A Motivated Introduction to Modular Forms

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1 May 3, 2006 Outline of talk: I. Motivating questions II. Ramanujan s τ function III. Theta Series IV. Congruent Number Problem V. My Research

2 Old Questions... What can you say about the coefficients of products of binomials? What numbers can be represented as the sum of four squares? For those that can be represented in how many ways can it be done? Given an integer n is there a way to determine if there s a right triangle with rational sides and area n?

3 with a very modern answer... Headway can be made in all these questions with some knowledge of modular forms.

4 Ingredients to make modular forms an upper half-space a group acting on the upper half-space an arithmetic group a functional equation and an automorphy factor a Fourier expansion

5 Classical Modular Forms Half-plane: H = {z = x + iy C : y > 0} Group: SL 2 (R) Group action: For z H and γ SL 2 (R), ( ) a b az + b γ.z =.z = c d def cz + d Arithmetic group: Γ := SL 2 (Z)

6 Classical Modular Forms (cont.) Let k Z and f : H C be holomorphic on H and at. If f satisfies the functional equation ( ) f (γ.z) = (cz + d) k a b f (z) for all γ = Γ c d we call f a modular form of weight k.

7 Classical Modular Forms (cont.) (cz + d) k is the automorphy factor f is periodic (let γ = ( )) it has a Fourier expansion f (z) = n Z a n q n ( q = def e 2πiz) holomorphic at means a n = 0 for n < 0 if a 0 = 0, we call f a cuspform we denote the space of modular forms of weight k by M k and the space of cusp forms by S k

8 Ramanujan s τ function (2π) 12 (z) = q n 1 (1 q n ) 24 = def τ(n)q n n 1 is the unique element of S 12 normalized so that τ(1) = 1 Its definition is similar to that of other arithmetic functions. E.g., the partition function: p(n)x n = def (1 x n ) 1 n 0 n 1

9 Ramanujan s τ function (cont.) τ(n) is multiplicative τ(n) σ 11 (n) (mod 691), τ(n) nσ 9 (n) (mod 5),... Lehmer s Conjecture: τ(n) > 0 for all n Ramanujan s Conjecture: τ(n) < σ 0 (n)n 11/2 τ(n) = 8000 [(σ 3 σ 3 ) σ 3 ] (n) 147 [σ 5 σ 5 ] (n)

10 Hecke operators Hecke defined a family of commuting operators T (i) (i Z 0 ) with the following properties: T (i) : Mk M k, T (i) : S k S k T (m)t (n) = T (mn) if (m, n) = 1 S k has a basis of Hecke eigenforms if f is a Hecke eigenform, a(1) = 1, f T (p) = λ p f, then a(p) = λ p

11 Lagrange s Four Squares Theorem In 1770, Lagrange proved Every integer can be written as the sum of 4 squares. In 1813, Cauchy proved the more general Every integer can be written as the sum of n n-gonal numbers.

12 Jacobi s Formulae In 1834 Jacobi proved that the number r 4 (n) of ways to represent n as the sum of four squares was { 8σ 1 (n) if n is odd r 4 (n) = 24σ 1 (n 0 ) if n = 2 r n 0 even, 2 n 0 Proof uses Θ 4 (z) = (x,y,z,w) Z 4 q x2+y 2+z2+w 2 which is a modular form of weight 2 and level 4.

13 Modular Forms with Level We will rewrite the functional equation in terms of the slash operator. Let α = ( ) a b c d GL + 2 (R) then (f k α)(z) = def (det α) k/2 (cz + d) k f (α.z) We change the arithmetic group in the definition of modular forms: {( Γ(N) = a b ) def c d SL2 (Z) : a d 1, b c 0 (mod N) }

14 Modular Forms with Level (cont.) A congruence subgroup of level N is any Γ so that Γ(N) Γ. A holomorphic modular form of weight k and level N is an f : H C so that 1. f (γ.z) = (cz + d) k f (z) for all γ Γ 2. if γ 0 SL 2 (Z), then (f k γ 0 )(z) has a Fourier expansion of the form n n γ0 a γ0 (n)q n N where q N = def e 2πiz/N 3. f is holomorphic on H and all its cusps

15 Congruent Numbers n Z 0 is congruent if there is a right triangle with rational side and area n. 1. Can you give an example of a congruent number? 2. How many congruent numbers are there? 3. What is the smallest congruent number?

