MODULAR FORMS AND THE FOUR SQUARES THEOREM. Contents 1. Introduction 1

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1 MODULAR FORMS AND THE FOUR SQUARES THEOREM AARON LANDESMAN Contents. Introduction 2. Definition of modular Forms 2.. Preliminaries A First Definition of Modular Forms The action of SL 2 (Z) on H Three Other Ways to Think about Modular Forms 5 3. The Space of modular Forms Eisenstein Series Notation for Spaces of Modular Forms A Bound on the Dimension of Modular Forms of a Fixed Weight A Complete Characterization of Modular Forms 2 4. Modular Forms of Higher Levels Congruence Subgroups A More Refined Definition of Modular Forms 4 5. The space M 2 (Γ 0 (4)) Computing the Fourier Series of G 2 (τ) The almost-invariance of G An Almost Invariant SL 2 (Z) function from G A Bounding Theorem The elements of M 2 (Γ 0 (4)) Theta Functions The Sum of Four Squares Theorem 27 References 28. Introduction The main goal of this paper is to introduce modular forms and to obtain an explicit formula for the number of ways to write a positive integer as a sum of four squares. Along the way, we shall also explicitly describe all modular forms with respect to the principal congruence subgroup SL 2 (Z). 2. Definition of modular Forms Modular forms are naturally viewed in several different contexts: as holomorphic functions satisfying transformation equations, as sections of line bundles on Riemann surfaces, and as functions on lattices. In the first section, we shall describe

2 2 AARON LANDESMAN these different perspectives for thinking about modular forms and explain how they relate. But first we must build up the necessary background. 2.. Preliminaries. Definition 2... Let F be a ring. The nth general linear group over F, GL n (F ) is the group of invertible n n matrices with coefficients in F, viewed as a group via multiplication and inversion of matrices. Definition Again, with F a ring, the nth special linear group over F, notated SL n (F ) is the group of n n matrices with coefficients in F with determinant. Definition Let F be a ring and let Z(n, F ) be the subgroup of diagonal matrices of GL n (F ). The Projective linear group of dimension n over F, P GL n (F ) = GL n (F )/Z(n, F ). Remark Let I n denote the n n identity matrix. Note that in the case that F = R or C, we have Z(n, F ) = F I n, is formed by diagonal matrices. In the further case that n = 2, which we shall mostly be dealing with in this paper, we have that P GL 2 (R) = GL 2 (R)/Z(2, R), P GL 2 (C) = GL 2 (C)/Z(2, C), since the matrices I 2, I 2 are the only two elements of Z(2, F ) with determinant. Definition Let Ĉ denote the Riemann sphere, also known as the one point compactification of C. The group of fractional linear transformations, notated az+b F LT, are rational functions f : Ĉ Ĉ which are of the form f(z) = cz+d. They form a group under composition. This is a group because it is the image of the natural surjection GL 2 (C) F LT. Note that the point of Ĉ not in C is labeled and any FLT of the form ax+b 0 is labeled. Lemma The group of fractional linear transformations is naturally isomorphic to P GL 2 (C) Proof. Observe ( ) that we have a natural map φ : GL 2 (C) F LT, mapping the a b matrix to the function f(z) = c d az+b cz+c. Clearly Z(2, C) ker φ. Hence, we have an induced map φ : P GL 2 (C) = GL 2 (C)/Z(2, C) F LT. This map is clearly surjective by choosing the same (a, b, c, d) for both groups. It only remains to check only the identity acts trivially, which holds because the identity is the only element mapping to the fractional linear transformation with a zero at z = 0 and a pole at z =. Remark As described above, we obtain that P GL 2 (C) acts naturally on the Riemann sphere by fractional linear transformations. Lemma Let Aut(Ĉ) denote the set of invertible meromorphic maps f : Ĉ Ĉ. We have Aut(Ĉ) = F LT. Proof. Clearly all fractional linear transformations are automorphisms, with their inverse given by their inverse in the group of fractional linear transformations, which we saw above was P GL 2 (C). Conversely, suppose T Aut(Ĉ). Then, T must have precisely one pole, and precisely zero. Suppose the pole is p and the zero is q. Note that we must have p, q distinct. Then, the meromorphic function S = T z p z q, has no zeros and no poles. Note this is written as a product of meromorphic

3 MODULAR FORMS AND THE FOUR SQUARES THEOREM 3 functions, with removable singularities filled in. That is, in S(z), we cancel off common factors from the numerator and denominator. Then, S is a function on the Riemann sphere with no poles or zeros. Hence, by Liouville s theorem, it must be constant, and so S = c, which implies T = cz cp z q, which can be rewritten in the form of a linear transformation by multiplying the numerator and denominator by the same constant so that the resulting determinant will be. Notation 2... Let Im(z) denote the imaginary part of z. Definition The upper half plane is the subset H = {z C Im(z) > 0}. ( ) a b Lemma Let g = SL c d 2 (R). Then, for z BC, Im(gz) = Im(z) cz+d. 2 Proof. Observe that Hence, Im(gz) = Im(z) cz+d 2. g(z) = az + b cz + d (az + b)(d + cz) = cz + d 2 = bd + ac z 2 + Re(z)(ad + bc) + i(ad bc)im(z) cz + d 2 = bd + ac z 2 + Re(z)(ad + bc) + iim(z) cz + d 2. Corollary 2... Let Aut(H) denote the group of meromorphic functions sending H to H with a meromorphic inverse on H. The restriction of fractional linear transformations to H define a subgroup of Aut(H). Proof. Clearly, and FLT maps R { } R { }, and by the previous lemma, it maps i to something with positive imaginary part. Therefore, it must map all of H H, and since it is an automorphism of Ĉ it restricts to an automorphism of H A First Definition of Modular Forms. The main idea of modular forms is that they have a certain invariance under composition with fractional linear transformations. Definition Let Γ = SL 2 (Z) act on the upper half plane via fractional linear transformations. Suppose γ Γ, γ(z) = az+b cz+c. A weakly modular function of weight k is a complex meromorphic function f : H H, satisfying f(γ(z)) = (cz + d) k f(z). Proposition The only weakly modular function of odd weight k is 0. Proof. Let f be such a weakly modular function. Then, applying the matrix I, we see f(z) = ( ) k f(z) = f(z) if k is odd. This implies f(z) = 0 identically on H. We d like to say that a modular function is simply a weakly modular function that is also meromorphic at infinity, where infinity is thought of as lying very far in the imaginary direction. We can also translate H to the unit disk by a linear

