Infinite Products and Number Theory 1

Transcription

1 Infinite Products and Number Theory We shall review the following subjects: Takashi Ichikawa 2 Abstract Basic theory of elliptic functions and modular forms; Relationship between infinite products and number theory; Processes of creation by great mathematicians. Contents. Introduction.. Aim of this lecture.2. Infinite product of sinx.3. Proof of pentagonal number theorem 2. Function theory 2.. Holomorphic functions and Cauchy s theorem 2.2. Residue theorem and Liouville s theorem 3. Elliptic functions 3.. History from Gauss 3.2. Weierstrass -function 3.3. Abel s theorem and elliptic curves 3.4. Jacobi s theta function and the triple product 4. Modular forms 4.. Eisenstein series 4.2. Discriminant of elliptic curves 4.3. Enumeration of modular forms 4.4. Application to number theory 5. Infinite product in modern mathematics 5.. Moonshine conjecture and Borcherds product 5.2. Proof of Borcherds product 5.3. Modular forms and Borcherds product 5.4. Borcherds lifts Appendices References Available at 2 Department of Mathematics, Graduate School of Science and Engineering, Saga University, Saga , Japan ichikawn@cc.saga-u.ac.jp

2 n=. Introduction.. Aim of this lecture. We explain a basic theory of elliptic functions and modular forms based on the infinite products by Euler and Jacobi: Euler: sin x = x x2 n 2 π 2. Euler s pentagonal number theorem: q n = n= m= Jacobi s triple product identity: m q m3m /2 = + q + q 5 q q 2 + q 7 q 5 + }{{}}{{} m>0 m<0, 5, 2,... are pentagon numbers. q 2n + q 2n ζ + q 2n ζ = n= m= q m2 ζ m. Further, we introduce a recent development called Borcherds products on this theory. Contribution of elliptic functions and modular forms to number theory. A theorem of Jacobi states that for any positive integer n, { } 4 m, m 2, m 3, m 4 Z 4 m 2 i = n = 8 d, i= 0<d n,4 d where S denotes the number of elements of a finite set S. This theorem is obtained by comparing the coefficients of q n in the following formula m Z q m24 = 32 4 n σ n/4q n + 8 σ nq n, where σ n denotes the sum of positive divisors of n. This formula was proved by Jacobi using his theory of elliptic functions, and later was understood as an equality between modular forms of weight 2 and level 2. 2 A conjecture of Fermat or Fermat s last theorem states that if n is an integer 3, then there is no positive integer a, b, c such that a n + b n = c n. This conjecture was finally proved in the paper of Wiles at 995 as follows. Assume that n= 2

3 there are a prime number p 5 and positive integers a, b, c such that a p + b p = c p. Then the equation y 2 = xx b p x c p defines an elliptic curve E over Q which is a geometric counterpart of elliptic functions. By proving a conjecture of Shimura and Taniyama substantially, Wiles showed that the zeta function of E becomes the Mellin transform of a special modular form, called a cusp form, of weight 2. By the fact that the discriminant of E is abc 2p and Frey-Ribet s theorem, the level of this cusp form actually becomes 2 which yields a contradiction since there is no cusp form of weight 2 and level 2. Actually, the proofs of Wiles and Ribet use algebraic geometry of modular curves which are geometric counterparts of modular forms. Two viewpoints of the above formulas. Identities between functions. 2 Identities between formal power series. Recall Riemann s zeta function ζs is defined as ζs def = n= Applying 2 to the infinite product of sinx, n s Res >. sin x = x x3 3! + x5 5! x7 7! + = m x2m+ ; Taylor expansion 2m +! m=0 = x x2 π 2 x2 2 2 π 2 x2 3 2 π 2. Therefore, comparing the coefficients of x 3 in the both sides, 3! = π π π 2 +. and hence we can compute the special value of ζs at s = 2 as ζ2 def = Exercise... Using ζ2 2 = ζ4 + 2 n<m def n 2, show ζ4 = m2 n= n= n 4 = π4 90. Remark. The Bernoulli numbers B k are defined by the power series expansion x e x = x k B k k!. k=0 3 n 2 = π2 6.

4 Then B k are seen to be rational numbers, and for any even positive integer 2n, Therefore, ζ2n/π 2n is a rational number. ζ2n = 22n 2n! B 2nπ 2n. Exercise..2. Using the pentagon number theorem, prove that where q /24 n= q n = am q m2 /24, m= if m, mod2, am = if m 5, 7 mod2, 0 otherwise. = 2 : Jacobi symbol. m Remark. The followings are important examples of infinite sum = infinite product. Rogers-Ramanujan s identity given by Rogers, Ramanujan and Schur in n= n= q n2 q q n q n2 +n q q n = = n= n= q 5n 4 q 5n, q 5n 3 q 5n 2. This formula is important in combinatorics and representation theory. Mock theta conjecture by Ramanujan on the 5th order case. Put Then n=0 n=0 a; q 0 =, a; q n = n k=0 aq k, a; q = q n2 = q5 ; q 5 q 5 ; q 0 q; q n q; q 5 q 4 ; q q n2 +n q; q n = q5 ; q 5 q 5 ; q 0 q 2 ; q 5 q 3 ; q 5 2 q k=0 aq k. q 0m2 q 2 ; q 0 m=0 m+ q 8 ; q 0, m q 0m2 + q 4 ; q 0 m+ q 6 ; q 0. m In 988, Hickerson proved this conjecture by technical calculation. In 2008, Bringmann, Ono, Rhoades, and Folsom gave this conceptual proof using Zwegers result in 2002 that mock theta functions become the holomorphic parts of harmonic modular forms. m=0 4

5 .2. Infinite product of sinx. We explain Euler s proof of the following formula: Since e x = cos x + sin x, sinx = x n= x2 n 2 π 2. sin x = e x e x 2, and hence by e z = lim n + z/n n, sin x = { 2 lim + n x n n n } x n. Putting n = 2m + and ζ = e 2π /n, X n Y n = n X ζ k Y k=0 = X Y = X Y Putting X = + t and Y = t, + t n t n = 2t = 2t = 2 n t m k= m k= m k= m X ζ k Y X ζ n k Y k= m k= X 2 2XY cos 2πk { 2πk + t 2 2 t 2 cos πk 4 sin 2 n πk m sin 2 n k= Comparing the coefficients of t in this both sides, 2n = 2 n m k= n πk + t 2 sin 2 n πk sin 2, n + Y 2. n + t 2 cot 2 πk n + t 2 }. and hence + t n t n = 2nt m k= πk + t 2 cot 2. n 5

6 Then putting t = x/n and letting n, by, Since we have and more precisely, Furthermore, k= infinite product x 2 sin x = lim x m x2 n n 2 cot2 k= cot 2 πk n = cos2 πk n sin 2 πk n x 2 lim n n 2 cot2 x 2 n 2 cot2 πk. n n2 π 2 n, k2 πk = x2 n π 2 k 2, πk = x2 n π 2 k 2 + O n 2. π 2 is absolutely convergent, and hence by Proposition. below, the k2 x2 π 2 k 2 is also absolutely convergent. Therefore, we have k= m k= and it follows that x2 n 2 cot2 sin x = x lim m n k= πk n = x2 n 2 cot2 m x2 m π 2 k 2 + O n 2 k= x2 π 2 k 2 n, k= πk n = x k= x2 π 2 k 2. This completes the proof. Remark. This formula inspired the infinite product formula of elliptic functions by Gauss and Abel cf

