Riemann s Zeta Function

Riemann s Zeta Function Lectures by Adam Harper Notes by Tony Feng Lent 204

Preface This document grew out of lecture notes for a course taught in Cambridge during Lent 204 by Adam Harper on the theory of the Riemann zeta function. There are likely to be errors, which are solely fault of the scribe. If you find any, please contact me at tonyfeng009@gmail.com.

Contents Preface i Analytic Theory of ζ(s). Introduction................................2 Basic theory of ζ(s)............................3 Chebyshev s function and Perron s formula.............. 4.4 Meromorphic continuation of ζ(s)................... 9.4. Some Fourier Analysis..................... 2 The Prime Number Theorem 6 2. The Prime Number Theorem..................... 6 2.2 The classical zero-free region...................... 2 2.3 Additional properties of ζ(s)...................... 28 3 Vinogradov s Method for Zero-Free Regions 30 3. Estimating ζ sums........................... 30 3.2 Bilinear Forms............................. 32 3.3 Vinogradov s Mean Value Theorem.................. 38 3.4 Second estimate on estimating zeta sums............... 46 3.5 The Vinogradov-Korobov zero-free region............... 49 4 Primes in Short Intervals 53 4. The explicit formula.......................... 53 4.2 Counting zeros............................. 55 4.3 Counting zeros with large real part.................. 56 5 Example Sheet 63 6 Example Sheet 2 72 7 Example Sheet 3 83 ii

Chapter Analytic Theory of ζ(s). Introduction The prime numbers are a fundamental object of interest in number theory. There are natural questions that we can ask about their distribution, such as: how many primes are x? how many primes are in the interval [x, x + x 0.99 ]? The course is about the Riemann zeta function, especially its applications to questions about the distribution of the primes such as the above. The connection between the zeta function and the distribution of the primes is not obvious. The Riemann zeta function is morally the Fourier transform of the prime numbers. This analogy can be made precise, and though we shall not do so here, it will crop up again and again in our investigation. In the first part of the notes, we develop some basic analytic theory of ζ(s) and use it to prove the Prime Number Theorem. In the second part, we describe Vinogradov s method, which leads to the widest known zero-free region (and hence the strongest error term in the Prime Number Theorem). In the last part, we describe applications to counting primes in short intervals. There are some notable omissions. Unfortunately, we did not have time to explain the functional equation of ζ(s), which is a fundamental part of the story, or applications such as the Explicit Formula..2 Basic theory of ζ(s). Definition.2.. For s C with Re (s) >, we define the Riemann zeta function ζ(s) = n s. n= Note that this series is absolutely convergent.

Analytic Theory of ζ(s) 2 More generally, any series of the form L(s) = n= a n n s, (a n) C is called a Dirichlet series (or an L-series). If only finitely many of the a n are non-zero, then the resulting finite sum a n n N n s is called a Dirichlet polynomial. The reason is that if we let s = σ + it, the sum can be rewritten as a n n = a n s n σ e it log n. n N n N If the e it log n were replaced with e itn, the sum would be a trigonometric polynomial. This may appear a bit far-fetched right now, but we will see that there is indeed a strong analogy between Dirichlet polynomials and trigonometric polynomials. Lemma.2.2. For any s with Re s > and any x N, we have ζ(s) = n x n s + x s s + O( s x σ ). The basic idea is that you can approximate a sum with an integral: ζ(s) = [ ] n z s s z dz = = s s s. However, this isn t very good because the integral doesn t approximate the sum very well for small n, where the summand decreasing more rapidly. The integral approximation turns out to be better for the later terms in the sum, which vary slowly. Proof. Note that so n>x 0 dw w s+ = n = s s n>x n [ w s s dw w s+ = s ] = s n x n s, ( x<n w ) dw w s+. (The interchange of summation and integration is justified because everything is absolutely convergent.) This tail can be bounded as s x ( x<n w ) dw w s+ = s = s x x ( w x) dw w s+ (w x) dw w s+ + O ( s x ) dw w s+

Analytic Theory of ζ(s) 3 since w w. Now, one simply computes that s x (w x) dw x s = ws+ s. Remark.2.3. Arguments such as the above are more naturally phrased using the apparatus of Riemann-Stieltjes integration, which we will do in the future. For an example of how it can be applied in this case, observe that x n s = = x t s d t = x s s t s d(t {t}) x t s d{t} = x s s {x}x s s x {t}t s+ dt. Corollary.2.4. If σ > and t 2, then Proof. Choose x = t. Then ζ(σ + it) = O(log t ). ζ(σ + it) = n + x s s s + O( s x σ ) n t = ( ) n + O + O() s t n t and now note that n s n t n t n log t. Remark.2.5. From the proof, we see that the Lemma also holds for small t if σ is bounded away from. The zeta function is defined in terms of n s, for all n N. Since each n has a unique prime factorization, we might hope to express ζ(s) only in terms of p s, for p prime.

Analytic Theory of ζ(s) 4 Lemma.2.6 (Euler product). For any s with Re s >, we have ζ(s) = ( ). p s By this infinite product we mean Proof. For any prime p, lim p prime P p prime P ( ) = p s ( ). p s k=0 p ks. Since we can rearrange and multiply out a finite product of absolutely convergent series, we see by unique prime factorization that p P ( ) = p s where a n = if all prime factors of n are at most P, and 0 otherwise. Therefore, ζ(s) ( ) p s n s n, σ n>p n>p p prime P which vanishes as P. We have seen that, when Re s >, ζ(s) has representations as a series and as an infinite product over primes. We can use analysis of the series representation to deduce facts about the distribution of primes. Most research on ζ(s) is about developing similar properties for analytic continuations of ζ(s), and playing them off each other to get information about ζ(s) or about primes. n= a n n s.3 Chebyshev s function and Perron s formula In this section, we will see how to leverage information about ζ(s) to deduce information about the prime numbers. Riemann introduced the function π(x) = p prime x to discuss the distribution of primes. It is standard to work instead with a different, technically more natural counting function.

