Graduate Texts in Mathematics 41Editorial BoardJ.H. Ewing F.W. Gehring P.R. HalmosGraduate Texts in Mathematics1 TAKEUTIIZARING. Introduction to Axiomatic Set Theory. 2nd ed.2 OXTOBY. Measure and Category. 2nd ed.3 SCHAEFFER. Topological Vector Spaces.4 HILTON/STAMMBACH. A Course in Homological Algebra.5 MAC LANE. Categories for the Working Mathematician.6 HUGHES/PIPER. Projective Planes.7 SERRE. A Course in Arithmetic.8 TAKEUTIlZARING. Axiomatic Set Theory.9 HUMPHREYS. Introduction to Lie Algebras and Representation Theory.10 COHEN. A Course in Simple Homotopy Theory.11 CONWAY. Functions of One Complex Variable. 2nd ed.12 BEALS. Advanced Mathematical Analysis.13 ANDERSON/FuLLER. Rings and Categories of Modules.14 GOLUBITSKy/GUILLEMIN. Stable Mappings and Their Singularities.15 BERBERIAN. Lectures in Functional Analysis and Operator Theory.16 WINTER. The Structure of Fields.17 ROSENBLATT. Random Processes. 2nd ed.18 HALMOS. Measure Theory.19 HALMOS. A Hilbert Space Problem Book. 2nd ed., revised.20 HUSEMOLLER. Fibre Bundles. 2nd ed.21 HUMPHREYS. Linear Algebraic Groups.22 BARNES/MACK. An Algebraic Introduction to Mathematical Logic.23 GREUB. Linear Algebra. 4th ed.24 HOLMES. Geometric Functional Analysis and its Applications.25 HEWITT/STROMBERG. Real and Abstract Analysis.26 MANES. Algebraic Theories.27 KELLEY. General Topology.28 ZARISKIISAMUEL. Commutative Algebra. Vol. I.29 ZARISKIISAMUEL. Commutative Algebra. Vol. II.30 JACOBSON. Lectures in Abstract Algebra I: Basic Concepts.31 JACOBSON. Lectures in Abstract Algebra II: Linear Algebra.32 JACOBSON. Lectures in Abstract Algebra III: Theory of Fields and Galois Theory.33 HIRSCH. Differential Topology.34 SPITZER. Principles of Random Walk. 2nd ed.35 WERMER. Banach Algebras and Several Complex Variables. 2nd ed.36 KELLEy/NAMIOKA et al. Linear Topological Spaces.37 MONK. Mathematical Logic.38 GRAUERT/FRITZSCHE. Several Complex Variables.39 ARVESON. An Invitation to C*-Algebras.40 KEMENy/SNELL/KNAPP. Denumerable Markov Chains. 2nd ed.41 ApoSTOL. Modular Functions and Dirichlet Series in Number Theory. 2nd ed.42 SERRE. Linear Representations of Finite Groups.43 GILLMAN/JERISON. Rings of Continuous Functions.44 KENDIG. Elementary Algebraic Geometry.45 LOEVE. Probability Theory I. 4th ed.46 LOEVE. Probability Theory II. 4th ed.47 MOISE. Geometric Topology in Dimensions 2 and 3.continued after IndetTom M. ApostolModular Functionsand Dirichlet Seriesin Number TheorySecond EditionWith 25 IllustrationsSpringer-VerlagNew York Berlin HeidelbergLondon Paris Tokyo Hong KongTom M. ApostolDepartment of MathematicsCalifornia Institute of TechnologyPasadena, CA 91125USAEditorial BoardJ.H. EwingDepartment ofMathematicsIndiana UniversityBloomington, IN 47405USAF.W. GehringDepartment ofMathematicsUniversity of MichiganAnn Arbor, MI 48109USAP.R. HalmosDepartment ofMathematicsSanta Clara UniversitySanta Clara, CA 95053USAAMS Subject ClassificationsIOA20, IOA45, 10045, IOH05, 10HIO, IOJ20, 30AI6Library of Congress Cataloging-in-Publication DataApostol, Tom M.Modular functions and Dirichlet series in number theory/Tom M. Apostol.-2nd ed.p. cm.-(Graduate texts in mathematics; 41)Includes bibliographical references.ISBN 0-387-97127-0 (alk. paper)I. Number theory. 2. Functions, Elliptic. 3. Functions, Modular. 4. Series,Dirichlet. I. Title. II. Series.QA241.A62 1990512'.7-dc20 89-21760Printed on acid-free paper. 1976, 1990 Springer-Verlag New York, Inc.All rights reserved. This work may not be translated or copied in whole or in part without thewritten permission of the publisher (Springer-Verlag New York, Inc., 17,:, Fifth Avenue, NewYork, NY 10010, USA), except for brief excerpts in connection with reviews or scholarlyanalysis. Use in connection with any form of information and retrieval, electronic adaptation,computer software, or by sirrlilar or dissimilar methodology now known or hereafter developedis forbidden.The use of general descriptive names, trade names, trademarks. etc., in this publication, even ifthe former are not especially identified, is not to be taken as a sign that such names, asunderstood by the Trade Marks and Merchandise Marks Act, may accordingly be used freelyby anyone.Typeset by Asco Trade Typesetting Ltd., Hong Kong.Printed and bound by Edwards Brothers, Inc., Ann Arbor, Michigan.Printed in the United States of America.9 8 7 6 5 432 IISBN 0-387-97 J27-0 Springer-Verlag New York Berlin HeidelbergISBN 3-540-97127-0 Springer-Verlag Berlin Heidelberg New YorkPrefaceThis is the second volume of a 2-volume textbook* which evolved from acourse (Mathematics 160) offered at the California Institute of Technologyduring the last 25 years.The second volume presupposes a background in number theory com-parable to that provided in the first volume, together with a knowledge ofthe basic concepts of complex analysis.Most of the present volume is devoted to elliptic functions and modularfunctions with some of their number-theoretic applications. Among themajor topics treated are Rademacher's convergent series for the partitionfunction, Lehner's congruences for the Fourier coefficients of the modularfunctionj(r), and Heeke's theory of entire forms with multiplicative Fouriercoefficients. The last chapter gives an account of Bohr's theory of equivalenceof general Dirichlet series.Both volumes of this work emphasize classical aspects of a subject whichin recent years has undergone a great deal of modern development. It ishoped that these volumes will help the nonspecialist become acquaintedwith an important and fascinating part of mathematics and, at the sametime, will provide some of the background that belongs to the repertory ofevery specialist in the field.This volume, like the first, is dedicated to the students who have takenthis course and have gone on to make notable contributions to numbertheory and other parts of mathematics.T.M.A.January, 1976* The first volume is in the Springer-Verlag series Undergraduate Texts in Mathematics underthe title Introduction to Analytic Number Theory.vPreface to the Second EditionThe major change is an alternate treatment of the transformation formula forthe Dedekind eta function, which appears in a five-page supplement to Chap-ter 3, inserted at the end of the book (just before the Bibliography). Other-wise, the second edition is almost identical to the first. Misprints have beenrepaired, there are minor changes in the Exercises, and the Bibliography hasbeen updated.T.M. A.July, 1989Chapter 1Elliptic functions1.1 Introduction1.2 Doubly periodic functions1.3 Fundamental pairs of periods1.4 Elliptic functions1.5 Construction of elliptic functions1.6 The Weierstrass g;J function1.7 The Laurent expansion of g;J near the origin1.8 Differential equation satisfied by g;J1.9 The Eisenstein series and the invariants 9z and 931.10 The numbers eb ez, e31.11 The discriminant .11.12 Klein's modular function J(T)1.13 Invariance of J under unimodular transformations1.14 The Fourier expansions of 9z(T) and 93(T)1.15 The Fourier expansions of .1(T) and J(1:)Exercises for Chapter 1Chapter 2The Modular group and modular functions2.1 Mobius transformations2.2 The modular group r2.3 Fundamental regions2.4 Modular functionsContents1124691111121314151618202326283034vii2.5 Special values of J 392.6 Modular functions as rational functions of J 402.7 Mapping properties of J 402.8 Application to the inversion problem for Eisenstein series 422.9 Application to Picard's theorem 43Exercises for Chapter 2 44Chapter 3The Dedekind eta function3.1 Introduction3.2 Siegel's proof of Theorem 3.13.3 Infinite product representation for .1(r)3.4 The general functional equation for YJ(r)3.5 Iseki's transformation formula3.6 Deduction of Dedekind's functional equation from Iseki'sformula3.7 Properties of Dedekind sums3.8 The reciprocity law for Dedekind sums3.9 Congruence properties of Dedekind sums3.10 The Eisenstein series G2(r)Exercises for Chapter 3Chapter 4Congruences for the coefficients of the modular function j4.1 Introduction4.2 The subgroup r o(q)4.3 Fundamental region of r o(p)4.4 Functions automorphic under the subgroup r o(P)4.5 Construction of functions belonging to r o(P)4.6 The behavior of fp under the generators of r4.7 The function R (z - OJ)converges absolutely and uniformly in the disk Iz I R.PROOF. We will show that there is a constant M (depending on R and such that, if '1, we have1 M--- R and all z with Iz I R. Then we invoke Lemma 1 toprove Lemma 2. Inequality (1) is equivalent toIz: WI" To exhibit M we consider all OJ in n with IOJ I > R. Choose one whosemodulus is minimal, say IOJ I = R + d, where d > O. Then if Iz I s RandIOJ I R + d we haveIz: Wi = 11- 1 1 - R d'81.6: The Weierstrass tJ functionand hencewhere(R )-a.M= l-R +d .This proves (2) and also the lemma. DAs mentioned earlier, we could try to construct the simplest ellipticfunction by using a series of the form1L (Z-W)2'WEnThis has the appropriate principal part near each period. However, theseries does not converge absolutely so we use, instead, a series with theexponent 2 replaced by 3. This will give us an elliptic function of order 3.Theorem 1.9. Let f be defined by the series1f(z) = L ( )3'WEn z - OJThenfis an ellipticfunction with periods OJ1, OJ2 and with a pole oforder 3ateach period OJ in Q.PROOF. By Lemma 2 the series obtained by summing over IOJ I > R convergesuniformly in the disk Iz I s R. Therefore it represents an analytic functionin this disk. The remaining terms, which are finite in number, are alsoanalytic in this disk except for a 3rd order pole at each period OJ in the disk.This proves thatfis meromorphic with a pole of order 3 at each OJ in Q.Next we show thatfhas periods OJ1 and OJ2 For this we take advantageof the absolute convergence of the series. We have1f(z + wd = L ( )3'WEn z + OJ1 - OJBut OJ - OJ1 runs through all periods in Q with OJ, so the series for f(z + OJ1)is merely a rearrangement of the series for f(z). By absolute convergence wehave f(z + OJ1) = f(z). Similarly, f(z + OJ2) = f(z) so f is doubly periodic.This completes the proof. 