ﬁnite product, hence a constant, and the second product is o(1) by choice of N. Therefore L(s, f˜) converges absolutely for Re(s) > σ + 1. For such s,

Y Y X pα Y L(s, f˜) = L (s, f˜) = = L (s, f) = L(s, f) p pαs p p p α≥0 p so for Re(s) suﬃciently large, the L-functions are equal, whence by Lemma 8.2.4 f˜ = f, and so f is multiplicative since f˜ is multiplicative by construction. Date: January 30th, 2019. EULER PRODUCTS 26

Corollary 10.1.1. If f is completely multiplicative with σf < ∞, then for Re(s) > σf , −1 Y f(p) L(s, f) = 1 − . ps p Proof. This follows immediately from f(pα) = f(p)α when f is completely multiplicative, so

−1 X f(pα) X f(p)α f(p) L (s, f) = = = 1 − , p pαs psα ps α≥0 α≥0 the last sum being geometric. We call such a product representation an Euler product.

Proposition 10.1.2. Let f be multiplicative with σf < ∞. Let σ > σf . Then there exists P > 0 such that

Y L(s, f) Lp(s, f) = Q Lp(s, f) p>P p≤P does not vanish in Re(s) ≥ σ. Thus the zeros of L(s, f) in Re(s) ≥ σ are exactly the zeros of the local factors Lp(s, f), p ≤ P .

Proof. Let σ > σf . By the previous Theorem 9.2.1, Y L(s, f) = Lp(s, f) p for Re(s) ≥ σ. The right-hand side is interpreted as a convergent limit, so there exists some P > 0 such that for p > P , 1 |L (s, f) − 1| < p 2 for Re(s) ≥ σ. Hence Lp(s, f) 6= 0 for p > P , Re(s) ≥ σ, since it is bounded away from zero. Example 10.1.3. For Re(s) > 1,

∞ −1 X 1 Y 1 ζ(s) = = 1 − , ns ps n=1 p which is never 0, so ζ(s) 6= 0 for Re(s) > 1. Correspondingly

1 Y 1 = 1 − ζ(s) ps p is also nonvanishing in Re(s) > 1. N EULER PRODUCTS 27

Example 10.1.4. For Re(s) > 1,

∞ −2 X d(n) Y 1 L(s, d) = = ζ(s)2 = 1 − . ns ps n=1 p

Let f be the ?-inverse of d. Then

2 1 Y 1 Y 2 1 L(s, f) = = 1 − = 1 − + . ζ(s)2 ps ps p2s N p p

Example 10.1.5. Let ϕ be the Euler ϕ-function. We have

ζ(s − 1) Y 1 − p−s L(s, ϕ) = = ζ(s) 1 − p1−s p for Re(s) > 2. N Let us now explore a slightly more esoteric example, but that serves a point in ﬁnding zeros between σf and σg, where f and g are each other’s ?-inverses.

Example 10.1.6. Let µs be the multiplicative function f deﬁned by  1, if α = 0  α 0, if α ≥ 2 µ2(p ) = −1, if α = 1, p 6= 2  −4, if α = 1, p = 2.

In other words, it is the usual M¨obiusfunction, except we’ve modiﬁed it specifically on n = 2. Note that |µ2(n)| ≤ 4, so L(s, µ2) converges absolutely for

Re(s) > 1; σµ2 ≤ 1. Moreover at s = 1,

∞ X |µ2(n)| X 1 |L(1, µ )| = ≥ = ∞, 2 ns p n=1 p6=2

so in fact σµ2 = 1. Now a natural question to ask is what the zeros of L(s, µ2) are for Re(s) > 1. We compute the local factors

α X µ2(2 ) 4 L (s, µ ) = = 1 − , 2 2 2αs 2s α≥0 and for p 6= 2, α X µ2(p ) 1 L (s, µ ) = = 1 − , p 2 pαs ps α≥0 meaning that 4 Y 1 L(s, µ ) = 1 − 1 − , 2 2s ps p6=2 where the ﬁrst factor is 0 if s = 2, and the remaining factors are never zero for Re(s) > 1. Hence the only zero of L(s, µ2) on Re(s) > 1 is s = 2. PRIMES IN ARITHMETIC PROGRESSIONS 28

(−1) Now let µ2 be the ?-inverse of µ2. It is multiplicative, since µ2 is, and ( 1, if p 6= 2 µ(−1)(pα) = 2 4α, if p = 2.

Let σ > 2. For Re(s) > σ,

(−1) −1 −1 σ (−1) X |µ (n)| 4 Y 1 1 − 1/2 |L(s, µ )| ≤ 2 = 1 − 1 − = ζ(σ). 2 nσ 2s ps 1 − 4/2σ n≥1 p6=2 We chose σ > 2 because (−1) α α α (−1) X µ (2 ) X 4 X 4 L (s, µ ) = 2 = = , 2 2 2αs 2sα 2s α≥0 α≥0 α≥0 which is geometric and converges for Re(s) < 2, so σ (−1) ≤ 2, and moreover µ2 (−1) α X |µ (n)| X 4 2 ≥ = ∞, n2 22 n≥1 α≥0 so σ (−1) = 2. µ2 (−1) Hence L(s, µ2)L(s, µ2 ) = 1 for Re(s) > 2, and between σµ2 = 1 and σ (−1) = 2 we ﬁnd the one zero at s = 2. µ2 N

Lecture 11 Primes in arithmetic progressions

11.1 Arithmetic progressions

Deﬁnition 11.1.1. An arithmetic progression is an inﬁnite subset of Z satisfying the property that there exists an integer q > 0 such that the distance between any two consecutive integers of this subset is q. This integer q is called the modulus of the arithmetic progression. Hence an arithmetic progression of modulus q is of the form

Lq,a = a + qZ, with a ∈ Z.

Note that if a ≡ b (mod q), then Lq,a = Lq,b. Therefore the arithmetic progressions of modulus q are indexed by the residue classes modulo q, i.e. Z/qZ. Theorem 11.1.2 (Dirichlet). Let a, q ∈ Z, q > 0, with (a, q) = 1. Then the set ℘q,a = ℘ ∩ Lq,a is infinite, i.e. there exist infinitely many primes p such that p ≡ a (mod q). Remark 11.1.3. The condition (a, q) = 1 is necessary. Indeed if (a, q) > 1, then there exists at most one prime p ≡ a (mod q), and such a prime p = (a, q). This is easy to see: everywhing in the arithmetic progression is of the form a + kq, but out of all of them we can factor (a, q) by definition. Hence the residue classes containing infinitely many primes are (Z/qZ)×. Date: February 1st, 2019. CHARACTERS 29

As with the inﬁnitude of the primes leading to us wondering about the Prime number theorem, a natural question to ask here is this: what is the density of the set ℘q,a? In the same vein, let

π(x; q, a) = #(℘q,a ∩ [1, x]) = #{ p ≤ x | p ≡ a (mod q) }.

Theorem 11.1.4 (Landau). Let a, q > 0 with (a, q) = 1. Then

1 π(x) X x π(x; q, a) = π(x)(1 + o(1)) ∼ ϕ(q) ϕ(q) ϕ(q) log x as x → ∞, the last being the Prime number theorem.

Now in particular, by deﬁnition, |(Z/qZ)×| = ϕ(q), so the primes are equidis- tributed amongst the residue classes modulo q.

Lecture 12 Characters

12.1 Analogue of Merten’s theorem Theorem 12.1.1. Let a, q > 0, (a, q) = 1. Then we have

X Λ(n) 1 (i) = log x + O(1). n ϕ(q) n≤x n≡a (mod q)

X log p 1 (ii) = log x + O(1). p ϕ(q) p≤x p≡a (mod q)

X 1 1 (iii) = log log x + O(1). p ϕ(q) p≤x p≡a (mod q) We will prove (i) later. Moreover (i) implies (ii) implies (iii) in exactly the same way as in Merten’s theorem. Remark 12.1.2. Note that this theorem implies Dirichlet’s theorem, since for weighted sums over the primes in arithmetic progressions to diverge, there must be inﬁnitely many primes in them.

12.2 Characters of ﬁnite abelian groups Let G be a ﬁnite abelian group with identity denoted by 1.

Deﬁnition 12.2.1. A character χ of G is a group homomorphism χ: G → C∗. Remark 12.2.2. Such a character χ has the following properties: (i) χ(ab) = χ(a)χ(b) for every a, b ∈ G since it is a homomorphism. (ii) χ(1) = 1.

Date: February 4th, 2019. CHARACTERS 30

(iii) χ(a)m = χ(am), which in turn is χ(1) = 1 if m = |G|. Hence every χ(a) is an mth root of unity, and |χ(a)| = 1. Thus χ: G → S1 ⊂ C, the unit circle. This also means that χ(a) = χ(a−1). Let Gˆ denote the set of all characters of G. Then Gˆ is a group with multi- plcation defined by χ1 · χ2(a) = χ1(a)χ2(a). We call Gˆ the dual group of G, and its identity element is the trivial character χ0(a) = 1 for all a ∈ G. Proposition 12.2.3. Let H be a subgroup of the finite abelian group G. Then every character of H extends to a character of G. Proof. We prove it by induction on the index [G : H] = |G/H|. If [G : H] = 1, then G = H and we are done. Assume that it is true for every subgroup H0 of G with [G : H0] < [G : H]. Take g ∈ G \ H, and let n be the smallest positive integer such that gn ∈ H (which exists since gn = 1 in G/H for some n). Then we have our character χ: H → C, but what should we define χ(g) to be? That’s hard to say, but certainly χ(gn) = t ∈ C is well-defined, since gn ∈ H. Let H0 = hH, gi. Now wn = t for some w (actually n options), so pick one of them and define χ0 : H0 → C∗ by χ0(h) = χ(h) for h ∈ H, and χ0(g) = w. For each h0 ∈ H0, we have h0 = hgk for some h ∈ H and k ∈ Z, so χ0(h0) = χ(h)wk. This is well-defined, and clearly it is a character on H0 and restricts to χ on H. Now H < H0 < G, with the first subgroup strict, so [G : H0] < [G : H], so by our inductive hypothesis, χ0 extends to G.

Remark 12.2.4. The restriction ρ: Gˆ → Hˆ deﬁned by χ 7→ χ|H is a homomorphism. The proposition implies that ρ is surjective. It’s kernel is, by deﬁnition,

ker ρ = { χ ∈ Gˆ | χ|H = χ0 }, which is the same as G/H[ , so we have the short exact sequence

ρ { 1 } G/H[ Gˆ Hˆ { 1 }.

Proposition 12.2.5. Let G be a ﬁnite abelian group. Then |Gˆ| = |G|. Proof. If G = hgi is cyclic, then χ ∈ Gˆ is determined by χ(g). Since χ(g) is a |G|th root of unity, and each |G|th root of unity determines a character χ, |Gˆ| = |G|. Suppose G is not cyclic. Then there exists a nontrivial cyclic subgroup H of G, and as above we have the short exact sequence

ρ { 1 } G/H[ Gˆ Hˆ { 1 }.

Hence |Gˆ| = |G/H[ | · |Hˆ |. Now suppose the proposition is true for all ﬁnite abelian groups of order less than n = |G|. Then

|Gˆ| = |G/H[ | · |Hˆ | = |G/H| · |H| = |G|, FOURIER ANALYSIS ON FINITE ABELIAN GROUPS 31 where in the penultimate step we use the inductive hypothesis for G/H and our argument about H being cyclic for the second term. Remark 12.2.6. One can show that there is an isomorphism between G and Gˆ, however it is not canonical—it depends on the choice of generators. That said, if G is isomorphic to its dual Gˆ, then similarly Gˆ must be isomor- ˆ phic to its dual Gˆ, and it turns out there is a canonical isomorphism between that and G.

Lecture 13 Fourier analysis on ﬁnite abelian groups

13.1 Orthogonality of characters

Let G be a ﬁnite abelian group. Let g ∈ G. Then g deﬁnes a character of Gˆ by

∗ gˆ: Gˆ → C ˆ by gˆ(χ) = χ(g). So we have a homomorphism ε: G → Gˆ defined by g 7→ gˆ. ˆ Proposition 13.1.1. The homomorphism ε: G → Gˆ is an isomorphism. ˆ Proof. We have |G| = |Gˆ| = |Gˆ|. Hence it suffices to show that ε is injective, i.e. if g 6= 1, then gˆ 6= 1, so gˆ is not the trivial character on Gˆ. In other words, there exists some χ ∈ Gˆ such that gˆ(χ) 6= 1, i.e. χ(g) 6= 1. Now let H = hgi ⊂ G. Define χ: H → C∗ by χ(g) = ξ 6= 1, with ξ an |H|th root of unity. Now extending χ to G finishes the proof. Theorem 13.1.2. Let G be a finite abelian group of order n. Then ( X n if χ = χ0 (i) χ(g) = , 0 if χ 6= χ g∈G 0 ( X n if g = 1 (ii) χ(g) = . 0 if g 6= 1 χ∈Gˆ

Proof. For the ﬁrst one, if χ = χ0, then X X χ0(g) = 1 = n. g∈G g∈G

If χ 6= χ0, then there exists some a ∈ G with χ(a) 6= 1. Multiplying the sum by χ(a) we have X X X X χ(a) χ(g) = χ(a)χ(g) = χ(ag) = χ(g) g∈G g∈G g∈G g∈G since χ is a homomorphism, and moreover ag, as g runs over G, is just permuting the arguments. Hence X (χ(a) − 1) χ(g) = 0, g∈G

Date: February 6th, 2019. FOURIER ANALYSIS ON FINITE ABELIAN GROUPS 32 but by choice of a the parenthesis is nonzero, so the sum is 0. For the second orthogonality relation, note that X X χ(g) = gˆ(χ), χ∈Gˆ χ∈Gˆ so it reduces to the ﬁrst case.

13.2 Fourier analysis on ﬁnite abelian groups

Let G be a ﬁnite abelian group. Let C(G) = { f : G → C }, which is a vector space over C. A basis for C(G) is given by B = { δg | g ∈ G } where ( 1 if a = g, δg(a) = 0 if a 6= g.

Hence dimC C(G) = |G|, and we can write any f ∈ C(G) as X f(a) = f(g)δg(a). g∈G We can also deﬁne the Hermitian inner product h·, ·i on C(G) by 1 X hf, hi = f(g)h(g). |G| g∈G Then B is an orthogonal basis since ( 1 X 1 if g = a hδ , δ i = δ (h)δ (h) = |G| g a |G| g a h∈G 0 if g 6= a. Remark 13.2.1. All of these properties hold for any ﬁnite set G. In other words, we have not used the group structure of G.

In order to instead take advantage of the group structure of G, let χ1, χ2 ∈ Gˆ. Then ( 1 X 1 if χ1 = χ2 hχ1, χ2i = χ1(g)χ2(g) = |G| 0 if χ 6= χ g∈G 1 2 by the orthogonality relation, since χ1(g)χ2(g) = (χ1χ2)(g) is another character, which is trivial if and only if they’re the same. ˆ ˆ Now |G| = |G| = dimC C(G) so G is an orthonormal set spanning C(G), and hence an orthonormal basis. This means that we also have a Fourier transform: for f ∈ C(G), fˆ: Gˆ → C∗ by 1 X fˆ(χ) = hf, χi = f(g)χ(g). |G| g∈G By orthogonality we then have the Fourier expansion X f(g) = fˆ(χ)χ(g). χ∈Gˆ DIRICHLET CHARACTERS 33

Moreover we have the Parseval-Plancherel formula 1 kfk2 = kfˆk2. |G|

13.3 Characters on cyclic groups

Let G = hgi be a ﬁnite cyclic group of order n. Note that a character χ ∈ Gˆ is completely determined by χ(g); i.e., χ(g) = ξ is an nth root of unity and each nth root of unity determined a character.

Proposition 13.3.1. Let G = hgi be a ﬁnite cyclic group of order n. Let µn be the set of all nth roots of unity. Then ϕ: µn → Gˆ deﬁned by ξ 7→ χξ, where χξ(g), is an isomorphism.

Example 13.3.2. A ﬁnite cyclic group G of order n is isomorphic to Z/nZ = h1i, seen as an additive group. The characters of G are ψm where

2πim/n ψm(1) = e , for m = 0, 1, 2, . . . , n − 1, noting that µn is a cyclic group of order n generated 2πi/n by ξ = e . N

13.4 Dirichlet characters Deﬁnition 13.4.1. Let q ≥ 1 be an integer. The characters of the abelian group (Z/qZ)× are called Dirichlet characters of modulus q. The trivial × ∗ character is denoted by ξ0. This deﬁnes χ:(Z/qZ) → C ; we extend it to Z/qZ and then to Z by taking χ(a) = 0 if (a, q) 6= 1. In other words ∗ χ: Z → Z/qZ → C by ( χ(n mod q) if (n, q) = 1 χ(n) = 0 if (n, q) 6= 1.