Analytic Theory of ζ(s) 5 Definition.3.. The von Mangoldt function Λ : N R is defined by { log p n = p k for some prime p Λ(n) :=. 0 otherwise Definition.3.2. Chebyshev s ψ function by ψ(x) = n x Λ(n). We can rewrite ψ(x) = p prime x log p + log p k=2 p x /k Note that log p = O( x log x). p x /k and only the first O(log x) values k make a contribution, since x / log x = e. We can conclude that ψ(x) = log p + O( x log 2 x). p x So Chebyshev s function is basically a weighted prime counting function. Why is the von Mangoldt function easier to work with? The reason is that it arises naturally from the ζ function, as its logarithmic derivative. Indeed, from the Euler product expansion ζ(s) = ( ) p s p we see that log ζ(s) = p Differentiating, we find that log( p s ) = p k p ks k ζ (s) ζ(s) = p log p k p ks = n Λ(n)n s. If we had looked at p, we would have ended up with something like log ζ(s), p s which is nasty because of issues with branch cuts. You might worry about the ζ(s) in the denominator. Since ζ(s) is an absolutely convergent product when Re s >, no factor of which is zero, the product itself

Analytic Theory of ζ(s) 6 does not vanish. This is a first glimpse of the connection between primes and the vanishing of ζ(s). We can think of Λ(n) n= as a bit like a Fourier transform of {Λ(n)}, since if n s s = σ + it then n σ + it = log n e it nσ and e it log n is like the phase e itn in a Fourier series. So maybe we can formulate some kind of Fourier inversion theorem in this setting, in which we get information about ψ(x) = n x Λ(n) by integrating Λ(n) n. This turns out to be a fruitful n σ+it idea. Lemma.3.3. Let y, c, T > 0. Define 0 0 < y <, δ(y) = y =, 2 y >. Then δ(y) c+it 2πi c it y s ds s { y c min{, T log y } y min{, c T } y =. Proof. The obvious idea is to try and use the residue theorem. We consider the cases separately. Suppose first that 0 < y <. Then y s s y σ σ 0 uniformly as σ = Re s, so we want to shift the contour to the right. Fixing T, we can shift the right side to, so the residue theorem implies that 0 = c+it c it y s ds +it s + y s ds c it c+it s + y s ds it s. Tony: [insert a picture?] Now, we can bound the two horizontal integrals as +it y s ds c+it s y σ dσ [ ] y σ c T = y c T log y c T log y. This gives half the bound; we still need to prove the bound y c. To do this, we consider instead consider completing the contour with the arc of a circle centered at the origin. Applying the residue theorem again, 0 = c+it c it y s ds s + y s ds Γ R s

Analytic Theory of ζ(s) 7 where Γ R the arc of a circle centered at the origin, cut off by the points c + it and c it, hence whose radius is R = c 2 + T 2. Parametrizing s = Re iθ, y s ds Γ T s yc πy c. dθ The case y > is similar, but with all contours shifted to the left. The case y = 0 is a direct computation. These are handled in Example Sheet, Question 4. We introduce the notation n x to mean that if x N, then a x should be replaced by a 2 x. Lemma.3.4 (Truncated Perron formula). Let x, c, T > 0, and suppose that a n n= a n n c is convergent. Then n x a n = ( c+it ) a n 2πi c it n s n= Proof. Note that ( x s s ds + O x c n x a n = n n= a n δ(x/n). ) a n n min{, c T log(x/n) }. Applying Lemma.3.3, a n δ(x/n) = [ c+it ( x ) ( )] s ds x c a n 2πi n n c it n s + O n min{, c T log(x/n) }. Since x s a n n is convergent (by the hypothesis), it is possible to swap the summation and integration. Remark.3.5. In fact, Lemma.3.4 is true with n x a n replaced by n x a n, since if x N, then a x is part of the big Oh. Remark.3.6. The integral on the right hand side isn t so nice because decays too s slowly. This is analogous to what happens with Fourier series, where the problem n s ds s

Analytic Theory of ζ(s) 8 is resolved by introducing the Féjer kernel. You can play a similar game here, weighting the a n by some slowly decaying function. See, for instance, Example Sheet Question 7 or Example Sheet 3 Question. Applying Perron s formula to the Chebyshev ψ-function, we get that for any < c 2, x >, and < T x, ψ(x) = c+it ( ) ( ζ (s) x s ) 2πi c it ζ(s) s ds + O x c Λ(n) min{, n c T log x/n }. We can simplify the error term. The T log x/n term will be smaller except when n is pretty close to x, so we should separate out that case. Doing so, we obtain the bound x c n= Λ(n) n c min{, T log x/n } xc T n= n= Λ(n) n c +x c x 2 <n<2x Λ(n) n c min{, T log x/n }. In the second sum we have xc = O() by the choice of range of summation. So we n c can substitute this in to get the error bound O xc log n + log 2 x min{, T n c n= x T log x/n }. 2 <n<2x By approximating the first sum with an integral, we obtain x c T n= log n n c ( = O x c T (c ) 2 For the second sum, note that tere are x x terms with n x <. For these terms, T 2T we use the bound on the summand. We split the rest of the sum into sums over dyadic intervals, with increasing length as n gets away from x. x min{, x 2 <n<2x T log x/n } x log 2 T + T + k=0 ). 2 k x n x < 2k x T T T log(x/n). By Taylor expansion, log x ( n = log + x n ) n x n n x n x since n x.

Analytic Theory of ζ(s) 9 Substituting this above, we obtain log x 2 T + T + k=0 2 k x n x < 2k x T T T log(x/n) x log 2 T + T + k=0 x log 2 T + T + k= n x < 2k x T x T (2 k x/t ) x T x T + x (log T ). T To summarize, we have obtained the two error terms and X T log2 T. We choose the parameter c to balance them, which leads to c = +. With this log x choice, we have proved x c T (c ) 2 Proposition.3.7. For any x >, if T x and < c < 2 then we have Ψ(x) = c+it ( ) ( ζ (s) x s x log 2 ) 2πi c it ζ(s) s ds + O x T For similar analyses of other arithmetic functions, it may be useful to record our technical result. Proposition.3.8. For any x >, if T x and < c < 2 then we have x c n min{, c T log(x/n) } x log x. T n x At this point, we could try to get very precise estimates for ζ( + + it) and logx ζ ( + + it), and plug them in. Instead, we will follow the classical approach, log x and try to estimate the integral by Cauchy s Residue Theorem. To do this, we must extend ζ(s) to the entire right half-plane..4 Meromorphic continuation of ζ(s) We have defined the zeta function in the region Re s > 0. The interesting arithmetic about prime numbers really comes from knowledge of ζ(s) in the critical strip Re s [0, ]. So we first have to analytically continue it. In fact, we have already done this in Remark.2.3! The expression there defines an analytic function for Re s > 0, since the integral converges (all other terms are obviously holomorphic). Therefore, we are justified in the following definition. Definition.4.. For each s C such that Re s > 0, except for s =, and for any x > 0, we define ζ(s) = n + x s s s + {x} {w} s dw, x s n x x ws+

Analytic Theory of ζ(s) 0 where {x} = x x is the fractional part of x. Remark.4.2. In Example Sheet 2, Question 2 you are asked to find a further analytic continuation to the region Re s >, using similar tricks. In fact, ζ admits a meromorphic continuation to all of C, with only a pole at s =. To prove this, one need to develop the functional equation. Lemma.4.3. For any t such that t is sufficiently large, and any σ > 00 log t (say), we have ζ(σ + it) = O(log t ). On the same range, we also have ζ (σ + it) = O(log 2 t ). Proof. If t is sufficiently large then certainly 00 log t Definition.4.. Again, we choose x = t, so we find ζ(σ + it) = n t n σ+it + O ( ) t 00/ log t t > 0, so we can apply ( ) + O + O ( σ + it t ). t σ σ Note that t / log t = e, so all the error terms are O(). We then get But n σ+it = n σ ζ(σ + it) = + O(). nσ+it n t n since n t, so n σ = O(log t ). n t To prove the estimate for ζ (σ +it), just differentiate and apply the same analysis. Remark.4.4. Although we stated this for sufficiently large t, the proof shows that it holds for small t as well as long as s is bounded away from, just as in Remark.2.5. Remark.4.5. We will often want to control ζ and ζ for small t and s near. To do this, one can use the pole at s = to establish that in the region 0.0 < Re s < 2 and Im s < C, one has ζ(s) s and ζ (s) s 2 You are asked to prove this in Example Sheet 2, Question. If we are to do analysis on ζ(s) left of the line Re s =, we need to have decent estimates on it. We will now develop an extremely good approximation to ζ(s), due to Hardy and Littlewood.