01.6 The Weierstrass 80 functionNow we use the function of Theorem 1.9 to construct an elliptic functionor order 2. We simply integrate the series for f(z) term by term. This gives usa principal part - (z - OJ) - 2/2 near each period, so we multiply by - 2 to91: Elliptic functionsget the principal part (z - w) - 2. There is also a constant of integration toreckon with. It is convenient to integrate from the origin, so we remove theterm z - 3 corresponding to w = 0, then integrate, and add the term z - 2.This leads us to the function1 fZ -22 + L ( )3 dt.z 0 1)*0 t - wIntegrating term by term we arrive at the following function, called theWeierstrass g;J function.Definition. The Weierstrass g;J function is defined by the series1 {II}.f.J(z) = 2 + L ( )2 - 2 .z w*o Z - W WTheorem 1.10. The function g;J so defined has periods Wt and w2 It is analyticexcept for a double pole at each period w in Q. Moreover g;J(z) is an evenfunction of z.PROOF. Each term in the series has modulus11 lIIW2-(Z-W)21IZ(2W-Z)1(z - W)2 - w2= w2(z - W)2 = w2(z - W)2 .Now consider any compact disk Iz I ::; R. There are only a finite number ofperiods W in this disk. If we exclude the terms of the series containing theseperiods we have, by inequality (1) obtained in the proof of Lemma 2,where M is a constant depending only on R. This gives us the estimateIz(2w - z) I MR(2Iwl + R) MR(2 + R/lwl) 3MR< < 0 anda( - m) =1= 0 we say that f has a pole of order m at ioo. If m ~ 0 we say f isanalytic at ioo. Condition (c) states that f has at worst a pole of order m atioo.The function J is a modular function. It is analytic in H with a first orderpole at ioo. Later we show that every modular function can be expressed asa rational function of J. The proof of this depends on the following propertyof modular functions.Theorem 2.4. Iff is modular and not identically zero, then in the closure of thefundamental region Rr , the number of zeros off is equal to the number ofpoles.Note. This theorem is valid only with suitable conventions at the boundarypoints of R r . First of all, we consider the boundary of R r as the union offour edges intersecting at four vertices p, i, P + 1, and ioo, where p = e 21ti/3(see Figure 2.3). The edges occur in equivalent pairs (1), (4) and (2), (3).Iffhas a zero or pole at a point on an edge, then it also has a zero or poleat the equivalent point on the equivalent edge. Only the point on the leftmostedge (1) or (2) is to be counted as belonging to the closure of Rr.The order of the zero or pole at the vertex p is to be divided by 3; the orderat i is to be divided by 2; the order at ioo is the order of the zero or pole atx = 0, measured in the variable x = e 21tit342.4: Modular functionsooi- - ~ - -(1)p./"//(2) (3)Figure 2.3......",(4)p + 1PROOF. Assume first that j has no zeros or poles on the finite part of theboundary of Rr . Cut Rr by a horizontal line, Im(r) = M, where M > 0 istaken so large that all the zeros or poles ofj are inside the truncated regionwhich we call R. [Ifjhad an infinite number of poles in Rr they would havean accumulation point at ioo, contradicting condition (c). Similarly, since jis not identically zero, j cannot have an infinite number of zeros in Rr .]Let oR denote the boundary of the truncated region R. (See Figure 2.4.)Let Nand P denote the number of zeros and poles ofjinside R. Then1 i f'(r) 1 {f f f f f}N - P = - - dr = - + + + +2ni oR f(r) 2ni (1) (2) (3) (4) (5)where the path is split into five parts as indicated in Figure 2.5. The integralsalong (1) and (4) cancel because of periodicity. They also cancel along (2)and (3) because (2) gets mapped onto (3) with a reversal of direction under-} + iM ,..------------, ! + iMpRFigure 2.4tp + 1352: The modular group and modular functions(5)(2)~(3)~Figure 2.5the mapping u = S(r) = -l/r, or r = S-l U = S(u). The integrand remainsunchanged because f[S(u)] = f(u) implies f'[S(u)]S'(u) = f'(u) sof'(r) dr = f'[S(u)] S'(u) du = f'(u) duof(r) f[S(u)] f(u)Thus we are left withN - P = _1 f f'(r) dr.2ni (5) f(r)We transform this integral to the x-plane, x = e21tit As r varies on thehorizontal segment r = u + iM, -t ~ u ~ t, we haveso x varies once around a circle K of radius e- 21tM about x = 0 in the negativedirection. The points above this segment are mapped inside K, so f has nozeros or poles inside K, except possibly at x = O. The Fourier expansiongives us( a-m (f r) = --;n + ... = F x),xsay, withHencef'(r) = F'(x) ~ : ' f'(r) dr = F'(x) dx.f(r) F(x)N - P = ~ f f'(r) dr = - ~ , [ F'(x) dx = -(NF - PF) = PF - NF ,2nl (5) f(r) 2nl JK F(x)where NF and PF are the number of zeros and poles of F inside K.362.4: Modular functionsIf there is a pole of order m at x = then m > 0, NF = 0, PF = m soPF - N F = m, andN = P + m.Thereforejtakes on the value in Rr as often as it takes the value 00.If there is a zero of order n at x = 0, then m = - n so PF = 0, N F = n,henceN + n = P.Again,jtakes the value in Rr as often as it takes the value 00. This provesthe theorem ifJhas no zeros or poles on the finite part of the boundary of Rr .Ifjhas a zero or a pole on an edge but not at a vertex, we introduce detoursin the path of integration so as to include the zero or pole in the interior of R,as indicated in Figure 2.6. The integrals along equivalent edges cancel asbefore. Only one member of each pair of new zeros or poles lies inside the newregion and the proof goes through as before, since by our convention onlyone of the equivalent points (zero or pole) is considered as belonging to theclosure of Rr .1Figure 2.6If j has a zero or pole at a vertex p or i we further modify the path ofintegration with new detours as indicated in Figure 2.7. Arguing as above wefind1 {(II) 1 J,-1 /2+iM}j"(7:)N - P = - + + + - d7:2ni Ct C3 C2 1/2 + iM j'(7:)= _1 {(I + 1)+ 1}J'(7:) d7: + m2ni Ct C3 C2 j(7:) ,where x-mis the lowest power of x occurring in the Laurent expansion nearx = 0, x = e21tit.372: The modular group and modular functions-1 + iM ,..------------, t + iM1'.p + 1Figure 2.7Near the vertex p we writef(r) = (r - p)kg(r), where g(p) # o.The exponent k is positive iffhas a zero at p, and negative iffhas a pole at p.On the path C1 we write r - p = rei8where r is fixed and rx ~ e~ nl2where rx depends on r. Thenf'(r) = _k_ + g'(r)f(r) r - p g(r)and1 f f'(r) 1 fa (k g'(p + reiO)) iO'- --dr = - -. + . re 1 de2ni Ct f(r) 2ni 1t/2 re'Og(p + re'O)_ -krx' r fa g'(p + rei8) i6 , _ ~ _- -- + - '0 e dO, where ex - 2 ex.2n 2n 1t/2 g(p + ref )As r -+ 0, the last term tends to 0 since the integrand is bounded. Also,rx' -+ n/3 as r -+ 0 so. 1 f f'(r) khm- --dr = --.r ~ O 2ni Ct f(r) 6Similarly,lim _1 f f'(r) dr = - ~r ~ O 2ni C3 f(r) 6382.5: Special values of Jsolim-1(J. + J. )f'(r) dr = - ~r-+O 2ni Ct C3 f( r) 3Similarly, near the vertex i we writef(r) = (r - i)'h(r), where h(i) # 0and we find, in the same way,lim _1 J. f'(r) dr = - i.r-'O 2rci C2 f( r) 2Therefore we get the formulak IN - P = m -"3 - 2Iffhas a pole at x = 0, and zeros at p and i, then m, k and I are positive andwe havek IN +"3 + 2= P + m.The left member counts the number of zeros offin the closure of Rr (with theconventions agreed on at the vertices) and the right member counts thenumber of poles. Iff has a zero of order n at x = 0 then m = - n and theequation becomesk IN + n + 3+ 2= P.Similarly, if f has a pole at p or at i the corresponding term k/3 or 1/2 isnegative and gets counted along with P. This completes the proof. DTheorem 2.5. If f is modular and not constant, then for every complex c thefunctionf - c has the same number of zeros as poles in the closure of Rr.In other words,ftakes on every value equally often in the closure of Rr.PROOF. Apply the previous theorem tof - c.Theorem 2.6. Iff is modular and bounded in H then f is constant.PROOF. Sincefis bounded it omits a value sofis constant.DD2.5 Special values of JTheorem 2.7. The function J takes every value exactly once in the closure ofRr. In particular, at the vertices we haveJ(p) = 0, J(i) = 1, J(ioo) = 00.There is afirst order pole at ioo, a triple zero at p, and J(r) - 1has a doublezero at r = i.392: The modular group and modular functionsPROOF. First we verify that 92(P) = 0 and 93(i) = o. Since p3= 1 andp2 + P + 1 = 0 we have~ (p) _ " ,,1 _ ~ " 160 92 - tn (m + np)4 tn (mp3 + np)4 - p4 tn (mp2 + n)41 1 1 1 1= PJ,:n (n - m - mpt = PM ~ N (N + Mp)4 = 60p g2(P),so g2(P) = O. A similar argument shows that g3(i) = O. ThereforeJ( ) = 923(P) = 0 and J(.) = 923(i) = 1p ~ ( p ) 1 923(i) .The multiplicities are a consequence of Theorem 2.4.2.6 Modular functions as rational functionsof JDTheorem 2.8. Every rational function of J is a modular function. Conversely,every modular function can be expressed as a rational function of J.PROOF. The first part is clear. To prove the second, suppose f has zeros atZ l' ... , Zn and poles at PI' ... , Pn with the usual conventions about multi-plicities. Letg(7:) = Ii J(7:) - J(Zk)k= 1 J (r) - J (pk)where a factor 1 is inserted whenever Zk or Pk is 00. Then 9 has the same zerosand poles asfin the closure of Rr , each with proper multiplicity. Thereforef /9 has no zeros or poles and must