4 4 AARON LANDESMAN fractional transformation, which translates R { } to the boundary of the unit disk. Now we develop a tiny bit a Fourier analysis to make this notion precise. Notation We shall use D to denote the complex unit disk D = {z C z < }, and D to denote the punctured complex unit disk, D = D {0}. Lemma Any function f : H C satisfying f(z) = f(z + ) can be written as f(z) = g(e 2πiz ) for g : C holomorphic. Proof. Since the upper half plane is simply connected, we have a well defined logarithm log : H D C. Locally, log has an inverse. Define the coordinate q(z) = 2πi log z. Then, at least locally, we can write z = e2πiq. Therefore, locally we can write g(q) = f(e 2πiq ) = f(z). This is well defined on the strip 0 Re(z) <. However, since f(z) = f(z + ), we can holomorphically extend this function by defining f(z) = f(z z ). So, g(q) is holomorphic because it is locally holomorphic. Notation Let S = ( ) 0, T = 0 ( ) as elements of SL 0 2 (Z). Remark Note that any weakly modular function f satisfies f(z) = f(z + ) because f(z +) = f(t z) = k f(z) = f(z) by definition of weakly modular. Therefore, we can write f(z) = f(e 2πiq ) = g(q) : D C. Because g(q) is holomorphic, on D and meromorphic at 0, we can expand it as a Laurent series around 0, g(q) = i= a nq n. Definition A function f : H C satisfying f(z) = f(z +) is meromorphic at infinity if we write f(z) = g(q) = i= a nq n as above, we have a n = 0 for all n N, for some N BZ. Equivalently, we can say that g(q) has a pole or a removable singularity at 0. Definition A function f : H C satisfying f(z) = f(z + ) is holomorphic at infinity if when we write f(z) = g(q) = i= a nq n as above, we have a n = 0 for all n < 0. Equivalently, we can say that g(q) has a removable singularity at 0. Definition A modular function is a weakly modular function f : H C that is meromorphic at infinity. Definition A modular form is a weakly modular function f : H C which is holomorphic on H, and is holomorphic at infinity The action of SL 2 (Z) on H. The purpose of the next several theorems is to gain a better understanding of the group P GL 2 (Z). Definition A fundamental domain for a group action Γ on H is an open set D H so that the closure of D, notated D satisfies ΓD = H, no two points in D are in the same Γ orbit, and for any point p D = D D, p is not in the same orbit as any point in D and p is only in the same orbit as finitely many other points in D. Theorem The set D = {z C z >, 2 < Re(z) < 2 } is a fundamental domain for P GL 2 (Z) acting as fractional linear transformations. Proof. Let the subgroup of SL 2 (Z) generated S, T be denoted by H. We will first show that if z C, it is in the same orbit as some element in D. To see this, by

5 MODULAR FORMS AND THE FOUR SQUARES THEOREM 5 Lemma 2..0, we may note that Im(gz) = Im(z) cz+d 2. However, there are only finitely many γ SL 2 (Z) for which cz + d. Therefore, there are only finitely many such γ H. But, for one of these finitely many γ, Im(γ(z)) is maximized over all γ H, and therefore there is some g such that Im(g(z)) attains its maximum over g H. We may apply an appropriate power of T so that 2 < Re(T k g(z)) 2. Of course, this has the same imaginary part as g(z). Note that we must have g(z) because if g(z) <, and g is such that Im(gz) is maximized, then Im(Sg(z)) = Im(g(z)) g(z) 2 < Im(z), contradicting maximality of g. All that remains to check is that no two elements of D are in the same SL 2 (Z) orbit. Suppose we have z D. Suppose g(z) D. We may assume Im(g(z)) Im(z), but by lemma 2..0, this implies that cz + d. This can only happen when c = ±, 0. This is because Im(z) 2, and so if c 2, Im(cz + d), and so cz + d >. In the case c = 0, we must have d = ±. This implies a =, and since the real part of z is constrained by Re(z) < 2, the only way it can be in the same orbit as another point is if Re(z) = ± 2 and z D. Next, if c = ±, we can assume without loss of generality c =, Then we need z + d. If d = 0, then z =, so b =. In the special case that z is a cube root of unity, that is Re(z) = 2, we may have a = ±, 0, but in all these cases it relates z to other points on the boundary of D. Otherwise, we must have a = 0, and in this case it relates z = α + βi, to α + βi, which again lies on the boundary. Therefore, no points in the interior are in the same SL 2 (Z) orbit. Corollary Let H be the subgroup of SL(2, Z) generated by S, T. Then, H = P SL(2, Z). Proof. Let z D. For a given g P GL 2 (Z), take w = gz. Then, as we showed in the proof of Theorem 2.3.2, there exists some h H with hw D. Therefore, since D is a fundamental domain, we must have hw = z. which means h = g, and hence g = h H. Therefore H = P SL(2, Z). Definition (Modular Form). A modular Form of weight k is a holomorphic function on the upper half plane satisfying f(z + ) = f(z) and f( z ) = zk f(z), which is also holomorphic at infinity. Lemma Definition is equivalent to our original definition Proof. Since S, T generate P GL 2 (Z), having f(γ(z)) = (cz + d) k f(z) for all k is equivalent to having this hold just for γ = S, T, as we are assuming in this definition Three Other Ways to Think about Modular Forms. The following definition perhaps illuminates the reason for calling it a modular form instead of simply a function. Since all modular forms of weight 2k + are 0, at least when our definition involves relations for all of SL 2 (Z), we can restrict our attention to even weight modular forms. Therefore, we can make the following definition. Definition 2.4. (Modular Form). A modular form of weight 2k is a holomorphic form ω = f(z)dz k defined on H that is SL 2 (Z) invariant, and f(z) is holomorphic at infinity. Remark In the above definition, dz k should be read as (dz) k and not d(z k ), as it should really be thought of as a section of the line bundle Ω k H.