16 Congruent Numbers (cont.) 6 is the area of a right triangle, so it is a congruent number. There are infinitely many congruent numbers since there are infinitely many Pythagorean triples. 5 is congruent since it is the area of a triangle with sides 3 2, 20 3 and 41 6.

17 A (somewhat) open problem The only open problem left from antiquity is the congruent number problem: Given an n, how can you determine if n is congruent. The problem is equivalent to Given an n, can you determine the number of rational points on the elliptic curve E n : y 2 = x(x n)(x + n)

18 The problem is hard (picture is Karl Rubin s)

19 More on E n : y 2 = x 3 n 2 x By Taniyama-Shimura, E n is associated to a modular form f En of weight 2 and level N En. a p = p + 1 #E n (F p ) L(E n, s) = p N En 1 1 a pp s +p 1 2s

20 Connection to Modular Forms n is congruent #E n (Q) = L(E n, 1) = 0 a n = 0 for a particular modular form

21 n is congruent #E(Q) = There is a 1-1 correspondence between rational solutions X, Y, Z to 1 2 XY = n and X 2 + Y 2 = Z 2 and rational numbers x, x + n, x n that are squares of rational numbers. The map is (X, Y, Z) (Z/2) 2 (X ± Y ) 2 = Z 2 ± 4n Divide both sides by four: ( X 2 ± Y ) 2 ± n = 2 ( ) 2 Z 2 Take-home Show that the map is injective Theorem E n has positive rank

22 #E n (Q) = L(E n, 1) = 0 A deep result of Coates and Wiles The converse is a deep ($ ) conjecture of Birch and Swinnerton-Dyer

23 L(E n, 1) = 0 a n = 0 for a particular modular form Let k 3 be odd. Shimura showed there is a nice injective map from M k/2 of level N to M k 1 for some level N. Tunnell constructed two forms of weight 3 2, f = a n q n and f = a nq n so that L(E, 1) = 0 iff a n = 0 (n odd) or a n/2 = 0 (n even) Moreover, Shimura(f ) = Shimura(f ).

24 Congruent Number Problem If n is squarefree and odd and n is the area of a right triangle with rational sides, then # { x, y, z Z : n = 2x 2 + y z 2} = # { x, y, z Z : n = 2x 2 + y 2 + 8z 2} Converse holds if BSD is true Proof follows from matching up images under Shimura Similar formula holds for even n. SAGE implementation of solution online

25 Integer vs half integer weight Half-space: H Group: Let G = {(α, φ(z)) : α GL 2 (Q), φ : H C hol.} where (α, φ(z))(β, ψ(z)) = (αβ, φ(βz)ψ(z)) Arithmetic group: Let Γ Γ 0 (4) Γ = { (γ, j(γ, z)) : γ Γ }

26 Integer vs half integer weight (cont.) Functional Equation: Let ξ = (α, φ(z)) G f (z) [ξ] k/2 = f (αz)φ(z) k Fourier Expansion: Complicated, but do-able

27 Siegel Modular Forms Comparison Sp g (Q) vs SL 2 (Q), H g vs H, clean Fourier expansion indexed by quadratic forms Θ series count number of ways of representing one quadratic form by another My interests Hecke theory computations for g = 2 Compute forms for g = 3

28 Function Field Modular Forms Comparison H = PSL 2 (K )/PSL 2 (O ), transform like Θ functions, finite Fourier expansion index by integers Θ series count number of ways of representing one polynomial over F q by another My interests I have developed a Hecke Theory Asymptotics for representation numbers

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