Moonshine of finite groups

EMS Series of Lectures in Mathematics Edited by Andrew Ranicki (University of Edinburgh, U.K.) EMS Series of Lectures in Mathematics is a book series aimed at students, professional mathematicians and scientists. It publishes polished notes arising from seminars or lecture series in all fields of pure and applied mathematics, including the reissue of classic texts of continuing interest. The individual volumes are intended to give a rapid and accessible introduction into their particular subject, guiding the audience to topics of current research and the more advanced and specialized literature. Previously published in this series: Katrin Wehrheim, Uhlenbeck Compactness Torsten Ekedahl, One Semester of Elliptic Curves Sergey V. Matveev, Lectures on Algebraic Topology Joseph C. Várilly, An Introduction to Noncommutative Geometry Reto Müller, Differential Harnack Inequalities and the Ricci Flow Eustasio del Barrio, Paul Deheuvels and Sara van de Geer, Lectures on Empirical Processes Iskander A. Taimanov, Lectures on Differential Geometry Martin J. Mohlenkamp, María Cristina Pereyra, Wavelets, Their Friends, and What They Can Do for You Stanley E. Payne and Joseph A. Thas, Finite Generalized Quadrangles Masoud Khalkhali, Basic Noncommutative Geometry Helge Holden, Kenneth H. Karlsen, Knut-Andreas Lie, Nils Henrik Risebro, Splitting Methods for Partial Differential Equations with Rough Solutions

Koichiro Harada

“Moonshine” of Finite Groups

Author: Prof. em. Koichiro Harada Department of Mathematics The Ohio State University 231 West 18th Avenue Columbus, OH 43210-1174 USA E-mail: [email protected]

2010 Mathematics Subject Classification: 20B05, 11F03 Key words: Monster simple group, congruence groups, modular functions, eta function

ISBN 978-3-03719-090-6 The Swiss National Library lists this publication in The Swiss Book, the Swiss national bibliography, and the detailed bibliographic data are available on the Internet at http://www.helveticat.ch. This work is subject to copyright. All rights are reserved, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, re-use of illustrations, recitation, broadcasting, reproduction on microfilms or in other ways, and storage in data banks. For any kind of use permission of the copyright owner must be obtained. © 2010 European Mathematical Society Contact address: European Mathematical Society Publishing House Seminar for Applied Mathematics ETH-Zentrum FLI C4 CH-8092 Zürich Switzerland Phone: +41 (0)44 632 34 36 Email: [email protected] Homepage: www.ems-ph.org Typeset using the author’s TEX files: I. Zimmermann, Freiburg Printing and binding: Druckhaus Thomas Müntzer GmbH, Bad Langensalza, Germany ∞ Printed on acid free paper 987654321

Preface This is an almost verbatim copy of the lecture notes from a two-quarter course I gave on Moonshine of Finite Groups at the Ohio State University in 1983–84. My original motivation was to understand the moonshine phenomena of the Monster simple group M. A quarter of a century later since, now around the year 2010, it does not seem that we understand them very well, although the main problem that existed then is now a theorem. Namely, the Conway–Norton conjecture has been solved by Richard Borcherds. The idea of his proof has created a new area of mathematics, now called the theory of vertex algebras. This algebraic structure has since become a flourishing area of mathematics for many researchers around the world. On the other hand, the interest generated by researchers in the original mystery of moonshine of the Monster simple group seems to have faded away somewhat, due perhaps to the difficulty of solving it. This is one reason why I am now making the lecture notes available for a wider body of readers. Over the last 25 years since the typed version first appeared, requests for copies of the original lecture notes became fewer but did not die. The original version of these lecture notes included the classification of 0 .N / such that the corresponding Riemann surface 0 .N /nH  (H the upper half plane) has genus 0. This was a home-work project of one of my former students, Kamal Narang, in the class of 1983–84. It was a well-known result and therefore it has been excluded from this published version. I considered the option of including some of the later results in order to make this publication more up-to-date and thus more valuable. However, I finally decided to keep the notes essentially the same as in the original version. As one can see in the literature, for example Gannon’s book (2006) Moonshine beyond the Monster and its bibliography (nearly 600 entries) [10], the horizon encircling ‘moonshine’ is vast, long and deep. It may even appear that we have yet to see a clear direction in which the truth of the moonshine will be found. But I have concluded that this kind of lecture notes should neither risk overwhelming readers with too much expansion, nor misdirect them with too little. It should be noted that many exercises are given in the text, but those are mostly theorems, propositions, or lemmas without proof. Although I did not update the actual mathematical contents of these notes, I did make the Bibliography somewhat longer. My purpose there was to list names of the researchers in this field, but not to pick up their papers in any exhaustive way, far from it as a matter of fact. I apologize if names of important contributors to the field are not mentioned in the Bibliography. I would like to express my sincere gratitude to C. H. Lam who scanned the original typed manuscript and painstakingly converted it to a TEX manuscript, and also to

vi

Preface

N. Chigira who read the manuscript and found many typographical errors. Chigira made also an improvement of one of the theorems in these notes. I would also like to thank the secretaries who typed these notes from my hand written manuscript back in the early 1980s, but their names have now been all but lost from my memory. August 2010

Koichiro Harada

Contents

Preface 1

v

Modular functions and modular forms 1.1 Linear fractional transformations . . . . . . . . . . . . 1.2 Fundamental domains, invariant measures . . . . . . . 1.3 Riemann surfaces associated with Fuchsian groups . . 1.4 Modular functions and modular forms . . . . . . . . . 1.5 Congruence subgroups . . . . . . . . . . . . . . . . . 1.6 Cusps of 0 .N /nH  . . . . . . . . . . . . . . . . . . 1.7 The normalizer of 0 .N / . . . . . . . . . . . . . . . . 1.8 The genus of 0 .N /nH  . . . . . . . . . . . . . . . . 1.9 The genus of nH  , where  D h0 .N /; We ; Wf ; : : :i 1.10 The subgroup njh C e; f; : : : . . . . . . . . . . . . . .

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17 17 18 23 25

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27

3

“Moonshine” of finite groups 3.1 Generalized partitions . . . . . . . . . . . . . . . . . . . . . . . . . 3.2 Harmonies . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3 Symmetric and alternating products of representations . . . . . . . .

31 31 44 47

4

Multiplicative product of  functions

53

2

Dedekind eta function 2.1 The Dedekind eta function .z/ . . . . . . . . 2.2 The Poisson Sum Formula and applications . . 2.3 Theta transformation formula . . . . . . . . . . 2.4 Transformation formula for .t/ . . . . . . . . 2.5 Quadratic reciprocity law, quadratic characters, and Petersson constants . . . . . . . . . . . . .

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Appendix. Genus zero discrete groups

65

Bibliography

67

1 Modular functions and modular forms 1.1 Linear fractional transformations Let GL2 .R/ be the group of all 2  2 real nonsingular matrices. Define GL2 .R/C D f˛ 2 GL2 .R/ j det.˛/ > 0g; SL2 .R/ D f˛ 2 GL2 .R/ j det.˛/ D 1g;   SL2 .Z/ D f˛ 2 SL2 .R/ j ˛ D ac db ; a; b; c; d 2 Zg;  ³ ² a 0 ˇˇ  a 2 S  R  ¹0º D R ; fS g D 0 a ˚  PGL2 .R/C D GL2 .R/C = R ; PSL2 .R/ D SL2 .R/= f˙1 g ; and PSL2 .Z/ D SL2 .Z/= f˙1 g : Let P D C [ f1g be the Riemann sphere. We define the action of GL2 .R/ on P as follows     az C b a b a b zD 2 GL2 .R/C ; z 2 P ; c d c d cz C d with the standard behavior for z D 1. We compute that      .ad  bc/ Im.z/ a b a b z D z ¤ 1: Im ; if c d c d jcz C d j2 Here jzj denotes the absolute value of the complex number z and Im.z/ is the imaginary part. Define H D fz 2 C j Im.z/ > 0g: By the action defined above, we see that H is invariant under GL2 .R/C . GL2 .R/ is a topological group with standard topology induced from R4 , and so SL2 .R/ is also a topological group. We are interested in discrete subgroups of SL2 .R/. PGL2 .R/C Š x denotes the image of PSL2 .R/ acts faithfully on H . For a subset   GL2 .R/C ,  C  in PGL2 .R/ . The bar notation may sometimes be omitted. We will often identify PGL2 .R/C D PSL2 .R/  PSL2 .Z/.

2

Chapter 1. Modular functions and modular forms

1.1 Definition. A subgroup  of SL2 .R/ is discrete if the induced topology of  from SL2 .R/ is discrete. A discrete subgroup of SL2 .R/ is called a Fuchsian group also. 1.2 Exercise. Let ˛ 2 GL2 .R/C . Suppose that ˛ is not a scalar matrix and F .˛/ is the set of fixed points on P by the action of ˛. Then one of the following holds: (1) F .˛/ D fz0 ; zN0 g; z0 2 H , where zN0 is the complex conjugate of z0 , (2) F .˛/ D fxg with x 2 R [ f1g, or (3) F .˛/ D fx; yg, x ¤ y, x; y 2 R [ f1g. ˛ is called elliptic, parabolic, or hyperbolic, according as (1) , (2) or (3) holds. 1.3 Exercise. Show that ˛ 2 GL2 .R/C is parabolic if and only if ˛ is conjugate in    GL2 .R/C to a0 a1 or a0 1 a for some a 2 R . 1.4 Definition. x 2 R [ f1g is a cusp of a Fuchsian group  if x is a fixed point of a parabolic element of . 1.5 Exercise. Let x 2 R [ f1g be a cusp of a Fuchsian group . Show that if x is the stabilizer of x in , then x f˙1g=f˙1g Š Z. Furthermore if ˛ 2 SL2 .R/ and ˛x D 1, then there exists h > 0 such that ˛x ˛ 1 f˙1g D f˙ 10 mh j m 2 Zg. 1

1.2 Fundamental domains, invariant measures We next describe the structure of the quotient space nH where  is a Fuchsian group. 1.6 Definition. A connected subset D of H is called a fundamental domain of  if S (i) H D  2 D, (ii) if U is the set of interior points of D, then D D Ux (closure of U ), (iii) U \ U D  for all  2   f˙1g. 1.7 Example. Let D D fz 2 H j  12  Re.z/  12 ; jzj  1 g. Then D is a fundamental domain of  D SL2 .Z/ (see Serre [23, p. 78, Theorem 1]).

1.2. Fundamental domains, invariant measures

3

1.8 Exercise. Let z be a variable on H , and let x D Re.z/, y D Im.z/. Show that .dx 0 /2 C .dy 0 /2 .dx/2 C .dy/2 D .y 0 /2 y2 and

dx ^ dy dx 0 ^ dy 0 D ; y2 .y 0 /2 0

where z 0 D ˛z for ˛ 2 GL2 .R/C and x D Re.z 0 /, y 0 D Im.z 0 /. A continuous map .t/ from the interval Œ0; 1  R to H is called a curve if .t/ is differentiable except possibly at finitely many points of .0; 1/. By abuse of notation the image C D .Œ0; 1/ is also called a curve. In fact, the term curve will be used more often in the latter sense. If C is a curve, we define Z 1p dt `.C / D xP 2 C yP 2 ; .t/ D x.t / C y.t/i; y 0 and call it the length of C , where the dots denote the derivatives with respect to t. Let z1 , z2 be points on H . We define the distance d.z1 ; z2 / between z1 and z2 by d.z1 ; z2 / D inff`.C / j C ranges over all curves from z1 to z2 g: 1.9 Exercise. Let z1 ; z2 2 H . Show that (1) `.C / does not depend on choice of ; (2) there is ˛ 2 SL2 .R/ such that ˛z1 D i, ˛z2 D i ( > 1), and the line segment between i and i is the unique curve that attains d.i; i/; (3) there exists a unique curve C D .Œ0; 1/ between z1 and z2 such that `.C / D d.z1 ; z2 /. 1.10 Definition. The unique curve C obtained in Exercise 1.9 is called the geodesic between z1 and z2 . 1.11 Exercise. Let  be a Fuchsian group. Show that for each z 2 H , there exists a neighborhood Uz of z such that for  2 , either .Uz / \ Uz D  or z D z holds (we say  acts discontinuously on H ). Use this result to show that the set F D fz 2 H j there exists ˛ D ˛z 2 nf˙1g D   f˙1g such that ˛z D zg is discrete. 1.12 Theorem. Let  be a Fuchsian group and z0 be a point in H not fixed by any element in nf˙1g. Define D D fz 2 H j d.z; z0 / < d.z; z0 / for all  2 g: Then the following holds:

4

Chapter 1. Modular functions and modular forms

S (1) H D  2 D, U \ U D ; if  2 nf˙1g, where U is the set of interior points of D. (2) D is a polygon enclosed by (possibly infinitely many) geodesics. (3) If C is any compact set of H , then C intersects only finitely many geodesics which are the sides of D. (4) D is a fundamental domain for . 

Proof. See [9], [24], [26]. 1.13 Definition. Define the volume of the region nH by Z dxdy : v.nH / D v.D/ D 2 D y

If v.D/ < 1, we say that  is a Fuchsian group of the first kind or a Fuchsian group of finite volume. 1.14 Exercise. Show that v.SL2 .Z/nH / D

 . 3

1.15 Theorem. Let D be a fundamental domain for a Fuchsian group  defined in Theorem 1.12. Suppose that v.nH / < 1. Then D D D0 [ V .x1 / [ [ V .x t / ( finite union) satisfying: (1) The closure of D0 is compact. (2) fx1 ; : : : ; x t g is a set of cusps of  and every cusp of  is equivalent under the action of  to one and only one of the set fx1 ; : : : ; x t g: (3) If gi .1/ D xi for gi 2 SL2 .R/, then V .xi / D V .xi ; i ; / D gi fz 2 H j jRe.z/j <

i 2

; Im.z/ >  0g:

gi

Proof. See [9], [24].