6 6 AARON LANDESMAN Lemma Definition 2.4. is equivalent to Definition Proof. Observe that d az+b dz cz+d = (a(cz+d) c(az+b) (cz+d) 2 = ad bc (cz+d) 2 = (cz+d) 2. Therefore, f(gz) = (cz + d) k f(z) if and only if f(gz)d(gz) k = (cz + d) 2k f(z)d(gz) k = (cz + d) 2k f(z)(cz + d) 2k dz k = f(z)dz k. Similarly, this invariance precisely implies f transforms as in the definition of modular form by the same computation. So, ω is SL 2 (Z) invariant if and only if f(z) is a modular form of weight 2k. Definition A modular Form of weight 2k is a holomorphic form ω = f(z)dz k defined on H/SL 2 (Z), and f(z) is holomorphic at infinity. Lemma Definition is equivalent to Definition 2.4. Proof. By definition, forms on H which are G invariant are equivalent to forms on H/G. In addition, we can also view modular forms as acting on the set of lattices. To justify this, we shall now see that the space of lattices is equivalent to H/SL 2 (Z). Definition Let V be a finite dimensional real vector space. A lattice Γ V is a discrete subgroup under addition of finite rank, which generates V as an R vector space. Notation Denote R to be the set of lattices in C. Let L denote pairs of nonzero complex numbers (ω, ω 2 ) (C ) 2 such that Im(ω /ω 2 ) > 0. Lemma The set R is in bijection with L/SL 2 (Z), where SL 2 (Z) acts on these two dimensional vectors of R by left multiplication. Proof. First, define the map F : L R. sending (ω, ω 2 ) Zω Zω2. First note that all elements of the image are lattices because Im(ω /ω 2 ) 0, or equivalently, (ω, ω 2 ) generates C as an R vector space. We have to show that this map is surjective, and that any two elements mapping to the same lattice are in the same SL 2 (Z) orbit. First, let us show surjectivity. Given any lattice, we can find two elements ω, ω 2 that generate it as a Z module. These elements clearly must be independent, and hence, after possibly reordering, we can arrange so that Im(ω /ω 2 ) > 0. Finally, we just have to check that two pairs mapping to the same set are in the same SL 2 (Z) orbit. First, it s clear that any two pairs which are in the same SL 2 (Z) orbit generate the same lattice, since multiplication by an element of SL 2 (Z) is simply a change of Z basis, with inverse given by its inverse in SL 2 (Z). However, suppose we have (ω, ω 2 ) and (η, η 2 ) both of which generate the same lattice. We must have matrices g, h with integer entries, mapping each lattice to the other. In particular, g and h must be mutual inverses. But since all integer matrices have integer determinants, this implies both g and h must have unit determinant. However, the fact that Im(ω /ω 2 ) > 0, Im(η /η 2 ) > 0 implies that the change of basis matrix must have positive determinant, and hence g, h SL 2 (Z). Corollary Denoting Λ = R/C where C acts on L by λ(ω, ω 2 ) = (λω, λω 2 ), we obtain a bijection between H = Λ. Proof. Clearly R/C = H, where C acts by scalar multiplication, since the point (ω, ω 2 )/C can be identified with the quotient ω /ω 2, which by assumption has

7 MODULAR FORMS AND THE FOUR SQUARES THEOREM 7 positive imaginary part, and hence lies in the upper half plane. Therefore, quotienting both sides the bijection from the previous lemma by the action of C, we obtain the corollary. Definition (Modular Form). Let Γ R, λ C. A modular Function of weight k is a function F : R C satisfying F (λγ) = λ k F (Γ), where F is holomorphic as a function from H after the identification Λ = H as in the previous corollary. Lemma Definition is equivalent to Definition Proof. (Proof of equivalence to our previous definitions of modular forms). Strictly speaking this is not precisely the same definition as the modular forms we have seen before, since this has a different domain and range, but we shall show that these two notions are equivalent. First, note that (ω, ω 2 ) is SL 2 (Z) invariant, as we showed in Lemma Therefore, to F we can associate the function f : H C, defined by f(ω /ω 2 ) = ω k 2 F (ω, ω 2 ). Then, observe that f is a modular function as defined in Definition We have f(γ(ω /ω 2 )) = F (γ(ω /ω 2 ), )) = F ( aω+bω2 cω +dω 2, ) = (cω /ω 2 + d) k F (aω /ω 2 + b, cω /ω 2 + d) = (cω /ω 2 + d) k F (ω /ω 2, ) = (cω /ω 2 + d) k f(ω /ω 2 ), which shows that f is a modular function as defined above. Similarly, we see that if f is modular in the sense defined above, then F is a modular function in this sense. These equivalence classes should be thought of as points in the moduli space of isomorphism classes of elliptic curves. That is, any elliptic curve can be represented by the quotient of C by a lattice Λ, and two elliptic curves are isomorphic if and only if their lattices are scalar multiples of each other. This allows us to view modular functions as functions on isomorphism classes of elliptic curves. 3. The Space of modular Forms In this section, we will describe the surprisingly simple complete classification of all modular forms on H. Essentially, all modular forms are given by Eisenstein series, which we will now define. Furthermore, the Eisenstein series of higher weights are generated by the Eisenstein series of weights 2 and Eisenstein Series. Definition 3... The Eisenstein series of weight k is a function R C, sending a lattice Γ to G k (Γ) = γ Γ,γ 0 γ k. Lemma For k > 2, the Eisenstein series converges absolutely Proof. First, note that given a lattice Γ, we can find a constant c Γ such that there are at most r 2 c Γ points γ Γ with γ < r. This implies that there is another constant k Γ so that there are at most rk Γ points with r < γ < r +. Then, in order to n< γ n+ show absolute convergence, we can write γ Γ,γ 0 γ k = n=0, γ k and it suffices to show the right hand series converges. Clearly, to show this converges, it suffices to show converges, as the n = 0 term is finite. n= n< γ n+ However, n< γ n+ γ k n= n< γ n+ γ k < γ k n< γ n+ < nk n k Γ = k n k γ. Then, n k n= k Γ n k, which converges so long as k > 2.

8 8 AARON LANDESMAN Notation 3... We shall now abuse notation, notating G k both as a function on lattices and on the upper half plane under the correspondence from the previous section. We shall write G k (z) to view it as a function on H, and G k (Γ) to view it as a function on lattices. Lemma Under the correspondence of functions on lattices with functions on H, G k (Γ) becomes the function G k (z) = k m,n Z,(m,n) (0,0). (mz+n) Proof. Let a lattice Γ = ω Z ω 2 Z. Then, under the correspondence from the previous section, by letting z = ω ω 2, we obtain the function G k (z) = ω2 k G k (ω Z ω 2 Z) = ω2 k γ k = ω k 2 = = γ Γ,γ 0 m,n Z,(m,n) (0,0) m,n Z,(m,n) (0,0) m,n Z,(m,n) (0,0) (mω + nω 2 ) k (m ω ω 2 + n) k (mz + n) k. Definition The Riemann Zeta Function is defined to be ζ(z) = n z. n>0,n Z Note that it is absolutely convergent for k C, Re(k) >. In particular it is convergent on reals greater than. Theorem The Eisenstein Series G k (z) is a modular function of weight k, for k 3, and G k ( ) = 2 ζ(k) for k even, and 0 for k odd. Proof. First, let us check it satisfies the transformation property. Clearly the Eisenstein series is preserved under the map z z +. For the other invariance, we need to check G k ( z ) = z k G k (z). Indeed, G k ( z ) = m,n Z,(m,n) (0,0) = z k (m z m,n Z,(m,n) (0,0) = z k = z k G k (z) m,n Z,(m,n) (0,0) + n)k (m + nz) k (mz + n) k Next, we shall check G k (z) is holomorphic in H. Defining f t = (mz + n) k, m,n Z, m <t, n <t,(m,n) (0,0)