be constant, so f is a rational functionofJ. D2.7 Mapping properties of JTheorem 2.7 shows that J takes every value exactly once in the closureof the fundamental region Rr . Figure 2.8 illustrates how Rr is mapped byJ onto the complex plane.The left half of Rr (the shaded portion of Figure 2.8a) is mapped onto theupper half-plane (shaded in Figure 2.8b) with the vertical part of the boundarymapping onto the real interval ( - 00, 0]. The circular part of the boundarymaps onto the interval [0, 1], and the portion of the imaginary axis v > 1,u = 0 maps onto the interval (1, +(0). Points in Rr symmetric about theimaginary axis map onto conjugate points in J(Rr). The mapping is con-formal except at the vertices r = i and r = p where angles are doubledand tripled, respectively.402.7: Mapping properties of JJ../P "-I"/ \I \I \\u-1 0 10= J(p) 1 = J(i)(a) (b)Figure 2.8These mapping properties can be demonstrated as follows. On theimaginary axis in Rr we have r = iv hence x = e 21tit= e- 21tv > 0, so theFourier seriesshows that. J(iv) is real. Since J(i) = 1 and J(iv) ~ +00 as v ~ +00 theportion of the imaginary axis 1 ~ v < +00 gets mapped onto the real axis1 ~ J(r) < +00.On the left boundary of Rr we have r = -1 + iv, hence x = e 21tit=e-21tve-1ti = _e-21tv< O. For large v (small x) we have J( -1 + iv) < 0 soJ maps the line u = -1 onto the negative real axis. Since J(p) = 0 andJ( (0) = 00, the left boundary of Rr is mapped onto the line - 00 < J(r) ~ O.As the boundary of Rr is traversed counterclockwise the points inside Rrlie on the left, hence the image points lie above the real axis in the imageplane.Finally, we show that J takes conjugate values at points symmetric aboutthe imaginary axis, that is,J(r) = J(-i).To see this, write r = u + iv. Thenx = e 21tit = e 21ti(u + iv) = e - 21tve 21tiuandThus rand - i correspond to conjugate points x and X, but the Fourierseries for J has real coefficients so J(r) and J( - i) are complex conjugates.412: The modular group and modular functionsIn particular, on the circular arc Ti = 1 we have - i = - liT, henceJ( - i) = J( -liT) = J(T) so J is real on this arc.2.8 Application to the inversion problem forEisenstein seriesIn the Weierstrass theory of elliptic functions the periods Wt, W2 determinethe invariants g2 and g3 according to the equations(4)A fundamental problem is to decide whether or not the invariants g2 and g3can take arbitrary prescribed values, subject only to the necessary conditiong2 3- 27g32=1= O. This is called the inversion problem for Eisenstein seriessince it amounts to solving the equations in (4) for Wt and W2in terms of g2and g3. The next theorem shows that the problem has a solution.Theorem 2.9. Given two complex numbers a2 and a3 such that a23- 27a32=1= O.Then there exist complex numbers Wt and W2 whose ratio is not real suchthatandPROOF. We consider three cases: (1) a2 = 0; (2) a3 = 0; (3) a2a3 =1= O.Case 1. If a2 = 0 then a3 =1= 0 since a23- 27a32 =1= O. Let Wt be anycomplex number such thatand let W2 = PWt, where p = e21ti/3. We know that g3(1, p) =1= 0 becauseg2(1, p) = 0 and ~ ( 1 , p) = g23- 27g32=1= O. Then1g2(Wt, w2) = g2(Wb wtp) = -4g2(1, p) = 0 = a2WtandCase 2. If a3 = 0 then a2 =1= 0 and we take W t to satisfy42(5)2.9: Application to Picard's theoremand let W2 = iwt . Then1g2(Wb w2) = g2(Wb iwt ) = -4g2(1, i) = a2Wtand1g3(Wb w2) = g3(Wb iwt ) = -6g3(1, i) = 0 = a3'WtCase 3. Assume a2 =1= 0 and a3 =1= O. Choose a complex r with 1m r > 0such thata23J(r) = 3 27 2'a2 - a3Note that J(r) =1= 0 since a2 =1= 0 and thatJ(r) - 1 27a32J(r) ~ .For this r choose Wt to satisfyand let W 2 = rwt . Theng2(Wt , w2) _ W t -4g2(1, r) _ 2 g2(1, r) _ a2-g-3(-W-t ,-w-2-) - Wt -6g3(1, r) - W t-g3-(1-,-r) - a3'so(6)But we also haveJ(r) - 1J(r)27g32(W t , w2)g23(Wb W2)27(a3/a2)2g22(Wb w2)g23(Wb W2)Comparing this with (5) we find that g2(Wt , W2) = a2 and hence by (6) wealso have g3(Wb W2) = a3' This completes the proof. D2.9 Application to Picard's theoremThe modular function J can be used to give a short proof of a famous theoremof Picard in complex analysis.Theorem 2.10. Every nonconstant entire function attains every complex valuewith at most one exception.Note. An example is the exponential function f(z) = eZwhich omitsonly the value O.432: The modular group and modular functionsPROOF. We assume f is an entire function which omits two values, say aand b, a =1= b, and show thatfis constant. Let( ) _ f(z) - ag Z - b .-aThen g is entire and omits the values 0 and 1.The upper half-plane H is covered by the images of the closure of thefundamental region Rr under transformations of r. Since J maps the closureof Rr onto the complex plane, J maps the half-plane H onto an infinite-sheeted Riemann surface with branch points over the points 0, 1 and 00(the images of the vertices p, i and 00, respectively). The inverse function J- 1maps the Riemann surface back onto the closure of the fundamental regionRr . Since J'(r) =1= 0 if r =1= p or r =1= i and since J'(p) == J'(i) == 0, each single-valued branch of J - 1 is locally analytic everywhere except at 0 == J(p),1 = J(i), and 00 = J( (0). For each single-valued branch of J-1the compositefunctionh(z) = J-1[g(z)]is a single-valued function element which is locally analytic at each finitez since g(z) is never 0 or 1. Therefore h is arbitrarily continuable in the entirefinite z-plane. By the monodromy theorem, the continuation of h exists as asingle-valued function analytic in the entire finite z-plane. Thus h is an entirefunction and so too isqJ(z) = eih(z).But 1m h(z) > 0 since h(z) E H soIqJ(z) I == e-Imh(z) < 1.Therefore qJ is a bounded entire function which, by Liouville's theorem,must be constant. But this implies h is constant and hence g is constant sinceg(z) = J[h(z)]. Thereforefis constant sincef(z) == a + (b - a)g(z). DExercises for Chapter 2In these exercises, r denotes the modular group, Sand T denote its gen-erators, S(r) = -l/r, T(r) = r + 1, and I denotes the identity element.1. Find all elements A of r which (a) commute with S; (b) commute with ST.2. Find the smallest integer n > 0 such that (ST)" = I.3. Determine the point r in the fundamental region Rr which is equivalent to(a) (8 + 60/(3 + 20; (b) (10i + 11)/(6i + 12).4. Determine all elements A of r which leave i fixed.5. Determine all elements A of r which leave p = e21ti/3fixed.44Exercises for Chapter 2QUADRATIC FORMS AND THE MODULAR GROUPThe following exercises relate quadratic forms and the modular group r. Weconsider quadratic forms Q(x, y) = ax2+ bxy + cy2 in x and y with realcoefficients a, b, c. The number d = 4ac - b2is called the discriminant ofQ(x, y).6. If x and yare subjected to a unimodular transformation, say(1) x = lXX' + Py', y = yx' + 8y', where (; ~ ) E r,prove that Q(x, y) gets transformed to a quadratic form Ql(X', y') having the samediscriminant. Two forms Q(x, y) and Ql(X', y') so related are called equivalent. Thisequivalence relation separates all forms into equivalence classes. The forms in a givenclass have the same discriminant, and they represent the same integers. That is, ifQ(x, y) = n for some pair of integers x and y, then Ql(X', y') = n for the pair ofintegers x', y' given by (1).In Exercises 7 thru 10 we consider forms ax2+ bxy + cy2 with d > 0,a > 0, and c > O. The associated quadratic polynomialf(z) = az2+ bz + chas two complex roots. The root T with positive imaginary part is called therepresentative of the quadratic form Q(x, y) = ax2+ bxy + cy2.7. (a) If d is fixed, prove that there is a one-to-one correspondence between the setof forms with discriminant d and the set of complex numbers r with Im(r) > o.(b) Prove that two quadratic forms with discriminant d are equivalent if and only iftheir representatives are equivalent under r.Note. A reduced form is one whose representative TERr. Thus, tworeduced forms are equivalent if and only if they are identical. Also, eachclass of equivalent forms contains exactly one reduced form.8. Prove that a form Q(x, y) = ax2+ bxy + cy2 is reduced if, and only if, either-a < b s a < cor 0 S b s a = c.9. Assume now that the form Q(x, y) = ax2+ bxy + cy2 has integer coefficientsa, b, c. Prove that for a given d there are only a finite number of equivalence classeswith discriminant d. This number is called the class number and is denoted by h(d).Hint: Show that 0 < a S Jdi3 for each reduced form.10. Determine all reduced forms with integer coefficients a, b, c and the class numberh(d) for each d in the interval 1 s d s 20.CONGRUENCE SUBGROUPSThe modular group r has many subgroups of special interest in numbertheory. The following exercises deal with a class of subgroups called con-gruence subgroups. Letand452: The modular group and modular functionsbe two unimodular matrices. (In this discussion we do not identify a matrixwith its negative.) If n is a positive integer writeA == B (mod n) whenever a == rJ., b == {3, c == y and d == {) (mod n).This defines an equivalence relation with the property thatimpliesAlBl == A2 B2 (mod n)Henceandand Al - l== A2 -1 (mod n).A == B (mod n) if, and only if, AB- l== I (mod n),where I is the identity matrix. We denote by r(n) the set of all matrices in rcongruent modulo n to the identity. This is called the congruence subgroupof level n (stufe n, in German).Prove each of the following statements:11. r(n) is a subgroup of r. Moreover, if BE r(n) then A-1BA E r(n) for every A in r.That is, r 0, and,. E H, we have(10) '1(:: : ~ ) = s(a, b, c, d){ - i(cr + dW/2'1(r)wheres(a, b, c, d) = exp{ n{al;cd + s( -d, C))}and(11) k-l r (hr [hrJ 1)s(h, k) = L - - - - - - .r= 1 k