Modular Forms Ting Gong May 18, 2020 1 Dimension of Modular Forms Definition 1.1. A modular form of weight k for SL(2; Z) is a holomorphic function f on H az + b a b satisfying f = (cz + d)kf(z) where 2 SL(2; ), and is holomorphic at the cz + d c d Z cusp 1. 1 1 Since 2 SL(2; ), f(z + 1) = f(z). Then we have a Fourier expansion f(z) = 0 1 Z 1 X 2πinz 2πinz ane . Below we denote q = e . n=−∞ Definition 1.2. A modular form that vanishes at 1 is called a cusp form. We denote the space of modular forms of weight k for Γ(1) = SL(2; Z) as Mk(Γ(1)) and the space of cusp forms as Sk(Γ(1)). Definition 1.3. An automorphic function for Γ is a meromorphic function f on H and at az + b 1 such that f = f(z) cz + d Then f is a meromorphic function on the compact Riemann surface Γ(1)=H∗. If f doesn't have a pole, then by Louiville theorem and maximal modulus principle, f is constant. Notice if f1; f2 2 Mk(Γ(1)), then k f1 az + b (cz + d) f1(z) f1 = k = f2 cz + d (cz + d) f2(z) f2 Therefore, f1=f2 is automophic. Proposition 1.4. Let X be a compact Riemann surface, P1;:::;Pn 2 X, let r1; : : : ; rn be positive integers. Let V be the vector space of meromorphic functions on X, which are holomorphic besides possibly at Pm, and which are holomorphic or else have poles of order at most rm at Pm. Then the space V has dimension at most r1 + ::: + rm + 1. 1 Proof. Let r = r1 + ::: + rm, pick a coordinate function t = tj in a neighborhood of Pj with −rj respect to which Pj is the origin. If φ 2 V , it has Laurent expansion, φ(t) = aj;−rj t + −rj +1 r aj;−rj +1t + :::. We associate φ with v 2 C whose entries are the Taylor coefficients. If P P φ1; : : : ; φN 2 V , N > r, then c1; : : : ; cN are not all zero with cjvj = 0. Thus cjφj has no poles. Then since above is meromorphic on a compact Riemann surface, it is constant. Thus any vector subspace of V having dimension greater than r contains a constant function. Thus dim V ≤ r + 1. Proposition 1.5. The space Mk(Γ(1)) is finite dimensional. Proof. Let f0 2 Mk(Γ(1)) be nonzero. Let X be the compactification of Γ(1)=H. Let P1;:::;Pm be zeroes of f0, let r1; : : : ; rm be the orders of zeroes of f0 at these points. If f 2 Mk(Γ(1)), then by our remark before, f=f0 is automorphic. Moreover, f 7! f0 is an isomorphism of Mk(Γ(1)) and V in the last proposition. Thus Mk(Γ(1)) is finite dimensional. 2 Jacobi's Triple Product Formula Definition 2.1. Let k be even, k ≥ 4, the Eisenstein series is defined as 1 X E (z) = (mz + n)−k k 2 m;n2Z;(m;n)6=(0;0) We notice that the Eisenstein series is absolutely convergent since Z 1 Z 1 Z 1 Z 1 −k −k Ek(z) ≤ (mz + n) dmdn = 4 (mz + n) dmdn < 1 −∞ −∞ 0 0 since k ≥ 4, and after integration, the k − 2 ≥ 2. Definition 2.2. Let r 2 C, the divisor sum is defined as X r σr(n) = d djn Proposition 2.3. The Eisenstein series is a modular form. Proof. We first show the first condition. az + b 1 X az + b E = (m + n)−k k cz + d 2 cz + d m;n2Z;(m;n)6=(0;0) 1 X = (cz + d)k (m(az + b) + n(cz + d))−k 2 m;n2Z;(m;n)6=(0;0) 1 X = (cz + d)k ((am + cn)z + (mb + nd))−k 2 m;n2Z;(m;n)6=(0;0) 2 Since c; d are coprime, (m; n) 7! (ma + nc; mb + nd) permutes Z × Z. Thus we see Ek(z) satisfies the first condition of a modular form. It suffices to show that it is holomorphic at 1. To do this, we compute its Fourier expansion. When m = 0, Ek(z) = ζ(k). When m 6= 0, since k is even, ±1 contributes equally. Thus, we only consider m > 0. Z 1 f^(n) = (mz + n)−ke2πinz −∞ Then by the residue theorem, 2πi f^(n) = 2πires(e2πinz(mz − n)−k) = nk−1e2πimnz (k − 1)! Then by Poisson Summation Formula, k 1 k n 1 (2πi) X (2π) (−1) 2 X E (z) = ζ(k) + nk−1e2πimnz = ζ(k) + σ (n)qn k (k − 1)! (k − 1)! k−1 n=1 n=1 where q = e2πiz. Therefore, we see that it is holomorphic at 1. For a given k, either Sk(Γ(1)) = Mk(Γ(1)) or dim Sk(Γ(1)) + 1 = dim Mk(Γ(1)), since if these is a modular form of weight k, either the constant is zero, or we can substract by a multiple. For k ≥ 4, we see that there is an Eisenstein series with nonzero constant term. Therefore, dim Mk(Γ(1)) = dim Sk(Γ(1)) + 1. We observe that the modular forms form a graded ring. It is easy to show that if f 2 Mk(Γ(1)) and g 2 Ml(Γ(1)), then fg 2 Mk+l(Γ(1)). Example 2.4. We construct example below: let 1 1 X n X n G4(z) = 1 + 240 σ3(n)q ; and G6(z) = 1 − 504 σ5(n)q n=1 n=1 Clearly, G4 has weight 4 and G6 has weight 6. Then we define 1 ∆(z) = (G3 − G2) = q − 24q2 + 252q3 − 1472q4 + ::: 1728 4 6 which becomes a cusp form of weight 12. Theorem 2.5 (Jacobi's Triple Formula). 1 1 X 2 Y qn xn = (1 − q2n)(1 + q2n−1x)(1 + q2n−1x−1) n=−∞ n=1 Proof. Let 1 X 2 ν(z; w) = qn xn; q = e2πiz; x = e2πiw n=−∞ 3 1 1 1 X 2 X 2 X 2 ν(z; w + 2z) = qn (xq2)n = qn +2nxn = (qx)−1 q(n+1) xn+1 = (qx)−1ν(z; w) n=−∞ n=−∞ n=−∞ And we let 1 Y P (z; w) = (1 + q2n−1x)(1 + q2n−1x−1) n=1 1 Y P (z; w + 2z) = (1 + q2n+1x)(1 + q2n−3x−1) = (qx)−1P (z; w) n=1 ν(z;w) Therefore, let Λ ⊂ C be the lattice f2mz + njm; n 2 Zg and f(w) = P (z;w) , then f(z) is an elliptic function over Λ. Assume P (z; w) = 0, for fixed z. Then some factor of P is zero. Namely, q2n−1x = 0 or q2n−1x−1 = 0, for some n. Then 2πiz(2n − 1) ± 2πiw = kπi, where k 1 is odd. Therefore, w = ±z + λ + 2 , where λ 2 Λ. Thus these w are zeroes of P (z; w). We show that these w are also zeroes of ν(z; w). Since n2(2πiz) + n(2πiw) mod 2 = πi(2zn2 + ±2nz + 2nλ + n) mod 2 = nπi mod 2. Thus it is a series permuting between −1 and 1. And the sum gives 0. Therefore, we see that f(w) doesn't have a pole. Hence, f(w) is a constant, say φ(q). Thus ν(z; w) = φ(q)P (z; w). Q1 2n Next, it suffices to show φ(q) = n=1(1 − q ). Consider 1 1 1 X n 4n2 X nπi n2 1 ν(4z; ) = (−1) q = e 2 q = ν(z; ) 2 4 n=−∞ n=−∞ 1 P (4z; 2 ) 4 Thus, we divide the two equations and we have φ(q) = 1 φ(q ), now we compute P (z; 4 ) 1 P (4z; 1 ) Y 2 = (1 − q4n−2)(1 − q8n−4) P (z; 1 ) 4 n=1 Q1 2n Then as q ! 0 φ(q) ! 1. Therefore, φ(q) = n=1(1 − q ). 3 Dimension of Cusp Forms 3 −1 We use Jacobi's triple product formula, replacing q with q 2 and x with −q 2 . Then 1 1 2 2 X n (6n+1) 1 X n 3n +n 1 Y 3n 3n−1 3n−2 Y n (−1) q 24 = q 24 (−1) q 2 = q 24 (1−q )1 − q )(1 − q ) = (1 − q ) n=−∞ n=−∞ n=1 n=1 Definition 3.1. The Dedekind eta function is defined as 1 2 1 Y n X n η(z) = q 24 (1 − q ) = χ(n)q 24 −∞ Where χ(n) = 1 if n ≡ ±1 mod 12, −1 if n ≡ ±5 mod 12 and 0 otherwise. 4 az + b Proposition 3.2. If γ 2 Γ(1) then there exists a 24th root of unity (γ) such that η = cz + d 1 (γ)(cz + d) 2 η(z). 1 1 0 −1 Proof. Since Γ(1) is spanned by S = and T = , it suffices to check these 0 1 1 0 p 1 2πi 1 −iz 24 24 24 two. T (q ) = e q . Then we are done. Since η(z) = θχ( 12 ), we have τ(χ) = 2 3 and p 1 N = 12. Thus −izη(z) = η(− 2 ). Thus, we are done. Therefore, it is reasonable to get rid of the root of unity by rising to 24th power. Let 2 Q1 n 24 ∆(z) = η(z) 4 = q n=1(1 − q ) , then ∆(z) is a cusp form of weight 12.