9 MODULAR FORMS AND THE FOUR SQUARES THEOREM 9 we know we have convergence {f t } G k. We wish to show this convergence is uniform on every compact set. Note that a small open set U containing D intersects only finitely many translates of the fundamental domain, and since we can cover any set in H by translates of this open set U, we can cover any compact set by finitely many translates. So, to show uniform convergence on any compact set, it suffices to show uniform convergence on the fundamental domain. But this follows simply from the estimate that mz + n 2 = m 2 z z + 2Re(z)mn + n 2 m 2 z z n2 = m 2 z z mn + n 2 = mw + n 2, where w is a primitive cube root of unity, and z lies in the fundamental domain D = {z H 2 Re(z) < 2, z }. Now, to check uniform convergence, we need that for all ɛ > 0, T t > T if f t (x) f(x) < ɛ. Using the above estimate, if we choose T such that it is satisfied at w, (which can clearly be done since the function converges at w,) then the same T works for all other points x D because f(x) f t (x) = m,n t m,n t < ɛ. mz+n k mx+n k Therefore, we have the {f t } G k (z) uniformly, and since all f t are holomorphic on H, so is G k (z). Finally, to check G k (z) is holomorphic at infinity, by Riemann s removable singularity theorem, it suffices to check it is bounded. However, again using the estimate from the previous paragraph, we have that it is bounded in the fundamental domain, by the value at w, a cube root of unity. Then, using the fact that G k (z) = G k (z + ), we have that the value at infinity is bounded, and so it must be holomorphic. Finally, we just have to check the value G k ( ). We know since the series is uniformly convergent, its value at is the term by term limit. For z 0, we have lim Im(z) (mz+n) k = 0. For even k, we have, G k ( ) = lim = Im(z) n,m,(n,m) (0,0) lim Im(z) n,m,(n,m) (0,0) = n k = 2 n 0 n 0 = 2ζ(k) n k (mz + n) k (mz + n) k In the case that k were odd, we would reach 0 at the penultimate step in the above computation. Corollary For odd k, G k = 0. Proof. It is a modular form of odd weight, hence 0. Of course, we could also see this by observing explicitly that the sum is 0, by pairing off a point in the lattice with its negative Notation for Spaces of Modular Forms. Notation Denote the vector space of modular forms of weight k by M k. Note that it is a vector space since linear combinations of modular forms of weight k are again obviously modular forms of weight k. Definition A modular form f is called a cusp form if a 0 = 0, where a 0 is the constant term in the fourier expansion of f at.

10 0 AARON LANDESMAN Definition Denote the vector space of cusp forms of weight k by M 0 k. Again, this too is a vector space, because M k is a vector space, and linear combinations of functions which vanish at infinity also vanish at infinity. Definition Denote the modular form algebra M = k Z M k, and the cusp form ideal M 0 = k Z M 0 k. Notation Denote = 60 3 (G 4 (z)) (G 6 (z)) 2. Lemma M 0 2. Proof. Clearly G 4 (z) 3 is a modular form of weight 2 = 4 3, because it is the cube of a modular form of weight 4, and (G 6 (z)) 2 is also a modular form of weight 2 because it is the square of a modular form of weight 6. Therefore, is indeed a modular form of weight 2. To check it is also a cusp form, we need to check that its value at infinity is 0. However, ( ) = 60 3 (2ζ(4)) (2ζ(6)) 2 = 60 3 ( π4 45 ) ( 2π6 945 )2 = 0. Lemma M k = 0 for k 2Z + Proof. This is simply a restatement of the fact that there are no nonzero odd weight modular forms A Bound on the Dimension of Modular Forms of a Fixed Weight. Above we have described the Eisenstein series of even weights. It will turn out that products of these account for all modular forms. We shall go about showing this by first obtaining an upper bound on the dimensions of modular forms of varying weights, and then showing that the Eisenstein Series generate a space of exactly this dimension. Definition The vanishing order of a holomorphic function f at a point p, denoted v p (f) is the maximum integer n for which we can write f(z) = g(z) (z p) n for g holomorphic. Lemma Note that the vanishing order is a valuation. That is, v p (fg) = v p (f) + v p (g) and v p (f + g) v p (f) + v p (g). Proof. After writing out Taylor series for f and g, the lemma is immediate. Lemma Let f be a modular form. If p = γ(q) then v p (f) = v q (f). Therefore, we have a well defined value v t (f) for t H/SL 2 (Z). Proof. Since f(γ(z)) = (cz+d) k f(z), and since v p (cz+d) = 0 for all p H, we have v γ (p)(f) = v p (f). Thus, we naturally have an induced valuation on the quotient H/SL 2 (Z). Definition Let G be a group acting on H. For p H define e p to be the order of the stabilizer of p under the G action. Remark Under the action of G = SL 2 (Z), by 2.3.2, we understand that in the fundamental domain, e p = 2 for p = i, e p = 3 for p = w, with w a sixth root of unity, and otherwise e p =. Theorem For f a nonzero modular function of weight 2k, v (f) + v p (f) = k e p 2. p H/SL 2(Z)

11 MODULAR FORMS AND THE FOUR SQUARES THEOREM Proof. To prove this, we shall integrate the function f (z) f(z). First observe that since f is a nonzero modular form, there is some neighborhood of, call it N, at which it has no zeros. Observe that the zeros of f are exactly the poles of f (z) f(z) and the residue of f (z) f(z) is the multiplicity of the zeros. Since D N is compact, f can only have finitely many zeros in that region. Define F to be a set containing D N and of the form D {z H Im(z) < k} for some real constant K. We shall then compute the contour integral f (z) F f(z) dz in two different ways. Therefore, the sum above is a finite sum, hence well defined. Define A, C to be the sixth root of unity and third root of unity, respectively, on our fundamental domain, and define B to be the point i. Next, we have to deal with poles on the boundary. If there are poles on the boundary, other than the points A, B, C, they will come in pairs. which are reflections around the real axis. We can then simply integrate along small semicircles around them, and take the limit as the radius approaches 0. Note that f (z) f(z) has the same residue at both of them, because the order of f is well defined on points that are equivalent under the SL 2 (Z) action. Hence, we can assume there are no points on the boundary. Next, we take care of poles at A, B, C. If there is a pole at B, then if we replace the contour going through B by a circle of small radius around B, in the limit that radius goes to 0, the circle approaches a half circle, and hence the integration around it is 2 Res B( f f ) = 2 v Bf. Next, we know there is a pole at A if and only if there is a pole at C, and if we replace the contour going through A and C by small circles around A and C, in the limit their radii go to 0, the each approach 6 of a circle, hence they contribute together ( 6 Res C( f f ) + 6 Res A( f f ) = 3 Res A( f f ) = v A(f). Now, using the residue theorem and adding the poles inside the contour, we have f (z) D f(z) dz = 2πi p H/SL 2(Z) e p v p (f). So, it remains to show that the integral along this region F is equal to 2πi( v (f) + k 2 ). Indeed, Let us rewrite F as the union of C curves CD, DE, EA, AB, BC, where D = 2 + Ki, E = 2 + Ki, where we defined K in the first paragraph. Then, observe that ED = 2πiv (f), because when we transform the upper half plane to the unit circle by taking a logarithm, the line segment ED travels around the origin exactly once, and since there are no other residues in the region, the integral is exactly 2πiRes ( f f ) = f f 2πiRes (f). Hence, DE = 2πiRes (f). So, our remaining goal is to show f EA,AB,BC,CD f = 2πi k 2. First, observe that because f(z + ) = f(z), we have f EA f = f DC f and so, f EA+CD f = 0. So, we just need to show f AB,BC f = 2πi k 2. To complete this, note f( z ) = zk f(z), and so making the change of variable y = z, we see f f B A f (z) B f(z) dz = = = A B C B C kz k f( z ) + zk f ( z ) z k f( z ) z 2 dz f (y)y k ky k f(y) y k dy f(y) f (y) f(y) + C B k y dy.