Analytic Theory of ζ(s) Theorem.4.6 (Hardy-Littlewood, 92). If s = σ + it for any σ > 0, t R, and if x > t π, then ζ(σ + it) = n x n s + x s s + O(x σ ). The point is that the implied constant is independent of s. Note that the error term O(x σ ) is much better than what we got in Lemma.2.2 (athough we do have the additional condition x t ). π The main ingredient in proving this is a lemma of Van der Corput, which says that certain sums are well-approximated by integrals, provided that the sum doesn t oscillate too quickly. Definition.4.7. We write e(θ) := exp(2πiθ). Lemma.4.8 (Special case of Van der Corput, 92). Let f(x) be a real-valued function on an interval [a, b] R. Suppose that f (x) is continuous and monotone on [a, b] and suppose that f (x) < δ for some δ <. Then a<n b e(f(n)) = b a ( ) e(f(x)) dx + O. δ Before giving the proof, we need to develop some Fourier analysis..4. Some Fourier Analysis If f is a function defined on [0, ], it is sometimes convenient to consider a trigonometric expansion for f of the form f(x) = n a n e(nx). Towards this end, ecall that we define the Fourier transform of f to be ˆf(n) = 0 e( xn)f(x) dx So that if there is an expansion as above, then (at least formally) we have ˆf(n) = a n. We would like to say that f(x) = n= ˆf(n)e(nx) but this is obviously impossible in certain situations. For instance, the right hand side is continuous even if the left hand side is not.

Analytic Theory of ζ(s) 2 Theorem.4.9 (Dirichlet). Let f be a piecewise-differentiable function. Then lim N N n= N ˆf(n)e(nx) f(x +) + f(x ). 2 This is not terribly hard to prove, but somewhat tangential to the course, so we will just assume it. Proof of Lemma.4.8. The idea is to use (a finite version of) Poisson summation. It will turn out that the zero Fourier mode produces the main term, while all the others contribute to the error term. By Theorem.4.9, we have the identity e(f(n)) + e(f(n + )) 2 = lim N N k= N u n (k), where u n (k) = e(f(n + x) kx) dx. 0 Notice that u n (0) = n+ n e(f(x)) dx. If k 0 then we expect u n (k) to be comparatively small since the integrand e(f(n + x) kx) is the product of the slowly varying function e(f(n + x)) and a more rapidly oscillating function e kx, producing a lot of cancellation. A general principle is that one should perform integration by parts in order to capture this cancellation. u n (k) = e(f(n + x) kx) dx 0 [ ] e(f(n + x))e( kx) = + 2πik 0 k e(f(n)) e(f(n + )) = + 2πik k 0 n+ n f (n + x)e(f(n + x) kx) dx f (x)e(f(x) kx) dx since e( kx) = e( k(n + x)). T first term can be ignored when we take the sum, because the k term cancels the k term. Therefore, e(f(n)) = n (0) + a<n b [a]+ n [b] (u u n (k)) + O() k 0 = b a e(f(x)) dx + O() + k 0 k b a f (x)e(f(x) kx) dx

Analytic Theory of ζ(s) 3 Here the error term integrals have rapidly oscillating integrands, so we should integrate by parts again. Doing so, we obtain a<n b e(f(n)) = b a e(f(x)) dx + O() + k 0 2πik f (x) d e(f(x) kx) dx. f (x) k dx f (x) f (x) k Notice that since f (x) is monotonic and f (x) <, the function is also monotonic for each fixed k 0. Now we are morally done. In general, if h(x) is monotonic, then h(x)g(x) is essentially bounded by max h(x) g(x). In our case, the g is O() since g is itself a derivative of a bounded function. Furthermore, max h(x) for all k k, so that part of the sum contributes. The remaining error comes k 2 from k =, which contributes. δ Lemma.4.0 (Abel). If c c 2... c N and if d,..., d N R are arbitrary, then N N N c n d n c N d n + (c N c N ) d n +... + (c c 2 ) d n= n= n= N (c c N + c N ) max d N N n. Proof. Recall the Abel summation formula: if D n = n k= d n, then Therefore, N c n d n = n= = n= N c n (D n D n ) n= N c n D n n= N n=0 c n+ D n N = c N D N + (c n c n+ )D n. N c n d n c N D N + n= which is the desired conclusion. n= N n= (c c N + c N ) max m c n c n+ D n D m

Analytic Theory of ζ(s) 4 find b a Approximating the integrals by Riemann sums and using Lemma.4.0, we f (x) f (x) k d e(f(x) kx) dx max dx x [a,b] since e(x). Therefore, a<n b e(f(n)) = b a k δ f (x) f (x) k max x [a,b] e(f(x)) dx + O() + O ( k 0 x [a]+ d e(f(x) kx) dx dx ). k ( k δ) Separating into the cases k and k =, as discussed above, we find that the last term is O ( δ ). We are now ready to prove the Hardy-Littlewood approximation. The point is that Lemma.4.8 gives us precise enough control of the Dirichlet polynomial part of ζ(s). Proof of Theorem.4.6. Let N x be a large parameter. By Definition.4., ( ) s. ζ(s) = n N n s + N s s + O(N σ ) + O N dw w σ+ In particular, if N is large enough (in terms of x, σ, t) then both big O terms are O(x σ ). So we need to show that the first two terms are not significantly affected by changing x to N, i.e. ( n N But the left hand side is x<n N n s N x ) n + N s s s w s dw = x<n N ( n x ) n + x s = O(x σ ). s s N n σ e it log n x w σ e it log w dw. (.) This is almost of the form of Lemma.4.8, but the exponential sums are weighted. We use summation by parts to get rid of the weights. Using the identity y σ = y x σ dv v σ+ + x σ

The Prime Number Theorem 5 we can rewrite the right hand side of (.) as = x<n N ( n = x σ ( x x<n N σ dv v σ+ e it log n + N x + x σ N x σ dv v σ+ ) N e it log n e it log w dw ( v<n N x ) e it log n ( w x N v σ v dv + ) e it log w dw σ+ x σ e it log w dw We have isolated terms comparing the exponential sums and integrals, which allows us to apply the previous lemma (after checking that the hypotheses hold). Note that e it log n = e(f(n)) where f(n) = t log n. We have f (x) = t, 2π 2πx which is indeed monotonic. Furthermore, f (z) t 2πx 2 by our assumption x t. So by Lemma.4.8 applied with δ =, we obtain that π 2 (.) is ( N ) (O(x σ σ ) + O dv = O(x σ ) x vσ+ ) Theorem.4.6 says that we can approximate ζ(σ+it) by a Dirichlet polynomial with t terms. It turns out that one can understand Dirichlet polynomials of length t, evaluated where Im s t quite well (as we will see in Chapter 3).