k k 2(13)Note. The sum s(h, k) in (11) is called a Dedekind sum. Some of its propertiesare discussed later in this chapter.We will prove Theorem 3.4 through a sequence of lemmas. First we notethat Dedekind's formula is a consequence of the following equation, obtainedby taking logarithms of both members of (10),( aT + b) (a + d )(12) log '1 cr + d = log '1(r) +ni ----u;;- + s( - d, c) +1 log{- i(cr + d)}.From the definition of l1(T) as a product we haveniT 00 niT 00log l1(T) = - + L log(1 - e21tlnt) = - - L A( -inT),12 n=l 12 n=lwhere A(X) is defined for Re(x) > 0 by the equation(14)00 e-21tmxA(X) = -log(1 - e-21tx) = L --.m=l mEquations (12) and (13) give usLemma 1. Equation (12) is equivalent to the relation(15) 00 00 ( aT + b) ni ( aT + b)LA(-inT) = LA -in-- + - T ---n =1 n =1 cT + d 12 CT + d( a + d )+ ni ----u;;- + s( -d, c) +! log{ - i(cr + d)}.We shall prove (15) as a consequence of a more general transformationformula obtained by Sh6 Iseki [17] in 1957. For this purpose it is convenientto restate (15) in an equivalent form which merely involves some changesin notation.523.5: Iseki's transformation formulaLemma 2. Let z be any complex number with Re(z) > 0, and let h, k and H beany integers satisfying (h, k) = 1, k > 0, hH == -1 (mod k). Then Equation(15) is equivalent to the formula(16) f (z - ih)} = f - iH)}n= 1 k n= 1 k z1 n ( 1)+ 2 log z - 12k z - + nis(h, k).PROOF. Given (: in r, with c > 0, and given r with Im(r) > 0, choosez, h, k, and H as follows:k = c, h = -d, H = a, z = -i(cr + d).Then Re(z) > 0, and the condition ad - bc = 1 implies - hH - bk = 1, so(h, k) = 1 and hH == -1 (mod k). Now b = -(hH + l)jk and iz = cr + d,soiz - d iz + hr=--=--c kand henceb iz + h hH + 1 iz ( i)ar + = H -k- - k = k H + -; .Therefore, since cr + d = iz, we havear + b = (H + cr + d k zConsequentlyar + b 1 i ( 1) a + d i ( 1)r - cr + d = k(h - H) + k z - = - -c- + k z - so (r -::: = -n{a1;Cd) - (z -DSubstituting these expressions in (15) we obtain (16). In the same way wefind that (16) implies (15). D3.5 Iseki's transformation formulaTheorem 3.5 (Iseki's formula). If Re(z) > and (1 1, f3 1, let00(17) 1\((1, {3, z) = L {A((r + (1)z - i{3) + A((r + 1 - (1)z + i{3)}.r=O533: The Dedekind eta functionThen if either 0 :::; a :::; 1 and 0 < f3 < 1, or 0 < a < 1 and 0 :::; f3 :::; 1,we haveNote. The sum on the right of (18), which contains Bernoulli polynomialsBn(x), is equal toPROOF. First we assume that 0 < a < 1 and 0 < f3 < 1. We begin with thefirst sum appearing in (17) and use (14) to write(19)00 00 00 e21timpL,1((r + a)z - if3) = L L --e-21tm(r+a)z.r=O r=Om=l mNow we use Mellin's integral for e-xwhich states that(20) 1 fC+ ooie-x= -2. r(s)x-Sds,nl c- ooiwhere c > 0 and Re(x) > O. This is a special case of Mellin's inversion formulawhich states that, under certain regularity conditions, we havefoo 1 fC + ooiq>(s) = XS-1tjJ(x) dx if, and only if, tjJ(x) = -2. q>(s)x-Sds.o nl c- ooiIn this case we take q>(s) to be the gamma function integral,ns) = LX' xS- le-Xdxand invert this to obtain (20). (Mellin's inversion formula can be deducedfrom the Fourier integral theorem, a proof of which is given in [3]. See also[49], p. 7.) Applying (20) with x = 2nm(r + a)z and c = 3/2 to the lastexponential in (19) and writing f(c) for f ~ = ~ ~ we obtain00 00 00 e21timP 1 fL ,1((r + a)z - if3) = L L ---2' r(s){2nm(r + a)z} -s dsr=O r=O m=l m nl (3/2)1 f r(s) 00 1 00 e21timp= -. --S L S L ----r+s ds2nl (3/2) (2nz) r=O (r + a) m=l m1 f r(s)= -2. -(2)S '(s, a)F(f3, 1 + s) ds.nl (3/2) nz543.5: Iseki's transformation formulaHere ((s, a) is the Hurwitz zeta function and F(x, s) is the periodic zeta functiondefined, respectively, by the seriesoc 1((s, tX) = L ( + )"r=O r a and00 e21timxF(x, s) = L -S-m=l mwhere Re(s) > 1, 0 < a S 1, and x is real. In the same way we findoc 1 f r(s)L A((r + 1 - a)z + i{3) = -2. -(2)S ((s, 1 - a)F(1 - {3, 1 + s) ds,r= 0 1tl (3/2) 1rZso (1 7) becomes(21)whereA(a, {3, z) = -21. f z - s(a, {3, s) ds,1tl (3/2)r(s)(22) (tX, 13, s) = (211:)' {((s, tX)F(f3, 1 + s) + ((s, 1 - tX)F(1 - 13, 1 + s)}.Now we shift the line of integration from c = ~ to c = -1. Actually, weapply Cauchy's theorem to the rectangular contour shown in Figure 3.2,! + iT ! + iT~~01---..t - iT ! - iTFigure 3.2and then let T ---. 00. In Exercise 8 we show that the integrals along thehorizontal segments tend to 0 as T ---. 00, so we getI3/2) - I-3/2) + Rwhere R is the sum of the residues at the poles of the integrand inside therectangle. This gives us the formulaA(a, {3, z) = -21. f z-s(a, {3, s) ds + R.1rl (- 3/2)55(23)3: The Dedekind eta functionIn this integral we make the change of variable u = - s to get it back inthe form of an integral along the 1line. This gives us1\(0:, {3, z) = -21. f zU(o:, {3, - u) du + R.n1 (3/2)Now the function satisfies the functional equation(24) (0:, {3, - s) = ( 1 - {3, 0:, s).This is a consequence of Hurwitz's formula for ((s, 0:) and a proof is outlinedin Exercise 7. Using (24) in (23) we find that(25) 1\(0:, {3, z) = 1\( 1 - {3, 0:, Z - 1) + R.To complete the proof of Iseki's formula we need to compute the residuesumR.Equation (22) shows that (0:, {3, s) has a first order pole at each of thepoints s = 1, and -1. Denoting the corresponding residues by R(1),R(O) and R( - 1) we findr( 1) 1 oc (2Trinp - 2Trinp)R(1) = -2- {F({3, 2) + F(1 - {3,2)} = -2 L ~ + _e-2-nz nz n= 1 n n1 oc e2Trinp 1 - (2n02n= -2 L -2- = -2 2' B2({3) = - B2({3),nz n = - oc n nz. zn*Owhere we have used Theorem 12.19 of [4J to express the Fourier series as aBernoulli polynomial.To calculate R(O) we recall that ((0,0:) = 1- 0:. Hence ((0, 1 - 0:) = 0: - 1sooc e2Trinp _ e - 2TrinpR(O) = ((0, o:)F({3, 1) + ((0, 1 - o:)F(l - {3, 1) = (1 - 0:) L ----n= 1 n00 e2Trinp oc e2Trinp= (1 - 0:) L -- = -B1(0:) L -- = 2niB1(0:)B1({3),n=-oo n n=-oc nn*O n*Owhere again we have used Theorem 12.19 of [4]. To calculate R( -1) wewriteR( -1) = Res z-s(o:, {3, s) = lim (s + 1)z-s(0:, {3, s)s=-l s ~ - l= lim( - s + 1)zs(0:, {3, - s).s ~ lUsing the functional equation (24) we findR(-1) = lim(1 - s)zs(1 - {3, 0:, s) = -Res zS(1 - {3, 0:, s).s= 1563.5: Iseki's transformation formulaNote that this is the same as R(I) == Ress = 1 z-S(o:, {3, s), except that z isreplaced by - z - 1, 0: by 1 - {3, and {3 by 0:. Hence we haveR( -1) == - nzB2(0:).ThusR = R( -1) + R(O) + R(l) = -nznto G)(iZ)-nB1-n(IX)Bi{3).This proves Iseki's formula under the restriction a < 0: < 1, a < {3 < 1.Finally, we use a limiting argument to show it is valid if a ~ 0: ~ 1 anda< {3 < 1, or if a~ {3 ~ 1 and a< 0: < 1. For example, consider the series00 ex.; oc e2nimf3L A((r + ex)z - i{3) == L L --e-2nm(r+a)zr=O r=Om=l mex 2nimf3 00== I _e__ e-2nmaz L e-2nmrzm=l m r=Osay, where1 e- 2nmaz.fa(m) == - 1 - 2nmzm -eAs m ~ 00, .fa(m) ~ auniformly in 0: if a s 0: S 1. Therefore the series00L e2nimf3fa(m)m=lconverges uniformly in 0: if a :::; 0: :::; 1, provided a < {3 < 1, so we can passto the limit 0: ~ a+ term by term. This gives us00 ex.;lim L A((r + o:)z - i{3) == L A(rz - i{3).a-O+ r=O r=OTherefore, if a < {3 < 1 we can let 0: ~ a+ in the functional equation. Theother limiting cases follow from the invariance of the formula under thefollowing replacements:0: ~ 1 - 0:, { 3 ~ 1 - { 30: ~ {3, 1{3 ~ 1 - 0:, z ~ -Z0: ~ 1 - {3, {3 ~ 0:,1D z ~ - .z57(26)(27)3: The Dedekind eta function3.6 Deduction of Dedekind's functionalequation from Iseki's formulaNow we use Iseki's formula to prove Equation (16) of Lemma 2. This, in turn,will prove Dedekind's functional equation for 11(r).Equation (16) involves integers hand k with k > O. First we treat the casek = 1 for which Equation (16) becomesI A{n(z - ih)} = I - iH)} + log z - (z - n= 1 n= 1 Z 2 12 zSince A(X) is periodic with period i this can be written as00 00 (n) 1 re ( 1)L A(nz) = L A - + - log z - - z - - .n=1 n=1 Z 2 12 zWe can deduce this from Iseki's formula (18) by taking f3 = 0 and lettinga ---+ 0+. Before we let a ---+ 0+ we separate the term r = 0 in the first termof the series on the left of (18) and in the second term of the series on the rightof (18). The difference of these two terms is A(az) - A(ia). Each of these tendsto 00 as a ---+ 0+ but their difference tends to a finite limit. We compute thislimit as follows:. 1 - A(CXZ) - A(iCX) = log(l - e-Z"IlX) - log(l - e-z"",Z) = log 1 _ e-z"",z'By L'Hopital's rule,1 - 2ni(X 21 - e 1 rel l1m _ 2n(Xz = 1m - = -(X-+O + 1 - e (X-+O 2rez zsolim (A(CXZ) - A(icx)) = log = ni - log Z.(X-+O+ Z 2Now when a ---+ 0+ the remaining terms in each series in (18) double upand we obtain, in the limit,(28) rei 00 00 (r) rez re rei- - log z + 2 L A(rz) = 2 L A - - - + - + -.2 r = 1 r = 1 Z 6 6z 2This reduces to (27) and proves (16) in the case k = 1.Next we treat the case k > 1. We choose rational values for a and f3 inIseki's formula (18) as follows. TakeJla = k' where 1 Jl k - 1583.6: Deduction of Dedekind's functional equation from Iseki's formulaand writehJ1 = qk + v, where 1 ~ v ~ k - 1.Now letNote that v == hJ1 (mod k) so - Hv == - HhJ1 == J1 (mod k), and therefore-Hv/k == J1/k (mod 1). Hence lJ. = J1/k == -Hv/k (mod 1) and f3 = v/k ==hJ1/k (mod 1). Substituting in Iseki's formula (18) and dividing by 2 we getRewrite this as follows:= ~ 00 { I(rk + V)G - iH)) (rk + k- V)G - iH))}2 Jo A\ k + A kNow sum both sides on J1 for J1 = 1, 2, ... , k - 1 and note that{rk + J1:r = 0,1,2, ... ; J1 = 1,2, ... ,k - I} = {n:n =1= (modk)}and similarly for the set of all numbers rk + k - J1. Also, since v == hJ1 (mod k),as J1 runs through the numbers 1, 2, ... , k - 1 then v runs through the same593: The Dedekind eta functionset of values in some other order. Hence we getCX) (n .) CX) (n (1 .)) n(1 )k- 1 J12L A - (z - zh) = L A -: - - zH + - - - z L 2n= 1 k n= 1 k z 2 z p.= 1 kn 0 (mod k) n 'i 0 (mod k)n (1 )k -1 J1 n (1 )- - - - z L - + - - - z (k - 1)2 Z p.=1 k 12 zk - 1J1 (V 1) ni k - 1V ni+ ni L - - - - - - L - + - (k - 1)p.= 1 k k 2 2 p.= 1 k 4((k - 1) (2k - 1) ) k - 1 J1 (V 1)X - 3(k - 1) + (k - 1) + ni L - - - -k p.= 1 k k 2I - iH)) + (z - - + l/f !!. - n= 1 k z 12 z k p.= 1 k k 2n$.O (mod k)But V was defined by the equation hJ1 = qk + V, so we haveThereforehJ1 Vk = q + k' = hJ1 _ [hJ1]k k kk-1 J1 (V 1) h-1 J1(hJ1 [hJ1] 1)L - - - - = L - - - - - - = s(h, k).p.= 1 k k 2 p.= 1 k k k 2Therefore we have proved that(29)