Note two things: there are precisely ϕ(q) Dirichlet characters to the modulus q, and for the record (Z/qZ)× is not usually cyclic: it is cyclic only when q is 2, 4, pk for p 6= 2, or 2pk (this is when there exist primitive roots modulo q).

Lecture 14 Dirichlet characters

14.1 L-functions attached to Dirichlet characters Remark 14.1.1. Dirichlet characters χ are completely multiplicative, i.e., χ(mn) = χ(m)χ(n) for all m, n ∈ Z. Hence the Dirichlet series associated with χ has an Euler product ∞ −1 X χ(n) Y χ(p) L(s, χ) = = 1 − ns ps n=1 p for Re(s) > 1 since |χ(n)| ≤ 1. Date: February 8th, 2019. DIRICHLET CHARACTERS 34

Proposition 14.1.2. Let χ (mod q) be a Dirichlet character.

(i) If χ = χ0, we have for Re(s) > 1,

Y 1 L(s, χ ) = 1 − ζ(s) 0 ps p|q

which has an analytic continuation to Re(s) > 0 with a simple pole at s = 1.

(ii) If χ 6= χ0, then L(s, χ) converges uniformly on compact subsets in Re(s) > 0 and thus it is a holomorphic function in Re(s) > 0. Moreover we have for Re(s) = σ > 0 and X ≥ 2,

X χ(n) q|s| L(s, χ) = + O x−σ , ns σ n≤X with the convergence here being conditional.

Proof. (i) If χ = χ0, then −1 Y χ0(p) L(s, χ ) = 1 − , 0 ps p where χ0(p) is 1 on primes (p, q) = 1 and otherwise 0, so

−1 −1 Y 1 Y 1 Y 1 L(s, χ ) = 1 − = 1 − 1 − , 0 ps ps ps (p,q)=1 p|q p where the last factor is ζ(s). The first product is finite and defined for all s ∈ C. Hence the analytic continuation and the simple pole follows from same of ζ(s).

(ii) If χ 6= χ0, by orthogonality X χ(n) = 0, a≤n 0, the ﬁrst term is

Mχ(X) q ≤ → 0 Xs Xσ DIRICHLET CHARACTERS 35 as X → ∞. For the integral, consider the tail Z ∞ |s|q M (t)t−s−1 dt ≤ → 0 χ σ X σX as X → ∞. Hence L(s, χ) converges uniformly on compact subsets in Re(s) > 0. Remark 14.1.3. Note that this is essentially the proof of Dirichlet’s test for convergence; if

X an < M n≤N uniformly, and bn decreases monotonically to 0, then X anbn n≥0 converges. With this we are ﬁnally equipped to prove Merten’s theorem for arithmetic progressions. Proof. We want to prove X Λ(n) 1 = log X + O(1). n ϕ(q) n≤X n≡a (mod q) Now to ﬁlter the terms n ≡ a (mod q) out of the sum over all n, we average the sum over all characters χ (mod q), since ( X 0 if b 6≡ 1 (mod q) χ(b) = ϕ(q) if b ≡ 1 (mod q). χ (mod q) Hence X Λ(n) X Λ(n) 1 X = χ(a)χ(n), n n ϕ(q) n≤X n≤X χ (mod q) n≡a (mod q) since the latter sum is ϕ(q) if a ≡ n (mod q), and 0 otherwise. We rewrite this sum as 1 X X Λ(n)χ(n) 1 X χ(a) = χ(a)S (X). ϕ(q) n ϕ(q) χ χ (mod q) n≤X χ (mod q)

We split this into two sums, one for χ0 and one for χ 6= χ0. If χ = χ0,

X Λ(n)χ0(n) X Λ(n) X Λ(n) X X 1 S (X) = = = − log p . χ0 n n n pα n≤X n≤X n≤X p|q α≥1 (n,q)=1 pα≤X Looking carefully at the last sum, the innermost sum is geometric and bounded by 1/(p(1 − 1/p)) ≤ 1, and with that bound in mind the sum over p | q becomes bounded by log q by bringing the sum inside the logarithm as a product. Hence

Sχ0 (X) = log X + O(1) + O(log q) = log X + Oq(1) MERTENS’ THEOREM FOR ARITHMETIC PROGRESSIONS 36 by Merten’s regular theorem. Now it remains to show that for χ 6= χ0,

X Λ(n)χ(n) S (X) = = O (1). χ n q n≤X

Lecture 15 Mertens’ theorem for arithmetic progressions

15.1 Proof ﬁnished

Proof continued. Carrying on from last time, we wish to show that for χ 6= χ0, Sχ(X) = Oq(1). To this end, consider

X χ(n) T (X) = (log n) . χ n n≤X

By Dirichlet’s test for convergence, this is Oq(1). Recall that X X log n = Λ(d) = Λ(a). d|n ab=n This means X X χ(n) X Λ(a)χ(a) X χ(b) T (X) = Λ(n) = . χ n a b n≤X ab=n a≤X b≤X/a

Looking at these carefully, we recognise the ﬁrst sum as Sχ(X) and the inner sum is L(1, χ) + O(qa/X) by Proposition 14.1.2. Hence the above is   X Λ(a)χ(a) qa T (X) = S (X)L(1, χ) + O , χ χ  a X  a≤x and the error term is q X q Λ(a) X = q X X a≤X where the second step is Chebyshev’s theorem. Hence Tχ(X) = Sχ(X)L(1, χ) + Oq(1), where the left-hand side is Oq(1) too, so if L(1, χ) 6= 0, then Sχ(X) = Oq(1). Date: February 13th, 2019. MERTENS’ THEOREM FOR ARITHMETIC PROGRESSIONS 37

15.2 Dirichlet L-functions at 1 This is one of Dirichlet’s results: Theorem 15.2.1 (Dirichlet). Let χ (mod q) be a nontrivial Dirichlet character. Then L(1, χ) 6= 0. There are essentially two proofs of this, one using real analysis and one using complex analysis. We will discuss the second here. Proof. Consider Y Y Lq(s) = L(s, χ) = L(s, χ0) L(s, χ).

χ (mod q) χ6=χ0 Note that this is a ﬁnite product since there are ﬁnitely many Dirichlet characters. If we enumerate those Dirichlet characters as χ0, χ1, . . . , χϕ(q)−1, we have in other words Lq(s) = L(s, χ0 ? χ1 ? ··· ? χϕ(q)−1), so ∞ X aq(n) L (s) = q ns n=1 with aq(n) = χ0 ?χ1 ?···?χϕ(q)−1(n), with abscissa of convergence σq ≤ 1 since it is bounded by the maximum of σχ for all χ, all of which are 1. We will show two properties later, which we will use here:

1. aq(n) ≥ 0 for all n ≥ 1, and ϕ(q) 2. aq(n ) ≥ 1 for all (n, q) = 1.

Evaluating our Lq(s) at σ = 1/ϕ(q), we get

∞ ϕ(q) X aq(n) X aq(n ) X 1 ≥ ≥ = ∞, nσ n n n=1 (n,q)=1 (n,q)=1 so σq ≥ 1/ϕ(q). Hence 1 ≤ σ ≤ 1. ϕ(q) q This contradicts the following lemma of Landau. Lemma 15.2.2 (Landau). Let X f(n) L(s) = n n≥1 be a Dirichlet series with σf < ∞ and f(n) ≥ 0 for all n ∈ N. Then L(s) does not admit a holomorphic continuation in a neighbourhood of s = σf .

The contradiction is as follows: By Landau’s theorem, Lq(s) does not admit a holomorphic continuation in some neighbourhood of s = σf . But by Propo- sition 14.1.2, Lq(s) admits a meromorphic continuation to Re(s) > 0 with the only possible simple pole at s = 1 (from χ0). However if we suppose L(1, χ) = 0 for some χ 6= χ0, then we no longer have a pole at s = 1, so Lq(s) is holomorphic on Re(s) > 0, which contradicts with Landau’s lemma. LANDAU’S LEMMA 38

Lecture 16 Landau’s lemma

We’ll ﬁnish up the proof of Dirichlet’s theorem by proving the three outstanding details: Landau’s lemma, and our two conditions on the coeﬃcients aq(n).

16.1 Landau’s lemma Lemma 16.1.1 (Landau). Let

X f(n) L(s) = ns n≥1 be a Dirichlet series with σf < ∞, and f(n) ≥ 0 for all n ∈ N. Then L(s) does not admit a holomorphic continuation in a neighbourhood of s = σf .

Proof. Without loss of generality, assume σf = 0 (else replace L(s) by L(s−σf )). Suppose that L(s) has a holomorphic continuation in an open disk D centred at σf = 0. Thus L(s) and all of its derivatives are holomorphic in D ∪ { s | Re(s) > 0 }. We will show that L(σ) converges for some σ ∈ D with σ < 0, which contradicts σf = 0, since f(n) ≥ 0 means that this convergence is absolute at σ. First, we claim that X f(n) = L(0). n≥1 In other words, what we naively would like the L(0) to be is, in fact, correct. Since f(n) ≥ 0, X f(n) L(σ) = nσ n≥1 is increasing as σ → 0+, and it is also bounded above by L(0). So by the monotone convergence theorem, L(σ) → L(0) as σ → 0+. Now consider

∞ X X f(n) X f(n) f(n) = lim ≤ lim = L(0). σ→0+ nσ σ→0+ nσ n≤n n≤N n=1

On the other hand, for σ > 0,

X f(n) X L(σ) = ≤ f(n), nσ n≥1 n≥1 and as σ → 0+, the left-hand side goes to L(0), so putting this together X L(0) ≤ f(n) ≤ L(0), n≥1 so they’re equal. Secondly, we claim that L(σ) converges for some σ < 0, σ ∈ D.

Date: February 20th, 2019. LANDAU’S LEMMA 39

First, by assumption, L is holomorphic here. For Re(s) > 0,

X f(n)(log n)k L(k)(s) = (−1)k . ns n≥1

Now importantly f(n)(log n)k ≥ 0, so the above argument again means that X L(k)(0) = (−1)k f(n)(log n)k. n≥1 For σ ∈ D with σ < 0, we have the Taylor expansion

∞ ∞ X L(k)(0) X (−1)k P f(n)(log n)k L(σ) = σk = σ . k! k! k k=0 k=0 Switching the order of summation, which we can do since −σ ≥ 0, so we have absolute convergence and can rearrange the sum, we get

X X (−σ log n)k X X f(n) L(σ) = f(n) = f(n) exp(−σ log n) = < ∞. k! nσ n≥1 k≥0 n≥1 n≥1

With this done, we need to establish two things about the coeﬃcients aq(n) of ∞ Y X aq(n) L (s) = L(s, χ) = . q ns χ (mod q) n=1

The two items it remains to show is that ﬁrst, aq(n) ≥ 0 for all n, and second, ϕ(q) aq(n ) ≥ 1 for (n, q) = 1. Since aq(n) is multiplicative (being the convolution of many (completely) k multiplicative characters χ), it suﬃces to show aq(p ) ≥ 0 for all primes p, k ≥ 0. We have −1 k Y Y χ(p) X aq(p ) (L ) (s) = L (s, χ) = 1 − = . q p p ps pks χ (mod q) χ k≥0

Now if p | q¡ then (Lq)p(s) = 1, so ( k 1 if k = 0 aq(p ) = 0 if k 6= 0, which is deﬁnitely nonnegative. If (p, q) = 1, then for notational convenience let z = p−s. We have |z| < 1/p for Re(s) > 1. Let

Y −1 X k k E(z) = (1 − χ(p)z) = aq(p )z . χ k≥0 Taking logarithms,

X X X χ(p)kzk log E(z) = − log(1 − χ(p)z) = k χ χ k≥1 RIEMANN’S MEMOIR 40 where we’ve Taylor expanded the logarithm, since |z| < 1/p < 1. Switching the order of summation, this is

X zk X X zk log E(z) = χ(pk) = ϕ(q) . k k k≥1 χ k≥1 pk≡1 (mod q)

Note that this is a power series with nonnegative coeﬃcients, so taking exponentials to retrieve E(z), we get

(log E(z))2 E(z) = exp(log E(z)) = 1 + log E(z) + + ..., 2!

k whence the coeﬃcients of E(z) are also nonnegative, so aq(p ) ≥ 0. m mϕ(q) For the second property, consider n = p . We want aq(p ) ≥ 1, with × ϕ(q) (p, q) = 1. Let ep be the order of p in (Z/qZ) , i.e. p ≡ 1 (mod q). Then k p ≡ 1 (mod q) if and only if ep | k, by Lagrange’s theorem. Hence ep | ϕ(q). In the above computation, then,

ϕ(q) X zepk ϕ(q) log E(z) = = (− log(1 − zep )) = log (1 − zep )−ϕ(q)/ep , ep k ep k so

ϕ(q)/ep 1 ϕ(q)/e E(z) = (1 − zep )−ϕ(q)/ep = = 1 + zep + z2ep + ... p . 1 − zep

k The exponent is an integer, and ep | mϕ(q), so if h = mϕ(q), aq(p ) is the kth k coeﬃcient, and hence aq(p ) ≥ 1.

Lecture 17 Riemann’s memoir

17.1 Riemann’s memoir In 1860 Riemann published his one and only paper on number theory. In it he proved two results:

1. The Riemann zeta function ζ(s) has a meromorphic continuation to all of C. It is holomorphic except for s = 1, which is a simple pole with residue 1. 2. ζ(s) satisﬁes the functional equation − 1 s s 1 (1−s) 1 π 2 Γ ζ(s) = π 2 Γ (1 − s) ζ(1 − s), 2 2

which is invariant under the transformation s 7→ 1 − s. Riemann further made several remarkable conjectures:

Date: February 22nd, 2019. RIEMANN’S MEMOIR 41

1. ζ(s) has inﬁnitely many zeros in the critical strip 0 < Re(s) < 1. Note that the distribution of zeros is symmetric with respect to the real axis and Re(s) = 1/2. 2. The number N(T ) of zeros of ζ(s) in the critical strip with 0 < Im(s) < T satisﬁes T T T N(T ) = log − + O(log T ), 2π 2π 2π proved by von Mangoldt in 1895. We will show N(T ) ∼ T log T . 3. The integral function

− 1 s s ζ (s) = s(s − 1)π 2 Γ ζ(s) 0 2 has the product representation

Y s ζ (s) = eA+Bs 1 − es/ρ, 0 ρ ρ

where A and B are constants and ρ runs through the zeros of ζ(s) in the critical strip. This was proved by Hadamard in 1893, and is a special case of the Hadamard factorisation theorem. (Note that s − 1 comes from the pole, and s comes from the wish to make it invariant under s 7→ 1 − s.) 4. There exists an explicit formula for π(x) − Li(x) with x > 1, involving a sum over the zeroes of ζ(s). This was proved by von Mangoldt in 1895. 5. Finally, and most importantly, the Riemann hypothesis: the zeros of ζ(s) in the critical strip all lie on the critical line Re(s) = 1/2. This, of course, remains unproven. There has been some progress, the most important of which is: Hardy, in 1914, showed that there are in- ﬁnitely many zeros on Re(s) = 1/2. (His argument also generalises to L(s, χ), and moreover to GL2 L-functions, but for higher dimensions it remains unknown.) Then, in 1942, Selberg showed that there is a positive proportion of zeros of ζ(s) in the critical strip lying on the critical line. This is where the (very powerful) molliﬁer method comes from.

17.2 Review of Fourier analysis Deﬁnition 17.2.1. Let (G, +) be an abelian topological group, i.e. (G, +) is an abelian group, G is a topological space, and moreover

G × G → G (a, b) 7→ a + b and

G → G a 7→ −a POISSON SUMMATION 42 are continuous. A character ψ of G is a continuous group homomorphism ψ : G → C×. Let Hom(G, C×) denote the set of all such characters. It is an abelian group under the multiplication of functions. Deﬁnition 17.2.2. A character ψ is called unitary if ψ : G → S1 = { z ∈ C× | |z| = 1 }. Let Gˆ denote the subgroup of Hom(G, C×) consisting of unitary characters. Remark 17.2.3. If G is compact, then every character is unitary.