1.3. Riemann surfaces associated with Fuchsian groups

5

1.3 Riemann surfaces associated with Fuchsian groups Let  be a Fuchsian group of finite volume and S be the set of all cusps of . x D f˙1g=f˙1g. Consider the quotient space Put H  D H D H [ S and   R D nH . R can be equipped with a compact Riemann surface structure according to the following procedure. The details may be found in some books (see, e.g., [11], [24]). As a set, R may be viewed equal to the set D D D [ fx1 ; : : : ; x t g with some sides of D identified. Occasionally, however, it may be easier to work with nH  directly. Let be the canonical map from H  onto R D nH  . As  acts discontinuously on H (Exercise 1.11), for each t 2 H , there exists a neighborhood U t of t such that .U t / \ U t D ; or  2  t . If  t ¤ 1, then t is an elliptic point and  t is conjugate to a subgroup of the rotation group SO2 .R/ (rotation group about the point t D i 2 H ). This implies that U t can be chosen to also satisfy  t .U t / D U t . x t D 1. Then t is an ordinary point of  and U t is homeomorphic Suppose that  to .U t /. Hence take f .U t /; z  tg as a local coordinate system for t. We have written t for .t / for simplicity. x t ¤ 1. Then t is an elliptic point and  t is a finite cyclic group. Suppose that  x U t is split into j t j equivalent regions by the rotation group  t .

U t0 

zt ztN

n

t

Let U t0 be one of the regions shown above. Identify two sides via the action of a genera x t . Then .U t0 / is homeomorphic to the resulting cone. Take f .U t0 /; zt n g, tor of  N zt x t j, as a local coordinate system for t. n D j Let t D xi be a cusp of . Let V .xi / be the set described in Theorem 1.15. Put V .xi / D V .xi / [ fxi g. Identify two sides of V .xi / emanating from xi via a xx . Then the resulting surface is of the shape below: generator of  i

6

Chapter 1. Modular functions and modular forms

radius D e 2=i

e 2

p 1gi1 .z/=i

which is homeomorphic to .V .xi / /. If xi D 1, then take f .V .1/ /; e 2 iz=i g as a local coordinate system for xi D 1. Observe that .z/ D e 2 iz=i maps V .1/ D fz 2 H j j Re.z/j < 2 ; Im.z/ > g [ f1g onto an open disc D fz 2 C j jzj < e 2=i g homeomorphically with .1/ D 0. Suppose that xp i ¤ 1. Then 2 1gi1 .z/=i  g choose gi 2 SL2 .R/ such that gi .1/ D xi and take f .V .xi / /; e as a local coordinate system of xi . We have given a local coordinate system fUzt ;  t g to each point t 2 nH  . We need to show that if Uzt \ Uzt 0 ¤ ;, then the product t 0  t1 is a homeomorphism from  t .Uzt \ Uzt 0 / to  t 0 .Uzt \ Uzt 0 /. This is almost obvious and the details may he found in the books [11], [24]. Compactness of R is also obvious by Theorem 1.15. Let k.R / be the function field of the compact Riemann surface R . k.R / is generated by two elements x and y; k.R / D C.x; y/, and there is a polynomial f .x; y/ such that f .x; y/ D 0. The genus of R is 0 if and only if k.R / D C.x/ for some x 2 k.R /. 1.16 Proposition. Let  be a Fuchsian group with finite volume. Let k D k.R / be the function field of the compact Riemann surface R . Then f˙1g D f 2 SL2 .R/ j f B  D f for all f 2 kg: Proof. Without loss of generality we may assume   f˙1g. By definition, k is the set of all meromorphic functions on H  invariant under . We have 1 D f 2 SL2 .R/ j f B  D f for all f 2 kg  . Suppose that 1 © . Let 2 be a subgroup of 1 such that 2   and Œ2 W  D r < 1. Then 2 is also a Fuchsian group with finite volume and so R2 is a compact Riemann surface. Let k2 D k.R2 / be its function field. Then the field index is Œk W k2  D r, since R is an r-fold covering of R2 . By assumption, the elements in k are 1 -invariant and so 2 -invariant. Thus k D k2 and r D 1. This proves that there is no subgroup 2 such that 1  2 ©  and Œ2 W  < 1. Let 1 act on nH D fz j z 2 H g by .z/ D z. Suppose that for some z 2 H the length of the orbit containing z under the action of 1 is finite, and let 2 be the stabilizer of z. We have 2   and 2 .z/ D z. Thus 2 acts on the discrete set z  H .

1.4. Modular functions and modular forms

7

Let .2 /z be the stabilizer of z in 2 . Then 2 D .2 /z . Since Œ2 W  D 1, Œ.2 /z W  \ .2 /z  D 1 must hold. On the other hand,  \ .2 /z is finite. By taking a suitable conjugate, we may assume that z D i and so .2 /z is an infinite subgroup of SO2 .R/ where  ³ ² cos  sin  j 0   < 2 : SO2 .R/ D  sin  cos  Since SO2 .R/ is a rotation group about i, if we choose a suitable neighborhood V of i in the fundamental domain of , then .2 /z y \ V is an infinite set for every y 2 V n fig. Now let f be a nonconstant function of k. By assumption f is constant on all points of .2 /z y \ V , and so f cannot be meromorphic in a small neighborhood of i . This proves that  D 1 . 

1.4 Modular functions and modular forms Let  be a Fuchsian group with finite volume and R D nH  be the corresponding Riemann surface. Let f be an element of k.R /. Then f is a meromorphic function on R . This implies, by definition, that f is a meromorphic function of  t on Uzt for every t 2 R , where fUzt ;  t g is a local coordinate system at t. Put fN D f B

where is the canonical mapping H  ! nH  . 1.17 Exercise. Show: (1) fN.z/ D fN.z/ for all  2 , z 2 H  . (2) fN is a meromorphic function of z, z 2 H . (3) If xi D 1 is a cusp and fQi .q/ D fN.z/ with q D e 2 iz=i then fQi is a meromorphic function of q near 0. (4) Suppose that xi ¤ 1 is a cusp. Then choose gi 2 SL2 .R/ such that gi .1/ D p 2 1gi1 .z/=i Q N a meromorphic function xi . Then fi .q/ D f .z/ with q D e of q near 0. Conversely, show that if a function fN of H  satisfies (1) – (4), then there exists f 2 k.R / such that fN D f B . For a function f .z/ on H [ R [ f1g and an integer k, define k

.f jk ˛/.z/ D .det ˛/ 2 .cz C d /k f .˛z/ where ˛ D

a b  c d

2 GL2 .R/C .

8

Chapter 1. Modular functions and modular forms

1.18 Definition. A function f on H  is a modular function of weight k (with respect to a Fuchsian group ) if the following holds: (1) f jk  D f for all  2 . (2) f is meromorphic on H . (3) If xi D 1 is a cusp, then fQi .q/ D f .z/, q D e 2 function of q for some i > 0.

p

1z=i

, is a meromorphic

Q (4) If xp i ¤ 1 is a cusp and gi .1/ D xi , then fi .q/ D .f jk gi /.z/, q D 1 e 2 1gi .z/=i , is a meromorphic function of q for some i > 0. Moreover, if f and all fQi are holomorphic, then f is a modular form. In addition, if fQi .0/ D 0 at all cusps xi , then f is a cusp form. Remark. Let f .z/ be a modular function of weight 2 with respect to a Fuchsian group  of the first kind. Let ! D f .z/dz and   a b 2 : D c d We have ! B  D f .z/d.z/



 az C b D f .z/.cz C d / d cz C d 2 D f .z/.cz C d / .cz C d /2 dz D f .z/dz D !: 2

This implies that ! D f .z/dz may be viewed as a (meromorphic) differential on nH  . Next suppose that f is a cusp form. If z 2 H , then f .z/dz is obviously a holomorphic differential near z. Let us discuss the holomorphy of f .z/dz at the cusps. p Suppose that xi D 1 is a cusp. Then writing f .z/ D fQi .q/ with q D exp.2 1z= i /, we have ! D f .z/dz D fQi .q/

i p

1 dq: 2 1 q

Since fQi .q/ is a power series in q such that fQi .0/ D 0, fQi .q/=q is holomorphic and so ! is p holomorphic near xi D 1. Next suppose xi ¤ 1 is a cusp. Since q D exp.2 1gi1 .z/= i / is the local coordinate at xi , we have  1 1 dgi .z/ i 1 dq: ! D f .z/dz D f .z/ p q 2 1 dz

1.5. Congruence subgroups

Choose gi D Then

 xi 1

 1 ; 0

gi1 D



9

 0 1 : 1 xi

d.gi1 .z// D .z C xi /2 : dz

Hence ! D f .z/ By assumption

1 i p .z C xi /2 dq: q 2 1

.f j2 gi /.z/ D z 2 f .gi .z//

is a holomorphic function of q 0 D e 2

p

1z=i

near z D 1 and so

.gi1 .z//2 f .z/ D .z C xi /2 f .z/ D fN.q/ is a holomorphic function of q D e 2

p

1gi1 .z/=i

near xi . Since

.f jgi /.1/ D 0; we have fN.0/ D 0 and so ! is holomorphic near xi also. By the argument above, we conclude that the space of all cusp forms of weight 2 with respect to  is “equal” to the space of all holomorphic differentials on the corresponding compact Riemann surface R D nH  .

1.5 Congruence subgroups For each natural number N we define the following subgroups of SL2 .Z/. 1.19 Definition.

 ² ³ a b 2 SL2 .Z/ j c 0 .mod N / ; 0 .N / D c d  ² ³ a b 2 SL2 .Z/ j c 0; a d 1 .mod N / ; 1 .N / D c d  ² ³ a b 2 SL2 .Z/ j b c 0; a d 1 .mod N / : .N / D c d

Obviously, 0 .N /  1 .N /  .N /. (Replacing c 0 by b 0, groups  0 .N / and  1 .N / are also defined, although they are not used in these notes.)

10

Chapter 1. Modular functions and modular forms

1.20 Exercise. Show: (1) SL2 .Z/ F .N /, SL2 .Z/=.N / Š SL2 .Z=.N //, Q ŒSL2 .Z/ W .N / D N 3 pjN .1  p 2 /. (2) 0 .N / F 1 .N /, 0 .N /=1 .N / Š .Z=.N // . Q (3) Œ0 .N / W .N / D .N /N D N 2 pjN .1  p 1 /. Q (4) ŒSL2 .Z/ W 0 .N / D N pjN .1 C p 1 /.

1.6 Cusps of 0 .N /nH  Let D D D be the standard fundamental domain for  D SL2 .Z/ and D  D D [ f1g D fz 2 H j jRe.z/j  12 ; jzj  1g [ f1g:

All cusps of SL2 .Z/ are equivalent to 1. Let  D 0 .N /g1 [ 0 .N /g2 [ [ 0 .N /gr be the coset decomposition of  D SL2 .Z/ by 0 .N /. We have r D ΠW 0 .N /:

1.6. Cusps of 0 .N /nH 

11

S If E D riD1 gi D, then E is a fundamental domain of 0 .N / and every cusp of 0 .N / is equivalent to an element of fg1 .1/; : : : ; gr .1/g. Suppose there exists g 2 0 .N / such that ggi .1/ D gj .1/ for some i, j . Then gj1 ggi 2 1 and so ggi 2 gj 1 where 1 is the stabilizer of 1 in  D SL2 .Z/, and so  ³ ²  1 b jb2Z : 1 D ˙ 0 1 Hence 0 .N /gi 1 D 0 .N /gj 1 . Arguing the reverse way, one concludes that if 0 .N /gi 1 D 0 .N /gj 1 , then the cusps gi .1/ and gj .1/ are equivalent under 0 .N /. Thus the number of inequivalent cusps of 0 .N / is j0 .N /n= 1 j: A set of double coset representatives can readily be obtained by the double coset decomposition of . The procedure is as follows. Let N N dN / 2 Z=.N /  Z=.N / j .c; N dN / D 1g MN D f.c; D f.c; N dN / 2 Z=.N /  Z=.N / j .c; N dN / is of order N g: Define an equivalence relation on MN as follows: N 2 .Z=.N // and .c; N dN / .cN 0 ; dN 0 / if and only if there exist m nN 2 Z=.N / such that cN 0 D m N cN and dN 0 D m N dN C nN c: N Put MN D MN = and let .c; N dN / be the equivalence class containing .c; N dN /. We  define the following map  from 0 .N /n=1 to MN :     a b 1 D .c; N dN / :  0 .N / c d 1.21 Exercise. Show that (1) is an equivalence relation on MN , and (2)  is well defined and  is one to one and onto.   If ac db is a representative of a double coset of 0 .N /n=1 , then the double   coset corresponds to a cusp ac db 1 D ac (if c D 0, ac D 1). The number of inequivalent cusps of 0 .N / is equal to the cardinality of MN . One can get jMN j pairs .c; d / 2 Z  Z such that they constitute a complete set of representatives of MN in the following procedure.

12

Chapter 1. Modular functions and modular forms

1.22 Exercise. Consider a set S of pairs .c; d / satisfying (1) c > 0, c jN , 1 < d < c, .c; d / D 1, and (2) if .c; d /; .c; d1 / 2 S and d1 d .mod .c; Nc //, then d D d1 . P Show that f.c; N dN / j .c; d / 2 S g D MN , jSj D jMN j D cjN ..c; Nc //. Let S be the set described in Exercise 1.22. Pick a such that ad 1 .mod c/. Then the set o na j .c; d / 2 S; ad 1 .mod c/ c is a complete set of representatives of all cusps of 0 .N /. If we choose 1  a < c, then a a1 .mod .c; Nc // implies d d1 .mod .c; Nc // and vice versa. Therefore the set f.c; a/ j .c; d / 2 S; ad 1 .mod c/; 1  a < cg satisfies the condition for S also. Therefore we have proved: 1.23 Proposition. Let S0 .N / be the set of pairs .c; a/ satisfying (0) .1; 0/ 2 S0 .N / ; (1) c > 1, c jN , 1  a < c, .c; a/ D 1, and (2) if .c; a/; .c; a1 / 2 S0 .N / and a1 a .mod .c; Nc //, then a D a1 . Then the set f ac j .c; a/ 2 S0 .N / g is a complete representatives of all inequivalent cusps of 0 .N /. 1.24 Example. Take 0 .18/. Then c D 1, 2, 3, 6, 9, or 18. Hence S0 .N / D f.1; 0/; .2; 1/; .3; 1/; .3; 2/; .6; 1/; .6; 5/; .9; 1/; .18; 1/g 1 g is a complete set of representatives of inequivalent and so f0; 12 ; 13 ; 23 ; 16 ; 56 ; 19 ; 18 cusps of 0 .18/.

