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Analytic Number Theory A course in Analytic Number Theory Taught by Barry Mazur Spring 2012 Last updated: January 19, 2014 1 Contents 1. January 244 2. January 268 3. January 31 11 4. February 2 15 5. February 7 18 6. February 9 22 7. February 14 26 8. February 16 30 9. February 21 33 10. February 23 38 11. March 1 42 12. March 6 45 13. March 8 49 14. March 20 52 15. March 22 56 16. March 27 60 17. March 29 63 18. April 3 65 19. April 5 68 20. April 10 71 21. April 12 74 22. April 17 77 Math 229 Barry Mazur Lecture 1 1. January 24 1.1. Arithmetic functions and first example. An arithmetic function is a func- tion N ! C, which we will denote n 7! c(n). For example, let rk;d(n) be the number of ways n can be expressed as a sum of d kth powers. Waring's problem asks what this looks like asymptotically. For example, we know r3;2(1729) = 2 (Ramanujan). Our aim is to derive statistics for (interesting) arithmetic functions via a study of the analytic properties of generating functions that package them. Here is a generating function: 1 X n Fc(q) = c(n)q n=1 (You can think of this as a Fourier series, where q = e2πiz.) You can retrieve c(n) as a Cauchy residue: 1 I F (q) dq c(n) = c · 2πi qn q This is the Hardy-Littlewood method. We can understand Waring's problem by defining 1 X nk fk(q) = q n=1 (the generating function for the sequence that tells you if n is a perfect kth power). Then d fk (q) is the generating function of the thing we want: d X nk d X n q = fk = rk;d(n)q Take the case when k = 1; you get a geometric series. If k = 2, then Y 2m 2m−1 2m+1 f2(q) = (1 − q )(1 + q )(1 − q ) by a theorem of Jacobi. Another example is the partition function: p(n) is the number of ways that n can be expressed as a sum of positive integers: 1 1 X Y 1 p(n)qn = 1 − qm n=1 m=1 A third example is Y X !(q) = q · (1 − qm)2(1 − q11m)2 = a(n)qn 2 3 2 If n 6= 11 is a prime, then a(p) = 1 + p − jE(Fp)j where E = y + y = x − x (here jE(Fp)j 2πiz is the number of points on the curve with values in Fp). When you replace q = e −1 there's a symmetry where you can replace z 7! 11z . 4 Math 229 Barry Mazur Lecture 1 1.2. Formal Dirichlet Series. 1 X c(n) D (s) = c ns n=1 Suppose we have two arithmetic functions n 7! a(n) and n 7! b(n). Define a product a ∗ b = c where X n c(n) = a(d)b d djn (Divisors are assumed always to be positive.) Why do we care? If you have Dirichlet series P a(n) P b(n) Da(s) = ns and Db(s) = ns then X a ∗ b(n) D = = D (s)D (s) a∗b ns a b The star product is commutative, associative, and has the identity " where ( 1 n = 1 "(n) = 0 else This is not the identity function 1 : n 7! 1. The identity function is invertible, and the inverse is the Moebius function ( 0 if n is not square-free µ(n) = (−1)# of prime factors of n else Exercise: check that X µ(d) · 1 = " djn so that X µ(n) · ζ(s) = 1 : ns Definition 1.1. An arithmetic function is called multiplicative if c(u · v) = c(u) · c(v) if (u; v) = 1. So v Y ki c(n) = c(pi ) i=1 Qv ki if n = i=1 pi . Suppose n 7! c(n) is multiplicative, and let p be prime. Let 1 X c(pm) D (s) = c;p pms m=0 X c(n) D (s) = c ns Formally Y (1.1) Dc(s) = Dc;p(s) p 5 Math 229 Barry Mazur Lecture 1 BUT, even if Dc(s) and Dp(s) converge, you may not have the formula, because the P1 −s product may not converge. If n=1 jc(n)n j converges, then everything works. 1 X X n−iy X e−log(n)iy c(n) · n−s = c(n) = c(n) nx nx n=1 If it converges at a point, then it converges at the entire vertical line through that point. 1.3. Riemann ζ function. Using the arithmetic function n 7! 