Proof. Let ψ : G → C× be a character. Then its image ψ(G) is compact since G is compact and ψ, by deﬁnition, is continuous. Hence ψ(G) is closed and bounded, since this characterises compact sets in C. Now suppose there exists some ψ(g) = z with |z| 6= 1. Then ψ(gn) = ψ(g)n = zn ∈ ψ(G). Since |z|= 6 1, we have two options: if it exceeds 1, then as n increases, zn diverges to inﬁnity, which contradicts ψ(G) being bounded. If |z| < 1, then zn converges to 0, and this contradicts ψ(G) being closed. In the sequel we will write e(x) := exp(2πix).

× Theorem 17.2.4. The map ϕ:(C, +) → Hom(R, C ) deﬁned by y 7→ ϕy, with ϕy(x) := e(yx) is an isomorphism of groups. The restriction of ϕ to R gives an isomoprhism of groups (R, +) =∼ Rˆ.

Lecture 18 Poisson summation

18.1 Fourier analysis Proof. That this is a homomorphism is clear by deﬁnition. To see that it is injective, let y ∈ ker ϕ. Then ϕy(x) = e(xy) = 1 for every x ∈ R, which naturally requires y = 0. For surjectivity, given ψ ∈ Hom(R, C×) we need to show that ψ(x) = e(xy) for some y ∈ C. To this end, let Z x Ψ(x) = ψ(t) dt. 0 Then Z x+y Z x Z x+y Ψ(x + y) = ψ(t) dt = ψ(t) dt + ψ(t) dt. 0 0 x In the latter integral, shift t 7→ t + x, so Z x+y Z y Z y ψ(t) dt = ψ(t + x) dt = ψ(t)ψ(x) dt x 0 0 since ψ is a homomorphism. Hence the above becomes Z x Z y Ψ(x + y) = ψ(t) dt + ψ(x) ψ(t) dt. 0 0 Date: February 25th, 2019. POISSON SUMMATION 43

In other words, Ψ(x + y) = Ψ(x) + ψ(x)Ψ(y). We now ﬁx y such that Ψ(y) 6= 0. This is possible since Ψ0(t) = ψ(t) isn’t identically zero. Taking derivatives with respect to x, this results in

ψ(x + y) = ψ(x) + ψ0(x)Ψ(y), where again we use the fact that ψ is a homomorphism to rewrite this as

ψ(x)ψ(y) = ψ(x) + ψ0(x)Ψ(y).

Rearranging this becomes

Ψ(y)ψ0(x) + (1 − ψ(y))ψ(x) = 0, which is a ﬁrst order ordinary diﬀerential equation, and

ψ(x) = C exp(ux) for some constants C, u ∈ C. Now ψ(0) = 1 so C = 1, i.e. ψ(x) = exp(ux), so we have surjectivity!

Note that ϕy(x) is unitary if and only if y ∈ R, since e(xy) = exp(2πixy), and x ∈ R. Corollary 18.1.1. The map

× (Z, +) → Hom(R/Z, C ) = Rd/Z deﬁned by n 7→ ϕn, ϕn(x) = e(nx), is an isomorphism of groups.

Proof. First, note that Hom(R/Z, C×) = Rd/Z since R/Z is compact. The injectivity part is identical to what we did for the previous theorem. For the onto part, consider ψ : R/Z → C×. Extend it to ψ : R → R/Z → C× periodically, i.e. ψ ≡ 1. Using the previous theorem., ψ = ϕy with y ∈ by Z R unitarity. Moreover ψ(1) = ϕy(1) = e(y) = 1 since 1 ∈ Z, so y ∈ Z.

Deﬁnition 18.1.2. A function f : R → C is said to be in the Schwartz class S(R) if f and all of its derivatives are rapidly decaying, meaning that for all A ≥ 0 and j ≥ 0, f (j)(x) (1 + |x|)−A. In other words, f and its derivatives go to 0 very quickly.

Deﬁnition 18.1.3. Let f ∈ S(R). The Fourier transform of f is the function Z fˆ(y) = f(x) e(−yx) dx, R for y ∈ R. POISSON SUMMATION 44

Remark 18.1.4. Because of the rapid decay, fˆ(y) converves absolutely and uniformly for all y ∈ R. Hence fˆ is inﬁnitely diﬀerentiable with Z fˆ(j)(y) = (−2πix)jf(x)e(−yx) dx. R Integrating this by parts, for y 6= 0, we have

1 Z 1 1 j fˆ(j)(y) = f 0(x)e(−yx) dx = fb0(y) = fd(j)(y). 2πiy 2πiy 2πiy R So for |y| ≥ 1, Z fˆ(y) |y|−j |f (j)(x)| dx |y|−j R for every j ≥ 0. Hence

Proposition 18.1.5. The Fourier transform is a linear map from S(R) to S(R). Let us note a few basic properties of the Fourier transformation. Let f ∈ S(R).

1. Let h ∈ R, and g(x) = f(x + h). Theng ˆ(y) = e(−hy)fˆ(y).

2. fd(j)(y) = (2πiy)jfˆ(y).

1 ˆ 3. Let λ ∈ R, λ 6= 0, and g(x) = f(λx). Theng ˆ(y) = |λ| f(y/λ). ˆ 4. The Fourier inversion formula says fˆ(x) = f(−x).

2 5. The Fourier transform of the Gaussian: if f(x) = e−πx , then fˆ(y) = f(y).

One particular and very important property is

Theorem 18.1.6 (Poisson summation formula). Let f ∈ S(R). Then for x ∈ R, X X f(x + n) = fˆ(n) e(nx). n∈Z n∈Z Proof. Let X g(x) = f(x + n). n∈Z Since f ∈ S(R), g(x) converges absolutely and uniformly for all x ∈ R. Hence we can rearrange terms, diﬀerentiate or integrate under the summation, and so on, and g ∈ C∞(R), and it is periodic with period 1 by construction. Hence g(x) has a Fourier expansion, X g(x) = a(n) e(nx), n∈Z THE FUNCTIONAL EQUATION FOR ζ(S) 45 with

Z 1 Z 1 X a(n) = g(x) e(−nx) dx = f(x + m) e(−nx) dx 0 0 m∈Z X Z 1 X Z m+1 = f(x + m) e(−n(x + m)) dx = f(t) e(−nt) dt 0 m m∈Z m∈Z Z = f(t) e(−nt) dt = fˆ(n). R Hence X g(x) = fˆ(n) e(nx). n∈Z

Corollary 18.1.7. For f ∈ S(R), X X f(n) = fˆ(n). n∈Z n∈Z Proof. Take x = 0 in the previous theorem.

This result is true in general for averaging over any lattice.

Lecture 19 The functional equation for ζ(s)

19.1 Mellin transformation An important corollary to the Poisson summation formula is the following, doing the same kind of averge but over an arithmetic sequence,

Corollary 19.1.1. Let q, a ∈ Z with q ≥ 1. For f ∈ S(R),

X 1 X n qn f(n) = fˆ e . q q q n≡a (mod q) n∈Z Proof. First note that X X f(n) = f(qk + q). n≡q (mod q) n∈Z

Let g(x) = f(qx + a), an aﬃne shift. By the properties of the Fourier transformation, 1 y ay gˆ(y) = fˆ e , q q q so X X X 1 X n an f(qk + q) = g(n) = gˆ(n) = fˆ e . q q q n∈Z n∈Z n∈Z n∈Z Date: February 27th, 2019. THE FUNCTIONAL EQUATION FOR ζ(S) 46

∞ Deﬁnition 19.1.2. Let f(x) ∈ C (R≥0) with rapid decay as x → ∞. The Mellin transform of f is Z ∞ dx f˜(s) = f(x)xs , 0 x for s ∈ C, Re(s) > 0. The integral converges uniformly on compact subsets of { s ∈ C | Re(s) > 0 } since f decays rapidly, and hence f˜(s) is a holomorphic function in Re(s) > 0. Moreover, note that if f(x) = O(xN ) as x → 0, then f˜(s) deﬁnes a holomorphic function in the domain Re(s) > −N, by just studying the power of x in the integrand. Like the Fourier transform, the Mellin transform has an inversion formula: Suppose f ∈ S(R). For any y > 0, 1 Z f(y) = f˜(s)y−s ds 2πi (σ) for any σ > 0, where (σ) means the line σ + it for −∞ < t < ∞. Remark 19.1.3. The Mellin inversion formula and the Fourier inversion formula are equivalent.

19.2 Gamma function The Gamma function is deﬁned as the Mellin transform of e−x, i.e. Z ∞ dx Γ(s) := e−xxs , 0 x for Re(s) > 0. This function, which turns out to be remarkably important, has many fas- cinating properties. For instance,

1.Γ( s) admits a meromorphic continuation to C, which has simple poles at s = −n, n = 0, 1, 2,... with residues (−1)n/n!, and no other poles; 2.Γ( s + 1) = sΓ(s);

π 3.Γ( s)Γ(1 − s) = sin(πs) ; √ 1 2 π 4.Γ( s)Γ(s + 2 ) = 22s Γ(2s); 1 √ 5.Γ( 2 ) = π, and Γ(k + 1) = k! for k = 0, 1, 2,... (which is a consequence of property 2);

6.Γ( s) has no zeros, i.e. Γ(s) 6= 0 for all s ∈ C; and ﬁnally (and incredibly usefully), 7. Sterling’s formula: For any s > 0, −π + δ < arg s < π − δ, 1 1 1 log Γ(s) = s − log s − s + log(2π) + O 2 2 |s| as |s| → ∞. THE FUNCTIONAL EQUATION FOR ζ(S) 47

19.3 The functional equation for ζ(s) Theorem 19.3.1. Let − s s ξ(s) = π 2 Γ ζ(s) 2 for Re(s) > 1. Then ξ(s) has a meromorphic continuation to C, holomorphic on C \{ 0, 1 }, and has simple poles as s = 0 ans s = 1 with residues −1 and 1 respectively. Moreover it satisﬁes the functional equation

ξ(s) = ξ(1 − s).

Proof. By the deﬁnition of Γ, Z ∞ s −t 1 t dt Γ = e t 2 . 2 0 t Let t = n2πx. Then Z ∞ − 1 s s −s −n2πx 1 s dx π 2 Γ n = e x 2 . 2 0 x Summing this over n, so as to develop a ζ(s) on the left-hand side, this becomes

∞ Z ∞ Z ∞ ∞ ! − s s X −n2πx 1 s dx X −n2πx 1 x dx π 2 Γ ζ(s) = e x 2 = e x 2 , 2 x x n=1 0 0 n=1 where switching the sum and integration is permitted since the rapid decay of the exponential gives us uniform convergence. Let ∞ X 2 w(x) = e−n πx, n=1 so that Z ∞ Z 1 Z ∞ 1 s dx 1 s dx 1 s dx ξ(s) = w(x)x 2 = w(x)x 2 + w(x)x 2 . 0 x 0 x 1 x Riemann’s trick was splitting this integral into two, and now performing the change of variables x 7→ 1/x in the ﬁrst one Z Z ∞ 1 − 1 s dx 1 s dx ξ(s) = w x 2 + w(x)x 2 . 1 x x 1 x

In order to make sense of w(x), we wish to extend it to a sum over all n ∈ Z, rather than n ≥ 1, so that we may use Poisson summation on it. To this end, Deﬁnition 19.3.2. The theta series θ(x) is deﬁned by

X 2 X 2 θ(x) = e−n πx = 1 + 2 e−n πx = 1 + 2w(x), n∈Z n≥1 so 1 w(x) = (θ(x) − 1). 2 THE FUNCTIONAL EQUATION FOR L-FUNCTIONS 48

We’ll prove the following lemma for the theta series in a moment: 1 1/2 Lemma 19.3.3. θ x = x θ(x) for x > 0. Taking the lemma for granted just now, we then get

1 1 1 1 1 w = θ − 1 = (x1/2θ(x) − 1) = (x1/2(1 + 2w(x)) − 1) x 2 x 2 2 1 1 = − + x1/2 + x1/2w(x). 2 2 Hence Z ∞ Z ∞ 1 − 1 s dx 1 1 1/2 1/2 − 1 s dx w x 2 = − + x + x w(x) x 2 1 x x 1 2 2 x Z ∞ 1 1 1 (1−s) dx = − + + w(x)x 2 . s s − 1 1 x Therefore Z ∞ 1 1 1 (1−s) 1 s dx ξ(s) = − + + w(x) x 2 + x 2 . s s − 1 1 x Staring at this carefully we see that the first term above is defined for s 6= 0, the second is defined for s 6= 1, and the integral is defined for all s ∈ C since w(x) decays rapidly. Hence ξ(s) has meromorphic continuation to C from the right-hand side, and ξ(s) = ξ(1−s) since the right-hand side is symmetric under s 7→ 1 − s.

Lecture 20 The functional equation for L-functions

20.1 Theta series It remains, form the last result, to prove

1 1/2 Lemma 20.1.1. θ( x ) = x θ(x) for x > 0.

2 Proof. Let f(y) = e−y πx. Then

1 2 fˆ(u) = √ e−πu /x, x so X X 1 X 2 1 1 θ(x) = f(n) = fˆ(n) = √ e−πn /x = √ θ , x x x n∈Z n∈Z n∈Z by Poisson summation. Corollary 20.1.2. The Riemann zeta function ζ(s) has analytic continuation to C\{ 1 }, has a simple pole at s = 1 with residue 1, and satisﬁes the functional equation πs ζ(1 − s) = 2(2π)−sΓ(s) cos ζ(s). 2 Date: March 1st, 2019. THE FUNCTIONAL EQUATION FOR L-FUNCTIONS 49

Proof. Recall how − s s ξ(s) = π 2 Γ ζ(s) 2 is holomorphic except at s = 0 and s = 1. s At s = 0, it has a simple pole. Moreover, Γ( 2 ) has a simple pole at s = 0, so ζ(s) must not have a pole there, since otherwise we have a double pole. For the same reason, ζ(0) 6= 0. 1 √ At s = 1, ξ(s) has a simple pole with residue 1, but Γ( 2 ) = π does not, so ζ(s) must have a simple pole at s = 1. Moreover

− 1 1 1 = Res ξ(s) = π 2 π 2 Res ζ(s), s=1 s=1 so the residue of ζ(s) at s = 1 is 1. Finally ξ(s) = ξ(1 − s) implies that

s 1 Γ( ) πs 2 −s 2 −s ζ(1 − s) = π 1−s ζ(s) = 2(2π) Γ(s) cos ζ(s) Γ( 2 ) 2 by manipulating the Gamma factors.

20.2 Trivial zeros of ζ(s) Since − s s ξ(s) = π 2 Γ ζ(s) 2 s is holomorphic in Re(s) < 0, and since Γ( 2 ) has simple poles as s = −2, −4, −6,..., it means that ζ(s) = 0 for s = −2, −4, −6,.... We call those the trivial zeros of ζ(s).