Chapter 2 The Prime Number Theorem 2. The Prime Number Theorem Theorem 2.. (Hadamard, de la Vallée Poussin, 896). As x, we have Ψ(x) x. More precisely, ( ) Ψ(x) = x + O x exp( c log /0 x). Hadamard and de la Vallée Poussin didn t originally obtain this quantitative error term at first, but a few years afterwards de la Vallée Poussin obtained a better error term O(x exp( c log x)), which is called the classical error term. We will develop this later. The usual version of the Prime Number theorem says something about the distribution of the primes. Corollary 2..2. As x, π(x) x 2 dt log t. The function x dt is called the logarithmic integral Li(x), and one can use 2 log t integration by parts to expand Li(x) as x log x + x log 2 x + O ( ) x log 3. x So π(x) Li(x) x, but the Li(x) is a better estimate. The relation between log x Ψ and π is investigated in Example Sheet, Question 3. The key ingredient to proving the prime number theorem is establishing that the Riemann zeta function has no zeroes on the line Re s =. The intuition is that if ζ has a zero at ρ, then ζ has a pole at ρ. The Euler product gives us good ζ 6

The Prime Number Theorem 7 control over ζ ζ for Re > s >. Therefore, if ζ in the region Re s > ; in particular, we know that it is bounded ζ had a pole on or near the line Re s =, it would have to decrease extremely rapidly. The basic zero-free region arguments are guided by a simple but convincing heuristic. Suppose there was a zero at ρ = σ + it. Then for small δ the Laurent expansion implies that ζ ζ (ρ + δ) = δ. On the other hand, we saw from the Euler product that ζ ζ (s) = p k= log p p ks. For σ > + ɛ, the contribution from prime powers is bounded, so 2 Re ζ ζ (ρ + δ) p log p cos(kt log p) p +δ. If this is to be close to for small δ, then it must be the case that cos(kt log p) δ for many values of k. But then cos(2kt log p) for many values of k, so we expect that Re ζ ( + 2it + δ), suggesting that there is a pole of ζ at + 2it. ζ δ Of course, we know from our meromorphic continuation that the only pole (in the region Re s > 0, for now) is at s =. This argument was made rigorous de la Vallée Poussin. A more elegant formulation was given by Mertens using the identity 3 + 4 cos θ + cos 2θ = 2(cos θ + ) 2 > 0 to make precise the idea that if cos θ is large and negative, then cos 2θ is large and positive. This is the key identity in the following proof. Theorem 2..3 (Weak zero-free region). There exists a small absolute constant c c > 0 such that the following is true. For any t R and any σ, we log 9 ( t +2) have ζ(σ + it) = O(log7 ( t + 2)). In particular, ζ(σ + it) 0. Proof. We assume throughout that σ is less than a small positive constant, since otherwise the result is trivial ( will in fact be O()). We may also assume ζ that t is greater than a small positive constant, since otherwise we are close to the pole at s =, in which case is small. ζ

The Prime Number Theorem 8 Let σ > be a number to be chosen later in terms of σ, t. Observe that for t R, we have ζ(σ + it ) = exp (Re log ζ(σ + it )) ( = exp Re ( log ) ) p σ +it p ( = exp Re ) kp k(σ +it ) p k ( = exp Re ) cos(t k log p) kp kσ p The key idea is to use the identity applied as above to the product 3 + 4 cos θ + cos 2θ = 2(cos θ + ) 2 > 0 ζ(σ ) 3 ζ(σ + it) 4 ζ(σ + 2it). Indeed, we have ζ(σ ) 3 ζ(σ + it) 4 ζ(σ + 2it) = exp 3 + 4 cos(kt log p) + cos(2kt log p) p k = exp. k kp kσ p k 2( + cos(kt log p)) 2 kp kσ Therefore, the only way that ζ(σ + it) can be small is if ζ(σ ) or ζ(σ + 2it) are very large. More precisely, ζ(σ + it) 3ζ(σ ) 3/4 ζ(σ + 2it) /4 (σ ) 3/4 log /4 ( t + 2) (2.) using our bounds from earlier. Now observe that ζ(σ + it) ζ(σ + it) σ σ ζ (r + it) dr. Note that this makes sense because we have assumed that t is bounded away from zero, so the integral avoids the pole and is well-defined. Using Lemma.4.3 (if t

The Prime Number Theorem 9 is large) and Remark.4.4 (if t is small and s is bounded away from ), we have ζ(σ + it) ζ(σ + it) σ σ ζ (r + it) dr (2.2) = ζ(σ + it) + O( σ σ log 2 ( t + 2)). (2.3) We will choose σ = + δ for some small δ so that we can use the above equation to deduce a lower bound on ζ(σ + it). By (2.) we have ζ(σ + it) δ3/4, log /4 ( t +2) so we should choose δ to balance the difference O( σ σ log 2 ( t + 2)) error term. That is, δ log 2 δ 3/4 ( t + 2) log /4 ( t + 2) = δ log9 ( t + 2). Therefore we set σ = + If σ + c log 9 ( t +2) c, so (2.) implies log 9 ( t +2) ζ(σ + it) c 3/4 log 7 ( t + 2). also, then we are already done. Otherwise, we have c log 9 ( t + 2) σ < + c log 9 ( t + 2) so σ σ log 2 c ( t + 2). If c is sufficiently small, then this is small log 7 ( t +2) compared to ζ(σ + it), and we have by (2.2) that ζ(σ + it) c 3/4 log 7 ( t + 2). We are now ready to prove Theorem 2... The trick is to use Perron s formula, and then shift the contour. The error term comes from the other integrals involved in the contour shift, and it is better the further we can shift to the left. This argument will come up again and again. Proof of Theorem 2... Recall from Lemma.3.4 (and the subsequent discussion) that if x is large and < T x, we have Ψ(x) = + log x +it ( ) ζ (s) x s 2πi + log x it ζ(s) s ds + O ( x log 2 x We will use Cauchy s residue theorem to estimate the integral, and then choose T to balance all the error terms. c Let δ = and σ log 9 (T +2) 0 = δ, σ = +,. We have proved that ζ has log x no zeros or poles in the rectangle with vertices (σ 0, ±it ) and (σ, ±it ) except at T ).