7 .3. Proof of pentagonal number theorem. The following formula called the pentagon number theorem was found in 74 and proved in 750 by Euler. q n = n= m= m q m3m /2. Euler s proof cf. [W, Chapter III, XXI]. Put P 0 = n= qn and Then we will show First, P n = q in q n q n+ q n+i n > 0. i=0 P n = q 2n+ q 3n+2 P n+ n 0. q q 2 P = q q 2 { q + q q q 2 + q 2 q q 2 q 3 + } = q q 2 q q 3 q q 2 q 4 q q 2 q 3 = q q 2 q 3 q q 2 q 4 q q 2 q 3 = q q 2 q 3 q 4 q q 2 q 3 = = q n, n= and hence holds when n = 0. Second, if n > 0, then P n = q n + q n q n q n+ + = q 2n q 2n+ + q 3n+ + q in q n q n+ q n+i i=2 q in q n+ q n+i i=2 q in+n q n+ q n+i i=2 = q 2n+ + q in q n+ q n+i i=2 } {{ } q in+n q n+ q n+i i= } {{ } 2 7

8 since q in+n q n+ q n+i i= = q 2n q 3n+. Then putting i = j + 2 in, and i = j + in 2, we have P n = q 2n+ + q jn+2n q n+ q n+j+ q n+j+2 j=0 q jn+2n q n+ q n+j+ j=0 = q 2n+ + q jn+2n q n+ q n+j+ q n+j+2 j=0 = q 2n+ q 3n+2 q jn+ q n+ q n++j j=0 = q 2n+ + q 3n+2 P n+, and hence holds when n > 0. Therefore, P 0 = q q 2 P = q q 2 q 3 q 5 P 2 = q q 2 + q 5 + q 2+5 P 2 = = f n q + n q n P n = f n q + n q an P n = f n q + n q an q 2n+ q 3n+2 P n+, where an = n = 3n2 + n, and f n q is a polynomial of q with 2 degree < an. Hence A n q = f n q + n q an satisfies A n+ q A n q = n q an q 2n+ q 3n+2 = n+ q a n + q an+. Therefore, q n = P 0 = lim A nq = n n= Remark. We call e π τ/2 n= n q an = n= n q 3n2 n/2. e 2π nτ = q /24 q n ; q = e 2π τ n= Dedekind s η-function which becomes a modular form of weight /2. 8 n=

9 Using this theorem and function theory, we will show in 3.4 Jacobi s triple product n= q n q n /2 ζ q n /2 ζ = m= q m2 /2 ζ m. Convergence of infinite sums. Recall that for a series n= a n of complex numbers is absolutely convergent, namely n= a n is convergent, then n= a n is convergent to the same value independent of the orders of this sum. Definition of convergence of infinite products. If a sequence {a n } satisfies that each + a n is not 0 and that subproducts m n= + a n converges to a nonzero number c, then the infinite product n= + a n is called to be convergent to c. Proposition.. The sum n= a n is convergent if and only if the infinite product n= + a n is convergent, and then n= + a n is convergent to the same value independent of the orders of this product In this case, n= + a n is called absolutely convergent. Proof. The first assertion follows from that m a n n= m + a n n= m m exp a n = exp a n, n= and that if a n /2, then m m + a n + an + a n 2 + n= n= n= n= m m + 2 a n exp 2 a n. n= The second one follows from the above property of absolute convergent sums. Exercise.3.. For any x C, using Proposition. show that the infinite product is absolutely convergent. n= x2 n 2 π 2 Exercise.3.2. Using Proposition. show that the infinite product is absolutely convergent if q <. q n n= 9

10 2. Function theory 2.. Holomorphic functions and Cauchy s theorem. Definition. A complex function is a complex valued function of a complex variable. For a C, fz is a holomorphic function at z = a if there is a neighborhood U of a, i.e., an open subset of C containing a such that fz is a complex function defined over U, fz fa The limit lim exists and is independent of ways of approaching z U,z a z a z a. This limit is called the derivative of fz at z = a, and denoted by f a. Remark. A holomorphic function at z = a is continuous at z = a. If fz, gz are holomorphic functions at z = a, then fz ± gz, fzgz are holomorphic at z = a, and fz/gz is holomorphic at z = a when ga 0. Example. The following functions are holomorphic on C, namely holomorphic at any point on C. Polynomial function: a n z n + + a z + a 0, Exponential function: e z = e x cos y + sin y; z = x + y, Trigonometric function: cos z = e z + e z Cauchy-Riemann s relation. if and only if fz satisfies 2, sin z = e z e z 2. A differentiable function fz is holomorphic at z = a f f a = x y a. Proof. We prove the only if part by considering the two approaches a + x a x +0 and a + y a y +0. Therefore, by definition f x a = f fa + y fa a = lim = f y 0 y y a. This completes the proof. Definition. Let D be a domain, namely connected and open subset of C, and C be an oriented path in D. Then for a holomorphic function on D, its line integral along C is defined as N fzdz = lim fz n z n z n, C N n= 0

11 where z 0 and z N are the starting and end points of C respectively. Cauchy s integral theorem actually founded by Gauss. Let D C be a domain, and C D be an counterclockwise oriented closed path such that the interior is contained in D. Then for any holomorphic function on D, fzdz = 0. C Proof. Let U be the interior of C. Then the boundary U is C, and hence by Green s formula, fzdz = dfzdz. Furthermore, dfzdz = C U f f dx + x y dy dx + f dy = x f dx dy y which is equal to 0 by Cauchy-Riemann s relation. Therefore, C fzdz = 0. Application of Cauchy s theorem. Let the notation be as in Cauchy s theorem. Cauchy s integral formula. For a point a in the interior of C, fa = 2π fz z a dz. 2 Taylor expansion. For points a, z D such that z a < da, D, fz = n=0 C f n a z a n. n! Proof. First, we prove. Let Ca; r C be a circle with center a and radius r > 0 which is contained in the interior of C. Then fz/z a is holomorphic outside Ca; r, and hence by Cauchy s theorem, 2π C fz z a dz = 2π This limit under r 0 is equal to = 2π = 2π f 2π fa 2π 0 2π 0 Ca;r 2π 0 fz z a dz f a + re θ re θ dθ re θ a + re θ dθ. dθ = fa

12 which implies. Second, we prove 2. If ζ Ca; r, then z a/ζ a <. Hence by, fz = 2π fζ ζ z dζ Therefore, if we put then = = 2π 2π c n = Ca;r Ca;r Ca;r n=0 z a dζ ζ a fζ ζ a z a n fζ ζ a 2π fζ dζ, Ca;r ζ a n+ fz = c n z a n, n=0 n+ dζ. and hence by taking the nth derivative and putting z = a of the both sides, we have This completes the proof. Example. c n = f n a. n! e z = + z + 2! z2 + 3! z3 +, cos z = 2! z2 + 4! z4, sin z = z 3! z3 + 5! z5. Meromorphic function. Let fz be a complex function. For a positive integer m, fz is called to have a zero of order m at z = a if around z = a, where c m 0. fz = c n z a n n=m For a positive integer m, fz is called to have a pole of order m at z = a if fz = c n z a n ; Laurent expansion of fz n= m around z = a, where c m 0. 2