1.7 The normalizer of 0 .N / For a pair of natural numbers r and s, we write r jjs if r js and .r; s=r/ D 1. 1.25 Exercise. Show: (1) Let a, d be integers and h be a divisor of 24. Then ad 1 .mod h/ if and only if a d .mod h/ and .ad; h/ D 1. (2) Let h be a natural number. Then the following conditions are equivalent:

13

1.7. The normalizer of 0 .N /

(i) ad 1 .mod h/ if and only if a d .mod h/ for all a; d 2 Z, and (ii) h is a divisor of 24. 1.26 Exercise. Let h be a divisor of 24, h2 j N and e jj hN2 . Let ˛ be a matrix of the following shape

ae b= h ; ˛D cN= h de where a; b; c; d; e 2 Z, and ade 2  cN b= h2 D e. Show that ˛ 1 0 .N /˛ D 0 .N /. 1.27 Definition. Let e jjN . Define

ae b ; We D cN de

a; b; c; d 2 Z; det.We / D e:

We is called an Atkin–Lehner involution of 0 .N /. By Exercise 1.26, we have We1 0 .N /We D 0 .N /. Remark. Since .e; N=e/ D 1, there are integers A, B such that Ae C BN=e D 1 or Ae 2 C BN D e. Writing Ae 2 D a0 e e and B D b 0 c 0 , we see that

0 a e b0 0 We D 0 cN e is also an Atkin–Lehner involution belonging to the same coset of 0 .N / (modulo the center of GL2 .R/C .) In other words, we can assume d D 1 or c > 0 if so desired. 1.28 Exercise. Show that We2 1;

We Wf Wf We Wg .mod 0 .N //;

where g D ef =.e; f /2 :

1.29 Definition. Define 0 .N /C to be the subgroup of GL2 .R/C generated by 0 .N / and all Atkin–Lehner involutions We of 0 .N /. 0 .N /C D h0 .N /; We ; Wf ; : : : i;

e jjN; f jjN; : : : :

1.30 Exercise. Show that the quotient group 0 .N / C =0 .N / is an elementary abelian group of order 2r where r is the number of distinct primes dividing N . 1.31 Definition. N C e; f; : : : ; g D h0 .N /; We ; Wf ; : : : ; Wg i.1 1

The alternative notation N C e C f C    C g may also be used. See Appendix.

14

Chapter 1. Modular functions and modular forms

Remark. One can define the Atkin–Lehner involution We to be p p

a p e b=p e : We D cN= e d e Then det.We / D 1 and so the bar notation such as 0 .N / may be avoided. Usually the action of We or 0 .N / on H is the issue and so the difference between the two definitions and the bar notation is immaterial. 1.32 Proposition. Let  D h0 .N /; We ; Wf ; : : : i. Then the cusps ² ³ 1 xe ; W xf ; : : : i xc 2 hW j there exists W N=c are all the cusps equivalent to 1 among the set of representatives na o j .a; c/ 2 S0 .N / c described in Proposition 1.23. Moreover, f ac j .a; c/ 2 S0 .N /g with denominator Nc .

1 N=c

is the only element in the set

xe ; W xf ; : : : i is written as W xc . Pick any W xc 2 Proof. We know every element of hW xe ; W xf ; : : : i. We may choose hW 

c Wc D N

b dc



1 and hence the first for some b; c 2 Z with cd  b Nc D 1. We have Wc .1/ D N=c N statement holds. Since .c; c / D 1, the condition (2) of Proposition 1.23 implies the second statement. This completes the proof. 

1.8 The genus of 0 .N /nH  Let H  D H [ Q [ f1g and 0 .N /nH  be the corresponding Riemann surface. The genus of 0 .N /nH  is given as follows (see Ogg [20] or Shimura [25]). 1.33 Theorem. If g0 .N / is the genus of 0 .N /nH  then g0 .N / D 1 C

.N / i .N / 0 .N / .N /    12 4 3 2

1.9. The genus of nH  , where  D h0 .N /; We ; Wf ; : : :i

where

Here

15

 Y 1 .N / D N 1C ; p primeI p pjN ´ 0 if 4jN; 4 i .N / D Q .1 C / otherwiseI pjN p ´ 0 if 2jN or 9jN; 3 0 .N / D Q pjN .1 C p / otherwiseI X .N / D ..d; N // where  is the Euler function. d

 

d jN

is the Legendre symbol, which will be briefly discussed in Section 2.5.

1.34 Theorem. g0 .N / D 0 if and only if N D 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 12, 13, 16, 18 or 25. This theorem must have been known for ages.

1.9 The genus of nH  , where  D h0 .N /; We ; Wf ; : : : i The genus of nH  with  D h0 .N /; We ; Wf ; : : : i may be obtained by Table 5 of [3] if N  300. 1.35 Example. N D 100 D 22 52 . There are five groups to consider: 0 .100/, h0 .100/; W4 i, h0 .100/; W25 i, h0 .100/, W4 W25 i, and h0 .100/; W4 ; W25 i. The genus of nH  is equal to the dimension of the space of all holomorphic differentials (differential forms) on H invariant under . The third column of Table 5 of [3] is the W -split of all forms. For N D 100, it is f1; 1; 3; 2g. First of all, this means that 1 C 1 C 3 C 2 D 7 is the genus xe2 2 0 .N /, We acts as an involution on all forms invariant of 0 .100/nH  . Since W under 0 .N /. Hence the eigenvalues of We are 1, 1. The dimensions of the eigenspace corresponding to the pairs of eigenvalues ."1 ; "2 / of W4 , W25 are listed in the order ."1 ; "2 / D .1; 1/, .1; 1/, .1; 1/, .1; 1/. Thus the genus of nH  is 1 C 1 D 2 if  D h0 .100/; W4 i; 1 C 3 D 4 if  D h0 .100/; W25 i; 1 C 2 D 3 if  D h0 .100/; W4 W25 i; 1 if  D h0 .100/; W4 ; W25 i:

16

Chapter 1. Modular functions and modular forms

In the Appendix one finds a list of all possible  D h0 .N /; We ; Wf ; : : : i such that the genus of nH  is 0, where N  300. There are exactly 123 possibilities.

1.10 The subgroup njh C e; f; : : : Let N be a natural number and let h be a positive integer such that h2 jN and hj24. Let n D Nh . Then nh D hN2 is an integer. 1.36 Definition.

1     n h 0 h 0 0 0 1 0 1 h  ² a b= h j a; b; c; d 2 Z; ad  D cn d 

0 .njh/ D

By Exercise 1.26, 0 .njh/  NGL2 .R/ .0 .N //. involution of 0 . nh /. Define 1   h 0 h we D We 0 1 0 

If We D then

ae b cn= h de

bcn h

³ D1 :

Let We be an Atkin–Lehner 

0 1

 with det We D e; e jj nh

 ae b= h : we D cn de 

1.37 Definition. njh C e1 ; e2 ; : : : ; ek D h0 .njh/; we1 ; we2 ; : : : wek i. This expression is not unique, since 30j1 C 3; 5 D 30j1 C 3; 5; 15 for example. However, it always has a unique expression njh C f1 ; : : : ; f` such that any pair of fi , fj (i ¤ j ) is coprime. We will adopt the following convention. (1) Omit h when h D 1; (2) write njhC if all e jj nh are present; and (3) write njh if no e is present. We remark that the genus of nH  with  D njh C e; f; : : : is obviously equal to the genus of 1 nH  where 1 D h0 . nh /; We ; Wf ; : : : ; i, and so it can be found in Table 5 of [3] if N  300.

2 Dedekind eta function 2.1 The Dedekind eta function .z/ For z 2 H , we define .z/ by .z/ D e

 iz 12

1 Y

.1  e 2 i nz /

nD1

An infinite product of complex numbers such as .1 C a1 /.1 C a2 / : : : .1 C an / : : : D

1 Y

.1 C an /

nD1

is evaluated by taking the limit of the partial product Pn D .1 C a1 /.1 C a2 / : : : .1 C an /. It is said to converge to the value P D limn!1 Pn if this 1imit exists and is different from zero. Remark. More generally, limits of infinite products may be defined by eliminating finitely many terms with 1 C an D 0. Q P1 2.1 Theorem. The infinite product 1 nD1 .1Cjan j/ converges if and only if nD1 jan j converges. Q Q1 It is easy to see if 1 nD1 .1 C jan j/ converges, then nD1 .1 C an / also converges (assuming 1 C an 6D 0 for all n). Q 2.2 Theorem. The infinite product of P functions 1 nD1 .1 C jfn .z/j/ converges uniformly on compact sets if and only if 1 jf .z/j converges absolutely and unin nD1 formly. Q1 Q1Again the convergence of nD1 .1 C jfn .z/j/ implies the same assertion for nD1 .1 C fn .z//. The following version will be the easiest to use. Namely, P if functions fn .z/ are defined on a set E and satisfy jf .z/j  M with n n n Mn Q1 converging, then nD1 .1 C fn .z// also converges uniformly on E. For the proof of these theorems, see Ahlfors [1, pp. 191–192] or standard textbooks on complex analysis.  iz Q 2 i nz Now let us come back to the Dedekind .z/ D e 12 1 /. Since nD1 .1  e z 2 H , no factors in the expression of .z/ is zero. Hence the uniform convergence of .z/ follows from the following exercise.

18

Chapter 2. Dedekind eta function

2.3 Exercise. Show that on compact sets.

P1 nD1

e 2 i nz , z 2 H , converges absolutely and uniformly

 iz Q Since the partial product Pn D e 12 nkD1 .1  e 2 izk / is a holomorphic function of z, .z/ is also holomorphic by uniform convergence.

2.4 Theorem. .z/ is holomorphic and nonzero on H . In the next section, we shall prove the transformation formula of .z/ under the action of elements of SL2 .Z/. It is trivial to prove that i

.z C 1/ D e 12 .z/: In the next section, we show . z1 / D e

 i 4

1

z 2 .z/;

where, for a real number r, z r D jzjr e ir arg.z/ ,  < arg.z/  .

2.2 The Poisson Sum Formula and applications Let f .x/ be a continuous function for all real x. We want to evaluate SD

1 X

f .n/:

nD1

There are several versions of the Poisson Sum Formula. We choose the following version and the main body of the argument given here is taken from Rademacher ([22, p. 71, Theorem A]). 2.5 Theorem. Let f .x/ be a twice continuously differentiable function on 1 < x < 1 and assume that Z 1 Z 1 f .x/ dx and jf 00 .x/j dx 1

1

exist. Then

1 X

SD

f .n/

nD1

converges, and we have SD

1 X kD1

Ak

2.2. The Poisson Sum Formula and applications

where

Z Ak D

1

19

f .t /e 2 ikt dt:

1

Proof. Step 1. limx!˙1 f .x/ D limx!˙1 f 0 .x/ D 0. Proof. For a fixed a 2 R, we have Z 1 Z f 00 .x/ dx D lim a

D D Existence of

R1 1

f 00 .x/ dx x!1 a lim Œf 0 .x/  f 0 .a/ x!1 lim f 0 .x/  f 0 .a/: x!1

jf 00 .x/j dx now implies that Z 1 f 00 .x/ dx lim f 0 .x/ D f 0 .a/ C x!1

exists. Existence of

x

R1 1

a

f .x/ dx will then imply Z

xC1

lim

x!1 x

and hence

Z lim

x!1 x

f .x/ dx D 0

xC1

.f .x C 1/  f .x// dx D 0:

By the mean value theorem for integration applied to a function g.x/ D f .x C 1/  f .x/, we obtain Z xC1 .f .x C 1/  f .x// dx D g.c/ D f .c C 1/  f .c/ x

for some c with x < c < x C 1. Using the mean value theorem for differentiation we have lim f 0 . / D 0

x!1

where c < < c C 1 (and so x < < x C 2). Since limx!1 f 0 .x/ exists, we must have lim f 0 .x/ D 0: x!1

Similarly, we obtain

lim f 0 .x/ D 0:

x!1

20

Chapter 2. Dedekind eta function

Next we shall show lim f .x/ D 0:

x!1

From

Z

xC1

f .x/dx D 0;

lim

x!1 x

we get lim f .cx / D 0

x!1

where cx is a number between x and x C 1. Since f .x/ D f 0 . x /.x  cx / C f .cx / for some x such that x < x < cx , we have lim f .x/ D lim .f 0 . x /.x  cx / C f .cx // D 0;

x!1

x!1

since limx!1 f 0 . x / D 0, jx  cx j < 1, and limx!1 f .cx / D 0. Similarly we have lim f .x/ D 0: x!1

Step 2. T D

P1 nD1

f 0 .n/ converges.