The Prime Number Theorem 20 s =, and the integral over the rectangle of ζ (s) ζ(s) Cauchy s residue theorem. So x s s is the residue at s = by ( Ψ(x) = Res s= ζ (s) ζ(s) ) x s + ( σ0 +it σ +it σ it ( ) ) + ζ (s) x s s 2πi σ 0 it σ 0 +it σ 0 it ζ(s) s ds. Finally, we can easily check that the residue is x since ζ has a simple pole at s =, so Ψ(x) = x + ( σ0 +it σ +it σ it ( ) ) + ζ (s) x s 2πi σ 0 it σ 0 +it σ 0 it ζ(s) s ds. We now bound the integrals. First we tackle the short horizontal integrals. On c the line segment I = [ + it, + + it ], we crudely bound each log 9 (T +2) log x factor: (s) x s ζ ζ(s) s x+/ log x max ζ (s) max T I I ζ(s) Lemma.4.3 implies that max ζ log 2 T and Theorem 2..3 implies that max ζ log 7 T, so we have (s) x s ζ ζ(s) s x T (log2 T )(log 7 T ) x T log9 x. The same is true on the other line segment, since all of these estimates were symmetric with ( respect ) to the y-axis. Therefore, the contribution from the short x log integrals is O 9 x (since the line of integration has length ). T It remains to bound the integral σ0 +it σ 0 it ζ (s) x s ζ(s) s ds. We again use a crude, factor-by-factor estimate (s) x s ζ ζ(s) s x δ max ζ (s) max ζ(s) s I I x δ s max ζ (s) log 7 T. When estimating max ζ (s), we want to use Lemma.4.3 but we must be careful because the some parts of the integral are quite close to the pole at s =. For these parts, we use the asymptotic expansion near the pole as in Remark.4.5 (since σ ) to deduce that log 9 T max ζ (s) log 8 T.

The Prime Number Theorem 2 Putting this together, and promising that we will later choose T x, we have ( T ) ( ψ(x) = x + O x δ log 25 x log 9 ) x T + t dt x + O. T The integral can be bounded by O(log T ), so the error term is ( x log 9 ) O(x δ log 26 x x + O. T We can crudely bound this popping out a factor of x log 9 x x log 26 x to get c log x O(x log 26 x(e log 9 (T +2) + e log T )). Notice that the two exponential terms change in opposite directions with T. We can balance them out by making them equal, so it s natural to set T = exp(log /0 x). That gives us an error term of O(x log 26 xe c log/0 x ). By increasing c, we can take out the log 26 x, which gives the form stated in the theorem. Remark 2..4. From now on, we shall always absorb logarithmic factors (as in the last line of the above proof) in order to obtain cleaner bounds. 2.2 The classical zero-free region The error term O(xe c log/0 x ) in (our version of) the prime number theory isn t very good. For example, suppose that we wanted to investigate the distribution of Γ(n) in fairly long intervals, e.g. [x, x + xe log/4 x ]. The obvious approach is to note that Λ(n) = ψ(x + xe log/4 x ) ψ(x) x<n x+xe log(/4 x and then apply the prime number theorem, which for us says ψ(x + xe log/4 x ) ψ(x) = xe log/4 x + O(xe c log/0 x ). But this is useless, since the main term is smaller than the error term. Our immediate goal is to improve the error term in Theorem 2... What we saw in the proof was that the quality of the error term depended on how far we could shift the line of integration to the left. To get a better zero-free region, we shall prove the following important technical theorem.

The Prime Number Theorem 22 Theorem 2.2. (Landau, 924). Let φ(t) and w(t) be non-decreasing functions such that φ(t) as t. Also let t 0 0 be any fixed constant. Suppose that w(t) = O(e φ(t)/2 ) and ζ(σ + it) = O(e φ(t) ) for all w(t) < σ 2, t t 0. Then there exists a constant c > 0, depending only on the implicit constants, such that ζ(σ + it) 0 for all σ c φ(2t + )w(2t + ) Corollary 2.2.2. There exists a constant c > 0 such that ζ(σ + it) 0 when σ c. log( t +2) Remark 2.2.3. This improves on the weak zero-free region earlier of σ > c log 9 ( t +2). Proof. We apply Theorem 2.2. with w(t) := 2, φ(t) = log t, and t 0 = 3 (say). We just need to check that ζ(σ + it) = O(t) in this region. To get this, use Theorem.4.6 with x = t: ζ(σ + it) = n t n + t (σ+it) σ+it σ + it + O(t σ ). Since σ t (σ+it), the error terms + 2 σ+it O(t σ ) are O() and the sum n t = n σ+it O( t). c So the theorem implies ζ(σ + it) 0 when σ. If 0 < t 3, then log(2t+) the theorem follows from the weaker result in Theorem 2..3 since this region is bounded in t. Remark 2.2.4. It is not necessary to use Theorem.4.6. In this bounded region, one can work directly with Definition.4.. Also, by choosing t 0 smaller and considering the pole at s =, we can avoid using Theorem 2..3. The proof of Theorem 2.2. has two ingredients. One of them is the key identity from before, ζ(σ ) 3 ζ(σ + it) 4 ζ(σ + 2it) if σ >, as we have seen before. The new ingredient is the following Lemma. Lemma 2.2.5. Let r, M > 0 and let z 0 C. Suppose that f(z) is a holomorphic function on the disc z z 0 r, that f(z 0 ) 0, and that M on the disk f(z) f(z 0 )

The Prime Number Theorem 23 z z 0 r. Then if f(z) 0 on the right half of the disc, we have ( ) f (z 0 ) Re 8 log M + Re f(z 0 ) r z 0 ρ. ρ:f(ρ)=0 ρ z 0 r/2 Proof. Let Z denote the multiset of all zeros of f in the small disc z z 0 r, 2 counted with multiplicity. (This is a finite set since f is not identically zero.) Define a function g(z) on the large disc z z 0 r to be g(z) = f(z) ρ Z z ρ. (A priori, this is not defined for ρ Z but it has removeable singularities at all such points.) By the maximum principle, max g(z) z z 0 r g(z 0 ) = max g(z) z z 0 =r g(z 0 ) max f(z) z z 0 =r f(z 0 ) max z 0 ρ z z 0 =r z ρ M where the last inequality follows from z ρ r and z 2 0 ρ r for all ρ. 2 Next, we can define ( ) g(z) h(z) = log. g(z 0 ) since g is a nonvanishing function on a simply-connected domain. This is holomorphic on the small disc z z 0 r, where we have h(z 2 0) = 0 and Re h(z) = log g(z) g(z 0 ) log M. Now we are in a situation to apply the Borel-Carathéeodory theorem, which gives a bound on the modulus of a holomorphic function at a point given a bound for its real part in a surrounding disc. Theorem 2.2.6 (Borel-Carathédory). If h(z) is a holomorphic function on the disc z R and if h(0) = 0 and r < R, then max h(z) z r Proof. See Example Sheet 2, Question 5. 2r R r max Re h(z). z R Since h(0) = 0, the Borel-Caratheodory Theorem implies that for any r < r 2, max h(z) 2r log M. z z 0 =r (r/2) r ρ