12 2 AARON LANDESMAN Therefore, AB,BC as we wanted to show. f C f = f (y) B f(y) f (y) f(y) + k C y dy = k dy B y = 3.4. A Complete Characterization of Modular Forms. Corollary For k < 0, M k = 0. iπ 2 iπ 3 kdz = 2πi Proof. If there were some form of negative weight, by it would have to have some v p (f) < 0. However, since it is holomorphic, we must have v p (f) 0. So, there are no negative weight forms. Corollary M 2 = 0. Proof. From 3.3.6, v (f) + p H/SL 2(Z) e p v p (f) = 2. However, since all v p 0, v p (f) Z and all e p 6, the above formula has no solutions, hence there cannot be any modular forms of weight 2. Lemma For k 2Z, if M k 0 0, then M k = M 0 k C Gk Proof. First, define the functional ξ : M k C, ξ(f) = f( ). This is clearly linear, and by definition Mk 0 = ker ξ. Since the image of ξ. In the case M 0 k 0, dim(im(ξ)) =, and so by rank nullity, dim M k = + dim Mk 0. Lemma Recall defined in Notation In Lemma we have shown is a cusp form of weight 2. There is an isomorphism M k 2 = M 0 k, defined by f f. Proof. Clearly multiplication by maps F : M k 2 Mk 0, f f but since is a cusp form, it follows that the product of with any modular form is a cusp form, and so the image lies in Mk 0. By we have v ( ) + p H/SL 2(Z) e p v p ( ) =. Since it has a 0 at, it cannot have any other 0 s. Therefore, we can define an inverse map division by, G : Mk 0 M k 2, g g. This is well defined because for any g Mk 0, we know g( ) = 0, and since has a zero of order at, g/ cannot have a pole at. Furthermore, since is holomorphic and non-vanishing on H, g/ is also holomorphic on H. Therefore, g/ is holomorphic on H, holomorphic at infinity, and is the quotient of two modular forms, hence it is itself a modular form, as desired. Clearly, G, F are mutual inverses, and so F defines an isomorphism. Theorem For k odd, and for k < 0, and k = 2, we have M k = 0. In the case k = 0, we have M 0 = C. For k = 4, 6, 8, 0, Mk = C Gk. And for k 2, M k = C Gk Mk 2. Proof. We have already proved that for k odd, k < 0, and k = 2 that M k = 0 in the above lemmas. For k = 0, 4, 6, 8, 0, we know Mk 0 = M k 2 = 0, and so we obtain dim M k, by However, we have explicitly produced modular functions of weight k, namely the constants C if k = 0, and G k (z) otherwise. Therefore, dim M k =, and it is generated by C if k = 0, and G k otherwise. It only remains to prove that for k 2, M k = C Gk Mk 2. By the previous lemma, we obtain that M0 k = M k 2. Further, by Lemma we k 2,

13 MODULAR FORMS AND THE FOUR SQUARES THEOREM 3 know that there is at most one linearly independent non-cusp form of any given weight. However, once again, we have produced such a modular form G k (z) which is not a cusp form because ζ(k) 0 for k > 0, k 2Z. Therefore, M k = C Gk M k 2. k/2 if k 2 mod 2 Corollary For k > 0, dim M k = k/2 + if k 0, 4, 6, 8, 0 mod 2. 0 otherwise Proof. The above theorem tells us this is the case for k < 2, and induction and the last statement of the theorem tells us the dimension counting holds for k 2. We are now able to completely characterize modular forms on C in terms of G 4 and G 6. Corollary M = C[G 4 (z), G 6 (z)] Proof. We clearly have a homomorphism φ : C[G 4 (z), G 6 (z)] M, since G 4 (z), G 6 (z) are modular forms, and linear combinations of modular forms are modular forms. We wish to show φ is an isomorphism. First, let us show φ is a surjection. Note that C[G 4 (z), G 6 (z)], since it is a linear combination of powers of G 4, G 6, so by the theorem, it suffices to show that G k C[G 4 (z), G 6 (z)]. However, we may note that for all k > 4, k 2Z we can write a form G a 4G b 6, which is a form of weigh 4a + 6b. Clearly, for any even k 4, we can find nonnegative integers a, b so that 4a+6b = k, and so this gives us some modular form of weight k. However, it is not a cusp form because G 4 and G 6 are not cusp forms. Hence, by our structure theorem above, we can write it uniquely as f + g where g Mk 0, f CG k. Inducting on k, we may assume that g C[G 4 (z), G 6 (z)]. This implies that f C[G 4 (z), G 6 (z)]. However, by our structure theorem, f is a constant multiple of G k (z). Therefore, G k (z) C[G 4 (z), G 6 (z)], completing the proof. To complete the proof, we just need to show φ is also an injection. Since φ is a map of graded algebras, it suffices to show φ is an injection on the kth graded component. However, by Corollary??, we precisely know the dimension of M k, and it is simple to see that C[G 4 (z), G 6 (z)] has the same dimension. Since the two dimensions are the same, and φ is surjective on graded components, it must also be injective on graded components. Therefore, φ is also injective, and hence an isomorphism. 4. Modular Forms of Higher Levels It seems above we have completely solved the subject of Modular Forms. However, we have made the stringent requirement that these forms transform properly under all of SL 2 (Z). The next step is to slightly loosen the requirements, and only necessitate that the functions transformed properly under certain finite index subgroups in SL 2 (Z). It turns out that there are many more rich ideas but first we have to specify the correct subgroups, called Congruence Subgroups. We will then use the more general modular forms to deduce the number of ways to write a number as a sum of four squares.