Proof. T D

1 X

Z

0

1

f .n/ D

f .x/ dx C 1

nD1

Z

1

D

f 0 .x/ dx C

1

Since

R1 1

1 X

0

1 Z X

0

f .n/ 

nD1 n1

nD1 1  X

f 0 .n/ 

Z

n n1

nD1

f 0 .x/ dx D Œf .x/1 1 D 0 by Step 1, we obtain T D D D D

1 Z X

n

nD1 n1 1 Z n X nD1 n1 1 Z n X nD1 n1 1 Z n X nD1 n1

.f 0 .n/  f 0 .x// dx Z

n



00

f .y/ dy Z

x y

00

n

f .y/ dx

dx  dy

n1

.y  n C 1/f 00 .y/ dy:

f 0 .x/dx

 f 0 .x/ dx :

21

2.2. The Poisson Sum Formula and applications

xDy n n1

n1

n

R1 Since 0  y  n C 1  1, jT j  1 jf 00 .y/j dy < 1. Thus T converges. P Step 3. S D 1 nD1 f .n/ converges. Proof. We have SD

1 X

Z

1

Since

1

Z

n

f .n/ 

f .x/ dx C

nD1

R1

1 ² X

1

f .n/ D

³ f .x/ dx :

n1

nD1

f .x/ dx < 1, S converges if and only if 1 ² X

S1 D

Z

³

n

f .n/ 

f .x/ dx n1

nD1

converges. As in the case of T , we obtain 1 Z X

S1 D

n

.y  n C 1/f 0 .y/ dy:

nD1 n1

We did not assume 2S1 D D

R1 1

f 0 .x/ dx < 1, and so we need to use integration by parts.

1 ² X

Œ.y  n C 1/ f

nD1 1 ² X

2

0

Z

Z .y/nn1

n



00

³

.y  n C 1/ f .y/ dy 2

n1

³ .y  n C 1/ f .y/ dy :

n

f .n/ 

nD1

0

2

00

n1

P 0 Since 1 nD1 f .n/ converges and 0  y  n C 1  1, we conclude S1 and hence S converges. Replacing x by x C u with u 2 R in the preceding argument, we conclude that 1 X nD1

f .n C u/

22

Chapter 2. Dedekind eta function

and 1 X

f 0 .n C u/

nD1

converge. The uniform convergence for the first sum depends on the same condition for 1 Z n X .y  n C 1/f 00 .y C u/ dy: nD1 n1

We compute 1 Z ˇ X ˇ ˇ

n

nD1 n1 1 ˇ X

ˇ Dˇ

2

1 Z X

.y  n C 1/f 00 .y C u1 / dy  Z

n

nD1 n1

n

nD1 n1 Z 1 00

ˇ ˇ .y  n C 1/f 00 .y C u2 / dy ˇ

ˇ ˇ .y  n C 1/.f 00 .y C u1 / dy  f 00 .y C u2 // dy ˇ

jf .y/j dy:

1 1 X

Similarly we conclude that the sum

f 0 .n C u/ converges uniformly on u. In

nD1

particular, if we put 1 X

S.u/ D

f .n C u/

nD1

then the derivative S 0 .u/ is obtained by termwise differentiation. Thus 1 X

S 0 .u/ D

f 0 .n C u/:

nD1

Since S 0 .u/ is also uniformly convergent, S 0 .u/ is continuous also. S is periodic with a period 1, and so we have 1 X

S.u/ D

Ak e 2 iku

kD1

with

Z Ak D 0

1

S.v/e 2 ikv dv

()

23

2.3. Theta transformation formula

As S.v/ converges uniformly, Z 1 X 1 f .n C v/e 2 ikv dv Ak D 0 nD1 1 Z 1 X

D D

f .n C v/e 2 ikv dv

nD1 0 1 Z nC1 X

f .w/e 2 ikw dw; w D n C v;

nD1 n 1

Z

f .w/e 2 ikw dw:

D nD1

Put u D 0 in () to complete the theorem.



The following is also a Poisson Sum Formula under a weaker condition. We will not need this theorem. 2.6 Theorem. P Suppose that f .t / is a continuous function for 1 < t < 1. Suppose also that 1 mD1 f .t C m/ is uniformly convergent for 0  t  1. Then 1 X

1 X

f .t C m/ D

mD1

e 2 imt

Z

1

f .x/e 2 imx dx 1

mD1

for all t such that the right-hand side converges. 

Proof. See [16].

2.3 Theta transformation formula 2.7 Theorem. For all complex numbers z and t with Re.t/ > 0, we have 1 X

e .nCz/

2t

1 Dp

nD1

Proof. Put .t; z/ D

1 X

t

e

 n2 C2 i nz t

:

nD1

1 X

2

e .nCz/ t :



nD1

2.8 Exercise. Show: 2

2

(1) There exist M; N > 0 such that je .nCz/ t j  e M n if n > N .

24

Chapter 2. Dedekind eta function

(2)

.t; z/ converges for all t; z 2 C such that Re.t/ > 0, and analytic function with respect to t and z.

Assume now that t; z 2 R (hence t > 0) and put f .x/ D e .xCz/ z fixed.

.t; z/ is an 2t

with t and

2.9 Exercise. Show: (1) f .x/ is continuous for 1 < x < 1 and f .x/ is twice continuously differentiable. R1 R1 (2) 1 f .x/ dx and 1 jf 00 .x/j dx exist. By Exercise 2.9, f .x/ satisfies all conditions of Theorem 2.5. Hence 1 X

f .n/ D

nD1

Z Ak D

(1) Ak D e 2 ikz

k 2 t

e

.nCz/2 t

nD1

where

2.10 Exercise. Show:

1 X

R1 1

1

D

1 X

Ak ;

kD1

2

e .xCz/ t e 2 ikx dx:

1

2

e  t w dw, where w D x C z C

ik . t

(2) Use Cauchy’s Theorem and let s ! 0 on the circuit C below and show that Z 1 Z 1 2  t w 2 e dw D e kt v dv D p : t 1 C k=s

_s

s

By Exercise 2.10, we have 1 k 2 Ak D p e 2 ikz t ; t

z; t 2 R; t > 0:

25

2.4. Transformation formula for .t/

Hence

1 X

e .nCz/

1 1 X  k2 C2 ikz Dp e t : t kD1

2t

nD1

A simple algebraic calculation now implies that e z .t; z/ D p t

2t

. 1t ; izt/;

z; t 2 R; t > 0:

()

By Exercise 2.8, we know that both sides of () are analytic functions of t; z 2 C if Re.t/ > 0 (hence Re. 1t / > 0 also). Thus by analytic continuation ./ holds for all t; z 2 C, Re.t/ > 0.

2.4 Transformation formula for .t/ 2.11 Theorem. For z 2 H , we have 1

1

. z1 / D .i/ 2 z 2 .z/: Proof. We use the following identity due to Euler: 1 Y

1 X

.1  x / D n

 iz 12

n.3nC1/ 2

for jxj < 1:

nD1

nD1

Since .z/ D e

.1/n x

Q1

nD1 .1

.z/ D e De De

 e 2 i nz /,  2 H , we have

 iz 12

 iz 12

 iz 12

1 X

.1/n e  izn.3nC1/

nD1 1 X nD1 1 X

.1/n e  izn.3n1/ 1

1

e 3 izfn 6 .1 z /g

2   iz .1 1 /2 z 12

nD1

De

 i 12z

C 6i

1 X

1

1

2

e 3 izfn 6 .1 z /g :

nD1

2.12 Exercise. Use Theorem 2.7 to show 1 X nD1

1

1

2

e 3 izfn 6 .1 z /g D p

1 3iz

1 X nD1

e

 i n2  3i n .1 3z

 z1 /

:

26

Chapter 2. Dedekind eta function

Hence i

.z/ D e  12z C

i 6

p

1 X

1 3iz

e

1  i n2 in 3z  3 .1 z /

:

nD1

Now we break up the summation on n according to mod 3: 1 X

D g0 .z/ C g1 .z/ C g2 .z/

nD1

where gk .z/ D

1 X

e

 i.3nCk/2 3z

e

 i.3nCk/ .1 z1 / 3

;

k D 0; 1; 2:

nD1

2.13 Exercise. Show 2 n/ P n  i.3n z (1) g0 .z/ D 1 ; nD1 .1/ e  i

(2) g1 .z/ D e 3 g0 .z/; P1 n n.nC1/ (3) D 0 for jxj < 1; nD1 .1/ x (4) g2 .z/ D 0. Thus  i

1  i .g0 .z/ C e 3 g0 .z// p 3iz 1 1 i i g0 .z/.e 6 C e  6 / p p 3 iz

.z/ D e 12z C  i

D e 12z Dp Dp Dp Dp Dp 1

1

1 iz 1 iz 1 iz 1 iz 1 iz

Hence . z1 / D .i/ 2 z 2 .z/.

i 6

 i

e 12z g0 .z/ e

 i 12z

1 X

 i

nD1 1 X

 i

nD1 1 Y

e 12z e 12z

.1/n e .1/n e

.1  e

 i z

.3n2 n/

 i.3n2 n/ z

2n i z

/

(Euler)

nD1

. z1 /: 

2.5. Quadratic reciprocity law, quadratic characters, and Petersson constants

27

2.5 Quadratic reciprocity law, quadratic characters, and Petersson constants Let a, m be integers such that .a; m/ D 1. If there is an integer solution to the equation x 2 a .mod m/; we say that a is a quadratic residue mod m; otherwise we say that a is a quadratic nonresidue mod m. 2.14 Definition. (1) Let p be an  odd prime and a be an integer such that .a; p/ D 1. We define Legendre’s symbol pa by   ´ a 1 if a is a quadratic residue mod p, D p 1 otherwise. (2) Let b bean odd positive integer and a be an integer such that .a; b/ D 1. If b D 1, we put a1 D 1, otherwise write b D p1 p2 : : : ps where pi are prime, not necessarily distinct. Define       a a a D ::: ; p1 ps b     where pai ’s are Legendre’s symbols. This (extended) symbol ab is called Jacobi’s symbol. Obviously Jacobi’s symbol and Legendre’s symbol coincide if b is an odd prime. 2.15 Theorem. Let m, m0 be odd positive integers and n, n0 be integers satisfying .n; m/ D .n0 ; m/ D .n; m0 / D .n0 ; m0 / D 1. Then:  n  n0  D m if n n0 .mod m/. (a) m  n  n   n   n n0  nn0  D mm0 ; m m D m . (b) m m0 (c)

1 m

D .1/

m1 2

;

2 m

D .1/

m2 1 8

.

(d) (Law of Quadratic Reciprocity) If m; n > 0 with n odd, then    n m n1 m1 D .1/ 2  2 : m n 2.16 Corollary. If m and n are odd integers and .m; n/ D 1, then    sgn.m/1 sgn.n/1 m n m1 n1 D .1/ 2  2 C 2  2 : jmj jnj

28

Chapter 2. Dedekind eta function

2.17 Definition. Let c, d be integers such that .c; d / D 1 and d is odd. Define     c c D d jd j     sgn.c/1 sgn.d /1 c c .1/ 2  2 : D d  jd j

and

Furthermore define



0 ˙1



 D 1;

0 ˙1

 D 1: 

Now we state the following theorem (Petersson [21]). 2.18 Theorem. Let .z/ be the Dedekind eta function and M D Then 1 .M z/ D v .M /.cz C d / 2 .z/

a b  c d

2 SL2 .Z/.

where

´  d exp¹ 12i Œ.a C d /c  bd.c 2  1/  3cº if c is odd, v .M / D  cc  i 2 exp¹ 12 Œ.a C d /c  bd.c  1/ C 3d  3  3cd º if c is even: d 

Remark. If c is even then d is odd (ad  bc D 1.) Moreover, if cd D 0, then ad  bc D 1 implies, jcj D 1 or jd j D 1. Hence Definition 2.17 covers all necessary cases. See also Rademacher [22]. The following lemma is useful. 2.19 Lemma. Suppose that c; d are both odd integers. Then the first expression of v .M / and the second are equal. Hence in this case   n i o d exp Œ.a C d /c  bd.c 2  1/  3c v .M / D c 12   n o

i c exp Œ.a C d /c  bd.c 2  1/ C 3d  3  3cd  : D 12 d 

Proof. Suppose that c; d are both odd. We need only to show   n i n i o c o d exp exp .3c/ D .3d  3  3cd / : c 12 12 d 

29

2.5. Quadratic reciprocity law, quadratic characters, and Petersson constants

We know

c

c  d  d  c

D

jd j

.1/

sgn.c/1 sgn.d /1  2 2

D .1/

d 

c1 d 1 2  2

jcj

by Corollary 2.16. Hence  ˚ d  o n i exp 12i .3c/ c1 d 1 c ˚ i  D .1/ 2  2 exp c  .c  1/.d  1/ 4 exp 12 .3d  3  3cd / d  D .1/2

c1 d 1 2  2

D 1;

as desired.

p



We shall next discuss the quadratic characters (see p. 237 of [4]). Let K D Q. d / be a quadratic field with a square free integer d ¤ 1, and let D be the discriminant of K (i.e., D D d if d 1 .mod 4/ and D D 4d if d 2; 3 .mod 4/.) Let x be an integer such that .x; D/ D 1. Define a function  from Z to f1; 1g as follows. 2.20 Definition.

8  x ˆ ˆ < jd j x1   x .x/ D D .x/ D .1/ 2 jd j ˆ 2 ˆ d 0 1  :.1/ x 81 C x1 2  2

if d 1 .mod 4/; 

x jd 0 j

if d 3 .mod 4/; if d D 2d 0 :

Note that if d 2; 3 .mod 4/, then D 0 .mod 4/ and so x must be odd. 2.21 Exercise. Show: (1) If x x 0 mod jDj, then .x/ D .x 0 /. (2) The induced map N W .Z=.D// to the group f1; 1g is a homomorphism. Now define D .x/ D 0 if .x; D/ ¤ 1 Then D is ap Dirichlet character modulo jDj. D is called the quadratic character of the field Q. d /. 2.22 Exercise. Let a be an odd positive integer such that .a; d / D 1. Show that   d D .a/ D : a

3 “Moonshine” of finite groups 3.1 Generalized partitions As mentioned in the Introduction, let F be the set of functions f .z/ satisfying the following conditions: (1) f .z/ is a modular function with respect to some discrete subgroup  of SL2 .R/ that contains 0 .N / for some N . (2) The genus of nH  is 0 and its function field is equal to k.nH  / D C.f /. (Recall that H is the upper half plane and the  notation indicates: cusps are adjoined.) (3) In a neighborhood of 1, f .z/ is expressed in the form 1

1 X an q n ; f .z/ D C q nD0

q D e 2 iz ; z 2 H; an 2 C:

We say a pair .G; / is a moonshine for a finite group G (see Kondo [17]) if  is a function from G to F and if, for 2 G, 1 X 1  .z/ D C a0 . / C an . /q n ; q nD1

q D e 2 iz ;

then the mapping ! an . / from G to C is a generalized character of G. In particular,  is a class function of G. Finding or constructing a moonshine .G; / for a given group G involves some nontrivial work. For each element of G, we have to find a natural number N and a Fuchsian group  containing 0 .N / in such a way that its function field k. nH  / is equal to C. / and that the coefficient an . / of the expansion of  .z/ at 1 are generalized characters of the finite group G for all n  1. In the Appendix, we provide a list of all Fuchsian groups  of genus 0 such that 0 .N /    0 .N / C for some N . There are exactly 123 possible ’s. Some subgroups, and some conjugates of those 123 ’s are, in practice, the only Fuchsian groups that could be used for a moonshine of a finite group G.

