Ramanujan s first letter to Hardy: 5 + = 1 + e 2π 1 + e 4π 1 +

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1 Ramanujan s first letter to Hardy: e 2π/ = 1 + e 2π e 4π 1 + e π/ 1 e π = e 2π 1 + Hardy: [These formulas ] defeated me completely. I had never seen anything in the least like this before. A single look at them is enough to show they could only be written down by a mathematician of the highest class. They must be true because no one would have the imagination to invent them. Where do these formulae come from, and why should they be true? 1

2 Let τ H and q = e 2πiτ. We are lead to define ( F (τ) def = q 1/ 1 1+ What is known about F (τ)? q q ). Rogers-Ramanujan: F = q 1/ n=1 (1 q n 4 )(1 q n 1 ) (1 q n 2 )(1 q n 3 ). Many singular values. For example, F (i), F (i/2), and F (i/ 10) = ( 90( + 2 ) 18 ) 1/. Modular equations: If x = F (τ) and y = F (3τ) then (x y 3 )(1 + xy 3 ) = 3x 2 y 2. Arithmetic: If K is an imaginary quadratic field and τ H K then F (τ) is a unit. 2

3 How are these things proved? Ramanujan s theory of modular functions (mostly identities). Clever manipulation of q series. Modular forms and multiplier systems. Kronecker s Limit Formula These methods are unsatisfactory as they do not provide any structure or framework in which to place the function F (τ). One cannot expect to prove more general results about F (τ) using these methods. 3

4 In this talk we will show: def j = F 1 is a modular function of full level, and hence an element of the function field of the modular curve X(). The function field C(X()) is rational, generated over C by j. This gives us the powerful interpretation of j (equivalently F ) as coordinate on the genus 0 modular curve X(). Using this viewpoint, we prove: x = j (τ) and y = j (nτ) satisfy a polynomial F n Z[X, Y ]. When K is an imaginary quadratic field and τ K H then j (τ) is a unit. The polynomial F n (X, X) Z[X] satisfies simple congruences modulo primes p. 4

5 X() as Riemann Surface Any subgroup Γ of SL 2 (Z) acts on H = H P 1 (Q) by fractional linear transformations. The quotient space X(Γ) def = Γ\H admits the structure of a compact Riemann surface. We consider congruence in particular subgroups, and Γ(N) def = {α SL 2 (Z) : α 1 mod N} and the associated Riemann surface X(N) def = X(Γ(N)). For any inclusion f : H G of congruence subgroups we get a field extension K(X(H))/K(X(G)) of degree [Ḡ : H]. For 1 N the genus of X(N) is 0, and hence C(X(N)) C(x).

6 Klein Forms L C a lattice with fixed Z basis ω 1, ω 2. Put W = ( ω 1 ω 2 ) and for a Q 2 set z = a W. Weierstrass σ function: σ(z, L) def = z ω L {0} ( 1 z ω ) e z/ω+1 2 (z/ω)2. Weierstrass η function: defined on L by σ σ σ (z + ω, L) = (z, L) + η(ω, L). σ Define the Klein form κ a (W ) def = e η(z,l)z/2 σ(z, L). κ a (W ) depends on the choice of basis of L. When W = ( τ 1 ) for τ H, we write κa (τ) for κ a (W ). 6

7 We have κ a (λw ) = λκ a (W ) for any λ C. For any b = (b 1, b 2 ) Z 2 and a = (a 1, a 2 ) Q 2, where κ a+b (τ) = ɛe πi(a 1b 2 b 1 a 2 ) κ a (τ), ɛ = 1 1 if b W 2L otherwise. For α SL 2 (Z), κ a (αw ) = κ aα (W ). Let q = e 2πiτ and q z = e 2πiz. Then κ a (τ) = q1/2(a2 1 a 1) e πia 2(a 1 1) (1 q z ) 2πi (1 q n q z )(1 q n /q z ) (1 q n ) 2. n=1 7

8 Constructing Functions on X(N) for Odd N Use Klein forms to construct functions on H invariant under Γ(N). Theorem 1[Kubert-Lang]: Let S Z 2 {0} and A = N 1 S. To each a = (a 1, a 2 ) A, associate an integer m(a) and let f def = a A κ m(a) a. Then f is invariant under the action of Γ(N) if and only if 1. a A m(a) = 0 2. a A m(a)n 2 a i a j 0 mod N for each pair (i, j) {(1, 1), (1, 2), (2, 2)}. 8

9 A Generator for C(X()). Theorem 2: Let ζ = e 2πi/ and define j def = ζ 1 4 k=0 κ ( 2, k ) κ ( 1, k ). Then j is a modular function of level and j = 1/F. Theorem 3: j as a function on X() has a single simple pole at. Corollary 4: For any number field K, we have K(X()) = K(j ). We have j (τ + 1) = ζ 1 j (τ) j ( 1/τ) = j (τ) j (τ)

10 Proofs For Theorem 2, use Theorem 1 with N =, A = {( j, k ) : j = 1, 2, 0 k 4}, and m(( j, )) = ( 1)j. To prove Theorem 3, compute the order of j at each cusp of X(). The product expansion makes it clear that j is holomorphic on H. Sample: The matrix α = ( ) SL2 (Z) takes the cusp represented by 2/9 to. Using the properties of Klein forms: j α 1 = ζ 2κ (0, 2 )κ ( 2,0)κ ( 1,1 )κ ( 1,2 )κ ( 2,1 ) κ (0, 1 ) κ ( 1,0)κ ( 2,2 )κ ( 1,3 )κ ( 2,4 ), and expanding this as a q series: j α 1 = (1+ζ 4 )+(1+3ζ +ζ 2 )q 1/ +..., so the order at the cusp represented by 2/9 is 0. 10