1, then define X 1 ζ(s) = D (s) = 1 ns This converges absolutely if Re(s) > 1, by calculus. There is an infinite product expansion 1 X 1 X 1 Y X = = ( p−ms) ns 1 − 1 p prime ps p prime m=0 (This is in the HW.) Theorem 1.2. X µ(n) ζ(s)−1 = = D (s) ns µ Proof. Dµ(s)ζ(s) = Dµ · D1 = D" = 1 This is analytically true when Re(s) > 1. Basic reason for interest in Dc(s) for any c : n 7! c(n): If we know enough, we will P be able to give very good estimates for p≤X c(p) := πc(X). In particular, when c = 1, we get estimates for π(X), the number of primes ≤ X. P 1 Theorem 1.3 (Euler). p≤X p diverges as X ! 1. Q 1 −1 Q P1 −m Proof. Define λ(X) = p≤X (1 − p ) . Write this as p≤X m=0 p . But X 1 ≤ λ(X) n n≤X 6 Math 229 Barry Mazur Lecture 1 1 P 1 because each n is a term in the expansion of λ. But n diverges (which immediately proves that there are infinitely many primes). Take the log of the infinite product: X 1 log λ(X) = − log 1 − p p≤X But 1 1 1 − log 1 − 1 = + + + ··· p p 2p2 3p3 1 1 1 ≤ + + + ··· p p2 p3 1 1 1 = + 2 · 1 p p 1 − p | P {z } n2 ∗ convergesN for some c constant. So we have X 1 lim log λ(X) = + something convergent X!1 p We showed above that λ(X) goes to 1, which means its log does too; if the LHS diverges, then so does the RHS. 1.4. Arithmetic functions related to the ζ-function. a will be an arithmetic P P n function, and let A(n) = djn a(d). Equivalently, a = µ ∗ A so a(n) = µ(d)A d . In terms of Dirichlet series, DA(s) = ζ(s) · D(a) Define A as the series where X k n 7! σk(n) = d djn and a : n 7! nk. Then X σk(n) D (s) = = ζ(s)D (s) = ζ(s) · ζ(s − k) A ns a X nk X 1 D (s) = = = ζ(s − k) a ns ns−k If k = 0 then define σ0(n) = ν(n) which counts the number of divisors of n. Then X ν(n) = ζ(s)2 ns P We call a = Φ, the Euler Phi function; then A(n) = djn Φ(d); you can check that A(n) = n. ζ(s − 1) D (s) = Φ ζ(s) th Recall rk;d(n), the number of ways that n is expressible as a sum of d k powers, and recall that when k = 2 we had a nice formula for this. Let R2;d(n) be the number of ways 7 Math 229 Barry Mazur Lecture 2 that n is expressible as the sum of d squares, counting (+α)2 and (−α)2 as different, and including 0. Also, order matters here! For example, R2;24(2) = 1104 Slightly more generally, 16 R2;24(n) = σ11(n) + e(n) 691 p where e(n) is the error term: up to taking logs, it is < other term. We have a complete description of the error term. 2. January 26 Define 1 Y ∆ = q (1 − qn)24 n=1 and recall X n2 f2 = q : It turns out that ∆ ≡ f2 (mod 2) : RECALL, a Dirichlet series is a formal expression that looks like D(s) = P a(n)n−s. We also had arithmetic functions n 7! a(n) 2 C. L-functions will be a special case. 2.1. Some properties to learn about. • Domain of definition. For example, ζ(s) doesn't make sense as a power series to the left of the line Re(s) = 1. However, there is a meromorphic continuation, with pole only at s = 1. (We have good control to the right of the line, but the most interesting information comes from the other half.) Also, there are symmetries: usually a relationship between L(s) and L(1 − s) (i.e. you can sort of reflect across the line). P • Poles and Zeroes. One way (the only way) of understanding p≤X a(p) is to under- P stand the poles and zeroes. In the case of ζ(s), this gives information about p≤X 1 = π(X), the number of primes ≤ X. The Riemann hypothesis says that the zeroes of the 1 ζ-function are at the negative even integers, on the line Re(s) = 2 , and nowhere else. If you're interested in statistics, consider P w(p)a(p) (i.e. a weighted sum). A good thing P to study is p≤X log(p).
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