32

Chapter 3. “Moonshine” of finite groups

Once  is chosen in some ‘judicious’ way, then we will need to find f .z/ 2 F such that k.nH  / D C.f /. In the most known cases, f .z/ is expressed as a product of some Dedekind  functions, for instance f .z/ D .2z/3 .3z/9 =.z/3 .6z/9 ;  D 0 .6/: Q 3.1 Definition. A symbol D t t r t is a generalized partition (generalized permuQ tation in [17]) if t 2 N, r t 2 Z for all t and t t r t is a product of finitely many t P 1P for finitely many t). Define deg. / D r t, wt. / D r (i.e., r t D 0 except t t t t 2 Q and sgn. / D t .1/r t 1 . Q 3.2 Definition. For a generalized partition D t t r t , define Y  .z/ D .tz/r t : t

3.3 Lemma. Suppose that D condition holds.

Q t

t r t is a generalized partition. Then the following

(1)  .z/ is holomorphic and nonzero on H .   (2) If ac db 2 SL2 .Z/ and c is divisible by every t in the expression of , then 



az C b cz C d

where C D



D C.cz C d /wt./  .z/

Y

 v

t

a c=t

bt d

r t :

Proof. Since .z/ is holomorphic and nonzero on H by Theorem 2.4, the same property holds for  .z/. This is (1). As for (2), we first note that v is the (Petersson’s) constant given in Theorem 2.18. We compute   Y   az C b r t az C b   t D cz C d cz C d t r t  Y atz C tb D  .c=t/tz C d t 

r t  Y 1 a bt 2 .cz C d / .tz/ D v c=t d t

1

D C.cz C d / 2 where C D



Q t

v

a bt c=t d

r t

P

rt

 .z/;

: This completes the proof.



33

3.1. Generalized partitions

To deal with cases with t not dividing c for some t , we need a lemma.   3.4 Lemma. Let A D ac db 2 SL2 .Z/, and t 2 Z. Then there exists A t D   at bt 2 SL2 .Z/ and ˛ t ; ˇ t 2 N, ı t 2 Z such that ct dt    at a b t 0 D c d 0 1 ct



bt dt



 ˇt : ıt

˛t 0

In particular, t D ˛ t ı t , c D c t ˛ t , c t ˇ t C d t ı t D d . Moreover, ˛ t D .t; c/ holds. Proof. Let ˛ D .t; c/ > 0 and v be a number such that cv ˛ .mod t/. Then   ta  a b D ˛c c d ˛



˛b  adv d˛  cdv t t

t 0 0 1



 ˛ dv : 0 ˛t 

Since v can be chosen so that dv > 0, the proof is complete. 3.5 Lemma. Let A D partition. Then 



az C b cz C d

a b  c d



2 SL2 .Z/ and let D

1

D C.cz C d / 2

P

 rt

exp

Q t

t r t be a generalized



iz X .t; c/2 r t g.z/; 12 t t

where C is a constant, and g.z/ is a holomorphic function of e 2 iz= h for some integer h in some neighborhood of z D 1. Proof. This result follows from a direct computation using Lemma 3.4. We have 



az C b cz C d



Y  az C b r t  t cz C d t    r t Y at bt ˛t ˇt D  z ct dt 0 ıt t ´

12  Y ˛ t ˇ t  ˛t 0 DC z C dt  ct 0 ıt 0 D

t

where C 0 D

Q

 t v

 ct

˛t 0

at bt ct dt

 μr t ˇt z ıt

r t . We have

 ˛t ct z C ˇt ct C dt ıt cz C d ˇt D : z C dt D ıt ıt ıt

34

Chapter 3. “Moonshine” of finite groups

Therefore, 



az C b cz C d



00

D C .cz C d /

1 2

1

D C 00 .cz C d / 2 where

P

P

rt

rt

Y  ˛ t z C ˇ t r t  ıt t  X

i ˛t z C ˇt exp r t g.z/; 12 t ıt

   r t 1  Y Y ˛t z C ˇt 1  exp 2 i n : g.z/ D ıt t nD1

Let h > 0 be the least common multiple of all ı t . Then g.z/ is a holomorphic function of e 2 iz= h in some neighborhood of z D 1. On the other hand, 



i X ˛t z C ˇt

iz X ˛ t2 000 r t D C exp rt exp 12 t ıt 12 t t

D C 000 exp





iz X .t; c/2 rt : 12 t t

This completes the proof. Q 3.6 Theorem. Let D t t r t be a generalized partition. Suppose that P (1) t r t D 2k, k 2 Z, P (2) t t r t 0 .mod 24/. Let f be a (unique) square free integer such that Q (3) t t jr t j D m2 f , m a natural number. Choose a natural number N satisfying (4) r t D 0 if t −N , P N (5) t t r t 0 .mod 24/, (6) N 0 .mod 4/ if f .1/kC1 .mod 4/, N 0 .mod 8/ if f 2 .mod 4/.  a b Then, if A D cN 2 0 .N /, we have d  .Az/ D .d /.cN z C d /k  .z/; p where  is the quadratic character of Q. "f / with " D .1/k defined mod N . Suppose furthermore,



35

3.1. Generalized partitions

(7)

P t

.t;c/2 rt t

 0.

Then  .z/ is holomorphic at the cusp then  .z/ vanishes at the cusp ac .

a c

of 0 .N / and if  is replaced by >,

Remark. The conditions (3) and (4) imply that f j N . The condition (6) implies p 4"f j N if "f 2; 3 .mod 4/, and so D j N if D is the discriminant of Q. "f /. Therefore, the character  is defined mod N also. Let Y  D .1/k t rt : p p   Then Q. "f / D Q. / and .d / D for all odd positive integers d such that d .d; N / D 1. In this case we may write    .cN z C d /k  .z/:  .Az/ D d 3.7 Definition. The level of  .z/ is the smallest integer N satisfying the conditions (4), (5) and (6) of the previous theorem. Proof of the theorem. Let be a generalized partition satisfying (1), (2), and assume that N is chosen to satisfy (4), (5), and (6). By Lemma 3.3, we know  .Az/ D C.cN t C d /k  .z/ where C D

Y t

 v

a cN=t

bt d

r t :

p We shall show that C D .d / where  is the quadratic character of Q. "f / with " D .1/k . Case 1. c D 0 (hence a D d D ˙1). n i o

Y  0 r t exp Œbtd C 3d  3r t C D 12 d  t i h i 3.d  1/2k D exp 12 ´ 1 if d D 1; D k if d D 1: .1/ Case 2. cN odd (hence cN=t odd for all t).

36

Chapter 3. “Moonshine” of finite groups

We have # Y  d  r t C D cN=t t n i h X io XN t rt Œa C d  bdcN  3c r t C bd exp 12 t t t Y  d  jr t j Y  d jr t j D D cN=t jcj N=t t t "   # "  #  Y d jr t j Y d jr t j D jcj N=t t t     Y d jr t j Y d jr t j  d  : D D D t Nt f t t "

Case 3. cN even (hence d odd). We have 

Y cN=t r t C D exp A d  t where AD

X X i XN

i h t r t C .3d  3/ rt : .a C d  bdcN  3d /c r t C bd 12 t t t t

Therefore, we have



d 1 d 1 exp A D exp i k D .1/ 2 k : 2 On the other hand, Y  cN=t r t d

t



D

Y  cN=t r t

.1/ jd j Y  cN t jr t j  f  : D D jd j jd j t  C D

f  1

.1/

11 2 k

D 1 and



t

Thus

Since

sgn.c/1 sgn.d /1  r t 2 2



f jd j

 .1/

11 f  .1/ 2 k j1j

d 1 2 k

:

D .1/k , the last case actually covers

3.1. Generalized partitions

37

the case c D 0. Summarizing, we have obtained 8  d ˆ if cN is odd, ˆ <f  f if cN is even, keven, C D jd j ˆ   ˆ d 1 f : .1/ 2 if cN is even, kodd. jd j

Therefore, p we need only to show that C D .d / where  is the quadratic character of Q. "f / with " D .1/k . Suppose that k is odd and f 1 .mod 4/. Then f 3 .mod 4/ and so N 0 .mod 4/. By Definition 2.20, we have       f f f 1 d 1 d d 1 d 1 d 1  .d / D .1/ 2 D .1/ 2 .1/ 2 2 D .1/ 2 D C: jf j jd j jd j  3.8 Exercise. Show that C D .d / for all other cases as well. In order to complete the proof of Theorem 3.6, we must show that if satisfies the  condition (7) then  .z/ is holomorphic, or vanishes at the cusp ac . Let A D ac db 2 SL2 .Z/. Then Lemma 3.5 implies that . jk A/.z/ is holomorphic with respect to P .t;c/2 r t > 0. e 2 iz= h for some h and also . jk A/.1/ D 0 if t 3.9 Corollary (M. Newman [19]). Let D that P (1) r t D 0, P (2) t r t 0 .mod 24/, Q jr t j is a square. (3) t

Q

t r t be a generalized partition. Suppose

Choose a positive integer N satisfying (4) r t D 0 if t jN , PN r t 0 .mod 24/. (5) t Then  .z/ is invariant under 0 .N /. 3.10 Exercise. Use Corollary 3.9 to show that  .z/ is invariant under 0 .N / where fN; g are listed below:

38

Chapter 3. “Moonshine” of finite groups

N



2

124 =224

3

112 =312

4

18 =48

5

16 =56

6

23 39 =13 69 , 28 34 =14 68 , 15 3=2 65

7

14 =74

8

14 42 =22 84

9

13 =93

10

2 55 =1 105 , 24 52 =12 104 , 13 5=2 103

12

44 62 =22 124 , 33 4=1 123 , 13 4 62 =22 3 123

13

12 =132

16

12 8=2 162

18

6 93 =3 183 , 22 9=1 182 , 12 6 9=2 3 182

25

1=25

 ae b  is an Atkin–Lehner involution of 0 .N / if a; b; c; We recall that We D cN de d; e 2 Z, det We D e, and .e; N=e/ D 1. Q Corollary 3.9 is useful when we wish to show that  .z/ D .tz/r t is invariant under 0 .N /. However, there are not many N such that 0 .N /nH  is of genus 0 (see Theorem 1.34). Therefore, we need to investigate a larger class of Fuchsian groups. Typically we deal with a group  D h0 .N /; We ; Wf ; : : : i where We , Wf , : : : are Atkin–Lehner involutions of 0 .N / (or conjugates of such ). We will give a set of conditions on in order for  .z/ to be invariant under 0 .N / and We . Firstly, we have 3.11 Definition. Let be a generalized partition and e be a natural integer. Define a new generalized partition  e by 

eD

Y  et t

.e; t/2

r t

39

3.1. Generalized partitions

 ae b  be an Atkin–Lehner involution of 0 .N /. Let t 3.12 Exercise. Let We D cN de be a positive integer such that t jN and z 2 H . Show 

aez C b  t cN z C de





cN z C de D v .M / ı

where ı D .e; t/ and

" M D

aı cN ı et

bt ı de ı

1  2

et  2 z ı



# 2 SL2 .Z/:

3.13 Exercise. Show that  .z/ with D 14 54 24 104 is invariant under h0 .10/; W5 i. 3.14 Theorem. Suppose that a generalized partition satisfies the following condition: P (1) t r t D 0, P (2) t t r t 0 .mod 24/, Q jr t j (3) t t is a square. Moreover, suppose that natural numbers e, N with e jjN satisfy (4)  e D , Q (5) t .e; t/r t D 1, (6) r t D 0 if t −N , P N (7) t t r t 0 .mod 24/, P e (8) if e is odd, t .e;t/ r t 0 .mod 8/, (9) if e is even, (10) if e is odd,

P t

r t 0 .mod 8/,

 N.e;t/ r t

Q

(11) if e is even,

N.e;t/ et et e .e;t/

t



Q t

e .e;t/ N.e;t/ et

r t

D 1, D 1,

  where  is the Jacobi symbol. Then  .z/ is invariant under h0 .N /; We i.