13 For a domain D C, fz is called meromorphic on D if fz is holomorphic or fz has a pole of finite order at each point on D. Exercise 2... Prove that if fz and gz have zeros resp. poles of order m and n respectively at z = a, then fzgz has a zero resp. pole of order m + n at z = a. Prove that if fz has a zero resp. pole of order m at z = a, then /fz has a pole resp. zero of order m at z = a. 3

14 2.2. Residue theorem and Liouville s theorem. Residue theorem. Let D C be a domain, and C D be a counterclockwise oriented closed path such that the interior of C is contained in D. Then for points a,..., a k in the interior of C and a holomorphic function on D {a,..., a k }, C fzdz = 2π k Res aj fz. Here Res aj fz = c j is called the residue of fz at z = a j, where denotes the Laurent expansion. fz = n Z j= c j n z a j n Proof. For each j =,..., k, let C j C be a circle with center a j and small radius such that C j and its interior are disjoint to each other and contained in the interior of C. Then fz is holomorphic outside C j, and hence by Cauchy s theorem, n C fzdz = k j= C j fzdz. Furthermore, by the following exercise, fzdz = c n j z a j n dz = c j n z a j n dz = 2π c j, C j C j n C j and hence This completes the proof. C fzdz = 2π k Res aj fz. Exercise Let r be a positive number and a be a point on C. Then for an integer n, show that { z a n 0 n, dz = 2π n =. z a =r j= A bounded holomorphic function on C becomes a constant func- Liouville s theorem. tion. Proof. Let fz be a bounded holomorphic function on C. Then for each a C, fz = fa + f az a + f a z a 2 +, 2 4

15 and hence z =R fz z a 2 = fa z a 2 + f a z a + f a +. 2 Since fz/z a 2 is holomorphic except z = a, for any R > a, by the residue theorem, fz fz z a 2 dz = 2π Res a z a 2 = 2π f a. By the assumption on fz, there is a constant M such that fz < M for any z C, and hence f a = 2π fz z =R z a 2 dz 2π max fz z =R z a 2 2πR M2πR 0 R. 2πR a 2 Therefore, f a = 0 for any a C, and hence fz is a constant function. Fundamental theorem of algebra first proved by Gauss. over C of positive degree has at least a solution in C. Any polynomial equation Proof. Assume, on the contrary, that there is a polynomial fz = a n z n + + a z + a 0 a i C, a n 0 of degree n > 0 such that fz = 0 has no root in C. Then gz = /fz becomes a holomorphic function on C. If z 0, then fz = a n z n + a n a n z + + a a n z n + a 0 a n z n, and hence lim z fz a n z n = lim z + a n a n z + + a a n z n + a 0 a n z n =. This implies that lim gz = z lim z fz = lim z a n z n = 0, and hence there is a positive constant M such that gz < if z > M. Furthermore, { gz z M} is compact and hence bounded since it is the image by the continuous function gz of the compact set {z C z M}. Therefore, gz is bounded on 5

16 the whole C, and hence by Liouville s theorem, gz becomes a constant function. This implies that fz is also a constant function which is a contradiction. Maximum modulus principle. Let fz be a holomorphic function on a connected domain D. If there exists a point a in the interior of D such that then fz becomes a constant function. fa = max z D fz, Proof. Take a point a satisfying the assumption, and put M = fa = max z D fz, F = {z D fz = M}. Since fz is a continuous function, F is a closed subset of D. First, we will prove that F = D. Assume, on the contrary that, F D. Then there is a point b on the boundary of F which is contained in the interior of D, and hence there is a positive number r such that If 0 < ρ < r, then {z C z b < r} D. fz = fb + f bz b + z b ρ, and hence fz z b = fb z b + f b +. Therefore, by the residue theorem, fb = Res b fz z b = Since b F and fz M for any z D, M = fb = fz 2π z b dz z b =ρ 2π z b =ρ 2π z b =ρ fz z b dz. M ρ dz = M. Then is an equality, and hence by the continuity of fz, fz = M if z b = ρ. Therefore, fz = M if z b < r. Since b belongs to the boundary of F, there is a point z on D such that z b < r and z F, and hence fz < M which is a contradiction. Therefore, F = D which means that fz = M for any z D. Second, we will prove that fz is a constant function. For the above a, gz = fz + fa 6

17 is a holomorphic function on D. If z D, then fz = M, and hence Since ga = 2fa = 2M, we have gz fz + fa = 2M. max gz = ga = 2M. z D Therefore, gz satisfies the assumption of this theorem, and hence for any z D, gz = fz + fa = 2M. Since fz = fa = M, fz = fa for any z D. Therefore, fz is a constant function. 7

18 3. Elliptic functions 3.. History from Gauss. cf. [T] In 797, as an extension of dx x=sin θ cos θ dθ = = θ = sin x; the inverse function of sin, x 2 cos θ Gauss found that the inverse function of dx x 4 has a double periodicity as a complex function. More precisely, Gauss considered the Lemniscate curve defined as x 2 = cos 2θ in the polar coordinates x, θ which is equivalent to that the product of the distances from ±/ 2, 0 is equal to /2. Put X = x cos θ, Y = x sin θ which is the regular coordinates, and denote by u the distance along this Lemniscate curve between the origin 0, 0 and X, Y with X, Y > 0. Then and hence u = x 0 u def = dx 2 + xdθ 2 = X 0 x 0 dx 2 + dy 2, + x dθ 2 x dx dx = dx 0 x 4. This inverse function is extended to a smooth function x = su of u R. Put ω = 0 dx x 4. Therefore, by the geometric meaning of the Lemniscate curve, su has the periodicity su+4ω = su. Furthermore, using s u = su and the addition law of su, Gauss extended su for a complex variable u. Then su has the another periodicity s u + 4ω = su, and { su = 0 u = 2m + 2n ω, su = u = { 2m + + 2n + } ω for integers m, n. Therefore, as an analogy to Euler s infinite product presentation of sin x, Gauss obtained an infinite product presentation of su as u 4 u m + n 4 2ω m,n=0,,2,..., m 0 su = u 4 m + n 4 ω m,n=,3,5,... 8

19 which is a meromorphic function represented as the quotient of holomorphic functions called theta functions. Exercise 3... Prove the above. Elliptic function. A meromorphic function fz on C is called an elliptic function if fz has double periodicity, namely there exist ω, ω 2 C which are linearly independent over R such that fz + ω = fz, fz + ω 2 = fz for any z C. The complex numbers ω, ω 2 are called periods of fz. Exercise Let fz be an elliptic function with periods ω, ω 2 C which are linearly independent over R. Then show the followings. The lattice in C L def = {mω + nω 2 m, n Z} is a subgroup of C under the addition of numbers. For each a L, we have fz + a = fz z C. Let Then D = {sω + tω 2 0 s, t }. {fz z C} = {fz z D}. Theorem 3.. If an elliptic function is holomorphic on the whole complex plane C, then this is a constant function. Proof. Let fz be a elliptic function with periods ω, ω 2, and assume that fz is holomorphic on C. Put D = {sω + tω 2 0 s, t }. Then by Exercise 3..2, {fz z C} = {fz z D} whose right hand side is the image of the compact set D by the continuous function fz. Therefore, this image is compact, and hence is bounded in C. By Liouville s theorem, fz is a constant function. Exercise Prove Theorem 3. using the maximum modulus principle. 9