The Prime Number Theorem 24 Combining this bound with Cauchy s integral formula, we have h h(z) (z 0 ) = 2πi (z z 0 ) dz 2 8 log M. r Finally, just note that Taking real parts, we are done. z z 0 = r 4 f (z 0 ) f(z 0 ) = d dz log f(z) z=z 0 = d dz (log g(z)) z=z 0 + ρ Z = h (z 0 ) + z 0 ρ. ρ Z z 0 ρ Remark 2.2.7. Note that Re = Re z 0 ρ z 0 ρ z 0 > 0 for each term in the sum. This ρ 2 means that we can get a lower bound by throwing away any terms in the sum. Now we are ready to tackle Landau s theorem. Before diving into the proof, it may be helpful to outline the broad strategy. As discussed before, the intuition is that ζ (s) wants to be large near a zero ρ = σ + it of the zeta function, since it has ζ(s) a pole with residue at ρ. Lemma 2.2.5 quantifies this behavior, giving a lower bound on Re ζ (s) in terms of nearby zeros. On the other hand, if ζ is large and ζ(s) ζ positive near ρ, then it should be large and negative near σ + 2it, by the heuristic discussed earlier. This was quantified by the key identity we used in the proof of the weak zero-free region. But we know that ζ is small just to the right of the line ζ Re s =. Therefore, any zeros cannot be located too close to this line. The proof is just a matter of carefully managing the parameters to quantify this intuition. Proof of Theorem 2.2.. Let ρ = σ + it be a zero of the Riemann zeta function. We wish to prove that σ < c w(2t + )φ(2t + ) where c is a constant that may depend on the implicit constants in the hypotheses of the theorem. By our work in Theorem 2..3, we may assume that σ < and t t 0. Tony: [assumed t 0 in class, is this necessary?] Let σ > be a number to be chosen later in terms of σ and t. We will carry out the strategy outlined above by comparing upper and lower bounds for ζ. ζ The lower bound, as before, comes from the key identity 3 + cos θ + cos 2θ 0.

The Prime Number Theorem 25 By applying this to the logarithmic derivative of the Euler product, we have Therefore, 3 ζ (σ ) ζ(σ ) 4Re ζ (σ + it) ζ(σ + it) Re ζ (σ + 2it) ζ(σ + 2it) 0. Re ζ (σ + it) ζ(σ + it) 3 ζ (σ ) 4 ζ(σ ) + 4 Re ζ (σ + 2it) ζ(σ + 2it). If we choose σ to be sufficiently close to, then we can control using the ζ asymptotic expansion near the pole. We want to beat the trivial lower bound Re ζ (σ +it) that comes from the asymptotic, so we choose ζ(σ +it) σ σ small enough so that ζ (σ ) 5/4, hence ζ(σ ) σ 3 ζ (σ ) 4 ζ(σ ) 5/6 σ. For the second term, we use Lemma 2.2.5. If M is such that ζ(s) ζ(σ + 2it) M for all s such that s σ 2it r where r is to be chosen later, then Lemma 2.2.5 implies that 4 Re ζ (σ + 2it) ζ(σ + 2it) 2 log M. r Let s summarize: if σ is sufficiently close to and M, r satisfy ζ(s) ζ(σ + 2it) M for all s such that s σ 2it r (2.4) then we have Lower bound: Re ζ (σ + it) ζ(σ + it) 5/6 σ 2 log M r Now let s work towards an upper bound. By Lemma 2.2.5, if M, r satisfy ζ(s) ζ(σ + it) M for all s such that s σ it r (2.5) then (dropping the contribution from all other zeros as in Remark 2.2.7) ζ (σ + it) ζ(σ + it) 8 log M r + σ σ.. ζ

The Prime Number Theorem 26 Combining this with our lower bound, we see that if M, r satisfy (2.4) and (2.5), then either σ < σ r/2, or We can simplify this to 5/6 σ 2 log M r 8 log M r σ σ. σ σ 5/6 σ + 8 log M r. (2.6) Let σ = + δ. To balance the right hand side, we should have δ r. For log M r good measure, we set δ = c min{, } where c is small enough to satisfy the log M requirements for the lower bound, and so that the right hand side of (2.6) above can be combined to + δ σ 3 32δ. We have thus established that if (2.4) and (2.5) are satisfied, then σ < + δ r 2 or σ δ 3. r Since δ = c min{, } we may decrease c further (if necessary) to conclude log M uniformly that σ < cδ. (2.7) It only remains to specify r, M. By the hypothesis, if we take r = w(2t+) M = e Cφ(2t+) for sufficiently large C, then ζ(s) M in the region s σ + it r. Using the Euler product, ζ(σ + it) = p σ +it ( + ) p σ p p n ζ(σ ) σ n σ = max{, log Mw(2t + )} c and similarly for, so after increasing C if necessary (2.4) and (2.5) are ζ(σ +2it) c satisfied with this choice. Then (2.7) implies that σ <, as desired. φ(2t+)w(2t+) and Having established the classical zero-free region (Corollary 2.2.2), we are almost ready to prove the Prime Number Theorem with classical error term O(xe c log x ). To do this we need a bound for ζ (s) inside the classical zero-free region. ζ(s)

The Prime Number Theorem 27 Lemma 2.2.8. There exists a small constant c > 0 such that the following is true: for any t, if σ c then log( t +2) ζ (σ + it) ζ(σ + it) log( t + 2). Proof sketch. Tony: [this is non-examinable] Examining the proof of Lemma 2.2.5, we see that we can in fact conclude Re f ( ) (z) M f(z) O + Re r z ρ ρ: f(ρ)=0 ρ z 0 r/2 Re ρ<re z 0 in a smaller ball, say z z 0 r/4. (Basically because the Cauchy integral formula gives a bound for f (z) on B D in terms of a bound on f(z) on D, as long as B is bounded away from D.) Applying this with M = O( t ) and r = (say), 2 and using our new zero-free region, for σ > zeros deduce that Re ζ (s) ζ(s) c log t O(log( t + 2)). Applying the Borel-Carathéodory theorem then gives the result. Theorem 2.2.9. For all x 2, we have ψ(x) = x + O(xe c log x ). we can drop the sum over the Proof. The proof is exactly similar to the proof of Theorem 2.., but using the line of integration further to the left, using the classical zero-free region. Let c σ 0 = and σ log(t +2) = +. By the same argument as before (using log x Perron s formula, shifting the contour into our new zero-free region, and applying the residue theorem) we obtain ψ(x) = x + ( σ0 +it σ +it σ it ( ) ) + ζ (s) x s 2πi σ 0 it σ 0 +it σ 0 it ζ(s) s ds + O ( x log 2 x On the horizontal lines I = [σ 0 ± it, σ ± T ], we have (using Lemma 2.2.8) (s) x s ζ ζ(s) s x+/ log x max ζ (s) T I ζ(s) x+/ log x log(t + 2) T x log x T T ).