20.3 Functional equations of L-functions Deﬁnition 20.3.1. Let χ be a Dirichlet character modulo q. Then there exists ∗ ∗ ∗ ∗ a minimal q | q such that χ = χ0 · χ , with χ a Dirichlet character modulo q and χ0 the trivial character modulo q. This character χ∗ is uniquely determined by χ, and q∗ is called the conduc- tor of χ. If q∗ = q, then χ is called a primitive character. We think of it in terms of the following commutative diagram:

Z Z ∗ qZ q Z χ χ∗

C× Proposition 20.3.2. The number of primitive characters modulo q is

2 X q Y 2 Y 1 ϕ∗(q) = ϕ ? µ(q) = µ(d)ϕ = q 1 − 1 − . d p p d|q pkq p2kq THE FUNCTIONAL EQUATION FOR L-FUNCTIONS 50

Proof. The number of Dirichlet characters modulo q is ϕ(q). Now for d | q, consider the diagram χ∗ Z Z × qZ dZ C

Hence for every d | q, if we find the primitive characters modulo d and extend them (as we know we can) to characters modulo q. In other words, X ϕ(q) = ϕ∗(d), d|q i.e. ϕ = ϕ∗ ? 1. By Möbiusinversion, convolving with µ, we get ϕ∗ = ϕ ? µ. Remark 20.3.3. Note that because of the first factor above, ϕ∗(q) = 0 if and only if 2 | q and 22 - q, i.e. q ≡ 2 (mod 4). Hence primitive characters χ modulo q exist only when q 6≡ 2 (mod 4). Proposition 20.3.4. For (n, q) = 1, we have X∗ X q χ(n) = ϕ(d)µ , d χ (mod q) d|(n−1,q) where by P∗ we mean that we sum only over primitive characters. Example 20.3.5. For q = 4, ϕ∗(4) = 1, so there is only one primitive Dirichlet n−1 character modulo 4, namely χ4(n) = (−1) 2 if 2 - n, else 0. Specifically χ4(1) = −1, χ4(3) = 1, and otherwise it is 0. N Example 20.3.6. For q = 8, ϕ∗(8) = 2, so there are two primitive characters. (n−1)(n+1) The first one is χ8(n) = (−1) 8 if 2 - n, and the second one is χ4 ·χ8(n) = (n−1)(n+5) (−1) 8 if 2 - n. N

20.4 Gauss sums Deﬁnition 20.4.1. Let χ (mod q) be a Dirichlet character. Then X b τ(χ) = χ(b) e q b (mod q) is called the Gauss sum of χ. Note that this sum only sums over b ∈ (Z/qZ)× since otherwise χ(b) = 0. By orthogonality, for (a, q) = 1, a 1 X e = χ(a)τ(χ). q ϕ(q) χ (mod q) This formula is very useful since it transforms from an additive character to a multiplicative character. Note also that it is the Fourier expansion of the additive character on (Z/qZ)×. Proposition 20.4.2. Let χ be a primitive character modulo q. Then X ab χ(b) e = χ(a)τ(χ). q b (mod q) THE FUNCTIONAL EQUATION FOR L-FUNCTIONS, CONTINUED 51

Lecture 21 The functional equation for L-functions, continued

21.1 Gauss sums Proposition 21.1.1. Let χ be a primitice character modulo q. Then

X ab χ(b) e = χ(a)τ(χ). q b (mod q)

Proof. We have two cases to consider: when (a, q) = 1, and when they’re not coprime. Let us start with when they are. Then X ab X ab X aba−1 χ(b) e = χ(b) e = χ(ba−1) e q q q × × b (mod q) b∈(Z/qZ) b∈(Z/qZ) X b = χ(b)χ(a−1) e = χ(a)τ(χ), q × b∈(Z/qZ) where we sum only over units since otherwise the charcater is 0 anyway, and we perform the change of variable b 7→ ba−1 since that just permutes the terms we’re summing over. If (a, q) 6= 1, then χ(a) = 0, so we need to show that

X ab χ(b) e = 0. q b (mod q)

Letting d = (a, q) and writing a = da1, q = dq1, with (a1, q1), so that X ab X a1b χ(b) e = χ(b) e q q1 b (mod q) b (mod q)   X  X  a1r =  χ(b) e .   q1 r (mod q1) b (mod q) b≡r (mod q1)

We claim, and will prove momentarily, that there exists some c ≡ q (mod q1) with (c, q) such that χ(c) 6= 1. In that case, X X X χ(b) = χ(bc) = χ(c) χ(b) b (mod q) b (mod q) b (mod q) b≡r (mod q1) b≡r (mod q1) b≡r (mod q1) by changing b 7→ bc, since c is a unit modulo q. Then X (1 − χ(c)) χ(b) = 0, b (mod q) b≡r (mod q1)

Date: March 4th, 2019. THE FUNCTIONAL EQUATION FOR L-FUNCTIONS, CONTINUED 52 so the sub over the characters must be zero, since χ(c) 6= 1. Now let us prove the claim. Suppose there is no such c, i.e., χ(c) = 1 for × all c ≡ 1 (mod q1) with (c, q) = 1. Then for any m, n ∈ (Z/qZ) with n ≡ m (mod q1), we have nm¯ ≡ 1 (mod q1), where bym ¯ we mean the multiplicative inverse modulo q1. Thus χ(nm¯ ) = 1, implying that χ(n) = χ(m), meaning that χ factors modulo q1, so it is not primitive. Proposition 21.1.2. Let χ be a primitive character modulo q. Then |τ(χ)| = √ q. Proof. We write 2 2 X ab X a(b1 − b2) χ(a)τ(χ) = χ(b) e = χ(b1)χ(b2) e . q q b (mod q) b1 (mod q) b2 (mod q) Now summing this over a (mod q), we note that the right-hand side is nonzero only for b1 − b2 = 0 since by orthogonality ( X am 0 if q m, e = - q q if q | m. a (mod q) Hence X X |τ(χ)|2 |χ(a)|2 = q |χ(b)|2, a (mod q) b (mod q) where the sums are nonzero, so dividing through we get |τ(χ)|2 = q. The proof of the functional equation for L(s, χ) is now very similar to that of the Riemann zeta function. Theorem 21.1.3. Let χ be a primitive character modulo q. Let

Λ(s, χ) = L∞(s, χ)L(s, χ) be the completed L-function, where q s/2 s + κ L (s, χ) = Γ ∞ π 2 is called the archimedean factor of the (local) factor at inﬁnity, and ( 1 0 if χ is even, κ = (1 − χ(−1)) = 2 1 if χ is odd.

A character χ being even means χ(−1) = 1, so that χ(−n) = χ(n), and odd correspondingly means χ(−1) = −1, so χ(−n) = −χ(n). Then Λ(s, χ) has a holomorphic continuation to all s ∈ C and satisﬁes the functional equation Λ(s, χ) = ε(χ)Λ(1 − s, χ), where τ(χ) ε(χ) = iκ √ q is called the root number of χ. THE FUNCTIONAL EQUATION FOR L-FUNCTIONS, CONTINUED 53

Note that by the previous proposition, |ε(χ)| = 1. Proof. We separate the case where χ is even and where it’s odd. If it is even, we use the deﬁnition of Γ(s), from which we have

Z ∞ 2 − 1 s 1 s 1 −s − n πt 1 s dt π 2 q 2 Γ s n = e q t 2 2 0 t for Re(s) > 0. (Cf. ζ(s) with q = 1.) Twisting by χ(n) and summing over n, we get   Z ∞ 2 − 1 s 1 s 1 X − n πt 1 s dt π 2 q 2 Γ s L(s, χ) = χ(n)e q t 2 . 2   t 0 n≥1

We switch the order of integration and summation since the exponential decay guarantees us absolute and uniform convergence for Re(s) > 1. Now let ∞ 2 X − n πt θχ(t) = χ(n)e q , n=−∞ so that Z ∞ Z 1 Z ∞ 1 1 t dt 1 1 t dt 1 1 t dt Λ(s, χ) = θχ(t)t 2 = θχ(t)t 2 + θχ(t)t 2 2 0 t 2 0 t 2 1 t Z ∞ Z ∞ 1 1 − 1 t dt 1 1 t dt = θχ t 2 + θχ(t)t 2 2 1 t t 2 1 t We claim, and will prove momentarily, that

1 t 1/2 θ = τ(χ)θ (t). χ t q χ

With that as writ, Z ∞ Z ∞ 1 τ(χ) − 1 (1−s) dt 1 1 s dt Λ(s, χ) = √ θχ(t)t 2 + θχ(t)t 2 , 2 q 1 t 2 1 t and Z ∞ Z ∞ 1 τ(χ) − 1 s dt 1 1 (1−s) dt Λ(1 − s, χ) = √ θχ(t)t 2 + θχ(t)t 2 , 2 q 1 t 2 1 t and comparing these we see that

Λ(s, χ) = ε(χ)Λ(1 − s, χ), √ where in this case, for even χ, ε(χ) = τ(χ)/ q. Note that as part of the computation, τ(χ)τ(χ) = τ(χ)τ(χ) = |τ(χ)|2 = q. FUNCTIONS OF FINITE ORDER 54

Lecture 22 Functions of ﬁnite order

22.1 Proof concluded We left the proof with two outstanding issues: we had only considered χ beign even, and we had a claim we hadn’t yet proved.

2 X − n πt Claim. With θχ(t) = χ(n)θχe q , we have n∈Z

1 t 1/2 θ = τ(χ)θ (t). χ t q χ Proof. By deﬁnition,

2 X b X − n πt τ(χ)θ (t) = χ(b) e χ(n)e q . χ q b (mod q) n∈Z Combining the exponentials and using Poisson summation, this becomes

b 2 2 1/2 (n+ ) πq X X − n πt + 2πibn X X q − q χ(b) e q 1 = χ(b) e t . t b (mod q) n∈Z b (mod q) n∈Z Multiplying and dividing the exponent by q, we get −(qn+b)2π/(qt), where the parenthesis then ranges over all integers n, so this becomes

1/2 2 1/2 q X − n π q 1 χ(n)e qt = θ . t t χ t n∈Z Hence, for χ even, we are done. Now let us tackle the case where χ is odd, i.e. χ(−1) = −1. Proof. Note that the method used above no longer works, since by χ being odd, θχ(t) is identically 0 for all t. Instead what we’ll do is replace s by s + 1 in the expression we play with for Γ, so

Z ∞ 2 − 1 (s+1) 1 (s+1) 1 −s − n πt 1 (s+1) dt π 2 q 2 Γ (s + 1) n = ne q t 2 , 2 0 t where we’ve moved one of the n from n−(s+1) over to the other side, so as to make the exponential, together with n, an odd function in n. We then proceed as usual: we twist by χ(n) and sum over n ≥ 1, getting   Z ∞ 2 − 1 (s+1) 1 (s+1) 1 X − n πt 1 (s+1) dt π 2 q 2 Γ (s + 1) L(s, χ) = χ(n)ne t t 2 , 2   t 0 n≥1 where the summand is now an even function in n. Deﬁning

n2πt 0 X − q θχ(t) = χ(n)ne , n∈Z Date: March 6th, 2019. FUNCTIONS OF FINITE ORDER 55

1 0 the sum is 2 θχ(t), and as before we write this as the integral from 0 to 1 plus the integral from 1 to ∞, then make a change of variables t 7→ 1/t in the former,

∞ 1 ∞ 1 Z 1 dt 1 Z 1 dt 1 Z 1 dt 0 2 (s+1) 0 2 (s+1) 0 2 (s+1) Λ(s, χ) = θχ(t)t = θχ(t)t + θχ(t)t 2 0 t 2 0 t 2 1 t 1 ∞ 1 Z 1 1 dt 1 Z 1 dt 0 − 2 (s+1) 0 2 (s+1) = θχ t + θχ(t)t . 2 0 t t 2 1 t By the same kind of computation as before, with Poisson summation and all, 1 θ0 = −iq−1/2t3/2τ(χ)θ0 (t), χ t χ by which

1/2 Z ∞ Z ∞ π −iτ(χ) 0 − 1 (s+1) dt 1 0 1 (s+1) dt Λ(s, χ) = θ (t)t 2 + θ (t)t 2 χ 1/2 χ 2 q 1 t q 1 t which is the same as ε(χ)Λ(1 − s, χ).

22.2 The Hadamard factorisation theorem

The goal of the coming discussion is to express a holomorphic function on C as an inﬁnite product indexed by its zeros. To start this, we need some preliminaries.

Deﬁnition 22.2.1. A holomorphic function f : C → C is of ﬁnite order if there exists a constant α > 0 such that for any ε > 0, we have

|f(s)| exp |s|α+ε for all s ∈ C, with the implied constant depending on ε and f. In this case we say that f is of order ≤ α. We say that f is of order α if it is of order ≤ α but not of order ≤ β for any β < α.

n n−1 Example 22.2.2. Let Pn(s) = ans + an−1s + ... + a0, ai ∈ C and an 6= 0. This is of order 0. N Example 22.2.3. The function f(s) = exp(s) is of order 1. N

Example 22.2.4. The function fn(s) = exp(Pn(s)) is of order n = deg Pn. N

Note also that fn(s), as deﬁned above, is a function of order n which does not vanish in C. This is not a coincidence: Proposition 22.2.5. A function f of order ≤ α which does not vanish on C is of the form f(s) = exp(p(s)) for some polynomial p(s) of degree ≤ α. Hence f is of integral order deg p.

Proof. Since f(x) does not vanish on C, we can take its logarithm, i.e., g(s) := log f(s) can be deﬁned and is holomorphic on C. Hence it has a Taylor expansion at s = 0, i.e. ∞ X n g(s) = cns , n=0 JENSEN’S FORMULA 56

with cn ∈ C. We claim that cn = 0 for n > α, from which the result follows. To prove the claim, note that since f is of order ≤ α,

α+ε α+ε Re g(s) = log|f(s)| ≤ Cε,f |s| = Cε,f R ,

2πix calling |s| = R. Writing cn = an + ibn, with an, bn ∈ R, and letting s = Re , we have

2πix X n X Re g Re = anR cos(2πnx) − i bnR6n sin(2πnx). n≥0 n≥1

Hence (R 1 2πix n 0 Re(g(Re )) dx if n = 0, anR = R 1 2πix 2 0 Re(g(Re )) cos(2πnx) dx if n ≥ 1 and Z 1 n 2πix bnR = 2 Re(g(Re )) sin(2πinx) dx, 0 whence Z 1 n 2πix α+ε |an|R ≤ 2 |Re(g(Re ))| · 1 dx ≤ 2Cε,f R 0 by bounding cos by 1. Hence for n > α,

α−n+ε |an| ≤ 2Cε,f R → 0 as R → ∞, so an = 0. By the same argument, bn is 0 for n > α too.

Lecture 23 Jensen’s formula

23.1 Jensen’s formula Remark 23.1.1. Note that in the proof of Proposition 22.2.5, it suﬃces to study ∞ a sequence of positive real numbers { Rn }n=1 with Rn → ∞ such that for all n ≥ 0, ε > 0, and |s| = Rn,

α+ε f(s) f,ε exp(|s| ).

This is because of the maximal principle, having control on the boundary of a circle implies having control inside the circle. In order to state Jensen’s theorem we need a bit of set up. Let f : C → C. For R > 0, we deﬁne

Z(f, R) := { ρ ∈ C | f(ρ) = 0, |ρ| ≤ R } and Z(f) := { ρ ∈ C | f(ρ) = 0 }. We will also take sums and products over these sets—in this case we add or multiply with multiplicity.

Date: March 18th, 2019. JENSEN’S FORMULA 57

Theorem 23.1.2 (Jensen’s formula). Let R > 0. Let f be a holomorphic function in a neighbourhood of DR = { s ∈ C | |s| < R }. Suppose f does not vanish at s = 0 nor on CR = { s ∈ C | |s| = R }. Then

Z 1 2πit f(Re ) Y R X R log dt = log = log . f(0) |ρ| |ρ| 0 ρ∈Z(f,R) ρ∈Z(f,r)

Note that like usual residue calculus with Cauchy’s redisue theorem on f 0/f, this also counts zeros, but not with weight 1. Proof. Write Y f(s) = F (s) (s − ρ) ρ∈Z(f,R) where F (s) is a holomorphic function that doesn’t vanish in DR. Hence

f(s) F (s) X s − ρ log = log + log . f(0) F (0) ρ ρ∈Z(f,R)

Looking at the ﬁrst term,

F (s) F (s) log = Re log . F (0) F (0)

The inside of this real part is a holomorphic function in a neighbourhood of DR, and vanishes at s = 0, so taking Re2πit = z,

Z 1 F (Re2πit) 1 Z F (z) dz log dt = log = 0 0 F (0) 2πi CR F (0) z since the integrand is a holomorphic function, so we can contract the integration to a point. For the second term, note that

2πit Re − ρ R 2πit ρ R ρ 2πit = e − = 1 − e ρ |ρ| R |ρ| R meaning that

2πit Re − ρ R ρ 2πit R ρ 2πit log = log + log 1 − e = log + Re log 1 − e . ρ |ρ| R |ρ| R As above, the latter is a holomorphic function vanishing at z = 0, so

Z 1 ρ log 1 − e2πit dt = 0, 0 R so when all the dust settles only the sum of log(R/|ρ|) remains. JENSEN’S FORMULA 58

23.2 Application of Jacobi’s formula Theorem 23.2.1. Let f be a holomorphic function of order ≤ α. Let X N(f, R) = 1, ρ∈Z(f,R) i.e., the number of zeros of f of modulus ≤ R, with multiplicity. (i) For all R > 0, ε > 0, we have ( α+ε 1, if f(0) = 0, N(f, R) f,ε R + 0, if f(0) 6= 0.

(ii) The series X 1 1 + |ρ|α+ε ρ∈Z(f) converges.

Proof. Suppose f(0) 6= 0 and f(s) 6= 0 on C2R. For ρ ∈ Z(f, R), log(2R/ρ) ≥ log 2, so

Z 1 2πit X 2R X 2R f(2Re ) (log 2)N(f, R) ≤ log ≤ log = log dt |ρ| |ρ| f(0) ρ∈Z(f,R) ρ∈Z(f,2R) 0 Z 1 (2R)α+ε dt = (2R)α+ε Rα+ε, 0 where for the integral we use Jensen’s formula, and the bound is by the order. 0 Note that if f(s) = 0 for some s ∈ C2R, then we consider R < R < R + δ for some δ > 0, such that f(s) 6= 0 for all s ∈ C2R0 . This is possible since f is a holomorphic function, meaning that if it has an accumulation point of zeros, then it is identically zero. Thus

N(f, R) ≤ N(f, R0) (R0)α+ε Rα+ε.