40

Chapter 3. “Moonshine” of finite groups

Proof. Firstly we note that  .z/ is invariant under 0 .N / by Corollary 3.9 and our conditions (1), (2), (3), (6), and (7). By the remark on p. 13 right after Definition 1.27, we assume that our Atkin–Lehner involution We satisfies d D 1. Using Exercise 3.12 and by (4), (1), and (5), we compute    r t Y cN z C e 1=2 etz v .M t /  .e; t/ .e; t/2 Y 1 1 D .v .M t /r t .cN z C e/ 2 r t .e; t/ 2 r t / .z/  Y D v .M t /r t  .z/;

 .We z/ D

where

" Mt D

a.e; t/ cN.e;t/ et

bt .e;t/ e .e;t/

# :

Suppose first that e is odd. Then Y

.v .M t //

rt

D

Y

cN.e;t/ !r t et e .e;t/ 

exp

h i i A : 12

We compute AD

X t

D

X ² acN.e; t/2 et

t

D acN

ac



cN.e; t/ bt e  et .e; t/ .e; t/ cN.e; t/ e e 33 rt C3 .e; t/ et .e; t/ e a.e; t/ C .e; t/

X .e; t/2

cN.e; t/ et

2

 1

³ e cN bc 2 N 2 bte cN C3 C  C 33 rt t et .e; t/2 .e; t/ t

X et bc 2 N X N rt rt C b et t e t .e; t/2 t t t t X e X XN C3 r t  3c rt  3 rt .e; t/ t t t t

XN t



t

rt C c

rt C b

0 .mod 24/:

X t

XN

rt 

t r t .mod 24/;

41

3.1. Generalized partitions

Here the conditions (1), (2), (4), (7) and (8) are repeatedly used. We next evaluate Y

cN.e;t/ et e .e;t/

t

!r t D 

D

Y t

cN.e;t/ et e .e;t/

Y

c

t

e .e;t/

D

Q

!r t

!

c

N.e;t / et e .e;t/

!

ert t .e;t/r t

Y t

!!r t

N.e;t / et e .e;t/

!r t

D 1 by (1), (5), (10): This completes the proof that  .z/ is invariant if e is odd. Assume next that e is even. Then ae  cNe b D 1 implies that b t Therefore, cN.e;t/ D cN = .e;t/ is odd also. Thus et e Y

v .M t /r t D

e .e;t/

Y

!r t exp.

cN.e;t/ et

cN e

is odd.

i B/; 12

where BD

X ² acN.e; t/2 et

t

D acN

X .e; t/2 et

t

Cb

X t

ac

XN t

t

³ cN bc 2 N 2 bte cN.e; t/ C  C rt 3 t et .e; t/2 et

rt C c

XN t

et r t  3c .e; t/2

rt C b

X

rt 

bc 2 N X N rt e t t

t X N.e; t / t

et

rt

t rt ;

t

0 .mod 24/ (by (2), (4), (7), (9).) Finally we have arrived at the computation of the remaining factor. For our Atkin–Lehner involution We , we may further assume c > 0 and so the asterisk  is not necessary.

42

Chapter 3. “Moonshine” of finite groups

As before, we compute Y t

e .e;t/ cN.e;t/ et

!r t D

!r t e .e;t/ cN.e;t/ et ! e .e;t/

Y t

D

Y t

Q

D D1

c ert t .e;t/r t

c

!!r t e .e;t/ N.e;t / et ! !r t e Y .e;t/ N.e;t / t et

by (1), (5), (11):

This completes the proof of the theorem.



Remark. The author expresses his gratitude to Naoki Chigira who has shown an improvement of the previous theorem. The author’s original version did not include: if e is odd in (8), if e is even in (9), if e is odd in (10), and if e is even in (11). This will have an effect on the computation of the following exercise. 3.15 Exercise. Use Theorem 3.14 and show that  .z/ is invariant under  where f; g is any of the following pairs. 



10 C 2

12 22 =52 102

10 C 5

14 54 =24 104

10 C 10

26 56 =16 106

20 C 4

12 42 102 =22 52 202

20 C 20

42 52 =12 202

20 C

28 108 =14 44 54 204

30 C 15

3 5=2 30, 1 62 102 15=22 3 5 302 , 12 6 10 152 =22 3 5 302

30 C 6; 10

13 63 103 153 =23 33 53 303

30 C 3; 5

1 3 5 15=2 6 10 30

30 C 2; 15

3 5 6 10=1 2 15 30

30 C 5; 6

22 32 102 152 =12 52 62 302

3.1. Generalized partitions

50 C 50

2 25=1 50

60 C 12; 15

1 12 15 20=3 4 5 60

60 C 4; 15

2 3 5 12 20 30=1 4 6 10 15 60

60 C

22 62 102 302 =1 3 4 5 12 15 20 60

70 C 10; 14

1 10 14 35=2 5 7 70

43

Note. The list above is taken from Table 3 of Conway–Norton [6]. This is only a partial list from [6]. An expression such as 30 C 6; 10 is not unique. In [6], it is listed as 30 C 6; 10; 5. The preceding results describe conditions under which a given function  .z/ is invariant under 0 .N / for some N , or groups generated by 0 .N / and some of its Atkin–Lehner involutions. The genera of the Riemann Surfaces nH  for the Fuchsian groups  appearing in Exercises 3.10 and 3.15 are all 0. Hence their function fields k.nH  / are generalized by just one function: k.nH  / D C.f /. We wish to prove that the  .z/ in Exercises 3.10 and 3.15 are ‘the’f for the corresponding . 3.16 Theorem. Let be a generalized partition. Assume that P (1) t t r t D 24, (2)  .z/ is invariant under the action of a discrete subgroup  of SL2 .R/ containing 0 .N / for some N ,   n (3) 1 D f˛ 2  j ˛.1/ D 1g is equal to f ˙1 0 ˙1 j n 2 Zg, (4) z D 1 is the unique pole of  .z/ among all inequivalent cusps of : Then  .z/ 2 F and the function field of nH  is equal to C. .z//. Proof. Condition (1) implies that  .z/ has a Fourier expansion of the form 1

1 X  .z/ D C an q n ; q nD0

q D e 2 iz ; z 2 H:

Condition (3) implies that 1 is a cusp and that V D fjzj < 12 ; Im.z/ > rg for a suitable r may be taken as a neighborhood of 1. Hence q D e 2 iz can be taken as a local parameter of z D 1. Since .z/ ¤ 0 on H ,  .z/ has no poles in H . Condition (4) implies that z D 1 is the unique pole on the Riemann surface nH  to C. Since  .z/ is an

44

Chapter 3. “Moonshine” of finite groups

open mapping, the image of  .z/ is open and compact, and hence  .z/ is onto and so  .z/ is a covering map from nH  to C [ f1g. Since  .z/ D 1 implies z D 1,  .z/ is one to one. This proves that nH  is analytically isomorphic to  C [ f1g. Hence the genus of nH  is 0. 3.17 Theorem. Let be a generalized partition of degree 24 and suppose that  .z/ is invariant under  D 0 .N / or  D h0 .N /; We1 ; We2 ; : : : ; Wer i for some N . Let N1 be the set of integers c such that 1  c < N , c j N , and c ¤ Ne if xe 2 hW xe1 ; : : : ; W xer i. Suppose that W X .t; c/2 rt  0 t for all c 2 N1 . Then  .z/ is a generator of k.nH  / In particular, the genus of  is zero. Proof. We have to prove conditions (1), (2), (3) and (4) of Theorem 3.16. (1) and (2) are already satisfied. Since the coset 0 .N /We does not contain an element of the form   ˛  ; 0 ˇ the condition (3) also holds. As for the condition (4), we first observe that all cusps of  are equivalent to ac where the pairs .c; a/ are in a set S0 .N / described in Proposition 1.23. Lemma 3.5, as in Theorem 3.6, implies that  .z/ is holomorphic at all cusps except at 1 (1 is equivalent to Na ; a 2 Z) if no ei is present: i.e.,  D 0 .N /. In the contrary case, Proposition 1.32 is applicable to get the desired result. This completes the proof.  3.18 Exercise. Show that each  .z/ is a generator of k.nH  / where f; g is one of the pairs in Exercises 3.10 and 3.15.

3.2 Harmonies Let h be a positive divisor of a natural integer n. By definition, nh Ce; f; : : : is equal to h0 . nh /; We ; Wf ; : : : i, the subgroup of GL2 .R/C generated by 0 . nh /; We ; Wf ; : : : , where We , Wf , : : : , are Atkin–Lehner involutions of 0 . nh /. 3.19 Definition. Define 1   h 0 h n njh C e; f; : : : D h0 . h /; We ; Wf ; : : : i 0 1 0  ² a b= h j a; b; c; d 2 Z; ad  D cn d

 0 1 ³ bcn D 1 hwe ; wf ; : : : i; h

3.2. Harmonies

45

  where we , wf , : : : are the conjugates of We , Wf , : : : by h0 01 . The group njh C e; f; : : : is called the r-th harmonics of nh C e; f; : : : D h0 . nh /; We ; Wf ; : : : i. Q Let D t r t be a generalized partition and r be a positive integer dividing all r t involved in . Define Y

jr D .rt/r t =r :

jr is called the r-th harmonics of . For some cases appearing in the moonshine for the Monster simple group M, we need to consider the function field of a subgroup of index h in njh C e; f; : : : . Suppose that a function f .z/ on H is invariant under  D nh C e; f; : : : , then f .hz/ is invariant under 1 D njh C e; f; : : : . Hence if k is the function field for nH  , then the function field of njh C e; f; : : : contains all f .hz/ where f 2 k. On the other hand, if g.z/ is in the function field for 1 , then g. hz / is invariant under . Therefore, if we know one of the function fields of  or of 1 then we will know the other. Suppose that  D nh C e; f; : : : is of genus 0. Then k.nH  / D C.f / for some f . Define g.z/ D f .hz/1= h (taking a suitable branch of f .hz/1= h ). Then ŒC.g.z// W C.f .hz// D h and so the group 2 corresponding to the function field C.g.z// is of index h in njh C e; f; : : : . 3.20 Proposition. Let be a generalized partition and r > 1 be a positive integer dividing r t for every t. Moreover assume (1)  .z/1 2 F, and Q r t =r is a rational square. (2) t Then jr .z/1 2 F where jr is the r-th harmonic of . Proof. Let f .z/ D  .z/1 ; g.z/ D jr .z/1 . Then g.z/r D f .rz/. Now let  be a group for f .z/ 2 F: i.e., k.nH  / D C.f .z//. We may assume   f˙1g. Then  is uniquely determined by Proposition 1.16. Set  ³   1   ²  a b=r ˇˇ a b r 0 r 0 2 : D  1 D c d cr d 0 1 0 1 It is immediate that 1 stabilizes f .rz/. On the other hand, suppose a matrix   ˛ ˇ 2 SL2 .R/  ı

46

Chapter 3. “Moonshine” of finite groups

stabilizes f .rz/. Then it holds that 

˛z C ˇ f r z C ı



D f .rz/:

Replacing z by zr , we obtain 

˛z C ˇr f z=r C ı 

and so we conclude

˛ =r

ˇr ı



D f .z/  2 :

Therefore 1 is the stabilizer group for f .rz/. By definition we have g.z/r D f .rz/, and so if 2 is the stabilizer group for g.z/, then 2  1 . Since g.M z/ D ıg.z/ for M 2 1 with ı being a r-th root of unity in C, we have that the group index Œ1 W 2  divides r. On the other hand, we have    Y   1   r 0 1 1 1 1=r r 0 1=r r t =r D .rtz/ f .rz/ j0 j0 0 1 0 1 0 1 0 1 D e 2 i=r f .rz/1=r and so





g zC

1 D e 2 i=r g.z/: r

Thus Œ1 W 2  D r: Since 2 nH  is an r-fold covering of 1 nH  , Œk.2 nH  / W k.1 nH  / D r. That ŒC.g.z// W C.f .rz// D r and k.2 nH  /  C.g.z// implies k.2 nH  / D C.g.z//. We have thus shown that 2 is of genus 0 and g.z/ is a generator of its function field. We need now only to show that 2  0 .N 0 / for some N 0 . Put N 0 D N r 2 , then N 0 satisfies (1), (2), (3), and (5) (also (4) trivially) of Corollary 3.9, for 0 .N 0 / as shown below. Put r t0 D rrt and t 0 D rt . It then holds: P 0 P rt (1) rt D D 0, r P 0 0 P (2) t r t D rt rrt D 24, Q Q Q rt 0 (3) .t 0 /r t D .rt/r t =r D t t is a square, P r t0 0 P r t =r P rt (5) N D N r2 D N 0 .mod 24/. t0 rt t Therefore, we have 2  0 .N r 2 / as desired.



3.3. Symmetric and alternating products of representations

47

3.3 Symmetric and alternating products of representations Let G be a finite group and V be a G-module of finite dimension over a field k of characteristic 0. Define the tensor product over k V .n/ D V ˝k V ˝k ˝k V

.n times/

and the wreath product G .n/ D G o †n ; z where †n is the symmetric group of degree n. G .n/ contains a normal subgroup G z isomorphic to G  G   G (n tuples) and †n acts on G by permuting the direct factors. Let z g 2 Gg: G0 D f.g; g; : : : ; g/ 2 G; Then G0 Š G. For simplicity we identify G with G0 . Obviously then the direct product G  †n acts on V .n/ . 3.21 Definition. S n .V / D fv 2 V .n/ j v  D v for all 2 †n gI An .V / D fv 2 V .n/ j v  D sgn. /v for all 2 †n g: Elements such as v1 ˝ v2 C v2 ˝ v1 are in S 2 .V / and v1 ˝ v2  v2 ˝ v1 2 A2 .V / (n D 2). S n .V / and An .V / are invariant under G and S n .V / \ An .V / D 0 if n  2. Bases for S n .V / and A n .V /. Let fe1 ; : : : ; em g be a basis of V where m D dim V . Then the set n X o .ei1 ˝ ei2 ˝ ˝ ein / j 1  i1  i2   in  m  2†n

is a basis for S n .V /. Thus the dimension of S n .V / is the number of ways to choose n elements (multiplicity allowed) from a set of m elements. Hence   mCn1 .m C n  1/Š n : D dim S .V / D nŠ.m  1/Š n Similarly we see that An .V / is spanned by the elements of the form X sgn. /.ei1 ˝ ei2 ˝ ˝ ein / : 2†n

48

Chapter 3. “Moonshine” of finite groups

3.22 Exercise. Show that if ij D ik for some 1  ij , ik  m, j ¤ k, then X sgn. /.ei1 ˝ ei2 ˝ ˝ ein / D 0: 2†n

Using Exercise 3.22, we see that the set o n X sgn. /.ei1 ˝ ei2 ˝ ˝ ein / j 1  i1 < i2 < < in  m 2†n

is a basis for An .V /, and so



dim A .V / D n

m n

 D

mŠ nŠ.m  n/Š

.n  m/:

3.23 Lemma. Let G be a finite group and V be an m-dimensional G-module over C, the complex number field. Let f"i . / j 1  i  mg be the eigenvalues of 2 G on V . Then the traces of on S n .V / and on An .V / are given by X "i1 . /"i2 . / : : : "in . / 1i1 i2 in m

and by

X

"i1 . /"i2 . / : : : "in . /;

1i1
respectively. 3.24 Definition. Let  be the characters of G on V . S n ./ and An ./ denote the characters of G on S n .V / and on An .V /, respectively. 3.25 Exercise. Show that S 2 ./. / D

Œ. /2 C . 2 / ; 2

A2 ./. / D

Œ. /2  . 2 / : 2

3.26 Lemma. Let G be a finite group and  be an m-dimensional representation of G over C. Let f"i . / j 1  i  mg be the eigenvalues of . /; 2 G. Define a. /, and b. / as follows: Œ.1  "1 . /q/.1  "2 . /q/ : : : .1  "m . /q/

1

D

1 X

ak . /q k

kD0

and .1 C "1 . /q/.1 C "2 . /q/ : : : .1 C "m . /q/ D

1 X kD0

bk . /q k :

3.3. Symmetric and alternating products of representations

49

Then the functions ak . / and bk . / from G to C are the characters of S k ./, and of Ak ./, respectively. Proof. Since 1 D 1 C x C x2 C : : : ; 1x we have

jxj < 1;

1 D 1 C "i . /q C "i . /2 q 2 C : : : ; 1  "i . /q

jqj < 1:

We see that a0 . / D 1 for all 2 G and X ak . / D "i1 . /"i2 . / : : : "in . /: 1ii i2 in m

On the other hand, we see

X

bk . / D

"i1 . /"i2 . / : : : "in . /:

1ii


In view of Lemma 3.23, this is what we wanted to prove.