20 3.2. Weierstrass -function. -function. For a lattice L = {mω + nω 2 m, n Z} of C, we define the associated Weierstrass -function as z = L z = z 2 + z a 2 a 2. Therorem 3.2. a L {0} z is a meromorphic function on C which is holomorphic outside L, and has poles of order 2 at points on L. 2 z is an even function, namely z = z, and for any a L, z+a = z. 3 z = z 2 + 2n + z 2n at z = 0. 4 Put g 2 = 60 n= a L {0} a 2n+2 a L {0} a 4, g 3 = 40 a L {0} Remark. If we put z = x, then by 4, d z x = z = dz = z = and hence z is the inverse function of an elliptic integral. a 6. Then z 2 = 4 z 3 g 2 z g 3. dx 4x 3 g 2 x g 3, Proof. Let R be a positive real number. If z R, then there is a constant c > 0 such that for any a C with a > 2R, z a 2 a 2 = 2za z 2 z a 2 a 2 < c a 3. Furthermore, there is a constant r > 0 such that {z C z r} is contained in {sω + tω 2 s, t }. Then a L {0} a 3 = k= max{ m, n }=k mω + nω 2 3 k= 8k kr 3 < +, and hence a L, a >2R z a 2 a 2 20

21 is absolutely convergent uniformly on z with z R. Therefore, the assertion follows since z 2 + z a 2 a 2 a L {0} a L,0< a 2R is a rational function of z with poles in L of order 2. 2 By definition, z = z 2 + z + a 2 b= a a 2 = z 2 + and hence z is an even function. Since z = 2 z 3 a L {0} 2 b L {0} z a 3 = b L is absolutely convergent in the wider sense, for any a L, z + a = b L z b 2 b 2 = z, 2 z b 3 2 c=b a z + a b 3 = 2 z c 3 = z. c L Therefore, for each a L, z + a z is a constant independent of z. Since z is an even function, this constant is a 2 + a a a = 2 2 a = Taking the derivatives by z of we have a z = z za z2 = + + a a a a 2 +, z a 2 = a a + 2z a 3 + a 2 + 3z2 z = z 2 + n + n= = n + zn a n+2. n=0 a n+2 a L {0} z n. Since z is an even function, this coefficients of z n is 0 if n is odd, and hence 3 holds. 4 Put fz = z 2 4 z 3 g 2 z g 3. Then by and 2, fz is a meromorphic function on C which is holomorphic outside L and satisfies that fz + a = fz for any a L. By the following Exercise 3.2., fz is holomorphic at z = 0, and hence by, it is also holomorphic at each point on L. 2

22 Therefore, fz is holomorphic on C, and hence by Theorem 3., fz becomes a constant function on C. Furthermore, by Exercise 3.2., f0 = 0, and hence fz = 0. Exercise Show that LHS RHS of 4 has a Taylor expansion at z = 0 of the form c z + c 2 z 2 +. Exercise Calculating the Taylor expansion of LHS RHS of 4, show that a L {0} a 8 = 3 7 a 4 a L {0} 2. 22

23 3.3. Abel s theorem and elliptic curves. Aim. Following [MM], we review Abel s proof of the addition formula of z = L z: z + w = z w + z w 2. 4 z w Theorem 3.3. If complex numbers z i i 3 satisfies z + z 2 + z 3 = 0, then zi, z i i 3 belong to a straight line. Exercise Using this theorem, show the above addition formula. Proof. For x = z, y = z, consider the equation y 2 = 4x 3 g 2 x g 3 given in Theorem and a linear equation y = ax + b together. Then the solutions e i i 3 of these equations satisfy F x def = 4x 3 g 2 x g 3 ax + b 2 = 0. Moving a and b, we study e i as functions of a, b. Since df x = F F F dx + da + x a b db = 0 at x = e i, we have F e i dx = 2ax + bxda + db, and hence dx y = 2e ida + db F e i Since F x = 4x e x e 2 x e 3, 3 i= 2e i da + db F e i at x = e i We take z i C satisfying z i = e i. Since dx/y = d z/ z = dz, by we have 3 dz i = i= 3 i= = 0. 2e i da + db F e i and hence c = z + z 2 + z 3 is a constant as a function of a, b. If we put a = 0 and let b, then each e i tends to, namely z i tends to a point on L. Therefore, = 0, z + z 2 + z 3 = c L; Abel s theorem! and hence z + z 2 + z 3 c = 0. Then the three points zi, z i = e i, ae i + b i 3 23

24 belong to the line defined by y = ax + b, and this line is uniquely determined by the two points z, z, z 2, z 2. Since z 3, z 3 = z 3 c, z 3 c, we completes the proof. Elliptic curve. The -dimensional algebraic variety E = { x, y y 2 = 4x 3 } g 2 x g 3 { } = { X : Y : Z P 2 Y 2 Z = 4X 3 g 2 XZ 2 g 3 Z 3} becomes an abelian group with as its identity satisfying that if two or three points x i, y i belong to one straight line, then their sum is equal to the identity. We call E an elliptic curve over C. The meaning of Abel s theorem. The map C E given by z { z, z z L, z L is a homomorphism which gives a morphism φ : C/L E between complex manifolds. Since the image of φ is an open and closed subset of the connected space E, φ is a surjection. Furthermore, if each fiber of φ contains two points, then z has poles except 0 which is a contradiction. Therefore, φ is an injection, and hence is an isomorphism C/L E. Remark. The original version of Abel s theorem is a generalization of the above theorem which shows an addition formula on integrals of general algebraic functions. Arithmetic of elliptic curves. Mordell s theorem. Let E be an elliptic curve over Q defined by y 2 = fx, where fx is a polynomial over Q of degree 3 without multiple root. Then the abelian group EQ = { x, y Q Q y 2 = fx } { } of its Q-rational points is generated by finite elements. Birch and Swinnerton-Dyer s conjecture. This conjecture claims that for an elliptic curve E over Q, the rank of EQ is equal to the order of the L-function LE, s of E at s =. 24

25 Kume s example. The elliptic curve defined by y 2 = x x has rank 2 over Q, and Kume found its linearly independent Q-rational points as x, y =, ,, ; linearly independent 24, 7 6, 45 36, 6 5, 5 2, 353 ; right triangles with area Application to Cryptography. Elliptic curves over finite fields are applied to factorizing integers and constructing cryptography. 25

26 3.4. Jacobi s theta function and the triple product. Theta function. Fix τ C with positive imaginary part, and define Jacobi s theta function of z C as θz= θ 3 z = exp π m 2 τ + 2π mz m Z = q m2 /2 ζ m ; q /2 = e π τ, ζ = e 2π z. m Z This function is not an elliptic function, however it is a holomorphic function of z having the following property close to the double periodicity. Proposition 3.4. The function θz is a holomorphic function over C and satisfies Proof. Since q <, for positive integers m, θz + = θz, θz + τ = q /2 ζ θz. q m+2 /2 ζ m+ q q m2 /2 ζ m = m+/2 q m 2 /2 ζ m q ζ 0, q m2 /2 ζ m = m+/2 ζ 0 as m. Then the series defining θz is absolutely convergent in the wider sense, and hence is a holomorphic function of z C. If z z +, then ζ = e 2π z is invariant, and hence the function θz is invariant since it is a function of ζ. Furthermore, θz + τ = m Z q m2 /2 ζq m = m Z q m+2 /2 q /2 ζ m+ ζ = q /2 ζ θz which completes the proof. Theorem 3.4 Gauss, Jacobi. θz = n= q n + q n /2 ζ + q n /2 ζ. Proof. First, we review Gauss ingenious proof cf. [T] given in his note at 88. Put T n = + an a t + an a n a a a 2 t2 + + an a n a a n a n a a 2 a n t n. Then + a n tt n = + a n t + an a t + an a n t 2 + an a n a a a a 2 t2 + + an a n a n a a n a n a a 2 a n t n+ 26