11 Corollary 4 follows from the fact that j Z((q 1/ )). To prove the formulae for j ( 1/τ), use the strategy in the proof of Theorem 3: we have j ( 1/τ) = (ζ 2 +ζ 3 )+(3+ζ+ζ 4 )q 1/ +..., so that 3 + ζ + ζ 4 j ( 1/τ) + (ζ 2 + ζ 3 ) j (τ) is a function on X() with no poles and is therefore constant. Inspection of the q series shows this constant to be ζ 2 + ζ 3. In this way, we find value of j at each cusp. In particular, j has a simple zero at 2/. We can now prove Ramanujan s evaluation for F (i) = 1/j (i). Indeed, i is fixed by τ 1/τ so x = j (i) satisfies x 2 (1 + )x 1 = 0. 11

12 Ramanujan s continued fraction is coordinate on X(). Gives analogy between j and j. For n > 1 and prime to, j (τ) and j (nτ) satisfy F n (X, Y ) Z[X, Y ]. If n is squarefree, H n (X) = F n (X, X) is monic. Because level is : more structure. There exists α SL 2 (Z) such that j α = 1/j For K imaginary quadratic and τ K H, j (τ) is a unit. For p ±1 mod F p (X, Y ) (X p Y )(X Y p ) mod p. For p ±2 mod F p (X, Y ) (X p Y )(XY p + 1) mod p. 12

13 An Involution on X(). Let σ a SL 2 (Z) satisfy σ a ( ) a 1 a mod. Then j σ a = j a ±1 mod j 1 a ±2 mod. Observe that σ a is in the normalizer of Γ() in SL 2 (Z). We expect σ a to be an involution since for any a, σ 2 a ±Γ(). Proof is simple and follows from transformation properties of Klein forms. Reduce to case a 2 mod since 1 acts trivially on H. 13

14 Modular Equations for j Existence follows from algebraic geometry, but we proceed classically. (Double coset decomposition): Let A = { ( ) } σ a b a 0 d : (a, b, d) = 1, ad = n, 0 b d. Then Γ() ( ) n Γ() = Γ()α. α A Let s k be the k th symmetric polynomial on {j α : α A}. Then s k = P k (j )/j m k where P k C[X]. s k is invariant under Γ() action, so is a rational function of j. Key is that (n, ) = 1 so j α has poles only at, 2/. 14

15 Since j, 1/j Z((q 1/ )) we find j α Z[ζ]((q 1/ )). Galois action on coeffs. of q series permutes the set {j α : α A}. Hence P k (X) Z[X] by Hasse Principle. Put f n (X) = j m 0 α A (X j α). Have shown f n (X) Z[X, j ], so let F n (X, Y ) be such that F n (X, j ) = f n (X). Hasse Principle shows F n (X, Y ) Z[X, Y ]. When n is squarefree, lead term in q series of j j α is a unit. Follows that H n (X) = F n (X, X) is monic. 1

16 H n (j (τ)) = 0 iff j (ατ) = j (τ) for some α A. For some β A and γ Γ(), we have σ 2 α = γβσ 2. Follows that j (σ 2 τ) = j (βσ 2 τ). Hence H n ( 1/j (τ)) = 0. Hence for any root z of H n (X), 1/z is also a root. By classical methods, if K is imaginary quadratic and for τ K H then j (τ) is a root of H n for some squarefree n. It follows that j (τ) is a unit. 16

17 F p (X, Y ) mod p. For a prime p, let α b = ( ) 1 b 0 p and ( ) α p = σ p 0 p 0 1. Then A = {αb : 0 b p}. Expanding as q series, find for 0 b < p j α 0 j α b 0 mod (1 ζ p ). Similarly j α p j p mod p if p ±1 mod and j α p 1 j p mod p if p ±2 mod. Finally, (j α 0 ) p j mod p. Since (1 ζ p ) is the unique prime of Z[ζ p ] over p, we can piece these congruences together. 17

18 Arithmetic of Singular Values Well known result [Shimura] that for K imaginary quadratic and τ K H the field K(ζ, f(τ)) is a certain class field of K, where we adjoin all values f(τ) for modular functions f of level with coefficients in Q. By Corollary 4, we know that K(ζ, j (τ)) is a class field. In fact, if τ = t 1 /t 2 with (t 1, t 2 ) O K then K(ζ, j (τ)) is the ray class field of K of conductor. 18

19 Computations Can compute F p (x, y) for small p using linear algebra. We find: F 2 (x, y) = yx 3 y 3 x 2 + x + y 2 F 3 (x, y) = yx 4 y 4 x 3 + 3y 2 x 2 + x y 3 F (x, y) = y (x 4 2x 3 + 4x 2 3x + 1) (x + 3x 4 + 4x 3 + 2x 2 + x) F 7 (x, y) = yx 8 + ( y 8 + 7y 3 )x 7 + 7y x 6 + ( 7y 7 + 7y 2 )x + 3y 4 x 4 + ( 7y 6 7y)x 3 7y 3 x 2 + (7y + 1)x y 7 F 11 (x, y) = x 12 +( y y 6 11y)x y 2 x y 3 x y 4 x 8 792y x 7 + (11y y 6 11y)x 6 792y 7 x + 49y 8 x 4 220y 9 x 3 +66y 10 x 2 +( 11y 11 11y 6 y)x+y 12 19

20 Questions Why are the coefficients of F n so small? Let K be imaginary quadratic, τ K H and L = K(ζ, j (τ)). Let O L be the full group of units in O L and let U be the subgroup of O L generated by all values j (τ) L with τ K H. What is the index [O L : U]? 20

21 Preprint and References bcais/papers.html 21

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