3.27 Definition. Let  be a finite dimensional representation of a finite group G over Q. Let x 2 G and .X/ be the characteristic polynomial of the linear transformation (matrix) .x/. .X/ is written uniquely in the form Y .X/ D .X t  1/r t ; where t is a positive and r t is a nonzero integer. Q rinteger

D x D t t is a generalized partition and is called the frame shape of x (with respect to ). 3.28 Lemma. Let Q G be a finite group and  an m-dimensional representation of G over Q. Let  D t r t be the frame shape of an element of G. Define the function n . / as follows: m

Π.z/1 D Π.z/1 D q  24

1 X

 k . /q k ;

q D e 2 iz :

kD0

Then the k . /’s are characters of G for all k. Proof. By definition, Œ .z/

1

Dq

m  24

1 Y Y t

nD1

nt r t

.1  q /

Dq

m  24

1 Y Y nD1

t

.1  q nt /r t :

50

Chapter 3. “Moonshine” of finite groups

Put

.q/ D Since and,

Q

hY i1 : .1  q t /r t t

t .X

t

 1/

rt

is the characteristic polynomial of . /, we have Y t

Setting X D

1 , q

m Y

.X t  1/r t D

P

t rt D m

.X  "i . //:

iD1

we obtain

.q/ D

m hY

.1  "i . /q/

i1

D

iD1

1 X

ak . /q k ;

kD0

where ak . / is the character of S k ./. Since

.q n / D

1 X

ak . /q k n ;

kD0

Q P1 0 n k the coefficients of q m=24  .z/1 D 1 nD1 .q / D kD0 ak . /q are characters 0  of G. Now define k . / D ak . / to complete the proof. 3.29 Exercise. Show det . / D .1/deg 

P

rt

:

3.30 Lemma (Kondo [17], Lemma 4.5). Let d be aQdivisor of 24 and  be a d dimensional representationQof G over Q. Let  D t r t be the frame shape of with respect to  and D hdh be a ( fixed ) generalized partition of degree 24=d . Put YY .htz/dh r t : j .z/ D t

Then

j .z/

h

has a Fourier expansion 1

j .z/ D

1 X C ak . /q k ; q kD0

where ak . / are generalized characters of G. Moreover, if is a (true) permutation (i.e., dh > 0 for all h), then all ak . / are (true) characters. P   Q Q t r t D 24. Hence Proof. We first note that t n .htdh r t / D t 24 d j .z/ D

1 C (a power series in q): q

3.3. Symmetric and alternating products of representations

51

Q For a fixed h, we showed 3.28 that t .htz/r t has coefficients which are Q Qin Lemma characters and so h . t .htz/r t dh / has coefficients of generalized characters. If  all dh > 0, then the coefficients are characters. This proves the lemma.

4 Multiplicative product of  functions Let .z/ be the Dedekind  function and g D .k1 ; k2 ; k3 ; : : : ; ks / be a tuple of positive integers ki , k1  k2   ks  1, 1  i  s. Define g .z/ D

s Y

.ki z/:

iD1

McKay conjectured that g .z/ is a primitive cusp form1 (of some level N and some character ) if and only if the following conditions are satisfied: (a) k1 is a multiple of all ki , 1  i  s, (b) k1 ks D ki ksC1i for all 1  i  s, Ps (c) i D1 ki D 24, and (d) s is even. This conjecture was shown to hold by Dummit, Kisilevsky, and McKay. and an alternative proof was given by M. Koike. 4.1 Theorem (Dummit, Kisilevsky, McKay [8], Koike [15]). Let g D .k1 ; : : : ; ks / be a tuple of positive integers satisfying the conditions (a)–(d) above. Then g .z/ is a primitive cusp form of weight s=2, of level N D k1 ks and with some Dirichlet character  which is described in Table 4.2 below. Conversely, if g .z/ is a primitive cusp form (of some weight, level, and character), then g must satisfy (a)–(d). Proof. Following [8] (and [15]), let us first treat the ‘if’ part of the conjecture. Collecting ki of the same value, we write n times

n times

n times

‚ 1…„ ƒ ‚ 2…„ ƒ ‚ `…„ ƒ g D .k1 ; k2 ; : : : ; ks / D . t1 ; : : : ; t1 ; t2 ; : : : ; t2 ; : : : ; t` ; : : : ; t` / and so we may also write g as a partition (with all ni positive) gD

` Y

n

ti i ;

t1 > t2 > t3 > > t` > 0:

iD1 1

An excellent introduction to primitive forms can be found in §4.6 of Miyake [18].

54

Chapter 4. Multiplicative product of  functions

P We have as in Definition 3.1, wt.g/ D 12 `iD1 ni D s=2 D s.g/=2, and we define N D N.g/ D k1 ks D t1 t` . Obviously there are only finitely many g D .k1 ; k2 ; : : : ; ks / satisfying the conditions (a)–(d). Since the partition number of 24 is 1575 (i.e., p.24/ D 1575), there are (at most) that many products of  functions (-products). Primitive forms do have the multiplicative property and it is stated in [8] that by examining the first few coefficients (by computer), of the 1575 partitions of 24, all but 30 partitions yield -products which are not multiplicative. Among these 30, only two partitions 24 and 83 have an odd s (s D 1 and s D 3) and it is well known that the corresponding -products .24z/ and .8z/3 have the multiplicative property. The remaining 28 partitions are listed below in Table 4.1. Table 4.1

s

g

2

23 1, 22 2, 21 3, 20 4, 18 6, 16 8, 122

4

15 5 3 1, 14 7 2 1, 12 6 4 2, 112 12 , 102 22 , 92 32 , 82 42 , 64

6

82 4 2 1 2 , 7 3 1 3 , 63 2 3 , 4 6

8

62 3 2 2 2 1 2 , 5 4 1 4 , 44 2 4 , 3 8

10

44 2 2 1 4

12

36 16 , 212

16

28 1 8

24

124

In [8], two theoretical ways are described to show the multiplicative property and the primitivity of the -products obtained from Table 4.1. The first one is by identifying the product as a modular function with specific weight, level, and character (“Nebentypus”) in a one-dimensional space of such forms and applying the theory of Hecke operators, and the other by identifying the associated Dirichlet series as an L-function. Following [8] and [15], we briefly describe how to show that, for each g of Table 4.1, g .z/ is a primitive cusp form of weight s=2 D wt.g/, of level N D t1 t` , with some Dirichlet character  mod N . We first treat the cases where s.g/ 0 .mod 4/. 4.2 Exercise (See [8] or Puse Theorem 2.18). Let m1 ; m2 ; : : : ; m t be non-zero integers, ei D sgn.mi /, 2k D tiD1 ei , and ni D ei mi D jmi j. Let N be a positive integer, divisible by ni for i D 1; 2; : : : ; t. Assume: Pt (a) i D1 mi 0 .mod 24/,

Chapter 4. Multiplicative product of  functions

55

Pt

N=mi 0 .mod 24/.  a b Q 2 0 .N /, then f .Az/ D Show that if f .z/ D tiD1 .ni z/ei and A D cN d .cN z C d /k f .z/ if k is even, and f .Az/ D .cN z C d /k .d /f .z/ if k is odd, where N  d  d 1 .d / D jd .1/ 2 if d is odd, and .d / D N if d is even. Note that since j P ad  bcN D 1, if d is even, then N must be odd. Moreover, if i mi > 0, then f is a cusp form (of weight k, of level N if k is even, and level N with character  if k is odd). Applying Exercise 4.2, we conclude (see also [15] and [13]): Let g be as in Table 4.1 and assume s.g/ 0 .mod 4/. Then g .z/ is a cusp form of weight s.g/=2 with respect to 0 .N / (with the trivial character). (b)

i D1

4.3 Exercise. Let g be as in Table 4.1 and assume s.g/ ¤ 0 .mod 4/. Show that g .z/ is a cusp form of weight s.g/=2 with respect top0 .N / with character , where  is the quadratic character (Definition 2.20) of Q. f / listed below in Table 4.2 (see [8]). (More precisely speaking, if 0 is the quadratic character of the corresponding field, then .d / D 0 .d / whenever both values are defined.) Table 4.2

f

g

1

122 , 46 , 44 22 14

2

16 8, 82 4 2 12

3

18 6, 63 23

5

20 4

7

21 3, 73 13

11

22 2

23

23 1

Thus we need only to show that g .z/ is primitive (hence the multiplicative property). The Hijikata’s trace formula [12] modified by Koike [14] is useful for the computation of the dimension of the space of cusp forms (see also [5].) 4.4 Theorem. If Sk .N; / is the space of cusp forms of weight k, level N , with character , then dim Sk .N; / D tv .k; N; /Ctp .k; N; /Cte1 .k; N; /Cte2 .k; N; /Ctı .k; N; /;

56

Chapter 4. Multiplicative product of  functions

where tv .k; N; / D te1 .k; N; / D

k  1 Y  1 p 1C 12 p pjN ´  3 k1 ! 0k1 Q 1  13 ! !! .p .!/ C p .! 0 // if 32 −N; 0 pjN 2 1 C p 0 otherwise,    Q .p .i/ C p .i// if 4−N and k even,  14 i k2 pjN 12 1 C 4 p

´ te2 .k; N; / D

0 otherwise, ´ 1 if k D 2 and  is trivial, tı .k; N; / D 0 otherwise, Y Y D p ; N D p  ; X 2 C X C 1 D .X  !/.X  ! 0 /; and pjN

tp .k; N; / D 

1 2N

pjN N X

Y

c.2; f; p/

f D1 pjN

with c.2; f; p/ being defined as follows: Pick f with 1  f  N . Let  D ordp f for a prime p dividing N , and let p m be the conductor of p . Put AQ D fx 2 Z j .x  1/2 0 mod p C2 ; 2x 2 mod p g; BQ D fx 2 AQ j .x  1/2 0 mod p C1C2 g: Q respectively, mod Let A or B be a complete system of representatives of AQ or B, C

p . Define X X c.2; f; p/ D p .x/ C p .2  y/: x2A

y2B

If  D 2n, n 2 Z, then we have 8 n n1 ˆ if m  n C ;

n C  C 1: If  D 2n C 1, n 2 Z, then we have ´ 2p n if m  n C  C 1; c.2; f; p/ D 0 if m > n C  C 1: Proof. See Koike [14] for the proof of this theorem and the details for tp .k; N; / and c.2; f; p/. We note that ! 6D ! 0 are cubic roots of unity and the notation tp .k; N; / is used both in [12] and [14]. Since p is a variable in the definition of tp .k; N; /, this may not be a proper notation. But it was kept here also. 

Chapter 4. Multiplicative product of  functions

57

4.5 Exercise. Let g be as in Table 4.1. Assume that s.g/  4. Show that dim Sk .N; / D dim Sk0 .N; / D 1; and hence g .z/ is primitive. Here Sk0 .N; / is the space of so-called new forms (see [15], and p. 162 of [18] for the definition of new forms). It now remains to show that g .z/ is primitive if s D s.g/ D 2. 4.6 Exercise. Assume that s D 2 and refer to Tables 4.1 and 4.2. Show that the partitions g, f , N D t1 t` satisfy the conditions (1)–(6) of Theorem 3.6. In order to identify g .z/ in our case s.g/ D 2, we need the theory of Hecke’s L-functions over the quadratic number fields. First we set up the following notation: K W a quadratic number field over Q; O W the ring of all algebraic integers of K over Q; I W the ideal group of K; P W the principal ideal group of K; m W an ideal of O; I.m/ W fa 2 I j a D bc ; .b; m/ D .c; m/ D 1 for some b; c 2 Og;

P .m/ W f.a/ 2 P j a 1 mod mg:

4.7 Definition. Let W I.m/ ! C be a character. Suppose that for all .a/ 2 P : 9 .1/ There is an integer u such that > > >  u > > > a > >

..a// D if K is imaginary. > > = jaj 0 () .2/ For some u; u 2 f0; 1g; and v 2 R; it holds that> > > 0 a > > >

..a// D .sgn.a//u .sgn.a0 //u j 0 jiv if K is real > > a > > ; 0 .a is the conjugate of a 2 K over Q/: Then is called a Hecke character of K mod m. Remark. Here the Hecke characters are defined only for quadratic number fields. See p. 90–91 of [18] for the Hecke characters of general cases. The notation mod denotes the multiplicative congruence and is defined as follows. Let .a/ 2 I.m/. Then there exist b; c 2 O such that a D bc with .b; m/ D .c; m/ D 1. We write a 1 mod m if in addition b c mod m holds.