A(-kn (z'- ih)) = f: - iH))n= 1 k zn 1- 0 (mod k) n 1= 0 (mod k)Add this to Equation (27) which corresponds to the case k = 1:CX) CX) (m) n( 1) 1L A(mz) = L A - - - z - - + -log z.m= 1 m= 1 Z 12 z 2This accounts for the missing terms in (29) with n == 0 (mod k), if we writen = mk. When (27) is combined with (29) we getf (z - ih)) = f - iH)) - (z - + log z + nis(h, k).n =1 k n = 1 k z 12k z 260(30)3.7: Properties of Dedekind sumsThis proves (16) which, in tum, completes the proof of Dedekind's functionalequation for Tl(T). For alternate proofs see p. 190 and [18], [35], and [45]. D3.7 Properties of Dedekind sumsThe Dedekind sums s(h, k) which occur in the functional equation for YJ(r)have applications to many parts of mathematics. Some of these are describedin an excellent monograph on Dedekind sums by Rademacher and Grosswald[38]. We conclude this chapter with some arithmetical properties of the sumss(h, k) which will be needed later in this book. In particular, Theorem 3.11plays a central role in the study of the invariance of modular functions undertransformations of certain subgroups of r, a topic discussed in the nextchapter.Note. Throughout this section we assume that k is a positive integer andthat (h, k) = 1.Dedekind sums are defined by the equations(h, k) = kt1!.- (hr - [hrJ _~ ) .r= 1 k k k 2First we express these sums in terms of the function ((x)) defined by( )_{x - [x] -! if xis not an integer,(x) - 0 fI X IS an Integer.This is a periodic function of x with period 1, and (( - x)) = - ((x)). Actually,((x)) is the same as the Bernoulli periodic function B1(x) discussed in [4],Chapter 12. Since ((x)) is periodic and odd we find thatL ( ( ~ ) ) - 0r mod k kand, more generally,L ((hr)) - 0 for (h, k) = 1.r mod k kSincethe Dedekind sums can now be represented as follows:(31) s(h,k) = r m ~ k ((i))((h:)).This representation is often more convenient than (30) because we can exploitthe periodicity of ((x)).613: The Dedekind eta functionTheorem 3.6(a) If h' == h (mod k), then s(h', k) = s(h, k), lvith the same sign asin the congruence. Similarly, lve have:(b) If hh == 1 (mod k) then s(h, k) = s(h, k).(c) ~ f h2+ 1 == 0 (mod k), then s(h, k) = O.PROOF. Parts (a) and (b) follow at once from (31). To prove (c) we note thath2+ 1 == 0 (mod k) implies h == -h (mod k), where h is the reciprocal ofh mod k, so from (a) and (b) we get s(h, k) = -s(h, k) = O. DFor small values of h the sum s(h, k) can be easily evaluated from itsdefinition. For example, when h = 1 we findk - 1 r(r 1) 1k - 1 2 1k - 1s(l, k) = L - - - - = 2 L r - - L rr = 1 k k 2 k r = 1 2k r = 1(k - 1)(2k - 1) k - 1 (k - 1)(k - 2)---6k 4 12kSimilarly, the reader can verify that(2 k) = (k - 1)(k - 5) if k is odd.s , 24kIn general there is no simple formula for evaluating s(h, k) in closed form.However, the sums satisfy a remarkable reciprocity law which can be usedas an aid in calculating s(h, k).3.8 The reciprocity law for Dedekind sumsTheorem 3.7 (Reciprocity law for Dedekind sums). If h > 0, k > 0 and(h, k) = 1 we have12hks(h, k) + 12khs(k, h) = h2+ k2- 3hk + 1.PROOF. Dedekind first deduced the reciprocity law from the functionalequation for log '1(7:). We give an arithmetic proof of Rademacher andWhiteman [39], in which the sum L ~ = 1 ((hr/k))2 is evaluated in two ways.First we have(32)62k ((hr))2 ((hr))2 ((r))2 k-l (r 1)2L - = L - = L - = L ---r = 1 k r mod k k r mod k k r = 1 k 2 .3.8: The reciprocity law for Dedekind sumsWe can also writeI ((hr))2 = kil (hr _ [hrJ _ ~ ) 2r= 1 k r= 1 k k 2= ki1( h 2 ~ 2 + [hrJ2 + ~ _ hr + [hrJ _ 2hr [hrJ)r= 1 k k 4 k k k k= 2hkf !:.- (hr _ [hrJ _ ~ )r= 1 k k k 2k-1 [hrJ([hrJ ) h2k-1 2 1 k-1+L - - +1 -2Lr +-2:1.r= 1 k k k r= 1 4 r= 1Comparing this with (32) and using (30) we obtaink - 1 [hrJ ([hrJ ) h2+ 1 k - 11k- 1(33) 2hs(h,k) + ~ -k -k + 1 = -k2- ~ r2- -k ~ r.r-1 r-1 r-1In the sum on the left we collect those terms for which [hr/k] has a fixed value.Since 0 < r < k we have 0 < hr/k < h and we can write(34) [ ~ J= v - 1, where v = 1, 2, ... , h.For a given v let N(v) denote the number of values of r for which [hr/k] =v - 1. Equation (34) holds if, and only ifhrv-1 0 so that Iarg zI nl2 - J, and showthat if s = (J + it where (J we haveIz-si = O(e1tl (1t/2-b),where the constant implied by the O-symbol depends on z.(c) If s = (J + it where (J and It I 1, show that1 (e-1t1tl)Is sin nsl = 0 -I-t-I 'and thatIe1tiS/21 = O(e1tltl/2), Ie-1tis/21 = O(e1tltl/2).(d) If (J and It I 1 obtain the estimate I((s, a) I = O( ItiC) for some c > 0(see [4J, Theorem 12.23) and use (b) to deduce thatIz-s(a, [3, s)1 = O(/tI2C-le-/tlt)).This shows that the integral of [3, s) along the horizontal segments of therectangle in Figure 3.2 tends to 0 as T 00.PROPERTIES OF DEDEKIND SUMS9. If k 1 the equations(h, k) = k ((i))((h:))is meaningful even if h is not relatively prime to k and is sometimes taken as thedefinition of Dedekind sums. Using this as the definition of s(h, k) prove thats(qh, qk) = s(h, k) if q > O.10. If p is prime prove thatp- 1(p + l)s(h, k) = s(ph, k) + L s(h + mk, pk).m=O11. For integers r, h, k with k 1 prove that we have the finite Fourier expansion((hr)) I k - 1 . 2nhrv nv- = - - L SIn--cot-k 2k \'= 1 k kand derive the following expression for Dedekind sums:1 k- 1 nhr nrs(h, k) = - L cot - cot-.4k r= 1 k k12. This exercise relates Dedekind sums with the sequence {lI(Il)] of Fibonacci numbers1, 1,2,3,5,8, ... , in which u(l) = u(2) = 1 and u(n + 1) = u(n) + u(n - 1).(a) If h = u(2n) and k = u(2n + 1) prove that s(h, k) = o.(b) If h = u(2n - 1) and k = u(2n) prove that 12hks(h, k) = h2+ k2- 3hk + 1.72Exercises for Chapter 3FORMULAS FOR EVALUATING DEDEKIND SUMSThe following exercises give a number of formulas for evaluating Dedekindsums in closed form in special cases. Assume throughout that (h, k) = 1,k ~ 1, h ~ 1.13. If k == r (mod h) prove that the reciprocity law implies12hks(h, k) = k2- {12s(r, h) + 3}hk + h2+ 1.Use the result of Exercise 13 to deduce the following formulas:14. If k == 1 (mod h) then 12hks(h, k) = (k - l)(k - h2- 1).15. If k == 2 (mod h) then 12hks(h, k) = (k - 2)(k - !(h 2+ 1)).16. If k == -1 (mod h) then 12hks(h, k) = k2+ (h 2- 6h + 2)k + h2+ 1.17. If k == r (mod h) and if h == t (mod r) where r ~ 1 and t = 1, thenh2- t(r - l)(r - 2)h + r 2+ 112hks(h,k) = k2- k + h2+ 1.rThis formula includes those of Exercises 14 and 15 as special cases.18. Show that the formula of Exercise 17 determines s(h, k) completely when r = 3and when r = 4.19. If k == 5 (mod h) and if h == t (mod 5), where t = 1 or 2, thenh k) k2 h2+ 4t(t - 2)(t + 2)h + 26 212hk s(, = - 5 k + h + 1.20. Assume 0 < h < k and let ro, r 1, ... , rn+ 1 denote the sequence of remainders in theEuclidean algorithm for calculating the gcd (h, k), so thatProve that1 n ~ 1 { "+ 1 r/ + rj_1 2+ I} (-1)" + 1s(h,k)=- ~ (-I)J -.12 j =1 rjrj-1 8This also expresses s(h, k) as a finite sum, but with fewer terms than the sum in theoriginal definition.73Congruences for the coefficientsof the modular function j4.1 IntroductionThe function j(r) = 123J(r) has a Fourier expansion of the form1 00 j(r) = - + L c(n)xn, (x = e2m!)x n=Owhere the coefficients c(n) are integers. At the end of Chapter 1 we mentioneda number of congruences involving these integers. This chapter shows howsome of these congruences are obtained. Specifically we will prove thatc(2n) == 0 (mod 211),c(3n) == 0 (mod 35),c(5n) == 0 (mod 52),c(7n) == 0 (mod 7).The method used to obtain these congruences can be illustrated for themodulus 52. We consider the function00f5(r) = L c(5n)xnn=1obtained by extracting every fifth coefficient in the Fourier expansion of j.Then we show that there is an identity of the form(1)where the ai are integers and (r) has a power series expansion in x = e 21ti!with integer coefficients. By equating coefficients in (1) we see that eachcoefficient of fs(r) is divisible by 25.Success in this method depends on showing that such identities exist.How are they obtained?744.2: The subgroup r o(q)Theorem 2.8 tells us that every modular function f is a rational functionof j. Sometimes this rational function is a polynomial in j with integer co-efficients, giving us an identity of the formf(r) = a1j(r) + a2j2(r) + ... + akjk(r).However, the function fs(r) is not invariant under all transformations ofthe modular group r and cannot be so expressed in terms of j(r). But weshall find that fs(r) is invariant under the transformations of a certainsubgroup of r, and the general theory enables us to express fs(r) as a poly-nomial in another basic function (r) which plays the same role as j(r)relative to this subgroup. This representation leads to an identity such as(1) and hence to the desired congruence property.The subgroup in question is the set of all unimodular matrices (: : )with c == 0 (mod 5). More generally we shall consider those matrices in rwith c == 0 (mod q), where q is a prime or a power of a prime.4.2 The