Now suppose f(0) = 0, with the zero of order m. Then

f(s) N(f, R) = N ,R + m Rα+ε + 1. sm f For the series,

X 1 Z R 1 = dN(f, r) 1 + |ρ|α+ε 1 + rα+ε ρ∈Z(f,R) 0 R Z R α−1+ε N(f, r) r = α+ε − N(f, r) α+ε 2 dr 1 + r 0 0 (1 + r ) N(f, R) Z R rα−1+ε = α+ε − N(f, r) α+ε 2 dr. 1 + R 0 (1 + r ) By the ﬁrst part, this is O(1), and letting R → ∞ we are done. HADAMARD FACTORISATION THEOREM 59

Lecture 24 Hadamard factorisation theorem

24.1 Hadamard factorisation Theorem 24.1.1. Let f be a holomorphic function of order at most 1 such that f(0) 6= 0. Then we have

Y s f(s) = A exp(bs) 1 − es/ρ ρ ρ∈Z(f) where A = f(0), b = f 0(0)/f(0), and the product converges uniformly on compact subsets of C. Proof. Let K ⊂ C be a compact subset. Then K ∩ Z(f) is a ﬁnite set (since otherwise f = 0, since a holomorphic function with an accumulation point of zeros must be zero everywhere). For s ∈ K and ρ ∈ Z(f) \ K, we have

s 1 1 − es/ρ = 1 + O ρ K |ρ|2 by expanding the exponential as a Taylor series. Q Note that an inﬁntie product (1 + an), with an ≥ 0, converges if and only P if an converges, and since X 1 |ρ|2 ρ∈Z(f) converges by Jensen’s formula,

Y s 1 − es/ρ ρ ρ∈Z(f) converges uniformly on K (since our estimate above is independent of s). Set Y s h(s) = 1 − es/ρ. ρ ρ∈Z(f) This is holomorphic and has exactly the same zeros as f(s). Hence f(s)/h(s) is holomorphic and does not vanish on C. We claim that f(s)/h(s) has order ≤ 1. Then, by Proposition 22.2.5, f(s) = exp(a + bs) h(s) for some a, b ∈ C. Now for all ρ ∈ Z(f), s s/ρ 1 − e = 1, ρ s=0 so h(0) = 1, whence f(0) = exp(a) = A.

Date: March 20th, 2019. THE INFINITE PRODUCT FOR ζ(S) 60

Similarly, d s s/ρ 1 − e = 0 ds ρ s=0 so h0(0) = 0, so b = f 0(0)/f(0). It remains to prove the claim, meaning we need to show

f(s) 1+ε log = O(|s| ). h(s) Proof. Writing the logarithm as

log|f(s)| − log|h(s)|, we note ﬁrst that f(s) is order at most 1 by assumption, so

log|f(s)| R1+ε on |s| = R for all R > 0. It remains to show that − log|h(s)| R1+ε on a sequence of R → ∞, |s| = R. The intelligent idea is this: Since

X 1 < ∞, |ρ|2 ρ∈Z(f) the total length of all intervals

Lecture 25 The inﬁnite product for ζ(s)

25.1 Hadamard factorisation ﬁnished The missing piece from our proof of Hadamard factorisation is to show that for

Y s h (s) = 1 − es/ρ, 3 ρ |ρ|>2R we have 1+ε − log|h3(s)| R for R = |s|. Write this as X s s/ρ X 2 − log|h3(s)| = − log 1 − e = − log 1 + O(|s/ρ| ) ρ |ρ|>2R |ρ|>2R since 2!! s s/ρ s s s 2 1 − e = 1 − 1 + + O = 1 + O(|s/ρ| ), ρ ρ ρ ρ

Date: March 25th, 2019. THE INFINITE PRODUCT FOR ζ(S) 61 by expanding the exponential as a Taylor series. Now expanding this logarithm as a Taylor series in turn, we have

2 1−ε X s X 1 R 1+ε 1+ε − log|h3(s)| = R R ρ |ρ|1+ε |ρ|1−ε |ρ|>2R |ρ|>2R since the ﬁrst term in the latter sum converges since the order is ≤ 1, and the second term is bounded by (1/2)1+ε since |ρ| > 2R. Corollary 25.1.1. Let f be a holomorphic function of order at most 1 such that f(0) 6= 0. For s ∈ C \ Z(f), we have d f 0(s) X 1 1 (log f(s)) = = b + − , ds f(s) ρ ρ − s ρ∈Z(f) with b = f 0(0)/f(0). Moreover

X 1 1 − ρ ρ − s ρ∈Z(f) converges uniformly on compact subsets K ⊂ C \ Z(f). Proof. The ﬁrst part is Hadamard factorisation on f 0(s)/f(s). For the second part, note that if s ∈ K, 1 1 −s 1 − = = O ρ ρ − s ρ(ρ − s) K |ρ|2 since s is in a bounded set.

25.2 The inﬁnite product for ζ(s) and the explicit formula Let − s s ξ (s) = s(1 − s)ζ(s) = s(1 − s)π 2 Γ ζ(s). 0 2

Proposition 25.2.1. ξ0(s) has order at most 1.

Proof. Since ξ(s) = ξ(1 − s), we also have ξ0(s) = ξ0(1 − s), so by symmetry we may assume Re(s) ≥ 1/2. Bounding each factor above one at a time, we have

− s − s log π 1+ε |π 2 | = e 2 exp |s| .

Next, s Γ exp |s|1+ε 2 by Stirling’s formula. Finally, s Z ∞ {t} ζ(s) = − s 1+s dt s − 1 1 t for Re(s) > 0. The ﬁrst term is bounded, with the pole at s = 1 cancelled by the factor (1 − s) from ξ0(s). The integral is convergent since Re(s) > 0. ε Hence s(1 − s)ζ(s) exp(|s| ), hence ξ0(s) has order at most 1. THE INFINITE PRODUCT FOR ζ(S) 62

Y s s Corollary 25.2.2. (i) ξ (s) = A exp(bs) 1 − e ρ , 0 ρ ρ∈Z(ξ0) X 1 (ii) < ∞. |ρ|1+ε ρ∈Z(ξ0)

Remark 25.2.3. The zeros of ξ0(s) are precisely the nontrivial zeros of ζ(s), since s sΓ( 2 ) has no zeros and the zero of (1 − s) is cancelled by the pole of ζ(s) at s = 1.

25.3 Inﬁnite product for L(s, χ) Let χ (mod q) be a primitive character, and

1 q 2 (s+κ) s + κ Λ(s, χ) = Γ L(s, χ), π 2 satisfying Λ(s, χ) = ε(χ)Λ(1 − s, χ) with ( 0, if χ is even, κ = 1, if χ is odd, and |ε(χ)| = 1. Proposition 25.3.1. Λ(s, χ) has order at most 1. Proof. By essentially the same proof as for ζ(s), except here we have

Z ∞ S(t) L(s, χ) = s 2+1 dt 1 t for Re(s) > 0, with X S(t) = χ(n), n≤t and |S(t)| ≤ q. This integral is convergent, so exp(|s|ε).

25.4 Application to counting zeros of ζ(s) By Corollary 25.1.1,

0 0 ξ0 1 1 ζ X 1 1 (s) = + 0 (s) + (s) = b + − ξ s s − 1 ζ∞ ζ ρ ρ − s 0 ζ ∞ ρ∈Z(ξ0) converges uniformly on compact subsets of C \ Z(f). Note that ξ0(s) takes real values on the real line, and it is holomorphic, so by Schwartz reﬂection principle ξ0(s) = ξ0(s). Hence Z(ξ0) is invariant under complex conjugation, so

X 1 1 1 X 1 1 1 1 − = + − − . ρ ρ − s 2 ρ ρ ρ − s ρ − s ρ∈Z(ξ0) ρ∈Z(ξ0) COUNTING ZEROS 63

Writing ρ = β + iγ, 0 ≤ β ≤ 1, γ ∈ R, and 1 1 2β + = , ρ ρ |ρ|2 we have X 1 1 + < ∞, ρ ρ ρ∈Z(ξ0) whereby 1 X 1 1 + . 2 s − ρ s − ρ ρ∈Z(ξ0) converges uniformly on compact subsets of C \ Z(f).

Proposition 25.4.1. For every s ∈ Z(ξ0),

ξ0 1 X 1 1 0 (s) = + + O(1) ξ0 2 s − ρ s − ρ ρ∈Z(ξ0) and

ζ0 1 1 1 X 1 1 ζ0 − (s) = + − + + ∞ (s) + O(1). ζ s s − 1 2 s − ρ s − ρ ζ∞ ρ∈Z(ξ0)

Lecture 26 Counting zeros

26.1 Application to counting zeros of ζ(s)

Corollary 26.1.1. Let N(T ) := #{ ρ = β + iγ ∈ Z(ξ0) | |γ| ≤ T }. For T ≥ 1, we have

(a) N(T + 1) − N(T ) log(2 + T ), (b) N(T ) T log(2 + T ). Proof. We have

ζ0 1 1 ζ0 1 X 1 1 − (s) = + + ∞ (s) − + + O(1). ζ s s − 1 ζ∞ 2 s − ρ s − ρ ρ∈Z(ξ0)

Letting s = 2 + iT , we have the series representation

∞ ζ0 X Λ(n) − (s) = < ∞ ζ ns n=1 since we are in the region of absolute convergence. Also,

ζ0 ∞ (s) log(2 + T ) ζ∞

Date: March 27th, 2019. COUNTING ZEROS 64 by Stirling’s formula. Hence 1 X 1 1 + log(2 + T ). 2 s − ρ s − ρ ρ∈Z(ξ0) Now writing ρ = β + iγ, with 0 ≤ β ≤ 1, we have

1 2 − β 1 1 1 Re = ≥ ≥ ≥ δ . s − ρ |s − ρ|2 (2 − β)2 + |γ − T |2 4 + |γ − T |2 5 |γ−T |≤1

Therefore

1 X X 1 1 1 ≤ Re + log(2 + T ), 5 s − ρ s − ρ |γ±T |≤1 ρ∈Z(ξ0) where the left-hand side sum, of course, is N(T + 1) − N(T − 1), which in turn is an upper bound for N(T + 1) − N(T ), so

N(T + 1) − N(T ) log(2 + T ).

For the second part, simply write N(T ) as a telescoping sum,

T X N(T ) = (N(n) − N(n − 1)) T log(2 + T ). n=1 Remark 26.1.2. From the proof we note, in particular, that

X 1 log(2 + T ), 4 + |γ ± T |2 ρ∈Z(ξ0) where ρ = β + iγ. Secondly, moving things over,

ξ0 1 X 1 1 0 (s) = + + O(log(2 + |T |)), ξ0 2 s − ρ s − ρ |γ±T |≤1 where s = σ + iT 6∈ Z(ξ0), −1 ≤ σ ≤ 2.

26.2 Applications to countinjg zeros of L(s, χ) We wish to perform the same sort of estimates for L(s, χ). 1 Let χ be a primitive character modulo q ≥ 1. Let κ = 2 (1 − χ(−1)), and

− s π 2 s + κ L (s, χ) := Γ . ∞ q 2

Then Λ(s, χ) = L∞(s, χ)L(s, χ) = ε(χ)Λ(1 − s, χ). Note that

Z(Λ(s, χ)) ⊂ { s ∈ C | 0 ≤ Re(s) ≤ 1 } is exactly the set of nontrivial zeros of L(s, χ). COUNTING ZEROS 65

By Hadamard factorisation,

Λ0 L0 L0 X 1 1 (s, χ) = ∞ (s, χ) + (s, χ) = b(χ) + − , Λ L∞ L ρ ρ − s ρ∈Z(Λ(s,χ)) where Λ0 b(χ) = (0, χ). Λ This causes trouble, since unlike for ζ it depends on a sum over zeros. Note that Λ(s, χ) = ε(χ)Λ(1 − s, χ) = εχΛ(1 − s, χ), so if ρ ∈ Z(Λ(s, χ)), then 1 − ρ ∈ Z(Λ(s, χ)). Thus

Λ0 Λ0 X 1 1 b(χ) = (0, χ) = − (1, χ) = −b(χ) − − . Λ Λ ρ0 ρ0 − 1 ρ0∈Z(Λ(s,χ))

Writing ρ = 1 − ρ0 ∈ Z(Λ(s, χ)), this becomes

X 1 1 −b(χ) − + . 1 − ρ ρ ρ∈Z(Λ(s,χ))

Combining this we have

1 X 1 1 Re b(χ) = − + 2 1 − ρ ρ ρ∈Z(Λ(s,χ)) since b(χ) + b(χ) = 2 Re b(χ), so

1 X 1 1 X 1 Re b(χ) = − Re + = − Re . 2 ρ ρ ρ ρ∈Z(Λ(s,χ)) ρ∈Z(Λ(s,χ))

Thus Λ0 X 1 Re (s, χ) = Re . Λ s − ρ ρ∈Z(Λ(s,χ)) Summarising,

Proposition 26.2.1. For s 6∈ Z(Λ(s, χ)),

Λ0 X 1 (a) Re (s, χ) = , and Λ s − ρ ρ∈Z(Λ(s,χ))

L0 X 1 L0 (b) Re (s, χ) = Re − Re ∞ (s, χ). L s − ρ L∞ ρ∈Z(Λ(s,χ)) WEIL’S EXPLICIT FORMULA 66

Lecture 27 Weil’s explicit formula

27.1 Application to counting zeros of L(s, χ) Corollary 27.1.1. Let N(T, χ) := #{ ρ = β + iγ ∈ Z(Λ(s, χ)) | |γ| ≤ T }. For T ≥ 1, we have (a) N(T + 1, χ) − N(T, χ) log(q(2 + T )), and

(b) N(T, χ) T log(q(2 + T )). Proof. For s = 2+iT , we have the (absolutely) convergent series representation

∞ L0 X Λ(n)χ(n) (s, χ) = < ∞, L ns n=1 and L0 ∞ (s, χ) log(q(2 + T )) L∞ by Stirling’s formula. By Proposition 26.2.1, we hence have

X 1 Re log(q(2 + T )). s − ρ ρ∈Z(Λ(s,χ))

Note that 1 σ − β 1 1 Re = ≥ ≥ δ , s − ρ |s − ρ|2 4 + |T − γ|2 5 |T −γ|≤1 writing ρ = β +iγ. For s = 2−iT we get the same result, but |T −γ| is replaced by |T + γ|. Hence

1 X X 1 1 ≤ Re log(q(2 + T )), 5 s − ρ |T ±γ|≤1 ρ so

N(T + 1, χ) − N(T, χ) ≤ N(T + 1, χ) − N(T − 1, χ) log(q(2 + T )).

The second part, as before, is just a telescoping sum. Remark 27.1.2. We have X 1 log(q(2 + T )). 4 + |γ ± T |2 ρ∈Z(Λ(s,χ))

Moreover, Λ0 X 1 (s, χ) = + O(log(q(2 + T ))) Λ s − ρ |γ±T |≤1 for s = σ + iT 6∈ Z(Λ(s, χ)), −1 ≤ σ ≤ 2.

Date: March 29th, 2019. WEIL’S EXPLICIT FORMULA 67

Proof. The second part is a little bit more delicate in this case than for ζ(s). For s = 2 + iT , Λ0 X 1 1 (s, χ) = b(χ) + + , Λ ρ 2 + iT − ρ ρ where the left-hand side is log(q(2 + T )). Thus Λ0 X 1 1 (s, χ) = − + O(log(q(2 + T ))) Λ s − ρ (2 + iT ) − ρ ρ X 1 X 1 X 1 1 = − + − + O(log(q(2 + T ))). s − ρ 2 + iT − ρ s − ρ 2 + iT − ρ |γ±T |≤1 |γ±T |≤1 |γ±T |>1 Now estimating the sums, X 1 X 1 log(q(2 + T )), 2 + iT − ρ |γ±T |≤1 |γ±T |≤1 and since 1 1 s − σ 1 − = . s − ρ 2 + iT − ρ (s − ρ)(2 + iT − ρ) |T ± γ|2 Summing both sides, we get log(q(2 + T )), meaning that all of the sums we don’t care about get baked into the same error term.