58

Chapter 4. Multiplicative product of  functions

Usually, is extended to all I by defining

.a/ D 0 Note. If K is imaginary, then

a jaj

if a … I.m/:

D e i ,  D arg.a/. If K is real, then

ˇ a ˇiv 0 ˇ ˇ ˇ 0 ˇ D e iv.log jajlog ja j/ : a Let be a Hecke character of K mod m and let  ˇ ˇ there exits a principal ideal .a/ 2 P .n/ \ I.m/ : M D n W an ideal of O ˇ such that ./ in the definition above holds for .a/ We have m 2 M and the G.C.D. of all ideals of M is the conductor of the character .

is said to be primitive if m is equal to the conductor of . We quote the following two theorems without proofs. See p. 182–85 in [18] or p. 204–07 in [7]. p 4.8 Theorem. Let K D Q. d /, d < 0, d the discriminant of K. Let be a Hecke character mod m. Suppose that there exists an integer u  0 such that  u a for a 1 mod m:

..a// D jaj Define f .z/ D

X

.a/e 2 iNK .a/

u=2 e 2 iNK .a/z

a

where a ranges over all ideals in O. Then, if is not the principal character, then f .z/ 2 SuC1 .N; / where N D jd jNK .m/, .n/ D d .n/ ..n// for n 2 Z. Moreover, if is a primitive character, then f .z/ is also primitive. p 4.9 Theorem. Let K D Q. d /, d > 0, d the discriminant of K. Let be a Hecke character of K mod m. Suppose that one of the following two possibilities holds: (1) ..a// D sgn.a/ for a 1 mod m, or (2) ..a// D sgn.a0 / for a 1 mod m (a0 is the conjugate of a in K=Q). Define f .z/ D

X

.a/e 2 iNK .a/z ;

a

where a ranges over all ideals of O. Then f .z/ 2 S1 .N; /, where N D dNK .m/, .n/ D d .n/ ..n// for n 2 Z. Moreover, if is primitive, then f .z/ is primitive as well.

Chapter 4. Multiplicative product of  functions

59

p 4.10 Example.  d  g D 23 1, N D t1 t2 D 23, f D 23. Put K D Q. 23/ and let .d / D 23 be the quadratic character of K. The class number of K is 3. Let be a Hecke character of I =P . By definition, ..a// D 1 for all .a/ 2 P , i.e., u D 0, m D 1 in ./ of Definition 4.7. Therefore, by Theorem 4.9, the function X

.a/e 2 iNK .a/z f .z/ D a

is an element of S1 .23; /, since 23NK .1/ D 23, .d / D

d  23

.

The following list taken from Koike [14] is our choice ofpdiscriminant dg , a character g of conductor fg of the quadratic field Kg D Q. dg / for each g of Table 4.1 with s.g/ D 2. Some additional information is also given. Table 4.3 2

g

Ng

dg

fg

23:1

23

23

1

"g .d / d 

11

2

d 

7

3

d 

22:2 21:3

44 63

20:4

80

5

18:6

108

3

16:8 122

128 144

8 4

4p1 6 4 6

5

3

11

3

7

4 d 1 2

.1/

.1/

2 3

3

2 jd j

23

.1/ d 

jd j

order of g

d 1 2

d 1 2

4 4

4.11 Exercise. Show that fg .z/ 2 S1 .Ng ; g / for each g of Table 4.3. (Note that fg is a conductor and fg .z/ is a function of z.) By definition, fg .z/ D

X

g .a/e 2 iNKg .a/z :

a

Since g is primitive for each g, fg .z/ is a primitive cusp form by Theorem 4.8. We now need only to show that fg .z/ D g .z/. It is known that fg .z/ D g .z/ holds 2 Koike informed the author that the combination fdg D 5; fg D 2; order of g D 4g will also work for the case g D 20  4.

60

Chapter 4. Multiplicative product of  functions

Q if their coefficients agree up to N12k pjN .1 C p1 / terms. (Note that D ŒSL2 .Z/ W Q 0 .N / D N pjN .1C p1 / and that an automorphic form f 2 Ak .0 .N // possesses k zeroes 12 fg .z/ and

in its fundamental domain.) So we need to compute the coefficients of g .z/ up to r terms where: g

23 1

22 2

21 3

20 4

18 6

16 8

122

r

2

6

8

12

18

16

24

The coefficient of g .z/ may be directly computed. 4.12 Example. g D 21 3 g .z/ D .21z/.3z/ 1 1 Y Y 21n Dq .1  q / .1  q 3n /; nD1

q D e 2 iz

nD1

q.1  q 3 /.1  q 6 / .mod q 9 / q  q 4  q 7 .mod q 9 /: Next we compute fg .z/ D f213 .z/, where X fg .z/ D

.a/q NK .a/ ;

q D e 2 iz :

p We have K D Q. 7/, .fg / D .3/ D m, I.m/=P .m/ Š Z4 , and g is a character of degree 4 of I.m/=P .m/. We need only to compute fg .z/ modulo q 9 . Let a be an ideal with norm N.a/  8. If fi is a prime divisor of a, then fi is above the ideals (2), (3), (5), or (7). D D 7 is the discriminant of K and by Theorem 2.15, we have            7 1 7 2 1 2 D D 1; D D D .1/3 D 1; 3 3 5 5 5 5 and .2/ D f1 f2 and .7/ D f32 . Hence a D f3 or a D f1a1 f2a2 where a1 C a2  3. Thus N.a/ D 1; N.a/ D 2;

a D .1/; a D f1 ; f2 ;

N.a/ D 4; N.a/ D 7;

a D f12 ; f1 f2 ; f22 ; a D f3 ;

N.a/ D 8;

a D f1 ; f12 f2 ; f1 f22 ; f23 :

61

Chapter 4. Multiplicative product of  functions

p p Since K D Q. 7/ and 7 1 .mod r/, we have O D Z C Z 1C 2 7 . Moreover, D D discriminant D 7, units D f˙1g, and decompositions of primes are

.7/ D f32 ; .2/ D f1 f2 We also have

and





7 3 7 5



 D



 D

1 3 2 5

.since  7 1 .mod 8//:

 D .1/ 

 D

1 5

31 2

D 1;

  2 2 D .1/.5 1/=8 D 1: 5

Moreover, we have the following exact sequences, where f D .3/, i.Kf / D h.a/ j a 2 K; .a; f/ D 1i, i.Kf;1 / D h.a/ j a 2 K; a 1 mod fi, and CK the class group of K: 1 !

i.Kf / If ! ! CK ! 1 i.Kf;1 / i.Kf;1 /

(exact)

and 1 !

Kf If .I /Kf;1 ! CK ! 1 ! ! Kf;1 Kf;1 i.Kf;1 / ˇ ˇ Kf ˇ ˇK

We have

f;1

and so

If i.Kf;1 /

(exact).

ˇ ˇ ˇ ˇ ˇ D ˇ.O=.3// ˇ D 8 ˇ

' Z4 .

We compute    1  7i 1 C 7i D f1 f2 ; .2/ D 2 2

 f12

D

3 C 7i 2



 ;

f14

D

I

1  21i 2



f and 121i 2 .˙1/Kf;1 . Therefore f1 is of order 4 in i.Kf;1 . We also have f3 D 2 / p If 2 . 7/, f3 D .7/ 2 .˙1/Kf;1 , and so f3 is of order 2 in i.Kf;1 / . We then conclude

p p fg .z/ q 1 C . 1  1/q 2 C .1 C 1  1/q 4 C .1/q 7 C .i C i  i C i/q 8 q  q 4  q 7 .mod q 9 /: Therefore, g .z/ D fg .z/ holds.

62

Chapter 4. Multiplicative product of  functions

4.13 Exercise. Show that g .z/ D fg .z/ for all the remaining g with s.g/ D 2. This completes the proof of the first part (if part) of Theorem 4.1. Next we show, following Koike [15], that if g .z/ is a primitive cusp form, then g D .k1 ; k2 ; : : : ; ks / (k1  k2   ks  1) satisfies the condition (a)–(d) of Theorem 4.1. Again it is convenient to write g as a partition: gD

` Y

n

ti i ;

t1 > t2 > t3 > > t` > 0:

iD1

Let us first observe that if Q is a generalized permutation and  .z/ is a constant, then rt ¤ ; and t0 is the least integer of the expression of

D ;. For, suppose

D P Q t Q1 0

. Then r t D 0 and t¤t0 nD1 .1  q n /jr t j would be expressed in terms of q t with t 0 > t0 , which is obviously false. Thus the property ‘ .z/ D constant’ implies

D ;. This in turn impliesQ that  .z/ D (constant)  0 .z/ forces D 0 . is a primitive cusp form of the space of new Now suppose g .z/ D t .tz/r tP forms Sk0 .N; /. Then a1 D 1 and so t r t D 24. This is (c).  a b 2 0 .N /. Then by Lemma 3.3, we have Let A D cN d s

g .t/jk A ´ .cN t C d /k g .At/ D (constant) .cN t C d /k .cN t C d / 2 g .t/: Hence s D 2k even. This is (d). To prove the other conditions, we make use ofAtkin–Lehner involutions of 0 .N /. Let   ae b ; e jjN; det.We / D e; a; b; c; d; e 2 Z: We D cN de Put e t D .t; e/. Then by Exercise 3.12, we have Y et g .z/jk We D (constant) . 2 z/r t : et

(A)

On the other hand, by [2] with e D N , we have P1

g .z/jk WN D (constant) N g .z/;

(B)

where D nD1 aN n q n (aN n denotes the complex conjugate of an ) if g .z/ D P1 N g .z/ n N g .z/ D g .z/. Thus (A), nD1 an q . In our cases, obviously all an are real and so  (B) and the observation mentioned above imply that Y Y  N t r t Y  N r t rt t D D ; t .N; t/2 t t t and therefore it holds that holds.

N ti

D t`C1i and ni D n`C1i . Hence the condition (b)

Chapter 4. Multiplicative product of  functions

63

Since We acts on the space Sk0 .N; / of new forms, g .z/jk We must be a constant multiple of some primitive cusp form. In particular, X eti i

ei2

D 24

()

must hold for every e such that e jjN . Q Let D liD1 ti ni be a partition (t1 > t2 > > t`  1, ni > 0 for all i) satisfying (b) if N D t1 t` , then (c) (d)

Pl

i D1 ti ni

Pl

i D1

N ti

D t`C1i for all i,

D 24, and

ni even.

Obviously there are only finitely many (less than p.24/ D 1575) possibilities for . One can find (perhaps with the help of a computer) that only the g’s in Table 4.1 satisfy the conditions (b)–(d) together with the condition ./ mentioned above. Therefore (a) also holds by inspection of Table 4.1. This completes the proof of Theorem 4.1. 

Appendix Genus zero discrete groups  such that 0 .N /    0 .N /C (compiled by M.-L. Lang)

The following is the list of all discrete subgroups  of SL2 .R/ specified in the title. There are exactly 123 such groups. For notation, see Definition 1.31. 1,

19 C 19,

2, 2 C 2,

20 C 4, 20 C 20, 20 C 4 C 5,

3, 3 C 3,

21 C 3, 21 C 21, 21 C 3 C 7,

4, 4 C 4,

22 C 11, 22 C 2 C 11,

5, 5 C 5,

23 C 23,

6, 6 C 2, 6 C 3, 6 C 6, 6 C 2 C 3,

24 C 8, 24 C 24, 24 C 8 C 3,

7, 7 C 7,

25, 25 C 25,

8, 8 C 8,

26 C 26, 26 C 2 C 13,

9, 9 C 9,

27 C 27,

10, 10 C 2, 10 C 5, 10 C 10, 10 C 2 C 5,

28 C 7, 28 C 4 C 7,

11 C 11,

29 C 29,

12, 12 C 4, 12 C 3, 12 C 12, 12 C 4 C 3,

30 C 15, 30 C 5 C 6, 30 C 2 C 15, 30 C 3 C 5, 30 C 6 C 10, 30 C 2 C 3 C 5,

13, 13 C 13, 14 C 7, 14 C 14, 14 C 2 C 7, 15 C 5, 15 C 15, 15 C 3 C 5, 16, 16 C 16, 17 C 17, 18, 18 C 2, 18 C 9, 18 C 18, 18 C 2 C 9,

31 C 31, 32 C 32, 33 C 11, 33 C 3 C 11, 34 C 2 C 17, 35 C 35, 35 C 5 C 7, 36 C 4, 36 C 36, 36 C 4 C 9,

66

Appendix

38 C 2 C 19, 39 C 39, 39 C 3 C 13, 41 C 41, 42 C 3 C 14, 42 C 6 C 14, 42 C 2 C 3 C 7, 44 C 4 C 11, 45 C 9 C 5, 46 C 23, 46 C 2 C 23, 47 C 47, 49 C 49, 50 C 50, 50 C 2 C 25, 51 C 3 C 17,

60 C 4 C 15, 60 C 20 C 15, 60 C 4 C 3 C 5, 62 C 2 C 31, 66 C 11 C 6, 66 C 2 C 3 C 11, 69 C 3 C 23, 70 C 10 C 14, 70 C 2 C 5 C 7, 71 C 71, 78 C 26 C 39, 78 C 2 C 3 C 13, 87 C 3 C 29, 92 C 4 C 23, 94 C 2 C 47,

54 C 2 C 27,

95 C 5 C 19,

55 C 5 C 11,

105 C 3 C 5 C 7,

56 C 8 C 7,

110 C 2 C 5 C 11,

59 C 59,

119 C 7 C 17.

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