14 4 AARON LANDESMAN 4.. Congruence Subgroups. Definition 4... The Principal Congruence Subgroup of level N is denoted {( ) ( ) ( ) a b a b a b Γ(N) = SL c d c d 2 (Z), c d where congruence mod N is taken entry-wise. Lemma The index [SL 2 (Z) : Γ(N)] <. ( ) 0 0 } mod N Proof. Observe Γ(N) is the kernel of the map SL 2 (Z) SL 2 (Z/NZ), where the latter is a finite group. Hence, the kernel Γ(N) has finite index. Definition A Congruence Subgroup of level N is a group Γ with Γ(N) Γ SL 2 (Z). Example 4... The subgroups of SL 2 (Z) defined by {( ) ( ) ( ) a b a b a b Γ (N) = SL c d c d 2 (Z), c d ( ) b 0 } mod N {( ) ( ) ( ) a b a b a b Γ 0 (N) = SL c d c d 2 (Z), c d ( ) a b 0 d } mod N are principle congruence subgroups of level N with Γ(N) Γ (N) Γ 0 (N) SL 2 (Z) A More Refined Definition of Modular Forms. We are now ready to define Modular Forms with respect to arbitrary principal congruence subgroups. Essentially, it will be the same as before, but they only need transform properly with respect to these subgroups. ( ) a b Definition For γ = SL c d 2 (Z), τ H the factor of automorphy, is denoted j(γ, τ) = cτ + d. Definition A weight k operator is a map [γ] k : Hom(H, C) Hom(H, C) with γ SL 2 (Z) defined by (f[γ] k )(τ) = j(γ, τ) k f(γ(τ)). Here we are denoting [γ] k (f) = f[γ] k. Definition For Γ a set of matrices in SL 2 (Z), a meromorphic function f : H C is invariant of weight k with respect to Γ if f[γ] k = f, γ Γ. Definition For Γ a set of matrices in SL 2 (Z), a meromorphic function f : H C is weakly modular of weight k with respect to Γ if is invariant of weight k with respect to Γ and f is holomorphic. Lemma For γ, γ SL 2 (R), the following statements hold () j(γγ, τ) = j(γ, γ (τ))j(γ, τ) (2) (γγ )(τ) = γ(γ(τ)) (3) [γγ ] k = [γ] k [γ ] k (4) Im(γ(τ)) = Im(τ) j(γ,τ) 2

15 MODULAR FORMS AND THE FOUR SQUARES THEOREM 5 Proof. We have already seen the fourth and second statement above. The other two can be seen by direct computation. That is, simply write out matrices for γ, γ write out their corresponding factors of automorphy, multiply them, and verify the equalities. Corollary If a function f is weakly modular with respect to a set of matrices Γ is weakly modular with respect to the group generated by Γ. Lemma For all congruence ( subgroups ) Γ SL 2 (Z) there exists a number h h Z so that the matrix m h = Γ. 0 Proof. Every congruence subgroup by definition contains ( some) principal congruence N subgroup Γ(N), which contains elements of the form. 0 Notation Throughout this section we shall simply use h = h(γ) to be the minimal positive h from the previous lemma. Proposition For f a weakly modular function of weight k with respect to Γ, f is hz periodic and there exists a function g : D {0} C so that f(τ) = g(e 2πiτ/h ). Proof. The proof is the same as Definition A weakly modular function f is holomorphic at if g has a removable singularity at q = 0, or equivalently f has a Fourier expansion of the form f(τ) = n=0 a ne 2πiτ/h. Definition Let Γ SL 2 (Z) be a congruence subgroup. A function f : H C is a modular form of weight k with respect to Γ if f is holomorphic, f is weakly modular of weight k with respect to Γ and f[α] k is holomorphic at for all α SL 2 (Z). Definition With Γ, f as above, we say f is a cusp form of weight k with respect to Γ if f is a modular form such that a 0 = 0 in the Fourier expansion of f[α] k for all α SL 2 (Z). Notation We shall denote the space of modular forms of weight k with respect to Γ by M k (Γ) and the space of cusp forms by S k (Γ). Then, the spaces of all modular forms is M(Γ) = k M k(γ) and the space of cusp forms is S(Γ) = k S k(γ). 5. The space M 2 (Γ 0 (4)) In this section we describe the space of modular forms of weight 2 on the principal congruence subgroup Γ 0 (4). This section will be devoted to showing that two explicit independent forms exist. However, it is a somewhat deeper result, involving Riemann surface theory, that these are the full space of modular forms, and so we shall omit the proof. The reason for studying this space is that one of its elements turns out to count the number of ways to write a positive integer as a sum of four squares, as shall be discussed in the next section.

16 6 AARON LANDESMAN 5.. Computing the Fourier Series of G 2 (τ). We shall first develop some standard trigonometric identities in order to go between the conditionally convergent Eisenstein series G 2 (τ) and its Fourier series. Definition 5... The Divisor Function is denoted σ(n) = d n,d>0 d. Lemma We have the following identities for π cot πz: () π cot πz = z + ( ) d=. τ d + τ+d (2) π cot πz = πi 2πi m=0 e2πizm where the first sum converges absolutely and the second sum converges conditionally. Proof. (Sketch) The first identity a standard complex analysis argument, so I will cos πz only give a sketch of the proof. It is easy to see that π cot πz = π sin πz has poles exactly where it s denominator has zeros. That is, its poles are exactly at the integer points, and we can see that its residues are all. The residues are easy to compute because by periodicity they are all the same, and at the origin the residue can be calculated by computing d z cos πz dz π sin πz πz sin πz z=0 = sin 2 πz z=0 =. This tells us that the principal parts at the poles of both sides of π cot πz = z + ( ) d= τ d + τ+d agree. Then, consider the function f(z) = π cot πz ( z + d= ( τ d + ) ). τ + d We wish to show it is 0. We would like to show it is bounded on the strip 0 Re(z) <, because it will then follow that it is bounded everywhere. This boundedness in the limit Im(z) + can be established by writing π cot πz = cos πz π sin πz = iπ e iπz +e iπz e iπz e, and noting that both numerator and the denominator approach e πz as z i. Therefore, both π cot πz and f(z) π cot πz are bounded on iπz the strip 0 Re(z), and their difference, f(z) is bounded. This tells us f(z) is constant by Liouville s Theorem, and checking the value of f(z) at any single point tells us it is 0. The easiest point to check is the limit as z i. The second identity is much more straightforward. As before we write cos πz π cot πz = π sin πz = iπ e iπz + e iπz e iπz e iπz = iπe2πiz + e 2πiz = πi 2πi e 2πizm, by expanding this as a geometric series. Lemma Let Z c = Then, for all τ H, the function { Z if c 0 Z {0} if c = 0. G 2 (τ) = c Z converges conditionally and satisfies d Z c G 2 (τ) = 2ζ(2) 8π 2 (cτ + d) 2 n= σ(n)e 2πiτn. m=0