27.2 Weil’s explicit formula ∞ Theorem 27.2.1. Let f ∈ Cc ((0, ∞)), (i.e., inﬁnitely diﬀerentiable with compact support) and let Z ∞ dx f˜(s) = f(x)xs 0 x be its Mellin transform. Let 1 1 fˇ(x) = f . x x We have X 1 Z ζ0 ζ0 X (f(n)+fˇ(n))Λ(n) = f˜(1)+f˜(0)+ ∞ (s) + ∞ (1 − s) f˜(s) ds− f˜(ρ). 2πi ( 1 ) ζ∞ ζ∞ n≥1 2 ρ∈Z(ξ0) Proof. First we claim that 1 Z ξ0 1 Z 1 1 ζ0 X f˜(s) 0 (s) ds = + + ∞ (s) f˜(s) − Λ(n)f(n). 2πi 3 ξ0 2πi 3 s s − 1 ζ∞ ( 2 ) ( 2 ) n≥1

Since ξ0(s) = s(1 − s)ζ∞(s)ζ(s), and ζ0 X Λ(n) − (s) = ζ ns n≥1 in the region of absolute convergence, 1 Z ξ0 1 Z 1 1 ζ0 X 1 Z f˜(s) 0 (s) ds = f˜(s) + + ∞ (s) ds− Λ f˜(s)n−s ds. 2πi 3 ξ0 2πi 3 s s − 1 ζ∞ 2πi 3 ( 2 ) ( 2 ) n≥1 ( 2 ) The ﬁnal integral is the Mellin inversion of f(n), whence we have our claim. WEIL’S EXPLICIT FORMULA 68

Lecture 28 Weil’s explicit formula

28.1 Proof continued We have X 1 Z ξ0 1 Z ξ0 f˜(ρ) = f˜(s) 0 (s) ds − f˜(s) 0 (s) s. 2πi ( 3 ) ξ0 2πi (− 1 ) ξ0 ρ∈Z(ξ0) 2 2 By making the change of variables s 7→ 1 − s in the second integral, we switch ξ0 ξ0 ξ0 from f˜(s) 7→ f˜(1 − s), 0 (s) 7→ 0 (1 − s) = − 0 (s), and ﬁnally the line of ξ0 ξ0 ξ0 3 integration moves to 2 . Hence by the claim this becomes 1 Z 1 1 ζ0 X + + ∞ (s) f˜(s) ds − Λ(n)f(n)+ 2πi 3 s s − 1 ζ∞ ( 2 ) n≥1 1 Z 1 1 ζ0 X + + ∞ (s) f˜(1 − s) ds − Λ(n)fˇ(n), 2πi 3 s s − 1 ζ∞ ( 2 ) n≥1 where we’ve used 1 Z ∞ dx 1 Z ∞ 1 1 dx fˇ˜(s) = f˜(x)xs = f xs , 2πi 0 x 2πi 0 x x x

1 which if we switch x 7→ t becomes Z ∞ dt tf(t)t−s = f˜(1 − s). 0 t

3 1 Moving the line of integration from ( 2 ) to ( 2 ) we pick up a pole at s = 1, with residues f˜(1) in the ﬁrst integral and f˜(0) in the second, so our expression becomes 1 Z 1 1 + (f˜(s) + f˜(1 − s)) ds = 2πi 3 s s − 1 ( 2 ) 1 Z 1 1 f˜(1) + f˜(0) + + (f˜(s) + f˜(1 − s)) ds. 2πi 1 s s − 1 ( 2 ) This last integral is 0, since if we switch s 7→ 1 − s, we get negative the same integral, so it is equal to negative itself. Theorem 28.1.1. Let χ be a Dirichlet character modulo q. Then, with fˇ as above, X (χ(n)f(n) + χ(n)fˇ(n))Λ(n) = n≥1 1 Z L0 L0 X ∞ (s, χ) + ∞ (1 − s, χ) f˜(s) ds − f˜(ρ). 2πi ( 1 ) L∞ L∞ 2 ρ∈Z(Λ(s,χ))

Proof. The proof is the same as above, except with L∞ in place of ζ∞. Date: April 1st, 2019. WEIL’S EXPLICIT FORMULA 69

Theorem 28.1.2 (Hadamard, de la Vall´ee-Poussin). There exist an absolute constant c > 0 such that ζ(s) 6= 0, s = σ + it, in the region c σ ≥ 1 − . log(2 + |t|)

We will prove this later on. For now, remark that the value of c is actually computable. For instance, Habiba and Kadin computed 1 c = 5.69693 in 2005. People are often improving this ever so slightly.

∞ Corollary 28.1.3. There exists a constant c > 0 such that for f ∈ Cc ((0, ∞)), X ≥ 2 large,

X n Z ∞ Λ(n)f = X f(t) dt + O X exp(−cplog X) . X f n≥1 0

We call this a smooth sum since f is smooth. One can ‘unsmooth’ the sum, obtaining X Λ(n) = X + O(X exp(−c0plog X)) n≤X for some c0 > 0, whence P Λ(n) ∼ X, which is equivalent to the Prime number theorem. n Remark 28.1.4. If supp f = [1, 2], we get 1 ≤ X ≤ 2, i.e., X ≤ n ≤ 2X. t ˜ s Proof of Corollary 28.1.3. Let g(t) = f( X ). Theng ˜(s) = f(s)X , so 1 1 1 1 gˇ(t) = g = f , t t t tX

1 which is zero for t ≥ 1 and X large since f has compact support, and tX ≈ 0. Thus by Weil’s explicit formula

X 1 Z ζ0 ζ0 X Λ(n)g(n) =g ˜(1)+˜g(0)+ ∞ (s) − ∞ (1 − s) g˜(s) ds− g˜(ρ), 2πi ( 1 ) ζ∞ ζ∞ n≥1 2 ρ∈Z(ξ0) whence X n Λ(n)f = X n≥1 1 Z ζ0 ζ0 X f˜(1)X + f˜(0) + ∞ (s) − ∞ (1 − s) f˜(s)Xs ds − f˜(ρ)Xρ. 2πi ( 1 ) ζ∞ ζ∞ 2 ρ∈Z(ξ0)

The ﬁrst term is the integral we want, with the second term being Of (1). For the sum over zeros,

X X X − c f˜(ρ)Xρ ≤ |f˜(ρ)|Xβ ≤ X |f˜(ρ)|X log(2+|γ|) ρ ρ=β+iγ ρ THE THEOREM OF HADAMARD AND DE LA VALLEE-POUSSIN´ 70 by the zero free region, where the exponential in the end is

x log X exp − . log(2 + |γ|)

√We attack this ﬁnal sum by splitting√ the sum over zeros with log(2 + |γ|) ≤ log X and those with log(2 + |γ|) > log X. The former becomes X exp(−cplog X) |f˜(ρ)| ρ wherein the sum converges by exponential decay of the Mellin transform. The latter is bounded by X |f˜(ρ)| ρ and the summand is 1/|γ|s by rapid decay—we can take any exponent. Such a sum converges since the L-function is of order ≤ 1.

Lecture 29 The theorem of Hadamard and de la Vall´ee-Poussin

29.1 Zero free region Surprisingly and remarkably, most of the following discussion is a consequence of the following elementary trigonometric inequality: for all θ ∈ R, 3 + 4 cos(θ) + cos(2θ) ≥ 0.

To see that this is true, note simply that the left-hand side is equal to 2(1 + cos(θ))2, which is of course nonnegative. To recover the proper zero free region, we need to ﬁrst establish that there are no zeros on the line (1).

Theorem 29.1.1. ζ(1 + it) 6= 0 for all t ∈ R. Proof. For s = σ + it, t 6= 0, σ > 1, we have

∞ X X 1 log ζ(s) = e−itn log p. npnσ p n=1

Taking real parts,

∞ X X 1 Re log ζ(s) = cos(tn log p). npnσ p n=1

Calling tn log p = θ, we apply the inequality, getting

3 Re log ζ(σ) + 4 Re log ζ(σ + it) + Re log ζ(σ + 2it) ≥ 0.

Date: April 3rd, 2019. THE THEOREM OF HADAMARD AND DE LA VALLEE-POUSSIN´ 71

More to the point,

3 log|ζ(σ)| + 4 log|ζ(σ + it)| + log|ζ(σ + 2it)| ≥ 0.

Taking exponentials,

|ζ(σ)|3|ζ(σ + it)|4|ζ(σ + 2it)| ≥ 1.

1 + Now ζ(σ) = σ−1 + O(1) as σ → 1 . Suppose ζ(σ + it) = 0 for some 4 4 + t 6= 0. This implies |ζ(σ + it)| = Ot((σ − 1) ) as σ → 1 , and also that + |ζ(σ + 2it)| = Ot(1) as σ → 1 . Hence

|ζ(σ)|3|ζ(σ + it)|4|ζ(σ + 2it)| → 0 as σ → 1+, which is a contradiction to it being bounded below by 1. Hence there can’t be any zeros on Re(s) = 1. Theorem 29.1.2. There exists a constant c > 0 such that c Z(ξ0) ⊂ s ∈ C Re(s) ≤ 1 − , log(2 + |Im(s)|) i.e., ζ(s) 6= 0 for c Re(s) > 1 − . log(2 + |Im(s)|) Proof. For σ > 1, we have the series representation

∞ ζ0 X Λ(n) − (σ + it) = , ζ nσ+it n=1 where as above we write the imaginary exponent n−it = e−it log n. Hence taking real parts, we have

∞ ζ0 X Λ(n) Re − (σ + it) = cos(t log n). ζ nσ n=1 Hence, in our trigonometric inequality,

ζ0 ζ0 ζ0 3 − (σ) + 4 Re − (σ + it) + Re − (σ + 2it) ≥ 0. ζ ζ ζ

Recall that by Hadamard factorisation, for s = σ + it ∈ Z(ξ0),

ζ0 1 1 X 1 1 ζ0 − (s) = + + b − + + ∞ (s). ζ s s − 1 s − ρ ρ ζ∞ ρ∈Z(ξ0)

For σ > 1, ζ0 1 − (σ) ≤ + C ζ σ − 1 1 because of the pole. THE THEOREM OF HADAMARD AND DE LA VALLEE-POUSSIN´ 72

For s = σ + it, σ > 1, t 6= 0, we have

ζ0 X 1 1 − Re (σ + it) ≤ C log(2 + |t|) − Re + , ζ 2 s − ρ ρ ρ where the logarithm is from the arhimedian factor, bounded with Stirling’s formula. In particular, 1 σ − β 1 Re = ≤ , s − ρ |s − ρ|2 |s − ρ|2 so the sum is bounded. Applying this for σ + it, we get

ζ0 − Re (σ + it) ≤ C log(2 + |t|), ζ 2 and for σ + 2it, ζ0 − Re (σ + 2it) ≤ C log(2 + |t|) ζ 3 as well. However we don’t want the ﬁrst of those two estimates above quite as-is, as it gets rid of the information from the zeros. Instead let us detect zeros one at a time, by ﬁxing ρ0 = β0 +iγ0 ∈ Z(ξ0), and taking s = σ +it with t = γ0. Then 1 1 1 Re = Re = , s − ρ σ + iγ0 − (β0 + iγ0) σ − β0 so if we get rid of the sum over zeros, except keeping only this one term from ρ0, we have ζ0 1 − Re (σ + it) ≤ C2 log(2 + |t|) − . ζ σ − β0 Thus our basic inequality becomes

1 1 3 + C1 + 4 C2 log(2 + |t|) − + C3 log(2 + |t|) ≥ 0 σ − 1 σ − β0 for t = γ0 and all σ > 0. Thus 4 3 − ≤ C4 log(2 + |t|) σ − β0 σ − 1

δ for all σ > 1. Taking in particular σ = 1 + log(2+|t|) , for some δ > 0, we get

δ 4δ c β0 ≤ 1 + − ≤ 1 − log(2 + |t|) (δC4 + 3) log(2 + |t|) log(2 + |t|) for some c > 0 by taking δ small enough. EXCEPTIONAL ZEROS 73

Lecture 30 Exceptional zeros

30.1 Zero free region for L(s, χ) Theorem 30.1.1. There exista a constant c > 0 such that if χ is a Dirichlet character modulo q, then L(s, χ) has no zeros in a region deﬁned by c σ > 1 − , log(q(2 + |t|)) writing s = σ + it, unless χ is a real quadratic character in which case L(s, χ) has at most one (necessarily real) zero β < 1 in this region. Such a zero is called an exceptional zero or a Siegel zero. Proof. The strategy is essentially the same, we just need to take a little bit of extra care sometimes. Letting s = σ + it with σ > 1 we again have the series representation ∞ L0 X Λ(n) − (s, χ) = χ(n)e−it log n. L nσ n=1 Note that |χ(n)e−it log n| = 1 for (n, q) = 1 (else 0), so let us write χ(n)e−it log n = eiθ, so that cos(θ) = Re(χ(n)e−it log n), and correspondingly e2iθ = χ2(n)e−i2t log n, and so cos(2θ) = Re(χ2(n)e−2it log n). We also need to be a little bit careful with the constant 3 in our basic triginometric inequality: we’ll write 1 = χ0(n) for (n, q) = 1 (and we don’t care about other n, since they contribute nothing). Hence, multiplying by the appropriate factors and summing, the inequality becomes L0 L0 L0 −3 (σ, χ ) + 4 Re − (σ + it, χ) + Re − (σ + 2it, χ2) ≥ 0. L 0 L L

As before we want an upper bount for this. Note ﬁrst of all that if χ = χ0, then Y L(s, χ) = ζ(s) (1 − p−s), p|q so the result follows from Theorem 29.1.2 for ζ(s). Now consider the case where χ is complex, in which case χ2 is nontrivial, so for σ > 1,

0 ∞ 0 L X χ0(n)Λ(n) ζ 1 − (σ, χ ) = ≤ − (σ) ≤ + C . L 0 nσ ζ σ − 1 1 n=1 Secondly, by Hadamard factorisation,

L0 1 q 1 Γ0 s + κ X 1 1 − (s, χ) = log + − b(χ) − − . L 2 π 2 Γ 2 s − ρ ρ 0≤Re(ρ)≤1

We showed before that X 1 Re b(χ) = − , ρ ρ

Date: April 5th, 2019. EXCEPTIONAL ZEROS 74 so taking real parts we have

L0 1 q 1 Γ0 s + κ X 1 Re − (s, χ) = log + Re − Re . L 2 π 2 Γ 2 s − ρ ρ

By Stirling’s formula for the Gamma factors and by nonnegativity of the last sum, we have L0 Re − (σ + 2it, χ2) ≤ C log(q(2 + |t|)) L 2 2 for χ 6= χ0. For the middle term, as before, we isolate a zero: given ρ0 = β0 + iγ0, take s = σ + it with σ > 0 and t = γ0. Then keeping precisely that term from the above sum, L0 1 Re − (σ + it, χ) ≤ C2 log(q(2 + |t|)) − . L σ − β0 Inserting these and going through the motions, c β ≤ 1 − , 0 log(q(2 + |t|)) for some c > 0 and t = γ0. Secondly, and more intricately, we need to consider the case where χ is real. 2 The above bound works only for χ 6= χ0, but now we in fact have precisely 2 χ = χ0, so we need to do something else. There are three cases to consider now. First, we are dealing with zeros at great height, say |γ0| ≥ 6(1 − β0). Take s = σ + it with t = γ0 again, and estimate 0 L 2 1 σ − 1 Re − (σ + 2it, χ ) ≤ C3 log(q(2+|t|))+Re ≤ C3 log(q(2+|t|))+ 2 2 L (σ + 2it) − 1 (σ − 1) + 144(1 − β0)

1 where the last term comes from the term s−1 , and the estimate stems from 1 σ − 1 σ − 1 Re = 2 2 ≤ 2 2 . (σ + 2it) − 1 (σ − 1) + 4γ0 (σ − 1) + 144(1 − β0) Therefore 1 1 σ − 1 3 + C1 +4 C2 log(q(2 + |t|)) − + 2 2 + C3 log(q(2 + |t|)) ≥ 0. σ − 1 σ − β0 (σ − 1) + 144(1 − β0)

Taking σ = 1 + 6(1 − β0), it eventually follows that c β ≤ 1 − . 0 log(q(2 + |t|))

Secondly, suppose we are concerned with a low-lying zero, say 0 < |γ0| < 6(1 − β0). Since χ is rael, ρ0 = β0 + iγ0 being a zero implies ρ0 = β0 − iγ0 is one too. Taking s = σ since the height t is quite small now, we have

0 L 1 1 −2(σ − β0) − (σ, χ) ≤ − Re −Re +C5 log q = 2 2 +C5 log q. L σ − ρ0 σ − ρ0 (σ − β0) + γ0 LANDAU’S THEOREM 75

2 Since t = 0 and χ real, χ = χ0, our inequality becomes simply L0 L0 − (σ, χ ) − (σ, χ) ≥ 0. L 0 L For the ﬁrst term the old bound works just ﬁne, and for the second term we use the above. Taking σ = 1 + δ(1 − β0), and then later δ = 13 (for example), we get the result we want. Finally, suppose γ0 = 0, so ρ0 = β0 ∈ R. This is where we might get an exceptional zero. Suppose there are two real zeros, say, 0 ≤ β0 ≤ β1 < 1. Then L0 1 1 2 − (σ, χ) ≤ − − + C5 log q ≤ − + C5 log q, L σ − β0 σ − β1 σ − β0 so 1 2 + C1 − + C5 log q ≥ 0. σ − 1 σ − β0

Taking σ = 1 + δ(1 − β0),

1 1 2 + + C6 log q ≥ 0, 1 − β0 δ δ + 1 which if we take δ = 2 gives us c β ≤ 1 − , 0 log q for some c > 0. Now this gives the same sort of zero free region, except for the notable hitch that it bounds β0—we might still have the hypothetical β1 to the right of it. Note that there can only ever be one exceptional zero, since otherwise we can repeat this argument with β1 in place of β0.