27 = + an+ a t + an+ a n+ a a a 2 t an+ a n+ a a n+ a n a a 2 a n+ = T n +. Therefore, since T = + t, we have t n+ T n = + t + at + a 2 t + a n t. Let n = 2m be an even positive integer. Then by the above two expressions of T n, + an a t + an a n a a a 2 t2 + + an a n a a n a m a a 2 a m t m + + an a n a a n a n a a 2 a n t n = + t + at + a 2 t + a n t. Putting a = α 2, t = α n ζ and dividing the above formula by a n a n a a n a m a a 2 a m we have t m = α2n α 2n 2 α n+2 α 2 α 4 α n α m n ζ m, + αn α n+2 αζ + ζ + αn α n 2 α n+2 α n+4 α4 ζ 2 + ζ αn α n 2 α 2 α n+2 α n+4 α 2n αn/22 ζ n/2 + ζ n/2 = + αζ + αζ + α 3 ζ + α 3 ζ + α n ζ + α n ζ α 2 α 4 α n α n+2 α n+4 α 2n. Therefore, as an equality between formal power series of α whose coefficients are polynomials of ζ ±, α m2 ζ m = + αζ + ζ + α 4 ζ 2 + ζ 2 + α 9 ζ 3 + ζ 3 + m Z = + αζ + αζ + α 3 ζ + α 3 ζ = α 2 α 4 α 6 α 2n + α 2n ζ + α 2n ζ. n= Then putting α = q /2, we complete the proof. 27

28 Second, we give this proof using the pentagon number theorem and the function theory. Since q <, the infinite product fz = n= + q n /2 ζ + q n /2 ζ is absolutely convergent in the wider sense, and hence is a holomorphic function of z. Note that fz = 0 ζ = q ±n /2 n N z = ± n τ + m + n N, m Z 2 2 z = n τ + m + ζ = q n /2 2 2 n, m Z, and that for any n, m Z, d + q n /2 e 2π z z=n dz 2τ+m+ 2 = q n /2 2π e 2π z z=n 2τ+m+ 2 = 2π 0. Therefore, fz has a zero of order at n τ + m n, m Z. Furthermore, for any n Z, θz ζ= q n /2 = m q m2 /2+mn /2 m Z = m q {m+n /22 n /2 2 }/2 m Z = l q l2 /2+ln /2 l Z by putting l = m + 2n, we have θz ζ= q n /2 = 0. Hence the quotient function gz = θz/fz is a holomorphic function on C. Further, gz + = θz + fz + = θz fz = gz, 28

29 and by Proposition 3.4, since fz + τ = gz + τ = θz + τ fz + τ = q /2 ζ θz q /2 ζ fz = gz + q n+/2 ζ + q n 3/2 ζ n= = + q /2 ζ + q /2 ζ = q /2 ζ fz. + q n /2 ζ + q n /2 ζ n= Therefore, gz is a double periodic holomorphic function on C, and hence is a constant C by Theorem 3.. Putting ζ = q /6, the Euler s pentagon theorem implies that and Therefore, θz ζ= q /6 = m Z m q /3 m3m /2 = fz ζ= q /6 = which completes the proof. q n 2/3 q n /3 = n= gz = C = q n n= q n/3, n= n= q n/3 q n. 29

30 4. Modular forms 4.. Eisenstein series. Exercise 4... Show that { SL 2 Z = a c b d a, b, c, d Z, ad bc = }. becomes a group under the product of matrices, and that SL 2 Z is not a commutative group. This group is called the modular group. Exercise Let H = {τ C Imτ > 0} be the Poincaré upper half plane. Then show the followings. a b For any τ H and γ = SL 2 Z, put γτ def = aτ + b c d cτ + d. Then Im γτ = Imτ cτ + d 2, and τ γτ gives a bijective map from H onto H. For any γ, γ 2 SL 2 Z, γ γ 2 τ = γ γ 2 τ τ H. Definition. Eisenstein series are functions of τ H which appear in the coefficients of the Laurent expansion of Z+Zτ z at z = 0. More precisely, for an even integer k 4, we define the Eisenstein series of weight k as Theorem 4.. E k τ = m,n Z 2 {0,0} mτ + n k τ H. The series E k τ is uniformly absolutely convergent in the wider sense, and hence is a holomorphic function of τ. 2 The function E k τ of τ H satisfies that aτ + b E k γτ := E k = cτ + d k E k τ cτ + d γ = a c b d SL 2 Z. 3 The function E k τ is represented as a nonnegative power series of q = e 2π τ. 30

31 4 Denote by ζk = n= n k, the value at k of the Riemann zeta function, and by σ k n = d k 0<d n the sum over the k th powers of positive divisors of a positive integer n. Then E k τ = 2ζk + 2 2π k σ k nq n. k! Definition. A modular form for SL 2 Z of weight k is defined as a function of τ H satisfying the above conditions 3. Proof of Theorem 4.. Let L = {mτ + n m, n Z} be the lattice of C generated by and τ H. Since k 4, as is shown in the proof of Theorem 3.2, there is a positive number r such that α L {0} α k 8n nr k = 8 r k n= n= n= <, nk and hence the assertion follows. a b Next, we show 2. For γ = SL 2 Z, the map φ γ : L L defined as c d φ γ mτ + n = ma + ncτ + mb + nd is a bijection since it has the inverse map given by m τ + n m d n cτ + m b + n a. Therefore, aτ + b cτ + d k E k = cτ + d φ γ α k = φ γ α k = E kτ α L {0} φ γα L {0} since the series E k τ is absolutely convergent. Finally, we show 4 from which 3 immediately follows. By taking the logarithmic derivative of the both sides of Euler s theorem shown in. sin x = x x2 n 2 π 2, n= we have under Imx > 0, d logsin x LHS = = cos x dx sin x = e2 x + e 2 x = 2 e 2 nx, n=0 RHS = d log x dx + d dx log x2 n 2 π 2 = x + x + nπ +. x nπ n= 3 n=

32 Then by putting x = πτ and multiplying the both sides by π, we have π 2π q n = τ + n=0 m= τ + n +, τ n and hence by taking the k th derivative of the both sides, Therefore, 2π k m Z Since k is even, by the above formlura, E k τ = = 2 = 2 n= n k q n = k k! m Z τ + m k = k! 2π k m,n Z {0,0} n= n= n k + 2 mτ + n k m= n Z n k + 2 m= = 2ζk + 22π k k! = 2ζk + 22π k k! mτ + n k τ + m k. n k q n. n= k! 2π k m= n= n k q nm σ k lq l l= n k q nm n= by putting l = nm. This completes the proof. Example. E 4 τ = 2 π π4 6 n= 2π 6 E 6 τ = E 8 τ = E 0 τ = 2π π π σ 5 nq n, σ 3 nq n = n= σ 7 nq n, n= σ 9 nq n. n= σ 3 nq n, n= 32