17 MODULAR FORMS AND THE FOUR SQUARES THEOREM 7 Proof. Using the above lemma 5..2 we have z + ( z d + ) = πi 2πi z + d d= e 2πizm, and so taking the derivative of both sides (as is justified by uniform convergence) yields the identity (z + d) 2 = 4π2 me 2πimz, d= where the doubly infinite sum is interpreted as summing in the order of increasing absolute value. Therefore, (cτ + d) 2 = (0τ + d) 2 + (cτ + d) 2 + (cτ + d) 2 c Z d Z c d Z c<0 0 d Z c c>0 d Z c = d (cτ + d) 2 d 0 c>0 d Z = 2 π π 2 c>0 = π π2 c= m=0 m=0 me 2πimcτ m=0 me 2πimcτ Now, note that c= m=0 me2πimcτ is absolutely convergent because τ H so, e 2πimcτ <. Therefore, for a fixed c, we can bound c m=0 m e2πimcτ = c ( e 2πiτc ), whose tail is bounded above by 2 2 c e 4πiτc 2 = e, which 4πiτc surely converges as e 4πiτc <. This tells us that based on the initial order of summation the sum is absolutely convergent, although this does not mean our original sum was absolutely convergent. However, since this sum is absolutely convergent, we can rearrange its terms to obtain m=0 Therefore, this tells us c Z d Z c me 2πimcτ = c= m=0 (cτ + d) 2 = π π2 and the sum is conditionally convergent. n= d>0,d n de 2πiτn. n= d>0,d n de 2πiτn, 5.2. The almost-invariance of G 2. The next step on the way to producing elements of M 2 (Γ 0 (4)) is to create a function which is invariant. It turns out G 2 isn t quite invariant, and we shall see how invariance fails in this subsection. Not to worry though, with a few tricks, actual invariant functions can be found. ( [( )] ) Lemma G 2 (τ) = G 0 2 (τ). 2

18 8 AARON LANDESMAN Proof. Using the previous lemma 5..3, we have the explicit Fourier expansion G 2 (τ) = π π2 de 2πiτn, and so G 2 (τ +) = π2 3 +8π2 Lemma Proof. First, note that n= d>0,d n n= d>0,d n de 2πi(τ+)n = π2 3 +8π2 ( [( )] ) 0 G 2 (τ) = G 0 2 (τ) 2πi τ. 2 n= d>0,d n ( [( )] ) 0 G 2 (τ) = τ 2 G 0 2 ( 2 τ ) = τ 2 (c c Z d Z c τ + d)2 = ( c + dτ) 2 c Z d Z c (cτ d) 2 = d Z c Z d = (cτ + d) 2 d Z c Z d = d 2 + d 0 d Z c 0 = 2 π2 6 + d Z c 0 (cτ + d) 2 (cτ + d) 2 Next, observe that we have a telescoping series (cτ + d)(cτ + d + ) = cτ + d cτ + d + = 0. d Z d Z Hence, we can add this to G 2 to obtain G 2 (τ) = π2 3 + (cτ + d) c 0 d Z = π2 3 + (cτ + d) 2 (cτ + d)(cτ + d + ) c 0 d Z c 0 d Z = π2 3 + (cτ + d) 2 (cτ + d)(cτ + d + ) c 0 d Z = π2 3 + (cτ + d) 2 (cτ + d + ) c 0 d Z de 2πiτn = G 2 (τ).

19 MODULAR FORMS AND THE FOUR SQUARES THEOREM 9 Now, we may observe that this double sum is on the order of d,c (cτ+d), and 3 therefore by the same argument as in 3..5 this converges absolutely. We may therefore rearrange terms in the series, to obtain G 2 (τ) = π2 3 + d Z c 0 = π2 = π2 3 + d Z c 0 (cτ + d) 2 (cτ + d + ) (cτ + d) 2 (cτ + d)(cτ + d + ) 3 + (cτ + d) 2 d Z c 0 c 0 = τ 2 G 2 ( τ ) d Z c 0 Hence, to prove the claim, it suffices to show Or equivalently, lim d Z c 0 N N d= N c 0 d Z (cτ + d)(cτ + d + ) (cτ + d)(cτ + d + ) = 2πi τ (cτ + d)(cτ + d + ) (cτ + d)(cτ + d + ) = 2πi τ Note that for N fixed, this sum converges absolutely. So, reversing the orders of the summations gives a telescoping sum, which works out to N d= N c 0 (cτ + d)(cτ + d + ) = c 0 = c 0 N d= N N d= N (cτ + d)(cτ + d + ) cτ + d cτ + d + = cτ N + cτ + N c 0 c 0 The next step is to employ our identities for cotangent from Recall Therefore, π cot πz = z + cτ N + cτ + N = c 0 c 0 c 0 d= ( τ d + ). τ + d τ N τ c + τ N τ + c = τ 2π cot π N τ 2 N.

20 20 AARON LANDESMAN Finally, lim N N d= N c 0 (cτ + d)(cτ + d + ) = lim N 2π τ cot π N τ 2 τ N = lim N τ 2π cot π N τ = τ + 2π lim τ e 2πi N τ = 2πi τ, as we wanted to show. ( ) a b Lemma (G 2 [γ] 2 )(τ) = G 2 (τ) 2πic cτ+d with γ = SL c d 2 (Z). ( ) ( ) 0 Proof. By we have that, generate SL (Z) action by Linear Fractional Transformations. We have just seen above that this lemma holds for the special case of the two generators. Therefore, to prove this lemma, it suffices to show that if this lemma is satisfied for γ, η, it is also ( satisfied ) for γ( and) γ η. a b e f First, let us show that it holds for γ η. Write γ =, η =. Using c d g h the general fact prove in that f[γ] 2 [η] 2 = f[γ η] 2, we would like to show that f[γ η] 2 (τ) = f(τ) 2πi(ce+dg) (ce+dg)τ+cf+dh. Indeed, since f[γ] 2[η] 2 = f[γ η] 2, we have N ie2πi N f[γ] 2 [η] 2 (τ) = (f(τ) 2πic cτ + d )[η] 2 = f(τ) 2πig 2πic (gτ + h) 2 gτ + h c eτ+f gτ+h + d = f(τ) 2πic 2πig(c(eτ + f) + d(gτ + h)). (gτ + h)(c(eτ + f) + d(gτ + h)) Therefore, to show multiplicativity holds, we need to show or equivalently, 2πic + 2πig(c(eτ + f) + d(gτ + h)) (gτ + h)(c(eτ + f) + d(gτ + h)) = 2πic + 2πig(c(eτ + f) + d(gτ + h)) gτ + h Clearing the denominator, we need to show only 2πi(ce + dg) (ce + dg)τ + cf + dh, = 2πi(ce + dg) 2πic + 2πig(c(eτ + f) + d(gτ + h)) = 2πi(ce + dg)(gτ + h) After expanding and cancelling matching terms, we obtain that it suffices to show 2πi(c + gcf) = 2πiceh, which holds because η SL 2 (Z) and so + gf = eh, as its determinant is. To complete the proof of the lemma, it suffices to show this law is closed under inversion. To do this, it suffices to check that the inverses of the generators S, T both satisfy this law, because any element of SL 2 (Z) can be written as products of