A question one now asks is this: For which q ∈ N is there a real quadratic character χ (mod q) such that L(s, χ) has an exceptional zero? There are no known examples. Moreover, if we assume the Riemann hypothesis for L(s, χ), then such q cannot exist. Indeed, Landau shows that such a q is very, very rare if it exists.

Lecture 31 Landau’s theorem

31.1 Exceptional zeros

Theorem 31.1.1 (Landau). There exists a constant c > 0 such that if χ1 (mod q1) and χ2 (mod q2) are real quadratic characters such that χ1χ2 is nontrivial, then L(s, χ1)L(s, χ2) has at most one real zero β such that c 1 − < β < 1. log q1q2 Date: April 8th, 2019. LANDAU’S THEOREM 76

Proof. We have

1 + χ1(n) + χ2(n) + χ1χ2(n) = (1 + χ1(n))(1 + χ2(n)) ≥ 0, so for σ > 1, multiply by Λ(n)/nσ and sum over n, getting

ζ0 L0 L0 L0 − (σ) − (σ, χ ) − (σ, χ ) − (σ, χ χ ) ≥ 0. ζ L 1 L 2 L 1 2

Now assume L(s, χi) has an exceptional zero βi, and assume β1 ≤ β2. We want an upper bound for the left-hand side above, and we have some from our previous discussion: ζ0 1 − (σ) ≤ + c , ζ σ − 1 1 as well as L0 1 − (σ, χi) ≤ − + c2 log qi, L σ − βi and ﬁnally L0 − (σ, χ χ ) ≤ c log q q L 1 2 3 1 2 since by assumption χ1χ2 6= χ0. Hence 1 1 1 − − + c4 log q1q2 ≥ 0. σ − 1 σ − β1 σ − β2

Since β1 ≤ β2, we have moreover 1 2 − + c4 log q1q2 ≥ 0. σ − 1 σ − β1

Taking σ = 1 + 2(1 − β1), we get c β1 ≤ 1 − , log q1q2 for some c > 0.

Note that this means that if β1 = β2, then neither zero is in this special region. Corollary 31.1.2. There exists a constant c > 0 such that Y L(s, χ) χ (mod q) has at most one zero in the region c σ > 1 − , log q(2 + |t|) for s = σ + it. If such a zero exists, then it is real and the associated character χ is quadratic. LANDAU’S THEOREM 77

Corollary 31.1.3 (Page). There exista a constant c > 0 such that for every Q ≥ 1, Y Y∗ L(s, χ), q≤Qχ (mod q) where by the asterisk we mean a product over primitive characters, has at most one zero in the region c σ < 1 − , log Q(2 + |t|) for s = σ + it. If such a zero exists, then it’s real and the associated character is quadratic.

31.2 Siegel’s theorem Siegel’s theorem is about bounding the zero-free region away from 1, and essentially the idea is to leverage upper bounds of the L-function at 1.

Theorem 31.2.1. Let χ 6= χ0 be a quadratic character modulo q. Then L(1, χ) q−1/2. Proof. Deﬁne τ(n) := 1?χ(n). We have showed before (in past homework) that τ(n) ≥ 0 for all n and τ(n) ≥ 1 if n is a perfect square. Consider X − n τ(n)e x . n≥1 We can easily get a lower bound for this:

2 2 X − n X 2 − m X − m 1/2 τ(n)e x ≥ τ(m )e x ≥ e x x n≥1 m≥1 m≥1 by comparing with the integral. To get asymptotics for it, note that by Mellin inversion 1 Z e−x = Γ(s)X−s ds, 2πi (2) so   Z X − n 1 X τ(n) s τ(n)e x = Γ(s)X ds, 2πi  ns  n≥1 (2) n≥1 where the inner sum is ζ(s)L(s, χ) by the deﬁnition of τ(n). Shifting the line 1 of integration from (2) to (− 2 ) picks up residues from ζ(s) at s = 1, namely XL(1, χ), and from Γ(s) at s = 0, namely ζ(0)L(0, χ). Hence the sum is 1 Z L(1, χ)X + ζ(0)L(0, χ) + ζ(s)L(s, χ)Γ(s)Xs ds. 2πi 1 (− 2 )

1 We have ζ(0) = − 2 , and L(0, χ) ≥ 0 since it’s positive at 1, and we translate by the functional equation. By studying the completed L-function for L(s, χ) at 3/2, using Stirling, and then using the functional equation we translate to SIEGEL’S THEOREM 78

−1/2 and get that L(s, χ) q|t| on this line. Hence by exponential decay, the entire integral is qX−1/2. Thus X1/2 L(1, χ)X + O(qX−1/2). Taking X = q, we deduce L(1, χ) q−1/2.

Corollary 31.2.2. There exists a cosntant c > 0 such that if χ (mod q) is a quadratic character with L(s, χ) having an exceptional zero β, then c β ≤ 1 − . q1/2(log q)2

Proof. We can show L0(σ, χ) (log q)2 for 1 1 − ≤ σ ≤ 1. log q By the mean value theorem,

q−1/2 L(1, χ) = L(1, χ) − L(β, χ) = L0(σ, χ)(1 − β) (1 − β)(log q)2, so c β ≤ 1 − , q1/2(log q)2 for c > 0. Theorem 31.2.3 (Siegel’s theorem). For any ε > 0, there exists a constant C(ε) > 0 such that C(ε) L(1, χ) ≥ qε for all primitive quadratic characters χ (mod q). Remark 31.2.4. This constant is ineﬀective—we don’t know how to compute a numerical value for ε < 1/2.

Lecture 32 Siegel’s theorem

32.1 Proof

Proof. Let χ1 (mod q1) and χ2 (mod q2) be two primitive characters. Deﬁne

τ(n) = 1 ? χ1 ? χ2 ? χ1χ2(n), so that L(s, τ) = ζ(s)L(s, χ1)L(s, χ2)L(s, χ1χ2). We have (essentially) showed before that τ(n) ≥ 0 for all n. We claim that for any ε > 0, there exists a primitive quadratic character χ1 (mod q1) and β with 1 − ε < β < 1 such that L(β, τ) ≤ 0 for all primitive quadratic characters χ2 (mod q2), with q2 6= q1.

Date: April 10th, 2019. SIEGEL’S THEOREM 79

Remark 32.1.1. We do not know how to compute this q1, which is what makes this result ineﬀective. We prove this in two cases. First, suppose there are no zeros on [1 − ε, 1] of L(s, χ) for any quadratic character χ. Then take any χ1 (mod q1). We have L(β, χ) > 0 since L(1, χ) > 0 for quadratic characters χ 6= χ0. Likewise L(s, χ2) > 0, and L(s, χ1χ2) > 0 since χ1χ2 6= χ0 because q1 6= q2. Moreover ζ(β) < 0 since it is approximately 1/(s − 1) near 1, so L(β, τ) < 0. Secondly, suppose there exists χ (mod ()q) with a real zero β ∈ [1 − ε, 1]. Take χ1 = χ, so that L(β, χ1) = 0. Then of course L(β, τ) = 0 too. With this in hand we are ready to prove Siegel’s theorem. Take χ1, q1, and β as in the claim. Let

λ = Res L(s, τ) = L(1, χ1)L(1, χ2)L(1, χ1χ2). s=1 Then Z 1 X τ(n) − n 1 s ≤ e x = L(s + β, τ)Γ(s)X ds e nβ 2πi n≥1 (2) by Mellin inversion. Moving the line of integration from (2) to (−β), we pick up poles at s = 1 − β and s = 0 with residues λΓ(1 − β)X1−β and L(β, τ) respectively. Thus 1 1 Z ≤ λΓ(1 − β)X1−β + L(β, τ) + L(s + β, τ)Γ(s)Xs ds, e 2πi (−β) where by choice of β, L(β, τ) ≤ 0 by the claim. We have L(it, χ) ((2 + |t|)q)1/2+ε for χ (mod q), and ζ(it) (2 + |t|)1/2+ε, so 2+ε 1+ε L(it, τ) (2 + |t|) (q1q2) . Note that −β is close to −1, and Γ(s) has a simple pole at s = −1, so Γ(−β) = O(1/(1 − β)), meaning that the integral above is

(q q )1+εX−β O 1 2 , 1 − β and therefore 1 (q q )1+εX−β ≤ λΓ(1 − β)X1−β + O 1 2 , e 1 − β

1−β 1+ε −β having dropped L(β, τ) since it’s nonpositive. Now if λX ≥ (q1q2) X , 1+ε so λX ≥ (q1q2) , so since

−1 λ = L(1, χ1)L(1, χ2)L(1, χ1χ2) (q1q2) , we have X 1+ε > (q1q2) , q1q2 THE PRIME NUMBER THEOREM IN ARITHMETIC PROGRESSIONS 80 meaning that λX1−β 1 1 − β 2+ε provided (q1q2) X, guaranteeing that

(q q )1+εX−β λΓ(1 − β)X1−β ≥ O 1 2 1 − β so that we have the asymptotics we want. Now for χ 6= χ0 (mod q), L(1, χ) log q, so

λ L(1, χ2)(log q1)(log q1q2).

2+ε Take X = (q1q2) , which may be a new epsilon, and insert into the bound on λ above. Then

1 −(2+ε)(1−β) L(1, χ2) · (q1q2) (1 − β) (log q1)(log q1q2)

Now q1 and β depend on ε, so we get

−(2+ε)(1−β) −1 L(1, χ2) ≥ C(ε)q2 (log q2) . Moreover, (1 − β) < ε, so (2 + ε)(1 − β) < 3ε, and log q qε, so

C(ε) L(1, χ ) ≥ . 2 qε

Corollary 32.1.2. For any ε > 0, there exists a constant C(ε) > 0 such that if χ (mod q) is a quadratic character such that L(s, χ) has an exceptional zero β, then C(ε) β ≤ 1 − . qε Proof. The same as Corollary 31.2.2.

Lecture 33 The Prime number theorem in Arith- metic progressions

33.1 Prime number theorem Recall how we proved the ordinary Prime number theorem by x X X π(x) ∼ ⇔ ϑ(x) = log p ∼ x ⇔ ψ(x) = Λ(n) ∼ x. log x p≤x n≤x

We prove this last step by smoothing the sum over all n with a smoothing function, using the explicit formula to write in terms of the Zeta function, and then use the zero-free region of ζ(s).

Date: April 12th, 2019. THE PRIME NUMBER THEOREM IN ARITHMETIC PROGRESSIONS 81

The strategy for arithmetic progressions is largely the same. For a and q with (a, q) = 1, X π(x) π(x; q, a) = 1 ∼ ϕ(q) p≤x p≡a (mod q) is equivalent to X x ϕ(x; q, a) = log p ∼ , ϕ(q) p≤x p≡a (mod q) is equivalent to X x ψ(x; q, a) = Λ(n) ∼ , ϕ(q) n≤x n≡a (mod q) which we will prove. ∞ Let f ∈ Cc ((0, ∞)), (a, q) = 1. Deﬁne X n ψ (x; q, a) := Λ(n)f . f x n≤x n≡a (mod q)

By the orthogonality relations,

1 X ψ (x; q, a) = χ(a)ψ (x, χ), f ϕ(q) f χ (mod q) where X n ψ (x, χ) := Λ(n)χ(n)f . f x n≥1

For χ = χ0 we get X n X n X n ψ (x, χ ) = Λ(n)f = Λ(n)f − Λ(n)f . f 0 x x x (n,q)=1 n≥1 (n,q)>1

Studying each sum separately, X n X X X X Λ(n)f Λ(n) = log p (log x) log p = (log x)(log q). x (n,q)>1 nx p|q pα<1

On the other hand,

X n Z ∞ Λ(n)f = x f(t) dt + O (x exp(−cplog x)) x f n≥1 0 for some c > 0 by Corollary 28.1.3. Hence ∞ x Z 1 X 1 p ψf (x; q, a) = f(t) dt+ χ(a)ψf (x, χ)+Of (x exp(−c log x) + log x log q) . ϕ(q) 0 ϕ(q) ϕ(q) χ6=χ0 (mod q)

1 √ Proposition 33.1.1. Suppose q exp( 2 log x). BOMBIERI-VINOGRADOV 82

(a) If χ 6= χ0 such that L(s, χ) has no exceptional√ zeros, then there exists a constant c > 0 such that ψf (x, χ) f x exp(−c log x) uniformly in χ.

(b) If χ 6= χ0 such that L(s, χ) has an exceptional zero β ∈ R, then there exists a constant c > 0 such that ˜ β p ψf (x, χ) = −f(β)X + Of (x exp(−c log x)) uniformly in χ.

1 Proof. Choose x large enough such that f( nx ) = 0 for all n ≥ 1. By the explicit formula, Z 0 0 1 L∞ L∞ ˜ s X ˜ ρ ψf (x, χ) = ( (s, χ) − (1 − s, χ))f(s)x ds − f(ρ)x . 2πi ( 1 ) L∞ L∞ 2 ρ∈Z(Λ(s,χ))

The integral is x1/2. By Theorem 30.1.1,

X0 c log x |f˜(ρ)| exp − , log q(2 + γ) ρ∈Z(Λ(s,χ)) with Im(ρ) = γ. Then X0 |f˜(ρ)| exp(−cplog x),

ρ∈Z(Λ(s,χ√)) log(q(2+|γ|))≤ log x and X0 X X 1 X 1 1 |f˜(ρ)| |f˜(ρ)| . |γ|2 |γ|3/2 |γ|1/2 ρ∈Z(Λ(s,χ√)) log(q(2+|γ|))> log x

1 √ Corollary 33.1.2. Suppose q exp( 2 log x). Then x Z ∞ 1 X ψ(x; q, a) = f(t) dt − χ(a)f˜(β)xβ + O (x exp(−cplog x)) ϕ(q) ϕ(q) f 0 χ for some c > 0, where the sum is over χ with exceptional zero β.

Lecture 34 Bombieri-Vinogradov

34.1 Prime number theorem in arithmetic progressions We can now finish the prime number theorem: Theorem 34.1.1 (Siegel-Walfisz theorem). Let C < 0 (effective) as in Corol- lary 33.1.2. Then for all A > 0, q (log x)A, (a, q) = 1, and X ≥ 2, we have ∞ x Z p ψf (x; q, a) = f(t) dt + Of,A(x exp(−c log x)), ϕ(q) 0 where the implied constant depending on A and f is ineffective.

Date: April 15th, 2019. BOMBIERI-VINOGRADOV 83

The ineﬀectiveness comes from us using Siegel’s theorem in the proof. Proof. We have

xβ = x exp(−(1 − β) log x) ≤ x exp(−C(ε)q−ε log x) by Siegel’s theorem. Since q (log x)A,

xβ x exp(−C(ε)(log x)1−Aε).

Taking ε = 1/(3A), we get

xβ x exp(−C(ε)(log x)2/3).

Thus also: Theorem 34.1.2 (Siegel-Walﬁsz). Suppose A > 0. For q (log x)A, (a, q) = 1, x ≥ 2, we have x ψ(x; q, a) = + O (exp(−cplog x)), ϕ(q) A for some c > 0, which is equivalent with the Prime number theorem in arithmetic progressions. Another way of saying this: Theorem 34.1.3 (Siegel-Walﬁsz, variant). Suppose A > 0. For q (log x)A and χ (mod q), we have p ψ(x, χ) − δ(χ)x A x exp(−c log x) for some c > 0, where ( 1, if χ = χ , δ(χ) = 0 0, if χ 6= χ0.