33 4.2. Discriminant of elliptic curves. Definition. The Poincaré upper half plane is defined as H = {τ C Imτ > 0} which gives a model of non-euclidean geometry. Let k be an integer. Then a function fτ on H is called a modular form of weight k for SL 2 Z if fτ satisfies the following conditions Holomorphic condition. fτ is a holomorphic function of τ H. a b 2 Automorphic condition. For any τ H and γ = SL 2 Z, c d fγτ := f aτ + b = cτ + d k fτ. cτ + d 3 Cusp condition. fτ is expanded to a nonnegative power series of q = e 2π τ as fτ = a n q n n=0 which is called the Fourier expansion of fτ. Denote by M k the space of modular forms of weight k. Exercise We define the linear combination of fτ, gτ M k as af + bgτ = afτ + bgτ a, b C. Then show that M k becomes a vector space over C. We define the product of fτ M k and gτ M l as Remark. fgτ = fτgτ. Then show that fgτ belongs to M k+l. 33

34 n Let fτ be a modular form of weight k. Then for any n Z, taking γ = 0 the automorphic condition 2 implies that fτ + n = fτ. On the other hand, τ e 2π τ gives a surjective map φ : H D def = {q C 0 < q < } such that φτ = φτ if and only if τ τ + Z. Therefore, fτ is regarded as a holomorphic function on D, and hence has a Laurent expansion of q. The cusp condition 3 says that fτ has actually a Taylor expansion of q. The space M k consists of global sections of a certain automorphic line bundle on the compactified modular curve H/SL 2 Z { }, and hence by Riemann-Roch s theorem, M k is seen to be finite dimensional. In 4.3, we will show the dimension formula of M k using the following theorems. The space M 0 of modular forms of weight 0 consists of constant func- Theorem 4.2. tions. Theorem 4.3. The modular form τ of weight 2 defined as τ = 60E 4τ E 6 τ 2 2π 2 = q 24q 2 + has no zero on H, namely τ 0 for any τ H. Proof of Theorem 4.2. First, we show that { D def = τ H 2 Reτ } 2, τ is a fundamental domain for SL 2 Z, especially γd = H. γ SL 2 Z It is clear that the left hand side is contained in the right hand side. For τ H, { } { } a b c, d Z cτ + d SL 2 Z cτ + d, c d c, d 0, 0 and hence there exists the value { m = min cτ + d a c b d SL 2 Z } 34

35 as a positive real number. Take an element a 0 b 0 γ 0 = SL 2 Z c 0 d 0 such that m = c 0 τ + d 0. By Exercise 4..2, Imγτ = Imτ cτ + d 2 γ = a c b d SL 2 Z, and hence we have Imγ 0 τ = max {Imγτ γ SL 2 Z}. Put T = 0 and let n be an integer satisfying, S = 0 0 SL 2 Z, n 2 Reγ 0τ n + 2. If we assume that T n γ 0 τ <, then by Exercise 4..2, Im ST n γ 0 τ = Im T n γ 0 τ T n γ 0 τ > Im T n γ 0 τ = Im γ 0 τ. which contradicts to. Therefore, T n γ 0 τ, and hence T n γ 0 τ D. This implies that τ is contained in γ0 T n D γd, γ SL 2 Z and hence holds. Second, we prove that M 0 consists of constant functions. For an element fτ M 0, by the automorphic condition 2 and by, { } 3 {fτ τ H} = {fτ τ D} = fτ Imτ. 2 Let fq denote fτ which is regarded as a function of q. Then by the above, { } {fq 0 < q < } = fq 0 < q e 3π. Furthermore, by the cusp condition 3, fq is holomorphic at q = 0, and hence { } {fq q < } = fq q e 3π 35

36 { is a compact set as } the image by the continuous function fq of the compact set q C q e 3π. Therefore, there exists a complex number q 0 with q 0 e 3π such that fq 0 = max {fq q < }, and hence by the maximum modulus principle in 2.2, fq is a constant function. Proof of Theorem 4.3. For an element τ of H, let L = {m + nτ m, n Z} be the lattice in C generated by, τ, and let F x, y be the polynomial y 2 φx, where φx = 4x 3 60E 4 τx 40E 6 τ. Then by Abel s theorem in 3.3, the map { L z, L z z C L, z z L gives an isomorphism from the complex torus C/L to the elliptic curve {x, y C C F x, y = 0} { }. Since C/L is a smooth manifold, if F x, y = 0, then F x = φ x, F y = 2y have no common zero. By the fact that φ x = 0, y = 0 φx = φ x = 0 x is a double root of φ = 0, φx = 0 has no double root, and hence the discriminant of φx, which is defined as the square of differences of its three roots, is not 0. Therefore, this discriminant given by { 40E6 τ } 60E4 τ 3 = 2π τ is not 0 for any τ H. Fundamental domain for Γ2 given by Gauss. 36

37 4.3. Enumeration of modular forms. Aim. Using Theorems 4.2 and 4.3, we show the following dimension formula. Theorem 4.4. Denote by M k the space of of modular forms of weight k. Then we have 0 k: odd, 0 k < 0, k = 0, dim C M k = 0 k = 2, k: even, 4 k 0, dim C M k 2 + k: even, k 2. Example. dim C M 00 = dim C M 88 + = = dim C M = dim C M = 9. Proof. 0 k: odd. For fτ M k, if γ = SL 2 Z, then 0 τ + 0 fτ = f = fγτ = k fτ = fτ, 0 and hence fτ = 0 for any τ H. Therefore, dim C M k = 0. k = 0. By Theorem 4.2, M k consists of constant functions, and hence dim C M k =. k < 0. For fτ M k, by Exercise 4.2., fτ 2 τ k M 2 k k 2 = M 0 = {constant functions}. If the Fourier expansion of fτ is represented as α n q n + with nonzero α n, then fτ 2 τ k = α 2 n q 2n k + is not a constant function since αn 2 0 and 2n k > 0. Therefore, fτ = 0 for any τ H, and hence dim C M k = 0. k: even, 4 k 0. Since the Eisenstein series E k τ belongs to M k, C E k τ def = {ae k τ a C} M k. The Fourier expansion of E k has the positive constant term 2ζk = 2 37 n= n k,

38 and hence for fτ M k with Fourier expansion α 0 + α q +, fτ α 0 2ζk E kτ M k has the Fourier expansion as β q +. As is shown in Theorem 4.3, τ has no zero on H, and hence the quotient fτ α 0 2ζk E kτ τ = β + belongs to the space M k 2 which becomes {0} since k 2 < 0. Therefore, fτ = α 0 2ζk E kτ C E k τ, and hence M k = C E k τ has dimension. k: even, k 2. If fτ = α 0 + α q + M k, then as is shown above fτ α / 0 2ζk E kτ τ M k 2. where fτ α 0 2ζk E kτ + τ M k 2, τ M k 2 = { τ gτ gτ M k 2 }. Therefore, τ M k 2 + C E k τ = M k. Further, by the following Exercise 4.3., dim C M k = dim C τ M k 2 + dim C C E k τ = dim C M k 2 +. Exercise Prove that τ M k 2 C E k τ = {0}. k = 2. Let fτ = α 0 + α q + be an element of M 2. If α 0 = 0, then fτ/ τ M 2 2 = M 0 = {0}, and hence fτ is identically 0. If α 0 0, then fτ/α 0 2 M 4 = C E 4 τ, fτ/α 0 3 M 6 = C E 6 τ, and hence there are constants c, c 2 such that + α /α 0 q + 2 = c +240q+, + α /α 0 q + 3 = c 2 504q+. This implies that c = c 2 = and that 2α /α 0 = 240, 3α /α 0 = 504 which is a contradiction. Therefore, M 2 = {0} has dimension 0. 38