21 MODULAR FORMS AND THE FOUR SQUARES THEOREM 2 S, T, and their inverses. This is easy to see, since S is actually self inverse and the same argument as given in Lemma 5.2. shows T satisfies this law An Almost Invariant SL 2 (Z) function from G 2.. ( ) a b Lemma For γ = SL c d 2 (Z), The function G 2 (τ) π I(τ) satisfies ( G 2 (τ) π ) ( [γ] 2 = G 2 (τ) π ) I(τ) I(τ) Proof. We showed in the previous lemma 5.2.3, that (G 2 [γ] 2 )(τ) = G 2 (τ) 2πic So, to complete this proof, it suffices to show π I(τ) [γ] 2 = π I(τ) 2πic cτ+d To see this, observe (cτ + d) 2 π π Im( aτ+b = (cτ + d) 2 cτ+d ) Im( (aτ+b)(cτ+d) (cτ+d)(cτ+d) ) = (cτ + d) 2 π I(aτd+bcτ) (cτ+d)(cτ+d) = (cτ + d) 2 π I(τ)(ad bc) (cτ+d)(cτ+d) = (cτ + d) 2 π I(τ) (cτ+d)(cτ+d) π(cτ + d) = (cτ + d)i(τ) So, it suffices to show that Note that So, we need to show Or equivalently, π(cτ + d) (cτ + d)i(τ) = π I(τ) 2πic cτ + d π I(τ) 2πic (cτ + d)π 2πicI(τ) = cτ + d I(τ)(cτ + d) π(cτ + d) (cτ + d)π 2πicI(τ) = (cτ + d)i(τ) I(τ)(cτ + d) π(cτ + d) = (cτ + d)π 2πicI(τ) cτ+d. Subtracting (cτ +d)π from both sides, we obtain that it suffices to show πc(τ τ) = 2πiI(τ), which is of course true, completing the proof A Bounding Theorem. We have above produced an SL 2 (Z) invariant function, but it is not holomorphic. We will later show that G 2,N (τ) = G 2 (τ) NG 2 (Nτ) satisfies G 2,N M 2 (Γ 0 (N)) are holomorphic on the upper half plane, which will deal with the issue of holomorphicity away from. To deal with the problem of holomorphicity, we will first need a crucial theorem, stating that if we can asymptotically bound the Fourier coefficients of a function by a polynomial, then it is holomorphic at infinity.

22 22 AARON LANDESMAN Lemma Let f : H C be holomorphic on H, satisfying f(z + ) = f(z). If there exist positive constants C, r, such that the Fourier expansion of f satisfies f(τ) = n=0 a ne 2πinτ/N, with a n < Cn r for all n, then ( ) f(τ) C 0 + C t r e 2πty/N dt + C t=0 y r. Proof. Write τ = x + iy. Note that f(τ) = a n e 2πiτ/N a n e 2πiτ/N < Cn r e 2πny/N n=0 n=0 n=0 Defining the function g(t) = e 2πty/N, we may observe that g (t) > 0, t (0, rn 2πy, g (t) < 0, t > rn 2πy. We can see this because g (t) = t r e ( 2πty/N r + t 2πy ) N. Let k = rn 2πy. Since g (t) is increasing on (0, k) it follows that k k n r e 2πny/N = g(n) < n= n= k t=0 g(t)dt, and additionally since g(t) is decreasing on (k +, ), it follows that n r e 2πny/N = g(n) < g(t)dt. n= n=k+ Combining these two facts yields that n=k+ k n r e 2πny/N = (k ) r e 2π(k )y/n + k r e 2πky/N + n r e 2πny/N + t=k n= k < (k ) r e 2π(k )y/n + k r e 2πky/N + = (k ) r e 2π(k )y/n + k r e 2πky/N + = (k ) r e 2π(k )y/n + k r e 2πky/N + t=0 k t=0 t=0 g(t)dt + g(t)dt + g(t)dt n=k+ t=k t=k g(t)dt g(t)dt n r e 2πny/N Since k > rn 2πy, and e2πky/n e rn2πy N2πy = e r, which is a constant. So, choosing C 0 large enough to encompass this factor of e r, it follows that ) ) ( f(τ) a 0 + C g(t)dt + O(y r ) t=0 ( < C 0 + C t r e 2πty/N dt t=0 + C y r. Theorem Let Γ be a principal congruence subgroup of level N and f : H C such that f is holomorphic on H, is meromorphic at, and f satisfies f[γ] k = f for all γ Γ. If there exist positive constants C, r such that the Fourier expansion of f satisfies f(τ) = n=0 a ne 2πinτ/N, with a n < Cn r for all n, then f M k (Γ).

23 MODULAR FORMS AND THE FOUR SQUARES THEOREM 23 Proof. To prove this, by the definition of modular form 4.2.0, we only have to show that f[α] k is holomorphic at infinity for all α SL 2 (Z), since we are assuming it is holomorphic elsewhere and translation invariant. We are in the situation of 5.4. and so we may apply the result that ( ) f(τ) C 0 + C t r e 2πty/N dt + C t=0 y r. Next, since f is holomorphic, and fractional linear transformations are holomorphic, it follows that for α SL 2 (Z), f[α] k (τ) = (cτ + d) k f(α(τ)) is holomorphic as well. Furthermore, it is weight k invariant after composing with any element of αγα because f was Γ invariant. Therefore, we have a Laurent series expansion f[α] k (τ) = n Z b ne 2πiτ/N. To complete the proof, we need to show this is holomorphic at infinity, or equivalently the b n = 0, n < 0. This is equivalent to the condition that lim qn 0(f[α] k )(τ) e 2πiτ/N = 0, because this precisely means there are no negative terms in the Laurent series expansion, as if there were this limit would be infinite. First, let us note that using the previous lemma 5.4. lim (f[α] k)(τ) = (cτ + d) k f(α(τ)) q N 0 = O(τ k )O(I(α(τ) r ) = O(y k )O(y r ) = O(y r k ). This implies that for D a constant, lim (f[α] k)(τ) e 2πiτ/N D y r k e 2πiτ/N = D y r k e 2πy/N, q N 0 which clearly goes to 0 as y, since the exponential term dominates the polynomial term. This means the Laurent series is actually a Fourier series, and so f is indeed holomorphic at infinity The elements of M 2 (Γ 0 (4)). We are finally ready to produce two independent elements of M 2 (Γ 0 (4)). Explicitly, these elements are G 2 (τ) 2G 2 (2τ), G 2 (τ) 4G 2 (4τ), as we shall soon show. Lemma The function G 2,N (τ) = G 2 (τ) NG 2 (Nτ) satisfies G 2,N M 2 (Γ 0 (N)) for all N > 0. ( ) ( ) a b a Nb Proof. Next, note that for γ Γ 0 (N), writing γ =, η = Nc d c d SL 2 (Z) we can write Nγ(τ) = anτ+bn cnτ+d = η(nτ). This tells us that for γ Γ 0(N),