subgroup r o(q)Definition. If q is any positive integer we define r o(q) to be the set of allmatrices (: ~ ) in r with c == 0 (mod q).It is easy to verify that r o(q) is a subgroup of r. The next theorem gives away of representing each element of r in terms of elements of r o(p) whenp is prime. In the language of group theory it shows that r o(p) is of finiteindex in r.Theorem 4.1. Let Sr = -l/r and Tr = r + 1 be the generators of the fullmodular group r, and let p be any prime. Then for every V in r, V ~ r o(P),there exists an element P in r o(P) and an integer k, 0 ~ k < p, such thatV = PSTkPROOF. Given V = ( ~ ~ ) where C =1= 0 (mod p). We wish to findp = G:). with c == 0 (mod p),and an integer k, 0 ~ k < p, such that754: Congruences for the coefficients of the modular function jAll matrices here are nonsingular so we can solve for ( : ) to get(: k to be that solution of the congruencekC == D (mod p) with 0 k < p.This is possible since C =1= 0 (mod p). Now takec = kC - D, a = kA - B, b = A, d = C.Then c == 0 (mod p) so PEro(p). This completes the proof. o4.3 Fundamental region of r o(p)As usual we write Sr = -l/r and Tr = r + 1, and let Rr denote the funda-mental region of r.Theorem 4.2. For any prime p the setp-lRr u USTk(Rr )k=Ois a fundamental region of the subgroup r o(p).This theorem is illustrated for p = 3 in Figure 4.1.PROOF. Let R denote the setp-lR = Rr u USTk(Rr ).k=OWe will prove(i) if r E H, there is a V in r o(p) such that Vr belongs to the closure of R, and(ii) no two distinct points of R are equivalent under r o(p).To prove (i), choose r in H, choose r 1 in the closure of Rr and chooseA in r such that Ar = r1. Then by Theorem 4.1 we can writeA-I = PWwhere PEro(p) and W = I or W = STkfor some k, 0 k p - 1. ThenP = A-IW- 1and p-I= WA. Let V = P- 1 Then VEro(p) andVr = WAr = Wri .Since W = I or W = STk'l this proves (i).764.3: Fundamental region of r o(p)-1 o 12"TFigure 4.1 Fundamental region for r 0(3)Next we prove (ii). Suppose!1E R, ! 2E R and v!1 = ! 2 for some V inr o(P). We will prove that!1 = ! 2' There are three cases to consider:(a) !1 ERr '!2 ERr In this case!1 =!2 since VEr.(b) ! 1 ERr, ! 2 ESTk(Rr ).(c) ! 1 E STkl(Rr ), ! 2 E STk2(Rr ).In case (b), !2 = STk!3 where !3 ERr. The equationimpliesk (0V = ST = 1This contradicts the fact that V E r o(p).Finally, consider case (c). In this case-1)k .andwhere !1' and !2' are in Rr . Since V!1 = !2 we have VSTkl!I' = STk2!2' soVSTkl = STk2,774: Congruences for the coefficients of the modular function iSince VE ro(p) this requires k2 == ki (mod pl. But both kl , k2 are in theinterval [0, p - IJ, so k2 = ki . ThereforeV = SToS = S2 = Iand r I = r 2. This completes the proof. DWe mention (without proof) the following theorem of Rademacher [34]concerning the generators of r o(p). (This theorem is not needed in thelater work.)Theorem 4.3. For any prime p> 3 the subgroup ro(p) has 2[p112J + 3generators and they may be selected from the following elements:where Tr = r + 1, Sr = -l/r, andk -k' ( k'v" = ST ST S = -(kk' + 1)where kk' == -1 (mod pl. The subgroup ro(2) has generators T and V I ~the subgroup r 0(3) has generators T and V2 .Here is a short table of generators:p 2 3 5 7 11 13 17 19Generators: T T T T T T T TVI V2 V2 V3 V4 V4 V4 VsV3 Vs V6 Vs V7 V8V8 Vg VI2VIO VI3 VI34.4 Functions automorphic urlder thesubgroup r o(p)We recall that a modular function f is one which has the following threeproperties:(a) f is meromorphic in the upper half-plane H.(b) f(Ar) = f(r) for every transformation A in the modular group r.(c) The Fourier expansion of f has the form00f(r) = L ane21tint.n= -m784.4: Functions automorphic under the subgroup r o(p)If property (b) is replaced by(b') f(Vr) = f(r) for every transformation V in r o(P),thenfis said to be automorphic under the subgroup ro(p). We also say thatf belongs to r o(P)The next theorem shows that the only bounded functions belonging tor o(p) are constants.Theorem 4.4. If f is automorphic under r o(p) and bounded in H, then f isconstant.PROOF. According to Theorem 4.1, for every V in r there exists an elementP in r o(P) and an integer k, ~ k ~ p, such thatV = PAk ,where Ak = STkif k < p, and Ap = I. For each k = 0, 1, ... , p, letrk = {PAk : PEro(p)}.Each set r k is called a right coset of r o(P). Choose an element ~ from thecoset rk and define a function fk on H by the equationfk(r) = f ( ~ r).Note that fp(r) = f(Pr) = f(r) since P E ro(P) and f is automorphic underro(P). The function value A(r) does not depend on which element ~ waschosen from the coset r k becauseA(r) = f ( ~ r ) = f(PAkr) = f(Akr)and the element Ak is the same for all members of the coset rk How does fk behave under the transformations of the full modular group?If V E r thenfk(Vr) = f ( ~ Vr).Now ~ V E r so there is an element Qin r o(P) and an integer m, ~ m ~ p,such thatTherefore we haveMoreover, as k runs through the integers 0, 1, 2, ... , p so does m. In otherwords, there is a permutation (J of {O, 1,2, ... , p} such thatfk(Vr) = fa(k)(r) for each k = 0,1, ... ,p.Now choose a fixed w in H and letpcp(r) = n{fk(r) - f(w)}.k=O794: Congruences for the coefficients of the modular function jThen if V E r we havep pcp(Vr) = n {fk(Vr) - f(w)} = n {f(j(k)(r) - f(w)} = cp(r),k=O k=Oso cp is automorphic under the full group r. Now cp is bounded in H (sinceeach fk is). T h e r e f o r e ~ cp omits some value hence, by Theorem 2.5, cp isconstant, so cp(r) = cp(w) for all r. But cp(w) = 0 becausepcp(w) = n {fk(W) - f(w)}k=Oand the factor with k = p vanishes since fp = f Therefore cp(r) = 0 for all r.Now take r = i. Thenpo= n {A(i) - f(w)}k=Ohence some factor is o. In other words, f(w) = A(i) for some k. But w wasarbitrary so f can take only the values fo(i), ... ,fp(i). This implies that f isconstant. 04.5 Construction of functions belongingto ro(p)This section shows how to construct functions automorphic under thesubgroup r o(p) from given functions automorphic under r.Theorem 4.5. If f is automorphic under r and if p is prime, letfir) =! Pff(r + A).p ),=0 PThenfp is automorphic under r o(p). Moreover, iff has the Fourier expansion00f(r) = L a(n)e21tinrn= -mthen fp has the Fourier expansion00fp(r) = L a(np)e21tinr.n= - [m/p]PROOF. First we prove the statement concerning Fourier expansions. We have1 p- 1 00fp(r) = - L L a(n)e21tin(r+).)/Pp ),=0 n=-m1 00 p- 1= - L a(n)e21tinr/p Le21tin)./p.p n= -m ),=0804.5: Construction of functions belonging to r o(p)Butsop-I {OL e21tin;"/p =;"=0 Pif p,(nif pin00 00fp('r) = L a(n)e21tinr/p = L a(np)e21tinr.n= -m n= -[m/p]pinThis shows that fp has the proper behavior at the point r = ioo. Also,fp is clearly meromorphic in H because it is a linear combination of functionsmeromorphic in H.Next we must show thatfp(Vr) = fp(r) whenever V E r o(p).For this we use a lemma.Lemma 1. If V E ro(p) and if ~ A ~ p - 1, let T;.,r = (r + A)jp. Thenthere exists an integer j1, 0 ~ j1 ~ p - 1 and a transformation ~ inr O(p2) such thatMoreover, as Aruns through a complete residue system modulo p, so does j1.First we use the lemma to complete the proof of Theorem 4.5, then wereturn to the proof of the lemma.If V E r o(p) we have1 P-1(Vr+A) 1 P-1fp(Vr) = - L f = - L f(T;., Vr).P ;"=0 P, P ;"=0Now we use the lemma to write the last sum as1 p- I 1 p- I- L f ( ~ ~ r) = - L f ( ~ r) = fp( r).PJL=o PJL=oThis proves that fp is invariant under all transformations in r o(p), so fp isautomorphic under r o(p). DPROOF OF LEMMA 1. Let V = (: :). where c == 0 (mod p), and let A. begiven, 0 ~ A ~ p - 1. We are to find an integer j1, 0 ~ j1 ~ p - 1 and atransformation ~ = ( ~ ~ ) such that ~ E r O(P2) andT;.,V = WJLTJL.814: Congruences for the coefficients of the modular function jSince TA. = G~ ) we must satisfy the matrix equationor( a + AC b + Ad) = (A AJl + BP)pc pd C CJl + Dpwith C == 0 (mod p2). Equating entries we must satisfy the relations{A = a + AC(2)C = pc(3) {AJl + Bp = b + AdCJl + Dp = pdwithC == 0 (mod p2) and AD - BC = 1.Now (2) determines A and C. Since pic, we have C == 0 (mod p2). Substi-tuting these values in (3) we must satisfy(4) {(a + AC)Jl + Bp = b + AdCpJl + Dp = pd.Choose Jl to be that solution of the congruenceJla == b + Ad (mod p)which lies in the interval 0 ~ Jl ~ P - 1. This is possible because ad - bc = 1and pic imply p,ta. Note that distinct values of Amod p give rise to distinctvalues of Jl mod p. Then, since pic we haveJla +JlAc == b + Ad (mod p)or(a + AC)Jl == b + Ad (mod p).Therefore there is an integer B such that(a + AC)Jl + Bp = b + Ad.Therefore the first relation in (4) is satisfied. The second relation requiresD = d - CJl. Thus, we have found integers Jl, A, B, C, D such thatClearly AD - BC = 1 since all matrices in this equation have determinant1 or p. This completes the proof of the lemma. D82or4.6: The behavior under the generators of r4.6 The behavior off p under the generatorsofrLet Tr = r + 1 and Sr = -l/r be the generators of r. Since T E r o(p) wehave fp(Tr) = fp(r). The next theorem gives a companion result for fp(Sr).Theorem 4.6. If f is automorphic under r and if p is prime, then (- = f, (r) + f(pr) - p r p p p PTo prove this we need another lemma.Lemma 2. Let T). r = (r + A)/p. Thenfor each Ain the interval 1 S ASP - 1there exists an integer J.1 in the same interval and a transformation V inr o(p) such thatT).S = VTl1 Moreover, as Aruns through the numbers 1,2, ... ,p - 1, so does J.1.PROOF OF LEMMA 2. We wish to find (: in ro(p) such thatG = G G- = (: )-Take a = A, c = p and let J.1 be that solution of the congruence,1,J.1 == -1 (mod p)in the interval 1 S J.1 S P - 1. This solution is unique and J.1 runs through areduced residue system mod p with ,1,. Choose b to be that integer such thataJ.1 + bp = - 1, and take d = - J.1. Then CJ.1 + dp = 0 and the proof iscomplete. DPROOF OF THEOREM 4.6. We havePfp( - = Pi! f(sr + A) = f(sr) + Pf f(T;.Sr)r ),=0 p P).