39 Exercise Prove that any element of M k is represented as a polynomial of E 4 τ and E 6 τ. For τ H, put q = e 2π τ. Then τ is equal to Ramanu- Theorem 4.5 Jacobi. jan s delta function q q n 24. n= Proof see [K] for its detail. Put L = Z + Zτ, and z = L z. As is shown in 4-2, if we put z 2 = 4 z e z e 2 z e 3, then 2π 2 τ = {4e e 2 e e 3 e 2 e 3 } 2. For α = /2, τ/2, + τ/2, 2α L, namely α α + L, and hence α = α. Since z is an odd function satisfying α = α, we have α = 0. Hence α {e, e 2, e 3 }, and actually {e, e 2, e 3 } = { /2, τ/2, + τ/2}. Therefore, τ is expressed as { 6 2π 2 2 τ } 2 { τ 2 } 2 { τ + τ 2 2 } 2. Put ζ = e 2π z, and define a theta function as an absolutely convergent product: q n ζ q n ζ θζ = ζ q n 2 Then θζ is invariant under z z + ζ ζ and satisfies that under z z + τ. Therefore, n= θqζ = ζ θζ ζ 2 θζ ζ 2 θζ ζ 2 θζ 2 θζ 2 2 is invariant under ζ qζ, ζ 2 qζ 2. Let ζ be a point on C q. Then the function of ζ 2 C becomes an elliptic function on C / q = C/Z + Zτ which has zeros of order at ζ 2 = ζ ± qn n Z, and has poles of order 2 at ζ 2 = q n n Z. On the other hand, for z Z + Zτ, the function z z 2 2 of ζ 2 = exp 2π z 2 is an elliptic function on C / q = C/Z + Zτ which has zero at ζ 2 = ζ ± qn n Z, and has poles of order 2 at ζ 2 = q n n Z, where ζ = exp 2π z. Therefore, the function Cz, z 2 = 2 39

40 of z 2 is a holomorphic elliptic function, and hence is a constant function. In a similar way, one can see that Cz, z 2 is also a constant function of z. Since lim z 2 0 z2 2 = θζ θζ z2 2 θζ 2 lim z2 0 e = 2π z 2 2 2π, 2 lim z 2 0 z2 2 2 = lim z 2 2 z2 0 z2 2 = which imply that Cz, z 2 = 2π 2 = 2π 2. Therefore, we have the formula: z z 2 = 2π 2 ζ 2 θζ ζ 2 θζ ζ 2 θζ 2 θζ 2 2 which was shown in [Si, Chapter V, Proposition.3] using Weierstrass σ-function. Substituting this formula to, we have This completes the proof. Corollary. weight 2. τ = 6 2π 2 2π2 q 3 θ q2 θ q /2 2 θq /2 2 θ 6 θq /2 6 θ q /2 6 = 6q θ 4 θq /2 4 θ q /2 4 = q q n 24. n= The infinite product q n= q n 24 q = e 2π τ Remark. Hurwitz, Siegel and Weil proved this fact directly. Remark. The function ητ = e π τ/2 n= e 2π nτ = τ /24 is a modular form of is called Dedekind s eta function which becomes a modular form of half integral weight /2. By Euler s pentagon number theorem, ητ = e π τ/2 n= n e π 3n 2 nτ. 40

41 4.4. Application to number theory. Example. By Theorem 4.4, M 8 has dimension, and hence is C E 4 τ 2 and C E 8 τ. Since 2π 4 E 4 τ = σ 3 nq n, 5 n= 2π 8 E 8 τ = σ 7 nq n, 7 there is a nonzero constant c such that σ 7 nq n = c σ 3 nq n. n= n= n= By comparing their constant terms, we have c =, and hence σ 7 nq n = σ 3 nq n n= = n= σ 3 nq n n= Therefore, comparing their coefficients of q, we have σ 7 n = σ 3 n + 20 n m= n n= m= σ 3 mσ 3 n m. σ 3 mσ 3 n mq n. Remark. Without using Theorem 4.4, we have E 8 τ = 3 7 E 4τ 2 by Exercise Exercise In a similar way as above, show using σ 9 n = 2σ 5 n 0σ 3 n E 4 τ = E 6 τ = E 0 τ = 2π π π n m= σ 3 mσ 5 n m σ 3 nq n, n= σ 5 nq n, n= 264 σ 9 nq n. n= 4

42 Example 2. Jacobi proved the following formula m Z q m24 = 32 4 n σ n/4q n + 8 σ nq n which is regarded as an equality between modular forms of weight 2 for Γ2. By this formula, Jacobi showed that for any positive integer n, { } 4 m, m 2, m 3, m 4 Z 4 m 2 i = n = 8 d > 0, where S denotes the number of elements of a finite set S. i= n= 0<d n,4 d Example 3. Euler proved that a prime p decomposes in Q as follows: if p = 2, then 2 = 2. if p mod4, then p = a + b a b for certain integers a and b. if p 3 mod4, then p does not decompose and remains prime. This result is an example of class field theory by Hilbert and Takagi which describes the decomposition rule of prime ideals in abelian extensions in terms of the congruence condition on these prime ideals. Serre gives an example of nonabelian class field theory which describes the decomposition of prime ideals in nonabelian extensions. More precisely, let L be the decomposition field of x 3 x = 0. Then L is a Galois extension over Q whose Galois group is isomorphic to the symmetric group of degree 3, and L contains K = Q 23. Let fτ = q q n q 23n n= = 2 m,n Z q m2 +mn+6n 2 m,n Z q 2m2 +mn+3n 2 ; q = e 2π τ. It is known that fτ is a modular form of weight with level 23. Denote the Fourier expansion of fτ by n= anqn. Then the ideal p of Q generated by a prime p 23 is unramified in L and its decomposition rule is described as follows: ap = 2 if and only if p is decomposed completely in K and L, and hence p is decomposed to the product of distinct 6 prime ideals of L. ap = if and only if p is decomposed to the product of distinct 2 ideals of K which remain prime in L. 42

43 ap = 0 if and only if p remains a prime ideal of K which is decomposed to the product of distinct 3 prime ideals of L. Ramanujan conjecture. For the coefficients τm of Ramanujan s delta function defined as τmq m = q q n 24, m= Ramanujan conjectured the followings n= If m and n are coprime, then τmn = τmτn. 2 If p is a prime number, then τ p n+ = τpτ p n p τ p n. 3 If p is a prime number, then τp 2p /2. Later Mordell proved and 2 using a theory of Hecke Mordell? operators, and Deligne proved 3 using the Weil conjecture which was also proved by Deligne. More precisely, Deligne applied the Weil conjecture to the modular curve X Γ = H/Γ {cusps} = H P Q /Γ associated with a congruence subgroup Γ SL 2 Z and to its fiber space whose fibers are products of elliptic curves. Example. Picture of X SL2 Z. Shimura-Taniyama conjecture proved by Wiles and others. For each elliptic curve E over Q, there exists a positive integer N and a surjective morphism φ : X Γ0 N E, where { } a b Γ 0 N = SL 2 Z c d c 0 modn is a congruence subgroup of SL 2 Z. As its applications, we have The analytic continuation and functional equation of the L-function of E: LE, s = ap Ep s + p 2s p: good prime ; a p E = p + EZ/pZ. The Gross-Zagier formula contributes the Birch and Swinnerton-Dyer conjecture: Proof of the Fermat conjecture see.. Rank of EQ = Order of LE, s at s =. 43