The Prime Number Theorem 28 since we will choose x T. On the vertical lines [σ 0 it, σ 0 + it ], we have (s) x s ζ ζ(s) s x c/ log(t +2) max ζ (s) s I ζ(s) x c/ log(t +2) log(t + 2) s x c/ log(t +2) log x s using Lemma 2.2.8 and the asymptotic expansion near the pole to get the second line. Putting these togther, ( ψ(x) = x + O = x + O x c T log(t +2) log x ( x log 2 x Choosing T = exp( log x) gives the result. T ) dt + t (e c log x log(t +2) + e log T ( x log 2 ) x + O T )). 2.3 Additional properties of ζ(s) Riemann proved that ζ(s) has a meromorphic continuation to all of C, with only a simple pole at s =. This is achieved by the following theorem. Theorem 2.3.. For all s C, we have π s/2 ζ(s)γ( s ( ) s 2 ) = π ( s)/2 ζ( s)γ. 2 This functional equation has a certain symmetry around the critical line Re (s) =. Now Γ(s) is meromorphic and non-vanishing, with simple poles at 0,, 2,.... 2 Therefore, ζ(s) = 0 when s = 2, 4, 6,.... These are called the trivial zeros. The functional fits into a much larger story. Riemann s original proof was by exploiting a symmetry of the theta function n πx e n2, coming from the fact that the e x2 is essentially its own Fourier transform. This was later re-interpreted by Hecke in the theory of modular forms. Hecke proved that a large class of L- functions (now called Hecke L-functions), vastly generalizing the Riemann zeta function, possess functional equations and analytic continuations. This result was, in turn, re-interpreted by Tate in his seminal thesis, in terms of Fourier analysis on adeles. As an alternative to repeatedly applying Perron s formula with the lines of integration in different positions, one can use an explicit formula that directly links ψ(x) to the zeros of ζ(s).

Vinogradov s Method for Zero-Free Regions 29 Theorem 2.3.2 (Explicit formula). For any 2 T x, we have ψ(x) = x ρ: ζ(ρ)=0 Im ρ T x ρ ρ + O ( x log 2 x Proof. (Sketch.) Apply Perron s formula, then move the line of integration very far to the left, picking up residues from all the zeros of ζ, and estimate the contour integral using the functional equation. T ).

Chapter 3 Vinogradov s Method for Zero-Free Regions 3. Estimating ζ sums Our machine for producing zero-free regions is Landau s Theorem (2.2.), which takes as input a bound ζ(σ + it) = O(e φ(t) ) in some region to the left of Re s = and outputs a zero-free region (approximately) of the form {Re s > }. φ(t) Therefore, we can obtain better zero-free regions from better bounds on ζ(s). In this chapter, we will prove a very strong bound for ζ(σ + it) due to Vinogradov and Korobov in 958, and use it to deduce the best-known zero-free region. If t and σ, then Hardy and Littlewood s approximation to ζ(s) (Theorem.4.6) with the choice x = t yields: ζ(σ + it) = n + t (σ+it) σ+it σ + it + O(t σ )) n t = + O(). nσ+it n t By partial summation, bounding ζ(σ+it) is basically equivalent to bounding sums of the form N n N+M n it where M N t. These are called zeta sums. They are highly oscillatory sums (since t is large compared to N), and as always in analytic number theory, the goal is account for as much cancellation as possible. When we proved Theorem 2.2., we used Fourier analysis (Lemma.4.8) to show that N+M n it w it dw when N t. N<n N+M N We don t know how to do that efficiently when N is smaller than t, so we will instead bound the sums directly, using more combinatorial arguments. The summands n it don t appear to have much useable structure, so we introduce structure 30

Vinogradov s Method for Zero-Free Regions 3 by creating auxiliary terms and Taylor expanding. Lemma 3... Suppose that N is large and M N t. Set r = 5.0 log t Then ( ) n it U(n) = O M max N<n 2N N + N 4/5 + Mt /500 4/5 where U(n) := N n N+M x N 2/5 y N 2/5 e(α xy + α 2 x 2 y 2 +... α r x r y r ) Proof. First write the seemingly empty equation N<n N+M n it = N 2/5 2 x N 2/5 y N 2/5 N<n N+M α j := ( )j t 2πjn j. n it. log N. We now shift the summand n n + xy. For fixed x and y, this shifts the inner sum to a translation of the original region by at most N 4/5, so the above is ( ) = (n + xy) it + O(N 4/5 ) N 2/5 2 = N<n N+M x N 2/5 y N 2/5 n it N 2/5 2 N<n N+M x N 2/5 y N 2/5 ( + xy n ) it + O(N 4/5 ). Since the shift xy is n, we can apply Taylor expansion. We have ( log + xy ) = n = ( ) j j= j r ( ) j j= j ( xy ) j n ( xy n ) j + O ( (xy n ) r+ ). Since r + 5.0 log t log N this estimate in ( + xy n and xy n N /5, the last error term is O(t +/500 ). Using ) ( it = e it t log( + xy/n) log(+xy/n) = e 2π ),

Vinogradov s Method for Zero-Free Regions 32 we find that x N 2/5 y N 2/5 ( + xy n ) it = x N 2/5 y N 2/5 = e x N 2/5 y N 2/5 [ ( r ) ] e α j x j y j e(o(t /500 )) j= ( r ) α j x j y j + O(N 4/5 t /500 ). Dividing by N 4/5 and inserting this into the zeta sum gives the result. The exact choices of the parameters are not important. What is important is that the degree of the polynomial is log t. It will turn out that we can only log n handle the case where r isn t too big compared with N. This will ultimately set the limit of the Vinogradov-Korobov method. You might wonder why we introduced two shift variables x, y. We could have applied Taylor expansion with just one shift variable. However, it is very often a good idea to have two independent variables. 3.2 Bilinear Forms In this section we will think about the general problem of bounding the exponential sum e(αx y), x X y Y where α R and X, Y are sets of r-vecors. First we will consider a simple problem, and then return to the sums U(n). Proposition 3.2. (Toy proposition). Let α = a q + θ q 2, where q, (a, q) =, θ. Let N N be large. Then p prime N p prime N j= e(αpp ) N max { N q, q} log(q + ). Note that the trivial bound is π(n) 2 N 2, so we beat the trivial bound log 2 N provided that q is isn t too big or too small. For this (and later work), we will need a small technical lemma. Lemma 3.2.2. Let α, N be as in the statement of Proposition 3.2., and let β, U 0 be arbitrary. Let x denote the distance from x R to the nearest integer. Then { } ( ) N min U, αn + β q + (U + q log q). n N were the implicit constant does not depend on β.