34.2 The Bombieri-Vinogradov theorem There are two questions we want to ask about all of this. • In the proof we use that there is no real zeros of L(s, χ) near 1 (i.e., Siegel’s theorem). What is the error term if we assume the Generalised Riemann Hypothesis (GRH), i.e., all zeros of L(s, χ) in 0 ≤ Re(s) ≤ 1 lie 1 on Re(s) = 2 . • Can we enlarge the range of q? Under GRH we can deduce that x ψ(x; q, a) = + O(x1/2(log x)2) ϕ(q) uniformly for all q. To have this make asymptotic sense, note that ϕ(q) ∼ q, so x > x1/2(log x)2 q BOMBIERI-VINOGRADOV 84 implies x1/2 q < , (log x)2 meaning q < x1/2. A remarkable result is that we can achieve the GRH error term on average: Theorem 34.2.1 (Bombieri-Vinogradov). For any A > 0, Q ≤ X1/2,

X y −A 1/2 4 max max ψ(y; q, a) − A X(log X) + X Q(log X) . (a,q)=1 y≤X ϕ(q) q≤Q

Remark 34.2.2. (i) The implied constant is ineﬀective since we use Siegel’s theorem in the proof. However, in 2013, Akbary and Hambrook managed to prove Bombieri-Vinogradov without relying on Siegel’s theorem. (ii) Taking X1/2(log X)−A−4 ≤ Q ≤ X1/2, then the second term in the theorem is larger, so the sum is bounded by X1/2(log X)4. The theorem says that if we average over q, this bound is as good as GRH for all χ (mod q), q ≤ X1/2(log X)−A−4. (iii) For Q > X1/2, we use the trivial bound

X X ψ(X; q, a) = Λ(n) ≤ (log X) + 1 , q n≤X n≡a (mod q)

so the sum is X X + 1 log X X(log X)(log Q) + Q(log X), q q≤Q

which is a better bound than the theorem. Proof. By orthogonality, 1 X ψ(X; q, a) = χ(a)ψ(X, χ). ϕ(q) χ (mod q)

Hence

y 1 X ψ(y; q, a) − = χ(a)ψ(y, χ) − δ(χ)y . ϕ(q) ϕ(q) χ (mod q) Getting rid of the character, since it is of modulus 1, we may potentially lose some cancellation. We get 1 X ≤ |ψ(y, χ) − δ(χ)y|. ϕ(q) χ (mod q)

So it suﬃces to bound X 1 X max|ψ(y, χ) − δ(χ)y|. ϕ(q) y≤X q≤Q χ (mod q) BOMBIERI-VINOGRADOV, CONTINUED 85

We can moreover reduce this to primitive characters. If χ (mod q) is induced by a primitive character χ∗ (mod q∗), q∗ | q, then  

∗ X X  ψ(y, χ) = ψ(y, χ ) + O  (log p) 1 .   p|q k≤ log y ∗ log p p-q To see this, note that χ∗ Z ∗ Z q Z C χ

Z qZ Evaluating at n = pk, p | q, p - q∗, we have χ(n) = 0 but χ∗(n) 6= 0, and the log p accounts for Λ(pk), and the inner sum counts how many of them there are.

Lecture 35 Bombieri-Vinogradov, continued

35.1 The trivial case The remainder at the end of the last discussion is of order (log y)(log q), so the sum is X 1 X X∗ max|ψ(y, χ∗) − δ(χ∗)y| + O(Q(log q)(log X)). ϕ(q) y≤X q≤Q q∗|qχ∗ (mod q∗) Writing q = q∗r, X X 1 X∗ max|ψ(y, χ∗) − δ(χ∗)y|. (35.1.1) ϕ(q∗r) y≤X q∗≤Q r≤Q/q∗ χ∗ (mod q∗) Note that ϕ(q∗r) ≥ ϕ(q∗)ϕ(r). Moreover X 1 log Q ϕ(q) q≤Q since ∞ ∞ ! X 1 Y X 1 Y X 1 ≤ = 1 + ϕ(q) ϕ(pn) pn − pn−1 q≤Q p≤Q n=0 p≤Q n=1 −1 Y 1 1 = 1 − 1 + log Q. p p(p − 1) p≤Q Thus for (35.1.1) we have the bound X log Q X∗ max|ψ(y, χ) − δ(χ)y|. ϕ(q) y≤X q≤Q χ (mod q)

Date: April 17th, 2019. BOMBIERI-VINOGRADOV, CONTINUED 86

Let R = (log X)A+4. We will splot the range of q according to whether q ≤ R, for which we can use Siegel’s theorem, and R < q ≤ Q, for which we need new machienery, namely the Large Sieve and Polya’s inequality. Claim. With deﬁnitions as above, X log Q X∗ p max|ψ(y, χ) − δ(χ)y| A R(log X)X exp(−c log X) ϕ(q) y≤X q≤R χ (mod q) for some c > 0. Note that this moreover is X(log X)−K for any K > 0. √ √ Proof.√ We consider two cases: y ≤ X, and X < y ≤ X. First, when y ≤ X, √ |ψ(y, χ) − δ(χ)y| y ≤ X trivially since we can bound the left-hand side by ψ(y) ∼ y. Hence X log Q X∗ √ √ √ X (log Q) XR (log X) XR. ϕ(q) q≤R χ (mod q) √ Second, when X < y ≤ X, q ≤ R = (log X)A+4 (log y)A+4, so by Siegel-Walﬁsz, p p |ψ(y, χ) − δ(χ)y| A y exp(−c log y) X exp(−c log X). Hence X log Q X∗ X exp(−cplog X) (log Q)X exp(−cplog X)R, ϕ(q) q≤R χ (mod q) and log Q log X.

35.2 The nontrivial part We now need to deal with X log Q X∗ max|ψ(y, χ) − δ(χ)y|. ϕ(q) y≤X R 1, there are no trivial primitive characters modulo q, so δ(χ) = 0, and we really only need to deal with X 1 X∗ log Q max|ψ(y, χ)|. (35.2.1) ϕ(q) y≤X R

We will prove this at some point, but for now let us assume it true. Let 1 X∗ f(q) = max|ψ(y, χ)|, ϕ(q) y≤X χ (mod q) the inner summand above, so that (35.2.1) is X X 1 (log Q) f(q) = (log Q) f(q)q. q R

Z Q 1 X (log Q) d f(q)q , t R q≤t where the weight is T (X, t). Integrating by parts gives us

Q Z Q ! 1 1 (log Q) T (X, t) + 2 T (X, t) dt t R R t Using the Mean value theorem above, this is Z Q ! (log Q) Q−1(X + X5/6Q + X1/2Q2)(log XQ)3 + t−2(X + X5/6t + X1/2t2)(log Xt)3 dt R and by bounding log Q log X and log Xt log XQ, Q−1(X +X5/6Q+X1/2Q2)(log X)4 +(XR−1 +X5/6 log Q+X1/2Q)(log X)4, which we can ﬁnally bound as X(log X)−A + X1/2Q(log X)4 since R = (log X)A+4.

Theorem 35.2.2 (Polya-Vinogradov inequality). Let χ 6= χ0 (mod q), with q > 1. Then M+N X 1/2 ≤ 2q (log q). n=M+1 Note that trivially the sum is ≤ q, but we know that if we sum over a full set of residues it is 0, so we should expect a lot of cancellation. This theorem reveals just how much!

Lecture 36 The Large Sieve

36.1 The Large siece inequality For N, Q > 1, we have

M+N 2 M+N X q X∗ X X a χ(n) (N + Q2)1/2 |a |2 ϕ(q) n n q≤Q χ (mod q) n=M+1 n=M+1

Date: April 19th, 2019. THE LARGE SIEVE 88

for any { an } ⊂ C. A consequence of this is Theorem 36.1.1. M N X q X∗ X X max ambnχ(mn) ϕ(q) y q≤Q χ (mod q) m=1 n=1 mn≤y

M !1/2 N !1/2 2 1/2 2 1/2 X 2 X 2 (log 2MN)(M + Q ) (N + Q ) |am| |bn| m=1 n=1 Note that without the condition mn ≤ y, the sum splits (since χ(mn) = χ(m)χ(n) by complete multiplicativity), and by Cauchy’s inequality

M N X q X∗ X X a b χ(mn) ϕ(q) m n q≤Q χ (mod q) m=1 n=1 1/2 1/2  M 2  N 2 X q X∗ X X q X∗ X ≤ a χ(m) b χ(n)  ϕ(q) m   ϕ(q) n  q≤Q χ (mod q) m=1 q≤Q χ (mod q) n=1 We can then use the Large sieve at each factor separately, getting

M !2 N !2 2 1/2 X 2 2 1/2 X 2 (M + Q ) |am| (N + Q ) |bn| m=1 n=1 So dropping the condition mn ≤ y gives us something quite close. We need a good way to detect this condition mn ≤ y.

Lemma 36.1.2. For any T > 0, β > 0, α ∈ R, and β 6= |α|, we have Z T itα sin(tβ) 1 e dt = δβ(α) + O , −T πt T |β − |α|| where ( 1, if |α| < β, δβ(α) = 0, if |α| > β. Proof sketch. First, Z ∞ itα sin(tβ) e dt = δβ(α). −∞ πt This is essentially a consequence of studying Z ds ys (c) s for c > 0, and then letting c → 0. Secondly, by integration by parts Z ∞ sin(tβ) 1 eitα dt , T πt T |β − |α|| and similarly for the lower tail. THE LARGE SIEVE 89

Now to use this lemma to detect mn ≤ y, we take β = log y and α = log mn, except the lemma only applies if β 6= |α|. We can get around this: since mn ∈ Z, 1 without loss of generality we can take y = k + 2 , k ∈ Z, so 0 ≤ k ≤ MN. Thus Z T it log(mn) sin(t log y) 1 δβ(log mn) = e dt + O −T πt T |β − |α|| Z T sin(t log y) 1 = (mn)it dt + O . −T πt T |β − |α|| Note that the term (mn)it is of modulus 1, which will come in handy soon. Note also that

mn |β − |α|| = |log y − log mn| = log . y Hence

M N M N X X X X ambnχ(mn) = ambnχ(mn)δβ(log mn) m=1 n=1 m=1 n=1 mn≤y Z T M ! N ! M N −1! X it X it sin(t log y) −1 X X mn = amm bnn dt + O T |ambn| log . πt y −T m=1 n=1 m=1 n=1 We need a few estimates. First, |log x| min{ 1, |x − 1| } for all x > 0, and therefore 1 mn mn y + 2 1 1 log min 1, − 1 min 1, − 1 . y y y y MN Also, |sin(t log y)| ≤ min{ 1, |t log y| } ≤ min{ 1, |t| log 2MN }, where we’ve taken 2MN to avoid the possible issue when M = N = 1. Therefore

M N Z T M N X X X −it X −it 1 max ambnχ(mn) amm χ(m) bnn χ(n) min , log 2MN dt y |t| m=1 n=1 −T m=1 n=1 mn≤y M N ! −1 X X +O T MN |ambn| , m=1 n=1 and by Cauchy the sums in the error term is

M !1/2 N !1/2 X 2 1/2 X 2 1/2 |am| M |bn| N . m=1 n=1 Therefore the ﬁnal error term is

 M !1/2 N !1/2 −1 3/2 3/2 X 2 X 2 O T M N |am| |bn|  m=1 n=1 VAUGHAN’S IDENTITY 90 and we’ll pick an appropriate T toward the end. Now average over X q X∗ , ϕ(q) q≤Q χ (mod q) bring it into the integral, since only the sums with χ in them depend on q, and so by the large sieve

M N X q X∗ X X max ambnχ(mn) ϕ(q) y q≤Q χ (mod q) m=1 n=1 mn≤y

M !1/2 N !1/2 X X Z T 1 (M + Q2)1/2(N + Q2)1/2 |a |2 |b |2 min , log 2MN dt+ m n |t| m=1 n=1 −T M !1/2 N !1/2 X q X∗ X X + T −1(MN)3/2 |a |2 |b |2 . ϕ(q) m n q≤Q χ (mod q) m=1 n=1

This last term is

M !1/2 N !1/2 −1 3/2 X 2 X 2 X T (MN) |am| |bn| q, m=1 n=1 q≤Q with the ﬁnal sum being Q2, so taking T = (MN)3/2 we get the error we want. Finally the remaining integral is

Z T 1 min , log 2MN dt log(2MN) −T |t| since T = (MN)3/2. Note that therefore it is important we didn’t take T → ∞ at ﬁrst to det just δβ(α), with no error, since then our upper bound is just inﬁnity.

Lecture 37 Vaughan’s Identity

37.1 Basic Mean Value Theorem, again Lemma 37.1.1 (Vaughan’s Identity). Suppose u > 0, v > 0, y ≥ 2, and f : N → C. Then X Λ(n)f(n) = S1 − S2 − s3 + s4 n≤y

Date: April 22nd, 2019. VAUGHAN’S IDENTITY 91 where X X S1(f) = µ(m) f(mn) log(n), m≤u y n≤ m X X X S2(f) = Cm f(mn), where Cm = Λ(k)µ(`), m≤uv y k≤u n≤ m `≤v kl=m   X X X  S3(f) =  Λ(k) µ(n)f(mn),   m>u n>v k|m mn≤y k>u X S4(f) = Λ(n)f(n). n≤v P Remark 37.1.2. Note how S1(f) and S2(f) are bilinear forms of the type m,n amcmn. If the length of the sum over m isn’t too big, then we can handle it. So we want u and v small compared to y. P Next, S3(f) is a bilinear form of the type m,n ambncmn. Here we can use the Large Sieve. Finally, S4(f) is of the same form as the original sum, but it’s shorter. Proof. Consider the purely algebraic identity

ζ0 1 − (s) = G(s)(−ζ0(s)) − F (s)G(s)ζ(s) − (−ζ0(s) − F (s)ζ(s))(G(s) − ) + F (s), ζ ζ(s) where X Λ(n) F (s) = ns n≤u and X µ(n) G(s) = . ns n≤v

Call the ﬁrst terms D1(s), D2(s), D3(s), and D4(s), respectively, and write them as Dirichlet series ∞ X aj(n) D (s) = j ns n=1 for j = 1, 2, 3, 4. Since the left-hand side above has coeﬃcients Λ(n), we get the identity Λ(n) = a1(n) − a2(n) − a3(n) + a4(n), VAUGHAN’S IDENTITY 92 where consequently X a1(n) = µ(d) log n, md=n d≤v X a2(n) = Λ(m)µ(d), mdr=n m≤u d≤v X X a3(n) = Λ(m) µ(d), mk=u d|k m>u d≤v k>1 ( Λ(n), n ≤ u a4(n) = 0, n > u.

Adding these up and weighing by f(n), we get X X X X X Λ(n)f(n) = a1(n)f(n) − a2(n)f(n) − a3(n)f(n) + a4(n)f(n), which is the identity sought. We have an outstanding result ot prove. We want to show that X q X∗ T (X,Q) := max|ψ(y, χ)| (X + X5/6Q + X1/2Q2)(log XQ)3, ϕ(q) y≤X q≤Q χ (mod q) where X ψ(y, χ) = Λ(n)χ(n). n≤y

Proof of Basic mean value theorem. We have two cases. First, if Q2 > X, then apply the Large Sieve with M = 1, a1 = 1, N = bXc, and bn = Λ(n), getting

1/2 bXc  2 1/2 X s T (X,Q) (log X)Q(X + Q )  Λ(n)  n=1

The penultimate term is (Q2)1/2 = Q, and the ﬁnal term we can bound by

 1/2 X 2 2 1/2  (log q)  (log X) X . pα≤X

So in all, this expression is bounded by

(log X)2Q2X1/2.

For the second case, Q2 ≤ X, we split into y ≤ u2 and u2 < y ≤ X. If we cheat a bit and let u = v = min{ Q2,X1/3,XQ−2 } (which are properly acquired VAUGHAN’S IDENTITY 93 by doing the calculations with u and v and then optimising), we get, for y ≤ u2, via the Large Sieve,

 1/2 2 2 1/2 2 X 2 Q(u + Q ) (log u )  Λ(n)  n≤u2 Q(u2 + Q2)1/2(log u2)(log(u2)u)1/2 Q(uQ)(log u)2u (QX2/3 + Q2X1/3)(log X)2.

Finally for u2 < y ≤ X, apply Vaughan’s theorem with ( χ(n), n ≤ y f(n) = 0, n > y.

Then it suﬃces to bound X q X∗ Tj := max |Sj(χ)|. ϕ(q) u2