44 5. Infinite products in modern mathematics 5.. Moonshine nonsense? conjecture and Borcherds product. Infinite product on the modular function. Since 45 π 4 E 4τ = σ 3 nq n, τ = q q n 24 q = e 2πiτ n= are modular forms of weight 4, 2 respectively, jτ = 45 π 4 E 4 τ 3 τ = q n= 3 σ 3 nq n n= n= + q n + q 2n + 24 = q q q q q q q q 7 + is a modular function, namely satisfies that j aτ + b = jτ for any cτ + d a c b d SL 2 Z. Remark. jτ is equal to an invariant called the j-invariant of the isomorphism classes containing the elliptic curves y 2 = 4x 3 60E 4 τx 40E 6 τ. Theorem 5.. Put jq = n= cnqn. Then for independent variables p and q, B jp jq = p p m q n cmn. q m,n= This formula was proved independently in the 980 s by Koike, Norton and Zagier. Calculation of B. Multiplying the both sides of B by we have /p /q = pq p q, B cpq c2 p 2 q + pq 2 c3 p 3 q + p 2 q 2 + pq 3 cn p n q + p n q pq n = pq c p 2 q c2 pq 2 c2 p 3 q c3 pq 3 c3 p 4 q c4 p 2 q 2 c4 pq 4 c4 p 5 q c5 pq 5 c5. 44

45 Therefore, by comparing the coefficients of p 2 q 2 in c3 p 3 q + p 2 q 2 + pq 3 = pq c p 2 q 2 c4 cc = cpq + p 2 q we have c4p 2 q 2 +, cc c3 = c , , , 299, 970 = 20, 245, 856, Exercise 5... By comparing the coefficients of p 3 q 2 in the both sides of B, show c4 = c6 + cc2. 20, 245, 856, 256 = 4, 252, 023, 300, , 884 2, 493, 760. Theorem 5.2 Recurrence of cn. Assume that Theorem 5. holds. Then for any n 6, cn is expressed as a polynomial over Z of cm 0 < m < n. Consequently, for any n, any cn is expressed as a polynomial over Z of c, c2,..., c5. Proof. First, comparing the coefficients of p n q 2 n 2 in B, cn + = c2n + a polynomial over Z of cm 0 < m < n, and hence when n 2 2n 4, c2n is expressed as a polynomial over Z of cm 0 < m < 2n. Furthermore, comparing the coefficients of p 2n q 2, p n q 4 and of p m q 3 0 < m < n in B, we have c2n + = c4n + a polynomial over Z of cm 0 < m < 2n, cn + 3 = c4n + a polynomial over Z of cm, c2m, c3m 0 < m < n, cm + 2 = c3m + a polynomial over Z of cl, c2l 0 < l < m, and hence c2n + = cn a polynomial over Z of { cm c2l, cl < m < 2n, 0 < l < n. Therefore, when n 3 2n + 7, c2n + is expressed as a polynomial over Z of cm 0 < m < 2n +. Monster group. Finite simple groups are shown to be one of the followings: Cyclic groups of prime order, 45

46 Alternative groups of degree 5 Galois, Certain matrix groups over finite fields Chevalley, Suzuki,..., 26 Sporadic groups which are not the above groups. The monster group is the sporadic simple group with maximal order 54 bits, and the sizes of its irreducible representations are d =, d 2 = 96883, d 3 = ,... Then Mckay and Thompson found that d n are related with certain Fourier coefficients cn of jτ as c = d + d 2, c2 = d + d 2 + d 3,... Moonshine conjecture proved by Borcherds 992 based on Conway-Norton s work. There is a sequences {H n } of representations of the monster group M such that for each g M, J g τ = Trace g Hn e 2π n n= is a modular function. Especially, if g =, then J τ = jτ + constant, and hence dim H n = cn. Idea of the proof. By the denominator formula of the characters of an infinite dimensional Lie algebra associated to M, we have the same infinite product on J τ as for jτ. Then by the recurrence of J τ and jτ, J τ = jτ + constant. 46

47 5.2. Proof of Borcherds product. Proposition 5.3. For a positive integer n, put T n = { a c b d a, b, c, d Z, ad bc = n }. Then we have the followings: For any γ SL 2 Z, φ γ α = αγ α T n gives a bijection φ γ : T n T n. 2 Define a subset of T n as { n = a b 0 d T n a, b > 0, 0 b < d }. Then any element α of T n is uniquely expressed as α = γ β β n, γ SL 2 Z, namely, there exists a unique element β, γ n SL 2 Z satisfying α = γ β. 3 Let γ be an element of SL 2 Z. Then for any β n, there exists an element β, γ n SL 2 Z such that βγ = γ β. Furthermore, β β gives a bijection ψ γ : n n. Exercise Prove. Proof. First, we show the uniqueness in 2. Assume that β, β n and γ, γ SL 2 Z satisfy γβ = γ β. Then γ γ SL 2 Z and γ γβ = β. Put γ x y a b γ =, β =, β a b = z w 0 d 0 d. Then ax az bx + dy bz + dw = a b 0 d, and hence az = 0, ax = a, bz + dw = d. Since a, a, d, d > 0, we have z = 0 and x > 0. Therefore, the integers x, w satisfy xw =, hence x = w =, a = a and d = d. Then b + dy = b and d y = b b. Since 0 b, b < d and y Z, we have y = 0, and hence This implies γ = γ and β = β. γ γ =

48 Second, we show the existence in 2. Put s t a b α =, β = u v 0 d, γ = x z y w. Since β n, γ SL 2 Z and α = γβ, we may find integers a, b, d, x, y, z, w satisfying A a, d > 0, ad = n, 0 b < d. B xy yz =. s t C = u v ax az bx + dy bz + dw. By B, x, z are coprime, and hence a = gcdax, az C = gcds, u which is a divisor of sv tu = n. Therefore, d, x, z are determined as d A = n a, x C = s a, z C = u a. Since x, z are coprime, there are integers y, w such that xw y z = from which we will construct integers y, w satisfying C. Multiplying this equality by n = ad, we have axdw dy az = n, and hence by C, sdw dy u = n. Therefore, from the fact that sv tu = n, we have sv dw = ut dy which implies xv dw = zt dy by C. Since x, z are coprime, there is an integer m satisfying v dw = mz and t dy = mx, and hence one can take integers q, b such that m = qd + b 0 b < d. Then t = mx + dy = bx + dy + qx and v = mz + dw = bz + dw + qz. Therefore, the above a, b, d, x, z and y = y + qx, w = w + qz satisfy A C. Finally, we prove 3. By 2, we have T n = and hence by, for any γ SL 2 Z, Therefore, T n = φ γ T n = β n and hence ψ γ is a bijection. β n β n SL 2 Z β, SL 2 Z βγ = SL 2 Z ψ γ β = T n = 48 β n β n SL 2 Z ψ γ β. SL 2 Z β,