=1= f(- + Pi! f(VTI1r) = f(rp) + Pi! f(TIL r) - rp 11=1 11=0 P= f(rp) + pfp(r) - fG)- 0834: Congruences for the coefficients of the modular function.i4.7 The function (r) = ifr E H. Then qJ is automorphic under ro(q). Moreover, the Fourier expansion ofqJhas theformwhere the bn are integers and x = e21tit844.7: The function (r) = = xq- 11+ 1r(n + 1)xnq= xq- 1(1 + f b xn) r) 1 + 1 r(n + 1)xnn = 1 nwhere the bn are integers.Now qJ is clearly meromorphic in H, and we will prove next that qJ isinvariant under r o(q).If V = (: E ro(q) then c = c1q for some integer c1 Hence Vr) = (cr + d) 12 r) = (c 1qr + d) 12 r).On the other hand,aT + b a(qr) + bqqVr = q--d = () d = W(qr),cr + C1 qr +wherew = (a bq).C1 dBut WE r because det W = ad - bc1q = ad - bc = 1. Hence = = (c1(qr) + so(Vr) = = (c1qr + = (r).qJ Vr) (c1qr + d) 12 r ) qJThis completes the proof. DNow qJ has a zero of order q - 1 at 00 and no further zeros in H. Next weshow that qJ does not vanish at the vertex r = 0 of the fundamental regionof r o(q). In fact, we show that qJ(r) -+ 00 as r -+ O.Theorem 4.8. Ifr E H we have(-1) 1qJ = q12qJ(r)Hence qJ(r) -+ 00 as r -+ O.85(5)4: Congruences for the coefficients of the modular function jPROOF. Since = we have -:r) = so4.8 The univalent function (r)The function cp has a zero of order q - 1 at 00 and no further zeros so itsvalence is q - 1. We seek a univalent function automorphic under r o(q)and this suggests that we consider cpa, where r:J. = 1/(q - 1). The Fourierexpansion of cpa need not have integer coefficients, since = x(1 + bnxnrOn the other hand we have the product representation00 = (2n)12xn(1 - Xn)24n=1sowhere the coefficients dq(n) are integers. Therefore if a = 1/(q - 1) we have(00 )24a = x 1 + din)xnand the Fourier series for cpa(r) will certainly have integer coefficients if24a is an integer, that is, if q - 1 divides 24. This occurs when q = 2, 3, 4, 5,7, 9, 13, and 25.Definition. If q - 1 divides 24 let a = 1/(q - 1) and r = 24a. We define thefunction (r) = = 1](r) 86(6)4.9: Invariance of ( r) under transformations of r o(q)The function so defined is analytic and nonzero in H. The Fourier seriesfor in (5) shows that has a first order zero at 00 and that1 1(r) = + I(x),where I(x) is a power series in x with integer coefficients.Since qJ is automorphic under r o(q) we have qJ(Vr) = qJ(r) for everyelement V of r o(q). Hence, extracting roots of order q - 1, we have( Vr) = E( r )where Eq- 1 = 1. The next theorem shows that, in fact, E = 1 whenever24/(q - 1) is an even integer and q is prime. This occurs when q = 2, 3, 5, 7,and 13. For these values of q the function is automorphic under r o(q).4.9 Invariance of (r) under transformationsof r o(q)The properties of Dedekind sums proved in the foregoing chapter lead to asimple proof of the invariance of the univalent function (r).Theorem 4.9. Let q = 2,3,5,7, or 13, and let r = 24/(q - 1). Then thefunction( r) = (11(qr))r11(r)is automorphic under the subgroup r o(q).PROOF.Ifq = 2wehaver = 24 and (r) = was already proved in Theorem 4.7. Therefore we shall assume that q 3.Let V = (: be any element of r o(q). Then ad - be = 1 andc == 0 (mod q). We can suppose that c O. If c = 0 then V is a power ofthe translation Tr = r + 1, and since 11(r + 1) = e1ti/1211(r) we find(r + 1) = (11(qr + q))r = e"ir(Q- 1l/12(r) = (r).11(r + 1)Therefore we can assume that c > 0 and that c = c1q, where C1 > O.Dedekind's functional equation for 11(r) gives us(7)where(8)874: Congruences for the coefficients of the modular function.iWe also have(a(qr) + bq)I](qVr) = I] c1(qr) + d = I](V1qr)where = (a bq)I CI dSince VI E r we havewhich, together with (7), gives us k + k1> N + 1 > !i1 - fi - fi fi'so (10) and (12) imply (11). The length of the chord is ::::; Iz11 + Iz21. D5.7 Rademacher's convergent series for pen)Theorem 5.10. If n ~ 1 the partition function p(n) is represented by theconvergent series1 (X) dp(n) = M LAk(n)jk -dny 2 k= 1 n1045.7: Rademacher's convergent series for p(n)whereA (n) ~ eftis(h,k)-2ftinh/k.k = LJO ~ h < k(h,k)= 1PROOF. We have(13) 1 I F(x)p(n) = -2. n+f dx'Ttl eX00 00where F(x) = n(1 - Xm)-l = L p(n)xn;m=l n=OC is any positively oriented closed curve surrounding x = 0 and lying insidethe unit circle. The change of variablemaps the unit disk Ix I ~ 1 onto an infinite vertical strip of width 1 in ther-plane, as shown in Figure 5.7. As x traverses counterclockwise a circle ofx-planeot-planeFigure 5.7radius e- 2ft with center at x = 0, the point r varies from i to i + 1 along ahorizontal segment. We replace this segment by the Rademacher path P(N)composed of the upper arcs of the Ford circles formed for the Farey seriesFN. Then (13) becomesi+ 1p(n) = r F(e2nh)e- 2ninr dr = r F(e2nit)e- 2nint dr.Ji Jp(N)In this discussion the integer n is kept fixed and the integer N will later beallowed to approach infinity. We can also writefP(N) ktl O " ~ < k L(h'k) = t,.;, l(h'kl(h,k)= 1where y(h, k) denotes the upper arc of the circle C(h, k), and Lh,k is anabbreviation for the double sum over hand k.1055: Rademacher's series for the partition functionNow we make the change of variableso thath izt = k + k2 Theorem 5.8 shows that this maps C(h, k) onto a circle K of radius! aboutz = ! as center. The arc y(h, k) maps onto an arc joining the points zl(h, k)and z2(h, k) in Figure 5.6. We now havep(n) = LJZ2(h. k) F(exp(21rih _ 2 ~ Z ) ) ~ e- 2ninh/k e2nnz/k2 dzh,k zt{h,k) k k k= L ik-2e-2ninh/k JZ2(h.k) e2nnZ/k2F(exp(21rih _ 2 ~ Z ) ) dz.h,k zt{h,k) k kNow we use the transformation formula for F (Theorem 5.1) which statesthat( Z)1/2 ( 1r 1rZ )F(x) = w(h, k) k exp 12z - 12k2 F(x'),whereand(21rih 21rZ)x = exp -k- - k2 ' , (21riH 21r)x = exp -k- - -;- ,w(h, k) = enis(h,k), hH == -1 (mod k), (h, k) = 1.Denote the elementary factor ZI/2 exp[,,-/(12z)- 1rz/{12k2)J by \fJk{Z) andsplit the integral into two parts by writingF(x') = 1 + {F(x') - I}.We then obtainp(n) = L ik- 5/2W(h, k)e-2ninh/k(11(h, k) + 12(h,k))h, kwhereJZ 2(h'k)11(h, k) = qJk(z)e2nnz/k2 dzzdh,k)andJZ2(h, k) {( (21riH 21r)) }I 2(h, k) = qJk(Z) F exp -k- - - - 1 e2nnz/k2 dz.zt{h, k) ZWe show next that 12 is small for large N. The path of integration in thez-plane can be moved so that we integrate along the chord joining z1(h, k)and z2(h, k). (See Figure 5.8.) We have already estimated the length of this1065.7: Rademacher's convergent series for p(n)oKFigure 5.8chord; it does not exceed 2J2k/N. On the chord itself we have Izi max{lzll, IZ21} < J2k/N. Note also that the mapping w = l/z mapsthe disk bounded by K onto the half-plane Re(w) 1. Inside and on thecircle K we have 0 < Re(z) 1 and Re(l/z) 1, while on K itself we haveRe(l/z) = 1.Now we estimate the integrand on the chord. We have= Iz1 1/2exp{ ReG) - Re(Z)}x e2nltRe 1 this reduces to a formula of J. Lehnerand M. Newman [25],(15)r-1L f(x, y) = f(l, 1) + L L {f(k, r) + f(r, k) - f(k, r - k)}. r=2 k=1(k,r)= 1This relates a sum involving Farey fractions to one which does not.7. Let1S = Ln (b, d) E bd(b + d)(a) Use Exercise 5 to show that 1/(2n - 1) S Sn S 1/(n + 1).(b) Choose f(x,y) = 1/(xy(x + y)) in (15) and show that3 n r 1Sn = - - 2 L L 2 .2 r =1 k =1 r (r + k)(k,r)= 1When n CJJ this gives a formula of Gupta [12],00 r 1 3L L 2 =-.r= 1 k= 1 r (r + k) 4(k,r)= 18. Exercise 7(a) shows that Sn 0 as n Cf:.;. This exercise outlines a proof of theasymptotic formula(16) _ 12 log 2 (log n)Sn - 2 + 0 2n n nobtained by Lehner and Newman in [25].1115: Rademacher's series for the partition functionLetso thatA =rr 1Lk=l r2(r + k)(k,r)= 1I /(d) ,k= 1 dl(r,k) r (r + k)(a) Show thatand deduce thatr>n=" ~ _d_J.l(r_/d_)Ar L. L. 3dl r h = 1 r (h + d)qJ(r) (1 )Ar = log 2 -3 + 0 3 L IJ.l(d) I .r r dlr(b) Show that L ~ = l Ldlr IJ.l(d)/ = O(n log n) and deduce that1 (lOg n)L 3LIJ.l(d)/ = 0 -2 .r> n r dlr n(c) Use the formula Lr ~ n qJ(r) = 3n2/n2+ O(n log n) (proved in [4J, Theorem 3.7) todeduce that(d) Use (a), (b), and (c) to deduce (16).112Modular forms withmultiplicative coefficients6.1 IntroductionThe material in this chapter is motivated by properties shared by the discrimi-nant L\(r) and the Eisenstein series1G2k(r) = L 2k'(m,n)::(O, 0) (m + nr)where k is an integer, k 2. All these functions satisfy the relation(1) f(:: : = + d)' where r is an integer and (: is any element of the modular group r.The function satisfies (1) with r = 12, and G2k satisfies (1) with r = 2k.Functions satisfying (1) together with some extra conditions concerninganalyticity are called modular forms. (A precise definition is given in thenext section.)Modular forms are periodic with period 1 and have Fourier expansions.For example, we have the Fourier expansion,00 = (2n)12 L r(n)e21tint,n=lwhere r(n) is Ramanujan's function, and2(2ni)2k 00 = 2(2k) + (2k _ 1)! (n)e2,,,nt,where (Ja(n) is the sum of the cxth powers of the divisors of n.1136: Modular forms with multiplicative coefficientsBoth r(n) and O"(X(n) are multiplicative arithmetical functions; that is, wehave(2) r(m)r(n) = r(mn) andThey also satisfy the more general multiplicative relations(3) r(m)r(n) = L dl(m,n)and(4) O"(X(m)O"(X(n) = L dl(m,n)for all positive integers m and n. These reduce to (2) when (m, n) = 1.The striking resemblance between (3) and (4) suggests the problem ofdetermining all modular forms whose Fourier coefficients satisfy a multi-plicative property encompassing (3) and (4). The problem was solved byHecke [16J in 1937 and his solution is discussed in this chapter.6.2 Modular forms of weight kIn this discussion k denotes an integer (positive, negative, or zero), H denotesthe upper half-plane, H = {r: Im(r) > O}, and r denotes the modular group.Definition. A function f is said to be an entire modular form of weight k if itsatisfies the following conditions:(a) f is analytic in the upper half-plane H.