Vinogradov s Method for Zero-Free Regions 33 Proof of Proposition 3.2.. It s very difficult to estimate sums over primes, so the crucial step is to complete the sums over all integers. By the Prime Number Theorem, we expect to lose only factors of log N in doing this, which isn t too bad. Using Cauchy-Schwarz, we have e(αpp ) e(αpn) p N p N n N p N N e(αpn) 2. n N p N Now we have replaced a sum over primes by a sum over all integers, which is much easier to work with. We have 2 e(αpn) = ( ) ( ) e(αpn) e( αpn) n N p N n N p N p N = e(α(p p )n) p N n N p N Let s think about what this sum looks like. In general, e(βn) = e(β(n + )) e(β) e(β) 2 e(β/2) e( β/2) = sin(πβ) 2 β. n N Here, as before, β is the distance to the nearest integer. So e(αpn) n N p N 2 p N N p N 0 n N { min } 2 α(p p ), N { min 2 αn, N where the last line follows since p p ranges over the integers between N and N, hitting each integer at most N times (in fact, at most O(N/ log N) times by the Prime Number Theorem). Finally, Lemma 3.2.2 implies that 0 n N { } min 2 αn, N ( N q + Putting these ingredients together, we are done. } ) (N + q log q) max {q, N 2 q } log(q + ). Remark 3.2.3. If you look carefully, you see that this duplication of variables here was slightly artificial. We could have obtained a better bound by simply

Vinogradov s Method for Zero-Free Regions 34 summing the geometric series at the first time; we would have ended up with min(n, αp p N and then we can immediately apply the lemma. Proof of Lemma 3.2.2. It will suffice to show that { } min U, U + q log q αn + β q 2 <n q 2 since we can break up the range for n into N + sums of length at most q, and q apply this result to each interval (changing β each time). We can also assume that q 2 since the assertion is trivial for q =. Note that for all q < n q, we have 2 2 αn an q = θn q 2 2q because α = a + θ for θ. Since (a, q) =, the numbers an vary over all q q 2 residue classes mod q as n does. So at most O() of the numbers αn lie in each interval [ r /2, r+/2 q q numbers αn + β in each interval [ r /2 If r = 0 we use the bound If r 0, then ]. Translating by β, we see that there are at most O() of the min Putting these estimators together, { } min U, αn + β q 2 <n q 2 q, r+/2 q ]. { } U, U. αn + β { } min U, q αn + β r. U + r q 2 q r U + q log q. Note that in the above proofs, the idea was to replace a sum over primes, which is hard to analyze because the distribution of the primes is so irregular, with a sum over all integers. We lose a bit here because the primes are sparse, but they aren t very sparse. For the sums U(n) from Lemma 3.., we are summing over vectors of the form (x, x 2,..., x r ) and (y, y 2,..., y r ) and these are very sparse in the box containing

Vinogradov s Method for Zero-Free Regions 35 them, so we need some new idea. The new idea is duplication of variables: we put in lots of copies of this sum in order to fill up the box. Lemma 3.2.4. [Duplication of variables] Let U(n) be the sum U(n) = e(α xy + α 2 x 2 y 2 +... + α r x r y r ) x N 2/5 y N 2/5 α j := ( )j t 2πjn j. Then for any natural number k, we have U(n) N 4/5 N 8k/5 (J k,r(n 2/5 )) 2 r min j= kn 2j/5 µ j kn 2j/5 { 3kN 2j/5, } /4k 2 α j µ j where J k/r (N 2/5 ) is the number of solutions (x,..., x 2k ) of the simultaneous equations k 2k x j i = j r i= with x i N 2/5 integers. i=k+ x j i The proof is like the argument of Toy Proposition 3.2., but with the application of Cauchy- Schwarz replaced by two applications of Hölder s inequality. This has the effect of producing a lot (specifically 2k) duplicate copies of the summands (x, x 2,..., x r ), (y,..., y r ) whose sums fill up the containing box more uniformly. So we see that this really similar in spirit to what we did before, and the proof starts out analogously. Proof. To simplify the notation, set Z = N 2/5. By Hölder s inequality, we have U(n) 2k Z 2k e(α xy +... + α r x r y r ) x Z y Z Z ( ( k 2k e α x y i x Z y,...,y 2k Z i= 2k 2k i=k+ y i ) +... + α r x r ( k i= If we let J k,r (λ,..., λ r ; Z) denote the number of solutions (x,..., x 2k ) of k x j i = i= 2k i=k+ x j i + λ j x i Z, j r, y r i 2k i=k+ y r i ))

Vinogradov s Method for Zero-Free Regions 36 then we can rewrite the above inequality by lumping together terms with the same values of k i= yj i 2k i=k+ yj i, which has the effect of decoupling the exponentials. U(n) 2k Z 2k... J k,r (λ,..., λ r, Z)e(α xλ +... + α r x r λ r ) x Z kz λ kz kz r λ r kz r Z 2k... J k,r (λ,..., λ r, Z) e(α xλ +... + α r x r λ r ) kz λ kz kz r λ r kz r x Z In the sequel, we will omit the range of summation to ease notation. The range will always be the same as in the above expressions: in particular, λ j ranges from kz j and kz j. At this point in the proof of Toy Proposiiton 3.2., we were now basically done because we could evaluate the inner sum as a geometric progression. We can t do that here, because of the coupling going on in the exponentials, so we use Hölder s inequality again: U(n) (2k)2 Z 2k(2k ) ( λ... λ r J k,r (λ,..., λ r ; Z) x Z e(α xλ +... + α r x r λ r ) Z 2k(2k ) ( λ... λ r J k,r (λ,..., λ r ; Z) 2k 2k ) 2k... e(α xλ +... + α r x r λ r ) λ λ r x Z To bound the first bracketed term, note that λ J k,r ( λ; Z) 2k 2k 2k max J k,r ( λ; Z) 2k... J k,r ( λ; Z) λ λ λ r Z 2k max J k,r ( λ) 2k. λ 2k ) 2k because the sum... J k,r (λ,..., λ r ; Z) λ λ r

Vinogradov s Method for Zero-Free Regions 37 counts all vectors (x,..., x 2k ) with x i integral in [, Z], which is trivially Z 2k. We still have to bound the maximum. For any (λ,..., λ r ) we have J k,r ( λ, Z) = k #{(x,..., x k ): x j i = L j} #{(x,..., x k ): L,...,L r Z i= #{(x,..., x k ): x i Z, L,...,L r Z i= k x j i = L j} 2 k x j i = L j λ j } i= where this last line follows from applying Cauchy-Schwarz and noting that the terms are equal. But this is just J k,r (0,..., 0, Z) =: J k,r (Z). Putting these pieces together, we have established. U(n) (2k)2 Z 4k(2k ) J k,r (Z)... e(α xλ +... + α r x r λ r ) λ λ r Expanding the inner sum as before, and grouping terms to decouple the exponentials, we obtain U(n) (2k)2 Z 4k(2k ) J k,r (Z) J k,r ( µ r ; Z)e(α µ λ +... + α r µ r λ r ) λ µ x Z Z 4k(2k ) J k,r (Z) J k,r ( µ; Z) e(α λ µ )... e(α r µ r λ r ) µ λ λ r Z 4k(2k ) (J k,r (Z)) r 2 e(α j λ j µ j ) µ j= λ j r Z 4k(2k ) (J k,r (Z)) 2 min{3kz, α µ j µ j } j j= where the last line follows by summing a geometric progression, as in the proof of Proposition 3.2. (the constant 3 comes from the lazy bound (2k + )Z 3kZ). 2k. In the proof of Lemma 3.2.4 we had to switch the order of the sums several times and duplicate variables. This shows the power of introducing two independent variables x, y.