(b) f(:: : :) = (cr + d)kf(r) whenever (: :) E r.(c) The Fourier expansion of f has the form00f(r) = L c(n)e21tint.n=ONote. The Fourier expansion of a function of period 1 is its Laurentexpansion near the origin x = 0, where x = e21tit. Condition (c) states thatthe Laurent expansion of an entire modular form contains no negativepowers of x. In other words, an entire modular form is analytic everywherein H and at ioo.The constant term c(O) is called the value of f at ioo, denoted by f(ioo).If c(O) = 0 the function f is called a cusp form (" Spitzenform" in German),and the smallest r such that c(r) i= 0 is called the order of the zero of f at i 00.It should be noted that the discriminant is a cusp form of weight 12 witha first order zero at i 00. Also, no Eisenstein series G2k vanishes at i 00.1146.3: The weight formula for zeros of an entire modular formWarning. Some authors refer to the weight k as the "dimension - k"or the "degree -k." Others write 2k where we have written k.In more general treatments a modular form is allowed to have poles inH or at i00. This is why forms satisfying our conditions are called entireforms. The modular function J is an example of a nonentire modular form ofweight 0 since it has a pole at ioo. Also, to encompass the Dedekind eta func-tion there are extensions of the theory in which k is not restricted to integervalues but may be any real number, and a factor e(a, b, c, d) of absolutevalue 1 is allowed in the functional equation (b). This chapter treats onlyentire forms of integer weight with multiplier e = 1.The zero function is a modular form of weight k for every k. A nonzeroconstant function is a modular form of weight k only if k = O. An entiremodular form of weight 0 is a modular function (as defined in Chapter 2) andsince it is analytic everywhere in H, including the point ioo, it must be constant.Our first goal is to prove that nonconstant entire modular forms existonly if k is even and ~ 4. Moreover, they can all be expressed in terms of theEisenstein series G4 and G6. The proof is based on a formula relating theweight k with the number of zeros of f in the closure of the fundamentalregion of the modular group.6.3 The weight formula for zeros of an entiremodular formWe recall that the fundamental region Rr has vertices at the points p, i,P + 1 and ioo. If f has a zero of order r at a point p we write r = N(p).Theorem6.1. Let f be an entire modularform ofweight k which is not identicallyzero, and assume f has N zeros in the closure of the fundamental regionRr , omitting the vertices. Then we have the formula(5) k = 12N + 6N(i) + 4N(p) + 12N(ioo).PROOF. The method of proof is similar to that of Theorem 2.4 where weproved that a modular function has the same number of zeros as poles inthe closure of Rr . Since f has no poles we can writeN = _1 r f'(7:) d7:.2ni JaR f(7:)The integral is taken along the boundary of a region R formed by truncatingthe fundamental region by a horizontal line y = M with sufficiently large M.The path 8R is along the edges of R with circular detours made around thevertices i, p and p + 1 and other zeros which might occur on the edges. By115(6)6: Modular forms with multiplicative coefficientscalculating the limiting value of the integral as M -+ 00 and the circulardetours shrink to their centers we find, as in the proof of Theorem 2.4,k 1 1N = 12 -"2 N(i) - 3N(p) - N(ioo).The only essential difference between this result and the correspondingformula obtained in the proof of Theorem 2.4 is the appearance of the termk112. This comes from the weight factor (cr + d)k in the functional equationf(A(r)) = (cr + d)kf(r),where A(r) = (ar + b)/(cr + d). Differentiation of this equation gives usf'(A(r))A'(r) = (cr + d)kf'(r) ,+ kc(cr + d)k-lfer)from which we findf'(A(r))A'(r) = f'(r) + ~ .f(A(r)) fer) cr + dConsequently, for any path y not passing through a zero we have_1 f f'(u) du = _1 f f'(r) dr + _1 f ~ dr.2ni A(y) feu) 2ni y f(r) 2ni y cr + dTherefore the integrals along the arcs (2) and (3) in Figure 2.5 do not cancelas they did in the proof of Theorem 2.4 unless k = O. Instead, they make acontribution whose limiting value is equal to-k Ii dr -k -k (ni 2ni) k2ni p ~ = 2ni (log i-log p) = 2ni 2 - 3 = 12"The rest of the proof is like that of Theorem 2.4 and we obtain (6), whichimplies (5). DFrom the weight formula (5) we obtain the following theorem.Theorem 6.2(a) The only entire modular forms of weight k = 0 are the constantfunctions.(b) Ifk is odd, ifk < 0, or ifk = 2, the only entire modular form of weight kis the zero function.(c) Every nonconstant entire modular.form has weight k ~ 4, where k is even.(d) The only entire cusp form of weight k < 12 is the zero function.PROOF. Part (a) was proved earlier. To prove (b), (c) and (d) we simply referto the weight formula in (5). Since each integer N, N(i), N(p) and N(ioo) isnonnegative, k must be nonnegative and even, with k ~ 4 if k =1= O. Also,if k < 12 then N(ioo) = 0 so f is not a cusp form unless f = O. D1166.4: Representation of entire forms in terms of G4 and G66.4 Representation of entire forms in termsofG4 and G6In Chapter 1 it was shown that every Eisenstein series Gk with k > 2 is apolynomial in G4 and G6 . This section shows that the same is true of everyentire modular form. Since the discriminant is a polynomial in G4 andG6 ,L\ = g23- 27g32= (60G4)3 - 27(140G6)2,it suffices to show that all entire forms of weight k can be expressed in termsof Eisenstein series and powers of The proof repeatedly uses the fact thatthe product fg of two entire forms f and g of weights WI and W2' respectively,is another entire form of weight WI + W2' and the quotient f /g is an entireform of weight WI - W2 if g has no zeros in H or at ioo.Notation. We denote by Mkthe set of all entire modular forms of weight k.Theorem 6.3. Let f be an entire modular form of even weight k 0 and defineGo(r) = 1 for all r. Then f can be expressed in one and only one way as asum of the type(7)[k/12]f = L r=Ok-12r*2where the ar are complex numbers. The cusp forms of even weight k arethose sums with ao = O.PROOF. If k < 12 there is at most one term in the sum and the theorem canbe verified directly. If f has weight k < 12 the weight formula (5) impliesN = N(ioo) = 0 so the only possible zeros of f are at the vertices p and i.For example, if k = 4 we have N(p) = 1 and N(i) = O. Since G4 has thispropertY,f/G4 is an entire modular form of weight 0 and therefore is constant,so f = aOG4 . Similarly, we find f = aOGk if k = 6,8 or 10. The theoremalso holds trivially for k = 0 (since f is constant) and for k = 2 (since thesum is empty). Therefore we need only consider even k 12.We use induction on k together with the simple observation that everycusp form in Mk can be written as a product L\h, where hEMk-12'Assume the theorem has been proved for all entire forms of even weight 1 prove that-S/2r(S) -s foo -rrn2x s/2- 1 dn -n = e x x2 0and use this to derive the representationn-S/ 2rG)c(s) = faror/J(x)xs/2- 1dx,where 21/J(x) = 9(x)-1.(b) Use (a) and (57) to obtain the representationn-S/2r(:)((s) = __1_ + joo(XS/2-1 + X(1-s)/2-1)I/J(X) dx2 s(s - 1) 1for (J > 1.(c) Show that the equation in (b) gives the analytic continuation of ((s) beyond theline (J = 1 and that it also implies the functional equation (56).141Kronecker's theoremwith applications7.1 Approximating real numbers by rationalnumbersEvery irrational number fJ can be approximated to any desired degree ofaccuracy by rational numbers. In fact, if we truncate the decimal expansionof fJ after n decimal places we obtain a rational number which differs fromfJ by less than 10-n However, the truncated decimals might have very largedenominators. For example, iffJ = n - 3 = 0.141592653 ...the first five decimal approximations are 0.1, 0.14, 0.141, 0.1415, 0.14159.Written in the form alb, where a and b are relatively prime integers, theserational approximations become1 7 141 283 1415910' 50' 1000' 2000' 100,000On the other hand, the fraction 1/7 = 0.142857 ... differs from fJ by less than2/1000 and is nearly as good as 141/1000 for approximating fJ, yet its denomi-nator 7 is very small compared to 1000.This example suggests the following type of question: Given a realnumber fJ, is there a rational number h/k which is a good approximation tofJ but whose denominator k is not too large?This is, of course, a vague question because the terms" good approxima-tion" and "not too large" are vague. Before we make the question moreprecise we formulate it in a slightly different way. If fJ - h/k is small, then(kfJ - h)/k is small. For this to be small without k being large the numeratorkfJ - h should be small. Therefore, we can ask the following question:1427.2: Dirichlet's approximation theoremGiven a real number fJ and given e > 0, are there integers hand k such thatIkfJ - hi < e?The following theorem of Dirichlet answers this question in the affirma-tive.7.2 Dirichlet's approximation theoremTheorem7.1. Given any real fJ and any positive integer N, there exist integershand k with 0 < k ~ N such that(1) 1IkfJ - hi 11a, for example,N = 1 + [l/a]. Then liN < a. Applying Theorem 7.2 with this N we obtaina pair of integers ha